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On functions

  1. Does there exist an entire f:\mathbb{C} \rightarrow \mathbb{C} such that f \left( \frac{1}{n} \right) =\frac{1}{n} for all n \in \mathbb{N}?
  2. Does there exist a holomorphic function f on the unit disk such that f \left( \frac{1}{n} \right) =\frac{1}{n!} for all n \in \mathbb{N}?

Solution

  1. Yes. It follows immediately from the identity theorem.
  2. Νο, there is no such function. If there was, its Taylor series centered at 0 would be of the form

        \[f(z) = a_kz^k+a_{k+1}z^{k+1}+\cdots\]

    for some k \in \mathbb{N} and a_k \neq 0. But then

        \[\lim_{n \rightarrow + \infty}\frac{\left\lvert f\left(\frac1n\right)\right\rvert}{\left\lvert\frac{a_k}{n^k}\right\rvert}=1\]

    In particular

        \[\lim_{n \rightarrow + \infty}\frac{\left\lvert f\left(\frac1n\right)\right\rvert}{\frac1{n^k}} \neq 0\]

    But

        \[\lim_{n \rightarrow +\infty}\frac{\frac1{n!}}{\frac1{n^k}}=0\]

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