Is this a valid proof for the general Leibniz rule?

Question

I want to prove the following formula by induction: $$\sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} f^{(k)}g^{(n-k)}$$

The base case is trivial but I am not sure if my reasoning is valid for the inductive step. Here is what I have got so far:

$$(fg)^{(n+1)}=\frac{d}{dx}(fg)^{(n)}=\frac{d}{dx} \left(\sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} f^{(k)}g^{(n-k)} \right) \\ = \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} f^{(k+1)}g^{(n-k)}+\sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix}f^{(k)}g^{(n+1-k)}\\ =\sum_{k=1}^{n+1} \begin{pmatrix}n \\ k-1\end{pmatrix}f^{(k)}g^{(n+1-k)}+\sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix}f^{(k)}g^{(n+1-k)} \\ = \sum_{k=0}^{n+1} \begin{pmatrix}n \\ k-1\end{pmatrix}f^{(k)}g^{(n+1-k)}-\color{red}{\begin{pmatrix}n \\ -1\end{pmatrix}f^{(0)}g^{(n+1)}}+\sum_{k=0}^{n+1} \begin{pmatrix}n \\ k\end{pmatrix}f^{(k)}g^{(n+1-k)}-\color{red}{\begin{pmatrix}n \\ n+1\end{pmatrix} f^{(n+1)}g^{(0)}}$$

My goal here was to change the starting point for the first sum from $k=1$ to $k=0$ and to change the end point for the second sum from $n$ to $n+1$. In order to compensate for that I subtracted the terms I added to the sums. I am just not sure about the binomial coefficients of the red parts. Wolfram alpha tells me that $\begin{pmatrix}n \\ n+1\end{pmatrix}=\begin{pmatrix}n \\ -1\end{pmatrix}=0$. Is that correct? Are binomial coefficients even defined for negative $k$? If this is valid then I think I can complete the proof:

$$\implies \sum_{k=0}^{n+1} \begin{pmatrix}n \\ k-1\end{pmatrix}f^{(k)}g^{(n+1-k)}+\sum_{k=0}^{n+1} \begin{pmatrix}n \\ k\end{pmatrix}f^{(k)}g^{(n+1-k)} \\ =\sum_{k=0}^{n+1} \left[ \begin{pmatrix}n \\ k\end{pmatrix}+\begin{pmatrix}n \\ k-1\end{pmatrix} \right]f^{(k)}g^{(n+1-k)} \\ =\sum_{k=0}^{n+1} \begin{pmatrix}n+1 \\ k\end{pmatrix}f^{(k)}g^{(n+1-k)} \space \space \space \square$$

Answer 1

I have no idea what $\binom n{-1}$ is. But note that\begin{multline}\sum_{k=1}^{n+1}\binom n{k-1}f^{(k)}g^{(n+1-k)}+\sum_{k=0}^n\binom nkf^{(k)}g^{(n+1-k)}=\\=f^{(n+1)}g^{(0)}+\sum_{k=1}^n\binom n{k-1}f^{(k)}g^{(n+1-k)}+\sum_{k=1}^n\binom nkf^{(k)}g^{(n+1-k)}+f^{(0)}g^{(n+1)}\end{multline}Now, you can use these facts:

• $\displaystyle\binom n{k-1}+\binom nk=\binom{n+1}k$ (as you did);
• $\displaystyle f^{(n+1)}g^{(0)}=\binom{n+1}{n+1}f^{(n+1)}g^{(0)}$;
• $\displaystyle f^{(0)}g^{(n+1)}=\binom{n+1}0f^{(0)}g^{(n+1)}$.

Answer 2

This is not an answer, but I don't have enough reputation to comment, so there you go. First off: the fact about combinatorial numbers that Wolfram Alpha told you is indeed valid, and as far as I could see your proof is correct.

If you want a quick intuition for $$\begin{pmatrix}n \\ n+1\end{pmatrix}=\begin{pmatrix}n \\ -1\end{pmatrix}=0$$ think of the Tartaglia triangle (aka Pascal triangle) written in terms of these numbers. The numbers in that triangle can be thought of as surrounded by zeroes for the sum of two contiguous numbers to equal the number below even at the sides of the triangle. In "more formal" terms, you can define $$\begin{pmatrix}n \\ n+1\end{pmatrix}=\begin{pmatrix}n \\ -1\end{pmatrix}=0$$ by imposing that the condition $$\begin{pmatrix}n \\ k\end{pmatrix}=\begin{pmatrix}n-1 \\ k-1\end{pmatrix} + \begin{pmatrix}n-1 \\ k\end{pmatrix}$$ holds.