For every $R>0$, prove that there is a positive integer $n$ such that $1 + z + \frac {z^2}{2!} +...+\frac {z^n}{n!}$ has no zeros in $|z|

Question

The problem is exactly that of the title.

I attempted to apply Rouché's theorem, by setting another function $f(z)=e^z$ and comparing the modulus of $f(z)$ and $f(z)-P_n (z)$ at the circle $|z|=R$, but I failed to proceed anymore.
Is my approach correct? Or is there another solution?

@zhw, it's good practice on Math.SE to make the body of the post self-contained. — Antonio Vargas, Jan 17 '17 at 14:54

score 3 · Accepted Answer · answered Jan 16 '17 at 00:02

Let $R >0$. We know $\exp(z) \neq 0$ for all $z \in B_R(0)$. So, there is an $M>0$ such that $|\exp(z)| \geq M > 0$ for all $z \in B_R(0)$. Now, there exist $N \in \mathbb{N}$ such that for all $n \geq N$ we have $$ \frac{M}{2}>|\exp(z)-P_n(z)| \geq ||\exp(z)|-|P_n(z)||,$$ and thus $$ \frac{M}{2} > |\exp(z)|-|P_n(z)| > -\frac{M}{2}.$$

Therefore, $$ \frac{M}{2}+|\exp(z)| > |P_n(z)| > |\exp(z)|-\frac{M}{2}\geq \frac{M}{2} > 0$$ for all $n \geq N$.

score 1 · Answer 2 · answered Jan 16 '17 at 00:05

1

Suppose that exist $r>0 $such that $p_n$has zeros in $U:=\{|z|<r\} $. Let $K\subset U$ compact. Prove that $p_n\to e^z$ over $K$ and obtain an absurd because $e^z$hasn't zeros

answered Jan 16 '17 at 00:05

Martín Vacas Vignolo

4,710

For every $R>0$, prove that there is a positive integer $n$ such that $1 + z + \frac {z^2}{2!} +...+\frac {z^n}{n!}$ has no zeros in $|z|

2 Answers2