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.
Theorem 13.8. Differentiation and Integration of Power Series.
A power series may be differentiated or integrated term by term. The resulting series converges to the derivative or integral of the function represented by the original series within the same circle of convergence as the original series.
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*}
Theorem 13.10. Substitution in Power Series.
One series may be substituted into another provided that the values of the substituted series are inside the circle of convergence of the other series.
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 \(z=\frac{\pi}{2}\text{.}\)
Theorem 13.13. Adding, Subtracting, and Multiplying Power Series.
Two power series of like powers may be added, subtracted, or multiplied. The resulting series converges at least within the common circle of convergence.
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?
Theorem 13.16. Dividing Power Series.
Two power series expanded around the same point may be divided. If the leading term(s) of the denominator series is not zero, or if the zero(s) is canceled by the numerator, then the resulting series converges within some circle. If the radius of convergence of the numerator and denominator series are \(r_1\) and \(r_2\text{,}\) respectively, and the distance from the origin of the circles to the nearest zero of the denominator series is \(s\text{,}\) then the quotient series converges at least inside the smallest of the three circles of radii \(r_1\text{,}\) \(r_2\text{,}\) and \(s\text{.}\)
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.
Theorem 13.18. Power Series Around Points Other than Zero.
The series expansions for most functions recorded in books are expansions around the point \(z=0\text{.}\) To expand around a point \(a\neq 0\) write every \(z\) which appears in the function as \((z-a) +a\text{,}\) simplify creatively, and use Theorem 13.10.
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*}
