mm-1063 === Subject: How to find this joint distribution of function of random variables? I am wondering how can I find the joint pdf of random variables X, Y. where X=cos(THETA), Y=sin(THETA), where THETA is uniformly distributed over [0, 2*pi]. It is difficult to me because all the tricks I know for this kind of problem is the Jacobian. But this is one R.V (THETA) mapping to two R.V(X, Y)... so I dont know how to write out the Jacobian...anybody knows how? Maybe I should define dummy variable? Please help me! === Subject: Re: How to find this joint distribution of function of random variables? > I am wondering how can I find the joint pdf of random variables X, Y. > where X=cos(THETA), Y=sin(THETA), where THETA is uniformly distributed over > [0, 2*pi]. How about I[x^2 + y^2 = 1]/(2 pi), where I[.] is the indicator function that returns 1 if its argument is true, and 0 otherwise. === Subject: Re: How to find this joint distribution of function of random variables? posting-account=tyq2IQ0AAAAuNErVkOWXc866QgLV9Cn9 > I am wondering how can I find the joint pdf of random variables X, Y. > where X=cos(THETA), Y=sin(THETA), where THETA is uniformly distributed over > [0, 2*pi]. > It is difficult to me because all the tricks I know for this kind of problem > is the Jacobian. X and Y are not jointly continuous, hence they do not have a joint pdf. === Subject: How to find the autocorrelation function of this random process? Hi all, The random process is Y(t)=(-1)^(X(t)), where X(t) is a Poisson process with rate lamda. The R.P. Y(t) starts at Y(0)=1. This is called semirandom telegraph signal because Y(0)=1 is not random initial value... The problem asks for the autocorrelation of this R.P. I did the following: E(Y(t)Y(s))=E((-1)^(X(t)+X(s)))=1*P(X(t)+X(s)=even number)-1*P(X(t)+X(s)=odd number) since X(t) is Poission distribution with rate lamda*t, X(s) is Poission distribution with rate lamda*s. X(t)+X(s) is Poission distribution with rate lamda*(t+s). Then I found the P(X(t)+X(s)=even number)-P(X(t)+X(s)=odd number)=exp(-2*lamda*(t+s)) hence this E(Y(t)Y(s)) depends on both s and t. But the solution gives E(Y(t)Y(s))=exp(-2*lamda*(|t-s|))... which is wierd... please help me! Anybody sees whats wrong with my derivation? === Subject: Re: How to find the autocorrelation function of this random process? > Hi all, > The random process is Y(t)=(-1)^(X(t)), > where X(t) is a Poisson process with rate lamda. The R.P. Y(t) starts > at Y(0)=1. This is called semirandom telegraph signal because > Y(0)=1 is not random initial value... > The problem asks for the autocorrelation of this R.P. > I did the following: > E(Y(t)Y(s))=E((-1)^(X(t)+X(s)))=1*P(X(t)+X(s)=even > number)-1*P(X(t)+X(s)=odd number) > since X(t) is Poission distribution with rate lamda*t, X(s) is > Poission distribution with rate lamda*s. > X(t)+X(s) is Poission distribution with rate lamda*(t+s). No. This is only true if you are working with a case where X(t), X(s) refer to independent realisations, or independent sections of the same process. For example, consider t=s. Then X(t)=X(s) always, so X(t)+X(t) is twice a Poisson r.v. with rate lamda*t, NOT a Poisson with rate lamda*(2t). You can probably make progress by considering, if s>t, X(t)=X(t)-X(0) X(s)={X(s)-X(t)} + {X(t)-X(0)} where the two terms in curly brackets are independent. David Jones === Subject: re:How to find the autocorrelation function of this random proc I havent done the calculation, but your answer just looks wrong. If s=t, the autocorrelation has to be 1. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: The Size of Grahams Number posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L x*(y+1) = x + (x*y) multiplication x^(y+1) = x * (x^y) exponentiation x^^1 = x x^^(y+1) = x ^ (x^^y) tetration x^^^1 = x x^^^(y+1) = x ^^ (x^^^y) hyper-tetration 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987 3^^4 = 3^(3^^3) = 3^(3^27) = 3^(7,625,597,484,987) [Has about 8.4 x 10^12 digits.] Grahams number is between 2^^^66 and 3^^^131. Herc === Subject: Re: The Size of Grahams Number format=ßowed; > x*(y+1) = x + (x*y) multiplication > x^(y+1) = x * (x^y) exponentiation > x^^1 = x > x^^(y+1) = x ^ (x^^y) tetration > x^^^1 = x > x^^^(y+1) = x ^^ (x^^^y) hyper-tetration > 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987 > 3^^4 = 3^(3^^3) = 3^(3^27) = 3^(7,625,597,484,987) > [Has about 8.4 x 10^12 digits.] > Grahams number is between 2^^^66 and 3^^^131. No, Grahams number is the 64th number in a rapidly growing sequence whose first element already far exceeds 3^^^131. In the hierarchy youve indicated above, let x{k} denote the operator with k ^s , e.g. x^^^y = x{3}y. Then Grahams number is g_64 in the sequence defined by g_1 = 3{4}3 g_n = 3{g_(n-1)}3 (n>1). Thus, g_1 = 3{4}3 g_2 = 3{3{4}3}3 g_3 = 3{3{3{4}3}3}3 ... g_64 = 3{3{...3{4}3...}3}3 (64 pairs of {}) = Grahams number Note that the g-sequence *starts* with the number g_1 = 3{4}3 = 3^^^^3 = 3^^^(3^^^3) >> 3^^^131. --r.e.s. === Subject: Re: The Size of Grahams Number posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L Completely different definition in Dave Renfros reference. http://mathforum.org/discuss/sci.math/m/394644/394645 Here is how Smorynski defines Grahams number (p. 149) ---> Let N be the natural numbers. First, we define K: N^2 --> N. K(0,n) = n^n K(m+1, n) = K(m, K(m,n)) Next, we define G: N --> N using K. G(0) = K(3,3) G(n+1) = K(G(n), 3) Smorynski then defines Grahams number to be G(64). Will the real Grahams number please stand up! Herc === Subject: Re: The Size of Grahams Number format=ßowed; > Completely different definition in Dave Renfros reference. > http://mathforum.org/discuss/sci.math/m/394644/394645 > Here is how Smorynski defines Grahams number (p. 149) --- Let N be the natural numbers. First, we define K: N^2 --> N. > K(0,n) = n^n > K(m+1, n) = K(m, K(m,n)) > Next, we define G: N --> N using K. > G(0) = K(3,3) > G(n+1) = K(G(n), 3) > Smorynski then defines Grahams number to be G(64). > Will the real Grahams number please stand up! Renfros page discusses both definitions, and apparently none of his other references use Smorynskis version. (Regrettably, I didnt notice, in his section How Big is Grahams Number?, essentially the content of my posting.) It would be nice to know for sure which definition is given in the original paper (to which, unfortunately, I have no ready access): Graham, R. L. and Rothschild, B. L. Ramseys Theorem for n-Parameter Sets. Trans. Amer. Math. Soc. 159, 257-292, 1971. --r.e.s. === Subject: Re: The Size of Grahams Number format=ßowed; reply-type=response >> Will the real Grahams number please stand up! > Renfros page discusses both definitions, and apparently none > of his other references use Smorynskis version. (Regrettably, > I didnt notice, in his section How Big is Grahams Number?, > essentially the content of my posting.) It would be nice to > know for sure which definition is given in the original paper > (to which, unfortunately, I have no ready access): > Graham, R. L. and Rothschild, B. L. > Ramseys Theorem for n-Parameter Sets. > Trans. Amer. Math. Soc. 159, 257-292, 1971. Update: I checked on this myself ... Apparently, practically all online sources have got the definition wrong. In the original paper by Graham & Rothchild, the number now known as Grahams number is defined as G = F(F(F(F(F(F(F(12,3),3),3),3),3),3),3) where F(m,n) is defined recursively as F(1,n) = 2^n, F(m,2) = 4, m >= 1, n >= 2 F(m,n) = F(m-1,F(m,n-1)), m >= 2, n >= 3. Please post any replies to the separate thread Ive started: Grahams Number -- Practically Everyone Has It Wrong!? --r.e.s. === Subject: Re: The Size of Grahams Number >The number of digits it would take to write down Grahams number is just >as unfathomable as the number itself. phil had a good point though - couldnt Grahams number concievably be certain precision in a certain volume V? Or is Grahams number so huge the precision of the Planck length is less than Grahams number? === Subject: Re: The Size of Grahams Number >>The number of digits it would take to write down Grahams number is just >>as unfathomable as the number itself. > phil had a good point though - couldnt Grahams number concievably be > certain precision in a certain volume V? Or is Grahams number so huge > the precision of the Planck length is less than Grahams number? 1 light year ~ 10^16m ~ 10^51 Planck lengths Approximate the universe as a cube of side 10^10 ly = 10^61 Planck lengths bigger than a googolplex but nowhere near Grahams number. === Subject: Re: The Size of Grahams Number >1 light year ~ 10^16m ~ 10^51 Planck lengths >Approximate the universe as a cube of side 10^10 ly = 10^61 Planck lengths >bigger than a googolplex but nowhere near Grahams number. I thought Grahams Number was some sort of (significant) upper bound to a problem in combinatorics. If the number of combinatoric solutions positions in the universe (i.e. the ultimate combinatorics problem) then what possible relevance can it have? (Excuse my literalness, I know maths as such is uninterested in the real world, it just interests me how you can pose a problem that theoretically could have so many solutions.) === Subject: Re: The Size of Grahams Number <9e6de$41b0d29b$5397ce19$23738@nf2.news-service.com> <271ff$41b1bcd7$5397ce19$29009@nf2.news-service.com> posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L >I thought Grahams Number was some sort of (significant) upper bound >to a problem in combinatorics. If the number of combinatoric solutions >positions in the universe (i.e. the ultimate combinatorics problem) >then what possible relevance can it have? (Excuse my literalness, I >know maths as such is uninterested in the real world, it just >interests me how you can pose a problem that theoretically could have >so many solutions.) Right! in actuality all these big numbers are from the same class of functions, exponential growth. (in between exponential growth and non computable functions whose growth cannot be considered at all). But 10^64 is hardly the largerst. What is useful are some constants to measure different exponential growths, how sharp they shoot upwards. Its not so much the height of the curve but the shape that determines different behaviours. 10^64 does not have the property that taking itself to its own power 10^64 times has relatively little change to the result, so in that respect its small. Whats bigger : G(64) or BusyBeaver(64) ? Herc === Subject: Re: The Size of Grahams Number <9e6de$41b0d29b$5397ce19$23738@nf2.news-service.com> <271ff$41b1bcd7$5397ce19$29009@nf2.news-service.com> posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L >Whats bigger : G(64) or BusyBeaver(64) ? It would be close, G is a small algorithm to code but would it fit into a 64 state TM? It could easily be coded in a few hundred states, so BB(500) > G(500). Busy Beaver is the BIGGEST POSSIBLE size of output from a fixed sized computer program. All you have to do is program your G function in say 400 states, then (pseudo) input the number 64, this will require a small TM to run first and put 64 1s on the tape, only about 50 more states to do this, and you have Grahams number with 450 states. Obviosly a TM with 451 states could output a larger number than a TM with 450 states. The fact tesselation is simple to code is one reason why the noncomputable BB function grows as fast as it does. Herc === Subject: Re: The Size of Grahams Number <9e6de$41b0d29b$5397ce19$23738@nf2.news-service.com> <271ff$41b1bcd7$5397ce19$29009@nf2.news-service.com> posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L >Whats bigger : G(64) or BusyBeaver(64) ? It would be close, G is a small algorithm to code but would it fit into a 64 state TM? It could easily be coded in a few hundred states, so BB(500) > G(500). Busy Beaver is the BIGGEST POSSIBLE size of output from a fixed sized computer program. All you have to do is program your G function in say 400 states, then (pseudo) input the number 64, this will require a small TM to run first and put 64 1s on the tape, only about 50 more states to do this, and you have Grahams number with 450 states. Obviosly a TM with 451 states could output a larger number than a TM with 450 states. The fact tesselation is simple to code is one reason why the noncomputable BB function grows as fast as it does. Herc === Subject: Re: The Size of Grahams Number <9e6de$41b0d29b$5397ce19$23738@nf2.news-service.com> <271ff$41b1bcd7$5397ce19$29009@nf2.news-service.com> posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L tetration not tesselation. Herc === Subject: Re: Abstract Algebra > Im missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. > Hint: If [G:H] = 3, then either H is normal in G, or H has a > subgroup of index 2 thats normal in G. > Ive never seen this before - Im not sure I believe it - and > I dont think anything like it is necessary for the problem at hand. Well, of course its not *necessary*. It just makes the problem easy to solve. And its certainly true. The action of G on the set of cosets of H gives a surjective homomorphism from G onto a transitive degree-3 subgroup of S_3, whose kernel lies in H (since the latter is the point stabilizer of the coset H in this action). S_3 has only two transitive subgroups in its natural action, so either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 (with the latter 3 cosets squaring to K). -- Jim Heckman === Subject: Re: Abstract Algebra Im missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. Hint: If [G:H] = 3, then either H is normal in G, or H has a > subgroup of index 2 thats normal in G. > Ive never seen this before - Im not sure I believe it - and > I dont think anything like it is necessary for the problem at hand. > Well, of course its not *necessary*. It just makes the problem > easy to solve. > And its certainly true. The action of G on the set of cosets of > H gives a surjective homomorphism from G onto a transitive > degree-3 subgroup of S_3, whose kernel lies in H (since the > latter is the point stabilizer of the coset H in this action). > S_3 has only two transitive subgroups in its natural action, so > either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 > with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 > (with the latter 3 cosets squaring to K). But. I suspect that anyone having trouble with the original problem is not going to understand any of what youve written here. Knowing that a problem on the second ßoor is easy to solve from the perspective of the 12th ßoor doesnt help the guy who has been struggling to reach the mezzanine. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Abstract Algebra > Im missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. [...] > And its certainly true. The action of G on the set of cosets of > H gives a surjective homomorphism from G onto a transitive > degree-3 subgroup of S_3, whose kernel lies in H (since the > latter is the point stabilizer of the coset H in this action). > S_3 has only two transitive subgroups in its natural action, so > either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 > with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 > (with the latter 3 cosets squaring to K). > But. > I suspect that anyone having trouble with the original problem > is not going to understand any of what youve written here. > Knowing that a problem on the second ßoor is easy to solve > from the perspective of the 12th ßoor doesnt help the guy > who has been struggling to reach the mezzanine. Actually, Im having trouble thinking of a way to solve the original problem that *doesnt* essentially use the action of G on the cosets of H to show that G has a surjective homomorphism to C_3 or S_3 whose kernel lies in H. I notice the OP said he figured it out, so I wonder just what ßoor hes on. Or maybe Im missing some more elementary approach... -- Jim Heckman === Subject: Re: Abstract Algebra [...] Im missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with > 1000 elements. > If x is in G, show that either x^2 or x^3 is in H. [...] > Actually, Im having trouble thinking of a way to solve the > original problem that *doesnt* essentially use the action of G > on the cosets of H to show that G has a surjective homomorphism > to C_3 or S_3 whose kernel lies in H. I notice the OP said he > figured it out, so I wonder just what ßoor hes on. Or maybe > Im missing some more elementary approach... obvious elementary approach. Doh! -- Jim Heckman === Subject: Re: Abstract Algebra > Actually, Im having trouble thinking of a way to solve the > original problem that *doesnt* essentially use the action of G > on the cosets of H to show that G has a surjective homomorphism > to C_3 or S_3 whose kernel lies in H. I notice the OP said he > figured it out, so I wonder just what ßoor hes on. Or maybe > Im missing some more elementary approach... Well, I dont know about a method that avoids cosets entirely, but heres an outline of what I came up with: Any two cosets aH and bH are either disjoint or equal. There is also a bijection between them (making them the same size if H is finite). We can see that if aH = bH, a^-1b must be in H. Then consider the 4 cosets H = x^0H, xH = x^1H, x^2H, x^3H. Since G is 3 times the size of H, the pigeonhole principle forces at least one pair of equal cosets; call them x^mH and x^nH where m < n. Since x^mH = x^nH, we have that x^(n-m) is in H. If n-m is 2 or 3, we are done. If n-m is 1, we use the fact that H is a group one last time to get the desired result. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Abstract Algebra Originator: grubb@lola > Im missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. >Actually, Im having trouble thinking of a way to solve the >original problem that *doesnt* essentially use the action of G >on the cosets of H to show that G has a surjective homomorphism >to C_3 or S_3 whose kernel lies in H. I notice the OP said he >figured it out, so I wonder just what ßoor hes on. Or maybe >Im missing some more elementary approach... Its easy to do this on a case by case basis: If x is in H, then x^2 is in H, so were done. If x is not in H, then H and xH are disjoint. Now consider x^2. If x^2 is in H, were done. If x^2 is in xH, then x is in H and were done. If x^2 is in neither, then x^2H is disjoint from both H and xH. Furthermore, since [G:H]=3, these three cosets cover G. Now look at x^3. If x^3 is in H, were done. If it is in xH, then x^2 is in H and were done. If x^3 is in x^2H, then x is in H and were done. --Dan Grubb === Subject: Re: How to measure randomness of a deck of cards? posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L take the difference between consecutive cards, step 2 take the sum of squares. compare it to a shufßed deck. also do it for suits, 0 1 2 3 in mod 4. could also do it for every second card, every 3rd card. or a neural net could be trained to give you a yes or no if its shufßed. theres no absolute answer unless youre using a naive interpretation of random, say the deck is 314159, looks stacked to me! step 2: subtract that from the expected difference (I think) Herc === Subject: Re: How to measure randomness of a deck of cards? > take the difference between consecutive cards, > step 2 > take the sum of squares. > compare it to a shufßed deck. > also do it for suits, 0 1 2 3 in mod 4. > could also do it for every second card, every 3rd card. > or a neural net could be trained to give you a yes or no if its > shufßed. > theres no absolute answer unless youre using a naive interpretation > of random, > say the deck is 314159, looks stacked to me! > step 2: subtract that from the expected difference (I think) > Herc What is this method called? And what is its main purpose? === Subject: Re: How to measure randomness of a deck of cards? posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L The sum of squares (of the differences) is standard for measuring total error of a sample. Im not sure how it will go using the heuristic of an Ôexpected difference between consecutive cards, but you should be able to detect a fresh deck from a shufßed one. Say you calculate the average difference between consecutive cards is 7. then a fresh deck will have differences of Cards = < 1,2,3,4,5,6,7,8,9,10,11,12,13,1,2,3,4,5,6,7,8,9,10,11,12,13,1, 2..> Delta = < 1,1,1,1,1,1,1,1,1,1,1,1,1,1,13,1,1,1,1,1,1,1,1,1,1,1,1,1,13,1, 1....> Errors = <6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,...> A random deck will have Delta = <2,4,1,10,12,4,2,6...> Errors = <5,1,3,5,1,5,1....> The standard deviation will pick it up (on Delta), thats a function of the sum of squares, so you dont need an expected difference or Errors. heres another technique : use GPS (General Problem Solver AKA Search algorithm) to sort the cards *into* order count the number of card swaps and you have a measure of the *distance* from an ordered list you could weight the cards swaps, *distance, *sqrt(distance), or just use adjacent swaps. Deck one took 5 card swaps to be ordered, deck two took 20, so deck two was more shufßed. Some problems... a reverse ordered deck will appear shufßed! Herc === Subject: Re: How to measure randomness of a deck of cards? > say the deck is 314159, looks stacked to me! Actually, the digits of pi look not too bad when you do PRNG stats tests... Not as good as SHA-1, but not too bad... === Subject: random 1 to 3 to random 1 to 5 One of my friend asked me this question which he was asked during a job interview. Couldnt figure it out and very curious about the solution (or this just mpossible? ) The question is, given a function that returns evenly distributed random number from 1 to 5, design a new function based on this function, and return evenly distributed random number from 1 to 3. === Subject: Re: random 1 to 3 to random 1 to 5 > One of my friend asked me this question which he was asked during a > job interview. Couldnt figure it out and very curious about the > solution (or this just mpossible? ) > The question is, given a function that returns evenly distributed > random number from 1 to 5, design a new function based on this > function, and return evenly distributed random number from 1 to 3. I would invoke the thing three times in a row and associate the first selection with the numbers 1-5, the second selection with 6-10 an the third with 11-15 and divide the result by three. OK, I figured this out in a minute but would NEVER have been able to do so during a job interview with mr/mrs personnel officer watching me, waiting for my defeat. There is no end to the abuse employers will go these days: I have had to sign a PISS form prior to even being interviewed. Well maybe I am an alcoholic! do you really care if I solve your problem? I view all of this as abuse: and it doesnt matter for the outcome anyway since the personnel person will hire people whom he(she) likes anyway but has to rationalize his(her) decisions with some pseudo-scientific rationalization. In the end he/she will hire: people ;ike him/her self! These would not be programmers of course but they would make great personnel officers. Now to the core of the matter: Is the person who can solve puzzles like the above in a minute the better candidate? I dont think so. Maybe this person has trained in goat wolf and cabbage puzzles all his life! Or maybe he is a member of Mensa which means that he can deduce the general term of any series with complete confidence from a mere three terms! I will never forget the scathing sarcasm of my algebra professor when he ridiculed certain puzzles in the newspaper asking the reader to complete a certain given series! I have heard that MICROSOFT uses puzzles like the above to evaluate potential job candidates so I assume they can cross any river without problems, maybe thats why their software is so popular. Now, there is such a thing as deep thinking and it requires time and an environment free of disturbances, much different from the environment of the job interview. There are also KNOWLEDGE and EXPERIENCE which one can actually measure in a scientific manner - provided one has some oneself. Finally, how a person will behave or fit in within a group cannot be deduced from superficial interpersonal behaviour in any way. Jentje Goslinga === Subject: Re: random 1 to 3 to random 1 to 5 > Now to the core of the matter: Is the person who can solve > puzzles like the above in a minute the better candidate? I believe this trend got started because Microsoft interviews programmers with logic puzzle, why are manhole covers round, and the like. But this is bad logic. Microsoft makes terrible software! What people should do is find out how Microsoft interviews their MARKETING people, and emulate that! === Subject: Re: random 1 to 3 to random 1 to 5 ETAsAhRige9JUWtymOvTm2G2qmtDPVFBJgIUUUFNAcz2vfIa4AVGxplK5kUosE E= uh, f(x) =(x+1)/2? --OL === Subject: Re: random 1 to 3 to random 1 to 5 > uh, f(x) =(x+1)/2? That will upset the ditribution. In the simplest example, if the 1-5 outputs all occur at equal frequency over a sufficiently long time interval, then the frequencys of the 1-3 outputs will be 1: 40% 2: 40% 3: 20% === Subject: Re: random 1 to 3 to random 1 to 5 ETAtAhUAxBJIWxM51G2i0uLGKaCpuJshQ2UCFHBkr2Qp/ 3fEVg4RMWkdEskPHtQo rphenry, I could not reach you via email. See what I subsequently posted for the discrete case. --OL === Subject: Re: random 1 to 3 to random 1 to 5 Having read Robert Vienneaus and Peter Webbs solution, I am getting interested in another similar question. The question is: Given a function F() that returns evenly distributed random number from {1,2,3,4,5} in constant time, design a new function G() that returns evenly distributed random number from {1,2,3}in constant time. I coudnt design such G(). === Subject: Re: random 1 to 3 to random 1 to 5 Not long ago there was a problem of designing a fair 7-sided die. One oof the solutions found for this was to roll a 6-sded die repeatedly, add up the numbers and take the residue mod 7. The result is the fairest possible division for the given number of rolls. To design your G() function: Start with a seed value, then define recursively: G() = mod(G()+F(),3)+1 As time progresses the probability distribution for G() approaches the uniform one. The function defined in this way is autocorrelated. When G() is 3 the next iterate is more likely to be 2 or 3 than 1. You can reduce the autocorrelation by storing the last several values of F. Calculate the sum modulo 3, calling it S(), then iterate according to: G() = mod(G()+S(), 3)+1 --OL === Subject: Re: random 1 to 3 to random 1 to 5 >Having read Robert Vienneaus and Peter Webbs solution, I am getting >interested in another similar question. >The question is: >Given a function F() that returns evenly distributed random number from >{1,2,3,4,5} in constant time, design a new function G() that returns evenly >distributed random number from {1,2,3}in constant time. Something similar was discussed here a while ago, except it was about contructing a uniform distribution in {1,2,3} from a uniform distribution in {1,2}: -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: random 1 to 3 to random 1 to 5 > One of my friend asked me this question which he was asked during a > job interview. Couldnt figure it out and very curious about the > solution (or this just mpossible? ) > The question is, given a function that returns evenly distributed > random number from 1 to 5, design a new function based on this > function, and return evenly distributed random number from 1 to 3. rand1to3 = (rand1to5 - 1)/2 + 1 === Subject: Re: random 1 to 3 to random 1 to 5 > One of my friend asked me this question which he was asked during a > job interview. Couldnt figure it out and very curious about the > solution (or this just mpossible? ) > The question is, given a function that returns evenly distributed > random number from 1 to 5, design a new function based on this > function, and return evenly distributed random number from 1 to 3. > rand1to3 = (rand1to5 - 1)/2 + 1 Im guessing the OP meant to say integer rather than number (otherwise its a kind of silly question). === Subject: Re: random 1 to 3 to random 1 to 5 > One of my friend asked me this question which he was asked during a > job interview. Couldnt figure it out and very curious about the > solution (or this just mpossible? ) > The question is, given a function that returns evenly distributed > random number from 1 to 5, design a new function based on this > function, and return evenly distributed random number from 1 to 3. Let f() be the first function. The new function is defined by the following pseudo-code: x := f(); while ( x > 3 ) do x := f(); end; return x; -- Mostly economics: r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Some Propositions Of Mathematical Economics 1. Because of price Wicksell effects, the interest rate is not equal, in equilibrium, to the marginal product of capital. 