mm-300 === Subject: : Re: please help with equationDon't worry about it, I recalled the rule. The ratio of the two frequencies has to be a rational number in order for f1-f2 to be periodic.Thanks for the reply.Nick> Referring to f1(t)=a*sin(w*t+p)and f2(t)=a*sin(c*w*t+p):>>Also, c is not tied to>>integer values only, it can be any positive real number. Functions f1>>and f2 are periodic signals, therefore the f1-f2 is periodic as well,> If that were correct, the problem might be easier. But if c is irrational> (and a is nonzero), then f1-f2 is not periodic.> David Cantrell === Subject: : Re: please help with equationI don't get that :). Isn't the sum (difference) of two periodic functions a periodic function? Please do explain, my maths are getting rusty.The problem applies to a real application, so the signals f1 and f2 are measured within some precision limits, which I guess could lead to the assumption that c is a rational number (a fraction of integers but not strictly an integer). How would that change things?Thanks.Nick> Referring to f1(t)=a*sin(w*t+p)and f2(t)=a*sin(c*w*t+p):>>Also, c is not tied to>>integer values only, it can be any positive real number. Functions f1>>and f2 are periodic signals, therefore the f1-f2 is periodic as well,> If that were correct, the problem might be easier. But if c is irrational> (and a is nonzero), then f1-f2 is not periodic.> David Cantrell === Subject: : Re: please help with equation> I don't get that :). Isn't the sum (difference) of two periodic> functions a periodic function? Please do explain, my maths are getting> rusty.No. Take a look athttp://webpages.acs.ttu.edu/tkarp/EE3323/chapter2.pdfunder almost periodic signals. However, if you replace the irrational numberin the example [I think it is sin(t) + sin(sqrt(2)t) ] by a rational number itbecomes periodic. Note also, in the graph they provide the sum of the two sinefuctions occassionally hits two (or close). That is what I was getting at inthe original question.Bill> The problem applies to a real application, so the signals f1 and f2 are> measured within some precision limits, which I guess could lead to the> assumption that c is a rational number (a fraction of integers but not> strictly an integer). How would that change things?> Thanks.> Nick> Referring to f1(t)=a*sin(w*t+p)and f2(t)=a*sin(c*w*t+p):>Also, c is not tied to>>integer values only, it can be any positive real number. Functions f1>>and f2 are periodic signals, therefore the f1-f2 is periodic as well,>If that were correct, the problem might be easier. But if c is irrational> (and a is nonzero), then f1-f2 is not periodic.David Cantrell === Subject: : Re: Al's Incessant Irritating Bark .> Hi Hanson ,> Of Al , You say :> He is a Southland dude of CA. ... Be careful . If you say so , I don't know much about him .I'm sure that his incessant irritating bark> is much worse than his bite .> I think what Jeff is saying is> that posters ought to say to Al,> Bite me!> when he barks at them.What I want to know is, how Google's target ad software came upwith fertility tests to accompany this thread. Maybe they are abetter way of telling when your bitch is in heat, other than herincessant barking. === Subject: : Re: using / applying Chebychev polynomial fitAnother link:http://www.cs.bham.ac.uk/~slb/sem307/g84.htmlRgds.... Celeste> Hi folks,> I need some serious interpolation help. I am working on taking a> general> distribution of data and trying to uniformly distribute it. This is> necessary for the learning phase of a machine learning algorithm.> I found a text speaking of a Chebychev polynomial fit (see text exerpt> below). My question is, 'Does anyone know what this Chebychev> polynomial' methodology is?' I would like to learn about it in order> to apply it to my application.> Any help would be apprecia. If you could point me to the right> text and maybe even software this would help.> Thanks> Anthony> === Subject:> Distribution Density Function (DDF) Conversion> This conversion function builds a distribution histogram from the raw> input data stored in the training file and constructs a custom> conversion function using a Chebychev polynomial fit. This is the most> general purpose conversion function and produces a reasonable level of> symmetry (a.k.a. uniformly distribu) for most input distribution> formats. This function is also the default used for stimulus> conversion.> theta = DDF(c*(s - m)/z)> where: theta = complex phase angle> c = conversion coefficient> s = raw input data> m = mean of input distribution> z = standard deviation of input distribution> DDF = Chebychev Polynomial> The sigmoid coefficient c modifies the size of the bins within the> distribution histogram and thus effects the span and degree of> resolution for the resultant polynomial fit. The size of the bins are> modified in linear proportion to the conversion coefficient. The> default value for the conversion> coefficient is 1.0, however it may be modified to any value within the> range of 0.1 to 10. Below is illustra the effect of varying the> conversion coefficient on an example of Gaussian distribu data.> (Some illlustrations in the text are here which show the distribution> histogram of the raw data, the 'sigmoidal' transformation funtion, I> guess the Chebychev polynomial, and finally, the resultant> 'uniformally distribu' data.> Similar to the other illustrations the horizontal axis of the DDF> conversion function represents the raw input values, plot as> standard deviations from the mean. The vertical axis of the DDF> conversion plot shows the phase angle output. As you can see from the> above plots the modification of the conversion coefficient has a> subtler effect upon the conversion function and it's efficiency. In> all of the above cases the conversion is adequate, however on closer> inspection one finds that the results from coefficient 1.0 are> slightly better (symmetry rating of 9.6 vs. approximately 9.2 for> coefficient values of 2.0 and 0.5). In most cases a conversion> coefficient of 1.0 will produce more than adequate results. === Subject: : Re: using / applying Chebychev polynomial fithttp://www.uni-koeln.de/REDUCE/numeric/section3_7.htmlRgds. ..Celeste> Hi folks,> I need some serious interpolation help. I am working on taking a> general> distribution of data and trying to uniformly distribute it. This is> necessary for the learning phase of a machine learning algorithm.> I found a text speaking of a Chebychev polynomial fit (see text exerpt> below). My question is, 'Does anyone know what this Chebychev> polynomial' methodology is?' I would like to learn about it in order> to apply it to my application.> Any help would be apprecia. If you could point me to the right> text and maybe even software this would help.> Thanks> Anthony> === Subject:> Distribution Density Function (DDF) Conversion> This conversion function builds a distribution histogram from the raw> input data stored in the training file and constructs a custom> conversion function using a Chebychev polynomial fit. This is the most> general purpose conversion function and produces a reasonable level of> symmetry (a.k.a. uniformly distribu) for most input distribution> formats. This function is also the default used for stimulus> conversion.> theta = DDF(c*(s - m)/z)> where: theta = complex phase angle> c = conversion coefficient> s = raw input data> m = mean of input distribution> z = standard deviation of input distribution> DDF = Chebychev Polynomial> The sigmoid coefficient c modifies the size of the bins within the> distribution histogram and thus effects the span and degree of> resolution for the resultant polynomial fit. The size of the bins are> modified in linear proportion to the conversion coefficient. The> default value for the conversion> coefficient is 1.0, however it may be modified to any value within the> range of 0.1 to 10. Below is illustra the effect of varying the> conversion coefficient on an example of Gaussian distribu data.> (Some illlustrations in the text are here which show the distribution> histogram of the raw data, the 'sigmoidal' transformation funtion, I> guess the Chebychev polynomial, and finally, the resultant> 'uniformally distribu' data.> Similar to the other illustrations the horizontal axis of the DDF> conversion function represents the raw input values, plot as> standard deviations from the mean. The vertical axis of the DDF> conversion plot shows the phase angle output. As you can see from the> above plots the modification of the conversion coefficient has a> subtler effect upon the conversion function and it's efficiency. In> all of the above cases the conversion is adequate, however on closer> inspection one finds that the results from coefficient 1.0 are> slightly better (symmetry rating of 9.6 vs. approximately 9.2 for> coefficient values of 2.0 and 0.5). In most cases a conversion> coefficient of 1.0 will produce more than adequate results. === Subject: : Re: What's the difference between 1/n, and n/1> 1/n - n and n - 1/n have a simple numerical relationship,> no matter whether n <> 1.Suposin' it's zero Brian?> Cut 1/n - n and n - 1/n have a simple numerical relationship,> no matter whether n <> 1.Suposin' it's zero Brian?Cut<> For n - 1/n, when n = 0,> my hand calculator gives 0 - [Error],> the mainframe computer gives 0 - [divide exception],> and the old mechanical adding machine gives 0 - [it's still cranking onit!].> Double-AYes it will crank maybe forever; as would by Casio Mini: That's 'cuz it willtake them forever to come up with the answer: Which is that 1/0 is infinite!I think.Modern computers nip that answer in the bud; 'cuz they ain't got time, orare programed not to waste it on the impossible goal of a final answer toinfinity; or sumpin. === Subject: : Re: Irony> Looking at a book on fractal image compression.> Towards the start he says something about closed and bounded> subsets of the plane - he's trying to be friendly to people who know> no math, so in a footnote on that page he says that closed and> bounded are just technicalities needed to make the proofs work.> He states> (i) the reason for the words closed and bounded is to minimize> complaints from mathematicians.> Then a little later he gives the definition (not making this up):> (ii) a metric space is compact if it's closed and bounded.> Looks like those words are not doing what they're intended to do...> ************************>I guess I have to admire your restraint in not pointingout that the so-called definition is... in-complete!-- chip === Subject: : Re: Irony>> Looking at a book on fractal image compression.>> Towards the start he says something about closed and bounded>> subsets of the plane - he's trying to be friendly to people who know>> no math, so in a footnote on that page he says that closed and>> bounded are just technicalities needed to make the proofs work.>> He states>> (i) the reason for the words closed and bounded is to minimize>> complaints from mathematicians.>> Then a little later he gives the definition (not making this up):>> (ii) a metric space is compact if it's closed and bounded.>> Looks like those words are not doing what they're intended to do...>> ************************>I guess I have to admire your restraint in not pointing>out that the so-called definition is... in-complete!Totally.>-- chip************************ === Subject: : Re: Extroverts on Usenet ?In sci.physics, mathedman<3f79a084.2097617@ netnews.worldnet.att.net>:>>Hi Hanson ,>> You call Usenet a : Choirs for the extrover . >>That's an interesting concept ... Usenet extroverts .>> Extrovert , Noun , >> A person concerned more with practical realities >> than with inner thoughts and feelings . > Help me out .> . I missed the mathematics here.Not to mention the physics and chemistry.-- #191, ewill3@earthlink.net -- although there can be a certainchemistry between two peopleIt's still legal to go .sigless. === Subject: : Re: Extroverts on Usenet ?(snip)>> . I missed the mathematics here.>Not to mention the physics and chemistry.Quite ... but don't help to continue it ... do this :)Bruce------------------------------------------------------- -------------Oook ! === Subject: : Re: Subfield of algebraics?>Dave, >>What is that supposed to mean? You're exact, > Yes.>you're implying that the>product of infinitely many countable sets may be uncountable. > No, I didn't imply that. It's true, of course, but doesn't follow from> anything I said.>Is it>not always? For example, consider NxNxNx.... There's a map from N>over NxN like there's a map from R over RxR, express the number as a>binary sequence a1 a2 a3 a4 and map that to the element (a1 a3 ... ,>a2 a4...). Similarly there's a map from non-negative Q (the set of>ordered pairs of elements of N, or NxN) over NxNxNxN, for each>rational p/q express the coordinate (p1 p3 ..., p2 p4 ..., q1 q3 ...,>q2 q4 ...). Similarly there's a map from N to NxNxNxNx...xN for>finitely many dimensions. If I can map N over the product of finitely>many instances (copies) of N, why can't I map Q over the product of>infinitely many copies of N?> Curiously, I'm a little confused by what you write here - are you> claiming that NxNx... is countable or uncountable? In fact it's> uncountable, and that's why you can't map Q _onto_ it.>Are you trying to tell me that R doesn't map to the product of>infinitely many copies of R? > No, why would you think I was trying to tell you that?> What about RxR? Does |RxRx...xR| thus>equal aleph_2? Would the product of that product with itself>countably infinitely many times thus have a cardinality to equal>aleph_3?>>How does R biject with NxNxNx...xN for countably infinitely many>dimensions? Does it?>Stephen, excuse me for writing on your thread about the algebraics, Iwas just writing to something Dave here has pos. I enjoy readingyour posts.Dave, what's an injection from RxRx...xRx... to R?Calling the unit interval of reals R, It's simple to determine aninjection from RxRx...xR, the product of finitely many copies of R, toR, simply for each binary coordinate (a1a2..., b1b2..., ..., z1z2...)assign it to map to a1b1...z1a2b2...z2a3.... I'm wondering thoughwhat is an injection from the product of infinitely many copies of Rto R.How might it be possible to diagonalize a function from R to R^N? Howmight it be possible to illustrate a bijection from R^N to P(R)?I consider these things, for at least five minutes, and have not asufficient answer to these questions.By the same token, I haven't thought of an injection from R^N to R. Now I have, heh heh heh heh. But seriously, What's a function thatinjects the product of infinitely many copies of R to R?The product of infinitely many copies of the reals is analogous to aninfinite-dimensional real vector space.Ross === Subject: : Re: Subfield of algebraics?>>Dave, >>What is that supposed to mean? You're exact, >> Yes.>>you're implying that the>>product of infinitely many countable sets may be uncountable. >> No, I didn't imply that. It's true, of course, but doesn't follow from>> anything I said.>>Is it>>not always? For example, consider NxNxNx.... There's a map from N>>over NxN like there's a map from R over RxR, express the number as a>>binary sequence a1 a2 a3 a4 and map that to the element (a1 a3 ... ,>>a2 a4...). Similarly there's a map from non-negative Q (the set of>>ordered pairs of elements of N, or NxN) over NxNxNxN, for each>>rational p/q express the coordinate (p1 p3 ..., p2 p4 ..., q1 q3 ...,>>q2 q4 ...). Similarly there's a map from N to NxNxNxNx...xN for>>finitely many dimensions. If I can map N over the product of finitely>>many instances (copies) of N, why can't I map Q over the product of>>infinitely many copies of N?>> Curiously, I'm a little confused by what you write here - are you>> claiming that NxNx... is countable or uncountable? In fact it's>> uncountable, and that's why you can't map Q _onto_ it.>>Are you trying to tell me that R doesn't map to the product of>>infinitely many copies of R? >> No, why would you think I was trying to tell you that?>> What about RxR? Does |RxRx...xR| thus>>equal aleph_2? Would the product of that product with itself>>countably infinitely many times thus have a cardinality to equal>>aleph_3?>>How does R biject with NxNxNx...xN for countably infinitely many>>dimensions? Does it?>Stephen, excuse me for writing on your thread about the algebraics, I>was just writing to something Dave here has pos. I enjoy reading>your posts.>Dave, what's an injection from RxRx...xRx... to R?Let phi be a bijection from R onto the power set P(N). LetA_1, A_2, ... be disjoint infinite subsets of N with union equal to N.Let f_j be a bijection from N onto A_j. For S a subset of N definef(S) = {f(x) : x in S}, as usual.Then (S1, S2, ...) |-> union(f_j(S_j))defines a bijection from P(N)^N to P(N), and hence (x1, x2, ...) |-> phi^(-1)(union(f_j(phi(x_j)))defines a bijection from R^N to R.>Calling the unit interval of reals R, It's simple to determine an>injection from RxRx...xR, the product of finitely many copies of R, to>R, simply for each binary coordinate (a1a2..., b1b2..., ..., z1z2...)>assign it to map to a1b1...z1a2b2...z2a3.... I'm wondering though>what is an injection from the product of infinitely many copies of R>to R.>How might it be possible to diagonalize a function from R to R^N? How>might it be possible to illustrate a bijection from R^N to P(R)?>I consider these things, for at least five minutes, and have not a>sufficient answer to these questions.>By the same token, I haven't thought of an injection from R^N to R. >Now I have, heh heh heh heh. But seriously, What's a function that>injects the product of infinitely many copies of R to R?>The product of infinitely many copies of the reals is analogous to an>infinite-dimensional real vector space.>Ross************************ === Subject: : Re: Matrix multiplication orderOf course not. The estimate for x will depend on the relationship betweenthe range and null space of A and B.Try a simple MATLAB example.....v=[3 4]';w=[-4 3]';A=2*v*v+1.e-5*w*w';e1=[1 0]';e2=[0 1]';B=5*e1*e1'+1.e-5*e2*e2';C1=AB;C2=BA;and you want to calc x such that for a given y , y=Ci*x where i=1,2x estima from C1, will always be in the direction of the e1, becauseinformation in the direction of e2 are destroyed upon multiplication of Bwhile x estima from C2 will always be in the direction of v, you can seethat by a similar arguement.I hope that helps.Alien+> I know, both systems are bad; but will both of them> give _equal_ performance on average ?> There is no reason, one should favor either approach (1) or (2), ineither> cases part of the information content in x is destroyed after being> transformed by A and B.You may use regularization to find a robust solution for x.> Google on Keyword Per Christian Hansen for his matlab toolbox thatoffer> techniques to obtain regularized solutions for ordinary least squares> problems.Alien+>>I have a question regarding the order of multiplication> for two systems.The first one is given as y=ABx+n (1)> while the second one is y=BAx+n (2).n is a noise vector with gaussian distribu entries, with> a mean of zero and unit variance. The elements of A also> follow the same identical gaussian distribution while B> is a correlation matrix.To find x I can invert (AB) or (BA), and obtain a noisy> estimate from y.The problem is with the correlation matrix B. If it's> ill-condtioned then with (1) _some_ elements of the estima x> are distor quite heavily.> In contrast, with (2) the noise-enhancement is typically> spread across _all_ elements.However, on average, it looks to me that the noise enhancement> due to B is the same for both systems.> Is it possible to show that in some way or am I quite wrong here ?Should I expect that, on average, with (2) my estimates of x will> be much worse than under (1) ? Ie upper bounded by the sum> of noise distortions inv(B)n ?Any help will be apprecia. === Subject: : Re: A Tensor problem>> So, if we are using the summation convention then>> 'i' is just a dummy index, it could be any letter>> I.e.>> a_ii = a_ll = a_jj = a_ff = a_pp>> which are all = a_11 + a_22 + a_33>> adam>Unfortunately, it doesn't work - in terms of giving the answer>required, e.g to computing the form:> a_ij - 2/5 [Delta_ij * a_ll]>Where Delta_ij is the Kronecker delta. E.g. on computing this form>(subst. in values of the respective matrices) one obtains:>(1.2 0 3)>(5 0.6 2)>(4 5 4.2)>However, the text gives:>(-2 0 3)>(5 -3 2)>(4 5 3)>So, exactly what is it that I'm doing wrong?>Or can we say with 100% confidence the text is in error?>Thanks in advance.What is the problem? And by that I mean, tell me what themath problem is that you're trying to solve. It's reallynot clear.adam === Subject: : Re: Groups with 16 elementsin message <70ae81fd.0309300430.382828cf@posting.google.com>:> in message <70ae81fd.0309281726.5e94cf58@posting.google.com>:> Consider the case of G genera by x and y, where x^8 = 1 = y^8.> Let X = , Y = , and yxy^(-1) = x^(-1) ; xyx^(-1) = y^(-1)Stop! This is already impossible. Remember, y must commute with> x^2. This limits the possible values of r in yxy^{-1} = x^r.Yes. r = 5. But this gives y^4 = 1, so I wonder if its possible to> have 2 cylic groups of order 8? It seems to lead to a contradiction,> unless I am making another error.?? How does it give y^4 = 1 ? Since y^2 = x^2, y^4 = x^4. And> yes it's possible. We've been over this before. There are 2 such> groups of order 16, the abelian group C_8 x C_2 = y^2 = 1> and the modular group Mod_{16} with the additional> relation yxy^{-1} = x^5. In both cases the 2 cyclic subgroups of> order 8 are and . That both groups do indeed exist can> be seen for example by constructing them as permutation groups:C_8 x C_2 = <(1,2,3,4,5,6,7,8), (9,10)Mod_{16} = <(1,2,3,4,5,6,7,8), (1,5)(3,7)>(These are their smallest-degree faithful permutation> representations.)> The Mod_{16} = group has, as you say,> x^8 = y^2 = 1 ; yxy^{-1} = x^5.> But it has only 1 cyclic subgroup of order 8.> Note the xyxy = x^6 y^2 = x^8 = 1, so (xy)^2 = 1.> So is not of order 8.Um, no. Since y^2 = 1, not x^2, therefore (xy)^2 = x^6. I seenow that I probably confused you by switching mid-paragraph froma presentation in which y^2 = x^2 to one in which y^2 = 1. The yin the first presentation corresponds to {x^3}y or {x^7}y in thesecond. It might help to go back and check all the y's andchange the relevant ones to, say, z's. That is, check everywhereyou or I have defined y^2, and if it's = 1, change all y's toz's (until someone redefine's y^2 again).> This group is like D_4 (or D_n), the dihedral group, which has> only 1 cyclic subgroup of order n. (This groups has> yx^2y^(-1) = x^2, but otherwise it is like D_8--I think.)> For n = 8, there are other> subgroups of order 8, but they aren't cyclic.Mod_{16} = has precisely 3subgroups of order 8. Two are cyclic C_8, namely and ,and one is the abelian direct product C_4 x C_2, namely .> I am looking for something like quaternions with order 8> instead of 4.There's no group of order 16 with more than 2 subgroupsisomorphic to C_8. The generalization of the quaternion group tohigher powers of 2 that finite group theorists usually use is. Forn>=2, all 2^n elements not in have order 4. There are 3subgroups of index 2; for n>=3, 1 is cyclic and 2 arequaternion.> Do quaternions have a rep. as a subgroup of S_4?