mm-4819 === Subject: Re: Polynomial into e^ix expansion ? > Let f(x) be any polynomial, > I can go for Taylor's polynomial expansion > for that function at x values > say x = p,q,r , I can get Taylor's Polynomials Is there any method or procedure to > express the same function in polynomial using > e^ix instead of using only x, > why I'm asking this particular one is > for higher powers of x its very hard to compute > where as if we use e^ix it will eventually > become e^inx for big 'n' values. Huh? What will become e^inx? What does whether > n is big or small have to do with it? And what's > hard to compute about powers of x? /*Question in a Single Line*/ How to go for a polynomial expansion for a function say F(x) > at given point using e^ix instead of using only x.? Look up Fourier series. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada ----------- > look if this is the problem, Let > T(p), T(q) & T(r) be the Taylor's polynomials > for the function f(x) at x = p,q,r I don't have the function f(x) > but data points as this x = 1 2 3 4 > y = 2 5 8 7 There is exactly one polynomial y = f(x) that matches the data (the so- > called Lagrange polynomial---seehttp://en.wikipedia.org/wiki/Lagrange polynomial > ), but there are infinitely many non-polynomials that are an exact fit > to the data. No amount of wishful thinking or arm-waving or > incomprehensible lunatic raving will ever change that fact. You just > are not thinking, and worse still, you never seem to learn and never > seem to want to learn. It seems to me that what he wants to find is an expansion for f that > looks > like this: f(x) = a 0 + a 1 (e^(ix)) + a 2 (e^(ix))^2 + a 3 (e^(ix))^3 + ... i.e. a polynomial with e^ix replacing x. That is a Fourier series, > and I > don't think it exists for non-periodic functions like he wants (i.e. > if f is a > polynomial function.). But it could be an approximation on the interval spanning the given > data points, which is possibly all he wants. That could be. One can pass an e^(ix) expansion through a given > set of data points. Consider 2 points, (x 1, y 1) and (x 2, y 2). We have a 0 + a 1 e^(ix 1) = y 1 > a 0 + a 1 e^(ix 2) = y 2 Then this is just linear equations for a 0 and a 1, giving a 0 = y 1 - (e^(ix 1) (y2 - y1))/(e^(ix 2) - e^(ix 1)) > a 1 = (y 2 - y 1)/(e^(ix 2) - e^(ix 1)). Then the interpolation is f(x) = a 0 + a 1 e^(ix). =-=-=-=-=-=-=- > apologized for late reply. > how to form a differential equation thread > we will continue here. > -=-=-=-=-=-=-=- Generally we will write a polynomial > by taking x P = a0 + a1x + a2x^2 + a3x^3 + a4x^4+............+ an x^n (1) here we can form a polynomial depending up on our wish > may be fitting roots like this (x-a)(x-b)(x-c).....(x-n) > or an interpolation polynomial like 'Lagrangian' form > but the central part of everything is 'x' > Whole my doubt is instead of using 'x'. > I have to have use 'w' Let, > omega = w = e^ix = cisx. > In eq(1) instead of going for x, I need to go for omega P = a0 + a1 (w) + a2 (w)^2 + a3 (w)^3 +......+... here whether we can go for fitting roots > (w-a)(w-b)(w-c).........or > going for interpolation polynomial like Lagrangian > form, what ever it is, /**/ Which ever polynomial I deal, replacing 'x' > with omega(w) that's exactly my problem /**/ If your known data points are (x 1,y 1), (x 2,y 2), ... (x n,y n) then you can set up simultaneous equations like this: a 0 + a 1*e^(i*x 1) + a 2*e^(2*i*x 1) + ... + a {n-1}*e^((n - 1) *i*x 1) = y 1 a 0 + a 1*e^(i*x 2) + a 2*e^(2*i*x 2) + ... + a {n-1}*e^((n - 1) *i*x 2) = y 2 ... and solve for the a's in the usual way (here I'm writing e^(n*i*x) for (e^(i*x))^n). The resulting function y = a 0 + a 1*e^(i*x) + a 2*e^(2*i*x) + ... + a {n-1}*e^((n - 1) *i*x) will pass through the data points, but there's no reason to expect it will do exactly what you want between them, and in general it won't be real-valued for real x other than x 1, x 2, ... x n. Alternatively -- amounting to exactly the same thing but not explicitly giving you the values of the a's -- you can use Lagrange- style interpolation except that everywhere you have an x or an x i in the Lagrange interpolation formula you use e^(i*x) and e^(i*x i) instead. You realise that if instead you had y = ... + b 2*e^(-2*i*x) + b 1*e^(-i*x) + a0 + a 1*e^(i*x) + a 2*e^ (2*i*x) ... then you'd have a standard fourier series? Then you could find the coefficients to approximate to a known function using standard methods. I'm not sure if there's a way to adapt these methods to give you a series without the negative powers. Maybe someone else will. === Subject: Re: Polynomial into e^ix expansion ? > You realise that if instead you had y = ... + b 2*e^(-2*i*x) + b 1*e^(-i*x) + a0 + a 1*e^(i*x) + a 2*e^ > (2*i*x) ... > I should have written it more tidily as y = ... + a {-2}*e^(-2*i*x) + a {-1}*e^(-i*x) + a 0 + a 1*e^(i*x) + a 2*e^(2*i*x) ... === Subject: Re: commutator question > assume x,y in a group G both commute with [x,y] (commutator). > prove that for all natural n, (xy)^n=x^n y^n [y,x]^(n(n-1)/2). it looks easy but i am stuck... Prove by induction. By definition x^0 = e, where e is the group the identity is true for some n. Note that if x commutes with a then x commutes with a^{-1} (since if xa = ax then xa^{-1}a = x and a^{-1}xa = a^{-1}ax = x, and so xa^{-1} = a^{-1}x), and also that [y,x] = [x,y]^ {-1}, so x and y both commute with [y,x]. Then (xy)^{n + 1} = (xy)^n (xy) = x^n y^n [y,x]^{n(n - 1)/2} xy = x^n y^n xy [y,x]^{n(n - 1)/2} Now note that y^n x = x y^n [y,x]^n (prove this by induction, using yx = [y,x] xy, together with the fact that both x and y commute with [y,x]). So we have (xy)^{n + 1} = x^n x y^n y [y,x]^n [y,x]^{n(n - 1)/2} = x^{n + 1} y^{n + 1} [y,x]^{(n + 1}n/2} as required. === Subject: Re: Desc set theory - Projections of product spaces > an example. Let me take a closed set in X*w -- X*w - U*w, > where U is open in X. Now, its projection is just the > set X - U. Its just a closed set. But can it always be > written as countably infinite union of closed sets?. There is no requirement in these considerations (descriptive > set theory) that the sets in a countable union (or a countable > intersection) be disjoint. In fact, you can even use the same > set more than once. Also, if countable means countably > infinite or finite (which it usually does, but some authors > don't and some authors don't tell you which way they intend > for it to mean), then you can just take the index set for the > countable union to have cardinality 1, and thus if C is a closed > set, then C is the union of the set {C}. Something like this has > to be assumed (either assume countable can include finite, > or assume the sets can be taken to be the same -- see [1]), > otherwise you can't deduce things like the empty set is > an F sigma set. [1] One possible mathematical grammar problem with this > (allowing the sets to be the same but using countable > to mean countably infinite), however, is that you > need to index over well ordered sets, since, as sets, > {C, C, ... } is equal to {C}. Thus, you're actually > taking the union of a sequence of closed sets, > not the union of a countable set of closed sets. Dave L. Renfro === Subject: Re: Probability & combination question > I'm having trouble knowing which methods I should be using to solve > this question: > (assume any other factors have no bearing on these simple > probabilities) having blonde hair = 0.2 & P(not blonde) = 0.8 If I pick a random group of 10 people, how to I work out the > probability that exactly 1 or 2... etc. to 10 will have blonde hair. I think that for all 10 to be blonde the probability is 0.2^10, and > that for only 1 to be blonde it is 0.8^9 (i.e. the probability that 9 > of them will not have blonde hair) No. P{10 blonde} = (0.2)^10 is correct, but P{1 blonde} is *not* equal > to (0.8)^9. Look at a group of 10 people, which group happens to have > only 1 blonde. For the order person 1, person 2, ..., person 10 and > using N = not blonde, B = blonde, we could have: BNNNNNNNNN or > NBNNNNNNNN or NNBNNNNNNN or ... or NNNNNNNNNB. Altogether, there are > 10 possible sample points, each having probability (0.2)*(0.8)^9, so > P{1 blonde} = 10*(0.2)*(0.8)^9. What about the case of 2 blondes? We > could have BBNNNNNNNN or BNBNNNNNNN or ... or NNNNNNNNBB, each point > having probability (0.2)^2 * (0.8)^8. How many points are there? Well, > you can enumerate them all and count them, but a neater way is to note > that the number is just 10*9/2, the number of ways of choosing 2 > things from 10 things. Why is this true? Think about it this way: we > have 10 empty boxes and want to but B into two of them. How many > distinct outcomes will there be? The first 'B' can go into any box, so > there are 10 ways to place the first 'B'. Then the second 'B' can go > into any of the remaining 9 boxes. Altogether, there are 10*9 ways of > placing the two 'B's. However, the arrangements in which B1 goes into > box 3 and B2 into box 7 is the same as that where B2 goes into box 3 > and B1 into box 7. In other words, all permutations of B1 and B2 give > the same final outcome, so the number of distinct outcomes is only > 10*9 / 2! In general, P{k blondes} = C(10,k)*(0.2)^k *(0.8)^(10-k), where C(M,j) > = binomial coefficient = number of combinations of j things chosen > from M things = number of subsets of size j = M!/[j! * (M-j)!] = M* > (M-1)* ... *(M-j+1)/j! = number of distinct arrangements of k 'B's in > M boxes. R.G. Vickson I know how to calculate combinations = n!/r!(n-r)!, but I'm having > difficulty visualizing how to apply it in this situation. > Peter in to a spreadsheet, then I'll know if I come unstuck. > Steve You are working with the so-called Binomial distribution. If you have > EXCEL (or most other decent spreadsheets) the binomial distribution is > built in. However, if you would like to program it yourself, you > should avoid direct computation of the formula and use instead an > iterative method. In the following, put n = 10 and p = 0.2 to get the > case you are dealing with. The general case of k blondes among n > people with blonde probability = p per person gives P{k blondes} = b > (k) = C(n,k)*p^k* q^(n-k), where q = 1-p and k = 0, 1, 2, ..., n. > Here, b(k) is the so-called binomial probability distribution. We have > b(0) = q^n, and b(k+1) = [(n-k)/(k+1)]*(p/q)*b(k) for k = > 0,1,2,...,n-1. Thus, you get b(1)=n*(p/q)*b(0), b(2) = [(n-1)/2]*b(1), > etc. Oops! This last one should be b(2) = [(n-1)/2]*(p/q)*b(1). RGV R.G. Vickson Nice... it works AND I understand it now. :-) I was going to ask > what happens if we choose from a small population of say, 100, as the > probability would then change each time.(if the first is blonde then > the second would have p = 19/99, I think). Yes. That is the problem of sampling without replacement. The > distribution of the number of blondes in this case is harder. It is > called the *hypergeometric distribution*. In my version of EXCEL it is > the built-in function HYPGEOMDIST. Basically, in a population of N things, N1 of type 1 and N2 of type 2 > (N = N1 + N2), the hypergeometric distribution with parameters N1, N2 > and M describes the probability P(k) of getting k objects of type 1 > when a sample of size M is drawn without replacement. In your case, > type 1 = blond, type 2 = other, N1 = 20, N2 = 80 (N = 100), and M = > 10. For more details on the hypergeometric distribution, see, eg.,http://en.wikipedia.org/wiki/Hypergeometric distributionorhttp://stat.... I am not sure how > well it is implemented in EXCEL, since EXCEL is well-known to be > inaccurate and unreliable for statistical computations. See, eg.,http://www.cs.uiowa.edu/~jcryer/JSMTalk2001.pdforhttp://www.coventry.... > (especially the links therein). R.G. Vickson I want to program a general function for this in C, so using the > iterative method you showed me I can just adjust the probability > each time through the loop. I'm not sure what I'm going to do when n! > becomes very large as I guess it will overflow even a 64bit type. Is > this why you said to avoid direct computation? Yes. However, if the population (sample size M) is very large, even > the recursion needs to be approached carefully, because starting from b > (0) = q^M might fail (inaccuracy, or perhaps even underflow). A better > method would be to start near the mean, say at k = k0 = [p*M] ([ ] = > integer part). Iterate upward using b(k+1) = [(M-k)/(k+1)]*(p/q)*b(k) > for k >= k0 and iterate downward using b(k-1) = c(M,k)*b(k) for k <= > k0; you can easily work out what the coefficient c(M,k) must be. > get the array of probabilities b(k) as [..., r(k0-2),r(k0-1),1,r > (k0+1),r(k0+2)...]*B, where the r(j) are what you would get if you had > put b(k0) = 1. Now determine B so that the b(j) sum to 1: B = 1/sum(r > (j),j=0..M), and finally re-scale the array to get b(k) = r(k)*B = r > (j)/sum(r). For large M, as you get very far from the center you may > start getting r(j) that are so small as to be negligible, maybe even > close to underflow. In that case, just stop the recursions and just > use sum(r) to be the sum of all the ones computed before stopping. That method will work even in cases where the direct method fails, and > will in all cases be at least as accurate, if not more accurate, than > the direct method. > Steve when it is necessary to take replacement/no-replacemnet in to account; > intuitively I would have thought that when the population is very > large it wouldn't have made much difference, so to get a feel for this > I concocted some examples: (assuming that all factors are independent) Population = 400,000 > NumBlondes = 150,000 > NumPeopleCalledDave = 100 > P(blonde) = 150000/400000 = 0.375 Instead of picking a group at random we pick the Daves (100) exactly 38 of these 100 Daves will be blonde = 0.08153 > ---------- Now flip the problem: > P(Dave) = 100/400000 = 0.00025 Now as our sample group we pick all the blondes. > exactly 38 of these 150,000 Blondes will be called Dave = 0.06437 Are you using the binomial distribution in both cases? If so, we can > see what is happening by changing the scenarios' descriptions > somewhat. In the first case we have 150K blondes and 250K others, all > as yet unnamed. Now we randomly throw in the label Dave 100 times. > The hypergeometric applies if we don't allow more than one label per > person (so once named, a person is essentially removed), while the > binomial applies if we all the label applications are independent (so > a person could have more than one Dave label sticking to him). It is > apparent that since there are so few labels compared to both sub- > populations N1 = 150K and N2 = 250K, the chance that anyone will have > more than one label is small; that is why the binomial gives a good > approximation to the hypergeometric in this case. In the second case > we have N1 = 100 'Daves' and N2 = 399,900 others, all without hair > color at the moment. Now we select 150,000 people at random and dye > them blonde. In this case there is a real difference between the > hypergeometric and the binomial. The hypergeometric case applies if we > don't dye anyone more than once, while the binomial applies if we pick > the 150,000 randomly one-by-one and dye them, irrespective of whether > or not they have been picked and dyed already. Now there is a good... read more é >It is >apparent that since there are so few labels compared to both sub- >populations N1 = 150K and N2 = 250K, the chance that anyone will have >more than one label is small; that is why the binomial gives a good >approximation to the hypergeometric in this case. In the second case >we have N1 = 100 'Daves' and N2 = 399,900 others, all without hair >color at the moment. Now we select 150,000 people at random and dye >them blonde. In this case there is a real difference between the >hypergeometric and the binomial. I can really visualize it now. is working well for the binomial, even with very large N. (and bailing out when values become negligible as suggested) method for just calculating C(n,k) also. Using that I'm calculating the hypergeometric distribution (avoiding having to calculate factorials). I'm happy to report I'm getting the same hypergeometric results as the calculator on the stattrek page you mentioned :-) (I haven't yet implemented the method from http://dissertations.ub.rug.nl/FILES/faculties/eco/2005/e.talens/c4.pdf) Steve === Subject: Re: POLL : e4 or d4 Given the choice of e4 or d4, I play e4; but I > play > b4 (the Sokolsky > Opening) more often than either d4 or e4. b4 ???? i wouldnt recommend that actually. the a pawn usually get isolated , leading to better > chances for black. and you give away the center and or initiative. i call b4 'the orangutan'. which it is also really called. tommy1729 I do not recall ever having a problem with an > isolated a-pawn in this > opening. This might be issue in the line 1 b4 e5 2 > Bb2 Bxb4 3 Bxe5, > but there are more aggressive responses to 2...Bxb4, > which I prefer to > play, like the Kucharkowski-Meybohm gambit. meybohm -> giving another pawn. btw at first sight , i might just respond with 1 b4 c5 or similar just to isolate the a pawn. In any case, at least > against the opponents I usually play, 2...d6 or > 2...f6 is a more > common response than 2...Bxb4, and in those lines, > White does not > exchange the b-pawn; usually it is pushed to b5 and > then supported > with pawns on a4 and c4, making a strong queenside > attack possible. that is often the idea yes ... I do not claim the Sokolsky opening (also called the > Orangutan or > Polish opening, in some sources) is as strong as 1 > e4, at the master > level; but I am no master, and among amateurs, it > seems to do passably > well for me, and I enjoy playing it--it seems to lead > to very > exciting, attacking games; Fischer's old games with > it are great > examples, like this one: >http://www.chessgames.com/perl/chessgame?gid=1255134 So what are the results of your poll, so far? not enough replies yet. === === Subject: Re: POLL : e4 or d4 > Given the choice of e4 or d4, I play e4; but I > play > b4 (the Sokolsky > Opening) more often than either d4 or e4. b4 ???? i wouldnt recommend that actually. the a pawn usually get isolated , leading to better > chances for black. and you give away the center and or initiative. i call b4 'the orangutan'. which it is also really called. tommy1729 I do not recall ever having a problem with an > isolated a-pawn in this > opening. This might be issue in the line 1 b4 e5 2 > Bb2 Bxb4 3 Bxe5, > but there are more aggressive responses to 2...Bxb4, > which I prefer to > play, like the Kucharkowski-Meybohm gambit. meybohm -> giving another pawn. btw at first sight , i might just respond with 1 b4 c5 or similar just to isolate the a pawn. In any case, at least > against the opponents I usually play, 2...d6 or > 2...f6 is a more > common response than 2...Bxb4, and in those lines, > White does not > exchange the b-pawn; usually it is pushed to b5 and > then supported > with pawns on a4 and c4, making a strong queenside > attack possible. that is often the idea yes ... I do not claim the Sokolsky opening (also called the > Orangutan or > Polish opening, in some sources) is as strong as 1 > e4, at the master > level; but I am no master, and among amateurs, it > seems to do passably > well for me, and I enjoy playing it--it seems to lead > to very > exciting, attacking games; Fischer's old games with > it are great > examples, like this one: >http://www.chessgames.com/perl/chessgame?gid=1255134 So what are the results of your poll, so far? not enough replies yet. > tommy1729 considering: http://psychos.here.ws if i would have to bother, i actually would prefer > teaching a little boy or girl, > go fish, > instead of how to play chess. i dunno why just somehow think its more appropriate. love and peace, > and, > peace and love,kirk- Hide quoted text - - Show quoted text - p.s. aw! five minute chess, mind if i vote for this move, moons ago, some up and coming master was kind enough to give me 4.45 minutes of time and i honored him the white pieces. i think he played e4, PKB4, and i thought and thought and thought, but gee maybe i could not think so all of a sudden i had only 15 seconds left and i grabed his king, ran out the door, jumped in my junk car, and sped away! with trepidation, i meekly appeared back at the club, only to find its owner, was rather upset because, it was his set that he had for weeks to play with, with NO KING! === Subject: Estimating the mean of the cumulative hypergeometric? I'm trying to save some computation when calculating the cumulative probability of the hypergeometric distribution by first estimating the mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 then just add the values from k = 1001 to 1100. (In tandem with skipping the many values close to zero, I'm typically only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse or skewed for this to be accurate, but even in the example above with large m, I'm getting a significant error: i.e when I brute-force add up all the values to the estimated mean I'm getting 0.509 instead of 0.5. Therefore, if I used this shortcut , I'd be out by .009. for any value of k. Is assuming mean = n * (m/N) always a no-no, or are there times when it's a reliably accurate estimate? Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Did you mean [ mCk ] [ N-mCn-k ] / [ NCn ]? (Or maybe you're just designating the variables differently from what I was expecting...) mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 > then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I'm sure you already know this, but just in case: you realise that you don't need to calculate the binomial coefficients afresh for each value of k, right? You can just multiply what you had for k - 1 by the appropriate ratio, thus saving a huge amount of computation. === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=K5_wDwoAAADH6PVGp0Ky-_BsyD7txqZY AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) > I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Did you mean [ mCk ] [ N-mCn-k ] / [ NCn ]? (Or maybe you're just > designating the variables differently from what I was expecting...) mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 > then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I'm sure you already know this, but just in case: you realise that you > don't need to calculate the binomial coefficients afresh for each > value of k, right? You can just multiply what you had for k - 1 by the > appropriate ratio, thus saving a huge amount of computation. Yes, I'd copied the notation incorrectly. (m = number successes in N, k = number successes in sample(n) method Ray showed me. I've got a single hypergeometric calculation working very fast now, it's just the speed of calculating the cumulative probability that's slowing me down, especially when n and k are large. >Sum k=0^q h(N,m,n,k) ~= phi((q + 1/2 - mu)/sigma) hypergeometric first to satisfy myself that it's correct, and then I'll compare the approximation using my typical data sizes to see how it compares. The errors you were getting would probably be acceptable. Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? This is my routine for calculating the cumulative _Binomial_ distribution, with relacement, given n and p: double q = 1-p, c = pow(q,n); // i.e. when k=0 double cp = c; // cumulative probability for(double k=1; k<=n; k++) { c = ((n-(k-1))/k)*c*(p/q); cp += c; } It's fast, and (for me) clear to understand, and as it's iterative, the cost of calculating the cumulative probability is little more than for calculating a specific probability. So... I REALLY thought I would be able to modify this to work with no- replacement, after all the only difference is that p&q are changing as a result of N and m being reduced. (m = number successes in N; k = number successes in sample(n)) Surely, in the loop above, if I could recalculate p&q by how N and m are changing, then this would give the same result as the hypergeometric?? But after much trying, I can't get it to work. I've trawled the internet too, and there seems to be no shortcut other than using h(N, m,n,k) = C(m,k)*C(N-m, n-k)/C(N,n) and summing over 0->k. I have found some sieving optimisations, but it's basically the same method. (sorry, that was all more of a rant than a question ;-) ) If only there was a reliable way to start anywhere near the middle of the distribution; working backward or forward from this point would mean I could skip the lower/upper ends where the value soon becomes almost zero. One last real world example: N = total number of people that have rented DVDs = 400,000 m = num people have rented a specific movie 'A': 0 < m 200,000 n = num people have rented a specific movie 'B': 0 < n < m k = num people that have rented both (As there is symmetry, it's always quicker to make m the larger of A or B, and n the smaller of A and B) What is the probability that k people have rented both? Then: h(N, m,n,k) = C(NumRentedA,k)*C(NumHaveNotRentedA, NumRentedA-k)/ C(N, NumRentedB) What is the probability that B and A are highly correlated in some unknown way? e.g. if A = 20,000, B = 5,000, the probability that >=294 have seen both is so small as to make it likely that there is an underlying reason: cumulativeP = 1 - SUM(k=0, k=310)[ C(20000,k)*C(380000, 20000- k)/C(N, 5000) ] = .0005 The median & mean are approx the same in this case: when k = median = mean = 250, P = 0.5 When k = 208: P = .0005. Intuitively, doesn't it feel astonishing that for a range of k=0 to k=5000, it's 99.9% likely that k will be between 208 to 294!! Unless, of course, this is all screwed! Given that N is fixed, and N > m*2, and m > n can anyone spot any optimisations? I've been testing this with lots of values to find when the median is almost equal to the mean. Can anyone suggest, from the limitations on the numbers given, when the WORST case scenario might be? Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > This is my routine for calculating the cumulative Binomial > distribution, with relacement, given n and p: double q = 1-p, c = pow(q,n); // i.e. when k=0 > double cp = c; // cumulative probability > for(double k=1; k<=n; k++) > { > c = ((n-(k-1))/k)*c*(p/q); > cp += c; } It's fast, and (for me) clear to understand, and as it's iterative, > the cost of calculating the cumulative probability is little more than > for calculating a specific probability. > So... I REALLY thought I would be able to modify this to work with no- > replacement, after all the only difference is that p&q are changing as > a result of N and m being reduced. > (m = number successes in N; k = number successes in sample(n)) > Surely, in the loop above, if I could recalculate p&q by how N and m > are changing, then this would give the same result as the > hypergeometric?? I don't see that this is an efficient way to approach it. p and q are changing, sure, but they're changing in a way that depends on the history of the previous selections, and you don't want to be tracking all those possibilities individually. Your loop code for the hypergeometric should not be that different from the binomial. You keep a running count of the product of the binomial coefficients involving k: this is what I was alluding to earlier. Each iteration involves multiplying by two ratios, something like c = c*(m - k)/(k + 1)*(n - k)/(N - m - n + k + 1), depending on exactly how you implement it. Of course, you can pre-calculate N - m - n + 1. You do also have the overhead of initialising this multiplier to (N - m)C(n) and calculating the denominator Ncn. === Subject: Re: Estimating the mean of the cumulative hypergeometric? This is my routine for calculating the cumulative Binomial > distribution, with relacement, given n and p: double q = 1-p, c = pow(q,n); // i.e. when k=0 > double cp = c; // cumulative probability > for(double k=1; k<=n; k++) > { > c = ((n-(k-1))/k)*c*(p/q); > cp += c; } It's fast, and (for me) clear to understand, and as it's iterative, > the cost of calculating the cumulative probability is little more than > for calculating a specific probability. > So... I REALLY thought I would be able to modify this to work with no- > replacement, after all the only difference is that p&q are changing as > a result of N and m being reduced. > (m = number successes in N; k = number successes in sample(n)) > Surely, in the loop above, if I could recalculate p&q by how N and m > are changing, then this would give the same result as the > hypergeometric?? I don't see that this is an efficient way to approach it. p and q are > changing, sure, but they're changing in a way that depends on the > history of the previous selections, and you don't want to be tracking > all those possibilities individually. Your loop code for the hypergeometric should not be that different > from the binomial. You keep a running count of the product of the > binomial coefficients involving k: this is what I was alluding to > earlier. Each iteration involves multiplying by two ratios, something > like c = c*(m - k)/(k + 1)*(n - k)/(N - m - n + k + 1), depending on > exactly how you implement it. Of course, you can pre-calculate N - m - > n + 1. You do also have the overhead of initialising this multiplier > to (N - m)C(n) and calculating the denominator Ncn. Oh, and you'll need to do some range-checking too, to make sure the k's you're summing are all in-range, and adjust the initialisation and loop termination as necessary, but that's only a tiny overhead. === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) > This is my routine for calculating the cumulative _Binomial_ > distribution, with relacement, given n and p: double q = 1-p, c = pow(q,n); // i.e. when k=0 > double cp = c; // cumulative probability > for(double k=1; k<=n; k++) > { > c = ((n-(k-1))/k)*c*(p/q); > cp += c; } It's fast, and (for me) clear to understand, and as it's iterative, > the cost of calculating the cumulative probability is little more than > for calculating a specific probability. > So... I REALLY thought I would be able to modify this to work with no- > replacement, after all the only difference is that p&q are changing as > a result of N and m being reduced. > (m = number successes in N; k = number successes in sample(n)) > Surely, in the loop above, if I could recalculate p&q by how N and m > are changing, then this would give the same result as the > hypergeometric?? > But after much trying, I can't get it to work. I've trawled the > internet too, and there seems to be no shortcut other than using > h(N, m,n,k) = C(m,k)*C(N-m, n-k)/C(N,n) > and summing over 0->k. > I have found some sieving optimisations, but it's basically the same > method. > (sorry, that was all more of a rant than a question ;-) ) If only there was a reliable way to start anywhere near the middle of > the distribution; working backward or forward from this point would > mean I could skip the lower/upper ends where the value soon becomes > almost zero. For the hypergeometric distribution with N = whole population, m = marked objects (like the blondes before), n = sample size (like M before) and k = number of marked in sample, the probabilities p(k) obey the recursion: p(k+1) = r(k)*p(k), where r(k) = [(m-k)/(k+1)]* [(n-k)/(N-m-n+k+1)]. So, you can iterate upward using p(k+1) = r(k)*p (k) for k >= k0 and downward using p(k-1) = p(k)/r(k-1) for k <= k0, starting as though p(k0) = 1. Stop when the resulting p(k) are negligible, let S = sum(p), then re-scale as p(k) <-- p(k)/S. If n is only a few hundred to a few thousand, that should be quick and accurate. One last real world example: N = total number of people that have rented DVDs = 400,000 > m = num people have rented a specific movie 'A': 0 < m 200,000 > n = num people have rented a specific movie 'B': 0 < n < m > k = num people that have rented both > (As there is symmetry, it's always quicker to make m the larger of A > or B, and n the smaller of A and B) What is the probability that k people have rented both? This question cannot be answered unless you specify a probability model. For example, there might be some weird reason that the only people who rent B are left-handed female redheads older than 40 years who rented A, or right-handed males under 30 who did not rent A. Then, to answer the question we would need to know the composition of the people who rented A and the composition of those who did not rent A. OK, this is an artificial illustration, but it makes a valid point: you need to specify a probability model. > Then: h(N, m,n,k) = C(NumRentedA,k)*C(NumHaveNotRentedA, NumRentedA-k)/ > C(N, NumRentedB) Shouldn't this be C(NumRentedA,k)*C(NumHaveNotRentedA, NumRentedB-k)/ C(N, NumRentedB)? Your use of the hypergeometric ASSUMES that the renters of B are a random sample from the whole population. Why should that be the case? What is the probability that B and A are highly correlated in some > unknown way? Surely that must depend on the probability model used. For example, if I specify that P{rent A|rent B} = 0.5, then on average one half of the renters of B will see both. However, I guess you are assuming a hypergeometric model, and in that case the number renting both is the number of A's in a random sample of NumRentedB people drawn from a population of size N and in which NumRentedA have rented A. > e.g. if A = 20,000, B = 5,000, the probability that >=294 have seen > both is so small as to make it likely that there is an underlying > reason: cumulativeP = 1 - SUM(k=0, k=310)[ C(20000,k)*C(380000, 20000- > k)/C(N, 5000) ] = .0005 Where on Earth does the number '310' come from? The median & mean are approx Here you say approx the same, then below you say they are exactly equal. You can't have both! > the same in this case: when k = median = > mean = 250, P = 0.5 > When k = 208: P = .0005. Intuitively, doesn't it feel astonishing that for a range of k=0 to > k=5000, it's 99.9% likely that k will be between 208 to 294!! Unless, of course, this is all screwed! Given that N is fixed, and N > m*2, and m > n can anyone spot any > optimisations? Do you mean to ask whether the code can be improved? Improved compared to what? What method have you used so far to do the computations? For example, there are may ways to compute binomial coefficients; which one (if any) did you use? Anyway, I have already suggested a method of starting in the middle and iterating up and down, then re-scaling, and I would bet that in any moderate-to-large problem it beats most direct method hands down. R.G. Vickson I've been testing this with lots of values to find when the median is > almost equal to the mean. Can anyone suggest, from the limitations on > the numbers given, when the WORST case scenario might be? > Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? > I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Did you mean [ mCk ] [ N-mCn-k ] / [ NCn ]? (Or maybe you're just > designating the variables differently from what I was expecting...) mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 > then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I'm sure you already know this, but just in case: you realise that you > don't need to calculate the binomial coefficients afresh for each > value of k, right? You can just multiply what you had for k - 1 by the > appropriate ratio, thus saving a huge amount of computation. If speed of computation is of the essence then another possibility is to approximate the hypergeometric distribution with a normal distribution of the same mean and variance, i.e. mu = n*m/N sigma^2 = n*m/N*(1 - m/N)*(N - n)/(N - 1) Then Sum k=0^q h(N,m,n,k) ~= phi((q + 1/2 - mu)/sigma) where phi is the standard cumulative normal distribution function, for which fast approximations are available that are accurate enough for most practical purposes. For very small probabilities the relative error is likely to be large (but the absolute error small), but for larger probabilities, and for suitably large N, m and n, the error may be acceptable. A random example that I tried, with N = 200, m = 70, n = 110, q = 37, gives the actual value as 0.382395... and the approximation as 0.383138... With with q = 34, otherwise the same, I get actual = 0.116702..., approximation = 0.117221... There is still the problem, though, of figuring out how the accuracy varies with N, m and n, so as to gauge whether the approximation is acceptable for a given set of parameters. === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Just for clarity: is kCm equal to m choose k (= m!/k!/(m-k)!) or is it equal to k choose m (= k!/m!/(k-m)!)? mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 The median (the 0.5 probability point) may not equal the mean. > then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. The computations MAY be OK, because the median need not equal the mean. > Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I don't understand the question. The formula mean = n * (m/N) is 100% accurate. The issue is whether the median is close to the mean, and how to know the cases in which that is more-or-less true. R.G. Vickson > Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? > I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Just for clarity: is kCm equal to m choose k (= m!/k!/(m-k)!) or is > it equal to k choose m (= k!/m!/(k-m)!)? mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 The median (the 0.5 probability point) may not equal the mean. then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. The computations MAY be OK, because the median need not equal the > mean. Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I don't understand the question. The formula mean = n * (m/N) is 100% > accurate. The issue is whether the median is close to the mean, and > how to know the cases in which that is more-or-less true. R.G. Vickson > Steve First of all, I should have written h(N, m,n,k) = C(m,k)*C(N-m, n-k)/C (N,n), I copied the other notation from a web page as i was unsure of the correct notation. Either way, I had tunnel-vision, mixing the mean and median up, so everything else I'd asked was pointless. Sorry. Back to the drawing board. Steve === Subject: test posting-account=QIM_hQoAAAD_XpQRD11KLkrLJctO0MFC Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) This is a test. === Subject: Re: test > This is a test. Great, does it have a solution? === Subject: Re: A simple pseudorandom number generator. posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Do you make any claim about the distribution of the numbers produced? Dave This is an improved version with an example that responds the Dave's question. A SIMPLE PSEUDORADOM NUMBERS GENERATOR 1.- Fix D, the number of digits of random numbers. 2.- Input an integer seed So . So >10^6 ;; N = So ; C = int(sqr (So)) ; n = 0 3.- Sum n to that number: N1 = N + n 4.- Multiply it by 4 and subtract 1 . N2 = 4*N - 1. 5.- Extract the square root: Q = sqr(N2) 6.- Determine the fractional part: F = Q - int(Q) 7.- Random number R = First D digits of F ; n = n + C 8.- Number of digits of initial seed : T = int((ln So)/ln 10)) 9.- Multiply R by 10^T: N1 = 10^T*R ; GO TO 3 It is based in the supposition that the digits of the fractional part of a square root (Not a perfect square) are pseudorandom. The mean of Random Numbers must be 0.5 The frequency of each R.N. must be uniform distributed. If the number of digits is D and the number of R.N produced is NN then the mean of frequencies of R.N.is, naturally = NN / 10^D. The frequencies of that frequencies must be Normal distributed. By the 7th condition ( n = n + C), the sequence never fall in a loop. Example: Let D = 3 Let be So = 1234567 ; NN = 10000 Results: Distribution: Uniform Mean of R.Ns = 0.499 Mean of frequencies = 10 RANGE FREQUENCY NORMAL. CHI^2 OF FREC. DISTR. TEST 0 - 1 30 30.47 0.48 2 [CapitalEth] 3 116 94.95 4.91 4 - 5 200 180.95 2.0 6 - 7 223 239.54 1.14 8 - 9 210 219.12 0.37 10 - 11 129 138.52 0.65 12 - 13 64 60.51 0.2 14 - 15 22 18.27 0.76 16 - 17 4 3.81 ------ 18 - 19 1 0.55 ------ 20 - 21 1 0.05 0.57 ------- -------- ---------- TOTAL 1000 989.85 11.08 === Subject: Re: A simple pseudorandom number generator. posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Do you make any claim about the distribution of the numbers produced? Dave This is an improved version with an example that responds the Dave's > question. A SIMPLE PSEUDORADOM NUMBERS GENERATOR 1.- Fix D, the number of digits of random numbers. > 2.- Input an integer seed So . So >10^6 ;; N = So ; C = int(sqr > (So)) ; n = 0 > 3.- Sum n to that number: N1 = N + n > 4.- Multiply it by 4 and subtract 1 . N2 = 4*N - 1. > 5.- Extract the square root: Q = sqr(N2) > 6.- Determine the fractional part: F = Q - int(Q) > 7.- Random number R = First D digits of F ; n = n + C > 8.- Number of digits of initial seed : T = int((ln So)/ln 10)) > 9.- Multiply R by 10^T: N1 = 10^T*R ; GO TO 3 It is based in the supposition that the digits of the fractional > part of a square root (Not a perfect square) are pseudorandom. > The mean of Random Numbers must be 0.5 > The frequency of each R.N. must be uniform distributed. > If the number of digits is D and the number of R.N produced is NN > then the mean of frequencies of R.N.is, naturally = NN / 10^D. > The frequencies of that frequencies must be Normal distributed. > By the 7th condition ( n = n + C), the sequence never fall in a loop. Example: > Let D = 3 > Let be So = 1234567 ; NN = 10000 Results: > Distribution: Uniform > Mean of R.Ns = 0.499 > Mean of frequencies = 10 RANGE FREQUENCY NORMAL. CHI^2 > OF FREC. DISTR. TEST > 0 - 1 30 30.47 0.48 > 2 [CapitalEth] 3 116 94.95 4.91 > 4 - 5 200 180.95 2.0 >6 - 7 223 239.54 1.14 >8 - 9 210 219.12 0.37 > 10 - 11 129 138.52 0.65 > 12 - 13 64 60.51 0.2 > 14 - 15 22 18.27 0.76 > 16 - 17 4 3.81 ------ > 18 - 19 1 0.55 ------ > 20 - 21 1 0.05 0.57 ------- ------- ------ > TOTAL 1000 989.85 11.08 === Subject: Re: A simple pseudorandom number generator. posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) This is the definitive version with the columns of the example in its proper place. A SIMPLE PSEUDORADOM NUMBERS GENERATOR 1.- Fix D, the number of digits of random numbers. 2.- Input an integer seed So . So >10^6 ; N = So ; C = int(sqr (So)) ; n = 0 3.- Sum n to that number: N1 = N + n 4.- Multiply it by 4 and subtract 1 . N2 = 4*N1 - 1. 5.- Extract the square root: Q = sqr(N2) 6.- Determine the fractional part: F = Q - int(Q) 7.- Random number R = First D digits of F ; n = n + C 8.- Number of digits of initial seed : T = int((ln So)/ln 10)) 9.- Multiply R by 10^T: N = 10^T*R ; GO TO 3 It is based in the supposition that the digits of the fractional part of a square root (Not a perfect square) are pseudorandom. The mean of Random Numbers must be 0.5 The frequency of each R.N. must be uniform distributed. If the number of digits is D and the number of R.N produced is NN then the mean of frequencies of R.N.is, naturally = NN / 10^D. The frequencies of that frequencies must be Normal distributed. By the 7th condition ( n = n + C), the sequence never fall in a loop. Example: Let D = 3 Let be So = 1234567 ; NN = 10000 Results: Distribution: Uniform Mean of R.Ns = 0.499 Mean of frequencies = 10 RANGE FREQUENCY NORMAL. CHI^2 OF FREC. DISTR. TEST 0 - 1 30 30.47 0.48 2 [CapitalEth] 3 116 94.95 4.91 4 - 5 200 180.95 2.0 6 - 7 223 239.54 1.14 8 - 9 210 219.12 0.37 10 - 11 129 138.52 0.65 12 - 13 64 60.51 0.2 14 - 15 22 18.27 0.76 16 - 17 4 3.81 ------ 18 - 19 1 0.55 ------ 20 - 21 1 0.05 0.57 ------- ------- ------ TOTAL 1000 989.85 11.08 === Subject: bijection from sequences of binary numbers to Reals posting-account=JTrNRgoAAABuGIzwXYnlAgkRTDpv8_sI .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How to get a bijection from sequences of binary numbers onto Reals? === Subject: Re: bijection from sequences of binary numbers to Reals > How to get a bijection from sequences of binary numbers onto Reals? Construct injections in each direction, and apply the Schroeder-Bernstein theorem. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: bijection from sequences of binary numbers to Reals posting-account=JTrNRgoAAABuGIzwXYnlAgkRTDpv8_sI .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > How to get a bijection from sequences of binary numbers onto Reals? Construct injections in each direction, and apply the Schroeder-Bernstein > theorem. > I actually need an explicit bijection. Binary representation of real numbers could work, but that would only work for real numbers in [0,1]. And there are duplicate representations. === Subject: Re: bijection from sequences of binary numbers to Reals days. My association with the Department is that of an alumnus. > How to get a bijection from sequences of binary numbers onto Reals? > Construct injections in each direction, and apply the Schroeder-Bernstein > theorem. I actually need an explicit bijection. Then construct explicit injections. Schroeder-Bernstein is constructive. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Sum(1/(p(i)*log(p(i))), i=1...infinity)=Pi/2 ???? bubu a .8ecrit : > Raymond Manzoni a .8ecrit : > Raymond Manzoni a .8ecrit : > Raymond Manzoni a .8ecrit : > Maria Povolotsky a .8ecrit : > I came across the following in > http://www.research.att.com/~njas/sequences/A000040 > I conjecture that > Sum(1/(p(i)*log(p(i))))=Pi/2=1.570796327... > .... > A numerical value was computed in 1998 by Henri Cohen : > 1.63661632335126086856965800392186367118159707613129... > found here http://pi.lacim.uqam.ca/piDATA/plogp.txt The more general sum : sum_{p prime} 1/(p^s*log(p)) > was studied in 2008 by Richard Mathar in 'Twenty digits of the > Erd.9as-Lebensold constant' http://arxiv.org/pdf/0811.4739 The table 2.4 gives the numerical results for the first integer > values of s. > A last link : the 1991 paper of Cohen 'High-precision > calculation of Hardy-Littlewood constants' > http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi providing very useful > methods for numerical evaluation of these sums as well as zeta and > other Euler series! > Raymond > but what are p(i) ? I think p(i) point out prime numbers Yes it seems to be the ith prime number here http://www.research.att.com/~njas/sequences/A000040 (that's at least what we supposed... to prove the initial conjecture wrong!) Raymond === Subject: Need a simple formula to calculate average percentage I am writing a little program to help me practice typing skills. The program generates a sequence of characters which I am to type. It compares what I type with the target sequence to determine which keys I am having more trouble with. Those keys are then presented more frequently in the future. The program is working but I am not happy with the algorithm I am using to decide which keys to drill. I would like some suggestions for better algorithms. Here's what I've tried and where they have problems. 1. Simple Average. Keep track of how many times each character is drilled (N) and how many times it was typed correctly (C). The average (C/N) is a measure of how difficult that key is. This works OK for awhile, but N quickly gets large and then C/N changes very little, so it's a relatively insensitive measure. 2. Consecutive Correct Keystrokes (CCK). Each time a character is typed correctly, add 1. Each time it is typed incorrectly, reset it to 0. This works much better than the simple average and, at first, appeared to be a pretty good measure of how difficult a key is. It is especially sensitive to recent data, which is a good thing. The problem is that it is too sensitive to recent date, especially misses. A key that is missed after having been typed correctly 1,000 (or 100,000) times is treated the same as one that is missed frequently. 3. Rolling Average. It seems like a good measure might be the average over the last N (100? 300?) drills. I haven't tried that because it requires that I keep track of the last N results. That's not that hard to do, but I hoping for a more elegant solution. 4. Weighted Average. It occurred to me that a weighted average might be a more general solution than a rolling average and may be easier to calculate (at least fewer data points). By weighted average, I mean an average where the most recent data point gets weight 1, the N-1 data point gets discounted by some factor (F, 0 What are the conditions for w + 1/w being rational, where w is real? Set w + 1/w = q, with q in Q. Solving for w, > If w must be real, then it must be |q| >= 2. It's just a special case of the arithmetic-geometric mean inequality: HINT w real>0 <=> w+1/w real>=2, special case v = 1/w of --Bill Dubuque === Subject: Re: caculate zeta function s->1 posting-account=W967OgoAAAASYsvrgrqxW7fN8WeVhaXp icafe8),gzip(gfe),gzip(gfe) I find the original form of zeta function converge in 0 Hi. I read this:http://en.wikipedia.org/wiki/Lambert%27s W function#Generalizations > and this:http://arxiv.org/PS cache/math-ph/pdf/0607/0607011v2.pdf Let's say then, that x = OMEGA (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q; u, v) > (on paper, one would write r 1, r 2, ... r p, then a bar under it, > then s 1, s 2, ..., s q, then that thing followed by ; u, v, > centered on the same row as the bar, or even just as it is written > above if that seems too odd, i.e. a big slash sign. I just came up > with this notation because it resembles that of the hypergeometric and > G-function, and OMEGA was the symbol used in the paper. They didn't > show a notation for the general case.) means that e^(-vx) = u((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x - > s p)). Then, why not also try out the obvious complement, where: x = omega (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q; u, v) means that log(-vx) = u((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x > - s p)). ? Then we can solve the Keplerian equation cos(x) = x in closed > form: x = -i ln(omega (2, 1)(-i, i / 0; i/2, -1)) Note that this is many-valued, as omega and ln are many-valued, it > actually has an infinite number of values (solutions). However there > is only one real solution, x ~ 0.73908513321516. By using different > branches of omega and ln we can get other (complex) solutions. The general Keplerian equation M = E - epsilon sin(E) has solution(s) E = -i ln(omega (2,1)(-iM/epsilon + sqrt(1 - (M^2)/(epsilon^2)), -iM/ > epsilon - sqrt(1 - (M^2)/(epsilon^2)) / 0; epsilon/2, -1)). (for the physical equation, one would choose the branches of omega and > ln to yield real solutions.) What do you think? Note that little-omega also generalizes W: W(x) = - > ln(omega (1,0)(0 / ; -1/x, -1)). Now what I'm wondering about is this: Can the generalizations given > above be used to invert the tower functions with towers greater than > 2? We know that the inverse for f(x) = ^2 x is tetroot 2(x) = e^(W(ln > (x))), is it possible to use an OMEGA or omega function to invert, > say, f(x) = ^3 x? (here, ^n x is x tetrated to the nth tower.) I also noticed that, at least for finite p, q, one can use e^x = H ((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x - s q)) and ln(x) = H ((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x - s q)) to give slightly simpler forms that are just as general: x = OMEGA (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q ; H) for the e^x eqn, and omega (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q ; H) for the ln(x) eqn. Then for cos(x) = x, x = -i ln(omega (2, 1)(-i, i / 0; i/2)) and for the Keplerian equation E = -i ln(omega (2,1)(-iM/epsilon + sqrt(1 - (M^2)/(epsilon^2)), -iM/ epsilon - sqrt(1 - (M^2)/(epsilon^2)) / 0; epsilon/2)). Note also that we can get the really simple identities log(x) = OMEGA (0, 0)( / ; x) exp(x) = omega (0, 0)( / ; x). with no parameters but the variable x. === Subject: Re: Generalization of Lambert W Function? posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.5) Gecko/2008120121 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Hi. I read this:http://en.wikipedia.org/wiki/Lambert%27s W function#Generalizations > and this:http://arxiv.org/PS cache/math-ph/pdf/0607/0607011v2.pdf Let's say then, that x = OMEGA (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q; u, v) > (on paper, one would write r 1, r 2, ... r p, then a bar under it, > then s 1, s 2, ..., s q, then that thing followed by ; u, v, > centered on the same row as the bar, or even just as it is written > above if that seems too odd, i.e. a big slash sign. I just came up > with this notation because it resembles that of the hypergeometric and > G-function, and OMEGA was the symbol used in the paper. They didn't > show a notation for the general case.) means that e^(-vx) = u((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x - > s p)). Then, why not also try out the obvious complement, where: x = omega (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q; u, v) means that log(-vx) = u((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x > - s p)). ? Then we can solve the Keplerian equation cos(x) = x in closed > form: x = -i ln(omega (2, 1)(-i, i / 0; i/2, -1)) Note that this is many-valued, as omega and ln are many-valued, it > actually has an infinite number of values (solutions). However there > is only one real solution, x ~ 0.73908513321516. By using different > branches of omega and ln we can get other (complex) solutions. The general Keplerian equation M = E - epsilon sin(E) has solution(s) E = -i ln(omega (2,1)(-iM/epsilon + sqrt(1 - (M^2)/(epsilon^2)), -iM/ > epsilon - sqrt(1 - (M^2)/(epsilon^2)) / 0; epsilon/2, -1)). (for the physical equation, one would choose the branches of omega and > ln to yield real solutions.) What do you think? Note that little-omega also generalizes W: W(x) = - > ln(omega (1,0)(0 / ; -1/x, -1)). Now what I'm wondering about is this: Can the generalizations given > above be used to invert the tower functions with towers greater than > 2? We know that the inverse for f(x) = ^2 x is tetroot 2(x) = e^(W(ln > (x))), is it possible to use an OMEGA or omega function to invert, > say, f(x) = ^3 x? (here, ^n x is x tetrated to the nth tower.) i have posted about multisected lambert W before but let me describe it briefly here the multisected lambert W is defined as n W(x) g ( W(x)) = x n 0 n n where 0 g n is the generalised coshinus then just as in the standard method you can calculate the inversion which rests on calculating terms j-1 / d 1 | | ----- ------ | | j-1 g (x) / |x=0 dx 0 n these can be expressed either in terms of my generalised berneuler numbers (which should be obvious by extracting taylor coefficients) or interestingly using the standard faa di bruno techniques in terms of j --- n-1 0's repeat pattern k -----/----- --/-- / (-1) (1) B (0, 0, ..., 0, 1, 0, ...) --- k j,k k=1 these are also related to the generalise bell numbers that arise in multisected u-tetration the multisected lamberts immediately solve the keplerian and provide a basis for similar iteration problems over the higher generalised trigonometrics -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: strange almost identity >Can anyone explain why (1/2)*Pi*sin(x) / (cos(x) + 1 - cos(x)/(sin(x)+cos(x)+1)) agrees so well with x on the interval [0,Pi/2]? It's exact for 0, Pi/ >6, Pi/4, and Pi/2, gets as high as about x + 0.02 and as low as about >x - 0.001 (it's less than x for x between Pi/6 and Pi/4). Perhaps the answer to why is just that the proposed formula is complicated enough to get a good approximation. Under the half-angle substitution sin(x)=2u/(1+u^2), cos(x)=(1-u^2)/(1+u^2), the question is equivalent to asking why u/( (u+1)*(u^2-2*u+3) ) agrees so well with arctan(u)/Pi on [0,1]. As you note, it matches perfectly when (u, arctan(u)/Pi) is one of (0,0) (2-sqrt(3), 1/12) (sqrt(2)-1, 1/8) (1, 1/4) It's not surprising that once the two smooth functions match perfectly at several points, they will behave similarly at intermediate points. Rational functions with rational coefficients which match these values will also send algebraic conjugates to 1/12 and 1/8 respectively, so in effect you are trying to match 6 rational points, which you should expect to be able to do as long as you have 6 free parameters, such as with a polynomial of degree 5, or a rational function of bidegree 2,3 . Fitting a polynomial to these data in the usual way gives u*(31-3*u-5*u^2+u^3) / 96 which actually fits the actangent curve a little better than the rational function you proposed. I'm not quite sure why both this polynomial and your rational function have degrees smaller (by one) than my heuristic would suggest, and I concede the rational function has especially small coefficients, but otherwise I think the proposed approximation is just complicated enough to be part of a general family of functions which is flexible enough to account for the good approximation. dave === Subject: Re: strange almost identity Can anyone explain why (1/2)*Pi*sin(x) / (cos(x) + 1 - cos(x)/(sin(x)+cos(x)+1)) agrees so well with x on the interval [0,Pi/2]? ?It's exact for 0, Pi/ >6, Pi/4, and Pi/2, gets as high as about x + 0.02 and as low as about >x - 0.001 (it's less than x for x between Pi/6 and Pi/4). Perhaps the answer to why is just that the proposed formula is > complicated enough to get a good approximation. Under the half-angle substitution > ? sin(x)=2u/(1+u^2), cos(x)=(1-u^2)/(1+u^2), > the question is equivalent to asking why > ? u/( (u+1)*(u^2-2*u+3) ) > agrees so well with ?arctan(u)/Pi ?on [0,1]. As you note, it matches perfectly when ?(u, arctan(u)/Pi) is one of > ?(0,0) > ?(2-sqrt(3), 1/12) > ?(sqrt(2)-1, 1/8) > ?(1, 1/4) > It's not surprising that once the two smooth functions match perfectly at > several points, they will behave similarly at intermediate points. Rational functions with rational coefficients which match these > values will also send algebraic conjugates to 1/12 and 1/8 respectively, > so in effect you are trying to match 6 rational points, which you > should expect to be able to do as long as you have 6 free parameters, such > as with a polynomial of degree 5, or a rational function of bidegree 2,3 . > Fitting a polynomial to these data in the usual way gives > ? u*(31-3*u-5*u^2+u^3) / 96 > which actually fits the actangent curve a little better than > the rational function you proposed. I'm not quite sure why both this polynomial and your rational function > have degrees smaller (by one) than my heuristic would suggest, and > I concede the rational function has especially small coefficients, > but otherwise I think the proposed approximation is just complicated > enough to be part of a general family of functions which is flexible > enough to account for the good approximation. dave One can try to approximate arctan(u) in a different way. Since it has a zero at 0 and an asymptote, one can try to approximate it by Pi/2 u/ (u+1) times a function g(x) (the factors u/(u+1) are roughly the heuristic behind the first part of the 'amateur''s reasoning). Approximating g(x) by a quadratic using least squares over [0,1] gives 1.53897 - 1.12859 u + 0.611177 u^2 which is close enough to (3-2u+u^2)/ 2 as claimed. === Subject: Re: strange almost identity >It's not surprising that once the two smooth functions match >perfectly at several points, they will behave similarly at >intermediate points. That's not quite right. The functions sin(10*x) and cos(10*x) match at quite a few points on the interval [0,1] but they don't behave similarly at intermediate points. -- Rouben Rostamian === Subject: Re: strange almost identity >It's not surprising that once the two smooth functions match >perfectly at several points, they will behave similarly at >intermediate points. That's not quite right. The functions sin(10*x) and cos(10*x) >match at quite a few points on the interval [0,1] but they >don't behave similarly at intermediate points. Certainly they do, for sufficiently weak values of similarly. Seriously: similarly is not precisely defined. Neither is smooth, at least in the sense in which I suspect he meant the term; smooth and similarly raise the questions how smooth? and how similarly?, respectively. If smooth means has small derivative and similar means difference is uniformly small then it's true that if f and g match at two points and are _sufficiently_ smooth in between then they behave _arbitrarily_ similarly in between... smooth in between then the values David C. Ullrich Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to. (John Jones, My talk about Godel to the post-grads. in sci.logic.) === Subject: Re: Most radical and stupid thing I ever said. Also the most correct. <18895165.1231244878749.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) > I am a Gentleman, said the Three Mouseketeers in harmony. > Mousieur Pascal at your service! > Je suis M. Desaurges, mon plesieur! > M. Brianchon, ici; nous avons trois deomonstrations > de la theorem dernier du Fermat, simulatamousement > dans trois columns. > Pascal dit, Gentile-hommes, > nous-sommes proceeder.... I have no idea what you are trying to communicate, but let me try this on you: There is a difference between triviality and nonsense. Triviality is not the same as nonsense. In physics, the smallest possible length is 1 Plancklength. My claim is that anything from 0 to 1 Plancklength is not nonsensical, but trivial. And there is a difference. We attempt to understand that difference on philosophical and analytical grounds, and that is what we do because we are right. Your fear of philosophy is a bit ironic, as mathematics is nothing more than just that. Philosophy. That is what this thread is all about. I have no idea what you are trying to communicate, but if you can explain a little further please....maybe I can understand. === Subject: Re: JSH: Algebraic integer problem, another try at explaining > On the complex plane consider an unknown factorization: > 175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2) If you're wondering why THAT particular quadratic which I keep using > over and over again, it's for historical reasons. Now I also multiply by 7 for historical reasons ... Tombstones of real men of genius: > Archimedes : sphere contained in minimal cylinder > Boltzmann: S = kLogW > Einstein: E=mc^2 William Claude Dukenfield: All things considered, I would rather be in Philadelphia. > Harris: 7*(175x^2 - 15x + 2) = 7*(5c_1(x) + 1)(5c_2(x)+ 2) -- Michael Press === Subject: Re: JSH: Algebraic integer problem, another try at explaining > Who to you sounds like the scientist, and who like people fighting to > hold on to dogma? Not much of a fight. You want a fight, I will let loose my dogma. -- Michael Press === Subject: Re: JSH: Algebraic integer problem, another try at explaining > Math people fighting me on this issue are always coming to one > conclusion: nothing to see here, not important, don't worry about it, > just go with what you were taught, trust the textbooks. The I will just move along. -- Michael Press === Subject: Re: JSH: Algebraic integer problem, another try at explaining posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) I find it hard to believe that even physicists would not get very >interested in this result if they believed it true, even though it >seems on the surface to only matter about some abstract number theory, >so here's yet another try at explaining this esoteric math issue that >takes out Galois Theory. James, could you please try to explain this: In the integers let: 2 * f(x) = (x + 1)(x + 2) At x = 0 we have: 2 * f(0) = 1 * 2 and the 2 is in the second term. At x = 1 we have: 2 * f(1) = 2 * 3 and the 2 is in the first term. At x = 2 we have: 2 * f(2) = 3 * 4 and the 2 is in the second term. At x = 3 we have: 2 * f(3) = 4 * 5 and the 2 is in the first term. In this example the 2 can appear either in the first term or in the > second term on the right hand side depending on the value of x. How does that happen? am putting forward here. rossum JSH responded completely predictably to this. He knows that f(x) = (x + 1)*(x + 2)/2 = x^2/2 + 3x/2 + 1 i.e., its definition requires division by 2. This is in contrast to his favorite, 7*P(x) = 7*(175x^2 - 15x + 2), where the function P(x) has only integer coefficients. even though your f(x) takes on only integer values for integer x. The fact that the factor 2 in your example jumps around like a kangaroo between (x + 1) and (x + 2) signifies nothing to him, even though in 7*(175x^2 - 15x + 2) = (5a_1(x) + 7)*(5a_2(x) + 7), the 7 splits up in various ways that jump around like a kangaroo, or more like a hyperactive schizophrenic, for various values of x, in the algebraic-integer factorization. JSH doesn't care that you have found something analogous in the integers. He nails you here not because there are essential differences, but because there are differences in ***appearance***. It is worth noting that the similarities are more than superficial. In your example, you can define 2 = c1(x) * c2(x), where c1(x) = GCD(2, x + 1) and c2(x) = GCD(2, x + 2), which results in c1(x) = 1 if x is even and 2 if odd, and c2(x) = 2 if x is even and 1 if x is odd, and (x + 1)/c1(x) and (x + 2)/c2(x) are both always integers for x an integer. The factorization of 2 is dependent on x. In the Harris example, one defines 7 = c1(x) * c2(x), where c1(x) = GCD(a_1(x), 7) and c2(x) = GCD(a_2(x), 7). and just as with your example, (5a_1(x) + 7)/c1(x) and (5a_2(x) + 7)/c2(x) are both always algebraic integers for x any integer. Again, the factorization of 7 is dependent on x. Harris's central error is that he refuses to consider this possibility. Marcus. === Subject: Re: JSH: Algebraic integer problem, another try at explaining > The mystery should resolve with a leap for some of you that of course > the one thing that must be invalid on the ring of algebraic integers, > as all the steps look like ok algebra is the start!!! How sure of this are you? You only used three exclamation points. If you had used seven I might be more inclined to take you at your word. -- Michael Press === Subject: Re: JSH: Algebraic integer problem, another try at explaining I find it hard to believe that even physicists would not get very >interested in this result if they believed it true, even though it >seems on the surface to only matter about some abstract number theory, >so here's yet another try at explaining this esoteric math issue that >takes out Galois Theory. >James, could you please try to explain this: >In the integers let: > 2 * f(x) = (x + 1)(x + 2) >At x = 0 we have: > 2 * f(0) = 1 * 2 and the 2 is in the second term. >At x = 1 we have: > 2 * f(1) = 2 * 3 and the 2 is in the first term. >At x = 2 we have: > 2 * f(2) = 3 * 4 and the 2 is in the second term. >At x = 3 we have: > 2 * f(3) = 4 * 5 and the 2 is in the first term. >In this example the 2 can appear either in the first term or in the >second term on the right hand side depending on the value of x. >How does that happen? >am putting forward here. >rossum > Post f(x). You read as badly as you do math. F(x) is defined above. Constrast with: 175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2) So the explanation is simple: you can't express f(x) in general in the > integers. His f(x) is an integer function. You do not seem to be able to read so the point of his demonstration was totally lost on you. You really should learn some math. I suggest though, that you test your intellect, and try. I dare you. It is funny to see you start to throw a tantrum. Hint: real mathematicians do not use dares to get proofs. That is the sign of a spoiled child. Post f(x). === Subject: Re: JSH: Algebraic integer problem, another try at explaining >I find it hard to believe that even physicists would not get very >interested in this result if they believed it true, even though it >seems on the surface to only matter about some abstract number theory, >so here's yet another try at explaining this esoteric math issue that >takes out Galois Theory. >James, could you please try to explain this: >In the integers let: > 2 * f(x) = (x + 1)(x + 2) >At x = 0 we have: > 2 * f(0) = 1 * 2 and the 2 is in the second term. >At x = 1 we have: > 2 * f(1) = 2 * 3 and the 2 is in the first term. >At x = 2 we have: > 2 * f(2) = 3 * 4 and the 2 is in the second term. >At x = 3 we have: > 2 * f(3) = 4 * 5 and the 2 is in the first term. >In this example the 2 can appear either in the first term or in the >second term on the right hand side depending on the value of x. >How does that happen? >am putting forward here. >rossum > Post f(x). > You read as badly as you do math. F(x) is defined above. > Constrast with: 175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2) > So the explanation is simple: you can't express f(x) in general in the > integers. > His f(x) is an integer function. You do not seem to be able to read > so the point of his demonstration was totally lost on you. You really > should learn some math. You're an idiot. The correct answer given 2 * f(x) = (x + 1)(x + 2), >is to solve for f(x): f(x) = (x+1)(x+2)/2 Agreed. and then substitute x = 2y, to get (don't let one quick move fool >you): No James. Remember at the beginning I stated that we were working in the integers so you are only allowed to use the integers. You cannot substitute x = 2y for *all* integers x if y is also in the integers. For example if x = 3 then what is the *integer* value of y? You are better off using the following: for x even use x = 2y for x odd use x = 2y + 1 That way all integers x are covered and y is always an integer. Remember we are working in the integers here James - nothing else allowed. f(y) = (2y+1)(y). Already corrected elsethread to: f(2y) = (2y + 1)(y) THAT is the solution in integers. Trivial. You people are math >infants. THAT is *half* of the solution in the integers. Not as trivial as you thought it was. So, looking at the case of x odd, we have x = 2y + 1 Subststute this in the original equation: 2f(2y + 1) = (2y + 2)(2y + 3) dividing out the two from the FIRST term this time gives: f(2y + 1) = (y + 1)(2y + 3) THAT is the other half of the solution in the integers. Trivial, but you missed it James. rossum === Subject: Re: JSH: Algebraic integer problem, another try at explaining posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > The correct answer given 2 * f(x) = (x + 1)(x + 2), > is to solve for f(x): f(x) = (x+1)(x+2)/2 and then substitute x = 2y, to get (don't let one quick move fool > you): f(y) = (2y+1)(y). That should be f(2y) = (2y+1)(y+1) Increment the OOPS counter. - William Hughes === Subject: Re: JSH: Algebraic integer problem, another try at explaining posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > The correct answer given 2 * f(x) = (x + 1)(x + 2), > is to solve for f(x): f(x) = (x+1)(x+2)/2 and then substitute x = 2y, to get (don't let one quick move fool > you): f(y) = (2y+1)(y). That should be f(2y) = (2y+1)(y+1) Increment the OOPS counter. - William Hughes They're equivalent Put y=2 in f(y) = (2y+1)(y) to get f(2)=(2*2+1)*2=10 put y=1 in f(2y)= (2y+1)(y+1) to get f(2)=(2*1 +1)(1+1) = 6 put x=2 in f(x) = (x+1)(x+2)/2 to get f(2) = (2+1)*(2+2)/2 = 6 - William Hughes === Subject: Re: JSH: Algebraic integer problem, another try at explaining f(x) = (2x+1)(x+1) >See? I can admit when I'm wrong. Why can't you? problem is that you, JSH are ALWAYS WRONG. we can count on that. === Subject: Re: JSH: Algebraic integer problem, another try at explaining posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > The correct answer given 2 * f(x) = (x + 1)(x + 2), > is to solve for f(x): f(x) = (x+1)(x+2)/2 and then substitute x = 2y, to get (don't let one quick move fool > you): f(y) = (2y+1)(y). That should be f(2y) = (2y+1)(y+1) Increment the OOPS counter. - William Hughes They're equivalent Put y=2 in f(y) = (2y+1)(y) to get f(2)=(2*2+1)*2=10 put y=1 in f(2y)= (2y+1)(y+1) to get f(2)=(2*1 +1)(1+1) = 6 put x=2 in f(x) = (x+1)(x+2)/2 to get f(2) = (2+1)*(2+2)/2 = 6 - William Hughes Damn. You're right. So the correct answer to the trivial problem is: f(x) = (2x+1)(x+1) put x=2 in f(x) = (2x+1)(x+1) we get f(2) = (2*2 +1)(2+1) = 15 (Hint, f(x) is not a polynomial with integer coefficients.) Probably what you want is f(x) = (2y+1)(y+1), y = x/2 However, y is only an integer for even x. (For the case of odd x, see rossum's post) - William Hughes === Subject: Re: JSH: Algebraic integer problem, another try at explaining I find it hard to believe that even physicists would not get very >interested in this result if they believed it true, even though it >seems on the surface to only matter about some abstract number theory, >so here's yet another try at explaining this esoteric math issue that >takes out Galois Theory. >James, could you please try to explain this: >In the integers let: > 2 * f(x) = (x + 1)(x + 2) >At x = 0 we have: > 2 * f(0) = 1 * 2 and the 2 is in the second term. >At x = 1 we have: > 2 * f(1) = 2 * 3 and the 2 is in the first term. >At x = 2 we have: > 2 * f(2) = 3 * 4 and the 2 is in the second term. >At x = 3 we have: > 2 * f(3) = 4 * 5 and the 2 is in the first term. >In this example the 2 can appear either in the first term or in the >second term on the right hand side depending on the value of x. >How does that happen? >am putting forward here. >rossum >Post f(x). >You read as badly as you do math. F(x) is defined above. >Constrast with: 175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2) >So the explanation is simple: you can't express f(x) in general in the >integers. >His f(x) is an integer function. You do not seem to be able to read >so the point of his demonstration was totally lost on you. You really >should learn some math. > You're an idiot. This from the fool who has been fighting reality for years. You have done nothing correct but you persist in your delusions. Professional treatment might help you. Studying math might also help The correct answer given 2 * f(x) = (x + 1)(x + 2), > is to solve for f(x): f(x) = (x+1)(x+2)/2 Wow. James can divide by two!!!!!! This is after not realizing that f(x) was defined above. and then substitute x = 2y, to get (don't let one quick move fool > you): f(y) = (2y+1)(y). James pulls another stupid move to try to hide his ignorance of math. You are a pretty sad person to have to live with your delusions. THAT is the solution in integers. Trivial. You people are math > infants. No, you are the math fool. You keep trying to find some imaginary problem in math but only show yourself to be monumentally stupid. Keep it up though as we enjoy laughing at you flailing and failing. === Subject: Re: factor phone numbers by hand calc? EASY. Fermat or trial. > don.lo...@paradise.net.nz said: > pc fc dw mcm nzmm, > ttfn Pardon.... what's that? wyw -- [mail]: Chuck F (cbfalconer at maineline dot net) [page]: Try the download section. === Subject: Re: zero knowledge protocols. An early protocol to flip a coin by mail. Ralph generates two large secret primes p and q such that p.q > almost certainly cannot be factored within the time to > execute the protocol, and tells Sarah pq. Sarah picks a secret number x and tells Ralph xx modulo pq. Ralph generates the square roots of xx modulo pq, picks > one, say y, and tells Sarah. If Ralph tells Sarah x or -x, then he wins the toss. > Otherwise he loses. If Sarah tells Ralph that Ralph lost the toss, then she > must tell Ralph x and it must not be the case that x = y > or x = -y modulo pq. I came up with a similar one once, that involves simpler computation, > and less exchanges of information. However, it is based on a problem > that, as far as I know, has not been proven to be hard. (In fact, I've > not heard of anyone even studying the problem). 1. Ralph generates two large primes, p and q, with p = 1 mod 4, and q = > 3 mod 4. Ralph gives N=pq to Sarah. 2. Sarah guesses whether p < q or q < p. She tells Ralph her guess. 3. Ralph reveals p and q. Sarah wins if her guess as to the order of p > and q was correct. Sarah can verify that Ralph revealed the correct p > and q by making sure p and q are prime and pq does equal N. The security of this is based on this problem: Given an integer N = 3 mod 4, where N is a product of two primes, > determine if the smaller of the two primes is equals 1 mod 4. I have no idea how hard this problem is. Then it is not as secure as the square root protocol. > Here's another coin flip protocol that is even less work: 1. Ralph generates two unequal numbers, A and B, and sends to Sarah > h(A) and h(B), where h is a cryptographically secure hash function > agreed upon in advance. 