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Math Help - P(s,t)=\sum_{0}^{\infty}\sum_{0}^{\infty}a_{m,n}s^ mt^n

  1. #1
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    P(s,t)=\sum_{0}^{\infty}\sum_{0}^{\infty}a_{m,n}s^ mt^n

    If \displaystyle\mbox{P(s,t)}=\sum_{m=0}^{\infty}\sum  _{n=0}^{\infty}a_{m,n}s^mt^n where the coefficients a_{m,n} satisfy the recurrence relation

    (m+1)a_{m+1,n}+5(n+1)a_{m,n+1}=0 \ \ m,n\geq 0,

    show that P satisfies the equation

    P_s+5P_t=0

    I am not sure where to begin with this one.
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  2. #2
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    \displaystyle P_s+5P_t = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n + 5\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ns^m  t^{n-1}

    \displaystyle = \sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n + 5\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{m,n}ns^m  t^{n-1}

    \displaystyle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}(m+1)  s^mt^n + 5\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}(n+1  )s^mt^n

    \displaystyle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}(a_{m,n}(m+1  )s^mt^n + 5a_{m,n}(n+1)s^mt^n)
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  3. #3
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    Quote Originally Posted by snowtea View Post
    \displaystyle P_s+5P_t = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n + 5\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ns^m  t^{n-1}

    \displaystyle = \sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n + 5\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{m,n}ns^m  t^{n-1}

    \displaystyle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}(m+1)  s^mt^n + 5\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}(n+1  )s^mt^n

    \displaystyle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}(a_{m,n}(m+1  )s^mt^n + 5a_{m,n}(n+1)s^mt^n)
    Shouldn't the indexing change after the derivative?

    Never mind, I see it in line two.
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  4. #4
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    Quote Originally Posted by snowtea View Post
    \displaystyle = \sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n + 5\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{m,n}ns^m  t^{n-1}

    \displaystyle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}(m+1)  s^mt^n + 5\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}(n+1  )s^mt^n
    I am sorry but I don't see how you went from line 2 to 3 getting (m+1) and (n+1), respectively.
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  5. #5
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    Snowtea,

    Never mind I figured it out but you forget to a_{m+1,n} \ \mbox{and} \ a_{m,n+1}
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    Snowtea,

    Never mind I figured it out but you forget to a_{m+1,n} \ \mbox{and} \ a_{m,n+1}
    Yup, that's what I get for rushing (and copy paste).
    It's the idea that counts
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