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Math Help - Find power series

  1. #1
    Super Member angel.white's Avatar
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    Find power series

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

    f(x) = \frac {x^3}{(x-2)^2}

    This is what I did:
    = x^3 \left( \frac {1}{x-2} \right)^2

    = x^3 \left( -\frac12 * \frac {1}{1-\frac x2}\right)^2

    = x^3 \left( -\frac12 * \sum_{n=0}^\infty \left( \frac x2 \right)^n\right)^2

    So radius = 2

    But I don't know where to go from here, the answer says:


    \sum_{n=3}^\infty \frac {n-2}{2^{n-1}} x^n
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  2. #2
    Moo
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    Hello,

    I've learnt to do this way :

    \frac{x^3}{(x-2)^2}=\sum_{n=0}^{+\infty} a_n x^n

    And then determine a_n by induction.

    I don't have time to see if it works here though...
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by angel.white View Post
    Find a power series representation for the function and determine the radius of convergence:

    f(x) = \frac {x^3}{(x-2)^2}

    This is what I did:
    = x^3 \left( \frac {1}{x-2} \right)^2

    = x^3 \left( -\frac12 * \frac {1}{1-\frac x2}\right)^2

    = x^3 \left( -\frac12 * \sum_{n=0}^\infty \left( \frac x2 \right)^n\right)^2

    So radius = 2

    But I don't know where to go from here, the answer says:


    \sum_{n=3}^\infty \frac {n-2}{2^{n-1}} x^n
    Consider this \frac{1}{(x-2)^2}=\frac{-D[\frac{1}{x-2}]}{dx} find the power series for \frac{1}{x-2} differentiate it adn take the opposite
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by angel.white View Post
    Find a power series representation for the function and determine the radius of convergence:

    f(x) = \frac {x^3}{(x-2)^2}

    This is what I did:
    = x^3 \left( \frac {1}{x-2} \right)^2

    = x^3 \left( -\frac12 * \frac {1}{1-\frac x2}\right)^2

    = x^3 \left( -\frac12 * \sum_{n=0}^\infty \left( \frac x2 \right)^n\right)^2

    So radius = 2

    But I don't know where to go from here, the answer says:


    \sum_{n=3}^\infty \frac {n-2}{2^{n-1}} x^n
    Given A(x)=\sum_{n=0}^{\infty}{a_n\cdot{x^n}} and B(x)=\sum_{n=0}^{\infty}{b_n\cdot{x^n}}

    We have: A(x)\cdot{B(x)}=\sum_{n=0}^{\infty}{\left(\sum_{k=  0}^n{a_k\cdot{b_{n-k}}}\right)\cdot{x^n}}

    Applying this: \left(\sum_{k=0}^{\infty}{\frac{x^n}{2^n}}\right)^  2= \sum_{n=0}^{\infty}{\left(\frac{n+1}{2^n}\right)\c  dot{x^n}}


    To do this exercise you can consider that: \frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}{n\cdot{x^n}} when |x|<1
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