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I've got a basic problem here with deriving the poisson distribution, where in part this summation is needed to show the expectation of the distribution is a parameter of the distribution's function, but while I can see from the derivation that this is the end result, I cannot see how it would be.

$$\sum_{x=0}^{\infty }{\frac{a^{x}}{x!}\; =\; e^{a}}$$

If I expand this, I get:

$$\lim_{n \to \infty} \frac{a^{0}}{0!}+\frac{a^{1}}{1!}+\frac{a^{2}}{2!}+...+\frac{a^{n}}{n!}$$

...but I'm not seeing it beyond this. Any pointers?

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  • $\begingroup$ This is the Taylor series expansion for the function $f(x) = e^x$. $\endgroup$ – Rolf Hoyer May 17 '15 at 5:10
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    $\begingroup$ That is the definition of $e^{a}$. Also, what is your definition of $e^{x}$? $\endgroup$ – user222031 May 17 '15 at 5:12
  • $\begingroup$ Some people would define $e^x$ exactly by that power sereis $\endgroup$ – Zach Effman May 17 '15 at 5:12
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    $\begingroup$ How do you define $\exp$? $\endgroup$ – copper.hat May 17 '15 at 5:25
  • $\begingroup$ I guess in my limited way I've relied on it as just the limits of (1+1/n)^n, but haven't had the number theory experience to see it as a sum of 1/n!...thats still hazy to me, but the answers make some sense given this current understanding of mine. $\endgroup$ – Topher May 17 '15 at 15:33
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As an addition (maybe not very necessary here) to @Paul's answer:
Provided you definition of $\exp$ is (otherwise you may just skip my answer) $$e:=\lim_{n\to\infty}\Big(1+\frac1n\Big)^n$$ which is more common as far as I know. Then $\forall z\in \Bbb R^+$, we have $$\lim_{n\to\infty}\Big(1+\frac zn\Big)^n=\lim_{n/z\to\infty}\Big[(1+\frac{1}{n/z})^{n/z}\Big]^z=\Big[\lim_{n/z\to\infty}(1+\frac{1}{n/z})^{n/z}\Big]^z=:e^z$$ And then for the $z<0$ case, note that $$\lim_{n\to\infty}\Big(1-\frac1n\Big)^n=\lim_{n\to\infty}\Big(1-\frac{1}{n+1}\Big)^{n+1}=\lim_{n\to\infty}\Big(1+\frac1n\Big)^{-1}\frac{1}{\Big(1+\frac1n\Big)^n}=:\frac1e$$ Take it from here, and you should be able to show $\forall z\in\Bbb R^{-}$ $$\lim_{n\to\infty}(1+\frac zn)^n=:e^z$$ And $z=0$ case is too trivial for discussion here.


The rest is all in @Paul's answer.

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If you define the exponential function as $$e^z=\lim_{n\to\infty}\left(1+\frac {z} {n}\right)^n$$ Then $$e^z=\lim_{n\to\infty}\left(1+\frac {z} {n}\right)^n=\lim_{n\to\infty}\sum_{k=0}^n\frac1{n^k}{n\choose k}z^k$$ Now $$\lim_{n\to\infty}\frac{1}{n^k}{n\choose k}=\lim_{n\to\infty}\frac{1}{n^k}\frac{n!}{k!(n-k)!}=\frac{1}{k!}\lim_{n\to\infty}\frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k}=\frac1{k!}$$ The limit equals $1$ because both the top and bottom of the fraction are polynomials of degree $k$, so the limit is the ratio of their leading coefficients ($1$ and $1$). Hence $$e^z=\lim_{n\to\infty}\left(1+\frac {z} {n}\right)^n=\lim_{n\to\infty}\sum_{k=0}^n\frac1{n^k}{n\choose k}z^k=\sum_{k=0}^\infty\frac{z^k}{k!}$$

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  • $\begingroup$ It's obvious what you mean, but if you're going to make a remark about the limit in the third line, you should clarify it equals one because they are both polynomials of same degree with the same leading coefficient. $\endgroup$ – MCT May 17 '15 at 6:27
  • $\begingroup$ I've clarified it now, as you suggested $\endgroup$ – Pauly B May 17 '15 at 11:35

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