How to Approximate a Partial Exponential Sum?

[Nurdan]

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

 

[djeitnstine]

recheck your code please

 

[Tedjn]

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 e^N. 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.

 

[adriank]

You should use [tex ][/tex] tags.

It is known that
$$\sum_{k = 0}^\infty {\frac{{N^k }}<br /> {{k!}}} = e^N$$


My question is

$$\sum_{k = 0}^M {\frac{{N^k }}<br /> {{k!}}} = ?$$
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?

 

[Nurdan]

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.

 

[adriank]

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 $$\Gamma(a) = \Gamma(a, 0)$$.

 

[Nurdan]

It is ok. I know the proof of this but still i need some approximations