Power series

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Infinite series can represent functions.

If we refuse to truncate a Taylor polynomial, and instead allow it to be a series (an infinite sum) we call it a power series.

A power series is a series of the form $\sum _{k=0}^\infty a_k(x-c)^k$ where the $a_k$ ’s are the coefficients and $c$ is the center.

Which of the following are power series functions?

$f(x) = 0$ $f(x) = -9$ $f(x) = 3x+1$ $f(x) = x^{1/2}-x +8$ $f(x) = -4x^{-3}+5x^{-1}+7-18x^2$ $f(x) = x^{-3}+x^{-2}+x^{-1}+1+x+x^2+x^3 +\cdots$ $f(x) = \frac {x^2 - 3x + 2}{x-2}$ $f(x) = x^7-32x^6-\pi x^3+45/84$ $f(x) = x^{10} + x^{20} + x^{30} + \cdots$

Every polynomial is a power series.

Here are four basic power series (centered at zero) that every mathematician knows.

$\begin{align*} e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots &|x|< \infty\\ \sin(x) &= x - \frac{x^3}{3!} + \frac{x^5}{5!} -\frac{x^7}{7!} + \cdots &|x|< \infty\\ \cos(x) &= 1-\frac{x^2}{2!} + \frac{x^4}{4!} -\frac{x^6}{6!} + \cdots &|x|< \infty\\ \frac{1}{1-x} &= 1+ x+ x^2 + x^3 + \cdots &|x|< 1 \end{align*}$ Next to each of the series, we list an interval which will correspond to the domain of the series. Using power series we can “read-off” properties of functions. Here are some examples.

We can easily see that $e^0 =1$ , $\sin (0)=0$ , and $\cos (0) =1$ .
Since every power of $x$ in the power series for sine is odd, we can see that sine is an odd function. Likewise, since every power of $x$ in the power series for cosine is even, we can see cosine is an even function.
Limits like $\lim _{x\to 0}\frac {\sin (x)}{x} = 1\qquad \text {and}\qquad \lim _{x\to 0} \frac {\cos (x)-1}{x} = 0$ are “easy” to compute, since they can be rewritten as follows. $\begin{align*} \lim_{x\to 0}\frac{\sin(x)}{x} &=\lim_{x\to 0} \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} -\frac{x^7}{7!} + \cdots}{x}\\ &=\lim_{x\to 0} \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} -\frac{x^6}{7!} + \cdots\right)\\ &=1, \end{align*}$ and $\begin{align*} \lim_{x\to 0} \frac{\cos(x)-1}{x}&=\lim_{x\to 0} \frac{1-\frac{x^2}{2!} + \frac{x^4}{4!} -\frac{x^6}{6!} + \cdots-1}{x}\\ &=\lim_{x\to 0} \left(-\frac{x}{2!} + \frac{x^3}{4!} -\frac{x^5}{6!} + \cdots\right)\\ &=0. \end{align*}$
Power series give us methods to actually compute values for these functions.

Convergence of power series

You may have noticed a small caveat above. $\frac {1}{1-x} = 1+ x+ x^2 + x^3 + \cdots \qquad |x|< 1$ The caveat is the “ $|x|<1$ ,” which we referred to as something like the domain of the function. This restriction is required because if our formula is true, then for any number $r$ , provided that $|r|<1$ , we have $\frac {1}{1-r} = 1+ r+ r^2 + r^3 + \cdots ,$ and the expression on the right-hand side of the equation above is a geometric series! As we’ve learned, geometric series only converge when the common ratio (in this case $r$ ) is between $-1$ and $1$ noninclusive. If we look at a graph of $y = \frac {1}{1-x}$ along with a graph of $y = 1+ x+ x^2 + x^3 + \cdots$ we see

True or False: $\frac {4}{3} = \frac {1}{1-(1/4)} = 1 + \left (\frac {1}{4}\right )+ \left (\frac {1}{4}\right )^2+ \left (\frac {1}{4}\right )^3 + \cdots$

true false

True or False: $\frac {1}{-3} = \frac {1}{1-4} = 1 + 4+ 4^2+ 4^3 + \cdots$

true false

Our next theorem tells us what possible scenarios we could encounter when investigating convergence of power series.

Convergence of Power Series Consider the power series $\sum _{n=0}^\infty a_n(x-c)^n.$ Exactly one of the following is true:

(a): The series converges only at $x=c$ .
(b): There is an $R>0$ such that the series converges for all $x$ in $(c-R,c+R)$ and diverges for all $x<c-R$ and $x>c+R$ .
(c): The series converges for all $x$ .