2. Sarah guess whether A < B or B < A. 3. Ralph reveals A and B. Sarah wins if she guessed right in #2. This one, I suspect, can be proven to be as secure as your hash function. hagman suggests that Ralph find A and A' such that h(A) = h(A'), then transmit a B that is between A and A'. I am entirely unpracticed in these matters. The square root protocol is more clearly fair to my eye than the two you propose. I also prefer heapsort because it is far easier to code than quicksort even though it runs twice as long on average; quicksort requires some engineering to guard against O(n^2) cases where heapsort is immune. Therefore my predispositions are of little weight. -- Michael Press === Subject: Re: bounded derivative ==> bounded function <8981945.1231692247321.JavaMail.jakarta@nitrogen.mathforum.org>, > let f is function with bounded derivative in (a,b). > prove that f is bounded in (a,b). > I know that bounded derivative in (a,b) implies that > f is uniformly continuous in (a,b), can it help ? Look at the mean value theorem. The mean value theorem is the bees knees. Took me a long time to figure out that it is the bridge between the differential calculus and everyday intuitive consequences of the DC. -- Michael Press === Subject: Re: radius of the universe > The usual analogy to describe the geometry of spacetime > is an expanding balloon. The surface is 2-D, closed, warped > in a 3rd dimension, inaccessible to the balloonists. However, > the ballon has a center - in the 3rd dimension. > Now, extend this picture to our expanding 3-D universe; > can we compute the 'radius', the distance to the center, > in the 4th space dimension? Analogous to the balloon > model, it should be the same for all observers. > And that would educe a circumference, would it not? > No Center > http://www.astro.ucla.edu/~wright/nocenter.html > http://www.astro.ucla.edu/~wright/infpoint.html This elaborates that we can locate no center in our > observable 3-D universe. It tells nothing about whether > a center exists in some 4th spatial dimension, in which > our universe is embedded. A 4th uncurled-up spacial dimension would render orbits unstable. Not only that, I could not keep my shoe laces tied. -- Michael Press === Subject: Re: I am greater than Jesus > Two thousand years ago Jesus got enough people to listen to him that a > church was started that survives today. I, on the other hand, > encounter only apathy or hostility. Yet my message is much more > important than his: I say things that actually are important, while > Jesus said mainly nonsense. > Oh greater thou than Jesus, for thee we give a bigger cross and meaner nails and modern medical aid to prolong your moment of greatness. I'm sure you mother will gladly give blood. > My beliefs are my religion, just as for Jesus and so many other > religious leaders. I have no need to take anyone else's beliefs > because I know better (and I also know what I don't know). God has > made me the most important person of my time but Satan is winning! Andrew Usher > === Subject: Re: I am greater than Jesus posting-account=n8pQvgkAAAAR2icJiQ21GsmYtflsTgld Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > My beliefs are my religion, just as for Jesus and so many other > religious leaders. I have no need to take anyone else's beliefs > because I know better (and I also know what I don't know). God has > made me the most important person of my time but Satan is winning! God made you double-post. DB === Subject: Re: I am greater than Jesus posting-account=504E-QkAAAA2v90r8nGnJKpfySa_yBSU 5.1),gzip(gfe),gzip(gfe) Dumb question I realize, but what connection has this crap with physics? After all, who the really cares about what some nutcase believes? Still, unfortunately that's what most of today's posts on sci.physics consist. Oddly enough, a few of the remaining sane readers of this newsgroup are here to discuss science, and physics in particular. Such threads today are rare indeed! Harry C. === Subject: Re: I am greater than Jesus posting-account=p0JNqwkAAAChY16-5zbk2O2xWfBB6K-z Gecko/20061206 Firefox/1.5.0.9,gzip(gfe),gzip(gfe) Two thousand years ago Jesus got enough people to listen to him that a > church was started that survives today. I, on the other hand, > encounter only apathy or hostility. Yet my message is much more > important than his: I say things that actually are important, while > Jesus said mainly nonsense. My beliefs are my religion, just as for Jesus and so many other > religious leaders. I have no need to take anyone else's beliefs > because I know better (and I also know what I don't know). God has > made me the most important person of my time but Satan is winning! Andrew, it's time for you to get help. I know Jesus, Jesus is a good > friend of mine, you're no Jesus No Jesus was a simple man from Palestine that dies around 2000 years ago. So I am not Jesus. > What I don't get is that you are totally unreligious but here you are > saying you are Jesus. It reminds me of Nietzche. He's the guy that > made a career out of saying God is Dead and then on his death bed he > thought he was God. Atheists, what a confused lot they are. I used Jesus as a useful comparison. I don't even know if I am an atheist because it really doesn't matter what's out there if our world doesn't interact with it. And Nietzsche really was crazy I think. Andrew Usher === Subject: Re: I am greater than Jesus posting-account=ktC8lwkAAACziLpj2cVSiK4oEjnF0N3x Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Two thousand years ago Jesus got enough people to listen to him that a > church was started that survives today. I, on the other hand, > encounter only apathy or hostility. Yet my message is much more > important than his: I say things that actually are important, while > Jesus said mainly nonsense. My beliefs are my religion, just as for Jesus and so many other > religious leaders. I have no need to take anyone else's beliefs > because I know better (and I also know what I don't know). God has > made me the most important person of my time but Satan is winning! Andrew, it's time for you to get help. I know Jesus, Jesus is a good > friend of mine, you're no Jesus No Jesus was a simple man from Palestine that dies around 2000 years > ago. So I am not Jesus. What I don't get is that you are totally unreligious but here you are > saying you are Jesus. It reminds me of Nietzche. He's the guy that > made a career out of saying God is Dead and then on his death bed he > thought he was God. Atheists, what a confused lot they are. I used Jesus as a useful comparison. I don't even know if I am an > atheist because it really doesn't matter what's out there if our world > doesn't interact with it. And Nietzsche really was crazy I think. I read his biography and kind of identified with him though I have never been an atheist. He was probably suffering from dementia toward the end plus he had a very difficult life. Those who do have tough lives do get closer to God...don't you think Andrew? Tom > Andrew Usher === Subject: Re: I am greater than Jesus posting-account=p0JNqwkAAAChY16-5zbk2O2xWfBB6K-z Gecko/20061206 Firefox/1.5.0.9,gzip(gfe),gzip(gfe) > I read his [Nietzsche's] biography and kind of identified with him though I have > never been an atheist. He was probably suffering from dementia toward > the end plus he had a very difficult life. Those who do have tough > lives do get closer to God...don't you think Andrew? Yes, I suppose it's often true. But guys like me can only get angry because I don't know how to get closer to God. I wish I did but all I know is this world and trying to make a difference here. And that's why I need people to join my wiki as you know - I''ve given all the arguments before. Andrew Usher === Subject: Re: I am greater than Jesus posting-account=ktC8lwkAAACziLpj2cVSiK4oEjnF0N3x Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) I read his [Nietzsche's] biography and kind of identified with him though I have > never been an atheist. He was probably suffering from dementia toward > the end plus he had a very difficult life. Those who do have tough > lives do get closer to God...don't you think Andrew? Yes, I suppose it's often true. But guys like me can only get angry > because I don't know how to get closer to God. I wish I did but all > I know is this world and trying to make a difference here. And that's > why I need people to join my wiki as you know - I''ve given all the > arguments before. I hear you and God does too. If you listen carefully you will hear God too. He loves it when one get it. Smitty > Andrew Usher === Subject: Re: MINIMAL POLYNOMIALS FOR TANGENTS It's all true! Proposition 2 on page 5 states the consolidated result, and the proof given contains the point I was missing (it's steeped in Galois theory, so brace yourself if you're not familiar with that). I hope to rephrase it in more mundane terms, and publish the final result as stated before. === Subject: Re: MINIMAL POLYNOMIALS FOR TANGENTS The assertion was that the algebraic degrees of these values are phi(n) unless n is a multiple of 4 (in which case it is [phi(n)]/2). Polynomials T(sub n) of degree phi(n) have been constructed which have these tangents as their root sets. By restricting the angles to the interval (0, pi), we see they are uniquely represented, and clearly there are phi(n) distinct such values. When n is a multiple of 4, the tangents can be partitioned to form two sets (according to whether k is congruent to 1 or to 3 (mod 4)), each containing half of them, and these have been shown to be the root sets of two irreducible polynomials over the integers whose product is T(sub n). We also have an inductive way to construct these factors; meaning we actually have access to the minimal polynomials in this case. The only remaining point to be rigorously established (yes, I am convinced it's true) is that T(sub n) is irreducible if n is not a multiple of 4. It will suffice to show T(sub n) is irreducible if n is odd, because in this case irreducibility of T(sub 2n) is an elementary consequence. I do have some nice closed forms and recursive relationships for these T(sub n) (and the polynomials themselves for n<128), but so far none of these have produced inspirations that lead to irreducibility. I am of the opinion that it is a peculiar property of the tangent function, rather than some Galois-theoretic property of polynomials in general, that will be the key. I am also convinced that this nagging detail may be the reason the problem may in fact be unsolved (meaning literature searches have not revealed any published solution). If you have any ideas, please let me know. You may post here or send message to my college email which is williamh(at)wcjc(dot)edu. Assuming we get it all done, I will publish the full paper on my website (with URL given in an announcement on this string). Oh, yeah: T(sub n)(x) = [1+x^2]^{[phi(n)]/2}* C(sub n)([1-x^2]/[1+x^2]), where C(sub n) is the minimal polynomial for cos(2k*pi/n). === Subject: Re: algebra: beginners question regarding factoring complete terms. <20090112071528.F7743@agora.rdrop.com> <496b68dd$0$5390$e4fe514c@dreader21.news.xs4all.nl> To be honest, I guess not. And that is because I'm not a native enlgish > speaker. Which means my english vocabulary is limited to everyday > 'chitchat' english but lacks when it comes to for example math or other > specialist activities. But I'm learning.. !!! Don't apologize for your English. It's better than a lot of the drivel > we see from the native speakers, and for a second language, I would > say it's excellent. > I concur. His English did not indicated to me that English is not his first language. Should I have taken clue from his name? Often it's apparent that the difficult English is because it's not the writer's first language. In such cases I try to help the poster both with the math and with the math terminology and without any grump. Grumps I reserve for those who know better and yet prefer deliberately remedial writing. As there's a large volume of blatantly illiterate writers, I tend to be too practice at perfecting grumps and at accumulating a choice collection of grumplets. As it is, it appears he's getting some bad habits from American brats who are in desperate need of taking English as a _first_ language. Yet he is to be commended for his self learning, even if it's from second rate citizens. I have some experience helping people to translate technical material into English and am willing to help if I can, those needing help expressing math in English. BTW, proper names, such as names of countries and their derivatives are capitalized. For example, England and English. Riddle of the day. What now will I do, now that the greatest inspiration of my life for satire and irony has left the White House? === Subject: non-integer winding number posting-account=NZALAQgAAAC1WZDjR6i1CsTqsen37btL SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) On complex analysis the argument principle is applicable only when the order of zero point or pole is integer. I tried to extend the argument principle to the case when the order or multiplicity of zero point or pole is not integer in the following site; http://hecoaustralia.fortunecity.com/argument/argument.htm === Subject: Re: non-integer winding number > On complex analysis the argument principle is applicable only when the > order of zero point or pole is integer. I tried to extend the argument > principle to the case when the order or multiplicity of zero point or > pole is not integer in the following site; http://hecoaustralia.fortunecity.com/argument/argument.htm The problem is that functions like f(z) = z^(2.3) are not analytic in a (punctured) neighborhood of a ... This is not called a zero of order 2.3 but rather a branch point. === Subject: Re: Non-integrability (Riemann) of f_xy(x,y). Additional Comment. I guess I should have also stated that f is real-valued. === Subject: Re: Product of pairwise differences between n different numbers (Pigeonhole Principle?) <5at9m4tehqmorrvsf7euebbdvm4fm87pn0@4ax.com> posting-account=RapriAoAAAAkZmTTP4ys2MiqlJPqfJgA Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) --------------------------------------------------------------- 1. Proof via Lagrange-related transformation matrices:- WLOG, the a i are distinct, else Pi {i b.87C(a+1,2) + (p-b).87C(a,2) = p.87C(a+1,2) + (p-b).87[C(a,2) - C(a+1,2)] = p.87C(a+1,2) + (p-b).87C(a,1) = p.87C(a+1,2) + ab - ap > p.87C(a+1,2) + ab = # of p appearing in 1!2!...(n-1)! This being true for each prime factor p, therefore .a6 is a multiple of 1!2!...(n-1)!. ### === Subject: Re: causion: dont read posting-account=W967OgoAAAASYsvrgrqxW7fN8WeVhaXp cafe8),gzip(gfe),gzip(gfe) > pay attention to the zeta function defined in wiki, not to the > continueation but the classical part where near the ambiguous line re > (s)=>1 not even real, contrary to many person's imaging the value is > infinit. I mistake. But the definition of zeta is confusing me. === Subject: Re: statistics question I would rather use the variance of the random vector (x,y), that is the mean of the distances of the various tips from the mean of these tips. H === Subject: What is a functional Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) I have seen many examples with vector spaces that mean little to me (an engineer). eg f(x)=x^+3x+2 is a function now if x is in turn a function of say z f(x(z))=x(z)^2 +3x(z)+2 is this a functional - or can you give me a simple example. H. === Subject: Re: What is a functional I have seen many examples with vector spaces that mean little to me > (an engineer). ... can you give me a simple example. Integral from a to b of f(x) dx is an example. If we just consider real f's for which the integral exists, the definite integral is a map from a set of functions to the real numbers and thus is a functional. -- But you see, I can believe a thing without understanding it. It's all a matter of training. --Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_ === Subject: Re: What is a functional > I have seen many examples with vector spaces that mean little to me > (an engineer). eg f(x)=x^+3x+2 is a function now if x is in turn a function of say z f(x(z))=x(z)^2 +3x(z)+2 is this a functional - or can you give me a simple example. > H. > If functional means a function whose domain is a set of functions, then yes, this is a functional. === === Subject: sum ln(1+1/n^2) How do find the sum of the serie sum ln(1+1/n^2) i.e. lim N->+infty sum_{n=1}^{N} ln(1+1/n^2) === Subject: Re: sum ln(1+1/n^2) Firefox/1.5.0.12,gzip(gfe),gzip(gfe) How do find the sum of the serie sum ln(1+1/n^2) > i.e. lim N->+infty sum_{n=1}^{N} ln(1+1/n^2) It's the logarithm of the infinite product product_{n=1}^infinity (1 + 1/n^2). One can compute this by evaluating the classical sine product sin(pi z) = pi z product_{n=1}^infinity (1 - z^2/n^2) at z = i. === Subject: Re: sum ln(1+1/n^2) > How do find the sum of the serie sum ln(1+1/n^2) > i.e. lim N->+infty sum {n=1}^{N} ln(1+1/n^2) It's the logarithm of the infinite product > product {n=1}^infinity (1 + 1/n^2). > One can compute this by evaluating the classical > sine product > sin(pi z) = pi z product {n=1}^infinity (1 - z^2/n^2) > at z = i. Well, we may also use the hyperbolic sine : sinh(x)/x =prod(1+x^2/((pi)^2*i^2) ) x = Pi and sum = ln(sinh(Pi)/Pi) , Alain === Subject: Re: sum ln(1+1/n^2) <130120090751545019%edgar@math.ohio-state.edu.invalid> posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) On 13 jan, 13:51, G. A. Edgar One can compute this by evaluating the classical > sine product > sin(pi z) = pi z product {n=1}^infinity (1 - z^2/n^2) > at z = i. Well, we may also use the hyperbolic sine : > sinh(x)/x =prod(1+x^2/((pi)^2*i^2) ) > x = Pi and sum = ln(sinh(Pi)/Pi) , Alain That's what he said: sin(i*x) = i*sinh(x) so the two functions are > really the same thing (just rotated in the complex plane). -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Bonjour G. A. Edgar , Of course these two ways are 'equivalent' and give the same result. It is just a matter of strictly remaining in R to calculate a real value, Alain === Subject: Dense images of Z^2 in R posting-account=JHgY7AoAAAB5Z10Fvw_ZmGANZtaQQoiP Gecko/20080410 SUSE/2.0.0.14-0.1 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in R. Now I wonder: What is known about necessary and sufficient conditions for the image of a function f : Z^2->R to be dense in R? If nothing can be said about general functions, what about some special function classes like injective functions? Rolf Z: Integers R: Real numbers === Subject: Re: Dense images of Z^2 in R > I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? > If D is dense subset X, f:X -> Y continuous surjection, then f(D) dense subset Y. D dense subset X when cl D = X. Exercise. D dense subset R iff D dense in R. Exercise. If multi-point S is a dense linear order, then D dense subset S iff D dense in S In general dense subset and dense in are not equivalent. === Subject: Re: Dense images of Z^2 in R <20090113052721.V10469@agora.rdrop.com> posting-account=JHgY7AoAAAB5Z10Fvw_ZmGANZtaQQoiP Gecko/2008120122 Firefox/3.0.5 Ubiquity/0.1.4,gzip(gfe),gzip(gfe) > I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? If D is dense subset X, f:X -> Y continuous surjection, > then f(D) dense subset Y. D dense subset X when cl D = X. Exercise. D dense subset R iff D dense in R. > Exercise. If multi-point S is a dense linear order, then > D dense subset S iff D dense in S In general dense subset and dense in are not equivalent. when is the image of f : Z^2 -> R a dense subset of R? === Subject: Re: Dense images of Z^2 in R posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? If D is dense subset X, f:X -> Y continuous surjection, > then f(D) dense subset Y. D dense subset X when cl D = X. Exercise. D dense subset R iff D dense in R. > Exercise. If multi-point S is a dense linear order, then > D dense subset S iff D dense in S In general dense subset and dense in are not equivalent. when is the image of f : Z^2 -> R a dense subset of R? As David said, at that level of generality you'll not get anything interesting: just that such a function has dense image it if has a dense image. -- m === Subject: Re: Dense images of Z^2 in R > I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? Rolf Z: Integers > R: Real numbers If f is an additive function (as in the example f(n,m) = n+sqrt(2)*m ), then the condition is that f(1,0) and f(0,1) be incommensurable. === Subject: Re: Dense images of Z^2 in R <130120090754213855%anniel@nym.alias.net.invalid> posting-account=JHgY7AoAAAB5Z10Fvw_ZmGANZtaQQoiP Gecko/2008120122 Firefox/3.0.5 Ubiquity/0.1.4,gzip(gfe),gzip(gfe) I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? Rolf Z: Integers > R: Real numbers If f is an additive function (as in the example > f(n,m) = n+sqrt(2)*m ), then the condition is that f(1,0) and f(0,1) > be incommensurable. interested in. === Subject: Re: Returning Fire: Vector Calculus posting-account=U4daugkAAABoPdCIBNuqtBcdKEss-z_p Gecko/2008121622 Ubuntu/8.10 (intrepid) Firefox/3.0.5,gzip(gfe),gzip(gfe) > A shot is fired and 3 microphones intercept the conical shock wave. With > electronics the arrival time differences between microphones is obtained. > between them, the location of both the source and destination of the > projectile is determined. You are still wrong, and trivially wrong at that. It is not even necessary to look at your solution to show it. Ian === Subject: Re: Elementary Derivation of Formula for Pi Here's a sketch of an elementary argument for an infinite product > expansion of pi. The details can be found at > http://www.ep.liu.se/ea/lsm/2005/002/lsm05002.pdf [neat sketch snipped.] This is known as Wallis' formula. Who knows how Wallis figured it out. In D J Struik, 'A Source Book in Mathematics, 1200-1800', Harvard UP, 1969 there is an English translation with notes. -- But you see, I can believe a thing without understanding it. It's all a matter of training. --Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_ === Subject: Partial Differential Equations posting-account=bIzc2QoAAAAijBTuofCklemWbVRps_8t Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; .NET CLR 2.0.50727; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) i am a civil engineer. I want to start learning PDEs theoretically. I would like to know which books you would recommend for someone like me, with only calculus (two variables), matrix algebra, analytic geometry, and ODEs as background. I want to know which path to fallow, Victor === Subject: Re: Partial Differential Equations > i am a civil engineer. I want to start learning PDEs theoretically. I > would like to know which books you would recommend for someone like > me, with only calculus (two variables), matrix algebra, analytic > geometry, and ODEs as background. I want to know which path to fallow, which hasn't had many replies. -- But you see, I can believe a thing without understanding it. It's all a matter of training. --Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_ === Subject: Re: Well ordering of reals Nntp-Posting-Host: hera.cwi.nl ... > Given a finite subset of R, we can certainly construct (many) well > orderings for it. We can even do it for many infinite subsets. > > This, however, is a ->very<- far cry from constructing well orderings > for ALL the finite subsets of R, which implies that you can do it > for all finite subsets simultaneously. I must be missing something. For every finite subset of R, the standard ordering forms a well-ordering. So do we not in a sense have a well-ordering for *all* fonite subsets of R? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Well ordering of reals > I must be missing something. For every finite subset of R, the > standard ordering forms a well-ordering. So do we not in a sense > have a well-ordering for *all* fonite subsets of R? Yes. Any linearly ordered finite set is well-ordered. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A MOD B NO MATH HAHAHAHA !! Nntp-Posting-Host: hera.cwi.nl > I have also never seen a keyboard with that symbol [.85, division sign, > U=+00F7]. On the computer where we used it it was formed by > overprinting a colon and a minus sign. BTW, the computer language > was Algol 60. > > Not APL? Surely as a mathematician close to my age, you > would have had the pleasure and heartache of using APL at > some time in the 1970s? Nope, never used APL. By the time I could have gotten access to APL I did already use Algol 68. Moreover, nowhere in my neighbourhood have I ever seen a keyboard that did show the APL graphics. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: A MOD B NO MATH HAHAHAHA !! > > I have also never seen a keyboard with that symbol [.85, division sign, > > U=+00F7]. On the computer where we used it it was formed by > > overprinting a colon and a minus sign. BTW, the computer language > > was Algol 60. > > Not APL? Surely as a mathematician close to my age, you > > would have had the pleasure and heartache of using APL at > > some time in the 1970s? Nope, never used APL. By the time I could have gotten access to APL I did > already use Algol 68. Moreover, nowhere in my neighbourhood have I ever > seen a keyboard that did show the APL graphics. It's sad that you have missed Another Programming Language :-) Han de Bruijn === Subject: Re: A MOD B NO MATH HAHAHAHA !! <4568100.1231630716322.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > <4568100.1231630716322.JavaMail.jaka...@nitrogen.mathforum.org>, > you got it wrong too ! > it was about mod 1 !! > sigh ... > that doesnt match your interpretation ! > Do you really mean mod 1? > For the mod operator , n mod 1 = 0 for any integer n, > and for the mod relation, n == 0 (mod 1) for any integer n. > Which means that mod 1 is of damn all use. By mod 1, Virgil is referring to the ring Z/Z, which is obviously isomorphic to the ring 1 = 0 (i.e., the trivial ring with only a single element). But when Plouffe refers to mod 1, he doesn't mean Z/Z, but R/Z. And R/Z is not the trivial ring. === Subject: Re: A MOD B NO MATH HAHAHAHA !! <4568100.1231630716322.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080614i-0etch1),gzip(gfe),gzip(gfe) <4568100.1231630716322.JavaMail.jaka...@nitrogen.mathforum.org>, > you got it wrong too ! > it was about mod 1 !! > sigh ... > that doesnt match your interpretation ! > Do you really mean mod 1? > For the mod operator , n mod 1 = 0 for any integer n, > and for the mod relation, n == 0 (mod 1) for any integer n. > Which means that mod 1 is of damn all use. By mod 1, Virgil is referring to the ring Z/Z, which > is obviously isomorphic to the ring 1 = 0 (i.e., the > trivial ring with only a single element). But when Plouffe refers to mod 1, he doesn't mean > Z/Z, but R/Z. And R/Z is _not_ the trivial ring. I already know that my own thinking is askance to your thinking here. But I have another means by which the mod 1 system is of importance. Upon generalizing sign the real number takes the form s x where s is sign and x is magnitude. For the real number there are two signs and the modulo arithmetic takes its place in the arithmetic product of two values ( s1 x1 )( s2 x2 ) whose resultant is (s1 + s2) x1 x2 where this sign summation is mod 2. Clearly when we step downward to the one-signed numbers we simply have one sign type with few dynamics. However, since this sign type is tied to a magnitude arithmetic can still be performed. These one-signed numbers carry the strongest congruence to time that I know of. They are unidirectional. The special time 'now' is embodied by them. While arithmetic can be performed with them in terms of geometrical dimension they are zero dimensional. This satisfies the physical behaviors of time, whereby no actual degree of freedom is observed; this in contrast with the three dimensions of space whose freedom is demonstrable. To appreciate the dimensional behavior a fuller description is necessary and so stepping up to the three-signed numbers (P3) would be wise. P3 are the complex numbers though their format is foreign to most. The complex numbers follow directly from the identical rules that build the real numbers under the polysign paradigm. They are merely P2(the reals) and P3(the complex numbers) tobe followed by P4, P5, etc. All of these polysign systems obey the commutative and associative behaviors under product and superposition. They are still limited by similar constraints that hampered Hamilton and Grassman and Clifford but they(polysign) are cleaner. The modern mathematics culture has come to obey the real number as fundamental and the polysign construction challenges this. Nearby are the field requirements and there we see a small exception in the division by zero which has been swallowed hook, line, and sinker. Multiplication by zero provides a dimensional collapse which takes a more general form in the high dimensional(P4+) systems of polysign. Such dimensional collapses will not be reversible. I suggest that the field requirements should be relaxed and generalized to have no exception within the division clause and instead observe this as dimensional collapse by specific values which are interesting behaviors and ought not be a cause of rejection. Furthermore these behaviors allow polysign math to claim support for spacetime from pure mathematics. That the modulo behavior is capable of providing a spacetime basis ought to be of interest to both physicists and mathematicians. - Tim === Subject: Re: About t^4-z^100=2^10*k posting-account=fPUT4goAAABJ1knpUmWe7fFdWT7SyGS2 Gecko/2008120122 Firefox/2.0.0.12;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) Also I have question about solutions existence of x^100+y^100=t^4-1 with all >0. === Subject: Re: About t^4-z^100=2^10*k >Also I have question about solutions existence of x^100+y^100=t^4-1 >with all >0. There are always the 2 trivial solutions: (x,y,t) = (0,0,1) (x,y,t) = (0,0,-1) I'm not sure if there are any other solutions. An easy necessary condition is that x,y must both be multiples of 10, and t must be odd but not a multiple of 5. quasi === Subject: Re: collections of well-ordered closed (or else F_sigma) subsets of the unit interval [0, 1] be represented as the well-ordering of a closed subset of [0, 1] , with > the ordering given by <. (Prop. A) That is correct. Proposition. > Every countable linear order can be order embedded into Z[1/2], > hence into Q and [0,1]. Let beta be countable compact ordinal, ie eta is a countable non-limit > ordinal. Let h:beta -> [0,1] be an order embedding of eta. Define f:beta -> [0,1] by transfinite induction. > If eta is successor ordinal in beta. let f(xi) = h(xi). > If eta is limit ordinal, let f(eta) = sup{ f(xi) | xi < eta } Show f is a supremum preserving order embedding. Since beta is compact, it is a complete order. > Thus also h(beta) and f(beta). Being complete orders, h(beta) > and f(beta) are compact within their intrinsic order topologies. The hard part is to apply the theorem that shows > the subspace topology of f(beta) is the same as the > intrinsic order topology for f(beta). Thus f(beta) > is a subspace of [0,1] and being compact, it's closed. [Assuming Prop. A is true below ] > If so, then countable unions of closed subsets of [0, 1] (or F_sigma > sets) could be the order type of any countable ordinal under > inclusion. That is ambiguous. Would you clarify? > Either way I read it, I've a counter example. An F_sigma set is by definition a union of up to countably many closed > sets in some topological space, usually the reals R, R^n, > a compact subset of R^n, etc. We can't get omega as a closed > subset of [0, 1] . But we can define By omega do you mean omega_0, in contrast to omega_1? > B_n = { 1 - 1/k : 1<=k<=n} for n>= 1. > They are closed subsets (finite) of [0, 1], and their countable union, > i.e. union_{n in N^*} B_n is an infinite set, an F_sigma > since each B_n is closed. But union_{n in N^*} B_n is not closed > in [0, 1] , the closure has the extra point 1. If a closed subset A of [0, 1] represents beta + 1 with <, beta an infinite > countable ordinal, [0, alpha) intersect A will sometimes give a closed set > in [0, 1] representing an ordinal <= beta+1. If it's not closed, > I still think the intersection is F_sigma (up to countable union > of closed subsets of [0, 1] ). It seems every ordinal < beta+1 > can be represented by [0, alpha) intersect A, for alpha > in [0, 1], and the sets are F_sigma. So it seems to every > ordinal < beta + 1 there corresponds naturally an F_sigma > subset of A. If beta is a countable infinite limit ordinal, > B subset [0, 1] which is F_sigma can, I think, represent > beta, with intersections with [0, alpha) , alpha in [0, 1] > representing smaller ordinals (or subsets which are F_sigma). > Hm. You're claiming every countable ordinal can be order embedded and/or topologically embedded into [0,1] as a F_sigma set? Since every countable subset of [0,1] is F_sigma, what's the significance? > So the question is: can we find aleph_1 F_sigma subsets > of [0, 1] which, under the inclusion U on its side, > have the order type aleph_1 ? > Yes. Well order [0,1]. [0,1] = { r_xi | xi < omega_1 }. For all eta < omega_1, let R_eta = { r_xi | xi < eta }. For all eta < omega_1, R_eta is F_sigma. For all psi,chi < omega_1, R_psi subset R_chi iff psi <= chi. ---- === Subject: Re: collections of well-ordered closed (or else F_sigma) subsets of the unit interval [0, 1] > I believe any infinite countable ordinal which is not a limit > ordinal can be represented as the well-ordering of a closed subset > of [0, 1] , with the ordering given by <. (Prop. A) > That is correct. Proposition. > Every countable linear order can be order embedded into Z[1/2], > hence into Q and [0,1]. Let beta be countable compact ordinal, ie eta is a countable > non-limit ordinal. Let h:beta -> [0,1] be an order embedding of eta. Define f:beta -> [0,1] by transfinite induction. > If eta is successor ordinal in beta. let f(xi) = h(xi). > If eta is limit ordinal, let f(eta) = sup{ f(xi) | xi < eta } Show f is a supremum preserving order embedding. Since beta is compact, it is a complete order. > Thus also h(beta) and f(beta). Being complete orders, h(beta) > and f(beta) are compact within their intrinsic order topologies. The hard part is to apply the theorem that shows > the subspace topology of f(beta) is the same as the > intrinsic order topology for f(beta). Thus f(beta) > is a subspace of [0,1] and being compact, it's closed. > [Assuming Prop. A is true below ] > If so, then countable unions of closed subsets of [0, 1] (or F_sigma > sets) could be the order type of any countable ordinal under inclusion. > That is ambiguous. Would you clarify? > Either way I read it, I've a counter example. > An F_sigma set is by definition a union of up to countably many closed > sets in some topological space, usually the reals R, R^n, > a compact subset of R^n, etc. We can't get omega as a closed > subset of [0, 1] . But we can define By omega do you mean omega_0, in contrast to omega_1? Yes, omega_0, i.e. the smallest infinite ordinal, omega_0 = {0, 1, 2, 3, .... } in von Neumann's way of defining the ordinals. > B_n = { 1 - 1/k : 1<=k<=n} for n>= 1. They are closed subsets (finite) of [0, 1], and their countable union, > i.e. union_{n in N^*} B_n is an infinite set, an F_sigma > since each B_n is closed. But union_{n in N^*} B_n is not closed > in [0, 1] , the closure has the extra point 1. > If a closed subset A of [0, 1] represents beta + 1 with <, beta an > infinite > countable ordinal, [0, alpha) intersect A will sometimes give a closed > set > in [0, 1] representing an ordinal <= beta+1. If it's not closed, > I still think the intersection is F_sigma (up to countable union > of closed subsets of [0, 1] ). It seems every ordinal < beta+1 > can be represented by [0, alpha) intersect A, for alpha > in [0, 1], and the sets are F_sigma. So it seems to every > ordinal < beta + 1 there corresponds naturally an F_sigma > subset of A. If beta is a countable infinite limit ordinal, > B subset [0, 1] which is F_sigma can, I think, represent > beta, with intersections with [0, alpha) , alpha in [0, 1] > representing smaller ordinals (or subsets which are F_sigma). > Hm. You're claiming every countable ordinal can be order embedded > and/or topologically embedded into [0,1] as a F_sigma set? Since every countable subset of [0,1] is F_sigma, what's the significance? You're right, every countable subset is an F_sigma, when singleton sets {p} are closed, as happens in any metric space W, p a point in W... I missed that basic fact. > So the question is: can we find aleph_1 F_sigma subsets > of [0, 1] which, under the inclusion U on its side, > have the order type aleph_1 ? > Yes. Well order [0,1]. [0,1] = { r_xi | xi < omega_1 }. > For all eta < omega_1, let R_eta = { r_xi | xi < eta }. > For all eta < omega_1, R_eta is F_sigma. > For all psi,chi < omega_1, > R_psi subset R_chi iff psi <= chi. Yes, very good. Your sets R_eta are all F_sigma sets, and 0<=eta <20090112202538.H48240@agora.rdrop.com> posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Since every countable subset of [0,1] is F_sigma, > what's the significance? > You're right, every countable subset is an F_sigma, > when singleton sets {p} are closed, as happens in > any metric space W, p a point in W... Each subset of the reals that is well ordered under the usual ordering for real numbers is also a G_delta set, by the way. The conditions of being countable and of being G_delt, when taken together, impose a fairly strengent requirement on a set (although not enough to force the set to be well ordered). For example, no such set can be dense in any interval (being dense and G_delta in an interval implies uncountability -- this is essentially the Baire category theorem for the interval). Indeed, no such set can be dense in any perfect set, even very thin Cantor-like sets with Hausdorff dimension zero that are far from being an interval (and thus are sets in which far fewer points are needed to be dense in), again being essentially the Baire category theorem (but for the perfect set as a complete metric space this time). TECHNICAL NIT-PICK: Note that not being dense in any interval is equivalent to not being dense in any closed (non-degenerate) interval, so the perfect set version really does imply the interval version. It's also easy to see that the perfect set version is strictly stronger than the interval version -- the endpoints of the complementary intervals of the middle thirds Cantor set form a set that is dense in the middle thirds Cantor set but the set of these endpoints is not dense in any interval. Dave L. Renfro === Subject: Re: collections of well-ordered closed (or else F_sigma) subsets of the unit interval [0, 1] Since every countable subset of [0,1] is F_sigma, > what's the significance? > You're right, every countable subset is an F_sigma, > when singleton sets {p} are closed, as happens in > any metric space W, p a point in W... Each subset of the reals that is well ordered under the > usual ordering for real numbers is also a G_delta set, > by the way. The conditions of being countable and of > being G_delt, when taken together, impose a fairly strengent > requirement on a set (although not enough to force the > set to be well ordered). For example, no such set can be > dense in any interval (being dense and G_delta in an interval > implies uncountability -- this is essentially the Baire > category theorem for the interval). Indeed, no such set > can be dense in any perfect set, even very thin Cantor-like > sets with Hausdorff dimension zero that are far from being > an interval (and thus are sets in which far fewer points > are needed to be dense in), again being essentially the > Baire category theorem (but for the perfect set as a > complete metric space this time). This is interesting. It's hard to visualize what a typical countable G_delta set for [0, 1] looks like ... David Bernier > TECHNICAL NIT-PICK: Note that not being dense in any > interval is equivalent to not being dense in any closed > (non-degenerate) interval, so the perfect set version > really does imply the interval version. It's also easy > to see that the perfect set version is strictly stronger > than the interval version -- the endpoints of the > complementary intervals of the middle thirds Cantor > set form a set that is dense in the middle thirds > Cantor set but the set of these endpoints is not dense > in any interval. Dave L. Renfro === Subject: Solutions manua to Engineering Mechanics - Statics (11th )by R.C.HIBBELER posting-account=jBP0yQoAAADkjbQMT90jR5JPojXnsRCF Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) Solution manual solutions manual solution manual I am a solutions manual collector, I offer solutions manual services Note: all solutions manual in soft copy that mean in Adobe Acrobat Reader (PDF ) format. if you want any book not just solutions just contact with us.81B to get the solution manual you want .81Cplease send message to happyren2008@hotmail.com .81Chappyren2008(at)hotmail.com.81C replace (at) to @ ,please email to me . 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(To search click in keyboard Ctrl+F) Solutions manua to Financial Accounting 6e by horngren Harrison Solutions manual to Advanced Accounting, 9th edition by Hoyle, Schaefer, & Doupnik Solutions manual to Complex Variables with Applications (Pie) by A.David Wunsch Solutions manual to Computer Design Fundamentals4E by Mano and Kime. 4th Solutions manua to Computer Networks Systems Approach 3ed by davie peterson Solutions manual to COMPUTER ORGANIZATION AND ARCHITECTURE Solutions manual to Cost Accounting, 13/e 13e by Horngren SM.zip Solutions manua to Data and Computer Communications, 7th Edition By Stallings Solutions manual to Differential Equations and Linear Algebra by Penney and Edwards, 2nd edition Solutions manual to Elementary Differential Equations and Boundary Value Problems, by Boyce andDiprima Solutions manua to Elements of engineering electromagnetics (6/e) by N.N.RAO Solutions manual to Engineering electromagnetics (7/e) by HAYT Solutions manual to Engineering Fluid Mechanics, 7th, By Clayton T. 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Martin(chapter 1 to chapter15) solution manual for Probability and Statistical Inference ( 7th edition by Hogg & Tanis) solution manual for Fundamentals of Communication Systems by John G. Proakis ,Masoud Salehi solution manual for Materials and Processes in Manufacturing,9thBy Degarmo === Subject: A statement about Rsa2048. Assuming there are only two prime factors of equal length in rsa2048 these two primes below would represent the smallest possible and largest possible prime factors. The smallest factor of 309 digits--- The range -- from 1000000000000000000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000010 4 59 to sqrt(rsa2048) The largest factor of 309 digits--- The range -- from sqrt(rsa2048) too -- 2519590847565789349402718324004839857142928212620403202777713783604366202070 7595556264018525880784406918290641249515082189298559149176184502808489120072 8 4499268739280728777673597141834727026189637501497182469116507761337985909570 0 0973304597488084284017974291006424586918171951187461215151726546322822168673 5 23 Knowing this and assuming the two factors are of equal length, neither one of these two primes could ever represent one of the factors in rsa2048. I have a simple proof. Dan === Subject: Re: A statement about Rsa2048. > Why do you assume that the two factors have the same number of decimal > digits? >Yes OK, if they are both 1024 bits long, then they must >each have 309 >digits. And be strictly within your lower and upper >bounds. Yes! To find these prime pairs of equal length that form these composites, the composites must reside on the same triangle number index as rsa2048 as does the composite resulting from the two factors I show. It is difficult to find these discrete paired prime composites that will only appear within this index. The number of these special composites is probably uncountable within the given index but still hard to find. Dan === Subject: Re: A statement about Rsa2048. posting-account=T7Gd-QoAAACeQajv7mi_Za6uPu3TpBXy AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.43 Safari/525.19,gzip(gfe),gzip(gfe) > Assuming there are only two prime factors of equal length > in rsa2048 these two primes below would represent the > smallest possible and largest possible prime factors. The smallest factor of 309 digits--- > The range -- > from > 100000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000010459 > to sqrt(rsa2048) The largest factor of 309 digits--- > The range -- > from > sqrt(rsa2048) > too -- > 251959084756578934940271832400483985714292821262040320277771378360436620207 075955562640185258807844069182906412495150821892985591491761845028084891200 728449926873928072877767359714183472702618963750149718246911650776133798590 957000973304597488084284017974291006424586918171951187461215151726546322822 168673523 Knowing this and assuming the two factors are > of equal length, neither one of these two primes > could ever represent one of the factors in rsa2048. I have a simple proof. Dan Why do you assume that the two factors have the same number of decimal digits? === Subject: Re: A statement about Rsa2048. posting-account=T7Gd-QoAAACeQajv7mi_Za6uPu3TpBXy AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.43 Safari/525.19,gzip(gfe),gzip(gfe) Assuming there are only two prime factors of equal length > in rsa2048 these two primes below would represent the > smallest possible and largest possible prime factors. The smallest factor of 309 digits--- > The range -- > from > 100000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000010459 > to sqrt(rsa2048) The largest factor of 309 digits--- > The range -- > from > sqrt(rsa2048) > too -- > 251959084756578934940271832400483985714292821262040320277771378360436620207 075955562640185258807844069182906412495150821892985591491761845028084891200 728449926873928072877767359714183472702618963750149718246911650776133798590 957000973304597488084284017974291006424586918171951187461215151726546322822 168673523 Knowing this and assuming the two factors are > of equal length, neither one of these two primes > could ever represent one of the factors in rsa2048. I have a simple proof. Dan Why do you assume that the two factors have the same number of decimal > digits? Yes OK, if they are both 1024 bits long, then they must each have 309 digits. And be strictly within your lower and upper bounds. === Subject: Help needed on combinatorial interpretation posting-account=JYIhQwoAAADfOqRKeWW3k_OkLnLoycin Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) Masters of Math, may I ask for your help to identify the combinatorial significance of this little formula: Sum( v=0..m, binomial(m,v) m! (v+2)^m ) === Subject: Re: Fwd: Planets Gather on May 5 and May 17, 2000 posting-account=plvunQoAAAAsFAHWouUDJEZVb5_cH_gI ---------- Forwarded message ---------- === Subject: Re: Planets Gather on May 5 and May 17, 2000 Hi Marcus, 15,000 flying bombs landed on London in 1945, World War II. (V1 and V2) 15,000 flying bombs then landed on Antwerp in 1945. 15,000 postings, cross posted to sci.chem and three other newsgroups, had the subject line Jeehad needs scientists. The blow by blow of what happened, when I pointed Big Bertha Thing and pulled the trigger, is contained on swnet.sci.astro, instead of sci.chem I have been the primary user of swnet.sci.astro since 1998. Would you honestly want me somewhere else? Everyone should have a home. Tony Lance tonylance@myinternetuk.com I am not interested in liitle bugs, just the gross misconduct sort. > Vice-presidents heads should roll. Don't bother even trying to understand. I don't think you'll ever will. > -- Marcus m9370@abc.se ++++++++++++++++++++++++++++++++++++++++++ > Hi there, > Taking the 2065 AD event as my staring point, I went looking for a > Ring Cycle of Uranus and > and found the one below. === Subject: Re: JSH problem posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) [David C. Ullrich] > I haven't been paying close attention. Sure you want to admit that? James has already warned that you > (academics in general) are legally responsible for reporting his > discoveries (although to whom they must report remains unclear), and > that anything you say on Usenet will be used against you as evidence in > court. I'm puzzled about the supposed problem with the algebraic > integers. That's easy enough: James dislikes them intensely, because he believes > they should behave in such a way instead that /one/ of the broken > foundations of his FLT proof stops being obviously broken even to him. This is something that has always puzzled me. James does not seem to have a > logical understanding of factor coprime divisible by etc.. Obviously > all these terms only make sense in the context of a specific ring, whereas > James seems to believe e.g. that only one of the roots *actually* has 7 as > a factor - i.e. that terms like factor have some absolute meaning (no > doubt based on the visual form of some expression). Given this outlook, how did James ever pick up on the algebraic integers > specifically? Hmm...not sure, but I remember him being incensed > that people secretly knew about algebraic integers > but didn't tell him and left him to twist in the > wind. (Or in fact with any ring? - this all just seems to be beyond > him...) This must have happened years ago, as when I started following he > was already fixated on the algebraic integers. Back in 2001, he had a group called The Algebra > of Factorizations and set up a chat room for > discussion. He had exactly one chat room discussion and, > typical for him, deleted the group. Here's a transcript of that single discussion in > which James demostrates that he knows nothing > about the subject of rings which he had been > simultaneously making his usual blowhard posts > on sci.math on the very subject. (James is using > the pseudonym Flatrings in this discussion as > he had been saying rings must be flat on sci.math). > This was posted here by Wilma Scranton and can be > found in a Google Group search. I think the algebraic integers came later. A transcript of that conversation follows: Wow. -- m === Subject: Re: JSH problem ... > So what is the object ring? You have to ask James ;-) No need to ask James. What it comes down to is that given a quadratic with two, non-integral roots, one of the roots is in the object ring and the other is not. This must be the case for every quadratic. The problem, obviously, but James does not see it, is that such a ring does not exist. That is, it is not possible to make a consistent choice for every pair of roots of the quadratics. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH problem > ... > > So what is the object ring? You have to ask James ;-) No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). Then x = (-p +/- sqrt(p^2 - 4q))/2 . Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. So (-p - sqrt(p^2 - 4q))/2 = q/a . Both roots are algebraic integers, and the extension field Q(a) contains q/a, which is an algebraic integer. So the ring of integers of Q(a) contains the other root, namely q/a . So if x^2 + px + q = 0 has non-rational roots, one of them being a, then the ring of integers in Q(a) also contains a/q, the other root. (x - a)(x - q/a) = x^2 + x*(-q/a - a) + q = x^2 + x*(-qa'/(a*a') -a ) + q = x^2 + x*(-a -a' ) +q = x^2 + px + q . a + a' = -p. So a' = -p - a. So any extension ring of Q containing a must contain the conjugate root a' . (case p, q in Q) Also, if a is an algebraic integer, we may assume p, q in Z and the an extension ring of Z containing a must contain the conjugate root a', which is also an algebraic integer. Mhh.. David Bernier === Subject: Re: JSH problem posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) > ... > > So what is the object ring? You have to ask James ;-) No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). > Then x = (-p +/- sqrt(p^2 - 4q))/2 . Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. So (-p - sqrt(p^2 - 4q))/2 = q/a . > Both roots are algebraic integers, and the extension > field Q(a) contains q/a, which is an algebraic integer. > So the ring of integers of Q(a) contains the other root, > namely q/a . So if x^2 + px + q = 0 has non-rational roots, one > of them being a, then the ring of integers in Q(a) > also contains a/q, the other root. (x - a)(x - q/a) = x^2 + x*(-q/a - a) + q = x^2 + x*(-qa'/(a*a') -a ) + q > = x^2 + x*(-a -a' ) +q = x^2 + px + q . a + a' = -p. So a' = -p - a. So any extension ring of Q containing a > must contain the conjugate root a' . (case p, q in Q) > Also, if a is an algebraic integer, Clearly it is. > we may assume p, q in Z > and the an extension ring of Z containing a must contain > the conjugate root a', which is also an algebraic integer. Mhh.. David Bernier All true I think, but that's not the question. Assume that q = r*s, where (say) r is prime. Then a Harrisian object ring would contain exactly one of a/r or a'/r. In general Harris has said that the object ring would also contain all the algebraic integers. It is not clear as Dik suggests that there is one object ring which contains one of a/r or a'/r for all irreducible quadratics that have r as a factor of the constant term. But it may be that something like A[a/r] is sufficient for the Harrisian oeuvre, where A is the ring of algebraic integers. He has been somewhat noncommittal beyond his defective definition of the object ring. Marcus. === Subject: Re: JSH problem > ... > > So what is the object ring? You have to ask James ;-) > No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. > Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). > Then x = (-p +/- sqrt(p^2 - 4q))/2 . > Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] > then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. > So (-p - sqrt(p^2 - 4q))/2 = q/a . > Both roots are algebraic integers, and the extension > field Q(a) contains q/a, which is an algebraic integer. > So the ring of integers of Q(a) contains the other root, > namely q/a . > So if x^2 + px + q = 0 has non-rational roots, one > of them being a, then the ring of integers in Q(a) > also contains a/q, the other root. > (x - a)(x - q/a) = x^2 + x*(-q/a - a) + q = x^2 + x*(-qa'/(a*a') -a ) + q > = x^2 + x*(-a -a' ) +q = x^2 + px + q . > a + a' = -p. So a' = -p - a. So any extension ring of Q containing a > must contain the conjugate root a' . (case p, q in Q) > Also, if a is an algebraic integer, Clearly it is. we may assume p, q in Z > and the an extension ring of Z containing a must contain > the conjugate root a', which is also an algebraic integer. > Mhh.. > David Bernier All true I think, but that's not the question. Assume that > q = r*s, where (say) r is prime. Then a Harrisian object ring > would contain exactly one of a/r or a'/r. In general Harris has said > that the object ring would also contain all the algebraic > integers. It is not clear as Dik suggests that there is > one object ring which contains one of a/r or a'/r for all > irreducible quadratics that have r as a factor of the constant term. > But it may be that something like A[a/r] is sufficient > for the Harrisian oeuvre, where A is the ring of algebraic > integers. He has been somewhat noncommittal beyond his > defective definition of the object ring. I appreciate that Arturo Magidin and you commented on what I like that, it's perhaps a consequence of the conclusions he wants. I remember there was a lot of discussion about specific examples (algebraic numbers of degree 2 I think) which disproved something James had claimed. Could you elaborate a bit on the counterexamples to James's claims regarding factors or whatever it was in algebraic number theory? David === Subject: Re: JSH problem posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) > ... > > So what is the object ring? You have to ask James ;-) > No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. > Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). > Then x = (-p +/- sqrt(p^2 - 4q))/2 . > Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] > then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. > So (-p - sqrt(p^2 - 4q))/2 = q/a . > Both roots are algebraic integers, and the extension > field Q(a) contains q/a, which is an algebraic integer. > So the ring of integers of Q(a) contains the other root, > namely q/a . > So if x^2 + px + q = 0 has non-rational roots, one > of them being a, then the ring of integers in Q(a) > also contains a/q, the other root. > (x - a)(x - q/a) = x^2 + x*(-q/a - a) + q = x^2 + x*(-qa'/(a*a') -a ) + q > = x^2 + x*(-a -a' ) +q = x^2 + px + q . > a + a' = -p. So a' = -p - a. So any extension ring of Q containing a > must contain the conjugate root a' . (case p, q in Q) > Also, if a is an algebraic integer, Clearly it is. > we may assume p, q in Z > and the an extension ring of Z containing a must contain > the conjugate root a', which is also an algebraic integer. > Mhh.. > David Bernier All true I think, but that's not the question. Assume that > q = r*s, where (say) r is prime. Then a Harrisian object ring > would contain exactly one of a/r or a'/r. In general Harris has said > that the object ring would also contain all the algebraic > integers. It is not clear as Dik suggests that there is > one object ring which contains one of a/r or a'/r for all > irreducible quadratics that have r as a factor of the constant term. > But it may be that something like A[a/r] is sufficient > for the Harrisian oeuvre, where A is the ring of algebraic > integers. He has been somewhat noncommittal beyond his > defective definition of the object ring. I appreciate that Arturo Magidin and you commented on what I > like that, it's perhaps a consequence of the conclusions > he wants. I remember there was a lot of discussion about > specific examples (algebraic numbers of degree 2 I think) > which disproved something James had claimed. Could you elaborate a bit on the counterexamples to > James's claims regarding factors or whatever it was > in algebraic number theory? David Succinctly, Harris thinks he has a proof that one of the roots of, e.g., a^2 - 6a + 35 is divisible by 7 in the ring of algebraic integers and the other root is coprime to 7 in that same ring. This contradicts well-established theory at a very low level. For example, assume that the root which is divisible by 7 is 3 + sqrt(-26). Thus u = (3 + sqrt(-26))/7 would be an algebraic integer. You conclude that 49u^2 - 42 u + 9 = -26, or 7u^2 - 6u + 5 = 0. This is an irreducible nonmonic primitive polynomial with leading coefficient 7, which means that, contrary to the Harris proof, it cannot be an algebraic integer. An almost identical argument shows that (3 - sqrt(-26))/7 similarly cannot be an algebraic integer. Harris waffles on this somewhat, saying e.g. that his result APPEARS to contradict known facts, or that one of the roots SHOULD be divisible by 7 and the other coprime to 7. But his argument does not invoke any ring properties that are not possessed by the ring of algebraic integers, which means that if his argument were valid, it would be valid in that ring. But clearly it is not. The root of all this is, he is making an assumption about how 7 distributes across (5a_1(x) + 7)*(5a_2(x) + 7). He assumes that 7 distributes across the two factors in a way that does not depend on x. Then he notes that when x = 0, 7 factors out of one of the two factors. He does not state this assumption explicitly and may not be aware, or want to admit, that he is making it. Logically, the contradiction that he encounters should make him say, Oh - one of my assumptions must be wrong. But instead of reaching that conclusion, he says Oh - the theory of algebraic integers must be wrong. Or, Oh - Galois theory must be mis- interpreted. Etc. Since he does not admit that he is making the assumption above, he does not conclude that assumption is the root of the contradiction. And that is where he has been for the last 5-6 years. Marcus. Marcus. === Subject: Re: JSH problem days. My association with the Department is that of an alumnus. > ... > > So what is the object ring? You have to ask James ;-) No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). The roots of such a quadratic would be integral. By non-integral, Dik means that the quadratic must be a constant multiple of a non-monic, primitive, irreducible quadratic. >Then x = (-p +/- sqrt(p^2 - 4q))/2 . Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. So (-p - sqrt(p^2 - 4q))/2 = q/a . >Both roots are algebraic integers, Indeed: and therefore they are integral. Dik said non-integral, not non-integer. > and the extension >field Q(a) contains q/a, which is an algebraic integer. >So the ring of integers of Q(a) contains the other root, >namely q/a . Now consider the case of tx^2 + px + q, with t,p,q in Z, gcd(t,p,q)=1, and t=/= 1,-1. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: microwave station posting-account=9GlOXwoAAAAlAbPpB4XhR7GNOEPVSRWt Trident/4.0; FunWebProducts; ub2; InfoPath.2; Zango 10.3.36.0; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How would i start to build a workable microwave station? === Subject: Re: microwave station posting-account=W7RQ6gkAAACLDC5JWapQU1UV8ot8KkAS .NET CLR 1.0.3705; Dealio Toolbar 3.1.1),gzip(gfe),gzip(gfe) > How would i start to build a workable microwave station? Microwave stations are so hard to be build, that's one of the reasons people invented Holograms, Licences, and Optical Computers though. === Subject: Re: microwave station > How would i start to build a workable microwave station? just buy one from ebay. === Subject: Re: microwave station Wrong group, you Google-posting AOL using twat. > How would I brane working ???????????? === Subject: Re: microwave station > Wrong group, you Google-posting AOL using twat. Off with your head, you top-posting OE-using twit! -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: microwave station > How would i start to build a workable microwave station? I hate to admit it, but _this_ is the main reason I read this group. B. -- Cheerfully resisting change since 1959. === Subject: subgroup of alternating group A_5 Hello teacher, How does one show that the alternating group A_5 has no subgroup of order 30. I know that A_5 has order 60. So it might be possible, as 30 divides 60. I also know that A_5 is a simple group. === Subject: Re: subgroup of alternating group A_5 Piet Vanraad schreef: > Hello teacher, How does one show that the alternating group A_5 has no subgroup of > order 30. I know that A_5 has order 60. So it might be possible, as 30 divides 60. > I also know that A_5 is a simple group. for a contradiction in this direction, but forgot about that theorem! === Subject: Re: subgroup of alternating group A_5 days. My association with the Department is that of an alumnus. >Hello teacher, How does one show that the alternating group A_5 has no subgroup of >order 30. I know that A_5 has order 60. So it might be possible, as 30 divides 60. >I also know that A_5 is a simple group. If you know that A_5 is a simple group, then show that a subgroup of order 30 would necessarily be normal. HINT: There is a theorem that talks about subgroups of index 2. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: subgroup of alternating group A_5 posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Hello teacher, How does one show that the alternating group A 5 has no subgroup of > order 30. I know that A 5 has order 60. So it might be possible, as 30 divides 60. > I also know that A 5 is a simple group. A subgroup of order 30 would be normal, because its index would be 2, and A 5 is simple. -- m === Subject: Please help - dynamic programming or robust optimization? posting-account=cewOKQoAAAAlnpN-CpPojngZgjG_pN_6 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) I'm wondering can you help me with this 9hope i'm i nthe right place, humble apologies if I'm not). I have certain process that occurs every day. To say the least it is an uncertain process. I think that can be said for sure. I am unsure if it fits any known distribution. Every day I take 2 numbers from this process. Once in the morning and 1 in the afternoon. I will call those numbers A and B for day n where n at the moment is day 1, the first day. I also have 3 sets of 2 numbers that represent % ranges. They are labelled: LLn = lower limit for day n ULn = upper limit for day n LLn+1 = lower limit for day n + 1 ULn+1 = upper limit for day n + 1 LLn+2 = lower limit for day n + 2 ULn+2 = upper limit for day n + 2 Every day I make 3 decisions (depending upon the values of A and B relative to the limits). Decision A: If on the first day (B < A - LLn) I exit and calculate (B - A) which will be negative. I store this number and repeat the process the next day. Decision B: If on the first day (B > A + LLn) I proceed to the next day n+1 and and repeat the process using the new lower and upper limits repectively LLn+1 and ULN+1 I proceed to the next day and repeat the process. Eventually at some stage decision A will occur. Like before I store this number, in this case it would be Bn+m - An, where m is the number of days the process continued for. Decision C: If on the first day (B <= A + UPn) AND (B >= A - LLn) I proceed to the next day n+1 and repeat the process using the previous day's upper and lower limits respectively: ULn and LLn I proceed to the next day and repeat the process. Eventually at some stage decision A will occur. Like before I store this number, in this case it would be Bn+m - An, where m is the number of days the process continued for. At the end of a certain time period I add up the outcomes to obtain a net total. My objective is to maximise this total. The only variables that I can control upper and lower limits respectively. Now I *believe that either the following can help me optimise this problem: 1. Dynamic programming with recourse 2. Robust optimization I feel that robusy optimization is the way to go, but am unsure how to do it. Can anyone out there lend me some assistance and take me under their wing please? Either in LP or QP, whichever is easier! Would greatly appreciate any advice/help/suggestions/mentoring on this one. Michael. === Subject: Re: Please help - dynamic programming or robust optimization? posting-account=gpERugkAAAB5_qKVhbO9UpGpOXFNrIYf CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) [snip long confused description of problem] > Can anyone out there lend me some assistance and take me > under their wing please? Either in LP or QP, whichever is easier! Would greatly appreciate any advice/help/suggestions/mentoring on this > one. Sure! ^ ^ What do you pay? Socks === Subject: Origin of Gunfire This solution obtains the origin of gunfire using three microphones. http://mypeoplepc.com/members/jon8338/math/id22.html I have posted this before but I corrected a mistake I made. This correction fixes the problem of the source of the gunfire originating from either side of the plane of the three microphones. There are two solutions. A fourth microphone comparing times determines which side of the bull's horns the shot originates. There's a link on the site to an Excel 97 workbook that calculates the solution for different inputs. === Subject: Re: A new definition for Life posting-account=BKLKygoAAACKhQOBleq3odJ7v98y_w-j Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > The second law of thermodynamics, states that the total entropy of any > isolated thermodynamic system tends to increase over time, approaching > a maximum value; in more general terms, systems move from a state of > higher complexity and activity, to one of lower complexity and > activity. This applies to inanimate systems, inanimate matter. Quite > clearly, for living systems and beings, for animate objects, the > opposite applies. Life is matter in reverse entropy: life > spontaneously and eternally moves from lower to higher levels of > complexity and structure. This definition subsumes Darwin's concept > of Evolution, and reconciles it with a sense of eternal purpose in the > Universe. It is funny how you got it exactly in reverse. The very reason Life exist, and if you will its purpose is to accelerate the entropy increase. Life is the most powerful entropy increase catalyst, because it uses part of the released energy to maintain its own complexity. And it is this complexity allows it to be such an efficient catalyst! The reason that Life's complexity is increasing with evolution is that it becomes more efficient entropy increase catalyst, while accounting for more and more new variations of external environments and involving new entropy increase energy flows (for example nuclear in addition to chemical) into this catalysis. Early life was only using chemical reactions (such as Fe3+ reduction) to increase entropy, later life found a way to increase entropy using sun-light, modern life (us) found a way to increase entropy of nuclear decomposition reactions as well as nuclear synthesis reactions. Future of Life is in expanding into new entropy increase processes (those increasing its own complexity as it does so), or being wiped out if existing entropy increase process is exhausted. No wonder Life choses the first and its complexity is increasing. Well, some form of Life did chose the second, and that is why they are dead and we don't know anything about them.... force of life complexity increase, but just a mechanism of optimization. Just like genetic algorithm is optimizing parameters of a function to satisfy certain criteria, Darvinian evolution optimizes Life for criteria maximal entropy increase efficiency over maximal possible time interval and maximal possible external conditions. Our future is in accelerating more energetic entropy increase processes - in stars, in black-holes, in quasars! Forget about your water, bacteria and sugar - Life is about the future of the Universe! Yevgen === Subject: Re: A new definition for Life > The second law of thermodynamics, states that the total > entropy of any isolated thermodynamic system tends to > increase over time, approaching a maximum value; in more > general terms, systems move from a state of higher > complexity and activity, to one of lower complexity and > activity. This applies to inanimate systems, inanimate > matter. Quite clearly, for living systems and beings, > for animate objects, the opposite applies. Life is > matter in reverse entropy: life spontaneously and > eternally moves from lower to higher levels of complexity > and structure. This definition subsumes Darwin's concept > of Evolution, and reconciles it with a sense of eternal > purpose in the Universe. Ok. === Subject: Re: A new definition for Life > - >There are numerous example of extinct species throughout history. >Species of man being among them. Was the divinity just learning how >to do it right? >The other example is that people are not evolving. The less intelligent >have more children so the average intelligence is decreasing. Also >medicine is allowing more of us to survive and pass on our various >defects. Not much of a divine plan. > You're failing to grasp my concept of divinity, Doug. It's not an all > powerful Pixie in the sky with an immutable plan. It's just a general > direction to things. That's all. That's all I see. The laws of physics set what happens in the universe. That is the only divine guidance. But, I do see > that. Which, is more than no divinity, at all. > The laws of physics are quite a divinity. === Subject: Re: A new definition for Life > If you are going to make up new concepts, then make new words. By > co-opting old words you just jam up communication. I know how it feels to want to invent new words. === Subject: Re: A new definition for Life >Evolution seems to involve the progressive development of >systematically more complex organisms. That is not correct. It is a random walk. Over time, things will increasingly diverge from the starting point. However, individual evolutionary events may be toward less or more complexity. Certainly it has no direct relationship to Darwinian evolution, which is based on fitness, not complexity. There is no general relationship between fitness and complexity. As to entropy... Consider the following analogy... Imagine a room, with screened windows on opposite sides. There are some balloons in the room. Their position is random = high entropy. Now turn on a wind machine, so that wind flows thru the room from one side to the other. The balloons will now tend to be all on one side, where the wind exits; low entropy. This is a simple example showing how the flow of energy can create (local) order. It is very much like a living system in that regard. Obviously, life cannot be defined simply in terms of entropy. Bob === Subject: Re: A new definition for Life <8senm4tvf3g9uf2m2cpts6aoths15f07sg@4ax.com> posting-account=5ayZ-goAAABGZmmwx8zZEwz6gU2OuVSd CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) >Evolution seems to involve the progressive development of >systematically more complex organisms. That is not correct. It is a random walk. Over time, things will > increasingly diverge from the starting point. However, individual > evolutionary events may be toward less or more complexity. Certainly > it has no direct relationship to Darwinian evolution, which is based > on fitness, not complexity. There is no general relationship between > fitness and complexity. Darwin was wrong. Evolution is not a random walk. More complex and sophisticated organisms will, invevitably, have more capacity to survive, reproduce, and thrive, in general. A more complex and sophisticated organism will have a better capacity to control its evironment, and therefore to survive within it. It won't have to adapt. It can control. === Subject: Re: A new definition for Life >Evolution seems to involve the progressive development of >systematically more complex organisms. That is not correct. It is a random walk. Over time, things will > increasingly diverge from the starting point. However, individual > evolutionary events may be toward less or more complexity. Certainly > it has no direct relationship to Darwinian evolution, which is based > on fitness, not complexity. There is no general relationship between > fitness and complexity. As to entropy... Consider the following analogy... Imagine a room, > with screened windows on opposite sides. There are some balloons in > the room. Their position is random = high entropy. Now turn on a wind > machine, so that wind flows thru the room from one side to the other. > The balloons will now tend to be all on one side, where the wind > exits; low entropy. This is a simple example showing how the flow of > energy can create (local) order. It is very much like a living system > in that regard. Obviously, life cannot be defined simply in terms of > entropy. Here you say it: wind MACHINE. A device created by intelligence. So this is a simple example showing how a flow designed by intelligence can make up local order. And life in the end. Han de Bruijn === Subject: Re: A new definition for Life <7p8nm4929s4kvauv5l45fb0eh454m85t11@4ax.com> posting-account=5ayZ-goAAABGZmmwx8zZEwz6gU2OuVSd CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > in > *Some* life has gotten more complex, but the vast majority of life > on Earth is single celled. Nevertheless, Matt, we see more complex life forms developing systematically with time. Quite systematically. Higher life now is far more intelligent and more socially complex than 100 million years ago or 500 million years ago. And, more complex than 100 years ago, for that matter. Pretty good proof of God, I'd say. === Subject: Re: A new definition for Life posting-account=5PxZDwoAAABv9p271mvbXvIpZRL9I54K CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > in > *Some* life has gotten more complex, but the vast majority of life > on Earth is single celled. Nevertheless, Matt, we see more complex life forms developing >systematically with time. Quite systematically. Higher life now is >far more intelligent and more socially complex than 100 million years >ago or 500 million years ago. And, more complex than 100 years ago, >for that matter. Pretty good proof of God, I'd say. Huh? Come again? Care to fill in some of the details of that > proof? -- > Angus Rodgers I must emphasize, I'm not talking about an anthropomorphic God, simply a meaningful direction to the Universe as a whole. What more meaningful direction could there be, but towards increasing sophistication, complexity and power? Which is clearly the direction of biological evolution. More God-like organisms automatically and spontaneously develop over time, everywhere. === Subject: Re: A new definition for Life > I must emphasize, I'm not talking about an anthropomorphic God, simply > a meaningful direction to the Universe as a whole. What more > meaningful direction could there be, but towards increasing > sophistication, complexity and power? Which is clearly the direction > of biological evolution. More God-like organisms automatically and > spontaneously develop over time, everywhere. Your remark reminds me of a problem I feel in considering possibilities of doing foolish things, even though I choose not to do them. Banging my head against a wall, or breaking things, or vandalizing property. Not to scare anyone -- I certainly don't recommend these actions -- but it seems like anyone could refute any argument about human nature by positing them. Now, why would you do them, except to win arguments and gain social status? Why would you *want* that social status? So that people would look up to you as an intelligent philosopher, admire you, be friends with you, perhaps marry and have children with you? Might you then be reversing the second law of thermodynamics after all, caring for your family and maintaining the orderliness of your environment? === Subject: Re: A new definition for Life in >I must emphasize, I'm not talking about an anthropomorphic God, simply >a meaningful direction to the Universe as a whole. This is called an equivocation and is dishonest. You are come up with some new concept, but you use an old word so you can get support from people who actually disagree with you. >What more >meaningful direction could there be, but towards increasing >sophistication, complexity and power? Meaningful to us, sure. But the Universe is, to good approximation, empty space. Not all that complex. >Which is clearly the direction >of biological evolution. Yet most life, overwhelmingly most life, is not complex. >More God-like organisms automatically and >spontaneously develop over time, everywhere. -- Matt Silberstein Do something today about the Darfur Genocide http://www.beawitness.org http://www.darfurgenocide.org http://www.savedarfur.org Darfur: A Genocide We can Stop === Subject: Re: A new definition for Life > in I must emphasize, I'm not talking about an anthropomorphic God, simply >a meaningful direction to the Universe as a whole. > > This is called an equivocation and is dishonest. You are come up with > some new concept, but you use an old word so you can get support from > people who actually disagree with you. > >What more >meaningful direction could there be, but towards increasing >sophistication, complexity and power? > > Meaningful to us, sure. But the Universe is, to good approximation, > empty space. Not all that complex. > >Which is clearly the direction >of biological evolution. > > Yet most life, overwhelmingly most life, is not complex. Wrong. Even a single cell is tremendously complex. >More God-like organisms automatically and >spontaneously develop over time, everywhere. Han de Bruijn === Subject: Re: A new definition for Life > This is called an equivocation and is dishonest. You are come up with > some new concept, but you use an old word so you can get support from > people who actually disagree with you. To be fair, to equivocate is not to use a word with several definitions but to deduce consequences of one definition from another. === Subject: Re: A new definition for Life > in > *Some* life has gotten more complex, but the vast majority of life > on Earth is single celled. > Nevertheless, Matt, we see more complex life forms developing > systematically with time. Quite systematically. Higher life now is > far more intelligent and more socially complex than 100 million years > ago or 500 million years ago. And, more complex than 100 years ago, > for that matter. Pretty good proof of God, I'd say. > Huh? Come again? Care to fill in some of the details of that > proof? > -- > Angus Rodgers I must emphasize, I'm not talking about an anthropomorphic God, simply > a meaningful direction to the Universe as a whole. What more > meaningful direction could there be, but towards increasing > sophistication, complexity and power? Which is clearly the direction > of biological evolution. More God-like organisms automatically and > spontaneously develop over time, everywhere. > It is quite clear that you know very little about biological evolution. === Subject: Radius of largest n+1 balls in n-dimensional unit cube? posting-account=WzP9FgoAAAANyEt4wx0YVvhakkQXYd72 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) I would appreciate your help or comment about the following problem. Consider a N dimensional space. If I want to put N+1 balls (all with the same radius R), within the unit hypercube such that: 1. The balls do not cut through each other 2. One of the balls is at the center of the cube, i.e. at (0.5 , 0.5, 0.5, ...., 0.5) Then what is the maximum possible R in terms of N? If not easy, a good approximation will be helpful too. Golabi === Subject: Re: Radius of largest n+1 balls in n-dimensional unit cube? I would appreciate your help or comment about the following problem. >Consider a N dimensional space. If I want to put N+1 balls (all with >the same radius R), within the unit hypercube such that: 1. The balls do not cut through each other >2. One of the balls is at the center of the cube, i.e. at (0.5 , 0.5, >0.5, ...., 0.5) Then what is the maximum possible R in terms of N? If not easy, a good >approximation will be helpful too. It seems intuitive that the maximum r for your problem occurs when the other n balls are in the corners, tangent to the central ball and tangent to the corner faces. For that configuration, r = sqrt(n) / (2*(2+sqrt(n))) But note, the unit cube in R^n has 2^n corners, so you could just as easily have (2^n)+1 balls with the same radius as above, rather than only n+1. quasi === Subject: Re: Radius of largest n+1 balls in n-dimensional unit cube? posting-account=WzP9FgoAAAANyEt4wx0YVvhakkQXYd72 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) derive the formula you proposed? The only piece I understand is that sqrt(n) is maximal distance between two corners of the cube, but have no idea how it entered into your formula. >I would appreciate your help or comment about the following problem. >Consider a N dimensional space. If I want to put N+1 balls (all with >the same radius R), within the unit hypercube such that: 1. The balls do not cut through each other >2. One of the balls is at the center of the cube, i.e. at (0.5 , 0.5, >0.5, ...., 0.5) Then what is the maximum possible R in terms of N? If not easy, a good >approximation will be helpful too. It seems intuitive that the maximum r for your problem occurs when the > other n balls are in the corners, tangent to the central ball and > tangent to the corner faces. For that configuration, r = sqrt(n) / (2*(2+sqrt(n))) But note, the unit cube in R^n has 2^n corners, so you could just as > easily have (2^n)+1 balls with the same radius as above, rather than > only n+1. quasi === Subject: Re: Radius of largest n+1 balls in n-dimensional unit cube? >derive the formula you proposed? The only piece I understand is that >sqrt(n) is maximal distance between two corners of the cube, but have >no idea how it entered into your formula. Ok, but please don't top post. Instead, either bottom post or intersperse portions of your reply after the relevant parts of the prior message (as I am doing here). That's the standard in sci.math. For the configuration I described, r can be calculated as follows: Place a little cube of edge length r in the corner of the unit cube. Then the distance from the center of the ball in that corner to the corner vertex is the length of the diagonal of the little cube, which is r*sqrt(n) (by proportionality to the unit cube). Then (draw a diagram to see it), r + r + r*sqrt(n) = (1/2)*sqrt(n) which yields r = sqrt(n) / (2*(2+sqrt(n))) >I would appreciate your help or comment about the following problem. >Consider a N dimensional space. If I want to put N+1 balls (all with >the same radius R), within the unit hypercube such that: >1. The balls do not cut through each other >2. One of the balls is at the center of the cube, i.e. at (0.5 , 0.5, >0.5, ...., 0.5) >Then what is the maximum possible R in terms of N? If not easy, a good >approximation will be helpful too. > It seems intuitive that the maximum r for your problem occurs when the > other n balls are in the corners, tangent to the central ball and > tangent to the corner faces. For that configuration, > r = sqrt(n) / (2*(2+sqrt(n))) > But note, the unit cube in R^n has 2^n corners, so you could just as > easily have (2^n)+1 balls with the same radius as above, rather than > only n+1. quasi === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem Angus Rodgers a .8ecrit : One bit of my sleepy mind says > that this makes some sense, while another bit says it is a bit odd Without having meant to think about it any more for the moment, > I am becoming a bit more sure that always switch is the right > answer. If I try not to think too much about it, my mind says: > On the one hand, /if/ you have chosen the door with the prize, > then Monty's choice is telling you nothing about whether the > prize is behind your door or the other one. On the other hand, > /if/ you have chosen the wrong door, then Monty's choice is tell- > ing you which the right door is - and it's the other one! Since > these two events are a priori equally likely, > DFont you mean * not a priori*? on balance Monty's > choice is telling you to go for the other door. Right. Anyway, I wanted to tell oyu more about this, especially as you > seem to have a very low opinion on yoursezlf on htose probability > matters, while I have not zseen you mke any mistake yet. So 1) dont > underestimate yourself. 