1

Here is the setup:

For all positive integers $n$, let define: $$P_n(z):=\sum_{k=0}^n\frac{z^k}{k!}.$$ Let $R_n$ be the minimum of the absolute values of the zeros of $P_n$.

I am asked to show that $R_n$ goes to infinity when $n\to+\infty$.

My thoughts.

  • $(P_n)$ is uniformly locally convergent towards $\exp$. Since, $\exp$ does not vanish on $\mathbb{C}$, it is sain and intuitive to have $R_n\underset{n\to+\infty}{\longrightarrow}+\infty$.

  • Using Rouché theorem, I am done if I am able to find a suitable $(r_n)_n\in(0,+\infty)^{\mathbb{N}}$ which goes to infinity and such that: $$\forall z\in\mathbb{C},|z|=r_n\Rightarrow |P_n(z)-\exp(z)|<|\exp(z)|.$$

  • Using the argument principle, one has: $$\forall n\in\mathbb{N}\setminus\{0\},\frac{1}{2i\pi}\int_{C(0,R_n)}\frac{P_{n-1}(z)}{P_n(z)}\,\mathrm{d}z=0.$$ But I have no clue how to use that $P_n$ converge toward $\exp$.

Any help will be greatly appreciated!

C. Falcon
  • 19,042

1 Answers1

3

For $r>0$ fixed, $e^z-P_n(z)$ goes uniformly to zero on $\overline{D(0,r)}$ so for $n$ large enough the difference is smaller than $e^{-r}$ and you may apply e.g. Rouché.

H. H. Rugh
  • 35,236