Polynomial $W(z) \in \mathbb{C}$ is defined by equation $$ W(z)=1+\sum_{k=1}^{2019} \frac{z^{k}}{k !} $$ prove that if $z \in \mathbb{C}$ and $|z|<10,$ then $W(z) \neq 0$
I need help and some hint. Can you take me some of first step to solution?
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Blabla
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4Estimate $e^z - W(z)$ – Anatoliy R Dec 22 '19 at 02:34
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Per the answer in zeros of the polynomials $\sum_{k=0}^{n} z^k/k!$, inside balls, all roots of $p_n(z)$ are inside $\frac{n}{e^2}<|z|<n$, so that there are no roots inside a disk with radius $273$. To show this for the much smaller radius $10$ should be downright trivial. See also the animation of the scaled roots converging to the Szegő curve in Roots of incomplete gamma in still diagrams in properties of roots of the incomplete exponential. – Lutz Lehmann Dec 22 '19 at 22:58
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Of course $e^z$ does not vanish over $|z|<10$, so if we manage to prove that $|W(z)-e^z|<e^{-10}$ over $|z|<10$ we are done. $W(z)-e^z$ is an entire function, so by the maximum modulus principle we only need to check its values over $|z|=10$. In such a case
$$ \left|e^z-W(z)\right| = \left|\sum_{k\geq 2020}\frac{z^{k}}{k!}\right|=10^{2020}\left|\sum_{h\geq 0}\frac{z^h}{(2020+h)!}\right|\leq\frac{10^{2020}}{2020!}\sum_{h\geq 0}\frac{10^h}{2020^h}=\frac{10^{2020}}{2020!\left(1-\frac{1}{202}\right)} $$ so it is enough to prove $$ 10^{2020} e^{10} < 2020!\frac{201}{202}. $$ Since $ 2020!> 2020^{2020} e^{-2020} $ it is enough to prove $$ 201\cdot 202^{2019} > e^{2030} $$ or $$ 2019 \log(202)+\log(201) > 2030, $$ which is blatantly true.
Jack D'Aurizio
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@Blabla: because over the given region $|e^z|=e^{\text{Re}(z)}>e^{-10}$. – Jack D'Aurizio Dec 22 '19 at 18:37