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[math-fun] An intriguing continued fraction cubic
by MCKAY＠vax2.concordia.ca 28 Apr '04

28 Apr '04

The real root of x*x*x - 156*x + 817 has a very large term round about the 800th position. Is there any explanation for this? It is of the order of 10**7. John McKay

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Re: [math-fun] continued fraction of root of cubic
by MCKAY＠vax2.concordia.ca 28 Apr '04

28 Apr '04

No - this one has no biggies until the one I mention. There is no apparent connection with class number 1 however there is a class number of 11 deriving from a prime factor of the discriminant. John McKay From: IN%"math-fun(a)mailman.xmission.com" "math-fun" 27-APR-2004 21:31:30.21 To: IN%"math-fun(a)mailman.xmission.com" CC: Subj: [math-fun] continued fraction of root of cubic Return-path: <math-fun-bounces+mckay=vax2.concordia.ca(a)mailman.xmission.com> Received: from bonnie.concordia.ca (bonnie.Concordia.CA [132.205.7.81]) by vax2.concordia.ca (PMDF V6.2 #30759) with ESMTP id <01L9FTU4LH9W005JEU(a)vax2.concordia.ca> for mckay(a)vax2.concordia.ca (ORCPT mckay(a)vax2.concordia.ca) Tue, 27 Apr 2004 21:31:29 -0400 (EDT) Received: from mailman.xmission.com (mailman.xmission.com [198.60.22.29]) by bonnie.concordia.ca (8.12.10/8.12.10) with ESMTP id i3S1VJA0013479 for <mckay(a)vax2.concordia.ca>; Tue, 27 Apr 2004 21:31:19 -0400 Received: from localhost ([127.0.0.1] helo=mailman.xmission.com) by mailman.xmission.com with esmtp (Exim 3.36 #1 (Debian)) id 1BIduw-0001Ey-05 for <mckay(a)vax2.concordia.ca>; Tue, 27 Apr 2004 19:31:18 -0600 Received: from mgr8.xmission.com ([198.60.22.208] ident=mail) by mailman.xmission.com with esmtp (Exim 3.36 #1 (Debian)) id 1BIduo-0001En-00 for <math-fun(a)mailman.xmission.com>; Tue, 27 Apr 2004 19:31:10 -0600 Received: from [192.12.69.5] (helo=optima.cs.arizona.edu) by mgr8.xmission.com with esmtp (Exim 4.30) id 1BIdun-0000bh-Q6 for math-fun(a)mailman.xmission.com; Tue, 27 Apr 2004 19:31:09 -0600 Received: from baskerville.CS.Arizona.EDU (baskerville.cs.arizona.edu [192.12.69.35]) by optima.cs.arizona.edu (8.12.9p2/8.12.8) with ESMTP id i3S1V8G7037718 for <math-fun(a)mailman.xmission.com>; Tue, 27 Apr 2004 18:31:08 -0700 Received: (from rcs@localhost) by baskerville.CS.Arizona.EDU (8.8.8p2+Sun/8.8.8) id SAA19959; Tue, 27 Apr 2004 18:31:08 -0700 (MST) Date: Tue, 27 Apr 2004 18:31:08 -0700 (MST) From: Richard Schroeppel <rcs(a)CS.Arizona.EDU> Subject: [math-fun] continued fraction of root of cubic Sender: math-fun-bounces+mckay=vax2.concordia.ca(a)mailman.xmission.com To: math-fun(a)mailman.xmission.com Errors-to: math-fun-bounces+mckay=vax2.concordia.ca(a)mailman.xmission.com Reply-to: math-fun <math-fun(a)mailman.xmission.com> Message-id: <200404280131.SAA19959(a)baskerville.CS.Arizona.EDU> Precedence: list X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on mgr8.xmission.com X-Spam-Level: X-Spam-Status: No, hits=0.0 required=8.0 tests=none autolearn=no version=2.63 X-SA-Exim-Mail-From: rcs(a)CS.Arizona.EDU X-SA-Exim-Version: 3.1 (built Mon Jan 26 13:00:24 MST 2004) X-SA-Exim-Scanned: Yes X-BeenThere: math-fun(a)mailman.xmission.com X-Mailman-Version: 2.1.4 X-Scanned-By: MIMEDefang 2.35 List-Post: <mailto:math-fun@mailman.xmission.com> List-Subscribe: <http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>, <mailto:math-fun-request@mailman.xmission.com?subject=subscribe> List-Unsubscribe: <http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>, <mailto:math-fun-request@mailman.xmission.com?subject=unsubscribe> List-Archive: <http://mailman.xmission.com/cgi-bin/mailman/private/math-fun> List-Help: <mailto:math-fun-request@mailman.xmission.com?subject=help> List-Id: math-fun <math-fun.mailman.xmission.com> Original-recipient: rfc822;mckay(a)vax2.concordia.ca > The real root of x*x*x - 156*x + 817 has a very large term round about the 800th position. Is there any explanation for this? It is of the order of 10**7. I think that seeing occasional terms that are O(N^2) is to be expected, so this is somewhat unusual. If there's an analogy with the Brillhart/Stark situation, which I think <fade, fade> was for X^3 = aX+b with a,b in {6,8,10}, then one of (X+-c)^2 for small c (< 10 or 20) should have a much larger term, and it should appear earlier. Rich rcs(a)cs.arizona.edu _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] continued fraction of root of cubic
by Richard Schroeppel 27 Apr '04

