# Inverse function of summation

How to find the inverse function of the following: $f\left( x \right)=\sum_{i=0}^{x}{\frac{2^{i}}{i!}}$

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Why are you looking for an inverse? What do you need it for? It's good always to add some motivation, in order for others to help you in a better way. –  Andy Jan 4 '11 at 10:25
Well, it requires the inverse of an incomplete gamma function... –  non-expert Jan 4 '11 at 12:03
Try to find a closed form. –  Raphael Jan 4 '11 at 12:27
Don't expect to find a closed form. –  leonbloy Jan 4 '11 at 16:23
I want to find the inverse CDF of the poisson distribution .. I found how to solve it somehow .. but still find that question interesting!! Can we find the inverse function of summations?? –  Osama Gamal Jan 6 '11 at 12:51

In case you actually want to find the inverse cdf of the Poisson distribution (with parameter $\lambda=2$), there seem to be many statistical/mathematical softwares that can do the job for you. For example, see this.
EDIT: The following refers to the OP's question in a comment below (and above). Let $F$ and $F^{-1}$ denote a CDF and its inverse, respectively. Given $a>0$, define a function $F_a$ by $F_a (x) = aF(x)$. If $y=F_a (x)$, then $y=aF(x)$, hence, $F(x)=y/a$, and so $x=F^{-1}(y/a)$. It thus follows that $F_a^{ - 1} (y) = F^{ - 1} (y/a)$, where $F_a^{-1}$ denotes the inverse of $F_a$. Returning to the original question, take $F(x) = e^{ - \lambda } \sum\nolimits_{i = 0}^{\left\lfloor x \right\rfloor } {\frac{{\lambda ^i }}{{i!}}}$ and $a=e^\lambda$ (where $\lambda=2$). Then, $F_a (x) = \sum\nolimits_{i = 0}^{\left\lfloor x \right\rfloor } {\frac{{\lambda ^i }}{{i!}}}$. Hence we are done, by the relation $F_a^{ - 1} (y) = F^{ - 1} (y/a)$. Note: Here, $F$ is an increasing step function, hence not invertible in the usual sense. However, the inverse can be defined as described in the link above.
Of course, the factor of $e^{-\lambda}$ should be added. –  Shai Covo Jan 4 '11 at 13:16