# Taylor series of a function

I am puzzled with the following problem and I am not able to figure out the answer no matter how hard I try. Let's say we have a function $$f(x)$$ and we know its Taylor series is of the form $$\sum_{n=0}^\infty a_nx^n$$ (center $$x_0=0$$). We want to find the Taylor series of $$f(\sqrt[k] x)$$ (if it exists). We can write $$f(\sqrt[k] x)= \sum_{n=0}^\infty a_nx^{n/k}$$. If we differentiate $$f(\sqrt[k] x)$$ and find that it is not differential at $$x=0$$, does that mean that the Taylor series we obtained is wrong and it does not have one.

• How do you treat complex numbers, e.g. $x^{1/k}$ for even $k$ and negative $x$? What is the 'Taylor series' of $\sqrt{x}$ at $x=0,$ i.e. $f=1, k=2?$ – gammatester Dec 8 '18 at 13:40
• You didn't "obtain a Taylor series." In a Taylor series, the exponents are all nonnegative integers. – saulspatz Dec 8 '18 at 14:22
• thanks for your comments, they helped me a lot! :) – mxaxc Dec 8 '18 at 14:32