2) after reading the link I gave you > (http://en.wikipedia.org/wiki/Monty_Hall_problem ), search next for > other interesting probability paradoxes (here is a list : > http://en.wikipedia.org/wiki/Category:Probability_theory_paradoxes ) > and test yourself : you could have a nice surprise :-) [incorrect stuff elided] >Is this argument correct? No -- you inadvertantly used the wrong formula for P(AB). It should be P(AB) = P(A|B) P(B). Also, consider the conceptual meaning of P(A|B) = P(A). It means that the conditional probabiity of A, given B, is the same as > the absolute probability of A. Thus, the even B gives no information > about A -- i.e. doesn't affect absolute probabiity of A. Sorry for the trouble. Yes, p(A|B) = p(A) implies A and B are independent. p(AB)=P(A|B)p(B) =P(A)p(B). I also proved the proposition, with probability at least 0.35, but will not inflict it on you. Berkson's paradox: The result is that two independent events become conditionally dependent (negatively dependent) given that at least one of them occurs. 0 right... Of course, the other crankish behavior of the poster does > nothing to change my mind... What other crankish behavior of the poster? I may have missed > something, but all I've seen is a guy who (1) thinks he's got an > elementary proof of FLT, and (2) wants to find some way to make some > money from it. Neither of these is crankish. > One of those is is credentials attitude (the I have gone to school in John Baez list), another is his fear of stolen ideas... I insist, see the list === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem Michael Press a .8ecrit : Angus Rodgers a .8ecrit : > One bit of my sleepy mind says > that this makes some sense, while another bit says it is a bit odd > Without having meant to think about it any more for the moment, > I am becoming a bit more sure that always switch is the right > answer. If I try not to think too much about it, my mind says: > On the one hand, /if/ you have chosen the door with the prize, > then Monty's choice is telling you nothing about whether the > prize is behind your door or the other one. On the other hand, > /if/ you have chosen the wrong door, then Monty's choice is tell- > ing you which the right door is - and it's the other one! Since > these two events are a priori equally likely, > DFont you mean * not a priori*? > on balance Monty's > choice is telling you to go for the other door. > Right. Anyway, I wanted to tell oyu more about this, especially as you > seem to have a very low opinion on yoursezlf on htose probability > matters, while I have not zseen you mke any mistake yet. So 1) dont > underestimate yourself. 2) after reading the link I gave you > (http://en.wikipedia.org/wiki/Monty_Hall_problem ), search next for > other interesting probability paradoxes (here is a list : > http://en.wikipedia.org/wiki/Category:Probability_theory_paradoxes ) > and test yourself : you could have a nice surprise :-) I went to the page and picked the first alphabetcally: > Berkson's paradox; and I find the statement to be wrong. They write > 0 p(A|B) = p(A), i.e. they are independent. The statement that they are indedepent > is wrong if `they' refers to A and B. The wording is perhaps wrong... But see http://en.wikipedia.org/wiki/Statistical_independence > for we have p(AB) = P(A|B)p(A) = p(A)^2 = p(B|A)p(B) What??? p(AB) = P(A|B)p(B) ... Suppose p(A)=p(A|B) = 1/2, then p(AB)= p(B|A)p(B) = 1/4. > Suppose p(B|A) = 1 and p(B) = 1/4. > Then we have p(B|A)p(B) = 1/4, in accordance with the specifications. > Also p(A)p(B) = 1/8 =/= 1/4 = p(AB). > But A and B independent iff p(AB) = p(A)p(B). > Contradiction. > Is this argument correct? Certainly not... > === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem > nice expository on the generalizations, > using a sort of meta-triple, where F'L'T eqs. are > (k,1,2) (and > Pythagoreans are (2,1,2), from monsieur PT3: > http://www.geocities.com/titus_piezas/Timeline1.htm Nice area for professionals and for amateurs ! Infinite capacity for patient math. creations ! > surprised if it's not valid comes from the fact > that I would not be > surprised if one doesn't exist. When coming back to FLT, I think the only elementary and partial solutions are for Sophie Germain primes... If Mr. David Fabian claims something else, I'll recommend to him to check if his methods are solving following equation: t^n = 2*n^u abtp +a^n + b^n where a;b;t;p;n are natural numbers and of gcd=1 This is equivalent of X^n +Y^n = Z^n for X;Y;Z;n natural numbers and of gcd=1 once n>=3 for specially chosen parameters. Such clear and correct parametrization could be written: 2Z = X+Y +Z-Y +Z-X = t^n + b^n + a^n 2X = X+Y +Z-Y -Z+X = t^n + b^n - a^n 2Y = X+Y -Z+Y +Z-X = t^n - b^n + a^n But only for 1-st case of FLT. What should be written for X;Y or Z divided by n ? What for value represents X+Y-Z ? Nice and simple but far away from some proof ! Ro-Bin === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem > Zachary Turner a .8ecrit : > Having read most of this thread, I can say pretty confidently you have > two options: > 1) Publish it on the web. If it is determined to be correct you will > likely get a book deal out of it, you will probably be offered high > paying jobs either in the finance industry or by the government as > well. You may be invited to give talks as well, all of which would > bring income. If it is to be incorrect you can try again, and once > you are choose again from options 1 or 2 listed here. > Poster never responded to the possibility of parlaying his success > into a better position. Some posts later, he inadvertently reveals > why: he's in his 50's. > M Well it's pretty much just a fact of life that nobody's just going to > write a big fat check for a proof which may or may not be correct. > And showing it to someone to verify its correctness will raise issues > of its own, as if it turns out to be incorrect, even slightly, they > might build off the work and end up with something correct and publish > it free. Therefore regardless of said proof's correctness and > strategy used to release proof in a way conducive to money-making, > there is going to be some inherent level of risk involved in terms of > possibly not receiving the payout the OP desires. So the best > strategy is probably just to publish it, and assume the risk that you > may not get the book deal or whatever. At least this has a >0% > probability that money will be made, as opposed to a 0% probability by > just sitting on it. In any case, one thing I will say is that there's a lot of cranks > going around claiming to have the secrets of the world, and most of > them are just cranks. However, in this case I would actually not be > surprised if the OP does have a valid proof Not be surprised show you know very little about the problem ( > and perhaps about math in general) (although I would also not be surprised if he doesn't). Search around for the name a bit. Compared to some of the people on this forum I probably do know > relatively little about math in general. But then again, compared to > some of the people not on this forum, some of the people on this forum > also know relatively little about math. I'm not sure what your point > is. I have an undergrad + masters in math from a pretty strong > american university, and was almost top in my class. Although I still > fail to see how that's relevant. News flash: People who don't know much about math in general tend not > to like it, and people who don't like it don't hang around math forums > in their free time. Disagree. There are people posting here at length who do not like mathematics. My external, objective criteria for liking mathematics include knowledge of the workings; ability to follow and generate proofs; and desire and gratitude for instruction. -- Michael Press === Subject: Re: The modern mathematical concept of infinity is indefensible <87zlhxaif1.fsf@alatheia.dsl.inet.fi> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > But let's set that aside. How does the inconsistency of ZFC > follow from the falsity of the completenss theorem? Because the completeness theorem follows from ZFC. > Consistent theories may well prove all sorts of falsities. NO, NOT *all* sorts. In particular, they NEVER prove the sort of falsity that is the DENIAL of one of their THEOREMS. === Subject: Re: The modern mathematical concept of infinity is indefensible > Because the completeness theorem follows from ZFC. ZFC is inconsistent follows from the axioms of the theory ZFC + ZFC is incosistent. Assuming ZFC + ZFC is inconsistent is in fact consistent, or equivalently, that ZFC is consistent, how would you use this fact in showing that ZFC + ZFC is inconsistent (or in ZFC)? If you agree there is in fact no way of doing so, how exactly do you propose to do so in case of the completeness theorem, supposing its falsity? > In particular, they NEVER prove the sort of falsity that is the > DENIAL of one of their THEOREMS. The relevance of this escapes me. There was no suggestion that even if the completeness theorem is false there is a derivation of the negation of the completeness theorem in ZFC. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > ZFC is inconsistent follows from the axioms of the theory ZFC + ZFC > is incosistent. Assuming ZFC + ZFC is inconsistent is in fact > consistent, or equivalently, that ZFC is consistent, how would you use > this fact in showing that ZFC + ZFC is inconsistent (or in ZFC)? My apologies. This should read Assuming ZFC + ZFC is inconsistent is in fact consistent, or equivalently, that ZFC is consistent, how would you use this fact in showing that ZFC + ZFC is inconsistent (or in ZFC) is inconsistent? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > You will already have problems when tying to explain what it means for an > axiom to be true... Consider an axiom in (certain systems of) ZFC: ExAy(y !e x). There is a set without elements With other words, there is an empty set. Now what does it mean to claim that this axiom is true? Exactly the same as to claim there is a set without elements. As to whether there is a set without elements, a very natural attitude is to regard this as not something determined by any matters of fact, but as something up to stipulation. In case of other axioms and proposed axioms, such as the axiom of choice or various large cardinal axioms, such an attitude is very unnatural and subtler considerations are called for. > What does it mean (in this case) for a statement in our theory to be > true other than: either this is an axiom, or it can be derived from > the axioms (and definitions)? That's provability from the axioms, not truth. It is a mathematical result that for example the truth of the axiom of choice or the continuum hypothesis is not equivalent to the provability of this or that in any formal theory. It is also a mathematical theorem that if ZFC is consistent there are infinitely many true arithmetical statements not provable in ZFC. And so on. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > Not at all. Check out Vicious Circles by Barwise and Moss, an > excellent (if poorly proofread) book which I'm currently reading. It's > about a particular version of set theory without the axiom of > foundation, with a new axiom which says, roughly speaking, that > equations such as x = {x,p} have unique solutions. In this context one might mention Aczel's book _Non-well-founded sets_ which is available on-line, at http://standish.stanford.edu/pdf/00000056.pdf -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <874p03931c.fsf@alatheia.dsl.inet.fi > Not at all. Check out Vicious Circles by Barwise and Moss, an > excellent (if poorly proofread) book which I'm currently reading. It's > about a particular version of set theory without the axiom of > foundation, with a new axiom which says, roughly speaking, that > equations such as x = {x,p} have unique solutions. In this context one might mention Aczel's book _Non-well-founded sets_ > which is available on-line, at http://standish.stanford.edu/pdf/00000056.pdf Ah, so it is still out there. === Subject: Re: The modern mathematical concept of infinity is indefensible Not at all. Check out Vicious Circles by Barwise and Moss, an > excellent (if poorly proofread) book which I'm currently reading. It's > about a particular version of set theory without the axiom of > foundation, with a new axiom which says, roughly speaking, that > equations such as x = {x,p} have unique solutions. In this context one might mention Aczel's book _Non-well-founded sets_ > which is available on-line, at http://standish.stanford.edu/pdf/00000056.pdf > Don't know if this is relevant to the issue but I've wondered if there is a kind of set system that: (a) there are multiple epsilon-relations (symbols) (b) there's one usual epsilon-relation that is well-founded but there are others that are not? -- To discover the proper approach to mathematical logic, we must therefore examine the methods of the mathematician. (Shoenfield, Mathematical Logic) === Subject: Re: The modern mathematical concept of infinity is indefensible <1ac8qu5fk14wf.1006vfhkth1o2$.dlg@40tude.net> <54obm4500241eimqhaisfucoqqk3b4koba@4ax.com> <87hc48lbkf.fsf@phiwumbda.org> <87vdslai9g.fsf@alatheia.dsl.inet.fi> <87sknp85u0.fsf@alatheia.dsl.inet.fi> <1lidr5olvinmz.1pkaemdj9ngtq.dlg@40tude.net You will already have problems when tying to explain what it means > for an axiom to be true... Consider an axiom in (certain systems of) ZFC: ExAy(y !e x). There is a set without elements With other words, there is an empty set. Now what does it mean to claim that this axiom is true? That there > ACTUALLY is an empty set (somewhere)? That there is an empty set, yes. > Well, sure, the statement ExAy(y !e x) may be interpreted to say > just that. But I have to admit, sadly, that I don't believe in the > existence of mathematical objects (and hence especially not in the > existence of the empty set). Well, then we are off down another bunny trail entirely. :-) > That there is an empty set somewhere, obviously, no, as pure sets > (if there are any) have no spatial location. Right. I also don't think that pure sets are spacio-temporal objects. > But imho they are not just _nowhere_, but not at all. Not an unreasonable view. > And if it does NOT mean the latter, what does true then refer to > here? I mean how can a sentence be true, without being the case what > it -the sentence- states? Etc. etc. Pretty clearly, it cannot be. > Well it can (imho), if you adhere to a coherence theory of truth (in > contrast to a correspondence theory of truth). Well, I was going to add unless one has a quirky and untenable view of truth. ;-) === Subject: Re: The modern mathematical concept of infinity is indefensible Chris Menzel says... > Well it can (imho), if you adhere to a coherence theory of truth (in > contrast to a correspondence theory of truth). Well, I was going to add unless one has a quirky and untenable view of >truth. ;-) I don't have a completely coherent view of mathematical truth, but it does seem to me that something like the coherence theory is plausible. Suppose you never heard of any notion of number other than natural number. Then someone asks you Does there exist a number x such that x+1=0? (or such that x*2 = 1, or such that x*x = 2, or such that x*x = -1) The answer is not simply yes or no. There is no such x among the numbers that you are aware of, but we can coherently *posit* the existence of negative numbers, rationals, reals and complex numbers. It would be unreasonable for someone to speculate that while it is *consistent* that there are negative numbers, they don't *really* exist. I think for *definable* sets, at least, (such as the empty set) their existence is just a matter of stipulation. We *define* the empty set to be the set with no elements, in the same way that we define -1 to be the solution to x+1=0. There is nothing more to say about the empty set, and there certainly isn't any Platonic question of whether it really exists. -- Daryl McCullough Ithaca, NY === Subject: Re: The modern mathematical concept of infinity is indefensible > Chris Menzel says... >Well it can (imho), if you adhere to a coherence theory of truth (in >contrast to a correspondence theory of truth). >Well, I was going to add unless one has a quirky and untenable view of >truth. ;-) I don't have a completely coherent view of mathematical truth, > but it does seem to me that something like the coherence theory > is plausible. Suppose you never heard of any notion of number other than > natural number. Then someone asks you Does there exist a > number x such that x+1=0? (or such that x*2 = 1, or such that > x*x = 2, or such that x*x = -1) The answer is not simply yes or no. There is no such x among > the numbers that you are aware of, but we can coherently *posit* > the existence of negative numbers, rationals, reals and complex > numbers. It would be unreasonable for someone to speculate that > while it is *consistent* that there are negative numbers, they > don't *really* exist. Objection ! If natural numbers exist, then there exist a _construction_ by which you can actually _make_ negative numbers, rational numbers and complex numbers out of this naturals material. Negative numbers are just pairs of natural numbers ,with newly defined addition and multiplication and _equality_ for those pairs, such that as many properties of naturals as possible are preserved. Rational numbers are pairs of whole numbers, with newly defined addition and multiplication and _equality_ for those pairs, such that as many properties of the whole numbers as possible are preserved. Now guess what complex numbers are .. > I think for *definable* sets, at least, (such as the empty set) > their existence is just a matter of stipulation. We *define* > the empty set to be the set with no elements, in the same way > that we define -1 to be the solution to x+1=0. There is nothing > more to say about the empty set, and there certainly isn't > any Platonic question of whether it really exists. Objection ! The empty set is an empty box. If the empty set exist, then there exist a _construction_ by which you can actually _make_ all other (heredetary finite) sets, e.g. according to: http://hdebruijn.soo.dto.tudelft.nl/jaar2007/set_theory.pdf And the empty set is _implementable_ as computer hardware and software; it's a piece of technology; therefore its existence is not disputable. > -- > Daryl McCullough > Ithaca, NY Han de Bruijn === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Negative numbers are just > pairs of natural numbers ,with newly defined addition and multiplication > and equality for those pairs, If they are then you'd have to explain how the negative integer marked by the pair <1 2> is the same as the negative integer marked by the pair <2 3>. The ordinary solution is to take integers (positive and non-positive) not as pairs of natural numbers but rather as equivalenece classes of pairs of natural numbers, which, too bad for you, requires having the set of natural numbers. > Rational numbers are pairs of whole numbers, > with newly defined addition and multiplication and equality for those > pairs, such that as many properties of the whole numbers as possible are > preserved. If they are then you'd have to explain how the rational marked by the pair <1 2> is the same as the rational number marked by the pair <2 4>. The ordinary solution is to take rational numbers not as pairs of integers but rather as equivalenece classes of pairs of integers, which, too bad for you, requires having the set of natural numbers. > Now guess what complex numbers are .. Complex numbers, ordinarily, are pairs of real numbers (not equivalence classes of pairs of real numbers). But what are real numbers? You've not shown a construction of real numbers that does not require having the set of natural numbers. > Objection ! The empty set is an empty box. No, it's not. A box is a physical object. > If the empty set exist, then > there exist a construction by which you can actually make all other > (heredetary finite) sets Whatever the merits of that claim, from the axiom of separation and axiom of extensionality, it is provable that there is a unique object, call it '0', such that no object in the epsilon to 0. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible MoeBlee says... > Negative numbers are just > pairs of natural numbers ,with newly defined addition and multiplication > and _equality_ for those pairs, If they are then you'd have to explain how the negative integer marked >by the pair <1 2> is the same as the negative integer marked by the >pair <2 3>. The ordinary solution is to take integers (positive and >non-positive) not as pairs of natural numbers but rather as >equivalenece classes of pairs of natural numbers, which, too bad for >you, requires having the set of natural numbers. In some varieties of constructive type theory, one defines a new type A' from an old type A by redefining equality. So the type of integers is the type NxN together with the equality relation Q_Z iff x+y' Q_N x'+y. (Where Q_N and Q_Z are the equality relations for N and Z, respectively) So if type A' is obtained from type A by imposing a new equality relation, and B' is obtained from type B by imposing a new equality relation, and f is a function from A to B, then to prove that f is a function from A' to B', you have to show that, forall x1, x2, if x1 Q_A' x2, then f(x1) Q_B' f(x2). All this is equivalent in some sense to working with equivalence classes, but requires less of the heavy guns of full blown set theory. -- Daryl McCullough Ithaca, NY === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > MoeBlee says... > Negative numbers are just > pairs of natural numbers ,with newly defined addition and multiplication > and equality for those pairs, If they are then you'd have to explain how the negative integer marked >by the pair <1 2> is the same as the negative integer marked by the >pair <2 3>. The ordinary solution is to take integers (positive and >non-positive) not as pairs of natural numbers but rather as >equivalenece classes of pairs of natural numbers, which, too bad for >you, requires having the set of natural numbers. In some varieties of constructive type theory, I have no dispute as to the many possibilities of formulations in theories weaker than Z set theory or not comparable in strength with Z set theory. My point in my post to Han is that if one does eschew roomier theories such as Z, then one does have to show that the alternative works and be willing to pay its cost in cumbersomeness. In that regard, Han is not at all prepared for the technicalities of constructive type theory and, in that regard, his claim that all we need is ordered pairs of naturals is bluster, as it would be for me or anyone to make that claim if one had not actually worked through to understand such alternative formulations as constructive type theory. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) >no object in the epsilon to 0. I meant no object is in the epsilon relation to 0. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > If they are then you'd have to explain how the negative integer > marked by the pair <1 2> is the same as the negative integer marked > by the pair <2 3>. The ordinary solution is to take integers > (positive and non-positive) not as pairs of natural numbers but > rather as equivalenece classes of pairs of natural numbers, which, > too bad for you, requires having the set of natural numbers. Requires in what sense? All talk about the negative integers, rationals and so on and not involving infinite sets, can be translated into talk about the naturals and basic operations on naturals (and nothing) else in a straightforward though tedious manner. For example, there is no difficulty in formulating arguments usually presented in terms of integers, such as one might find in Euclid and modern variants thereof, in Peano arithmetic. Instead of equivalence classes we simply explicitly use the relevant equivalence relation, which in no way depends on the existence of any infinite sets. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <87bpubvzs5.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > If they are then you'd have to explain how the negative integer > marked by the pair <1 2> is the same as the negative integer marked > by the pair <2 3>. The ordinary solution is to take integers > (positive and non-positive) not as pairs of natural numbers but > rather as equivalenece classes of pairs of natural numbers, which, > too bad for you, requires having the set of natural numbers. Requires in what sense? In the sense of the ordinary set theoretic, axiomatic handling of the matter. That there may be some other axiomatic handling that does not require the set of natural numbers is not disputed by me, nor is it disputed by me that one may choose not to rise to the rigor or axiomatization. My immediate point here is that the poster has not given reference to such a handling, thus my remark you'd have to explain. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > My immediate point here is that the poster has not given reference > to such a handling, thus my remark you'd have to explain. Well, the fact such an explanation can be found in any number of standard texts makes one wonder why you should feel it necessary to make this point in this context. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <87zlhvukp3.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > My immediate point here is that the poster has not given reference > to such a handling, thus my remark you'd have to explain. Well, the fact such an explanation can be found in any number of > standard texts makes one wonder why you should feel it necessary to > make this point in this context. The point in this context may be that every explanation has a cost in cumbersomeness. An explanation of operations on integers (positive and non-positive) and rationals in a more restricted theory than set theory would likely be more cumbersome, as would be actually working with integers and rationals in the more restricted theory. So, in that sense, there is a cost for such as Han de Bruijn's claim that we can just take integers to be ordered pairs of naturals and rationals as ordered pairs of integers, which is a cost we can bet he is not personally willing to pay in his actual mathematical practice (he surely is not going to try to perform integer and rational operations as coded in Peano arithmetic (as well as, by the way, a definition of 'first order PA', ordinarily (again, that context I first mentioned) requires having the set of natural numbers just to set up (defined) the language of first order PA). So, in that practical sense, such simplifications by the likes of Han, as just ordered pairs of naturals are bluster. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > The point in this context may be that every explanation has a cost > in cumbersomeness. An explanation of operations on integers > (positive and non-positive) and rationals in a more restricted > theory than set theory would likely be more cumbersome, as would be > actually working with integers and rationals in the more restricted > theory. That depends on what we understand by working with and what theories we have in mind. In a suitable system, e.g. one allowing us to introduce new sets by introducing a notion of equality, as in constructive type theory, the approach need not be cumbersome at all. Han might well be sensible enough in this context to mention, say, the handling of arbitrary rationals (of any type in the Num typeclass) in Haskell, given his fondness of computer implementations. I don't think your point is very well taken. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <87ljtfujht.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Han might well be sensible enough in this context to mention, > say, the handling of arbitrary rationals (of any type in the Num > typeclass) in Haskell, given his fondness of computer > implementations. IF he were indeed taking rationals as just pairs and are thus constructed in such a context, then, granted, my retort was off- base. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > IF he were indeed taking rationals as just pairs and are thus > constructed in such a context, then, granted, my retort was off- > base. That would be the charitable interpretation, in any case. Whether you feel charitable towards Han and his various utterances is up to you, of course. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <877i4zui5n.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > IF he were indeed taking rationals as just pairs and are thus > constructed in such a context, then, granted, my retort was off- > base. That would be the charitable interpretation, in any case. Whether you > feel charitable towards Han and his various utterances is up to you, > of course. I don't wish to be unreasonable or overly restrictive as to what qualifies as a construction, and Han did mention not just pairs but re- defintion of equality too. So while it is still not clear to me in what sense his proposal constructs objects onto themselves (i.e., aside from operations), since I can't rule out that he has some reasonable context in mind, I had better back off my original retort. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > And the empty set is _implementable_ as computer hardware and software; > it's a piece of technology; therefore its existence is not disputable. Right. In contrast the existence of arbitrarily complex hereditarily finite sets is very much disputable. Extensive computer experiments are necessary to determine which of them exist and which are but figments of our collective imagination. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible And the empty set is _implementable_ as computer hardware and software; >it's a piece of technology; therefore its existence is not disputable. Right. In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. Extensive computer experiments > are necessary to determine which of them exist and which are but > figments of our collective imagination. A bit of common reasoning will be more helpful than all that. Han de Bruijn === Subject: Re: The modern mathematical concept of infinity is indefensible A bit of common reasoning will be more helpful than all that. > Since, as it seems, infinite collections of things aren't part of our usual experience of the world, finitistic common sense won't help in dealing with (the ideas of) such entities. The world of the infinite is ... open for exploration, but to embark we must unlearn our finitistic intuitions which instill fear and confusion by making some consistent and demonstrable results about the infinite literally counter-intuitive. [...] (Peter Suber, Infinite Reflections) Herb P.S. Btw. I guess with common reasoning you refer to those inconsistent (extremely idiotic) stuff concerning math at your homepage, right? === Subject: Re: The modern mathematical concept of infinity is indefensible > Since, as it seems, infinite collections of things aren't part of > our usual experience of the world, finitistic common sense won't > help in dealing with (the ideas of) such entities. Arbitrarily large finite collections aren't a part of our usual experience of the world either, so we need rely on some other source, perhaps some theoretical generalisation of that experience, to reason about finitary objects in general, in a finitistically meaningful and justified way or otherwise. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. Extensive computer > experiments are necessary to determine which of them exist and > which are but figments of our collective imagination. A bit of common reasoning will be more helpful than all that. Surely the existence of arbitrarily complex hereditarily finite set is intimately tied to the limits and possibilities of computer simulations. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. Extensive computer > experiments are necessary to determine which of them exist and > which are but figments of our collective imagination. > A bit of common reasoning will be more helpful than all that. > Surely the [alleged] existence of arbitrarily complex hereditarily > finite set is intimately tied to the limits and possibilities of > computer simulations. > Of course. Since computer resources are bound (assuming that our universe is finite), there are (or can be) only finitely many HFS. Simple as that. Oh... oh... another ultrafinitist ... Herb === Subject: Re: The modern mathematical concept of infinity is indefensible In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. Extensive computer > experiments are necessary to determine which of them exist and > which are but figments of our collective imagination. > A bit of common reasoning will be more helpful than all that. > Surely the [alleged] existence of arbitrarily complex hereditarily > finite set is intimately tied to the limits and possibilities of > computer simulations. > Of course. Since computer resources are bound (assuming that our universe > is finite), there are (or can be) only finitely many HFS. > Simple as that. Oh... oh... another ultrafinitist ... Do you think there's evidence for the infinite in the real world? I'm wondering about these things because in the opposing view to the infinite world, the other guy could say: << That's rubbish. You've only seen/measured very big things and very small things. It's all on a scale from 10^{-googolplex} to 10^{googolplex}, but you'll never know, of course. > What do you think? David Bernier === Subject: Re: The modern mathematical concept of infinity is indefensible > Oh... oh... another ultrafinitist ... > Do you think there's evidence for the infinite in the real world? > Well, actually, it's hard to say. Maybe there is, maybe not. But so what? As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality. (A. Einstein) I'm wondering about these things because in the opposing view > to the infinite world, the other guy could say: << That's rubbish. You've only seen/measured very big things and > very small things. It's all on a scale from > 10^{-googolplex} to 10^{googolplex}, but you'll never > know, of course. What do you think? > I'd agree. We simply don't know. Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > Do you think there's evidence for the infinite in the real world? > I'm wondering about these things because in the opposing view to the > infinite world, the other guy could say: << That's rubbish. You've only seen/measured very big things and > very small things. It's all on a scale from > 10^{-googolplex} to 10^{googolplex}, but you'll never > know, of course. What do you think? Thing like that have their appeal from a certain point of view. What is very difficult is to squeeze any coherent or interesting mathematics out of such a view. It is also often the case that people who say stuff like that in fact reason, when caught off-guard, in a way that presupposes it makes sense to speak of, say, arbitrarily complex formal proofs, formulas, hypothetical computers and such like. Consistent ultra-finitism, anti-realism about all of mathematics, etc. are heroic undertakings requiring almost super-human determination and effort. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Consistent ultra-finitism [is a] heroic undertakings requiring > almost super-human determination and effort. > Seems so: [Though] Ultrafinitism is a form of constructivism [...] even constructivists generally view the philosophy as unworkably extreme. The logical foundation of ultrafinitism is unclear; in his comprehensive survey Constructivism in Mathematics (1988), the constructive logician A. S. Troelstra dismissed it as no satisfactory development exists at present. This was not so much a philosophical objection as it was an admission that, in a rigorous work of mathematical logic, there was simply nothing precise enough to include. http://en.wikipedia.org/wiki/Ultrafinitism Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > Consistent ultra-finitism [is a] heroic undertakings requiring > almost super-human determination and effort. Seems so: [Though] Ultrafinitism is a form of constructivism [...] even > constructivists generally view the philosophy as unworkably extreme. > The logical foundation of ultrafinitism is unclear; in his comprehensive > survey Constructivism in Mathematics (1988), the constructive logician A. > S. Troelstra dismissed it as no satisfactory development exists at > present. This was not so much a philosophical objection as it was an > admission that, in a rigorous work of mathematical logic, there was simply > nothing precise enough to include. http://en.wikipedia.org/wiki/Ultrafinitism There is nothing objectionable in this quote from Wikipedia, but the lack of a satisfactory development, in the foundational sense, is merely a symptom of the inherent difficulty to being consistent about ultra-finitism in mathematics. It is very difficult to set aside our natural intuitions about the infinite progression of naturals, and consequently there is a dearth of any actual ultra-finitist arguments to formalise, arguments which might serve as guidance and raw material when unearthing basic principles and concepts of ultra-finitism. In contrast, with set theory, constructive mathematics, and so on, there was already a formidable body of mathematical reasoning which to consult, and in relation to which to evaluate the suitability of various formalisations and explications. A paper of interest in this connexion is /A Formalization of Essenin-Volpin's Proof Theoretical Studies by Means of Nonstandard Analysis/, James R. Geiser, The Journal of Symbolic Logic, Vol. 39, No. 1 (Mar., 1974), pp. 81-87 -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Do you think there's evidence for the infinite in the real world? >I'm wondering about these things because in the opposing view to the >infinite world, the other guy could say: ><< That's rubbish. You've only seen/measured very big things and > very small things. It's all on a scale from > 10^{-googolplex} to 10^{googolplex}, but you'll never > know, of course. >What do you think? Thing like that have their appeal from a certain point of view. What > is very difficult is to squeeze any coherent or interesting > mathematics out of such a view. It is also often the case that people > who say stuff like that in fact reason, when caught off-guard, in a > way that presupposes it makes sense to speak of, say, arbitrarily > complex formal proofs, formulas, hypothetical computers and such > like. Consistent ultra-finitism, anti-realism about all of > mathematics, etc. are heroic undertakings requiring almost super-human > determination and effort. Sure. Computational mathematics is a heroic undertaking requiring almost super-human determination and effort. Welcome to the 21'th century ! Han de Bruijn === Subject: Re: The modern mathematical concept of infinity is indefensible Sure. Computational mathematics is a heroic undertaking requiring > almost super-human determination and effort. [...] > Huh?! Man you are talking nonsense. (Well, not that uncommon for a crank.) Herb === Subject: Re: The modern mathematical concept of infinity is indefensible Consistent ultra-finitism, anti-realism about all of mathematics, > etc. are heroic undertakings requiring almost super-human > determination and effort. Sure. Computational mathematics is a heroic undertaking requiring > almost super-human determination and effort. What leads you to this pecular conclusion? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Surely the [alleged] existence of arbitrarily complex hereditarily > finite set is intimately tied to the limits and possibilities of > computer simulations. Of course. Since computer resources are bound (assuming that our > universe is finite), there are (or can be) only finitely many HFS. > Simple as that. Quite, if we are serious about the justification for the existence of such sets as being intimately tied to computer simulations. If we are prepared to accept the existence of arbitrarily complex hereditarily finite sets in our mathematical arguments and reasoning, and most of us are, without so much as blinking, this can't be because they can be simulated on any actual computer (since they in fact can't). Whatever computers are involved, and computers and computer implementations might well enter into our explanation of the world of hereditarily finite sets, serving as helpful illustrations perhaps, are thus just as theoretical, imaginary and dubious -- or just as innocuous -- as the arbitrarily complex hereditarily finite sets themselves. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible And the empty set is _implementable_ as computer hardware and software; > it's a piece of technology; therefore its existence is not disputable. Right. In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. > Let alone the existence of infinite sets!!! Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > And the empty set is _implementable_ as computer hardware and > software; it's a piece of technology; therefore its existence > is not disputable. Right. In contrast the existence of arbitrarily complex > hereditarily finite sets is very much disputable. Let alone the existence of infinite sets!!! Right! Their existence is in fact infinitely disputable. Incidentally, I seem to recall you killfiled me for being an asshole. Did something lead you to revise your opinion? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Right! Their existence is in fact infinitely disputable. Incidentally, > I seem to recall you killfiled me for being an asshole. Did something > lead you to revise your opinion? > Not really, but I've access to a different computer now, and I just didn't start to build up a killfile there. Still you will see that I'm not eager to engage in any (serious) discussion with you, I simply don't like your attitude(s). Herb === Subject: Re: The modern mathematical concept of infinity is indefensible Incidentally, I seem to recall you killfiled me for being an > asshole. Did something lead you to revise your opinion? Not really, but I've access to a different computer now, and I just > didn't start to build up a killfile there. Still you will see that I'm not eager to engage in any (serious) > discussion with you, I simply don't like your attitude(s). What sort of attitudes you or anyone finds congenial is of course entirely their business. Perhaps in the future we may both engage in and enjoy some entirely frivolous discussion. Is there something in particular in my attitude or attitudes that rubs you the wrong way? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > ... and there certainly isn't any Platonic question of > whether it really exists. > It isn't if we agree that math is just some sort of mythological story (i.e. a fiction). Stating ExAy(y !e x), as an axiom in ZFC (then) is just this: a statement in the (mathematical) story called ZFC. It's like stating Holmes had a friend called Watson in the context of one of the (fictional) stories by Sir Arthur Conan Doyle. On the other hand, there have been (and still are) mathematicians who are convinced that such a statement actually refers to a state of affairs in some reality (realm) ... Classes and concepts may, however, also be conceived as real objects, namely classes as pluralities of things or as structures consisting of a plurality of things and concepts as the properties and relations of things existing independently of our definitions and constructions. It seems to me that the assumption of such objects is quite as legitimate as the assumption of physical bodies and there is quite as much reason to believe in their existence. (K. G.9adel, Russell's Mathematical Logic) Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > It isn't if we agree that math is just some sort of mythological > story (i.e. a fiction). Stating ExAy(y !e x), as an axiom in ZFC (then) is just this: a statement in the > (mathematical) story called ZFC. It's like stating Holmes had a > friend called Watson in the context of one of the (fictional) > stories by Sir Arthur Conan Doyle. This is a very appealing idea for all sorts of reasons in case such claims as the existence of the empty set. In case of other statements, such as ZFC is inconsistent, every even natural greater than two is the sum of two primes and so on, the idea has considerably less appeal. In particular, we make use of such statements outside mathematics, e.g. in such non-mathematical claims as Even if ZFC is inconsistent, it might be impossible for us to ever find this out. Here ZFC is inconsistent is used, so to speak, as a factual hypothesis. In his dissertation Torkel refers to the idea that there is nothing more to the truth or falsity of claims of abstract set theory than what is inherent in the mathematical stories we tell as a distancing view. It is in practice very difficult to maintain such a view about /all/ of mathematics, and since abstract set theoretic principles have consequences we are naturally inclined to regard as determined as a matter of mathematical fact, such as whether ZFC is consistent or not, we are lead to ponder the interesting question of why we should accept consequences of fictional claims about fictional entities that are, so to speak, factual. It is also obscure just what reality purely abstract fictional entities not posited to ineract with us in any way could possess in addition to being parts of the conceptual stories we tell about them, and consequently just what is claimed in saying that, say, these or those mathematical objects don't actually exist. That is, just what it is we are denying in such claims? (In contrast, there is nothing obscure in noting that Sherlock Holmes or Captain Planet don't really exist.) -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > Suppose you never heard of any notion of number other than > natural number. Then someone asks you Does there exist a > number x such that x+1=0? (or such that x*2 = 1, or such that > x*x = 2, or such that x*x = -1) The answer is not simply yes or no. There is no such x among > the numbers that you are aware of, but we can coherently *posit* > the existence of negative numbers, rationals, reals and complex > numbers. It would be unreasonable for someone to speculate that > while it is *consistent* that there are negative numbers, they > don't *really* exist. Sure. But it is a rather odd notion that, say, the truth of Fermat's last theorem, consistency of ZFC, and so on, are a matter of coherence or stipulation, even though it makes perfect sense to suppose there is nothing more to the existence of the rationals or zero than our stipulation to that effect. Also, in case of the reals we seem to be introducing substantial ontological assumptions, about arbitrary sets or sequences of naturals, and so on; assumptions that have arithmetical consequences it would be odd to regard as open to stipulation, or determined merely by their coherence with these or those mathematical claims. (Merely introducing ontological assumptions need not be dubious, of course. I believe it was Kreisel who once cautioned against such doubts, remarking that they seem to be founded in thinking about mathematical objects as something akin to huge physical objects.) Of course, we must also note that the notion of consistency involved here is not the technical one applying to formal theories -- as you're well aware, consistency of a formal theory is insufficient to guarantee it describes any coherent mathematical picture -- and that this notion, whatever it amounts to exactly, is not philosophically unproblematic. > I think for *definable* sets, at least, (such as the empty set) > their existence is just a matter of stipulation. We *define* the > empty set to be the set with no elements, in the same way that we > define -1 to be the solution to x+1=0. There is nothing more to say > about the empty set, and there certainly isn't any Platonic question > of whether it really exists. If someone came forth and said the empty set exists or the empty set does not exist it would indeed be very odd to regard them as reporting a mathematical discovery, something they could be mistaken about. Naturally, we would take such assertions as stipulations, perhaps occurring in the context of introduction to an exposition of some theory of sets. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Supersedes: <1ax3y7zpprxre.eg5cegu645bl$.dlg@40tude.net> I mentioned the existence claim of the empty set as one of the axioms of ZFC: Ex(Ay(y !e x). Now you say: If someone came forth and said the empty set exists or the empty > set does not exist it would indeed be very odd to regard them as > reporting a mathematical discovery ... > Well: [...] despite their remoteness from sense experience, we do have something like a perception also of the objects of set theory, as is seen from the fact that the axioms force themselves upon us as being true. I don't see any reason why we should have less confidence in this kind of perception, and more generally, in mathematical intuition than in sense perception. (K. G.9adel) Herb === Subject: Re: The modern mathematical concept of infinity is indefensible I mentioned the existence claim of the empty set as one of the axioms of > ZFC: Ex(Ay(y !e x). Now you say: If someone came forth and said the empty set exists or the > empty set does not exist it would indeed be very odd to regard > them as reporting a mathematical discovery ... Well: [...] despite their remoteness from sense experience, we do > have something like a perception also of the objects of set theory, > as is seen from the fact that the axioms force themselves upon us as > being true. I don't see any reason why we should have less > confidence in this kind of perception, and more generally, in > mathematical intuition than in sense perception. On the conception of sets as arbitrary extensional collections in the cumulative hierarchy many axioms do force themselves upon us, in the sense that they seem evident and compelling. But before they can do anything like that we must have some idea of what we're talking about, some explanation. In saying that the existence of the empty set is a matter of stipulation I have in mind in part that such stipulation would be naturally regarded as a component in such an explanation of what we're talking about. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Supersedes: <10zwlr3t1erf1.1p2lyj0ihq5gw.dlg@40tude.net> I meantioned the existence claim of the empty set as one of the axioms of ZFC: Ex(Ay(y !e x). Now you say: If someone came forth and said the empty set exists or the empty > set does not exist it would indeed be very odd to regard them as > reporting a mathematical discovery ... > Well: [...] despite their remoteness from sense experience, we do have something like a perception also of the objects of set theory, as is seen from the fact that the axioms force themselves upon us as being true. I don't see any reason why we should have less confidence in this kind of perception, and more generally, in mathematical intuition than in sense perception. (K. G.9adel) Herb === Subject: Re: The modern mathematical concept of infinity is indefensible <873afofw9v.fsf@phiwumbda.org> <87tz84e8s8.fsf@phiwumbda.org> <1uwp3cfq5z0ox.1rebiiv9nmql7.dlg@40tude.net > I think you could make it work in Aczel set theory. > I don't think so. If 1 = {1} and 2 = {1,2}, then 1 and 2 are bisimilar and hence equal > in Aczel's theory. Sure? (Note that I know almost nothing about Aczel's theory). I mean 1 = {1} (in that theory) is the unique set x such that x = {x}. Now the unique set y such that y = {1, y} needn't be identical with 1, > imho. Well, since we know that (by definition) 1 = {1} and since {1} = {1,1}, we have 1 = {1,1} and so we know that 1 satisfies y = {1,y}. Hence, if, as you say, there is a *unique* such y, it must be identical to 1. I think what you are questioning is whether there *is* a unique such y. Aczel's anti-foundation axiom AFA asserts exactly that. But this is by no means self-evident. > After all 1 only has just one element, while this unique set, let's > call it 2, might have two elements; I mean it will iff 1 =/= 2. ;-) But maybe I'm just plainly wrong. Well, just a thought. You are not plainly wrong. AFA is only one of several reasonable anti-foundation alternatives and in fact on at least one, SAFA (Scott's anti-foundation axiom), there is a set satisfying y = {1,y} that is not identical to 1. It seems to me (without having put any thought into it) that the proposed definition of a series of n.w.f. finite ordinals satisfying 1 = {1}, 2 = {1,2}, ... is feasible under SAFA. Anyone? I'm very rusty here... === Subject: Re: The modern mathematical concept of infinity is indefensible > AFA is only one of several reasonable anti-foundation alternatives > and in fact on at least one, SAFA (Scott's anti-foundation axiom), > there is a set satisfying y = {1,y} that is not identical to 1. It > seems to me (without having put any thought into it) that the > proposed definition of a series of n.w.f. finite ordinals > satisfying 1 = {1}, 2 = {1,2}, ... is feasible under SAFA. Anyone? > I'm very rusty here... I am even rustier! Where can one learn more about Scott's only that it is a stenghtening of the axiom of extensionality... -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <873afofw9v.fsf@phiwumbda.org> <87tz84e8s8.fsf@phiwumbda.org> <1uwp3cfq5z0ox.1rebiiv9nmql7.dlg@40tude.net> <87eiz77m9i.fsf@alatheia.dsl.inet.fi> said: > AFA is only one of several reasonable anti-foundation alternatives > and in fact on at least one, SAFA (Scott's anti-foundation axiom), > there is a set satisfying y = {1,y} that is not identical to 1. It > seems to me (without having put any thought into it) that the > proposed definition of a series of n.w.f. finite ordinals > satisfying 1 = {1}, 2 = {1,2}, ... is feasible under SAFA. Anyone? > I'm very rusty here... I am even rustier! Where can one learn more about Scott's > anti-foundation axiom? Eh? Something about logic that you haven't already mastered? :-) Herb has given a good reference -- from a quick look, Smith's paper is basically a distillation of the first four chapters or so of Aczel's book. At one time the latter was out of print and available for free as a huge scanned PDF from CSLI, but it has been reprinted and appears once again to be available in hard copy only. > stengthening of the axiom of extensionality... Right. All of the anti-foundation axioms Aczel discusses can be so considered, and are needed (as you no doubt have surmised) because set identity isn't settled in general by extensionality in the non-well-founded realm. === Subject: Re: The modern mathematical concept of infinity is indefensible > said: I am even rustier! Where can one learn more about Scott's > anti-foundation axiom? Eh? Something about logic that you haven't already mastered? :-) As strange as it sounds there are such things. There are, so it seems, also many a thing I've managed to entirely forget about ever reading, some in books I've spent considerably time with, much to my shame no doubt. I have a vivid mental image of the notebooks MoeBlee keeps, in which he records, in tedious formal detail, everything he's learnt, every argument he's worked out in detail. Often I wish I had the patience to keep my notes in such meticulous order, but alas, I find I'm a very unsystematic reader, often having to revisit sources the identity of which I can't quite recall, going by nothing but the vague recollection that what I need was contained in a thick volume with yellow covers and a dent in the back. > Herb has given a good reference -- from a quick look, Smith's paper > is basically a distillation of the first four chapters or so of > Aczel's book. At one time the latter was out of print and available > for free as a huge scanned PDF from CSLI, but it has been reprinted > and appears once again to be available in hard copy only. It is, as you have now learned from a later post, still available. I did actually read it with some attention some time in the past, but apparently it has left nothing but vague traces clouded in time in my often not very relieable memory... > Right. All of the anti-foundation axioms Aczel discusses can be so > considered, and are needed (as you no doubt have surmised) because > set identity isn't settled in general by extensionality in the > non-well-founded realm. Indeed. Without some principled account of ill-founded sets it is simply not obvious whether e.g. there is a unique set x such that x = {x}, and so on, and how and on what grounds we are to approach such questions. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <87eiz77m9i.fsf@alatheia.dsl.inet.fi > I am even rustier! Where can one learn more about Scott's > anti-foundation axiom? > You will find something about it in http://bsmith7.asp.radford.edu/Hypersets.pdf Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > > I am even rustier! Where can one learn more about Scott's > anti-foundation axiom? You will find something about it in > http://bsmith7.asp.radford.edu/Hypersets.pdf Much obliged. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > But this means that sets can be, and often are, members of themselves, > which opens a different can of worms. [Virgil] In Aczel's set theory this is NOT a can of worms, but one of the distinct > features of that system. If Aczel's theory deals with it successfully, it only means that he has > a lid on that particular can of worms, not that the can does not exist. Huh? You are beginning to sound like a real crank. A: ...this means that sets can be infinite, which opens a can of worms. B: In ZFC this is NOT a can of worms, but one of the distinct features of > that system. A: If ZFC deals with it successfully, it only means that it has a lid on > that particular can of worms, not that the can does not exist. B: Well... (*argl*) P.S. Note that Aczel showed that his theory is consistent if ZFC is (iirc). > Actually his theory shows that allowing for unfounded sets does in no way > mean opening a can of worms (if dealt with in an appropriate framework, > of course). Hint: Aczel's theory is virtually identical with ZFC, just the axiom of > foundation is replaced with AFA (anti-foundation axiom). As far as I can tell (not far) axiom of foundation is more of an aesthetic judgement than a necessary foundational support for mathematics. What has the axiom of foundation done for me lately? According to wikipedia, Kuratowski ordered pairs are handier to work with: (a,b) = {a, {a,b}} rather than (a,b) = {{a}, {a,b}}. -- Michael Press === Subject: Re: The modern mathematical concept of infinity is indefensible It occurred to me a while ago that I've never seen a crank object to > Euclid's infinitude-of-primes argument. Oh, you don't know WM (M.9fckenheim) right? translation): This proof ignores the fact that for a set of 10^100 prime numbers no more > [prime] can be found, since there are no means available to store more than > 10^100 prime numbers. How to gauge the profound depth of such an insight? Chicken farmer: I have so many chickens, I don't know > how many there are, and the bookkeeping > seems hopeless. Have you got any advice > as a mathematician? > Matheologer: Try to use arithmetic. You apply numbers > to count the chicken, to decide about > each one's age etc. There are plenty of > numbers available to do all that stuff. > Chicken farmer: I don't know. I have darn'd many chickens, > and I doubt that arithmetic has enough > numbers to count all of them. > Matheologer: That is not a worry. No matter how many > chickens you have, the numbers will suffice. > The set of natural numbers is actually > infinite. This is guaranteed by Set Theory. > There is no bound to the amount of chickens > that it can deal with. > WM: Ha! I caught you now: Set Theory allows for > infinitely many chickens. But an argument > involving the age of the Universe and the > procreation method of chickens and some > arithmetic proves that there can only be a > finite number of chickens present at any time! > This contradiction proves that Set Theory is > dead wrong in everything, just as I thought! Stupidity is infinite. QED -- Michael Press === Subject: Re: The modern mathematical concept of infinity is indefensible You usually get an answer like start with 0 and add 1, and > just keep adding 1, until you do this an infinite number of > times, at which point you have reached an infinite natural. In case of an answer like that the obvious thing to do is to ask what > is meant by adding one an infinite number of times. at which point > as witnessed often on these groups > the immediate followup is to ask back > what does it mean to collect things an infinite number of times? In case of a follow-up like that the obvious thing is to suggest that > we set aside any worries or claims about infinite sets or set theory > in general and concentrate on the matter at hand. The point to these > tedious rejoinders is to try and make the confused student -- there is > no point in asking the crank anything -- explain what he means by an > infinite natural, by way of a clear definition in simple terms not > involving finitude or infinitude, the way we may explain that that > there are infinitely many naturals means just that for any natural n > there is a natural m > n, hopefully leading them to recognise their > confusion, that they do not in fact mean anything definite by the > locution. Sometimes this works and sometimes it doesn't. but often people coming in asking questions > if that is what they are doing > only have very vague notions they may not understand that the map from naturals > to certain collections > provides an answer to some of their intuitive notions they may not yet understand that there is an infinite ordinal > omega > which behaves in many the ways are expecting and they may not understand how it behaves differently by engaging in a discussion about infinite process What do you mean by infinite process? Seems that embarking on such a discussion with a student unpracticed enough to entertain the notion of an infinite natural number will only confuse the student. Better to stay with finite processes. Each natural is reached in a finite number of steps. That is a formulation for natural numbers. When that is cleared up we can address the question How many natural numbers are there? With the definition of natural number nailed down, it looks less feasible to speak about counting up the natural numbers one-by-one. And _now_ the only sane course is to consider the axiom EI(0 in I / Ax in I((x / {x}) in I)) in one way or another intelligible to the student. [...] -- Michael Press === Subject: Re: The modern mathematical concept of infinity is indefensible > What do you mean by infinite process? Seems that embarking on > such a discussion with a student unpracticed enough to entertain the > notion of an infinite natural number will only confuse the student. > Better to stay with finite processes. Each natural is reached in a > finite number of steps. That is a formulation for natural numbers. That may well be, but if you actually try that approach with a genuinely confused student of mathematics you'll find it is not particularly helpful. It is best to avoid all mention of finite processes, set theory, what not. We can formulate the claim that there are infinitely many naturals simply as for any given natural n, there is a natural m > n. The question now is: what is the analogous reformulation for the claim that there is an infinite natural, and why should that reformulation follow from the infinity of naturals, understood in the way explained above? For certainly the reformulation can't be there exists a natural n such that for all naturals m, n > m, which even the most confused student will usually grant is obviously false. Focusing on this question we needn't concern ourselves with any conundrums about sets, or even whether there /are/ infinitely many naturals. Dealing with one conglasses at a time is, I believe, the pedagogically preferable method. > When that is cleared up we can address the question How many > natural numbers are there? With the definition of natural number > nailed down, it looks less feasible to speak about counting up the > natural numbers one-by-one. And _now_ the only sane course is to > consider the axiom EI(0 in I / Ax in I((x / {x}) in I)) in one way or another intelligible to the student. Well, it is unlikely that a student confused about the infinity of naturals is much helped by abstract set theoretic reductions. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <873afofw9v.fsf@phiwumbda.org> <87tz84e8s8.fsf@phiwumbda.org> <1uwp3cfq5z0ox.1rebiiv9nmql7.dlg@40tude.net> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Let Omega be a set such that Omega = {Omega} (since the first of the > two above equations by itself is also a flat system of equations it > follows that such a set exists according to AFA). Then x = y = Omega > satisfies the above system, and since AFA says that the solution is > unique it follows that in Aczel's theory the y you define above is > indeed equal to x. > Thanx, that comment is rather helpful. Hence if you and Jesse F. Hughes are > right (which seems to be the case), the simple approach > 1 = {1} > 2 = {1, 2} > : > will not work in Aczel's theory. I've enjoyed reading this discussion about the theory Z+AFA, Aczel's Antifoundation Axiom. About a year ago, zuhair also wanted to come up with a theory in which one can have sets like 1 = {1}, and 2 = {1,2}, and so on. And he, just like Hughes and Rotwang, arrived at the exact same problem -- there's no way to prevent 1 = 2. We all agree that 1 = {1} is possible -- such a set is called a Quine atom. Now zuhair's axioms were: 1. Extensionality 2. Irregularity: Ax (xex) (so that not only the naturals, but every set would contain itself as an element) 3. Quine atom: Ex (Ay (yex <-> y=x)) So by Axiom 3, x = {x} exists (and we might as well let 1 be an instantiation of the axiom). Now zuhair wanted to form 2 = {1,2}, and guarantee that 2 is not 1. So he tried something like: 4. Successor: Az (Ex (~x=z & Ay (yex <-> (yez or y=x)))) Then applying the axiom to z = 1 gives: Ex (~x=1 & Ay (yex <-> ye1 or y=x)) so that x is a set distinct from 1 , whose elements are all of 1's elements along with x itself. So we can call this set 2, and have 2 = {1,2} distinct from 1. What doomed zuhair's theory is that he wanted to have a Comprehension Schema that he hoped would avoid Russell's Paradox, but someone showed him that the paradox would still occur anyway in a disguised form. > What a pity. I agree. I can see why Aczel would demand uniqueness in the solution to the system of equations: 1 = {1} 2 = {1,2} with 1 = 2 = Quine as the unique solution. Otherwise, as Rotwang points out, one couldn't guarantee that the Quine atom itself is unique. I've seen Aczel's axioms mentioned several times before, and I wouldn't mind seeing the axioms for myself. A web search for Z+AFA doesn't reveal any list of axioms, meaning that I'd have to buy a book to see the axioms. The book countdown is currently at 33 days, but this countdown is for a book about ZFC, not Z+AFA or an ill-founded or otherwise nonstandard theory. Books about nonstandard theories are even more elusive and more expensive than books about ZFC. === Subject: Re: The modern mathematical concept of infinity is indefensible sha1:670WY30I7foWn3uMK9m3igscfls= > About a year ago, zuhair also wanted to come up with > a theory in which one can have sets like 1 = {1}, and > 2 = {1,2}, and so on. And he, just like Hughes and > Rotwang, arrived at the exact same problem -- there's > no way to prevent 1 = 2. Sure there is! Just alter Aczel's axiom. His axiom says something like: every set of flat equations has a unique solution (pardon me if I don't state it quite right). Take away uniqueness, and you could well have distinct 1 and 2 defined as above. But the presumption that such solutions are unique make a much more interesting and useful theory. Without it, we have too few provable equalities. -- Many argue that its programmers have turned out shoddy programs, but [their] objective is to make profit, not superlative programs per se. By the profit criterion, Microsoft has been one of the greatest companies in the history of this country. -- ADTI defends Microsoft === Subject: Re: The modern mathematical concept of infinity is indefensible <873afofw9v.fsf@phiwumbda.org> <87tz84e8s8.fsf@phiwumbda.org> <1uwp3cfq5z0ox.1rebiiv9nmql7.dlg@40tude.net> posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) [...] I've seen Aczel's axioms mentioned several times before, > and I wouldn't mind seeing the axioms for myself. A web > search for Z+AFA doesn't reveal any list of axioms, > meaning that I'd have to buy a book to see the axioms. A Google search for anti foundation axiom led, via Wikipedia, to this: http://standish.stanford.edu/pdf/00000056.pdf I've just had a quick look and it appears that the version of the AFA given there is in terms of graphs, rather than flat systems of equations (the former being Aczel's original formulation). But it's the graph version of AFA applies: consider the graph G whose nodes are x, y and whose edge relation is a -> b iff in {,,} i.e. the graph which looks like this: _ _ / / | x <- y | | ^ ^ | -/ -/ (apologies for my piss-poor attempt at ascii art). Then a decoration of this graph is precisely a pair of sets x, y which satisfy the system of equations I gave earlier: x = {x} y = {x,y} so the existence and uniqueness of a decoration of G is equivalent to the existence and uniqueness of a solution to the flat system of equations. === Subject: Re: The modern mathematical concept of infinity is indefensible A Google search for anti foundation axiom led, via Wikipedia, to > this: http://standish.stanford.edu/pdf/00000056.pdf I've just had a quick look and it appears that the version of the AFA > given there is in terms of graphs, rather than flat systems of > equations (the former being Aczel's original formulation). > Maybe helpful: http://bsmith7.asp.radford.edu/Hypersets.pdf Herb === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) Again: The simple sketch O OO OOO OOOO OOOOO ... shows: the number (cardinality) of the natural numbers can't be larger as the natural numbers are able to number. And if there would be a number which is larger than any natural number, this number would be _too large_ to depict the cardinality of the natural numbers. Therefore, and because there is no largest natural number, there is no number which depicts the cardinality of the natural numbers (or any infinite totality). Infinity is just a mode, not a quantity. Infinitum actu non datur. Albrecht S. Storz === Subject: Re: The modern mathematical concept of infinity is indefensible > Again: The simple sketch O > OO > OOO > OOOO > OOOOO > ... shows: the number (cardinality) of the natural numbers can't be larger > as the natural numbers are able to number. It shows no such thing. And if there would be a > number which is larger than any natural number, this number would be > _too large_ to depict the cardinality of the natural numbers. There is and it isn't. At least outside of Albrecht's teeny world. Therefore, and because there is no largest natural number There is no smallest unit fraction, so that by Albrecht's argument, there cannot be a zero rational number.. Infinity is just a mode, not a quantity. Define mode and quantity. According to standard definitions there are everal infinite quantities. Infinitum actu non datur. Lying in latin does not help your case. Albrecht S. Storz === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > I guess, one of > the problems of this crank is to accept the (logical) possibility of > infinitely many finite natural numbers. > This is a common aliment among many of the sci.math cranks. > They can't accept that there are an infinite number of finite > naturals. I agree that many so-called cranks reject that there are infinitely many naturals. > Their personal take on it varies, though, including: > a) since there are an infinite number of naturals, there > must exist infinite naturals. This does sound a lot like the Transfer Principle. In some theories with a Transfer Principle, one can prove that any infinite set must necessarily contain a nonstandard element. Thus the infinite set of naturals must contain a nonstandard natural (colloquially called an infinite natural, since one can prove it to be greater than every standard finite natural). Therefore the Transfer Principle is one way to make this common claim of so many cranks rigorous. > b) the infinite set of all naturals does not exist. > c) potentially infinite sets may exist, but actual infinite > sets do not exist. > d) infinite sets do not have cardinalities. -drt === Subject: Re: The modern mathematical concept of infinity is indefensible <9cqvt0bupgxj$.n1gcnpcbbcig.dlg@40tude.net> <87ljthbxxh.fsf@alatheia.dsl.inet.fi> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > It is trivial though tedious exercise to formulate it in the language > of set theory. This is standard stuff. > OK. I consider the issue settled. Perhaps from now on, when I need to refer to a finitist theory, I'll write it as something like ZF-Infinity+Finite. Then I'll leave it open for interpretation whether Finite means the negation of Infinity, or Friedman's Finiteness Axiom, or whatever axiom is necessary to prove that in all models of the theory, all sets are finite, so the theory really is finitist. Of course, the point the others are trying to make here is that if it took several posts for Aatu, Bader, and MoeBlee to agree on what axiom is necessary to guarantee that every set is finite, then Albrecht never knew what the axiom was, and therefore Albrecht never knew what it takes to make a rigorous finitist theory. To think that it would take this much trouble just to work with finite objects! === Subject: Re: The modern mathematical concept of infinity is indefensible > To think that it would take this much trouble just > to work with _finite_ objects! The trouble has nothing to do with working with finite objects, but with arcane technicalities of formal theories, of no conceptual or philosophical significance. