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Re: Re: Some bugs in Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59288] Re: [mg59260] Re: Some bugs in Mathematica
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 4 Aug 2005 02:08:16 -0400 (EDT)
  • References: <dcn6ho$kbr$1@smc.vnet.net> <200508030520.BAA06414@smc.vnet.net> <9140CFBD-F453-4FC2-BE19-55CD5ED7C7B0@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

Just in case you decided to object that n was supposed to be >1, a  
trivial computation by hand shows that for n=2 the answer is

(Gamma[3/2]*Gamma[1/2])/(Gamma[1]*Gamma[1]) + (Gamma[1/2]*Gamma[3/2])/ 
(Gamma[0]*Gamma[2])

The first term is Pi/2 and the second is 0 so the answer comes to Pi/ 
2 and not 1.

What made you claim it is 1?

Andrzej Kozlowski


On 3 Aug 2005, at 23:09, Andrzej Kozlowski wrote:

>
> On 3 Aug 2005, at 07:20, akhmel at hotmail.com wrote:
>
>
>>>> From: "Alex Khmelnitsky" <akhmel at hotmail.com>
To: mathgroup at smc.vnet.net
>>>>
>>>>
>>>> Date: 2005/08/01 Mon AM 10:07:41 EDT
>>>> To: "Bob Hanlon" <hanlonr at cox.net>
>>>> Subject: [mg59288] [mg59260] Re:  Some bugs in Mathematica
>>>>
>>>> 1) You are wrong. My sum is equal to 1 for any n and the  
>>>> respectable
>>>>
>>>>
>>> program
>>>
>>>
>>>> like Mathematica claims to be should know it.
>>>>
>
> 1. This is obviously false. Take n=1. Than your sum is
>
> Sum[(Gamma[1/2 - k ]*Gamma[k + 1/2])/
>    (Gamma[ - k ]*Gamma[k + 1]), {k, 0, 0}]
>
> which is simply Gamma[1/2]*Gamma[1/2]/(Gamma[0]*Gamma[1])
>
> The issue is what is Gamma[0]? The Gamma function has a simple pole  
> at 0, so Mathematica correctly gives
>
>
> Gamma[0]
>
>
> ComplexInfinity
>
> That forces the answer to be 0, thus contradicting your claim that  
> the sum is always 1.
>
> 2. This group is not a place for reporting Mathematica bugs. If you  
> wan to report a bug you shoudl report it directly to WRI's  
> technical support.
> This is a Mathematica discussion group. No employees of Wolfram are  
> obliged to read any posting to this group or reply to them. Those  
> who choose to do so, do so voluntarily because they want to help  
> Mathematica users. Referring to them by name in your postings  is  
> not going to make them reply  and I would guess that if you make  
> yourself sound as if you were demanding rather than asking for an  
> answer they are more likely to ignore you. Which they are fully  
> entitled to do.
>
> Andrzej Kozlowski
>
>
>
>
>
>


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