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I found the following equation, to solve for $\{c_k\}$, with $0<x<1$: $$ \sum_{k=0}^{\infty} c_k x^k = \log x $$

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.

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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.

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  • $\begingroup$ 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! $\endgroup$
    – G Frazao
    Commented Jun 2, 2022 at 19:02

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