> I am aware of the rep in terms of Pauli matrices.No, the smallest faithful permutation rep of Q_8 is the regularone: with, say,x = (1,2,3,4)(5,6,7,8) and w = (1,5,3,7)(2,8,4,6).It's easy to see Q_8 can't be a subgroup of S_4 since Sylow'sTheorem tells us the latter has, up to isomorphism, only 1subgroup of order 8, and of course that's dihedral.> I recall the Dirac matrices (there are 16). I wonder if they> for a group, and if so what group they represent.Interesting question. This would appear to be just themultiplicative group genera by the usual basis vectors of theClifford algebra Cl(3,1), which has order 32. (You have to throwin their negatives, just as you do with quaternions, to get agroup of order 8, not 4.) And of course the commutationrelations mean no element has order more than 4. But there aresomething like a couple of dozen such groups of order 32, andI'm not familiar enough with them to know if there are anystandard naming conventions for the ones corresponding toClifford algebras.-- Jim Heckman === Subject: : Re: Question About Prime Numbers> You seem to be locking your AP's length to its first term.> In which case, the next few results can be found here:> http://listserv.nodak.edu/scripts/wa.exe?A2=ind0111&L=nmbrthry &P=R233> Thanks .Zakir Zeidov sent me this interesting message:[QUOTE]Interesting problem, Russell! For p=11, minimal d = 4911773580 (OEISA088430), and AP contains maximal number, 11, primes. For p=13, dshould be a factor of 2310. Who first find it (and then try17,19,...)? Zak BTW I guess that found d is indeed minimal not unique-there is no reason of absense of other larger d's.[END/QUOTE]So, for 11, minimal d = 4911773580 ...very interesting! 114911773591982354717114735320751196470943312455886791129470641 49134382415071392941886514420596223149117735811 === Subject: : Re: polysigned numbers>>I haven't checked any of the field properties, but I suspect that this >>gives us a field that is isomorphic to the complex numbers by the linear >>transformation:>>f((x,0,0)) = x*e^(i 2pi/3)>>f((0,x,0)) = x*e^(i 4pi/3)>>f((0,0,x)) = x>>Does this correspond with what you have in mind?> Hi Will.> The transform f that you mention above seems coherent.> The direct transform to C from Y that I have used goes as follows:> z(y) = s - ( m + p )/2 + i sqrt(3)( m - p )/2.> where s is the star component, m the minus, and p the plus> component of y.> I am sorry, but I do not see the need for star as a product operator> when parenthesis do just fine, especially when the star is being used> to denote a new sign. I'll agree with that. I was thinking in terms of abstract algebra, where a symbol for the operator is generally given and used explicitly. > I have stopped using the comma form of> three-signed values in favor of the pure sum, just as in complex math> people use a + bi instead of (a,b).> I am most interes in any qualms that you may have with this math,> or ideas that are applicable to it. I am happy with the notation as it> has been developed. I do understand your qualms with it, however, the> signs chosen are mnically correct and work backward to the reals> and up to four signed arithmetic with '#' as the fourth sign.Using # as your fourth sign, does that span R^3 or still R^2? Depending on which it is, four-signed might be *more* interesting to me than three-signed arithmetic (voting for R^3).I think it's interesting that you have effectively found a way to reduce the number of signs in the complex numbers from positive real, negative real, positive imaginary, negative imaginary or 4 signs down to 3. As far as applications, it might have some uses when doing fractal analysis, but I don't know for sure. I'm thinking about things like the Mandelbrot set where you are looking at the behavior of various points under an itera function. Whether it would be more useful than the e^z notation I don't know.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Please give some hints on Probability problems> However I don't know how to handle this one.> Let X and Y be independent r.v. having geometric densities with > parameters> p1 and p2 respectively.> Find:> a) P ( x >= y)I use the notation Sum( n = a, b, g(n) )to mean the sum from n = a to b of g(n).Let f1( x ) be the Probability Density Function (PDF) for Xand f2( y ) be the PDF for Y.Since X and Y are stochastically independent, the joint PDF ofX and Y is f1( x ) f2( y ).In any case, f1( x ) = p1 (1 - p1)^(x - 1), x = 1, 2, 3, ...for a geometric distribution.Then Pr( x >= y ) = Sum( y = 1, inf, Sum( x = y, inf, f1( x ) f2( y ) ) )where inf is how I write infinity. I believe you can evaluate thosesums in close form by some manipulation of geometric series.The point is to sum over integrate the joint PDF over both randomvariables in the region defined by the probability you want toevaluate.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html 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: Generator polynomial of a product code?Thanks a lotJaco> According to Lin and Costello ( Lin, Shu, and Daniel J. Costello, Jr.,> Error Control Coding: Fundamentals and Applications, Englewood Cliffs,> N.J., Prentice-Hall, 1983.) a product code has a generator polynomial> if n1 and n2 are relatively prime, where (n1,k1)C1 and (n2,k2)C2 are> the cyclic codes used to derive the product code. (p274 - 278).> The generator polynomial for the product code can then be compu by:> g(X) = GCD[X^(n1*n2) - 1,g1(X^b*n2)g2(X^a*n1)]> where a*n1 + b*n2 = 1 (since n1 and n2 are relatively prime).> What does the notation g1(X^b*n2) mean: do I simply multiply b*n2 with> the powers of the X's of g1(X)?> I presume it means g_1(X^{b n_2}). Just replace X by X^{b n_2}> in g_1(X).> As an example, I have two codes in GF(2^4):> 1. C1 is a (15,11) Reed-Solomon code with generator polynomial> g1(X) = alpha^10 + alpha^3X + alpha^6X^2 + alpha^13X^3+X^4> g1(X^-119) = alpha^10 + alpha^3X^-119 + alpha^6X^-238 +> alpha^13X^-357 + X^-476> Horrible notation: X^-119 indeed! As> g_1(X) = X^4 + alpha^13 X^3 + ... + alpha^10> then> g_1(X^{-119}) = X^{-476} + alpha^13 X^{-357} + ... + alpha^10.> Are these steps correct thus far? What bothers me is the fact that> after the multiplication of the two polynomials, we will have X's with> negative powers. > That doesn't really matter: as we are taking gcds with X^255 - 1,> only the congruence class of the exponents modulo 255 matters.> One can replace X^{-119} with X^{136} (as 119 + 136 = 255)> if that worries you (NB X^{-119} - X^{136} = -X^{-119})(X^255 - 1)).> Also, the polynomial after multiplication will have> larger powers than 255, the highest power of (X^255-1), or doesnt this> matter?> g(X) = gcd(X^255 - 1, ...).> For cyclic codes of length n, generator polynomials etc> really only exist modulo X^n - 1. Effectively> one is working in the group algebra over the cyclic group of> order n; in practice one can always replace X^k by X^{k-n} or> X^{k+n} without loss. === Subject: : Re: Change all f=ma; to f=wa/gIn sci.math, Donald G. Shead:> The usual equation for finding force is f = ma; but that is wrong: There's> more to that insignificant little symbol [m]: Like m = f/a, and m = w/g:> The mathematical equation for the quantity of matter in any object;> body or mass, is: m = f/a = w/g:> Where we can use f/a = w/g as the working formula; so that we> can solve f = wa/g - etc. - by> simple transposition.Yeah, so?Of course you do realize that f, g, w, and a are, strictlyspeaking, vector quantities, and the expressions f/aand w/g are therefore technically meaningless. (I saytechnically because most individuals will divide anyway,using the scalar forms instead. This works for the mostpart on throw a ball up in the air-type problems butmay not be all that useful for, say, Earth-moon freetrajectory approximations.)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: Change all f=ma; to f=wa/g> In sci.math, Donald G. Shead> :> The usual equation for finding force is f = ma; but that is wrong:There's> more to that insignificant little symbol [m]: Like m = f/a, and m = w/g:The mathematical equation for the quantity of matter in any object;> body or mass, is: m = f/a = w/g:Where we can use f/a = w/g as the working formula; so that we> can solve f = wa/g - etc. - by> simple transposition.Yeah, so?> Of course you do realize that f, g, w, and a are, strictly> speaking, vector quantities, and the expressions f/a> and w/g are therefore technically meaningless.In a hosse's neck: These expressions are vector quantities only in that theycan be represen in magnitude and direction by vectors:Not so strictly speaking: All directions and speeds are _relative_; to theobservers of them. (I say> technically because most individuals will divide anyway,> using the scalar forms instead. This works for the most> part on throw a ball up in the air-type problems but> may not be all that useful for, say, Earth-moon free> trajectory approximations.)> --> #191, ewill3@earthlink.net> It's still legal to go .sigless. === Subject: : Re: Change all f=ma; to f=wa/g> Sorry, typo; I ought to have said:> I also demand that y be changed to:> sqrt[(y-304)^2]-[3/(.1^2)-(-2)/sqrt(.25)]For heavens sake; as if anyone cares. === Subject: : Re: Change all f=ma; to f=wa/g> The usual equation for finding force is f = ma; but that is wrong:There's> more to that insignificant little symbol [m]: Like m = f/a, and m = w/g:The mathematical equation for the quantity of matter in any object; bodyor> mass, is: m = f/a = w/g:Where we can use f/a = w/g as the working formula; so that we can solvef => wa/g - etc. - by> simple transposition.Mathematics should be used to simplify things; properly used it will: Massis the quantity of matter in a body; as is determined by it's inertia; whichis determined from the ratio of the net force[f] to the acceleration [a] that it causes:A body's mass and/or it's inertia is a ratio; which must be written as m =f/a = w/g:The term ma; should be written as fg/a; the term mg should be written aswa/g. If you have to argue with a professor about it, do it their way(;^);unless you want to flunk. === Subject: : Re: Change all f=ma; to f=wa/g> Mathematics should be used to simplify things; properly used it will: Mass> is the quantity of matter in a body; as is determined by it's inertia; which> is determined from the ratio of the net force> [f] to the acceleration [a] that it causes:> A body's mass and/or it's inertia is a ratio; which must be written as m => f/a = w/g:> The term ma; should be written as fg/a; the term mg should be written as> wa/g. If you have to argue with a professor about it, do it their way(;^);> unless you want to flunk.I think you have those backwards, but I know what you meant to say.You've added more symbols to an equation that was already valid(in fact, probably more so, given the issue that force andacceleration are vectors). How is that simplification? === Subject: : Re: how to prove that all even numbers can be represen in binary?In sci.math, Joona I Palaste:> Will Twentyman scribbled the following:> how does one prove that all even numbers can be represen in binary?how does one prove that all even numbers can be represen in binary?how does one prove that all even numbers can be represen in binary?how does one prove that all even numbers can be represen in binary?>> All real numbers are representable in binary. The evens are a subset of >> the reals, therefor all evens are representable in binary.>> I suspect there is something about binary that you are not >> understanding, the question seems unusual.> Not only are all reals representable in binary, they are representable> in any base whose radix is an integer greater than 1.If one allows a transfinite number of digits, perhaps.It's basically the Z[1/2] =? R problem.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: how to prove that all even numbers can be represen in binary?The Ghost In The Machine scribbled the following:> In sci.math, Joona I Palaste> : >> Will Twentyman scribbled the following:>> how does one prove that all even numbers can be represen in binary?how does one prove that all even numbers can be represen in binary?how does one prove that all even numbers can be represen in binary?how does one prove that all even numbers can be represen in binary?> All real numbers are representable in binary. The evens are a subset of > the reals, therefor all evens are representable in binary.> I suspect there is something about binary that you are not > understanding, the question seems unusual.>> Not only are all reals representable in binary, they are representable>> in any base whose radix is an integer greater than 1.> If one allows a transfinite number of digits, perhaps.> It's basically the Z[1/2] =? R problem.Sorry, I don't understand what the notation Z[1/2] means. Pleaseexplain in detail.-- /-- Joona Palaste (palaste@cc.helsinki.fi) ---------------------------| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.fi/~palaste W++ B OP+ |----------------------------------------- Finland rules! ------------/Hasta la Vista, Abie! - Bart Simpson === Subject: : Re: algebra problem?? please| Z2 * Z2 * Z2 * Z3 * Z3 * Z5 = (2*2*2*3*3*5 verbatim)[...]| 2*6*30The next question might be, Why did they representthe group as Z/2 x Z/6 x Z/30 instead of Z/2 x Z/2 x Z/2x Z/3 x Z/3 x Z/5? Certainly the decomposition as aproduct of cyclic groups of prime-power order is natural.But it appears they are using a second natural convention,where you write a finite abelian group as a productZ/d1 x Z/d2 x Z/d3 x ... x Z/dn in which d_{i+1} dividesd_i for each i=1,...,n-1 (except that they've written thefactors in the reverse order). One reason to write itthis way is that it displays the exponent of the group,defined as the least m such that g^m is the identity foreach element of the group, as d1. I seem to rememberthat at least one usual proof of the structure theorem forfinite abelian groups (or modules over a principal idealdomain) is formula in terms of such a representationof the group (module). === Subject: : Re: algebra problem?? please> But it appears they are using a second natural convention,> where you write a finite abelian group as a product> Z/d1 x Z/d2 x Z/d3 x ... x Z/dn in which d_{i+1} divides> d_i for each i=1,...,n-1 (except that they've written the> factors in the reverse order). One reason to write it> this way is that it displays the exponent of the group,> defined as the least m such that g^m is the identity for> each element of the group, as d1. I seem to remember> that at least one usual proof of the structure theorem for> finite abelian groups (or modules over a principal ideal> domain) is formula in terms of such a representation> of the group (module).The buzzword is Smith normal form.-- === Subject: : Re: Non-Polynomial Time Algorithms|What are the top five or ten most famous NP algorithms, ie. algorithms that |will not complete in polynomial time?I think you mean problems rather than algorithms. Algorithms areparticular methods for solving problems. Plenty of algorithms will failto complete in polynomial time, even though the problem is easy.I also wonder whether you mean NP-complete rather than just NP. The classP consists of yes/no problems having polynomial-time solution algorithms.NP is a class containing P. A yes/no problem is in NP if there are twopolynomials P and Q and an algorithm, such that if the answer to a case Xof the problem has the answer yes, then there is a verification Y oflength |Y| <= Q(|X|) such that when X and Y are given to the algorithm, itaccepts within time P(|X|+|Y|), and if the answer is no there does notexist such a verification Y. Roughly, they are the problems where yesanswers always can be briefly proven. NP includes all the easiest problemsas well as some apparently hard ones.It's thought that NP properly contains P, but this is hard. (There's a$1,000,000 U.S. prize for proving it or disproving it.)The class of NP-complete problems are ones in NP to which all problems inNP can be reduced in a certain sense. If any of the NP-complete problemsare in P, then the reduction would show that all NP problems are in P. Orby the contrapositive, if P<>NP, then the NP-complete problems are not inP. The NP-complete problems are pretty well expec not to be in P.(If all problems in NP can be reduced to a given problem, it's known asNP-hard, whether or not it is itself in NP. The NP-complete problemsare the NP-hard problems in NP.)| I'm familiar with the traveling |salesman problem, the knapsack problemThese are NP-complete if you formulate the first one as a yes/no question,Is there are route of length <= N?.| and the problem of factoring large |integers,Factoring is not a yes/no question as usually sta, so technically it'snot in NP, but it's very closely rela to problems in NP. Given afactorization, you can in polynomial time verify that it's correct(now that we know primality testing is in P, a recent result). For a numberof years it was known that for each prime, there's a short certificate ofits being prime, which is enough to put it in NP.As far as we know these rela problems are not NP-complete. There arealgorithms for factoring which are between polynomial and exponential time.It's not clear to me that anybody has good evidence that factoring is notin P; it's just that they've worked hard looking for a good algorithm forit, and only gotten non-polynomial ones so far.|but I understand that this is such a common difficulty in |algorithm design that there are a VAST number of problems in this category. |What are the most famous examples?The book by Garey and Johnson is perhaps a little da (70s) but has anappendix with dozens of the best-known ones. The original one is calledsatisfiability. Given a collection of n boolean variables, and a list ofclauses of the form (a1 or a2 or a3 or ... or a_k) where each a_i is eithera variable or the negation of a variable, does there exist an assignment oftrue or false to the variables which makes all the clauses true? Thatwas refined into 3-SAT which restricts the clauses to only 3 variables ornegations of variables each. Those are both NP-complete. Certain kinds ofcircuit-design problems can be shown NP-complete by showing that 3-SAT orSAT reduces to them.2-matching is in P. 2-matching is also known as the marriage problem: giventwo lists of n elements and a table of feasible pairs consisting of oneelement from each list, does there exist a way to put the elements intoone-to-one correspondence so that all the pairs are on the feasible list?But the analogous 3-matching problem, where one has three lists, and a tableof compatible triples of elements, one from each list, is NP-complete. Anumber of scheduling problems can be reduced naturally either to 3-matchingor to knapsack.Hamilton circuit is NP complete: given a graph, is there a circuit thatvisits each vertex exactly once?Integer programming: given a set of linear inequalities, is there asolution in integers? Without the restriction to integers, there arenow polynomial-time algorithms, as well as the popular simplex methodwhich although not invariably polynomial-time is usually good. With therestriction to integers, the logic problems I mentioned can be reducedto it.Given a collection of first-order axioms, a proposition, and a string,is there a proof of the proposition from the axioms which is no longerthan the string? The reason for having the input include the string andnot just the length of the string, is that I want the polynomial-timebound to be in terms of the length of the string, and not the length ofthe representation of the length. That's NP complete too. So presumablyhunting for a proof of a theorem from given axioms can never be madevery trivial, even if one is given a limit on how long the proof can be.It seems likely that the time required to find a proof will in hardcases grow exponentially in the length of the proof. If P=NP were true,the time needed would grow only polynomially in the length of the proof!There are of course assor computable-in-principle problems which arenot in NP at all, which means they are also not in P. Some of the naturalproblems which are *probably* not in P are ones which are PSPACE complete.That means that they are in PSPACE, and every problem in PSPACE reduces tothem, where PSPACE is the class of problems that can be compu inpolynomial-bounded space. PSPACE contains NP, and presumably is bigger.Being PSPACE-complete is even better evidence that a problem is not in Pthan being NP-complete.Given an N by N grid filled with chess pieces, with White to play, doesWhite have a winning strategy? That's PSPACE-complete. So is the problemwith an arbitrary N by N position in GO, or checkers. Sadly the positionsare (of course) very contrived, so it remains possible that there's aneasier optimal strategy starting from the usual starting position.The more fundamental PSPACE-complete problem is parallel to satisfiability,but is called quantified boolean formula. In a quantified boolean formulayou are allowed boolean variables, the usual logical operators and, or,and not, and quantifiers for all p and there exists a p. Thesatisfiability problem I described above is a special case of the problemwhere you're only allowed existential quantifiers. You ask whether it'strue that there exist p1,p2,...,pn which make a given boolean formula true.In quantified boolean formula, you are allowed universal quantifiers too.It's harder because quantifiers can alternate. You can ask whether thereexists a p1 (i.e., a choice of assigning either true or false to p1) suchthat for every p2 (i.e., both ways of assigning true or false to p2), thereexists a p3 such that... a given boolean formula is true. These arenaturally rela to games, because this is equivalent to asking who hasthe winning strategy in a game where the players take turns setting thevalues of p1, p2, p3,... to be true or false, and who wins at the end isdetermined by the value of the boolean formula inside the quantifiers. === Subject: : Re: Non-Polynomial Time AlgorithmsGracious, Keith! Thank you *very, very* much. I never did quite understand the concepts of NP-complete and NP-hard, and you have explained them very succintly and clearly. Bravo!To get back to the OP's intent, I think, the simplex algorithm for linear programming is widely used and in practice usually efficient, even though it is not guaranteed to run in polynomial time. My (casual) understanding of the Karmakar (interior-point) algorithm for LP is the following: In its pure form it is guaranteed to finish in polynomial time. If one tries to implement that algorithm, it generally takes a lot longer than the simplex algorithm. To get it to run faster, one makes a slight alteration - and with this, the algorithm is no longer guaranteed to complete in polynomial time.