Thread: 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.

I'm stuck. Please help me! Thanks.

Differentiate and see what you get...

Apply the same to the sum you're given

Originally Posted by Deadstar

Differentiate 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?

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 is 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.