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related to this question:

Is there an easy closed-form term for

$$\sum_{j=k}^{\infty} \frac{x^j}{j!}e^{-x},$$

thus when the sum starts at a constant $k$ instead of $1$?

EDIT: Thanks for your help. Is there a Chance to solve this sum-term? Because this is not really what I expect, when I talk about a closed-form term.

A Little bit more of context might help, maybe:

I have $$f(n,p)=\sum_{j=k}^{\infty} \frac{(np)^j}{j!} e^{-np}$$ and it is meant that the partial Derivation is $$\frac{\delta f(n,p)}{\delta n}=\frac{p (np)^{k-1}}{(k-1)!}e^{-np}$$ but I have no idea how to get to this.

Because to me:

$$\frac{\delta f(n,p)}{\delta n}=\sum_{j=k}^{\infty} \left( \frac{p (np)^{j-1}}{j!} e^{-np} -\frac{p (np)^j}{j!} e^{-np} \right)$$ but then I am stuck.

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    $\begingroup$ See CDF of Poisson distribution. $\endgroup$ – Tunk-Fey Jul 7 '14 at 12:49
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    $\begingroup$ From $\displaystyle{\large\tt M}$: $\displaystyle{\large\frac{\Gamma (k)-\Gamma (k,x)}{\Gamma (k)}}$. $\endgroup$ – Felix Marin Jul 7 '14 at 17:03
  • $\begingroup$ @FelixMarin Could you specify what you mean with that? I know about the Gamma and the incomplete Gamma functions, but I can't see how to apply this here. $\endgroup$ – user136457 Jul 7 '14 at 17:05
  • $\begingroup$ I just put your expression in $\tt Mathematica$. $\endgroup$ – Felix Marin Jul 7 '14 at 18:53
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You just made a mistake when differentiating $$f(x)=\sum_{j=k}^{\infty} \frac{(xp)^j}{j!} e^{-xp}.$$ The actual derivative is $$f'(x)=\sum_{j=k}^{\infty} \left( \color{red}{j}\,\frac{p\,(xp)^{j-1}}{j!} e^{-xp} -\frac{p\,(xp)^j}{j!} e^{-xp} \right),$$ that is, provided $k\geqslant1$, using the change of indices $\ell=j-1$ in the first summation, $$f'(x)=\sum_{\ell=k-1}^{\infty}\frac{p(xp)^{\ell}}{\ell!} e^{-xp} -\sum_{j=k}^{\infty} \frac{p (xp)^j}{j!} e^{-xp}=\frac{p(xp)^{k-1}}{(k-1)!} e^{-xp}.$$

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$$ \sum_{j=k}^{\infty} \frac{x^j}{j!}e^{-x}= \left(\sum_{j=k}^{\infty} \frac{x^j}{j!}+\sum_{j=0}^{k-1} \frac{x^j}{j!}-\sum_{j=0}^{k-1} \frac{x^j}{j!}\right)e^{-x}\\ =\left(e^x-\sum_{j=0}^{k-1} \frac{x^j}{j!}\right)e^{-x}=1-e^{-x}\sum_{j=0}^{k-1} \frac{x^j}{j!} $$

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  • $\begingroup$ thanks for your answer. I stated above in the question the whole Problem, pointing out what I could mean with a closed form, because somehow I don't get how they got to this solution. $\endgroup$ – user136457 Jul 7 '14 at 12:49
  • $\begingroup$ @mookid correct but my sum's upper limit is $k-1$, or did I miss something... $\endgroup$ – draks ... Jul 7 '14 at 14:34
  • $\begingroup$ you are right. But I don't know if it is what OP was looking for.... $\endgroup$ – mookid Jul 7 '14 at 14:44
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Yes:

$$e^{-x}\sum_{j=k}^\infty \frac 1{j!} x^j = e^{-x}\left[\exp x - \sum_{j=0}^{k-1} \frac 1{j!} x^j \right] =e^{-x}\int_0^x \frac{(x-t)^{k-1}}{(k-1)!} e^{t} dt $$

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    $\begingroup$ $x$ as limit and inside the integral looks a little strange. Are you sure? $\endgroup$ – draks ... Jul 7 '14 at 12:50
  • $\begingroup$ x does not appear as a limit, does it? $\endgroup$ – mookid Jul 7 '14 at 13:02
  • $\begingroup$ $e^{-x}\int_0^{\color{red}x}$... $\endgroup$ – draks ... Jul 7 '14 at 14:32
  • $\begingroup$ there is no error, see en.wikipedia.org/wiki/… $\endgroup$ – mookid Jul 7 '14 at 14:43
  • $\begingroup$ @draks...: it would only seem strange to me if $x$ were the variable of integration, but $t$ is the variable of integration. $\endgroup$ – robjohn Jul 8 '14 at 15:51
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Incomplete Gamma function and Eq.2 in : http://mathworld.wolfram.com/IncompleteGammaFunction.html

enter image description here

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