-- === Subject: : Value of rational functions on a curve at infinityDear NG,Recently I have had problems understanding the value of a rational function defined on an elliptic curve at the point of infinity. Before all these, I had to understand the degree of a several variable polynomial, and I am not sure if my understanding is correct. Let our working field be K. Given an elliptic curve of the form f(x,y)=y^2-x^3-ax-b=0. And given a function g in K[x,y]/ (ie polynomials modulo f), I do not know exactly what is the degree of g. I do know for instance that if g were in K[x,y] then the degree is the smallest integer d so that z^d*g(x/z,y/z) is in K[x,y,z] (or please correct me if I am wrong). But then I get confused when I look at polynomial class modulo f. My professor told me that the degree of the function x in K[x,y]/ and y in K[x,y]/ are different (one is of degree 2 and the other of degree 3 ? or am I wrong here?). I would appreciate it if someone could clear this matter for me. I would like for instance know what are the degrees of x,y, x/y and y/x I needed this concept because I wan to know the value of a rational function at the point of infinity in the elliptic curve. So for a function g=p(x,y)/q(x,y) in K(x,y)/ where p and q are in K[x,y]/, .. how do I determine precisely its value at infinity. I know this to some point, but I am really unsure whether it is correct. For instance I know that if I somehow know the highest degree term of p and q (that is also one thing that results into confusion!) then if the degree of p and q are equal the value of p/q at infinity is the just the ratio of the coefficients of highest degree term of p and q respectively (Again one must know how this highest degree is precisely defined and why one can write p and q in such a way that they have only one highest degree term...)So in general I am confused of the definitions and wish to learn of a precise definition (than what I know) and why such a definition is used. Lastly I do also have a general question. Especially in the field of algebraic geometry in which I am just a beginner. Why do they want to extend the rational functions of a curve (which we usually know in the affine plane) into the projective plane. How does projective geometry help and why is it of interest here. I was hin that for my questions above all I have to do is to write all the rational functions in its form in projective space (i.e. f(x,y) -> F(x:y:z) by transforming x to x/z and y to y/z.. to have the homogenous function) and use (0:0:1) to find its value at infinity. But that also confuses me .. (for instance for x/y) and besides it still doesnt answer my question about degree.I would appreciate any comment or suggestion and thanks in advance.Sincerely,Jose Capco === Subject: : Re: Value of rational functions on a curve at infinity>Dear NG,>Recently I have had problems understanding the value of a rational function >defined on an elliptic curve at the point of infinity.This makes no sense . There is no at the point of infinity. There isno such point. infinity is not a number, or a point, or a place.Thank you for having the integrity to give a valid e-mail addr> Before all these, I had >to understand the degree of a several variable polynomial, and I am not sure if >my understanding is correct. >Let our working field be K. >Given an elliptic curve of the form f(x,y)=y^2-x^3-ax-b=0. And given a function >g in K[x,y]/ (ie polynomials modulo f), I do not know exactly what is the >degree of g. I do know for instance that if g were in K[x,y] then the degree >is the smallest integer d so that z^d*g(x/z,y/z) is in K[x,y,z] (or please >correct me if I am wrong). But then I get confused when I look at polynomial >class modulo f. My professor told me that the degree of the function x in >K[x,y]/ and y in K[x,y]/ are different (one is of degree 2 and the other >of degree 3 ? or am I wrong here?). >I would appreciate it if someone could clear this matter for me. I would like >for instance know what are the degrees of x,y, x/y and y/x >I needed this concept because I wan to know the value of a rational function >at the point of infinity in the elliptic curve. So for a function >g=p(x,y)/q(x,y) in K(x,y)/ where p and q are in K[x,y]/, .. how do I >determine precisely its value at infinity. I know this to some point, but I am >really unsure whether it is correct. For instance I know that if I somehow know >the highest degree term of p and q (that is also one thing that results into >confusion!) then if the degree of p and q are equal the value of p/q at >infinity is the just the ratio of the coefficients of highest degree term of p >and q respectively (Again one must know how this highest degree is precisely >defined and why one can write p and q in such a way that they have only one >highest degree term...)>So in general I am confused of the definitions and wish to learn of a precise >definition (than what I know) and why such a definition is used. >Lastly I do also have a general question. Especially in the field of algebraic >geometry in which I am just a beginner. Why do they want to extend the rational >functions of a curve (which we usually know in the affine plane) into the >projective plane. How does projective geometry help and why is it of interest >here. I was hin that for my questions above all I have to do is to write all >the rational functions in its form in projective space (i.e. f(x,y) -> F(x:y:z) >by transforming x to x/z and y to y/z.. to have the homogenous function) and >use (0:0:1) to find its value at infinity. But that also confuses me .. (for >instance for x/y) and besides it still doesnt answer my question about degree.>I would appreciate any comment or suggestion and thanks in advance.>Sincerely,>Jose Capco === Subject: : Re: Value of rational functions on a curve at infinity> Let our working field be K.> Given an elliptic curve of the form f(x,y)=y^2-x^3-ax-b=0. And given a> function g in K[x,y]/ (ie polynomials modulo f), I do not know exactly> what is the degree of g.Write g as h(x) + k(x)y(can eliminate higher powers of y using the defining equation).Then deg(g) = max(2 deg(h), 2 deg(k) + 3).It's the order of the pole of g at infinity. (Basically x and yhave degrees 2 and 3 resp.).> I needed this concept because I wan to know the value of a rational> function at the point of infinity in the elliptic curve. So for a function> g=p(x,y)/q(x,y) in K(x,y)/ where p and q are in K[x,y]/, .. how do I> determine precisely its value at infinity.Check p and q have the same degree. Write each as h(x) + k(x)y.Isolate the terms of highest degree in each, and take their quotient.-- === Subject: : Re: Value of rational functions on a curve at infinityrjc@ivorynospamtower.freeserve.co.uk says...>Write g as h(x) + k(x)y>(can eliminate higher powers of y using the defining equation).>Then deg(g) = max(2 deg(h), 2 deg(k) + 3).>It's the order of the pole of g at infinity. (Basically x and y>have degrees 2 and 3 resp.).Why is that so? This is using the weights of x and y if I understand correctly. But I don't still understand, why does x have weight (or degree) of 2 and y of 3. It would help if you could explain a bit more. Thanks a lot.>Check p and q have the same degree. Write each as h(x) + k(x)y.>Isolate the terms of highest degree in each, and take their quotient.Why can one have write them with only one highest degree term! I believe you meant the quotient of the coefficients. Jose Capco === Subject: : Re: Value of rational functions on a curve at infinity> rjc@ivorynospamtower.freeserve.co.uk says...>>Write g as h(x) + k(x)y>>(can eliminate higher powers of y using the defining equation).>>Then deg(g) = max(2 deg(h), 2 deg(k) + 3).>>It's the order of the pole of g at infinity. (Basically x and y>>have degrees 2 and 3 resp.).> Why is that so? This is using the weights of x and y if I understand> correctly. But I don't still understand, why does x have weight (or> degree) of 2 and y of 3.As I said, that is the order of the poles at infinity.Consider the uniformization by means of the Weierstrss P-function:basically x = P(z), y = P'(z) (up to some constant factors).P has double poles at the lattice points, P' has triplepoles. 2 and 3.Or if you prefer, a uniformizer at infinity is t = y/x. Onecan expand x as a Laurent series x = b_{-2}t^{-2} + b_{-1} t^{-1} + b_0 + b_1 t + ...and similarlyy = c_{-3} t^{-3} + ... .>>Check p and q have the same degree. Write each as h(x) + k(x)y.>>Isolate the terms of highest degree in each, and take their quotient.> Why can one have write them with only one highest degree term!The terms in h(x) have degrees 0, 2, 4, 6, ... etc.The terms in y k(x) have degrees 3, 5, 7, 9, ... etc.Isn't it obvious that there's a unique highest degree term?> I believe> you meant the quotient of the coefficients.Believe what you like. It makes no difference dividing the highest degreeterms or their coefficents (provided that p/q has neither polenor zero at infinity).-- === Subject: : Re: Why poles so named?>Zeros so named makes sense but what has infinity got to do with a pole?> This has been discussed here before, but I don't remember all the details.> Some engineers explain it as the modulus of a complex-valued function> depic as the height of a circus tent above the complex plane. At> the zeros the tent is nailed to the floor and at the (tent) poles the> height is infinite. That metaphor works for me. But, it's not surprising that that methaphor works in Aeronautical Engineering, since all AEs worry about is dynamic stabilty. But, try that methaphor in Signal Processing and you'll find that the tent collapses, just like the Physics it came from. === Subject: : Joe Uptaught (Was Re: David Ullrich on Identity)I notice you pos on the epsilon operator some time back.> I am wondering whether, where quanta are concerned, an> epsilon operator might be of some use in referring to> buggers that can't be named:> Yes, you are a funny dude.> It seems to me that all references to 'unlikely objects' are buggers.> I am deligh that you noticed my remarks about indefinite descriptions.> Your opinions are always, more than welcome.> The Boyz of the group notwithstanding.> It is disconcerting that the Boyz always resort to insults and flaming,> in one way or other. Franz is simply more blatent than David,> But, bad manners 'sucks', don't you think so?See below.> My expression of 'indefinite descriptions' is derived from Russell's 'On> Denoting'. And, I believe that it shows that Hilbert's 'epsilon> function' rendition is in error.> If you are interes, I will persue this point further.the Stanford Encyclopedia.> Please keep in mind that, I am not educa like you and most others> on this board are. I have not attended any logic courses at all. I> am self- decieved, (hahaha). Please show that I am wrong, slowly: so> that I can understand.Everybody makes mistakes. But you and I want to correct our mistakes;the unabashedly uptaught make a fetish of theirs.Which is one reason why it is perhaps more important(and more productive) for you and me (and others like us)to find out what sociologists of knowledge have written aboutthe mores of Joe Uptaught than it is to try to emulatemathie speak, mathie rigidity, or mathie arrogance. A propos, here are cites from Pierre Bourdieu's_Language & Symbolic Power_ (the titles are mine).Enjoy! CensorshipAmong the most effective and best concealed censorshipsare all those which consist in excluding different agentsfrom communication by excluding them from the groups whichspeak or the places which allow one to speak with authority.In order to explain what may or may not be said in a group,one has to take into account not only the symbolic relationsof power which become established within it and which deprivecertain individuals (e.g. women) of the possibility of speaking,but also the laws of group formation themselves (e.g. the logicof conscious or unconscious exclusion) which function like aprior censorship. (p. 138) There Oughta Be A LawAs you say, my good knight! There ought to be laws toprotect the body of acquired knowledge. Take one of our good pupils, for example: modestand diligent, from his earliest grammar classeshe's kept a little notebook full of phrases.After hanging on the lips of his teachers fortwenty years, he's managed to build up anintellectual stock-in-trade; doesn't it belongto him as if it were a house or money? (P. Claudel,_Le Soulier du Satin_; ci in _Language andSymbolic Power_, p. 43) Cool-Hand LukePerhaps from force of occupational habit, perhapsby virtue of the calm that is acquired by everyimportant man who is consul for his advice andwho, knowing that he will keep control over thesituation, sits back and lets his interlocutor flapand fluster, perhaps also in order to show to advantagethe character of his head (which he believed to beGrecian, in spite of his whiskers), while somethingwas being explained to him, M. de Norpois mantainedan immobility of expression as absolute as if you hadbeen speaking in front of some classical--and deaf--bustin a museum. (Marcel Proust, _A la recherche du tempsperdu_; ci in _Language and Symbolic Power_, p. 66)John === Subject: : Re: Fundamental Reason for High Achievements of Jews>And of course, as can be seen by the many posts>asserting that many people in America, and many posters in the>newsgroups are anti-Semetic, it is clear that many (Most?) Jews>sense that conflict exists.> Starting with punks like you.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.>I suggest that the root cause of the problem>lies in the fact that Jews are far more intensely>brainwashed with religious and ethnic dogma,>than Muslims, Catholics, Protestents, Buddhists, etc.> I suggest that the root cause is ignorant fools like you.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.>and that a solution would be for governments>to limit the amount of time that a child can be>subjec to religious/ethnic brainwashing.> As defined by whom? Certainly I wouldn't mind seeing you and yours> kept out of the classroom.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.> --> Shmuel (Seymour J.) Metz, SysProg and JOAT--Tom Potter http://tompotter.us === Subject: : Re: Fundamental Reason for High Achievements of Jews>> The question is, does the intense ethnic/religious conditioning>> of Jewish children, bring them into conflict with their>> communities, and other religious and ethnic groups?>> In the Uni States, Jewish children live in a peacful and lawabiding>> way among and with their Gentile neibhbors. What conflict? Theoccurence>> of violence and criminality is much lower among Jews, than in the>> country taken as a whole. What conflict?>> Bob Kolker>>Of course, one point in time and space does not>negate the fact that Jews have come into conflict with>their non-Jewish neighbors more often than not>trhoughout history.>>And of course, as can be seen by the many posts>asserting that many people in America, and many posters>in the newsgroups are anti-Semetic, it is clear that>many (Most?) Jews sense that conflict exists.>>The question that is of vital inportance to mankind>is why Jews come into conflict with theitr neighbors,>and what can be done to eliminate this problem.>>I suggest that the root cause of the problem>lies in the fact that Jews are far more intensely>brainwashed with religious and ethnic dogma,>than Muslims, Catholics, Protestents, Buddhists, etc.>and that a solution would be for governments>to limit the amount of time that a child can be>subjec to religious/ethnic brainwashing.>>Tom PotterI suggest that the problem is you're a bed-wetting little bigot.> I probably won't see any response to this post of> mine, since my access is through Google Groups and> finding a post that doesn't display my name as the> author is very difficult in long threads. Tho, if> any respondent would e-mail me in addition to posting> .... That said, here's a true, personal experience,> maybe less than two weeks old.> I'm walking north on 5th Avenue in NYC, on the east side> 33rd I pluck a styrofoam container that has some heft> to it, open it, and see that someone had eaten all> the rice but left a considerable amount of meat. As> I cross 33rd I do the obvious experiment and determine> that the meat is pork. As I stroll and nosh (or more> properly nosh and stroll -- in alphabetical order) I> find sitting about mid-block a large cleanly dressed> male with a sign hung from his neck. The sign said> xxxxx> HUNGRY> JEW> where xxxxx was a word I don't recall, perhaps referring> to someting associa with the religeon. Thinking that> here was a healthy-looking and apparently competent person> begging for food rather than looking for work or even gleaning> the trash for recyclable bottles and cans (as I do), I> literally pulled my head away from the direction I was> looking and walked on. I heard him speak and turned to> determine what it was that he was saying. He repea> ...something about what a sorry person I was. I approached> him and said something like hungry jew, I just picked this> food from a trash can, would you like to have it? He gave> the usual response, that he couldn't eat pork. I then went> on that he shouldn't be making disparaging remarks about> passers by. He responded by saying something like you> shouldn't be shaking your head like an asshole. I left> without further comment.> Is this clear to you...the hungry jew refuses food and> wants to pick a fight?> Mark (Didn't the prohibition of pork come about> because of trichinosis and now in modern> times the ban is unnecessary? Well, baaaa!> And can be trained to do math, play music> written by others, play doctor.... Most historians believe that Jews avoid pork,because the ancient Jews associa pigs with leprosy,and pigs and people with leprosy were unclean.the Jews expelled from Egypt had leprosy,and that they avoided pork as a preventative measure.I understand that science has found noassociation between pigs and leprosy.--Tom Potter http://tompotter.us === Subject: : Re: Fundamental Reason for High Achievements of Jews>The origins of delusions> are the drugs you've been taking. And you're repeating yourself.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.>and as can be clearly seen,> Many things can be clearly seen when you have DT. That doesn't make> them real.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.>Jewish conflict with their Palestinian neighbors has>been the root cause of 911,> Bin kalbah's interest came after he had launched the attack. The real> issue was the American presence in Saudi Arabia, which has to do with> oil, not Israel.>the movement of America toward>a police state, and the enormous encrease in the>cost of government in America.> You forgot sunspots and flying saucers.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.> --> Shmuel (Seymour J.) Metz, SysProg and JOATI suggest that the ***responses*** of Shmuel (Seymour J.) Metzto my post about why Jews have come into conflictwith peoples and nations throughout historypretty much makes my point thatmany people have been intensely brainwashedand are unable to address certain issuesin a rational, intelligent, MORAL way,and that they respond to this issuesby trying to make the messenger the issue.--Tom Potter http://tompotter.us === Subject: : Re: Fundamental Reason for High Achievements of Jews>And of course, as can be seen by the many posts>asserting that many people in America, and many posters in the>newsgroups are anti-Semetic, it is clear that many (Most?) Jews>sense that conflict exists.> Starting with punks like you.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.>I suggest that the root cause of the problem>lies in the fact that Jews are far more intensely>brainwashed with religious and ethnic dogma,>than Muslims, Catholics, Protestents, Buddhists, etc.> I suggest that the root cause is ignorant fools like you.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.>and that a solution would be for governments>to limit the amount of time that a child can be>subjec to religious/ethnic brainwashing.> As defined by whom? Certainly I wouldn't mind seeing you and yours> kept out of the classroom.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.> --> Shmuel (Seymour J.) Metz, SysProg and JOAT--Tom Potter http://tompotter.us === Subject: : Re: Fundamental Reason for High Achievements of Jews>And of course, as can be seen by the many posts>asserting that many people in America, and many posters in the>newsgroups are anti-Semetic, it is clear that many (Most?) Jews>sense that conflict exists.>>Starting with punks like you.> As can be seen, Shmuel (Seymour J.) Metz> uses the standard tactic of brainwashed people> who cannot address an issue,> and try to make the messenger the issue.>I suggest that the root cause of the problem>lies in the fact that Jews are far more intensely>brainwashed with religious and ethnic dogma,>than Muslims, Catholics, Protestents, Buddhists, etc.>>I suggest that the root cause is ignorant fools like you.> As can be seen, Shmuel (Seymour J.) Metz> uses the standard tactic of brainwashed people> who cannot address an issue,> and try to make the messenger the issue.>and that a solution would be for governments>to limit the amount of time that a child can be>subjec to religious/ethnic brainwashing.>>As defined by whom? Certainly I wouldn't mind seeing you and yours>>kept out of the classroom.> As can be seen, Shmuel (Seymour J.) Metz> uses the standard tactic of brainwashed people> who cannot address an issue,> and try to make the messenger the issue.>>-->> Shmuel (Seymour J.) Metz, SysProg and JOAT> --> Tom Potter http://tompotter.usWhat is abundantly clear is that Potter defines any rational response to his accusations as a result of the condition he insists are applicable to Jews but finds no evidence among those who are not Jews. Hence the wonderfully circular logic of Potter. The proof of his assertion is in the response that it elicits. Huh ?????FK === Subject: : Re: Fundamental Reason for High Achievements of Jews>And of course, as can be seen by the many posts>asserting that many people in America, and many posters in the>newsgroups are anti-Semetic, it is clear that many (Most?) Jews>sense that conflict exists.> Starting with punks like you.As can be seen, Shmuel (Seymour J.) Metzuses the standard tactic of brainwashed peoplewho cannot address an issue,and try to make the messenger the issue.>I suggest that the root cause of the problem>lies in the fact that Jews are far more intensely>brainwashed with religious and ethnic dogma,>than Muslims, Catholics, Protestents, Buddhists, etc.> I suggest that the root cause is ignorant fools like you.>and that a solution would be for governments>to limit the amount of time that a child can be>subjec to religious/ethnic brainwashing.> As defined by whom? Certainly I wouldn't mind seeing you and yours> kept out of the classroom.> --> Shmuel (Seymour J.) Metz, SysProg and JOAT> Unsolici bulk E-mail will be subject to legal action. I reserve> the right to publicly post or ridicule any abusive E-mail.> Reply to domain Patriot dot net user shmuel+news to contact me. Do> not reply to spamtrap@library.lspace.org === Subject: : Re: Fundamental Reason for High Achievements of Jews>Considering that Jews frown on Jews breeding>with non-Jews,> For most of the last two millenia non-Jews frowned on Jews marrying> non-Jews, tonto.>I think that the reason that Jews come into>conflict with their neighbors is caused by>the intense religious/ethnic brainwashing>that Jewish children are subjec to,> Herr G.9abells, you keep repeating the same lie. So tell me, faygeleh,> how many hours a week do Jews spend in religious instruction.> Catholics? Muslims? Protestants?> --> Shmuel (Seymour J.) Metz, SysProg and JOATI suggest that Shmuel (Seymour J.) Metz response to mypost tends to confirm my suggestion that many peopleare brainwashed to obfuscate and hide from discussionsof why Jews have come into conflict with so manyother groups and nations throughout history.Note that Shmuel (Seymour J.) Metzused the tactic common to people who have been brainwashedintently to some dogma, by attacking the messenger,rather than addressing the message in arational, intelligent, moral way.As can be seen, even Shmuel (Seymour J.) Metz thinks thatFor most of the last two millenia non-Jews frowned on Jews marryingnon-Jews.and perceives that non-Jews are hostile to Jews.If he thinks this, why not analyze why he thinks non-Jewswould not want to marry Jews?What is the root cause of these conflict?Is it nature?Is it nurture?Is it religious/ethnic brainwashing?Are there other groups thatsubject their children to intense religious/ethnic conditioning?If so, how do they interface with the folks around them?Do Muslims, Christians, Buddhists, or other religionsintegrate ethnicity with their religion?If so, how do they interface with the folks around them?