3
$\begingroup$

I'm hoping someone can help me with the following question from Stein and Shakarchi Volume 1 on Fourier Analysis (p. 24):

For $z \in \mathbb C$, we define the complex exponential by $$\sum_{n=0}^{\infty} \frac{z^n}{n!}.$$ Prove that the above definition makes sense, by showing the series converges for every complex number $z$. Moreover, show that the convergence is uniform on every bounded subset of $\mathbb C$.

I know that convergence is easily shown with the ratio test or an absolute convergence argument, but I'm trying to prove convergence formally, and keep getting stuck. So far, I've tried the following. A series converges if the sequence of partial sums $$\sum_{j=0}^{n} \frac{z^j}{j!}$$ converges as $n \to \infty$. So showing that $$\lim_{n \to \infty} \left| \sum_{j=0}^{n} \frac{z^j}{j!} - e^z \right| = 0$$ appears to be the goal. We could also use the $N-\epsilon$ definition, which may help, or show the sequence is Cauchy. But I can't seem to get any of these to work. With the uniform convergence, I seem to run into the same problems.

Any help or hints would be appreciated.

$\endgroup$
  • $\begingroup$ Note that $|e^z| \le e^{|z|}$, so the problem of convergence reduces to the real case. $\endgroup$ – copper.hat Nov 16 '16 at 3:42
1
$\begingroup$

HINT:

Note that for any fixed $z$, we have for $M>N>|z|$

$$\begin{align} \left|\sum_{n=N}^M \frac{z^n}{n!}\right|&\le \sum_{n=N}^M \frac{|z|^n}{n!}\\\\ &\le \sum_{n=N}^M \frac{|z|^n}{N^n}\\\\ &=\frac{|z/N|^N -|z/N|^{M+1}}{1-|z/N|} \end{align}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ By writing the sum to $\infty$, you are assuming that it converges. You need to make the upper limit $m$ and show that the sum converges as $m \to \infty$. $\endgroup$ – marty cohen Nov 16 '16 at 5:09
  • $\begingroup$ @martycohen Actually, the upper limit implies that a limit is taken. In fact, the last step actually evaluates that implied limit. Does that make sense? $\endgroup$ – Mark Viola Nov 16 '16 at 5:15
  • $\begingroup$ I would make the upper limit $m$, where $m > N$. The numerator of the fraction would then be $|z/N|^N-|z/N|^{m+1} = |z/N|^N(1-|z/N|^{m-N+1})$ and show that that expression goes to zero. $\endgroup$ – marty cohen Nov 16 '16 at 5:22
  • $\begingroup$ @martycohen Marty, I've edited to incorporate your suggestion. Much appreciative. -Mark $\endgroup$ – Mark Viola Nov 17 '16 at 4:50
  • $\begingroup$ Thanks a lot. I assume this goes to showing uniform convergence right? I just don't really see how... I may be missing a theorem somewhere... $\endgroup$ – K.Reeves Nov 18 '16 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.