Objectives
-
Learn the definition of a power series.
Section 1.5 Power Series
In this section, we will expand the idea of a series from Section 1.4 to that as a function that is written as a series. The simplest one is that of a power series in which each term is a power function.
Subsection 1.5.1 Power Series
Definition 1.5.1.
A power series is a series of the form:
\begin{equation*}
\sum_{n=0}^{\infty}c_{n} x^{n} = c_{0} + c_{1} x + c_{2} x^{2} + \cdots
\end{equation*}
where \(x\) is a variable and the \(c_{n}\)’s are constants called coefficients.
A power series is a function of \(x\) and resembles a polynomial with an infinite number of terms.
Definition 1.5.2.
The series:
\begin{equation*}
\sum_{n=0}^{\infty}c_{n} (x-a)^{n} = c_{0} + c_{1} (x-a) + c_{2} (x-a)^{2} + \cdots
\end{equation*}
is called a power series center at \(x=a\).
Example 1.5.3.
Consider the power series
\begin{equation*}
\sum_{n=0}^{\infty}x^{n} = 1+x+x^{2}+x^{3}+ x^{4} + \cdots
\end{equation*}
This is a geometric series with \(r=x\) and \(a=1\text{.}\) The series converges for \(|x| \lt 1\) and diverges for \(|x|\geq 1\text{.}\)
From what we know about geometric series, when the series converges, it converges to
\begin{equation*}
\frac{1}{1-x}
\end{equation*}
Example 1.5.4.
Find the center of
\begin{equation*}
\sum_{n=0}^{\infty}\frac{x^{n}}{n!}
\end{equation*}
and all values of \(x\) for which the series converges.
Solution.
First, the center is \(a=0\text{.}\) To find the values of \(x\) of convergence, we will use RATFACE and
\begin{equation*}
\lim_{n \rightarrow \infty}\biggl\vert \frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^{n}}{n!}}\biggr\vert = \lim_{n \rightarrow \infty}\frac{|x|}{n+1}=0
\end{equation*}
and since \(L \lt 1\) and does not depend on \(x\text{,}\) this series converges for all \(x\text{.}\)
Example 1.5.5.
Find the domain for the power series:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{(x-1)^{n}}{3^{n}\sqrt{n}}
\end{equation*}
Solution.
Find the domain is identical to find all values of \(x\) for which the series converges. Let’s use RATFACE again.
\begin{equation*}
\lim_{n \rightarrow \infty}\biggl\vert \frac{\frac{(x-1)^{n+1}}{3^{n+1}\sqrt{n+1}}}{\frac{(x-1)^{n}}{3^{n}\sqrt{n}}}\biggr\vert = \lim_{n \rightarrow \infty}\frac{|x-1|}{3}\frac{\sqrt{n}}{\sqrt{n+1}}= \frac{|x-1| }{3}
\end{equation*}
And this value is less than 1 (and thus converges) when \(|x-1| \lt 3\) or \(-2 \lt x \lt 4\) and diverges when \(|x-1|\gt 4\) or \(x \lt -2\) and \(x \gt 4\text{.}\) This test is inconclusive when \(|x-1|=1\) or at \(x=-2\) and \(x=4\text{.}\) We will thus need to test each of those separately.
- \(x=-2\)
- Plug this value into the series to get\begin{equation*} \sum_{n=1}^{\infty}\frac{(-3)^{n}}{3^{n}\sqrt{n}}= \sum_{n=0}^{\infty}\frac{(-1)^{n}}{\sqrt{n}} \end{equation*}and this is an alternating series.We will use the Alternating Series Test (Theorem 1.4.9) with \(b_{n}=1/\sqrt{n}\) and this is positive, decreasing and approaches zero, so it converges.\(x=4\)
- \(x=4\)
- Plug this value into the series to get:\begin{equation*} \sum_{n=1}^{\infty}\frac{(3)^{n}}{3^{n}\sqrt{n}} = \sum_{n=0}^{\infty}\frac{1}{\sqrt{n}} \end{equation*}which diverges as it is a \(p\)-series (see Theorem thm:p:test) with \(p=1/2\text{.}\) Thus the domain of this function is \(-2 \leq x \lt 4\text{.}\)
Theorem 1.5.6.