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <9cqvt0bupgxj$.n1gcnpcbbcig.dlg@40tude.net> <87ljthbxxh.fsf@alatheia.dsl.inet.fi> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) It is trivial though tedious exercise to formulate it in the language > of set theory. This is standard stuff. > OK. I consider the issue settled. Perhaps from now on, when I need to refer to a > finitist theory, I'll write it as something like > ZF-Infinity+Finite. Then I'll leave it open for > interpretation whether Finite means the negation > of Infinity, or Friedman's Finiteness Axiom, or > whatever axiom is necessary to prove that in all > models of the theory, all sets are finite, so the > theory really is finitist. So you are saying that you will purposefully make imprecise statements? -- m === Subject: Re: The modern mathematical concept of infinity is indefensible <87ljthbxxh.fsf@alatheia.dsl.inet.fi> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Perhaps from now on, when I need to refer to a > finitist theory, I'll write it as something like > ZF-Infinity+Finite. Then I'll leave it open for > interpretation whether Finite means the negation > of Infinity, or Friedman's Finiteness Axiom, or > whatever axiom is necessary to prove that in all > models of the theory, all sets are finite, so the > theory really is finitist. > So you are saying that you will purposefully > make imprecise statements? OK then, what would you (Mariano) consider to be a precise statement of a finitist theory -- one sufficiently precise that no one will question whether the theory really proves that every set is finite in the theory? In other words, how can I fill in the blank in: ZF-Infinity+ that is precise enough to guarantee that every set is finite, so that no one will claim that the theory could have a model with an infinite set. I don't want to make imprecise statements. I would very much love to write a precise theory which everyone will agree that it's finitist. But, as the discussion with Aatu, Bader, MoeBlee has shown, this is easier said than done. === Subject: Re: The modern mathematical concept of infinity is indefensible > I would very much _love_ to write a precise theory which everyone > will agree that it's finitist. But, as the discussion with Aatu, > Bader, MoeBlee has shown, this is easier said than done. The theory ZF without infinity but with the negation of infinity is not finitistically justified, and the complications we have been discussing are nothing but arcane technicalities. You'll have much more success in finding a finitistically justified theory if instead of set theory you concentrate your efforts on proof theory where you'll meet such exciting theories as primitive recursive arithmetic. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible It's only basic and elementar: natural numbers are both cardinal > numbers and ordinal numbers at the same time. There is no stage on the > ladder of natural numbers at which the cardinality exceed the > ordinality. This is entirely conformant with usual conventions. > It has nothing to do with your claim that the natural numbers number > themselves. > It also has nothing to do with any ?ladder. > The cardinality and the ordinality of 7 are both 7, but > 7= {0,1,2,3,4,5,6} , so BOTH of these DO exceed > EVERY one of the things they are numbering. Just an arbitrary thesis. With more relevance I say: 7 := 1, 2, 3, 4, 5, 6, 7 This is circular. You can't define 7 in this way. Exactly what aspect is circular here: X > XO > XOO > XOOO > XOOOO > XOOOOO > XXXXXXX > ? There is nothing defined here. There are only stated facts. That's why > there is nothing circular. The Seven is seven units long because it > is at the seventh position. You can read the definition > n := 1, 2, 3, ..., n > with the cardinal number on the left side of the := and ordinal > numbers on the right side. You are defining each natural in terms of itself, which is circular. In particular, 1 := 1 does not have a source for the 1 on the right hand side. When one starts with 0, a la von Neumann, there is no such problem, as each natural after 0 is defined solely in terms of prior naturals. === Subject: Re: Non-integrability (Riemann) of f_xy(x,y) What's my question: What warrants the conclusion by the author using as only assumption that y(x,t) is continuous, ( I included the full context of what the author derived) not only that y is twice differentiable, but that from: Del^2y/DeluDelv First that: Dely/Delv=g'(v) Then y=g(v)+h(u) Only conditions given by the author were that y(x,t) is continuous, together with (are these the conditions you wanted?. Only relevant mathematical information I thought was that y(x,t) is continuous) the conditions of the wave equation, which are: A tightly -stretched string between two fixed points 0 and L on the x-axis. The string is pulled back vertically a distance that is very small compared to the length L , and released at time t=0. Our problem is to determine the displacement y(x,t) that is x units away from the end 0, at any time t. Now, *Question* Do we just need that : Del^2y/DeluDelv be continuous (and therefore bounded in a rectangle, by compactness: what if instead of a rectangle, we have an open, simply-connected region of integration?, Would boundedness of Del^2y/DeluDelv be enough?). Also: I was hoping you would correct some of the comments I made, even with just a 'yes', or a short statement of why not: > You have to be much more careful about this > sort of thing. For example, the existence of > the iterated integral int_a^b (int_c^d g(x,y) dx) dy is _not_ the same as saying g is Riemann integrable > on the rectangle [c,d]x[a,b]; you need to decide > which notion you're interested in. > notion is there? A response here, please? And: Also, the existence of a derivative is not enough > to make the Fundamental Theorem of Calculus > work. > O.K, right, we need f' to be continuous on [a,b] too, > and then Int(a^b)(f')=f(b)-f(a). Just a comment. Yes/No, why not?. Please guive me a quick ref. of the relevant theorems, so I can re-review them. More: O.K, let me try to do that, piece-by-piece. > (For example, you assume that f_x and f_y > exist and then wonder why something about > f_xy does not follow. The answer to that > is that it doesn't even follow that f_xy > _exists_! And it would not exist because f_x may not be differentiable. Is this the crux? And a new one: If f_x (x,.) , and f_y(.,y) are continuous is that enough to guarantee a repeated appli- cation of the fundamental thm. of calc: Int(Int Del^y/DeluDelv)dudv)= Int[Int y_uvdu/delu)]delv)= Int[y_u]delv ??. Could you please tell me what conditions I need for the integrability of Int(Intf(x,y)dxdy)) other than the ones discussed above?.Does continuity and boundedness of f(x,y) guarantee the existence of the integral over, say, a simply-connected region with smooth or piecewise- smooth boundary? Do you suggest some sources? === Subject: Re: Non-integrability (Riemann) of f_xy(x,y) Hi: > Just trying to understand better why , given > f(x,y) > such that f_x, f_y exist , may not be > Riemann-Integrable , i.e., Int(Int )f_xydxdy , may > not exist . There's something missing from that sentence... I guess I would need to add that f_x is itself differentiable , or, equiv. that f itself is twice differentiable, to talk about f_xy, since if either of f_x or f_y are not continuous, then f is not differentiable, so that f_xy does not make sense. Is that it?. Yes, existence of f_x f_y alone does not guarantee differentiability of f, let alone existence of f_xy. (with f_xy = the partials of f with resp. to x and > to > y) (Over a bounded region) > > It would seem, naively, that we could apply the > fundamental thm. of calc twice, to get the > integral, > i.e: Int(Int f_xy dxdy)=Int[ Intf_x(x,.)dx]dy= (let the points a,b be the limits of integration for x ; let c,d be the limits for y) Int(f_y(a)-f_y(b))dy= f(a)-f(b)|_(y=c...d) You have to be much more careful about this > sort of thing. For example, the existence of > the iterated integral int_a^b (int_c^d g(x,y) dx) dy is _not_ the same as saying g is Riemann integrable > on the rectangle [c,d]x[a,b]; you need to decide > which notion you're interested in. > notion is there? > Also, the existence of a derivative is not enough > to make the Fundamental Theorem of Calculus > work. > O.K, right, we need f' to be continuous on [a,b] too, and then Int(a^b)(f')=f(b)-f(a). > If you state _all_ the hypotheses much more > carefully, state the non-conclusion more > carefully, then give an argument that seems > to you to show that the non-conclusion > actually follows, someone will point out > the error you're making. But in its present > form we really can't do that. > O.K, let me try to do that, piece-by-piece. > (For example, you assume that f_x and f_y > exist and then wonder why something about > f_xy does not follow. The answer to that > is that it doesn't even follow that f_xy > _exists_! Please see my second comment about the need for f_x itself to be differentiable. Is that it? Of course that doesn't answer your > real question, but it shows that the question > needs to be stated much more carefully. > I can't guess exactly what the real question > is, there are various things it could be. > If this is inspired by something you read, > transcribe what you read very carefully.) result used in the derivation of the D'Alembert solution to the wave equation in a Math Engineering book. The authors (Derrick and Grossman) just go on to conclude , after only assuming that y(x,t) (the displacement along the y-axis of a plucked string) is continuous, not only that : del^2y/(delu)(delv) (we used the C.Variables: u=x+ct and v=x-ct, c a constant ) exists, but that, given the above we can go from having del^2y/(delu)(delv)=0 To concluding that : dely/delv =g'(v), and from this, to concluding that y=g(v)+h(u) David C. Ullrich Understanding Godel isn't about following his formal > proof. > That would make a mockery of everything Godel was up > to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) === Subject: Re: Non-integrability (Riemann) of f_xy(x,y) <295pm4lkbtbrf7r4e3ngc38dt54e0sso4m@4ax.com> posting-account=fVOpuAkAAAB0gOUkQMH0DG_KdwTVgKXP Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) >Hi: > Just trying to understand better why , given > f(x,y) > such that f x, f y exist , may not be > Riemann-Integrable , i.e., Int(Int )f xydxdy , may > not exist . > There's something missing from that sentence... I wasn't talking about the math - the sentence > is simply not a sentence. The verb clause > may not be Riemann-Integrable has no > subject . I guess I would need to add that f x is itself > differentiable , or, equiv. that f itself is > twice differentiable, to talk about f xy, since > if either of f x or f y are not continuous, then > f is not differentiable, so that f xy does not > make sense. Is that it?. Yes, existence of f x > f y alone does not guarantee differentiability > of f, let alone existence of f xy. >(with f xy = the partials of f with resp. to x and > to > y) (Over a bounded region) > It would seem, naively, that we could apply the > fundamental thm. of calc twice, to get the > integral, > i.e: > Int(Int f xy dxdy)=Int[ Intf x(x,.)dx]dy= > (let the points a,b be the limits of integration > for x ; let c,d be the limits for y) > Int(f y(a)-f y(b))dy= f(a)-f(b)| (y=c...d) > You have to be much more careful about this > sort of thing. For example, the existence of > the iterated integral > int a^b (int c^d g(x,y) dx) dy > is not the same as saying g is Riemann integrable > on the rectangle [c,d]x[a,b]; you need to decide > which notion you're interested in. notion is there? > Also, the existence of a derivative is not enough > to make the Fundamental Theorem of Calculus > work. O.K, right, we need f' to be continuous on [a,b] too, > and then Int(a^b)(f')=f(b)-f(a). > If you state all the hypotheses much more > carefully, state the non-conclusion more > carefully, then give an argument that seems > to you to show that the non-conclusion > actually follows, someone will point out > the error you're making. But in its present > form we really can't do that. O.K, let me try to do that, piece-by-piece. > (For example, you assume that f x and f y > exist and then wonder why something about > f xy does not follow. The answer to that > is that it doesn't even follow that f xy > exists ! Please see my second comment about the need for > f x itself to be differentiable. Is that it? Of course that doesn't answer your > real question, but it shows that the question > needs to be stated much more carefully. > I can't guess exactly what the real question > is, there are various things it could be. > If this is inspired by something you read, > transcribe what you read very carefully.) result used in the derivation of the D'Alembert > solution to the wave equation in a Math Engineering > book. The authors (Derrick and Grossman) just go > on to conclude , after only assuming that y(x,t) > (the displacement along the y-axis of a plucked > string) is continuous, not only that : del^2y/(delu)(delv) (we used the C.Variables: u=x+ct and v=x-ct, c a constant ) exists, but that, given the above we can go from having del^2y/(delu)(delv)=0 To concluding that : dely/delv =g'(v), and from this, to concluding that y=g(v)+h(u) > Let us know if you ever decide what your question is. > David C. Ullrich > Understanding Godel isn't about following his formal > proof. > That would make a mockery of everything Godel was up > to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) David C. Ullrich Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) David, It's clear to me that f(x,y) is the entity that (according to the OP) may not be Riemann-integrable. I think that's obvious from the quote: i.e., Int(Int )f xydxdy , may not exist . Paul Epstein === Subject: Re: prime numbers formula please. <648574.1231877305848.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) miss.helpful a .8ecrit : > On 11 Jan., 13:34, Axel Vogt Not a prove ... but you may search for Jones' > formula. > Why? Is it the same formula? Jones paper does not > seem to > be openly accessible :-/ This link should help : http://www.joma.org/images/upload library/22/Ford/Jone > sSatoWadaWiens.pdf Well it's not your formula but interesting too! > Fine discovery, > Raymond I have a problem with this. It looks to me like the number must be composite or > 0, because it is of the > form (k+2) * (polynomial) with the variables ranging > over non-negative > integers, so the first term must be >= 2 What am I missing? well (k+2) * polynomial with polynomial of the form 1 - (..)^2 - (..)^2 - ... thus an upperbound of polynomial = 1. then k + 2 * poly = k + 2 = prime. > tommy1729 ps : note (..)^2 + .. are equivalent to simul diophantine equations. i suppose simul diophantines are always at the basis of prime representing polynomials ? disproof anyone ? disproof of *what*? -- m === Subject: Re: prime numbers formula please. > ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. > This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. > I hope you people could help me out - > I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. > Hope I came to right address, but if I didn't I'd appreciate > your help. > thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) Hi Ofer, thank you for bringing this interesting question to my >attention. I worked for the last 19 years on your problem, >now, finally, I found the solution. It is simple and uses >only the power function n^m and the factorial function n!. W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; Now take the numerator(1/W(n)). I have written it as 'Maple' code, but I think it is >easy to rewrite it with mathematical symbols. A first test showed this: seq(numer(1/W(i)),i=0..49); 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, >1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, >1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, Hope it helps! P.S. Perhaps some member of sci.math can supply a short proof? >Very interesting. >One minor suggestion ... >In your formula, replace every occurrence of n by (n-1). That way, the >indices won't be off by 1 when you produce the list. In other words, >define W by >W := proc(n) local v,k; >add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) >*(v+2)^(n-1),v=0..k),k=0..n-1) end; >Then your conjecture appears to be as follows ... >Let w(n) be the numerator of the fraction 1/W(n), when reduced to >lowest terms. Also, why the flip? Instead, perhaps say it this way: Let w(n) be the denominator of W(n), when reduced to lowest terms. >Conjecture (miss.helpful's Primality Test): >If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. I can prove half of the result, namely: If n is prime, then w(n)=n. It's cute. My proof uses both Wilson's theorem and Fermat's little theorem. quasi === Subject: Re: prime numbers formula please. > ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. I hope you people could help me out - I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. Hope I came to right address, but if I didn't I'd appreciate > your help. thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) >Hi Ofer, >thank you for bringing this interesting question to my >attention. I worked for the last 19 years on your problem, >now, finally, I found the solution. It is simple and uses >only the power function n^m and the factorial function n!. > W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; > Now take the numerator(1/W(n)). >I have written it as 'Maple' code, but I think it is >easy to rewrite it with mathematical symbols. >A first test showed this: > seq(numer(1/W(i)),i=0..49); >1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, >1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, >1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, >Hope it helps! >P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the >indices won't be off by 1 when you produce the list. In other words, >define W by W := proc(n) local v,k; >add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) >*(v+2)^(n-1),v=0..k),k=0..n-1) end; Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to >lowest terms. >Also, why the flip? Instead, perhaps say it this way: >Let w(n) be the denominator of W(n), when reduced to lowest terms. >Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. I can prove half of the result, namely: If n is prime, then w(n)=n. It's cute. My proof uses both Wilson's theorem and Fermat's little theorem. I have both halves now -- the other half was actually the easy one (and I already had it, but didn't realize it). quasi === Subject: Re: prime numbers formula please. > ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. > This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. > I hope you people could help me out - > I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. > Hope I came to right address, but if I didn't I'd appreciate > your help. > thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) Hi Ofer, thank you for bringing this interesting question to my >attention. I worked for the last 19 years on your problem, >now, finally, I found the solution. It is simple and uses >only the power function n^m and the factorial function n!. W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; Now take the numerator(1/W(n)). I have written it as 'Maple' code, but I think it is >easy to rewrite it with mathematical symbols. A first test showed this: seq(numer(1/W(i)),i=0..49); 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, >1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, >1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, Hope it helps! P.S. Perhaps some member of sci.math can supply a short proof? >Very interesting. >One minor suggestion ... >In your formula, replace every occurrence of n by (n-1). That way, the >indices won't be off by 1 when you produce the list. In other words, >define W by >W := proc(n) local v,k; >add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) >*(v+2)^(n-1),v=0..k),k=0..n-1) end; >Then your conjecture appears to be as follows ... >Let w(n) be the numerator of the fraction 1/W(n), when reduced to >lowest terms. Also, why the flip? Instead, perhaps say it this way: Let w(n) be the denominator of W(n), when reduced to lowest terms. >Conjecture (miss.helpful's Primality Test): >If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. >I can prove half of the result, namely: >If n is prime, then w(n)=n. >It's cute. >My proof uses both Wilson's theorem and Fermat's little theorem. I have both halves now -- the other half was actually the easy one >(and I already had it, but didn't realize it). Oops -- the other half doesn't yield so easily, it was an illusion. But I do have a proof of the first half (the case where n is prime). quasi === Subject: Re: prime numbers formula please. <6svcn6F8f8g9U1@mid.dfncis.de> posting-account=NbUMfwoAAADsMBfKflSkbPXwYcSgauQP Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Am 11.01.2009 18:09 schrieb quasi: > ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. I hope you people could help me out - I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. Hope I came to right address, but if I didn't I'd appreciate > your help. thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) > Hi Ofer, > thank you for bringing this interesting question to my > attention. I worked for the last 19 years on your problem, > now, finally, I found the solution. It is simple and uses > only the power function n^m and the factorial function n!. > W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; > Now take the numerator(1/W(n)). > I have written it as 'Maple' code, but I think it is > easy to rewrite it with mathematical symbols. > A first test showed this: > seq(numer(1/W(i)),i=0..49); > 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, > 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, > 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, > Hope it helps! > P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the > indices won't be off by 1 when you produce the list. In other words, > define W by W := proc(n) local v,k; > add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^(n-1),v=0..k),k=0..n-1) end; Another formulation: (Pari/GP) { W(n)=local(m=n-1,result); > result= m! * sum(k=0,m, 1/(k+1) * sum(v=0,k,binomial(k,v)*(v+2)^m) ); > return(result); } MH Primetest(n) = numerator(1/W(n)) if n is prime -> n > if n is composite -> 1 This way W(n) looks a bit more familiar.. Gottfried Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to > lowest terms. Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. quasi There are a number of useful binomial coefficient identities you can apply to this, although I haven't had enough time to play with it long enough to see if any yield an expression simpler than the original. But you can start by expanding (v+2)^m as a sum and then changing order of summations, and there's a number of directions to proceed from there. I suspect it can't be simplified too much, but still might be worth checking out. === Subject: Re: prime numbers formula please. Peter Webb a .8ecrit : This link should help : > http://www.joma.org/images/upload_library/22/Ford/JonesSatoWadaWiens.pdf > Well it's not your formula but interesting too! > Fine discovery, > Raymond I have a problem with this. It looks to me like the number must be composite or 0, because it is of > the form (k+2) * (polynomial) with the variables ranging over > non-negative integers, so the first term must be >= 2 What am I missing? If one or more of the squares is not zero the polynomial (and the result) will be 0 or negative so that all the squares must be 0. See here for a earlier discussion : I think this is the Wilson theorem in disguise! Hoping it helped, Raymond === Subject: Re: prime numbers formula please. <496a38b3$0$18380$ba4acef3@news.orange.fr> <496b0b39$0$3253$afc38c87@news.optusnet.com.au> <496bdd51$0$18371$ba4acef3@news.orange.fr> posting-account=JYIhQwoAAADfOqRKeWW3k_OkLnLoycin Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) Raymond Manzoni a .8ecrit : > I think this is the Wilson theorem in disguise! I doubt this. Why? No one came in here and said You are not Santa Claus, you are Papa! And this, I think, would have happened on sci.math if it is a simple disguise. On more formal ground: Both propositions are equivalent. So it is only a matter of effort and ten pages of paper to derive one formulation from the other. However, this does not qualify to name it a disguise. My definition of a disguise would be: One can give a three line proof deriving one proposition from the other. Till now no one had given one. === Subject: Re: prime numbers formula please. miss.helpful a .8ecrit : > Raymond Manzoni a .8ecrit : I think this is the Wilson theorem in disguise! I doubt this. Why? No one came in here and said > You are not Santa Claus, you are Papa! And this, I think, would have happened > on sci.math if it is a simple disguise. On more formal ground: Both propositions are > equivalent. So it is only a matter of effort > and ten pages of paper to derive one formulation > from the other. However, this does not qualify > to name it a disguise. My definition of a disguise would be: > One can give a three line proof deriving one > proposition from the other. Till now no one had given one. Except that... Peter Webb and myself are talking about Jones&all formula... http://www.joma.org/images/upload_library/22/Ford/JonesSatoWadaWiens.pdf === Subject: Re: prime numbers formula please. <496a38b3$0$18380$ba4acef3@news.orange.fr> <496b0b39$0$3253$afc38c87@news.optusnet.com.au> posting-account=JYIhQwoAAADfOqRKeWW3k_OkLnLoycin Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Raymond Manzoni a .8ecrit : > I think this is the Wilson theorem in disguise! > Except that... Peter Webb and myself are talking about Jones&all > formula...http://www.joma.org/images/upload library/22/Ford/JonesSatoWadaWiens.pdf Hi Raymond, you are disappointing me ;-) This old and complicated formula no one uses ;-)) But even more: you can stop to conjecture about the origin of the Jones et alia formula. They explicitly say on page 452: The fundamental tool in both constructions is Wilsons's theorem which characterizes the primes in terms of the factorial function. Voil.88! === Subject: Re: prime numbers formula please. ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. I hope you people could help me out - I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. Hope I came to right address, but if I didn't I'd appreciate > your help. thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) > Hi Ofer, > thank you for bringing this interesting question to my > attention. I worked for the last 19 years on your problem, > now, finally, I found the solution. It is simple and uses > only the power function n^m and the factorial function n!. > W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; > Now take the numerator(1/W(n)). > I have written it as 'Maple' code, but I think it is > easy to rewrite it with mathematical symbols. > A first test showed this: > seq(numer(1/W(i)),i=0..49); > 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, > 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, > 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, > Hope it helps! > P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the > indices won't be off by 1 when you produce the list. In other words, > define W by W := proc(n) local v,k; > add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^(n-1),v=0..k),k=0..n-1) end; Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to > lowest terms. Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. quasi Ok, but how do you reduce 'to lowest terms' (which is, what Maple probably does internally) without having the primes? Those numbers are quite large ... === Subject: Re: prime numbers formula please. <6t1qijF8pjhpU1@mid.individual.net> posting-account=JYIhQwoAAADfOqRKeWW3k_OkLnLoycin Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Ok, but how do you reduce 'to lowest terms' (which is, what Maple > probably does internally) without having the primes? > Those numbers are quite large ... First let me say that I use from now onward the enumeration proposed by quasi and helms, which is clearly the more natural one. Now back to your question. Axel, I give you a hint: Multiply the sum by 'n' (the input). What happens then? === Subject: Re: prime numbers formula please. ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. > This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. > I hope you people could help me out - > I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. > Hope I came to right address, but if I didn't I'd appreciate > your help. > thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) > Hi Ofer, thank you for bringing this interesting question to my > attention. I worked for the last 19 years on your problem, > now, finally, I found the solution. It is simple and uses > only the power function n^m and the factorial function n!. W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; Now take the numerator(1/W(n)). I have written it as 'Maple' code, but I think it is > easy to rewrite it with mathematical symbols. A first test showed this: seq(numer(1/W(i)),i=0..49); 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, > 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, > 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, Hope it helps! P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the > indices won't be off by 1 when you produce the list. In other words, > define W by W := proc(n) local v,k; > add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^(n-1),v=0..k),k=0..n-1) end; Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to > lowest terms. Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. quasi Ok, but how do you reduce 'to lowest terms' (which is, what Maple >probably does internally) without having the primes? Those numbers are quite large ... The Euclidean Algorithm makes finding the GCD very easy, and thereby, reducing to lowest terms very easy. Rob Johnson take out the trash before replying === Subject: Re: prime numbers formula please. ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. > This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. > I hope you people could help me out - > I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. > Hope I came to right address, but if I didn't I'd appreciate > your help. > thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) > Hi Ofer, thank you for bringing this interesting question to my > attention. I worked for the last 19 years on your problem, > now, finally, I found the solution. It is simple and uses > only the power function n^m and the factorial function n!. W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; Now take the numerator(1/W(n)). I have written it as 'Maple' code, but I think it is > easy to rewrite it with mathematical symbols. A first test showed this: seq(numer(1/W(i)),i=0..49); 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, > 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, > 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, Hope it helps! P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the > indices won't be off by 1 when you produce the list. In other words, > define W by W := proc(n) local v,k; > add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^(n-1),v=0..k),k=0..n-1) end; Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to > lowest terms. Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. quasi Ok, but how do you reduce 'to lowest terms' (which is, what Maple >probably does internally) without having the primes? Using GCDs (i.e. the Euclidean Algorithm). >Those numbers are quite large ... In any case, to say w(n)=1 is equivalent to saying that W(n) is an integer, which can be checked with a single division. But note: miss.helpful is not claiming that the test is algorithmically efficient, but rather, that it's a simple algebraic formula which, for some mysterious reason (which she will reveal soon), distinguishes primes from composites. numerical evidence, the formula seems valid, but if it is valid, why so? quasi === Subject: Re: prime numbers formula please. Am 12.01.2009 22:48 schrieb quasi: > numerical evidence, the formula seems valid, but if it is valid, why > so? quasi Something in the formula reminds me of the bernoulli/Clausen-von Staudt- theorem and also the wilson-prime-test should be examined, because we have (n-1)! in the formula for a test of isprime(n). Just ideas... Gottfried === Subject: Re: JSH: Research speaks for itself posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) > Through the years I've been certain that useful research will be > used. Seems like it makes sense but hey, it often seems like a crazy > world, and I have experience with failure with my research. For years > I actually *was* a math crackpot, Indeed you were. But that's all past. > with numerous failed attempts at > proving Fermat's Last Theorem. Lots of failure. Years of it. But I > believed that research speaks for itself so that if I were right, it > wouldn't be about me and failure, it would only be about the success. > It would be. If. > If I could ever succeed. > Yes. If. > The more astute of you may have noticed a sharp dichotomy in recent > exchanges on the newsgroups: ----> I emphasize web searches primarily on research results, citing > information as key. ----> Argumentative posters emphasized searches on me, and disdained > search results as important tools. I could keep going after failure because I found I just liked fiddling > around with equations, as it occupied my time, and in accepting that a > good reason for what I was doing was enjoyment and not the end itself, > as there is no end, I started getting real results, and found this > massive problem in abstract number theory which has tested my belief > that research speaks for itself. > Now you are doing what people love the most. You are bragging and preaching. People just love to listen to braggarts and windbags. Please continue to do more bragging. Not only is is pleasant to read, but also it is so informative. People here are not interested in mathematics. What they want most are the blitherings of an immodest, self-important and delusional windbag. We love it. > In the past I'd attack my own credibility considering it useless in > considering mathematical proofs. > Oh yes. The days when you tried to make yourself a Discredited Source. What a brilliant strategy. Or the time that you were going to attack your own work. Great, and it was so convincing. You had me going. You actually convinced me that you didn't know your rectum from a hole in the ground. > Now I have to convince that experts in the field of mathematics are > for some reason or other ignoring quite a few major discoveries > including a find of a massive error in some abstract number theory and > the credibility issue has taken a new twist. > That is what you have to do all right. The best way to do that is to brag and pontificate and whine. You are doing exactly what you should do. Your adoring public loves you for it. More bragging, more patting yourself on the back, more whining. Everyone loves it. Bring it on. It is surprising that Gauss didn't try the same approach. No wonder he is so underrated. > What is more credible? People or their results? > Must be a trick question. Posts like this one, with no math or results whatsoever, just a lot of hot air and bragging, are what people most want. That is what all the great mathematicians of the past have done. Give us more. > I and my opponents have taken two different paths in answering those > questions. > Not that different. You brag, they adulate. What else could they do? Hold their noses when you walk by? > Our two paths are revealed through Google searches: 1. I've emphasized research. Explaining, explaining, explaining. > I've worked at simplifying my prior results and worked at making more, > and even focused on their usefulness. My focus has been on research. > Much too modest. What you need to say here is, you have expressed yourself with complete mathematical rigor and maximal elegance. Your exposition is so simple that it can be understood by the brain of a mosquito. No one has laid a glove on your arguments. Then you top it all off with truly superlative bragging. That is what wows your huge loyal fan club. Give us more gushing, self-congratulatory bragging. Lay it on thick. We never get enough. Everybody loves you for it too. All the world loves a braying braggart. A lot of people actually sneer at Riemann because he was so modest. Dumb , never tried to tell people he was the King. At least you have avoided that mistake! > Results: Google search results linking to my research in key areas. > e.g. Google: define mathematical proof > That is proof, absolute proof positive. High ranking in Google searches is an infallible way to tell whether or not you are a crank. Point this out over and over again. Your disciples and worshippers just eat it up. They cannot get enough of your incessant bragging. It is much the same as the phenomenon of people smelling farts in the elevator. How sweet they are and how sweet are your ravings. > Or, Google: solving binary quadratic Diophantine > Better yet. The testimonials to your method just pop up everywhere. What is amazing is how many of those posts cited by Google say your method is the absolute best. Tippy top best, they all say. You should make sure to tell everyone that you are a genius and that they are just warts on a toad. They want that and need it. Please resume bragging to remind everyone, right now. > sanity to repeating over and over again that I'm wrong, or that even > if I'm right nothing I have is important, it has been a non-stop > attack on my worth. Their focus: personality. > What demeaning skullduggery. Clearly this is not about you. Never has been. When you brag you have done something, people are so appreciative to be informed, to learn some new math or physics. They love posts like this one because they are so dense with mathematics and new ideas, but most of all they just love to hear your brag. Who can blame them? > Results: Google searches on my name bring up a crank.net webpage > against me in the top 20 on major search engines, including Google and > Yahoo! which I just checked. On Google the flame page came up #20, on > Yahoo! it came up #18. > Shocking. Well, you should make the most of it. Let everyone know that you are a top-ranking crank. Not everyone can be, you know. Keep trying! Of course there are jealous people out there who want to out-crank you. Keep bragging and yelling about how good you are, and be assured you will move up in the crank ratings. And remember: bragging about being one of the greatest cranks of all time is what people most want to hear. I just cannot say it enough: People love bragging. And more is better. But you know that. > So the search engine results reveal two competing strategies. > Fascinating. In fact I am surprised that you didn't use the word 'fascinating' more often in this post, because it really is fascinating how these Google searches prove your worth. As your fellow Superior Being, I think I am entitled to say how fascinating you are, and further how fascinating it is that I am so very good at discerning how fascinating you are. No one is better at it than I. Not to brag, but Google searches prove that I am your top-ranked fan. When it comes to being fascinated by your magnificent accomplishments I am the best. > I have yet to find searches on my name bringing up ANY of my > research. Not any. Incredible. Just incredible. Not any, you say? I am floored. Of course the same is true with Gauss. Just do a search on his name. Nothing about what he actually did ever comes up. Never a mention of his work in number theory, analysis, algebra, geometry, physics, astronomy, etc.. Just nothing, just like you. The problem with Gauss - and with you, to be frank - is, you have not sufficiently tooted your own horn. Too modest. So was Gauss. Of course he didn't actually have that much to show for all his efforts. Pretty good at arithmetic apparently. But you, my man, you take the cake. You really MUST brag more to make that clear. Did I mention that people love it? Remember the old Doris Day song ... Every-body Loves a braggart I'm a braggart Everybody loves me, Yes they do! And I love every braggart Since started bragging to you! > So no, I can't just search on my name and come up > with my research on solving binary quadratic Diophantine equations, or > defining mathematical proof. > Told ya. You gotta brag more. Remember George Bush, and that big Mission Accomplished sign on the aircraft carrier? Now that is the kind of example you should follow! Get out that old U.S. Army uniform and get a big red flag that says, I am the World's Biggest Crank! Mission Accomplished! People need to know. They will fall down at your feet and kiss your bunion. > Their strategy of personal attacks worked in that sense. Their > continued use of that strategy can be seen in reply after reply after > reply on these newsgroups. > It's despicable. You brag, they try to tear you down. Remember, it's just a few rotten apples in the bottom of the barrel. Most sci.math readers really love you. Many of them would like to cuddle up with you. You can tell that there is a huge silent majority out there that loves you, because (1) they are VERY silent - it would take millions and millions of fans out there to build up such an enormous amount of silence, and (2) look at how high you rank in those Google searches. Amazing. (3) not one of them - NOT ONE - says you are right. That massive silence just proves how many people actually think you are right but they don't say anything, because they are silent! You should brag about this! > And my disdain for that strategy can be seen in my responses: I > believed and believe that research speaks for itself. > Make no mistake, it does. It really does. But it is so far above the head of the average Joe that it might as well be just a disgusting, smelly, putrid, fetid, stinking, load of . The only thing you can do to counter this reaction is to brag. Or, if you prefer, boast. That way people will want to suck up to you and they will understand how extremely good you really are. Just TELL them. Brag or boast, either one. Both work like a charm. > They clearly believe in the usefulness of personal attacks. Attack > the person to attack the research. > That's what they do! All the time! And if they are so stupid as to include actual mathematical content in what they say, you give them the old heave-ho and just ignore them. The ones you need to answer here are the Uncle Al's of this world, who just treat you derisively. Imagine that, your fellow physicist, and he ///seems/// to hold you in contempt. Probably just kidding of course. Make sure Uncle Al knows you are the best. Drop him a personal e-mail that includes some of your very best bragging. He will thank you for it. Please, do it right now. Uncle Al is waiting. Or better yet, find out where he lives and go visit him. Take along your extremely popular and successful best-selling book. Give him a free copy and see what he says. I bet he will just break down and cry from sheer gratitude. He puts on the gruff act here, but inside his evil shrivelled bigotted little black heart, there is another even littler heart that is even blacker and more shrivelled, and that is the true Uncle Al, who loves you. Please, go see him and give him a hug. He will thank you as only Uncle Al can thank someone. > But what kind of scientists would engage in such behavior? > Good question. Hmm. Short ones? Ones with red hair? Republicans? Scientists who watch World Wrestling Federation? Scientists with anal warts? I don't have a clue. Tell me and tell us all. But don't stint on the bragging. That is what we hunger for. > Oh, oops! My mistake. Talking about members of the mathematical > community, not scientists. > Certainly not. Witch-doctors, more like. > Behavior speaks for itself. > God knows. But it's not enough. You need to brag more so that people will (1) love you and (2) not overlook your amazing discoveries, e.g., your APPARENT proof that the algebraic integers are thoroughly buggered. > My situation is about time. I see Google search results as leading > indicators of world interest. I see the focus of search engine > results on my research and not on me as a brilliant demonstration that > it's not the person--it's their work. > Well you need not fear that there is a focus on you as brilliant. As for your work: I am mindful of the time I ate too much and drank too much and passed out for 18 hours without having a bowel movement. When I finally got to it, I produced a turd that was 2.8 feet long and thick as an ear of corn. I was so proud. And this is how you should boasting about it. That is what people love to hear you say. More! > I am very tempted to just kick back and relax, turning my problem > solving skills to more natural things, like pursuing extremely > beautiful women (now that takes a great deal of genius, a problem > worthy of my skills). > Absolutely. Women love a braggart even more than men do. Oh James - tell me again how you discovered Nonpolynomial Factorization - you are such a genius - and please, please James, tell me more about your gigantic turd! Truly, I just love to hear you brag! > However, I cannot simply walk away from the reality of some people > abusing their position to maintain their ability to teach wrong > mathematics, nor deny my responsibility to help in the end of the use > of failed ideas, so that correct ones can be used, which can further > the advancement of the entire human species. > Yes! Well put! You have a responsibility first and foremost to Yourself. You are not just the Best Mathematician of All Time. You are God. People yearn for salvation, and they can only be saved by Your suffering! > To the world---They are your children: both the ones being taught > false information now, and the ones yet to be born, who may need the > continuing pursuit of knowledge. > do. Verily you have straightened them out. Unto us a Gift is given. Ask not what your country can do for your. Good King Wenceslas. Able was I ere I saw Elba! Garbonzo, Roz? No, brag! James Harris is Lord. We three kings of Orient are. Psalm 38, James. Look it up. > Don't allow a world that wakes up in some distant future when the > reality of the math error and its importance are finally realized, > belatedly, only to find it is too far behind to catch up to the > solutions it needs for problems we cannot imagine today any more than > Galileo or Newton could imagine ours. > Yup, yup, yup. A sentence worthy of Henry James or Henry Ford. Can't decide which. Should be longer. > The fate of the world is not about personalities. Or who can be more > creative in slamming the other guy. > No indeed. > It's about the research that speaks for itself. If you listen. > (Listens intently for several hours) Hmm. Nothing coming through. You need to brag some more. > The future of our world depends on people asking questions. > It does? Marcus. > James Harris === Subject: Re: JSH: Research speaks for itself [...] > do. Verily you have straightened them out. Unto us a Gift > is given. Ask not what your country can do for your. Good > King Wenceslas. Able was I ere I saw Elba! Garbonzo, Roz? > No, brag! James Harris is Lord. We three kings of Orient > are. Psalm 38, James. Look it up. ----------------- 19:038:004 For mine iniquities are gone over mine head: as an heavy burden they are too heavy for me. 19:038:005 My wounds stink and are corrupt because of my foolishness. 19:038:006 I am troubled; I am bowed down greatly; I go mourning all the day long. 19:038:007 For my loins are filled with a loathsome disease: and there is no soundness in my flesh. 19:038:008 I am feeble and sore broken: I have roared by reason of the disquietness of my heart. ----------------- _That_ Psalm 38? -- Michael Press === Subject: Re: JSH: Research speaks for itself posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) > [...] do. Verily you have straightened them out. Unto us a Gift > is given. Ask not what your country can do for your. Good > King Wenceslas. Able was I ere I saw Elba! Garbonzo, Roz? > No, brag! James Harris is Lord. We three kings of Orient > are. Psalm 38, James. Look it up. ----------------- > 19:038:004 For mine iniquities are gone over mine head: as an heavy > burden they are too heavy for me. 19:038:005 My wounds stink and are corrupt because of my foolishness. 19:038:006 I am troubled; I am bowed down greatly; I go mourning all the > day long. 19:038:007 For my loins are filled with a loathsome disease: and there is > no soundness in my flesh. 19:038:008 I am feeble and sore broken: I have roared by reason of the > disquietness of my heart. > ----------------- _That_ Psalm 38? -- > Michael Press That's the one. Not all the Psalms are soothing little poems. Marcus. === Subject: Re: JSH: Research speaks for itself posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Or, Google: solving binary quadratic Diophantine Google: verbing nouns === Subject: Get instructor's solutions manual posting-account=EmCVswoAAACJGR09me6whGKmUcssb5cc Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) Do you suffer from a tough class? Are you looking for instructor solutions manual to do your homework? Just send me email with its name and edition and I may be able to help you in low price! It is my list, however if you don't find it here don't give up because it is only a list of some. Please , DO NOT REPLY HERE , instead send me email to : cartermath (at) gmail(dot)com instructor solution manual for A First Course in Differential Equations (7th ed.) 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Beer) instructor solution manual for Vector Mechanics for Engineers Statics & Dynamics (6th Ed., Ferdinand P. Beer) instructor solution manual for Wireless Communications Principles and Practice, 2nd Ed, by Rappaport === Subject: Solution manual for Introduction to functional Analysis by Kreyzig posting-account=z0CEbwoAAAC4Yfd6z5bsJTekd_ucAbBB Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Can somebody provide me with solution manual for Introduction to functional Analysis - Kreyzig === Subject: Re: Linear Algebra question about adjacent bases > Now, by choice if j, a_j =/= 0. So this is a nontrivial linear > combination of P_1,...,P_n equal to zero, which contradicts the > assumption that P_1,...,P_n forms a basis. This contradiction arises > from the assumption that Q was linearly dependent on > P_1,...,P_{j-1},P_{j+1},...,P_n. Hence, Q is not linearly dependent on > them, so the list > {P_1,...,P_{j-1},P_{j+1},....,P_n,Q} > is linearly independent, hence a basis (by size considerations) as was > to be proven. I get it now. They wanted to prove that Q was linearly independent >with respect to P_i excepting P_j, That last sentence still doesn't quite make sense, although it's closer. >so they just found a contradiction >between the alternate situation and the definition of linear >independence. This linear algebra stuff is so simple, yet it drives >me nuts! I guess my problem was that I was conceiving of the >definition of linear dependence as P_n = a1P1 + a2P2 + . . . + a_{n-1} >P_{n-1}, instead of all the P's on the left hand side and zero on the >right hand side. Or something. I don't know why this is so hard for >me. On the one hand a lot of students find this stuff hard. On the other hand, a lot of students, including you, are simply not _precise_ enough in the things they say about this stuff. Which means they're not precise enough in the things they _think_ about it, which makes it impossible. You still haven't _correctly_ stated what it is you're trying to prove! Here's the story. Say you were learning arithmetic. You need to calculate 2 + 2. You ask someone about 2 + 3, they tell you 2 + 3 = 5, you reply the answer's supposed to be 4, eventually you say that yeah, you meant to ask about 2 + 2 instead of 2 + 3, you thought that was clear... Of course that would be ridiculous, you'd never do that. But the point is this: The Point: If you _were_ being that sloppy about addition that would make it impossible for you to learn arithemtic. Right? Now suddenly in linear algebra you have to manipulate _concepts_ instead of manipulating _numbers_. As long as you're asking how to show Q is independent instead of what you meant to ask you're being just as sloppy as that hypothetical guy who asks about 2 + 3 when the problem is 2 + 2. And the result is the same. Then when you come back and say oops, you meant to ask how to how Q is independent with respect to .... it's like the arithmetic student saying oops, he meant to ask about 2 - 2. Until the guy starts being careful enough that he asks about 2 + 2 when the problem is 2 + 2 it's going to he hopeless. Exactly what is it you're trying to prove here? David C. Ullrich Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to. (John Jones, My talk about Godel to the post-grads. in sci.logic.) === Subject: Re: Linear Algebra question about adjacent bases posting-account=qiuOxgkAAABg9uMm_CgxOdnWf40hvxg7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) If you have a basis, P1,P2,...,Pj,...,Pm, and a vector Q that is a > linear combination of this basis, then you can replace one of those > basis vectors (call it Pj) with Q and the vectors would still be > linearly independent and therefore still be a basis. Now Pj can be > expressed in terms of this new basis, P1,P2,...,Q,...,Pm. This is not true without some restrictions of Q. > In order to be able to replace some Pj by Q it is necessary (and > sufficient) that the coefficient of Pj in the linear combination of the > basis vectors t form Q not be zero. I.e., If Q = a1 P1 + ... ak Pk + ... +am Pm then the replacement set > of vectors has the same span only if the corresponding ak is not zero. Yeah, I was thinking of this, but I didn't write it down because it > How would you prove that Q is not linearly dependent? If any ak is not zero, you can Solve Q = a1 P1 + ... ak Pk + ... +am Pm > for Pk, so that if Q were dependent only on the other basis vectors, so > is Pk, thus the original basis is not a basis. === Subject: Re: Linear Algebra question about adjacent bases If you have a basis, P1,P2,...,Pj,...,Pm, and a vector Q that is a > linear combination of this basis, then you can replace one of those > basis vectors (call it Pj) with Q and the vectors would still be > linearly independent and therefore still be a basis. Now Pj can be > expressed in terms of this new basis, P1,P2,...,Q,...,Pm. This is not true without some restrictions of Q. > In order to be able to replace some Pj by Q it is necessary (and > sufficient) that the coefficient of Pj in the linear combination of the > basis vectors t form Q not be zero. I.e., If Q = a1 P1 + ... ak Pk + ... +am Pm then the replacement set > of vectors has the same span only if the corresponding ak is not zero. Yeah, I was thinking of this, but I didn't write it down because it > How would you prove that Q is not linearly dependent? If any ak is not zero, you can Solve Q = a1 P1 + ... ak Pk + ... +am Pm > for Pk, so that if Q were dependent only on the other basis vectors, so > is Pk, thus the original basis is not a basis. > You're welcome. === Subject: Re: JSH: Leading authority check The twin paradox goes away when you take into account acceleration. The twin paradox is a special relativity problem. There is NO > acceleration in special relativity. The solution of the problem must > remain within the field of discussion. Your so-called solution is > unacceptable. Without acceleration, you can't even set up the twin paradox. > The twins have to start together, separate, and come back together. > This *cannot* be done without acceleration. Yes, it can be. -- Michael Press === Subject: Re: JSH: Leading authority check posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Minefield/3.0.5,gzip(gfe),gzip(gfe) > The twin paradox goes away when you take into account acceleration. The twin paradox is a special relativity problem. There is NO > acceleration in special relativity. The solution of the problem must > remain within the field of discussion. Your so-called solution is > unacceptable. Without acceleration, you can't even set up the twin paradox. > The twins have to start together, separate, and come back together. > This *cannot* be done without acceleration. Yes, it can be. > -- > Michael Press Not the classical twins paradox, but true enough. === Subject: Re: JSH: Leading authority check > The twin paradox goes away when you take into account acceleration. The twin paradox is a special relativity problem. There is NO > acceleration in special relativity. The solution of the problem must > remain within the field of discussion. Your so-called solution is > unacceptable. Without acceleration, you can't even set up the twin paradox. > The twins have to start together, separate, and come back together. > This *cannot* be done without acceleration. Yes, it can be. > Not the classical twins paradox, but true enough. It is mathematically equivalent to the classic scenario. This scenario can be enhanced. Let the three spaceships manufacture their clocks only when they are needed to be set in motion , record the transit times on pieces of paper, and destroy the clocks just as soon as they have measured their relevant time interval. No accelerated clocks. -- Michael Press === Subject: Re: JSH: Leading authority check The twin paradox goes away when you take into account acceleration. The twin paradox is a special relativity problem. There is NO > acceleration in special relativity. The solution of the problem must > remain within the field of discussion. Your so-called solution is > unacceptable. Without acceleration, you can't even set up the twin paradox. > The twins have to start together, separate, and come back together. > This *cannot* be done without acceleration. Yes, it can be. > Not the classical twins paradox, but true enough. It is mathematically equivalent to the classic scenario. This scenario can be enhanced. Let the three spaceships > manufacture their clocks only when they are needed to be > set in motion , record the transit times on pieces of > paper, and destroy the clocks just as soon as they have > measured their relevant time interval. No accelerated clocks. Here it is written out. -- Michael Press === Subject: Re: JSH: Leading authority check posting-account=3WPJYgoAAAA55VjhzK9i07RN8h8u8eEs Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Google search results are leading indicators of what people around the > world are doing. Really? Well what does you Google listing tell you that people are > doing James? > I mean, all those hits on your page, and yet you get no positive > responses. I know when I publish something (of course, unlike you, Why do you think that? > I've already given possible answers. But you still haven't answered this seemingly simple question on an issue in which you seem to attach such significance. Why is that? I don't get much feedback I'll admit, but I do get some. > Care to share? M === Subject: Re: JSH: Leading authority check If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine Now try Fermat crank I guess you are the world's leading FLT crank? You *guess*? JSH is the creme de la creme.[1] Footnotes: > [1] I'm probably losing my Okie cred using that cliche (not to > mention looking like a twit), so let's say instead that JSH is the > biggest dog in the yard. If you'd only used `cr.8fme de la creep' you could have retained your cred. It's a long way to Ponca City, and even further to get back. -- Michael Press === Subject: Re: JSH: Leading authority check If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: > solving quadratic residues > solving binary quadratic Diophantine > Now try Fermat crank I guess you are the world's leading FLT crank? > You *guess*? > JSH is the creme de la creme.[1] > Footnotes: > [1] I'm probably losing my Okie cred using that cliche (not to > mention looking like a twit), so let's say instead that JSH is the > biggest dog in the yard. If you'd only used `cr.8fme de la creep' > you could have retained your cred. It's a long way to Ponca City, > and even further to get back. > In OooooooooooooooooK! lahoma, I'd think the term would'a been Cheez de la Whiz Dale === Subject: Re: JSH: Leading authority check sha1:jvGFfe7gNYpZ3jTPgKU6VV3Onp4= If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine > Now try > Fermat crank > I guess you are the world's leading FLT crank? You *guess*? JSH is the creme de la creme.[1] Footnotes: > [1] I'm probably losing my Okie cred using that cliche (not to > mention looking like a twit), so let's say instead that JSH is the > biggest dog in the yard. If you'd only used `cr.8fme de la creep' > you could have retained your cred. Not if I put an accent on creme. Ain't you uppity! -- Jesse F. Hughes Part of the problem here, Peter, is that you are an idiot. -- Daryl McCullough gives a diagnosis === Subject: Re: JSH: Leading authority check > If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine > Now try > Fermat crank > I guess you are the world's leading FLT crank? You *guess*? JSH is the creme de la creme.[1] Footnotes: > [1] I'm probably losing my Okie cred using that cliche (not to > mention looking like a twit), so let's say instead that JSH is the > biggest dog in the yard. If you'd only used `cr.8fme de la creep' > you could have retained your cred. Not if I put an accent on creme. Ain't you uppity! I shop at Tarjay. -- Michael Press === Subject: Re: JSH: Leading authority check posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.43 Safari/525.19,gzip(gfe),gzip(gfe) > If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine I should come up in the top 10 for both searches, if not then consider > the statement above rebutted for your particular area, as it is > probably, at least for the moment, country specific, but that will > probably change as my mathematical research continues to rapidly take > over. How long does it take you to find those three or four magic > words to input to the Great Google? It took me about 15 minutes to come up with: Monte Carlo riemann-siegel (my magic words ...) David Bernier Meaningless jumble in comparison to: define mathematical proof === Subject: Re: JSH: Leading authority check posting-account=BVr-MgkAAABE4LRE1rHDnN9heo0IZZTk .NET CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine I should come up in the top 10 for both searches, if not then consider > the statement above rebutted for your particular area, as it is > probably, at least for the moment, country specific, but that will > probably change as my mathematical research continues to rapidly take > over. How long does it take you to find those three or four magic > words to input to the Great Google? It took me about 15 minutes to come up with: Monte Carlo riemann-siegel (my magic words ...) David Bernier Meaningless jumble in comparison to: define mathematical proof- Hide quoted text - - Show quoted text - So, what about that Solving binary quadratic Diophantine equations blog entry? I don't see any progress there. Got it programmed for test purposes yet? Here's an easy test problem: Given 2x^2 + 4x*y + y^2 - 37x + 3y - 19 = 0 Find integers x, y Answer: x = 7, y = 5 2*7^2 + 4*7*5 + 5^2 - 37*7 +3*5 - 19 = 0 98 + 140 + 25 -259 +15 -19 = 0 Enrico === Subject: Re: Leading authority check [...] > I have no doubt you will discount that fact and forget the information > again. There is no rationality left in you. Only the blindness of belief. -- Michael Press === Subject: Re: Leading authority check > The point here is non-rationality by people like you. There is no evidence that will convince people like you. I need readers to understand your behavior is about emotion, denial, > and a refusal to accept things like the mathematical proofs I do have, > which are driving the interest of the world. That is, they need to understand YOU are the cranks and crackpots. > Not me. Why? Because I need them to stop trusting people who are lying to > them. As until they do so, they'll keep paying them to teach students > false things. Those poor undergrads all over the world. As for fun searches, I recommend people try one more: intellectual foodchain. See who is on top. James Harris Yeah, but you forget I'm a degreed mathematician. It's BLANTANTLY OBVIOUS So? I have a degree in physics. Here is an undergraduate physics exercise. A rope is wrapped around a cylinder of radius r. The total angle of wrapping is a. The coefficient of friction between the rope and the surface of the cylinder is c. The tension on one end of the rope is T. The tension on the other end of the rope is T'. The rope is slowly slipping around the cylinder. Derive the relationship among a, c, r, T, and T'. -- Michael Press === Subject: Get instructor's solutions manual posting-account=EmCVswoAAACJGR09me6whGKmUcssb5cc Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) Do you suffer from a tough class? Are you looking for instructor solutions manual to do your homework? Just send me email with its name and edition and I may be able to help you in low price! It is my list, however if you don't find it here don't give up because it is only a list of some. Please , DO NOT REPLY HERE , instead send me email to : cartermath (at) gmail(dot)com instructor solution manual for A First Course in Differential Equations (7th ed.) Zill & Diferential Equations (5th ed.)Zill & Cullen instructor solution manual for A Course in Modern Mathematical Physics by Peter Szekeres instructor solution manual for A First Course in Abstract Algebra (7th Ed., John B. 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Beer) instructor solution manual for Wireless Communications Principles and Practice, 2nd Ed, by Rappaport === Subject: helllllp posting-account=BZnlawoAAACoV0jPNDt-90THeJ8F9FEu AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) 1. a recent survey was done to find out what types of financial goals recent university grads have. 100 recent uni grads were asked to think about what they hoped their financial situation would be like 10 years from now, and asked if they would own a house, a car, and/or a boat. there were only 3 who said they would have none of these things, while 4 said they would have a only a boat, 25 said only a house, and 30 said only a car. of those who felt they would own a house, 27 did not think they would have a car, while 42 didnt think they would have a boat. 45 of those surveyed said they would have a boat or a car, but not a house. how many of the 100 surveyed grads think they will have all three? 2. when peter was moving to his new house he labelled each of this boxes with kitchen, living room, office, spare room, bedroom garage. when he was done packing, he had filled and labelled 48 boxes. his friend removed all the labelled so when he moved he simply asked the moving company to place each box in any of the six locations. In how many ways can the boxes be distributed that result in no box being left in the correct room? 3. consider the permutaitons of the letters SUCCESSFUL a)how many permuations are there in total? b) how many permutations have a consonant as the first letter? 4. a particular website requires that users select a password with between 6 and 8 characters (inclusive). each character can be a lower- case letter, an upper-case letter, or one of the digits 0-9. a password must start with a letter, and must have at least 1 digit. a sample password is tIc4SS. how many passwords are possible? 5. twelve couples attend a party at which 4 identical door-prizes are to be given out to 4 different individuals. of all the ways to select 4 winers from the 24 people, how many outcomes result in no couple going home with more than one prize? === Subject: Re: Kinematics And Dynamics of Machinery third edition I'm looking for the solution manual for Kinematics And Dynamics of > Machinery third edition Charles E. Wilson & J. Peter Sadler I'm not sure why you're announcing as much in a mathematics newsgroup but good luck anyway. -- But you see, I can believe a thing without understanding it. It's all a matter of training. --Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_ === Subject: Re: Kinematics And Dynamics of Machinery third edition Hi there, === Subject: new list of solutions manual posting-account=WLvLsAoAAAAwFP1gs27vbetS0eUhA4m1 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) Hello here I leave my list of solutions manual that I have ,his list is more extensive. please contact: mbfix8117@gmail.com or luisquipu@hotmail.com *Embedded Microcomputer Systems: Real Time Interfacing Valvano *Chip Design for Submicron VLSI: CMOS Layout and Simulation, 1st Edition John P. Uyemura *Digital Signal Processing - A Modern Introduction, 1st Edition Ashok Ambardar *Digital Signal Processing Using MATLAB¬, 2nd Edition Vinay K. Ingle | John G. 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Hwang Modern Power System Analysis Kothari *Embedded Systems Kamal *Circuits and Networks Sudhakar *An Introduction to Signals and Systems, 1st Edition John Alan Stuller Hazardous Waste Management, 2nd Edition Michael D LaGrega, Phillip L Buckingham, Retired Jeffrey C Evans *Power Systems Analysis and Design, 4th Edition J. Duncan Glover , Mulukutla S. Sarma , Thomas Overbye *Chemistry for Environmental Engineering and Science, 5th Edition Clair N Sawyer,Perry L. McCarty, Gene F. Parkin, *Embedded Systems: Architechture, Programming and Design D. P. Kothari,I. J. Nagrath *Advanced Digital Logic Design Using Verilog, State Machines, and Synthesis for FPGA's, 1st Edition Sunggu Lee *Engineering Mechanics: Statics-Computational Edition, 1st Edition Robert W. Soutas-Little | Daniel J. Inman | Daniel S. Balint *Design of Fluid Thermal Systems, 2nd Edition William S. Janna *Principles of Heat Transfer, 6th Edition Frank Kreith *Foundations of Materials Science and Engineering, 4th Edition William F. Smith, Javad Hashemi, *Microwave Engineering Annapurna Das Introduction to Environmental Engineering, 4th Edition Mackenzie L Davis,David A Cornwell *An Introduction to Mechanical Engineering, 2nd Edition Jonathan Wickert *Mechanics of Materials , 6th Edition James M. Gere *Engineering Mechanics: Dynamics - Computational Robert W. Soutas-Little | Daniel J. Inman | Daniel S. Balint *Engineering Mechanics: Statics-Computational , 1st Edition Robert W. Soutas-Little | Daniel J. Inman | Daniel S. Balint *Structural Analysis , 3rd Edition Aslam Kassimali *The Science and Design of Engineering Materials, 2nd Edition James P Schaffer,Ashok Saxena, Stephen D. Antolovich,Thomas H. Sanders, Steven B Warner, *Traffic and Highway Engineering, 3rd, 3rd Edition Nicholas J. Garber | Lester A. Hoel *Modern Digital Electronics R.P. Jain *Embedded System Design with C805, 1st Edition Han-Way Huang *Chip Design for Submicron VLSI: CMOS Layout and Simulation, 1st Edition John P. Uyemura *Digital Communications, 5th Edition John Proakis,Massoud Salehi *Construction Planning, Equipment, and Methods, 7th Edition Robert L Peurifoy, Clifford J. Schexnayder, Aviad Shapira, Technion *Control Systems Madan Gopal *Modern Power System Analysis D. P. Kothari,I. J. Nagrath *Embedded Microcomputer Systems: Real Time Interfacing, 2nd Edition Jonathan W. Valvano *Introduction to Embedded Microcomputer Systems: Motorola 6811/6812 Simulations, 1st Edition Jonathan W. Valvano *Principles of Foundation Engineering, 6th Edition Braja M. Das *Engineering Fundamentals: An Introduction to Engineering, 3rd Edition Saeed Moaveni *Introduction to Scientific Computation and Programming, 1st Edition Daniel Kaplan *Engineering Economy, 6/e Leland Blank,Anthony Tarquin *Basics of Engineering Economy Leland Blank,Anthony Tarquin *Programmable Logic Controllers, 3rd Edition Frank D. Petruzella *Introduction to Chemical Processes: Principles, Analysis, Synthesis, 1st Edition Regina M. Murphy *physical chemistry 8/E P.Atkins-Trapp-Cady *Analytical chemistry D.Harvey *chemistry C.Housecroft *Plant Design and Economics for Chemical Engineers, 5th Edition Max S Peters , Klaus D Timmerhaus, Ronald E. West. *Conceptual Design of Distillation Systems Michael F. Doherty,Michael F. Malone *Fundamentals of Applied Electromagnetics, 5/E Fawwaz T. Ulaby *Unit Operations and Processes in Environmental Engineering, 2nd Edition Tom D. Reynolds | Paul Richards *Introduction to Environmental Engineering, 2nd Edition P. Aarne Vesilind | Susan M. Morgan *Introduction to Scientific Computation and Programming, 1st Edition Daniel Kaplan *An Introduction to Signals and Systems, 1st Edition John Alan Stuller *Introduction to Wireless and Mobile Systems, 2nd Edition Dharma P. Agrawal | Qing-An Zeng *Power Systems Analysis and Design, 4th Edition J. Duncan Glover | Mulukutla S. Sarma | Thomas Overbye *Introduction to Signal and System Analysis, 1st Edition Kaliappan Gopalan *Signals, Systems, and Transforms, 4/E Charles L Phillips oohn Parr Eve Riskin *Electrical Engineering: Principles and Applications, 4/E Allan R. Hambley *Introduction to Logic Design, 2nd Edition Alan B Marcovitz *Digital Principles and Design, 1st Edition Donald D. Givone Modern Digital Electronics, 1st Edition R.P. Jain Elements of Engineering Electromagnetics, 6/E Nannapaneni Narayana Rao Introduction to Microelectronic Fabrication: Volume 5 of Modular Series on Solid State Devices, 2/E Richard C. Jaeger Electrical Machines, Drives and Power Systems, 6/E Theodore Wildi Engineering Economy and the Decision-Making Process Joseph C. Hartman Interpreting and Analyzing Financial Statement, 4/E Karen P. Schoenebeck Financial Accounting, Reporting & Analysis: International Edition, 2/E Barry Elliott,Jamie Elliott Financial Statement Analysis: A Valuation Approach Leonard C. Soffer, Robin J. Soffer Interpreting and Analyzing Financial Statement, 4/E Karen P. Schoenebeck Effective Writing, 8/E Claire B. May,Gordon S. May Getting Started with Peachtree and Quickbooks 2006 Elaine Heldstab Getting Started with Peachtree 2005 Elaine Heldstab Advanced Accounting, 10/E Floyd A. BeamsRobin P. Clement, Joseph H. Anthony,Suzanne Lowensohn Financial Accounting 7/E Walter T. Harrison, Charles T. Horngren Financial Accounting: A Business Process Approach, 2/E Jane L. Reimers Financial and Managerial Accounting Charles T. Horngren,Walter T. Harrison Financial Accounting Jane L. Reimers Financial Accounting, Reporting & Analysis: International Edition, 2/E Barry Elliott, Jamie Elliott Cases in Management Accounting and Control Systems, 4/E Brandt R. Allen, E. Richard Brownlee, II, Mark E. Haskins, Luann J. Lynch, Jane W. Rotch Introduction to Management Accounting-Chapters 14/E Charles T. Horngren,Gary L. Sundem,William O. Stratton,Jeff Schatzberg,Dave Burgstahler Management Accounting: Analysis and Interpretation Cheryl McWatters,Jerold L Zimmerman, Dale Morse Management Accounting, 5/E Anthony A. Atkinson, Robert S. Kaplan,Ella Mae Matsumura, S. Mark Young Takeovers, Restructuring, and Corporate Governance, 4/E J. Fred Weston,Mark L. Mitchell, J. Harold Mulherin, Cases in Management Accounting and Control Systems, 4/E Brandt R. Allen,E. Richard Brownlee, Mark E. Haskins, Luann J. Lynch, Jane W. Rotch Governmental and Nonprofit Accounting: Theory and Practice, 9/E Robert J. Freeman,Craig D. Shoulders,Gregory S. Allison,Terry Patton,G. Robert Smith, Jr Introduction to Government and Non-for-Profit Accounting, 6/E Martin Ives,Joseph R. Razek,Gordon A. Hosch,Larry A. Johnson Prentice Hall's Federal Taxation 2009: Comprehensive, 22/E Thomas R. Pope,Kenneth E. Anderson,John L. Kramer Prentice Hall's Federal Taxation 2009: Individuals, 22/E Thomas R. Pope,Kenneth E. Anderson,John L. Kramer Accounting Information Systems, 9/E George H. Bodnar, William S. Hopwood Manual AIS Practice Set Frank A. Buckless, laura R. Ingraham, James G. Jenkins Tax Research, 4/E Barbara H Karlin Survey of Accounting: Making Sense of Business Katherene P. Terrell, Katherene P. Terrell, Robert L. Terrell Mathematics for Economics and Business, 5/E Ian Jacques Mathematics for Business, 8/E Stanley A. Salzman, Charles D. Miller,Gary Clendenen Foundations of Finance: The Logic and Practice of Financial Management, 6/E Art J Keown,John D Martin,john W Petty,David F Scott Financial Management: Principles and Applications, 10/E Arthur J. Keown,John D. 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Riedel === Subject: Re: Polynomial into e^ix expansion ? posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > Let f(x) be any polynomial, > I can go for Taylor's polynomial expansion > for that function at x values > say x = p,q,r , I can get Taylor's Polynomials Is there any method or procedure to > express the same function in polynomial using > e^ix instead of using only x, > why I'm asking this particular one is > for higher powers of x its very hard to compute > where as if we use e^ix it will eventually > become e^inx for big 'n' values. Huh? What will become e^inx? What does whether > n is big or small have to do with it? And what's > hard to compute about powers of x? /*Question in a Single Line*/ How to go for a polynomial expansion for a function say F(x) > at given point using e^ix instead of using only x.? Look up Fourier series. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada ----------- > look if this is the problem, Let > T(p), T(q) & T(r) be the Taylor's polynomials > for the function f(x) at x = p,q,r I don't have the function f(x) > but data points as this x = 1 2 3 4 > y = 2 5 8 7 There is exactly one polynomial y = f(x) that matches the data (the so- > called Lagrange polynomial---seehttp://en.wikipedia.org/wiki/Lagrange polynomial > ), but there are infinitely many non-polynomials that are an exact fit > to the data. No amount of wishful thinking or arm-waving or > incomprehensible lunatic raving will ever change that fact. You just > are not thinking, and worse still, you never seem to learn and never > seem to want to learn. It seems to me that what he wants to find is an expansion for f that > looks > like this: f(x) = a 0 + a 1 (e^(ix)) + a 2 (e^(ix))^2 + a 3 (e^(ix))^3 + ... i.e. a polynomial with e^ix replacing x. That is a Fourier series, > and I > don't think it exists for non-periodic functions like he wants (i.e. > if f is a > polynomial function.). But it could be an approximation on the interval spanning the given > data points, which is possibly all he wants. That could be. One can pass an e^(ix) expansion through a given > set of data points. Consider 2 points, (x 1, y 1) and (x 2, y 2). We have a 0 + a 1 e^(ix 1) = y 1 > a 0 + a 1 e^(ix 2) = y 2 Then this is just linear equations for a 0 and a 1, giving a 0 = y 1 - (e^(ix 1) (y2 - y1))/(e^(ix 2) - e^(ix 1)) > a 1 = (y 2 - y 1)/(e^(ix 2) - e^(ix 1)). Then the interpolation is f(x) = a 0 + a 1 e^(ix). =-=-=-=-=-=-=- > apologized for late reply. > how to form a differential equation thread > we will continue here. > -=-=-=-=-=-=-=- Generally we will write a polynomial > by taking x P = a0 + a1x + a2x^2 + a3x^3 + a4x^4+............+ an x^n (1) here we can form a polynomial depending up on our wish > may be fitting roots like this (x-a)(x-b)(x-c).....(x-n) > or an interpolation polynomial like 'Lagrangian' form > but the central part of everything is 'x' > Whole my doubt is instead of using 'x'. > I have to have use 'w' Let, > omega = w = e^ix = cisx. > In eq(1) instead of going for x, I need to go for omega P = a0 + a1 (w) + a2 (w)^2 + a3 (w)^3 +......