I came across a double recursion relation I want to solve
$a_{m,k}=a_{m+1,k-1}-a_{m,k-1}$ for $m,k\geq 0$.
Context
The first derivative of a function satisfies $f^{(1)}(\lambda)=f(\lambda+1)-f(\lambda)$. I want to find a formula for all its derivatives i.e. $f^{(k)}(\lambda)=?$. So I set $a_{m,k}=f^{(k)}(\lambda+m)$ and I want to compute $a_{0,k}$.
Any ideas?
Thanks
You can prove by induction on $n$ that
$$f^{(n)}(\lambda)=\sum_{k=0}^n\binom{n}k(-1)^kf(\lambda+n-k)\;.$$
This is clear for $n=0$, and
$$\begin{align*} f^{(n+1)}(\lambda)&=f^{(n)}(\lambda+1)-f^{(n)}(\lambda)\\ &=\sum_{k=0}^n\binom{n}k(-1)^kf(\lambda+1+n-k)-\sum_{k=0}^n\binom{n}k(-1)^kf(\lambda+n-k)\\ &=\sum_{k=0}^n\binom{n}k(-1)^kf(\lambda+1+n-k)+\sum_{k=1}^{n+1}\binom{n}{k-1}(-1)^kf(\lambda+n+1-k)\\ &=\sum_{k=0}^{n+1}\left(\binom{n}k+\binom{n}{k-1}\right)(-1)^kf(\lambda+n+1-k)\\ &=\sum_{k=0}^{n+1}\binom{n+1}k(-1)^kf(\lambda+n+1-k)\;. \end{align*}$$