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Math Help - Power Series Help

  1. #1
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    Power Series Help

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    Hey guys I need help with this Power Series question, I used expl differentiation first then I got stuck in anti differentiation, where do I go after?

    Thanks
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by chalk View Post
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    Hey guys I need help with this Power Series question, I used expl differentiation first then I got stuck in anti differentiation, where do I go after?

    Thanks
    I am not sure what you tried, but the usual method start like this

    Assume the solution have the form

    \displaystyle y=\sum_{n=0}^{\infty}a_nx^n

    Then take some derivatives to get
    \displaystyle y'=\sum_{n=0}^{\infty}na_nx^{n-1}

    and

    \displaystyle y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}

    Now plug these into the ode and work from here to solve for the a_n
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by chalk View Post
    Hey guys I need help with this Power Series question, I used expl differentiation first then I got stuck in anti differentiation, where do I go after?

    After writing y(x)=\sum_{n=0}^{+\infty}a_nx^n , what did you obtain?


    Edited : Sorry, I didn't see the previous post.
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  4. #4
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    Yeah I had those forms and subbed it back into the d.eq

    \displaystyle y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \displaystyle y'=\sum_{n=0}^{\infty}na_nx^{n-1} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0

    then

    \displaystyle y''=\sum_{n=0}^{\infty}(n+2)(n+1)a_nx^{n} + \displaystyle y'=\sum_{n=1}^{\infty}na_nx^{n} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0

    Right?
    My equation becomes 2a_2 + a_0 = 0 then to a_2 = -(1/2) a_0

    EDIT: Ah I dunno how to get the formula source code up :/
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by chalk View Post
    Yeah I had those forms and subbed it back into the d.eq

    \displaystyle y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \displaystyle y'=\sum_{n=0}^{\infty}na_nx^{n-1} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0

    then


    \displaystyle y''=\sum_{n=0}^{\infty}(n+2)(n+1)a_nx^{n} + \displaystyle y'=\sum_{n=1}^{\infty}na_nx^{n} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0

    Right?
    My equation becomes 2a_2 + a_0 = 0 then to a_2 = -(1/2) a_0

    EDIT: Ah I dunno how to get the formula source code up :/
    you need to wrap them in math tags
    [tex] [ /math] no space between the bracket and forward slash. [ /

    If you change your editor to advanced mode you can just click

    Tex button in the upper right


    \displaystyle \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \sum_{n=0}^{\infty}na_nx^{n-1} +\sum_{n=0}^{\infty}a_nx^n = 0

    \displaystyle \sum_{n=0}^{\infty}(n+2)(n+1)a_nx^{n} + \sum_{n=1}^{\infty}na_nx^{n} + \sum_{n=0}^{\infty}a_nx^n = 0

    I think this is what you typed, but you are missing the x and x^2 in front of the function and the first derivative
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  6. #6
    MHF Contributor chisigma's Avatar
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    Even if 'non reccommended' the most confortable approach is to find y as...

    \displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n} (1)

    The values of y(0) and y^{(1)} (0) are given by the 'initial conditions'...

    y(0)= 5

    y^{(1)}(0)= -2

    ... and the others are given by the DE...

    y^{(2)} (0) = -x^{2} y^{(1)} (0) + x y(0)= 0

    y^{(3)} (0) = y(0) -x y^{(1)} (0) -x^{2} y^{(2)}(0)= 5

    y^{(4)} (0) = -3 x y^{(2)} (0) -x^{2} y^{(3)}(0)= 0

    y^{(5)} (0) = -3 y^{(2)} (0) -5 x y^{(3)}(0) - x^{2} y^{(4)} (0) = 0

    y^{(6)} (0) = -8 y^{(3)} (0) -7 x y^{(4)}(0) - x^{2} y^{(5)} (0) = -40

    ... and so one, so that the power series is...

    \displaystyle y(x)= 5 -2 x + \frac{5}{6} x^{3} -\frac{1}{18} x^{6} + ... (2)

    Kind regards

    \chi \sigma
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