Example 5.52.
By Proposition 5.33, the power series
\begin{equation}
\sum_{n=0}^{\infty}
x^n = 1 + x+ x^2 + \cdots,\tag{5.3}
\end{equation}
(centered at \(0\)) represents the function \(\frac{1}{1-x}\) on \(x \in (-1,1)\text{.}\)
Section 5.5 Power Series
A power series in \(x\) about a point \(x_0\) is a series of the form
\begin{equation*}
\sum_{n=0}^{\infty} a_n(x-x_0)^n = a_0 + a_1(x-x_0) + a_2(x-x_0)^2
+ \cdots
\end{equation*}
The point \(x_0\) is called the center of the power series.
We may ask whether a power series in \(x\) is convergent at various values of \(x\text{.}\) A power series always converges at its center. So, by a convergent power series we mean a power series that converges at some point other than its center. We say that a convergent power series represents a function \(f\) on a set if it converges to the value of \(f\) at every point of the set.
Example 5.53.
Consider the power series
\begin{equation}
\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = x + \frac{x^2}{2} +
\frac{x^3}{3} + \cdots\tag{5.4}
\end{equation}
Since \(\frac{|x^{n+2}/(n+2)|}{|x^{n+1}/(n+1)|} =
\frac{n+1}{n+2}|x| \to |x|\text{,}\) as \(n\to \infty\text{.}\) Therefore, according to the ratio test, the power series converges when \(|x|
\lt 1\text{,}\) i.e. \(x \in (-1,1)\) and diverges when \(|x| \gt
1\text{.}\) When \(|x|=1\text{,}\) i.e. when \(x=\pm 1\text{,}\) the ratio test is inconclusive. We need further investigation to decide the convergence at those points.
When \(x = 1\text{,}\) the power series turns into the Harmonic Series
\begin{equation*}
\sum_{n=0}^{\infty} \frac{1}{n+1} = 1+ \frac{1}{2} + \frac{1}{3}
+ \cdots
\end{equation*}
and is divergent.
On the other hand, when \(x=-1\text{,}\) then the power series turns into the alternating series
\begin{equation*}
\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n+1} = 1- \frac{1}{2}
+ \frac{1}{3} - \cdots
\end{equation*}
and is convergent by the alternating series test. And we conclude that power series \(\sum_{n=0}^{\infty}
\frac{x^{n+1}}{n+1}\) converges on the interval \([-1,1)\text{.}\)
Example 5.54.
Consider the power series
\begin{equation}
\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} +
\frac{x^3}{3!} + \cdots\tag{5.5}
\end{equation}
Since \(\frac{|x^{n+1}/(n+1)!|}{|x^n/n!|} = |x|\frac{n!}{(n+1)!}
= \frac{|x|}{n+1} \to 0\) for any fixed \(x\text{,}\) the power series converges for all real numbers. We will see that the function represented by this power series is the exponential function \(e^x\text{.}\)
Example 5.55.
Consider the power series
\begin{equation}
\sum_{n=0}^{\infty} n!{x^n} = 1 + x + 2!{x^2} + 3!{x^3} + \cdots\tag{5.6}
\end{equation}
Since \(\frac{|(n+1)!x^{n+1}|}{|n!x^n|} = (n+1)|x| \to +\infty\) for any fixed \(x \neq 0\text{,}\) the power series is divergent.
More generally, we have the following key statement about convergence of power series:
Theorem 5.56.
For any sequence \((a_n)\text{,}\) there is an \(0 \le R \le \infty\) so that the power series \(\sum_{n=0}^{\infty} a_n(x-x_0)^n\text{,}\)
- converges absolutely on \(|x-x_0| \lt R\) (i.e. on \((x_0-R,x_0+R)\text{.}\)
- diverges on \(|x-x_0| \gt R\)
Remark 5.57.
- The \(R\) appears in the theorem is called the radius of convergence of the power series . When \(R = \infty\text{,}\) the power series converges everywhere. When \(R = 0\text{,}\) the power series is divergent.
