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I want to determine the radius of convergence of the series

\begin{equation*} \sum_0^\infty \frac{f^{k}(5)}{k!}(z-5)^k, \end{equation*}

where $f(z) = \frac{z^2}{e^{iz}-1}$.

In the solution of the problem it says that I need to find the singularity of $f$ that is closest to $z=5$. I need some help in making sense of that claim. In the series we're using $f^{k}(5)$, and I don't see how that value depends on the singularities of $f$. I find it hard therefore to understand why the radius of convergence of the series is determined by the singularities of $f$ (since $f$ has no singularities for $z=5$).

Really grateful for any help with this. Thankful for "simple", intuitive explanations, when possible.

Jarvi79
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2 Answers2

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You are asking about a very nice feature of holomorphic functions. If $\Omega \subseteq \mathbb{C}$ is an open subset and $f \colon \Omega \rightarrow \mathbb{C}$ is a holomorphic function then, given any point $z_0 \in \Omega$, the power series expansion of $f$ around $z = z_0$ converges to $f$ on any open disc $B(z_0, R)$ such that $B(z_0, R) \subseteq \Omega$. That is, the power series expansion of $f$ around $z = z_0$ is guaranteed to converge at least on the maximal open disc around $z_0$ that is contained in $\Omega$. You can find a proof in any textbook on complex analysis.

This is indeed somewhat surprising - the power series of $z = z_0$ is calculated using only local information (the derivatives $f^{k}(z_0)$ depend on the values of $f$ in an arbitrary small neighborhood of $z = z_0$) and yet the behavior of the power series depends on "how far" the function $f$ can be extended while remaining a well-defined, holomorphic function.

In your case, the natural domain of definition of $f$ is $\Omega = \mathbb{C} \setminus \{ 2\pi k \, | \, k \in \mathbb{Z} \}$. This guarantees that the power series expansion of $f$ around $z_0 = 5$ converges to $f$ on $B(z_0, 2\pi - 5)$. It cannot converge on any larger disk, for if it would, then the power series expansion would provide a holomorphic extension of $f$ across $2\pi$, showing that $z = 2\pi$ is a removable singularity for $f$. However, $$ \lim_{x \to 2\pi^{-}} f(x) = \lim_{x \to 2\pi^{-}} \frac{x^2}{e^{ix} - 1} = -\infty $$ so $2\pi$ cannot be a removable singularity of $f$.

levap
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The radius of convergence of a power series about a point $a$ is a real number $R>0$ such that the series converges in $G=\{ z: |z-a|<R\}$, this gives that the function $f$ equals its power series in $G$, that is the first singularity of $f$ appears on the boundary of $G$. Then of course a way to obtain $R$ is by finding the closest singularity to the point $a$. In your case $a=5$.

Alonso Delfín
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