27 Apr '04

> The real root of x*x*x - 156*x + 817 has a very large term round about the 800th position. Is there any explanation for this? It is of the order of 10**7. I think that seeing occasional terms that are O(N^2) is to be expected, so this is somewhat unusual. If there's an analogy with the Brillhart/Stark situation, which I think <fade, fade> was for X^3 = aX+b with a,b in {6,8,10}, then one of (X+-c)^2 for small c (< 10 or 20) should have a much larger term, and it should appear earlier. Rich rcs(a)cs.arizona.edu

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[math-fun] Re: zeta function and prime partitions
by Mike Stay 22 Apr '04

22 Apr '04

Ack! I'm not doing very well today. I wrote: (more or less) prime partitions -> Melzak -> zeta function But Melzak's transform leaves the sequence the same, and there's certainly more than one way to partition most numbers into sums of primes. I guess the transform I'm looking for in this case is one that takes prod (k in S) 1/(1-x^k) to prod(k in S) 1/(1-k^x) rather than sum (n in N) a_n x^n to sum(n in N) a_n n^x -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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[math-fun] zeta function and prime partitions
by M. Stay 22 Apr '04

22 Apr '04

Here's something I just noticed that I'm sure someone else must have seen before. The difference between an OGF f and a Dirichlet GF g is that you swap the free variable and the exponent: f(x) = sum(n) a_n x^n <=> g(x) = sum(n) a_n n^x The ways of partitioning an integer into elements of a set S which is a subset of N is p(x) = sum(n in N) p_n x^n = prod(k in S) 1/(1-x^k) If you swap, you get q(x) = sum(n in N) q_n n^x = prod(k in S) 1/(1-k^x) If S is the set of primes, then q(x) = zeta(-x) and q_n = 1 for all n. So the number of ways to partition an integer into primes is related to the zeta function by Melzak's integral that Marc LeBrun posted. g(s,x) = Ds p(x) = 1/Gamma(s) integral(z=0..inf) z^(s-1) p(x e^-z) dz g(s,1) = zeta(s) And p(x) >= 2 for even x is Goldbach's conjecture. -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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Re: [math-fun] how to find sequence associated with dirichletgenerating function?
by Dan Asimov 22 Apr '04

