Re: Why won't this sum evaluate?
- To: mathgroup at smc.vnet.net
- Subject: [mg120655] Re: Why won't this sum evaluate?
- From: Dana DeLouis <dana01 at me.com>
- Date: Tue, 2 Aug 2011 07:13:00 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
> In[122]:= Sum[c^n/(1 + c^(2*n)), {n, 1, Infinity}]
> Out[122]= -(1/4) + 1/4 EllipticTheta[3, 0, c]^2
Hi. I'll take a crack at it.
I'm not familiar with EllipticTheta, but I do notice what appears to be another issue.
If we look at your solution at values of c=1, and just to either side, we get the following:
f[c_]:=-(1/4)+1/4 EllipticTheta[3,0,c]^2
f[0.999]
784.7554\[VeryThinSpace]-9.953975*10^-14 I
f[1.0]
-(1/4)+1/4 EllipticTheta[3,0,1.]^2
f[1.001]
-(1/4)+1/4 EllipticTheta[3,0,1.001]^2
Notice that it is not evaluated at values of 1.0 or greater.
If we plot the equation at sums of 100, and 1000, we see that there is a spike at 1.
So, the value should be infinity at 1.0 when we sum to infinity.
equ = Sum[c^n/(1+c^(2*n)),{n,1,100}];
Plot[equ,{c,0,2},PlotRange->All]
equ = Sum[c^n/(1 + c^(2*n)), {n, 1, 1000}];
Plot[equ, {c, -2, 2}, PlotRange -> All]
The function SumConvergence also suggests that the value of c can not be 1
SumConvergence[c^n/(1+c^(2*n)),n] //FullSimplify
Abs[c]!=1
I think the program can not handle this sum.
Sometimes, a workaround is to use a symbolic upper limit instead of infinity, and then change it later.
For example, use z as the upper limit here...
Sum[c^n/(1+c^(2*n)),{n,1,z},Assumptions->Abs[c]!=1] //FullSimplify ;
Sometimes, using Limit as z -> infinity works, but it does not here. Instead, I just substituted...
%/.z->\[Infinity] //FullSimplify
Which gives this equation instead:
g[c_]:=(I (QPolyGamma[0,1-(I \[Pi])/(2 Log[c]),c]-QPolyGamma[0,1+(I\[Pi])/(2 Log[c]),c]))/(2 Log[c])
This gives a complex number, with the imaginary part very small. Just use Chop.
Notice that this does give indeterminate at 1, and a value when c > 1
g[.999]
784.7554\[VeryThinSpace]+0. I
g[1.0]
Indeterminate
g[1.001]
2357.122\[VeryThinSpace]+0. I
If we go back to your problem, and sum from zero...
Sum[c^n/(1+c^(2*n)),{n,0,z},Assumptions->c!=1] //FullSimplify ;
%/.z->\[Infinity] //FullSimplify
Which leads to this equation:
g[c_]:=(I (QPolyGamma[0,-((I \[Pi])/(2 Log[c])),c]-QPolyGamma[0,(I\[Pi])/(2 Log[c]),c]))/(2 Log[c])
g[0.999]
785.2554\[VeryThinSpace]+0. I
g[1.0]
Power::infy: Infinite expression 1/0. encountered. >>
Indeterminate
g[1.001]
2357.622\[VeryThinSpace]+0. I
Notice that at c = 0.999, the differences between the two values are indeed 0.5 as you pointed out.
Not an expert, but hopefully this will help. :>)
= = = = = = = = = = = =
Dana DeLouis
"8.0 for Mac OS X x86 (64-bit) (November 6, 2010)=94
On Jul 27, 6:20 am, PAR123 <reiser.p... at gmail.com> wrote:
> In[120]:= $Version
> Out[120]= "7.0 for Mac OS X x86 (32-bit) (January 30, 2009)"
>
> In[122]:= Sum[c^n/(1 + c^(2*n)), {n, 1, Infinity}]
> Out[122]= -(1/4) + 1/4 EllipticTheta[3, 0, c]^2
>
> In[123]:= Sum[c^n/(1 + c^(2*n)), {n, 0, Infinity}]
> Out[123]= (won't simplify)
>
> The only thing different in the two sums is that the second sum is from 0 to Infinity rather than 1 to Infinity. Clearly, the n=zero term is 1/2.
>
> I have tried various Regularizations and Methods, (not exhaustively) but none seem to work on either of the sums, much less the last.
>
> A side problem - Is there a way to determine what Regularization and Method were used when none were specified?
>
> Thanks