9.7 Representations of Functions as Power Series
Geometric Power Series
A power series representation for $f(x)=\frac{1}{1-x}$ can be obtained from the sum of a geometric series $\sum_{n=0}^{\infty} a r^n=\frac{a}{1-r}$, if you let $a=1$ and $r=x$.
Therefore $\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n=1+x+x^2+x^3+\cdots,|x|<1$.
Example1
- Find the first four nonzero terms and the general terms for the power series expansion of the function given by $f(x)=\frac{2}{1+x^2}$.
▶️Answer/Explanation
Solution
$f(x)=\frac{2}{1+x^2}=\frac{2}{1-\left(-x^2\right)}$ is a geometric series with $a=2$ and $r=-x^2$
Therefore, $f(x)=2-2 x^2+2 x^4-2 x^6+\cdots+(-1)^n 2 x^{2 n}+\cdots$.
Example 2
- Find the first four nonzero terms and the general terms for the power series expansion of the function given by $f(x)=\frac{x^2}{1+x}$.
▶️Answer/Explanation
Solution
$f(x)=\frac{x^2}{1+x}=\frac{x^2}{1-(-x)}$ is a geometric series with $a=x^2$ and $r=-x$
Therefore, $f(x)=x^2-x^3+x^4-x^5+\cdots+(-1)^n x^{n+2}+\cdots$.
Differentiation and Integration of Power Series
If the function given by
$
\begin{aligned}
f(x) & =\sum_{n=0}^{\infty} a_n(x-c)^n \\
& =a_0+a_1(x-c)+a_2(x-c)^2+\cdots+a_n(x-c)^n+\cdots
\end{aligned}
$
is differentiable, the derivative and antiderivative of $f$ are as follows.
1. $f^{\prime}(x)=\sum_{n=0}^{\infty} n a_n(x-c)^{n-1}$
$
=a_1+2 a_2(x-c)+3 a_3(x-c)^2+\cdots
$
2. $\int f(x)=C+\sum_{n=0}^{\infty} a_n \frac{(x-c)^{n+1}}{n+1}$
$
=C+a_0(x-c)+a_1 \frac{(x-c)^2}{2}+a_2 \frac{(x-c)^3}{3}+\cdots
$
The radius of convergence of the series obtained by differentiating or integrating a power series is the same as that of the original power series. However the interval of convergence may differ as a result of the behavior at the endpoints.
Example 3
- Let $f$ be a function given by $f(x)=\sum_{n=1}^{\infty} \frac{(-1)^n(x-2)^n}{n}$. Find the interval of convergence for each of the following.
(a) $f(x)$
(b) $f^{\prime}(x)$
(c) $\int f(x) d x$
▶️Answer/Explanation
Solution
(a)
$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim _{n \rightarrow \infty}\left|\frac{(-1)^{n+1}(x-2)^{n+1}}{n+1} \cdot \frac{n}{(-1)^n(x-2)^n}\right| \\
& =\lim _{n \rightarrow \infty}\left|(x-2) \frac{n}{n+1}\right|=|x-2|
\end{aligned}
$
By the Ratio Test, the series converges if $|x-2|<1$.
$
|x-2|<1 \Rightarrow 1<x<3 \text {. }
$
Now check the endpoints.
If $x=1, f(x)=\frac{(-1)^n(1-2)^n}{n}=\frac{1}{n}$, which diverges by the $p$-Series Test.
If $x=3, f(x)=\frac{(-1)^n(3-2)^n}{n}=\frac{(-1)^n}{n}$, which converges by the Alternating Series Test.
The interval of convergence is $(1,3]$.
