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Math Help - Power Series

  1. #1
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    Power Series

    Question:

    Find a power series representation for the function and determine the interval of convergence:

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

    My attempt:

    we know that

    \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n

    Therefore,

    \frac{1}{1-x^2} = \sum x^{2n}

    Now, if we multiply by x^2 we get:

    \frac{x^2}{1-x^2} = \sum x^{4n}

    Then we add: \frac{1}{1-x^2}

    So, \sum_{n=0}^{\infty} x^{2n} + x^{4n}

    Am I correct?
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  2. #2
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    Quote Originally Posted by calc101 View Post


    \frac{1}{1-x^2} = \sum x^{2n}

    Now, if we multiply by x^2 we get:

    \frac{x^2}{1-x^2} = \sum x^{4n}
    Nope. x^2 \cdot x^{2n} = x^{2n+2}
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  3. #3
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    Also note that x^{2n}+x^{2n+2}=(1+x^2)x^{2n}.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Also note that x^{2n}+x^{2n+2}=(1+x^2)x^{2n}.
    Therefore the correct answer is:

    \sum_{n=0}^{\infty}(1+x^2)x^{2n} ?
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  5. #5
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    Quote Originally Posted by calc101 View Post
    Question:

    Find a power series representation for the function and determine the interval of convergence:

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

    My attempt:

    we know that

    \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n

    Therefore,

    \frac{1}{1-x^2} = \sum x^{2n} (**)

    Now, if we multiply by x^2 we get:

    \frac{x^2}{1-x^2} = \sum x^{4n}

    Then we add: \frac{1}{1-x^2}

    So, \sum_{n=0}^{\infty} x^{2n} + x^{4n}

    Am I correct?
    It might be a little easier to recognize that

     <br />
\frac{1+x^2}{1-x^2} = \frac{2}{1-x^2} - 1.<br />
then use (**) above
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