- #1
randybryan
- 52
- 0
I need to prove that ez1 x ez2 = e(z1 + z2)
using the power series ez = (SUM FROM n=0 to infinity) zn/n!
(For some reason the Sigma operator isn't working)
In the proof I have been given, it reads
(SUM from 0 to infinity) z1n/n! x (SUM from 0 to infinity)z2m/m!
= (SUM n,m) z1nz2m/n!m!
and this is the step that I can't follow:
= (SUM from p=0 to infinity) x (SUM from q=0 to p) z1q z2(p-q)/q!(p-q)!
It may be easier to copy out onto paper using the sigma symbols rather than the ridiculous brackets, but I can't get the Sigma operator to work (like I say).
I just don't understand where the q and (p - q) have come from and how it can be split into a multiple of a sum from p=0 to infinity and q=0 to p.
If anyone can explain, I would be extremely grateful :)
= (SUM from p=0 to infinity) 1/p! (SUM from q=0 to p) p!/q! (p- q)! x z1q x z2(p-q)
= (SUM from p=0 to infinity) 1/p! (z1 + z2)p
using the power series ez = (SUM FROM n=0 to infinity) zn/n!
(For some reason the Sigma operator isn't working)
In the proof I have been given, it reads
(SUM from 0 to infinity) z1n/n! x (SUM from 0 to infinity)z2m/m!
= (SUM n,m) z1nz2m/n!m!
and this is the step that I can't follow:
= (SUM from p=0 to infinity) x (SUM from q=0 to p) z1q z2(p-q)/q!(p-q)!
It may be easier to copy out onto paper using the sigma symbols rather than the ridiculous brackets, but I can't get the Sigma operator to work (like I say).
I just don't understand where the q and (p - q) have come from and how it can be split into a multiple of a sum from p=0 to infinity and q=0 to p.
If anyone can explain, I would be extremely grateful :)
= (SUM from p=0 to infinity) 1/p! (SUM from q=0 to p) p!/q! (p- q)! x z1q x z2(p-q)
= (SUM from p=0 to infinity) 1/p! (z1 + z2)p