2. The wage is not determined by the value of the marginal product of labor. 3. Consider a so-called perverse switch point. Around such a point, a higher wage is associated with a cost minimizing technique in which firms hire more labor to produce a given level of net output. These propositions use technical terms, and I have explained those terms in different ways on many occasions. If you do not like my explanations, you can always look up some of the literature I have cited. For some reason, mainstream economists posting to Usenet have argued against these true statements. Of course, they are not consistent with certain erroneous teaching found in certain textbooks. -- Mostly economics: r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: talk science and mathematics, you faggots. >> i m supreme. i alone command you. > still boring... Lunacy is never boring. BTW CHAOS stands for Could (do to) Have Another Olanzapine Soon. === Subject: about semialgebraic set My definition of semialgebraic subset is: A subset V of R^n is called semi-algebraic if it admits some representation of the form V = cup_{i=1}^s cap_{j=1}^{r_i} {x in R^n | P_{i,j}(x) s_{ij} 0 }, where Ôcup stands for Ôunion Ôcap stands for Ôintersection s_{ij} in {>,=,<} P_{ij}(X) in R[X], X = (X_1,...,X_n). May you help me to show that the complementary in R^n of a semialgebraic set is a semialgebraic set itself, please? S.T. === Subject: Re: about semialgebraic set > My definition of semialgebraic subset is: > A subset V of R^n is called semi-algebraic > if it admits some representation of the form > V = cup_{i=1}^s cap_{j=1}^{r_i} {x in R^n | P_{i,j}(x) s_{ij} 0 }, > where > Ôcup stands for Ôunion > Ôcap stands for Ôintersection > > s_{ij} in {>,=,<} > P_{ij}(X) in R[X], X = (X_1,...,X_n). > May you help me to show that the complementary in R^n > of a semialgebraic set is a semialgebraic set itself, > please? Start by showing the intersection of two semialgebraic sets is semialgebraic. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Proving the cross product is orthogonal I hope I dont seem like one of those louts that only turn up at assignment time!! However I am struggling with a proof. My answer is not coming out as expected, and I think I am making a silly mistake somewhere. If anyone could have a glance over it, that would be appreciated. Q) a =(1, -2, 1) b =(3, 1, 0) Find a x b and prove that your cross-product vector is perpendicular to each of the vectors a and b. A) [I have worked out the cross-product, and I have proven that a is perpendicular to the cross-product, however I am struggling to prove that b is perpendicular to the cross-product] a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k = -3i + 2j + 7k Let w = -3i + 2j + 7k Prove vector a is perpendicular to w a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) cos&= (a.w)/(|a||w|) = 0 Prove vector a is perpendicular to w b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) I know that is wrong, can anyone see what my mistkae is?? Cassandra === Subject: Re: Proving the cross product is orthogonal >>I hope I dont seem like one of those louts that only turn up at >>assignment time!! However I am struggling with a proof. My answer is not >>coming out as expected, and I think I am making a silly mistake >>somewhere. If anyone could have a glance over it, that would be appreciated. >>Q) a =(1, -2, 1) b =(3, 1, 0) >>Find a x b and prove that your cross-product vector is perpendicular to >>each of the vectors a and b. >>A) [I have worked out the cross-product, and I have proven that a is >>perpendicular to the cross-product, however I am struggling to prove >>that b is perpendicular to the cross-product] > Do you know about the dot product? Yes, I like the dot product...much easier then the cross product I think! It turned out my main mistake was silly mistakes in my workings (ie 2*0 = 2 etc) I did this two places, I think it was only ßuke that my workings for a being perpendicular to a x b, after the silly errors where fixed both proofs worked. Cassandra === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE You simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess that you were writing too small print too tightly in the left upper corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); etc., etc. IHTH - Johan E. Mebius > I hope I dont seem like one of those louts that only turn up at > assignment time!! However I am struggling with a proof. My answer is > not coming out as expected, and I think I am making a silly mistake > somewhere. If anyone could have a glance over it, that would be > appreciated. > Q) a =(1, -2, 1) b =(3, 1, 0) > Find a x b and prove that your cross-product vector is perpendicular > to each of the vectors a and b. > A) [I have worked out the cross-product, and I have proven that a is > perpendicular to the cross-product, however I am struggling to prove > that b is perpendicular to the cross-product] > a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k > = -3i + 2j + 7k > Let w = -3i + 2j + 7k > Prove vector a is perpendicular to w > a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 > |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) > |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) > cos&= (a.w)/(|a||w|) = 0 > Prove vector a is perpendicular to w > b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 > |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) > |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) > cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) > I know that is wrong, can anyone see what my mistkae is?? > Cassandra === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE > You simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess > that you were writing too small print too tightly in the left upper > corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); > etc., etc. > IHTH - Johan E. Mebius >> I hope I dont seem like one of those louts that only turn up at >> assignment time!! However I am struggling with a proof. My answer is >> not coming out as expected, and I think I am making a silly mistake >> somewhere. If anyone could have a glance over it, that would be >> appreciated. >> Q) a =(1, -2, 1) b =(3, 1, 0) >> Find a x b and prove that your cross-product vector is perpendicular >> to each of the vectors a and b. >> A) [I have worked out the cross-product, and I have proven that a is >> perpendicular to the cross-product, however I am struggling to prove >> that b is perpendicular to the cross-product] >> a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k >> = -3i + 2j + 7k >> Let w = -3i + 2j + 7k >> Prove vector a is perpendicular to w >> a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 >> |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) >> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >> cos&= (a.w)/(|a||w|) = 0 >> Prove vector a is perpendicular to w >> b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 >> |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >> cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) >> I know that is wrong, can anyone see what my mistkae is?? >> Cassandra Much appreciated. I have been writing my equations in MS Word using the equation editor. Perhaps I should write them on paper Ôbig and large first!! Cassie === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE > Much appreciated. I have been writing my equations in MS Word using the > equation editor. Perhaps I should write them on paper Ôbig and large > first!! With a pencil. The trolls in this newsgroup post all kinds of nonsense, sometimes on purpose and sometimes because they dont check their work. === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE >>Much appreciated. I have been writing my equations in MS Word using the >>equation editor. Perhaps I should write them on paper Ôbig and large >>first!! > With a pencil. > The trolls in this newsgroup post all kinds of nonsense, sometimes on purpose > and sometimes because they dont check their work. Yes, one of my biggest downfalls in life is that I make lots of mistakes. I once had to do a speed & accuracy test as part of a pysch test. I scored full marks (not sure how they calculated it), but I remember the pysch saying something about me getting alot further through the test then most, but making more mistakes then most. I used to have a turtle on my desk to remind myself to slow down and double check things. I find it so difficult, but it is something I should really work on. I have made two very elementary mistakes in the course of this thread...I do need to improve. === Subject: Re: Proving the cross product is orthogonal >Q) a =(1, -2, 1) b =(3, 1, 0) >Find a x b and prove that your cross-product vector is perpendicular to >each of the vectors a and b. >A) [I have worked out the cross-product, and I have proven that a is >perpendicular to the cross-product, however I am struggling to prove >that b is perpendicular to the cross-product] >a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k > = -3i + 2j + 7k Huh? Check your algebra. -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny Infinite-dimensional only. And AC required. How about this: Banach spaces l^2 and l^1 are isomorphic as linear spaces (both having Hamel dimension c). -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny > Infinite-dimensional only. And AC required. How about this: > Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). one has to necessarily appeal to AC? In other words, is AC equivalent to that statement? I am asking that, since I have a somehow realated problem: It is well known that the dual of L^infty strictly contains L^1 (this can be directly argued from separability arguments; clearly, only in infinite dimension)). In spite of that, I am not aware af any example of a continuous linear functional on L^infty which is not L^1 without appealing to the Hahn-Banach theorem, which is to say, without appealing to AC. jenny === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny Infinite-dimensional only. And AC required. How about this: > Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). > one has to necessarily appeal to AC? In other words, is AC > equivalent to that statement? Equivalent, probably not, but in fact it cannot be proved in ZF alone. In Solovays model where every set of reals (and every set in a Polish space) has the property of Baire, it follows (proved by Christensen, I guess) that any linear map of separable Banach spaces is automatically continuous. So, in particular, if a given vector space admits two complete separable norms, then the identity is a homeomorphism. (And since, in metric spaces, a map is continuous if and only if its restriction to every separable subspace is, we can get rid of the separable assumptions I made above.) Thats in Solovays model. But there is some principle that goes with this saying any spaces and maps THAT YOU CAN ACTUALLY WRITE DOWN EXPLICITLY also work like this, merely using ZF. So, depending on what accessible means in the original question, the answer may be no. > I am asking that, since I have a somehow realated problem: > It is well known that the dual of L^infty strictly contains L^1 > (this can be directly argued from separability arguments; clearly, only > in infinite dimension)). > In spite of that, I am not aware af any example of a continuous linear > functional on L^infty which is not L^1 without appealing to > the Hahn-Banach theorem, which is to say, without appealing to AC. Again, this existence cannot be proved in ZF. But it can be proved using HB. It is known that HB is strictly weaker than AC, so this answers the equivalent question above. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: question about Banach spaces Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny > Infinite-dimensional only. And AC required. How about this: Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). > one has to necessarily appeal to AC? In other words, is AC > equivalent to that statement? > Equivalent, probably not, but in fact it cannot be proved in ZF alone. (also for David C. Ullrich) When you say that it cannot be proved what do you exactly mean? I mean, if it is provable with AC but not provable without AC why it is not automatically equivalent to AC. My point is: one cannot prove this within ZF because he/she is not able to, but in principle he/she could, or there is a way to show this impossibility? jenny === Subject: Re: question about Banach spaces >> Is there an accessible example of a linear >> space endowed with two non-equivalent >> complete norms? >> jenny >> Infinite-dimensional only. And AC required. How about this: >> Banach spaces l^2 and l^1 are isomorphic as linear spaces >> (both having Hamel dimension c). >> one has to necessarily appeal to AC? In other words, is AC >> equivalent to that statement? >> Equivalent, probably not, but in fact it cannot be proved in ZF alone. >(also for David C. Ullrich) >When you say that it cannot be proved what do you exactly mean? >I mean, if it is provable with AC but not provable without AC >why it is not automatically equivalent to AC. >My point is: one cannot prove this within ZF because he/she is >not able to, but in principle he/she could, or there is a way to >show this impossibility? There is a way to show this impossibility. There was a big hint how that works in the paragraph you snipped, where Edgar said something about a certain model of ZF. Any statement that can be proved from the axioms of ZF will be true in every model of ZF, so if there is a model of ZF in which P is false it follows that P _cannot_ be proved in ZF. You need to study a small bit of mathematical logic to really know exactly what that means. For now an analogy: Say GT is the axioms for a group: (ab)c = a(bc), etc. So for example (ab(cd) = (a(bc))d is a theorem of GT - it can be proved from the axioms. Now a model of GT is precisely the same thing as a _group_. There exists a group in which it is not true that ab = ba for all a, b, and the existence of such a group shows that ab = ba for all a, b cannot be proved from the axioms of GT. >jenny ************************ David C. Ullrich === Subject: Re: question about Banach spaces > (also for David C. Ullrich) > When you say that it cannot be proved what do you exactly mean? > I mean, if it is provable with AC but not provable without AC > why it is not automatically equivalent to AC. > My point is: one cannot prove this within ZF because he/she is > not able to, but in principle he/she could, or there is a way to > show this impossibility? An example is the Hahn-Banach theorem, HB. Assuming consistency when needed, it has been shown that: (1) ZF does not prove HB, (2) ZF+AC proves HB, (3) ZF+HB does not prove AC. In that sense, HB is beyond ZF, but is not as strong as AC. On the other hand, Zorns Lemma, ZL, is equivalent to AC, meaning: (4) ZF+AC proves ZL (5) ZF+ZL proves AC In most cases, a cannot be proved result is shown by exhibiting a model. For example, a model of ZF+HB where AC fails will show (3), above. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: question about Banach spaces >> Is there an accessible example of a linear >> space endowed with two non-equivalent >> complete norms? >> jenny >> Infinite-dimensional only. And AC required. How about this: >> Banach spaces l^2 and l^1 are isomorphic as linear spaces >> (both having Hamel dimension c). >one has to necessarily appeal to AC? In other words, is AC >equivalent to that statement? Not that I know the answer to either question, but Ill point out that the in other words isnt right - saying something requires AC is not the same as saying its equivalent to AC. If ZF does not prove P but ZFC does then P requires AC; saying P is equivalent to AC is a much stronger statement. (For example: If ZFC proves P but, say, ZF + AD (Axiom of Determinacy) proves not P then AC is required for P, although this doesnt say that P is equivalent to AC. This actually happens if P is every set of reals is Lebesgue measurable.) >I am asking that, since I have a somehow realated problem: >It is well known that the dual of L^infty strictly contains L^1 >(this can be directly argued from separability arguments; clearly, only >in infinite dimension)). >In spite of that, I am not aware af any example of a continuous linear >functional on L^infty which is not L^1 without appealing to >the Hahn-Banach theorem, which is to say, without appealing to AC. >jenny ************************ David C. Ullrich === Subject: Group presentations Im struggling to understand group presentations. For example, Mathworld gives a group presentation for the Dihedral Group D_2n (the symmetry group of the regular n-gon) as: D_2n = < x, y | x^2 = 1, y^2 = 1, (xy)^n = 1 > What are the xs and ys? I assume they are elements in the group of isometries of the plane since D_2n is a subgroup thereof? Can we thus pick *any* xs and ys which satisfy the given relations, for example, why cant we pick x = 1 and y = 1? They satisfy the relations but clearly dont generate D_2n. Has Mathworld missed something out? Am I missing the point? Richard Hayden. === Subject: Re: Group presentations days. My association with the Department is that of an alumnus. >Im struggling to understand group presentations. >For example, Mathworld gives a group presentation for the Dihedral Group >D_2n (the symmetry group of the regular n-gon) as: >D_2n = < x, y | x^2 = 1, y^2 = 1, (xy)^n = 1 What are the xs and ys? I assume they are elements in the group of >isometries of the plane since D_2n is a subgroup thereof? No. They are telling you that x and y are elemnts of D_2n, that every element of D_2n can be written as a product of xs, ys, and their inverses, and that the only thing you know about x and y are that x^2 is trivial, that y^2 is trivial, and that (xy)^n is trivial. So, for example, you know that y^{-1} = y, because y^2 = 1. You know that (xy)y = y(xy)^{-1}, because (xy)y = xyy = x1 = x, and y(xy)^{-1} = y(y^{-1}x^{-1}) = x^{-1} = x (the last because x^2 = 1, so x=x^{-1}. > Can we thus >pick *any* xs and ys which satisfy the given relations, for example, >why cant we pick x = 1 and y = 1? They satisfy the relations but >clearly dont generate D_2n. D_2n is the LARGEST group which contains elements x and y that satisfy these conditions. If you pick ANY group with elements a and b which satisfy these conditions, there will be a group homomorphism from D_2n to that group, mapping x to a and y to b. The ONLY thing you know about x and y in D_2n is the x^2 =1 , y^2=1, (xy)^n=1, and the things you can deduce from this. No equality which cannot be deduced from these holds in D_2n. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Group presentations > For example, Mathworld gives a group presentation for the Dihedral Group > D_2n (the symmetry group of the regular n-gon) as: > D_2n = < x, y | x^2 = 1, y^2 = 1, (xy)^n = 1 What are the xs and ys? I assume they are elements in the group of > isometries of the plane since D_2n is a subgroup thereof? The standard choice consists in taking x and y to be reßections whose axes are neighbours, i.e. form an angle of pi/n. The relations are then quite obvious (recall that the product of two reßections with intersecting axes is a rotation whose angle is twice the angle between these axes). This presentation is inspired by what you see in a kaleidoscope - and thats also why dihedral groups are called dihedral. > Can we thus > pick *any* xs and ys which satisfy the given relations, for example, > why cant we pick x = 1 and y = 1? They satisfy the relations but > clearly dont generate D_2n. The full answer to that entails the notion of a free group and quotients thereof. Roughly speaking, the idea is that x and y should be picked in such a way that (a) relations x^2=1, y^2=1 and (xy)^n=1 hold, *and* (b) any other relation between x and y should be an *algebraic* consequence of these three relations. LD === Subject: Re: Group presentations > Im struggling to understand group presentations. > For example, Mathworld gives a group presentation for the Dihedral Group > D_2n (the symmetry group of the regular n-gon) as: > D_2n = < x, y | x^2 = 1, y^2 = 1, (xy)^n = 1 What are the xs and ys? I assume they are elements in the group of > isometries of the plane since D_2n is a subgroup thereof? Can we thus > pick *any* xs and ys which satisfy the given relations, for example, > why cant we pick x = 1 and y = 1? They satisfy the relations but > clearly dont generate D_2n. > Has Mathworld missed something out? Am I missing the point? Theres two ways (equivalent, of course) to look at this. The first is that D_2n is the _maximal_ group satisfying those relations; there are no extra relations beyond those. Of course, this doesnt always preclude the group being trivial even so; it might happen that the relations given imply that x = y = 1. But in this case you can exhibit a nontrivial group that works. The other way is that D_2n is the quotient of the free group on two generators by the normal subgroup generated by all the words which appear in the relations (that is, x^2, y^2, (xy)^n = xyxy...xy). That is, the intersection of all normal subgroups containing those words. You can see that this gives no extra relationships; it uses just enough to make all those words 1 and the subset that remains a group. In either case, x and y are not any particular things; they are symbols. You can draw an isomorphism between D_2n as written here and a certain group of plane isometries, of course, but you can also just use the symbols (and there are other interpretations Im forgetting). -- Ryan Reich ryanr@uchicago.edu === Subject: BibTex why can LaTeX not process this entry, if i cite with cite{ENG82}? @ARTICLE{ENG82, AUTHOR = Robert Engle, TITLE = Autoregressive Conditional Heteroskedasticity with Estimates of the Variance of UK Inßation, JOURNAL = Econometrica, YEAR = 1982, volume = 50, pages = 987-1008 } === Subject: Re: BibTex 3QLpj-NoP*NzsIC,boYU]bQ]Hy<#4ga3$21: > why can LaTeX not process this entry, if i cite with cite{ENG82}? > @ARTICLE{ENG82, > AUTHOR = Robert Engle, > TITLE = Autoregressive Conditional Heteroskedasticity with > Estimates of the Variance of UK Inßation, > JOURNAL = Econometrica, > YEAR = 1982, > volume = 50, > pages = 987-1008 You need quotes around the pages, I think. Anyway you should be using an en-dash (--) not a hyphen (-) to separate the page numbers. -- David Eppstein Computer Science Dept., Univ. of California, Irvine http://www.ics.uci.edu/~eppstein/ === Subject: Re: BibTex > why can LaTeX not process this entry, if i cite with cite{ENG82}? > @ARTICLE{ENG82, > AUTHOR = Robert Engle, > TITLE = Autoregressive Conditional Heteroskedasticity with > Estimates of the Variance of UK Inßation, > JOURNAL = Econometrica, > YEAR = 1982, > volume = 50, > pages = 987-1008 This question should be posted at the comp.text.tex newsgroup. Besides, you should provide more details. For instance, what do you mean when you say that LaTeX cannot process this entry? (Dont reply! This is something you should say while posting at comp.text.tex.) Jose Carlos Santos === Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere