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Math Help - Calculus: Taylor Series

  1. #1
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    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.
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  2. #2
    Super Member Deadstar's Avatar
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    Differentiate \frac{1}{1-x} and see what you get...

    Apply the same to the sum you're given
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  3. #3
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    Quote Originally Posted by Deadstar View Post
    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?
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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Mathbo View Post
    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.
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