This theorem is an open invitation to collect a bag of cute tricks. It doesn't matter how you find a series for a function, once you have it, it is the series. The rest of the theorems in this section should be in your bag of cute tricks. Scientists use these theorems, rather that the method in Section 8.1 to find series expansions whenever possible. We give these theorems without proof, but include one or more examples of how each is used, in practice. It is worth taking the time to work carefully through each of these short examples.

\begin{align*} \sin z \amp = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!} \amp\forall z\\ \cos z = {d\over dz}\sin z \amp = \sum_{n=0}^{\infty} \frac{(-1)^n (2n+1) z^{2n}}{(2n+1)!} \nonumber\\ \amp = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} \amp\forall z \end{align*}
\begin{align*} \frac{1}{1+z} \amp = 1 - z + z^2 - z^3 + \dots = \sum_{n=0}^{\infty} (-z)^n \amp |z|\lt 1 \end{align*}
\begin{align*} \frac{1}{1+\sin z} \amp = 1 - \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!} + \left( \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!} \right)^2 \nonumber\\ \amp \qquad - \left( \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!} \right)^3 + \dots\\ \amp = 1 - z + z^2 + \left(\frac{1}{3!}-1\right)\, z^3 + \dots \amp |\sin z|\lt 1 \nonumber \end{align*}

What happens if you try this same trick to find a power series for $1/(1+\cos z)\text{?}$ Why?

\begin{align*} z-\frac{\pi}{2} \amp = z-\frac{\pi}{2}\amp\forall z \nonumber \end{align*}

Note: This is a very short power series with just two non-zero terms.

\begin{align*} \cos(z) \amp = -\sin (z-\frac{\pi}{2})\\ \amp = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} (z-\frac{\pi}{2})^{2n+1}} {(2n+1)!} \amp\forall z \nonumber \end{align*}

Note: Starting with a power series for $\sin(z)$ expanded around $z=0\text{,}$ we have obtained a power series for $\cos(z)$ expanded around $z=\frac{\pi}{2}\text{.}$

\begin{align*} \frac{2}{1-z^2} \amp = \frac{1}{1+z} + \frac{1}{1-z} \nonumber\\ \amp = (1 - z + z^2 -z^3 + \dots) + (1 + z + z^2 + z^3 + \dots) \nonumber\\ \amp = 2 (1 + z^2 + z^4 + \dots) \amp |z|\lt 1 \end{align*}

Compare this to the result you would get using the previous theorem. Which method is faster?

\begin{align*} \frac{\sin z}{1+z} \amp = \left( z - \frac{z^3}{3!} + \frac{z^5}{5!} + \dots \right) \left( 1 - z + z^2 - z^3 + z^4 - z^5 \right) \nonumber\\ \amp = z - z^2 + \left(-\frac{1}{3!}+1\right) z^3 + \left(\frac{1}{3!}-1\right) z^4 \nonumber\\ \amp \qquad + \left(\frac{1}{5!}-\frac{1}{3!}+1\right)z^5 +\dots \amp |z|\lt 1 \end{align*}

Compare this series to the series for the function $1-\frac{1}{1+\sin(z)}$ in Example 8.7.5. What can you conclude about the wisdom of assuming two series are the same if their first three terms are identical?

Try the previous example $\sin z/(1+z)$ using synthetic division, instead. Is this method easier or harder? Imagine what you would do if the denominator were a power series with an infinite number of non-zero terms.

Expand $\sin z$ around $z=\pi\text{.}$

\begin{align*} \sin z \amp = \sin[(z-\pi)+\pi] \nonumber\\ \amp = \sin(z-\pi)\cos\pi +\cos(z-\pi)\sin\pi \nonumber\\ \amp = -\sin(z-\pi) \nonumber\\ \amp = -\sum_{n=0}^{\infty} {(-1)^n (z-\pi)^{2n+1}\over (2n+1)!} \amp\forall z \end{align*}