True or False: A power series $\sum _{k=0}^\infty a_k(x-c)^k$ always converges when $x=c$ .

true false

If $x=c$ , then $\begin{align*} \sum_{k=0}^\infty a_k(x-c)^k &= \sum_{k=0}^\infty a_k(c-c)^k \\ &= \sum_{k=0}^\infty a_k(0)^k. \end{align*}$

True or False: If $f(x) = \sum _{k=0}^\infty a_k(x-c)^k,$ then $f(c) = 0$ .

true false

$\begin{align*} f(c) &= \sum_{k=0}^\infty a_k(c-c)^k \\ &= \sum_{k=0}^\infty a_k(0)^k\\ &= a_0 \end{align*}$ since $0^k = 0$ when $k\ne 0$ and $0^0 = 1$ .

Power series are both similar to and different from the series we’ve previously studied. When we fix some value for $x$ , we are working with the sort of series we’ve already studied - a series of numbers. In this way, we can use all of our previous tools for working with series. We can also let $x$ be a variable, and consider our power series as a function. Because power series can define functions, we no longer exclusively talk about convergence at a point, instead we talk about the radius and interval of convergence.

If a power series converges absolutely for all $x$ , then its radius of convergence is said to be $\infty$ and the interval of convergence is $(-\infty ,\infty )$ .
If a power series converges absolutely for all $x$ in $(c-R,c+R)$ and diverges for all $x<c-R$ and $x>c+R$ , then its radius of convergence is said to be $R$ and the interval of convergence is one of the following: $(c-R,c+R),\quad [c-R,c+R),\quad (c-R,c+R],\quad [c-R,c+R].$
If a power series converges only at one value $x=a$ , then its radius of convergence is said to be $0$ and the series does not have an interval of convergence.

In the previous definition, the interval of convergence depends on the series. We must separately consider the behavior of a power series at the endpoints of its interval of convergence. In other words, we plug in values for $x$ , and consider the series as a series of numbers!

Suppose you know that $\sum _{n=0}^\infty a_n (x-3)^n$ converges when $x =7$ and diverges when $x = -1$ . Must the series converge at $x=4$ ?

yes no there is not enough information

Since we know that every power series converges either exactly at a single point or on an interval, we see that this power series must converge with radius of convergence $R=4$ .

How do we check for radius of convergence? Two old friends can come to the rescue: the ratio and root tests.

Consider the power series: $\sum _{n=0}^\infty \frac {x^n}{n!} = 1 + x + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots \\$ Determine the radius and interval of convergence.

For this power series we will use the ratio test. Since the ratio test requires positive terms, we must look at the absolute values of the terms in the series. $\begin{align*} \lim_{n\to\infty} \frac{\frac{|x|^{n+1}}{(n+1)!}}{\frac{|x|^n}{n!}} &= \lim_{n\to\infty} \frac{|x|^{n+1}n!}{(n+1)!|x|^n}\\ &= \lim_{n\to\infty} \frac{|x|}{n}. \end{align*}$ Now, for any fixed value of $x$ , we have that $\lim _{n\to \infty } \frac {|x|}{n} = \answer [given]{0},$ since we recall that $|x|$ is a constant in this limit and its value does not affect the value of the limit. Hence, the radius of convergence for $\sum _{n=0}^\infty \frac {x^n}{n!}$ is $R=\infty$ , and the interval of convergence is $(-\infty , \infty )$ .

While the ratio and root test are good for determining the radius of convergence of a power series, they are useless for determining convergence at the end-points of the interval. Let’s see an example:

Consider the power series: $\sum _{n=1}^\infty \frac {(x-1)^n}{n \cdot 9^n} = \frac {(x-1)}{9} + \frac {(x-1)^2}{2\cdot 9^2} + \frac {(x-1)^3}{3\cdot 9^3} + \cdots \\$ Determine the radius and interval of convergence.

