The $n$-th derivative of $(x-c)^k$ (a compact formulation?) As part of our calculations, we are trying to compute the $n$-th derivative of $(x-c)^k$ for $0<x<c$, $c>0$ and any positive integer $k$.
From what K.K.McDonald proposed here, I think maybe we can write it as a series but it makes future analysis extremely complicated.
Is it possible to have a more compact and "easy-to-handle" formulation for the higher order derivatives of $(x-c)^k$?
EDIT
I think I should have mentioned this before.
We are trying to compute the $n$-th derivative of
$$f(x)=\sum_{k=1}^{\infty}a_k(x-c)^k.$$
So we wrote
$$f^{(n)}(x)=\sum_{k=1}^{\infty}a_k\frac{d^n}{dx^n}[(x-c)^k],$$
and got stuck!
 A: 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 $n>k$.
A: 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.
A: If $$f(x) = \sum_{k \geq 0} a_k (x-c)^k$$ (so in your case, $a_0 = 0$, this is a bit more general, but if you're taking derivatives, it doesn't really matter), we have $$f'(x) = \sum_{k \geq 1}  k a_k (x-c)^{k-1},$$ the term with just $a_0$ we can throw out, similarly $$f''(x) = \sum_{k \geq 2}  k (k-1) a_k (x-c)^{k-2}.$$ Again, the first term with $a_1 \times 1$ becomes zero so we throw it away. In general, by following the pattern, we suspect $$f^{(n)} (x) = \sum_{k \geq n} k (k-1) \cdots (k - (n-1)) a_k (x-c)^{k-n},$$ and this is true. One would usually prove this by induction.