--Tom Potter http://tompotter.us === Subject: : Re: Fundamental Reason for High Achievements of Jews> Tom Potter said> it is clear that many (Most?) > Jews sense that conflict exists.> Starting with punks like you.> Considering that Jews frown on > Jews breeding with non-Jews,> tell me, faygeleh, how many hours... do Jews spend> Shmuel (Seymour J.) Metz, SysProg and JOATMr. Potter, starting with punks like you, as this Shmuel faygeleh begs you for, means that he wishes to confess and profess his basal desire to perform his own faygelen and spend his ShMutz on you for many hours like a faygeleh that loves to make his foreplay byF. Art Ingatu PS: Disregard posters like him. He is not a Jew's Jew. Good Jews refer to punks like Metz as a Shmutz kike. === Subject: : Re: Fundamental Reason for High Achievements of Jews(snip)>Do Muslims, Christians, Buddhists, or other religions>integrate ethnicity with their religion?>If so, how do they interface with the folks around them?Can we take this elsewhere ? ... no relevance to any of the groups it's actually pos to.Thanks :)------------------------------------------------------------ --------Oook ! === Subject: : Re: Fundamental Reason for High Achievements of Jews>Apparently Another Wise Guy - Macon, GA USA>thinks I am wrong in my assessment.> More likely he knows that you are simply lying.>I suggest that he has been brainwashed,> I suggest that you have an ax to grind.>Here on Earth,>Jewish children are subjec to much more>religious/ethnic dogma,>than any other large group.> That, of course, is a bare-faced lie.As can be seen, the response of Shmuel (Seymour J.) Metzto my post regarding why Jews have come into conflict withgroups and nations throughout history,tends to confirm my suggestion that the root of the problemlies in the intense religious/ethnic brainwashing of Jewish children.Note that Shmuel (Seymour J.) Metz uses the standard,knee jerk, boilerplate response, common to people whohave been brainwashed, and note that he avoids thebasic issue, and tries to make the messenger the issue,in order to deflect the subject.As the real world facts cannot support their belief system,brainwashed people are unable to deal withcertain issues, and use what psychologists callavoiding the scene, and transferred aggression.Hopefully rational, intelligent, MORAL folkswho are interes in the welfare of the human racewill begin to examine the roots cause of why Jewscome into conflict with nations and groups of peopleand find a way to solve the problem, as it has beenthe root cause of much of mankind's suffering.To address Shmuel (Seymour J.) Metz effortsto make me the issue, rather than the central issuewhich is of great importance to mankind,I challenge him to elaborate on his diversionary statements:More likely he knows that you are simply lying.I suggest that you have an ax to grind.That, of course, is a bare-faced lie. Note that Shmuel (Seymour J.) Metzcompletely avoided the prime issue,and used the standard boilerplate tacticof brainwashed, closed-minded (Or immoral) people,of trying to make the messenger the issue.To paraphrase the Brown Bomber,You can run from the truth,but you can't hide from the truth.--Tom Potter http://tompotter.us === Subject: : Re: Fundamental Reason for High Achievements of Jews>>Apparently Another Wise Guy - Macon, GA USA>>thinks I am wrong in my assessment.>> More likely he knows that you are simply lying.>>I suggest that he has been brainwashed,>> I suggest that you have an ax to grind.>>Here on Earth,>>Jewish children are subjec to much more>>religious/ethnic dogma,>>than any other large group.>> That, of course, is a bare-faced lie.>As can be seen, the response of Shmuel (Seymour J.) Metz>to my post regarding why Jews have come into conflict with>groups and nations throughout history,>tends to confirm my suggestion that the root of the problem>lies in the intense religious/ethnic brainwashing of Jewish children.>Note that Shmuel (Seymour J.) Metz uses the standard,>knee jerk, boilerplate response, common to people who>have been brainwashed, and note that he avoids the>basic issue, and tries to make the messenger the issue,>in order to deflect the subject.>As the real world facts cannot support their belief system,>brainwashed people are unable to deal with>certain issues, and use what psychologists call>avoiding the scene, and transferred aggression.>Hopefully rational, intelligent, MORAL folks>who are interes in the welfare of the human race>will begin to examine the roots cause of why Jews>come into conflict with nations and groups of people>and find a way to solve the problem, as it has been>the root cause of much of mankind's suffering.>To address Shmuel (Seymour J.) Metz efforts>to make me the issue, rather than the central issueYou are the issue, just as your idol Adolf was. When evil rears its head, all good people must confront it, challenge it, and fight it. As Edmund Burke said, all that is necessary for the triumph of evil is for good men to do nothing. === Subject: : Re: Sum of k^(k+r) x^k /k!> If W(y) is the Lambert W-function, then it is known that> W(y) = sum{k=1 to infinity} k^(k-1) y^k/k!,> I think. (This equation above has NOT been found by me, however.)You are missing an extra (-1)^(n-1). The correct expansion is:LW(z) = Sum([(-1)*n]^(n-1)*z^n/n!, n=1..+oo)> But, I am wondering about the generalization:> w(r,y) = sum{k=1 to oo} k^(k+r) y^k /k!.> I think that: w(0,y) = -W(-y)/(1 +W(-y)).Your notation is confusing. The notation w(k,y) is usually used todenote the various branches of the LambertW. YOUR notation w(r,y) seemsto denote something entirely different.Try it again with the correct expansion. You are missing a factor of(-1)^n. Otherwise it looks right.> And, for r = integer >= 1,> w(r,y) = (sum{k=1 to r} (-W(-y))^k a(r,k))/(1 +W(-y))^(2r+1)> where a(r,k) is, recursively,> a(r+1,k) = a(r,k) k + a(r,k-1) (2r+2-k)> (I think.)> So,..> If, for instance, n = odd positive integer,> w((n-1)/2,-n exp(n)) is an integer divisible by n,> and this integer is congruent to -n (mod n^2).> (These number-theory results can be continued.)> What is the closed from for the a(r,k)s??>-- Ioannishttp://users.forthnet.gr/ath/jgal/_____________________ ______________________Eventually, _everything_ is understandable. === Subject: : Re: Equation differentielle: circuit RC> Salut!> j'ai qq problemes a resoudre le probleme suivant:> (c'est de l'electronique a la base je vous l'accorde mais les outils> de resolution sont des maths alors ...)> Supposons un circuit RC banal aliment.8e par une tension U0 continue.> Que'ce que c'est, un circuit RC banal? Est-il comme ca:On a en s.8erie un g.8en.8erateur + une Capa + une r.8esistance.> (U0) ---R---C--- (0)> La tension U0, est-elle continue ou constante?> la tension est constante et non continue comme je l'ai .8ecrit ci dessus:U(t)=U0> Michael. === Subject: : Re: Euler vs. =?ISO-8859-2?Q?Erd=F5s?=Greetings.> Rela:> I do not even know if Euler DIRECTLY collabora with anyone.> (Everyone knows that his results have been widely used, in any case.)Euler was known to have carried on lively discussions with his colleagues,such as Goldbach, Stirling, and d'Alembert. It's not unthinkable that someof these discussions may have resul in joint papers, though, as you say,I haven't been able to find any evidence of direct collaboration. Perhapsgood friend to the family.There has been at least some indirect or uncredi collaboration. In 1794(after Euler's death), Lagrange transla and significantly revisedEuler's two-volume series Anleitung zur Algebra. Also, as Euler wasblind for the latter part of his life, he employed several amanuenses whowere not mere secretaries but with whom he also discussed matters oftheory. Three of these men -- Albrecht, Krafft and Lexell -- receivedacknowledgment in his 1772 work on the moon, though I don't believe theyreceived an actual coauthorship credit. Nikolai Fuss was an amanuensis forunder Euler's direction, but again I am not aware of any work credijointly to both mathematicians.> If he did collaberate, I wonder if he had a finite Erdos-number. It is> very unlikely that he did have a finite EN (given the time-span); but> if he did, does anyone know it?I've written to Jerry Grossman, maintainer of the Erds number website, tosee if he has anything to say on the matter. I'll post any information Iget back.Tristan-- _ _V.-o Tristan Miller [en,(fr,de,ia)] >< Space is limi / |`-' -=-=-=-=-=-=-=-=-=-=-=-=-=-=-= <> In a haiku, so it's hard(7_ http://www.nothingisreal.com/ >< To finish what you === Subject: : Re: Naive Q: Set theory, logic - which comes first?> Personally I would say that the twin prime hypothesis is definitely either true> or false, because the intuitive notion of the natural numbers determines that> system of objects completely. That's only if natural numbers are positive numbers. The only thing intuition ever proved about negative numbers is that there are no negative prime numbers. With more abstract objects like sets, it is not> too surprising if the underlying intuitive concept fails to determine all the> details of the relevant system of objects (to borrow axiomatic terminology, the> theory is not intuitively categorical). If the continuum hypothesis follows> from propositions that should be axioms, then of course you have to accept it> as true, and the same goes for its negation. But if there are no such> propositions, it seems reasonable to remain agnostic about it, or to opt for one> or the other truth value based on considerations of convenience. The> completeness of the real number system, for example, seems to have convenience> as its main justification. (Or do most people have strong geometrical intuitions> about it? No. The only thing Geometrical about the real numbers is 0.Has anything been written about this question?) osphers star writing about it in 2000 BC. And haven't finished yet, Since many present day osophers mistaking believe that Goedel was clued in about things like DNA and evolution. === Subject: : Continuously Tracing the HyperPower function x^^yThere was a nice bonus on my construction for a Continuous Extension forthe HyperPower operator, which I discovered only two days ago.The definition that I use, seems to produce an infinity of new shapes ifone traces the hyperexponent continuously.For example, although (-1)^^n = -1, for all n in N, (so that thefunction (-1)^^y returns infinately often to -1) the behavior of (-1)^^yfor values in between natural numbers appears to be quite complex.A couple of .GIF trace samples, along with a short cross-referencedindirect proof that lim[n->+oo]i^^n exists (by considering the traces ofi^^y, y>=0), can be found at:in my Math section:Enjoy the read.-- Ioannishttp://users.forthnet.gr/ath/jgal/_____________________ ______________________Eventually, _everything_ is understandable. === Subject: : Re: Two coin flip/ clarification for C Bond>>System 3:>> A = we were told at least one is H>> That isn't a proper event. If A is an event in the sample space E then>> the complement of A must be also be a member of E.>The compliment of at least one head in a toss of 2 coins is 2 >tails.Yes, but the logical complement of X took place is X did not takeplace, not 'not X' took place.A = at least one head is a proper event.B = we were told X isn't.Look at it this way. Events A and B must be subsets of the permutationof all possible outcomes of the two-coin toss. But event B containsinformation that is not decidable based on the coin-toss alone. Howare the coins supposed to know what we were told, can they read? === Subject: : Re: Two coin flip/ clarification for C Bond>>System 3:A = we were told at least one is HThat isn't a proper event. If A is an event in the sample space E then> the complement of A must be also be a member of E.> The compliment of at least one head in a toss of 2 coins is 2 > tails.Not true. Flip two coins and look at one. If it's heads the at least one headsstatement is true. If it's tails the at least one tails statement istrue.To make the two tails statement, you must look at both coins. Thisis NOT complimentary.Eldon === Subject: : Re: Two coin flip/ clarification for C Bond>>System 3:A = we were told at least one is HThat isn't a proper event. If A is an event in the sample space E then> the complement of A must be also be a member of E.> The compliment of at least one head in a toss of 2 coins is 2 > tails.> Not true. > Flip two coins and look at one. If it's heads the at least one heads> statement is true. If it's tails the at least one tails statement is> true.> To make the two tails statement, you must look at both coins. This> is NOT complimentary.> EldonThe sample space for the two coin toss experiment mat be represen by S = {hh,ht,th,tt}.The event of at least one head is E = {hh,ht,th}, and the compliment, relative to S, of E is F = {tt}.The method by which you determine which outcome(s) occurred is irrelevant. === Subject: : Re: Two coin flip/ clarification for C Bond>> The compliment of at least one head in a toss of 2 coins is 2>> tails.> Not true.!> To make the two tails statement, you must look at both coins. This> is NOT complimentary.And I'm not going to be complimentary to you!-- === Subject: : Re: Two coin flip/ clarification for C Bond>>I'll toss in my 2 cents.> Thanx for your extreme sacrifice. The bad news is that with inflation,> it's already worth less.Dang, I'll dig up some nickels for next time.> I get criticized for posting on this question, however, this question> gets deep into the definition of probability itself. Probability, by> definition, must have many events, then we use a ratio.> But, we can have a single event, then reiterate that event many times.> The problem then being, we must iterate the given event. In this> question, we have a one time, or a first time event. We are> reiterating a first time event. That confuses some.> Two coins can be flipped four ways. HH, HT, TH, TT. When we flip> umpteen thousands of times, we should get that ratio. HH, or HT, or> TH, or TT could have occurred first. We can select any event in the> umpteen thousands and assume it to have been first.> We have our statements, Two coins were flipped and at least one is a> tail. What are the chances for two tails? and Two coins were flipped> and at least one is a head. What are the chances for two heads?> With HH, the heads statement is true. With TT, the tails statement is> true. With HT, or TH, either statement is true.> Our two bettors, Bill and Jane, only hear the statement, all they have> to do to make the proper bet is answer the question correctly. Bill> always bets for two. Jane covers the bet.All of these provide a wealth of context that was not presen in the original statement and may change the meaning of the discussion.>We have prior agreement, even with Dr. Ullrich that the following two>statements are the same:>Two coins were flipped and at least one is a head.>Two coins were flipped and at least one is a tail.>>Obviously they are different, but if you are talking about >>probabilities, you can substitute head for tail and tail for >>head everywhere in a problem and generate an equivalent problem.> We agree to that. They are mathematically equivalent. >If you agree to here, then, flip two identical coins. You can make>neither statement without some way inspecting at least one coin.>Inspect one and only one, if it's a head, make the heads statement, if>it's a tail, make the tails statement.>>Neither of your statements refers to inspecting one and only one coin. >>The statement above corresponds to:>>Two coins were flipped and the first one is a head. or>>Two coins were flipped and the second one is a head.> With HH, the at least one is a head statement is true, whether we> saw one, or two. We only have to have seen one to have made the> statement.Yes, but with HT, at least one is a head is true, but is not a statement I could make if I happened to look at the tail. This is where the issue comes in: at least one is a head without further discussion suggests that I have information about *both* coins. I have looked at a coin and it is a head suggests that I have information about only *one, specific* coin. WLOG, I can then think of the pairs in my sample space as being S = { (looked at)(not looked at) | each is either HT } = {HH,HT,TH,TT}.If that's not acceptable, then you must change the sample space to reflect how the coins flipped *and* which you looked at (1st or 2nd). Then S = {HH1,HH2,HT1,HT2,TH1,TH2,TT1,TT2}.What seems clear to me is that you are not choosing to take either of these approaches in analyzing your problem.>>Of course, you can substitute tail for head without affecting the >>nature of the problem.>The answer to the corresponding question is 1/2.>>Only after changing what you mean by at least one is a head to a >>specific coin (1st or 2nd) is a head.> Not true. I saw one and only one, don't know which one. The statement> is true. I saw a head, I said at least one is a head. I did not lie.> I saw a tail, I said, at least one is a tail. I did not lie.Then you must use the second sample space I gave.>Define a repeatable coin flip sequence which has correct answer 1/3.>You will have to inspect both coins.>>Which is what at least one suggests.> This is where you err. You are reading into the question. The at> least one suggests something to you, other than what it means. That's> why you get it wrong. With TT, at least one is a tail is a true> statement, whether I saw one, or two is moot, the statement is true.> Bill will bet for two tails, Jane will cover, Bill will win. You say> she should pay double, you're wrong.I have no idea what exactly Bill and Jane are doing in their betting. You have not clearly explained this.I do know that at least one is a tail and at least one is a head are of probability are written, it is you who is reading additional information into the problem.> In the above paragraph, substitute heads for tails, then Bill wins> with TT, and HH. Jane wins with HT, and TH. She wins half the time,> you have her paying double when she loses. Shame shame on you.You have made no statements about how this betting works within this thread. You are not giving me useful information.>There is no evidence in our question that the writer of the question>knew the outcome of both coins. Imagine that there is always two>bettors. One bets for two of the kind, the other covers the bet. On>HH, for the losing bettor to pay two to one, there must be some>assurance, in the statement, to the losing bettor, that there would>have been some kind of demur at TT.>>Here the distinction seems to come down to whether you are looking at a >>sequence of flips, or simultaneous flips of distinguishable coins. Is >>it not reasonable for me to take a dime and nickel, flip them (concealed >>from you) and announce that at least one is a head after inspecting >>both? Can you reasonably assume that I will be consistent as to whether >> I am informing you about the dime or the nickel?> As we can substitute 'heads' for 'tails' and vice versa, it doesn't> matter. Bill will win with HH and TT, Jane will win with HT and TH.> That's fifty fifty. It's an even money bet. The probability is 1/2.What are the rules of this game?>example: Two coins were flipped until at least one is a head. The>'until' tells the losing bettor to pay double on HH. It protects her>from also having to pay double on TT.>Eldon:)>>This is a completely different type of scenario. How you work out the >>unmentioned payments is up to the gamblers, but I would certainly hope >>they can work out a system of payments for 0, 1, or 2 heads that >>satisfies both.> The 'until' tells Jane that on HH, she pays double. It tells her that> with TT there would have been a reflip, and that she wouldn't have to> pay double there also. It shows that the writer of the question had> access to both coins.> To get an answer of 1/3 for our question, the answerer has to assume> the 'until' or some equivelant into our question. This is where they> err.> They are asked one question, they assume a different question.> Eldon MoritzEldon, may I suggest:1) telling us the rules of the game.2) determining what you think should happen.3) run an experiment to see if your prediction appears to be correct.I suspect at this point that what you have in mind will give a result of 1/2, but that you are not stating it in the standard way and are being misinterpret. If this is the case, conforming to how others state the situation would probably clear up the problem rather than trying to make everyone else switch to how you want to communicate.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Two coin flip/ clarification for C Bond> With HH, the at least one is a head statement is true, whether we> saw one, or two. We only have to have seen one to have made the> statement.> Yes, but with HT, at least one is a head is true, but is not a > statement I could make if I happened to look at the tail. This is where > the issue comes in: at least one is a head without further discussion > suggests that I have information about *both* coins. I have looked at > a coin and it is a head suggests that I have information about only > *one, specific* coin. WLOG, I can then think of the pairs in my sample > space as being S = { (looked at)(not looked at) | each is either HT } = > {HH,HT,TH,TT}.> If that's not acceptable, then you must change the sample space to > reflect how the coins flipped *and* which you looked at (1st or 2nd). > Then S = {HH1,HH2,HT1,HT2,TH1,TH2,TT1,TT2}.> What seems clear to me is that you are not choosing to take either of > these approaches in analyzing your problem.Actually, using upper case for looked at and lower case for not looked at, first coin a nickle and second one a dime, you could haveS = {HH,Hh,hH,hh, HT,Ht,hT,ht, TH,Th,Th,th, TT,Tt,tT,tt}, although you might want to eliminate hh,ht,th and tt as being indeterminate. === Subject: : Re: Two coin flip/ clarification for C Bond>>I'll toss in my 2 cents.> Thanx for your extreme sacrifice. The bad news is that with inflation,> it's already worth less.> Dang, I'll dig up some nickels for next time. > Eldon, may I suggest:> 1) telling us the rules of the game.> 2) determining what you think should happen.> 3) run an experiment to see if your prediction appears to be correct.