22 Apr '04

Hi, Mike. Using Google I found a paper at <http://www.cs.ust.hk/faculty/golin/pubs/MergeAv.ps> which proves how the Mellin-Perron transform can be used to get a recurrence fmla. for the Dirichlet coefficients from the function f(s) (assuming it converges for Re(s) > 2), which I think can be used to easily recover the coefficents under the convergence assumption. When you get to the paper, search for the text "recover the coefficients" -- or just go to page 6. Regards, Dan Asimov ---------------------------------------------------------------- > > > ...How does one find > > > the sequence a(n) for a Dirichlet generating function > > > f(s)=sum(n=1 to inf) a(n)/n^s ? > > > > This isn't directly the answer you seek, but I recalled Z. > > A. Melzak gives a transform to "Dirichletize" a Taylor > > series: > > > > If f(x) has Taylor series coefficients An then the > > operator Ds, defined by > > > > Ds f(x) = integral z^(s-1) f(x e^-z) dz / Gamma(s), > > z=0..inf > > > > gives the function with Taylor coefficients An/n^s. > > > > Then you let x=1 and you get the corresponding Dirichlet > > series function, call it g(s). > > > > > > Now if you can manage to invert that operator you could > > presumably start with g(s), transform it into the > > corresponding Taylor series, and then pick the > > coefficients off f(x) with the usual derivative operators. > > The compositions of the inverse with those operators > > would give you the operators yielding the Dirichlet > > sequence. > > > > > > Probably not much help, but maybe a start... > > > > > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > -- > Mike Stay > staym(a)clear.net.nz > http://www.cs.auckland.ac.nz/~msta039 > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] how to find sequence associated with dirichlet generating function?
by Mike Stay 21 Apr '04

21 Apr '04

For a standard or exponential generating function, you find the sequence by taking the Taylor series. How does one find the sequence a(n) for a Dirichlet generating function f(s)=sum(n=1 to inf) a(n)/n^s ? -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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Re: [math-fun] how to find sequence associated with dirichlet generating function?
by Mike Stay 21 Apr '04

21 Apr '04

Wow, well done! One question before I try to remember how to invert this thing: in Dirichlet series, the index starts at 1, whereas in the Taylor series, the index starts at zero. Does that operator shift the sequence or drop a term, or what? Thanks, Mike ----- Original Message Follows ----- > >=Mike Stay > > ...How does one find > > the sequence a(n) for a Dirichlet generating function > > f(s)=sum(n=1 to inf) a(n)/n^s ? > > This isn't directly the answer you seek, but I recalled Z. > A. Melzak gives a transform to "Dirichletize" a Taylor > series: > > If f(x) has Taylor series coefficients An then the > operator Ds, defined by > > Ds f(x) = integral z^(s-1) f(x e^-z) dz / Gamma(s), > z=0..inf > > gives the function with Taylor coefficients An/n^s. > > Then you let x=1 and you get the corresponding Dirichlet > series function, call it g(s). > > > Now if you can manage to invert that operator you could > presumably start with g(s), transform it into the > corresponding Taylor series, and then pick the > coefficients off f(x) with the usual derivative operators. > The compositions of the inverse with those operators > would give you the operators yielding the Dirichlet > sequence. > > > Probably not much help, but maybe a start... > > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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RE: [math-fun] how to find sequence associated with dirichletgenerating function?
by Dan Asimov 20 Apr '04

20 Apr '04

Very interesting question. Right off the bat, it's not even clear to me that the function f(s) defined by a convergent Dirichlet series (when one converges, it tends to be on some right half-plane) necessarily determines the coefficents a(n) uniquely! Assuming convergence at s = 0, it's easy to check that the Kth derivative of f(s) at s = 0 is sum(n=1 to oo) a(n) log(n)^K. (Call this sum b(K).) I'm out of my element with infinite-dimensional operators, but one could say that the (infinite x infinite) matrix M whose (K,L)th entry is given by M(K,L) = log(L)^K, when applied to the infinite vector A = (a(1), a(2), ... ), yields the known infinite vector B = (b(1), b(2), ... ). So I'd say, naively, that if this known matrix M (which is an infinite Vandermonde matrix) has a left inverse M^(-1), then applying M^(-1) to both sides of M A = B should yield A = M^(-1) B, thus solving for all the a(n)'s. --Dan ----------------------------------------------------------------------- Mike Stay asks: > For a standard or exponential generating function, you find > the sequence by taking the Taylor series. How does one find > the sequence a(n) for a Dirichlet generating function > f(s)=sum(n=1 to inf) a(n)/n^s ?

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RE: [math-fun] For sale on eBay: an Erdos number
by Dan Asimov 20 Apr '04

20 Apr '04

Thane, How do you *find* these things!?! --Dan

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