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MathGroup Archive 2007

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Re: question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75660] Re: [mg75622] question
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 8 May 2007 05:48:12 -0400 (EDT)
  • References: <200705070927.FAA27927@smc.vnet.net>

On 7 May 2007, at 18:27, dimitris wrote:

> Hi.
>
> I want to show that
>
> x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x])
>
> is equal to zero.
>
> In[36]:=
> Plot[Chop[x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x])], {x, -2,
> 2}, Axes -> False, Frame -> True]
>
> with Mathematica.
>
> Simplify, FullSimplify, FunctionExpand fail to do this task. Even
> replacing x by specific value does
> not yield better results.
>
> In[40]:=
> FullSimplify[3/Pi - (1/2)*3*(StruveH[-1, 3] + StruveH[1, 3])]
>
> Out[40]=
> 3/Pi - (3/2)*(StruveH[-1, 3] + StruveH[1, 3])
>
> On the contrary, in another CAS I took
>
>   simplify(3/Pi - (1/2)*3*(StruveH(-1, 3) + StruveH(1, 3)));
>   0
>
> Any ideas?
>
> A quick way to show that x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1,
> x]) is indeed
> zero is
>
> In[41]:=
> Integrate[x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x]), x]
>
> Out[41]=
> 0
>
> but I would really appreciate any other suggestions.
>
> Thanks
>
>

In Mathemaica 6.0 I get:

In[1]:= FullSimplify[x/Pi -
      (1/2)*x*(StruveH[-1, x] +
           StruveH[1, x])]
Out[1]= 0

Another reason to upgrade ;-)

Andrzej Kozlowski


  • References:
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      • From: dimitris <dimmechan@yahoo.com>
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