> I suspect at this point that what you have in mind will give a result of > 1/2, but that you are not stating it in the standard way and are being > misinterpret. If this is the case, conforming to how others state > the situation would probably clear up the problem rather than trying to > make everyone else switch to how you want to communicate.The rules? The rules of fair play prevail. Thou shalt not kill, thoushalt not covet, thou shalt not steal.......Thanx Will,Bill and Jane are imaginary bettors. They only hear what we tell them.I use them to illustrate that we only have the problem statement. Weshouldn't add, or subtract from it. As they only hear the problemstatement, that's all they have. As we only have the problemstatement, that's all we should use.Flip two coins, they landed HH, or TT, or TH, or HT. Notice the or's.Flip them umpteen thousand times, and those four are all you get. Theycome up one at a time, and anyone of them could have appeared first,hence the or's.TT landed first and we made the tails statement, or HH landed firstand we made the heads statement.Suppose that HH landed first and we made the statement, Two coinswere flipped and at least one is a head. What are the chances for twoheads? Bill bet for two heads, Jane covered the bet. Bill put up adollar and he will win. How much should Jane pay?OR:Suppose that TT landed first and we made the statement, Two coinswere flipped and at least one is a Tail. What are the chances for twotails? Bill bet for two tails, Jane covered the bet. Bill put up adollar and he will win. How much should Jane pay?Eldon:> === Subject: : Re: Two coin flip/ clarification for C Bond>>I'll toss in my 2 cents.>Thanx for your extreme sacrifice. The bad news is that with inflation,>it's already worth less.>>Dang, I'll dig up some nickels for next time.> Question: Why did you eliminate my analysis without comment? Do you consider it invalid? If so, why? If not, why did you respond with your further example?[examples snipped because they've been addressed seperately]-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Two coin flip/ clarification for C Bond>>I'll toss in my 2 cents.>Thanx for your extreme sacrifice. The bad news is that with inflation,>it's already worth less.>>Dang, I'll dig up some nickels for next time.>Eldon, may I suggest:>>1) telling us the rules of the game.>>2) determining what you think should happen.>>3) run an experiment to see if your prediction appears to be correct.>>I suspect at this point that what you have in mind will give a result of >>1/2, but that you are not stating it in the standard way and are being >>misinterpret. If this is the case, conforming to how others state >>the situation would probably clear up the problem rather than trying to >>make everyone else switch to how you want to communicate.> The rules? The rules of fair play prevail. Thou shalt not kill, thou> shalt not covet, thou shalt not steal.......> Thanx Will,> Bill and Jane are imaginary bettors. They only hear what we tell them.> I use them to illustrate that we only have the problem statement. We> shouldn't add, or subtract from it. As they only hear the problem> statement, that's all they have. As we only have the problem> statement, that's all we should use.> Flip two coins, they landed HH, or TT, or TH, or HT. Notice the or's.> Flip them umpteen thousand times, and those four are all you get. They> come up one at a time, and anyone of them could have appeared first,> hence the or's.> TT landed first and we made the tails statement, or HH landed first> and we made the heads statement.> Suppose that HH landed first and we made the statement, Two coins> were flipped and at least one is a head. What are the chances for two> heads? Bill bet for two heads, Jane covered the bet. Bill put up a> dollar and he will win. How much should Jane pay?1 dollar. She covered his bet. If he wins, she loses.> OR:> Suppose that TT landed first and we made the statement, Two coins> were flipped and at least one is a Tail. What are the chances for two> tails? Bill bet for two tails, Jane covered the bet. Bill put up a> dollar and he will win. How much should Jane pay?Same as before.> Eldon:>Which statement is made when TH comes up? How about HT? You still haven't explained how the betting works. There is a difference between whether the game is a *fair* game and how much Jane should pay for a *particular* game.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Two coin flip/ clarification for C Bond>> Thanx for your extreme sacrifice. The bad news is that with inflation,>> it's already worth less.>Dang, I'll dig up some nickels for next time.Here's a penny for your thoughts - got any change? === Subject: : Re: Two coin flip/ clarification for C Bond>Thanx for your extreme sacrifice. The bad news is that with inflation,>it's already worth less.>>Dang, I'll dig up some nickels for next time.> Here's a penny for your thoughts - got any change?delta-x-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Algebraic Integers>> In sci.physics, >> <3c65f87.0309291419.161d2dd9@posting.google.com>:>> I thought I was set. With a high I.Q. group set to publish my paper>> on factoring polynomials into non-polynomial factors I figured that>> now certainly I could push my agenda to fame and fortune. I was>> wrong. So here I am back again, humbled yet again, as there's no escaping it,>> there is no way I'll get anywhere pissing off mathematicians. So here are some concessions. If mathematician wish to dismiss my prime counting function.>> Dismiss, no.>> Blow away, yes. Christian Bau did a remarkably good job with>> his implementation at the higher numbers, although he had>> help from Meissel and Lehmer.>That's false. All Christian Bau even claimed was that he was>rehashing old work and I never claimed to have the fastest prime>counting algorithm.Not true. You eventually retrac that claim, replacing it withan assertion that you didn't care about speed. But you wereclaiming that your algorithm was the fastest possible forquite some time. (No, you never claimed that your _implementation_was the fastest possible.)>So you're caught in a falsehood as it's impossible for my position to>have been blown away by someone putting up someone else's work when my>position was never the one that you imply.>However, someone naive who actually *trus* you might have believed>you and the implication, which is that Christian Bau actually did>something, versus copy from what other researchers found, and that my>position was that I had the fastest prime counting algorithm, when>that was not my position.>But now they can put you in their list of sources not to be trus.>who looks for targets to attack to bring attention to yourself. You>probably figured I was a good and easy target, so you trot out your>bogus claim.>**************************As far as I'm concerend you're trying to wait until I die, so I figuremaybe you should die instead. How about that, eh? Wouldn't that be abetter twist?You refuse to follow the math, so the great Powers that controlreality and *speak* in mathematics decide to kill you instead of me.So what do you think about that, eh? Oh, can't hear Them talking?Well, I guess that's because you don't really understand Mathematics,the true language, which is THE language.They're talking about you now, and They agree with my assessment, andwill not penalize me as They allowed the others like Galois and Abelto be penalized.They will kill you instead. speaking on Weird factorization, genius === Subject: : Re: Algebraic Integers[...]|>(Does anyone know if the ring of algebraic integers is|>a principal ideal ring? It's clear that the algebraic|>integers, like their namesake the integers, form an|>integral domain.||It's a Bezout Domain: every finitely genera ideal is|principal. However, some infinitely genera ideals are not|principal; for example:||(2, 2^{1/2}, 2^{1/3}, 2^{1/5}, 2^{1/7}, ..., 2^{1/p}, ...)||is easily seen to non-principal, since it contains elements of|arbitrarily high degree over Q.The nontrivial principal ideals of O, the ring of algebraic integers,contain elements of arbitrarily high degree over Q. In (u), for u<>0,if uv has degree <=N then v=(uv)/u has degree at most N times the degreeof u, so the degree of uv goes to infinity as the degree of v does.So I think something more needs to be said. === Subject: : Re: Algebraic Integers Visiting Assistant Professor at the University of Montana.>[...]>|>(Does anyone know if the ring of algebraic integers is>|>a principal ideal ring? It's clear that the algebraic>|>integers, like their namesake the integers, form an>|>integral domain.>|>|It's a Bezout Domain: every finitely genera ideal is>|principal. However, some infinitely genera ideals are not>|principal; for example:>|>|(2, 2^{1/2}, 2^{1/3}, 2^{1/5}, 2^{1/7}, ..., 2^{1/p}, ...)>|>|is easily seen to non-principal, since it contains elements of>|arbitrarily high degree over Q.>The nontrivial principal ideals of O, the ring of algebraic integers,>contain elements of arbitrarily high degree over Q. In (u), for u<>0,>if uv has degree <=N then v=(uv)/u has degree at most N times the degree>of u, so the degree of uv goes to infinity as the degree of v does.>So I think something more needs to be said.Oops. Yeah, I guess you're right there. If (2, 2^{1/2}, 2^{1/3}, 2^{1/5}, 2^{1/7}, ..., 2^{1/p}, ...)=(u),note that u is not a unit as the ideal is not the total ideal. Note that since u divides 2^{1/p} for every prime p, it follows thatu^p divides 2 for every p>0. So u^n is a proper divisor of 2 for every n>0.Let K= Q(u) and let R be the ring of integers of K. Then the ideal (u)is a proper divisor of (2), and (u)^n is a proper divisor of 2 for alln; this contradicts unique factorization of ideals intoprimes. Therefore, no such u exists. === Subject: : Re: Algebraic Integers> Blow away, yes. Christian Bau did a remarkably good job with> his implementation at the higher numbers, although he had> help from Meissel and Lehmer.> That's false. All Christian Bau even claimed was that he was> rehashing old work and I never claimed to have the fastest prime> counting algorithm.> What I claimed was that _you_ are rehashing 200 year old work. I on the > other hand have taken twenty year old work as a starting point and > improved it. I am standing on the shoulders of giants; you are a dwarf.Get published on it then. Until then you're just a hack trying to getsomething off of Odlzyko, Lagarias and others.Now I have actually corresponded with both Odlyzko and Lagarias(though they may regret it, especially Odlyzko), but what about you?What do they say about your work, loudmouth? === Subject: : Re: Algebraic Integers> message > Blow away, yes. Christian Bau did a remarkably good job with> his implementation at the higher numbers, although he had> help from Meissel and Lehmer.That's false. All Christian Bau even claimed was that he was> rehashing old work and I never claimed to have the fastest prime> counting algorithm.> What I claimed was that _you_ are rehashing 200 year old work. I on the > other hand have taken twenty year old work as a starting point and > improved it. I am standing on the shoulders of giants; you are a dwarf.> Get published on it then. Until then you're just a hack trying to get> something off of Odlzyko, Lagarias and others.It is available from my webpage. Which, unlike yours, is available to everyone. And searching for my name on www.mathworld.com actually produces results. As long as they don't add a page of the worlds greatest mathematical crackpots there is no chance of you appearing there. > Now I have actually corresponded with both Odlyzko and Lagarias> (though they may regret it, especially Odlyzko), but what about you?> What do they say about your work, loudmouth?Why would they be interes? Their work was more than 20 years ago, they have moved on. If you search on the web for Odlyzko and Lagarias, you will find some very good stuff that has nothing to do with counting prime numbers, and some very good stuff that has nothing to do with mathematics. Miller is harder to find; there are just too many with that name. However, I have been asked to write an implementation for the PARI calculator. No promises, but it is quite likely to happen. The point is: 1. You found (or copied, nobody knows) a prime counting algorithm, claimed very loudly that it is the most brilliant invention under the sun, how it will revolutionise things and so on and so on. You were so proud because the algorithm was simple and fast. implementations. The first ones were dead slow, then they got faster and more obfusca. 3. All the time I got to the conclusion that you are just one asshole, and it would be worthwhile to demonstrate that you've got . For that purposeconsiderable simpler than your simplest implementation, uses less space, and is comparable in speed to your simple implementations. 1:0 for me. You don't have the simplest prime counting algorithm. blows yours out of the water both in speed and in memory consuption. 2:0 for me. You are _nowhere near_ the fastest prime counting algorithm. Six and a half years for your algorithm where mine takes eight hours!In other words, I demonstra that you are an incompetent jerk. It was a pleasure to do this. === Subject: : Re: Algebraic Integersmessage > Blow away, yes. Christian Bau did a remarkably good job with> his implementation at the higher numbers, although he had> help from Meissel and Lehmer.That's false. All Christian Bau even claimed was that he was> rehashing old work and I never claimed to have the fastest prime> counting algorithm.What I claimed was that _you_ are rehashing 200 year old work. I on the > other hand have taken twenty year old work as a starting point and > improved it. I am standing on the shoulders of giants; you are a dwarf.> Get published on it then. Until then you're just a hack trying to get> something off of Odlzyko, Lagarias and others.> It is available from my webpage. Which, unlike yours, is available to > everyone. And searching for my name on www.mathworld.com actually > produces results. As long as they don't add a page of the worlds > greatest mathematical crackpots there is no chance of you appearing > there. All that was necessary to verify is that the OP treating you as ifyou'd done something important was wrong.If it were worth anything, you'd publish it somewhere besides tossingit on a webpage. Or at least try. > Now I have actually corresponded with both Odlyzko and Lagarias> (though they may regret it, especially Odlyzko), but what about you?> What do they say about your work, loudmouth?> Why would they be interes? Their work was more than 20 years ago, > they have moved on. If you search on the web for Odlyzko and Lagarias, > you will find some very good stuff that has nothing to do with counting > prime numbers, and some very good stuff that has nothing to do with > mathematics. Miller is harder to find; there are just too many with that > name. Once again the OP's statements about you were clearly overra.You just used other people's work. === Subject: : Re: Algebraic Integers> All that was necessary to verify is that the OP treating you as if> you'd done something important was wrong.I kicked your ass, that's what I've done. No, it is not important, but it was fun. You know, and I know, that I can kick your ass anytime I want. And you know, and I know, that there are some people reading this who can kick your ass much much better than I can. Some are just too polite. ONE THOUSAND TIMES SLOWER!!!!!> If it were worth anything, you'd publish it somewhere besides tossing> it on a webpage. Or at least try.I don't have your obsessions. Narcisistic Personality Disorder, wasn't that what you are suffering from? What kind of medication do you have to take, anyway? > Once again the OP's statements about you were clearly overra.> You just used other people's work.I think I made quite clear what additions to the Extended Meissel-Lehmer algorithm I have made. I don't say its much, but it is more genuine mathematics than you have ever done in your life. === Subject: : Re: Algebraic Integers> message > Blow away, yes. Christian Bau did a remarkably good job with> his implementation at the higher numbers, although he had> help from Meissel and Lehmer.That's false. All Christian Bau even claimed was that he was> rehashing old work and I never claimed to have the fastest prime> counting algorithm.> What I claimed was that _you_ are rehashing 200 year old work. I on the > other hand have taken twenty year old work as a starting point and > improved it. I am standing on the shoulders of giants; you are a dwarf.> Get published on it then. Until then you're just a hack trying to get> something off of Odlzyko, Lagarias and others.> Now I have actually corresponded with both Odlyzko and Lagarias> (though they may regret it, especially Odlyzko), but what about you?> What do they say about your work, loudmouth?Remember when you used to tell yourself to just talk about the math? Think of it this way. Can it help your cause to get into personal fights with people?I heard a great quote on TV today, mentioned by of all people Tom McClintock, ultra-conservative candidate for California Governor. He said, Lincoln used to point out that you can't disprove Euclidean geometry by calling Euclid an idiot. === Subject: : Re: Algebraic IntegersIn sci.physics, <3c65f87.0309301233.fe76af8@posting.google.com >:>> In sci.physics, >> <3c65f87.0309291419.161d2dd9@posting.google.com>:>> I thought I was set. With a high I.Q. group set to publish my paper>> on factoring polynomials into non-polynomial factors I figured that>> now certainly I could push my agenda to fame and fortune. I was>> wrong. So here I am back again, humbled yet again, as there's no escaping it,>> there is no way I'll get anywhere pissing off mathematicians. So here are some concessions. If mathematician wish to dismiss my prime counting function.>> Dismiss, no.>> Blow away, yes. Christian Bau did a remarkably good job with>> his implementation at the higher numbers, although he had>> help from Meissel and Lehmer.> That's false. All Christian Bau even claimed was that he was> rehashing old work and I never claimed to have the fastest prime> counting algorithm.> So you're caught in a falsehood as it's impossible for my position to> have been blown away by someone putting up someone else's work when my> position was never the one that you imply.Oh, I see. Well, never mind then. What *was* your claim?Elegance? Simplicity? No memoization? The last at leastis verifiable and true (the other two being value judgements).> However, someone naive who actually *trus* youNobody trusts me. Nobody should.Trust No one. (*psst* Mulder. We have another one on the line. )This is, after all, Usenet, the denizens of quacks andcharlatans, the place of weird theories (some of which maybe true), the oddball hangout of the oddballs who hang out.Here, one might find truth and integrity. However, onewill also find various idiots whose sole purpose is totroll for amusement, who will argue with you just forthe sheer pleasure of it -- And the term argue one hasto take *very* loosely.(Not that I'm claiming either way what you are, James. Itdoesn't really matter. Mathematics can be verified, even ifone's identity cannot. But one does have to be somewhatcareful; even 1 + 1 need not always equal 2 here; 1 + 1could, for instance, equal 10 -- if done base 2.)> might have believed> you and the implication, which is that Christian Bau actually did> something, versus copy from what other researchers found, and that my> position was that I had the fastest prime counting algorithm, when> that was not my position.> But now they can put you in their list of sources not to be trus.Good for them, and they should. I do not claim to be amathematical authority; merely a hacker (in the good sense)who knows a bit about symbolic manipulations. If I'm luckyI might answer questions in a comprehensible fashion --if one's *real* lucky the answers are correct!> who looks for targets to attack to bring attention to yourself. You> probably figured I was a good and easy target, so you trot out your> bogus claim.Yes, of course, my apologies. Did you have aWebpage/PDF/foolscap/book [*] that put out a clear, cogent theory aboutthe divisibility/factorizability of roots of a certainpolynomial again? I'd like to see it, with some clarity,so that I *can* critique it (science relies on peerreview; no one should accept scientific pronouncements onblind faith alone). However, if you prefer I can comment(translation: lie like a rug) on how smelly your armpitsare, or how grotesquely weird your hair looked last Tuesdayas you stared out the bus going to the Mustang Ranch.Not that either would be all that truthful, of course,but many trolls do exactly that. (Someone's apparentlylet them out of alt.flame again; they're infesting,among other spots, comp.os.linux.advocacy, another groupwhich I happen to frequent on occasion. )The critiquing would be done on your posts alone, forthat's all I have. At best, you could post an ATCGsequence which I would have to replicate for a few yearsto gain an approximation of your physical appearance.(Why I would want to is another question. )In any event, if you're attempting to get us to accept,without any questions whatsoever, your stupendouslyfantastic results regarding whatever, you've obviouslycome to the wrong newsgroup(s).> [*] a foolscap is not automatically an item one wears on the 1st of April. One definition, which apparently stuck in Great Britain, is of a piece of paper, originally apparently with a fool's cap watermark.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: Algebraic Integers> In sci.physics, > :>> In sci.physics, >> <3c65f87.0309291419.161d2dd9@posting.google.com>:>> I thought I was set. With a high I.Q. group set to publish my paper>> on factoring polynomials into non-polynomial factors I figured that>> now certainly I could push my agenda to fame and fortune. I was>> wrong. So here I am back again, humbled yet again, as there's no escaping it,>> there is no way I'll get anywhere pissing off mathematicians. So here are some concessions. If mathematician wish to dismiss my prime counting function. Dismiss, no. Blow away, yes. Christian Bau did a remarkably good job with>> his implementation at the higher numbers, although he had>> help from Meissel and Lehmer.> That's false. All Christian Bau even claimed was that he was> rehashing old work and I never claimed to have the fastest prime> counting algorithm.> So you're caught in a falsehood as it's impossible for my position to> have been blown away by someone putting up someone else's work when my> position was never the one that you imply.> Oh, I see. Well, never mind then. What *was* your claim?> Elegance? Simplicity? No memoization? The last at least> is verifiable and true (the other two being value judgements).Yup, you're a critic troll. I said I'm dropping talking about primecounting, as I've realized people don't give a damn about it anyway.You must have seen that as an opening so you presen your falsehoodwhere you pumped up Christian Bau.Challenged on it, you decide to try and entertain.But hey, if you were any good at entertaining with your own stuff, youwouldn't need to be a critic troll looking for people to attack, nowwould you? === Subject: : Primitive Element TheoremPrimitive Element Theorem:If u,v are algebraic over F, a field of characteristic zero, then there's some algebraic w with F(u,v) = F(w).-- Here's an example of the primitive element theorem with proof.Q(sqr n, sqr m) = Q(sqr n + sqr m). Let a = sqr n + sqr ma^3 = n.sqr n + 3n.sqr m + 3m.sqr n + m.sqr m = (n+3m)sqr n + (3n+m)sqr ma^3 - (n+3m)a = 2(n-m)sqr m; a^3 - (3n+m)a = 2(m-n)sqr nsqr m = a(a^2 - n - 3m)/2(n-m) in Q(a)sqr n = a(a^2 - 3n - m)/2(m-n) in Q(a)(This example with division by n-m illustrates why the premise, field of characteristic zero, is needed.)Similar algebra for Q(u,v) = Q(u+v) where u^2 = au + b, v^2 = rv + sis more than twice as complex.Faced by such rising complexity, I see my feeble algebra is inadequate.Thus the question, how is the primitive element theorem approached?