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Let $r>0.$ In each problem below, answer the question "Is there a holomorphic $f$ in $D(0,r)$ that satisfies the given condition for infinitely many $n \in \mathbb{N}?$"

  1. $f(\frac{1}{n}) = \frac{1}{n^2}$

  2. $f(\frac{1}{n}) = \frac{1}{n^2-1}$

  3. $|f(\frac{1}{n})| \le \frac{1}{2^n}$

  4. $f(\frac{1}{n}) = \frac{(-1)^{n+1}}{n}$

I have a hunch that the following theorem should be useful here, but I'm not sure how it should be used: if two holomorphic functions $f$ and $g$ on a domain $D$ agree on a set $S$ which has an accumulation point $c$ in $D$ then $f = g$ on $D$.

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    The problem needs restating. It should be, in each case, "Is there a holomorphic function $f$ such that ..." – 2017-01-14
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    @zhw. Yes, exactly. I've edited the question. I hope the statement is clearer now. – 2017-01-14
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    I'd say your hunch is right. What do you think the answer is for 1.? There's an obvious function that sends $\frac{1}{n}\mapsto\frac{1}{n^2}$. – 2017-01-14

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Hints: 1. @juan arroyo has given you the heads-up.

  1. Think of it as $f(z) = \frac {1}{(1/z)^2-1}$ for certain values of $z.$

  2. $f\equiv 0$ comes to mind. You'll have a more interesting problem if "nonconstant" is part of the requirement.

  3. Think about $n$ odd.