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Given monotonic sequence $x_n$ of reals. Does there exist a subsequence $x_{n_k}$ and real-analytic function $f$ on $(1;+\infty)$ such that $f(n_k)=x_{n_k}$?

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    $\begingroup$ What have you tried? $\endgroup$ – David G. Stork Dec 23 '19 at 6:24
  • $\begingroup$ @DavidG.Stork: I have tried formal series $$ f(s) = a_0 + \sum_{n=1}^\infty a_n\cdot\frac{\prod\limits_{k=0}^{n-1}(s-i_k)}{\prod\limits_{k=0}^{n-1}(i_n-i_k)} $$ where $a_n$ could be determined by induction from equations $f(i_k)=x_{i_k}$. $\endgroup$ – user2935704 Dec 23 '19 at 8:19
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  • If $x_n\to \infty$, take $|x_{n_k}|> k^2$ $$h(z) =\prod_k (1-\frac{z}{x_{n_k}}), \qquad h_m(z)=\frac1{1-\frac{z}{x_{n_m}}}$$ are entire and uniformly bounded by $\prod_k (1+\frac{|z|}{|x_{n_k}|})$, take $g(z)$ entire growing fast enough on the real axis such that $$|\frac{f(x_{n_m})}{h_m(x_{n_m})g(x_{n_m})} |\le 1/m^2$$ then $$g(z)\sum_m \frac{f(x_{n_m})}{h_m(x_{n_m})g(x_{n_m})} h_m(z)$$ converges locally uniformly thus it is entire

  • If $x_n$ is bounded, if there exists a function analytic at $a=\sup x_n$ agreeing with $f(x_n)$ then it is the only one, and it is given by the Taylor series at $a$ which is found from the differences of the $f(x_n)$.

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  • $\begingroup$ sorry, but $h_m(x_{n_m})$ is not defined $\endgroup$ – user2935704 Dec 23 '19 at 7:50

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