-- conjecture. When u,v have the same degree, then Q(u,v) = Q(u+v)---- === Subject: : Re: Primitive Element Theorem> Primitive Element Theorem:> If u,v are algebraic over F, a field of characteristic zero,> then there's some algebraic w with F(u,v) = F(w). -- conjecture. When u,v have the same degree, then Q(u,v) = Q(u+v)So Q(sqrt(2), -sqrt(2)) = Q(0).-- === Subject: : Re: Primitive Element Theorem> -- conjecture. When u,v have the same degree, then Q(u,v) = Q(u+v)> So Q(sqrt(2), -sqrt(2)) = Q(0).Or Q(1 - sqr 3, 1 + sqr 3) = Q(2) ?? ThusWhen non-conjugate u,v have the same degree, then Q(u,v) = Q(u+v) === Subject: : Re: Primitive Element Theorem>> -- conjecture. When u,v have the same degree, then Q(u,v) = Q(u+v)>> So Q(sqrt(2), -sqrt(2)) = Q(0).> Or Q(1 - sqr 3, 1 + sqr 3) = Q(2) ?? Thus> When non-conjugate u,v have the same degree, then Q(u,v) = Q(u+v)Aha!Then Q(sqrt(2), 1-sqrt(2)) = Q(1)-- === Subject: : Still looking for rule dividing & multiplying by 1>snip<> As a rule:You can divide or multiply anything - including zero - by 1, without> changing its value; but you cannot divide anything by zero!Is that right?So is 1/m the inverse of m?>YES.Kostas.>I think the answer Kostas, is that you can't divide anything, includingone;> by anything other than itself; without changing its value:> Obviously, then, you can't divide two by one and get two. I do this> on my calculator and get two, so, in the future, doing that should be> forbidden.Obviously my last statement doesn't jibe with my as a rule statement!There _is_ some sort of rule; because 1/m is _not always_ the inverse of m:Is it? === Subject: : Re: Still looking for rule dividing & multiplying by 1> There _is_ some sort of rule; because 1/m is _not always_ the inverse of m:> Is it?m * 1 = m1 * m = mm / 1 = mm / m = 1The inverse of m is defined to be 1/m.1/m (the inverse of m) is not = to m unless m=1 or m=-1. === Subject: : Re: Still looking for rule dividing & multiplying by 1> There _is_ some sort of rule; because 1/m is _not always_ the inverse ofm:> Is it?> m * 1 = m> 1 * m = m> m / 1 = m> m / m = 1I think you're onto something there Spiro.> The inverse of m is defined to be 1/m.> 1/m (the inverse of m) is not = to m unless m=1 or m=-1.They say there's an exception to every rule: Is this the rule _I'm_ seeking? === Subject: : Re: Still looking for rule dividing & multiplying by 1 > Obviously my last statement doesn't jibe with my as a rule statement! > There _is_ some sort of rule; because 1/m is _not always_ the inverse of m: > Is it?What do you mean? In mathematics, 1/m is by *definition* the notation forthe inverse of m when m is a number.-- === Subject: : Re: Still looking for rule dividing & multiplying by 1> Obviously my last statement doesn't jibe with my as a rule statement!> There _is_ some sort of rule; because 1/m is _not always_ the inverseof m:> Is it?> What do you mean? In mathematics, 1/m is by *definition* the notation for> the inverse of m when m is a number.Ya vol Dik: But it's a different answer with each different number!My original claim was that 'a' is proportional to 'f ', but 'a' is NOTproportional to '1/m': Which it claims to be in my physics textbook, andprobably in your's too?Acceleration _is_ proportional to 1/(f/a), and 1/(w/g) however; so thatsimple transpositions will solve any variable: But 'a' is not proportionalto 'f/m', because 'm' wouldn't provide for simple transpositions to solveit, or the other variables.> --> 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: Still looking for rule dividing & multiplying by 1> Obviously my last statement doesn't jibe with my as a rule statement!> There _is_ some sort of rule; because 1/m is _not always_ the inverse> of m:> Is it?What do you mean? In mathematics, 1/m is by *definition* the notation for> the inverse of m when m is a number.> Ya vol Dik: But it's a different answer with each different number!Um, yeah, the inverse of different numbers is different. Areyou rediscovering third grade now?> My original claim was that 'a' is proportional to 'f ', but 'a' is NOT> proportional to '1/m': Which it claims to be in my physics textbook, and> probably in your's too?If you are talking about the relationship a = f/m, then a proportional to m means As f is held constant, if 1/m is multiplied by a constant, a is multiplied bythe same constant. Apply the same force to half themass, you get twice the acceleration.That is certainly true. You think not? === Subject: : Re: Still looking for rule dividing & multiplying by 1> Obviously my last statement doesn't jibe with my as a rulestatement!> There _is_ some sort of rule; because 1/m is _not always_ theinverse> of m:> Is it?What do you mean? In mathematics, 1/m is by *definition* the notationfor> the inverse of m when m is a number.Ya vol Dik: But it's a different answer with each different number!> Um, yeah, the inverse of different numbers is different. Are> you rediscovering third grade now?My original claim was that 'a' is proportional to 'f ', but 'a' is NOT> proportional to '1/m': Which it claims to be in my physics textbook, and> probably in your's too?> If you are talking about the relationship a = f/m, then> a proportional to m means As f is held constant,> if 1/m is multiplied by a constant, a is multiplied by> the same constant. Apply the same force to half the> mass, you get twice the acceleration.> That is certainly true. You think not?What I think Randy is that you have the right conclusion; but the wrongpremise: === Subject: : Re: Still looking for rule dividing & multiplying by 1 > Obviously my last statement doesn't jibe with my as a rule statement! > There _is_ some sort of rule; because 1/m is _not always_ the inverse > of m: > Is it? > What do you mean? In mathematics, 1/m is by *definition* the notation for > the inverse of m when m is a number. > Ya vol Dik: But it's a different answer with each different number!Of course for each different 'm' we get a different '1/m'. > My original claim was that 'a' is proportional to 'f ', but 'a' is NOT > proportional to '1/m': Which it claims to be in my physics textbook, and > probably in your's too?I have no physics textbook... > Acceleration _is_ proportional to 1/(f/a),Eh? I though 'f = m.a' or 'a = f/m'. This means simply that 'a' isproportional to 'f' *and* that 'a' is proportional to '1/m'.Increasing 'm' while leaving 'f' constant, decreases 'a'. So? > Acceleration _is_ proportional to 1/(f/a),Now, this is nonsense, because 'f/a' = 'm', so that would also meanthat it is proportional to '1/m', which you claim it is not.-- === Subject: : Re: Still looking for rule dividing & multiplying by 1G.> Obviously my last statement doesn't jibe with my as a rulestatement!> There _is_ some sort of rule; because 1/m is _not always_ theinverse> of m:> Is it?What do you mean? In mathematics, 1/m is by *definition* thenotation for> the inverse of m when m is a number.Ya vol Dik: But it's a different answer with each different number!> Of course for each different 'm' we get a different '1/m'.> My original claim was that 'a' is proportional to 'f ', but 'a' is NOT> proportional to '1/m': Which it claims to be in my physics textbook,and> probably in your's too?> I have no physics textbook...I'm surprised to hear that: How do you know so much about physics? Memory?> Acceleration _is_ proportional to 1/(f/a),> Eh? I though 'f = m.a' or 'a = f/m'. This means simply that 'a' is> proportional to 'f' *and* that 'a' is proportional to '1/m'.> Increasing 'm' while leaving 'f' constant, decreases 'a'. So?That sounds so right; but how can you increase m in that formula, withoutincreasing f?> Acceleration _is_ proportional to 1/(f/a),> Now, this is nonsense, because 'f/a' = 'm', so that would also mean> that it is proportional to '1/m', which you claim it is not.It sounds like you are saying that: As well as f/a being _equal_ to m: Thatf/a is also _proportional_ to 1/m. Can that be?> --> 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: Dividing & multiplying by 1As a rule:You can divide or multiply anything - including zero - by 1, without> changing its value; but you cannot divide anything by zero!Is that right?So is 1/m the inverse of m?>YES.Kostas.>I think the answer Kostas, is that you can't divide anything, includingone;> by anything other than itself; without changing its value:> Obviously, then, you can't divide two by one and get two. I do this> on my calculator and get two, so, in the future, doing that should be> forbidden.You're carrying this forbidden thing to an extreme: Obviously somethings notkosher with my reasoning. === Subject: : Re: how to find limit of...> escribi.97 en el mensaje>>Lim((1-n^2)^(1/3) + n^(3/2))?>> Seems clear that the limit (as n -> infinity) is 0.>> For example, say f(x) = x^(1/3). Then f'(x) = x^(-2/3)/3,>> so the Mean Value Theorem shows that>> |f(1-n^2) - f(-n^2)| <= c/n^(4/3)>> for large n.>> ************************>??Already retrac this - I misread the problem.>For large n, it is roughly n^(3/2) - n^(2/3), that got to 0 as n go to>infinite.************************ === Subject: : Re: how to find limit of...>> as n-->0 the limit = 0>WHAT? n---> infinity.Well, then *say* that. === Subject: : optimizationHello everybody, I'm a post-gratuate student from greece and since mymath and software knowledge is limi I seem to find it difficult tosolve the problem below.So this is the case:................After conducting an experiment I've collec some (323) values. I know that those values come from the equation:value = A*x^2 + B*y^2 + C*z^2 + 2*D*x*y + 2*E*y*z + 2*F*z*xx, y and z are known and different for each value.My aim is to calculate the optimum values of A, B, C, D, E and F for which the values that will come up by the use of the equation willbe as close they can get to the experimental.I know for a fact that A,B and C are between 0-1 and D,E and F between(-1)-1What I've done so far:..............x, y and z are limi numbers(mixed up cosines with sines and stafflike that)So I choose randomly a heaxad values for A-F and calculate the belowsumsum from i=1 to 323 ev(i)-value(equation)After that I choose a different hexad and calculate the new sum.Lets say, after 100 different hexads the one where the sum is minimumis repor.This story continues for 100 times so from 100x100 sums I collect thehexads of the 100 minimun.These hexads litim the values of A-F. For example the A of the ortimumhexads is now between 0.2-0.5 etc...I rerun the same program having change the limits where A-F take theirrandom values.I keep doing this until I get variation of A-F value under the thirddecomale.g. A [0.356 - 0.369]I want to know if there is a faster easier or more secure way tocalculate the optimum values of A-F.I used matlab to run the prementioned optimisation which is the onlyone I know (I've learned it only to solve this problem) so if someonehas a set solution of a program for matlab ot would be muchapprecia.I know it was to long a letter so even if you don't know the answer tomy problem I ought to thank you for your time.TIA spiros. === Subject: : Re: optimization >Hello everybody, I'm a post-gratuate student from greece and since my >math and software knowledge is limi I seem to find it difficult to >solve the problem below. >So this is the case:................ >After conducting an experiment I've collec some (323) values. >I know that those values come from the equation: >value = A*x^2 + B*y^2 + C*z^2 + 2*D*x*y + 2*E*y*z + 2*F*z*x >x, y and z are known and different for each value. >My aim is to calculate the optimum values of A, B, C, D, E and F >for which the values that will come up by the use of the equation will >be as close they can get to the experimental. >I know for a fact that A,B and C are between 0-1 and D,E and F between >(-1)-1 >What I've done so far:.............. >x, y and z are limi numbers(mixed up cosines with sines and staff >like that) >So I choose randomly a heaxad values for A-F and calculate the below >sum >sum from i=1 to 323 ev(i)-value(equation) >After that I choose a different hexad and calculate the new sum. >Lets say, after 100 different hexads the one where the sum is minimum >is repor. >This story continues for 100 times so from 100x100 sums I collect the >hexads of the 100 minimun. These hexads litim the values of A-F. For example the A of the ortimum >hexads is now between 0.2-0.5 etc... >I rerun the same program having change the limits where A-F take their >random values. >I keep doing this until I get variation of A-F value under the third >decomal >e.g. A [0.356 - 0.369] I want to know if there is a faster easier or more secure way to >calculate the optimum values of A-F. >I used matlab to run the prementioned optimisation which is the only >one I know (I've learned it only to solve this problem) so if someone >has a set solution of a program for matlab ot would be much >apprecia. >I know it was to long a letter so even if you don't know the answer to >my problem I ought to thank you for your time. >TIA spiros.I assume you have x,y,z,value in vectors of length 323. so this is now vectorized:A=[x.^2 2*x.*y 2*x.*z y.^2 2*y.*z z.^2 ] ;b=value;param=lsqlin(A,b,[],[],[],[],[0,-1,-1,0,-1,0],[ 1,1,1,1,1,1]);if you have no access to the optimization toolbox and the bounds are not active then simply useparam=Ab;hthpeter === Subject: : Re: optimization[F-up set to comp.soft-sys.matlab]> After conducting an experiment I've collec some (323) values. > I know that those values come from the equation:> value = A*x^2 + B*y^2 + C*z^2 + 2*D*x*y + 2*E*y*z + 2*F*z*x> x, y and z are known and different for each value.> My aim is to calculate the optimum values of A, B, C, D, E and F > for which the values that will come up by the use of the equation will> be as close they can get to the experimental.> I know for a fact that A,B and C are between 0-1 and D,E and F between> (-1)-1I think you'll want a constrained least-square solution. Imagine thatyou construct a matrix for the x^2, y^2, etc values:x1^2 y1^2 z1^2 2*x1*y1 2*y1*z1 2*z1*x1x2^2 y2^2 z2^2 2*x2*y2 2*y2*z2 2*z2*x2x3^2 y3^2 z3^2 2*x3*y3 2*y3*z3 2*z3*x3...Call it C. Now when you multiply this by a vector x, defined as[A B C D E F]', you get another vector of your values (call them y).x is unknown, so you want to minimize (C*x - y).^2. If there were noconstraints, you could just do x = Cy; But since you haveconstraints for x, you can use lsqlin instead, which allowsconstraints, again specified by a matrix and a vector. The first few rowsmight look like:[1 0 0 0 0 0] * x <= [1][-1 0 0 0 0 0] [1][0 1 0 0 0 0] [1]Give it a shot!-- Peter Boettcher MIT Lincoln LaboratoryMATLAB FAQ: http://www.mit.edu/~pwb/cssm/ === Subject: : Re: optimization> After conducting an experiment I've collec some (323) values.> I know that those values come from the equation:> value = A*x^2 + B*y^2 + C*z^2 + 2*D*x*y + 2*E*y*z + 2*F*z*x> x, y and z are known and different for each value.> My aim is to calculate the optimum values of A, B, C, D, E and F> for which the values that will come up by the use of the equation will> be as close they can get to the experimental.This is a typical problem called least squares approximation.> I know for a fact that A,B and C are between 0-1 and D,E and F between> (-1)-1That's nice to know (for checking the solution) but not needed in thefollowing.> What I've done so far:..............> x, y and z are limi numbers(mixed up cosines with sines and staff> like that)> So I choose randomly a heaxad values for A-F and calculate the below> sum> sum from i=1 to 323 ev(i)-value(equation)> After that I choose a different hexad and calculate the new sum.> Lets say, after 100 different hexads the one where the sum is minimum> is repor.> This story continues for 100 times so from 100x100 sums I collect the> hexads of the 100 minimun.> These hexads litim the values of A-F. For example the A of the ortimum> hexads is now between 0.2-0.5 etc...> I rerun the same program having change the limits where A-F take their> random values.> I keep doing this until I get variation of A-F value under the third> decomal> e.g. A [0.356 - 0.369]Wow, that's really unmathematical. Stop it. It sounds weird,time-consuming and not exact. (I didn't read on after 100 times.> I want to know if there is a faster easier or more secure way to> calculate the optimum values of A-F.Yep, there's a very fast, easy and stable algorithm to do this based onthe so-called QR decomposition of a matrix.The following code between the === Subject:== lines is valid LaTeX code. Justput it into a file and run LaTeX to get a nicer output, if you don'twant to read the code in text form. (If you don't have/know LaTeX, I cansend you a PDF file, if you give me your e-mail address.)=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:begin{document}Let's call your equation values $b_1,ldots,b_n$, where $n=323$ inyour case.Similarly let $x_1,ldots,x_n$ be the $x$ values, $y_1,ldots,y_n$ the$y$ values, and $z_1,ldots,z_n$ the $z$ values you obtained from yourexperiment.Then let $A$ be your data matrix$$A = left( begin{array}{cccccc} x_1^2 & y_1^2 & z_1^2 & x_1 y_1 & y_1 z_1 & x_1 z_1 vdots & vdots & vdots & vdots & vdots & vdots x_n^2 & y_n^2 & z_n^2 & x_n y_n & y_n z_n & x_n z_nend{array} right),$$let $b$ your vector of experimental values$$b = left( begin{array}{c} b_1 vdots b_nend{array} right),$$and let $x$ be the vector of values you are searching for$$x = left( begin{array}{c} A B C 2D 2E 2Fend{array} right).$$Then you can write your optimatzation problem as follows:$$min_x left| A x - b right|_2$$Remark: You can bring all parameter fitting problems into this form ifthe equation is linear in the parameters ($A$ to $2F$ in your case).There are several ways to solve this. Two of them are:begin{itemize}item Solve the linear equation$$ A^T A x = A^T b$$using the Gauss-Algorithm (LR decomposition of $A^T A$).item Solve the problem by using the QR decomposition of $A$$$ A = Q R$$which produces an orthogonal matrix $Q$ and an upper right matrix $R$.Then the minimazation problem is equivalent to the following linearequation for $x$$$ R x = Q^T b$$which can be easily solved, because $R$ has upper-right form.end{itemize}The first method is not recommended, since it uses more operations thennecessary and is not as stable as the second one.end{document}=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:> I used matlab to run the prementioned optimisation which is the only> one I know (I've learned it only to solve this problem) so if someone> has a set solution of a program for matlab ot would be much> apprecia.Here comes the MatLab code, assuming you built the A matrix and b vectoras explained above.For applying the second (stable) method, use the following commands:% perform QR decomposition of A[Q R] = qr(A);% calculate solution of R x = Q^T bx = R(Q'*b);The Matlab help to the QR algorithm proposes sth. based on a mixture ofthe two methods which should also be very reliable:% perform QR decomposition of A[Q R] = qr(A);% calculate the solution of R^T R x (= A^T A x) = A^T bx = R(R'(A'*b));So, you only have to build the matrix A, vector b and enter two commandlines to obtain your solution x=(A,B,C,2D,2E,2F).The 2 methods normally are working equally reliable and are very fastand accurate. You could even use much more data samples then 323.If you want to check the accuracy, you can (in both cases) perform acorrection step:% calculate residualr = b - A*x;% calculate error correction step (depending on which% method you prefer), which should be incredible smalle = R(Q'*b); ORe = R(R'(A'*r));% add correction to solution vectorx = x + e;A final note:I realized that your equation has a quadratic form. It can be written as(this time it's ASCII art for better understanding): |A D F| |x||x y z| * |D B E| * |y| |F E C| |z|Therefore I think, there could be an even better solution to theoptimization problem which uses the special structure of the equation.But I'm too lazy to think about it for the time being, sorry.-- AndreasFor replying, remove the fruit from my address. === Subject: : Re: optimization> Here comes the MatLab code, assuming you built the A matrix and bvector> as explained above.> For applying the second (stable) method, use the following commands:> % perform QR decomposition of A> [Q R] = qr(A);> % calculate solution of R x = Q^T b> x = R(Q'*b);> The Matlab help to the QR algorithm proposes sth. based on a mixtureof> the two methods which should also be very reliable:> % perform QR decomposition of A> [Q R] = qr(A);> % calculate the solution of R^T R x (= A^T A x) = A^T b> x = R(R'(A'*b));> So, you only have to build the matrix A, vector b and enter twocommand> lines to obtain your solution x=(A,B,C,2D,2E,2F).> The 2 methods normally are working equally reliable and are very fast> and accurate. You could even use much more data samples then 323.> If you want to check the accuracy, you can (in both cases) perform a> correction step:> % calculate residual> r = b - A*x;> % calculate error correction step (depending on which> % method you prefer), which should be incredible small> e = R(Q'*b);> OR> e = R(R'(A'*r));> % add correction to solution vector> x = x + e;I discovered in the thread minimization of complex argued functionwhich is currently active in the group comp.soft-sys.matlab, that onedoesn't even have to perform the QR decomposition manually. Thecommandlinex = Ab;is already sufficient, since Matlab detects wether the system isoverdetermined or not and uses the QR or LR decomposition, respectively.Great, isn't it?-- AndreasFor replying, remove the fruit from my address. === Subject: : Re: optimizationThis is a good starting place, but the problem is neither linear or normal,so if it does not lead to a satisfactory solution you may want to considerthe maximum likelihood estimation method.> After conducting an experiment I've collec some (323) values.> I know that those values come from the equation:> value = A*x^2 + B*y^2 + C*z^2 + 2*D*x*y + 2*E*y*z + 2*F*z*x> x, y and z are known and different for each value.> My aim is to calculate the optimum values of A, B, C, D, E and F> for which the values that will come up by the use of the equation will> be as close they can get to the experimental.> This is a typical problem called least squares approximation.> I know for a fact that A,B and C are between 0-1 and D,E and F between> (-1)-1> That's nice to know (for checking the solution) but not needed in the> following.> What I've done so far:..............> x, y and z are limi numbers(mixed up cosines with sines and staff> like that)> So I choose randomly a heaxad values for A-F and calculate the below> sum> sum from i=1 to 323 ev(i)-value(equation)> After that I choose a different hexad and calculate the new sum.> Lets say, after 100 different hexads the one where the sum is minimum> is repor.> This story continues for 100 times so from 100x100 sums I collect the> hexads of the 100 minimun.> These hexads litim the values of A-F. For example the A of the ortimum> hexads is now between 0.2-0.5 etc...> I rerun the same program having change the limits where A-F take their> random values.> I keep doing this until I get variation of A-F value under the third> decomal> e.g. A [0.356 - 0.369]> Wow, that's really unmathematical. Stop it. It sounds weird,> time-consuming and not exact. (I didn't read on after 100 times.> I want to know if there is a faster easier or more secure way to> calculate the optimum values of A-F.