The $n$-th derivative of $(x-c)^k$ (a compact formulation?)

Solution 1:

The important part of this question should actually be the following: Why can we exchange the differentiation $\frac{d}{dx}$ and the series $\sum_{k=1}^\infty$. There is a theorem in calculus which states that you can do this for $x$ inside the radius of convergence, namely if the radius of convergence of $(a_k)_k$ is $R>0$, $$\frac{d}{dx}\sum_{k=1}^\infty a_k(x-c)^k = \sum_{k=1}^\infty a_k \frac{d}{dx} (x-c)^k\quad \forall x:|x-c|<R. $$ This can be repeated for finitely many times (by induction) so that we can solve your problem with the $n$-th derivative. We find for $x$ with $|x-c|<R$ $$f^{(n)}(x) = \left(\frac{d}{dx}\right)^n\sum_{k=1}^\infty a_k(x-c)^k = \sum_{k=1}^\infty a_k \left(\frac{d}{dx}\right)^n (x-c)^k = \sum_{k=n}^\infty a_k \frac{k!}{(k-n)!}(x-c)^{k-n}.$$

Also have a look here: https://en.wikipedia.org/wiki/Power_series#Radius_of_convergence.

Solution 2:

So $$\frac{d^n}{dx^n}(x-c)^k=\frac{k!}{(k-n)!}(y-c)^{k-n}$$ if $n\le k$, which can easily confimed by induction.

It is $0$ if $n>k$.