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.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.