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Consider the complex-valued family of functions $$ f_n(z) = \sum_{k=0}^n \frac{1}{k!}z^k. $$

Is it possible to use Rouché's Theorem to prove that for any $R \in \mathbb R$ there exists some $n$ such that all roots of $f_n$ have modulus strictly greater than $R$?


The solutions that I've seen for this exploit the uniform convergence of the Taylor series for $e^z$, but I'm curious if it's possible to use Rouché's Theorem to show that there are no roots inside the circle of radius $R$ for sufficiently large $n$.

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  • $\begingroup$ Are you interested in a proof that uses uniform convergence and Rouché's theorem? $\endgroup$ – José Carlos Santos Jul 25 '18 at 17:30
  • $\begingroup$ @JoséCarlosSantos, I'm interested in a proof that uses Rouché's theorem in any useful way—it's fine if it uniform convergence too, but it doesn't need to. $\endgroup$ – Peter Kagey Jul 25 '18 at 19:07
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Take $R>0$ and let $M=\inf\left\{\bigl|e^z\bigr|\,\middle|\,|z|=R\right\}$. Since the series $\sum_{n=0}^\infty\frac{z^n}{n!}$ converges uniformly to $e^z$ on the circle $\{z\,|\,|z|=R\}$, there is a natural $N$ such that $\bigl|f_n(z)-e^z\bigr|<M$ for each $z$ there. In particular$$|z|=R\implies\bigl|f_n(z)-e^z\bigr|<|e^z|.$$Therefore, by Rouché's theorem, on the closed disk $D_R(0)$ the functions $f_n$ and $\exp$ have the same number of zeros — that is, no zeros.

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The radius of convergence of the power series for $e^x$ is $\infty,$ so $(f_n)_n$ converges uniformly to $e^z $ on any compact subset of $\Bbb C.$

The function $g(z)=|e^z|$ is continuous from $\Bbb C$ to $\Bbb R^+,$ and $d(R)=\{z:|z|\leq R\}$ is compact, so $m(R)=\min \{|e^z|:R\geq |z|\}$ exists (and is positive).

Take $n\in \Bbb N$ large enough that $\forall n\geq N\;(\sup \{|f_n(z)-e^z|: z\in d(R)\}<m(R)\;).$ Then $\forall n\geq N \;\forall z\in d(R)\;(|f_n(z)|>0).$

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