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Math Help - Power series solution - are these Recurrence relationships correct?

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    Power series solution - are these Recurrence relationships correct?

    I'm working with solving an ODE via power series, and the recurrence relations seem like they're more complicated than they should be. Can someone check what I've done so far and let me know if I've screwed up royally somewhere?


    (3 - x^2)y'' - 3xy' - y = 0 \qquad x_0 = 0

    y = \sum_{n=0}^{\infty} a_n x^n

    y' = \sum_{n=1}^{\infty} na_n x^{n - 1}

    y'' = \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2}


    (3 - x^2) \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2} - 3x \sum_{n=1}^{\infty} na_n x^{n - 1} - \sum_{n=0}^{\infty} a_n x^n = 0


    3 \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2} - \sum_{n=2}^{\infty} n(n - 1) a_n x^n - 3 \sum_{n=1}^{\infty} na_n x^n - \sum_{n=0}^{\infty} a_n x^n = 0

    3 \sum_{k=0}^{\infty} (k + 2)(k + 1)a_{k + 2} x^k - \sum_{k=2}^{\infty} k(k - 1) a_k x^k - 3 \sum_{k=1}^{\infty} ka_k x^k - \sum_{k=0}^{\infty} a_k x^k = 0


    3 \sum_{k=2}^{\infty} (k + 2)(k + 1)a_{k + 2} x^k - \sum_{k=2}^{\infty} k(k - 1) a_k x^k - 3 \sum_{k=2}^{\infty} ka_k x^k - \sum_{k=2}^{\infty} a_k x^k + 6a_2 + 18a_3x - 3a_1x - a_0 - a_1x = 0


    \sum_{k=2}^{\infty} \left[ x^k \left[ 3(k + 2)(k + 1)a_{k + 2} - k(k - 1) a_k - 3ka_k - a_k \right] \right] + 18a_3x + 6a_2 - 4a_1x - a_0 = 0


    18a_3x + 6a_2 - 4a_1x - a_0 = 0

    a_3 = \frac{-6a_2 + 4a_1x + a_0}{18x}


    3(k + 2)(k + 1)a_{k + 2} - k(k - 1) a_k - 3ka_k - a_k = 0

    a_{k + 2} = \frac{k(k - 1)a_k + 3ka_k + a_k}{3(k + 2)(k + 1)}


    It's that first recurrence relationship for a_3 that seems odd. The RR for the orphan terms has always evaluated to something simple, and certainly never with three levels of a terms, nor with x making an appearance.

    It makes me think that I made a mistake somewhere. Did I?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Power series solution - are these Recurrence relationships correct?

    Quote Originally Posted by Lancet View Post
    18a_3x + 6a_2 - 4a_1x - a_0 = 0
    a_3 = \frac{-6a_2 + 4a_1x + a_0}{18x}

    It should be: 18a_3x + 6a_2 - 4a_1x - a_0 = 0 \Leftrightarrow \begin{Bmatrix} 18a_3-4a_1=0\\6a_2-a_0=0\end{matrix}\Leftrightarrow \begin{Bmatrix} a_3=2a_1/9\\a_2=a_0/6\end{matrix}
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    Re: Power series solution - are these Recurrence relationships correct?

    Quote Originally Posted by FernandoRevilla View Post
    It should be: 18a_3x + 6a_2 - 4a_1x - a_0 = 0 \Leftrightarrow \begin{Bmatrix} 18a_3-4a_1=0\\6a_2-a_0=0\end{matrix}\Leftrightarrow \begin{Bmatrix} a_3=2a_1/9\\a_2=a_0/6\end{matrix}
    Fascinating!

    Why? How do you know which terms to form a sub-split with?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Power series solution - are these Recurrence relationships correct?

    What about the uniqueness of the series expansion?
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    Re: Power series solution - are these Recurrence relationships correct?

    Quote Originally Posted by FernandoRevilla View Post
    What about the uniqueness of the series expansion?
    I'm afraid I don't understand...
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Power series solution - are these Recurrence relationships correct?

    Quote Originally Posted by Lancet View Post
    I'm afraid I don't understand...
    f(x)=\sum_{n=0}^{+\infty}\alpha_nx^n=\sum_{n=0}^{+ \infty}\beta_nx^n for all x\in (-R,R) implies \alpha_n=\beta_n=\dfrac{f^{(n)}(0)}{n!} for all n .
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