+... here whether we can go for fitting roots > (w-a)(w-b)(w-c).........or > going for interpolation polynomial like Lagrangian > form, what ever it is, /**/ Which ever polynomial I deal, replacing 'x' > with omega(w) that's exactly my problem /**/ If your known data points are (x 1,y 1), (x 2,y 2), ... (x n,y n) then you can set up simultaneous equations like this: a 0 + a 1*e^(i*x 1) + a 2*e^(2*i*x 1) + ... + a {n-1}*e^((n - 1) *i*x 1) = y 1 a 0 + a 1*e^(i*x 2) + a 2*e^(2*i*x 2) + ... + a {n-1}*e^((n - 1) *i*x 2) = y 2 ... and solve for the a's in the usual way (here I'm writing e^(n*i*x) for (e^(i*x))^n). The resulting function y = a 0 + a 1*e^(i*x) + a 2*e^(2*i*x) + ... + a {n-1}*e^((n - 1) *i*x) will pass through the data points, but there's no reason to expect it will do exactly what you want between them, and in general it won't be real-valued for real x other than x 1, x 2, ... x n. Alternatively -- amounting to exactly the same thing but not explicitly giving you the values of the a's -- you can use Lagrange- style interpolation except that everywhere you have an x or an x i in the Lagrange interpolation formula you use e^(i*x) and e^(i*x i) instead. You realise that if instead you had y = ... + b 2*e^(-2*i*x) + b 1*e^(-i*x) + a0 + a 1*e^(i*x) + a 2*e^ (2*i*x) ... then you'd have a standard fourier series? Then you could find the coefficients to approximate to a known function using standard methods. I'm not sure if there's a way to adapt these methods to give you a series without the negative powers. Maybe someone else will. === Subject: Re: Polynomial into e^ix expansion ? posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > You realise that if instead you had y = ... + b 2*e^(-2*i*x) + b 1*e^(-i*x) + a0 + a 1*e^(i*x) + a 2*e^ > (2*i*x) ... > I should have written it more tidily as y = ... + a {-2}*e^(-2*i*x) + a {-1}*e^(-i*x) + a 0 + a 1*e^(i*x) + a 2*e^(2*i*x) ... === Subject: Re: commutator question posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > assume x,y in a group G both commute with [x,y] (commutator). > prove that for all natural n, (xy)^n=x^n y^n [y,x]^(n(n-1)/2). it looks easy but i am stuck... Prove by induction. By definition x^0 = e, where e is the group the identity is true for some n. Note that if x commutes with a then x commutes with a^{-1} (since if xa = ax then xa^{-1}a = x and a^{-1}xa = a^{-1}ax = x, and so xa^{-1} = a^{-1}x), and also that [y,x] = [x,y]^ {-1}, so x and y both commute with [y,x]. Then (xy)^{n + 1} = (xy)^n (xy) = x^n y^n [y,x]^{n(n - 1)/2} xy = x^n y^n xy [y,x]^{n(n - 1)/2} Now note that y^n x = x y^n [y,x]^n (prove this by induction, using yx = [y,x] xy, together with the fact that both x and y commute with [y,x]). So we have (xy)^{n + 1} = x^n x y^n y [y,x]^n [y,x]^{n(n - 1)/2} = x^{n + 1} y^{n + 1} [y,x]^{(n + 1}n/2} as required. === Subject: Re: Desc set theory - Projections of product spaces posting-account=LdgEVwoAAADzlB0-fAT_xkiYXMrEZ-Zc Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > an example. Let me take a closed set in X*w -- X*w - U*w, > where U is open in X. Now, its projection is just the > set X - U. Its just a closed set. But can it always be > written as countably infinite union of closed sets?. There is no requirement in these considerations (descriptive > set theory) that the sets in a countable union (or a countable > intersection) be disjoint. In fact, you can even use the same > set more than once. Also, if countable means countably > infinite or finite (which it usually does, but some authors > don't and some authors don't tell you which way they intend > for it to mean), then you can just take the index set for the > countable union to have cardinality 1, and thus if C is a closed > set, then C is the union of the set {C}. Something like this has > to be assumed (either assume countable can include finite, > or assume the sets can be taken to be the same -- see [1]), > otherwise you can't deduce things like the empty set is > an F sigma set. [1] One possible mathematical grammar problem with this > (allowing the sets to be the same but using countable > to mean countably infinite), however, is that you > need to index over well ordered sets, since, as sets, > {C, C, ... } is equal to {C}. Thus, you're actually > taking the union of a sequence of closed sets, > not the union of a countable set of closed sets. Dave L. Renfro === Subject: Re: Probability & combination question posting-account=K5_wDwoAAADH6PVGp0Ky-_BsyD7txqZY AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) > I'm having trouble knowing which methods I should be using to solve > this question: > (assume any other factors have no bearing on these simple > probabilities) having blonde hair = 0.2 & P(not blonde) = 0.8 If I pick a random group of 10 people, how to I work out the > probability that exactly 1 or 2... etc. to 10 will have blonde hair. I think that for all 10 to be blonde the probability is 0.2^10, and > that for only 1 to be blonde it is 0.8^9 (i.e. the probability that 9 > of them will not have blonde hair) No. P{10 blonde} = (0.2)^10 is correct, but P{1 blonde} is *not* equal > to (0.8)^9. Look at a group of 10 people, which group happens to have > only 1 blonde. For the order person 1, person 2, ..., person 10 and > using N = not blonde, B = blonde, we could have: BNNNNNNNNN or > NBNNNNNNNN or NNBNNNNNNN or ... or NNNNNNNNNB. Altogether, there are > 10 possible sample points, each having probability (0.2)*(0.8)^9, so > P{1 blonde} = 10*(0.2)*(0.8)^9. What about the case of 2 blondes? We > could have BBNNNNNNNN or BNBNNNNNNN or ... or NNNNNNNNBB, each point > having probability (0.2)^2 * (0.8)^8. How many points are there? Well, > you can enumerate them all and count them, but a neater way is to note > that the number is just 10*9/2, the number of ways of choosing 2 > things from 10 things. Why is this true? Think about it this way: we > have 10 empty boxes and want to but B into two of them. How many > distinct outcomes will there be? The first 'B' can go into any box, so > there are 10 ways to place the first 'B'. Then the second 'B' can go > into any of the remaining 9 boxes. Altogether, there are 10*9 ways of > placing the two 'B's. However, the arrangements in which B1 goes into > box 3 and B2 into box 7 is the same as that where B2 goes into box 3 > and B1 into box 7. In other words, all permutations of B1 and B2 give > the same final outcome, so the number of distinct outcomes is only > 10*9 / 2! In general, P{k blondes} = C(10,k)*(0.2)^k *(0.8)^(10-k), where C(M,j) > = binomial coefficient = number of combinations of j things chosen > from M things = number of subsets of size j = M!/[j! * (M-j)!] = M* > (M-1)* ... *(M-j+1)/j! = number of distinct arrangements of k 'B's in > M boxes. R.G. Vickson I know how to calculate combinations = n!/r!(n-r)!, but I'm having > difficulty visualizing how to apply it in this situation. > Peter in to a spreadsheet, then I'll know if I come unstuck. > Steve You are working with the so-called Binomial distribution. If you have > EXCEL (or most other decent spreadsheets) the binomial distribution is > built in. However, if you would like to program it yourself, you > should avoid direct computation of the formula and use instead an > iterative method. In the following, put n = 10 and p = 0.2 to get the > case you are dealing with. The general case of k blondes among n > people with blonde probability = p per person gives P{k blondes} = b > (k) = C(n,k)*p^k* q^(n-k), where q = 1-p and k = 0, 1, 2, ..., n. > Here, b(k) is the so-called binomial probability distribution. We have > b(0) = q^n, and b(k+1) = [(n-k)/(k+1)]*(p/q)*b(k) for k = > 0,1,2,...,n-1. Thus, you get b(1)=n*(p/q)*b(0), b(2) = [(n-1)/2]*b(1), > etc. Oops! This last one should be b(2) = [(n-1)/2]*(p/q)*b(1). RGV R.G. Vickson Nice... it works AND I understand it now. :-) I was going to ask > what happens if we choose from a small population of say, 100, as the > probability would then change each time.(if the first is blonde then > the second would have p = 19/99, I think). Yes. That is the problem of sampling without replacement. The > distribution of the number of blondes in this case is harder. It is > called the *hypergeometric distribution*. In my version of EXCEL it is > the built-in function HYPGEOMDIST. Basically, in a population of N things, N1 of type 1 and N2 of type 2 > (N = N1 + N2), the hypergeometric distribution with parameters N1, N2 > and M describes the probability P(k) of getting k objects of type 1 > when a sample of size M is drawn without replacement. In your case, > type 1 = blond, type 2 = other, N1 = 20, N2 = 80 (N = 100), and M = > 10. For more details on the hypergeometric distribution, see, eg.,http://en.wikipedia.org/wiki/Hypergeometric distributionorhttp://stat.... I am not sure how > well it is implemented in EXCEL, since EXCEL is well-known to be > inaccurate and unreliable for statistical computations. See, eg.,http://www.cs.uiowa.edu/~jcryer/JSMTalk2001.pdforhttp://www.coventry.... > (especially the links therein). R.G. Vickson I want to program a general function for this in C, so using the > iterative method you showed me I can just adjust the probability > each time through the loop. I'm not sure what I'm going to do when n! > becomes very large as I guess it will overflow even a 64bit type. Is > this why you said to avoid direct computation? Yes. However, if the population (sample size M) is very large, even > the recursion needs to be approached carefully, because starting from b > (0) = q^M might fail (inaccuracy, or perhaps even underflow). A better > method would be to start near the mean, say at k = k0 = [p*M] ([ ] = > integer part). Iterate upward using b(k+1) = [(M-k)/(k+1)]*(p/q)*b(k) > for k >= k0 and iterate downward using b(k-1) = c(M,k)*b(k) for k <= > k0; you can easily work out what the coefficient c(M,k) must be. > get the array of probabilities b(k) as [..., r(k0-2),r(k0-1),1,r > (k0+1),r(k0+2)...]*B, where the r(j) are what you would get if you had > put b(k0) = 1. Now determine B so that the b(j) sum to 1: B = 1/sum(r > (j),j=0..M), and finally re-scale the array to get b(k) = r(k)*B = r > (j)/sum(r). For large M, as you get very far from the center you may > start getting r(j) that are so small as to be negligible, maybe even > close to underflow. In that case, just stop the recursions and just > use sum(r) to be the sum of all the ones computed before stopping. That method will work even in cases where the direct method fails, and > will in all cases be at least as accurate, if not more accurate, than > the direct method. > Steve when it is necessary to take replacement/no-replacemnet in to account; > intuitively I would have thought that when the population is very > large it wouldn't have made much difference, so to get a feel for this > I concocted some examples: (assuming that all factors are independent) Population = 400,000 > NumBlondes = 150,000 > NumPeopleCalledDave = 100 > P(blonde) = 150000/400000 = 0.375 Instead of picking a group at random we pick the Daves (100) exactly 38 of these 100 Daves will be blonde = 0.08153 > ---------- Now flip the problem: > P(Dave) = 100/400000 = 0.00025 Now as our sample group we pick all the blondes. > exactly 38 of these 150,000 Blondes will be called Dave = 0.06437 Are you using the binomial distribution in both cases? If so, we can > see what is happening by changing the scenarios' descriptions > somewhat. In the first case we have 150K blondes and 250K others, all > as yet unnamed. Now we randomly throw in the label Dave 100 times. > The hypergeometric applies if we don't allow more than one label per > person (so once named, a person is essentially removed), while the > binomial applies if we all the label applications are independent (so > a person could have more than one Dave label sticking to him). It is > apparent that since there are so few labels compared to both sub- > populations N1 = 150K and N2 = 250K, the chance that anyone will have > more than one label is small; that is why the binomial gives a good > approximation to the hypergeometric in this case. In the second case > we have N1 = 100 'Daves' and N2 = 399,900 others, all without hair > color at the moment. Now we select 150,000 people at random and dye > them blonde. In this case there is a real difference between the > hypergeometric and the binomial. The hypergeometric case applies if we > don't dye anyone more than once, while the binomial applies if we pick > the 150,000 randomly one-by-one and dye them, irrespective of whether > or not they have been picked and dyed already. Now there is a good... read more é >It is >apparent that since there are so few labels compared to both sub- >populations N1 = 150K and N2 = 250K, the chance that anyone will have >more than one label is small; that is why the binomial gives a good >approximation to the hypergeometric in this case. In the second case >we have N1 = 100 'Daves' and N2 = 399,900 others, all without hair >color at the moment. Now we select 150,000 people at random and dye >them blonde. In this case there is a real difference between the >hypergeometric and the binomial. I can really visualize it now. is working well for the binomial, even with very large N. (and bailing out when values become negligible as suggested) method for just calculating C(n,k) also. Using that I'm calculating the hypergeometric distribution (avoiding having to calculate factorials). I'm happy to report I'm getting the same hypergeometric results as the calculator on the stattrek page you mentioned :-) (I haven't yet implemented the method from http://dissertations.ub.rug.nl/FILES/faculties/eco/2005/e.talens/c4.pdf) Steve === Subject: Re: POLL : e4 or d4 <22985642.1231025152287.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=beI85QkAAACAAUTQbCemu04CsboPXCD4 Given the choice of e4 or d4, I play e4; but I > play > b4 (the Sokolsky > Opening) more often than either d4 or e4. b4 ???? i wouldnt recommend that actually. the a pawn usually get isolated , leading to better > chances for black. and you give away the center and or initiative. i call b4 'the orangutan'. which it is also really called. tommy1729 I do not recall ever having a problem with an > isolated a-pawn in this > opening. This might be issue in the line 1 b4 e5 2 > Bb2 Bxb4 3 Bxe5, > but there are more aggressive responses to 2...Bxb4, > which I prefer to > play, like the Kucharkowski-Meybohm gambit. meybohm -> giving another pawn. btw at first sight , i might just respond with 1 b4 c5 or similar just to isolate the a pawn. In any case, at least > against the opponents I usually play, 2...d6 or > 2...f6 is a more > common response than 2...Bxb4, and in those lines, > White does not > exchange the b-pawn; usually it is pushed to b5 and > then supported > with pawns on a4 and c4, making a strong queenside > attack possible. that is often the idea yes ... I do not claim the Sokolsky opening (also called the > Orangutan or > Polish opening, in some sources) is as strong as 1 > e4, at the master > level; but I am no master, and among amateurs, it > seems to do passably > well for me, and I enjoy playing it--it seems to lead > to very > exciting, attacking games; Fischer's old games with > it are great > examples, like this one: >http://www.chessgames.com/perl/chessgame?gid=1255134 So what are the results of your poll, so far? not enough replies yet. > tommy1729 considering: http://psychos.here.ws if i would have to bother, i actually would prefer teaching a little boy or girl, go fish, instead of how to play chess. i dunno why just somehow think its more appropriate. love and peace, and, peace and love, kirk === Subject: Re: POLL : e4 or d4 <22985642.1231025152287.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=beI85QkAAACAAUTQbCemu04CsboPXCD4 Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; http://bsalsa.com) ; FDM),gzip(gfe),gzip(gfe) > Given the choice of e4 or d4, I play e4; but I > play > b4 (the Sokolsky > Opening) more often than either d4 or e4. b4 ???? i wouldnt recommend that actually. the a pawn usually get isolated , leading to better > chances for black. and you give away the center and or initiative. i call b4 'the orangutan'. which it is also really called. tommy1729 I do not recall ever having a problem with an > isolated a-pawn in this > opening. This might be issue in the line 1 b4 e5 2 > Bb2 Bxb4 3 Bxe5, > but there are more aggressive responses to 2...Bxb4, > which I prefer to > play, like the Kucharkowski-Meybohm gambit. meybohm -> giving another pawn. btw at first sight , i might just respond with 1 b4 c5 or similar just to isolate the a pawn. In any case, at least > against the opponents I usually play, 2...d6 or > 2...f6 is a more > common response than 2...Bxb4, and in those lines, > White does not > exchange the b-pawn; usually it is pushed to b5 and > then supported > with pawns on a4 and c4, making a strong queenside > attack possible. that is often the idea yes ... I do not claim the Sokolsky opening (also called the > Orangutan or > Polish opening, in some sources) is as strong as 1 > e4, at the master > level; but I am no master, and among amateurs, it > seems to do passably > well for me, and I enjoy playing it--it seems to lead > to very > exciting, attacking games; Fischer's old games with > it are great > examples, like this one: >http://www.chessgames.com/perl/chessgame?gid=1255134 So what are the results of your poll, so far? not enough replies yet. > tommy1729 considering: http://psychos.here.ws if i would have to bother, i actually would prefer > teaching a little boy or girl, > go fish, > instead of how to play chess. i dunno why just somehow think its more appropriate. love and peace, > and, > peace and love,kirk- Hide quoted text - - Show quoted text - p.s.s. too not to! sorry if my grammar upset you, i just thought if i offended a few people, i could be, a polish-american president! kalbassa anyone? just kidding! === Subject: Re: POLL : e4 or d4 <22985642.1231025152287.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=beI85QkAAACAAUTQbCemu04CsboPXCD4 Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; http://bsalsa.com) ; FDM),gzip(gfe),gzip(gfe) > Given the choice of e4 or d4, I play e4; but I > play > b4 (the Sokolsky > Opening) more often than either d4 or e4. b4 ???? i wouldnt recommend that actually. the a pawn usually get isolated , leading to better > chances for black. and you give away the center and or initiative. i call b4 'the orangutan'. which it is also really called. tommy1729 I do not recall ever having a problem with an > isolated a-pawn in this > opening. This might be issue in the line 1 b4 e5 2 > Bb2 Bxb4 3 Bxe5, > but there are more aggressive responses to 2...Bxb4, > which I prefer to > play, like the Kucharkowski-Meybohm gambit. meybohm -> giving another pawn. btw at first sight , i might just respond with 1 b4 c5 or similar just to isolate the a pawn. In any case, at least > against the opponents I usually play, 2...d6 or > 2...f6 is a more > common response than 2...Bxb4, and in those lines, > White does not > exchange the b-pawn; usually it is pushed to b5 and > then supported > with pawns on a4 and c4, making a strong queenside > attack possible. that is often the idea yes ... I do not claim the Sokolsky opening (also called the > Orangutan or > Polish opening, in some sources) is as strong as 1 > e4, at the master > level; but I am no master, and among amateurs, it > seems to do passably > well for me, and I enjoy playing it--it seems to lead > to very > exciting, attacking games; Fischer's old games with > it are great > examples, like this one: >http://www.chessgames.com/perl/chessgame?gid=1255134 So what are the results of your poll, so far? not enough replies yet. > tommy1729 considering: http://psychos.here.ws if i would have to bother, i actually would prefer > teaching a little boy or girl, > go fish, > instead of how to play chess. i dunno why just somehow think its more appropriate. love and peace, > and, > peace and love,kirk- Hide quoted text - - Show quoted text - p.s. aw! five minute chess, mind if i vote for this move, moons ago, some up and coming master was kind enough to give me 4.45 minutes of time and i honored him the white pieces. i think he played e4, PKB4, and i thought and thought and thought, but gee maybe i could not think so all of a sudden i had only 15 seconds left and i grabed his king, ran out the door, jumped in my junk car, and sped away! with trepidation, i meekly appeared back at the club, only to find its owner, was rather upset because, it was his set that he had for weeks to play with, with NO KING! mercy, kirk p.s. you guys are cool! boys will be boys! (:-D roflol peace and love, whatever way you want it, what me worry? kirk === Subject: Estimating the mean of the cumulative hypergeometric? posting-account=K5_wDwoAAADH6PVGp0Ky-_BsyD7txqZY AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) I'm trying to save some computation when calculating the cumulative probability of the hypergeometric distribution by first estimating the mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 then just add the values from k = 1001 to 1100. (In tandem with skipping the many values close to zero, I'm typically only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse or skewed for this to be accurate, but even in the example above with large m, I'm getting a significant error: i.e when I brute-force add up all the values to the estimated mean I'm getting 0.509 instead of 0.5. Therefore, if I used this shortcut , I'd be out by .009. for any value of k. Is assuming mean = n * (m/N) always a no-no, or are there times when it's a reliably accurate estimate? Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Did you mean [ mCk ] [ N-mCn-k ] / [ NCn ]? (Or maybe you're just designating the variables differently from what I was expecting...) mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 > then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I'm sure you already know this, but just in case: you realise that you don't need to calculate the binomial coefficients afresh for each value of k, right? You can just multiply what you had for k - 1 by the appropriate ratio, thus saving a huge amount of computation. === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=K5_wDwoAAADH6PVGp0Ky-_BsyD7txqZY AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) > I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Did you mean [ mCk ] [ N-mCn-k ] / [ NCn ]? (Or maybe you're just > designating the variables differently from what I was expecting...) mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 > then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I'm sure you already know this, but just in case: you realise that you > don't need to calculate the binomial coefficients afresh for each > value of k, right? You can just multiply what you had for k - 1 by the > appropriate ratio, thus saving a huge amount of computation. Yes, I'd copied the notation incorrectly. (m = number successes in N, k = number successes in sample(n) method Ray showed me. I've got a single hypergeometric calculation working very fast now, it's just the speed of calculating the cumulative probability that's slowing me down, especially when n and k are large. >Sum k=0^q h(N,m,n,k) ~= phi((q + 1/2 - mu)/sigma) hypergeometric first to satisfy myself that it's correct, and then I'll compare the approximation using my typical data sizes to see how it compares. The errors you were getting would probably be acceptable. Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=K5_wDwoAAADH6PVGp0Ky-_BsyD7txqZY AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) This is my routine for calculating the cumulative _Binomial_ distribution, with relacement, given n and p: double q = 1-p, c = pow(q,n); // i.e. when k=0 double cp = c; // cumulative probability for(double k=1; k<=n; k++) { c = ((n-(k-1))/k)*c*(p/q); cp += c; } It's fast, and (for me) clear to understand, and as it's iterative, the cost of calculating the cumulative probability is little more than for calculating a specific probability. So... I REALLY thought I would be able to modify this to work with no- replacement, after all the only difference is that p&q are changing as a result of N and m being reduced. (m = number successes in N; k = number successes in sample(n)) Surely, in the loop above, if I could recalculate p&q by how N and m are changing, then this would give the same result as the hypergeometric?? But after much trying, I can't get it to work. I've trawled the internet too, and there seems to be no shortcut other than using h(N, m,n,k) = C(m,k)*C(N-m, n-k)/C(N,n) and summing over 0->k. I have found some sieving optimisations, but it's basically the same method. (sorry, that was all more of a rant than a question ;-) ) If only there was a reliable way to start anywhere near the middle of the distribution; working backward or forward from this point would mean I could skip the lower/upper ends where the value soon becomes almost zero. One last real world example: N = total number of people that have rented DVDs = 400,000 m = num people have rented a specific movie 'A': 0 < m 200,000 n = num people have rented a specific movie 'B': 0 < n < m k = num people that have rented both (As there is symmetry, it's always quicker to make m the larger of A or B, and n the smaller of A and B) What is the probability that k people have rented both? Then: h(N, m,n,k) = C(NumRentedA,k)*C(NumHaveNotRentedA, NumRentedA-k)/ C(N, NumRentedB) What is the probability that B and A are highly correlated in some unknown way? e.g. if A = 20,000, B = 5,000, the probability that >=294 have seen both is so small as to make it likely that there is an underlying reason: cumulativeP = 1 - SUM(k=0, k=310)[ C(20000,k)*C(380000, 20000- k)/C(N, 5000) ] = .0005 The median & mean are approx the same in this case: when k = median = mean = 250, P = 0.5 When k = 208: P = .0005. Intuitively, doesn't it feel astonishing that for a range of k=0 to k=5000, it's 99.9% likely that k will be between 208 to 294!! Unless, of course, this is all screwed! Given that N is fixed, and N > m*2, and m > n can anyone spot any optimisations? I've been testing this with lots of values to find when the median is almost equal to the mean. Can anyone suggest, from the limitations on the numbers given, when the WORST case scenario might be? Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > This is my routine for calculating the cumulative Binomial > distribution, with relacement, given n and p: double q = 1-p, c = pow(q,n); // i.e. when k=0 > double cp = c; // cumulative probability > for(double k=1; k<=n; k++) > { > c = ((n-(k-1))/k)*c*(p/q); > cp += c; } It's fast, and (for me) clear to understand, and as it's iterative, > the cost of calculating the cumulative probability is little more than > for calculating a specific probability. > So... I REALLY thought I would be able to modify this to work with no- > replacement, after all the only difference is that p&q are changing as > a result of N and m being reduced. > (m = number successes in N; k = number successes in sample(n)) > Surely, in the loop above, if I could recalculate p&q by how N and m > are changing, then this would give the same result as the > hypergeometric?? I don't see that this is an efficient way to approach it. p and q are changing, sure, but they're changing in a way that depends on the history of the previous selections, and you don't want to be tracking all those possibilities individually. Your loop code for the hypergeometric should not be that different from the binomial. You keep a running count of the product of the binomial coefficients involving k: this is what I was alluding to earlier. Each iteration involves multiplying by two ratios, something like c = c*(m - k)/(k + 1)*(n - k)/(N - m - n + k + 1), depending on exactly how you implement it. Of course, you can pre-calculate N - m - n + 1. You do also have the overhead of initialising this multiplier to (N - m)C(n) and calculating the denominator Ncn. === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) This is my routine for calculating the cumulative Binomial > distribution, with relacement, given n and p: double q = 1-p, c = pow(q,n); // i.e. when k=0 > double cp = c; // cumulative probability > for(double k=1; k<=n; k++) > { > c = ((n-(k-1))/k)*c*(p/q); > cp += c; } It's fast, and (for me) clear to understand, and as it's iterative, > the cost of calculating the cumulative probability is little more than > for calculating a specific probability. > So... I REALLY thought I would be able to modify this to work with no- > replacement, after all the only difference is that p&q are changing as > a result of N and m being reduced. > (m = number successes in N; k = number successes in sample(n)) > Surely, in the loop above, if I could recalculate p&q by how N and m > are changing, then this would give the same result as the > hypergeometric?? I don't see that this is an efficient way to approach it. p and q are > changing, sure, but they're changing in a way that depends on the > history of the previous selections, and you don't want to be tracking > all those possibilities individually. Your loop code for the hypergeometric should not be that different > from the binomial. You keep a running count of the product of the > binomial coefficients involving k: this is what I was alluding to > earlier. Each iteration involves multiplying by two ratios, something > like c = c*(m - k)/(k + 1)*(n - k)/(N - m - n + k + 1), depending on > exactly how you implement it. Of course, you can pre-calculate N - m - > n + 1. You do also have the overhead of initialising this multiplier > to (N - m)C(n) and calculating the denominator Ncn. Oh, and you'll need to do some range-checking too, to make sure the k's you're summing are all in-range, and adjust the initialisation and loop termination as necessary, but that's only a tiny overhead. === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) > This is my routine for calculating the cumulative _Binomial_ > distribution, with relacement, given n and p: double q = 1-p, c = pow(q,n); // i.e. when k=0 > double cp = c; // cumulative probability > for(double k=1; k<=n; k++) > { > c = ((n-(k-1))/k)*c*(p/q); > cp += c; } It's fast, and (for me) clear to understand, and as it's iterative, > the cost of calculating the cumulative probability is little more than > for calculating a specific probability. > So... I REALLY thought I would be able to modify this to work with no- > replacement, after all the only difference is that p&q are changing as > a result of N and m being reduced. > (m = number successes in N; k = number successes in sample(n)) > Surely, in the loop above, if I could recalculate p&q by how N and m > are changing, then this would give the same result as the > hypergeometric?? > But after much trying, I can't get it to work. I've trawled the > internet too, and there seems to be no shortcut other than using > h(N, m,n,k) = C(m,k)*C(N-m, n-k)/C(N,n) > and summing over 0->k. > I have found some sieving optimisations, but it's basically the same > method. > (sorry, that was all more of a rant than a question ;-) ) If only there was a reliable way to start anywhere near the middle of > the distribution; working backward or forward from this point would > mean I could skip the lower/upper ends where the value soon becomes > almost zero. For the hypergeometric distribution with N = whole population, m = marked objects (like the blondes before), n = sample size (like M before) and k = number of marked in sample, the probabilities p(k) obey the recursion: p(k+1) = r(k)*p(k), where r(k) = [(m-k)/(k+1)]* [(n-k)/(N-m-n+k+1)]. So, you can iterate upward using p(k+1) = r(k)*p (k) for k >= k0 and downward using p(k-1) = p(k)/r(k-1) for k <= k0, starting as though p(k0) = 1. Stop when the resulting p(k) are negligible, let S = sum(p), then re-scale as p(k) <-- p(k)/S. If n is only a few hundred to a few thousand, that should be quick and accurate. One last real world example: N = total number of people that have rented DVDs = 400,000 > m = num people have rented a specific movie 'A': 0 < m 200,000 > n = num people have rented a specific movie 'B': 0 < n < m > k = num people that have rented both > (As there is symmetry, it's always quicker to make m the larger of A > or B, and n the smaller of A and B) What is the probability that k people have rented both? This question cannot be answered unless you specify a probability model. For example, there might be some weird reason that the only people who rent B are left-handed female redheads older than 40 years who rented A, or right-handed males under 30 who did not rent A. Then, to answer the question we would need to know the composition of the people who rented A and the composition of those who did not rent A. OK, this is an artificial illustration, but it makes a valid point: you need to specify a probability model. > Then: h(N, m,n,k) = C(NumRentedA,k)*C(NumHaveNotRentedA, NumRentedA-k)/ > C(N, NumRentedB) Shouldn't this be C(NumRentedA,k)*C(NumHaveNotRentedA, NumRentedB-k)/ C(N, NumRentedB)? Your use of the hypergeometric ASSUMES that the renters of B are a random sample from the whole population. Why should that be the case? What is the probability that B and A are highly correlated in some > unknown way? Surely that must depend on the probability model used. For example, if I specify that P{rent A|rent B} = 0.5, then on average one half of the renters of B will see both. However, I guess you are assuming a hypergeometric model, and in that case the number renting both is the number of A's in a random sample of NumRentedB people drawn from a population of size N and in which NumRentedA have rented A. > e.g. if A = 20,000, B = 5,000, the probability that >=294 have seen > both is so small as to make it likely that there is an underlying > reason: cumulativeP = 1 - SUM(k=0, k=310)[ C(20000,k)*C(380000, 20000- > k)/C(N, 5000) ] = .0005 Where on Earth does the number '310' come from? The median & mean are approx Here you say approx the same, then below you say they are exactly equal. You can't have both! > the same in this case: when k = median = > mean = 250, P = 0.5 > When k = 208: P = .0005. Intuitively, doesn't it feel astonishing that for a range of k=0 to > k=5000, it's 99.9% likely that k will be between 208 to 294!! Unless, of course, this is all screwed! Given that N is fixed, and N > m*2, and m > n can anyone spot any > optimisations? Do you mean to ask whether the code can be improved? Improved compared to what? What method have you used so far to do the computations? For example, there are may ways to compute binomial coefficients; which one (if any) did you use? Anyway, I have already suggested a method of starting in the middle and iterating up and down, then re-scaling, and I would bet that in any moderate-to-large problem it beats most direct method hands down. R.G. Vickson I've been testing this with lots of values to find when the median is > almost equal to the mean. Can anyone suggest, from the limitations on > the numbers given, when the WORST case scenario might be? > Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Did you mean [ mCk ] [ N-mCn-k ] / [ NCn ]? (Or maybe you're just > designating the variables differently from what I was expecting...) mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 > then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I'm sure you already know this, but just in case: you realise that you > don't need to calculate the binomial coefficients afresh for each > value of k, right? You can just multiply what you had for k - 1 by the > appropriate ratio, thus saving a huge amount of computation. If speed of computation is of the essence then another possibility is to approximate the hypergeometric distribution with a normal distribution of the same mean and variance, i.e. mu = n*m/N sigma^2 = n*m/N*(1 - m/N)*(N - n)/(N - 1) Then Sum k=0^q h(N,m,n,k) ~= phi((q + 1/2 - mu)/sigma) where phi is the standard cumulative normal distribution function, for which fast approximations are available that are accurate enough for most practical purposes. For very small probabilities the relative error is likely to be large (but the absolute error small), but for larger probabilities, and for suitably large N, m and n, the error may be acceptable. A random example that I tried, with N = 200, m = 70, n = 110, q = 37, gives the actual value as 0.382395... and the approximation as 0.383138... With with q = 34, otherwise the same, I get actual = 0.116702..., approximation = 0.117221... There is still the problem, though, of figuring out how the accuracy varies with N, m and n, so as to gauge whether the approximation is acceptable for a given set of parameters. === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Just for clarity: is kCm equal to m choose k (= m!/k!/(m-k)!) or is it equal to k choose m (= k!/m!/(k-m)!)? mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 The median (the 0.5 probability point) may not equal the mean. > then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. The computations MAY be OK, because the median need not equal the mean. > Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I don't understand the question. The formula mean = n * (m/N) is 100% accurate. The issue is whether the median is close to the mean, and how to know the cases in which that is more-or-less true. R.G. Vickson > Steve === Subject: Re: Estimating the mean of the cumulative hypergeometric? posting-account=K5_wDwoAAADH6PVGp0Ky-_BsyD7txqZY AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) > I'm trying to save some computation when calculating the cumulative > probability of the hypergeometric distribution by first estimating the > mean and working forwards/backwards from there. h(N, m, n, k) = [ kCm ] [ N-kCn-m ] / [ NCn ] Just for clarity: is kCm equal to m choose k (= m!/k!/(m-k)!) or is > it equal to k choose m (= k!/m!/(k-m)!)? mean = n * (m/N) So for instance, if mean = 1000 and I wanted to know the cumulative P > for k <= 1100, I'd start with a cumulative value of 0.5 for k=1000 The median (the 0.5 probability point) may not equal the mean. then just add the values from k = 1001 to 1100. > (In tandem with skipping the many values close to zero, I'm typically > only having to compute about 25% of the values I would have to.) I understand why for small values of m the distribution is too coarse > or skewed for this to be accurate, but even in the example above with > large m, I'm getting a significant error: i.e when I brute-force add > up all the values to the estimated mean I'm getting 0.509 instead of > 0.5. The computations MAY be OK, because the median need not equal the > mean. Therefore, if I used this shortcut , I'd be out by .009. for any > value of k. Is assuming mean = n * (m/N) always a no-no, or are there times > when it's a reliably accurate estimate? I don't understand the question. The formula mean = n * (m/N) is 100% > accurate. The issue is whether the median is close to the mean, and > how to know the cases in which that is more-or-less true. R.G. Vickson > Steve First of all, I should have written h(N, m,n,k) = C(m,k)*C(N-m, n-k)/C (N,n), I copied the other notation from a web page as i was unsure of the correct notation. Either way, I had tunnel-vision, mixing the mean and median up, so everything else I'd asked was pointless. Sorry. Back to the drawing board. Steve === Subject: test posting-account=QIM_hQoAAAD_XpQRD11KLkrLJctO0MFC Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) This is a test. === Subject: Re: test > This is a test. Great, does it have a solution? === Subject: Re: A simple pseudorandom number generator. posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Do you make any claim about the distribution of the numbers produced? Dave This is an improved version with an example that responds the Dave's question. A SIMPLE PSEUDORADOM NUMBERS GENERATOR 1.- Fix D, the number of digits of random numbers. 2.- Input an integer seed So . So >10^6 ;; N = So ; C = int(sqr (So)) ; n = 0 3.- Sum n to that number: N1 = N + n 4.- Multiply it by 4 and subtract 1 . N2 = 4*N - 1. 5.- Extract the square root: Q = sqr(N2) 6.- Determine the fractional part: F = Q - int(Q) 7.- Random number R = First D digits of F ; n = n + C 8.- Number of digits of initial seed : T = int((ln So)/ln 10)) 9.- Multiply R by 10^T: N1 = 10^T*R ; GO TO 3 It is based in the supposition that the digits of the fractional part of a square root (Not a perfect square) are pseudorandom. The mean of Random Numbers must be 0.5 The frequency of each R.N. must be uniform distributed. If the number of digits is D and the number of R.N produced is NN then the mean of frequencies of R.N.is, naturally = NN / 10^D. The frequencies of that frequencies must be Normal distributed. By the 7th condition ( n = n + C), the sequence never fall in a loop. Example: Let D = 3 Let be So = 1234567 ; NN = 10000 Results: Distribution: Uniform Mean of R.Ns = 0.499 Mean of frequencies = 10 RANGE FREQUENCY NORMAL. CHI^2 OF FREC. DISTR. TEST 0 - 1 30 30.47 0.48 2 [CapitalEth] 3 116 94.95 4.91 4 - 5 200 180.95 2.0 6 - 7 223 239.54 1.14 8 - 9 210 219.12 0.37 10 - 11 129 138.52 0.65 12 - 13 64 60.51 0.2 14 - 15 22 18.27 0.76 16 - 17 4 3.81 ------ 18 - 19 1 0.55 ------ 20 - 21 1 0.05 0.57 ------- -------- ---------- TOTAL 1000 989.85 11.08 === Subject: Re: A simple pseudorandom number generator. posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Do you make any claim about the distribution of the numbers produced? Dave This is an improved version with an example that responds the Dave's > question. A SIMPLE PSEUDORADOM NUMBERS GENERATOR 1.- Fix D, the number of digits of random numbers. > 2.- Input an integer seed So . So >10^6 ;; N = So ; C = int(sqr > (So)) ; n = 0 > 3.- Sum n to that number: N1 = N + n > 4.- Multiply it by 4 and subtract 1 . N2 = 4*N - 1. > 5.- Extract the square root: Q = sqr(N2) > 6.- Determine the fractional part: F = Q - int(Q) > 7.- Random number R = First D digits of F ; n = n + C > 8.- Number of digits of initial seed : T = int((ln So)/ln 10)) > 9.- Multiply R by 10^T: N1 = 10^T*R ; GO TO 3 It is based in the supposition that the digits of the fractional > part of a square root (Not a perfect square) are pseudorandom. > The mean of Random Numbers must be 0.5 > The frequency of each R.N. must be uniform distributed. > If the number of digits is D and the number of R.N produced is NN > then the mean of frequencies of R.N.is, naturally = NN / 10^D. > The frequencies of that frequencies must be Normal distributed. > By the 7th condition ( n = n + C), the sequence never fall in a loop. Example: > Let D = 3 > Let be So = 1234567 ; NN = 10000 Results: > Distribution: Uniform > Mean of R.Ns = 0.499 > Mean of frequencies = 10 RANGE FREQUENCY NORMAL. CHI^2 > OF FREC. DISTR. TEST > 0 - 1 30 30.47 0.48 > 2 [CapitalEth] 3 116 94.95 4.91 > 4 - 5 200 180.95 2.0 >6 - 7 223 239.54 1.14 >8 - 9 210 219.12 0.37 > 10 - 11 129 138.52 0.65 > 12 - 13 64 60.51 0.2 > 14 - 15 22 18.27 0.76 > 16 - 17 4 3.81 ------ > 18 - 19 1 0.55 ------ > 20 - 21 1 0.05 0.57 ------- ------- ------ > TOTAL 1000 989.85 11.08 === Subject: Re: A simple pseudorandom number generator. posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) This is the definitive version with the columns of the example in its proper place. A SIMPLE PSEUDORADOM NUMBERS GENERATOR 1.- Fix D, the number of digits of random numbers. 2.- Input an integer seed So . So >10^6 ; N = So ; C = int(sqr (So)) ; n = 0 3.- Sum n to that number: N1 = N + n 4.- Multiply it by 4 and subtract 1 . N2 = 4*N1 - 1. 5.- Extract the square root: Q = sqr(N2) 6.- Determine the fractional part: F = Q - int(Q) 7.- Random number R = First D digits of F ; n = n + C 8.- Number of digits of initial seed : T = int((ln So)/ln 10)) 9.- Multiply R by 10^T: N = 10^T*R ; GO TO 3 It is based in the supposition that the digits of the fractional part of a square root (Not a perfect square) are pseudorandom. The mean of Random Numbers must be 0.5 The frequency of each R.N. must be uniform distributed. If the number of digits is D and the number of R.N produced is NN then the mean of frequencies of R.N.is, naturally = NN / 10^D. The frequencies of that frequencies must be Normal distributed. By the 7th condition ( n = n + C), the sequence never fall in a loop. Example: Let D = 3 Let be So = 1234567 ; NN = 10000 Results: Distribution: Uniform Mean of R.Ns = 0.499 Mean of frequencies = 10 RANGE FREQUENCY NORMAL. CHI^2 OF FREC. DISTR. TEST 0 - 1 30 30.47 0.48 2 [CapitalEth] 3 116 94.95 4.91 4 - 5 200 180.95 2.0 6 - 7 223 239.54 1.14 8 - 9 210 219.12 0.37 10 - 11 129 138.52 0.65 12 - 13 64 60.51 0.2 14 - 15 22 18.27 0.76 16 - 17 4 3.81 ------ 18 - 19 1 0.55 ------ 20 - 21 1 0.05 0.57 ------- ------- ------ TOTAL 1000 989.85 11.08 === Subject: bijection from sequences of binary numbers to Reals posting-account=JTrNRgoAAABuGIzwXYnlAgkRTDpv8_sI .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How to get a bijection from sequences of binary numbers onto Reals? === Subject: Re: bijection from sequences of binary numbers to Reals > How to get a bijection from sequences of binary numbers onto Reals? Construct injections in each direction, and apply the Schroeder-Bernstein theorem. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: bijection from sequences of binary numbers to Reals posting-account=JTrNRgoAAABuGIzwXYnlAgkRTDpv8_sI .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > How to get a bijection from sequences of binary numbers onto Reals? Construct injections in each direction, and apply the Schroeder-Bernstein > theorem. > I actually need an explicit bijection. Binary representation of real numbers could work, but that would only work for real numbers in [0,1]. And there are duplicate representations. === Subject: Re: bijection from sequences of binary numbers to Reals days. My association with the Department is that of an alumnus. > How to get a bijection from sequences of binary numbers onto Reals? > Construct injections in each direction, and apply the Schroeder-Bernstein > theorem. I actually need an explicit bijection. Then construct explicit injections. Schroeder-Bernstein is constructive. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Sum(1/(p(i)*log(p(i))), i=1...infinity)=Pi/2 ???? bubu a .8ecrit : > Raymond Manzoni a .8ecrit : > Raymond Manzoni a .8ecrit : > Raymond Manzoni a .8ecrit : > Maria Povolotsky a .8ecrit : > I came across the following in > http://www.research.att.com/~njas/sequences/A000040 > I conjecture that > Sum(1/(p(i)*log(p(i))))=Pi/2=1.570796327... > .... > A numerical value was computed in 1998 by Henri Cohen : > 1.63661632335126086856965800392186367118159707613129... > found here http://pi.lacim.uqam.ca/piDATA/plogp.txt The more general sum : sum_{p prime} 1/(p^s*log(p)) > was studied in 2008 by Richard Mathar in 'Twenty digits of the > Erd.9as-Lebensold constant' http://arxiv.org/pdf/0811.4739 The table 2.4 gives the numerical results for the first integer > values of s. > A last link : the 1991 paper of Cohen 'High-precision > calculation of Hardy-Littlewood constants' > http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi providing very useful > methods for numerical evaluation of these sums as well as zeta and > other Euler series! > Raymond > but what are p(i) ? I think p(i) point out prime numbers Yes it seems to be the ith prime number here http://www.research.att.com/~njas/sequences/A000040 (that's at least what we supposed... to prove the initial conjecture wrong!) Raymond === Subject: Need a simple formula to calculate average percentage I am writing a little program to help me practice typing skills. The program generates a sequence of characters which I am to type. It compares what I type with the target sequence to determine which keys I am having more trouble with. Those keys are then presented more frequently in the future. The program is working but I am not happy with the algorithm I am using to decide which keys to drill. I would like some suggestions for better algorithms. Here's what I've tried and where they have problems. 1. Simple Average. Keep track of how many times each character is drilled (N) and how many times it was typed correctly (C). The average (C/N) is a measure of how difficult that key is. This works OK for awhile, but N quickly gets large and then C/N changes very little, so it's a relatively insensitive measure. 2. Consecutive Correct Keystrokes (CCK). Each time a character is typed correctly, add 1. Each time it is typed incorrectly, reset it to 0. This works much better than the simple average and, at first, appeared to be a pretty good measure of how difficult a key is. It is especially sensitive to recent data, which is a good thing. The problem is that it is too sensitive to recent date, especially misses. A key that is missed after having been typed correctly 1,000 (or 100,000) times is treated the same as one that is missed frequently. 3. Rolling Average. It seems like a good measure might be the average over the last N (100? 300?) drills. I haven't tried that because it requires that I keep track of the last N results. That's not that hard to do, but I hoping for a more elegant solution. 4. Weighted Average. It occurred to me that a weighted average might be a more general solution than a rolling average and may be easier to calculate (at least fewer data points). By weighted average, I mean an average where the most recent data point gets weight 1, the N-1 data point gets discounted by some factor (F, 0 What are the conditions for w + 1/w being rational, where w is real? Set w + 1/w = q, with q in Q. Solving for w, > If w must be real, then it must be |q| >= 2. It's just a special case of the arithmetic-geometric mean inequality: HINT w real>0 <=> w+1/w real>=2, special case v = 1/w of --Bill Dubuque === Subject: Re: caculate zeta function s->1 posting-account=W967OgoAAAASYsvrgrqxW7fN8WeVhaXp icafe8),gzip(gfe),gzip(gfe) I find the original form of zeta function converge in 0 Hi. I read this:http://en.wikipedia.org/wiki/Lambert%27s W function#Generalizations > and this:http://arxiv.org/PS cache/math-ph/pdf/0607/0607011v2.pdf Let's say then, that x = OMEGA (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q; u, v) > (on paper, one would write r 1, r 2, ... r p, then a bar under it, > then s 1, s 2, ..., s q, then that thing followed by ; u, v, > centered on the same row as the bar, or even just as it is written > above if that seems too odd, i.e. a big slash sign. I just came up > with this notation because it resembles that of the hypergeometric and > G-function, and OMEGA was the symbol used in the paper. They didn't > show a notation for the general case.) means that e^(-vx) = u((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x - > s p)). Then, why not also try out the obvious complement, where: x = omega (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q; u, v) means that log(-vx) = u((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x > - s p)). ? Then we can solve the Keplerian equation cos(x) = x in closed > form: x = -i ln(omega (2, 1)(-i, i / 0; i/2, -1)) Note that this is many-valued, as omega and ln are many-valued, it > actually has an infinite number of values (solutions). However there > is only one real solution, x ~ 0.73908513321516. By using different > branches of omega and ln we can get other (complex) solutions. The general Keplerian equation M = E - epsilon sin(E) has solution(s) E = -i ln(omega (2,1)(-iM/epsilon + sqrt(1 - (M^2)/(epsilon^2)), -iM/ > epsilon - sqrt(1 - (M^2)/(epsilon^2)) / 0; epsilon/2, -1)). (for the physical equation, one would choose the branches of omega and > ln to yield real solutions.) What do you think? Note that little-omega also generalizes W: W(x) = - > ln(omega (1,0)(0 / ; -1/x, -1)). Now what I'm wondering about is this: Can the generalizations given > above be used to invert the tower functions with towers greater than > 2? We know that the inverse for f(x) = ^2 x is tetroot 2(x) = e^(W(ln > (x))), is it possible to use an OMEGA or omega function to invert, > say, f(x) = ^3 x? (here, ^n x is x tetrated to the nth tower.) I also noticed that, at least for finite p, q, one can use e^x = H ((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x - s q)) and ln(x) = H ((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x - s q)) to give slightly simpler forms that are just as general: x = OMEGA (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q ; H) for the e^x eqn, and omega (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q ; H) for the ln(x) eqn. Then for cos(x) = x, x = -i ln(omega (2, 1)(-i, i / 0; i/2)) and for the Keplerian equation E = -i ln(omega (2,1)(-iM/epsilon + sqrt(1 - (M^2)/(epsilon^2)), -iM/ epsilon - sqrt(1 - (M^2)/(epsilon^2)) / 0; epsilon/2)). Note also that we can get the really simple identities log(x) = OMEGA (0, 0)( / ; x) exp(x) = omega (0, 0)( / ; x). with no parameters but the variable x. === Subject: Re: Generalization of Lambert W Function? posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.5) Gecko/2008120121 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Hi. I read this:http://en.wikipedia.org/wiki/Lambert%27s W function#Generalizations > and this:http://arxiv.org/PS cache/math-ph/pdf/0607/0607011v2.pdf Let's say then, that x = OMEGA (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q; u, v) > (on paper, one would write r 1, r 2, ... r p, then a bar under it, > then s 1, s 2, ..., s q, then that thing followed by ; u, v, > centered on the same row as the bar, or even just as it is written > above if that seems too odd, i.e. a big slash sign. I just came up > with this notation because it resembles that of the hypergeometric and > G-function, and OMEGA was the symbol used in the paper. They didn't > show a notation for the general case.) means that e^(-vx) = u((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x - > s p)). Then, why not also try out the obvious complement, where: x = omega (p,q)(r 1, r 2, ..., r p / s 1, s 2, ..., s q; u, v) means that log(-vx) = u((x - r 1)(x - r 2)...(x - r p))/((x - s 1)(x - s 2)...(x > - s p)). ? Then we can solve the Keplerian equation cos(x) = x in closed > form: x = -i ln(omega (2, 1)(-i, i / 0; i/2, -1)) Note that this is many-valued, as omega and ln are many-valued, it > actually has an infinite number of values (solutions). However there > is only one real solution, x ~ 0.73908513321516. By using different > branches of omega and ln we can get other (complex) solutions. The general Keplerian equation M = E - epsilon sin(E) has solution(s) E = -i ln(omega (2,1)(-iM/epsilon + sqrt(1 - (M^2)/(epsilon^2)), -iM/ > epsilon - sqrt(1 - (M^2)/(epsilon^2)) / 0; epsilon/2, -1)). (for the physical equation, one would choose the branches of omega and > ln to yield real solutions.) What do you think? Note that little-omega also generalizes W: W(x) = - > ln(omega (1,0)(0 / ; -1/x, -1)). Now what I'm wondering about is this: Can the generalizations given > above be used to invert the tower functions with towers greater than > 2? We know that the inverse for f(x) = ^2 x is tetroot 2(x) = e^(W(ln > (x))), is it possible to use an OMEGA or omega function to invert, > say, f(x) = ^3 x? (here, ^n x is x tetrated to the nth tower.) i have posted about multisected lambert W before but let me describe it briefly here the multisected lambert W is defined as n W(x) g ( W(x)) = x n 0 n n where 0 g n is the generalised coshinus then just as in the standard method you can calculate the inversion which rests on calculating terms j-1 / d 1 | | ----- ------ | | j-1 g (x) / |x=0 dx 0 n these can be expressed either in terms of my generalised berneuler numbers (which should be obvious by extracting taylor coefficients) or interestingly using the standard faa di bruno techniques in terms of j --- n-1 0's repeat pattern k -----/----- --/-- / (-1) (1) B (0, 0, ..., 0, 1, 0, ...) --- k j,k k=1 these are also related to the generalise bell numbers that arise in multisected u-tetration the multisected lamberts immediately solve the keplerian and provide a basis for similar iteration problems over the higher generalised trigonometrics -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: strange almost identity >Can anyone explain why (1/2)*Pi*sin(x) / (cos(x) + 1 - cos(x)/(sin(x)+cos(x)+1)) agrees so well with x on the interval [0,Pi/2]? It's exact for 0, Pi/ >6, Pi/4, and Pi/2, gets as high as about x + 0.02 and as low as about >x - 0.001 (it's less than x for x between Pi/6 and Pi/4). Perhaps the answer to why is just that the proposed formula is complicated enough to get a good approximation. Under the half-angle substitution sin(x)=2u/(1+u^2), cos(x)=(1-u^2)/(1+u^2), the question is equivalent to asking why u/( (u+1)*(u^2-2*u+3) ) agrees so well with arctan(u)/Pi on [0,1]. As you note, it matches perfectly when (u, arctan(u)/Pi) is one of (0,0) (2-sqrt(3), 1/12) (sqrt(2)-1, 1/8) (1, 1/4) It's not surprising that once the two smooth functions match perfectly at several points, they will behave similarly at intermediate points. Rational functions with rational coefficients which match these values will also send algebraic conjugates to 1/12 and 1/8 respectively, so in effect you are trying to match 6 rational points, which you should expect to be able to do as long as you have 6 free parameters, such as with a polynomial of degree 5, or a rational function of bidegree 2,3 . Fitting a polynomial to these data in the usual way gives u*(31-3*u-5*u^2+u^3) / 96 which actually fits the actangent curve a little better than the rational function you proposed. I'm not quite sure why both this polynomial and your rational function have degrees smaller (by one) than my heuristic would suggest, and I concede the rational function has especially small coefficients, but otherwise I think the proposed approximation is just complicated enough to be part of a general family of functions which is flexible enough to account for the good approximation. dave === Subject: Re: strange almost identity posting-account=HLRthgoAAADkf0S91nMQVx8QWMLWTqUm Can anyone explain why (1/2)*Pi*sin(x) / (cos(x) + 1 - cos(x)/(sin(x)+cos(x)+1)) agrees so well with x on the interval [0,Pi/2]? ?It's exact for 0, Pi/ >6, Pi/4, and Pi/2, gets as high as about x + 0.02 and as low as about >x - 0.001 (it's less than x for x between Pi/6 and Pi/4). Perhaps the answer to why is just that the proposed formula is > complicated enough to get a good approximation. Under the half-angle substitution > ? sin(x)=2u/(1+u^2), cos(x)=(1-u^2)/(1+u^2), > the question is equivalent to asking why > ? u/( (u+1)*(u^2-2*u+3) ) > agrees so well with ?arctan(u)/Pi ?on [0,1]. As you note, it matches perfectly when ?(u, arctan(u)/Pi) is one of > ?(0,0) > ?(2-sqrt(3), 1/12) > ?(sqrt(2)-1, 1/8) > ?(1, 1/4) > It's not surprising that once the two smooth functions match perfectly at > several points, they will behave similarly at intermediate points. Rational functions with rational coefficients which match these > values will also send algebraic conjugates to 1/12 and 1/8 respectively, > so in effect you are trying to match 6 rational points, which you > should expect to be able to do as long as you have 6 free parameters, such > as with a polynomial of degree 5, or a rational function of bidegree 2,3 . > Fitting a polynomial to these data in the usual way gives > ? u*(31-3*u-5*u^2+u^3) / 96 > which actually fits the actangent curve a little better than > the rational function you proposed. I'm not quite sure why both this polynomial and your rational function > have degrees smaller (by one) than my heuristic would suggest, and > I concede the rational function has especially small coefficients, > but otherwise I think the proposed approximation is just complicated > enough to be part of a general family of functions which is flexible > enough to account for the good approximation. dave One can try to approximate arctan(u) in a different way. Since it has a zero at 0 and an asymptote, one can try to approximate it by Pi/2 u/ (u+1) times a function g(x) (the factors u/(u+1) are roughly the heuristic behind the first part of the 'amateur''s reasoning). Approximating g(x) by a quadratic using least squares over [0,1] gives 1.53897 - 1.12859 u + 0.611177 u^2 which is close enough to (3-2u+u^2)/ 2 as claimed. === Subject: Re: strange almost identity >It's not surprising that once the two smooth functions match >perfectly at several points, they will behave similarly at >intermediate points. That's not quite right. The functions sin(10*x) and cos(10*x) match at quite a few points on the interval [0,1] but they don't behave similarly at intermediate points. -- Rouben Rostamian === Subject: Re: strange almost identity >It's not surprising that once the two smooth functions match >perfectly at several points, they will behave similarly at >intermediate points. That's not quite right. The functions sin(10*x) and cos(10*x) >match at quite a few points on the interval [0,1] but they >don't behave similarly at intermediate points. Certainly they do, for sufficiently weak values of similarly. Seriously: similarly is not precisely defined. Neither is smooth, at least in the sense in which I suspect he meant the term; smooth and similarly raise the questions how smooth? and how similarly?, respectively. If smooth means has small derivative and similar means difference is uniformly small then it's true that if f and g match at two points and are _sufficiently_ smooth in between then they behave _arbitrarily_ similarly in between... smooth in between then the values David C. Ullrich Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to. (John Jones, My talk about Godel to the post-grads. in sci.logic.) === Subject: Re: Most radical and stupid thing I ever said. Also the most correct. <18895165.1231244878749.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) > I am a Gentleman, said the Three Mouseketeers in harmony. > Mousieur Pascal at your service! > Je suis M. Desaurges, mon plesieur! > M. Brianchon, ici; nous avons trois deomonstrations > de la theorem dernier du Fermat, simulatamousement > dans trois columns. > Pascal dit, Gentile-hommes, > nous-sommes proceeder.... I have no idea what you are trying to communicate, but let me try this on you: There is a difference between triviality and nonsense. Triviality is not the same as nonsense. In physics, the smallest possible length is 1 Plancklength. My claim is that anything from 0 to 1 Plancklength is not nonsensical, but trivial. And there is a difference. We attempt to understand that difference on philosophical and analytical grounds, and that is what we do because we are right. Your fear of philosophy is a bit ironic, as mathematics is nothing more than just that. Philosophy. That is what this thread is all about. I have no idea what you are trying to communicate, but if you can explain a little further please....maybe I can understand. === Subject: Re: JSH: Algebraic integer problem, another try at explaining > On the complex plane consider an unknown factorization: > 175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2) If you're wondering why THAT particular quadratic which I keep using > over and over again, it's for historical reasons. Now I also multiply by 7 for historical reasons ... Tombstones of real men of genius: > Archimedes : sphere contained in minimal cylinder > Boltzmann: S = kLogW > Einstein: E=mc^2 William Claude Dukenfield: All things considered, I would rather be in Philadelphia. > Harris: 7*(175x^2 - 15x + 2) = 7*(5c_1(x) + 1)(5c_2(x)+ 2) -- Michael Press === Subject: Re: JSH: Algebraic integer problem, another try at explaining > Who to you sounds like the scientist, and who like people fighting to > hold on to dogma? Not much of a fight. You want a fight, I will let loose my dogma. -- Michael Press === Subject: Re: JSH: Algebraic integer problem, another try at explaining > Math people fighting me on this issue are always coming to one > conclusion: nothing to see here, not important, don't worry about it, > just go with what you were taught, trust the textbooks. The I will just move along. -- Michael Press === Subject: Re: JSH: Algebraic integer problem, another try at explaining posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) I find it hard to believe that even physicists would not get very >interested in this result if they believed it true, even though it >seems on the surface to only matter about some abstract number theory, >so here's yet another try at explaining this esoteric math issue that >takes out Galois Theory. James, could you please try to explain this: In the integers let: 2 * f(x) = (x + 1)(x + 2) At x = 0 we have: 2 * f(0) = 1 * 2 and the 2 is in the second term. At x = 1 we have: 2 * f(1) = 2 * 3 and the 2 is in the first term. At x = 2 we have: 2 * f(2) = 3 * 4 and the 2 is in the second term. At x = 3 we have: 2 * f(3) = 4 * 5 and the 2 is in the first term. In this example the 2 can appear either in the first term or in the > second term on the right hand side depending on the value of x. How does that happen? am putting forward here. rossum JSH responded completely predictably to this. He knows that f(x) = (x + 1)*(x + 2)/2 = x^2/2 + 3x/2 + 1 i.e., its definition requires division by 2. This is in contrast to his favorite, 7*P(x) = 7*(175x^2 - 15x + 2), where the function P(x) has only integer coefficients. even though your f(x) takes on only integer values for integer x. The fact that the factor 2 in your example jumps around like a kangaroo between (x + 1) and (x + 2) signifies nothing to him, even though in 7*(175x^2 - 15x + 2) = (5a_1(x) + 7)*(5a_2(x) + 7), the 7 splits up in various ways that jump around like a kangaroo, or more like a hyperactive schizophrenic, for various values of x, in the algebraic-integer factorization. JSH doesn't care that you have found something analogous in the integers. He nails you here not because there are essential differences, but because there are differences in ***appearance***. It is worth noting that the similarities are more than superficial. In your example, you can define 2 = c1(x) * c2(x), where c1(x) = GCD(2, x + 1) and c2(x) = GCD(2, x + 2), which results in c1(x) = 1 if x is even and 2 if odd, and c2(x) = 2 if x is even and 1 if x is odd, and (x + 1)/c1(x) and (x + 2)/c2(x) are both always integers for x an integer. The factorization of 2 is dependent on x. In the Harris example, one defines 7 = c1(x) * c2(x), where c1(x) = GCD(a_1(x), 7) and c2(x) = GCD(a_2(x), 7). and just as with your example, (5a_1(x) + 7)/c1(x) and (5a_2(x) + 7)/c2(x) are both always algebraic integers for x any integer. Again, the factorization of 7 is dependent on x. Harris's central error is that he refuses to consider this possibility. Marcus. === Subject: Re: JSH: Algebraic integer problem, another try at explaining > The mystery should resolve with a leap for some of you that of course > the one thing that must be invalid on the ring of algebraic integers, > as all the steps look like ok algebra is the start!!! How sure of this are you? You only used three exclamation points. If you had used seven I might be more inclined to take you at your word. -- Michael Press === Subject: Re: JSH: Algebraic integer problem, another try at explaining I find it hard to believe that even physicists would not get very >interested in this result if they believed it true, even though it >seems on the surface to only matter about some abstract number theory, >so here's yet another try at explaining this esoteric math issue that >takes out Galois Theory. >James, could you please try to explain this: >In the integers let: > 2 * f(x) = (x + 1)(x + 2) >At x = 0 we have: > 2 * f(0) = 1 * 2 and the 2 is in the second term. >At x = 1 we have: > 2 * f(1) = 2 * 3 and the 2 is in the first term. >At x = 2 we have: > 2 * f(2) = 3 * 4 and the 2 is in the second term. >At x = 3 we have: > 2 * f(3) = 4 * 5 and the 2 is in the first term. >In this example the 2 can appear either in the first term or in the >second term on the right hand side depending on the value of x. >How does that happen? >am putting forward here. >rossum > Post f(x). You read as badly as you do math. F(x) is defined above. Constrast with: 175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2) So the explanation is simple: you can't express f(x) in general in the > integers. His f(x) is an integer function. You do not seem to be able to read so the point of his demonstration was totally lost on you. You really should learn some math. I suggest though, that you test your intellect, and try. I dare you. It is funny to see you start to throw a tantrum. Hint: real mathematicians do not use dares to get proofs. That is the sign of a spoiled child. Post f(x). === Subject: Re: JSH: Algebraic integer problem, another try at explaining >I find it hard to believe that even physicists would not get very >interested in this result if they believed it true, even though it >seems on the surface to only matter about some abstract number theory, >so here's yet another try at explaining this esoteric math issue that >takes out Galois Theory. >James, could you please try to explain this: >In the integers let: > 2 * f(x) = (x + 1)(x + 2) >At x = 0 we have: > 2 * f(0) = 1 * 2 and the 2 is in the second term. >At x = 1 we have: > 2 * f(1) = 2 * 3 and the 2 is in the first term. >At x = 2 we have: > 2 * f(2) = 3 * 4 and the 2 is in the second term. >At x = 3 we have: > 2 * f(3) = 4 * 5 and the 2 is in the first term. >In this example the 2 can appear either in the first term or in the >second term on the right hand side depending on the value of x. >How does that happen? >am putting forward here. >rossum > Post f(x). > You read as badly as you do math. F(x) is defined above. > Constrast with: 175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2) > So the explanation is simple: you can't express f(x) in general in the > integers. > His f(x) is an integer function. You do not seem to be able to read > so the point of his demonstration was totally lost on you. You really > should learn some math. You're an idiot. The correct answer given 2 * f(x) = (x + 1)(x + 2), >is to solve for f(x): f(x) = (x+1)(x+2)/2 Agreed. and then substitute x = 2y, to get (don't let one quick move fool >you): No James. Remember at the beginning I stated that we were working in the integers so you are only allowed to use the integers. You cannot substitute x = 2y for *all* integers x if y is also in the integers. For example if x = 3 then what is the *integer* value of y? You are better off using the following: for x even use x = 2y for x odd use x = 2y + 1 That way all integers x are covered and y is always an integer. Remember we are working in the integers here James - nothing else allowed. f(y) = (2y+1)(y). Already corrected elsethread to: f(2y) = (2y + 1)(y) THAT is the solution in integers. Trivial. You people are math >infants. THAT is *half* of the solution in the integers. Not as trivial as you thought it was. So, looking at the case of x odd, we have x = 2y + 1 Subststute this in the original equation: 2f(2y + 1) = (2y + 2)(2y + 3) dividing out the two from the FIRST term this time gives: f(2y + 1) = (y + 1)(2y + 3) THAT is the other half of the solution in the integers. Trivial, but you missed it James. rossum === Subject: Re: JSH: Algebraic integer problem, another try at explaining posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > The correct answer given 2 * f(x) = (x + 1)(x + 2), > is to solve for f(x): f(x) = (x+1)(x+2)/2 and then substitute x = 2y, to get (don't let one quick move fool > you): f(y) = (2y+1)(y). That should be f(2y) = (2y+1)(y+1) Increment the OOPS counter. - William Hughes === Subject: Re: JSH: Algebraic integer problem, another try at explaining posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > The correct answer given 2 * f(x) = (x + 1)(x + 2), > is to solve for f(x): f(x) = (x+1)(x+2)/2 and then substitute x = 2y, to get (don't let one quick move fool > you): f(y) = (2y+1)(y). That should be f(2y) = (2y+1)(y+1) Increment the OOPS counter. - William Hughes They're equivalent Put y=2 in f(y) = (2y+1)(y) to get f(2)=(2*2+1)*2=10 put y=1 in f(2y)= (2y+1)(y+1) to get f(2)=(2*1 +1)(1+1) = 6 put x=2 in f(x) = (x+1)(x+2)/2 to get f(2) = (2+1)*(2+2)/2 = 6 - William Hughes === Subject: Re: JSH: Algebraic integer problem, another try at explaining f(x) = (2x+1)(x+1) >See? I can admit when I'm wrong. Why can't you? problem is that you, JSH are ALWAYS WRONG. we can count on that. === Subject: Re: JSH: Algebraic integer problem, another try at explaining posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > The correct answer given 2 * f(x) = (x + 1)(x + 2), > is to solve for f(x): f(x) = (x+1)(x+2)/2 and then substitute x = 2y, to get (don't let one quick move fool > you): f(y) = (2y+1)(y). That should be f(2y) = (2y+1)(y+1) Increment the OOPS counter. - William Hughes They're equivalent Put y=2 in f(y) = (2y+1)(y) to get f(2)=(2*2+1)*2=10 put y=1 in f(2y)= (2y+1)(y+1) to get f(2)=(2*1 +1)(1+1) = 6 put x=2 in f(x) = (x+1)(x+2)/2 to get f(2) = (2+1)*(2+2)/2 = 6 - William Hughes Damn. You're right. So the correct answer to the trivial problem is: f(x) = (2x+1)(x+1) put x=2 in f(x) = (2x+1)(x+1) we get f(2) = (2*2 +1)(2+1) = 15 (Hint, f(x) is not a polynomial with integer coefficients.) Probably what you want is f(x) = (2y+1)(y+1), y = x/2 However, y is only an integer for even x. (For the case of odd x, see rossum's post) - William Hughes === Subject: Re: JSH: Algebraic integer problem, another try at explaining I find it hard to believe that even physicists would not get very >interested in this result if they believed it true, even though it >seems on the surface to only matter about some abstract number theory, >so here's yet another try at explaining this esoteric math issue that >takes out Galois Theory. >James, could you please try to explain this: >In the integers let: > 2 * f(x) = (x + 1)(x + 2) >At x = 0 we have: > 2 * f(0) = 1 * 2 and the 2 is in the second term. >At x = 1 we have: > 2 * f(1) = 2 * 3 and the 2 is in the first term. >At x = 2 we have: > 2 * f(2) = 3 * 4 and the 2 is in the second term. >At x = 3 we have: > 2 * f(3) = 4 * 5 and the 2 is in the first term. >In this example the 2 can appear either in the first term or in the >second term on the right hand side depending on the value of x. >How does that happen? >am putting forward here. >rossum >Post f(x). >You read as badly as you do math. F(x) is defined above. >Constrast with: 175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2) >So the explanation is simple: you can't express f(x) in general in the >integers. >His f(x) is an integer function. You do not seem to be able to read >so the point of his demonstration was totally lost on you. You really >should learn some math. > You're an idiot. This from the fool who has been fighting reality for years. You have done nothing correct but you persist in your delusions. Professional treatment might help you. Studying math might also help The correct answer given 2 * f(x) = (x + 1)(x + 2), > is to solve for f(x): f(x) = (x+1)(x+2)/2 Wow. James can divide by two!!!!!! This is after not realizing that f(x) was defined above. and then substitute x = 2y, to get (don't let one quick move fool > you): f(y) = (2y+1)(y). James pulls another stupid move to try to hide his ignorance of math. You are a pretty sad person to have to live with your delusions. THAT is the solution in integers. Trivial. You people are math > infants. No, you are the math fool. You keep trying to find some imaginary problem in math but only show yourself to be monumentally stupid. Keep it up though as we enjoy laughing at you flailing and failing. === Subject: Re: factor phone numbers by hand calc? EASY. Fermat or trial. > don.lo...@paradise.net.nz said: > pc fc dw mcm nzmm, > ttfn Pardon.... what's that? wyw -- [mail]: Chuck F (cbfalconer at maineline dot net) [page]: Try the download section. === Subject: Re: zero knowledge protocols. An early protocol to flip a coin by mail. Ralph generates two large secret primes p and q such that p.q > almost certainly cannot be factored within the time to > execute the protocol, and tells Sarah pq. Sarah picks a secret number x and tells Ralph xx modulo pq. Ralph generates the square roots of xx modulo pq, picks > one, say y, and tells Sarah. If Ralph tells Sarah x or -x, then he wins the toss. > Otherwise he loses. If Sarah tells Ralph that Ralph lost the toss, then she > must tell Ralph x and it must not be the case that x = y > or x = -y modulo pq. I came up with a similar one once, that involves simpler computation, > and less exchanges of information. However, it is based on a problem > that, as far as I know, has not been proven to be hard. (In fact, I've > not heard of anyone even studying the problem). 1. Ralph generates two large primes, p and q, with p = 1 mod 4, and q = > 3 mod 4. Ralph gives N=pq to Sarah. 2. Sarah guesses whether p < q or q < p. She tells Ralph her guess. 3. Ralph reveals p and q. Sarah wins if her guess as to the order of p > and q was correct. Sarah can verify that Ralph revealed the correct p > and q by making sure p and q are prime and pq does equal N. The security of this is based on this problem: Given an integer N = 3 mod 4, where N is a product of two primes, > determine if the smaller of the two primes is equals 1 mod 4. I have no idea how hard this problem is. Then it is not as secure as the square root protocol. > Here's another coin flip protocol that is even less work: 1. Ralph generates two unequal numbers, A and B, and sends to Sarah > h(A) and h(B), where h is a cryptographically secure hash function > agreed upon in advance. 