- When \(\lim |a_{n+1}|/|a_n|\) exists or is \(+\infty\text{,}\) it is \(1/R\) (understood with the convention \(1/0 = \infty\) and \(1/\infty = 0\)). Same conclusion if the limit is replaced by \(\lim|a_n|^{1/n}\text{.}\)
- The convergence of the power series at the points \(x_0-R, x_0+R\) needs to be determine on a case-by-case bases (see Example 5.53). The interval on which the power series converges is called its interval of convergence.
Example 5.58.
To find the interval of convergence of the power series
\begin{equation*}
\sum_{n=1}^{\infty} \frac{8^n(x-5)^n}{n+5},
\end{equation*}
first note that the center of this power series is \(5\text{.}\) Note that the coefficient of \((x-5)^n\) is \(a_n = 8^n/(n+5)\text{.}\) We have \((n+5)^{1/n} \to 1\) as \(n \to \infty\) (see Exercise 5.7.4), so
\begin{equation*}
|a_n|^{1/n} = 8/(n+5)^{1/n} \to 8
\end{equation*}
and hence the \(R\text{,}\) the radius of convergence is \(1/8\text{.}\) At the left endpoint \(x=5-1/8=39/8\text{,}\) the alternating series test applies and the power series is convergent. At the right endpoint \(x=5+1/8 =
41/8\text{,}\) the power series becomes \(\sum 1/(n+5)\text{.}\) By comparing it with the Harmonic series using the limit comparison test, we see that the series is divergent as well. Hence the interval of convergence of the power series is \([39/8, 41/8)\text{.}\)
Within the interval of convergence the function represented by a power series behaves nicely. This is the key reason why power series is a powerful tool in studying functions.
Theorem 5.59.
Let \(\sum_{n=0}^{\infty} a_n(x-x_0)^n\) be a power series with radius of convergence \(R \gt 0\text{.}\) Then on the open interval \((x_0-R,
x_0 +R)\) the function \(f(x) = \sum a_n(x-x_0)^n\) is both differentiable and has an anti-derivative. Moreover,
- \(\displaystyle f'(x) = \sum_{n=1}^{\infty} na_n (x-x_0)^{n-1}.\)
- \(\displaystyle \int_{x_0}^x f(t)dt = \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-x_0)^{n+1}.\)
and these power series all have radius of convergence \(R\text{.}\) Also, if the power series is convergent at an end-point, i.e. \(x_0+R\) or \(x_0-R\text{,}\) then \(f(x)\) is continuous at that end-point.
Remark 5.60.
- A function \(f\) is analytic at a point \(x_0\) in its domain if it is represented by some convergent power series with center at \(x_0\text{.}\) A function is analytic on a set \(D\) if if it analytic at every point in \(D\text{.}\) Most functions studied in a Calculus course are analytic on some open interval and hence can be studied via power series.
- The theorem above guarantees that an analytic function is both differentiable and possesses an anti-derivative. Moreover, both can be found easily---on a term-by-term basis. In fact, by repeat application of the theorem, we see that an analytic function has derivative of all orders.
- Suppose a function \(f(x)\) is analytic at \(x_0\text{,}\) i.e.\begin{equation*} f(x)=\sum_{n=0}^{\infty} a_n(x-x_0)^n =a_0 + a_1(x-x_0) + a_2(x-x_0)^2 + a_3(x-x_0)^3 + \cdots \end{equation*}on some open interval \(I\) containing \(x_0\text{.}\) Then \(a_0 = f(x_0)\text{.}\) Since, according to the theorem above,\begin{equation*} f'(x) = a_1 + 2a_2(x-x_0) + 3a_3(x-x_0)^2 + 4a_4(x-x_0)^2 + \cdots \end{equation*}\(a_1 = f'(x_0)\text{.}\) Differentiating one more time and set \(x=x_0\) yields \(2 a_2 = f''(x_0)\text{.}\) By repeat differentiation, we see that\begin{equation*} f^{(n)}(x) = n(n-1)(n-2)\cdots 1 a_n + \text{terms with factor} (x-x_0). \end{equation*}Thus, \(f^{(n)}(x_0) = n!a_n\text{.}\) In other words, \(a_n = \dfrac{f^{(n)}(x_0)}{n!}\text{.}\) This tells us that the coefficients \(a_n\)’s are completely determined by \(f\text{.}\) In other words, if \(f\) is analytic at \(x_0\) then it is represented by a unique power series, namely\begin{equation*} f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \end{equation*}on an open interval containing \(x_0\text{.}\) This series is called the Taylor series of \(f\) at \(x_0\text{.}\)
- Two things to keep in mind. The Taylor series of \(f\) at \(x_0\) exists if \(f\) is smooth (i.e. having derivatives of all orders) at \(x_0\text{.}\) However, 1) the Taylor series maybe divergent and 2) even if the Taylor series is convergent there is no guarantee that it represent \(f\) on any open interval containing \(x_0\text{.}\) In other words, a function being smooth at a point does not imply it is analytic at that point.
