Fall 2013 - Problem 2

calculation

Show that there is no function $f$ that is holomorphic near $0 \in \C$ and satisfies

f\p{\frac{1}{n^2}} = \frac{n^2 - 1}{n^5}

for all large $n \in \N$ .

Suppose $f$ is holomorphic, and write

f\p{z} = \sum_{n=0}^\infty a_n z^n

in some small neighborhood of $0$ . Then

f'\p{0} = \lim_{n\to\infty} \frac{f\p{\frac{1}{n^2}} - f\p{0}}{1/n^2} = \frac{n^2 - 1 - n^5f\p{0}}{n^3} = \frac{1}{n} - \frac{1}{n^3} - n^2f\p{0}.

If $f\p{0} \neq 0$ , then this sequence is unbounded, which is impossible, so we require $f\p{0} = 0$ . This also implies that $f'\p{0} = 0$ , and so

\begin{aligned} f\p{z} = a_2z^2 + z^3\sum_{n=3}^\infty a_n z^{n-3} &\implies \frac{n^2 - 1}{n^5} = \frac{a_2}{n^4} + \frac{1}{n^6}\sum_{k=3}^\infty \frac{a_k}{n^{2\p{k-3}}} \\ &\implies \frac{n^2 - 1}{n} = a_2 + \frac{1}{n^2} \sum_{k=3}^\infty \frac{a_k}{n^{2\p{k-3}}}. \end{aligned}

But sending $n \to \infty$ , we see that $a_2 = \infty$ , which is impossible. Thus, no function $f$ could have existed to begin with.