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I am about to solve the recurrence

$a(n,m)=\frac{1}{2}a(n,m-1)+\frac{1}{2}a(n-1,m)$ with boundary conditions $a(0,m)=(\frac{1}{2})^m, \forall m\in\mathbb N$ and $a(n,0)=(\frac{1}{2})^n, \forall n\in\mathbb N$ where $a:\mathbb N^2 \to \mathbb R$.

My questions are

  1. For this specific problem, I found the solution $a(n,m)=(1/2)^{n+m}\binom{n+m}{n}$ using graphics (see my sketch here). Is there an algebraic way to solve this problem?
  2. For "1D" linear constant-coefficient difference equation, we know that a geometric sequence with proper base is a fundamental solution to the difference equation. Can we generalise this notion to "2D difference equations"?
  3. Is there a closed-form general solution to the "2D difference equation" without boundary condition?

** Side note **

I found this problem as I was analysing a simplified Swiss tournament model. $a(n,m)$ represents the portion of teams which won $n$ games and lost $m$ games after $n+m$ games.

user8253063
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2 Answers2

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We can get rid of the coefficients by setting

$$ b(n, m) = 2^{n+m}a(n, m) $$

for all $n, m \in \mathbb{N}$, giving us the recurrence

$$ b(n, m) = b(n, m - 1) + b(n - 1, m). $$

(Adding this factor is not just motivated by our prior knowledge of the solution, it's a standard method when solving recurrence relations in one variable.) In this form it has already been solved here, where they use generating functions in two variables. I believe they also answer some of your other questions there.

user23571113
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Hint.

We can use the characteristic function technique. From

$$a(n,m)=\frac{1}{2}a(n,m-1)+\frac{1}{2}a(n-1,m)$$

after multiplication by $x^ny^m$ we have

$$ a(n,m)x^n y^m-\frac{1}{2}a(n,m-1)x^n y^m-\frac{1}{2}a(n-1,m)x^n y^m $$

and summing up

$$ S(x,y) -\frac y2 S(x,y)-\frac x2 S(x,y) = \phi_0(x,y) $$

or

$$ S(x,y) = \frac{\phi_0(x,y)}{1-\frac 12(x+y)} $$

Here $\phi_0(x,y)=\sum_k\left(\frac x2\right)^k + \sum_j\left(\frac x2\right)^j$. Now calling $u = \frac x2, u = \frac y2$ we follow with

$$ Z(u,v) = \frac{\sum_k u^k + \sum_j v^j}{1-(u+v)} $$

or

$$ Z(u,v) = \sum_i\left(u+v\right)^i\left(\sum_k u^k + \sum_j v^j\right) $$

so the polynomial $Z(u,v)=\sum_n\sum_m b(n,m)u^nv^m$ coefficients are related to $a(n,m)$

Cesareo
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