We often invoke the uniqueness to find the series that represents an analytic function instead of computing the coefficients by differentiation. Let us illustrate this by an example.
Example 5.61.
Let us show that \(\dfrac{1}{1+x^2} \) is analytic at \(0\) and find its power series representation. Recall that
\begin{equation*}
\frac{1}{1-x} = 1 + x + x^2 + \cdots
\end{equation*}
for all \(x\) with \(|x| \lt 1\text{.}\) We conclude that
\begin{align*}
\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} & = 1 + (-x^2) +
(-x^2)^2 +
\cdots\\
& = 1-x^2 + x^4 - \cdots = \sum_{n=0}^{\infty} (-1)^nx^{2n}
\end{align*}
for all \(x\) with \(|(-x^2)| = |x|^2 \lt 1\text{,}\) i.e. with \(|x|
\lt 1\text{.}\)
Now with the help of Theorem 5.59 we can find the representation of \(\arctan(x)\)
Example 5.62.
\begin{align*}
\arctan(x) & = \int_0^x \frac{1}{1+t^2} dt = \int_0^x (1 -
t^2 + t^4 - \cdots) dt\\
& = \int_0^{x} \sum_{n = 0}^{\infty} (-1)^{n} t^{2n} dt =
\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}\\
& = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots
\end{align*}
for all \(|x| \lt 1\text{.}\)
Note that by the alternating series test, the power series converges at \(\pm 1\text{.}\) So by continuity, we have
\begin{equation*}
\frac{\pi}{4} = \arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} -
\frac{1}{7} + \cdots
\end{equation*}
This gives us a way to estimate \(\pi\text{.}\)
Here is an explanation of this equality without using Calculus.Example 5.63.
We have
\begin{equation*}
\frac{1}{1+x} = \frac{1}{1-(-x)} = \sum_{n=0}^{\infty} (-1)^n x^n =
1-x + x^2 - x^3 + \ldots
\end{equation*}
on \(|-x| = |x| \lt 1\text{.}\) By Theorem 5.59, we have
\begin{equation*}
\ln(1+x) = \int_0^x \frac{1}{1+t} dt = \sum_{n=0}^{\infty} (-1)^n
\frac{x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}
+ \ldots
\end{equation*}
The series converges at \(x=1\) by the alternating series test and diverges to \(-\infty\) at \(x=-1\) by the \(p\)-test. So, we have
\begin{equation*}
\ln(2) = \ln(1+1) = 1-\frac{1}{2} +\frac{1}{3} - \frac{1}{4} +
\cdots
\end{equation*}
Example 5.64.
Later we will show that for any \(x \in \mathbb{R}\)
\begin{equation*}
\sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} =
x-\frac{x^3}{3!} +\frac{x^5}{5} - \frac{x^7}{7!} + \cdots
\end{equation*}
Thus, by Theorem 5.59 again, we have for any \(x
\in \mathbb{R}\)
\begin{equation*}
\cos{x} = \frac{d}{dx}\sin(x) = \sum_{n=0}^{\infty} (-1)^n
\frac{x^{2n}}{(2n)!} = 1- \frac{x^2}{2!} + \frac{x^4}{4!} -
\frac{x^6}{6!} + \cdots
\end{equation*}
Next we show how to find the power series expansions for various analytic functions from a few fundamental ones (Appendix C).