For a given power series
\begin{equation*}
\sum_{n=0}^{\infty}c_{n}(x-a)^{n}
\end{equation*}
there are only three possibilities:
-
The series converges only when \(x=a\text{.}\)
-
The series converges for all \(x\text{.}\)
-
There is a positive number \(R\) such that the series converges if \(|x-a| \lt R\) and diverges for \(|x-a| \gt R\text{.}\)
We call the number \(R\) the radius of convergence and in the 2nd and 3rd cases, the interval of convergence.The center of the interval is always the point \(x=a\) and the interval stretches from \(a-R\) to \(a+R\text{.}\)
Note that the theorem says nothing in the 3rd case about the points \(x=a+R\) and \(x=a-R\text{.}\) These are the endpoints of the interval, and must be tested for individually.
Example 1.5.7.
List the center, the radius of convergence and the interval of convergence of the series: \(\displaystyle \sum_{n=0}^{\infty}x^{n}\text{,}\) \(\displaystyle \sum_{n=1}^{\infty}\frac{(x-1)^{n}}{3^{n}\sqrt{n}}\) and \(\displaystyle \sum_{n=0}^{\infty}\frac{x^{n}}{n!}\)
Solution.
\begin{equation*}
\sum_{n=0}^{\infty}x^{n}
\end{equation*}
In this case, above, we found that the series converges for \(|x| \lt 1\) and diverges for \(|x| \geq 1\text{.}\) Thus the radius of convergence is 1 and the interval of convergence is \((-1,1)\text{.}\)
\begin{equation*}
\sum_{n=0}^{\infty}\frac{x^{n}}{n!}
\end{equation*}
In this case, we found that it converged for all \(x\text{,}\) so the radius of convergence is \(R=\infty\) and the interval of convergence is \((-\infty, \infty)\text{.}\)
\begin{equation*}
\sum_{n=1}^{\infty}\frac{(x-1)^{n}}{3^{n}\sqrt{n}}
\end{equation*}
In this case we found that the radius of convergence was \(R=3\) and the interval of convergence was \([-2,4)\text{.}\)
Subsection 1.5.2 Representations of Functions as Power Series
Power series are extremely important in many ways.
-
How do you evaluate integrals that are either extremely difficult to find an antiderivative or one may not exist?
-
Some differential equations are very difficult to solve. One method to solve them is to use power series.
In short, power series represent functions that we may need to approximate or can’t find another representation.
We saw the power series
\begin{equation*}
\frac{1}{1-x} = \sum_{n=0}^{\infty}x^{n}
\end{equation*}
in the previous section. Recall that it is a geometric series that converges when \(|x| \lt 1\text{.}\)
Example 1.5.8.
Express \(1/(1+x^{2})\) as a power series using the power series above.
Example 1.5.9.
Create the power series of the function \(f(x)=\frac{x^{3}}{4+x}\) using the power series of \(\frac{1}{1-x}\text{.}\) What is its radius of convergence?
Solution.
If we write it as
\begin{equation*}
\begin{aligned} \frac{x^{3}}{4+x} \amp = \frac{x^{3}}{4(1+x/4)}\\\amp= \frac{x^{3}}{4}\cdot \frac{1}{1+x/4} \end{aligned}
\end{equation*}
now replace \(x\) in the power series for \(1/(1-x)\) with \(-x/4\text{,}\) to get
\begin{equation*}
\begin{aligned} \frac{x^{3}}{4+x} \amp = \frac{x^{3}}{4}\sum_{n=0}^{\infty}\biggl(-\frac{x}{4}\biggr)^{n} \\\amp= \frac{x^{3}}{4}\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n}}{4^{n}}\\\amp= \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n+3}}{4^{n+1}} \end{aligned}
\end{equation*}
The radius of convergence could be found using RATFACE, however, it is easier than that.
Recall that the radius of convergence for \(1/(1-x)\) is \(R=1\text{,}\) since this required that \(|x| \lt 1\text{.}\) Since we replaced \(x\) with \(-x/4\text{,}\) then the series converges when \(|x/4| \lt 1\) so the radius is \(R=4\text{.}\)
Subsubsection 1.5.2.1 Differentiation and Integration of Power Series
Theorem 1.5.10.
If a power series of the form \(\sum c_{n} (x-a)^{n}\) has a radius of convergence of \(R\text{,}\) then the function \(f\) defined as
\begin{equation*}
f(x)= \sum_{n=0}^{\infty}c_{n} (x-a)^{n}
\end{equation*}
is differentiable on the interval \((a-R,a+R)\) and
\begin{equation*}
f'(x) = \sum_{n=1}^{\infty}n c_{n} (x-a)^{n-1}
\end{equation*}
and
\begin{equation*}
\int f(x) \, dx = C+\sum_{n=0}^{\infty}\frac{c_{n}}{n+1}(x-a)^{n+1}
\end{equation*}
both have a radius of convergence \(R\text{.}\) Also, the derivative and the integral of \(f\) is performed term by term.