> Yep, there's a very fast, easy and stable algorithm to do this based on> the so-called QR decomposition of a matrix.> The following code between the === Subject:== lines is valid LaTeX code. Just> put it into a file and run LaTeX to get a nicer output, if you don't> want to read the code in text form. (If you don't have/know LaTeX, I can> send you a PDF file, if you give me your e-mail address.)> === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:> begin{document}> Let's call your equation values $b_1,ldots,b_n$, where $n=323$ in> your case.> Similarly let $x_1,ldots,x_n$ be the $x$ values, $y_1,ldots,y_n$ the> $y$ values, and $z_1,ldots,z_n$ the $z$ values you obtained from your> experiment.> Then let $A$ be your data matrix> $$> A = left( begin{array}{cccccc}> x_1^2 & y_1^2 & z_1^2 & x_1 y_1 & y_1 z_1 & x_1 z_1 > vdots & vdots & vdots & vdots & vdots & vdots > x_n^2 & y_n^2 & z_n^2 & x_n y_n & y_n z_n & x_n z_n> end{array} right),> $$> let $b$ your vector of experimental values> $$> b = left( begin{array}{c}> b_1 vdots b_n> end{array} right),> $$> and let $x$ be the vector of values you are searching for> $$> x = left( begin{array}{c}> A B C 2D 2E 2F> end{array} right).> $$> Then you can write your optimatzation problem as follows:> $$> min_x left| A x - b right|_2> $$> Remark: You can bring all parameter fitting problems into this form if> the equation is linear in the parameters ($A$ to $2F$ in your case).> There are several ways to solve this. Two of them are:> begin{itemize}> item Solve the linear equation> $$> A^T A x = A^T b> $$> using the Gauss-Algorithm (LR decomposition of $A^T A$).> item Solve the problem by using the QR decomposition of $A$> $$> A = Q R> $$> which produces an orthogonal matrix $Q$ and an upper right matrix $R$.> Then the minimazation problem is equivalent to the following linear> equation for $x$> $$> R x = Q^T b> $$> which can be easily solved, because $R$ has upper-right form.> end{itemize}> The first method is not recommended, since it uses more operations then> necessary and is not as stable as the second one.> end{document}> === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:> I used matlab to run the prementioned optimisation which is the only> one I know (I've learned it only to solve this problem) so if someone> has a set solution of a program for matlab ot would be much> apprecia.> Here comes the MatLab code, assuming you built the A matrix and b vector> as explained above.> For applying the second (stable) method, use the following commands:> % perform QR decomposition of A> [Q R] = qr(A);> % calculate solution of R x = Q^T b> x = R(Q'*b);> The Matlab help to the QR algorithm proposes sth. based on a mixture of> the two methods which should also be very reliable:> % perform QR decomposition of A> [Q R] = qr(A);> % calculate the solution of R^T R x (= A^T A x) = A^T b> x = R(R'(A'*b));> So, you only have to build the matrix A, vector b and enter two command> lines to obtain your solution x=(A,B,C,2D,2E,2F).> The 2 methods normally are working equally reliable and are very fast> and accurate. You could even use much more data samples then 323.> If you want to check the accuracy, you can (in both cases) perform a> correction step:> % calculate residual> r = b - A*x;> % calculate error correction step (depending on which> % method you prefer), which should be incredible small> e = R(Q'*b);> OR> e = R(R'(A'*r));> % add correction to solution vector> x = x + e;> A final note:> I realized that your equation has a quadratic form. It can be written as> (this time it's ASCII art for better understanding):> |A D F| |x|> |x y z| * |D B E| * |y|> |F E C| |z|> Therefore I think, there could be an even better solution to the> optimization problem which uses the special structure of the equation.> But I'm too lazy to think about it for the time being, sorry.> --> Andreas> For replying, remove the fruit from my address. === Subject: : Re: optimization> This is a good starting place, but the problem is neither linear ornormal,> so if it does not lead to a satisfactory solution you may want toconsider> the maximum likelihood estimation method.Sorry, but I can't see any nonlinearity in the unknow parameters A, ...,2F. Where do you see them?And yes, parameter fitting is quite normal. Thousands of people do itevery day.And why do you want to make things more comlica than they are? Thisalgorithm is one of the most stable algorithm known in mathematics. I'mnot sure, because my statistics lectures are far away, but I think asolution to maximum likelihood estimation can be obtained by the samealgorithms? (I may be wrong in this point.)-- AndreasFor replying, remove the fruit from my address. === Subject: : Re: optimization> Sorry, but I can't see any nonlinearity in the unknow parameters A, ...,> 2F. Where do you see them?large values of x, y, z contribute disproportionately> And yes, parameter fitting is quite normal. Thousands of people do it> every day.I meant normal in the sense of Gaussian.> And why do you want to make things more comlica than they are?Because you may get a different answer, if the constraints are viola thenit maybe worth looking into.I agree LS will probably do the job, but as I don't know what the OP istrying to achieve I provided the information so they could choose.>I'm> not sure, because my statistics lectures are far away, but I think a> solution to maximum likelihood estimation can be obtained by the same> algorithms? (I may be wrong in this point.)for y=a+bx and normal data it can. === Subject: : Re: optimization> Sorry, but I can't see any nonlinearity in the unknow parameters> A, ..., 2F. Where do you see them?> large values of x, y, z contribute disproportionatelySo what? The parameters are A, ..., F. And the paramter-dependingfunction that has to be fittet may even include e^x and 1/x. Thesolution obtained using the qr decomposition is optimal in the 2-normsense.> And yes, parameter fitting is quite normal. Thousands of> people do it every day.> I meant normal in the sense of Gaussian.You star this normal thing. I used the word typical and youdisagreed with that.> And why do you want to make things more comlica than they are?> Because you may get a different answer, if the constraints areviola then> it maybe worth looking into.> I agree LS will probably do the job, but as I don't know what the OPis> trying to achieve I provided the information so they could choose.It will definitely do the job when one doesn't take the constraints intoaccount. Because from the original text, I thought the given intervalswere meant to be estimations rather than constraints. Maybe tzavalascould tell us, wether he has constraints or estimations for theparameters.> I'm not sure, because my statistics lectures are> far away, but I think a solution to maximum> likelihood estimation can be obtained by the> same algorithms? (I may be wrong in this point.)> for y=a+bx and normal data it can.Aha, that's always good to know. Thanks.-- AndreasFor replying, remove the fruit from my address. === Subject: : Re: optimizationA short question, before I post my answer: Do you know LaTeX notation?That would make an answer much easier.-- AndreasFor replying, remove the fruit from my address. === Subject: : Help with algorithm equationI'm not too experienced with this (as you can tell) and I don't understandthe following statement: mY0:=1/m SIGMA (Pi-(Y1)Si = (Y2)(Si)^2 ) i=1where: m=number of points (Si,Pi), Y1,Y2 are already transformed 'outer' points.What does the = signify in (Pi-(Y1)Si = (Y2)(Si)^2 ) ?I'm trying to understand/get an algorithm to work, which transforms aquadratic regression curve to be unimodal using Bezier control points. Thealgorithm works out the transform.I can post the whole algorithm if its needed to put things in context (itsnot very long), but what I'm particularly stuck on is the above equation.The text states that:this statement evaluates the best least squares estimate of Y0 for theparabola with new values of Y2 and Y1 resulting from the first adjustment.The second adjustment, a vertical translation of the B.8ezier-reshapedparabola, occurs only when and Y1 and Y2 have been altered.In fact, I'm not sure that the steps in the algorithm I've done before thisequation are right either:(So ANY help apprecia.PS I have access to matlabTIA === Subject: : What are the equations calculate what dates are New Moon, Full Moon?The === Subject: says its all.If you know what books, papers, URL sites have this information,please follow up and reply to me in email.Thank Q in advance! === Subject: : Re: What are the equations calculate what dates are New Moon, Full Moon?> The === Subject: says its all.> If you know what books, papers, URL sites have this information,> please follow up and reply to me in email.> Thank Q in advance!If you read Lisp try Calendrical Calculations E M Reingold & NDershowitz, CUP.-- G.C. === Subject: : Re: What are the equations calculate what dates are New Moon, Full Moon?> The === Subject: says its all.> If you know what books, papers, URL sites have this information,> please follow up and reply to me in email.> Thank Q in advance!Your question would be better placed in sci.astro, butsince you're here...Get hold of the book Astronomical Algorithms byJean Meeus. It has what you're looking for. === Subject: : Re: What are the equations calculate what dates are New Moon, Full Moon?T = moon's rotation period around the Earth, at the phase point relative to thesun, that is from a noon to a noon.T2 = moon's rotation period from midnight to midnight.T = T2?Any way New Moon is every T.Starting with any known New Moon moment in the past, preferedably the latestone, at date X seconds from The beginning of Time, like Gregorian,next date.i = X + T*iYours, Goncz (at aol dot com)Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/I will be signing an amended Affidavit soon. === Subject: : Re: What are the equations calculate what dates are New Moon, Full Moon?> The === Subject: says its all.> If you know what books, papers, URL sites have this information,> please follow up and reply to me in email.> T = moon's rotation period around the Earth, at the phase point > relative to the sun, that is from a noon to a noon....> Starting with any known New Moon moment in the past, preferedably the latest> one, at date X seconds from The beginning of Time, like Gregorian,> next date.i = X + T*i...Slightly-more-accurate formulae include an i*i term; eg, D = 5.597661 + 29.5305888610*i + (102.026*10^-12)*i*i, relative to 2000-01-01 00:00:00, as shown under Approximate formula in http://www.wikipedia.org/wiki/New_moon , whichalso has 2 literature refs. It also says The true new moon may differ from this by over 14 hours due to periodicperturbations.http://www.moshier.net/ has a link to newmoontab.zip withand a link to the program that calcula the table.-jiw === Subject: : Re: What are the equations calculate what dates are New Moon, Full Moon?> T = moon's rotation period around the Earth, at the phase point relative to the> sun, that is from a noon to a noon.> T2 = moon's rotation period from midnight to midnight.> T = T2?> Any way New Moon is every T.The moon rotate around the earth, sometimes is 29 days, sometimes is 30days. How do I know what period is 29 days, what period is 30 days?Chinese calendar is unlike Gregorian calendar.Gregorian calendar always have31, 28/29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 days (365/3666)each year.But Chinese calendar is very irregular. It depends on the period of moonaround the earth. Sometime you may see continue two, even three monthshave 30 days.> Starting with any known New Moon moment in the past, preferedably the latest> one, at date X seconds from The beginning of Time, like Gregorian,> next date.i = X + T*iI can only find the UTC time (Jan. 1st, 1970 00:00:00 is 0)in seconds.Or I can get the Julian Day (Jan. 1st is 1; Dec. 31st is 365/366)in seconds.in seconds. === Subject: : Proof of Fourier integral theoremI have read several book of math analysis which contains the Fourierintegral theorem and its proof. Quite distressed did I find that everybook access to this theorem by the analogous theorem, i.e. firstchange the order of integration, then add and cancel the term f(x+0)and f(x-0), and last use the Riemann Lebesgue lemma. Are there otherpossible method to get the proof of this theorem?One of these book I read, did suggest that we can take the limit ofthe periodic Fourier series (of period 2l, and let l -->infinity). Butconsidering the evaluation of error in the possible deduction Iretreat.Can anyone help me out? === Subject: : Re: Proof of Fourier integral theorem>I have read several book of math analysis which contains the Fourier>integral theorem and its proof. Quite distressed did I find that every>book access to this theorem by the analogous theorem, i.e. first>change the order of integration, then add and cancel the term f(x+0)>and f(x-0), and last use the Riemann Lebesgue lemma. Are there other>possible method to get the proof of this theorem?First show that if f and g are in L^1 then int f g^ = int f^ g.This is immediate from Fubini's theorem. It follows that f * (g^) = (f^ g)^(that may be off by a minus sign in the exponent), where* denotes convolution.Now suppose that f and f^ are both in L^1. Replace g aboveby a suitable approximate identity g_n and see what happensto both sides as n -> infinity.>One of these book I read, did suggest that we can take the limit of>the periodic Fourier series (of period 2l, and let l -->infinity). But>considering the evaluation of error in the possible deduction I>retreat.>Can anyone help me out?************************ === Subject: : Q: Product code generator polynomialIf I have two codes C1 and C2, which are both cyclic, and the lengthof C1 (n1) and C2 (n2) are relatively prime, according to Lin andCostello (Lin, Shu, and Daniel J. Costello, Jr.,Error Control Coding:Fundamentals and Applications, Englewood Cliffs, N.J., Prentice-Hall,1983.) these two codes can be used to form a product code, and furtherthe product code will also be a (n1n2,k1k2) cyclic code. Also, theminimum distance of the code will be d1*d2, where d1 and d2 is theminimum distance of codes C1 and C2 respectively.Is this an upperbound or an existence theorem, that is, if aboveconditions hold, the best will be the product code with the givenparameters, or given n1 and n2 (relatively prime), we will always beable to find a code C1 and a code C2 such that the product code willhave the properties sta above?Also, how do I read the (n1n2,k1k2) cyclic code from the product code?If I have a product code of the form:P_{1,1} d_{1,1} d_{1,2} ... d_{1,k1} P_{1,2} d_{2,1} d_{2,2} ... d_{2,k1}..P_{1,k2} d_{k2,1) d_{k2,2} ... d_{k2,k1}P_{1,k2+1} P_{2,1} P_{2,2} ... P_{2,k1}(With P_{1,i} genera using the encoding of C1, and P_{2,i} usingthe encoding of C2.)how can I form the (n1n2,k1k2) cyclic code of the form:c_{1}, c_{2}, c_{3}, ... ,c_{n1n2}(Do I read row for row or column for column, or diagonal?)Your time, effort and suggestions will be greatly appreciaJaco Versfeld === Subject: : Re: Q: Product code generator polynomial> If I have two codes C1 and C2, which are both cyclic, and the length> of C1 (n1) and C2 (n2) are relatively prime, according to Lin and> Costello (Lin, Shu, and Daniel J. Costello, Jr.,Error Control Coding:> Fundamentals and Applications, Englewood Cliffs, N.J., Prentice-Hall,> 1983.) these two codes can be used to form a product code, and further> the product code will also be a (n1n2,k1k2) cyclic code. Also, the> minimum distance of the code will be d1*d2, where d1 and d2 is the> minimum distance of codes C1 and C2 respectively.OK> Is this an upperbound or an existence theorem, that is, if above> conditions hold, the best will be the product code with the giventhe best what?> Also, how do I read the (n1n2,k1k2) cyclic code from the product code?Chinese remainder theorem. Ponder this diagram illustratingn_1 = 3, n_2 = 5. 0 11 7 3 14 5 1 12 8 4 10 6 2 13 9-- === Subject: : Re: An optimization problemhi,> How do you determine that a function> space is convex? Such a determination> requires some kind of metric and a > corresponding ordering on the functions. let's see. i was thinking of convexness in the following sense:let X be a space (can be a function space) in which the operation of'+' and multiplication by a scalar (in R or C) is defined then X isconvex ifgiven any two elements x1 and x2 in X and any nonnegative pair ofnumbers a,b such that a+b=1 then a*x1+b*x2 is again an element of X. this looks like a fine definition, and it becomes the usualdefinition of convexness if X is a subspace of a Euclidean space. as far as i know, to check for convexness one does not require thepresence of a metric or even a corresponding ordering on the functions(this can be seen from the definition itself). all you need is adefinition of the operations of + and multiplication by a scalar. doyou think that R^n (n>1) has an ordering defined on it? even withoutthe Euclidean metric one can still check for the convexness of asubset of R^n.> You can lead a horse's ass to knowledge, but you can't make him think.what is this suppose to mean? J. === Subject: : Re: Halmos Set Theory Text> There are apparently several editions of this book, some from around> 1960 all the way to about 1990. Is any particular edition superior> from a self-teaching standpoint to the others? All advice> apprecia.Several copies are lis at www.bookfinder.com and among them, theonly edition mentioned is the first. Others mention no edition atall. As mentions, there are various publication andcopyright dates. However, these do not necessarily indicate anythingmore than printing dates and assertions of intellectual propertyrights.David Ames === Subject: : Re: unit element in Ring>>Hi guys,>>One of the characteristics of a Ring is the absence of an inverse element>>(regarding multiplication).> No. The characteristics of a Ring is that it has two binary> associative operations that satisfy the ring axioms. The existence of> inverse elements with respect to multiplication are neither required> nor banned.>>How is it possible then to define a so called 'Unit' in a Ring. > A unit is a NEUTRAL element for multiplication; in order to define> multiplicative inverses you need a neutral element, but it is possible> to define neutral element without having talked about multiplicative> inverses. (The neutral element, however, will always have a> multiplicative inverse: itself).Terminology varies a bit here.Sometimes the neutral element w.r.t. multiplication is called identityand the invertible elements are called units.I am sure the OP had this terminology in mind.Perhaps that usage stems from translating Einheit into English?Marc === Subject: : Re: unit element in RingIn sci.math, Bouwman P.:> Hi guys,> One of the characteristics of a Ring is the absence of an inverse element> (regarding multiplication).> How is it possible then to define a so called 'Unit' in a Ring. A Unit u is> an element for which an inverse v exists so that u.v=1. Also sta as u> divides 1.Not all rings contain such units; the ring of even integersis one example. However, a ring need not exclude inverses.http://mathworld.wolfram.com/Ring.html> Peter-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: unit element in RingTo everybody who answered. Thanks a lot.I've been set straight again.Peter> Richard,> I love to play semantic games but be serious.> If an multiplicative inverse is excluded from the definition of a ring,it> means (should mean?) that that inverse is not allowed in a ring.> It is not excluded. It is only not a requirement for a ring, so it is> not in the definition. Every field is a ring, but not every ring is a> field. A ring is a field if and only if every element except 0 has a> multiplicative inverse.> --> 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: Why is Sum (1/p) divergent>>http://www.utm.edu/research/primes/infinity.shtml>>Given that 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 ... diverges, it would be>>very surprising if one of>>1/2 + 1/5 + 1/11 ... and>>1/3 + 1/7 + 1/13 ...>>was finite and the other was not.>>Perhaps the standard proof could be modified to easily show this?>> It's immediate from the fact that the full series diverges and>> has decreasing terms that both sub-series diverge:>> 1/5 + 1/11 ... < 1/3 + 1/7 + 1/13 ... < 1/2 + 1/5 + 1/11 ...>???The inequalities show that either both subseries converge or theyboth diverge. But they can't both converge because the originalseries diverges.>I reached the same conclusion via a different route. Both cant converge>because then they would be both absolutely convergent and could be added to>get a divergent series. The only possibility is the smaller 1/3+1/7.. is>convergent and 1/2+1/5.. is divergent. But because there is prime between>p_n and 2p_n we can write 1/2+1/5+... <2/3+2/7+2/13+... giving both>convergent - a contradiction. Hence the only possibility is both diverge.>This process can be repea on the two divergent series, can it be asser>that the sum 1/p_an is divergent? Yes, by the argument I gave - consider the n series sum 1/p_{an}, sum 1/p_{an+1}, ... sum 1/p_{an + (n-1)}. They satisfy inequalities as above, so either they all converge or they all diverge.>Or more interestingly is sum 1/p_p>divergent?The notation makes no sense. I imagine you mean to ask aboutthe series sum 1/p_{p_n}, where p_n is the n-th prime.Let's see. As no, the n-th prime is roughly n log(n), by theprime number theorem. So p_{p_n} is roughly p_n log(p_n),which is roughly (n log(n)) log(n log(n)), so 1/p_{p_n}is something like 1/(n log(n) log(n log(n))). This is less than1/(n log(n)^2) and that sum converges (by the integraltest).************************ === Subject: : Re: Why is Sum (1/p) divergent>http://www.utm.edu/research/primes/ infinity.shtml>>Given that 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 ... diverges, it wouldbe>>very surprising if one of>>1/2 + 1/5 + 1/11 ... and>>1/3 + 1/7 + 1/13 ...>>was finite and the other was not.>>Perhaps the standard proof could be modified to easily show this?>> It's immediate from the fact that the full series diverges and>> has decreasing terms that both sub-series diverge:>> 1/5 + 1/11 ... < 1/3 + 1/7 + 1/13 ... < 1/2 + 1/5 + 1/11 ...>>???> The inequalities show that either both subseries converge or they> both diverge. But they can't both converge because the original> series diverges.>I reached the same conclusion via a different route. Both cant converge>because then they would be both absolutely convergent and could be addedto>get a divergent series. The only possibility is the smaller 1/3+1/7.. is>convergent and 1/2+1/5.. is divergent. But because there is prime between>p_n and 2p_n we can write 1/2+1/5+... <2/3+2/7+2/13+... giving both>convergent - a contradiction. Hence the only possibility is both diverge.>>This process can be repea on the two divergent series, can it beasser>that the sum 1/p_an is divergent?> Yes, by the argument I gave - consider the n series sum 1/p_{an},> sum 1/p_{an+1}, ... sum 1/p_{an + (n-1)}. They satisfy inequalities> as above, so either they all converge or they all diverge.>Or more interestingly is sum 1/p_p>divergent?> The notation makes no sense. I imagine you mean to ask about> the series sum 1/p_{p_n}, where p_n is the n-th prime.