Power Series to Functional Notation?

Question

I need some help with my calculus homework. I'm trying to convert the power series x^(n+2) / 2^(n+1)(n+1) to its functional notation.

Answer 1

Ok, reverse-engineering the Taylor Series:
The general form of the TS is:
f(x) = Σ_n=0..∞ df^n(a)/dx^n / n! * (x-a)^n

Firstly you must have a=0 since it's just x^(n+2)
Substituting a=0 and equating coefficients of x^i we get:

df^n(x)/dx^n / n! * x^n = x^(n+2) / 2^(n+1)(n+1)
df^n(x)/dx^n / n! = x² / 2^(n+1)(n+1)
df^n(x)/dx^n = x² n! / 2^(n+1)(n+1)
df^n(x)/dx^n = x² (n+1)! / 2^(n+1)

Integrating n times (and ignoring all constants of integration):
f(x) = x^(n+2) (n+1) / 2^(n+1)
= 2(n+1) (x/2)^(n+2)

Hmm... looks like an arithmetic progression-geometric progression to me.
Not sure but I hope that's right.

Answer 2

It is the series for x ln [x/2] about x = 0 if that is what you want..F(t) = ln [1 + t ] has F ' = 1 / [1 + t ] = sigma n=0..infinity t^[n] and upon integrating the terms are t^[n+1} / [n+1]..replace t with x/2 and you have the series except for a factor of "x".