Note: if \(f\) has a particular interval of convergence, the interval of convergence for \(f'(x)\) and \(\int f \, dx\) may change, however only at the endpoints.
Example 1.5.11.
Solution.
Note that
\begin{equation*}
\begin{aligned} \frac{1}{(1-x)^{2}}\amp= - \frac{d}{dx}\frac{1}{1-x}\\\amp= - \frac{d}{dx}\sum_{n=0}^{\infty}x^{n} \\\amp= - \sum_{n=1}^{\infty}n x^{n-1}\\\amp= \sum_{n=1}^{\infty}(-n) x^{n-1} \end{aligned}
\end{equation*}
And the radius of convergence of this series is \(R=1\text{.}\)
Example 1.5.12.
Solution.
Recall that from the last chapter we found that
\begin{equation*}
\frac{1}{1+x^{2}} = \sum_{n=0}^{\infty}(-1)^{n} x^{2n}.
\end{equation*}
Now, we’ll integrate both sides of this:
\begin{equation*}
\begin{aligned} \tan^{-1}x\amp= \int \frac{1}{1+x^{2}}\, dx \\\amp= \int \sum_{n=0}^{\infty}(-1)^{n} x^{2n}\, dx \\\amp= C + \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n+1}}{2n+1} \end{aligned}
\end{equation*}
to determine the value of \(C\text{,}\) we will substitute in \(x=0\text{:}\)
\begin{equation*}
\tan^{-1}(0) = C = 0
\end{equation*}
so this leads to the power series for arctangent:
\begin{equation*}
\tan^{-1}x = \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n+1}}{2n+1}.
\end{equation*}
Example 1.5.13.
Example 1.5.14.
Use power series to approximate:
\begin{equation*}
\int_{0}^{1/2}\frac{1}{1+x^{5}}\, dx
\end{equation*}
Solution.
If you use a CAS (like Maple) to find the antiderivative of this function, then you get a solution that is a mess. Instead, we’ll use the power series of \(1/(1+x^{5})\) to find the value of the integral to 16 digits.
Since
\begin{equation*}
\frac{1}{1-x} = \sum_{n=0}^{\infty}x^{n}
\end{equation*}
then if we replace \(x\) with \(-x^{5}\text{,}\) we get:
\begin{equation*}
\begin{aligned} \frac{1}{1+x^{5}}\amp= \sum_{n=0}^{\infty}(-x^{5})^{n} \\\amp= \sum_{n=0}^{\infty}(-1)^{n} x^{5n} \end{aligned}
\end{equation*}
Next, if we integrate this:
\begin{equation*}
\begin{aligned} \int_{0}^{1/2}\frac{1}{1+x^{5}}\, dx\amp= \int_{0}^{1/2}\sum_{n=0}^{\infty}(-1)^{n} x^{5n}\, dx \\\amp= \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{5n+1}}{5n+1}\biggr\vert_{0}^{1/2}\\\amp= \sum_{n=0}^{\infty}(-1)^{n} \frac{(1/2)^{5n+1}}{5n+1} \end{aligned}
\end{equation*}
To find an approximation, we’ll use a partial sum instead of the full series. By the Alternating Series Remainder Theorem, the remainder is bounded by the \(b_{n+1}\) term, that is
\begin{equation*}
\frac{(1/2)^{5n+6}}{5n+6} \lt 10^{-16}
\end{equation*}
If we substitute in values of \(n\) until this is reached, then we get \(n=9\text{,}\) so if we find \(s_{9}\) using the series above, then we will successfully find the appropriate approximation.
\begin{equation*}
\begin{aligned} \int_{0}^{1/2}\frac{1}{1+x^{5}}\,dx\amp\approx s_{9} = \sum_{n=0}^{9}(-1)^{n} \frac{(1/2)^{5n+1}}{5n+1}\\\amp= 0.49743929101160003470 \end{aligned}
\end{equation*}
Example 1.5.15.
Show that
\begin{equation*}
y(x)= \sum_{n=0}^{\infty}\frac{x^{n}}{n!}
\end{equation*}
satisfies the differential equation
\begin{equation*}
y'(x) = y(x).
\end{equation*}
Solution.
\begin{equation*}
\begin{aligned} y'(x)\amp= \sum_{n=1}^{\infty}n \frac{x^{n-1}}{n!}= \sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!}\\\amp= \sum_{k=0}^{\infty}\frac{x^{k}}{k!}\\\amp= \sum_{n=0}^{\infty}\frac{x^{n}}{n!}\\\amp= y(x) \end{aligned}
\end{equation*}
so the differential equation is satisfied.