Here, let’s start with the root test. Again, we must first use the absolute value of the terms in the series: $\begin{align*} \lim_{n\to\infty}\sqrt[n]{\frac{|x-1|^n}{n \cdot 9^n}} &= \lim_{n\to\infty}\sqrt[n]{\frac{|x-1|^n}{n \cdot 9^n}}\\ &= \lim_{n\to\infty} \frac{|x-1|}{\sqrt[n]{n} \cdot 9}\\ &= \frac{|x-1|}{\sqrt[n]{n} \cdot 9} \lim_{n\to\infty} n^{-1/n}. \end{align*}$ Using logarithms and L’Hôpital’s rule, we can show that $\lim _{n\to \infty } n^{-1/n} = \answer [given]{1}.$ Hence $\lim _{n\to \infty }\sqrt [n]{\frac {|x-1|^n}{n \cdot 9^n}} = \frac {|x-1|}{9}.$ The root test gives us convergence when this limit is between $-1$ and $1$ . In other words, the series converges absolutely when $\begin{align*} \frac{|x-1|}{9} &<\answer[given]{1}\\ |x-1| &< \answer[given]{9}. \end{align*}$ However, $|x-1| < 9 \qquad \text {means that}\qquad \answer [given]{-9} < x-1 < \answer [given]{9}$ and so adding $1$ to all sides of the inequality, we need $x$ such that $\answer [given]{-8} < x < \answer [given]{10}.$ Since our power series is centered at $x=1$ , the radius of convergence is $R=9$ . However, the root test (and ratio test) is inconclusive at the end points $x=-8$ and $x=10$ . For this, we need to investigate separately the following two series, found by plugging in $x = -8$ and $x=10$ . $\sum _{n=1}^\infty \frac {(-8-1)^n}{n \cdot 9^n}\qquad \text {and}\qquad \sum _{n=1}^\infty \frac {(10-1)^n}{n \cdot 9^n}$ For the first, where $x=-8$ , note that $\begin{align*} \sum_{n=1}^\infty \frac{(-8-1)^n}{n \cdot 9^n} &= \sum_{n=1}^\infty \frac{(-9)^n}{n \cdot 9^n}\\ &= \sum_{n=1}^\infty \frac{(-1)^n}{n}. \end{align*}$ This is the alternating harmonic series, which we know converges. So our power series converges at $x= -8$ . For the second, where $x=10$ , note that $\begin{align*} \sum_{n=1}^\infty \frac{(10-1)^n}{n \cdot 9^n} &= \sum_{n=1}^\infty \frac{(9)^n}{n \cdot 9^n}\\ &= \sum_{n=1}^\infty \frac{1}{n}. \end{align*}$ This is the harmonic series, which we know diverges. So our power series diverges at $x= 10$ . Hence the interval of convergence for $\sum _{n=1}^\infty \frac {(x-1)^n}{n \cdot 9^n}$ must include everything between $-8$ and $10$ , as well as $-8$ , but does not include $10$ . In other words, the interval of convergence is $[-8,10)$ .

Let’s work through an example of a power series that only converges at a single point.

Consider the power series: $\sum _{n=0}^\infty n!(x+7)^n = 1 + (x+7) + 2(x+7)^2 + 6(x+7)^3 + \cdots .$ Determine the radius and interval of convergence.

Here we’ll use the ratio test, looking at the absolute value of the terms in the series. $\lim _{n\to \infty } \frac {(n+1)!|x+7|^{n+1}}{n!|x+7|^n}= \lim _{n\to \infty } \answer [given]{(n+1)}|x+7|$ This limit diverges unless $x=-7$ , the center of the power series. The the radius of convergence is $R=0$ , and there is no interval of convergence, since the series only converges at a single point.

New power series from old

With the basic power series above, we can produce new power series via algebraic manipulation.

Algebra of Power Series Let $\begin{align*} f(x) &= \sum_{n=0}^\infty a_nx^n\\ g(x) &= \sum_{n=0}^\infty b_nx^n \end{align*}$ converge absolutely for $|x|<R$ , and let $h(x)$ be a continuous function.

$f(x)\pm g(x) = \sum _{n=0}^\infty (a_n\pm b_n)x^n$ for $|x|<R$ .
$\begin {aligned}[t] f(x)g(x) &= \left (\sum _{n=0}^\infty a_nx^n\right )\left (\sum _{n=0}^\infty b_nx^n\right )\\ &= \sum _{n=0}^\infty \big (a_0b_n+a_1b_{n-1}+\dots + a_nb_0\big )x^n \end {aligned}$ for $|x|<R$ .
$f\big (h(x)\big ) = \sum _{n=0}^\infty a_n\big (h(x)\big )^n$ for $|h(x)|<R$ .

In our first example we will derive Euler’s famous formula $e^{ix} = \cos (x) + i \sin (x),$ where $i$ is the number $i^2=-1$ .

Use power series to show $\cos (x) + i \sin (x) = e^{ix}$

We start by writing the relevant power series for cosine $\cos (x) = 1-\frac {x^2}{2!} + \frac {x^4}{4!} -\frac {x^6}{6!} + \cdots$ and now we consider $i \sin (x) = ix - \frac {ix^3}{3!} + \frac {ix^5}{5!} -\frac {i x^7}{7!} + \cdots .$ Adding these power series (and ordering the terms by degree) we find $1+ ix -\frac {x^2}{2!} - \frac {ix^3}{3!} + \frac {x^4}{4!} + \frac {ix^5}{5!} -\frac {x^6}{6!} -\frac {i x^7}{7!} + \cdots .$ Since $i^0=\answer [given]{1}$ , $i^1= \answer [given]{i}$ , $i^2 = \answer [given]{-1}$ , $i^3 = \answer [given]{-i}$ , $i^4 = \answer [given]{1}$ , and so on with a repeating pattern we may now write $\begin{align*} 1+ ix &+\frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} +\frac{(ix)^6}{6!} +\frac{(i x)^7}{7!} + \cdots\\ &= e^{ix}. \end{align*}$

