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I am trying to show that if $$f_n(z)=1+z+\frac{z^2}{2!}+...+\frac{z^n}{n!}$$ Then $f_n(z)$ don’t have zeros inside the unitary disk. I have tryied to use Rouche’s theorem or use that in the limit the polinomial converges to the exponential, but i dont get hoy to do this.

TeemoJg
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  • It converges locally uniformly to an entire function $f$. So if $f$ has no zeros it is finished. Otherwise assume $f$ has a zero of order $k$ at $z=a$, $f(z) \sim C (z-a)^k$. Fix $\epsilon$ very small, there is $N$ such that every $f_n,n \ge N$ have $k$ zeros on $|z-a| < \epsilon$ and $f_n(z) = h_n(z)\prod_{l=1}^k (z-a_{l,n})=(C+O(z-a))\prod_{l=1}^k (z-a_{l,n})$, $\frac{f_n'}{f_n}(z) = O(z-a) + \sum_{l=1}^k \frac{1}{z-a_{l,n}}$.

    But $\frac{f_n'}{f_n}(z) = 1-\frac{\frac{z^n}{n!}}{f_n(z)}=1-\frac{a^n+O(z-a)}{n! (C+O(z-a))} \frac{1}{\prod_{l=1}^k (z-a_{l,n})}$, a contradiction.

     – reuns Nov 18 '18 at 04:45
  • @reuns: Unless I am mistaken, your argument shows only that $f_n$ has no zeros in the unit disk for sufficiently large $n$. – Martin R Nov 18 '18 at 08:31

1 Answers1

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You have all the facts you need. Here is an outline of what you need to do:

Use Rouche's theorem with $f_n$ and $z^n\over n!$ to show that if some $f_N$ has a zero within the unit disk, then every $f_n$ with $n\gt N$ has a zero in the unit disk. Now since there are an infinite number of zeros in the unit disk, the set of points mapping to zero under our $f_n$'s has at least one accumulation point. So we have a sequence of pairs, $f_i, p_i$ with $f_i(p_i)=0$ for each i, with a limit point $f, p$ (as the $f_n$'s also converge). For the sake of contradiction, we want to show $f(p)=0$ as we already know that f is the exponential and so has no zeros.

To do this use an epsilon delta argument: let $\epsilon \gt 0$, then find $N_0$ such that $f_n(x)$ is within $\epsilon \over 3$ of f(x) for all x in the unit disk when $n\gt N_0$, then find $\delta$ such that when $|x-x_0|\lt\delta$, then $|f(x)-f(x_0)| \lt$ $ \epsilon \over 3$. Lastly find $N_1$ so that $n\gt N_1$ implies $|p-p_n| \gt \delta$. Use the triangle inequality to show that when $n>max(N_0,N_1)$ we have $|f(p)-f_n(p_n)|<\epsilon$. This completes the contradiction, as it implies $f(x)=e^x$ has a zero in the unit disk.

Mark
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