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MathGroup Archive 2007

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Re: question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75661] Re: question
  • From: Peter Pein <petsie at dordos.net>
  • Date: Tue, 8 May 2007 05:48:45 -0400 (EDT)
  • References: <f1mrii$rb4$1@smc.vnet.net>

dimitris schrieb:
> 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
> 
> 
Hi Dimitris,

I don't get it in the usual way (showing, that the derivative is identical to
zero _and_ expression at x==0 is zero (the latter is easy)).

My way is similar to your approach via integration:

ee = x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x]);
InverseLaplaceTransform[LaplaceTransform[ee, x, s], s, x]

--> 0

and calculations like Series[ee, {x, 0, 50}] --> O[x]^51 are no proof.

Playing around with SeriesTerm[] leads to a funny result:

e2 = Sum[SeriesTerm[ee, {x, 0, k}]x^(k)*k!^(3/2), {k, 0, â??}]

gives two warnings "Sum does not converge" and the "result" x/Pi.


But really astonishing is:

e3 = Sum[SeriesTerm[ee, {x, 0, k}]*(x^k/k), {k, 0, Infinity}]
--> x/Pi - (1/(9*Pi))*(x^3*HypergeometricPFQ[{1}, {5/2, 5/2}, -(x^2/4)])
    -(1/2)*StruveH[0, x]
ZeroQ[e3]
--> False

*but:*

FullSimplify[e3]
--> 0

I understood ZeroQ as if it tries transformations, which FullSimplify uses too??

But using SeriesTerm, we get a proof:

First FullSimplify the even terms, then the odd terms with k>=3 and at last
the coefficient with k==1:

ak = SeriesTerm[ee, {x, 0, k}];

FullSimplify[ak /. k -> 2*j, j \[Element] Integers && j >= 0]
--> 0
FullSimplify[ak /. k -> 2*j + 1, j \[Element] Integers && j > 0]
--> 0

and finally
FullSimplify[ak /. k -> 1]
--> 0


Cheers,
Peter


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