2
$\begingroup$

For this question, it is defined that $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$, $\forall x\in\mathbb{C}$.

Though I know why $\lim_{n\rightarrow \infty}(1+x/n)^n=e^x$ pointwise for any real $x$, I feel difficult to show that the sequence of functions converges uniformly on each bounded subset in $\mathbb{C}$. This question comes up in an undergraduate analysis class, but I have little experience working with the complex number. Could you help me on that? Thank you!

$\endgroup$
0
$\begingroup$

Consider the power series for $\log$ which converges in the complex plane if $|z| < 1:$

$$\log(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \ldots.$$

Hence, if $|z| < n$ we have

$$\log(1+z/n) = \frac{z}{n} - \frac{z^2}{2n^2} + \frac{z^3}{3n^3} - \frac{z^4}{4n^4} + \ldots,$$

and, using the triangle inequality

$$\left|n\log(1+z/n) - z\right| \leqslant \frac{|z|^2}{2n} + \frac{|z|^3}{3n^2} + \frac{|z|^4}{4n^3} + \ldots.$$

In a bounded subset $S \subset \mathbb{C}$, we have a positive real number $R$ such that $|z| \leqslant R$. For all sufficiently large $n$, we have $n > 2R \geqslant 2|z| \implies |z|/n \leqslant 1/2$ for all $z \in S.$

Hence,

$$\left|n\log(1+z/n) - z\right| \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{|z|}{3n} + \frac{|z|^2}{4n^2} + \ldots\right] \\ \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 2^2} + \ldots\right] \\ \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\right] \\ = \frac{|z|^2}{n} \\ \leqslant \frac{R^2}{n}.$$

Thus,

$$\lim_{n \to \infty} n\log(1+z/n) = \lim_{n \to \infty} \log(1+z/n)^n = z,$$

uniformly for $z \in S$.

The function $z \mapsto \exp(z)$ is uniformly continuous on the compact set $\{z : |z| \leqslant R \},$ and it follows that

$$\lim_{n \to \infty} (1+z/n)^n = e^z, $$

uniformly for $z \in S$.

$\endgroup$

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.