ETAtAhUApQL6jYNcW0hWib4EVPgc7y0QsTUCFDITqMFsndm0APzGwrAYok6z6K eo Its actually a closed curve, which seems very differentfrom what we get on the plane. Imagine that AB is fixed on the Equator of a spherical Earth, say in South America with A in northern Ecuador and B in far nothern Brazil. Put C initally in the Carribean Sea and start moving it west, adjusting its latitude so that the triangle keeps the same area. Eventually, angle A reaches 180 degrees and C accordingly reaches the Equator -- but the triangle does not collapse into a straight line. Conserving the nonzero area has forced the formation of a lune with B and C as the poles (C is now in eastern Indonesia, opposite B). Sides AB and AC together form a 180-degree arc on the Equator, while side BC is an alternate great circular arc that byapsses the Equator to the north. Now you can ßip this lune through line AB, so that BC is now in the Southern Hemisphere, and continue to rotate C around the base AB. C goes through Bolivia and Brazil, all the way around to western Indonesia, opposite A, where another lune is formed. Flip the lune again to complete tracing out the locus. The net result is a closed curve, symmetric about AB, with cusps a the points opposite A and B. If you specify a smnall lenght ofr AB and a small are for the triangle, the locus in the vicinity of AB looks like the pair of parallel lines you get on the plane. --OL === Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere > Its actually a closed curve, which seems very differentfrom what we get > on the plane. > Imagine that AB is fixed on the Equator of a spherical Earth, say in > South America with A in northern Ecuador and B in far nothern Brazil. > Put C initally in the Carribean Sea and start moving it west, adjusting > its latitude so that the triangle keeps the same area. Eventually, > angle A reaches 180 degrees and C accordingly reaches the Equator -- but > the triangle does not collapse into a straight line. Conserving the > nonzero area has forced the formation of a lune with B and C as the > poles (C is now in eastern Indonesia, opposite B). Sides AB and AC > together form a 180-degree arc on the Equator, while side BC is an > alternate great circular arc that byapsses the Equator to the north. > Now you can ßip this lune through line AB, so that BC is now in the > Southern Hemisphere, and continue to rotate C around the base AB. C > goes through Bolivia and Brazil, all the way around to western > Indonesia, opposite A, where another lune is formed. Flip the lune > again to complete tracing out the locus. The net result is a closed > curve, symmetric about AB, with cusps a the points opposite A and B. OK, ready on our tracks for a symbolic quantificating trigonometry... === Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere ETAuAhUAl7G/ DoN3ibTTTtFAwwKX0m7AnQcCFQCUOVX9ZFzKEYVRs9Nd4y62DEDKNg== OK, Ill start that. Let LtA be the latitude of A, north positive, LoA be the longitude of A, east positive, with similar nomenclature for points B and C. Set N to quantities we have: cos(AB) = sin(LtA)sin(LtB)+cos(LtA)cos(LtB)cos(LoA-LoB) Likewise draw Triangles NAC and NBC and get analogous expressions for cos(AC) and cos(BC). With the sides of Triangle ABC thus characterized, obtain the angles in that triangle by using the Law of Cosines solved for a vertex angle; for example: cos C = (cos(AB)-cos(AC)cos(BC))/(sin(AC)sin(BC)) where angle C is taken to be within Triangle ABC (no longer involving N). To get the sines in the denominator you must take square roots (sin x = (+/-) sqrt(1-cos^2 x)); give the square roots positive signs because the arcs measure 180 degrees or less. Take the inverse cosines of your angles, add them up and set the sum to the desired value. The locus may be a simple closed curve, but its algebra is not very simple! --OL === Subject: Re: Delta-eps proof Let q be a real number between 0 and 1, that is, let 0oo. the slope goes to oo as x->0, this suggest that we have to separate the domain [0,oo) into [0,r] and (r,oo), where r is a small real. The slope is bounded on (r,oo) and so f is uniformly continuous on (r,oo). For [0,r], the slope is too big, but luckily, the values of f are small on [0,r], so we can use this fact to overcome the difficulty of infinite slope. A bit of computation shows that we have to take r to be the delta. So we are separating the region [0,oo) into [0,delta] and [delta,oo) in the following proof. @ a Formal Proof @ Theorem: f(x)=x^q is uniformly continuous on the domain [0,oo) Proof: (For convenience of the writing, we are going to use d and e, instead of delta and epsilon.) Let e>0 be arbitrarily given. Then set d to be the real number such that (2d)^q=e, d>0. Suppose x and y are arbitrary reals such that |x-y|=0 and y>=0. Now, We are going to show that | f(y)-f(x) | < e. There are two cases. Case I: one or both of x and y is in [0,d]. For this case, since |x-y|d. Therefore z^(q-1) < d^(q-1). (Note that (q-1) is *negative*) So | f(y)-f(x) | = q z^(q-1) |y-x| < d^(q-1) |y-x| < d^(q-1) d < d^q < e. So in both cases, | f(y)-f(x) | < e QED. === Subject: Re: Delta-eps proof > I need to show that x^(1/m), m in pos integers, is uniformly continuous on > its domain. That is and for x =c in domain ,I must show that given eps >0 > there exist delta > 0 s/t > | x^(1/m) - a^(1/m) | < eps whenever 0< |x - a| < delta. > <--> |x - a| |x^(1/m - 1) +x^(1/m - 2)a +..+x a^(1/m -2) + a ^(1/m -1)| < > eps whenever 0< |x - a| < delta > I do not see how to bound (x^(1/m - 1) +x^(1/m - 2)a +..+x a^(1/m -2) + a > ^(1/m -1)) > Any hints....please. sqr x = x^(1/2) is not uniformly continuous on its domain [0,oo). === Subject: Re: Delta-eps proof >> I need to show that x^(1/m), m in pos integers, is uniformly continuous on >> its domain. That is and for x =c in domain ,I must show that given eps >0 >> there exist delta > 0 s/t >> | x^(1/m) - a^(1/m) | < eps whenever 0< |x - a| < delta. >> <--> |x - a| |x^(1/m - 1) +x^(1/m - 2)a +..+x a^(1/m -2) + a ^(1/m -1)| < >> eps whenever 0< |x - a| < delta >> I do not see how to bound (x^(1/m - 1) +x^(1/m - 2)a +..+x a^(1/m -2) + a >> ^(1/m -1)) >> Any hints....please. >sqr x = x^(1/2) is not uniformly continuous on its domain [0,oo). Proof? Hint: Its uniformly continuous on [0,1] by compactness. And its derivative is bounded on [1,infinity), making it uniformly continuous there as well. ************************ David C. Ullrich === Subject: Re: Delta-eps proof Proof? > Hint: Its uniformly continuous on [0,1] by compactness. And Well that settles the problem about the derivative being oo at 0. > its derivative is bounded on [1,infinity), making it uniformly > continuous there as well. === Subject: Re: Delta-eps proof > sqr x = x^(1/2) is not uniformly continuous on its domain [0,oo). Yes, it is! Take delta > 0 and take x and y such that |x - y| < delta^2. Then |x^(1/2) - y^(1/2)|*(x^(1/2) + y^(1/2)) < delta^2, and therefore at least one of the numbers |x^(1/2) - y^(1/2)| and x^(1/2) + y^(1/2) is smaller than delta. In the first case, were done. Otherwise, if x^(1/2) + y^(1/2) < delta, then you also have |x^(1/2) - y^(1/2)| < < delta. Jose Carlos Santos === Subject: Re: Delta-eps proof <31dkrtF3ad2qvU1@individual.net sqr x = x^(1/2) is not uniformly continuous on its domain [0,oo). > Yes, it is! Take delta > 0 and take x and y such that |x - y| < delta^2. > Then |x^(1/2) - y^(1/2)|*(x^(1/2) + y^(1/2)) < delta^2, and therefore at > least one of the numbers |x^(1/2) - y^(1/2)| and x^(1/2) + y^(1/2) is > smaller than delta. In the first case, were done. Otherwise, if > x^(1/2) + y^(1/2) < delta, then you also have |x^(1/2) - y^(1/2)| < > < delta. Hm, an example where a divergent derivative doesnt ruin uniformity. The great tragedy of science - the slaying of beautiful hypothesis by an ulgy fact. -- Thomas H. Huxley. === Subject: Re: toplogically equevalent... > hello.....doctor~ > Let (X,d) be a metric space. > Define a function d :XxX -> (R+) U {0} by > d(x,y) = min {1, d(x,y)}, the minumum of 1 and d(x,y), > for all x,y in X. > show that > (1) d is a bounded metric for X. > (2) The metric space (X,d) is topologically equivalent to > the bounded metric space (X,d) > ---------------------------------------------------- > i can do (1) by metric conditions. > but...in the (2) > let x in X and e>0 > then y in B_d_(x,e) => d(x,y) < e > => d(x,y) <= d(x,y) < e > => y in B_d_(x,e) > y in B_d_(x,e) => d(x,y) < e > =>min {1, d(x,y)} < e > => 1 < e or d(x,y) < e > if d(x,y) < e => y in B_d_(x,e) > else 1 < e => (***) > i cant deduce y in B_d_(x,e) in the (***) step. > so, i need your advice. > thank you very much for your advice. Let N_e(x) = {y in X | d(x,y)0, B_L = { N_1/n(x) | x in X, n in N, 1/N Let (X,d) be a metric space. > Define a function d :XxX -> (R+) U {0} by > d(x,y) = min {1, d(x,y)}, the minumum of 1 and d(x,y), > for all x,y in X. > show that > (1) d is a bounded metric for X. > (2) The metric space (X,d) is topologically equivalent to > the bounded metric space (X,d) Exercise: Start with a metric d and f:R -> R, f(x) = 0 iff x = 0, f(x+y) <= f(x) + f(y) Show fd = f o d is a metric. Show when f is ascending and continuous at 0 that f is uniformly continuous and d,fd are equivalent metrics For your problem f(x) = min{ 1,x } and d = fd. > but...in the (2) > let x in X and e>0 > then y in B_d_(x,e) => d(x,y) < e > => d(x,y) <= d(x,y) < e > => y in B_d_(x,e) > y in B_d_(x,e) => d(x,y) < e > =>min {1, d(x,y)} < e > => 1 < e or d(x,y) < e > if d(x,y) < e => y in B_d_(x,e) > else 1 < e => (***) > i cant deduce y in B_d_(x,e) in the (***) step. I think you want to show y in B_d(x, min(1,r)) ==> y in B_d(x,r) Heres from my notes for the exercise: Let F(a,r) = { x | fd(a,x) < r } F(a,f(r)) subset B(a,r). If x in F(a,f(r)): fd(a,x) < f(r) if r <= d(a,x): f(r) <= fd(a,x) < f(r) which cannot be d(a,x) < r; x in B(a,r) some s > 0 with B(a,s) subset F(a,r). Some s > 0 with f(s) < r if x in B(a,s): d(a,x) < s; fd(a,x) <= f(s) < r; x in F(a,r) ---- === Subject: Re: need proof!!! ok, so why is the oppside function exist??? > A partial answer to some of your questions: > The set R-Q is a G_delta, i.e. a countable intersection > of open subsets of R. This is because > R-Q= cap_{qin Q} (R-{q}). > Here the complement of a singleton is obviously open, > and as Q is countable, so is this intersection. Obviously > this generalizes to the statement that the complement > of a countable set is a G_delta. > Q is not a G_delta as explained by William Elliott > (the complement of a G_delta is an F_sigma, i.e. a countable > union of closed sets). > I dont know, whether being a G_delta is a sufficient > condition for a set to be equal to the points of contuinity > of some function. It is relatively easy to see that it is > a necessary condition, though. This is an exercise (together > with some hints) in Roydens book Real Analysis. Im sure > many other first year graduate texts in real analysis have > related material as well. > Jyrki Lahtonen, Turku, Finland but i have one more problem you are based on the fact that The set of points at which f is discontinuous must be an F_sigma set and thats o.k but i only know that: a function f(x) in a single variable x is said to be continuous at point y if lim x-->y f(x) = f(y) can you proof basing on the lim of a function????? === Subject: Re: need proof!!! > but i have one more problem > you are based on the fact that The set of points at which f is > discontinuous must be an F_sigma set and thats o.k > but i only know that: > a function f(x) in a single variable x is said to be continuous at point y if > lim x-->y f(x) = f(y) > can you proof basing on the lim of a function????? I did this as an exercise as a first year graduate student. It was good for me to work it out myself. Im sure its good for you as well. If you get stuck, look at Dennis Mays reply:) Jyrki === Subject: Re: need proof!!! > I dont know, whether being a G_delta is a sufficient > condition for a set to be equal to the points of contuinity > of some function. It is relatively easy to see that it is > a necessary condition, though. This is an exercise (together > with some hints) in Roydens book Real Analysis. Im sure > many other first year graduate texts in real analysis have > related material as well. It is also sufficient. The book Counterexamples in Analysis shows a construction of a function which is discontinuous at precisely the points of an arbitrary F_sigma set. === Subject: Re: need proof!!! > i resently asked the following question > Im looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... > but if you dont know just tell me why Suppose f:R->R is an arbitrary real function. For n in N, n>=1 and y in R let A_n(y) = f^-1[(y-1/n, y+1/n)] U_n(y) = int A_n(y) where int = topological interior U_n = UNION {U_n(y) | y in R} C = INTERSECTION {U_n | n in N, n>=1} If f is continuous at x: x in int(f^-1[O]) whenever O is open and f(x) in O so x in U_n(f(x)) for all n so x in U_n for all n so x in C If x in C: For all n, x in U_n So for each n we can find y_n in R with x in U_n(y_n) So for all n, x in int f^-1[(y_n-1/n, y_n+1/n)] Thus for all n, |f(x)-y_n| < 1/n and also for all n there is d_n>0 such that |x-z| < d_n => |f(z)-y_n| < 1/n Suppose e>0 and pick N such that 1/N < e/2 |x-z| < d_N => |f(z)-f(x)| <= |f(z)-y_N| + |y_N-f(x)| < 1/N + 1/N < e Hence f is continuous at x. So C is precisely the set of points at which f is continuous. Each U_n is open since it is a union of open sets and so C is a G_delta set as required. If Q were a G_delta set, R-Q would be an F_sigma set, so there would be a sequence of closed sets F_0, F_1, ... with R-Q = UNION {F_n | n in N} Each F_i contains no rational and so is nowhere dense. Let (q_n) be an enumeration of Q. Each {q_n} is also a nowhere dense set. Then R = F_0 U F_1 U ... U q_0 U q_1 U ... is a countable union of nowhere dense sets, which contradicts Baires category theorem. Hence Q is not a G_delta set. === Subject: Re: need proof!!! > i resently asked the following question > Im looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... > but if you dont know just tell me why Why do you ask? When youve asked your previous question, Dennis May discontinuous must be an F_sigma set. The irrationals are not an F_sigma set. If thats not a formal proof, then I have no clue about what a formal proof is. Of course, if you had written that you do not understand it, that would have been different. But its a proof. And, please, stop putting those exclamation marks at the subjects of your posts. Jose Carlos Santos === Subject: Re: Subbases, compactness === Subject: Re: Subbases, compactness > Let S be a subbase for a topological space X. Suppose every cover > of X by sets in S has a finite subcollection which covers X. Is > X necessarily compact? >> Alexanders Theorem. >> Requires the Axiom of Choice for the proof. > Rather than a laborious set theory proof, is there a simpler filter proof? > The proof in Kelleys General Topology does not seem so laborious > to me. It is a rather nifty application of Tuckeys Lemma, which > is a version of Zorns lemma. Whats Tuckeys Lemma? Zorns Lemma for P(S) with subset order? > The theorem we use here is more easily found as Alexanders Subbase > Theorem; sometimes you will see Sub-base hyphenated. Well as included below, I did find an EasyProofOfTychonoff http://www.theoryandpractice.org/kyle/Wiki/ EasyProofOfTychonoff covering both Alexs Lemma and Tychonovs Theorem. However for all its clarity, it is lacking a crucial step where it is claimed G = H / S covers X (marked **). Im not even seeing why a maximal cover H of X with no finite subcover, would contain any subbase sets S, much less why all the subbase sets of H would cover X. What steps are needed to show this? In the Tychonov section I corrected a couple of apparent typos ... x_i in X - / G_i for every i. ... and x in X_i - / G_i. and seeing no need, removed the one time occurrence of bar{U} = -- Alexander subbasis lemma If X is a topological space and S is a subbasis of X, then X is compact if and only if every S-cover of X has a finite subcover. Proof: Let B be the collection of all open covers of X that do not have finite subcovers. Assume, by way of contradiction, that B is nonempty. Partial order B by set inclusion. Let C be a chain in B. Let D be the union of all elements of C. If D has a finite subcover, then those finitely many open sets would be contained in finitely many elements of the chain C. Every finite subset of the chain has a maximum element, and so each of those finitely many open sets that together cover X are contained in that maximum element. This is impossible since every element of C is an element of B, and thus has no finite subcover. Thus D has no finite subcover, so is in B, and is an upperbound for C inside B. Since C was an arbitrary chain in B, the conditions of Zorns lemma are satisfied, and thus B has a maximal element, call it H. ** Consider G = H / S. We show that G covers X, so that by hypothesis it has a finite subcover, but that finite subcover is a subcover of H as well. This is a contradiction, and so the assumption that B is nonempty is untenable. Thus every open cover of X has a finite subcover, and X is compact. -- Theorem (Tychonoff): Suppose that I is a set and for every i in I suppose X_i is a compact topological space. Endow X = prod_{i in I} X_i with the product topology. X is compact. Proof of Tychonoff: Let S be the standard subbasis of the product topology, S = { pi_i^{-1}(U) : U is open in X_i }. By Alexanders lemma it suffices to consider S-covers. Let G be an S-cover, that is let G consist of sets pi_i^{-1}(U) for various i in I and U open in X_i. Define G_i = { U : U is open in X_i, pi^{-1}_i(U) in G }. Assume BWOC, that for every i in I that / G_i /= X_i. Using the axiom of choice, choose x_i in X_i - / G_i for every i. Since x = (x_i)_{i in I} is in X, x in pi_i^{-1}(U) in G for some i in I and U open in X_i. This is a contradiction since x in U in G_i and x_i in X_i - / G_i. Therefore there must exist an i in I such that / G_i = X_i. Since X_i is compact, choose a finite subcover, U_1,... U_n. Let C = { pi_i^{-1}(U_1),... pi_i^{-1}(U_n) }. C is a finite subcover of X. Since G was an arbitrary S-cover, the conditions of Alexanders subbase lemma are satisfied, and X is compact. ---- === Subject: Re: JSH: Look at it backwards >snip >There is only one variable Nora Baron . . . Here I agree with Mr. Harris. I once posted that there are evidently a number of real people living in the United States that really are named Nora Baron, but those individuals are constant while our Nora Baron may indeed be the only one that is a variable -- although that property has not yet been established for sure. === Subject: Re: JSH: Look at it backwards >snip >There is only one variable Nora Baron . . . > Here I agree with Mr. Harris. I once posted that there are evidently > a number of real people living in the United States that really are > named Nora Baron, but those individuals are constant while our Nora > Baron may indeed be the only one that is a variable -- although that > property has not yet been established for sure. LOL. Ive been having a good time today. Lots of fun stuff, and some neat accomplishments. And sci.math posters are actually getting a LOT more entertaining with some of their replies. Its a hoot. The real work though is in working through the details of *explaining* my result. Which is an interesting problem, which is also why I need so many posts as I just throw a bunch of ideas at the problem until I find something that sticks. Its basically how I found my math results, brainstorming, but with expository style. I think Im almost done. Ive got that mathematicians ignoring my work are passive-aggressive, and using a passive-aggressive style. I have that the work itself resolves down to acceptance of some VERY BASIC mathematics known for thousands of years. And I have the explanation for why some posters obsessively reply to my posts as theyve learned that simply disagreeing is their most potent tactic. The information can be played out in many different combinations, and as each one plays out, I use various tools to look for impact. Eventually Ill get the right combination. The key is what Im looking for, and Im hot on the hunt now. Its just so, terribly exciting! James Harris === Subject: Re: JSH: Look at it backwards > The key is what Im looking for, and Im hot on the hunt now. Its > just so, terribly exciting! Be sure and keep your zipper up. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: JSH: Without a trace Some of you may have noticed that Im talking about abstractions now at higher and higher levels, which I think should help. For instance, given a multiple of a polynomial it has been well accepted that you can divide that multiple off without leaving a trace. Consider P(x) = 5(x+1)(x+2) = 5x^2 + 15x + 10 and it is true that you can divide 5 from that factorizaton giving (x+1)(x+2) = x^2 + 3x + 2 without leaving a trace. That is, there is no indication left that the polynomial was ever multiplied by 5, as how could there be? Rationally there are an *infinity* of potential multiples that you could use, so why mathematically should a factorization of x^2 + 3x + 2, have traces of one particular multiple, like traces of 5. Now thats obvious with polynomial factors but some sci.mathers clearly have many of you convinced that things change if you have P(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10 and NOW divide the 5 off, as many of you seem convinced that NOW if the as and bs are somehow complex, or weird, or otherwise different from what you get with polynomial then maybe, hmmm, possibly, you know? Maybe there IS a trace, right? But how? If I divide the 5 off, then I have a factorization of x^2 + 3x + 2 just as before, and there is no rational reason to suppose that a trace of 5 is left, as why 5? Why not 7? Or 293874983? Logically, it doesnt matter how complicated that as and bs are in that example, when the 5 is divided off, it goes--without a trace. The situation now where posters argue with me, with arguments that have to boil down to some trace being left by a multiple, so that they can argue that the multiple divides through dependent on some variable, is not unlike an argument between a scientist and people who dont believe in evolution, but worse. In my case I have precedent from thousands of years of mathematics that a multiple can be divided off, absolute logic, and just the plain oddity of the notion that a multiple has to leave a trace when its divided off, but STILL people have argued with me quite successfully, and I figure many of you, despite what I say here, remain unconvinced that Im right. And you are no different than Creationists arguing with scientists against evolution. Or people who dont believe man landed on the moon. You are no different, and in fact worse, as here its mathematics, with absolute proof. You people who cant accept mathematics are no different from those other people who cant accept science. You may think you like or even love mathematics, but you cannot when you refuse to accept even the most basic concepts in mathematics, with a result that clearly you dont like. I understand the *desire* to have certain things be true, but in mathematics that doesnt matter. At the end of the day, you put away your emotions, and go with whats true--if you truly value mathematics. