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## #2424 Simplification/Infinite sum

open
nobody
5
2012-06-16
2012-06-15
No

Enter in Maxima:

simplify_sum(sum(n^2/(2*n)!,n,1,inf));

Maxima returns:

(sqrt(%pi)*(sqrt(2)*bessel_i(3/2,1)+2^(3/2)*bessel_i(1/2,1)))/8

Maxima should return: %e/4

build_info("5.27.0","2012-04-24 08:52:03","i686-pc-mingw32","GNU Common Lisp (GCL)","GCL 2.6.8")

Regards

Chris

## Discussion

• Raymond Toy
2012-06-15

• status: open --> pending

• Raymond Toy
2012-06-15

expand(exponentialize(ev(%,besselexpand=true)))
-> %e/4

I only knew to try this because bessel_i with half integer orders have representations in elementary functions, which you get by setting besselexpand to true.

Marking as pending/wontfix

• status: pending --> open

Yes, if you are a specialist, it’s quite simple. If not, you have a problem.
First you notice that all simplifications in the pull down menu of wx Maxima fail.
Then you have to read a chapter about Bessel functions until you find “besselexpand”.
After applying “besselexpand”, Maxima returns a sum of hyperbolic functions.
Now you need “exponentialize” to get a sum of e functions and finally you see %e/4.

Why do I have to do all these things?
If I enter sum(n^2/(2*n)!,n,1,inf) in Wolfram Alpha I immediately get the correct result.

<marking as pending/wontfix>
Ok.

Thanks again

Chris

• Raymond Toy
2012-06-20

I think it's very hard in general to know what the right answer should be. Yes %e/4 is a very simple answer. But sometimes it's also nice to know that the sum can be expressed in terms of bessel_i, which might lead to insight into other similar sums. If maxima simplified to %e/4, you wouldn't know about bessel_i, possibly missing out on the insight.

But it's also nice to know that maxima can simplify the result to %e/4, for the case where you don't care about the insight. :-)