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From: rancid moth on 1 Jul 2010 23:07 "Rob Johnson" <rob(a)trash.whim.org> wrote in message news:20100701.154233(a)whim.org... > 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 these and all the other posts are most helpful. thank you. cheers moth |