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THE GEOMETRY OF MATHEMATICAL METHODS

Section 13.10 Theorems about Power Series

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 13.2 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.

Example 13.9.

\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*}

Example 13.11.

\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?

Example 13.12.

\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 \(z=\frac{\pi}{2}\text{.}\)

Example 13.14.

\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?

Example 13.15.

\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 13.11. What can you conclude about the wisdom of assuming two series are the same if their first three terms are identical?

Example 13.17.

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.

Example 13.19.

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*}