I am trying to simplify this expression $$ \sum_{k\geq 0} \frac{(k+x)!}{k!} \frac{b^{s+k+x}}{(s+k+x)!} $$ to an expression with finite summation. I am able to do for x = 0 and 1, but not able to reduced for x>1. If anyone has similar experience in solving this or is able to reduced, can you please help to share.
Asked
Active
Viewed 33 times
0
-
are $s,x,b$ real, integer, positive, ....? – G Cab Mar 27 '17 at 18:37
-
yes .. real ($s$ and $x$ are positive integers and $b$ is positive but may not be integers). – S. Bha Mar 28 '17 at 16:14
1 Answers
1
Assuming that $s$ is the shift parameter and is thus always a non-negative integer, you can always express this sum in terns of powers and exponentials of $b$ and incomplete gamma functions.
For example, when $x=2$ the expression is $$ \frac{b^{s+2}}{s+1)!} \left[1+b-s+\frac{e^b}{b^{s+2}}\left(b^2 -2bs+s(s+1)\right) \left(\Gamma(s+2)-\gamma(s+2,b)\right)\right] $$
Since even for $x=0$ your expression involves an incomplete gamma function, this is as good as you are going to get.
Mark Fischler
- 41,743