Let f be analytic on D=zinmathbf C/z<3 such that f!(1+frac1n)=2n+1/n+1, n=1,2,dots. Then f(z)=

Q26Section II (Mathematics)26 of 180

Recovering $f$ from its values on a convergent sequence

Let $f$ be analytic on $D=\{z\in\mathbf C/|z|<3\}$ such that $f\!\left(1+\frac1n\right)=\dfrac{2n+1}{n+1}, n=1,2,\dots$ . Then $f(z)=$

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