Determine whether there exist functions \(f\) and \(g\), analytic at \(0\), such that for every positive integer \(n\), \[ f(1/n)=f(-1/n)=1/n^2, \] and \[ g(1/n)=g(-1/n)=1/n^3. \]
Solution
1. A function \(f\) satisfying the first interpolation condition exists; one may take\[ \boxed{f(z)=z^2}. \]
The polynomial \(f(z)=z^2\) is analytic at \(0\), and for every positive integer \(n\), \[ f(1/n)=\frac1{n^2}=f(-1/n). \]
2. No function \(g\) analytic at \(0\) can satisfy the second interpolation condition.
Suppose such a function \(g\) existed. There is a disk \(D\) centered at \(0\) on which \(g\) is analytic. For every sufficiently large positive integer \(n\), the point \(1/n\) lies in \(D\), and \[ g(1/n)=\frac1{n^3}=\left(\frac1n\right)^3. \] Thus the analytic function \[ h(z)\coloneqq g(z)-z^3 \] has infinitely many zeros \(1/n\) in \(D\) accumulating at the interior point \(0\). By the identity theorem, \(h\equiv0\) on \(D\), so \(g(z)=z^3\) on \(D\).
For all sufficiently large \(n\), also \(-1/n\in D\), and hence \[ g(-1/n)=\left(-\frac1n\right)^3=-\frac1{n^3}, \] contradicting the required equality \(g(-1/n)=1/n^3\).
3. Q.E.D.
Step 1 settles the existence of \(f\), and step 2 proves the nonexistence of \(g\).