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I'm a student and I'm looking for a solution for the following finite power series:

$$ \sum_{n=0}^m \frac{1}{n!} x^n $$

By "solution" I meant expansion of the series and finding a closed form representation of it (without the sum). Thanks to everyone for the replies! The solution with the Taylor remainder was what I needed.

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  • $\begingroup$ If the upper limit $m\to\infty$, then the series is the Maclaurin expansion of $e^x$ $\endgroup$
    – Prasun Biswas
    Commented May 8, 2015 at 16:53
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    $\begingroup$ Welcome to math.se! What do you mean by "solution"? Also, you're more likely to receive good answers if you post your work to this point. $\endgroup$
    – Travis Willse
    Commented May 8, 2015 at 16:53
  • $\begingroup$ In agreement with @Travis, I’d point out that ordinarily, it’s only equations that have solutions, and an equation requires the verb “$=$”. Maybe you meant a description of the finite sum? $\endgroup$
    – Lubin
    Commented May 8, 2015 at 16:57
  • $\begingroup$ What is the question? $\endgroup$
    – copper.hat
    Commented May 8, 2015 at 16:58
  • $\begingroup$ @Lubin I think that's probably the intention, in which case OP might find illuminating the question I linked in the previous comment. $\endgroup$
    – Travis Willse
    Commented May 8, 2015 at 17:06

1 Answer 1

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I only think you can get a "closed" for the finite sum by using the Taylor reminder. That is if the real function $f$ is $m+1$ times differentiable at the point $a$ then $$ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \cdots + \frac{f^{(m)}(a)}{m!}(x-a)^m + \frac{f^{(m+1)}(\xi)}{(m+1)!}(x-a)^{m+1} $$ where $\xi$ is obtained by the mean value theorem and belongs between $a$ and $x$.

Then in your case, clearly $f(x)= e^x$ and $a=0$, which gives $$ e^x = \sum_{n=0}^m \frac{1}{n!}x^n + \frac{e^{\xi}}{(m+1)!}x^{m+1} $$ For a $\xi$ such that $|\xi|<|x|$ ( which of course gives that $\xi$ depends entirely on $x$). Thus $$ \sum_{n=0}^m \frac{1}{n!} x^n= e^x - \frac{e^{\xi}}{(m+1)!}x^{m+1} $$

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  • $\begingroup$ Thanks a lot. It helped me solve my problem. I learned a lot about Taylor series and the remainder. $\endgroup$
    – Mapl
    Commented May 12, 2015 at 11:44
  • $\begingroup$ That is great @Alf , I am glad I could help !! $\endgroup$
    – Alonso Delfín
    Commented May 12, 2015 at 12:34

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