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Thread: 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 $\displaystyle \frac{1}{(1+x)^2}=\frac{-D[\frac{1}{1+x}]}{dx}$

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

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

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

    so $\displaystyle a_3=(-1)^{4}{3}x^{3-1}=3x^2$
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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 $\displaystyle f(x)=\frac{1}{(1+x)^2}$

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

    So first you have to compute $\displaystyle f''(0)$.
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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 $\displaystyle \frac{1}{(1+x)^2}=\frac{-D[\frac{1}{1+x}]}{dx}$

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

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

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

    so $\displaystyle a_3=(-1)^{4}{3}x^{3-1}=3x^2$
    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 $\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n}$

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

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