About $e^{i z} = \cos z + i \sin z$ in Michael Spivak "Calculus 3rd Edition".

Question

I am reading "Calculus 3rd Edition" by Michael Spivak.

The author wrote as follows (p. 555):

Moreover, if we replace $z$ by $i z$ in the series for $e^z$, and make a rearrangement of the terms (justified by absolute convergence), something particularly interesting happens:

$$e^{i z} = 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots \\ =1 + iz - \frac{z^2}{2!} - \frac{i z^3}{3!} + \frac{z^4}{4!} + \frac{i z^5}{5!} + \cdots \\ = (1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots) + i (z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots),$$
so $$e^{i z} = \cos z + i \sin z.$$

But I think the author didn't use a rearrangement of the terms at all.

Am I right?

Let $$c_n := 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots + (-1)^n \frac{z^{2 n}}{(2 n)!},$$
$$s_n := z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots + (-1)^n \frac{z^{2 n + 1}}{(2 n + 1)!},$$
$$e_n := 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots + \frac{(iz)^n}{n!}.$$

Then, $$e_{2 n + 1} = c_n + i s_n,$$
$$\lim_{n \to \infty} e_{2 n + 1} = e^{i z},$$
$$\lim_{n \to \infty} c_n + i s_n = \lim_{n \to \infty} c_n + i \lim_{n \to \infty} s_n = \cos z + i \sin z,$$
so $$e^{i z} = \cos z + i \sin z.$$

Answer 1

Your statement:

But I think the author didn't use a rearrangement of the terms at all.

He did re-arrange the terms as said...See:

Answer 2

Technically a rearrangement of a series $\sum_{n=0}^\infty a_n$ is understood to be any series $\sum_{n=0}^\infty a_{f(n)}$ where $f : \mathbb N_0 \to \mathbb N_0$ is a bijection. Here $\mathbb N_0$ is the set of nonnegative integers.

In that sense no rearrangement of $e^{iz} = \sum_{n=0}^\infty \frac{(iz)^n}{n!}$ can produce the desired equation $e^{iz} = \cos z + i\sin z$. The "trick" is that for any strictly increasing $g : \mathbb N_0 \to \mathbb N_0$ we have $$\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty a'_n \text{ with } a'_n = \begin{cases} a_m & n = g(m) \\ 0 & n \notin g(\mathbb N_0) \end{cases}$$ Let us write $\sum_{n=0}^\infty a'_n = [\sum_{n=0}^\infty a_n]*g$. Applying this to $a_n = (-1)^n\frac{z^{2n}}{(2n)!}$ with $g(m) = 2m$ and $b_n = (-1)^n\frac{z^{2n+1}}{(2n+1)!}$ with $h(m) = 2m+1$ yields the result: We get $$\left[\sum_{n=0}^\infty (-1)^n\frac{z^{2n}}{(2n)!}\right]*g + i\left[\sum_{n=0}^\infty (-1)^n\frac{z^{2n+1}}{(2n+1)!}\right]*h = e^{iz} .$$

Answer 3

You are right that absolute convergence, while present as "security blanket", is not really used. The only re-arrangement that happens is that the exponential series is organized in groups of two subsequent terms, so that then the addition rule for convergent series can be applied:

$\sum_{n=0}^\infty a_n+\sum_{n=0}^\infty b_n=\sum_{n=0}^\infty(a_n+b_n)$, if two of the series converge, then so does the third.

This is also what you traced with the partial sums.

In this grouping procedure one reduces the sequence of partial sums of the exponential series to the sub-sequence of even indices. For this to be unproblematic requires that the terms of the series converge to zero, which for this power series is true for every $z$.