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From: TefJlives on 9 Aug 2010 23:56 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
From: achille on 10 Aug 2010 00:34 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)) ?
From: Rob Johnson on 10 Aug 2010 07:41 In article <d0f8a489-3e5e-4eea-b4a9-ce24d7631cc9(a)f20g2000pro.googlegroups.com>, achille <achille_hui(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 <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: Noone on 10 Aug 2010 12:33 On Mon, 9 Aug 2010 20:56:12 -0700 (PDT), TefJlives <gmarkowsky(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 If you don't need a worked out solution, at wolframalpha.com copy/paste the following: sum z^n/(2n+1)!! from n=0 to infinity
From: Gottfried Helms on 10 Aug 2010 17:05
Am 10.08.2010 13:41 schrieb Rob Johnson: > In article <d0f8a489-3e5e-4eea-b4a9-ce24d7631cc9(a)f20g2000pro.googlegroups.com>, > achille <achille_hui(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 <rob(a)trash.whim.org> Very well! Gottfried Helms |