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Apr 9, 2019 at 9:28 history closed Pietro Majer
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Alexandre Eremenko
Sean Lawton
Wolfgang
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Apr 9, 2019 at 5:20 comment added ar.grig Sorry for the question. Obviously, $g$ doesn't exist, because must have all derivatives equal to zero in $t=0$. And it was explained earlier in other question.
Apr 8, 2019 at 23:58 comment added Alexandre Eremenko Real-analytic WHERE?
Apr 8, 2019 at 22:31 comment added Igor Khavkine To expand on the comment by Christian Remling, computing divided differences of the values of $g(1/n_k)$ gives sequences that converge to $g^{(p)}(0)$ for any $p$. But by fast convergence of $x_n \to x$ we get $g^{(p)}(0)=0$ for $p>0$. So $g(t)$ cannot be analytic (the covergent sum of its Taylor series $x + 0t + 0t^2 + \cdots$) unless $g(t)$ is constant.
Apr 8, 2019 at 22:25 review Close votes
Apr 9, 2019 at 9:28
Apr 8, 2019 at 21:48 comment added Christian Remling So $(x_n-x)n^k\to 0$ for all $k$, making all derivatives zero at $t=0$.
Apr 8, 2019 at 20:22 history asked ar.grig CC BY-SA 4.0