# Calculus: Taylor Series

• Apr 27th 2010, 05:09 PM
Mathbo
Calculus: Taylor Series
Find the Taylor series for the function f(x) = 1/((1-x)^2) valid for x near zero, using the fact that 1/(1-x) = the sum from k=0 to infinity of x^k.

• Apr 27th 2010, 05:26 PM
Differentiate $\displaystyle \frac{1}{1-x}$ and see what you get...

Apply the same to the sum you're given
• Apr 27th 2010, 06:19 PM
Mathbo
Quote:

Originally Posted by Deadstar
Differentiate $\displaystyle \frac{1}{1-x}$ and see what you get...

Apply the same to the sum you're given

When I took the derivative, I got my function and when I took the derivative of the sum, I got f(x)=sum from K=0 to infinity of (K)x^K, but it seems like sum from K=0 to infinity of (1+K)x^K better describes my function. Is that the answer? Is it considered a Taylor Series when you don't write out all the terms, but write it as an infinite sum instead?
• Apr 28th 2010, 02:23 AM
The derivative of $\displaystyle x^n$ is $\displaystyle nx^{n-1}$ so yeah that 1+K one you got is better. (just let n=k+1)