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I have a recurrence relation with 2 variables similar to $$ F(n,m) = n\cdot F(n-1,m) + (n-m)\cdot F(n-1,m-1) $$ I want to know the steps required to get the generating Function for such recurences. I have gone through a similar post : Solving recurrence relation in 2 variables . But how do I solve when there is a multiplication as in the above equations ?

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  • $\begingroup$ Related to OEIS A078341, which does not give a generating function. $\endgroup$ – Henry Feb 22 '14 at 16:09
  • $\begingroup$ @Henry The recurrence relation that I am trying to solve is actually more complex than the one mentioned in my question. Which is why I am looking for the steps I can take to solve similar questions. $\endgroup$ – Kyuubi Feb 22 '14 at 16:17
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The strategy is to look for the generating function $$ G(x,y)=\sum_{n,m}F(n,m)x^ny^m. $$ To do so:

  • Multiply by $x^ny^m$ each term in the identity, getting $$ F(n,m)x^ny^m = n\cdot F(n-1,m) x^ny^m+ (n-m)\cdot F(n-1,m-1)x^ny^m. $$
  • The LHS terms sum to the generating function $G(x,y)$.
  • For the first terms on the RHS, use the identity $$ nx^ny^m=\left(x^2\cdot\frac{\partial}{\partial x}+x\right) x^{n-1}y^m. $$
  • For the second terms on the RHS, use the identity $$ (n-m)x^ny^m=xy\cdot\left(x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}\right) x^{n-1}y^{m-1}. $$
  • Deduce that $$ G(x,y)=x^2\cdot\frac{\partial G(x,y)}{\partial x}+xG(x,y)+x^2y\cdot\frac{\partial G(x,y)}{\partial x}-xy^2\frac{\partial G(x,y)}{\partial y}+H(x,y), $$ where $H(x,y)$ is a polynomial correction due to the terms of low degrees in the sums.

Whether this PDE can be solved is another story.

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  • $\begingroup$ How would I calculate $H(x,y)$ ? $\endgroup$ – Kyuubi Feb 22 '14 at 18:50
  • $\begingroup$ Looking at the terms of low degrees. $\endgroup$ – Did Feb 22 '14 at 19:06

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