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I'm very new to Mathematica, so excuse my innocence. I have the following expression:

$$ \left( \sum_{n=0}^r \frac{(-1)^n}{n!} y^n \right)^f $$

I would like Mathematica to expand out the expression in powers of $y$, as a polynomial, where $r$ and $f$ are arbitrary positive integers. Is this possible?

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  • $\begingroup$ Welcome! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$
    – user9660
    May 22, 2016 at 7:36
  • $\begingroup$ People here generally like users to post code as Mathematica code instead of images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. $\endgroup$
    – user9660
    May 22, 2016 at 7:37
  • $\begingroup$ Take the result from a given r, f (here called result), then Series[result, {y, Infinity, 0}] // Normal $\endgroup$
    – ciao
    May 22, 2016 at 7:41
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    $\begingroup$ That will give an expression in terms of specific $r$,$f$. I desire an expression for general $r$ and $f$...In other words ungiven. $\endgroup$
    – Anand Chotai
    May 22, 2016 at 8:10

1 Answer 1

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A bit complicated, this one:

With[{m = 5, r = 3}, 
     CoefficientList[Sum[(-x)^n/n!, {n, 0, m}]^r, x] == 
     Table[Sum[FactorialPower[r, k]
               BellY[n, k, Table[(-1)^i, {i, m}]],
               {k, 0, r}]/n!, {n, 0, m r}]]
   True

Recall that the partial Bell polynomials are a way to express Faà di Bruno's formula, which applies here since the coefficients of a polynomial are the same as the successive derivatives of a polynomial divided by an appropriate factorial. There may be a less cumbersome closed form; I'll keep trying to look.

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  • $\begingroup$ Thanks a lot for that! A very good idea, I will examine it more closely. $\endgroup$
    – Anand Chotai
    May 23, 2016 at 4:03

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