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Thread: Frobenius method

  1. #1
    Senior Member
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    Frobenius method

    Hi. I need some help with this problem. I have to solve

    $\displaystyle x(1-x)y''+2(1-2x)y'-2y=0$ (1)

    Using Frobenius method around zero. So proposing $\displaystyle y=\Sigma_{n=0}^\infty a_n x^{n+\alpha}$, differentiating and replacing in (1):

    $\displaystyle x(1-x)y''+2(1-2x)y'-2y=\Sigma_{n=0}^\infty a_n(n+\alpha)(n+\alpha-1)x^{n+\alpha-1}- \Sigma_{n=0}^\infty a_n(n+\alpha)(n+\alpha-1)x^{m+\alpha} + 2 \Sigma_{n=0}^\infty a_n(n+\alpha)x^{n+\alpha-1} -4 \Sigma_{n=0}^\infty a_n(n+\alpha)x^{n+\alpha}-2 \Sigma_{n=0}^\infty a_n n^{n+\alpha}=0$

    Regrouping and working a little bit this becomes:

    $\displaystyle a_0\alpha(\alpha+1)+\Sigma_{n=1}^\infty a_n(n+\alpha)(n+\alpha+1)x^{n+\alpha-1}- \Sigma_{n=0}^\infty a_n [(n+\alpha)^2+2]x^{n+\alpha}$
    I think this is fine, because the indicial equation I get gives the same roots from the other method: $\displaystyle \alpha(\alpha-1)+p_0\alpha+q_0=0$

    Then $\displaystyle \alpha_1=0,\alpha_2=-1$

    So I have:
    $\displaystyle a_0\alpha(\alpha+1)+\Sigma_{n=0}^\infty a_{n+1}(n+\alpha+1)(n+\alpha+2)x^{n+\alpha}- \Sigma_{n=0}^\infty a_n [(n+\alpha)^2+2]x^{n+\alpha}$

    And then $\displaystyle a_{n+1}(n+\alpha+1)(n+\alpha+2)-a_n[(n+\alpha)^2+2]=0$
    So for $\displaystyle \alpha_1=0$

    $\displaystyle a_{n+1}=\frac{a_n[n^2+2]}{(n+1)(n+2)}$

    The thing is I've tried some iterations but I can't get a expression for $\displaystyle a_k$ in terms of $\displaystyle a_0$. How should I do this?

    Bye there, thanks for your help.-
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  2. #2
    Junior Member
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    Just use repeated iterations to get to the next expression:

    $\displaystyle a_n = a_0 \frac{((n-1)^2+2)((n-2)^2+2)...2}{((n-1)+1)((n-1)+2)((n-2)+1)((n-2)+2)....1\cdot 2} $

    You can rearrange the expression to get something more compact.
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  3. #3
    Senior Member bkarpuz's Avatar
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    If you have a relation like
    $\displaystyle a_{n+1}=p_{n}a_{n}$ for $\displaystyle n\geq m$,
    which is a first-order difference equation, then
    $\displaystyle a_{n}=a_{m}\prod_{j=m}^{n-1}p_{j}$ for $\displaystyle n\geq m$,
    with the convention that the empty product is the unity.
    In your case, you have
    $\displaystyle a_{n}=a_{0}\prod_{j=0}^{n-1}\frac{{j^{2}+2}}{(j+1)(j+2)}$
    .....$\displaystyle =a_{0}\frac{1}{n!(n+1)!}\prod_{j=0}^{n-1}(j^{2}+2)$
    for $\displaystyle n\geq0$.
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