2. Sarah guess whether A < B or B < A. 3. Ralph reveals A and B. Sarah wins if she guessed right in #2. This one, I suspect, can be proven to be as secure as your hash function. hagman suggests that Ralph find A and A' such that h(A) = h(A'), then transmit a B that is between A and A'. I am entirely unpracticed in these matters. The square root protocol is more clearly fair to my eye than the two you propose. I also prefer heapsort because it is far easier to code than quicksort even though it runs twice as long on average; quicksort requires some engineering to guard against O(n^2) cases where heapsort is immune. Therefore my predispositions are of little weight. -- Michael Press === Subject: Re: bounded derivative ==> bounded function <8981945.1231692247321.JavaMail.jakarta@nitrogen.mathforum.org>, > let f is function with bounded derivative in (a,b). > prove that f is bounded in (a,b). > I know that bounded derivative in (a,b) implies that > f is uniformly continuous in (a,b), can it help ? Look at the mean value theorem. The mean value theorem is the bees knees. Took me a long time to figure out that it is the bridge between the differential calculus and everyday intuitive consequences of the DC. -- Michael Press === Subject: Re: radius of the universe > The usual analogy to describe the geometry of spacetime > is an expanding balloon. The surface is 2-D, closed, warped > in a 3rd dimension, inaccessible to the balloonists. However, > the ballon has a center - in the 3rd dimension. > Now, extend this picture to our expanding 3-D universe; > can we compute the 'radius', the distance to the center, > in the 4th space dimension? Analogous to the balloon > model, it should be the same for all observers. > And that would educe a circumference, would it not? > No Center > http://www.astro.ucla.edu/~wright/nocenter.html > http://www.astro.ucla.edu/~wright/infpoint.html This elaborates that we can locate no center in our > observable 3-D universe. It tells nothing about whether > a center exists in some 4th spatial dimension, in which > our universe is embedded. A 4th uncurled-up spacial dimension would render orbits unstable. Not only that, I could not keep my shoe laces tied. -- Michael Press === Subject: Re: I am greater than Jesus > Two thousand years ago Jesus got enough people to listen to him that a > church was started that survives today. I, on the other hand, > encounter only apathy or hostility. Yet my message is much more > important than his: I say things that actually are important, while > Jesus said mainly nonsense. > Oh greater thou than Jesus, for thee we give a bigger cross and meaner nails and modern medical aid to prolong your moment of greatness. I'm sure you mother will gladly give blood. > My beliefs are my religion, just as for Jesus and so many other > religious leaders. I have no need to take anyone else's beliefs > because I know better (and I also know what I don't know). God has > made me the most important person of my time but Satan is winning! Andrew Usher > === Subject: Re: I am greater than Jesus posting-account=n8pQvgkAAAAR2icJiQ21GsmYtflsTgld Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > My beliefs are my religion, just as for Jesus and so many other > religious leaders. I have no need to take anyone else's beliefs > because I know better (and I also know what I don't know). God has > made me the most important person of my time but Satan is winning! God made you double-post. DB === Subject: Re: I am greater than Jesus posting-account=504E-QkAAAA2v90r8nGnJKpfySa_yBSU 5.1),gzip(gfe),gzip(gfe) Dumb question I realize, but what connection has this crap with physics? After all, who the really cares about what some nutcase believes? Still, unfortunately that's what most of today's posts on sci.physics consist. Oddly enough, a few of the remaining sane readers of this newsgroup are here to discuss science, and physics in particular. Such threads today are rare indeed! Harry C. === Subject: Re: I am greater than Jesus posting-account=p0JNqwkAAAChY16-5zbk2O2xWfBB6K-z Gecko/20061206 Firefox/1.5.0.9,gzip(gfe),gzip(gfe) Two thousand years ago Jesus got enough people to listen to him that a > church was started that survives today. I, on the other hand, > encounter only apathy or hostility. Yet my message is much more > important than his: I say things that actually are important, while > Jesus said mainly nonsense. My beliefs are my religion, just as for Jesus and so many other > religious leaders. I have no need to take anyone else's beliefs > because I know better (and I also know what I don't know). God has > made me the most important person of my time but Satan is winning! Andrew, it's time for you to get help. I know Jesus, Jesus is a good > friend of mine, you're no Jesus No Jesus was a simple man from Palestine that dies around 2000 years ago. So I am not Jesus. > What I don't get is that you are totally unreligious but here you are > saying you are Jesus. It reminds me of Nietzche. He's the guy that > made a career out of saying God is Dead and then on his death bed he > thought he was God. Atheists, what a confused lot they are. I used Jesus as a useful comparison. I don't even know if I am an atheist because it really doesn't matter what's out there if our world doesn't interact with it. And Nietzsche really was crazy I think. Andrew Usher === Subject: Re: I am greater than Jesus posting-account=ktC8lwkAAACziLpj2cVSiK4oEjnF0N3x Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Two thousand years ago Jesus got enough people to listen to him that a > church was started that survives today. I, on the other hand, > encounter only apathy or hostility. Yet my message is much more > important than his: I say things that actually are important, while > Jesus said mainly nonsense. My beliefs are my religion, just as for Jesus and so many other > religious leaders. I have no need to take anyone else's beliefs > because I know better (and I also know what I don't know). God has > made me the most important person of my time but Satan is winning! Andrew, it's time for you to get help. I know Jesus, Jesus is a good > friend of mine, you're no Jesus No Jesus was a simple man from Palestine that dies around 2000 years > ago. So I am not Jesus. What I don't get is that you are totally unreligious but here you are > saying you are Jesus. It reminds me of Nietzche. He's the guy that > made a career out of saying God is Dead and then on his death bed he > thought he was God. Atheists, what a confused lot they are. I used Jesus as a useful comparison. I don't even know if I am an > atheist because it really doesn't matter what's out there if our world > doesn't interact with it. And Nietzsche really was crazy I think. I read his biography and kind of identified with him though I have never been an atheist. He was probably suffering from dementia toward the end plus he had a very difficult life. Those who do have tough lives do get closer to God...don't you think Andrew? Tom > Andrew Usher === Subject: Re: I am greater than Jesus posting-account=p0JNqwkAAAChY16-5zbk2O2xWfBB6K-z Gecko/20061206 Firefox/1.5.0.9,gzip(gfe),gzip(gfe) > I read his [Nietzsche's] biography and kind of identified with him though I have > never been an atheist. He was probably suffering from dementia toward > the end plus he had a very difficult life. Those who do have tough > lives do get closer to God...don't you think Andrew? Yes, I suppose it's often true. But guys like me can only get angry because I don't know how to get closer to God. I wish I did but all I know is this world and trying to make a difference here. And that's why I need people to join my wiki as you know - I''ve given all the arguments before. Andrew Usher === Subject: Re: I am greater than Jesus posting-account=ktC8lwkAAACziLpj2cVSiK4oEjnF0N3x Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) I read his [Nietzsche's] biography and kind of identified with him though I have > never been an atheist. He was probably suffering from dementia toward > the end plus he had a very difficult life. Those who do have tough > lives do get closer to God...don't you think Andrew? Yes, I suppose it's often true. But guys like me can only get angry > because I don't know how to get closer to God. I wish I did but all > I know is this world and trying to make a difference here. And that's > why I need people to join my wiki as you know - I''ve given all the > arguments before. I hear you and God does too. If you listen carefully you will hear God too. He loves it when one get it. Smitty > Andrew Usher === Subject: Re: MINIMAL POLYNOMIALS FOR TANGENTS It's all true! Proposition 2 on page 5 states the consolidated result, and the proof given contains the point I was missing (it's steeped in Galois theory, so brace yourself if you're not familiar with that). I hope to rephrase it in more mundane terms, and publish the final result as stated before. === Subject: Re: MINIMAL POLYNOMIALS FOR TANGENTS The assertion was that the algebraic degrees of these values are phi(n) unless n is a multiple of 4 (in which case it is [phi(n)]/2). Polynomials T(sub n) of degree phi(n) have been constructed which have these tangents as their root sets. By restricting the angles to the interval (0, pi), we see they are uniquely represented, and clearly there are phi(n) distinct such values. When n is a multiple of 4, the tangents can be partitioned to form two sets (according to whether k is congruent to 1 or to 3 (mod 4)), each containing half of them, and these have been shown to be the root sets of two irreducible polynomials over the integers whose product is T(sub n). We also have an inductive way to construct these factors; meaning we actually have access to the minimal polynomials in this case. The only remaining point to be rigorously established (yes, I am convinced it's true) is that T(sub n) is irreducible if n is not a multiple of 4. It will suffice to show T(sub n) is irreducible if n is odd, because in this case irreducibility of T(sub 2n) is an elementary consequence. I do have some nice closed forms and recursive relationships for these T(sub n) (and the polynomials themselves for n<128), but so far none of these have produced inspirations that lead to irreducibility. I am of the opinion that it is a peculiar property of the tangent function, rather than some Galois-theoretic property of polynomials in general, that will be the key. I am also convinced that this nagging detail may be the reason the problem may in fact be unsolved (meaning literature searches have not revealed any published solution). If you have any ideas, please let me know. You may post here or send message to my college email which is williamh(at)wcjc(dot)edu. Assuming we get it all done, I will publish the full paper on my website (with URL given in an announcement on this string). Oh, yeah: T(sub n)(x) = [1+x^2]^{[phi(n)]/2}* C(sub n)([1-x^2]/[1+x^2]), where C(sub n) is the minimal polynomial for cos(2k*pi/n). === Subject: Re: algebra: beginners question regarding factoring complete terms. <20090112071528.F7743@agora.rdrop.com> <496b68dd$0$5390$e4fe514c@dreader21.news.xs4all.nl> To be honest, I guess not. And that is because I'm not a native enlgish > speaker. Which means my english vocabulary is limited to everyday > 'chitchat' english but lacks when it comes to for example math or other > specialist activities. But I'm learning.. !!! Don't apologize for your English. It's better than a lot of the drivel > we see from the native speakers, and for a second language, I would > say it's excellent. > I concur. His English did not indicated to me that English is not his first language. Should I have taken clue from his name? Often it's apparent that the difficult English is because it's not the writer's first language. In such cases I try to help the poster both with the math and with the math terminology and without any grump. Grumps I reserve for those who know better and yet prefer deliberately remedial writing. As there's a large volume of blatantly illiterate writers, I tend to be too practice at perfecting grumps and at accumulating a choice collection of grumplets. As it is, it appears he's getting some bad habits from American brats who are in desperate need of taking English as a _first_ language. Yet he is to be commended for his self learning, even if it's from second rate citizens. I have some experience helping people to translate technical material into English and am willing to help if I can, those needing help expressing math in English. BTW, proper names, such as names of countries and their derivatives are capitalized. For example, England and English. Riddle of the day. What now will I do, now that the greatest inspiration of my life for satire and irony has left the White House? === Subject: non-integer winding number posting-account=NZALAQgAAAC1WZDjR6i1CsTqsen37btL SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) On complex analysis the argument principle is applicable only when the order of zero point or pole is integer. I tried to extend the argument principle to the case when the order or multiplicity of zero point or pole is not integer in the following site; http://hecoaustralia.fortunecity.com/argument/argument.htm === Subject: Re: non-integer winding number > On complex analysis the argument principle is applicable only when the > order of zero point or pole is integer. I tried to extend the argument > principle to the case when the order or multiplicity of zero point or > pole is not integer in the following site; http://hecoaustralia.fortunecity.com/argument/argument.htm The problem is that functions like f(z) = z^(2.3) are not analytic in a (punctured) neighborhood of a ... This is not called a zero of order 2.3 but rather a branch point. === Subject: Re: Non-integrability (Riemann) of f_xy(x,y). Additional Comment. I guess I should have also stated that f is real-valued. === Subject: Re: Product of pairwise differences between n different numbers (Pigeonhole Principle?) <5at9m4tehqmorrvsf7euebbdvm4fm87pn0@4ax.com> posting-account=RapriAoAAAAkZmTTP4ys2MiqlJPqfJgA Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) --------------------------------------------------------------- 1. Proof via Lagrange-related transformation matrices:- WLOG, the a i are distinct, else Pi {i b.87C(a+1,2) + (p-b).87C(a,2) = p.87C(a+1,2) + (p-b).87[C(a,2) - C(a+1,2)] = p.87C(a+1,2) + (p-b).87C(a,1) = p.87C(a+1,2) + ab - ap > p.87C(a+1,2) + ab = # of p appearing in 1!2!...(n-1)! This being true for each prime factor p, therefore .a6 is a multiple of 1!2!...(n-1)!. ### === Subject: Re: causion: dont read posting-account=W967OgoAAAASYsvrgrqxW7fN8WeVhaXp cafe8),gzip(gfe),gzip(gfe) > pay attention to the zeta function defined in wiki, not to the > continueation but the classical part where near the ambiguous line re > (s)=>1 not even real, contrary to many person's imaging the value is > infinit. I mistake. But the definition of zeta is confusing me. === Subject: Re: statistics question I would rather use the variance of the random vector (x,y), that is the mean of the distances of the various tips from the mean of these tips. H === Subject: What is a functional Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) I have seen many examples with vector spaces that mean little to me (an engineer). eg f(x)=x^+3x+2 is a function now if x is in turn a function of say z f(x(z))=x(z)^2 +3x(z)+2 is this a functional - or can you give me a simple example. H. === Subject: Re: What is a functional I have seen many examples with vector spaces that mean little to me > (an engineer). ... can you give me a simple example. Integral from a to b of f(x) dx is an example. If we just consider real f's for which the integral exists, the definite integral is a map from a set of functions to the real numbers and thus is a functional. -- But you see, I can believe a thing without understanding it. It's all a matter of training. --Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_ === Subject: Re: What is a functional > I have seen many examples with vector spaces that mean little to me > (an engineer). eg f(x)=x^+3x+2 is a function now if x is in turn a function of say z f(x(z))=x(z)^2 +3x(z)+2 is this a functional - or can you give me a simple example. > H. > If functional means a function whose domain is a set of functions, then yes, this is a functional. === === Subject: sum ln(1+1/n^2) posting-account=jXDJAgoAAACvCBNKtFyPLZBPxKdTyO2v 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) 1.0 localhost.localdomain:8080 (squid/2.6.STABLE5) How do find the sum of the serie sum ln(1+1/n^2) i.e. lim N->+infty sum_{n=1}^{N} ln(1+1/n^2) === Subject: Re: sum ln(1+1/n^2) Firefox/1.5.0.12,gzip(gfe),gzip(gfe) How do find the sum of the serie sum ln(1+1/n^2) > i.e. lim N->+infty sum_{n=1}^{N} ln(1+1/n^2) It's the logarithm of the infinite product product_{n=1}^infinity (1 + 1/n^2). One can compute this by evaluating the classical sine product sin(pi z) = pi z product_{n=1}^infinity (1 - z^2/n^2) at z = i. === Subject: Re: sum ln(1+1/n^2) > How do find the sum of the serie sum ln(1+1/n^2) > i.e. lim N->+infty sum {n=1}^{N} ln(1+1/n^2) It's the logarithm of the infinite product > product {n=1}^infinity (1 + 1/n^2). > One can compute this by evaluating the classical > sine product > sin(pi z) = pi z product {n=1}^infinity (1 - z^2/n^2) > at z = i. Well, we may also use the hyperbolic sine : sinh(x)/x =prod(1+x^2/((pi)^2*i^2) ) x = Pi and sum = ln(sinh(Pi)/Pi) , Alain === Subject: Re: sum ln(1+1/n^2) <130120090751545019%edgar@math.ohio-state.edu.invalid> posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) On 13 jan, 13:51, G. A. Edgar One can compute this by evaluating the classical > sine product > sin(pi z) = pi z product {n=1}^infinity (1 - z^2/n^2) > at z = i. Well, we may also use the hyperbolic sine : > sinh(x)/x =prod(1+x^2/((pi)^2*i^2) ) > x = Pi and sum = ln(sinh(Pi)/Pi) , Alain That's what he said: sin(i*x) = i*sinh(x) so the two functions are > really the same thing (just rotated in the complex plane). -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Bonjour G. A. Edgar , Of course these two ways are 'equivalent' and give the same result. It is just a matter of strictly remaining in R to calculate a real value, Alain === Subject: Dense images of Z^2 in R posting-account=JHgY7AoAAAB5Z10Fvw_ZmGANZtaQQoiP Gecko/20080410 SUSE/2.0.0.14-0.1 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in R. Now I wonder: What is known about necessary and sufficient conditions for the image of a function f : Z^2->R to be dense in R? If nothing can be said about general functions, what about some special function classes like injective functions? Rolf Z: Integers R: Real numbers === Subject: Re: Dense images of Z^2 in R > I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? > If D is dense subset X, f:X -> Y continuous surjection, then f(D) dense subset Y. D dense subset X when cl D = X. Exercise. D dense subset R iff D dense in R. Exercise. If multi-point S is a dense linear order, then D dense subset S iff D dense in S In general dense subset and dense in are not equivalent. === Subject: Re: Dense images of Z^2 in R <20090113052721.V10469@agora.rdrop.com> posting-account=JHgY7AoAAAB5Z10Fvw_ZmGANZtaQQoiP Gecko/2008120122 Firefox/3.0.5 Ubiquity/0.1.4,gzip(gfe),gzip(gfe) > I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? If D is dense subset X, f:X -> Y continuous surjection, > then f(D) dense subset Y. D dense subset X when cl D = X. Exercise. D dense subset R iff D dense in R. > Exercise. If multi-point S is a dense linear order, then > D dense subset S iff D dense in S In general dense subset and dense in are not equivalent. when is the image of f : Z^2 -> R a dense subset of R? === Subject: Re: Dense images of Z^2 in R posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? If D is dense subset X, f:X -> Y continuous surjection, > then f(D) dense subset Y. D dense subset X when cl D = X. Exercise. D dense subset R iff D dense in R. > Exercise. If multi-point S is a dense linear order, then > D dense subset S iff D dense in S In general dense subset and dense in are not equivalent. when is the image of f : Z^2 -> R a dense subset of R? As David said, at that level of generality you'll not get anything interesting: just that such a function has dense image it if has a dense image. -- m === Subject: Re: Dense images of Z^2 in R > I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? Rolf Z: Integers > R: Real numbers If f is an additive function (as in the example f(n,m) = n+sqrt(2)*m ), then the condition is that f(1,0) and f(0,1) be incommensurable. === Subject: Re: Dense images of Z^2 in R <130120090754213855%anniel@nym.alias.net.invalid> posting-account=JHgY7AoAAAB5Z10Fvw_ZmGANZtaQQoiP Gecko/2008120122 Firefox/3.0.5 Ubiquity/0.1.4,gzip(gfe),gzip(gfe) I recently saw an exercise asking to show that Z+sqrt(2)Z is dense in > R. > Now I wonder: What is known about necessary and sufficient conditions > for the image of a function f : Z^2->R to be dense in R? If nothing > can be said about general functions, what about some special function > classes like injective functions? Rolf Z: Integers > R: Real numbers If f is an additive function (as in the example > f(n,m) = n+sqrt(2)*m ), then the condition is that f(1,0) and f(0,1) > be incommensurable. interested in. === Subject: Re: Returning Fire: Vector Calculus posting-account=U4daugkAAABoPdCIBNuqtBcdKEss-z_p Gecko/2008121622 Ubuntu/8.10 (intrepid) Firefox/3.0.5,gzip(gfe),gzip(gfe) > A shot is fired and 3 microphones intercept the conical shock wave. With > electronics the arrival time differences between microphones is obtained. > between them, the location of both the source and destination of the > projectile is determined. You are still wrong, and trivially wrong at that. It is not even necessary to look at your solution to show it. Ian === Subject: Re: Elementary Derivation of Formula for Pi Here's a sketch of an elementary argument for an infinite product > expansion of pi. The details can be found at > http://www.ep.liu.se/ea/lsm/2005/002/lsm05002.pdf [neat sketch snipped.] This is known as Wallis' formula. Who knows how Wallis figured it out. In D J Struik, 'A Source Book in Mathematics, 1200-1800', Harvard UP, 1969 there is an English translation with notes. -- But you see, I can believe a thing without understanding it. It's all a matter of training. --Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_ === Subject: Partial Differential Equations posting-account=bIzc2QoAAAAijBTuofCklemWbVRps_8t Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; .NET CLR 2.0.50727; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) i am a civil engineer. I want to start learning PDEs theoretically. I would like to know which books you would recommend for someone like me, with only calculus (two variables), matrix algebra, analytic geometry, and ODEs as background. I want to know which path to fallow, Victor === Subject: Re: Partial Differential Equations > i am a civil engineer. I want to start learning PDEs theoretically. I > would like to know which books you would recommend for someone like > me, with only calculus (two variables), matrix algebra, analytic > geometry, and ODEs as background. I want to know which path to fallow, which hasn't had many replies. -- But you see, I can believe a thing without understanding it. It's all a matter of training. --Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_ === Subject: Re: Well ordering of reals Nntp-Posting-Host: hera.cwi.nl ... > Given a finite subset of R, we can certainly construct (many) well > orderings for it. We can even do it for many infinite subsets. > > This, however, is a ->very<- far cry from constructing well orderings > for ALL the finite subsets of R, which implies that you can do it > for all finite subsets simultaneously. I must be missing something. For every finite subset of R, the standard ordering forms a well-ordering. So do we not in a sense have a well-ordering for *all* fonite subsets of R? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Well ordering of reals > I must be missing something. For every finite subset of R, the > standard ordering forms a well-ordering. So do we not in a sense > have a well-ordering for *all* fonite subsets of R? Yes. Any linearly ordered finite set is well-ordered. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A MOD B NO MATH HAHAHAHA !! Nntp-Posting-Host: hera.cwi.nl > I have also never seen a keyboard with that symbol [.85, division sign, > U=+00F7]. On the computer where we used it it was formed by > overprinting a colon and a minus sign. BTW, the computer language > was Algol 60. > > Not APL? Surely as a mathematician close to my age, you > would have had the pleasure and heartache of using APL at > some time in the 1970s? Nope, never used APL. By the time I could have gotten access to APL I did already use Algol 68. Moreover, nowhere in my neighbourhood have I ever seen a keyboard that did show the APL graphics. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: A MOD B NO MATH HAHAHAHA !! > > I have also never seen a keyboard with that symbol [.85, division sign, > > U=+00F7]. On the computer where we used it it was formed by > > overprinting a colon and a minus sign. BTW, the computer language > > was Algol 60. > > Not APL? Surely as a mathematician close to my age, you > > would have had the pleasure and heartache of using APL at > > some time in the 1970s? Nope, never used APL. By the time I could have gotten access to APL I did > already use Algol 68. Moreover, nowhere in my neighbourhood have I ever > seen a keyboard that did show the APL graphics. It's sad that you have missed Another Programming Language :-) Han de Bruijn === Subject: Re: A MOD B NO MATH HAHAHAHA !! <4568100.1231630716322.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > <4568100.1231630716322.JavaMail.jaka...@nitrogen.mathforum.org>, > you got it wrong too ! > it was about mod 1 !! > sigh ... > that doesnt match your interpretation ! > Do you really mean mod 1? > For the mod operator , n mod 1 = 0 for any integer n, > and for the mod relation, n == 0 (mod 1) for any integer n. > Which means that mod 1 is of damn all use. By mod 1, Virgil is referring to the ring Z/Z, which is obviously isomorphic to the ring 1 = 0 (i.e., the trivial ring with only a single element). But when Plouffe refers to mod 1, he doesn't mean Z/Z, but R/Z. And R/Z is not the trivial ring. === Subject: Re: A MOD B NO MATH HAHAHAHA !! <4568100.1231630716322.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080614i-0etch1),gzip(gfe),gzip(gfe) <4568100.1231630716322.JavaMail.jaka...@nitrogen.mathforum.org>, > you got it wrong too ! > it was about mod 1 !! > sigh ... > that doesnt match your interpretation ! > Do you really mean mod 1? > For the mod operator , n mod 1 = 0 for any integer n, > and for the mod relation, n == 0 (mod 1) for any integer n. > Which means that mod 1 is of damn all use. By mod 1, Virgil is referring to the ring Z/Z, which > is obviously isomorphic to the ring 1 = 0 (i.e., the > trivial ring with only a single element). But when Plouffe refers to mod 1, he doesn't mean > Z/Z, but R/Z. And R/Z is _not_ the trivial ring. I already know that my own thinking is askance to your thinking here. But I have another means by which the mod 1 system is of importance. Upon generalizing sign the real number takes the form s x where s is sign and x is magnitude. For the real number there are two signs and the modulo arithmetic takes its place in the arithmetic product of two values ( s1 x1 )( s2 x2 ) whose resultant is (s1 + s2) x1 x2 where this sign summation is mod 2. Clearly when we step downward to the one-signed numbers we simply have one sign type with few dynamics. However, since this sign type is tied to a magnitude arithmetic can still be performed. These one-signed numbers carry the strongest congruence to time that I know of. They are unidirectional. The special time 'now' is embodied by them. While arithmetic can be performed with them in terms of geometrical dimension they are zero dimensional. This satisfies the physical behaviors of time, whereby no actual degree of freedom is observed; this in contrast with the three dimensions of space whose freedom is demonstrable. To appreciate the dimensional behavior a fuller description is necessary and so stepping up to the three-signed numbers (P3) would be wise. P3 are the complex numbers though their format is foreign to most. The complex numbers follow directly from the identical rules that build the real numbers under the polysign paradigm. They are merely P2(the reals) and P3(the complex numbers) tobe followed by P4, P5, etc. All of these polysign systems obey the commutative and associative behaviors under product and superposition. They are still limited by similar constraints that hampered Hamilton and Grassman and Clifford but they(polysign) are cleaner. The modern mathematics culture has come to obey the real number as fundamental and the polysign construction challenges this. Nearby are the field requirements and there we see a small exception in the division by zero which has been swallowed hook, line, and sinker. Multiplication by zero provides a dimensional collapse which takes a more general form in the high dimensional(P4+) systems of polysign. Such dimensional collapses will not be reversible. I suggest that the field requirements should be relaxed and generalized to have no exception within the division clause and instead observe this as dimensional collapse by specific values which are interesting behaviors and ought not be a cause of rejection. Furthermore these behaviors allow polysign math to claim support for spacetime from pure mathematics. That the modulo behavior is capable of providing a spacetime basis ought to be of interest to both physicists and mathematicians. - Tim === Subject: Re: About t^4-z^100=2^10*k posting-account=fPUT4goAAABJ1knpUmWe7fFdWT7SyGS2 Gecko/2008120122 Firefox/2.0.0.12;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) Also I have question about solutions existence of x^100+y^100=t^4-1 with all >0. === Subject: Re: About t^4-z^100=2^10*k >Also I have question about solutions existence of x^100+y^100=t^4-1 >with all >0. There are always the 2 trivial solutions: (x,y,t) = (0,0,1) (x,y,t) = (0,0,-1) I'm not sure if there are any other solutions. An easy necessary condition is that x,y must both be multiples of 10, and t must be odd but not a multiple of 5. quasi === Subject: Re: collections of well-ordered closed (or else F_sigma) subsets of the unit interval [0, 1] be represented as the well-ordering of a closed subset of [0, 1] , with > the ordering given by <. (Prop. A) That is correct. Proposition. > Every countable linear order can be order embedded into Z[1/2], > hence into Q and [0,1]. Let beta be countable compact ordinal, ie eta is a countable non-limit > ordinal. Let h:beta -> [0,1] be an order embedding of eta. Define f:beta -> [0,1] by transfinite induction. > If eta is successor ordinal in beta. let f(xi) = h(xi). > If eta is limit ordinal, let f(eta) = sup{ f(xi) | xi < eta } Show f is a supremum preserving order embedding. Since beta is compact, it is a complete order. > Thus also h(beta) and f(beta). Being complete orders, h(beta) > and f(beta) are compact within their intrinsic order topologies. The hard part is to apply the theorem that shows > the subspace topology of f(beta) is the same as the > intrinsic order topology for f(beta). Thus f(beta) > is a subspace of [0,1] and being compact, it's closed. [Assuming Prop. A is true below ] > If so, then countable unions of closed subsets of [0, 1] (or F_sigma > sets) could be the order type of any countable ordinal under > inclusion. That is ambiguous. Would you clarify? > Either way I read it, I've a counter example. An F_sigma set is by definition a union of up to countably many closed > sets in some topological space, usually the reals R, R^n, > a compact subset of R^n, etc. We can't get omega as a closed > subset of [0, 1] . But we can define By omega do you mean omega_0, in contrast to omega_1? > B_n = { 1 - 1/k : 1<=k<=n} for n>= 1. > They are closed subsets (finite) of [0, 1], and their countable union, > i.e. union_{n in N^*} B_n is an infinite set, an F_sigma > since each B_n is closed. But union_{n in N^*} B_n is not closed > in [0, 1] , the closure has the extra point 1. If a closed subset A of [0, 1] represents beta + 1 with <, beta an infinite > countable ordinal, [0, alpha) intersect A will sometimes give a closed set > in [0, 1] representing an ordinal <= beta+1. If it's not closed, > I still think the intersection is F_sigma (up to countable union > of closed subsets of [0, 1] ). It seems every ordinal < beta+1 > can be represented by [0, alpha) intersect A, for alpha > in [0, 1], and the sets are F_sigma. So it seems to every > ordinal < beta + 1 there corresponds naturally an F_sigma > subset of A. If beta is a countable infinite limit ordinal, > B subset [0, 1] which is F_sigma can, I think, represent > beta, with intersections with [0, alpha) , alpha in [0, 1] > representing smaller ordinals (or subsets which are F_sigma). > Hm. You're claiming every countable ordinal can be order embedded and/or topologically embedded into [0,1] as a F_sigma set? Since every countable subset of [0,1] is F_sigma, what's the significance? > So the question is: can we find aleph_1 F_sigma subsets > of [0, 1] which, under the inclusion U on its side, > have the order type aleph_1 ? > Yes. Well order [0,1]. [0,1] = { r_xi | xi < omega_1 }. For all eta < omega_1, let R_eta = { r_xi | xi < eta }. For all eta < omega_1, R_eta is F_sigma. For all psi,chi < omega_1, R_psi subset R_chi iff psi <= chi. ---- === Subject: Re: collections of well-ordered closed (or else F_sigma) subsets of the unit interval [0, 1] > I believe any infinite countable ordinal which is not a limit > ordinal can be represented as the well-ordering of a closed subset > of [0, 1] , with the ordering given by <. (Prop. A) > That is correct. Proposition. > Every countable linear order can be order embedded into Z[1/2], > hence into Q and [0,1]. Let beta be countable compact ordinal, ie eta is a countable > non-limit ordinal. Let h:beta -> [0,1] be an order embedding of eta. Define f:beta -> [0,1] by transfinite induction. > If eta is successor ordinal in beta. let f(xi) = h(xi). > If eta is limit ordinal, let f(eta) = sup{ f(xi) | xi < eta } Show f is a supremum preserving order embedding. Since beta is compact, it is a complete order. > Thus also h(beta) and f(beta). Being complete orders, h(beta) > and f(beta) are compact within their intrinsic order topologies. The hard part is to apply the theorem that shows > the subspace topology of f(beta) is the same as the > intrinsic order topology for f(beta). Thus f(beta) > is a subspace of [0,1] and being compact, it's closed. > [Assuming Prop. A is true below ] > If so, then countable unions of closed subsets of [0, 1] (or F_sigma > sets) could be the order type of any countable ordinal under inclusion. > That is ambiguous. Would you clarify? > Either way I read it, I've a counter example. > An F_sigma set is by definition a union of up to countably many closed > sets in some topological space, usually the reals R, R^n, > a compact subset of R^n, etc. We can't get omega as a closed > subset of [0, 1] . But we can define By omega do you mean omega_0, in contrast to omega_1? Yes, omega_0, i.e. the smallest infinite ordinal, omega_0 = {0, 1, 2, 3, .... } in von Neumann's way of defining the ordinals. > B_n = { 1 - 1/k : 1<=k<=n} for n>= 1. They are closed subsets (finite) of [0, 1], and their countable union, > i.e. union_{n in N^*} B_n is an infinite set, an F_sigma > since each B_n is closed. But union_{n in N^*} B_n is not closed > in [0, 1] , the closure has the extra point 1. > If a closed subset A of [0, 1] represents beta + 1 with <, beta an > infinite > countable ordinal, [0, alpha) intersect A will sometimes give a closed > set > in [0, 1] representing an ordinal <= beta+1. If it's not closed, > I still think the intersection is F_sigma (up to countable union > of closed subsets of [0, 1] ). It seems every ordinal < beta+1 > can be represented by [0, alpha) intersect A, for alpha > in [0, 1], and the sets are F_sigma. So it seems to every > ordinal < beta + 1 there corresponds naturally an F_sigma > subset of A. If beta is a countable infinite limit ordinal, > B subset [0, 1] which is F_sigma can, I think, represent > beta, with intersections with [0, alpha) , alpha in [0, 1] > representing smaller ordinals (or subsets which are F_sigma). > Hm. You're claiming every countable ordinal can be order embedded > and/or topologically embedded into [0,1] as a F_sigma set? Since every countable subset of [0,1] is F_sigma, what's the significance? You're right, every countable subset is an F_sigma, when singleton sets {p} are closed, as happens in any metric space W, p a point in W... I missed that basic fact. > So the question is: can we find aleph_1 F_sigma subsets > of [0, 1] which, under the inclusion U on its side, > have the order type aleph_1 ? > Yes. Well order [0,1]. [0,1] = { r_xi | xi < omega_1 }. > For all eta < omega_1, let R_eta = { r_xi | xi < eta }. > For all eta < omega_1, R_eta is F_sigma. > For all psi,chi < omega_1, > R_psi subset R_chi iff psi <= chi. Yes, very good. Your sets R_eta are all F_sigma sets, and 0<=eta <20090112202538.H48240@agora.rdrop.com> posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Since every countable subset of [0,1] is F_sigma, > what's the significance? > You're right, every countable subset is an F_sigma, > when singleton sets {p} are closed, as happens in > any metric space W, p a point in W... Each subset of the reals that is well ordered under the usual ordering for real numbers is also a G_delta set, by the way. The conditions of being countable and of being G_delt, when taken together, impose a fairly strengent requirement on a set (although not enough to force the set to be well ordered). For example, no such set can be dense in any interval (being dense and G_delta in an interval implies uncountability -- this is essentially the Baire category theorem for the interval). Indeed, no such set can be dense in any perfect set, even very thin Cantor-like sets with Hausdorff dimension zero that are far from being an interval (and thus are sets in which far fewer points are needed to be dense in), again being essentially the Baire category theorem (but for the perfect set as a complete metric space this time). TECHNICAL NIT-PICK: Note that not being dense in any interval is equivalent to not being dense in any closed (non-degenerate) interval, so the perfect set version really does imply the interval version. It's also easy to see that the perfect set version is strictly stronger than the interval version -- the endpoints of the complementary intervals of the middle thirds Cantor set form a set that is dense in the middle thirds Cantor set but the set of these endpoints is not dense in any interval. Dave L. Renfro === Subject: Re: collections of well-ordered closed (or else F_sigma) subsets of the unit interval [0, 1] Since every countable subset of [0,1] is F_sigma, > what's the significance? > You're right, every countable subset is an F_sigma, > when singleton sets {p} are closed, as happens in > any metric space W, p a point in W... Each subset of the reals that is well ordered under the > usual ordering for real numbers is also a G_delta set, > by the way. The conditions of being countable and of > being G_delt, when taken together, impose a fairly strengent > requirement on a set (although not enough to force the > set to be well ordered). For example, no such set can be > dense in any interval (being dense and G_delta in an interval > implies uncountability -- this is essentially the Baire > category theorem for the interval). Indeed, no such set > can be dense in any perfect set, even very thin Cantor-like > sets with Hausdorff dimension zero that are far from being > an interval (and thus are sets in which far fewer points > are needed to be dense in), again being essentially the > Baire category theorem (but for the perfect set as a > complete metric space this time). This is interesting. It's hard to visualize what a typical countable G_delta set for [0, 1] looks like ... David Bernier > TECHNICAL NIT-PICK: Note that not being dense in any > interval is equivalent to not being dense in any closed > (non-degenerate) interval, so the perfect set version > really does imply the interval version. It's also easy > to see that the perfect set version is strictly stronger > than the interval version -- the endpoints of the > complementary intervals of the middle thirds Cantor > set form a set that is dense in the middle thirds > Cantor set but the set of these endpoints is not dense > in any interval. Dave L. Renfro === Subject: Solutions manua to Engineering Mechanics - Statics (11th )by R.C.HIBBELER posting-account=jBP0yQoAAADkjbQMT90jR5JPojXnsRCF Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) Solution manual solutions manual solution manual I am a solutions manual collector, I offer solutions manual services Note: all solutions manual in soft copy that mean in Adobe Acrobat Reader (PDF ) format. if you want any book not just solutions just contact with us.81B to get the solution manual you want .81Cplease send message to happyren2008@hotmail.com .81Chappyren2008(at)hotmail.com.81C replace (at) to @ ,please email to me . (To search click in keyboard Ctrl+F) Solutions manua to Financial Accounting 6e by horngren Harrison Solutions manual to Advanced Accounting, 9th edition by Hoyle, Schaefer, & Doupnik Solutions manual to Complex Variables with Applications (Pie) by A.David Wunsch Solutions manual to Computer Design Fundamentals4E by Mano and Kime. 4th Solutions manua to Computer Networks Systems Approach 3ed by davie peterson Solutions manual to COMPUTER ORGANIZATION AND ARCHITECTURE Solutions manual to Cost Accounting, 13/e 13e by Horngren SM.zip Solutions manua to Data and Computer Communications, 7th Edition By Stallings Solutions manual to Differential Equations and Linear Algebra by Penney and Edwards, 2nd edition Solutions manual to Elementary Differential Equations and Boundary Value Problems, by Boyce andDiprima Solutions manua to Elements of engineering electromagnetics (6/e) by N.N.RAO Solutions manual to Engineering electromagnetics (7/e) by HAYT Solutions manual to Engineering Fluid Mechanics, 7th, By Clayton T. 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(To search click in keyboard Ctrl+F) Solutions manua to Financial Accounting 6e by horngren Harrison Solutions manual to Advanced Accounting, 9th edition by Hoyle, Schaefer, & Doupnik Solutions manual to Complex Variables with Applications (Pie) by A.David Wunsch Solutions manual to Computer Design Fundamentals4E by Mano and Kime. 4th Solutions manua to Computer Networks Systems Approach 3ed by davie peterson Solutions manual to COMPUTER ORGANIZATION AND ARCHITECTURE Solutions manual to Cost Accounting, 13/e 13e by Horngren SM.zip Solutions manua to Data and Computer Communications, 7th Edition By Stallings Solutions manual to Differential Equations and Linear Algebra by Penney and Edwards, 2nd edition Solutions manual to Elementary Differential Equations and Boundary Value Problems, by Boyce andDiprima Solutions manua to Elements of engineering electromagnetics (6/e) by N.N.RAO Solutions manual to Engineering electromagnetics (7/e) by HAYT Solutions manual to Engineering Fluid Mechanics, 7th, By Clayton T. 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Martin(chapter 1 to chapter15) solution manual for Probability and Statistical Inference ( 7th edition by Hogg & Tanis) solution manual for Fundamentals of Communication Systems by John G. Proakis ,Masoud Salehi solution manual for Materials and Processes in Manufacturing,9thBy Degarmo === Subject: A statement about Rsa2048. Assuming there are only two prime factors of equal length in rsa2048 these two primes below would represent the smallest possible and largest possible prime factors. The smallest factor of 309 digits--- The range -- from 1000000000000000000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000010 4 59 to sqrt(rsa2048) The largest factor of 309 digits--- The range -- from sqrt(rsa2048) too -- 2519590847565789349402718324004839857142928212620403202777713783604366202070 7595556264018525880784406918290641249515082189298559149176184502808489120072 8 4499268739280728777673597141834727026189637501497182469116507761337985909570 0 0973304597488084284017974291006424586918171951187461215151726546322822168673 5 23 Knowing this and assuming the two factors are of equal length, neither one of these two primes could ever represent one of the factors in rsa2048. I have a simple proof. Dan === Subject: Re: A statement about Rsa2048. > Why do you assume that the two factors have the same number of decimal > digits? >Yes OK, if they are both 1024 bits long, then they must >each have 309 >digits. And be strictly within your lower and upper >bounds. Yes! To find these prime pairs of equal length that form these composites, the composites must reside on the same triangle number index as rsa2048 as does the composite resulting from the two factors I show. It is difficult to find these discrete paired prime composites that will only appear within this index. The number of these special composites is probably uncountable within the given index but still hard to find. Dan === Subject: Re: A statement about Rsa2048. posting-account=T7Gd-QoAAACeQajv7mi_Za6uPu3TpBXy AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.43 Safari/525.19,gzip(gfe),gzip(gfe) > Assuming there are only two prime factors of equal length > in rsa2048 these two primes below would represent the > smallest possible and largest possible prime factors. The smallest factor of 309 digits--- > The range -- > from > 100000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000010459 > to sqrt(rsa2048) The largest factor of 309 digits--- > The range -- > from > sqrt(rsa2048) > too -- > 251959084756578934940271832400483985714292821262040320277771378360436620207 075955562640185258807844069182906412495150821892985591491761845028084891200 728449926873928072877767359714183472702618963750149718246911650776133798590 957000973304597488084284017974291006424586918171951187461215151726546322822 168673523 Knowing this and assuming the two factors are > of equal length, neither one of these two primes > could ever represent one of the factors in rsa2048. I have a simple proof. Dan Why do you assume that the two factors have the same number of decimal digits? === Subject: Re: A statement about Rsa2048. posting-account=T7Gd-QoAAACeQajv7mi_Za6uPu3TpBXy AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.43 Safari/525.19,gzip(gfe),gzip(gfe) Assuming there are only two prime factors of equal length > in rsa2048 these two primes below would represent the > smallest possible and largest possible prime factors. The smallest factor of 309 digits--- > The range -- > from > 100000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 000010459 > to sqrt(rsa2048) The largest factor of 309 digits--- > The range -- > from > sqrt(rsa2048) > too -- > 251959084756578934940271832400483985714292821262040320277771378360436620207 075955562640185258807844069182906412495150821892985591491761845028084891200 728449926873928072877767359714183472702618963750149718246911650776133798590 957000973304597488084284017974291006424586918171951187461215151726546322822 168673523 Knowing this and assuming the two factors are > of equal length, neither one of these two primes > could ever represent one of the factors in rsa2048. I have a simple proof. Dan Why do you assume that the two factors have the same number of decimal > digits? Yes OK, if they are both 1024 bits long, then they must each have 309 digits. And be strictly within your lower and upper bounds. === Subject: Help needed on combinatorial interpretation posting-account=JYIhQwoAAADfOqRKeWW3k_OkLnLoycin Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) Masters of Math, may I ask for your help to identify the combinatorial significance of this little formula: Sum( v=0..m, binomial(m,v) m! (v+2)^m ) === Subject: Re: Fwd: Planets Gather on May 5 and May 17, 2000 posting-account=plvunQoAAAAsFAHWouUDJEZVb5_cH_gI ---------- Forwarded message ---------- === Subject: Re: Planets Gather on May 5 and May 17, 2000 Hi Marcus, 15,000 flying bombs landed on London in 1945, World War II. (V1 and V2) 15,000 flying bombs then landed on Antwerp in 1945. 15,000 postings, cross posted to sci.chem and three other newsgroups, had the subject line Jeehad needs scientists. The blow by blow of what happened, when I pointed Big Bertha Thing and pulled the trigger, is contained on swnet.sci.astro, instead of sci.chem I have been the primary user of swnet.sci.astro since 1998. Would you honestly want me somewhere else? Everyone should have a home. Tony Lance tonylance@myinternetuk.com I am not interested in liitle bugs, just the gross misconduct sort. > Vice-presidents heads should roll. Don't bother even trying to understand. I don't think you'll ever will. > -- Marcus m9370@abc.se ++++++++++++++++++++++++++++++++++++++++++ > Hi there, > Taking the 2065 AD event as my staring point, I went looking for a > Ring Cycle of Uranus and > and found the one below. === Subject: Re: JSH problem posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) [David C. Ullrich] > I haven't been paying close attention. Sure you want to admit that? James has already warned that you > (academics in general) are legally responsible for reporting his > discoveries (although to whom they must report remains unclear), and > that anything you say on Usenet will be used against you as evidence in > court. I'm puzzled about the supposed problem with the algebraic > integers. That's easy enough: James dislikes them intensely, because he believes > they should behave in such a way instead that /one/ of the broken > foundations of his FLT proof stops being obviously broken even to him. This is something that has always puzzled me. James does not seem to have a > logical understanding of factor coprime divisible by etc.. Obviously > all these terms only make sense in the context of a specific ring, whereas > James seems to believe e.g. that only one of the roots *actually* has 7 as > a factor - i.e. that terms like factor have some absolute meaning (no > doubt based on the visual form of some expression). Given this outlook, how did James ever pick up on the algebraic integers > specifically? Hmm...not sure, but I remember him being incensed > that people secretly knew about algebraic integers > but didn't tell him and left him to twist in the > wind. (Or in fact with any ring? - this all just seems to be beyond > him...) This must have happened years ago, as when I started following he > was already fixated on the algebraic integers. Back in 2001, he had a group called The Algebra > of Factorizations and set up a chat room for > discussion. He had exactly one chat room discussion and, > typical for him, deleted the group. Here's a transcript of that single discussion in > which James demostrates that he knows nothing > about the subject of rings which he had been > simultaneously making his usual blowhard posts > on sci.math on the very subject. (James is using > the pseudonym Flatrings in this discussion as > he had been saying rings must be flat on sci.math). > This was posted here by Wilma Scranton and can be > found in a Google Group search. I think the algebraic integers came later. A transcript of that conversation follows: Wow. -- m === Subject: Re: JSH problem ... > So what is the object ring? You have to ask James ;-) No need to ask James. What it comes down to is that given a quadratic with two, non-integral roots, one of the roots is in the object ring and the other is not. This must be the case for every quadratic. The problem, obviously, but James does not see it, is that such a ring does not exist. That is, it is not possible to make a consistent choice for every pair of roots of the quadratics. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH problem > ... > > So what is the object ring? You have to ask James ;-) No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). Then x = (-p +/- sqrt(p^2 - 4q))/2 . Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. So (-p - sqrt(p^2 - 4q))/2 = q/a . Both roots are algebraic integers, and the extension field Q(a) contains q/a, which is an algebraic integer. So the ring of integers of Q(a) contains the other root, namely q/a . So if x^2 + px + q = 0 has non-rational roots, one of them being a, then the ring of integers in Q(a) also contains a/q, the other root. (x - a)(x - q/a) = x^2 + x*(-q/a - a) + q = x^2 + x*(-qa'/(a*a') -a ) + q = x^2 + x*(-a -a' ) +q = x^2 + px + q . a + a' = -p. So a' = -p - a. So any extension ring of Q containing a must contain the conjugate root a' . (case p, q in Q) Also, if a is an algebraic integer, we may assume p, q in Z and the an extension ring of Z containing a must contain the conjugate root a', which is also an algebraic integer. Mhh.. David Bernier === Subject: Re: JSH problem posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) > ... > > So what is the object ring? You have to ask James ;-) No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). > Then x = (-p +/- sqrt(p^2 - 4q))/2 . Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. So (-p - sqrt(p^2 - 4q))/2 = q/a . > Both roots are algebraic integers, and the extension > field Q(a) contains q/a, which is an algebraic integer. > So the ring of integers of Q(a) contains the other root, > namely q/a . So if x^2 + px + q = 0 has non-rational roots, one > of them being a, then the ring of integers in Q(a) > also contains a/q, the other root. (x - a)(x - q/a) = x^2 + x*(-q/a - a) + q = x^2 + x*(-qa'/(a*a') -a ) + q > = x^2 + x*(-a -a' ) +q = x^2 + px + q . a + a' = -p. So a' = -p - a. So any extension ring of Q containing a > must contain the conjugate root a' . (case p, q in Q) > Also, if a is an algebraic integer, Clearly it is. > we may assume p, q in Z > and the an extension ring of Z containing a must contain > the conjugate root a', which is also an algebraic integer. Mhh.. David Bernier All true I think, but that's not the question. Assume that q = r*s, where (say) r is prime. Then a Harrisian object ring would contain exactly one of a/r or a'/r. In general Harris has said that the object ring would also contain all the algebraic integers. It is not clear as Dik suggests that there is one object ring which contains one of a/r or a'/r for all irreducible quadratics that have r as a factor of the constant term. But it may be that something like A[a/r] is sufficient for the Harrisian oeuvre, where A is the ring of algebraic integers. He has been somewhat noncommittal beyond his defective definition of the object ring. Marcus. === Subject: Re: JSH problem > ... > > So what is the object ring? You have to ask James ;-) > No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. > Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). > Then x = (-p +/- sqrt(p^2 - 4q))/2 . > Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] > then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. > So (-p - sqrt(p^2 - 4q))/2 = q/a . > Both roots are algebraic integers, and the extension > field Q(a) contains q/a, which is an algebraic integer. > So the ring of integers of Q(a) contains the other root, > namely q/a . > So if x^2 + px + q = 0 has non-rational roots, one > of them being a, then the ring of integers in Q(a) > also contains a/q, the other root. > (x - a)(x - q/a) = x^2 + x*(-q/a - a) + q = x^2 + x*(-qa'/(a*a') -a ) + q > = x^2 + x*(-a -a' ) +q = x^2 + px + q . > a + a' = -p. So a' = -p - a. So any extension ring of Q containing a > must contain the conjugate root a' . (case p, q in Q) > Also, if a is an algebraic integer, Clearly it is. we may assume p, q in Z > and the an extension ring of Z containing a must contain > the conjugate root a', which is also an algebraic integer. > Mhh.. > David Bernier All true I think, but that's not the question. Assume that > q = r*s, where (say) r is prime. Then a Harrisian object ring > would contain exactly one of a/r or a'/r. In general Harris has said > that the object ring would also contain all the algebraic > integers. It is not clear as Dik suggests that there is > one object ring which contains one of a/r or a'/r for all > irreducible quadratics that have r as a factor of the constant term. > But it may be that something like A[a/r] is sufficient > for the Harrisian oeuvre, where A is the ring of algebraic > integers. He has been somewhat noncommittal beyond his > defective definition of the object ring. I appreciate that Arturo Magidin and you commented on what I like that, it's perhaps a consequence of the conclusions he wants. I remember there was a lot of discussion about specific examples (algebraic numbers of degree 2 I think) which disproved something James had claimed. Could you elaborate a bit on the counterexamples to James's claims regarding factors or whatever it was in algebraic number theory? David === Subject: Re: JSH problem posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) > ... > > So what is the object ring? You have to ask James ;-) > No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. > Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). > Then x = (-p +/- sqrt(p^2 - 4q))/2 . > Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] > then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. > So (-p - sqrt(p^2 - 4q))/2 = q/a . > Both roots are algebraic integers, and the extension > field Q(a) contains q/a, which is an algebraic integer. > So the ring of integers of Q(a) contains the other root, > namely q/a . > So if x^2 + px + q = 0 has non-rational roots, one > of them being a, then the ring of integers in Q(a) > also contains a/q, the other root. > (x - a)(x - q/a) = x^2 + x*(-q/a - a) + q = x^2 + x*(-qa'/(a*a') -a ) + q > = x^2 + x*(-a -a' ) +q = x^2 + px + q . > a + a' = -p. So a' = -p - a. So any extension ring of Q containing a > must contain the conjugate root a' . (case p, q in Q) > Also, if a is an algebraic integer, Clearly it is. > we may assume p, q in Z > and the an extension ring of Z containing a must contain > the conjugate root a', which is also an algebraic integer. > Mhh.. > David Bernier All true I think, but that's not the question. Assume that > q = r*s, where (say) r is prime. Then a Harrisian object ring > would contain exactly one of a/r or a'/r. In general Harris has said > that the object ring would also contain all the algebraic > integers. It is not clear as Dik suggests that there is > one object ring which contains one of a/r or a'/r for all > irreducible quadratics that have r as a factor of the constant term. > But it may be that something like A[a/r] is sufficient > for the Harrisian oeuvre, where A is the ring of algebraic > integers. He has been somewhat noncommittal beyond his > defective definition of the object ring. I appreciate that Arturo Magidin and you commented on what I > like that, it's perhaps a consequence of the conclusions > he wants. I remember there was a lot of discussion about > specific examples (algebraic numbers of degree 2 I think) > which disproved something James had claimed. Could you elaborate a bit on the counterexamples to > James's claims regarding factors or whatever it was > in algebraic number theory? David Succinctly, Harris thinks he has a proof that one of the roots of, e.g., a^2 - 6a + 35 is divisible by 7 in the ring of algebraic integers and the other root is coprime to 7 in that same ring. This contradicts well-established theory at a very low level. For example, assume that the root which is divisible by 7 is 3 + sqrt(-26). Thus u = (3 + sqrt(-26))/7 would be an algebraic integer. You conclude that 49u^2 - 42 u + 9 = -26, or 7u^2 - 6u + 5 = 0. This is an irreducible nonmonic primitive polynomial with leading coefficient 7, which means that, contrary to the Harris proof, it cannot be an algebraic integer. An almost identical argument shows that (3 - sqrt(-26))/7 similarly cannot be an algebraic integer. Harris waffles on this somewhat, saying e.g. that his result APPEARS to contradict known facts, or that one of the roots SHOULD be divisible by 7 and the other coprime to 7. But his argument does not invoke any ring properties that are not possessed by the ring of algebraic integers, which means that if his argument were valid, it would be valid in that ring. But clearly it is not. The root of all this is, he is making an assumption about how 7 distributes across (5a_1(x) + 7)*(5a_2(x) + 7). He assumes that 7 distributes across the two factors in a way that does not depend on x. Then he notes that when x = 0, 7 factors out of one of the two factors. He does not state this assumption explicitly and may not be aware, or want to admit, that he is making it. Logically, the contradiction that he encounters should make him say, Oh - one of my assumptions must be wrong. But instead of reaching that conclusion, he says Oh - the theory of algebraic integers must be wrong. Or, Oh - Galois theory must be mis- interpreted. Etc. Since he does not admit that he is making the assumption above, he does not conclude that assumption is the root of the contradiction. And that is where he has been for the last 5-6 years. Marcus. Marcus. === Subject: Re: JSH problem days. My association with the Department is that of an alumnus. > ... > > So what is the object ring? You have to ask James ;-) No need to ask James. What it comes down to is that given a quadratic with > two, non-integral roots, one of the roots is in the object ring and the other > is not. This must be the case for every quadratic. The problem, obviously, > but James does not see it, is that such a ring does not exist. That is, it > is not possible to make a consistent choice for every pair of roots of the > quadratics. Let's see: x^2 + px + q = 0 (field = Q, p, q in Z). The roots of such a quadratic would be integral. By non-integral, Dik means that the quadratic must be a constant multiple of a non-monic, primitive, irreducible quadratic. >Then x = (-p +/- sqrt(p^2 - 4q))/2 . Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...] then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q. So (-p - sqrt(p^2 - 4q))/2 = q/a . >Both roots are algebraic integers, Indeed: and therefore they are integral. Dik said non-integral, not non-integer. > and the extension >field Q(a) contains q/a, which is an algebraic integer. >So the ring of integers of Q(a) contains the other root, >namely q/a . Now consider the case of tx^2 + px + q, with t,p,q in Z, gcd(t,p,q)=1, and t=/= 1,-1. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: microwave station posting-account=9GlOXwoAAAAlAbPpB4XhR7GNOEPVSRWt Trident/4.0; FunWebProducts; ub2; InfoPath.2; Zango 10.3.36.0; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How would i start to build a workable microwave station? === Subject: Re: microwave station posting-account=W7RQ6gkAAACLDC5JWapQU1UV8ot8KkAS .NET CLR 1.0.3705; Dealio Toolbar 3.1.1),gzip(gfe),gzip(gfe) > How would i start to build a workable microwave station? Microwave stations are so hard to be build, that's one of the reasons people invented Holograms, Licences, and Optical Computers though. === Subject: Re: microwave station > How would i start to build a workable microwave station? just buy one from ebay. === Subject: Re: microwave station Wrong group, you Google-posting AOL using twat. > How would I brane working ???????????? === Subject: Re: microwave station > Wrong group, you Google-posting AOL using twat. Off with your head, you top-posting OE-using twit! -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: microwave station > How would i start to build a workable microwave station? I hate to admit it, but _this_ is the main reason I read this group. B. -- Cheerfully resisting change since 1959. === Subject: subgroup of alternating group A_5 Hello teacher, How does one show that the alternating group A_5 has no subgroup of order 30. I know that A_5 has order 60. So it might be possible, as 30 divides 60. I also know that A_5 is a simple group. === Subject: Re: subgroup of alternating group A_5 Piet Vanraad schreef: > Hello teacher, How does one show that the alternating group A_5 has no subgroup of > order 30. I know that A_5 has order 60. So it might be possible, as 30 divides 60. > I also know that A_5 is a simple group. for a contradiction in this direction, but forgot about that theorem! === Subject: Re: subgroup of alternating group A_5 days. My association with the Department is that of an alumnus. >Hello teacher, How does one show that the alternating group A_5 has no subgroup of >order 30. I know that A_5 has order 60. So it might be possible, as 30 divides 60. >I also know that A_5 is a simple group. If you know that A_5 is a simple group, then show that a subgroup of order 30 would necessarily be normal. HINT: There is a theorem that talks about subgroups of index 2. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: subgroup of alternating group A_5 posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Hello teacher, How does one show that the alternating group A 5 has no subgroup of > order 30. I know that A 5 has order 60. So it might be possible, as 30 divides 60. > I also know that A 5 is a simple group. A subgroup of order 30 would be normal, because its index would be 2, and A 5 is simple. -- m === Subject: Please help - dynamic programming or robust optimization? posting-account=cewOKQoAAAAlnpN-CpPojngZgjG_pN_6 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) I'm wondering can you help me with this 9hope i'm i nthe right place, humble apologies if I'm not). I have certain process that occurs every day. To say the least it is an uncertain process. I think that can be said for sure. I am unsure if it fits any known distribution. Every day I take 2 numbers from this process. Once in the morning and 1 in the afternoon. I will call those numbers A and B for day n where n at the moment is day 1, the first day. I also have 3 sets of 2 numbers that represent % ranges. They are labelled: LLn = lower limit for day n ULn = upper limit for day n LLn+1 = lower limit for day n + 1 ULn+1 = upper limit for day n + 1 LLn+2 = lower limit for day n + 2 ULn+2 = upper limit for day n + 2 Every day I make 3 decisions (depending upon the values of A and B relative to the limits). Decision A: If on the first day (B < A - LLn) I exit and calculate (B - A) which will be negative. I store this number and repeat the process the next day. Decision B: If on the first day (B > A + LLn) I proceed to the next day n+1 and and repeat the process using the new lower and upper limits repectively LLn+1 and ULN+1 I proceed to the next day and repeat the process. Eventually at some stage decision A will occur. Like before I store this number, in this case it would be Bn+m - An, where m is the number of days the process continued for. Decision C: If on the first day (B <= A + UPn) AND (B >= A - LLn) I proceed to the next day n+1 and repeat the process using the previous day's upper and lower limits respectively: ULn and LLn I proceed to the next day and repeat the process. Eventually at some stage decision A will occur. Like before I store this number, in this case it would be Bn+m - An, where m is the number of days the process continued for. At the end of a certain time period I add up the outcomes to obtain a net total. My objective is to maximise this total. The only variables that I can control upper and lower limits respectively. Now I *believe that either the following can help me optimise this problem: 1. Dynamic programming with recourse 2. Robust optimization I feel that robusy optimization is the way to go, but am unsure how to do it. Can anyone out there lend me some assistance and take me under their wing please? Either in LP or QP, whichever is easier! Would greatly appreciate any advice/help/suggestions/mentoring on this one. Michael. === Subject: Re: Please help - dynamic programming or robust optimization? posting-account=gpERugkAAAB5_qKVhbO9UpGpOXFNrIYf CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) [snip long confused description of problem] > Can anyone out there lend me some assistance and take me > under their wing please? Either in LP or QP, whichever is easier! Would greatly appreciate any advice/help/suggestions/mentoring on this > one. Sure! ^ ^ What do you pay? Socks === Subject: Origin of Gunfire This solution obtains the origin of gunfire using three microphones. http://mypeoplepc.com/members/jon8338/math/id22.html I have posted this before but I corrected a mistake I made. This correction fixes the problem of the source of the gunfire originating from either side of the plane of the three microphones. There are two solutions. A fourth microphone comparing times determines which side of the bull's horns the shot originates. There's a link on the site to an Excel 97 workbook that calculates the solution for different inputs. === Subject: Re: A new definition for Life posting-account=BKLKygoAAACKhQOBleq3odJ7v98y_w-j Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > The second law of thermodynamics, states that the total entropy of any > isolated thermodynamic system tends to increase over time, approaching > a maximum value; in more general terms, systems move from a state of > higher complexity and activity, to one of lower complexity and > activity. This applies to inanimate systems, inanimate matter. Quite > clearly, for living systems and beings, for animate objects, the > opposite applies. Life is matter in reverse entropy: life > spontaneously and eternally moves from lower to higher levels of > complexity and structure. This definition subsumes Darwin's concept > of Evolution, and reconciles it with a sense of eternal purpose in the > Universe. It is funny how you got it exactly in reverse. The very reason Life exist, and if you will its purpose is to accelerate the entropy increase. Life is the most powerful entropy increase catalyst, because it uses part of the released energy to maintain its own complexity. And it is this complexity allows it to be such an efficient catalyst! The reason that Life's complexity is increasing with evolution is that it becomes more efficient entropy increase catalyst, while accounting for more and more new variations of external environments and involving new entropy increase energy flows (for example nuclear in addition to chemical) into this catalysis. Early life was only using chemical reactions (such as Fe3+ reduction) to increase entropy, later life found a way to increase entropy using sun-light, modern life (us) found a way to increase entropy of nuclear decomposition reactions as well as nuclear synthesis reactions. Future of Life is in expanding into new entropy increase processes (those increasing its own complexity as it does so), or being wiped out if existing entropy increase process is exhausted. No wonder Life choses the first and its complexity is increasing. Well, some form of Life did chose the second, and that is why they are dead and we don't know anything about them.... force of life complexity increase, but just a mechanism of optimization. Just like genetic algorithm is optimizing parameters of a function to satisfy certain criteria, Darvinian evolution optimizes Life for criteria maximal entropy increase efficiency over maximal possible time interval and maximal possible external conditions. Our future is in accelerating more energetic entropy increase processes - in stars, in black-holes, in quasars! Forget about your water, bacteria and sugar - Life is about the future of the Universe! Yevgen === Subject: Re: A new definition for Life > The second law of thermodynamics, states that the total > entropy of any isolated thermodynamic system tends to > increase over time, approaching a maximum value; in more > general terms, systems move from a state of higher > complexity and activity, to one of lower complexity and > activity. This applies to inanimate systems, inanimate > matter. Quite clearly, for living systems and beings, > for animate objects, the opposite applies. Life is > matter in reverse entropy: life spontaneously and > eternally moves from lower to higher levels of complexity > and structure. This definition subsumes Darwin's concept > of Evolution, and reconciles it with a sense of eternal > purpose in the Universe. Ok. === Subject: Re: A new definition for Life > - >There are numerous example of extinct species throughout history. >Species of man being among them. Was the divinity just learning how >to do it right? >The other example is that people are not evolving. The less intelligent >have more children so the average intelligence is decreasing. Also >medicine is allowing more of us to survive and pass on our various >defects. Not much of a divine plan. > You're failing to grasp my concept of divinity, Doug. It's not an all > powerful Pixie in the sky with an immutable plan. It's just a general > direction to things. That's all. That's all I see. The laws of physics set what happens in the universe. That is the only divine guidance. But, I do see > that. Which, is more than no divinity, at all. > The laws of physics are quite a divinity. === Subject: Re: A new definition for Life > If you are going to make up new concepts, then make new words. By > co-opting old words you just jam up communication. I know how it feels to want to invent new words. === Subject: Re: A new definition for Life >Evolution seems to involve the progressive development of >systematically more complex organisms. That is not correct. It is a random walk. Over time, things will increasingly diverge from the starting point. However, individual evolutionary events may be toward less or more complexity. Certainly it has no direct relationship to Darwinian evolution, which is based on fitness, not complexity. There is no general relationship between fitness and complexity. As to entropy... Consider the following analogy... Imagine a room, with screened windows on opposite sides. There are some balloons in the room. Their position is random = high entropy. Now turn on a wind machine, so that wind flows thru the room from one side to the other. The balloons will now tend to be all on one side, where the wind exits; low entropy. This is a simple example showing how the flow of energy can create (local) order. It is very much like a living system in that regard. Obviously, life cannot be defined simply in terms of entropy. Bob === Subject: Re: A new definition for Life <8senm4tvf3g9uf2m2cpts6aoths15f07sg@4ax.com> posting-account=5ayZ-goAAABGZmmwx8zZEwz6gU2OuVSd CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) >Evolution seems to involve the progressive development of >systematically more complex organisms. That is not correct. It is a random walk. Over time, things will > increasingly diverge from the starting point. However, individual > evolutionary events may be toward less or more complexity. Certainly > it has no direct relationship to Darwinian evolution, which is based > on fitness, not complexity. There is no general relationship between > fitness and complexity. Darwin was wrong. Evolution is not a random walk. More complex and sophisticated organisms will, invevitably, have more capacity to survive, reproduce, and thrive, in general. A more complex and sophisticated organism will have a better capacity to control its evironment, and therefore to survive within it. It won't have to adapt. It can control. === Subject: Re: A new definition for Life >Evolution seems to involve the progressive development of >systematically more complex organisms. That is not correct. It is a random walk. Over time, things will > increasingly diverge from the starting point. However, individual > evolutionary events may be toward less or more complexity. Certainly > it has no direct relationship to Darwinian evolution, which is based > on fitness, not complexity. There is no general relationship between > fitness and complexity. As to entropy... Consider the following analogy... Imagine a room, > with screened windows on opposite sides. There are some balloons in > the room. Their position is random = high entropy. Now turn on a wind > machine, so that wind flows thru the room from one side to the other. > The balloons will now tend to be all on one side, where the wind > exits; low entropy. This is a simple example showing how the flow of > energy can create (local) order. It is very much like a living system > in that regard. Obviously, life cannot be defined simply in terms of > entropy. Here you say it: wind MACHINE. A device created by intelligence. So this is a simple example showing how a flow designed by intelligence can make up local order. And life in the end. Han de Bruijn === Subject: Re: A new definition for Life <7p8nm4929s4kvauv5l45fb0eh454m85t11@4ax.com> posting-account=5ayZ-goAAABGZmmwx8zZEwz6gU2OuVSd CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > in > *Some* life has gotten more complex, but the vast majority of life > on Earth is single celled. Nevertheless, Matt, we see more complex life forms developing systematically with time. Quite systematically. Higher life now is far more intelligent and more socially complex than 100 million years ago or 500 million years ago. And, more complex than 100 years ago, for that matter. Pretty good proof of God, I'd say. === Subject: Re: A new definition for Life posting-account=5PxZDwoAAABv9p271mvbXvIpZRL9I54K CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > in > *Some* life has gotten more complex, but the vast majority of life > on Earth is single celled. Nevertheless, Matt, we see more complex life forms developing >systematically with time. Quite systematically. Higher life now is >far more intelligent and more socially complex than 100 million years >ago or 500 million years ago. And, more complex than 100 years ago, >for that matter. Pretty good proof of God, I'd say. Huh? Come again? Care to fill in some of the details of that > proof? -- > Angus Rodgers I must emphasize, I'm not talking about an anthropomorphic God, simply a meaningful direction to the Universe as a whole. What more meaningful direction could there be, but towards increasing sophistication, complexity and power? Which is clearly the direction of biological evolution. More God-like organisms automatically and spontaneously develop over time, everywhere. === Subject: Re: A new definition for Life > I must emphasize, I'm not talking about an anthropomorphic God, simply > a meaningful direction to the Universe as a whole. What more > meaningful direction could there be, but towards increasing > sophistication, complexity and power? Which is clearly the direction > of biological evolution. More God-like organisms automatically and > spontaneously develop over time, everywhere. Your remark reminds me of a problem I feel in considering possibilities of doing foolish things, even though I choose not to do them. Banging my head against a wall, or breaking things, or vandalizing property. Not to scare anyone -- I certainly don't recommend these actions -- but it seems like anyone could refute any argument about human nature by positing them. Now, why would you do them, except to win arguments and gain social status? Why would you *want* that social status? So that people would look up to you as an intelligent philosopher, admire you, be friends with you, perhaps marry and have children with you? Might you then be reversing the second law of thermodynamics after all, caring for your family and maintaining the orderliness of your environment? === Subject: Re: A new definition for Life in >I must emphasize, I'm not talking about an anthropomorphic God, simply >a meaningful direction to the Universe as a whole. This is called an equivocation and is dishonest. You are come up with some new concept, but you use an old word so you can get support from people who actually disagree with you. >What more >meaningful direction could there be, but towards increasing >sophistication, complexity and power? Meaningful to us, sure. But the Universe is, to good approximation, empty space. Not all that complex. >Which is clearly the direction >of biological evolution. Yet most life, overwhelmingly most life, is not complex. >More God-like organisms automatically and >spontaneously develop over time, everywhere. -- Matt Silberstein Do something today about the Darfur Genocide http://www.beawitness.org http://www.darfurgenocide.org http://www.savedarfur.org Darfur: A Genocide We can Stop === Subject: Re: A new definition for Life > in I must emphasize, I'm not talking about an anthropomorphic God, simply >a meaningful direction to the Universe as a whole. > > This is called an equivocation and is dishonest. You are come up with > some new concept, but you use an old word so you can get support from > people who actually disagree with you. > >What more >meaningful direction could there be, but towards increasing >sophistication, complexity and power? > > Meaningful to us, sure. But the Universe is, to good approximation, > empty space. Not all that complex. > >Which is clearly the direction >of biological evolution. > > Yet most life, overwhelmingly most life, is not complex. Wrong. Even a single cell is tremendously complex. >More God-like organisms automatically and >spontaneously develop over time, everywhere. Han de Bruijn === Subject: Re: A new definition for Life > This is called an equivocation and is dishonest. You are come up with > some new concept, but you use an old word so you can get support from > people who actually disagree with you. To be fair, to equivocate is not to use a word with several definitions but to deduce consequences of one definition from another. === Subject: Re: A new definition for Life > in > *Some* life has gotten more complex, but the vast majority of life > on Earth is single celled. > Nevertheless, Matt, we see more complex life forms developing > systematically with time. Quite systematically. Higher life now is > far more intelligent and more socially complex than 100 million years > ago or 500 million years ago. And, more complex than 100 years ago, > for that matter. Pretty good proof of God, I'd say. > Huh? Come again? Care to fill in some of the details of that > proof? > -- > Angus Rodgers I must emphasize, I'm not talking about an anthropomorphic God, simply > a meaningful direction to the Universe as a whole. What more > meaningful direction could there be, but towards increasing > sophistication, complexity and power? Which is clearly the direction > of biological evolution. More God-like organisms automatically and > spontaneously develop over time, everywhere. > It is quite clear that you know very little about biological evolution. === Subject: Radius of largest n+1 balls in n-dimensional unit cube? posting-account=WzP9FgoAAAANyEt4wx0YVvhakkQXYd72 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) I would appreciate your help or comment about the following problem. Consider a N dimensional space. If I want to put N+1 balls (all with the same radius R), within the unit hypercube such that: 1. The balls do not cut through each other 2. One of the balls is at the center of the cube, i.e. at (0.5 , 0.5, 0.5, ...., 0.5) Then what is the maximum possible R in terms of N? If not easy, a good approximation will be helpful too. Golabi === Subject: Re: Radius of largest n+1 balls in n-dimensional unit cube? I would appreciate your help or comment about the following problem. >Consider a N dimensional space. If I want to put N+1 balls (all with >the same radius R), within the unit hypercube such that: 1. The balls do not cut through each other >2. One of the balls is at the center of the cube, i.e. at (0.5 , 0.5, >0.5, ...., 0.5) Then what is the maximum possible R in terms of N? If not easy, a good >approximation will be helpful too. It seems intuitive that the maximum r for your problem occurs when the other n balls are in the corners, tangent to the central ball and tangent to the corner faces. For that configuration, r = sqrt(n) / (2*(2+sqrt(n))) But note, the unit cube in R^n has 2^n corners, so you could just as easily have (2^n)+1 balls with the same radius as above, rather than only n+1. quasi === Subject: Re: Radius of largest n+1 balls in n-dimensional unit cube? posting-account=WzP9FgoAAAANyEt4wx0YVvhakkQXYd72 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) derive the formula you proposed? The only piece I understand is that sqrt(n) is maximal distance between two corners of the cube, but have no idea how it entered into your formula. >I would appreciate your help or comment about the following problem. >Consider a N dimensional space. If I want to put N+1 balls (all with >the same radius R), within the unit hypercube such that: 1. The balls do not cut through each other >2. One of the balls is at the center of the cube, i.e. at (0.5 , 0.5, >0.5, ...., 0.5) Then what is the maximum possible R in terms of N? If not easy, a good >approximation will be helpful too. It seems intuitive that the maximum r for your problem occurs when the > other n balls are in the corners, tangent to the central ball and > tangent to the corner faces. For that configuration, r = sqrt(n) / (2*(2+sqrt(n))) But note, the unit cube in R^n has 2^n corners, so you could just as > easily have (2^n)+1 balls with the same radius as above, rather than > only n+1. quasi === Subject: Re: Radius of largest n+1 balls in n-dimensional unit cube? >derive the formula you proposed? The only piece I understand is that >sqrt(n) is maximal distance between two corners of the cube, but have >no idea how it entered into your formula. Ok, but please don't top post. Instead, either bottom post or intersperse portions of your reply after the relevant parts of the prior message (as I am doing here). That's the standard in sci.math. For the configuration I described, r can be calculated as follows: Place a little cube of edge length r in the corner of the unit cube. Then the distance from the center of the ball in that corner to the corner vertex is the length of the diagonal of the little cube, which is r*sqrt(n) (by proportionality to the unit cube). Then (draw a diagram to see it), r + r + r*sqrt(n) = (1/2)*sqrt(n) which yields r = sqrt(n) / (2*(2+sqrt(n))) >I would appreciate your help or comment about the following problem. >Consider a N dimensional space. If I want to put N+1 balls (all with >the same radius R), within the unit hypercube such that: >1. The balls do not cut through each other >2. One of the balls is at the center of the cube, i.e. at (0.5 , 0.5, >0.5, ...., 0.5) >Then what is the maximum possible R in terms of N? If not easy, a good >approximation will be helpful too. > It seems intuitive that the maximum r for your problem occurs when the > other n balls are in the corners, tangent to the central ball and > tangent to the corner faces. For that configuration, > r = sqrt(n) / (2*(2+sqrt(n))) > But note, the unit cube in R^n has 2^n corners, so you could just as > easily have (2^n)+1 balls with the same radius as above, rather than > only n+1. quasi === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem Angus Rodgers a .8ecrit : One bit of my sleepy mind says > that this makes some sense, while another bit says it is a bit odd Without having meant to think about it any more for the moment, > I am becoming a bit more sure that always switch is the right > answer. If I try not to think too much about it, my mind says: > On the one hand, /if/ you have chosen the door with the prize, > then Monty's choice is telling you nothing about whether the > prize is behind your door or the other one. On the other hand, > /if/ you have chosen the wrong door, then Monty's choice is tell- > ing you which the right door is - and it's the other one! Since > these two events are a priori equally likely, > DFont you mean * not a priori*? on balance Monty's > choice is telling you to go for the other door. Right. Anyway, I wanted to tell oyu more about this, especially as you > seem to have a very low opinion on yoursezlf on htose probability > matters, while I have not zseen you mke any mistake yet. So 1) dont > underestimate yourself. 