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Is it possible to find a close form solution for $S_1$.

$S_1$ is defined as follows:

$S_1=\sum_{k=b}^{\infty}\frac{x^k}{k!}$ ; Where $0<x<b<\infty$

If $b=0$ then $S_2 = e^x$. But how do we solve $S_1$ for the condition given above? I think we can use upper incomplete gamma function but not sure how to get a solution for this.

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I don't think this is what you are after, but I would consider $$ S_1 = e^x - \sum_0 ^b \frac{x^k}{k!} $$ a closed form for your sum.

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  • $\begingroup$ How do I find the sum for $x^k/k!$ term ? Do you want me to see math.stackexchange.com/questions/1283/… for the second term? $\endgroup$ – Jenn Aug 2 '15 at 8:17
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    $\begingroup$ @Ria You could use the expression there, which here corresponds to $\sum_0 ^b \frac{x^k}{k!} = \frac{e^x \int_x ^\infty t^b e^{-t} d t }{ \Gamma (b+1) }$. I am not aware of any other one, however. I think it is generally accepted/defined that finite sums are closed forms, although perhaps not necessarily very elegant ones. In this case, we would get that $$ S_1 = e^{x} \left( 1 - \frac{\int_x ^\infty t^b e^{-t} d t }{ \Gamma (b+1) } \right) = \frac{ e^{x} }{ \Gamma (b+1) } \left( \Gamma (b+1) - \int_x ^\infty t^b e^{-t} d t \right) . $$ $\endgroup$ – izœc Aug 2 '15 at 22:37

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