# Can a power series of $x$ converge to $\log(x)$?

I found the following equation, to solve for $$\{c_k\}$$, with $$0: $$\sum_{k=0}^{\infty} c_k x^k = \log x$$

I know that $$\log x$$ does not have a Taylor series expansion around $$x=0$$. Does this mean that no $$\{c_k\}$$ can be found such that the equality holds?

Note that the equality is not required to hold for $$x=0$$, where $$\log x$$ is not defined.

Does this mean that no $$\{c_k\}$$ can be found such that the equality holds?
Exactly. If it would hold for some $$x \neq 0$$, then it would also hold for any $$z$$ with $$|z| < |x|$$. As it is a power series, this would mean that $$\ln$$ is analytic on the disc $$|z|<|x|$$ which is not the case because $$\ln x$$ has a singularity at $$x=0$$. Not even a radius of convergence of $$0$$ would work because $$\ln 0$$ is not defined.
And not even a Laurent series does work, because if such a series converged on some annulus around $$0$$, it would be analytic there. But you would hit a branch cut somewhere.
• For a proof of "converge for some $x\neq 0$, also converges $\forall z$ with $|z|<|x|$" see the proof of Note 10.1.1 in this link. This bit is exactly what I was looking for. Thanks! Commented Jun 2, 2022 at 19:02