Let $$f_n(z)=\sum_{k=0}^n \frac{z^k}{k!}$$ and let $$f(z)=\sum_{k=0}^\infty \frac{z^k}{k!}=e^z.$$ As a polynomial $f_n$ has $n$ roots in $\Bbb{C}$. Prove for every $R>0$ there's an integer $n_0>0$ such that if $n\geq n_0$ then $f_n$ has no zeroes in $B_R(0)$. So I want to appeal to Hurwitz's theorem that States if $f_n \rightarrow f$ uniformly on every compact subset of $\Bbb{C}$ which the $f_n$ do converge uniformly to $e^z$ on every compact subset, then if $f$ has a zero at $z_0$ of order $m$ so do the $f_n$. Then since $f(z)$ has no zeroes, can I conclude the $f_n$'s have no zeroes? Or do I need to find such an $n_0$.
Asked
Active
Viewed 82 times
1 Answers
2
Fix $R>0.$ Since $f_n$ tends uniformly to $e^z$ on the disc $|z|\le R,$ there exists $n_0$ such that for $$|f_n(z)-e^z|< e^{-2R},\ n\ge n_0,\ |z|\le R$$ Then $$|f_n(z)|\ge |e^z|- |f_n(z)-e^z|\\ \ge e^{-R}-e^{-2R}>0,\ n\ge n_0,\ |z|\le R$$
Remark No analytic function theory is needed. Almost uniform convergence to a function which does not vanish, suffices.
Ryszard Szwarc
- 30,319
-
how'd you go from $\vert f_n(z) - e^z \vert < e^{-2R}$ to saying that its greater than or equal to $e^{-2R}$ – homosapien Nov 05 '22 at 16:27
-
and how does this show $f_n$'s don't have zero? on ball of radius $R$? – homosapien Nov 05 '22 at 16:29
-
and I thought I had to use the fact that $e^z$ has no zeroes – homosapien Nov 05 '22 at 16:30
-
$|f_n|$ is greater or equal $e^{-R}-e^{-2R}>0.$ Thus $f_n$ does not vanish for $|z|\le R$ and $n\ge n_0.$ Concerning $e^z$ we have $|e^z|=e^{{\rm Re }z}\ge e^{-R}>0.$ In particular $e^z$ dos not vanish. – Ryszard Szwarc Nov 05 '22 at 16:41
-
also how'd you go from |()−|<−2 to saying that its greater than or equal to −2 – homosapien Nov 05 '22 at 16:49
-
I have used the triangle inequality$|a|\ge |b|-|a-b|.$ I didn"t claim it was greater or equal $e^{-2R}$ but greater or equal $e^{-R}-e^{-2R}.$ – Ryszard Szwarc Nov 05 '22 at 16:56
-
aha!! I just realized that after I asked it lol. – homosapien Nov 05 '22 at 16:56
-
so there was no need to invoke Hurwitz Theorem? – homosapien Nov 05 '22 at 19:58
-
No result from complex analysis theory is needed. The proof is valid for any sequence of continuous functions $f_n$ convergent to a nonvanishing function $f$ almost uniformly, i.e. uniformly on bounded sets. – Ryszard Szwarc Nov 05 '22 at 20:04