How to Approximate a Partial Exponential Sum?

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The discussion centers on approximating the partial sum of the exponential series, specifically \(\sum_{k = 0}^M \frac{N^k}{k!}\) for integer \(M \leq N\). It is noted that this sum serves as a Taylor polynomial approximation of \(e^N\), with accuracy improving as \(M\) increases. A participant mentions that when \(M = N\), the result approaches \(\frac{e^N}{2}\). Additionally, Mathematica's representation of the sum involves the incomplete gamma function, suggesting a more complex relationship for approximation. Overall, the conversation seeks asymptotic approaches for evaluating the sum as \(M\) approaches \(N\).
Nurdan
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It is known that
\[\sum\limits_{k = 0}^\infty {\frac{{N^k }}
{{k!}}} = e^N
\]


My question is

\[\sum\limits_{k = 0}^M {\frac{{N^k }}
{{k!}}} = ?
\]
where $M\leq N$ an integer.


This is not an homework
 
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recheck your code please
 
I do not understand the question. Whether M \leq N or not, as long as M is not infinity, all you have is a Taylor polynomial approximation of eN. The larger M is, the better your approximation.

Your LaTeX is fine, but the forum requires you to put code in between [ tex ][ /tex ] tags (no spaces) or [ itex ][ /itex ] tags (no spaces) for inline TeX.
 
You should use [tex ][/tex] tags.
Nurdan said:
It is known that
\sum_{k = 0}^\infty {\frac{{N^k }}<br /> {{k!}}} = e^N<br />


My question is

\sum_{k = 0}^M {\frac{{N^k }}<br /> {{k!}}} = ?<br />
where M\leq N an integer.
What kind of answer are you looking for? There isn't really a simpler expression, and why the restriction M ≤ N?
 
My empiric results show that if M=N the answer is (e^N)/2. I am looking for any asymptotic approach that gives the solution as M=N.
 
Yes, it does seem to approach that (although of course it's not exact).

Interestingly, according to Mathematica,
\sum_{k = 0}^{n} \frac{x^k}{k!} = e^x \frac{\Gamma(n + 1, x)}{\Gamma(n + 1)}
where \Gamma(a, x) is the incomplete gamma function
\Gamma(a, x) = \int_x^\infty t^{a - 1} e^t \,dt
and \Gamma(a) = \Gamma(a, 0).
 
It is ok. I know the proof of this but still i need some approximations
 

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