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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.

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  • $\begingroup$ 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. $f=1, k=2?$ $\endgroup$ – gammatester Dec 8 '18 at 13:40
  • $\begingroup$ You didn't "obtain a Taylor series." In a Taylor series, the exponents are all nonnegative integers. $\endgroup$ – saulspatz Dec 8 '18 at 14:22
  • $\begingroup$ thanks for your comments, they helped me a lot! :) $\endgroup$ – mxaxc Dec 8 '18 at 14:32

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