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Without a trace > Some of you may have noticed that Im talking about abstractions now > at higher and higher levels, which I think should help. > For instance, given a multiple of a polynomial it has been well > accepted that you can divide that multiple off without leaving a > trace. > Consider > P(x) = 5(x+1)(x+2) = 5x^2 + 15x + 10 > and it is true that you can divide 5 from that factorizaton giving > (x+1)(x+2) = x^2 + 3x + 2 > without leaving a trace. That is, there is no indication left that > the polynomial was ever multiplied by 5, as how could there be? > Rationally there are an *infinity* of potential multiples that you > could use, so why mathematically should a factorization of x^2 + 3x + > 2, have traces of one particular multiple, like traces of 5. > Now thats obvious with polynomial factors but some sci.mathers > clearly have many of you convinced that things change if you have > P(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10 > and NOW divide the 5 off, as many of you seem convinced that NOW if > the as and bs are somehow complex, or weird, or otherwise different > from what you get with polynomial then maybe, hmmm, possibly, you > know? Maybe there IS a trace, right? > But how? > If I divide the 5 off, then I have a factorization of > x^2 + 3x + 2 > just as before, and there is no rational reason to suppose that a > trace of 5 is left, as why 5? Why not 7? Or 293874983? > Logically, it doesnt matter how complicated that as and bs are in > that example, when the 5 is divided off, it goes--without a trace. > The situation now where posters argue with me, with arguments that > have to boil down to some trace being left by a multiple, so that they > can argue that the multiple divides through dependent on some > variable, is not unlike an argument between a scientist and people who > dont believe in evolution, but worse. Here yet again, lacking a proof, you try to convince people with an oversimplified toy example. Examples are not proofs. As usual, your example is a reducible polynomial. As you well know, at the heart of this controversy is the difference between reducible and irreducible polynomials. So consider an *irreducible* quadratic: a little simpler than your cubic example, but more complicated than the quadratic example you just gave. Q(x) = 3 * (4*x^2 + 2*x - 1). = 3 * (4*x^2 + 2*(x - 2) + 3). Consider factoring this in the form Q(x) = (2*a1(x) + 3)*(2*a2(x) + 3). I will call this a Harristotelian factorization: that is, you are factoring as if 2 is a polynomial variable. The as satisfy the equation a^2 - (x - 2)*a + 3 x^2 = 0. Note that the roots of this equation are always algebraic integers. For example, when x = 0, the as are a = 0 and a = -2. This gives the factorization Q(0) = 3 * (-4 + 3) = -3. The obvious way to factor 3 out of both sides of this is thus: Q(0)/3 = (2*0 + 3)*(2*(-2) + 3)/3 = (0/3 + 3/3)*(-4 + 3) = -1. That is, a1(0) = 0 and is divisible by 3, and a2(0) = -1 and is NOT divisible by 3. Now consider Q(1). In this case, the as satisfy the equation a^2 + a + 3 = 0. This is irreducible. The two roots are a_1(1) = (-1 + sqrt(-11))/2 and a_2(1) = (-1 - sqrt(-11))/2. You can easily check that a_1(1) * a_2(1) = 3, as it should. Therefore both a_1(1) and a_2(1) are not coprime to 3. Moreover, neither a_1(1) nor a_2(1) are divisible by 3. You can easily check that too. So how should Q(1) be factored? Recall from above that Q(1) = (2 a_1(1) + 3)*(2 a_2(1) + 3). Now I will divide both sides by 3. But I am NOT going to divide the first term by 3 and the second term by 1. That will not work, because 2 a_1(1)/3 is not an algebraic integer. Instead I am going to divide the first term by w_1(1) = a_1(1) and the second term by w_2(1) = a_2(1) ! Does this work? Lets see --- The first term becomes (2 a_1(1) + 3)/a_1(1) = (2 + a_2(1)). The second term becomes (2 a_2(1) + 3)/a_2(1) = (2 + a_1(1)). Both of these are algebraic integers! Thats good! But I know from the definition of Q(x) that Q(1) = 15, that is, Q(1)/3 = 5. So, is it true that (2 + a_2(1)) * (2 + a_1(1)) = 5? Check it out! It works! Everything comes out right! So whats the moral of all this? The moral is, contrary to what Harris says, the RIGHT way to divide both sides of such equations, so that you get algebraic integer factors after the division, has to be dependent on x . When x = 0, you divide the first and second terms respectively by 3 and 1. When x = 1, you divide the first and second terms respectively by w_1(1) = a_1(1) and w_2(1) = a_2(1), that is, by two DIFFERENT factors of 3. All the terms after the division are algebraic integers. Finally: no, examples are not proofs. Examples are disproofs. [non-math blather deleted] Nora B. === Subject: Re: JSH: Without a trace > Some of you may have noticed that Im talking about abstractions now > at higher and higher levels, which I think should help. > For instance, given a multiple of a polynomial it has been well > accepted that you can divide that multiple off without leaving a > trace. > Consider > P(x) = 5(x+1)(x+2) = 5x^2 + 15x + 10 > and it is true that you can divide 5 from that factorizaton giving > (x+1)(x+2) = x^2 + 3x + 2 > without leaving a trace. That is, there is no indication left that > the polynomial was ever multiplied by 5, as how could there be? > Rationally there are an *infinity* of potential multiples that you > could use, so why mathematically should a factorization of x^2 + 3x + > 2, have traces of one particular multiple, like traces of 5. Without a trace has no precise meaning, (and not much meaning in any sense). You are retreating into not even wrong territory. However, something that has disappeared without a trace is any mention of the constant term. This makes perfect sense. Each time you brought up the constant term, everyone would ask: isnt the constant term of (a(x)/w(x) + 7/w(x)) equal to 7? Unfortunately for you, you could not answer this question without aknowleging that your entire argument was fallacious. And too many people were asking the question for you to comfortably ignore it. So you could either admit you were wrong or retreat into not even wrong territory. No one is surprised by your choice. - William Hughes > Now thats obvious with polynomial factors but some sci.mathers > clearly have many of you convinced that things change if you have > P(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10 > and NOW divide the 5 off, as many of you seem convinced that NOW if > the as and bs are somehow complex, or weird, or otherwise different > from what you get with polynomial then maybe, hmmm, possibly, you > know? Maybe there IS a trace, right? > But how? > If I divide the 5 off, then I have a factorization of > x^2 + 3x + 2 > just as before, and there is no rational reason to suppose that a > trace of 5 is left, as why 5? Why not 7? Or 293874983? > Logically, it doesnt matter how complicated that as and bs are in > that example, when the 5 is divided off, it goes--without a trace. > The situation now where posters argue with me, with arguments that > have to boil down to some trace being left by a multiple, so that they > can argue that the multiple divides through dependent on some > variable, is not unlike an argument between a scientist and people who > dont believe in evolution, but worse. > In my case I have precedent from thousands of years of mathematics > that a multiple can be divided off, absolute logic, and just the plain > oddity of the notion that a multiple has to leave a trace when its > divided off, but STILL people have argued with me quite successfully, > and I figure many of you, despite what I say here, remain unconvinced > that Im right. > And you are no different than Creationists arguing with scientists > against evolution. Or people who dont believe man landed on the > moon. > You are no different, and in fact worse, as here its mathematics, > with absolute proof. > You people who cant accept mathematics are no different from those > other people who cant accept science. You may think you like or even > love mathematics, but you cannot when you refuse to accept even the > most basic concepts in mathematics, with a result that clearly you > dont like. > I understand the *desire* to have certain things be true, but in > mathematics that doesnt matter. At the end of the day, you put away > your emotions, and go with whats true--if you truly value > mathematics. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Without a trace > Some of you may have noticed that Im talking about abstractions now > at higher and higher levels, which I think should help. Im begging for enlightnment. > For instance, given a multiple of a polynomial it has been well > accepted that you can divide that multiple off without leaving a > trace. You are obviously referring to those wicked but rational criminal rings who go about stripping integer polynomials of their factors leaving no finger prints, DNA or other forensic evidence. Its always a shock when your polynomials are tampered with in this way. However, more brutal irrationals tend to leave polys hideously mutilated and because of the resistance of the victim the perpetrators blood and other bodily ßuids can be left at the scene. Lets hope the forces of law and order can at least capture these irrational miscreants. === Subject: Re: JSH: Without a trace > Some of you may have noticed that Im talking about abstractions now > at higher and higher levels, which I think should help. > For instance, given a multiple of a polynomial it has been well > accepted that you can divide that multiple off without leaving a > trace. Cant we just agree on this one bit of terminology? If you consider the number 12, say ... 3 is a FACTOR of 12, and 24 is a MULTIPLE of 12. Surely its not against your principles to use that bit of correct terminology? === Subject: Re: JSH: Without a trace posting-account=KR2cuw0AAACZ_86pfubjOKsQkAVb6Rpe If you really understood mathematics, youd see where your errors are. Youre just too pompous to see that. Dave === Subject: Re: JSH: Without a trace > If you really understood mathematics, youd see where your errors are. > Youre just too pompous to see that. > Dave Yet Ive made quite a few mistakes over the years trying out different ideas, and dropping those that failed. Here posters need only do one thing: prove me wrong. And its a sad lie for some of them to just repeat that they have or that I dont answer objections as Ive answered objections many, many, many times over a period of years. These people simpy will not listen to reason. What do they appear to accomplish? My guess is that leading mathematicians who are aware of my work, pay attention to them, as evidence that they can rely on the passive-aggessive strategy of waiting. So then sci.math may actually be having an impact after all. You may be convincing certain people who do know that my work is correct they have the luxury of waiting. James Harris === Subject: Re: JSH: Without a trace > Here posters need only do one thing: prove me wrong. > And its a sad lie for some of them to just repeat that they have or > that I dont answer objections as Ive answered objections many, many, > many times over a period of years. Yes, but your answers are consistently off base. First, you misrepresent the objections; second, you ignore the counter-examples or disproofs entirely; third, you repeat your own previously ßawed argument. In your mind, that disposes of the matter. Everyone else, however, sees that you have *not* answered the objection, but only substituted a loud noise. Of all the p-baked cranks, where Ôp equals 1/2, you take the q, where Ôq equals cake. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Unstoppable Force vs Immovable Object > Whats that supposed to mean, I address a physical concept in a > non-physical way? > I conclude that its logically impossible for an unstoppable force to > meet an immovable object. Do you disagree? Do you think its logically > possible? If you consider what the word force means you would conclude there is no unstopable (i.e. infinite) force in the physical universe. Likewise, an immovable object would have infinite momentum (physically impossible). In either case you are talking about things which not only dont exist, but cant exist. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object Whats that supposed to mean, I address a physical concept in a > non-physical way? I conclude that its logically impossible for an unstoppable force to > meet an immovable object. Do you disagree? Do you think its logically > possible? > If you consider what the word force means you would conclude there is no > unstopable (i.e. infinite) force in the physical universe. Likewise, an > immovable object would have infinite momentum (physically impossible). > In either case you are talking about things which not only dont exist, > but cant exist. > Bob Kolker So you conclude that its logically impossible for an unstoppable force or an immovable object to exist. Im not sure thats correct, but in any event its consistent with my conclusion. So whats to argue about? === Subject: Clothoid construction A Clothoid (Euler/Cornus spiral) has a constant arc rate of curvature change. How can the rate be determined either geometrically or by calculation when four points on it are given ? === Subject: Re: Clothoid construction ETAuAhUAsogYrSiC/xJdV6QhrdX0vKr4H14CFQCRNGGy/KyR0A6NC8FHL+ QGeH3C6Q== Well ... The curvature k isgiven by |d^2y/dx^2|/(ds/dx)^(3/2). Let k = cs, then c^2 s^2 = (d^2y/dx^2)^2/(ds/dx)^3 Differentiate this twice w.r.t. x, then combine the given equation with the first and second derivatives to eliminate c and s. Plug in ds = sqrt(dx^2+dy^2) and good luck solving the resulting fourth-order nonlinear diffeential equation. --OL === Subject: Cauchys Resume ... text? Would someone be kind enough to provide the French original and/or English translation of a portion of Cauchys text described below? Either email or posting to this newsgroup is OK (though email preferred, as this would not be of interest to the public). The text in question is: A.L. Cauchy: Resume des lecons donnees a lEcole Polytechnique sur le calcul infinitesimal (1823) lecture 37 (topic is Taylor-Maclaurin theorem) The first paragraph, where the definition of infinite series conversion is given. More specifically, the first sentence where (seemingly) the term series is defined. I ask this because the Japanese translation I have at hand is a bit obscure and awkward for this sentence (possibly a mistranslation), and am wondering what the original text says. - Yuzuru Hiraga (Univ. of Tsukuba, Japan) === Subject: Re: Cauchys Resume ... text? hiraga@slis.tsukuba.ac.jp (Yuzuru Hiraga) dixit: >Would someone be kind enough to provide the French original and/or >English translation of a portion of Cauchys text described below? >Either email or posting to this newsgroup is OK (though email preferred, >as this would not be of interest to the public). >The text in question is: >A.L. Cauchy: Resume des lecons donnees a lEcole Polytechnique > sur le calcul infinitesimal (1823) >lecture 37 (topic is Taylor-Maclaurin theorem) >The first paragraph, where the definition of infinite series conversion >is given. More specifically, the first sentence where (seemingly) >the term series is defined. >I ask this because the Japanese translation I have at hand >is a bit obscure and awkward for this sentence (possibly a >mistranslation), and am wondering what the original text says. >- Yuzuru Hiraga (Univ. of Tsukuba, Japan) Maybe you could look at http://gallica.bnf.fr/ click on recherche then in auteur field enter cauchy and click on the rechercher button, the result lists a lot of Oeuvres compl.8ftes dAugustin Cauchy but youll have to look at all the tables of contents to see if what you want is in there. === Subject: JSH: Fool all of the people, all of the time? So I can show that popular objections on sci.math boil down to weird things like arguing that dividing a multiple of a polynomial can occur as a function of some variable. Or explain over and over again that at its heart my work work depends on previously uncontroversial truths like the distributive property, and w/7 = 1, meaning that w=7. Ive mentioned before that in my contacts with leading mathematicians, and even math students like that Cornell math grad student, I dont get the objections that play out month after month on these newsgroups. A properly trained mathematician is just NOT going to seriously consider that a multiple of a polynomial leaves some trace after its divided off. A properly trained mathematician will not question the distributive property in a commutative ring. So, if I can contact leading mathematicians, and they dont give the same objections, why am I arguing on Usenet about what I claim is a very basic argument that is also one of the greatest discoveries in math history? Wouldnt some of these leading mathematicians say something? Why arent they giving press releases? How can business as usual continue in the math world? I now think part of it is you. After all, Ive argued on Usenet for years, and been VERY WRONG for years, with arguments that turned out to be false or wrong. It seems reasonable to conclude from a slanted perspective that it will be very difficult for me to get people to believe what I say now. And that has been true. Month after month Ive explained, and events have occurred like that weirdness with the APF paper, and youve absorbed it all. Its becoming increasingly clear that hey, maybe no one will EVER react, and business can continue as usual, professors can teach work that they probably know now is ßawed, and their world can continue with little change. Why would they do it? Good question. But to believe that my results havent traveled through mathematical society at this point you have to believe that a very basic argument, which I know I can explain in about an hour as I did it in-person at my alma mater Vanderbilt University, which shoots down one of the underpinnings of algebraic number theory could just ßoat out there, be argued about by me on Usenet for years, and never get heard of by leading mathematicians. Oh, but wait, Ive actually contacted leading mathematicians. I even got some commentary on the APF paper from Barry Mazur. So, you have to believe that those people are remarkably dense. I say theyre not dense, theyre media savvy. They watch you, and see me posting all the time, with people arguing with me with really dumb stuff, and they dont get feedback indicating that its not ok, and it seems easier to just go with the way things are. As a backup they have that I have a paper at a leading journal, so they can let the weight sit on that journal, and claim later that with such a controversial result, and such a controversial figure, it seemed prudent to wait. But, you have to believe that theres some question, when there is not. Ill tell you one of the weirdest things I face when dealing with mathematicians outside of Usenet, when Im sure they get it, is that they tend to walk away. Its bizarre. Its like some door just closes. After the explanations are done, any questions or concerns are handled, they quit talking to me. Then again, the mathematical research I have shatters their world in many ways though as I mentioned yesterday, its not really a loss. Its a gain. For the true mathematician, whats actually correct is whats valuable. People make mistakes. Better to live today and learn the truth, than to be one of those poor saps who died deluded, thinking they knew certain things that they just didnt. Thinking they had proofs that they didnt. Mathematicians today are lucky. They need to start counting their blessings. As make no mistake, you CANNOT make fools of all of the people, all of the time. Eventually the silliness of challenging mathematics at its base will play out. People will eventually get tired of hearing from posters who get caught in dumb lies, and there will be people who will ask more and more questions. And make no mistake, the benefit to the passive-aggessive strategy of sitting tight and waiting, maybe hoping that people will be fooled indefinitely, will not be lost on people. Many people may be intimidated by mathematics, but they understand money and power very well. So, yeah, I think Usenet has something to do with the continued foot-dragging, this passive-aggressive strategy of just keeping quiet, and, you know I also think it has to do with a contempt for the public that you can see at times traveling through intellectual circles. Maybe they just think that people arent being made fools of, but that they are fools, incapable of ever realizing the truth on their own. James Harris === Subject: Re: JSH: Fool all of the people, all of the time? > Ill tell you one of the weirdest things I face when dealing with > mathematicians outside of Usenet, when Im sure they get it, is that > they tend to walk away. Thats because what they get is that *you* are *never* going to get it. They walk away because theyve realized that reasoning with you is a hopeless task. > Its bizarre. Its like some door just closes. After the > explanations are done, any questions or concerns are handled, they > quit talking to me. Because, unlike many of us here in sci.math, they have neither the time nor the inclination to indulge an idiot with the attention he craves. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Fool all of the people, all of the time? > Good question. But to believe that my results havent traveled > through mathematical society at this point you have to believe that a > very basic argument, which I know I can explain in about an hour as I > did it in-person at my alma mater Vanderbilt University, which shoots > down one of the underpinnings of algebraic number theory could just > ßoat out there, be argued about by me on Usenet for years, and never > get heard of by leading mathematicians. Impressive bit of sentence construction. Incidentally, though: argued about by me on Usenet for years - which bit _is_ this? I know you seem to have been claiming to have found errors in Ôcore for years, but I thought they were different arguments. Since as you say yourself, youve been wrong a lot in the past, and the important bit is the current argument, which alone of course is Correct, how long has this bit been going? Id have thought less than years...? > ... ill play out. People will eventually get tired of hearing from > posters who get caught in dumb lies, and there will be people who will > ask more and more questions. Newsßash!! People are asking questions!! Heres the commonest one: What does Ôproperly a unit mean? Only you can answer. Brian Chandler http://imaginatorium.org === Subject: Re: JSH: Fool all of the people, all of the time? > Good question. But to believe that my results havent traveled > through mathematical society at this point you have to believe that a > very basic argument, which I know I can explain in about an hour as I > did it in-person at my alma mater Vanderbilt University, which shoots > down one of the underpinnings of algebraic number theory could just > ßoat out there, be argued about by me on Usenet for years, and never > get heard of by leading mathematicians. > Impressive bit of sentence construction. Incidentally, though: argued > about by me on Usenet for years - which bit _is_ this? I know you > seem to have been claiming to have found errors in Ôcore for years, > but I thought they were different arguments. Since as you say > yourself, youve been wrong a lot in the past, and the important bit > is the current argument, which alone of course is Correct, how long > has this bit been going? Id have thought less than years...? The full timeline is that back in December 1999 I first discovered an approach which would lead to the analysis tool of non-polynomial factorization. It took a while though to figure out what I was doing as I stumbled abot with various proofs of FLT that turned out to be bogus, but I kept working at it despite the failures, and time passed. I think it was around two years later when I finally was fully using non-polynomial factorization in my FLT research, as I definitely was by November 2001. I focused on non-polynomial factorization itself, in discussions around May 2002, when I also discovered my prime counting function. Its actually neat in that I discovered the prime counting function from trying to explain my FLT work! I was talking about simple polynomials, and I started thinking that hey, this could have something to do with counting prime numbers! It took me about two weeks from there to find the prime counting function. I took some time talking about the prime counting function, and fiddling with it, and at times would also talk about non-polynomial factorization, and eventually I kind of settled into working on either one or the other up until recently. Arguments on sci.math about non-polynomial factorization probably go back to before November 2001. Over time Ive pushed down quite a few of the early objections, which might surprise many of you who have a snapshot view, and dont realize how objections have changed over time. The earliest arguments in this area go back to December 1999. Its some of my most well-worked research, with almost every piece of it argued out over a period of years. Which is also why some of you may see me talking about having explained to various posters MANY times before, as I have. > ... ill play out. People will eventually get tired of hearing from > posters who get caught in dumb lies, and there will be people who will > ask more and more questions. > Newsßash!! People are asking questions!! Heres the commonest one: > What does Ôproperly a unit mean? > Only you can answer. Thats an old game of trying to cause major arguments over the use of some term or other, as I now simply shift from usage that sci.mathers find easily works to provoke confusion. What I find fascinating is that *clearly* some of you have worked rather hard to confuse the issue, hide the reality, and fight for arguments that just dont work, when I know that I learned years ago that its just futile in mathematics to do those things. And now some of you may finally be learning why, as I simply adjust explanations, and soon enough people will realize that you had to understand how it all worked to confuse them so well, and then they probably wont appreciate your efforts. After all, mathematics is objective in many ways. Sure some of you have personalized it, so that you can attack it as if it were mine. But its like if you hated Pythagoras and went after the Pythagorean Theorem. Youre not doing the world any favors, and fighting a battle you will lose. These things have happened before, history repeats itself, and for some reason there are people like some of you who step out to fight what is mathematically true, for personal reasons. James Harris === Subject: Re: JSH: Fool all of the people, all of the time? > Good question. But to believe that my results havent traveled > through mathematical society at this point you have to believe that a > very basic argument, which I know I can explain in about an hour as I > did it in-person at my alma mater Vanderbilt University, which shoots > down one of the underpinnings of algebraic number theory could just > ßoat out there, be argued about by me on Usenet for years, and never > get heard of by leading mathematicians. Impressive bit of sentence construction. Incidentally, though: argued > about by me on Usenet for years - which bit _is_ this? I know you > seem to have been claiming to have found errors in Ôcore for years, > but I thought they were different arguments. Since as you say > yourself, youve been wrong a lot in the past, and the important bit > is the current argument, which alone of course is Correct, how long > has this bit been going? Id have thought less than years...? The full timeline is that back in December 1999 I first discovered an > approach which would lead to the analysis tool of non-polynomial > factorization. Newsßash!! People are asking questions!! Heres the commonest one: What does Ôproperly a unit mean? Only you can answer. Thats an old game of trying to cause major arguments over the use of > some term or other, as I now simply shift from usage that sci.mathers > find easily works to provoke confusion. Sorry, Im a bit lost here. Youre accusing me of playing a game, just because I asked what a bit of your argument means? I dont understand what you mean when you claim that while something-or-other is not a unit in the ring of algebraic integers, it _is_ properly a unit. I guess it means something to you, but I surmise it means nothing