unknown Taylor series
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From: TefJlives on 12 Aug 2010 04:43 Thanks for all the replies, I've got it now. Best, Greg On Aug 10, 2:05 pm, Gottfried Helms <he...(a)uni-kassel.de> wrote: > Am 10.08.2010 13:41 schrieb Rob Johnson: > > > > > In article <d0f8a489-3e5e-4eea-b4a9-ce24d7631...(a)f20g2000pro.googlegroups.com>, > > achille <achille_...(a)yahoo.com.hk> wrote: > >> On Aug 10, 11:56 am, TefJlives <gmarkow...(a)gmail.com> wrote: > >>> Hello all, > > >>> Does anyone recognize this one? I'm looking for a closed form. > > >>> f(z) = 1/1 + z/(1*3) + z^2/(1*3*5) + z^3/(1*3*5*7) + ... > > >>> The denominator in the n-th term is the product of the odd integers up > >>> to 2n+1. Thanks. > > >>> Greg > >> f(z) = sqrt(pi/(2*t))*exp(t/2)*erf(sqrt(t/2)) ? > > > Indeed. > > > Define > > > x x^3 x^5 > > g(x) = - + --- + ----- + ... [1] > > 1 1*3 1*3*5 > > > Then f(x) = g(sqrt(x))/sqrt(x). Furthermore, > > > x^2 x^4 > > g'(x) = 1 + --- + --- + ... > > 1 1*3 > > > = 1 + x g(x) [2] > > > To solve the differential equation [2], we need an integrating > > factor of exp(-x^2/2): > > > (exp(-x^2/2) g(x))' > > > = exp(-x^2/2) (g'(x) - x g(x)) > > > = exp(-x^2/2) [3] > > > Integrating [3], we get > > > exp(-x^2/2) g(x) > > > = sqrt(pi/2) erf(x/sqrt(2)) [4] > > > Therefore, > > > g(x) = sqrt(pi/2) exp(x^2/2) erf(x/sqrt(2)) [5] > > > Thus, > > > f(x) > > > = g(sqrt(x))/sqrt(x) > > > pi x x > > = sqrt( -- ) exp( - ) erf(sqrt( - )) [6] > > 2x 2 2 > > > Rob Johnson <r...(a)trash.whim.org> > > Very well! > > Gottfried Helms |