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 Post subject: Uniform convergence
PostPosted: Tue Nov 22, 2016 8:46 pm 
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Joined: Mon Nov 09, 2015 1:52 pm
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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 convergence
PostPosted: Mon Dec 05, 2016 1:05 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 150
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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