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Q31Section II (Mathematics)31 of 180

Recovering ff from a convergent sequence of values inside a disk

Let G={z∈C:∣z−1∣<1}G=\{z\in\mathbf C:|z-1|<1\} and let f:G→Cf:G\to\mathbf C be an analytic function such that f(1−1/n)=2n−1n+1f(1-1/n)=\dfrac{2n-1}{n+1} for n=2,3,…n=2,3,\dots. Then:

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