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

1. 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.

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

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

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

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

3. 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.

4. 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.

5. Snowtea,

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

6. 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