# Math Help - Calculus: Taylor Series

1. ## 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.

2. Differentiate $\frac{1}{1-x}$ and see what you get...

Apply the same to the sum you're given

3. Originally Posted by Deadstar
Differentiate $\frac{1}{1-x}$ 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?

4. Originally Posted by Mathbo
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?
The derivative of $x^n$ is $nx^{n-1}$ so yeah that 1+K one you got is better. (just let n=k+1)

Yes it's still considered a Taylor series if you do that.