Timeline for Real-analytic function with given set of values [closed]
Current License: CC BY-SA 4.0
7 events
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Apr 9, 2019 at 9:28 | history | closed |
Pietro Majer user44191 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 |