Example 5.65.
Let us find the power series expansion of \(\ln(x)\) at \(x=5\) That is, to find the coefficients \(c_n (n \ge 0)\) so that
\begin{equation*}
\ln(x) = c_0 + c_1(x-5) + c_2(x-5)^2 + \cdots = \sum_{n=0}^\infty
c_n(x-5)^n
\end{equation*}
Since the power series expansions that we know are all about \(0\text{,}\) first let us translate the center from \(5\) to \(0\text{.}\) We achieve this by making the substitution \(x+5\text{:}\)
\begin{equation*}
\ln(x+5) = \sum_{n=0}^\infty c_n(x+5 -5)^n = \sum_{n=0}^{\infty}c_nx^n
\end{equation*}
So the task is now to find the power series expansion of \(\ln(x+5)\) about \(x=0\text{.}\) To do that we use a property of \(\ln(x)\) and write
\begin{equation*}
\ln(x+5) = \ln(5(1+x/5)) = \ln(5)+\ln(1+x/5)
\end{equation*}
Next, substituting \(x\) by \(x/5\) in the expansion of \(\ln(1+x)\text{,}\) we get for \(-1 \lt x/5 \le 1\text{,}\) i.e. \(-5 \lt x \le 5\text{:}\)
\begin{align*}
\ln(x+5) & = \ln(5) + \ln(1+x/5) \\
& = \ln(5) + \sum_{n=0}^{\infty} (-1)^{n}
\frac{(x/5)^{n+1}}{n+1}\\
& = \ln(5) + \sum_{n=0} \frac{(-1)^n x^{n+1}}{5^{n+1}(n+1)}
\end{align*}
So substituting \(x\) by \(x-5\) into the above, we get for \(-5
\lt x-5 \le 5\text{,}\) i.e. for \(0 \lt x \le 10\text{:}\)
\begin{align*}
\ln(x) & = \ln(5) + \sum_{n=0}^{\infty} \frac{(-1)^n}{5^{n+1}(n+1)}(x-5)^{n+1} \\
& = \ln(5) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n5^n}(x-5)^n
\end{align*}
That is \(c_0 = \ln(5)\) and \(c_n = (-1)^{n-1}/(n5^n)\) for \(n
\ge 1\text{.}\)
Checkpoint 5.66.
Find the power series expansions of \(f(x)=\dfrac{3}{5+x}\) about \(0\text{.}\)
Hint.
Note that
\begin{equation*}
f(x) = \frac{3}{5}\frac{1}{1+x/5}.
\end{equation*}
Then substitute \(u=-x/5\) into the expansion of \(1/(1-u)\text{.}\)
Checkpoint 5.67.
Find the power series expansions of \(\cos(x)\) at \(x_0=\pi/2\text{.}\)
Hint.
The task is to find the coefficients \(a_n\) in
\begin{equation*}
\cos(x) = \sum_{n=0}^{\infty} a_n(x-\pi/2)^n
\end{equation*}
Note that that
\begin{equation*}
\cos(x + \pi/2) = \sum_{n=0}^{\infty} a_n(x+\pi/2 - \pi/2)^n =
\sum_{n=0}^{\infty} a_n x^n
\end{equation*}
and \(\cos(x+\pi/2) = \cos(x)\cos(\pi/2)-\sin(x)\sin(\pi/2)
= -\sin(x)\text{.}\) Then use the power series expansion of \(sin(x)\) at \(x_0=0\) to find the \(a_n\)’s.
Example 5.68.
