Proving ƩP(n)/n! x^n = P(x)e^x for Polynomials

[aaaa202]

You have:

Ʃ(n+1)/n! x^n = (1+x)e^x

Is it in general true with a polynomium that:

ƩP(n)/n! x^n = P(x)e^x

?

 

[clamtrox]

That's easy to check. What is ƩP(n)/n! x^n for P(n) = n(n-1)?

 

[aaaa202]

uhhh I don't know. Is that a series I should know? :(

Edit: Wait if you let the factorials cancel out you get: Ʃx^n/(n-2)! but doesn't get me furhter

 

[Ray Vickson]

uhhh I don't know. Is that a series I should know? :(

Edit: Wait if you let the factorials cancel out you get: Ʃx^n/(n-2)! but doesn't get me furhter

Yes, it does (or should); you just need to write things out more explicitly:
e^x = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!} x^3 + \frac{1}{4!} x^4 + \cdots. Now write out $\Sigma\: n(n-1) x^n/n!$ in a similarly-detailed way.

Using the Ʃ notation saves time and writing after you are thoroughly familiar with the techniques, but until then you might do better to avoid relying on it.