2) after reading the link I gave you > (http://en.wikipedia.org/wiki/Monty_Hall_problem ), search next for > other interesting probability paradoxes (here is a list : > http://en.wikipedia.org/wiki/Category:Probability_theory_paradoxes ) > and test yourself : you could have a nice surprise :-) [incorrect stuff elided] >Is this argument correct? No -- you inadvertantly used the wrong formula for P(AB). It should be P(AB) = P(A|B) P(B). Also, consider the conceptual meaning of P(A|B) = P(A). It means that the conditional probabiity of A, given B, is the same as > the absolute probability of A. Thus, the even B gives no information > about A -- i.e. doesn't affect absolute probabiity of A. Sorry for the trouble. Yes, p(A|B) = p(A) implies A and B are independent. p(AB)=P(A|B)p(B) =P(A)p(B). I also proved the proposition, with probability at least 0.35, but will not inflict it on you. Berkson's paradox: The result is that two independent events become conditionally dependent (negatively dependent) given that at least one of them occurs. 0 right... Of course, the other crankish behavior of the poster does > nothing to change my mind... What other crankish behavior of the poster? I may have missed > something, but all I've seen is a guy who (1) thinks he's got an > elementary proof of FLT, and (2) wants to find some way to make some > money from it. Neither of these is crankish. > One of those is is credentials attitude (the I have gone to school in John Baez list), another is his fear of stolen ideas... I insist, see the list === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem Michael Press a .8ecrit : Angus Rodgers a .8ecrit : > One bit of my sleepy mind says > that this makes some sense, while another bit says it is a bit odd > Without having meant to think about it any more for the moment, > I am becoming a bit more sure that always switch is the right > answer. If I try not to think too much about it, my mind says: > On the one hand, /if/ you have chosen the door with the prize, > then Monty's choice is telling you nothing about whether the > prize is behind your door or the other one. On the other hand, > /if/ you have chosen the wrong door, then Monty's choice is tell- > ing you which the right door is - and it's the other one! Since > these two events are a priori equally likely, > DFont you mean * not a priori*? > on balance Monty's > choice is telling you to go for the other door. > Right. Anyway, I wanted to tell oyu more about this, especially as you > seem to have a very low opinion on yoursezlf on htose probability > matters, while I have not zseen you mke any mistake yet. So 1) dont > underestimate yourself. 2) after reading the link I gave you > (http://en.wikipedia.org/wiki/Monty_Hall_problem ), search next for > other interesting probability paradoxes (here is a list : > http://en.wikipedia.org/wiki/Category:Probability_theory_paradoxes ) > and test yourself : you could have a nice surprise :-) I went to the page and picked the first alphabetcally: > Berkson's paradox; and I find the statement to be wrong. They write > 0 p(A|B) = p(A), i.e. they are independent. The statement that they are indedepent > is wrong if `they' refers to A and B. The wording is perhaps wrong... But see http://en.wikipedia.org/wiki/Statistical_independence > for we have p(AB) = P(A|B)p(A) = p(A)^2 = p(B|A)p(B) What??? p(AB) = P(A|B)p(B) ... Suppose p(A)=p(A|B) = 1/2, then p(AB)= p(B|A)p(B) = 1/4. > Suppose p(B|A) = 1 and p(B) = 1/4. > Then we have p(B|A)p(B) = 1/4, in accordance with the specifications. > Also p(A)p(B) = 1/8 =/= 1/4 = p(AB). > But A and B independent iff p(AB) = p(A)p(B). > Contradiction. > Is this argument correct? Certainly not... > === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem > nice expository on the generalizations, > using a sort of meta-triple, where F'L'T eqs. are > (k,1,2) (and > Pythagoreans are (2,1,2), from monsieur PT3: > http://www.geocities.com/titus_piezas/Timeline1.htm Nice area for professionals and for amateurs ! Infinite capacity for patient math. creations ! > surprised if it's not valid comes from the fact > that I would not be > surprised if one doesn't exist. When coming back to FLT, I think the only elementary and partial solutions are for Sophie Germain primes... If Mr. David Fabian claims something else, I'll recommend to him to check if his methods are solving following equation: t^n = 2*n^u abtp +a^n + b^n where a;b;t;p;n are natural numbers and of gcd=1 This is equivalent of X^n +Y^n = Z^n for X;Y;Z;n natural numbers and of gcd=1 once n>=3 for specially chosen parameters. Such clear and correct parametrization could be written: 2Z = X+Y +Z-Y +Z-X = t^n + b^n + a^n 2X = X+Y +Z-Y -Z+X = t^n + b^n - a^n 2Y = X+Y -Z+Y +Z-X = t^n - b^n + a^n But only for 1-st case of FLT. What should be written for X;Y or Z divided by n ? What for value represents X+Y-Z ? Nice and simple but far away from some proof ! Ro-Bin === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem > Zachary Turner a .8ecrit : > Having read most of this thread, I can say pretty confidently you have > two options: > 1) Publish it on the web. If it is determined to be correct you will > likely get a book deal out of it, you will probably be offered high > paying jobs either in the finance industry or by the government as > well. You may be invited to give talks as well, all of which would > bring income. If it is to be incorrect you can try again, and once > you are choose again from options 1 or 2 listed here. > Poster never responded to the possibility of parlaying his success > into a better position. Some posts later, he inadvertently reveals > why: he's in his 50's. > M Well it's pretty much just a fact of life that nobody's just going to > write a big fat check for a proof which may or may not be correct. > And showing it to someone to verify its correctness will raise issues > of its own, as if it turns out to be incorrect, even slightly, they > might build off the work and end up with something correct and publish > it free. Therefore regardless of said proof's correctness and > strategy used to release proof in a way conducive to money-making, > there is going to be some inherent level of risk involved in terms of > possibly not receiving the payout the OP desires. So the best > strategy is probably just to publish it, and assume the risk that you > may not get the book deal or whatever. At least this has a >0% > probability that money will be made, as opposed to a 0% probability by > just sitting on it. In any case, one thing I will say is that there's a lot of cranks > going around claiming to have the secrets of the world, and most of > them are just cranks. However, in this case I would actually not be > surprised if the OP does have a valid proof Not be surprised show you know very little about the problem ( > and perhaps about math in general) (although I would also not be surprised if he doesn't). Search around for the name a bit. Compared to some of the people on this forum I probably do know > relatively little about math in general. But then again, compared to > some of the people not on this forum, some of the people on this forum > also know relatively little about math. I'm not sure what your point > is. I have an undergrad + masters in math from a pretty strong > american university, and was almost top in my class. Although I still > fail to see how that's relevant. News flash: People who don't know much about math in general tend not > to like it, and people who don't like it don't hang around math forums > in their free time. Disagree. There are people posting here at length who do not like mathematics. My external, objective criteria for liking mathematics include knowledge of the workings; ability to follow and generate proofs; and desire and gratitude for instruction. -- Michael Press === Subject: Re: The modern mathematical concept of infinity is indefensible <87zlhxaif1.fsf@alatheia.dsl.inet.fi> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > But let's set that aside. How does the inconsistency of ZFC > follow from the falsity of the completenss theorem? Because the completeness theorem follows from ZFC. > Consistent theories may well prove all sorts of falsities. NO, NOT *all* sorts. In particular, they NEVER prove the sort of falsity that is the DENIAL of one of their THEOREMS. === Subject: Re: The modern mathematical concept of infinity is indefensible > Because the completeness theorem follows from ZFC. ZFC is inconsistent follows from the axioms of the theory ZFC + ZFC is incosistent. Assuming ZFC + ZFC is inconsistent is in fact consistent, or equivalently, that ZFC is consistent, how would you use this fact in showing that ZFC + ZFC is inconsistent (or in ZFC)? If you agree there is in fact no way of doing so, how exactly do you propose to do so in case of the completeness theorem, supposing its falsity? > In particular, they NEVER prove the sort of falsity that is the > DENIAL of one of their THEOREMS. The relevance of this escapes me. There was no suggestion that even if the completeness theorem is false there is a derivation of the negation of the completeness theorem in ZFC. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > ZFC is inconsistent follows from the axioms of the theory ZFC + ZFC > is incosistent. Assuming ZFC + ZFC is inconsistent is in fact > consistent, or equivalently, that ZFC is consistent, how would you use > this fact in showing that ZFC + ZFC is inconsistent (or in ZFC)? My apologies. This should read Assuming ZFC + ZFC is inconsistent is in fact consistent, or equivalently, that ZFC is consistent, how would you use this fact in showing that ZFC + ZFC is inconsistent (or in ZFC) is inconsistent? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > You will already have problems when tying to explain what it means for an > axiom to be true... Consider an axiom in (certain systems of) ZFC: ExAy(y !e x). There is a set without elements With other words, there is an empty set. Now what does it mean to claim that this axiom is true? Exactly the same as to claim there is a set without elements. As to whether there is a set without elements, a very natural attitude is to regard this as not something determined by any matters of fact, but as something up to stipulation. In case of other axioms and proposed axioms, such as the axiom of choice or various large cardinal axioms, such an attitude is very unnatural and subtler considerations are called for. > What does it mean (in this case) for a statement in our theory to be > true other than: either this is an axiom, or it can be derived from > the axioms (and definitions)? That's provability from the axioms, not truth. It is a mathematical result that for example the truth of the axiom of choice or the continuum hypothesis is not equivalent to the provability of this or that in any formal theory. It is also a mathematical theorem that if ZFC is consistent there are infinitely many true arithmetical statements not provable in ZFC. And so on. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > Not at all. Check out Vicious Circles by Barwise and Moss, an > excellent (if poorly proofread) book which I'm currently reading. It's > about a particular version of set theory without the axiom of > foundation, with a new axiom which says, roughly speaking, that > equations such as x = {x,p} have unique solutions. In this context one might mention Aczel's book _Non-well-founded sets_ which is available on-line, at http://standish.stanford.edu/pdf/00000056.pdf -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <874p03931c.fsf@alatheia.dsl.inet.fi > Not at all. Check out Vicious Circles by Barwise and Moss, an > excellent (if poorly proofread) book which I'm currently reading. It's > about a particular version of set theory without the axiom of > foundation, with a new axiom which says, roughly speaking, that > equations such as x = {x,p} have unique solutions. In this context one might mention Aczel's book _Non-well-founded sets_ > which is available on-line, at http://standish.stanford.edu/pdf/00000056.pdf Ah, so it is still out there. === Subject: Re: The modern mathematical concept of infinity is indefensible Not at all. Check out Vicious Circles by Barwise and Moss, an > excellent (if poorly proofread) book which I'm currently reading. It's > about a particular version of set theory without the axiom of > foundation, with a new axiom which says, roughly speaking, that > equations such as x = {x,p} have unique solutions. In this context one might mention Aczel's book _Non-well-founded sets_ > which is available on-line, at http://standish.stanford.edu/pdf/00000056.pdf > Don't know if this is relevant to the issue but I've wondered if there is a kind of set system that: (a) there are multiple epsilon-relations (symbols) (b) there's one usual epsilon-relation that is well-founded but there are others that are not? -- To discover the proper approach to mathematical logic, we must therefore examine the methods of the mathematician. (Shoenfield, Mathematical Logic) === Subject: Re: The modern mathematical concept of infinity is indefensible <1ac8qu5fk14wf.1006vfhkth1o2$.dlg@40tude.net> <54obm4500241eimqhaisfucoqqk3b4koba@4ax.com> <87hc48lbkf.fsf@phiwumbda.org> <87vdslai9g.fsf@alatheia.dsl.inet.fi> <87sknp85u0.fsf@alatheia.dsl.inet.fi> <1lidr5olvinmz.1pkaemdj9ngtq.dlg@40tude.net You will already have problems when tying to explain what it means > for an axiom to be true... Consider an axiom in (certain systems of) ZFC: ExAy(y !e x). There is a set without elements With other words, there is an empty set. Now what does it mean to claim that this axiom is true? That there > ACTUALLY is an empty set (somewhere)? That there is an empty set, yes. > Well, sure, the statement ExAy(y !e x) may be interpreted to say > just that. But I have to admit, sadly, that I don't believe in the > existence of mathematical objects (and hence especially not in the > existence of the empty set). Well, then we are off down another bunny trail entirely. :-) > That there is an empty set somewhere, obviously, no, as pure sets > (if there are any) have no spatial location. Right. I also don't think that pure sets are spacio-temporal objects. > But imho they are not just _nowhere_, but not at all. Not an unreasonable view. > And if it does NOT mean the latter, what does true then refer to > here? I mean how can a sentence be true, without being the case what > it -the sentence- states? Etc. etc. Pretty clearly, it cannot be. > Well it can (imho), if you adhere to a coherence theory of truth (in > contrast to a correspondence theory of truth). Well, I was going to add unless one has a quirky and untenable view of truth. ;-) === Subject: Re: The modern mathematical concept of infinity is indefensible Chris Menzel says... > Well it can (imho), if you adhere to a coherence theory of truth (in > contrast to a correspondence theory of truth). Well, I was going to add unless one has a quirky and untenable view of >truth. ;-) I don't have a completely coherent view of mathematical truth, but it does seem to me that something like the coherence theory is plausible. Suppose you never heard of any notion of number other than natural number. Then someone asks you Does there exist a number x such that x+1=0? (or such that x*2 = 1, or such that x*x = 2, or such that x*x = -1) The answer is not simply yes or no. There is no such x among the numbers that you are aware of, but we can coherently *posit* the existence of negative numbers, rationals, reals and complex numbers. It would be unreasonable for someone to speculate that while it is *consistent* that there are negative numbers, they don't *really* exist. I think for *definable* sets, at least, (such as the empty set) their existence is just a matter of stipulation. We *define* the empty set to be the set with no elements, in the same way that we define -1 to be the solution to x+1=0. There is nothing more to say about the empty set, and there certainly isn't any Platonic question of whether it really exists. -- Daryl McCullough Ithaca, NY === Subject: Re: The modern mathematical concept of infinity is indefensible > Chris Menzel says... >Well it can (imho), if you adhere to a coherence theory of truth (in >contrast to a correspondence theory of truth). >Well, I was going to add unless one has a quirky and untenable view of >truth. ;-) I don't have a completely coherent view of mathematical truth, > but it does seem to me that something like the coherence theory > is plausible. Suppose you never heard of any notion of number other than > natural number. Then someone asks you Does there exist a > number x such that x+1=0? (or such that x*2 = 1, or such that > x*x = 2, or such that x*x = -1) The answer is not simply yes or no. There is no such x among > the numbers that you are aware of, but we can coherently *posit* > the existence of negative numbers, rationals, reals and complex > numbers. It would be unreasonable for someone to speculate that > while it is *consistent* that there are negative numbers, they > don't *really* exist. Objection ! If natural numbers exist, then there exist a _construction_ by which you can actually _make_ negative numbers, rational numbers and complex numbers out of this naturals material. Negative numbers are just pairs of natural numbers ,with newly defined addition and multiplication and _equality_ for those pairs, such that as many properties of naturals as possible are preserved. Rational numbers are pairs of whole numbers, with newly defined addition and multiplication and _equality_ for those pairs, such that as many properties of the whole numbers as possible are preserved. Now guess what complex numbers are .. > I think for *definable* sets, at least, (such as the empty set) > their existence is just a matter of stipulation. We *define* > the empty set to be the set with no elements, in the same way > that we define -1 to be the solution to x+1=0. There is nothing > more to say about the empty set, and there certainly isn't > any Platonic question of whether it really exists. Objection ! The empty set is an empty box. If the empty set exist, then there exist a _construction_ by which you can actually _make_ all other (heredetary finite) sets, e.g. according to: http://hdebruijn.soo.dto.tudelft.nl/jaar2007/set_theory.pdf And the empty set is _implementable_ as computer hardware and software; it's a piece of technology; therefore its existence is not disputable. > -- > Daryl McCullough > Ithaca, NY Han de Bruijn === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Negative numbers are just > pairs of natural numbers ,with newly defined addition and multiplication > and equality for those pairs, If they are then you'd have to explain how the negative integer marked by the pair <1 2> is the same as the negative integer marked by the pair <2 3>. The ordinary solution is to take integers (positive and non-positive) not as pairs of natural numbers but rather as equivalenece classes of pairs of natural numbers, which, too bad for you, requires having the set of natural numbers. > Rational numbers are pairs of whole numbers, > with newly defined addition and multiplication and equality for those > pairs, such that as many properties of the whole numbers as possible are > preserved. If they are then you'd have to explain how the rational marked by the pair <1 2> is the same as the rational number marked by the pair <2 4>. The ordinary solution is to take rational numbers not as pairs of integers but rather as equivalenece classes of pairs of integers, which, too bad for you, requires having the set of natural numbers. > Now guess what complex numbers are .. Complex numbers, ordinarily, are pairs of real numbers (not equivalence classes of pairs of real numbers). But what are real numbers? You've not shown a construction of real numbers that does not require having the set of natural numbers. > Objection ! The empty set is an empty box. No, it's not. A box is a physical object. > If the empty set exist, then > there exist a construction by which you can actually make all other > (heredetary finite) sets Whatever the merits of that claim, from the axiom of separation and axiom of extensionality, it is provable that there is a unique object, call it '0', such that no object in the epsilon to 0. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible MoeBlee says... > Negative numbers are just > pairs of natural numbers ,with newly defined addition and multiplication > and _equality_ for those pairs, If they are then you'd have to explain how the negative integer marked >by the pair <1 2> is the same as the negative integer marked by the >pair <2 3>. The ordinary solution is to take integers (positive and >non-positive) not as pairs of natural numbers but rather as >equivalenece classes of pairs of natural numbers, which, too bad for >you, requires having the set of natural numbers. In some varieties of constructive type theory, one defines a new type A' from an old type A by redefining equality. So the type of integers is the type NxN together with the equality relation Q_Z iff x+y' Q_N x'+y. (Where Q_N and Q_Z are the equality relations for N and Z, respectively) So if type A' is obtained from type A by imposing a new equality relation, and B' is obtained from type B by imposing a new equality relation, and f is a function from A to B, then to prove that f is a function from A' to B', you have to show that, forall x1, x2, if x1 Q_A' x2, then f(x1) Q_B' f(x2). All this is equivalent in some sense to working with equivalence classes, but requires less of the heavy guns of full blown set theory. -- Daryl McCullough Ithaca, NY === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > MoeBlee says... > Negative numbers are just > pairs of natural numbers ,with newly defined addition and multiplication > and equality for those pairs, If they are then you'd have to explain how the negative integer marked >by the pair <1 2> is the same as the negative integer marked by the >pair <2 3>. The ordinary solution is to take integers (positive and >non-positive) not as pairs of natural numbers but rather as >equivalenece classes of pairs of natural numbers, which, too bad for >you, requires having the set of natural numbers. In some varieties of constructive type theory, I have no dispute as to the many possibilities of formulations in theories weaker than Z set theory or not comparable in strength with Z set theory. My point in my post to Han is that if one does eschew roomier theories such as Z, then one does have to show that the alternative works and be willing to pay its cost in cumbersomeness. In that regard, Han is not at all prepared for the technicalities of constructive type theory and, in that regard, his claim that all we need is ordered pairs of naturals is bluster, as it would be for me or anyone to make that claim if one had not actually worked through to understand such alternative formulations as constructive type theory. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) >no object in the epsilon to 0. I meant no object is in the epsilon relation to 0. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > If they are then you'd have to explain how the negative integer > marked by the pair <1 2> is the same as the negative integer marked > by the pair <2 3>. The ordinary solution is to take integers > (positive and non-positive) not as pairs of natural numbers but > rather as equivalenece classes of pairs of natural numbers, which, > too bad for you, requires having the set of natural numbers. Requires in what sense? All talk about the negative integers, rationals and so on and not involving infinite sets, can be translated into talk about the naturals and basic operations on naturals (and nothing) else in a straightforward though tedious manner. For example, there is no difficulty in formulating arguments usually presented in terms of integers, such as one might find in Euclid and modern variants thereof, in Peano arithmetic. Instead of equivalence classes we simply explicitly use the relevant equivalence relation, which in no way depends on the existence of any infinite sets. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <87bpubvzs5.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > If they are then you'd have to explain how the negative integer > marked by the pair <1 2> is the same as the negative integer marked > by the pair <2 3>. The ordinary solution is to take integers > (positive and non-positive) not as pairs of natural numbers but > rather as equivalenece classes of pairs of natural numbers, which, > too bad for you, requires having the set of natural numbers. Requires in what sense? In the sense of the ordinary set theoretic, axiomatic handling of the matter. That there may be some other axiomatic handling that does not require the set of natural numbers is not disputed by me, nor is it disputed by me that one may choose not to rise to the rigor or axiomatization. My immediate point here is that the poster has not given reference to such a handling, thus my remark you'd have to explain. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > My immediate point here is that the poster has not given reference > to such a handling, thus my remark you'd have to explain. Well, the fact such an explanation can be found in any number of standard texts makes one wonder why you should feel it necessary to make this point in this context. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <87zlhvukp3.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > My immediate point here is that the poster has not given reference > to such a handling, thus my remark you'd have to explain. Well, the fact such an explanation can be found in any number of > standard texts makes one wonder why you should feel it necessary to > make this point in this context. The point in this context may be that every explanation has a cost in cumbersomeness. An explanation of operations on integers (positive and non-positive) and rationals in a more restricted theory than set theory would likely be more cumbersome, as would be actually working with integers and rationals in the more restricted theory. So, in that sense, there is a cost for such as Han de Bruijn's claim that we can just take integers to be ordered pairs of naturals and rationals as ordered pairs of integers, which is a cost we can bet he is not personally willing to pay in his actual mathematical practice (he surely is not going to try to perform integer and rational operations as coded in Peano arithmetic (as well as, by the way, a definition of 'first order PA', ordinarily (again, that context I first mentioned) requires having the set of natural numbers just to set up (defined) the language of first order PA). So, in that practical sense, such simplifications by the likes of Han, as just ordered pairs of naturals are bluster. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > The point in this context may be that every explanation has a cost > in cumbersomeness. An explanation of operations on integers > (positive and non-positive) and rationals in a more restricted > theory than set theory would likely be more cumbersome, as would be > actually working with integers and rationals in the more restricted > theory. That depends on what we understand by working with and what theories we have in mind. In a suitable system, e.g. one allowing us to introduce new sets by introducing a notion of equality, as in constructive type theory, the approach need not be cumbersome at all. Han might well be sensible enough in this context to mention, say, the handling of arbitrary rationals (of any type in the Num typeclass) in Haskell, given his fondness of computer implementations. I don't think your point is very well taken. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <87ljtfujht.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Han might well be sensible enough in this context to mention, > say, the handling of arbitrary rationals (of any type in the Num > typeclass) in Haskell, given his fondness of computer > implementations. IF he were indeed taking rationals as just pairs and are thus constructed in such a context, then, granted, my retort was off- base. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > IF he were indeed taking rationals as just pairs and are thus > constructed in such a context, then, granted, my retort was off- > base. That would be the charitable interpretation, in any case. Whether you feel charitable towards Han and his various utterances is up to you, of course. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <877i4zui5n.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > IF he were indeed taking rationals as just pairs and are thus > constructed in such a context, then, granted, my retort was off- > base. That would be the charitable interpretation, in any case. Whether you > feel charitable towards Han and his various utterances is up to you, > of course. I don't wish to be unreasonable or overly restrictive as to what qualifies as a construction, and Han did mention not just pairs but re- defintion of equality too. So while it is still not clear to me in what sense his proposal constructs objects onto themselves (i.e., aside from operations), since I can't rule out that he has some reasonable context in mind, I had better back off my original retort. MoeBlee === Subject: Re: The modern mathematical concept of infinity is indefensible > And the empty set is _implementable_ as computer hardware and software; > it's a piece of technology; therefore its existence is not disputable. Right. In contrast the existence of arbitrarily complex hereditarily finite sets is very much disputable. Extensive computer experiments are necessary to determine which of them exist and which are but figments of our collective imagination. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible And the empty set is _implementable_ as computer hardware and software; >it's a piece of technology; therefore its existence is not disputable. Right. In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. Extensive computer experiments > are necessary to determine which of them exist and which are but > figments of our collective imagination. A bit of common reasoning will be more helpful than all that. Han de Bruijn === Subject: Re: The modern mathematical concept of infinity is indefensible A bit of common reasoning will be more helpful than all that. > Since, as it seems, infinite collections of things aren't part of our usual experience of the world, finitistic common sense won't help in dealing with (the ideas of) such entities. The world of the infinite is ... open for exploration, but to embark we must unlearn our finitistic intuitions which instill fear and confusion by making some consistent and demonstrable results about the infinite literally counter-intuitive. [...] (Peter Suber, Infinite Reflections) Herb P.S. Btw. I guess with common reasoning you refer to those inconsistent (extremely idiotic) stuff concerning math at your homepage, right? === Subject: Re: The modern mathematical concept of infinity is indefensible > Since, as it seems, infinite collections of things aren't part of > our usual experience of the world, finitistic common sense won't > help in dealing with (the ideas of) such entities. Arbitrarily large finite collections aren't a part of our usual experience of the world either, so we need rely on some other source, perhaps some theoretical generalisation of that experience, to reason about finitary objects in general, in a finitistically meaningful and justified way or otherwise. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. Extensive computer > experiments are necessary to determine which of them exist and > which are but figments of our collective imagination. A bit of common reasoning will be more helpful than all that. Surely the existence of arbitrarily complex hereditarily finite set is intimately tied to the limits and possibilities of computer simulations. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. Extensive computer > experiments are necessary to determine which of them exist and > which are but figments of our collective imagination. > A bit of common reasoning will be more helpful than all that. > Surely the [alleged] existence of arbitrarily complex hereditarily > finite set is intimately tied to the limits and possibilities of > computer simulations. > Of course. Since computer resources are bound (assuming that our universe is finite), there are (or can be) only finitely many HFS. Simple as that. Oh... oh... another ultrafinitist ... Herb === Subject: Re: The modern mathematical concept of infinity is indefensible In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. Extensive computer > experiments are necessary to determine which of them exist and > which are but figments of our collective imagination. > A bit of common reasoning will be more helpful than all that. > Surely the [alleged] existence of arbitrarily complex hereditarily > finite set is intimately tied to the limits and possibilities of > computer simulations. > Of course. Since computer resources are bound (assuming that our universe > is finite), there are (or can be) only finitely many HFS. > Simple as that. Oh... oh... another ultrafinitist ... Do you think there's evidence for the infinite in the real world? I'm wondering about these things because in the opposing view to the infinite world, the other guy could say: << That's rubbish. You've only seen/measured very big things and very small things. It's all on a scale from 10^{-googolplex} to 10^{googolplex}, but you'll never know, of course. > What do you think? David Bernier === Subject: Re: The modern mathematical concept of infinity is indefensible > Oh... oh... another ultrafinitist ... > Do you think there's evidence for the infinite in the real world? > Well, actually, it's hard to say. Maybe there is, maybe not. But so what? As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality. (A. Einstein) I'm wondering about these things because in the opposing view > to the infinite world, the other guy could say: << That's rubbish. You've only seen/measured very big things and > very small things. It's all on a scale from > 10^{-googolplex} to 10^{googolplex}, but you'll never > know, of course. What do you think? > I'd agree. We simply don't know. Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > Do you think there's evidence for the infinite in the real world? > I'm wondering about these things because in the opposing view to the > infinite world, the other guy could say: << That's rubbish. You've only seen/measured very big things and > very small things. It's all on a scale from > 10^{-googolplex} to 10^{googolplex}, but you'll never > know, of course. What do you think? Thing like that have their appeal from a certain point of view. What is very difficult is to squeeze any coherent or interesting mathematics out of such a view. It is also often the case that people who say stuff like that in fact reason, when caught off-guard, in a way that presupposes it makes sense to speak of, say, arbitrarily complex formal proofs, formulas, hypothetical computers and such like. Consistent ultra-finitism, anti-realism about all of mathematics, etc. are heroic undertakings requiring almost super-human determination and effort. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Consistent ultra-finitism [is a] heroic undertakings requiring > almost super-human determination and effort. > Seems so: [Though] Ultrafinitism is a form of constructivism [...] even constructivists generally view the philosophy as unworkably extreme. The logical foundation of ultrafinitism is unclear; in his comprehensive survey Constructivism in Mathematics (1988), the constructive logician A. S. Troelstra dismissed it as no satisfactory development exists at present. This was not so much a philosophical objection as it was an admission that, in a rigorous work of mathematical logic, there was simply nothing precise enough to include. http://en.wikipedia.org/wiki/Ultrafinitism Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > Consistent ultra-finitism [is a] heroic undertakings requiring > almost super-human determination and effort. Seems so: [Though] Ultrafinitism is a form of constructivism [...] even > constructivists generally view the philosophy as unworkably extreme. > The logical foundation of ultrafinitism is unclear; in his comprehensive > survey Constructivism in Mathematics (1988), the constructive logician A. > S. Troelstra dismissed it as no satisfactory development exists at > present. This was not so much a philosophical objection as it was an > admission that, in a rigorous work of mathematical logic, there was simply > nothing precise enough to include. http://en.wikipedia.org/wiki/Ultrafinitism There is nothing objectionable in this quote from Wikipedia, but the lack of a satisfactory development, in the foundational sense, is merely a symptom of the inherent difficulty to being consistent about ultra-finitism in mathematics. It is very difficult to set aside our natural intuitions about the infinite progression of naturals, and consequently there is a dearth of any actual ultra-finitist arguments to formalise, arguments which might serve as guidance and raw material when unearthing basic principles and concepts of ultra-finitism. In contrast, with set theory, constructive mathematics, and so on, there was already a formidable body of mathematical reasoning which to consult, and in relation to which to evaluate the suitability of various formalisations and explications. A paper of interest in this connexion is /A Formalization of Essenin-Volpin's Proof Theoretical Studies by Means of Nonstandard Analysis/, James R. Geiser, The Journal of Symbolic Logic, Vol. 39, No. 1 (Mar., 1974), pp. 81-87 -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Do you think there's evidence for the infinite in the real world? >I'm wondering about these things because in the opposing view to the >infinite world, the other guy could say: ><< That's rubbish. You've only seen/measured very big things and > very small things. It's all on a scale from > 10^{-googolplex} to 10^{googolplex}, but you'll never > know, of course. >What do you think? Thing like that have their appeal from a certain point of view. What > is very difficult is to squeeze any coherent or interesting > mathematics out of such a view. It is also often the case that people > who say stuff like that in fact reason, when caught off-guard, in a > way that presupposes it makes sense to speak of, say, arbitrarily > complex formal proofs, formulas, hypothetical computers and such > like. Consistent ultra-finitism, anti-realism about all of > mathematics, etc. are heroic undertakings requiring almost super-human > determination and effort. Sure. Computational mathematics is a heroic undertaking requiring almost super-human determination and effort. Welcome to the 21'th century ! Han de Bruijn === Subject: Re: The modern mathematical concept of infinity is indefensible Sure. Computational mathematics is a heroic undertaking requiring > almost super-human determination and effort. [...] > Huh?! Man you are talking nonsense. (Well, not that uncommon for a crank.) Herb === Subject: Re: The modern mathematical concept of infinity is indefensible Consistent ultra-finitism, anti-realism about all of mathematics, > etc. are heroic undertakings requiring almost super-human > determination and effort. Sure. Computational mathematics is a heroic undertaking requiring > almost super-human determination and effort. What leads you to this pecular conclusion? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Surely the [alleged] existence of arbitrarily complex hereditarily > finite set is intimately tied to the limits and possibilities of > computer simulations. Of course. Since computer resources are bound (assuming that our > universe is finite), there are (or can be) only finitely many HFS. > Simple as that. Quite, if we are serious about the justification for the existence of such sets as being intimately tied to computer simulations. If we are prepared to accept the existence of arbitrarily complex hereditarily finite sets in our mathematical arguments and reasoning, and most of us are, without so much as blinking, this can't be because they can be simulated on any actual computer (since they in fact can't). Whatever computers are involved, and computers and computer implementations might well enter into our explanation of the world of hereditarily finite sets, serving as helpful illustrations perhaps, are thus just as theoretical, imaginary and dubious -- or just as innocuous -- as the arbitrarily complex hereditarily finite sets themselves. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible And the empty set is _implementable_ as computer hardware and software; > it's a piece of technology; therefore its existence is not disputable. Right. In contrast the existence of arbitrarily complex hereditarily > finite sets is very much disputable. > Let alone the existence of infinite sets!!! Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > And the empty set is _implementable_ as computer hardware and > software; it's a piece of technology; therefore its existence > is not disputable. Right. In contrast the existence of arbitrarily complex > hereditarily finite sets is very much disputable. Let alone the existence of infinite sets!!! Right! Their existence is in fact infinitely disputable. Incidentally, I seem to recall you killfiled me for being an asshole. Did something lead you to revise your opinion? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Right! Their existence is in fact infinitely disputable. Incidentally, > I seem to recall you killfiled me for being an asshole. Did something > lead you to revise your opinion? > Not really, but I've access to a different computer now, and I just didn't start to build up a killfile there. Still you will see that I'm not eager to engage in any (serious) discussion with you, I simply don't like your attitude(s). Herb === Subject: Re: The modern mathematical concept of infinity is indefensible Incidentally, I seem to recall you killfiled me for being an > asshole. Did something lead you to revise your opinion? Not really, but I've access to a different computer now, and I just > didn't start to build up a killfile there. Still you will see that I'm not eager to engage in any (serious) > discussion with you, I simply don't like your attitude(s). What sort of attitudes you or anyone finds congenial is of course entirely their business. Perhaps in the future we may both engage in and enjoy some entirely frivolous discussion. Is there something in particular in my attitude or attitudes that rubs you the wrong way? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > ... and there certainly isn't any Platonic question of > whether it really exists. > It isn't if we agree that math is just some sort of mythological story (i.e. a fiction). Stating ExAy(y !e x), as an axiom in ZFC (then) is just this: a statement in the (mathematical) story called ZFC. It's like stating Holmes had a friend called Watson in the context of one of the (fictional) stories by Sir Arthur Conan Doyle. On the other hand, there have been (and still are) mathematicians who are convinced that such a statement actually refers to a state of affairs in some reality (realm) ... Classes and concepts may, however, also be conceived as real objects, namely classes as pluralities of things or as structures consisting of a plurality of things and concepts as the properties and relations of things existing independently of our definitions and constructions. It seems to me that the assumption of such objects is quite as legitimate as the assumption of physical bodies and there is quite as much reason to believe in their existence. (K. G.9adel, Russell's Mathematical Logic) Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > It isn't if we agree that math is just some sort of mythological > story (i.e. a fiction). Stating ExAy(y !e x), as an axiom in ZFC (then) is just this: a statement in the > (mathematical) story called ZFC. It's like stating Holmes had a > friend called Watson in the context of one of the (fictional) > stories by Sir Arthur Conan Doyle. This is a very appealing idea for all sorts of reasons in case such claims as the existence of the empty set. In case of other statements, such as ZFC is inconsistent, every even natural greater than two is the sum of two primes and so on, the idea has considerably less appeal. In particular, we make use of such statements outside mathematics, e.g. in such non-mathematical claims as Even if ZFC is inconsistent, it might be impossible for us to ever find this out. Here ZFC is inconsistent is used, so to speak, as a factual hypothesis. In his dissertation Torkel refers to the idea that there is nothing more to the truth or falsity of claims of abstract set theory than what is inherent in the mathematical stories we tell as a distancing view. It is in practice very difficult to maintain such a view about /all/ of mathematics, and since abstract set theoretic principles have consequences we are naturally inclined to regard as determined as a matter of mathematical fact, such as whether ZFC is consistent or not, we are lead to ponder the interesting question of why we should accept consequences of fictional claims about fictional entities that are, so to speak, factual. It is also obscure just what reality purely abstract fictional entities not posited to ineract with us in any way could possess in addition to being parts of the conceptual stories we tell about them, and consequently just what is claimed in saying that, say, these or those mathematical objects don't actually exist. That is, just what it is we are denying in such claims? (In contrast, there is nothing obscure in noting that Sherlock Holmes or Captain Planet don't really exist.) -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > Suppose you never heard of any notion of number other than > natural number. Then someone asks you Does there exist a > number x such that x+1=0? (or such that x*2 = 1, or such that > x*x = 2, or such that x*x = -1) The answer is not simply yes or no. There is no such x among > the numbers that you are aware of, but we can coherently *posit* > the existence of negative numbers, rationals, reals and complex > numbers. It would be unreasonable for someone to speculate that > while it is *consistent* that there are negative numbers, they > don't *really* exist. Sure. But it is a rather odd notion that, say, the truth of Fermat's last theorem, consistency of ZFC, and so on, are a matter of coherence or stipulation, even though it makes perfect sense to suppose there is nothing more to the existence of the rationals or zero than our stipulation to that effect. Also, in case of the reals we seem to be introducing substantial ontological assumptions, about arbitrary sets or sequences of naturals, and so on; assumptions that have arithmetical consequences it would be odd to regard as open to stipulation, or determined merely by their coherence with these or those mathematical claims. (Merely introducing ontological assumptions need not be dubious, of course. I believe it was Kreisel who once cautioned against such doubts, remarking that they seem to be founded in thinking about mathematical objects as something akin to huge physical objects.) Of course, we must also note that the notion of consistency involved here is not the technical one applying to formal theories -- as you're well aware, consistency of a formal theory is insufficient to guarantee it describes any coherent mathematical picture -- and that this notion, whatever it amounts to exactly, is not philosophically unproblematic. > I think for *definable* sets, at least, (such as the empty set) > their existence is just a matter of stipulation. We *define* the > empty set to be the set with no elements, in the same way that we > define -1 to be the solution to x+1=0. There is nothing more to say > about the empty set, and there certainly isn't any Platonic question > of whether it really exists. If someone came forth and said the empty set exists or the empty set does not exist it would indeed be very odd to regard them as reporting a mathematical discovery, something they could be mistaken about. Naturally, we would take such assertions as stipulations, perhaps occurring in the context of introduction to an exposition of some theory of sets. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Supersedes: <1ax3y7zpprxre.eg5cegu645bl$.dlg@40tude.net> I mentioned the existence claim of the empty set as one of the axioms of ZFC: Ex(Ay(y !e x). Now you say: If someone came forth and said the empty set exists or the empty > set does not exist it would indeed be very odd to regard them as > reporting a mathematical discovery ... > Well: [...] despite their remoteness from sense experience, we do have something like a perception also of the objects of set theory, as is seen from the fact that the axioms force themselves upon us as being true. I don't see any reason why we should have less confidence in this kind of perception, and more generally, in mathematical intuition than in sense perception. (K. G.9adel) Herb === Subject: Re: The modern mathematical concept of infinity is indefensible I mentioned the existence claim of the empty set as one of the axioms of > ZFC: Ex(Ay(y !e x). Now you say: If someone came forth and said the empty set exists or the > empty set does not exist it would indeed be very odd to regard > them as reporting a mathematical discovery ... Well: [...] despite their remoteness from sense experience, we do > have something like a perception also of the objects of set theory, > as is seen from the fact that the axioms force themselves upon us as > being true. I don't see any reason why we should have less > confidence in this kind of perception, and more generally, in > mathematical intuition than in sense perception. On the conception of sets as arbitrary extensional collections in the cumulative hierarchy many axioms do force themselves upon us, in the sense that they seem evident and compelling. But before they can do anything like that we must have some idea of what we're talking about, some explanation. In saying that the existence of the empty set is a matter of stipulation I have in mind in part that such stipulation would be naturally regarded as a component in such an explanation of what we're talking about. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible Supersedes: <10zwlr3t1erf1.1p2lyj0ihq5gw.dlg@40tude.net> I meantioned the existence claim of the empty set as one of the axioms of ZFC: Ex(Ay(y !e x). Now you say: If someone came forth and said the empty set exists or the empty > set does not exist it would indeed be very odd to regard them as > reporting a mathematical discovery ... > Well: [...] despite their remoteness from sense experience, we do have something like a perception also of the objects of set theory, as is seen from the fact that the axioms force themselves upon us as being true. I don't see any reason why we should have less confidence in this kind of perception, and more generally, in mathematical intuition than in sense perception. (K. G.9adel) Herb === Subject: Re: The modern mathematical concept of infinity is indefensible <873afofw9v.fsf@phiwumbda.org> <87tz84e8s8.fsf@phiwumbda.org> <1uwp3cfq5z0ox.1rebiiv9nmql7.dlg@40tude.net > I think you could make it work in Aczel set theory. > I don't think so. If 1 = {1} and 2 = {1,2}, then 1 and 2 are bisimilar and hence equal > in Aczel's theory. Sure? (Note that I know almost nothing about Aczel's theory). I mean 1 = {1} (in that theory) is the unique set x such that x = {x}. Now the unique set y such that y = {1, y} needn't be identical with 1, > imho. Well, since we know that (by definition) 1 = {1} and since {1} = {1,1}, we have 1 = {1,1} and so we know that 1 satisfies y = {1,y}. Hence, if, as you say, there is a *unique* such y, it must be identical to 1. I think what you are questioning is whether there *is* a unique such y. Aczel's anti-foundation axiom AFA asserts exactly that. But this is by no means self-evident. > After all 1 only has just one element, while this unique set, let's > call it 2, might have two elements; I mean it will iff 1 =/= 2. ;-) But maybe I'm just plainly wrong. Well, just a thought. You are not plainly wrong. AFA is only one of several reasonable anti-foundation alternatives and in fact on at least one, SAFA (Scott's anti-foundation axiom), there is a set satisfying y = {1,y} that is not identical to 1. It seems to me (without having put any thought into it) that the proposed definition of a series of n.w.f. finite ordinals satisfying 1 = {1}, 2 = {1,2}, ... is feasible under SAFA. Anyone? I'm very rusty here... === Subject: Re: The modern mathematical concept of infinity is indefensible > AFA is only one of several reasonable anti-foundation alternatives > and in fact on at least one, SAFA (Scott's anti-foundation axiom), > there is a set satisfying y = {1,y} that is not identical to 1. It > seems to me (without having put any thought into it) that the > proposed definition of a series of n.w.f. finite ordinals > satisfying 1 = {1}, 2 = {1,2}, ... is feasible under SAFA. Anyone? > I'm very rusty here... I am even rustier! Where can one learn more about Scott's only that it is a stenghtening of the axiom of extensionality... -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <873afofw9v.fsf@phiwumbda.org> <87tz84e8s8.fsf@phiwumbda.org> <1uwp3cfq5z0ox.1rebiiv9nmql7.dlg@40tude.net> <87eiz77m9i.fsf@alatheia.dsl.inet.fi> said: > AFA is only one of several reasonable anti-foundation alternatives > and in fact on at least one, SAFA (Scott's anti-foundation axiom), > there is a set satisfying y = {1,y} that is not identical to 1. It > seems to me (without having put any thought into it) that the > proposed definition of a series of n.w.f. finite ordinals > satisfying 1 = {1}, 2 = {1,2}, ... is feasible under SAFA. Anyone? > I'm very rusty here... I am even rustier! Where can one learn more about Scott's > anti-foundation axiom? Eh? Something about logic that you haven't already mastered? :-) Herb has given a good reference -- from a quick look, Smith's paper is basically a distillation of the first four chapters or so of Aczel's book. At one time the latter was out of print and available for free as a huge scanned PDF from CSLI, but it has been reprinted and appears once again to be available in hard copy only. > stengthening of the axiom of extensionality... Right. All of the anti-foundation axioms Aczel discusses can be so considered, and are needed (as you no doubt have surmised) because set identity isn't settled in general by extensionality in the non-well-founded realm. === Subject: Re: The modern mathematical concept of infinity is indefensible > said: I am even rustier! Where can one learn more about Scott's > anti-foundation axiom? Eh? Something about logic that you haven't already mastered? :-) As strange as it sounds there are such things. There are, so it seems, also many a thing I've managed to entirely forget about ever reading, some in books I've spent considerably time with, much to my shame no doubt. I have a vivid mental image of the notebooks MoeBlee keeps, in which he records, in tedious formal detail, everything he's learnt, every argument he's worked out in detail. Often I wish I had the patience to keep my notes in such meticulous order, but alas, I find I'm a very unsystematic reader, often having to revisit sources the identity of which I can't quite recall, going by nothing but the vague recollection that what I need was contained in a thick volume with yellow covers and a dent in the back. > Herb has given a good reference -- from a quick look, Smith's paper > is basically a distillation of the first four chapters or so of > Aczel's book. At one time the latter was out of print and available > for free as a huge scanned PDF from CSLI, but it has been reprinted > and appears once again to be available in hard copy only. It is, as you have now learned from a later post, still available. I did actually read it with some attention some time in the past, but apparently it has left nothing but vague traces clouded in time in my often not very relieable memory... > Right. All of the anti-foundation axioms Aczel discusses can be so > considered, and are needed (as you no doubt have surmised) because > set identity isn't settled in general by extensionality in the > non-well-founded realm. Indeed. Without some principled account of ill-founded sets it is simply not obvious whether e.g. there is a unique set x such that x = {x}, and so on, and how and on what grounds we are to approach such questions. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <87eiz77m9i.fsf@alatheia.dsl.inet.fi > I am even rustier! Where can one learn more about Scott's > anti-foundation axiom? > You will find something about it in http://bsmith7.asp.radford.edu/Hypersets.pdf Herb === Subject: Re: The modern mathematical concept of infinity is indefensible > > I am even rustier! Where can one learn more about Scott's > anti-foundation axiom? You will find something about it in > http://bsmith7.asp.radford.edu/Hypersets.pdf Much obliged. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible > But this means that sets can be, and often are, members of themselves, > which opens a different can of worms. [Virgil] In Aczel's set theory this is NOT a can of worms, but one of the distinct > features of that system. If Aczel's theory deals with it successfully, it only means that he has > a lid on that particular can of worms, not that the can does not exist. Huh? You are beginning to sound like a real crank. A: ...this means that sets can be infinite, which opens a can of worms. B: In ZFC this is NOT a can of worms, but one of the distinct features of > that system. A: If ZFC deals with it successfully, it only means that it has a lid on > that particular can of worms, not that the can does not exist. B: Well... (*argl*) P.S. Note that Aczel showed that his theory is consistent if ZFC is (iirc). > Actually his theory shows that allowing for unfounded sets does in no way > mean opening a can of worms (if dealt with in an appropriate framework, > of course). Hint: Aczel's theory is virtually identical with ZFC, just the axiom of > foundation is replaced with AFA (anti-foundation axiom). As far as I can tell (not far) axiom of foundation is more of an aesthetic judgement than a necessary foundational support for mathematics. What has the axiom of foundation done for me lately? According to wikipedia, Kuratowski ordered pairs are handier to work with: (a,b) = {a, {a,b}} rather than (a,b) = {{a}, {a,b}}. -- Michael Press === Subject: Re: The modern mathematical concept of infinity is indefensible It occurred to me a while ago that I've never seen a crank object to > Euclid's infinitude-of-primes argument. Oh, you don't know WM (M.9fckenheim) right? translation): This proof ignores the fact that for a set of 10^100 prime numbers no more > [prime] can be found, since there are no means available to store more than > 10^100 prime numbers. How to gauge the profound depth of such an insight? Chicken farmer: I have so many chickens, I don't know > how many there are, and the bookkeeping > seems hopeless. Have you got any advice > as a mathematician? > Matheologer: Try to use arithmetic. You apply numbers > to count the chicken, to decide about > each one's age etc. There are plenty of > numbers available to do all that stuff. > Chicken farmer: I don't know. I have darn'd many chickens, > and I doubt that arithmetic has enough > numbers to count all of them. > Matheologer: That is not a worry. No matter how many > chickens you have, the numbers will suffice. > The set of natural numbers is actually > infinite. This is guaranteed by Set Theory. > There is no bound to the amount of chickens > that it can deal with. > WM: Ha! I caught you now: Set Theory allows for > infinitely many chickens. But an argument > involving the age of the Universe and the > procreation method of chickens and some > arithmetic proves that there can only be a > finite number of chickens present at any time! > This contradiction proves that Set Theory is > dead wrong in everything, just as I thought! Stupidity is infinite. QED -- Michael Press === Subject: Re: The modern mathematical concept of infinity is indefensible You usually get an answer like start with 0 and add 1, and > just keep adding 1, until you do this an infinite number of > times, at which point you have reached an infinite natural. In case of an answer like that the obvious thing to do is to ask what > is meant by adding one an infinite number of times. at which point > as witnessed often on these groups > the immediate followup is to ask back > what does it mean to collect things an infinite number of times? In case of a follow-up like that the obvious thing is to suggest that > we set aside any worries or claims about infinite sets or set theory > in general and concentrate on the matter at hand. The point to these > tedious rejoinders is to try and make the confused student -- there is > no point in asking the crank anything -- explain what he means by an > infinite natural, by way of a clear definition in simple terms not > involving finitude or infinitude, the way we may explain that that > there are infinitely many naturals means just that for any natural n > there is a natural m > n, hopefully leading them to recognise their > confusion, that they do not in fact mean anything definite by the > locution. Sometimes this works and sometimes it doesn't. but often people coming in asking questions > if that is what they are doing > only have very vague notions they may not understand that the map from naturals > to certain collections > provides an answer to some of their intuitive notions they may not yet understand that there is an infinite ordinal > omega > which behaves in many the ways are expecting and they may not understand how it behaves differently by engaging in a discussion about infinite process What do you mean by infinite process? Seems that embarking on such a discussion with a student unpracticed enough to entertain the notion of an infinite natural number will only confuse the student. Better to stay with finite processes. Each natural is reached in a finite number of steps. That is a formulation for natural numbers. When that is cleared up we can address the question How many natural numbers are there? With the definition of natural number nailed down, it looks less feasible to speak about counting up the natural numbers one-by-one. And _now_ the only sane course is to consider the axiom EI(0 in I / Ax in I((x / {x}) in I)) in one way or another intelligible to the student. [...] -- Michael Press === Subject: Re: The modern mathematical concept of infinity is indefensible > What do you mean by infinite process? Seems that embarking on > such a discussion with a student unpracticed enough to entertain the > notion of an infinite natural number will only confuse the student. > Better to stay with finite processes. Each natural is reached in a > finite number of steps. That is a formulation for natural numbers. That may well be, but if you actually try that approach with a genuinely confused student of mathematics you'll find it is not particularly helpful. It is best to avoid all mention of finite processes, set theory, what not. We can formulate the claim that there are infinitely many naturals simply as for any given natural n, there is a natural m > n. The question now is: what is the analogous reformulation for the claim that there is an infinite natural, and why should that reformulation follow from the infinity of naturals, understood in the way explained above? For certainly the reformulation can't be there exists a natural n such that for all naturals m, n > m, which even the most confused student will usually grant is obviously false. Focusing on this question we needn't concern ourselves with any conundrums about sets, or even whether there /are/ infinitely many naturals. Dealing with one confusion at a time is, I believe, the pedagogically preferable method. > When that is cleared up we can address the question How many > natural numbers are there? With the definition of natural number > nailed down, it looks less feasible to speak about counting up the > natural numbers one-by-one. And _now_ the only sane course is to > consider the axiom EI(0 in I / Ax in I((x / {x}) in I)) in one way or another intelligible to the student. Well, it is unlikely that a student confused about the infinity of naturals is much helped by abstract set theoretic reductions. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <873afofw9v.fsf@phiwumbda.org> <87tz84e8s8.fsf@phiwumbda.org> <1uwp3cfq5z0ox.1rebiiv9nmql7.dlg@40tude.net> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Let Omega be a set such that Omega = {Omega} (since the first of the > two above equations by itself is also a flat system of equations it > follows that such a set exists according to AFA). Then x = y = Omega > satisfies the above system, and since AFA says that the solution is > unique it follows that in Aczel's theory the y you define above is > indeed equal to x. > Thanx, that comment is rather helpful. Hence if you and Jesse F. Hughes are > right (which seems to be the case), the simple approach > 1 = {1} > 2 = {1, 2} > : > will not work in Aczel's theory. I've enjoyed reading this discussion about the theory Z+AFA, Aczel's Antifoundation Axiom. About a year ago, zuhair also wanted to come up with a theory in which one can have sets like 1 = {1}, and 2 = {1,2}, and so on. And he, just like Hughes and Rotwang, arrived at the exact same problem -- there's no way to prevent 1 = 2. We all agree that 1 = {1} is possible -- such a set is called a Quine atom. Now zuhair's axioms were: 1. Extensionality 2. Irregularity: Ax (xex) (so that not only the naturals, but every set would contain itself as an element) 3. Quine atom: Ex (Ay (yex <-> y=x)) So by Axiom 3, x = {x} exists (and we might as well let 1 be an instantiation of the axiom). Now zuhair wanted to form 2 = {1,2}, and guarantee that 2 is not 1. So he tried something like: 4. Successor: Az (Ex (~x=z & Ay (yex <-> (yez or y=x)))) Then applying the axiom to z = 1 gives: Ex (~x=1 & Ay (yex <-> ye1 or y=x)) so that x is a set distinct from 1 , whose elements are all of 1's elements along with x itself. So we can call this set 2, and have 2 = {1,2} distinct from 1. What doomed zuhair's theory is that he wanted to have a Comprehension Schema that he hoped would avoid Russell's Paradox, but someone showed him that the paradox would still occur anyway in a disguised form. > What a pity. I agree. I can see why Aczel would demand uniqueness in the solution to the system of equations: 1 = {1} 2 = {1,2} with 1 = 2 = Quine as the unique solution. Otherwise, as Rotwang points out, one couldn't guarantee that the Quine atom itself is unique. I've seen Aczel's axioms mentioned several times before, and I wouldn't mind seeing the axioms for myself. A web search for Z+AFA doesn't reveal any list of axioms, meaning that I'd have to buy a book to see the axioms. The book countdown is currently at 33 days, but this countdown is for a book about ZFC, not Z+AFA or an ill-founded or otherwise nonstandard theory. Books about nonstandard theories are even more elusive and more expensive than books about ZFC. === Subject: Re: The modern mathematical concept of infinity is indefensible sha1:670WY30I7foWn3uMK9m3igscfls= > About a year ago, zuhair also wanted to come up with > a theory in which one can have sets like 1 = {1}, and > 2 = {1,2}, and so on. And he, just like Hughes and > Rotwang, arrived at the exact same problem -- there's > no way to prevent 1 = 2. Sure there is! Just alter Aczel's axiom. His axiom says something like: every set of flat equations has a unique solution (pardon me if I don't state it quite right). Take away uniqueness, and you could well have distinct 1 and 2 defined as above. But the presumption that such solutions are unique make a much more interesting and useful theory. Without it, we have too few provable equalities. -- Many argue that its programmers have turned out shoddy programs, but [their] objective is to make profit, not superlative programs per se. By the profit criterion, Microsoft has been one of the greatest companies in the history of this country. -- ADTI defends Microsoft === Subject: Re: The modern mathematical concept of infinity is indefensible <873afofw9v.fsf@phiwumbda.org> <87tz84e8s8.fsf@phiwumbda.org> <1uwp3cfq5z0ox.1rebiiv9nmql7.dlg@40tude.net> posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) [...] I've seen Aczel's axioms mentioned several times before, > and I wouldn't mind seeing the axioms for myself. A web > search for Z+AFA doesn't reveal any list of axioms, > meaning that I'd have to buy a book to see the axioms. A Google search for anti foundation axiom led, via Wikipedia, to this: http://standish.stanford.edu/pdf/00000056.pdf I've just had a quick look and it appears that the version of the AFA given there is in terms of graphs, rather than flat systems of equations (the former being Aczel's original formulation). But it's the graph version of AFA applies: consider the graph G whose nodes are x, y and whose edge relation is a -> b iff in {,,} i.e. the graph which looks like this: _ _ / / | x <- y | | ^ ^ | -/ -/ (apologies for my piss-poor attempt at ascii art). Then a decoration of this graph is precisely a pair of sets x, y which satisfy the system of equations I gave earlier: x = {x} y = {x,y} so the existence and uniqueness of a decoration of G is equivalent to the existence and uniqueness of a solution to the flat system of equations. === Subject: Re: The modern mathematical concept of infinity is indefensible A Google search for anti foundation axiom led, via Wikipedia, to > this: http://standish.stanford.edu/pdf/00000056.pdf I've just had a quick look and it appears that the version of the AFA > given there is in terms of graphs, rather than flat systems of > equations (the former being Aczel's original formulation). > Maybe helpful: http://bsmith7.asp.radford.edu/Hypersets.pdf Herb === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) Again: The simple sketch O OO OOO OOOO OOOOO ... shows: the number (cardinality) of the natural numbers can't be larger as the natural numbers are able to number. And if there would be a number which is larger than any natural number, this number would be _too large_ to depict the cardinality of the natural numbers. Therefore, and because there is no largest natural number, there is no number which depicts the cardinality of the natural numbers (or any infinite totality). Infinity is just a mode, not a quantity. Infinitum actu non datur. Albrecht S. Storz === Subject: Re: The modern mathematical concept of infinity is indefensible > Again: The simple sketch O > OO > OOO > OOOO > OOOOO > ... shows: the number (cardinality) of the natural numbers can't be larger > as the natural numbers are able to number. It shows no such thing. And if there would be a > number which is larger than any natural number, this number would be > _too large_ to depict the cardinality of the natural numbers. There is and it isn't. At least outside of Albrecht's teeny world. Therefore, and because there is no largest natural number There is no smallest unit fraction, so that by Albrecht's argument, there cannot be a zero rational number.. Infinity is just a mode, not a quantity. Define mode and quantity. According to standard definitions there are everal infinite quantities. Infinitum actu non datur. Lying in latin does not help your case. Albrecht S. Storz === Subject: Re: The modern mathematical concept of infinity is indefensible posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > I guess, one of > the problems of this crank is to accept the (logical) possibility of > infinitely many finite natural numbers. > This is a common aliment among many of the sci.math cranks. > They can't accept that there are an infinite number of finite > naturals. I agree that many so-called cranks reject that there are infinitely many naturals. > Their personal take on it varies, though, including: > a) since there are an infinite number of naturals, there > must exist infinite naturals. This does sound a lot like the Transfer Principle. In some theories with a Transfer Principle, one can prove that any infinite set must necessarily contain a nonstandard element. Thus the infinite set of naturals must contain a nonstandard natural (colloquially called an infinite natural, since one can prove it to be greater than every standard finite natural). Therefore the Transfer Principle is one way to make this common claim of so many cranks rigorous. > b) the infinite set of all naturals does not exist. > c) potentially infinite sets may exist, but actual infinite > sets do not exist. > d) infinite sets do not have cardinalities. -drt === Subject: Re: The modern mathematical concept of infinity is indefensible <9cqvt0bupgxj$.n1gcnpcbbcig.dlg@40tude.net> <87ljthbxxh.fsf@alatheia.dsl.inet.fi> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > It is trivial though tedious exercise to formulate it in the language > of set theory. This is standard stuff. > OK. I consider the issue settled. Perhaps from now on, when I need to refer to a finitist theory, I'll write it as something like ZF-Infinity+Finite. Then I'll leave it open for interpretation whether Finite means the negation of Infinity, or Friedman's Finiteness Axiom, or whatever axiom is necessary to prove that in all models of the theory, all sets are finite, so the theory really is finitist. Of course, the point the others are trying to make here is that if it took several posts for Aatu, Bader, and MoeBlee to agree on what axiom is necessary to guarantee that every set is finite, then Albrecht never knew what the axiom was, and therefore Albrecht never knew what it takes to make a rigorous finitist theory. To think that it would take this much trouble just to work with finite objects! === Subject: Re: The modern mathematical concept of infinity is indefensible > To think that it would take this much trouble just > to work with _finite_ objects! The trouble has nothing to do with working with finite objects, but with arcane technicalities of formal theories, of no conceptual or philosophical significance. