Solving recurrence relation with 2 variables

Question

If I have a recurrence relation like $$T(n,k)=\frac{T(n-1,k)+T(n,k-1)}{2}$$ with initial values $\forall n \quad T(n,0)=T_0$ and $\forall k \quad T(0,k)=0$. How can I solve it? By the way this came up when I was solving a physics problem

Answer 1

Here is a graphical representation of your recurrence relationship :

(which, now that you have settled correctly your initial data, isn't compulsory, but is interesting by itself because it shows its similarity with the ¨Pascal's triangle" (see below).

Some numerical computations on the first values of $T_{n,k}$ in the case of $T_0=1$ give the following first numerical results with denominators $2^{n+k-1}$.

(please note the diagonal values equal to $1/2$) :

Out of this array, we can build a simplified one by turning it $135°$clockwise in the "Pascal's triangle' manner and keeping only the numerators where the right diagonal, instead of being filled by "ones", is filled by successive powers of $2$ :

$$\begin{array}{ccccccccc} &&&&&1&&&&\\ &&&&\color{blue}{1}&&\color{red}{2}&&&&&\\ &&&\color{blue}{1}&&3&&\color{red}{4}&&&&&&&&\\ &&\color{blue}{1}&&4&&7&&\color{red}{8}&&&\\ &\color{blue}{1}&&5&&11&&15&&\color{red}{16}&&&\\ \color{blue}{1}&&6&&16&&26&&31&&\color{red}{32} \end{array}\tag{1}$$

We have simplified the problem because in this way only integers are managed, and (thanks to an indication by the OP) this is known in the litterature under the name "Bernoulli triangle" yielding the explicit formula for the coefficients in the previous "Pascal's like" array (1) (it is why we write $T'$ instead of $T$):

$$\displaystyle T'_{n,k}=\frac{1}{2^{n}}\sum _{p=0}^{k}{\binom {n}{p}} \ \ \text{for} \ \ k=0,1,\cdots n$$