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

**dwsmith** · January 11th 2011, 11:49 AM

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.

**snowtea** · January 11th 2011, 11:58 AM

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

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

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

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

**dwsmith** · January 11th 2011, 12:00 PM

Originally Posted by snowtea

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

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

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

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

Shouldn't the indexing change after the derivative?

Never mind, I see it in line two.

**dwsmith** · January 11th 2011, 12:05 PM

Originally Posted by snowtea

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

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

I am sorry but I don't see how you went from line 2 to 3 getting (m+1) and (n+1), respectively.

**dwsmith** · January 11th 2011, 12:42 PM

Snowtea,

Never mind I figured it out but you forget to $a_{m+1,n} \ \mbox{and} \ a_{m,n+1}$

**snowtea** · January 11th 2011, 12:46 PM

Originally Posted by dwsmith

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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