General Leibniz's rule, why cases k=0 and k=n are separated from the sum?

Question

1) Firstly, I don't understand why cases $n=0$ and $k=n$ are seperated from the sum?

2) Then if we put k=1 then why n on the top of the sum symbol still stays the same instead of changing to $n+1$?

3) Why after the first summation f is derived in k-th order instead of $(k+1)$ like before? And if the n on the sum symbol doesn't change, then why do function $g$ is derived at $(n+1-k)$ th order? ( You can correct my terminology.)

4) Why after the second summation symbol there is $k-1$?

Answer 1

The following might be helpful.

Ad (1)

Firstly, I don't understand why cases $n=0$ and $k=n$ are separated from the sum?

In fact it is not plausible to separate the cases $n=0$ and $k=n$ at this moment. It becomes more plausible when we change the order of the steps.

Note, splitting the summation was done in order to use the binomial identity

\begin{align*} \binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}\tag{2} \end{align*}

We start with the line commented with "splitting the summation" and obtain

\begin{align*} &\color{blue}{\left(f(x)g(x)\right)^{(n+1)}}\\ &\quad=\sum_{k=0}^n\binom{n}{k}f^{(k+1)}(x)g^{(n-k)}(x) +\sum_{k=0}^n\binom{n}{k}f^{(k)}(x)g^{(n+1-k)}(x)\\ &\quad=\sum_{k=\color{blue}{1}}^{\color{blue}{n+1}}\binom{n}{\color{blue}{k-1}}f^{(\color{blue}{k})}(x)g^{(n\color{blue}{+1-k})}(x) +\sum_{k=0}^n\binom{n}{k}f^{(k)}(x)g^{(n+1-k)}(x)\tag{3}\\ &\quad=\left(\sum_{k=1}^{n}\binom{n}{k-1}f^{(k)}(x)g^{(n+1-k)}(x)+\binom{n}{n}f^{(n+1)}(x)g(x)\right)\\ &\quad\qquad+\left(\sum_{k=1}^n\binom{n}{k}f^{(k)}(x)g^{(n+1-k)}(x)+\binom{n}{0}f(x)g^{(n+1)}(x)\right)\tag{4}\\ &\quad=\binom{n}{0}f(x)g^{(n+1)}(x)+\sum_{k=1}^{n}\binom{n+1}{k}f^{(k)}(x)g^{(n+1-k)}(x)\\ &\quad\qquad+\binom{n}{n}f^{(n+1)}(x)g(x)\tag{5}\\ &\quad\,\,\color{blue}{=\sum_{k=0}^{n+1}\binom{n+1}{k}f^{(k)}(x)g^{(n+1-k)}(x)}\tag{6} \end{align*}

Comment:

• In (3) we shift the index of the left-hand sum by one to start with $k=1$. We now have the convenient situation that we can use the binomial identity from (2). But the sums do not have the same index region.

  This is the reason why we separate the case $k=n+1$ from the left-hand sum and the case $k=0$ from the right-hand sum in (4).

• In (5) we can apply (2) since we have separated the two cases where the indices do not coincide.

• In (6) we merge the single terms back into the sum.

The above derivation should help to also clarify parts of the other questions. A somewhat more detailed look to the index-shift in (3):

\begin{align*} \sum_{k=0}^n f(k)&=f(0)+f(1)+\cdots+f(n)\\ \sum_{k=1}^{n+1}f(k-1)&=f(0)+f(1)+\cdots+f(n)\\ \sum_{k=-1}^{n-1}f(k+1)&=f(0)+f(1)+\cdots+f(n) \end{align*}

Note that an index shift by one implies an argument shift by one in the other direction.