> Let's see. As no, the n-th prime is roughly n log(n), by the> prime number theorem. So p_{p_n} is roughly p_n log(p_n),> which is roughly (n log(n)) log(n log(n)), so 1/p_{p_n}> is something like 1/(n log(n) log(n log(n))). This is less than> 1/(n log(n)^2) and that sum converges (by the integral> test).Is it valid to subsitute nlog(n) like that, in general if f(x) is asymptoticto g(x) why should f(f(x)) be asymptotic to g(g(x))?I think it would be better to use an inequality like:n (log n + log log n - 1.0073) < p(n) < n (log n + log log n - 0.9385) forn>8601.http://www.utm.edu/research/primes/howmany.shtml# 2Though I notice it gives the same result (the nloglog(n)-n1.0073 iseventually >0). === Subject: : Re: Why is Sum (1/p) divergent>>http://www.utm.edu/research/primes/infinity.shtml >>Given that 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 ... diverges, it would>be>very surprising if one of>1/2 + 1/5 + 1/11 ... and>1/3 + 1/7 + 1/13 ...>>was finite and the other was not.>>Perhaps the standard proof could be modified to easily show this?It's immediate from the fact that the full series diverges and> has decreasing terms that both sub-series diverge: 1/5 + 1/11 ... < 1/3 + 1/7 + 1/13 ... < 1/2 + 1/5 + 1/11 ...>???>> The inequalities show that either both subseries converge or they>> both diverge. But they can't both converge because the original>> series diverges.>>I reached the same conclusion via a different route. Both cant converge>>because then they would be both absolutely convergent and could be added>to>>get a divergent series. The only possibility is the smaller 1/3+1/7.. is>>convergent and 1/2+1/5.. is divergent. But because there is prime between>>p_n and 2p_n we can write 1/2+1/5+... <2/3+2/7+2/13+... giving both>>convergent - a contradiction. Hence the only possibility is both diverge.>>This process can be repea on the two divergent series, can it be>asser>>that the sum 1/p_an is divergent?>> Yes, by the argument I gave - consider the n series sum 1/p_{an},>> sum 1/p_{an+1}, ... sum 1/p_{an + (n-1)}. They satisfy inequalities>> as above, so either they all converge or they all diverge.>>Or more interestingly is sum 1/p_p>>divergent?>> The notation makes no sense. I imagine you mean to ask about>> the series sum 1/p_{p_n}, where p_n is the n-th prime.>> Let's see. As no, the n-th prime is roughly n log(n), by the>> prime number theorem. So p_{p_n} is roughly p_n log(p_n),>> which is roughly (n log(n)) log(n log(n)), so 1/p_{p_n}>> is something like 1/(n log(n) log(n log(n))). This is less than>> 1/(n log(n)^2) and that sum converges (by the integral>> test).>Is it valid to subsitute nlog(n) like that, in general if f(x) is asymptotic>to g(x) why should f(f(x)) be asymptotic to g(g(x))?You'll notice I was careful not to actually say I'd provedthat sum 1/p_{p_n} was convergent. The argumentabove suffices to convince me that the sum does in factconverge, but yes it is a little fuzzy.>I think it would be better to use an inequality like:>n (log n + log log n - 1.0073) < p(n) < n (log n + log log n - 0.9385) for>n>8601.Actually it may well be that just knowing the prime numbertheorem as usually sta (with no higher-order terms orerror estimates) is enough - I'd be very surprised if thiswere not so, but the argument would have to be a littlemore careful than what I did above. Yes, using anexplicit inequality as you suggest certainly makes iteasier to give a rigorous argument.>http://www.utm.edu/research/primes/howmany.shtml#2> Though I notice it gives the same result (the nloglog(n)-n1.0073 is>eventually >0).************************ === Subject: : Re: Why is Sum (1/p) divergent>>http://www.utm.edu/research/primes/infinity.shtml>> Given that 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 ... diverges, it wouldbe>very surprising if one of>1/2 + 1/5 + 1/11 ... and>1/3 + 1/7 + 1/13 ...>>was finite and the other was not.>>Perhaps the standard proof could be modified to easily show this?It's immediate from the fact that the full series diverges and> has decreasing terms that both sub-series diverge: 1/5 + 1/11 ... < 1/3 + 1/7 + 1/13 ... < 1/2 + 1/5 + 1/11 ...???> I reached the same conclusion via a different route. Both cant converge> because then they would be both absolutely convergent and could be addedto> get a divergent series. The only possibility is the smaller 1/3+1/7.. is> convergent and 1/2+1/5.. is divergent. But because there is prime between> p_n and 2p_n we can write 1/2+1/5+... <2/3+2/7+2/13+... giving both> convergent - a contradiction. Hence the only possibility is both diverge.> This process can be repea on the two divergent series, can it beasser> that the sum 1/p_an is divergent?Yes it is divergent. More generally taking every a'th term of any divergentseries will give a divergent series.>Or more interestingly is sum 1/p_p> divergent?I'm not sure, given that the n'th prime p_n is asymptotic to nlogn, is itpossible to say the p_n th prime is asymptotic to nlog(n) log(nlog(n))? Ithink I need to know the approximation to the next order. === Subject: : Re: Why is Sum (1/p) divergent>>http://www.utm.edu/research/primes/infinity.shtml >>Given that 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 ... diverges, it would>be>>very surprising if one of>>1/2 + 1/5 + 1/11 ... and>>1/3 + 1/7 + 1/13 ...>>was finite and the other was not.>>Perhaps the standard proof could be modified to easily show this?>> It's immediate from the fact that the full series diverges and>> has decreasing terms that both sub-series diverge:>> 1/5 + 1/11 ... < 1/3 + 1/7 + 1/13 ... < 1/2 + 1/5 + 1/11 ...>> ???>> I reached the same conclusion via a different route. Both cant converge>> because then they would be both absolutely convergent and could be added>to>> get a divergent series. The only possibility is the smaller 1/3+1/7.. is>> convergent and 1/2+1/5.. is divergent. But because there is prime between>> p_n and 2p_n we can write 1/2+1/5+... <2/3+2/7+2/13+... giving both>> convergent - a contradiction. Hence the only possibility is both diverge.>> This process can be repea on the two divergent series, can it be>asser>> that the sum 1/p_an is divergent?>Yes it is divergent. More generally taking every a'th term of any divergent>series will give a divergent series.Nope. True if you assume the terms are positive and decreasing.>>Or more interestingly is sum 1/p_p>> divergent?>I'm not sure, given that the n'th prime p_n is asymptotic to nlogn, is it>possible to say the p_n th prime is asymptotic to nlog(n) log(nlog(n))? I>think I need to know the approximation to the next order.************************ === Subject: : Re: Why is Sum (1/p) divergent> Yes it is divergent. More generally taking every a'th term of any> divergent series will give a divergent series.Aha!So taking the even terms of the divergent series1/1 + 1/1 + 1/2 + 1/4 + 1/3 + 1/9 + 1/4 + 1/16 + ...gives a divergent series-- === Subject: : Re: Why is Sum (1/p) divergentYes it is divergent. More generally taking every a'th term of any> divergent series will give a divergent series.Aha!> So taking the even terms of the divergent series> 1/1 + 1/1 + 1/2 + 1/4 + 1/3 + 1/9 + 1/4 + 1/16 + ...> gives a divergent seriesOK it does rather assume the terms capable of being expressed f(n) with fcontinuous non-negative and monotone decreasing on [1,oo). (Trivial changeof variable in the integral test). Testing for divergence means we canignore any finite number of initial terms and f(n) just needs to beasymptotic to the series we're testing.If sum(1/p(p(n))) diverges using n (log n + log log n - 1.0073) < p(n) forn>8601 and applying the integral test I suspect this fact could beestablished. Conversely using p(n) < n (log n + log log n - 0.9385) mightestablish convergence.Perhaps something more general holds:Hypothesis: If Sum f(n) is divergent then Sum f(p(n)) is divergent. Where fcontinuous non-negative monotone decreasing.... I can't think of acounterexample. === Subject: : Re: Why is Sum (1/p) divergent>> Yes it is divergent. More generally taking every a'th term of any>> divergent series will give a divergent series.>> Aha!>> So taking the even terms of the divergent series>> 1/1 + 1/1 + 1/2 + 1/4 + 1/3 + 1/9 + 1/4 + 1/16 + ...>> gives a divergent series>OK it does rather assume the terms capable of being expressed f(n) with f>continuous non-negative and monotone decreasing on [1,oo). Continuity of f is irrelevant here.>(Trivial change>of variable in the integral test). Testing for divergence means we can>ignore any finite number of initial terms and f(n) just needs to be>asymptotic to the series we're testing.>If sum(1/p(p(n))) diverges using n (log n + log log n - 1.0073) < p(n) for>n>8601 and applying the integral test I suspect this fact could be>established. Conversely using p(n) < n (log n + log log n - 0.9385) might>establish convergence.It converges.>Perhaps something more general holds:>Hypothesis: If Sum f(n) is divergent then Sum f(p(n)) is divergent. Where f>continuous non-negative monotone decreasing.... I can't think of a>counterexample.This is obviously false. The fact that sum 1/p(p(n)) converges requires some sort of careful estimate, but to show that thereexists f such that sum f(n) diverges while sum f(p(n)) divergesonly uses the fact that p(n)/n -> infinity.************************ === Subject: : Re: Why is Sum (1/p) divergentOr indeed0+1+0+1...The words strictly reducing were obviously missing.And Ullrichs proof clearly works with n sub-series.At least one must diverge. Its easy see its the first, but that's notneccessary. Let it be the ith. Then by chopping off the first i-j terms youcan permute it to be the jth series plus the constant terms chopped off thefront. So it too must diverge and so must them all.How about an example that proves that the selection of terms must be simplypicking each nth item - that it can't be generalised. How about the membersof one set is each term of the original the original series that is an exactpower of two in position, and the other set is all the others. Can you dothis to form the opposite of what the OP wan - a division into divergentand convergent infinite series? Or prove its impossible?Yes it is divergent. More generally taking every a'th term of any> divergent series will give a divergent series.Aha!> So taking the even terms of the divergent series> 1/1 + 1/1 + 1/2 + 1/4 + 1/3 + 1/9 + 1/4 + 1/16 + ...> gives a divergent series> --Needless to say, I had the last laugh.> Alan Partridge, _Bouncing Back_ (14 times) === Subject: : Re: Why is Sum (1/p) divergentAnswering my own question.Yes, it is possible to divide a strictly decreasing series into a convergentinfinite subset (and a divergent subset).Use the 1 + 1/2 + 1/3 + 1/4+Consider the sub-series1+1/2^1 + 1/2^2 + 1/2^3 ...this is a convergent sub-series of the original. The remaining terms ofcourse produce a divergent series.So the selection must be regular (every nth) to make all subsets divergent.> Or indeed> 0+1+0+1...> The words strictly reducing were obviously missing.> And Ullrichs proof clearly works with n sub-series.> At least one must diverge. Its easy see its the first, but that's not> neccessary. Let it be the ith. Then by chopping off the first i-j termsyou> can permute it to be the jth series plus the constant terms chopped offthe> front. So it too must diverge and so must them all.> How about an example that proves that the selection of terms must besimply> picking each nth item - that it can't be generalised. How about themembers> of one set is each term of the original the original series that is anexact> power of two in position, and the other set is all the others. Can you do> this to form the opposite of what the OP wan - a division intodivergent> and convergent infinite series? Or prove its impossible?>Yes it is divergent. More generally taking every a'th term of any> divergent series will give a divergent series.>Aha!So taking the even terms of the divergent series> 1/1 + 1/1 + 1/2 + 1/4 + 1/3 + 1/9 + 1/4 + 1/16 + ...> gives a divergent series--> Needless to say, I had the last laugh.> Alan Partridge, _Bouncing Back_ (14 times) === Subject: : Re: Why is Sum (1/p) divergent>Or indeed>0+1+0+1...>The words strictly reducing were obviously missing.>And Ullrichs proof clearly works with n sub-series.>At least one must diverge. Its easy see its the first, but that's not>neccessary. Let it be the ith. Then by chopping off the first i-j terms you>can permute it to be the jth series plus the constant terms chopped off the>front. So it too must diverge and so must them all.>How about an example that proves that the selection of terms must be simply>picking each nth item - that it can't be generalised. It generalizes this far: If s(n) is an increasing sequence of integers such that there exists N such that each interval [A, A + N] contains at least one s(n), then sum f(n) divergent, f as above, implies that sum f(s(n)) diverges.>How about the members>of one set is each term of the original the original series that is an exact>power of two in position, and the other set is all the others. Can you do>this to form the opposite of what the OP wan - a division into divergent>and convergent infinite series? Or prove its impossible?It's easy to show that if s(n)/n -> infinity then there exists a positive decreasing f such that sum f(n) diverges butsum f(s(n)) converges.At least I'm pretty sure it's easy to see that - I have to go towork soon, so I can't write down the details. It's certainlyeasy to see that if s(n) is a regular sequence like 2^nor something.>> Yes it is divergent. More generally taking every a'th term of any>> divergent series will give a divergent series.>> Aha!>> So taking the even terms of the divergent series>> 1/1 + 1/1 + 1/2 + 1/4 + 1/3 + 1/9 + 1/4 + 1/16 + ...>> gives a divergent series>> -- Needless to say, I had the last laugh.>> Alan Partridge, _Bouncing Back_ (14 times)************************ === Subject: : Re: polynom of power 12>> Greetings,>> since there are many good mathematicians amongst you, I hope that >> maybe somebody can help me with my problem.>> I want to solve an equation of the following type analytically, if >> possible.>> A*x^12 + B*x^6 + C^x + D = 0>> I want to solve for x, ABCD are constants. In my case, they are set so >> that only one real solution will exist. Both Maple and Mathematica >> fail to find analytical solutions. Numerically, the real solution >> would be easy to find, but I would prefer to have the solution in >> analytical form.>> Maybe somebody has an idea how to crack this fellow.>> Anton.> Polynomials of degree 5 or higher do not, in general, have analytic > solutions. Unless yours is a special case, it won't either.How can you be sure that this degree is 5? Why?br,Anton. === Subject: : Re: polynom of power 12>> Greetings,>> since there are many good mathematicians amongst you, I hope that >> maybe somebody can help me with my problem.>> I want to solve an equation of the following type analytically, if >> possible.>> A*x^12 + B*x^6 + C^x + D = 0>> I want to solve for x, ABCD are constants. In my case, they are set so >> that only one real solution will exist. Both Maple and Mathematica >> fail to find analytical solutions. Numerically, the real solution >> would be easy to find, but I would prefer to have the solution in >> analytical form.>> Maybe somebody has an idea how to crack this fellow.> Anton.> Polynomials of degree 5 or higher do not, in general, have analytic > solutions. Unless yours is a special case, it won't either.> How can you be sure that this degree is 5? Why?> br,> Anton.Expressions with radicals for the zeroes of polynomials of degree < 5 are known. For example, the quadratic polynomiala*x^2 + b*x + c has zeroes (-b +- sqrt(b^2 - 4*a*c))/(2*a).Abel (and Galois) showed that polynomials of degree >= 5 need not have zeroes expressible with radicals. === Subject: : Re: polynom of power 12 > A*x^12 + B*x^6 + C^x + D = 0>> Polynomials of degree 5 or higher do not, in general, have analytic >> solutions. Unless yours is a special case, it won't either. >How can you be sure that this degree is 5? Why?You misunderstood what he said.The degree is the highest power of the variable that occurs in the polynomial. If A <> 0 the degree of your polynomial is 12. If A = 0but B <> 0 the degree is 6. Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: : Re: polynom of power 12>> I want to solve an equation of the following type analytically, if >> possible. >> A*x^12 + B*x^6 + C^x + D = 0 I presume that's A*x^12 + B*x^6 + C*x + D = 0>> I want to solve for x, ABCD are constants. In my case, they are set so >> that only one real solution will exist. Both Maple and Mathematica fail >> to find analytical solutions. Numerically, the real solution would be >> easy to find, but I would prefer to have the solution in analytical form.>> Maybe somebody has an idea how to crack this fellow. >Polynomials of degree 5 or higher do not, in general, have analytic >solutions. Unless yours is a special case, it won't either.Be careful with your use of the term analytic. You'd be right if yousaid in terms of radicals. But quintics, and some others, can be solvedusing hypergeometric functions. I don't know about this one.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: : Re: How long must physics put up w/f=ma?(ghytrfvbnmju7654)>> What was all that stuff we learnt about 'dependent' and 'independent'>> variables?>> Is 'f' dependent on 'a' or 'a dependent on 'f'?>>Yes, of course.> Well? Is 'f' dependent on 'a' or is 'a' dependent on 'f'?What is more; aren't 'w' and 'g' dependent on each other? So that at anygiven location on Earth or some similar planet; a body's weight [w], dividedby the acceleration [a] at which it will free fall there is a constant; theconstant we call gravitational mass. === Subject: : Re: How long must physics put up w/f=ma?Expires: 28 days>(ghytrfvbnmju7654)> What was all that stuff we learnt about 'dependent' and 'independent'> variables?> Is 'f' dependent on 'a' or 'a dependent on 'f'?>>Yes, of course.>> Well? Is 'f' dependent on 'a' or is 'a' dependent on 'f'?>What is more; aren't 'w' and 'g' dependent on each other? So that at any>given location on Earth or some similar planet; a body's weight [w], divided>by the acceleration [a] at which it will free fall there is a constant; the>constant we call gravitational mass.Since the only way we can detect gravity is through its ability to acceleratemass, I supposethat is not as silly as it sounds.But can you generalize that idea to encompass ALL forces? === Subject: : Re: How long must physics put up w/f=ma?>>(ghytrfvbnmju7654)>What was all that stuff we learnt about 'dependent' and'independent'> variables?> Is 'f' dependent on 'a' or 'a dependent on 'f'?>>Yes, of course.>> Well? Is 'f' dependent on 'a' or is 'a' dependent on 'f'?>What is more; aren't 'w' and 'g' dependent on each other? So that at any>given location on Earth or some similar planet; a body's weight [w],divided>by the acceleration [a] at which it will free fall there is a constant;the>constant we call gravitational mass.Since the only way we can detect gravity is through its ability toaccelerate> mass, I supposethat is not as silly as it sounds.It doesn't sound the least bit silly to me.> But can you generalize that idea to encompass ALL forces?Oh but of course Henry; with a simple spring weight-scale: Either hang abody on one, or pull a body with one. === Subject: : Re: How long must physics put up w/f=ma?>:::aye, I cry tears of blood for thee, for this is sad:::>>Please let us not simplify this equation to just F=ma or a=F/m...>>You all miss an extremely important part of the relationship between>force and acceleration... Essentially, force has nothing to do with>acceleration as it does with fricken momentum...>>Force is simply the change in momentum over an interval of time... F = d(momentum)/d(time) or... F = d(mv)/dt (1)>>Try basic calculus before you try and limit this to only F=ma... You>will instead come to find... F = [dm/dt]*v + m*[dv/dt] (2)>>In Intro level physics you will only see the latter part of (2),>mass times the change in velocity per unit time... But for any sort of>propulsion problem or whatever, where mass is changing, this will also>induce and contribute to force... so please... F = [dm/dt]*v + ma> You are still defining F as the dependent variable.> This equation should not be use as a definition of force.>Essentially, there is no logical reason to define or refer to this>equation as a=F/m>-G00glyMinataur->>------------------------------------ --------->>Oh yes, and to whoever got his/her panties in a bunch with the crap>about a=F/m and the inverse of m... F*** you, sissy... its a>mathematical model that can be interpre either way...>>Nonetheless, I have a solution for you... leave mathematics and>physics and go pursue a career in social work or business... You>aren't cut out for any of this... :::aye, I cry tears of blood for thee, for this is sad::: it's not - andnever could be - f = ma, or about a=F/m and the inverse of m; because massis a ratio: Its got to be f = (f/a)a; w = (w/g)g, or f = wa/g - or since aand g are ratios themselves - any (small) number of other combinations! === Subject: : Re: How long must physics put up w/f=ma?In sci.math, Bob Pease:>> Surely you must all know how to work with algebraic parentheses;> therefore>> you must see that since m = f/a, and m = w/g; the equation f = ma is>> improper; but can be written as: f = wa/g.>> A simple lesson in algebra:>> ma = ma>> (f/a)a = ma>> fa/a = ma>> f = ma> Shead lives in his own world.> I would suggest googling him before continuing.I'd rather ogle Anna Kournikova. Of course onecould Google her, as well -- and find a fan pageor two. (Besides, AFAIK she's happily married.)As for the algebra: f = ma makes sense when f and aare replaced by vectors; m = f/a does not, as itraises rather embarrassing questions (like: what if fand a are not pointing in the same direction?[they always do, of course]).Apparently Mr. Shead's world is one-dimensional, as well,not requiring vectors.> Bob Pease-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: How long must physics put up w/f=ma?> In sci.math, Bob Pease> :> Surely you must all know how to work with algebraic parentheses;> therefore>> you must see that since m = f/a, and m = w/g; the equation f = ma is>> improper; but can be written as: f = wa/g.>> A simple lesson in algebra:>> ma = ma>> (f/a)a = ma>> fa/a = ma>> f = maShead lives in his own world.> I would suggest googling him before continuing.> I'd rather ogle Anna Kournikova. Of course one> could Google her, as well -- and find a fan page> or two. (Besides, AFAIK she's happily married.)> As for the algebra: f = ma makes sense when f and a> are replaced by vectors; m = f/a does not, as it> raises rather embarrassing questions (like: what if f> and a are not pointing in the same direction?> [they always do, of course]).> Apparently Mr. Shead's world is one-dimensional, as well,> not requiring vectors.Goully: If f and a are replaced by vectors; wouldn't m = f/a indicate thatm is a vector; or would that be like double jeopardy(;^? Ya danged old ghostyou.Bob Pease--> #191, ewill3@earthlink.net> It's still legal to go .sigless. === Subject: : Pi r^2 / 3Please look at my thread in: http://www.physicsforums.com/showthread.php?s=&threadid= 6539Thank you,Doron Shadmi (Organic) === Subject: : Re: Pi r^2 / 3>Please look at my thread in: >http://www.physicsforums.com/showthread.php?s=&threadid=6539> Thank you,>Doron Shadmi (Organic) why would we care about your thread in a ohysics group???