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I have the following question, and I don't even know where to begin:

Do functions $f$ or $g$ exist which are analytic at the point $z=0$ and satisfy the conditions: $f(\frac{1}{n})=f(\frac{-1}{n})=\frac{1}{n^2}$, and $g(\frac{1}{n})=g(\frac{-1}{n})=\frac{1}{n^3}$?

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$f$ certainly exists! $g$ I suspect of being trickier. – Ben Millwood Aug 15 '12 at 18:25
Ok...I assume $f(z)=z^2$? That's analytic at zero and satisfies the condition. – Frank White Aug 15 '12 at 18:27
up vote 4 down vote accepted

take $f(z)=z^2$. For $g$, define $h(z):=g(z)-z^3$. Then $h(n^{-1})=0$, and by uniqueness theorem, $g(z)=z^3$. But it doesn't match with the second condition, hence $g$ cannot exist.

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