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Re: question

  • To: mathgroup at
  • Subject: [mg75273] Re: [mg75255] question
  • From: Andrzej Kozlowski <akoz at>
  • Date: Mon, 23 Apr 2007 05:40:08 -0400 (EDT)
  • References: <> <>

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.

FullSimplify[EllipticK[z] == (Pi/2)*Hypergeometric2F1[1/2, 1/2, 1,  
z], Abs[z] < 1]


FullSimplify[Unevaluated[Gamma[1/4]^2/(4*Sqrt[Pi]) == (Pi/2) 
*Hypergeometric2F1[1/2, 1/2, 1, 1/2]]]


This leads us to


which is what I wrote above, was equivalent to the result you wanted  
proved. Let us demonstrate that. First, by definition of EllipticK


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

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      • From: dimitris <>
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