I was re-visiting the binomial series expansion, and noticed that one author took special pains to restrict the general form of expression like
$$ (a+x)^k, $$
saying that the results for $(1+x)^k$ can be extended to this general case using the identity
$$ (a+x)^k = a^k \; \big(1+ \tfrac{x}{a}\big )^k, $$
for $\boldsymbol{a > 0}$.
After checking a few other textbooks, I found that none of them addressed binomial series expansions of the form $(x-b^2)^k$, where $b, k\, \in \mathrm{I\! R}$.
Is there a standard way of handling these cases? Is there anything to watch out for, besides the domain?
I see why the Maclaurin series can't be defined for something like $(x-1)^k$, in general, because $x = 0$ may not be in the domain, if the function is real-valued;$^{\color{red}1}$ however, expansion about $c > 1$ would appear to work just fine, as far as I can tell. I would also expect there to be a general result about the radius of convergence being limited by the distance of the centre of expansion from the point of singularity / discontinuity, for the obvious reason that the function isn't (well) defined at those points.
$^{\color{red}1}$: For example, if $f(x) = \sqrt{x-2}$, then, for the expansion about $c = 0$ we have:
$$ \begin{array}{c|c} f(x) = \sqrt{x-2} \hfill & f(0) = \sqrt{-2} \hfill \\[3pt] \hline % f'(x) = \tfrac{1}{2}(x-2)^{-1/ 2} \hfill & f'(0) = \tfrac{1}{2}(-2)^{-1/ 2} \hfill \\[3pt] \hline % f''(x) = (\tfrac{1}{2})(-\tfrac{1}{2}) (x-2)^{-3/2} \hfill & f''(0) = (\tfrac{1}{2})(-\tfrac{1}{2}) (-2)^{-3/2} \hfill \\[3pt] \hline % f'''(x) = (\tfrac{1}{2})(-\tfrac{1}{2})(-\tfrac{3}{2})(x-2)^{-5/2} & f'''(0) = (\tfrac{1}{2})(-\tfrac{1}{2})(-\tfrac{3}{2})(-2)^{-5/2} \hfill \end{array} $$
so the Tylor coefficients $a_n = \tfrac{f^{(n)}(c )}{n!}$ would be complex-valued.