to anyone else. (That means you are speaking a private language, and the closest to fame you can hope for is the status of a Voynich manuscript in a half-millennium or two.) I find all this stuff about peering into polynomials looking for Ôfactors a bit confusing, too. I mean, of course in (a) below, theres a factor of 4 that can be divided off: (a): 4x^2 + 4x - 4 So if n is an integer - any integer - I know that 4n^2 + 4n -4 is a multiple of 4. However: (b): (x+1)(x+2) Does this polynomial have a factor of 2 in it? I can see a Ô2, but it doesnt look like a factor. And again I know that any integer n means that (n+1)(n+2) is a multiple of 2. So whats going on? Is there any difference from the case of (a). You seem to reject the idea that theres really any difference between talking about polynomials *as* polynomials and evaluations of polynomials, because thats voodoo math - have I got that right? And what ever should I think about (c) - absolutely no trace of a Ô3 anywhere: (c): x(x+1)(x+2) Another thing I wonder about: OK, suppose the dam bursts. Suppose suddenly you are on tv shows and whatnot. The mathematicians are all disgraced, ßung into debtors jails. Are you going to rewrite all the text books? Who will do it? FWIW, I have a copy of Hersteins Topics in [pre-Harrisian] Algebra here, and it only mentions algebraic integers very brießy, in the problems. Theres the usual definition, then the first problem reads: 10. If _a_ is any algebraic number, prove that there is a positive integer _n_ such that _na_ is an algebraic number. (I can type more if required.) Well, is Hersteins solution to this problem wrong? Brian Chandler http://imaginatorium.org === Subject: Re: JSH: Fool all of the people, all of the time? > Good question. But to believe that my results havent traveled > through mathematical society at this point you have to believe that a > very basic argument, which I know I can explain in about an hour as I > did it in-person at my alma mater Vanderbilt University, which shoots > down one of the underpinnings of algebraic number theory could just > ßoat out there, be argued about by me on Usenet for years, and never > get heard of by leading mathematicians. Impressive bit of sentence construction. Incidentally, though: argued > about by me on Usenet for years - which bit _is_ this? I know you > seem to have been claiming to have found errors in Ôcore for years, > but I thought they were different arguments. Since as you say > yourself, youve been wrong a lot in the past, and the important bit > is the current argument, which alone of course is Correct, how long > has this bit been going? Id have thought less than years...? The full timeline is that back in December 1999 I first discovered an > approach which would lead to the analysis tool of non-polynomial > factorization. > OK - Ill give you your arguing for years. So you will accept what is true. How wonderful for you. > Newsßash!! People are asking questions!! Heres the commonest one: What does Ôproperly a unit mean? Only you can answer. > Thats an old game of trying to cause major arguments over the use of > some term or other, as I now simply shift from usage that sci.mathers > find easily works to provoke confusion. > Sorry, Im a bit lost here. Youre accusing me of playing a game, > just because I asked what a bit of your argument means? I dont > understand what you mean when you claim that while something-or-other > is not a unit in the ring of algebraic integers, it _is_ properly a > unit. I guess it means something to you, but I surmise it means > nothing to anyone else. (That means you are speaking a private > language, and the closest to fame you can hope for is the status of a > Voynich manuscript in a half-millennium or two.) You take a phrase out of context, and make a big deal out of it, asking for some explanation, when in context when used the explanation was in what I said. Its an old tactic that sci.mathers have used for years. Pick a phrase, take it out of context, make a big deal out of it. And, you know? Lots of times I HAVE used certain phrase differently from standard usage, and misused terms I didnt understand or simply just didnt care a lot to get exactly right in informal discussions. Time after time, posters have reacted as if these posts are such important communications that perfection is a requirement. I reserve the right to at times just babble. Its not a big deal. This is Usenet. > I find all this stuff about peering into polynomials looking for > Ôfactors a bit confusing, too. I mean, of course in (a) below, > theres a factor of 4 that can be divided off: > (a): 4x^2 + 4x - 4 > So if n is an integer - any integer - I know that 4n^2 + 4n -4 is a > multiple of 4. However: > (b): (x+1)(x+2) > Does this polynomial have a factor of 2 in it? I can see a Ô2, but it > doesnt look like a factor. And again I know that any integer n means > that (n+1)(n+2) is a multiple of 2. So whats going on? Is there any > difference from the case of (a). You seem to reject the idea that > theres really any difference between talking about polynomials *as* > polynomials and evaluations of polynomials, because thats voodoo > math - have I got that right? Well heres a good chance to show how sci.mathers routinely try to mislead. With integer you have the any integer either is even or has a residue of 1. That is, given an integer x, either x = 0 mod 2, or x = 1 mod 2. So trivially you can put up something like (x+1)(x+2) and know it must be even. However, in the ring of algebraic integers, no such relations exist. That is, there is no non-unit in the ring of algebraic integers such that EVERY algebraic integer is either divisible by that number or has the same residue. You dont even have the even case with algebraic integers as in the ring of algebraic integers it is NOT true that (x+1)(x+2) must have 2 as a factor. Now the example (x+1)(x+2) in the ring of integers is childish mathematics at the most basic level, but posters have routinely used that example for years to try and claim that it shows how my research can be wrong. You people dont even try hard. > And what ever should I think about (c) - absolutely no trace of a Ô3 > anywhere: > (c): x(x+1)(x+2) Yet another childish example based on any *integer* x, either being divisible by 3 or having a residue of 1 or 2. That childish game can be played on and on with integers, but it has no meaning outside of the ring of integers, and it doesnt invalidate my research findings. > Another thing I wonder about: OK, suppose the dam bursts. Suppose > suddenly you are on tv shows and whatnot. The mathematicians are all > disgraced, ßung into debtors jails. Are you going to rewrite all the > text books? Who will do it? Mathematicians will NOT all be disgraced and ßung into jail. If there are some who end up prosecuted itd most likely be a choice few. And hey, not all mathematicians work in the area of algebraic number theory. Number theorists are the ones who are really going to take the hit. How might a prosecution work? Well, consider a federal prosecution that considers whether or not some number theorist knowingly continued both to teach what Id shown to be false, and to receive federal funds for their research. So theyd have tax dollars for two things: teaching and research The charge would be fraud. Evidence might include emails of that mathematician and conversations had with colleagues who would be pulled into grand juries and later into court to testify under oath. Likely punishment? A fine. Blockage from receipt of any more federal funds. Censure from their university, and probably removal from teaching position. > FWIW, I have a copy of Hersteins Topics in [pre-Harrisian] Algebra > here, and it only mentions algebraic integers very brießy, in the > problems. Theres the usual definition, then the first problem reads: > 10. If _a_ is any algebraic number, prove that there is a positive > integer _n_ such that _na_ is an algebraic number. > (I can type more if required.) Well, is Hersteins solution to this > problem wrong? Amazingly enough, most people actually care about whats TRUE. More than likely there will be a ßood of mathematicians into algebraic number theory, as it will be an opened up field with major opportunities for advancement. Think about it. In many areas a young mathematician can work for years and get nowhere. In algebraic number theory, they will have the opportunity to be a significant figure in the remaking of an entire discipline. When that finally gets out, it will be the hottest field in mathematics. James Harris === Subject: Re: JSH: Fool all of the people, all of the time? posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn > Good question. But to believe that my results havent traveled >> through mathematical society at this point you have to believe that a >> very basic argument, which I know I can explain in about an hour as I >> did it in-person at my alma mater Vanderbilt University, which shoots >> down one of the underpinnings of algebraic number theory could just >> ßoat out there, be argued about by me on Usenet for years, and never >> get heard of by leading mathematicians. >> Impressive bit of sentence construction. Incidentally, though: argued >> about by me on Usenet for years - which bit _is_ this? I know you >> seem to have been claiming to have found errors in Ôcore for years, >> but I thought they were different arguments. Since as you say >> yourself, youve been wrong a lot in the past, and the important bit >> is the current argument, which alone of course is Correct, how long >> has this bit been going? Id have thought less than years...? >> The full timeline is that back in December 1999 I first discovered an >> approach which would lead to the analysis tool of non-polynomial >> factorization. >> The earliest arguments in this area go back to December 1999. >> OK - Ill give you your arguing for years. >So you will accept what is true. How wonderful for you. >> Newsßash!! People are asking questions!! Heres the commonest one: >> What does Ôproperly a unit mean? >> Only you can answer. >> Thats an old game of trying to cause major arguments over the use of >> some term or other, as I now simply shift from usage that sci.mathers >> find easily works to provoke confusion. >> Sorry, Im a bit lost here. Youre accusing me of playing a game, >> just because I asked what a bit of your argument means? I dont >> understand what you mean when you claim that while something-or-other >> is not a unit in the ring of algebraic integers, it _is_ properly a >> unit. I guess it means something to you, but I surmise it means >> nothing to anyone else. (That means you are speaking a private >> language, and the closest to fame you can hope for is the status of a >> Voynich manuscript in a half-millennium or two.) >You take a phrase out of context, and make a big deal out of it, >asking for some explanation, when in context when used the explanation >was in what I said. When real mathematicians use a new term, they define it. Period. No dependency on Ôcontext. >Its an old tactic that sci.mathers have used for years. The tactic here is evasion. >Pick a phrase, take it out of context, make a big deal out of it. >And, you know? Lots of times I HAVE used certain phrase differently >from standard usage, and misused terms I didnt understand or simply >just didnt care a lot to get exactly right in informal discussions. Mathematics, unlike sociology, has to be precise. >Time after time, posters have reacted as if these posts are such >important communications that perfection is a requirement. There is no reason in mathematical discourse not to be explicit and precise. >I reserve the right to at times just babble. Its not a big deal. Then in return dont expect anyone to believe you. >This is Usenet. So? Are you trying to do math or not? It doesnt matter where you do it. >> I find all this stuff about peering into polynomials looking for >> Ôfactors a bit confusing, too. I mean, of course in (a) below, >> theres a factor of 4 that can be divided off: >> (a): 4x^2 + 4x - 4 >> So if n is an integer - any integer - I know that 4n^2 + 4n -4 is a >> multiple of 4. However: >> (b): (x+1)(x+2) >> Does this polynomial have a factor of 2 in it? I can see a Ô2, but it >> doesnt look like a factor. And again I know that any integer n means >> that (n+1)(n+2) is a multiple of 2. So whats going on? Is there any >> difference from the case of (a). You seem to reject the idea that >> theres really any difference between talking about polynomials *as* >> polynomials and evaluations of polynomials, because thats voodoo >> math - have I got that right? >Well heres a good chance to show how sci.mathers routinely try to >mislead. >With integer you have the any integer either is even or has a residue >of 1. >That is, given an integer x, either x = 0 mod 2, or x = 1 mod 2. >So trivially you can put up something like (x+1)(x+2) and know it must >be even. >However, in the ring of algebraic integers, no such relations exist. >That is, there is no non-unit in the ring of algebraic integers such >that EVERY algebraic integer is either divisible by that number or has >the same residue. So do you think, given (x + 1)(x + 2), where x is an algebraic integer, that its divisibility or coprimeness to 2 does not depend on x ? If it *is* divisible by 2, does 2 always factor out of it in the same way? >You dont even have the even case with algebraic integers as in the >ring of algebraic integers it is NOT true that (x+1)(x+2) must have 2 >as a factor. I would like to see you prove that. Can you? But in any case you are missing the point. (x + 1)*(x + 2) may be coprime or noncoprime to 2 for various values of x. But there can be no doubt that its coprimeness or noncoprimeness is a FUNCTION of x. And that is really the point that is relevant to your factorization of P(x) and P(x)/49, isnt it? How factors of 49 might divide out of a product of functions of x ? Does it necessarily always happen in the same way, or is it dependent on x ? >Now the example (x+1)(x+2) in the ring of integers is childish >mathematics at the most basic level, but posters have routinely used >that example for years to try and claim that it shows how my research >can be wrong. Its meant as an analogy, since you do not seem to be able to understand the real thing. >You people dont even try hard. >> And what ever should I think about (c) - absolutely no trace of a Ô3 >> anywhere: >> (c): x(x+1)(x+2) >Yet another childish example based on any *integer* x, either being >divisible by 3 or having a residue of 1 or 2. >That childish game can be played on and on with integers, but it has >no meaning outside of the ring of integers, and it doesnt invalidate >my research findings. >> Another thing I wonder about: OK, suppose the dam bursts. Suppose >> suddenly you are on tv shows and whatnot. The mathematicians are all >> disgraced, ßung into debtors jails. Are you going to rewrite all the >> text books? Who will do it? >Mathematicians will NOT all be disgraced and ßung into jail. >If there are some who end up prosecuted itd most likely be a choice >few. >And hey, not all mathematicians work in the area of algebraic number >theory. Number theorists are the ones who are really going to take >the hit. >How might a prosecution work? >Well, consider a federal prosecution that considers whether or not >some number theorist knowingly continued both to teach what Id shown >to be false, and to receive federal funds for their research. Are people jailed now for teaching false things? You are aware probably that some people still teach Bohrs model of the hydrogen atom. Are they being prosecuted? >So theyd have tax dollars for two things: teaching and research >The charge would be fraud. Evidence might include emails of that >mathematician and conversations had with colleagues who would be >pulled into grand juries and later into court to testify under oath. >Likely punishment? A fine. Blockage from receipt of any more federal >funds. Censure from their university, and probably removal from >teaching position. Might the blade cut both ways? You have tried very hard to convince large numbers of people that your proof is right, even in the face of counterexamples which you have not refuted. Prosecuting you for fraud would be trivial. There would be literally thousands of expert witnesses, and none of them for the defense. You have repeatedly accused people here of lying simply because they post arguments saying your proof is wrong. You have accused people of committing fraud. If you are wrong (AND YOU ARE!), YOU could be sued for libel. In fact you have previously accused people of lying about many of your former arguments, and then later you were forced to admit they were right. RIGHT NOW, your own postings which include (1) the accusations of lying and (2) the later admission that those you were accusing were correct, could be used against you in a lawsuit for libel. Who do you think would win? >> FWIW, I have a copy of Hersteins Topics in [pre-Harrisian] Algebra >> here, and it only mentions algebraic integers very brießy, in the >> problems. Theres the usual definition, then the first problem reads: >> 10. If _a_ is any algebraic number, prove that there is a positive >> integer _n_ such that _na_ is an algebraic number. Error there: last phrase should be Ôalgebraic integer. >> (I can type more if required.) Well, is Hersteins solution to this >> problem wrong? >Amazingly enough, most people actually care about whats TRUE. Which in this very case you overlooked entirely with your self-serving non-answer. >More than likely there will be a ßood of mathematicians into >algebraic number theory, as it will be an opened up field with major >opportunities for advancement. Delusion. >Think about it. In many areas a young mathematician can work for >years and get nowhere. In algebraic number theory, they will have the >opportunity to be a significant figure in the remaking of an entire >discipline. >When that finally gets out, it will be the hottest field in >mathematics. Hey, my offer of a bet of $100 that the Annals of Mathematics will reject your paper Advanced Polynomial Factorization still stands. If they publish it, I send you a one hundred dollar bill. If they reject it, you send $100 to the charity you did it. If you are confident enough to threaten lawsuits and prosecution for fraud, surely you must be confident that the worlds most prestigious math journal will get it right. How about it? Do you believe your own proof, or not ? . Nora B. >James Harris === Subject: Re: JSH: Fool all of the people, all of the time? posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn > Good question. But to believe that my results havent traveled >> through mathematical society at this point you have to believe that a >> very basic argument, which I know I can explain in about an hour as I >> did it in-person at my alma mater Vanderbilt University, which shoots >> down one of the underpinnings of algebraic number theory could just >> ßoat out there, be argued about by me on Usenet for years, and never >> get heard of by leading mathematicians. >> Impressive bit of sentence construction. Incidentally, though: argued >> about by me on Usenet for years - which bit _is_ this? I know you >> seem to have been claiming to have found errors in Ôcore for years, >> but I thought they were different arguments. Since as you say >> yourself, youve been wrong a lot in the past, and the important bit >> is the current argument, which alone of course is Correct, how long >> has this bit been going? Id have thought less than years...? >> The full timeline is that back in December 1999 I first discovered an >> approach which would lead to the analysis tool of non-polynomial >> factorization. >> The earliest arguments in this area go back to December 1999. >> OK - Ill give you your arguing for years. >So you will accept what is true. How wonderful for you. >> Newsßash!! People are asking questions!! Heres the commonest one: >> What does Ôproperly a unit mean? >> Only you can answer. >> Thats an old game of trying to cause major arguments over the use of >> some term or other, as I now simply shift from usage that sci.mathers >> find easily works to provoke confusion. >> Sorry, Im a bit lost here. Youre accusing me of playing a game, >> just because I asked what a bit of your argument means? I dont >> understand what you mean when you claim that while something-or-other >> is not a unit in the ring of algebraic integers, it _is_ properly a >> unit. I guess it means something to you, but I surmise it means >> nothing to anyone else. (That means you are speaking a private >> language, and the closest to fame you can hope for is the status of a >> Voynich manuscript in a half-millennium or two.) >You take a phrase out of context, and make a big deal out of it, >asking for some explanation, when in context when used the explanation >was in what I said. When real mathematicians use a new term, they define it. Period. No dependency on Ôcontext. >Its an old tactic that sci.mathers have used for years. The tactic here is evasion. >Pick a phrase, take it out of context, make a big deal out of it. >And, you know? Lots of times I HAVE used certain phrase differently >from standard usage, and misused terms I didnt understand or simply >just didnt care a lot to get exactly right in informal discussions. Mathematics, unlike sociology, has to be precise. >Time after time, posters have reacted as if these posts are such >important communications that perfection is a requirement. There is no reason in mathematical discourse not to be explicit and precise. >I reserve the right to at times just babble. Its not a big deal. Then in return dont expect anyone to believe you. >This is Usenet. So? Are you trying to do math or not? It doesnt matter where you do it. >> I find all this stuff about peering into polynomials looking for >> Ôfactors a bit confusing, too. I mean, of course in (a) below, >> theres a factor of 4 that can be divided off: >> (a): 4x^2 + 4x - 4 >> So if n is an integer - any integer - I know that 4n^2 + 4n -4 is a >> multiple of 4. However: >> (b): (x+1)(x+2) >> Does this polynomial have a factor of 2 in it? I can see a Ô2, but it >> doesnt look like a factor. And again I know that any integer n means >> that (n+1)(n+2) is a multiple of 2. So whats going on? Is there any >> difference from the case of (a). You seem to reject the idea that >> theres really any difference between talking about polynomials *as* >> polynomials and evaluations of polynomials, because thats voodoo >> math - have I got that right? >Well heres a good chance to show how sci.mathers routinely try to >mislead. >With integer you have the any integer either is even or has a residue >of 1. >That is, given an integer x, either x = 0 mod 2, or x = 1 mod 2. >So trivially you can put up something like (x+1)(x+2) and know it must >be even. >However, in the ring of algebraic integers, no such relations exist. >That is, there is no non-unit in the ring of algebraic integers such >that EVERY algebraic integer is either divisible by that number or has >the same residue. So do you think, given (x + 1)(x + 2), where x is an algebraic integer, that its divisibility or coprimeness to 2 does not depend on x ? If it *is* divisible by 2, does 2 always factor out of it in the same way? >You dont even have the even case with algebraic integers as in the >ring of algebraic integers it is NOT true that (x+1)(x+2) must have 2 >as a factor. I would like to see you prove that. Can you? But in any case you are missing the point. (x + 1)*(x + 2) may be coprime or noncoprime to 2 for various values of x. But there can be no doubt that its coprimeness or noncoprimeness is a FUNCTION of x. And that is really the point that is relevant to your factorization of P(x) and P(x)/49, isnt it? How factors of 49 might divide out of a product of functions of x ? Does it necessarily always happen in the same way, or is it dependent on x ? >Now the example (x+1)(x+2) in the ring of integers is childish >mathematics at the most basic level, but posters have routinely used >that example for years to try and claim that it shows how my research >can be wrong. Its meant as an analogy, since you do not seem to be able to understand the real thing. >You people dont even try hard. >> And what ever should I think about (c) - absolutely no trace of a Ô3 >> anywhere: >> (c): x(x+1)(x+2) >Yet another childish example based on any *integer* x, either being >divisible by 3 or having a residue of 1 or 2. >That childish game can be played on and on with integers, but it has >no meaning outside of the ring of integers, and it doesnt invalidate >my research findings. >> Another thing I wonder about: OK, suppose the dam bursts. Suppose >> suddenly you are on tv shows and whatnot. The mathematicians are all >> disgraced, ßung into debtors jails. Are you going to rewrite all the >> text books? Who will do it? >Mathematicians will NOT all be disgraced and ßung into jail. >If there are some who end up prosecuted itd most likely be a choice >few. >And hey, not all mathematicians work in the area of algebraic number >theory. Number theorists are the ones who are really going to take >the hit. >How might a prosecution work? >Well, consider a federal prosecution that considers whether or not >some number theorist knowingly continued both to teach what Id shown >to be false, and to receive federal funds for their research. Are people jailed now for teaching false things? You are aware probably that some people still teach Bohrs model of the hydrogen atom. Are they being prosecuted? >So theyd have tax dollars for two things: teaching and research >The charge would be fraud. Evidence might include emails of that >mathematician and conversations had with colleagues who would be >pulled into grand juries and later into court to testify under oath. >Likely punishment? A fine. Blockage from receipt of any more federal >funds. Censure from their university, and probably removal from >teaching position. Might the blade cut both ways? You have tried very hard to convince large numbers of people that your proof is right, even in the face of counterexamples which you have not refuted. Prosecuting you for fraud would be trivial. There would be literally thousands of expert witnesses, and none of them for the defense. You have repeatedly accused people here of lying simply because they post arguments saying your proof is wrong. You have accused people of committing fraud. If you are wrong (AND YOU ARE!), YOU could be sued for libel. In fact you have previously accused people of lying about many of your former arguments, and then later you were forced to admit they were right. RIGHT NOW, your own postings which include (1) the accusations of lying and (2) the later admission that those you were accusing were correct, could be used against you in a lawsuit for libel. Who do you think would win? >> FWIW, I have a copy of Hersteins Topics in [pre-Harrisian] Algebra >> here, and it only mentions algebraic integers very brießy, in the >> problems. Theres the usual definition, then the first problem reads: >> 10. If _a_ is any algebraic number, prove that there is a positive >> integer _n_ such that _na_ is an algebraic number. Error there: last phrase should be Ôalgebraic integer. >> (I can type more if required.) Well, is Hersteins solution to this >> problem wrong? >Amazingly enough, most people actually care about whats TRUE. Which in this very case you overlooked entirely with your self-serving non-answer. >More than likely there will be a ßood of mathematicians into >algebraic number theory, as it will be an opened up field with major >opportunities for advancement. Delusion. >Think about it. In many areas a young mathematician can work for >years and get nowhere. In algebraic number theory, they will have the >opportunity to be a significant figure in the remaking of an entire >discipline. >When that finally gets out, it will be the hottest field in >mathematics. Hey, my offer of a bet of $100 that the Annals of Mathematics will reject your paper Advanced Polynomial Factorization still stands. If they publish it, I send you a one hundred dollar bill. If they reject it, you send $100 to the charity you did it. If you are confident enough to threaten lawsuits and prosecution for fraud, surely you must be confident that the worlds most prestigious math journal will get it right. How about it? Do you believe your own proof, or not ? . Nora B. >James Harris === Subject: Re: JSH: Fool all of the people, all of the time? ... > That is, given an integer x, either x = 0 mod 2, or x = 1 mod 2. Yup. > So trivially you can put up something like (x+1)(x+2) and know it must > be even. Indeed. > However, in the ring of algebraic integers, no such relations exist. Eh? > That is, there is no non-unit in the ring of algebraic integers such > that EVERY algebraic integer is either divisible by that number or has > the same residue. This does not parse. But I would think that in the integers a number is divisible by 3 or has a residu of either 1 or 2 when divided by 3. So I see nothing like the same residu here. So, mod 3, in the integers, there are three residue classes. So I am not surprised when the same happens in a larger ring. > You dont even have the even case with algebraic integers as in the > ring of algebraic integers it is NOT true that (x+1)(x+2) must have 2 > as a factor. Indeed, because there are infinitely many residue classes mod 2 in the algebraic integers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Fool all of the people, all of the time? > Well, consider a federal prosecution that considers whether or not > some number theorist knowingly continued both to teach what Id shown > to be false, and to receive federal funds for their research. That would make Court TV forget all about Scott Peterson. === Subject: Re: JSH: Fool all of the people, all of the time? > FWIW, I have a copy of Hersteins Topics in [pre-Harrisian] Algebra Um, I think that should be pre-Harristotelian. HTH Gib === Subject: Re: JSH: Fool all of the people, all of the time? days. My association with the Department is that of an alumnus. >> FWIW, I have a copy of Hersteins Topics in [pre-Harrisian] Algebra >Um, I think that should be pre-Harristotelian. Im partial to the adjective Harrisible. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Fool all of the people, all of the time? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > FWIW, I have a copy of Hersteins Topics in [pre-Harrisian] Algebra >>Um, I think that should be pre-Harristotelian. > Im partial to the adjective Harrisible. Gesundheit. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Fool all of the people, all of the time? > What does Ôproperly a unit mean? > Only you can answer. > Thats an old game of trying to cause major arguments over the use of > some term or other, as I now simply shift from usage that sci.mathers > find easily works to provoke confusion. > What I find fascinating is that *clearly* some of you have worked > rather hard to confuse the issue, hide the reality, and fight for > arguments that just dont work, when I know that I learned years ago > that its just futile in mathematics to do those things. > And now some of you may finally be learning why, as I simply adjust > explanations, and soon enough people will realize that you had to > understand how it all worked to confuse them so well, and then they > probably wont appreciate your efforts. > After all, mathematics is objective in many ways. Sure some of you > have personalized it, so that you can attack it as if it were mine. > But its like if you hated Pythagoras and went after the Pythagorean > Theorem. > Youre not doing the world any favors, and fighting a battle you will > lose. > These things have happened before, history repeats itself, and for > some reason there are people like some of you who step out to fight > what is mathematically true, for personal reasons. The question was What does Ôproperly a unit mean? fundamental (remaining) problem is finding a value for your sample polynomials at x = 0, and then extending that result to all values of x. How have you answered that? === Subject: Re: JSH: Fool all of the people, all of the time? === >Subject: Re: JSH: Fool all of the people, all of the time? >> Good question. But to believe that my results havent traveled >> through mathematical society at this point you have to believe that a >> very basic argument, which I know I can explain in about an hour as I >> did it in-person at my alma mater Vanderbilt University, which shoots >> down one of the underpinnings of algebraic number theory could just >> ßoat out there, be argued about by me on Usenet for years, and never >> get heard of by leading mathematicians. >> Impressive bit of sentence construction. Incidentally, though: argued >> about by me on Usenet for years - which bit _is_ this? I know you >> seem to have been claiming to have found errors in Ôcore for years, >> but I thought they were different arguments. Since as you say >> yourself, youve been wrong a lot in the past, and the important bit >> is the current argument, which alone of course is Correct, how long >> has this bit been going? Id have thought less than years...? >The full timeline is that back in December 1999 I first discovered an >approach which would lead to the analysis tool of non-polynomial >factorization. >It took a while though to figure out what I was doing as I stumbled >abot with various proofs of FLT that turned out to be bogus, but I >kept working at it despite the failures, and time passed. >I think it was around two years later when I finally was fully using >non-polynomial factorization in my FLT research, It took you two years to learn what a ring is, Flatrings? Or do you still not understand? >as I definitely was >by November 2001. I focused on non-polynomial factorization itself, >in discussions around May 2002, when I also discovered my prime >counting function. > Newsßash!! People are asking questions!! Heres the commonest one: >> What does Ôproperly a unit mean? >> Only you can answer. > Thats an old game of trying to cause major arguments over the use of > some term or other, as I now simply shift from usage that sci.mathers > find easily works to provoke confusion. No ones asking what the conventional definition is. There isnt one. We just want to know what *you* mean when you use the term, since it apparently plays an important role in understanding the problem with the algebraic integers. You *do* mean *something* when you say such-and-such is properly a unit (but not a unit in the algebraic integers), right? So *what* do you mean? Is there any non-zero complex number which is *not* properly a unit? Is there any non-zero, non-integer in C which is *not* properly a unit? Note: An answer to the above two questions is nice, but it doesnt settle the issue. What is the definition of properly a unit? -- Yup, you guessed it. If worse comes to worse, I *will* turn to the Army to help me with mathematicians. And then mathematicians dont think the NSA or CIA can save your asses, as generals LIKE me. -- James Harriss latest foray into mathematical logic. === Subject: Re: JSH: Fool all of the people, all of the time? -- Min I blame the jelly === Subject: Re: Fool all of the people, all of the time? > So, if I can contact leading mathematicians, and they dont give the > same objections, why am I arguing on Usenet about what I claim is a > very basic argument that is also one of the greatest discoveries in > math history? Because the leading mathematicians arent interested in your ßuff, and they just say, Oh, thats nice or some such? Because your apparent megalomania isnt fueled by a once-a-year 15-minute talk with a person who really isnt interested in what you have to say? Because you need to see your name in print and no reputable journal is interested in your work but you can see JSH with the stroke of a key on the net? > Wouldnt some of these leading mathematicians say something? Why > arent they giving press releases? How can business as usual continue > in the math world? > I now think part of it is you. Ah, yes. Youve uncovered the most exciting discovery in the history of mathematics, and the mathematical world *would* have heralded you as the latest Fields Medal candidate, but they all ran and checked sci.math to see what *we* thought of you first. > After all, Ive argued on Usenet for years, and been VERY WRONG for > years, with arguments that turned out to be false or wrong. I definitely cannot argue with that statement. Lets leave it there, on a rare point of total agreement. Michael === Subject: Re: JSH: Fool all of the people, all of the time? === >Subject: JSH: Fool all of the people, all of the time? >I say theyre not dense, theyre media savvy. They watch you, and see >me posting all the time, with people arguing with me with really dumb >stuff, and they dont get feedback indicating that its not ok, and it >seems easier to just go with the way things are. Well, then obviously you should simply stop posting to sci.math. >James Harris -- Mensanator Ace of Clubs === Subject: Re: JSH: Fool all of the people, all of the time? > So I can show that popular objections on sci.math boil down to weird > things like arguing that dividing a multiple of a polynomial can occur > as a function of some variable. Or explain over and over again that > at its heart my work work depends on previously uncontroversial truths > like the distributive property, and w/7 = 1, meaning that w=7. [misprint in line above] > Ive mentioned before that in my contacts with leading mathematicians, > and even math students like that Cornell math grad student, I dont > get the objections that play out month after month on these > newsgroups. You know, continually mentioning that you have CONTACTED these people, with no evidence or documentation that they AGREED with you, is not exactly real compelling evidence that you might be right. > A properly trained mathematician is just NOT going to seriously > consider that a multiple of a polynomial leaves some trace after its > divided off. A properly trained mathematician will not question the > distributive property in a commutative ring. > So, if I can contact leading mathematicians, and they dont give the > same objections, why am I arguing on Usenet about what I claim is a > very basic argument that is also one of the greatest discoveries in > math history? Oh. So, since they didnt say you were wrong, that proves you must be right ??? > Wouldnt some of these leading mathematicians say something? Your paper, APF, because of its amateur style and syntax and its unmotivated reference to a highly specialized polynomial, is not something that people understand when they first read it. Remember, it contained an error for OVER A YEAR that you yourself did not notice until someone pointed it out here, even though you had argued about it endlessly and recomposed it several times and submitted to two or three different journals. APF is hard to read. A professional mathematician is going to glance at it, see it is some kind of crank nonsense, and forget it. He is not going to take the time to actually see what is wrong with it. So he doesnt point out the errors - he knows they are almost certainly there, but he doesnt waste his time trying to find them. He just ... sort of ... politely brushes you off. That is what happened. > Why > arent they giving press releases? How can business as usual continue > in the math world? This is simply delusional talk. > I now think part of it is you. Blame the victims? > After all, Ive argued on Usenet for years, and been VERY WRONG for > years, with arguments that turned out to be false or wrong. It seems > reasonable to conclude from a slanted perspective that it will be very > difficult for me to get people to believe what I say now. True. However that is not why people are still opposing what you say. This time, as in all previous instances, they are opposing what you say *because it is wrong*. > And that has been true. Month after month Ive explained, and events > have occurred like that weirdness with the APF paper, and youve > absorbed it all. > Its becoming increasingly clear that hey, maybe no one will EVER > react, and business can continue as usual, professors can teach work > that they probably know now is ßawed, and their world can continue > with little change. > Why would they do it? Do what? > Good question. Glad you asked! > But to believe that my results havent traveled > through mathematical society at this point you have to believe that a > very basic argument, which I know I can explain in about an hour as I > did it in-person at my alma mater Vanderbilt University, Yes ... Dr. McKenzie. Did he jump up and embrace you as a budding genius, and say you had made an astounding discovery ? Or did he just ... sort of ... brush you off to get rid of you? Why do you keep mentioning this episode? Is it supposed to be evidence that your argument is right? An appeal to Authority ? Except in this case the Authority didnt say anything, so you assume that silence means approval ? > which shoots > down one of the underpinnings of algebraic number theory could just > ßoat out there, be argued about by me on Usenet for years, and never > get heard of by leading mathematicians. > Oh, but wait, Ive actually contacted leading mathematicians. Oh, yes. Forgot that. Did they endorse your work? Did they say you were right, or just fail to say you were wrong? I mean, you must be mentioning this for SOME reason! > I even got some commentary on the APF paper from Barry Mazur. What did he say, exactly? Great original work here, Harris! This is astonishing! Im calling the Fields Medal people right now! Has he continued to correspond with you on this? Did he offer to help you get it published ? Or did he just ... sort of ... brush you off? > So, you have to believe that those people are remarkably dense. Or that they sense very quickly that you are a crank and just ... sort of ... politely brush you off. > I say theyre not dense, theyre media savvy. They watch you, and see > me posting all the time, with people arguing with me with really dumb > stuff, and they dont get feedback indicating that its not ok, and it > seems easier to just go with the way things are. More delusion. Barry Mazur probably never heard of sci.math. > As a backup they have that I have a paper at a leading journal, so > they can let the weight sit on that journal, and claim later that with > such a controversial result, and such a controversial figure, it > seemed prudent to wait. If they actually thought you had something and that it was important, they would be all over it. They would help you write it and help you get it published. The evidence is right there in front of you. You are misinterpreting every bit of it. And now, here, you try to use these total non-endorsements here as evidence that your proof is right. Pathetic! > But, you have to believe that theres some question, when there is > not. I agree, there is no question. > Ill tell you one of the weirdest things I face when dealing with > mathematicians outside of Usenet, when Im sure they get it, is that > they tend to walk away. Right. Cranks make people nervous. They know that arguing with them is hopeless. And they have the uneasy feeling that the crank might actually be a crazy person. > Its bizarre. Its like some door just closes. After the > explanations are done, any questions or concerns are handled, they > quit talking to me. Yes. And now you are trying to use the fact that you actually CONTACTED these people as evidence that your proof makes sense. Again, PATHETIC! > Then again, the mathematical research I have shatters their world in > many ways though as I mentioned yesterday, its not really a loss. > Its a gain. > For the true mathematician, whats actually correct is whats > valuable. True! > People make mistakes. True! Got THAT right! > Better to live today and learn the truth, than > to be one of those poor saps who died deluded, thinking they knew > certain things that they just didnt. Thinking they had proofs that > they didnt. Those poor saps! > Mathematicians today are lucky. They need to start counting their > blessings. > As make no mistake, you CANNOT make fools of all of the people, all of > the time. Eventually the silliness of challenging mathematics at its > base will play out. People will eventually get tired of hearing from > posters who get caught in dumb lies, and there will be people who will > ask more and more questions. > And make no mistake, the benefit to the passive-aggessive strategy of > sitting tight and waiting, maybe hoping that people will be fooled > indefinitely, will not be lost on people. Many people may be > intimidated by mathematics, but they understand money and power very > well. OK, lets see how confident you are. I will bet you $100 that the Annals of Mathematics rejects your paper, Advanced Polynomial Factorization. If they publish it I will mail you a one hundred dollar bill. If they reject it, you mail $100 to the charity of site. Agreed? Nora B. > So, yeah, I think Usenet has something to do with the continued > foot-dragging, this passive-aggressive strategy of just keeping quiet, > and, you know I also think it has to do with a contempt for the public > that you can see at times traveling through intellectual circles. > Maybe they just think that people arent being made fools of, but that > they are fools, incapable of ever realizing the truth on their own. > James Harris === Subject: Re: JSH: Fool all of the people, all of the time? until I read this sentence -- and I went no further than it -- Id thought that the header had formulated the penultimate question of the formulator ... because it could be a pyshcomath bot! > its unmotivated reference to a highly specialized polynomial, is not --Advice, 0.05; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: JSH: Fool all of the people, all of the time? > OK, lets see how confident you are. I will bet you $100 that > the Annals of Mathematics rejects your paper, Advanced Polynomial > Factorization. If they publish it I will mail you a one hundred > dollar bill. If they reject it, you mail $100 to the charity of site. > Agreed? I am also up for a bet with James Harris. But I will him 10 to 1 odds, just to sweeten the deal. If the Annals of Mathematics (of Princeton) publish it I will mail him a one hundred dollar bill. If they reject it, he can mail, if he chooses, only $10 to the charity of my choice. === Subject: Re: JSH: Fool all of the people, all of the time? > OK, lets see how confident you are. I will bet you $100 that > the Annals of Mathematics rejects your paper, Advanced Polynomial > Factorization. If they publish it I will mail you a one hundred > dollar bill. If they reject it, you mail $100 to the charity of site. > Agreed? > I am also up for a bet with James Harris. But I will him 10 to 1 > odds, just to sweeten the deal. If the Annals of Mathematics (of > Princeton) publish it I will mail him a one hundred dollar bill. > If they reject it, he can mail, if he chooses, only $10 to the > charity of my choice. Just so you know, Harris doesnt have a very good history of coming through on bets. See: === Subject: Re: JSH: Times up, and finally proof : > : > How about this, as for rules, Im still not quite ready to bet : > on the Case 2 proof, but I have $500 at ten to one on the : > Case 1 proof. : > : > That means I get $5,000 if Im right (which I am). : : We all know that your original bet was on the full FLT, not on : the easy Case I (which I as well as many others did as students : eons ago). What we now know is the following: : : (a) You do not currently have (or even feel confident that you : have) a proof of the full FLT -- contrary to your previous boasts : and offer of a ten to one wager on the validity of your proof. : : (b) You are welching on your original bet and trying to weasel : out of it by changing it to a different bet. In some parts of : the world, people take it *very seriously* when one welches on : a bet, with serious consequences to the person who reneges. Jim Burns === Subject: Re: JSH: Fool all of the people, all of the time? >[...] >Oh, but wait, Ive actually contacted leading mathematicians. >I even got some commentary on the APF paper from Barry Mazur. Why dont you ever tell us what this commentary _said_? >So, you have to believe that those people are remarkably dense. >I say theyre not dense, theyre media savvy. They watch you, and see >me posting all the time, with people arguing with me with really dumb >stuff, and they dont get feedback indicating that its not ok, and it >seems easier to just go with the way things are. Yeah, thats it. Leading mathematicians confronted with earth- shattering result, and they check sci.math to see whether they can get away with just ignoring it. Youre really sounding a little off today, you know. >As a backup they have that I have a paper at a leading journal, so >they can let the weight sit on that journal, and claim later that with >such a controversial result, and such a controversial figure, it >seemed prudent to wait. >But, you have to believe that theres some question, when there is >not. >Ill tell you one of the weirdest things I face when dealing with >mathematicians outside of Usenet, when Im sure they get it, is that >they tend to walk away. Doesnt seem weird to anyone but you - makes perfect sense to the rest of us. >Its bizarre. Its like some door just closes. After the >explanations are done, any questions or concerns are handled, they >quit talking to me. This has been explained to you many times. When you contact a leading mathematician he will eventually realize youre a lunatic and that theres no point in talking to you. You dont seem to get this, because you find people on sci.math dont just shut up. The reason for that is that the people here who continue to talk to you are doing it for _fun_ - those leading mathematians have work to do. >[...]Eventually the silliness of challenging mathematics at its >base will play out. That seems unlikely - as far as we can see you intend to continue the silliness of challenging mathematics at its base until you die. >People will eventually get tired of hearing from >posters who get caught in dumb lies, and there will be people who will >ask more and more questions. Oh, you were talking about _us_ challenging mathematics at its base. Never mind... >And make no mistake, the benefit to the passive-aggessive strategy of >sitting tight and waiting, maybe hoping that people will be fooled >indefinitely, will not be lost on people. Many people may be >intimidated by mathematics, but they understand money and power very >well. >So, yeah, I think Usenet has something to do with the continued >foot-dragging, this passive-aggressive strategy of just keeping quiet, >and, you know I also think it has to do with a contempt for the public >that you can see at times traveling through intellectual circles. >Maybe they just think that people arent being made fools of, but that >they are fools, incapable of ever realizing the truth on their own. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Fool all of the people, all of the time? > So I can show that popular objections on sci.math boil down to weird > things like arguing that dividing a multiple of a polynomial can occur > as a function of some variable. ??? > Or explain over and over again that > at its heart my work work depends on previously uncontroversial truths > like the distributive property, and w/7 = 1, meaning that w=7. No one has denied the distributive property or that w/7 = 1 means that w = 7. So the controvesy must be elsewhere. > Ive mentioned before that in my contacts with leading mathematicians, > and even math students like that Cornell math grad student, I dont > get the objections that play out month after month on these > newsgroups. Maybe they cant be bothered with cranks. In this newsgroup, your continued defense of false math is a scourge. Some readers have shown praiseworthy patience in trying to help you correct your arguments. > A properly trained mathematician is just NOT going to seriously > consider that a multiple of a polynomial leaves some trace after its > divided off. A properly trained mathematician will not question the > distributive property in a commutative ring. If I wanted to know what a Ôproperly trained mathematician does or does not do, I certainly wouldnt consult you about it. > So, if I can contact leading mathematicians, and they dont give the > same objections, why am I arguing on Usenet about what I claim is a > very basic argument that is also one of the greatest discoveries in > math history? Greatest? According to you. Your credibility, however, was shattered long ago. > Wouldnt some of these leading mathematicians say something? Why > arent they giving press releases? How can business as usual continue > in the math world? > I now think part of it is you. > After all, Ive argued on Usenet for years, and been VERY WRONG for > years, with arguments that turned out to be false or wrong. It seems > reasonable to conclude from a slanted perspective that it will be very > difficult for me to get people to believe what I say now. No kidding! But note that people didnt believe you BECAUSE YOU WERE WRONG. If you want people to believe you, correct your errors and post something right. [snip snivelling, tear-jerking, woe-is-me, self-serving crap] -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Fool all of the people, all of the time? Discussion, linux) > Wouldnt some of these leading mathematicians say something? Why > arent they giving press releases? How can business as usual continue > in the math world? > I now think part of it is you. Leading mathematician 1: What a startling and groundbreaking piece of research! Lets call the New York Times! And Oprah! Leading mathematician 2: Not so fast. Lets see if this guy posts on Usenet. Leading mathematician 1: Whoa. Good call. Look at how easily misled the masses are. Leading mathematician 2: Yes, let us continue as if nothing has happened. Besides, my mortgage is due. -- Jesse F. Hughes And a journal can beg me for the right to publish it [...] because Id rather see it in People magazine [...] --James Harris on his simple proof of Fermats last theorem === Subject: Re: Just checking line slope > I hope this doesnt sound dumb, but is the following an acceptable > formula for a straight line. I am sure it is, but just wanted to check. > y = x/6 Add or subtract any number to your x,or y or 6 in this equation and it is still a straight line. I first learnt it by visual detection: If the plot or graph is straight then it is linear, but if bent it is not so.There is only one way keeping it straight, so many ways to bend it. === Subject: Re: Just checking line slope <3Mcsd.58178$K7.34752@news-server.bigpond.net.au I hope this doesnt sound dumb, but is the following an acceptable > formula for a straight line. I am sure it is, but just wanted to check. > y = x/6 Yes. Checking your work is never dumb. In general the equation of a straight line in the plane is y = mx + b or x = a Are you familiar with this general equation? Your equation fits that general form for what values of m and b? === Subject: Re: Just checking line slope >>I hope this doesnt sound dumb, but is the following an acceptable >>formula for a straight line. I am sure it is, but just wanted to check. >>y = x/6 > Yes. Checking your work is never dumb. > In general the equation of a straight line in the plane is > y = mx + b or x = a > Are you familiar with this general equation? > Your equation fits that general form for what values of m and b? x/6 part. I have been struggling with some of those finer rules for what constitutes a linear equation. My textbook says this... The following *are* linear equations: 1) 3x + 2y = 7 2) x1-2x2 +10x3 + x4 = 0 3) 1/2x + y - (pi)z = sqrt(2) 4) (sin((pi)/2))x1 - 4x2 = e^2 The following *are not* linear equations: 5) xy + z = 2 6) e^x - 2y = 4 7) sin(x1) + 2x2 - 3x3 = 0 8) 1/x + 1/y = 4 Firstly my apologies for being unsure of ASCII syntax. Where I have written a number after the variable I am trying to indicate the number is a subscript, ie x1 is x subscript 1. What is the correct syntax? While I can understand why most of these are linear/not linear. Some have me confused. Why is 4 okay, but not 7? What is wrong with 8? Cassie === Subject: Re: Just checking line slope The following *are* linear equations: > 1) 3x + 2y = 7 > 2) x1-2x2 +10x3 + x4 = 0 > 3) 1/2x + y - (pi)z = sqrt(2) > 4) (sin((pi)/2))x1 - 4x2 = e^2 > The following *are not* linear equations: > 5) xy + z = 2 > 6) e^x - 2y = 4 > 7) sin(x1) + 2x2 - 3x3 = 0 > 8) 1/x + 1/y = 4 > Firstly my apologies for being unsure of ASCII syntax. Where I have > written a number after the variable I am trying to indicate the number > is a subscript, ie x1 is x subscript 1. What is the correct syntax? x_1 and x^n for n-th power of x. Best readable ascii style is to persist like youre doing for the most part with adequate spaces like x1 - 2x2 + 10x3 + x4 = 0 > While I can understand why most of these are linear/not linear. Some > have me confused. > Why is 4 okay, but not 7? sin pi/2 is a constant. sin x_1 isnt ax_1 for some a in R. > What is wrong with 8? It doesnt have the form ax + by = c, the most general equation for a line, hence the expression linear, line like. Indeed, y + x = 4xy Were you to graph the equations, what is linear and what is not would likely jump out at you for only lines are linear. === Subject: Re: Just checking line slope > Were you to graph the equations, what is linear and what is not > would likely jump out at you for only lines are linear. an equation looks linear on a graph, then it is linear. Aah, I enjoy maths so much...but it is also very hard at times...posibly why I like it so much! cassandra === Subject: Re: Big Bertha Thing blogs Big Bertha Thing pathos Cosmic Ray Series Possible Real World System Constructs http://web.onetel.com/~tonylance/pathos.html Access page JPG 12K Image Astrophysics net ring access site Newsgroup Reviews including uk.rec.cycling Detail from painting of captive musketeers. Caption:- Porthos took hold of a bar (foot rail) with both hands Twenty Years After by Alexandre Dumas Published by George G.Harrup & Co.Ltd., 1923 Reprinted 1929 (C) Copyright Tony Lance 1998 Distribute complete and free of charge to comply. Big Bertha Thing poem Some Days, Then Some by Tony Lance Ive had better days, he thought and said. When I could get my sorry butt out of bed. When I wasnt mistook for as good as dead. When they didnt fill my boots with all that lead. There are days sometimes, of sunshine on my head. Windswept shores viewed from along a beachy-head. Carefree larks, in a clearly blue sky, over-head. Then of course, I became a headmaster, the old man said. Tony Lance joehorace@big-bertha-thing.com Big Bertha Thing testament New Testament Rosary 1. The Angel Tells of the Virgin Birth. 2. The Blessing of John the Baptist in the Womb by the Unborn Christ.(13 weeks) 3. The Birth of Our Lord. 4. The Purification of Our Lady and the Presentation of Our Lord. 5. The Finding in the Temple. 6. The Baptism of Our Lord in the Jordan. (Luke 3) 7. The Wedding at Cana. (John 2) 9. The Transfiguration of Christ with Moses and Elias. (Matthew 17) 10. The Last Supper. (Matthew 26) 11. The Agony in the Garden. 12. The Scourging at the Pilar. 13. The Crowning With Thorns. 14. The Carrying of the Cross. 15. The Crucifixion. 16. The Resurrection. 17. The Ascension into Heaven. 18. The Coming Down of the Holy Ghost. 19. The Assumption into Heaven. 20. The Crowning of Our Lady Queen of Heaven. (1st five Joyfull, 2nd five Luminous, 3rd five Sorrowful, 4th five Glorious.) Old Testament Rosary by Tony Lance 1. The Promise of Hannah. (I Kings 1) Birth of Samuel. 2. The Three Angels Visit Abraham. (Genesis 18) Birth of Isaac. 3. The Betrothal of Ruth. (Ruth 3) Stranger and great-grandmother of David. 4. The Offering up of Isaac by Abraham. (Genesis 22) 5. The Finding of Joseph in Egypt by Jacob. (Genesis 46) 7. The Feast in the Temple for the Rechabites. (Jeremias 35) 8. The Covenant of Moses on Mount Sinai. (Exodus 19) 9. The Forty Day Walk of Elias. (III Kings 19) 10. The Writing on the Wall. (Daniel 5) The Fall of Empires. 11. The Water Dungeon of Jeremias. (Jeremias 38) The Babylonian Transmigration for 70 years. 12. The Deriding of David by the Kinsman of Saul. (II Kings 16) The Fall of Jerusalem. 14. The Ark is Carried for Forty Years. (Numbers 14) The Entry into the Promised Land. 15. The Execution of Jonah. (Jonah 1) The City of Nineveh Wears Sackcloth and Ashes. 16. The Raising to Life by the Tomb of Eleusius. (IV Kings 13) 17. The Taking up of Elias in the Fiery Chariot. (IV Kings 2) 18. The Spirit of Moses Comes Down on the Judges. (Numbers 11) The Law of Precident Cases. 19. The Slaying of Holofernes by Judith. (Judith 13) The Treasure to the 12 tribes. 20. The Crowning of Esther Queen of Susan. (Esther 2) The Decree of Pogrom is reversed. Rosary Intentions for each decade of St. Louis Marie de Montfort Luminous Mysteries by Tony Lance Lord through thy Holy Mother that we may:- 1. know true humility. 2. have charity one for another. 3. love poverty. 4. have purity of body and soul. 5. obtain Divine Wisdom. 6. be free from Original Sin.(Wet head saying ÔI baptise you in the name of the Father and of the Son and of the Holy Ghost. Amen) 7. give grace one to another. 8. enlighten one another. 9. obtain Union with Christ.(3 Nights of Soul) St.John of the Cross Battle vices and deny 5 physical senses. (Pictures, icons and Rock and Roll.) Battle ghosts of old vices and deny spiritual senses. (Old sins plague you for second time.) Visions, Locutions and tears. All else gone except the love of God. (Last of all the memory is perfected. It gets worse first.) 10. break bread together. 11. have perfect contrition. 12. know mortification. 13. have contempt for the world. 14. have patience in crosses. 15. obtain repentance for sinners, perserverance in grace for the just and reief for the Holy Souls in Purgatory. 16. love thee and be fervent in thy service. 17. love Heaven our true home. Annual Preparation. 12 days seek Contempt for World, 6 days each know self, Mary and Christ. 19. know the Holy Ghost. 20. obtain perserverance in grace for the just and the crown of Heavenly Glory. NB. Douai-Rheims cross reference links. The Sermon on the Mount. Mark 1,14:15 Matthew 4,12:17 Matthew 5,3 Luke 6,20 NB. Parallels between old and new testament rosary event readings. They had no wine, yet drank it. Others had wine and drank not. One was crowned with thorns and crucified. Another robed as a king and decreed to hang. One was scourged at the pillar. Another derided at the head of a whole column of soldiers. One was found with the wise men. Another was found, as the wisest man in Egypt. Elizabeth and Rachel were barren and both bore child. One was crucified. and descended into Hell (Limbo), for three days. Another walked the plank and was swallowed by a whale, for three days. One was born in a stable. Another betrothed in a barn. One was crowned Queen of Heaven. Another crowned Queen of an empire. One rose up into Heaven. Another was taken up in a fiery chariot. One was baptised in the Jordan. Another was saved from the bullrushes. That is 10 out of 20 direct parallels. The other 10 are more nuanced, but who can tell? NB. The Original Divine Office of the Church was Simply to Read Book Of Psalms Once a Week. (Divisions into days and hours can be the 15 minute segments as follows.) 3 by 7 Each psalm is followed by the prayer. Glory by to the Father and to the Son and to the Holy Ghost, as it was in the beggining, is now and ever shall be, world without end. Amen. 1-9 10-17 18-24 (25-32 33-37 38-44) 45-53 54-60 61-67 (68-72 73-77 78-84) 85-89 90-98 99-104 (105-109 110-117 118-118) 119-131 132-139 140-150 NB. On meditation and contemplation. You have to dig the field before you can watch the ßowers grow. Both are problem solving. Dread Visions Six Harms By Tony Lance (Based on the writings of St.John of the Cross; the Mystic Doctor of the Church.) Faith fails, replaced by a vision sensible. What we see is more real, than God invisible. Visions turn the spirit from its ßight Heavenward. Had Mary Magdallen touched the feet of the Lord. Then grounded in faith, would she never be. How can we hold this vision dearer than Thee. In nakedness of spirit, hold ourselves renouncible. Grace fails, imperceptible forsaken for perceptible. The gift becomes a right, which denies the gift. Decide the vision not from God, with Satan left. Seeking after visions is a sinfull occasion. Turning into diabolical, that divine vision. With the Devil disguised, as an angel of light, And man too ignorant, to even take fright. Grace cannot be rejected; into the soul infused. The taint of evil can be rejected, by vision refused. === Subject: Re: about semialgebraic set by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB4GMNU03889; >My definition of semialgebraic subset is: > A subset V of R^n is called semi-algebraic >if it admits some representation of the form > V = cup_{i=1}^s cap_{j=1}^{r_i} {x in R^n | P_{i,j}(x) s_{ij} 0 }, >where > Ôcup stands for Ôunion > Ôcap stands for Ôintersection > > s_{ij} in {>,=,<} > P_{ij}(X) in R[X], X = (X_1,...,X_n). >May you help me to show that the complementary in R^n >of a semialgebraic set is a semialgebraic set itself, >please? Its not hard to show that the collection of semi-algebraic sets is the closure under union and intersection of the class of sets defined by P(x) s 0 (to use your notation). Then, since complements of unions are intersections and vice versa, it comes down to showing that the complement of a set defined by P(x) s 0 is semi-algebraic, which is easy. Todd Trimble === Subject: Re: about semialgebraic set > Its not hard to show that the collection of semi-algebraic sets > is the closure under union and intersection of the class of sets > defined by P(x) s 0 (to use your notation). Then, since > complements of unions are intersections and vice versa, it comes > down to showing that the complement of a set defined by P(x) s 0 > is semi-algebraic, which is easy. > Todd Trimble Only a problem remains: Is it true that, for any given polynomial P in n real variables, i.e. P(x) in R[x], comp {x in |R^n : P(x) > 0} = {x in R^n : P(x) < 0 vel P(x) = 0} ?? Saem ___ comp denotes the operator which takes the complementary in R^n, i.e. for any given subset X in R^n, comp returns comp(X) = the complementary of X in |R^n. === Subject: Re: How to find this joint distribution of function of random variables? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB4GMPx03956; >Hi all, >I am wondering how can I find the joint pdf of random variables X, Y. >where X=cos(THETA), Y=sin(THETA), where THETA is uniformly distributed over >[0, 2*pi]. >It is difficult to me because all the tricks I know for this kind of problem >is the Jacobian. >But this is one R.V (THETA) mapping to two R.V(X, Y)... so I dont know how >to write out the Jacobian...anybody knows how? >Maybe I should define dummy variable? >Please help me! It is the uniform distribution on the circumference of the unit circle. For any point (X,Y) you get X^2+Y^2=1, and that it is uniform is also easy to see. Note that the result is still 1-dimensional. === Subject: Re: need help in understanding Torkels ZFC comment >>Ullrich> Thats not a formal proof from the axioms of ZFC. >Neither is the MetaMath proof of 2+2=4. >>Yes it is. If you believe otherwise then point out an axiom that is used >>in the proof but is not part of ZFC. >Very few of the terminal nodes of the proof tree are ZFC axioms, >because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set >is involved, so we need not consider the question of what sets exist >in general. 2+2=4 regardless of what ZFC says.) >At these nodes you will find other axioms, rules that allow >arbitrary wffs to be considered proven (unsound) and unverified >definitions (also unsound) which must be set up to verify new theorems >(a strict no-no in axiomatic systems.) There are also many nodes that >have no links or justification, such as ECOPRASS at >http://us.metamath.org/mpegif/ecoprass.html . What is the >justification for lines 23 and 24 (and the other lines labeled >ecoprass)? >>Looks like hypotheses of the theorem to me. > What are the hypotheses to 2+2=4? What does that have to do with anything? The theorem in question isnt 2 + 2 = 4. Now that you mention it, why dont you show us a complete listing of a formal proof that 2 + 2 = 4 in the Charlie Boo system, so that we can get an idea of how your formal proof compares to a Metamath formal proof. >An axiomatic system is one that derives everything from a fixed set of >axioms and rules (and perhaps definitions.) After it is set up, (1) >you dont have to add anything to the system to develop a theorem, and >(2) only actual theorems can be constructed. (Agree?) >Otherwise you have no more than a word processor with a Mathematical >font. >C-B > You know, this is kind of like deja vue. All these people yelling and > screaming and Im the one who (as is often the case) puts out the > formal definition to resolve technical issues - and they all go > running for the hills until time for the next skirmish. > What a life! -- Replace Roman numerals with digits to reply by email === Subject: FLT and the n-body Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB4GsRT06407; FLT AND THE n-BODY PROBLEM These two problems may seem remote but the resolution of the former provided the key to the solution of the latter. In 1992 I posed this question: why is FLT unresolved after 360 years? To resolve FLT these questions must be answered first: (1) does the problem make sense, that is, are the concepts and the formulation of the problem well-defined? (2) is the conjecture decidable? (3) is the conjecture true? The questions of decidability and truth are separate. To prove that FLT is true it is necessary to rule out the existence of triples of integers (x,y,z) satisfying Fermat.89s equation for n > 2. So far no one has found such triples and verifying it for all integers is both a mathematical and impractical impossibility because the integers are supposedly infinite. To prove it false it is necessary to construct a counterexample. At any rate, the answer to the first question is: no, because the underlying fields of FLT, namely, foundations, number theory and the real number system are defective, ill-defined. Therefore, I undertook their critique-rectification and the results are summarized in my post, .8bThe state-of-the-art in mathematics.8a. One of the major results of this critique-rectification is the characterization of undecidable propositions: a proposition is undecidable in a mathematical space if and only if it involves some ill-defined concept, explicitly or implicitly. It follows that a proposition is undecidable if it involves concepts from distinct spaces since they are independent and each is well-defined only by its axioms. Another major implication of this characterization theorem is that the present methodology of conventional modeling of physics is ßawed since it DESCRIBES nature in terms of some mathematical space. The two systems are independent and a problem or proposition in one is! not decidable in the other. In other words, there is no relation of necessity between them and solving a problem in one space in another amounts to reasoning by analogy. The remedy is dynamic modeling, i.e., modeling or explaining nature in terms of its laws. Consider the gravitational n-body problem. It says that given n bodies in the Cosmos with respective masses, positions and velocities subject to their gravitational attraction find their positions, velocities and paths at later time. Here, .8bbody.8a and .8bgravity.8a are ill-defined. To well-define .8bbody.8a it is necessary to know the basic constituent of matter. .8bGravity.8a requires a theory of gravitation to well-define and explain. A suitable set of laws of nature are needed to accomplish them. It took 11 initial laws of nature to do it in 1997. They constitute the initial natural laws of the ßux theory of gravitation that provide the specific solution of the gravitational n-body problems. I acknowledge quite a few regulars in Sci Math who are serious mathematicians. Among them are Don Palmer and Don Taylor. Don Taylor debated me by e-mail and I learned much from him. Together, we resolved a number of major mathematical issues. For the serious ones I recommend MathForge.net. There is less intellectual pollution there. It is less facetious and has fewer mathematicians from antiquity. E. E. Escultura === Subject: Re: FLT and the n-body Problem > FLT AND THE n-BODY PROBLEM Utter and total balderdash! Bob Kolker === Subject: Re: FLT and the n-body Problem > FLT AND THE n-BODY PROBLEM > These two problems may seem remote but the resolution of the former > provided the key to the solution of the latter. Why and how? Explanations and references - please. In 1992 I posed this question: why is FLT unresolved after 360 years? To resolve FLT these questions must be answered first: (1) does the problem make sense, that is, are the concepts and the formulation of the problem well-defined? Iff they are not well-defined, then references - please. (2) is the conjecture decidable? (3) is the conjecture true? The questions of decidability and truth are separate. To prove that FLT is true it is necessary to rule out the existence of triples of integers (x,y,z) satisfying Fermats equation for n > 2. So far no one has found such triples and verifying it for all integers is both a mathematical and impractical impossibility because the integers are supposedly infinite. To prove it false it is necessary to construct a counterexample. What is the counter example in the finite integers? > At any rate, the answer to the first question is: no, because the > underlying fields of FLT, namely, foundations, number theory and the real > number system are defective, ill-defined. References - please! Where and who has proofed or demonstrated that thing? Therefore, I undertook their critique-rectification and the results are summarized in my post, The state-of-the-art in mathematics. One of the major results of this critique-rectification is the characterization of undecidable propositions: a proposition is undecidable in a mathematical space if and only if it involves some ill-defined concept, explicitly or implicitly. It follows that a proposition is undecidable if it involves concepts from distinct spaces since they are independent and each is well-defined only by its axioms. Is it generally well-accepted by the mathematical community? In other case it is just a claim. Another major implication of this characterization theorem is that the present methodology of conventional modeling of physics is ßawed since it DESCRIBES nature in terms of some mathematical space. The two systems are independent and a problem or proposition in one is! This (above) is just the claim, but is this claim generally accepted? It is Your claim. > not decidable in the other. In other words, there is no relation of > necessity between them and solving a problem in one space in another > amounts to reasoning by analogy. The remedy is dynamic modeling, i.e., > modeling or explaining nature in terms of its laws. You have to explain it so plain that the most stupid people understand it! Including me. :-) > Consider the gravitational n-body problem. It says that given n bodies in > the Cosmos with respective masses, positions and velocities subject to > their gravitational attraction find their positions, velocities and paths > at later time. Here, body and gravity are ill-defined. OK, specify your references. In other case you just claim. :-) To well-define body it is necessary to know the basic constituent of matter. Gravity requires a theory of gravitation to well-define and explain. A suitable set of laws of nature are needed to accomplish them. It took 11 initial laws of nature to do it in 1997. They constitute the initial natural laws of the ßux theory of gravitation that provide the specific solution of the gravitational n-body problems. OK. This is your claim and very well done, but is it accepted by the scientific community? If not - forget it! > Every time this theory is applied to a new scientific problem suitable > natural laws are required to solve it. For example, 3 new laws of nature > were required to explain the Columbia space shuttle disaster last year. Who?, Where? and references - please. Fourteen biological and physio-psychological laws were needed to build the theories of intelligence, evolution and learning. To-date 42 natural laws, mathematical principles and a general law (that applies to the natural and on these subjects see the references I have made available. ??? Put those here: > I acknowledge quite a few regulars in Sci Math who are serious > mathematicians. Among them are Don Palmer and Don Taylor. Don Taylor > debated me by e-mail and I learned much from him. Together, we resolved a > number of major mathematical issues. For the serious ones I recommend > MathForge.net. There is less intellectual pollution there. It is less > facetious and has fewer mathematicians from antiquity. We prefer facts that are generally accepted - after referred discussions. Tapio > E. E. Escultura === Subject: Re: ladies In sci.math, Lord of Chaos(Suresh Devanathan) <31bp26F39llvmU1@individual.net>: > shut up you twirt Hes much more coherent than youll ever be. Answered my questions yet? :-) > blah blah blah ba ba ba mm m ohmm ohm ohh o o 0 i m supreme >> The adult answer is yada yada. Save the drama for yer mama. Get >> down and push, maggot! >> -- >> Uncle Al >> http://www.mazepath.com/uncleal/ >> (Toxic URL! Unsafe for children and most mammals) >> http://www.mazepath.com/uncleal/qz.pdf -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Circle construction problem Points A, B, and C lie in a straight line and are the centres of 3 circles. The circle centred on B touches those centred on A and C. Point O is the intersection between the straight line and a line tangent to all 3 circles. The problem is to construct 2 more circles, such that the first touches the circles centred on A and B, and the second touches those centred on B and C, and the 2 new circles also touch each other. Tim (For e-mail subeliminate) === Subject: Re: Circle construction problem posting-account=CfSJ5AwAAAD1yt3VP50q913IBHikxMCd > Points A, B, and C lie in a straight line and are the centres of 3 > circles. The circle centred on B touches those centred on A and C. > Point O is the intersection between the straight line and a > linetangent to all 3 circles. > The problem is to construct 2 more circles, such that the first > touches the circles centred on A and B,and the second touches > those centred on B and C, and the 2 new circles also touch each > other. Tim, The missing constraint is that << the line of centres of the 2 new circles should also pass through point O.