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From: Rob Johnson on 27 Jun 2010 23:01 In article <20100627.163137(a)whim.org>, Rob Johnson <rob(a)trash.whim.org> wrote: >In article <rbisrael.20100627222655$02d7(a)news.acm.uiuc.edu>, >Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: >>"rancid moth" <rancidmoth(a)yahoo.com> writes: >> >>> product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? >> >>Well, if p is a prime, by a theorem of Kummer the highest power of p in >>binomial(2k,k) is the number of times you "carry 1" in adding k + k in >>base p. >>If you call that C(k,p), your answer is >> >>product_p p^(8 sum_{k=1}^m (2k+1) C(k,p)) >> >>the product being over all primes <= 2m. > >The number of base p carries in adding k+k is (2 S_p(k) - S_p(2k))/p >where S_p(k) is the sum of the base p digits of k. Oops, that should be (2 S_p(k) - S_p(2k))/(p-1) Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: rancid moth on 30 Jun 2010 03:06 "Rob Johnson" <rob(a)trash.whim.org> wrote in message news:20100627.195425(a)whim.org... > In article <20100627.163137(a)whim.org>, > Rob Johnson <rob(a)trash.whim.org> wrote: >>In article <rbisrael.20100627222655$02d7(a)news.acm.uiuc.edu>, >>Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: >>>"rancid moth" <rancidmoth(a)yahoo.com> writes: >>> >>>> product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? >>> >>>Well, if p is a prime, by a theorem of Kummer the highest power of p in >>>binomial(2k,k) is the number of times you "carry 1" in adding k + k in >>>base p. >>>If you call that C(k,p), your answer is >>> >>>product_p p^(8 sum_{k=1}^m (2k+1) C(k,p)) >>> >>>the product being over all primes <= 2m. >> >>The number of base p carries in adding k+k is (2 S_p(k) - S_p(2k))/p >>where S_p(k) is the sum of the base p digits of k. > > Oops, that should be (2 S_p(k) - S_p(2k))/(p-1) > > Rob Johnson <rob(a)trash.whim.org> > take out the trash before replying > to view any ASCII art, display article in a monospaced font ok. i still can't get my head around it yet...so i obviously need to do more work before Kummer's result makes sense to me. However my real aim is to get an asymptotic formulae when m->oo. to this end i can show the following product(k=1..m) Binomial(2k,k)^(16k)) = H(m)* {product(k=2..m-1) G(m+2 -k)/G(m+3/2-k) }^(16) H(m) is a reasonably involved function of the Glashier constant along with, e, G(m) and pi. But what this representation does is put m into the function within the product...which i think then allows me to simply use the asymptotic formulae for G(z) z-->oo under the product. I'm sure that's all hunky dory. if using the Robert's representation, we let m-->oo can one obtain an asymptotic formulae?
From: David Bernier on 30 Jun 2010 05:51 rancid moth wrote: > > > "Rob Johnson" <rob(a)trash.whim.org> wrote in message > news:20100627.195425(a)whim.org... >> In article <20100627.163137(a)whim.org>, >> Rob Johnson <rob(a)trash.whim.org> wrote: >>> In article <rbisrael.20100627222655$02d7(a)news.acm.uiuc.edu>, >>> Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: >>>> "rancid moth" <rancidmoth(a)yahoo.com> writes: >>>> >>>>> product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? >>>> >>>> Well, if p is a prime, by a theorem of Kummer the highest power of p in >>>> binomial(2k,k) is the number of times you "carry 1" in adding k + k in >>>> base p. >>>> If you call that C(k,p), your answer is >>>> >>>> product_p p^(8 sum_{k=1}^m (2k+1) C(k,p)) >>>> >>>> the product being over all primes <= 2m. >>> >>> The number of base p carries in adding k+k is (2 S_p(k) - S_p(2k))/p >>> where S_p(k) is the sum of the base p digits of k. >> >> Oops, that should be (2 S_p(k) - S_p(2k))/(p-1) >> >> Rob Johnson <rob(a)trash.whim.org> >> take out the trash before replying >> to view any ASCII art, display article in a monospaced font > > ok. i still can't get my head around it yet...so i obviously need to do > more work before Kummer's result makes sense to me. > > However my real aim is to get an asymptotic formulae when m->oo. to this > end i can show the following > > product(k=1..m) Binomial(2k,k)^(16k)) = H(m)* {product(k=2..m-1) G(m+2 > -k)/G(m+3/2-k) }^(16) > > H(m) is a