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Math Help - Taylor Series Help

  1. #1
    TFD
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    Taylor Series Help

    So tomorrow I need to explain a problem to the class, I know the correct answer is 3, i just do not know how to show the work necessary to explain it.

    Here is the problem:

    What is the coefficient of x^2 in the Taylor series for 1/((1+x)^2) about x=0.

    (A) 1/6 (B) 1/3 (C) 1 (D)3 (E)6


    Thanks for the help.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TFD View Post
    So tomorrow I need to explain a problem to the class, I know the correct answer is 3, i just do not know how to show the work necessary to explain it.

    Here is the problem:

    What is the coefficient of x^2 in the Taylor series for 1/((1+x)^2) about x=0.

    (A) 1/6 (B) 1/3 (C) 1 (D)3 (E)6


    Thanks for the help.
    The power series for this would be found by seeing that \frac{1}{(1+x)^2}=\frac{-D[\frac{1}{1+x}]}{dx}

    So \frac{1}{1+x}=\sum_{n=1}^{\infty}(-1)^{n}x^{n}

    so the oppositve of the derivative would be -\sum_{n=0}^{\infty}n(-1)^{n}x^{n-1}

    so a_n=(-1)^{n+1}nx^{n}

    so a_3=(-1)^{4}{3}x^{3-1}=3x^2
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  3. #3
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    Quote Originally Posted by TFD View Post
    So tomorrow I need to explain a problem to the class, I know the correct answer is 3, i just do not know how to show the work necessary to explain it.

    Here is the problem:

    What is the coefficient of x^2 in the Taylor series for 1/((1+x)^2) about x=0.

    (A) 1/6 (B) 1/3 (C) 1 (D)3 (E)6


    Thanks for the help.

    Given f(x)=\frac{1}{(1+x)^2}

    The coefficient of x^2 is going to be \frac{f''(0)}{2!}.

    So first you have to compute f''(0).
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  4. #4
    TFD
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    Quote Originally Posted by Mathstud28 View Post
    The power series for this would be found by seeing that \frac{1}{(1+x)^2}=\frac{-D[\frac{1}{1+x}]}{dx}

    So \frac{1}{1+x}=\sum_{n=1}^{\infty}(-1)^{n}x^{n}

    so the oppositve of the derivative would be -\sum_{n=0}^{\infty}n(-1)^{n}x^{n-1}

    so a_n=(-1)^{n+1}nx^{n}

    so a_3=(-1)^{4}{3}x^{3-1}=3x^2
    Im a bit confused on what the oppositve derivative is?
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TFD View Post
    Im a bit confused on what the oppositve derivative is?
    Since \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n}

    and since the derivative of \frac{1}{1+x} is \frac{-1}{(1+x)^2} you can see that \frac{1}{(1+x)^2} is the opposite(or negative) of the derivative of \frac{1}{1+x}=\sum_{n=1}^{\infty}(-1)^{n}x^{n}

    so by differentiating we get \frac{D[\sum_{n=0}^{\infty}(-1)^{n}x^{n}]}{dx}=\sum_{n=1}^{\infty}n(-1)^{x}x^{n-1} and since we said that \frac{1}{(1+x)^2} was the opposite of the derivative the answer is -1\sum_{n=1}^{\infty}n(-1)^{n}x^{n}=\sum_{n=1}^{\infty}n(-1)^{n+1}x^{n-1}
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