Euler’s formula $e^{ix} = \cos (x) + i \sin (x)$ allows us to produce (by setting $x=\pi$ ) the amazing identity: $e^{i \pi } + 1 = 0.$ This identity combines the fundamental constants, $0$ , $1$ , $i$ , $\pi$ and $e$ , along with the fundamental operations of addition, multiplication, and exponentiation!

Use power series to give evidence (by looking at the first $4$ terms of a power series) for the double angle formula: $2\sin (x) \cos (x) = \sin (2x)$

Write with me. $\begin{align*} \sin(x) &= x - \frac{x^3}{3!} + \frac{x^5}{5!} -\frac{x^7}{7!} + \cdots &|x|< \infty\\ \cos(x) &= 1-\frac{x^2}{2!} + \frac{x^4}{4!} -\frac{x^6}{6!} + \cdots &|x|< \infty \end{align*}$ Multiplying we find

So $2\sin (x)\cos (x)$ $\begin{align*} &= 2x + \left(\answer[given]{\frac{-2^3}{3!}}\right)x^3 + \left(\answer[given]{\frac{2^5}{5!}}\right)x^5+\left(\frac{-2^7}{7!}\right)x^7 +\cdots\\ &= (2x) -\frac{(2x)^3}{3!} + \frac{(2x)^5}{5!}-\frac{(2x)^7}{7!} +\cdots\\ &= \sin(2x). \end{align*}$

Use the power series for $\frac {1}{1-x}$ to give a power series for $\frac {1}{1+x^2}$

We know that $\frac {1}{1-x}= 1+ x+ x^2 + x^3 + \cdots \qquad |x|< 1.$ Since $-x^2$ is continuous when $|x|<1$ , we can compose this with $\frac {1}{1-x}$ to see $\begin{align*} \frac{1}{1-(-x^2)} &=\frac{1}{1+x^2}\\ &= \answer[given]{1 - x^2 + x^4} - x^6 + \cdots &|x|< 1\\ &= \sum_{n=0}^\infty (-1)^n x^{2n} &|x|< 1 \end{align*}$

Derivatives and Indefinite Integrals of Power Series Let $f(x) = \sum _{n=0}^\infty a_n(x-c)^n$ be a function defined by a power series, with radius of convergence $R$ .

$f(x)$ is continuous and differentiable on $(c-R,c+R)$ .
$f'(x) = \sum _{n=1}^\infty a_n\cdot n\cdot (x-c)^{n-1}$ , with radius of convergence $R$ .
$\int f(x) dx = C+\sum _{n=0}^\infty a_n\frac {(x-c)^{n+1}}{n+1}$ , with radius of convergence $R$ .

A few notes about the theorem above:

The theorem states that differentiation and integration do not change the radius of convergence. It does not state anything about the interval of convergence. They are not always the same. Check the endpoints!
Notice how the summation for $f'(x)$ starts with $n=1$ . This is because the derivative of the constant term $a_0$ of $f(x)$ is $0$ .
Differentiation and integration are simply calculated term-by-term using the power rule.

Let’s see an example.

Use the power series for $\frac {1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots \qquad |x|< 1$ to find a power series for $\arctan (x)$ . Given the radius and interval of convergence.

Since $\int \frac {1}{1+x^2} dx = \arctan (x)+C$ we can find the desired power series by integrating. Write with me. $\int \left (1 - x^2 + x^4 - x^6 + \cdots \right )dx = \answer [given]{x - \frac {x^3}{3} + \frac {x^5}{5}} - \frac {x^7}{7} + \cdots +C$ Since $\arctan (0) = 0$ , $C=0$ , and we have our desired power series. Its radius of convergence is $R=1$ . However, we recall that the interval of convergence may be different from the original series, so we set out to check the endpoints. First note that our power series can be written in summation notation as $\sum _{n=0}^\infty \frac {(-1)^n x^{2n+1}}{2n+1}.$ If $x=1$ or $x=-1$ we can see that this sequence is $\sum _{n=0}^\infty \frac {(-1)^n}{2n+1}\qquad \text {or}\qquad \sum _{n=0}^\infty \frac {(-1)^{n+1}}{2n+1}$ In both cases, the series converges by the alternating series test. Hence the interval of convergence is $[\answer [given]{-1},\answer [given]{1}]$ .