reasonably involved function of the Glashier constant along > with, e, G(m) and pi. But what this representation does is put m into > the function within the product...which i think then allows me to simply > use the asymptotic formulae for G(z) z-->oo under the product. I'm sure > that's all hunky dory. > > if using the Robert's representation, we let m-->oo can one obtain an > asymptotic formulae? [...] I don't know if what follows can be of much help in finding asymptotic formulae, but hopefully the natural logs of the product found using PARI-gp are "close". Suppose we denote the finite product by BP(m). I used PARI-gp to evaluate BP(m) exactly for 1 <= m <= 65. As a check, I get BP(1) = 2^24 = 16,777,216. For each value of m in the range 1 to 65, I had PARI-gp print m along with the nearest integer to log(BP(m)), or in one notation: round(log(BP(m))) . David Bernier m round(log(BP(m))) ======================= 1 17 2 88 3 256 4 562 5 1049 6 1759 7 2736 8 4023 9 5663 10 7700 11 10178 12 13140 13 16630 14 20692 15 25369 16 30706 17 36745 18 43532 19 51110 20 59523 21 68815 22 79030 23 90212 24 102405 25 115653 26 130000 27 145490 28 162168 29 180077 30 199261 31 219764 32 241631 33 264906 34 289633 35 315856 36 343619 37 372966 38 403941 39 436589 40 470954 41 507079 42 545010 43 584789 44 626462 45 670073 46 715665 47 763284 48 812972 49 864775 50 918736 51 974901 52 1033312 53 1094014 54 1157051 55 1222468 56 1290309 57 1360618 58 1433439 59 1508817 60 1586795 61 1667418 62 1750730 63 1836776 64 1925599 65 2017245
From: Rob Johnson on 30 Jun 2010 18:16 In article <i0eqcc$qmo$1(a)news.eternal-september.org>, "rancid moth" <rancidmoth(a)yahoo.com> wrote: >"Rob Johnson" <rob(a)trash.whim.org> wrote in message >news:20100627.195425(a)whim.org... >> In article <20100627.163137(a)whim.org>, >> Rob Johnson <rob(a)trash.whim.org> wrote: >>>In article <rbisrael.20100627222655$02d7(a)news.acm.uiuc.edu>, >>>Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: >>>>"rancid moth" <rancidmoth(a)yahoo.com> writes: >>>> >>>>> product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? >>>> >>>>Well, if p is a prime, by a theorem of Kummer the highest power of p in >>>>binomial(2k,k) is the number of times you "carry 1" in adding k + k in >>>>base p. >>>>If you call that C(k,p), your answer is >>>> >>>>product_p p^(8 sum_{k=1}^m (2k+1) C(k,p)) >>>> >>>>the product being over all primes <= 2m. >>> >>>The number of base p carries in adding k+k is (2 S_p(k) - S_p(2k))/p >>>where S_p(k) is the sum of the base p digits of k. >> >> Oops, that should be (2 S_p(k) - S_p(2k))/(p-1) >> >> Rob Johnson <rob(a)trash.whim.org> >> take out the trash before replying >> to view any ASCII art, display article in a monospaced font > >ok. i still can't get my head around it yet...so i obviously need to do >more work before Kummer's result makes sense to me. > >However my real aim is to get an asymptotic formulae when m->oo. to this >end i can show the following > >product(k=1..m) Binomial(2k,k)^(16k)) = H(m)* {product(k=2..m-1) >G(m+2 -k)/G(m+3/2-k) }^(16) > >H(m) is a reasonably involved function of the Glashier constant along with, >e, G(m) and pi. But what this representation does is put m into the >function within the product...which i think then allows me to simply use the >asymptotic formulae for G(z) z-->oo under the product. I'm sure that's all >hunky dory. > >if using the Robert's representation, we let m-->oo can one obtain an >asymptotic formulae? I was responding to Robert's post, giving what, to me, seems an easy method to compute the number of carries occurs when adding k+k in base p. If I were to look for an asymptotic expansion, I would start with the asymptotic expansion for log(k!) log(k!) ~ k log(k) - k + log(k)/2 + log(2pi)/2 + 1/(12k) - 1/(360k^3) + 1/(1260k^5) - ... to get the asymptotic expansion of log(C(2k,k)) log(C(2k,k)) ~ k log(4) - log(k)/2 - log(pi)/2 - 1/(8k) + 1/(192k^3) - 1/(640k^5) + ... Multiplying the asymptotic expansion of log(C(2k,k)) by 16k+8, and summing with the Euler-Maclaurin Summation Formula, I would get m --- > (16k+8) log(C(2k,k)) --- k=1 ~ 16/3 log(4) m^3 - 4 m^2 log(m) + (12 log(4) + 2 - 4 log(pi)) m^2 - 8 m log(m) + (20/3 log(4) + 2 - 8 log(pi)) m - 11/3 log(m) - 11/(12m) + 67/(720m^2) + 19/(720m^3) - 107/(4032m^4) - 6.0818194771915516013508 The constant is gotten by plugging in m = 10000. By extending the asymptotic approximations above, the next term is -17/(8064m^5). Therefore, at m = 10000, the error is about 2.1e-23, so the constant should be good to 22 places after the decimal point. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: Rob Johnson on 1 Jul 2010 19:40
