Understanding $\sum_{n=0}^{m}\sum_{k=1}^\infty {m\choose n}g^{(m-n)}(x) a_k \frac{k!}{(n-k)!}(x-c)^{k-n}$ (application of general Leibniz rule)

Solution 1:

Remember that the actual formula for differentiating a power is $$\frac{d}{dx}(x-c)^k = k(x-c)^{k-1}$$ Applying this $n$ times gives $$\frac{d^n}{dx^n}(x-c)^k = k(k-1)(k-2)\cdots(k-n+1)(x-c)^{k-1}$$ This is the actual formula. What happens when $n > k$? Somewhere in that list of coefficients will be $k - k = 0$, which makes the entire thing $0$.

The expression $\dfrac{k!}{(n-k)!}$ is an attempt to express $k(k-1)(k-2)\cdots(k-n+1)$ more compactly. It is not a successful attempt, as it should be $\dfrac{k!}{(k-n)!}$ as the second answer to your other question points out. You can test this by noting when $n = 1$, it gives $$\dfrac{k!}{(k-1)!} = \dfrac{k(k-1)!}{(k-1)!} = k$$ as you know is the coefficient after a single differentiation, while the other formula is $\dfrac{k!}{(1-k)!}$ which does not reduce to $k$.

By the simple definition of the factorial, $0! = 1; m! = m(m-1)!$ for $m > 0$, this expression requires $k \ge n$ for it to even be defined. So the actual result would be $$k(k-1)\cdots(k-n+1) = \begin{cases}\dfrac{k!}{(k-n)!}&k \ge n\\0&k < n\end{cases}$$

Alternatively, factorials can be extended to all complex numbers (I won't go into the process). But the result is that for natural numbers $n > 0, (-n)! = \infty$ (in complex numbers, there is a single $\infty$ that is approached by increasing in all directions, closing the complex plane into a sphere). Thus when $n > k$, $$\dfrac{k!}{(k-n)!} = \dfrac{k!}{\infty} = 0$$