To find the power series expansion of \(f(x) = \dfrac{1}{(1-3x)^2}\) about \(0\text{,}\) first note that we can start with the one that we know: for \(|u| \lt 1\text{,}\)
\begin{equation*}
\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n = 1+u +u^2 + u^3 + \cdots
\end{equation*}
Differentiating both sides with respect to \(u\text{,}\) yields
\begin{equation*}
\frac{1}{(1-u)^2} = 1+ 2u + 3u^2 + \cdots = \sum_{n=1}^{\infty} nu^{n-1} =
\sum_{n=0}^{\infty} (n+1)u^n.
\end{equation*}
Substitute \(u\) by \(3x\text{,}\) we conclude that for \(|3x| \lt 1\text{,}\) i.e. for \(|x| \lt 1/3\text{,}\)
\begin{equation*}
\frac{1}{(1-3x)^2} = \sum_{n=0}^{\infty} (n+1)3^n x^n.
\end{equation*}
Subsection 5.5.1 Applications to summing series
The theory of power series can be used to extend the types of infinite sums that we can compute. Suppose a function \(f(x)\) is represented by a power series \(\ds \sum_{n=0}^{\infty} a_n x^n\) on \(|x| \lt R\) with \(R \gt 0\text{.}\) Then according to Theorem 5.59,
\begin{align*}
x\frac{d}{dx}f(x) & = x\frac{d}{dx}\sum_{n=0}^{\infty} a_n x^n\\
xf'(x) & = x\sum_{n=1}^{\infty} na_n x^{n-1} =
\sum_{n=0}^{\infty} na_n x^n.
\end{align*}
So the sum \(\sum_{n=0}^{\infty} na_n c^n\) is \(cf'(c)\) for \(|c|
\lt R\text{.}\) By iterating the application of \(\ds x\dfrac{d}{dx}\text{,}\) we can compute the sum of \(\sum_{n=0}^{\infty} n^ka_n c^n \) for any \(|c| \lt R\text{.}\)
Here are a few concrete examples.
Example 5.69.
To compute \(\sum_{n=0}^{\infty} \frac{n}{3^n}\text{,}\) note that it is the value of \(\sum_{n=0}^{\infty} nx^n\) at \(x = 1/3\text{.}\) Since
\begin{equation*}
\frac{1}{1-x} = \sum_{n=0}^\infty x^n \quad |x| \lt 1.
\end{equation*}
It follows from the discussion above that for \(|x| \lt 1\text{,}\)
\begin{equation*}
\sum_{n=0}^{\infty} nx^n = x\dfrac{d}{dx} \frac{1}{1-x} =
\frac{x}{(1-x)^2}.
\end{equation*}
Hence,
\begin{align*}
\sum_{n=0}^{\infty} \frac{n}{3^n} & = \left.
\frac{x}{(1-x)^2}\right|_{x=1/3} = \frac{1/3}{(1-1/3)^2} = \frac{3}{4}.
\end{align*}
Example 5.70.
Let us compute \(\sum_{n=0}^{\infty} \frac{n^2}{n!}\text{.}\) Since \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\) for all \(x \in
\mathbb{R}\text{.}\) Applying \(x\dfrac{d}{dx}\) twice to both sides yields,
\begin{align*}
\left( x\frac{d}{dx} \right)^2 e^x & = \left( x\frac{d}{dx}
\right)^2 \sum_{n=0}^{\infty} \frac{x^n}{n!} \\
x\frac{d}{dx}(xe^x) & = x\frac{d}{dx} \sum_{n=0}^{\infty}
\frac{nx^n}{n!}\\
(x^2 + x)e^x & = \sum_{n=0}^{\infty} \frac{n^2x^n}{n!}
\end{align*}
Evaluate both sides at \(x=1\) and we get \(\sum_{n=0}^{\infty}
\frac{n^2}{n!} = 2e.\)
It is easy to generalize the method in these examples to find the sum of series of the form \(\ds \sum_{n=0}^{\infty} \frac{P(n)}{b^n}\) or \(\ds
\sum_{n=0}^{\infty} \frac{P(n)}{n!}\) where \(P(x)\) is a polynomial in \(x\) and \(b \gt 1\text{.}\) In Exercise 5.7.8, we will look at a slightly different way of finding these sums.
