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Given an expression, for example, $$f(x,y)=\frac{x}{(x-1)(1-y)(xy+x-1)}$$

How can we find the general expression involving $m$ and $n$ for $$\lim_{x\to 0, y\to 0}{\frac{\partial^{n+m}{f}}{\partial{x}^{m}\partial{y}^{n}}}$$

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Have you considered expanding into power series of $x$ and $y$? –  Shai Covo Feb 17 '11 at 21:30
@Shai, yes, but how to do it in details? Can you please give the details as well as the final answer? –  Qiang Li Feb 17 '11 at 21:37
I just wanted to know whether you considered that; I haven't thought about the details... –  Shai Covo Feb 17 '11 at 21:44
Why is the final answer so important? Given the method, you could find the answer for this question, as well as for similar ones in the future. –  Yuval Filmus Feb 18 '11 at 6:16

1 Answer 1

up vote 1 down vote accepted

Suppose $f = \sum_{i,j \geq 0} c_{ij} x^i y^j$. The limit you're looking for is $m!n! c_{mn}$. Your function is better written as $$f(x,y) = \frac{x}{(1-x)(1-y)(1-x-xy)},$$ where the most interesting factor is $$\frac{1}{1-x-xy} = \sum_{t \geq 0} x^t(1 + y)^t = \sum_{t,s \geq 0} \binom{t}{s} x^t y^s.$$ Multiplying by a variable $1/(1-z)$ is like taking "running sums" across $z$. So $$\frac{x}{(1-x)(1-y)(1-x-xy)} = \sum_{t,s \geq 0} x^{t+1} y^s \left[ \sum_{T \leq t, S \leq s} \binom{T}{S} \right]. $$ So the value you're looking for is $$m!n!\sum_{T \leq m-1,S \leq n} \binom{T}{S}.$$

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