Is there a sequence of complex polynomials $p_n$ s.t. $p_n(0) = 1$ for every $n \in \mathbb{N}$ and $p_n(z) \to 0$ for each $z \in \mathbb{C} \setminus \{0\}$?
Any help with this would be great!
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Caleb Stanford
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user148029
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Consider $p_n(z):=1+ \frac {z^n} {n!}.$ – user64494 May 05 '14 at 16:34
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@user64494 That goes to $1$ for all $z$, not $0$. – Caleb Stanford May 05 '14 at 16:39
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1This is explicitly proved in "Pointwise Limits of Analytic Functions", Kennith Davidson, Amer. Math. Monthly, Vol 90, No. 6 (1983), as a consequence of Runge's theorem. – Omran Kouba May 05 '14 at 17:46
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@ Omran Kouba : it is strange: I find nothing neither in ZMATH nor in MRlookup, searching for "Pointwise Limits of Analytic Functions", Kenneth Davidson, Amer. Math. Monthly, Vol 90, No. 6 (1983) – user64494 May 05 '14 at 18:04
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@user64494, Nothing is strange, you didn't search hard, look here: http://webs.um.es/beca/Docencia/1112.ac/2975578.pdf – Omran Kouba May 05 '14 at 19:04
1 Answers
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Yes, such a sequence exists. Use the polynomials $Q_n$ defined in this answer to a very similar question. From that answer:
$Q_n$ is a polynomial that is less than $\frac1{n(n+1)}$ on $K_n^\times$ and such that $Q_n(0)=1$. Thus $Q_n(x)\to0$ pointwise on $\Bbb C^\times$.
I have been trying to come up with a simpler explicit construction but I have not been able to.
Caleb Stanford
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