>> (Another missing constraint is that each circle should touch any of the others externally only.) And yes, Id like to be able to construct this by straight edge and compass. When and where did you introduce the above << >> first? It appears first in reply to Philippe. You ought to have mentioned it a priori in OP, so it would nt gainlessly expend mutual times. Anyway your solution was with homothetically (similarly) placed circles with respect to pole O situation included in first hint. If R1 and R2 are circle radii and rho1 and rho2 distance to pole O, by similar triangles sin(TA)=(R2-R1)/(rho2-rho1) where TA is angle between tangents and line of centres of A,B,C. By algebra and trigonometry we get R2/R1=rho2/rho1=tan(45+TA/2)^2, a homothetic or similitude ratio of geometric progression away from pole O. A set of similarities are noted when you join tangent points to circle contact points as given by Philippe. Through tangent point of central circle B construct a mirror of inversion (I like to call) between circles A and C that was suggested in second hint. This is standard fare in inversions. About a half of circle B mirrors onto its second half mirrored about sqrt(power) mirror circle as well as circle A onto circle C. Next, find by Apollonius problem procedure how a contact circle among 3 circles A,B and mirror circle can be drawn.Its reßection behind the mirror automatically touches B and C. Important it is to realize that this mirrors not only A to C but also new circles 4 and 5. You would appreciate that if only two circles A,B are given, a set of infinite set of circles is generated by the problem you outlined. Also, if pi/TA is an integer, it would have full polar symmetry for other circumferentially generated copies. If there was a desription like homothetically placed contacting circles or the like, it could be shorter than the verbose problem statement. === Subject: Re: Circle construction problem posting-account=CfSJ5AwAAAD1yt3VP50q913IBHikxMCd > Points A, B, and C lie in a straight line and are the centres of 3 circles. > The circle centred on B touches those centred on A and C. Point O is the > intersection between the straight line and a line tangent to all 3 circles. > The problem is to construct 2 more circles, such that the first touches the > circles centred on A and B, and the second touches those centred on B and C, and the 2 new circles also touch each other. ... HINT: Line of centers and common tangents are all collinear at O. === Subject: Re: Circle construction problem > Points A, B, and C lie in a straight line and are the centres of 3 > circles. > The circle centred on B touches those centred on A and C. Point O is > the > intersection between the straight line and a line tangent to all 3 > circles. > The problem is to construct 2 more circles, such that the first > touches the > circles centred on A and B, and the second touches those centred on B > and C, and the 2 new circles also touch each other. > ... > HINT: Line of centers and common tangents are all collinear at O. I can see what youre getting at. The tangents are easy to construct once the 4th and 5th circles have been drawn. But how do you determine where those tangents will be drawn beforehand? Tim === Subject: Re: Circle construction problem posting-account=CfSJ5AwAAAD1yt3VP50q913IBHikxMCd > Points A, B, and C lie in a straight line and are the centres of 3 > circles. > The circle centred on B touches those centred on A and C. Point O is > the > intersection between the straight line and a line tangent to all 3 > circles. The problem is to construct 2 more circles, such that the first > touches the > circles centred on A and B, and the second touches those centred on B > and C, and the 2 new circles also touch each other. > ... > HINT: Line of centers and common tangents are all collinear at O. > I can see what youre getting at. The tangents are easy to construct once > the 4th and 5th circles have been drawn. But how do you determine where > those tangents will be drawn beforehand? > Tim ANOTHER HINT: Project by Inversion a set of hexagonally packed circles with respect to one of circle centers as pole O and using any radius for circle of inversion. === Subject: Re: Circle construction problem >> Points A, B, and C lie in a straight line and are the centres of 3 >> circles. The circle centred on B touches those centred on A and C. Point O >> is the intersection between the straight line and a line tangent to all 3 >> circles. >> >> The problem is to construct 2 more circles, such that the first touches >> the circles centred on A and B, and the second touches those centred on B >> and C, and the 2 new circles also touch each other. > ... > HINT: Line of centers and common tangents are all collinear at O. > >> I can see what youre getting at. The tangents are easy to construct once >> the 4th and 5th circles have been drawn. But how do you determine where >> those tangents will be drawn beforehand? >> Tim > ANOTHER HINT: Project by Inversion a set of hexagonally packed circles > with respect to one of circle centers as pole O and using any radius > for circle of inversion. However the original problem doesnt ask draw any 5 circles such and such but *given* 3 circles such and such, *find* 2 additional circles such and such. And also I wonder how you transform the 3 *given* circles by inversion in 3 *equal* circles ! (if they are not equal, the hexagonal trick doesnt work) Your notation of pole O is confusing as there is already a point O defined as intersection of the two common tangents of the 3 given circles. I shall name P the inversion pole, center of one of your hex lattice circles. I understand that P = B center of one of the circles. Inverting the two equal tangent circles to the one you choosed with pole = center results obviously into 2 equal circles. Given circles (A) (B) (C) are *unequal* as their common tangents intersect. So your trick of the set of hexagonally packed circles doesnt work. Hint : The 4th and 5th circles are undetermined. There is an infinite number of choices for the 4th circle. Once choosen, there are 2 possibilities for the 5th. You need an additional constraint on the choice of the 4th and 5th circles. Method : Choose and construct a 4th circle tangent to (A) and (B). As I said there are an infinite number of such circles. The easiest case is to arbitrarily choose the radius R4 of this 4th circle. Its center is then at intersect point of circles centered in A with radius RA + R4 and B with radius RB + R4. Other additional constraint are more difficult, for instance if (4) should go through a given point, or be tangent to some other given line or circle, or should be related to 5th circle, for instance find two *equal* additional circles ... and may be cant be constructed (with compass and straightedge). Now construct the 5th circle as the 2 only possibilities for a circle tangent to (B) (C) and (4) To do this, use an inversion with pole the contact point I of (B) and (4), and such as circle (C) is unchanged. The two circles (B) and (4) are transformed into two parallel lines (4) and (B), easy to construct. Now the image (5) of (5) is a circle tangent to those two parallel lines, and tangent to circle (C)=(C). This gives the two possibilities for circle (5). Construct the inverse of circle (5). OT Note : I couldnt read original post (not on news server) I had to search through Google groups to get it. Hopefully, this time Google groups was OK ... -- philippe (chephip at free dot fr) === Subject: Re: Circle construction problem [requote] > Points A, B, and C lie in a straight line and are the centres of 3 > circles. The circle centred on B touches those centred on A and C. Point > O is the intersection between the straight line and a line tangent to > all 3 circles. The problem is to construct 2 more circles, such that the first touches > the circles centred on A and B, and the second touches those centred on B > and C, and the 2 new circles also touch each other. >> ... ... > The 4th and 5th circles are undetermined. > There is an infinite number of choices for the 4th circle. > Once choosen, there are 2 possibilities for the 5th. > You need an additional constraint on the choice of the 4th and 5th circles. Warning : the message Im replying to here doesnt appear on >> The missing constraint is that the line of centres of the 2 >> new circles should also pass through point O. >> (Another missing constraint is that each circle should touch >> any of the others externally only.) And yes, Id like to >> be able to construct this by straightedge and compass. This seems much harder as construct just any 4th circle, then deduce the 5th as given in my previous method, but it is not. The difficult part is that it seems you have to construct the two additional circles simultaneously... However here also there is a nice inversion trick : Consider an inversion with pole O and such as circle (B) is invariant. The common tangents to circles (A), (B), (C) going through O, the inversion circle goes through the contact point T of circle (B) with the tangent. Lets name this circle (O). These tangents are also invariant because going through the inversion pole O. Hence circles (A) and (B) are just exchanged by the inversion. The 4th and 5th circles are hence also exchanged by this inversion as the center line O4-O5 goes through O. Hence the whole figure is invariant. The contact point between circles (4) and (5) lies hence on the inversion circle (O). Hence circles (4) and (5) are tangent to the inversion circle (O). Circle (4) is one of the circles tangent to (A), (B) and (O). There are 2 symetric non degenerated solutions. Construct it for instance with the method I gave previously : Lets name I the contact point of circles (A) and (B). Inversion with pole I and such as (O) is invariant, transforms circles (A) and (B) into two parallel lines (A) and (B). The transformed (4) of circle (4) is then easy to construct : tangent to two parallel lines and one circle. Similarly, the 5th circle is circle tangent to (B), (C) and (O). But it is also an homothetic circle of circle (4) with center O, and tangent to circle (4), which is easier to construct. Note : (I suppose O, A, B, C in that order) Proove that (B) is just the perpendicular to OB from O, and (A) is a perpendicular to OB at distance OH = 2*OB. Hint : power of I to (O) is the inversion constant. Hence construction is easier than in general case : (see http://chephip.free.fr/img/5circles.gif) T being the contact point of tangent from O to circle (B). Draw the perpendicular to OB from B. Circle centered in O with radius OT + OB intersects that perpendicular at U. U is the center of a circle (4) tangent to (O), (A) and (B). The parallel to OB from U intersects (B) at V which is the contact point of (4) and (B). IV intersects circle (B) at K, which is the contact point of circle (4) with circle (B). The center of circle (4) is hence the intersect point O4 of BK with IU. Draw circle (4) as centered in O4 and going through K. Circle (5) is homothetic of circle (4) : The parallel to B-O4 from C intersects O-O4 at O5 O-O4 intersects circle (4) at R, on the same side as circle (C). Circle (5) is centered in O5 and goes through R. Constructing a circle tangent to three given circles is the Apollonius problem. The general solution gives 8 such circles. This reduces if two of the given circles are tangent, wich is our case. However we could wonder where the additional solutions disappear... These additional solutions in our case degenerate into all circles which are tangent to both (A) and (B) precisely at their contact point I. This results into an infinitely many circles (4) and (5) with their centers on the ABC line. The intersection of O4-O5 and ABC line is then undefined and is no more point O but any point on ABC. These degenerate solutions are rejected : Another missing constraint is that each circle should touch any of the others externally only I found a wonderfull paper giving the general construction of the 8 Apollonius circles, by Gish and Ribando. (http://www.ajur.uni.edu/v3n1/Gisch%20and%20Ribando.pdf) This is rather complicated in the general case, when none of the given circles are tangent nor reduced to a single point. -- philippe (chephip at free dot fr) === Subject: Re: Circle construction problem >> Points A, B, and C lie in a straight line and are the centres of 3 >> circles. The circle centred on B touches those centred on A and C. Point O >> is the intersection between the straight line and a line tangent to all 3 >> circles. >> The problem is to construct 2 more circles, such that the first touches >> the circles centred on A and B, and the second touches those centred on B >> and C, and the 2 new circles also touch each other. > ... > HINT: Line of centers and common tangents are all collinear at O. > I can see what youre getting at. The tangents are easy to construct once >> the 4th and 5th circles have been drawn. But how do you determine where >> those tangents will be drawn beforehand? >> Tim > ANOTHER HINT: Project by Inversion a set of hexagonally packed circles > with respect to one of circle centers as pole O and using any radius > for circle of inversion. > However the original problem doesnt ask draw any 5 circles such and > such > but *given* 3 circles such and such, *find* 2 additional circles such > and such. > And also I wonder how you transform the 3 *given* circles by inversion > in 3 *equal* > circles ! (if they are not equal, the hexagonal trick doesnt work) > Your notation of pole O is confusing as there is already a point O > defined as > intersection of the two common tangents of the 3 given circles. > I shall name P the inversion pole, center of one of your hex lattice > circles. > I understand that P = B center of one of the circles. > Inverting the two equal tangent circles to the one you choosed with > pole = center > results obviously into 2 equal circles. > Given circles (A) (B) (C) are *unequal* as their common tangents > intersect. > So your trick of the set of hexagonally packed circles doesnt work. > Hint : > The 4th and 5th circles are undetermined. > There is an infinite number of choices for the 4th circle. > Once choosen, there are 2 possibilities for the 5th. > You need an additional constraint on the choice of the 4th and 5th > circles. > Method : > Choose and construct a 4th circle tangent to (A) and (B). > As I said there are an infinite number of such circles. > The easiest case is to arbitrarily choose the radius R4 of this 4th > circle. > Its center is then at intersect point of circles centered in A with > radius RA + R4 > and B with radius RB + R4. > Other additional constraint are more difficult, for instance if (4) > should go > through a given point, or be tangent to some other given line or > circle, or should > be related to 5th circle, for instance find two *equal* additional > circles ... > and may be cant be constructed (with compass and straightedge). > Now construct the 5th circle as the 2 only possibilities for a circle > tangent to > (B) (C) and (4) > To do this, use an inversion with pole the contact point I of (B) and > (4), and such > as circle (C) is unchanged. > The two circles (B) and (4) are transformed into two parallel lines > (4) and (B), > easy to construct. > Now the image (5) of (5) is a circle tangent to those two parallel > lines, > and tangent to circle (C)=(C). > This gives the two possibilities for circle (5). > Construct the inverse of circle (5). > OT Note : I couldnt read original post (not on news server) > I had to search through Google groups to get it. > Hopefully, this time Google groups was OK ... which I left out of the original problem. As you werent able to see that message, I repeat it here: Points A, B, and C lie in a straight line and are the centres of 3 circles. The circle centred on B touches those centred on A and C. Point O is the intersection between the straight line and a line tangent to all 3 circles. The problem is to construct 2 more circles, such that the first touches the circles centred on A and B, and the second touches those centred on B and C, and the 2 new circles also touch each other. The missing constraint is that the line of centres of the 2 new circles should also pass through point O. (Another missing constraint is that each circle should touch any of the others externally only.) And yes, Id like to be able to construct this by straightedge and compass. Can you help? Tim (For e-mail, sub-eliminate.) === Subject: Re: Circle construction problem >which I left out of the original problem. As you werent able to see that >message, I repeat it here: >Points A, B, and C lie in a straight line and are the centres of 3 circles. >The circle centred on B touches those centred on A and C. Point O is the >intersection between the straight line and a line tangent to all 3 circles. >The problem is to construct 2 more circles, such that the first touches the >circles centred on A and B, and the second touches those centred on B and C, >and the 2 new circles also touch each other. >The missing constraint is that the line of centres of the 2 new circles >should also pass through point O. (Another missing constraint is that each >circle should touch any of the others externally only.) And yes, Id like to >be able to construct this by straightedge and compass. >Can you help? >Tim >(For e-mail, sub-eliminate.) Are hunches permitted on Sci Math ? If the radii of the first three circles are a, b, c and the radii of the two required circles are d (touching a and b) and e (touching b and c) , then I suspect that a,d,b,e and c are in geometric progression, with ratio of SQRT(b/c) So radius d = SQRT(a*b) and radius e = SQRT(b*c) Both of which are easily constructed. Hope this helps Geoff G === Subject: Re: Circle construction problem >> Points A, B, and C lie in a straight line and are the centres of 3 circles. >> The circle centred on B touches those centred on A and C. Point O is the >> intersection between the straight line and a line tangent to all 3 circles. >> The problem is to construct 2 more circles, such that the first touches the >> circles centred on A and B, and the second touches those centred on B and C, >> and the 2 new circles also touch each other. >> The missing constraint is that the line of centres of the 2 new circles >> should also pass through point O. (Another missing constraint is that each >> circle should touch any of the others externally only.) And yes, Id like to >> be able to construct this by straightedge and compass. > Are hunches permitted on Sci Math ? > If the radii of the first three circles are a, b, c and the radii of > the two required circles are d (touching a and b) and e (touching b > and c) , then I suspect that a,d,b,e and c are in geometric > progression, with ratio of SQRT(b/c) > So radius d = SQRT(a*b) and radius e = SQRT(b*c) > Both of which are easily constructed. > Hope this helps Hunches are fine ! However they have to be prooved or disprooved one day... This one is true. Consider circles (A) (B) and (D)=(4) Contact points with the common tangent U on (A) and T on B I the contact point of (A) and B and J the contact point of (B) and (C) Homothecy ratio of (B) and (A) is OT/OU = OJ/OI By the inversion proof I gave, (D) is tangent to circle with radius OT and similarly, is also tangent to circle with radius OU Let S the contact point of the tangent from O to (D) The same arguments prooves that OS = OI : (A) and (B) tangent to circle with radius OS. Hence the similitude ratio of (D) and (A) is OI/OU Circles A and B being homothetic, OJ/OT = OI/OU Power of I relative to circle (B) is OT^2 = OI.OJ that is OJ/OT = OT/OI hence OT/OI = OI/OU = sqrt(OT/OU) QED. Construction is now easy as (D) is just equal to circle with diameter TU ! -- philippe (chephip at free dot fr) === Subject: New math for biology? ÔBack to the Future, the following sentence is stated at the bottom of the 2nd to last paragraph, To understand biology, is new mathematics required, the way that string theory requires new mathematics? Why is new math required for biology? Please provide any references to what the above for futher reading. Brett === Subject: Re: New math for biology? titled > ÔBack to the Future, the following sentence is stated at the bottom of the > 2nd to last paragraph, To understand biology, is new mathematics required, > the way that string theory requires new mathematics? Off topic, but while were on the subject, why does string theory require new mathematics. Specifically, what _is_ it a bout string theory that cannot be solved by mathematics today? I always figured it was some sort of PDE that we dont know how to solve, or something, but the above quote makes it sound like its something more. === Subject: New math for biology? ÔBack to the Future, the following sentence is stated at the bottom of the 2nd to last paragraph, To understand biology, is new mathematics required, the way that string theory requires new mathematics? Why is new math required for biology? Please provide any references to what the above for futher reading. Brett === Subject: Re: Products of Sines? Daryl McCullough escribi.97: > Does the following product have a nice closed-form solution? > sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) Let P(z) = (z^n - 1)/(z - 1) = 1 + z + z^2 + ... + z^(n-1) (#1) But also P(z) = Prod(z - z_k, k, 1, n-1) (#2) being z_k = cos(k*2pi/n) + i*sen(k*2pi/n), k = 1, 2, ..., n -1 the n-1 n-roots of unity distinct of 1. |P(1)| = |Prod(z - z_k, k, 1, n-1)| = Prod(|z - z_k|, k, 1, n-1) = Prod(d_k, k, 1, n-1) where d_k is the distance from (1, 0) to (cos(k*2pi/n), sen(k*2pi/n)). But d_k = sqrt((1 - cos(k*2pi/n))^2 + sen^2(k*2pi/n)) = sqrt(2 - 2cos(k*2pi/n)) = 2sqrt((1 - 1cos(k*2pi/n))/2) = 2*sen(kpi/n) Then |P(1)| = n = Prod(2sen(k*pi/n), k, 1, n-1) = 2^(n-1)*Prod(sen(k*pi/n), k, 1, Prod(sen(k*pi/n), k, 1, n-1) = n/2^(n-1) Curiously, it is the same that the probability of n points choosen at ramdom in a circle lie in the same semicircle. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Products of Sines? Robin Chapman says... >> Does the following product have a nice closed-form solution? >> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) >This is very well-known. Set w = exp(pi i/n). The product >is then >(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} - w^{-n+1}) >= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 - w^{-2n+2}). >Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) >= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}. Okay, Im puzzled by the step (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X - 1) -- Daryl McCullough Ithaca, NY === Subject: Re: Products of Sines? > Robin Chapman says... > Does the following product have a nice closed-form solution? sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) >>This is very well-known. Set w = exp(pi i/n). The product >>is then >>(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} - w^{-n+1}) >>= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 - w^{-2n+2}). >>Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) >>= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}. > Okay, Im puzzled by the step > (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X - 1) What is the factorization of X^n - 1 into linear factors? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: How to prove this for periodic functions? Hi all, I have an integral: f(t, s)=Integrate(g(t-u)g(t-s-u), w.r.t. u from 0 to 2) where the function g(t) is a periodic function with period 2 where t from -inf to +inf... I want to prove that the above integral is dependent on t, or the integral is independent of t... basically, I want to see if f(t, s) is a function of both t and s or it is a function of s only... I am wondering how to do that? I am also wondering if the shape of g(t) matters? say, for example, g(t) is a periodic sawtooth shape, g(0)=-1, g(2)=+1, then it immediately drop down so g(2+)=-1, g(4)=+1, g(4+)=-1, so on and so forth...? === Subject: Re: How to prove this for periodic functions? > Hi all, > I have an integral: > f(t, s)=Integrate(g(t-u)g(t-s-u), w.r.t. u from 0 to 2) > where the function g(t) is a periodic function with period 2 where t > from -inf to +inf... > I want to prove that the above integral is dependent on t, or the integral > is independent of t... basically, I want to see if f(t, s) is a function of > both t and s or it is a function of s only... > I am wondering how to do that? > I am also wondering if the shape of g(t) matters? say, for example, g(t) is > a periodic sawtooth shape, g(0)=-1, g(2)=+1, then it immediately drop down > so g(2+)=-1, g(4)=+1, g(4+)=-1, so on and so forth...? Independent of t. f(s,t_1)=int(u=0,u=2)[g(t_1-u)g(t_1-s-u)] f(s,t_2)=int(u=0,u=2)[g(t_2-u)g(t_2-s-u)]=int(u=t_2-t_1,u=t_2 -t_1+2)[g(t_2-u )g(t_2-s-u)]=int(u=0,u=2)[g(t_1-u)g(t_1-s-u)]=f(s,t_1)