Re: Infinite series problem

• To: mathgroup at smc.vnet.net
• Subject: [mg35958] Re: [mg35917] Infinite series problem
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Fri, 9 Aug 2002 05:18:06 -0400 (EDT)
• References: <200208081006.GAA12939@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Seung Soon Jang wrote:
>
> Hi, Group,
>
> I am just beginner to start using Mathematica to solve my problem.
> MY goal is to get some asymptotic value of the following series. But it
> doesn't give the answer and
> I have no clue how to resolve the error message I met.
>
> Please give me the idea to go through this hard thing.
>
> Thanks,
>
> Seung Soon
>
> In[9]:=
> Sum[1/(2^(0.5)*(Pi*k)^2*
>
> NIntegrate[(x/2)^(-0.5)*(1-(x/2))^(-0.5)*Cos[Pi*k*x],{x,0,1}]),{k,1,
>     Infinity}]
>
> NIntegrate::inum: Integrand 2.3094 Cos[1.5708 k] is not numerical at {x}
> = \
> {0.5}.
>
> NIntegrate::inum: Integrand 2.3094 Cos[1.5708 k] is not numerical at {x}
> = \
> {0.5}.
>
> NIntegrate::inum: Integrand 2.3094 Cos[1.5708 k] is not numerical at {x}
> = \
> {0.5}.
>
> General::stop: Further output of NIntegrate::inum will be suppressed
> during \
> this calculation.
>
> --
> ----------------------------------------------------------
>
> Seung Soon Jang, Ph.D.
> California Institute of Technology
> Chemistry 139-74
> Pasadena, CA 91125
> TEL) 1-626-395-8147
> FAX) 1-626-585-0917
> e-mail) jsshys at wag.caltech.edu
>
> ----------------------------------------------------------

One approach would be to break the sum into a piece from, say, 1 to 100,
and another from 101 to Infinity. Compute the first in roughly as you
now do. For the second, approximate by only integrating from 0 to 2/k
and throw away the slowly changing part of the denominator,
(1-x/2)^(1/2). Offhand I do not have an error estimate for how good an
approximation this is to your actual integral. But after some
experimentation I believe it is reasonable, perhaps three or so decimal
places.

In[126]:= ff[k_?NumberQ, opts___] :=
NIntegrate[(x/2)^(-1/2)*(1-(x/2))^(-1/2)*Cos[Pi*k*x], {x,0,1}, opts]

In[127]:= firstpart = Sum[1/(2^(N[1/2,40])*(Pi*k)^2*ff[k,
WorkingPrecision->40, PrecisionGoal->10]), {k,1,100}]
Out[127]= 0.17722511803

In[130]:= rest = N[Sum[1/(2^(1/2)*(Pi*k)^2*
Integrate[(x/2)^(-1/2)*Cos[Pi*k*x], {x,0,2/k}, Assumptions->k>0]),
{k,101,Infinity}], 40]
Out[130]= 0.01463711889688464408318349581501263184547

In[131]:= approx = firstpart + rest
Out[131]= 0.19186223693

Daniel Lichtblau
Wolfram Research

```

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