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible <9cqvt0bupgxj$.n1gcnpcbbcig.dlg@40tude.net> <87ljthbxxh.fsf@alatheia.dsl.inet.fi> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) It is trivial though tedious exercise to formulate it in the language > of set theory. This is standard stuff. > OK. I consider the issue settled. Perhaps from now on, when I need to refer to a > finitist theory, I'll write it as something like > ZF-Infinity+Finite. Then I'll leave it open for > interpretation whether Finite means the negation > of Infinity, or Friedman's Finiteness Axiom, or > whatever axiom is necessary to prove that in all > models of the theory, all sets are finite, so the > theory really is finitist. So you are saying that you will purposefully make imprecise statements? -- m === Subject: Re: The modern mathematical concept of infinity is indefensible <87ljthbxxh.fsf@alatheia.dsl.inet.fi> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Perhaps from now on, when I need to refer to a > finitist theory, I'll write it as something like > ZF-Infinity+Finite. Then I'll leave it open for > interpretation whether Finite means the negation > of Infinity, or Friedman's Finiteness Axiom, or > whatever axiom is necessary to prove that in all > models of the theory, all sets are finite, so the > theory really is finitist. > So you are saying that you will purposefully > make imprecise statements? OK then, what would you (Mariano) consider to be a precise statement of a finitist theory -- one sufficiently precise that no one will question whether the theory really proves that every set is finite in the theory? In other words, how can I fill in the blank in: ZF-Infinity+ that is precise enough to guarantee that every set is finite, so that no one will claim that the theory could have a model with an infinite set. I don't want to make imprecise statements. I would very much love to write a precise theory which everyone will agree that it's finitist. But, as the discussion with Aatu, Bader, MoeBlee has shown, this is easier said than done. === Subject: Re: The modern mathematical concept of infinity is indefensible > I would very much _love_ to write a precise theory which everyone > will agree that it's finitist. But, as the discussion with Aatu, > Bader, MoeBlee has shown, this is easier said than done. The theory ZF without infinity but with the negation of infinity is not finitistically justified, and the complications we have been discussing are nothing but arcane technicalities. You'll have much more success in finding a finitistically justified theory if instead of set theory you concentrate your efforts on proof theory where you'll meet such exciting theories as primitive recursive arithmetic. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is indefensible It's only basic and elementar: natural numbers are both cardinal > numbers and ordinal numbers at the same time. There is no stage on the > ladder of natural numbers at which the cardinality exceed the > ordinality. This is entirely conformant with usual conventions. > It has nothing to do with your claim that the natural numbers number > themselves. > It also has nothing to do with any ?ladder. > The cardinality and the ordinality of 7 are both 7, but > 7= {0,1,2,3,4,5,6} , so BOTH of these DO exceed > EVERY one of the things they are numbering. Just an arbitrary thesis. With more relevance I say: 7 := 1, 2, 3, 4, 5, 6, 7 This is circular. You can't define 7 in this way. Exactly what aspect is circular here: X > XO > XOO > XOOO > XOOOO > XOOOOO > XXXXXXX > ? There is nothing defined here. There are only stated facts. That's why > there is nothing circular. The Seven is seven units long because it > is at the seventh position. You can read the definition > n := 1, 2, 3, ..., n > with the cardinal number on the left side of the := and ordinal > numbers on the right side. You are defining each natural in terms of itself, which is circular. In particular, 1 := 1 does not have a source for the 1 on the right hand side. When one starts with 0, a la von Neumann, there is no such problem, as each natural after 0 is defined solely in terms of prior naturals. === Subject: Re: Non-integrability (Riemann) of f_xy(x,y) What's my question: What warrants the conclusion by the author using as only assumption that y(x,t) is continuous, ( I included the full context of what the author derived) not only that y is twice differentiable, but that from: Del^2y/DeluDelv First that: Dely/Delv=g'(v) Then y=g(v)+h(u) Only conditions given by the author were that y(x,t) is continuous, together with (are these the conditions you wanted?. Only relevant mathematical information I thought was that y(x,t) is continuous) the conditions of the wave equation, which are: A tightly -stretched string between two fixed points 0 and L on the x-axis. The string is pulled back vertically a distance that is very small compared to the length L , and released at time t=0. Our problem is to determine the displacement y(x,t) that is x units away from the end 0, at any time t. Now, *Question* Do we just need that : Del^2y/DeluDelv be continuous (and therefore bounded in a rectangle, by compactness: what if instead of a rectangle, we have an open, simply-connected region of integration?, Would boundedness of Del^2y/DeluDelv be enough?). Also: I was hoping you would correct some of the comments I made, even with just a 'yes', or a short statement of why not: > You have to be much more careful about this > sort of thing. For example, the existence of > the iterated integral int_a^b (int_c^d g(x,y) dx) dy is _not_ the same as saying g is Riemann integrable > on the rectangle [c,d]x[a,b]; you need to decide > which notion you're interested in. > notion is there? A response here, please? And: Also, the existence of a derivative is not enough > to make the Fundamental Theorem of Calculus > work. > O.K, right, we need f' to be continuous on [a,b] too, > and then Int(a^b)(f')=f(b)-f(a). Just a comment. Yes/No, why not?. Please guive me a quick ref. of the relevant theorems, so I can re-review them. More: O.K, let me try to do that, piece-by-piece. > (For example, you assume that f_x and f_y > exist and then wonder why something about > f_xy does not follow. The answer to that > is that it doesn't even follow that f_xy > _exists_! And it would not exist because f_x may not be differentiable. Is this the crux? And a new one: If f_x (x,.) , and f_y(.,y) are continuous is that enough to guarantee a repeated appli- cation of the fundamental thm. of calc: Int(Int Del^y/DeluDelv)dudv)= Int[Int y_uvdu/delu)]delv)= Int[y_u]delv ??. Could you please tell me what conditions I need for the integrability of Int(Intf(x,y)dxdy)) other than the ones discussed above?.Does continuity and boundedness of f(x,y) guarantee the existence of the integral over, say, a simply-connected region with smooth or piecewise- smooth boundary? Do you suggest some sources? === Subject: Re: Non-integrability (Riemann) of f_xy(x,y) Hi: > Just trying to understand better why , given > f(x,y) > such that f_x, f_y exist , may not be > Riemann-Integrable , i.e., Int(Int )f_xydxdy , may > not exist . There's something missing from that sentence... I guess I would need to add that f_x is itself differentiable , or, equiv. that f itself is twice differentiable, to talk about f_xy, since if either of f_x or f_y are not continuous, then f is not differentiable, so that f_xy does not make sense. Is that it?. Yes, existence of f_x f_y alone does not guarantee differentiability of f, let alone existence of f_xy. (with f_xy = the partials of f with resp. to x and > to > y) (Over a bounded region) > > It would seem, naively, that we could apply the > fundamental thm. of calc twice, to get the > integral, > i.e: Int(Int f_xy dxdy)=Int[ Intf_x(x,.)dx]dy= (let the points a,b be the limits of integration for x ; let c,d be the limits for y) Int(f_y(a)-f_y(b))dy= f(a)-f(b)|_(y=c...d) You have to be much more careful about this > sort of thing. For example, the existence of > the iterated integral int_a^b (int_c^d g(x,y) dx) dy is _not_ the same as saying g is Riemann integrable > on the rectangle [c,d]x[a,b]; you need to decide > which notion you're interested in. > notion is there? > Also, the existence of a derivative is not enough > to make the Fundamental Theorem of Calculus > work. > O.K, right, we need f' to be continuous on [a,b] too, and then Int(a^b)(f')=f(b)-f(a). > If you state _all_ the hypotheses much more > carefully, state the non-conclusion more > carefully, then give an argument that seems > to you to show that the non-conclusion > actually follows, someone will point out > the error you're making. But in its present > form we really can't do that. > O.K, let me try to do that, piece-by-piece. > (For example, you assume that f_x and f_y > exist and then wonder why something about > f_xy does not follow. The answer to that > is that it doesn't even follow that f_xy > _exists_! Please see my second comment about the need for f_x itself to be differentiable. Is that it? Of course that doesn't answer your > real question, but it shows that the question > needs to be stated much more carefully. > I can't guess exactly what the real question > is, there are various things it could be. > If this is inspired by something you read, > transcribe what you read very carefully.) result used in the derivation of the D'Alembert solution to the wave equation in a Math Engineering book. The authors (Derrick and Grossman) just go on to conclude , after only assuming that y(x,t) (the displacement along the y-axis of a plucked string) is continuous, not only that : del^2y/(delu)(delv) (we used the C.Variables: u=x+ct and v=x-ct, c a constant ) exists, but that, given the above we can go from having del^2y/(delu)(delv)=0 To concluding that : dely/delv =g'(v), and from this, to concluding that y=g(v)+h(u) David C. Ullrich Understanding Godel isn't about following his formal > proof. > That would make a mockery of everything Godel was up > to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) === Subject: Re: Non-integrability (Riemann) of f_xy(x,y) <295pm4lkbtbrf7r4e3ngc38dt54e0sso4m@4ax.com> posting-account=fVOpuAkAAAB0gOUkQMH0DG_KdwTVgKXP Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) >Hi: > Just trying to understand better why , given > f(x,y) > such that f x, f y exist , may not be > Riemann-Integrable , i.e., Int(Int )f xydxdy , may > not exist . > There's something missing from that sentence... I wasn't talking about the math - the sentence > is simply not a sentence. The verb clause > may not be Riemann-Integrable has no > subject . I guess I would need to add that f x is itself > differentiable , or, equiv. that f itself is > twice differentiable, to talk about f xy, since > if either of f x or f y are not continuous, then > f is not differentiable, so that f xy does not > make sense. Is that it?. Yes, existence of f x > f y alone does not guarantee differentiability > of f, let alone existence of f xy. >(with f xy = the partials of f with resp. to x and > to > y) (Over a bounded region) > It would seem, naively, that we could apply the > fundamental thm. of calc twice, to get the > integral, > i.e: > Int(Int f xy dxdy)=Int[ Intf x(x,.)dx]dy= > (let the points a,b be the limits of integration > for x ; let c,d be the limits for y) > Int(f y(a)-f y(b))dy= f(a)-f(b)| (y=c...d) > You have to be much more careful about this > sort of thing. For example, the existence of > the iterated integral > int a^b (int c^d g(x,y) dx) dy > is not the same as saying g is Riemann integrable > on the rectangle [c,d]x[a,b]; you need to decide > which notion you're interested in. notion is there? > Also, the existence of a derivative is not enough > to make the Fundamental Theorem of Calculus > work. O.K, right, we need f' to be continuous on [a,b] too, > and then Int(a^b)(f')=f(b)-f(a). > If you state all the hypotheses much more > carefully, state the non-conclusion more > carefully, then give an argument that seems > to you to show that the non-conclusion > actually follows, someone will point out > the error you're making. But in its present > form we really can't do that. O.K, let me try to do that, piece-by-piece. > (For example, you assume that f x and f y > exist and then wonder why something about > f xy does not follow. The answer to that > is that it doesn't even follow that f xy > exists ! Please see my second comment about the need for > f x itself to be differentiable. Is that it? Of course that doesn't answer your > real question, but it shows that the question > needs to be stated much more carefully. > I can't guess exactly what the real question > is, there are various things it could be. > If this is inspired by something you read, > transcribe what you read very carefully.) result used in the derivation of the D'Alembert > solution to the wave equation in a Math Engineering > book. The authors (Derrick and Grossman) just go > on to conclude , after only assuming that y(x,t) > (the displacement along the y-axis of a plucked > string) is continuous, not only that : del^2y/(delu)(delv) (we used the C.Variables: u=x+ct and v=x-ct, c a constant ) exists, but that, given the above we can go from having del^2y/(delu)(delv)=0 To concluding that : dely/delv =g'(v), and from this, to concluding that y=g(v)+h(u) > Let us know if you ever decide what your question is. > David C. Ullrich > Understanding Godel isn't about following his formal > proof. > That would make a mockery of everything Godel was up > to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) David C. Ullrich Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) David, It's clear to me that f(x,y) is the entity that (according to the OP) may not be Riemann-integrable. I think that's obvious from the quote: i.e., Int(Int )f xydxdy , may not exist . Paul Epstein === Subject: Re: prime numbers formula please. <648574.1231877305848.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) miss.helpful a .8ecrit : > On 11 Jan., 13:34, Axel Vogt Not a prove ... but you may search for Jones' > formula. > Why? Is it the same formula? Jones paper does not > seem to > be openly accessible :-/ This link should help : http://www.joma.org/images/upload library/22/Ford/Jone > sSatoWadaWiens.pdf Well it's not your formula but interesting too! > Fine discovery, > Raymond I have a problem with this. It looks to me like the number must be composite or > 0, because it is of the > form (k+2) * (polynomial) with the variables ranging > over non-negative > integers, so the first term must be >= 2 What am I missing? well (k+2) * polynomial with polynomial of the form 1 - (..)^2 - (..)^2 - ... thus an upperbound of polynomial = 1. then k + 2 * poly = k + 2 = prime. > tommy1729 ps : note (..)^2 + .. are equivalent to simul diophantine equations. i suppose simul diophantines are always at the basis of prime representing polynomials ? disproof anyone ? disproof of *what*? -- m === Subject: Re: prime numbers formula please. > ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. > This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. > I hope you people could help me out - > I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. > Hope I came to right address, but if I didn't I'd appreciate > your help. > thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) Hi Ofer, thank you for bringing this interesting question to my >attention. I worked for the last 19 years on your problem, >now, finally, I found the solution. It is simple and uses >only the power function n^m and the factorial function n!. W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; Now take the numerator(1/W(n)). I have written it as 'Maple' code, but I think it is >easy to rewrite it with mathematical symbols. A first test showed this: seq(numer(1/W(i)),i=0..49); 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, >1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, >1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, Hope it helps! P.S. Perhaps some member of sci.math can supply a short proof? >Very interesting. >One minor suggestion ... >In your formula, replace every occurrence of n by (n-1). That way, the >indices won't be off by 1 when you produce the list. In other words, >define W by >W := proc(n) local v,k; >add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) >*(v+2)^(n-1),v=0..k),k=0..n-1) end; >Then your conjecture appears to be as follows ... >Let w(n) be the numerator of the fraction 1/W(n), when reduced to >lowest terms. Also, why the flip? Instead, perhaps say it this way: Let w(n) be the denominator of W(n), when reduced to lowest terms. >Conjecture (miss.helpful's Primality Test): >If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. I can prove half of the result, namely: If n is prime, then w(n)=n. It's cute. My proof uses both Wilson's theorem and Fermat's little theorem. quasi === Subject: Re: prime numbers formula please. > ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. I hope you people could help me out - I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. Hope I came to right address, but if I didn't I'd appreciate > your help. thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) >Hi Ofer, >thank you for bringing this interesting question to my >attention. I worked for the last 19 years on your problem, >now, finally, I found the solution. It is simple and uses >only the power function n^m and the factorial function n!. > W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; > Now take the numerator(1/W(n)). >I have written it as 'Maple' code, but I think it is >easy to rewrite it with mathematical symbols. >A first test showed this: > seq(numer(1/W(i)),i=0..49); >1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, >1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, >1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, >Hope it helps! >P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the >indices won't be off by 1 when you produce the list. In other words, >define W by W := proc(n) local v,k; >add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) >*(v+2)^(n-1),v=0..k),k=0..n-1) end; Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to >lowest terms. >Also, why the flip? Instead, perhaps say it this way: >Let w(n) be the denominator of W(n), when reduced to lowest terms. >Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. I can prove half of the result, namely: If n is prime, then w(n)=n. It's cute. My proof uses both Wilson's theorem and Fermat's little theorem. I have both halves now -- the other half was actually the easy one (and I already had it, but didn't realize it). quasi === Subject: Re: prime numbers formula please. > ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. > This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. > I hope you people could help me out - > I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. > Hope I came to right address, but if I didn't I'd appreciate > your help. > thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) Hi Ofer, thank you for bringing this interesting question to my >attention. I worked for the last 19 years on your problem, >now, finally, I found the solution. It is simple and uses >only the power function n^m and the factorial function n!. W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; Now take the numerator(1/W(n)). I have written it as 'Maple' code, but I think it is >easy to rewrite it with mathematical symbols. A first test showed this: seq(numer(1/W(i)),i=0..49); 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, >1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, >1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, Hope it helps! P.S. Perhaps some member of sci.math can supply a short proof? >Very interesting. >One minor suggestion ... >In your formula, replace every occurrence of n by (n-1). That way, the >indices won't be off by 1 when you produce the list. In other words, >define W by >W := proc(n) local v,k; >add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) >*(v+2)^(n-1),v=0..k),k=0..n-1) end; >Then your conjecture appears to be as follows ... >Let w(n) be the numerator of the fraction 1/W(n), when reduced to >lowest terms. Also, why the flip? Instead, perhaps say it this way: Let w(n) be the denominator of W(n), when reduced to lowest terms. >Conjecture (miss.helpful's Primality Test): >If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. >I can prove half of the result, namely: >If n is prime, then w(n)=n. >It's cute. >My proof uses both Wilson's theorem and Fermat's little theorem. I have both halves now -- the other half was actually the easy one >(and I already had it, but didn't realize it). Oops -- the other half doesn't yield so easily, it was an illusion. But I do have a proof of the first half (the case where n is prime). quasi === Subject: Re: prime numbers formula please. <6svcn6F8f8g9U1@mid.dfncis.de> posting-account=NbUMfwoAAADsMBfKflSkbPXwYcSgauQP Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Am 11.01.2009 18:09 schrieb quasi: > ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. I hope you people could help me out - I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. Hope I came to right address, but if I didn't I'd appreciate > your help. thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) > Hi Ofer, > thank you for bringing this interesting question to my > attention. I worked for the last 19 years on your problem, > now, finally, I found the solution. It is simple and uses > only the power function n^m and the factorial function n!. > W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; > Now take the numerator(1/W(n)). > I have written it as 'Maple' code, but I think it is > easy to rewrite it with mathematical symbols. > A first test showed this: > seq(numer(1/W(i)),i=0..49); > 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, > 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, > 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, > Hope it helps! > P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the > indices won't be off by 1 when you produce the list. In other words, > define W by W := proc(n) local v,k; > add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^(n-1),v=0..k),k=0..n-1) end; Another formulation: (Pari/GP) { W(n)=local(m=n-1,result); > result= m! * sum(k=0,m, 1/(k+1) * sum(v=0,k,binomial(k,v)*(v+2)^m) ); > return(result); } MH Primetest(n) = numerator(1/W(n)) if n is prime -> n > if n is composite -> 1 This way W(n) looks a bit more familiar.. Gottfried Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to > lowest terms. Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. quasi There are a number of useful binomial coefficient identities you can apply to this, although I haven't had enough time to play with it long enough to see if any yield an expression simpler than the original. But you can start by expanding (v+2)^m as a sum and then changing order of summations, and there's a number of directions to proceed from there. I suspect it can't be simplified too much, but still might be worth checking out. === Subject: Re: prime numbers formula please. Peter Webb a .8ecrit : This link should help : > http://www.joma.org/images/upload_library/22/Ford/JonesSatoWadaWiens.pdf > Well it's not your formula but interesting too! > Fine discovery, > Raymond I have a problem with this. It looks to me like the number must be composite or 0, because it is of > the form (k+2) * (polynomial) with the variables ranging over > non-negative integers, so the first term must be >= 2 What am I missing? If one or more of the squares is not zero the polynomial (and the result) will be 0 or negative so that all the squares must be 0. See here for a earlier discussion : I think this is the Wilson theorem in disguise! Hoping it helped, Raymond === Subject: Re: prime numbers formula please. <496a38b3$0$18380$ba4acef3@news.orange.fr> <496b0b39$0$3253$afc38c87@news.optusnet.com.au> <496bdd51$0$18371$ba4acef3@news.orange.fr> posting-account=JYIhQwoAAADfOqRKeWW3k_OkLnLoycin Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) Raymond Manzoni a .8ecrit : > I think this is the Wilson theorem in disguise! I doubt this. Why? No one came in here and said You are not Santa Claus, you are Papa! And this, I think, would have happened on sci.math if it is a simple disguise. On more formal ground: Both propositions are equivalent. So it is only a matter of effort and ten pages of paper to derive one formulation from the other. However, this does not qualify to name it a disguise. My definition of a disguise would be: One can give a three line proof deriving one proposition from the other. Till now no one had given one. === Subject: Re: prime numbers formula please. miss.helpful a .8ecrit : > Raymond Manzoni a .8ecrit : I think this is the Wilson theorem in disguise! I doubt this. Why? No one came in here and said > You are not Santa Claus, you are Papa! And this, I think, would have happened > on sci.math if it is a simple disguise. On more formal ground: Both propositions are > equivalent. So it is only a matter of effort > and ten pages of paper to derive one formulation > from the other. However, this does not qualify > to name it a disguise. My definition of a disguise would be: > One can give a three line proof deriving one > proposition from the other. Till now no one had given one. Except that... Peter Webb and myself are talking about Jones&all formula... http://www.joma.org/images/upload_library/22/Ford/JonesSatoWadaWiens.pdf === Subject: Re: prime numbers formula please. <496a38b3$0$18380$ba4acef3@news.orange.fr> <496b0b39$0$3253$afc38c87@news.optusnet.com.au> posting-account=JYIhQwoAAADfOqRKeWW3k_OkLnLoycin Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Raymond Manzoni a .8ecrit : > I think this is the Wilson theorem in disguise! > Except that... Peter Webb and myself are talking about Jones&all > formula...http://www.joma.org/images/upload library/22/Ford/JonesSatoWadaWiens.pdf Hi Raymond, you are disappointing me ;-) This old and complicated formula no one uses ;-)) But even more: you can stop to conjecture about the origin of the Jones et alia formula. They explicitly say on page 452: The fundamental tool in both constructions is Wilsons's theorem which characterizes the primes in terms of the factorial function. Voil.88! === Subject: Re: prime numbers formula please. ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. I hope you people could help me out - I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. Hope I came to right address, but if I didn't I'd appreciate > your help. thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) > Hi Ofer, > thank you for bringing this interesting question to my > attention. I worked for the last 19 years on your problem, > now, finally, I found the solution. It is simple and uses > only the power function n^m and the factorial function n!. > W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; > Now take the numerator(1/W(n)). > I have written it as 'Maple' code, but I think it is > easy to rewrite it with mathematical symbols. > A first test showed this: > seq(numer(1/W(i)),i=0..49); > 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, > 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, > 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, > Hope it helps! > P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the > indices won't be off by 1 when you produce the list. In other words, > define W by W := proc(n) local v,k; > add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^(n-1),v=0..k),k=0..n-1) end; Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to > lowest terms. Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. quasi Ok, but how do you reduce 'to lowest terms' (which is, what Maple probably does internally) without having the primes? Those numbers are quite large ... === Subject: Re: prime numbers formula please. <6t1qijF8pjhpU1@mid.individual.net> posting-account=JYIhQwoAAADfOqRKeWW3k_OkLnLoycin Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Ok, but how do you reduce 'to lowest terms' (which is, what Maple > probably does internally) without having the primes? > Those numbers are quite large ... First let me say that I use from now onward the enumeration proposed by quasi and helms, which is clearly the more natural one. Now back to your question. Axel, I give you a hint: Multiply the sum by 'n' (the input). What happens then? === Subject: Re: prime numbers formula please. ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. > This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. > I hope you people could help me out - > I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. > Hope I came to right address, but if I didn't I'd appreciate > your help. > thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) > Hi Ofer, thank you for bringing this interesting question to my > attention. I worked for the last 19 years on your problem, > now, finally, I found the solution. It is simple and uses > only the power function n^m and the factorial function n!. W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; Now take the numerator(1/W(n)). I have written it as 'Maple' code, but I think it is > easy to rewrite it with mathematical symbols. A first test showed this: seq(numer(1/W(i)),i=0..49); 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, > 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, > 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, Hope it helps! P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the > indices won't be off by 1 when you produce the list. In other words, > define W by W := proc(n) local v,k; > add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^(n-1),v=0..k),k=0..n-1) end; Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to > lowest terms. Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. quasi Ok, but how do you reduce 'to lowest terms' (which is, what Maple >probably does internally) without having the primes? Those numbers are quite large ... The Euclidean Algorithm makes finding the GCD very easy, and thereby, reducing to lowest terms very easy. Rob Johnson take out the trash before replying === Subject: Re: prime numbers formula please. ofe...@cs.openu.ac.il (Ofer Luft) > Hello folks. > This is my first time here in the math news groups > and probably the last. Don't even know if it's the right group. > I hope you people could help me out - > I am looking for a certain formula that produces prime > numbers. I only know something like that exists but any sort > of formula will do. > Hope I came to right address, but if I didn't I'd appreciate > your help. > thanx. (This is also my first posting ever, so hope my sig. > turns out ok.) > Hi Ofer, thank you for bringing this interesting question to my > attention. I worked for the last 19 years on your problem, > now, finally, I found the solution. It is simple and uses > only the power function n^m and the factorial function n!. W := proc(n) local v,k; > add(add(n!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^n,v=0..k),k=0..n) end; Now take the numerator(1/W(n)). I have written it as 'Maple' code, but I think it is > easy to rewrite it with mathematical symbols. A first test showed this: seq(numer(1/W(i)),i=0..49); 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, > 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, > 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, Hope it helps! P.S. Perhaps some member of sci.math can supply a short proof? Very interesting. One minor suggestion ... In your formula, replace every occurrence of n by (n-1). That way, the > indices won't be off by 1 when you produce the list. In other words, > define W by W := proc(n) local v,k; > add(add((n-1)!*k!*((k+1)*v!*(k-v)!)^(-1) > *(v+2)^(n-1),v=0..k),k=0..n-1) end; Then your conjecture appears to be as follows ... Let w(n) be the numerator of the fraction 1/W(n), when reduced to > lowest terms. Conjecture (miss.helpful's Primality Test): If n is in N, then w(n)=n if n is prime, otherwise w(n)=1. quasi Ok, but how do you reduce 'to lowest terms' (which is, what Maple >probably does internally) without having the primes? Using GCDs (i.e. the Euclidean Algorithm). >Those numbers are quite large ... In any case, to say w(n)=1 is equivalent to saying that W(n) is an integer, which can be checked with a single division. But note: miss.helpful is not claiming that the test is algorithmically efficient, but rather, that it's a simple algebraic formula which, for some mysterious reason (which she will reveal soon), distinguishes primes from composites. numerical evidence, the formula seems valid, but if it is valid, why so? quasi === Subject: Re: prime numbers formula please. Am 12.01.2009 22:48 schrieb quasi: > numerical evidence, the formula seems valid, but if it is valid, why > so? quasi Something in the formula reminds me of the bernoulli/Clausen-von Staudt- theorem and also the wilson-prime-test should be examined, because we have (n-1)! in the formula for a test of isprime(n). Just ideas... Gottfried === Subject: Re: JSH: Research speaks for itself posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) > Through the years I've been certain that useful research will be > used. Seems like it makes sense but hey, it often seems like a crazy > world, and I have experience with failure with my research. For years > I actually *was* a math crackpot, Indeed you were. But that's all past. > with numerous failed attempts at > proving Fermat's Last Theorem. Lots of failure. Years of it. But I > believed that research speaks for itself so that if I were right, it > wouldn't be about me and failure, it would only be about the success. > It would be. If. > If I could ever succeed. > Yes. If. > The more astute of you may have noticed a sharp dichotomy in recent > exchanges on the newsgroups: ----> I emphasize web searches primarily on research results, citing > information as key. ----> Argumentative posters emphasized searches on me, and disdained > search results as important tools. I could keep going after failure because I found I just liked fiddling > around with equations, as it occupied my time, and in accepting that a > good reason for what I was doing was enjoyment and not the end itself, > as there is no end, I started getting real results, and found this > massive problem in abstract number theory which has tested my belief > that research speaks for itself. > Now you are doing what people love the most. You are bragging and preaching. People just love to listen to braggarts and windbags. Please continue to do more bragging. Not only is is pleasant to read, but also it is so informative. People here are not interested in mathematics. What they want most are the blitherings of an immodest, self-important and delusional windbag. We love it. > In the past I'd attack my own credibility considering it useless in > considering mathematical proofs. > Oh yes. The days when you tried to make yourself a Discredited Source. What a brilliant strategy. Or the time that you were going to attack your own work. Great, and it was so convincing. You had me going. You actually convinced me that you didn't know your rectum from a hole in the ground. > Now I have to convince that experts in the field of mathematics are > for some reason or other ignoring quite a few major discoveries > including a find of a massive error in some abstract number theory and > the credibility issue has taken a new twist. > That is what you have to do all right. The best way to do that is to brag and pontificate and whine. You are doing exactly what you should do. Your adoring public loves you for it. More bragging, more patting yourself on the back, more whining. Everyone loves it. Bring it on. It is surprising that Gauss didn't try the same approach. No wonder he is so underrated. > What is more credible? People or their results? > Must be a trick question. Posts like this one, with no math or results whatsoever, just a lot of hot air and bragging, are what people most want. That is what all the great mathematicians of the past have done. Give us more. > I and my opponents have taken two different paths in answering those > questions. > Not that different. You brag, they adulate. What else could they do? Hold their noses when you walk by? > Our two paths are revealed through Google searches: 1. I've emphasized research. Explaining, explaining, explaining. > I've worked at simplifying my prior results and worked at making more, > and even focused on their usefulness. My focus has been on research. > Much too modest. What you need to say here is, you have expressed yourself with complete mathematical rigor and maximal elegance. Your exposition is so simple that it can be understood by the brain of a mosquito. No one has laid a glove on your arguments. Then you top it all off with truly superlative bragging. That is what wows your huge loyal fan club. Give us more gushing, self-congratulatory bragging. Lay it on thick. We never get enough. Everybody loves you for it too. All the world loves a braying braggart. A lot of people actually sneer at Riemann because he was so modest. Dumb , never tried to tell people he was the King. At least you have avoided that mistake! > Results: Google search results linking to my research in key areas. > e.g. Google: define mathematical proof > That is proof, absolute proof positive. High ranking in Google searches is an infallible way to tell whether or not you are a crank. Point this out over and over again. Your disciples and worshippers just eat it up. They cannot get enough of your incessant bragging. It is much the same as the phenomenon of people smelling farts in the elevator. How sweet they are and how sweet are your ravings. > Or, Google: solving binary quadratic Diophantine > Better yet. The testimonials to your method just pop up everywhere. What is amazing is how many of those posts cited by Google say your method is the absolute best. Tippy top best, they all say. You should make sure to tell everyone that you are a genius and that they are just warts on a toad. They want that and need it. Please resume bragging to remind everyone, right now. > sanity to repeating over and over again that I'm wrong, or that even > if I'm right nothing I have is important, it has been a non-stop > attack on my worth. Their focus: personality. > What demeaning skullduggery. Clearly this is not about you. Never has been. When you brag you have done something, people are so appreciative to be informed, to learn some new math or physics. They love posts like this one because they are so dense with mathematics and new ideas, but most of all they just love to hear your brag. Who can blame them? > Results: Google searches on my name bring up a crank.net webpage > against me in the top 20 on major search engines, including Google and > Yahoo! which I just checked. On Google the flame page came up #20, on > Yahoo! it came up #18. > Shocking. Well, you should make the most of it. Let everyone know that you are a top-ranking crank. Not everyone can be, you know. Keep trying! Of course there are jealous people out there who want to out-crank you. Keep bragging and yelling about how good you are, and be assured you will move up in the crank ratings. And remember: bragging about being one of the greatest cranks of all time is what people most want to hear. I just cannot say it enough: People love bragging. And more is better. But you know that. > So the search engine results reveal two competing strategies. > Fascinating. In fact I am surprised that you didn't use the word 'fascinating' more often in this post, because it really is fascinating how these Google searches prove your worth. As your fellow Superior Being, I think I am entitled to say how fascinating you are, and further how fascinating it is that I am so very good at discerning how fascinating you are. No one is better at it than I. Not to brag, but Google searches prove that I am your top-ranked fan. When it comes to being fascinated by your magnificent accomplishments I am the best. > I have yet to find searches on my name bringing up ANY of my > research. Not any. Incredible. Just incredible. Not any, you say? I am floored. Of course the same is true with Gauss. Just do a search on his name. Nothing about what he actually did ever comes up. Never a mention of his work in number theory, analysis, algebra, geometry, physics, astronomy, etc.. Just nothing, just like you. The problem with Gauss - and with you, to be frank - is, you have not sufficiently tooted your own horn. Too modest. So was Gauss. Of course he didn't actually have that much to show for all his efforts. Pretty good at arithmetic apparently. But you, my man, you take the cake. You really MUST brag more to make that clear. Did I mention that people love it? Remember the old Doris Day song ... Every-body Loves a braggart I'm a braggart Everybody loves me, Yes they do! And I love every braggart Since started bragging to you! > So no, I can't just search on my name and come up > with my research on solving binary quadratic Diophantine equations, or > defining mathematical proof. > Told ya. You gotta brag more. Remember George Bush, and that big Mission Accomplished sign on the aircraft carrier? Now that is the kind of example you should follow! Get out that old U.S. Army uniform and get a big red flag that says, I am the World's Biggest Crank! Mission Accomplished! People need to know. They will fall down at your feet and kiss your bunion. > Their strategy of personal attacks worked in that sense. Their > continued use of that strategy can be seen in reply after reply after > reply on these newsgroups. > It's despicable. You brag, they try to tear you down. Remember, it's just a few rotten apples in the bottom of the barrel. Most sci.math readers really love you. Many of them would like to cuddle up with you. You can tell that there is a huge silent majority out there that loves you, because (1) they are VERY silent - it would take millions and millions of fans out there to build up such an enormous amount of silence, and (2) look at how high you rank in those Google searches. Amazing. (3) not one of them - NOT ONE - says you are right. That massive silence just proves how many people actually think you are right but they don't say anything, because they are silent! You should brag about this! > And my disdain for that strategy can be seen in my responses: I > believed and believe that research speaks for itself. > Make no mistake, it does. It really does. But it is so far above the head of the average Joe that it might as well be just a disgusting, smelly, putrid, fetid, stinking, load of . The only thing you can do to counter this reaction is to brag. Or, if you prefer, boast. That way people will want to suck up to you and they will understand how extremely good you really are. Just TELL them. Brag or boast, either one. Both work like a charm. > They clearly believe in the usefulness of personal attacks. Attack > the person to attack the research. > That's what they do! All the time! And if they are so stupid as to include actual mathematical content in what they say, you give them the old heave-ho and just ignore them. The ones you need to answer here are the Uncle Al's of this world, who just treat you derisively. Imagine that, your fellow physicist, and he ///seems/// to hold you in contempt. Probably just kidding of course. Make sure Uncle Al knows you are the best. Drop him a personal e-mail that includes some of your very best bragging. He will thank you for it. Please, do it right now. Uncle Al is waiting. Or better yet, find out where he lives and go visit him. Take along your extremely popular and successful best-selling book. Give him a free copy and see what he says. I bet he will just break down and cry from sheer gratitude. He puts on the gruff act here, but inside his evil shrivelled bigotted little black heart, there is another even littler heart that is even blacker and more shrivelled, and that is the true Uncle Al, who loves you. Please, go see him and give him a hug. He will thank you as only Uncle Al can thank someone. > But what kind of scientists would engage in such behavior? > Good question. Hmm. Short ones? Ones with red hair? Republicans? Scientists who watch World Wrestling Federation? Scientists with anal warts? I don't have a clue. Tell me and tell us all. But don't stint on the bragging. That is what we hunger for. > Oh, oops! My mistake. Talking about members of the mathematical > community, not scientists. > Certainly not. Witch-doctors, more like. > Behavior speaks for itself. > God knows. But it's not enough. You need to brag more so that people will (1) love you and (2) not overlook your amazing discoveries, e.g., your APPARENT proof that the algebraic integers are thoroughly buggered. > My situation is about time. I see Google search results as leading > indicators of world interest. I see the focus of search engine > results on my research and not on me as a brilliant demonstration that > it's not the person--it's their work. > Well you need not fear that there is a focus on you as brilliant. As for your work: I am mindful of the time I ate too much and drank too much and passed out for 18 hours without having a bowel movement. When I finally got to it, I produced a turd that was 2.8 feet long and thick as an ear of corn. I was so proud. And this is how you should boasting about it. That is what people love to hear you say. More! > I am very tempted to just kick back and relax, turning my problem > solving skills to more natural things, like pursuing extremely > beautiful women (now that takes a great deal of genius, a problem > worthy of my skills). > Absolutely. Women love a braggart even more than men do. Oh James - tell me again how you discovered Nonpolynomial Factorization - you are such a genius - and please, please James, tell me more about your gigantic turd! Truly, I just love to hear you brag! > However, I cannot simply walk away from the reality of some people > abusing their position to maintain their ability to teach wrong > mathematics, nor deny my responsibility to help in the end of the use > of failed ideas, so that correct ones can be used, which can further > the advancement of the entire human species. > Yes! Well put! You have a responsibility first and foremost to Yourself. You are not just the Best Mathematician of All Time. You are God. People yearn for salvation, and they can only be saved by Your suffering! > To the world---They are your children: both the ones being taught > false information now, and the ones yet to be born, who may need the > continuing pursuit of knowledge. > do. Verily you have straightened them out. Unto us a Gift is given. Ask not what your country can do for your. Good King Wenceslas. Able was I ere I saw Elba! Garbonzo, Roz? No, brag! James Harris is Lord. We three kings of Orient are. Psalm 38, James. Look it up. > Don't allow a world that wakes up in some distant future when the > reality of the math error and its importance are finally realized, > belatedly, only to find it is too far behind to catch up to the > solutions it needs for problems we cannot imagine today any more than > Galileo or Newton could imagine ours. > Yup, yup, yup. A sentence worthy of Henry James or Henry Ford. Can't decide which. Should be longer. > The fate of the world is not about personalities. Or who can be more > creative in slamming the other guy. > No indeed. > It's about the research that speaks for itself. If you listen. > (Listens intently for several hours) Hmm. Nothing coming through. You need to brag some more. > The future of our world depends on people asking questions. > It does? Marcus. > James Harris === Subject: Re: JSH: Research speaks for itself [...] > do. Verily you have straightened them out. Unto us a Gift > is given. Ask not what your country can do for your. Good > King Wenceslas. Able was I ere I saw Elba! Garbonzo, Roz? > No, brag! James Harris is Lord. We three kings of Orient > are. Psalm 38, James. Look it up. ----------------- 19:038:004 For mine iniquities are gone over mine head: as an heavy burden they are too heavy for me. 19:038:005 My wounds stink and are corrupt because of my foolishness. 19:038:006 I am troubled; I am bowed down greatly; I go mourning all the day long. 19:038:007 For my loins are filled with a loathsome disease: and there is no soundness in my flesh. 19:038:008 I am feeble and sore broken: I have roared by reason of the disquietness of my heart. ----------------- _That_ Psalm 38? -- Michael Press === Subject: Re: JSH: Research speaks for itself posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081203 Firefox/2.0.0.19,gzip(gfe),gzip(gfe) > [...] do. Verily you have straightened them out. Unto us a Gift > is given. Ask not what your country can do for your. Good > King Wenceslas. Able was I ere I saw Elba! Garbonzo, Roz? > No, brag! James Harris is Lord. We three kings of Orient > are. Psalm 38, James. Look it up. ----------------- > 19:038:004 For mine iniquities are gone over mine head: as an heavy > burden they are too heavy for me. 19:038:005 My wounds stink and are corrupt because of my foolishness. 19:038:006 I am troubled; I am bowed down greatly; I go mourning all the > day long. 19:038:007 For my loins are filled with a loathsome disease: and there is > no soundness in my flesh. 19:038:008 I am feeble and sore broken: I have roared by reason of the > disquietness of my heart. > ----------------- _That_ Psalm 38? -- > Michael Press That's the one. Not all the Psalms are soothing little poems. Marcus. === Subject: Re: JSH: Research speaks for itself posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Or, Google: solving binary quadratic Diophantine Google: verbing nouns === Subject: Get instructor's solutions manual posting-account=EmCVswoAAACJGR09me6whGKmUcssb5cc Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) Do you suffer from a tough class? Are you looking for instructor solutions manual to do your homework? Just send me email with its name and edition and I may be able to help you in low price! It is my list, however if you don't find it here don't give up because it is only a list of some. Please , DO NOT REPLY HERE , instead send me email to : cartermath (at) gmail(dot)com instructor solution manual for A First Course in Differential Equations (7th ed.) 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Beer) instructor solution manual for Vector Mechanics for Engineers Statics & Dynamics (6th Ed., Ferdinand P. Beer) instructor solution manual for Wireless Communications Principles and Practice, 2nd Ed, by Rappaport === Subject: Solution manual for Introduction to functional Analysis by Kreyzig posting-account=z0CEbwoAAAC4Yfd6z5bsJTekd_ucAbBB Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Can somebody provide me with solution manual for Introduction to functional Analysis - Kreyzig === Subject: Re: Linear Algebra question about adjacent bases > Now, by choice if j, a_j =/= 0. So this is a nontrivial linear > combination of P_1,...,P_n equal to zero, which contradicts the > assumption that P_1,...,P_n forms a basis. This contradiction arises > from the assumption that Q was linearly dependent on > P_1,...,P_{j-1},P_{j+1},...,P_n. Hence, Q is not linearly dependent on > them, so the list > {P_1,...,P_{j-1},P_{j+1},....,P_n,Q} > is linearly independent, hence a basis (by size considerations) as was > to be proven. I get it now. They wanted to prove that Q was linearly independent >with respect to P_i excepting P_j, That last sentence still doesn't quite make sense, although it's closer. >so they just found a contradiction >between the alternate situation and the definition of linear >independence. This linear algebra stuff is so simple, yet it drives >me nuts! I guess my problem was that I was conceiving of the >definition of linear dependence as P_n = a1P1 + a2P2 + . . . + a_{n-1} >P_{n-1}, instead of all the P's on the left hand side and zero on the >right hand side. Or something. I don't know why this is so hard for >me. On the one hand a lot of students find this stuff hard. On the other hand, a lot of students, including you, are simply not _precise_ enough in the things they say about this stuff. Which means they're not precise enough in the things they _think_ about it, which makes it impossible. You still haven't _correctly_ stated what it is you're trying to prove! Here's the story. Say you were learning arithmetic. You need to calculate 2 + 2. You ask someone about 2 + 3, they tell you 2 + 3 = 5, you reply the answer's supposed to be 4, eventually you say that yeah, you meant to ask about 2 + 2 instead of 2 + 3, you thought that was clear... Of course that would be ridiculous, you'd never do that. But the point is this: The Point: If you _were_ being that sloppy about addition that would make it impossible for you to learn arithemtic. Right? Now suddenly in linear algebra you have to manipulate _concepts_ instead of manipulating _numbers_. As long as you're asking how to show Q is independent instead of what you meant to ask you're being just as sloppy as that hypothetical guy who asks about 2 + 3 when the problem is 2 + 2. And the result is the same. Then when you come back and say oops, you meant to ask how to how Q is independent with respect to .... it's like the arithmetic student saying oops, he meant to ask about 2 - 2. Until the guy starts being careful enough that he asks about 2 + 2 when the problem is 2 + 2 it's going to he hopeless. Exactly what is it you're trying to prove here? David C. Ullrich Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to. (John Jones, My talk about Godel to the post-grads. in sci.logic.) === Subject: Re: Linear Algebra question about adjacent bases posting-account=qiuOxgkAAABg9uMm_CgxOdnWf40hvxg7 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) If you have a basis, P1,P2,...,Pj,...,Pm, and a vector Q that is a > linear combination of this basis, then you can replace one of those > basis vectors (call it Pj) with Q and the vectors would still be > linearly independent and therefore still be a basis. Now Pj can be > expressed in terms of this new basis, P1,P2,...,Q,...,Pm. This is not true without some restrictions of Q. > In order to be able to replace some Pj by Q it is necessary (and > sufficient) that the coefficient of Pj in the linear combination of the > basis vectors t form Q not be zero. I.e., If Q = a1 P1 + ... ak Pk + ... +am Pm then the replacement set > of vectors has the same span only if the corresponding ak is not zero. Yeah, I was thinking of this, but I didn't write it down because it > How would you prove that Q is not linearly dependent? If any ak is not zero, you can Solve Q = a1 P1 + ... ak Pk + ... +am Pm > for Pk, so that if Q were dependent only on the other basis vectors, so > is Pk, thus the original basis is not a basis. === Subject: Re: Linear Algebra question about adjacent bases If you have a basis, P1,P2,...,Pj,...,Pm, and a vector Q that is a > linear combination of this basis, then you can replace one of those > basis vectors (call it Pj) with Q and the vectors would still be > linearly independent and therefore still be a basis. Now Pj can be > expressed in terms of this new basis, P1,P2,...,Q,...,Pm. This is not true without some restrictions of Q. > In order to be able to replace some Pj by Q it is necessary (and > sufficient) that the coefficient of Pj in the linear combination of the > basis vectors t form Q not be zero. I.e., If Q = a1 P1 + ... ak Pk + ... +am Pm then the replacement set > of vectors has the same span only if the corresponding ak is not zero. Yeah, I was thinking of this, but I didn't write it down because it > How would you prove that Q is not linearly dependent? If any ak is not zero, you can Solve Q = a1 P1 + ... ak Pk + ... +am Pm > for Pk, so that if Q were dependent only on the other basis vectors, so > is Pk, thus the original basis is not a basis. > You're welcome. === Subject: Re: JSH: Leading authority check The twin paradox goes away when you take into account acceleration. The twin paradox is a special relativity problem. There is NO > acceleration in special relativity. The solution of the problem must > remain within the field of discussion. Your so-called solution is > unacceptable. Without acceleration, you can't even set up the twin paradox. > The twins have to start together, separate, and come back together. > This *cannot* be done without acceleration. Yes, it can be. -- Michael Press === Subject: Re: JSH: Leading authority check posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Minefield/3.0.5,gzip(gfe),gzip(gfe) > The twin paradox goes away when you take into account acceleration. The twin paradox is a special relativity problem. There is NO > acceleration in special relativity. The solution of the problem must > remain within the field of discussion. Your so-called solution is > unacceptable. Without acceleration, you can't even set up the twin paradox. > The twins have to start together, separate, and come back together. > This *cannot* be done without acceleration. Yes, it can be. > -- > Michael Press Not the classical twins paradox, but true enough. === Subject: Re: JSH: Leading authority check > The twin paradox goes away when you take into account acceleration. The twin paradox is a special relativity problem. There is NO > acceleration in special relativity. The solution of the problem must > remain within the field of discussion. Your so-called solution is > unacceptable. Without acceleration, you can't even set up the twin paradox. > The twins have to start together, separate, and come back together. > This *cannot* be done without acceleration. Yes, it can be. > Not the classical twins paradox, but true enough. It is mathematically equivalent to the classic scenario. This scenario can be enhanced. Let the three spaceships manufacture their clocks only when they are needed to be set in motion , record the transit times on pieces of paper, and destroy the clocks just as soon as they have measured their relevant time interval. No accelerated clocks. -- Michael Press === Subject: Re: JSH: Leading authority check The twin paradox goes away when you take into account acceleration. The twin paradox is a special relativity problem. There is NO > acceleration in special relativity. The solution of the problem must > remain within the field of discussion. Your so-called solution is > unacceptable. Without acceleration, you can't even set up the twin paradox. > The twins have to start together, separate, and come back together. > This *cannot* be done without acceleration. Yes, it can be. > Not the classical twins paradox, but true enough. It is mathematically equivalent to the classic scenario. This scenario can be enhanced. Let the three spaceships > manufacture their clocks only when they are needed to be > set in motion , record the transit times on pieces of > paper, and destroy the clocks just as soon as they have > measured their relevant time interval. No accelerated clocks. Here it is written out. -- Michael Press === Subject: Re: JSH: Leading authority check posting-account=3WPJYgoAAAA55VjhzK9i07RN8h8u8eEs Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Google search results are leading indicators of what people around the > world are doing. Really? Well what does you Google listing tell you that people are > doing James? > I mean, all those hits on your page, and yet you get no positive > responses. I know when I publish something (of course, unlike you, Why do you think that? > I've already given possible answers. But you still haven't answered this seemingly simple question on an issue in which you seem to attach such significance. Why is that? I don't get much feedback I'll admit, but I do get some. > Care to share? M === Subject: Re: JSH: Leading authority check If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine Now try Fermat crank I guess you are the world's leading FLT crank? You *guess*? JSH is the creme de la creme.[1] Footnotes: > [1] I'm probably losing my Okie cred using that cliche (not to > mention looking like a twit), so let's say instead that JSH is the > biggest dog in the yard. If you'd only used `cr.8fme de la creep' you could have retained your cred. It's a long way to Ponca City, and even further to get back. -- Michael Press === Subject: Re: JSH: Leading authority check If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: > solving quadratic residues > solving binary quadratic Diophantine > Now try Fermat crank I guess you are the world's leading FLT crank? > You *guess*? > JSH is the creme de la creme.[1] > Footnotes: > [1] I'm probably losing my Okie cred using that cliche (not to > mention looking like a twit), so let's say instead that JSH is the > biggest dog in the yard. If you'd only used `cr.8fme de la creep' > you could have retained your cred. It's a long way to Ponca City, > and even further to get back. > In OooooooooooooooooK! lahoma, I'd think the term would'a been Cheez de la Whiz Dale === Subject: Re: JSH: Leading authority check sha1:jvGFfe7gNYpZ3jTPgKU6VV3Onp4= If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine > Now try > Fermat crank > I guess you are the world's leading FLT crank? You *guess*? JSH is the creme de la creme.[1] Footnotes: > [1] I'm probably losing my Okie cred using that cliche (not to > mention looking like a twit), so let's say instead that JSH is the > biggest dog in the yard. If you'd only used `cr.8fme de la creep' > you could have retained your cred. Not if I put an accent on creme. Ain't you uppity! -- Jesse F. Hughes Part of the problem here, Peter, is that you are an idiot. -- Daryl McCullough gives a diagnosis === Subject: Re: JSH: Leading authority check > If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine > Now try > Fermat crank > I guess you are the world's leading FLT crank? You *guess*? JSH is the creme de la creme.[1] Footnotes: > [1] I'm probably losing my Okie cred using that cliche (not to > mention looking like a twit), so let's say instead that JSH is the > biggest dog in the yard. If you'd only used `cr.8fme de la creep' > you could have retained your cred. Not if I put an accent on creme. Ain't you uppity! I shop at Tarjay. -- Michael Press === Subject: Re: JSH: Leading authority check posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.43 Safari/525.19,gzip(gfe),gzip(gfe) > If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine I should come up in the top 10 for both searches, if not then consider > the statement above rebutted for your particular area, as it is > probably, at least for the moment, country specific, but that will > probably change as my mathematical research continues to rapidly take > over. How long does it take you to find those three or four magic > words to input to the Great Google? It took me about 15 minutes to come up with: Monte Carlo riemann-siegel (my magic words ...) David Bernier Meaningless jumble in comparison to: define mathematical proof === Subject: Re: JSH: Leading authority check posting-account=BVr-MgkAAABE4LRE1rHDnN9heo0IZZTk .NET CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) If you go by Google search results it appears I'm becoming the world's > leading authority in two key areas, which is a statement meant to be > challenged, so here are the two searches, web engine must be Google > and no quotes: solving quadratic residues solving binary quadratic Diophantine I should come up in the top 10 for both searches, if not then consider > the statement above rebutted for your particular area, as it is > probably, at least for the moment, country specific, but that will > probably change as my mathematical research continues to rapidly take > over. How long does it take you to find those three or four magic > words to input to the Great Google? It took me about 15 minutes to come up with: Monte Carlo riemann-siegel (my magic words ...) David Bernier Meaningless jumble in comparison to: define mathematical proof- Hide quoted text - - Show quoted text - So, what about that Solving binary quadratic Diophantine equations blog entry? I don't see any progress there. Got it programmed for test purposes yet? Here's an easy test problem: Given 2x^2 + 4x*y + y^2 - 37x + 3y - 19 = 0 Find integers x, y Answer: x = 7, y = 5 2*7^2 + 4*7*5 + 5^2 - 37*7 +3*5 - 19 = 0 98 + 140 + 25 -259 +15 -19 = 0 Enrico === Subject: Re: Leading authority check [...] > I have no doubt you will discount that fact and forget the information > again. There is no rationality left in you. Only the blindness of belief. -- Michael Press === Subject: Re: Leading authority check > The point here is non-rationality by people like you. There is no evidence that will convince people like you. I need readers to understand your behavior is about emotion, denial, > and a refusal to accept things like the mathematical proofs I do have, > which are driving the interest of the world. That is, they need to understand YOU are the cranks and crackpots. > Not me. Why? Because I need them to stop trusting people who are lying to > them. As until they do so, they'll keep paying them to teach students > false things. Those poor undergrads all over the world. As for fun searches, I recommend people try one more: intellectual foodchain. See who is on top. James Harris Yeah, but you forget I'm a degreed mathematician. It's BLANTANTLY OBVIOUS So? I have a degree in physics. Here is an undergraduate physics exercise. A rope is wrapped around a cylinder of radius r. The total angle of wrapping is a. The coefficient of friction between the rope and the surface of the cylinder is c. The tension on one end of the rope is T. The tension on the other end of the rope is T'. 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Beer) instructor solution manual for Wireless Communications Principles and Practice, 2nd Ed, by Rappaport === Subject: helllllp posting-account=BZnlawoAAACoV0jPNDt-90THeJ8F9FEu AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) 1. a recent survey was done to find out what types of financial goals recent university grads have. 100 recent uni grads were asked to think about what they hoped their financial situation would be like 10 years from now, and asked if they would own a house, a car, and/or a boat. there were only 3 who said they would have none of these things, while 4 said they would have a only a boat, 25 said only a house, and 30 said only a car. of those who felt they would own a house, 27 did not think they would have a car, while 42 didnt think they would have a boat. 45 of those surveyed said they would have a boat or a car, but not a house. how many of the 100 surveyed grads think they will have all three? 2. when peter was moving to his new house he labelled each of this boxes with kitchen, living room, office, spare room, bedroom garage. when he was done packing, he had filled and labelled 48 boxes. his friend removed all the labelled so when he moved he simply asked the moving company to place each box in any of the six locations. In how many ways can the boxes be distributed that result in no box being left in the correct room? 3. consider the permutaitons of the letters SUCCESSFUL a)how many permuations are there in total? b) how many permutations have a consonant as the first letter? 4. a particular website requires that users select a password with between 6 and 8 characters (inclusive). each character can be a lower- case letter, an upper-case letter, or one of the digits 0-9. a password must start with a letter, and must have at least 1 digit. a sample password is tIc4SS. how many passwords are possible? 5. twelve couples attend a party at which 4 identical door-prizes are to be given out to 4 different individuals. of all the ways to select 4 winers from the 24 people, how many outcomes result in no couple going home with more than one prize? === Subject: Re: Kinematics And Dynamics of Machinery third edition I'm looking for the solution manual for Kinematics And Dynamics of > Machinery third edition Charles E. Wilson & J. Peter Sadler I'm not sure why you're announcing as much in a mathematics newsgroup but good luck anyway. -- But you see, I can believe a thing without understanding it. It's all a matter of training. --Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_ === Subject: Re: Kinematics And Dynamics of Machinery third edition Hi there, === Subject: new list of solutions manual posting-account=WLvLsAoAAAAwFP1gs27vbetS0eUhA4m1 Gecko/2008120122 Firefox/3.0.5,gzip(gfe),gzip(gfe) Hello here I leave my list of solutions manual that I have ,his list is more extensive. please contact: mbfix8117@gmail.com or luisquipu@hotmail.com *Embedded Microcomputer Systems: Real Time Interfacing Valvano *Chip Design for Submicron VLSI: CMOS Layout and Simulation, 1st Edition John P. Uyemura *Digital Signal Processing - A Modern Introduction, 1st Edition Ashok Ambardar *Digital Signal Processing Using MATLAB¬, 2nd Edition Vinay K. Ingle | John G. 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Hwang Modern Power System Analysis Kothari *Embedded Systems Kamal *Circuits and Networks Sudhakar *An Introduction to Signals and Systems, 1st Edition John Alan Stuller Hazardous Waste Management, 2nd Edition Michael D LaGrega, Phillip L Buckingham, Retired Jeffrey C Evans *Power Systems Analysis and Design, 4th Edition J. Duncan Glover , Mulukutla S. Sarma , Thomas Overbye *Chemistry for Environmental Engineering and Science, 5th Edition Clair N Sawyer,Perry L. McCarty, Gene F. Parkin, *Embedded Systems: Architechture, Programming and Design D. P. Kothari,I. J. Nagrath *Advanced Digital Logic Design Using Verilog, State Machines, and Synthesis for FPGA's, 1st Edition Sunggu Lee *Engineering Mechanics: Statics-Computational Edition, 1st Edition Robert W. Soutas-Little | Daniel J. Inman | Daniel S. Balint *Design of Fluid Thermal Systems, 2nd Edition William S. Janna *Principles of Heat Transfer, 6th Edition Frank Kreith *Foundations of Materials Science and Engineering, 4th Edition William F. Smith, Javad Hashemi, *Microwave Engineering Annapurna Das Introduction to Environmental Engineering, 4th Edition Mackenzie L Davis,David A Cornwell *An Introduction to Mechanical Engineering, 2nd Edition Jonathan Wickert *Mechanics of Materials , 6th Edition James M. Gere *Engineering Mechanics: Dynamics - Computational Robert W. Soutas-Little | Daniel J. Inman | Daniel S. Balint *Engineering Mechanics: Statics-Computational , 1st Edition Robert W. Soutas-Little | Daniel J. Inman | Daniel S. Balint *Structural Analysis , 3rd Edition Aslam Kassimali *The Science and Design of Engineering Materials, 2nd Edition James P Schaffer,Ashok Saxena, Stephen D. Antolovich,Thomas H. Sanders, Steven B Warner, *Traffic and Highway Engineering, 3rd, 3rd Edition Nicholas J. Garber | Lester A. Hoel *Modern Digital Electronics R.P. Jain *Embedded System Design with C805, 1st Edition Han-Way Huang *Chip Design for Submicron VLSI: CMOS Layout and Simulation, 1st Edition John P. Uyemura *Digital Communications, 5th Edition John Proakis,Massoud Salehi *Construction Planning, Equipment, and Methods, 7th Edition Robert L Peurifoy, Clifford J. Schexnayder, Aviad Shapira, Technion *Control Systems Madan Gopal *Modern Power System Analysis D. P. Kothari,I. J. Nagrath *Embedded Microcomputer Systems: Real Time Interfacing, 2nd Edition Jonathan W. Valvano *Introduction to Embedded Microcomputer Systems: Motorola 6811/6812 Simulations, 1st Edition Jonathan W. Valvano *Principles of Foundation Engineering, 6th Edition Braja M. Das *Engineering Fundamentals: An Introduction to Engineering, 3rd Edition Saeed Moaveni *Introduction to Scientific Computation and Programming, 1st Edition Daniel Kaplan *Engineering Economy, 6/e Leland Blank,Anthony Tarquin *Basics of Engineering Economy Leland Blank,Anthony Tarquin *Programmable Logic Controllers, 3rd Edition Frank D. Petruzella *Introduction to Chemical Processes: Principles, Analysis, Synthesis, 1st Edition Regina M. Murphy *physical chemistry 8/E P.Atkins-Trapp-Cady *Analytical chemistry D.Harvey *chemistry C.Housecroft *Plant Design and Economics for Chemical Engineers, 5th Edition Max S Peters , Klaus D Timmerhaus, Ronald E. West. *Conceptual Design of Distillation Systems Michael F. Doherty,Michael F. Malone *Fundamentals of Applied Electromagnetics, 5/E Fawwaz T. Ulaby *Unit Operations and Processes in Environmental Engineering, 2nd Edition Tom D. Reynolds | Paul Richards *Introduction to Environmental Engineering, 2nd Edition P. Aarne Vesilind | Susan M. Morgan *Introduction to Scientific Computation and Programming, 1st Edition Daniel Kaplan *An Introduction to Signals and Systems, 1st Edition John Alan Stuller *Introduction to Wireless and Mobile Systems, 2nd Edition Dharma P. Agrawal | Qing-An Zeng *Power Systems Analysis and Design, 4th Edition J. Duncan Glover | Mulukutla S. Sarma | Thomas Overbye *Introduction to Signal and System Analysis, 1st Edition Kaliappan Gopalan *Signals, Systems, and Transforms, 4/E Charles L Phillips oohn Parr Eve Riskin *Electrical Engineering: Principles and Applications, 4/E Allan R. Hambley *Introduction to Logic Design, 2nd Edition Alan B Marcovitz *Digital Principles and Design, 1st Edition Donald D. Givone Modern Digital Electronics, 1st Edition R.P. Jain Elements of Engineering Electromagnetics, 6/E Nannapaneni Narayana Rao Introduction to Microelectronic Fabrication: Volume 5 of Modular Series on Solid State Devices, 2/E Richard C. Jaeger Electrical Machines, Drives and Power Systems, 6/E Theodore Wildi Engineering Economy and the Decision-Making Process Joseph C. Hartman Interpreting and Analyzing Financial Statement, 4/E Karen P. Schoenebeck Financial Accounting, Reporting & Analysis: International Edition, 2/E Barry Elliott,Jamie Elliott Financial Statement Analysis: A Valuation Approach Leonard C. Soffer, Robin J. Soffer Interpreting and Analyzing Financial Statement, 4/E Karen P. Schoenebeck Effective Writing, 8/E Claire B. May,Gordon S. May Getting Started with Peachtree and Quickbooks 2006 Elaine Heldstab Getting Started with Peachtree 2005 Elaine Heldstab Advanced Accounting, 10/E Floyd A. BeamsRobin P. Clement, Joseph H. Anthony,Suzanne Lowensohn Financial Accounting 7/E Walter T. Harrison, Charles T. Horngren Financial Accounting: A Business Process Approach, 2/E Jane L. Reimers Financial and Managerial Accounting Charles T. Horngren,Walter T. Harrison Financial Accounting Jane L. Reimers Financial Accounting, Reporting & Analysis: International Edition, 2/E Barry Elliott, Jamie Elliott Cases in Management Accounting and Control Systems, 4/E Brandt R. Allen, E. Richard Brownlee, II, Mark E. Haskins, Luann J. Lynch, Jane W. Rotch Introduction to Management Accounting-Chapters 14/E Charles T. Horngren,Gary L. Sundem,William O. Stratton,Jeff Schatzberg,Dave Burgstahler Management Accounting: Analysis and Interpretation Cheryl McWatters,Jerold L Zimmerman, Dale Morse Management Accounting, 5/E Anthony A. Atkinson, Robert S. Kaplan,Ella Mae Matsumura, S. Mark Young Takeovers, Restructuring, and Corporate Governance, 4/E J. Fred Weston,Mark L. Mitchell, J. Harold Mulherin, Cases in Management Accounting and Control Systems, 4/E Brandt R. Allen,E. Richard Brownlee, Mark E. Haskins, Luann J. Lynch, Jane W. Rotch Governmental and Nonprofit Accounting: Theory and Practice, 9/E Robert J. Freeman,Craig D. Shoulders,Gregory S. Allison,Terry Patton,G. Robert Smith, Jr Introduction to Government and Non-for-Profit Accounting, 6/E Martin Ives,Joseph R. Razek,Gordon A. Hosch,Larry A. Johnson Prentice Hall's Federal Taxation 2009: Comprehensive, 22/E Thomas R. Pope,Kenneth E. Anderson,John L. Kramer Prentice Hall's Federal Taxation 2009: Individuals, 22/E Thomas R. Pope,Kenneth E. Anderson,John L. Kramer Accounting Information Systems, 9/E George H. Bodnar, William S. Hopwood Manual AIS Practice Set Frank A. Buckless, laura R. Ingraham, James G. Jenkins Tax Research, 4/E Barbara H Karlin Survey of Accounting: Making Sense of Business Katherene P. Terrell, Katherene P. Terrell, Robert L. Terrell Mathematics for Economics and Business, 5/E Ian Jacques Mathematics for Business, 8/E Stanley A. Salzman, Charles D. Miller,Gary Clendenen Foundations of Finance: The Logic and Practice of Financial Management, 6/E Art J Keown,John D Martin,john W Petty,David F Scott Financial Management: Principles and Applications, 10/E Arthur J. Keown,John D. Martin,john W. Petty,David F. Scott Advanced Corporate Finance Joseph Ogden, Frank C. Jen, Philip F. O'Connor Modern Investment Theory, 5/E Robert A. Haugen Financial Economics, 2/E Zvi Bodie,Robert Merton,david Cleeton Advanced Corporate Finance Joseph Ogden, Frank C. Jen, Philip F. O'Connor Takeovers, Restructuring, and Corporate Governance, 4/E J. Fred Weston, Mark L. Mitchell, J. Harold Mulherin E conomic Growth, 2/E David N. Weil Introduction to Econometrics, 2/E James H. Stock, Mark W. Watson Engineering Economy and the Decision-Making Process Joseph C. Hartman, Lehigh University Microeconomics: Theory and Applications with Calculus Jeffrey M. Perloff Engineering Economy and the Decision-Making Process Joseph C. Hartman Managing Engineering and Technology, 4/E Lucy C. Morse,Dan L. Babcock Cost Analysis and Estimating for Engineering and Management Phillip F. Ostwald, Timothy S. McLaren Supply Chain Management, 3/E Sunil Chopra,Peter Meindl Manufacturing Processes for Engineering Materials, 5/E Serope Kalpakjian,Steven Schmid Automation, Production Systems, and Computer-Integrated Manufacturing, 3/E Mikell P. Groover Operations Research: An Introduction, 8/E Hamdy A. Taha Water-Resources Engineering, 2/E David A. Chin Water and Wastewater Technology, 6/E Mark J. Hammer Structural Steel Design, 4/E Jack C. McCormac Finite Element Analysis Theory and Application with ANSYS, 3/E Saeed Moaveni Introduction to Environmental Engineering and Science, 3/E Gilbert M. Masters,Wendell P. Ela Electric Circuits, 8/E James W. Nilsson,Susan Riedel Introductory Circuits for Electrical and Computer Engineering James W. Nilsson,Susan A. Riedel