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My question is as follows:

For $n\in\mathbb{N}$, let $$P_n(z)=\sum_{k=0}^n\frac{z^k}{k!},$$ where $z\in\mathbb{C}$.

Show that for every $R>0$, there exists $N>0$ such that $\forall n>N$ and $|z|<R$, $P_n(z)\neq 0$.

I tried to solve it using the fact that $P_n(z)$ converges to $e^z$ and $e^z$ is nonzero, but I didn't get the desired result.

How can I solve that problem?

Thank you in advance.

Sigma Algebra
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James
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  • Think about the type of convergence the sequence $(P_n)$ has. It converges uniformly towardz $ z \mapsto e^z$ on every compact subset. Using that $e^z$ is never zero should get you to an answer – Enguerrand Moulinier Oct 17 '24 at 08:14
  • One can find more insight under "incomplete Gamma function" instead of "partial sums of the exponential series". Among them https://math.stackexchange.com/questions/131479/complex-zeros-of-the-polynomials-sum-k-0n-zk-k-inside-balls and https://math.stackexchange.com/questions/109360/roots-of-the-incomplete-gamma-function – Lutz Lehmann Oct 17 '24 at 08:28

1 Answers1

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Assume the contrary. There exist $R>0$ and $z_n$ with $|z_n|<R$ such that $P_n(z_n)=0$ for infinitely many values of $n$. There is a subsequence $z_{n_k}$ thet converges to some point $z_0$. Now $P_n(z) \to e^{z}$ uniformly for $|z| \le R$. (I will add a proof of this if you ask for it).

This implies that $0=P_{n_k}(z_{n_k}) \to e^{z_0}$, a contradiction.

Sigma Algebra
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Kavi Rama Murthy
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