Exploring Infinite Series with Maxima: Finding Exact Sums with Analytic Programs

[kq6up]

Homework Statement

From Mary Boas: Math for Phys. Sci. Ch1.15.20

20. By computer or tables, find the exact sum of each of the following series.

a. $$\sum _{ n=1 }^{ \infty }{ \frac { { n } }{ { (4{ n }^{ 2 }-1) }^{ 2 } } }$$

Homework Equations

N/A. One is supposed to use an analytic program, or a table to find this.

The Attempt at a Solution

I was able to get Maxima to give me exact answers for some really simple sums like $$\sum _{ n=1 }^{ \infty }{ \frac { { 1 } }{ { n }^{ 2 } } }$$ Is this just too much to ask Maxima? Would mathematica handle this? It has been a long time since I have used Mathematica because I prefer to use GNU software when it is available.

 

[vela]

Mathematica can do it, but you could also do it by hand using partial fractions.

 

[kq6up]

Are you familiar with Maxima? Will it do it?

Chris

 

[Ray Vickson]

Are you familiar with Maxima? Will it do it?

Chris

Maple will do it, but I am not familiar with Maxima, so I cannot answer your direct question. However, people are trying to help you answer it for yourself; if you resort to a partial fraction expansion, you will get simpler sums that Maxima can undoubtedly handle.

 

[Bill Simpson]

http://www.wolframalpha.com/input/?i=Sum[n/(4+n^2+-+1)^2,+{n,+1,+Infinity}]

 

[kq6up]

Thanks, Bill. How powerful is Alpha compared to the actual Mathematica program?

Regards,
Chris

 

[Bill Simpson]

WolframAlpha uses much of the same underlying engine, but it won't let you define functions, won't let you use multi-character variable names and accepts a mostly-undocumented-pseudo-pidgin-english language for you to write your request in. It will accept a limited subset of Mathematica notation for input. But it is more forgiving about use of () versus {} versus [], etc. Paying for access to the Pro WolframAlpha does provide greater access to processing, but I haven't had time to do that. If I could get real documentation explaining exactly what it could do for me I'd probably try that.

I tried to get Maxima to do your problem, but unfortunately it has been too many years since I last did that and I wasn't able to find a way to get it to give me an answer. I expect it can do it, it just takes brain cells that I have misplaced. And my Macsyma (not Maxima) manuals are buried somewhere that I can't find.

 

[kq6up]

Thanks for trying, Bill. I am playing around with Sage now to see what I can do. I do know Mathematica (rusty), but I really like Maxima. I find it intuitive. I think think the lease license model of Mathematica is really stupid, so if I can find a viable alternative, I would much prefer it.

The bad thing about picking up a new symbolic math program is the time it takes to figure out how do the same easy thing you already knew how to do in the old one.

Wolfram Alpha seems like it might be a stop gap in doing what Maxima cannot.

Thanks,
Chris Maness

 

[kq6up]

Yay! Sage gives me the sum. Pretty cool for a free program.

Regards,
Chris Maness

 

[Dick]

Yay! Sage gives me the sum. Pretty cool for a free program.

Regards,
Chris Maness

Yes, it is! I've got Sage too and I'm a big fan. Mind telling me how you did that?

 

[kq6up]

Her is the command I used:

k=var('n'); sum(n/(4*n^2-1)^2, n, 1, oo)

Regards,
Chris Maness

 

[Dick]

Her is the command I used:

k=var('n'); sum(n/(4*n^2-1)^2, n, 1, oo)

Regards,
Chris Maness

Very nice! I'm not sure if Maxima can handle that on its own directly. Sage might be using sympy. Learn something new every day...

Yeah, here's the corresponding 'isympy' line:

summation(k/(4*k**2-1)**2,(k,1,oo))

Sage does a pretty good job of putting a lot of free software tools into a common wrapper. I continue to be impressed.

 

[Robert Dodier]

I've arrived late at the party, but for the record, here's how to find the sum with Maxima.

foo : 'sum(n/(4*n^2-1)^2, n, 1, inf);
load (simplify_sum);
simplify_sum (foo);
=> 1/8

 

[Dick]

I've arrived late at the party, but for the record, here's how to find the sum with Maxima.

foo : 'sum(n/(4*n^2-1)^2, n, 1, inf);
load (simplify_sum);
simplify_sum (foo);
=> 1/8

Thanks Robert Dodier! Ah, simplify_sum.

 

[kq6up]

I started to use sage, and found that it was better at finding the limits of infinite sums. Since starting at CSUF I found that we get Mathematica for free, and that is even more powerful yet.

Thanks,
Chris