In article <20100630.112308(a)whim.org>, Rob Johnson <rob(a)trash.whim.org> wrote: >In article <i0eqcc$qmo$1(a)news.eternal-september.org>, >"rancid moth" <rancidmoth(a)yahoo.com> wrote: >>"Rob Johnson" <rob(a)trash.whim.org> wrote in message >>news:20100627.195425(a)whim.org... >>> In article <20100627.163137(a)whim.org>, >>> Rob Johnson <rob(a)trash.whim.org> wrote: >>>>In article <rbisrael.20100627222655$02d7(a)news.acm.uiuc.edu>, >>>>Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: >>>>>"rancid moth" <rancidmoth(a)yahoo.com> writes: >>>>> >>>>>> product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? >>>>> >>>>>Well, if p is a prime, by a theorem of Kummer the highest power of p in >>>>>binomial(2k,k) is the number of times you "carry 1" in adding k + k in >>>>>base p. >>>>>If you call that C(k,p), your answer is >>>>> >>>>>product_p p^(8 sum_{k=1}^m (2k+1) C(k,p)) >>>>> >>>>>the product being over all primes <= 2m. >>>> >>>>The number of base p carries in adding k+k is (2 S_p(k) - S_p(2k))/p >>>>where S_p(k) is the sum of the base p digits of k. >>> >>> Oops, that should be (2 S_p(k) - S_p(2k))/(p-1) >>> >>> Rob Johnson <rob(a)trash.whim.org> >>> take out the trash before replying >>> to view any ASCII art, display article in a monospaced font >> >>ok. i still can't get my head around it yet...so i obviously need to do >>more work before Kummer's result makes sense to me. >> >>However my real aim is to get an asymptotic formulae when m->oo. to this >>end i can show the following >> >>product(k=1..m) Binomial(2k,k)^(16k)) = H(m)* {product(k=2..m-1) >>G(m+2 -k)/G(m+3/2-k) }^(16) >> >>H(m) is a reasonably involved function of the Glashier constant along with, >>e, G(m) and pi. But what this representation does is put m into the >>function within the product...which i think then allows me to simply use the >>asymptotic formulae for G(z) z-->oo under the product. I'm sure that's all >>hunky dory. >> >>if using the Robert's representation, we let m-->oo can one obtain an >>asymptotic formulae? > >I was responding to Robert's post, giving what, to me, seems an easy >method to compute the number of carries occurs when adding k+k in >base p. > >If I were to look for an asymptotic expansion, I would start with the >asymptotic expansion for log(k!) > > log(k!) ~ k log(k) - k + log(k)/2 + log(2pi)/2 > > + 1/(12k) - 1/(360k^3) + 1/(1260k^5) - ... > >to get the asymptotic expansion of log(C(2k,k)) > > log(C(2k,k)) ~ k log(4) - log(k)/2 - log(pi)/2 > > - 1/(8k) + 1/(192k^3) - 1/(640k^5) + ... > >Multiplying the asymptotic expansion of log(C(2k,k)) by 16k+8, and >summing with the Euler-Maclaurin Summation Formula, I would get > > m > --- > > (16k+8) log(C(2k,k)) > --- > k=1 > > ~ 16/3 log(4) m^3 > > - 4 m^2 log(m) > > + (12 log(4) + 2 - 4 log(pi)) m^2 > > - 8 m log(m) > > + (20/3 log(4) + 2 - 8 log(pi)) m > > - 11/3 log(m) > > - 11/(12m) > > + 67/(720m^2) > > + 19/(720m^3) > > - 107/(4032m^4) > > - 6.0818194771915516013508 > >The constant is gotten by plugging in m = 10000. By extending the >asymptotic approximations above, the next term is -17/(8064m^5). >Therefore, at m = 10000, the error is about 2.1e-23, so the constant >should be good to 22 places after the decimal point. I had meant to include that the asymptotic approximation given above for the logarithm of the product yields the following asymptotic approximation for the product in question e^{a m^3 + b m^2 + c m + d} --------------------------- [1] m^{4 m^2 + 8 m + 11/3} where a = 32/3 log(2) b = 24 log(2) + 2 - 4 log(pi) c = 40/3 log(2) + 2 - 8 log(pi) d = -6.0818194771915516013508 The ratio of [1] to the true product is approximately e^{11/(12m)}, which tends to 1 as m tends to infinity. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |