Re: question
- To: mathgroup at smc.vnet.net
- Subject: [mg75273] Re: [mg75255] question
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 23 Apr 2007 05:40:08 -0400 (EDT)
- References: <200704220910.FAA20298@smc.vnet.net> <A0A84097-6960-4D4E-9BF8-64F74237A957@mimuw.edu.pl>
On 22 Apr 2007, at 23:33, Andrzej Kozlowski wrote: > *This message was transferred with a trial version of CommuniGate > (tm) Pro* > On 22 Apr 2007, at 18:10, dimitris wrote: > >> Can somebody point me out a way to show that the >> Gamma[1/4]^2/(2*Sqrt[2*Pi]) is equal to 2*EllipticF[Pi/4, 2] or/and a >> series >> of steps from taking from the latter to the former and vice versa? >> > > I can't see how to do this but I can at least show that this is > equivalent to the following curious formula for Pi: > > Pi == Gamma[1/4]^4/(16*EllipticK[1/2]^2) > > So if one can prove (or find somewhere a proof) of this formula > then one will also obtain your derivation. > > Andrzej Kozlowski I think I can now prove it. We want to show that: Gamma[1/4]^2 == 4*EllipticF[Pi/4, 2]*Sqrt[2*Pi]) First, we shal start with some identities which Mathematica can prove. In[1]:= FullSimplify[EllipticK[z] == (Pi/2)*Hypergeometric2F1[1/2, 1/2, 1, z], Abs[z] < 1] Out[1]= True In[3]:= FullSimplify[Unevaluated[Gamma[1/4]^2/(4*Sqrt[Pi]) == (Pi/2) *Hypergeometric2F1[1/2, 1/2, 1, 1/2]]] Out[3]= True This leads us to Gamma[1/4]^2/(4*Sqrt[Pi])==EllipticK[1/2] which is what I wrote above, was equivalent to the result you wanted proved. Let us demonstrate that. First, by definition of EllipticK EllipticK[1/2]==EllipticF[Pi/2,1/2] Now we refer to page 593 of Abramowitz and Stegun: EllipticF[theta, m] == Sqrt[m]*EllipticF[phi, m^(-1)] where Sin[phi] == m^(1/2)*Sin[theta] this gives us EllipticK[1/2]==EllipticF[Pi/2,1/2]==Sqrt[2] * EllipticF[Pi/4,2] (Mathematica can't prove the last relationship because it always reduced EllipticF[Pi/2,a] to EllipticK[a] but you can do: N[EllipticF[Pi/2, 1/2] - Sqrt[2]*EllipticF[Pi/4, 2], 100] // Chop) which gives us: Gamma[1/4]^2/(4*Sqrt[Pi]) ==Sqrt[2] * EllipticF[Pi/4,2] which is what you wanted to prove. Andrzej Kozlowski
- References:
- question
- From: dimitris <dimmechan@yahoo.com>
- question