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 Post subject: Uniform convergencePosted: Tue Nov 22, 2016 8:46 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Examine if there exists a sequence $\displaystyle{\left(p_n(z)\right)_{n\in\mathbb{N}}}$ of

complex polynomials such that $\displaystyle{p_n(z)\to \dfrac{1}{z}}$ uniformly to

$\displaystyle{C_{r}=\left\{z\in\mathbb{C}: |z|=r\right\}}$ ($\displaystyle{r>0}$).

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 Post subject: Re: Uniform convergencePosted: Mon Dec 05, 2016 1:05 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 157
Location: Melbourne, Australia
Suppose that this is not the case. If such sequence existed then the convergence at the compact set $\left| z \right| = r$ would be uniform. However,

$$2 \pi i = \oint \limits_{\mathcal{C}_r} \frac{{\rm d}z}{z} = \oint \limits_{\mathcal{C}_r} \lim_{n \rightarrow +\infty} p_n (z) \, {\rm d}z = \lim_{n \rightarrow +\infty} \oint \limits_{\mathcal{C}_r} p_n (z) \, {\rm d}z = \lim_{n \rightarrow +\infty} 0 = 0$$

since the polynomials have no poles inside the circle thus the contour integral is zero. The last relation is an obscurity of course. Thus, no complex polynomials exist. This completes our arguement.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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