(b)
$
\begin{aligned}
& f^{\prime}(x)=\sum_{n=1}^{\infty} \frac{(-1)^n n(x-2)^{n-1}}{n}=\sum_{n=1}^{\infty}(-1)^n(x-2)^{n-1} \\
& \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim _{n \rightarrow \infty}\left|\frac{(-1)^{n+1}(x-2)^n}{(-1)^n(x-2)^{n-1}}\right|=|x-2| \\
& |x-2|<1 \Rightarrow 1<x<3 . \text { Check endpoints. }
\end{aligned}
$
If $x=1, f^{\prime}(x)=\sum_{n=1}^{\infty}(-1)^n(1-2)^{n-1}=\sum_{n=1}^{\infty}(-1)^{2 n-1}$, which diverges by the $n$ th-Term Test.
If $x=3, f^{\prime}(x)=\sum_{n=1}^{\infty}(-1)^n(3-2)^{n-1}=\sum_{n=1}^{\infty}(-1)^n$, which also diverges by the $n$ th-Term Test.
The interval of convergence is $(1,3)$.
(c)
$
\begin{aligned}
& \int f(x) d x=C+\sum_{n=1}^{\infty} \frac{(-1)^n(x-2)^{n+1}}{n(n+1)} \\
& \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim _{n \rightarrow \infty}\left|\frac{(-1)^{n+1}(x-2)^{n+2}}{(n+1)(n+2)} \cdot \frac{n(n+1)}{(-1)^n(x-2)^{n+1}}\right| \\
& =\lim _{n \rightarrow \infty}\left|\frac{n(x-2)}{n+2}\right|=|x-2| \\
& |x-2|<1 \Rightarrow 1<x<3 . \text { Check endpoints. }
\end{aligned}
$
If $x=1, \int f(x) d x=C+\sum_{n=1}^{\infty} \frac{(-1)^n(1-2)^{n+1}}{n(n+1)}=C+\sum_{n=1}^{\infty} \frac{(-1)^{2 n+1}}{n(n+1)}$,
which convergent because $\frac{1}{n(n+1)}<\frac{1}{n^2}$.
If $x=3, \int f(x) d x=C+\sum_{n=1}^{\infty} \frac{(-1)^n(3-2)^{n+1}}{n(n+1)}=C+\sum_{n=1}^{\infty} \frac{(-1)^n}{n(n+1)}$, which also converges by the Alternating Series Test.
The interval of convergence is $[1,3]$.
Example4
- The power series expansion for $\frac{1}{1-x}$ is $\sum_{n=0}^{\infty} x^n$. Which of the following is a power series expansion for $\frac{1}{1+x^3} ?$
(A) $1+x^2+x^4+x^6+\cdots$
(B) $1-x^3+x^6-x^9+\cdots$
(C) $1+\frac{x^3}{3}+\frac{x^6}{6}+\frac{x^9}{9}+\cdots$
(D) $1-\frac{x^3}{3}+\frac{x^6}{6}-\frac{x^9}{9}+\cdots$
▶️Answer/Explanation
Ans:B
Example5.
The power series expansion for $\frac{1}{1-x}$ is $\sum_{n=0}^{\infty} x^n$. Which of the following is a power series expansion for $\frac{1}{2-x}$ ?
(A) $1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}+\cdots$
(B) $1-\frac{x}{2}+\frac{x^2}{4}-\frac{x^3}{8}+\cdots$
(C) $\frac{1}{2}+\frac{x}{4}+\frac{x^2}{8}+\frac{x^3}{16}+\cdots$
(D) $\frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+\cdots$
▶️Answer/Explanation
Ans:C
Example6.
- If $f(x)=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{(x-2)^n}{n !}=(x-2)-\frac{(x-2)^2}{2 !}+\frac{(x-2)^3}{3 !}-\frac{(x-2)^4}{4 !}+\cdots$, which of the following represents $f^{\prime}(x) ?$
(A) $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{(x-2)^{n-1}}{n !}$
(B) $\sum_{n=1}^{\infty}(-1)^n \frac{(x-2)^{n-1}}{(n+1) !}$
(C) $\sum_{n=0}^{\infty}(-1)^{n+1} \frac{(x-2)^{n-1}}{n !}$
(D) $\sum_{n=0}^{\infty}(-1)^n \frac{(x-2)^n}{n !}$
▶️Answer/Explanation
Ans:D