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MathGroup Archive 2007

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Re: Re: question (2nd)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75300] Re: [mg75263] Re: question (2nd)
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 24 Apr 2007 03:27:34 -0400 (EDT)
  • References: <f0f8ud$k47$1@smc.vnet.net> <200704230934.FAA10305@smc.vnet.net>

It depends what you mean by "verification". I believe a numerical and  
not a symbolic method is used here, though possibly one that in fact  
amounts to a mathematical proof.

Andrzej Kozlowski

On 23 Apr 2007, at 18:34, dimitris wrote:

> I think I have just encountered the quickest verification...
>
> In[2]:=
> Timing[Developer`ZeroQ[Gamma[1/4]^2/(2*Sqrt[2*Pi]) - 2*EllipticF[Pi/4,
> 2]]]
>
> Out[2]=
> {0.07800000000000001*Second, True}
>
> Dimitris
>
>
> =CF/=C7 dimitris =DD=E3=F1=E1=F8=E5:
>> Hello.
>>
>> In Mathematica 5.2 I got
>>
>> In[73]:=
>> Integrate[1/Sqrt[Sin[u]], {u, 0, Pi/2}]
>>
>> Out[73]=
>> 2*EllipticF[Pi/4, 2]
>>
>> The result ic correct as a numerical integration indicates.
>>
>> In[74]:=
>> {Chop[N[%]], NIntegrate[1/Sqrt[Sin[u]], {u, 0, Pi/2}]}
>>
>> Out[74]=
>> {2.6220575244897986, 2.622057554312378}
>>
>> It can be proved that the definite integral under discussion is
>> given by Gamma[1/4]^2/(2*Sqrt[2*Pi]).
>>
>> In[75]:=
>> N[Gamma[1/4]]^2/(2*Sqrt[2*Pi])
>>
>> Out[75]=
>> 2.6220575542921196
>>
>> 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?
>>
>> Thanks
>> Dimitris
>
>



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