Lesson 19: Power Series and Functions & Properties of Power Series

Motivating question

What the heck does something like $\sum_{n=0}^\infty x^n$ even mean???

Something we should notice in Warmup exercise 1 is: not only is each series geometric, but the general term is of the form $r^n$ for each of them. The only difference among the three is the value of $r$.

Introduction to power series

Motivating example

Let’s do a deep dive on our motivating question: What the heck does something like $\sum_{n=0}^\infty x^n$ even mean??? Buckle up because we’ll be with this example for a while. It will be worth it because this example will really help us understand power series in general. We’ll take it slowly. Let’s start with what we know:

• We could write the first few terms of the expanded sum:\[\sum_{n=0}^\infty x^n = 1 + x +x^2+x^3+x^4 + \cdots.\] But we’ve just rewritten the infinite sum, it doesn’t really tell us what it means.
• Now that we have a definition, we can say that “$\sum_{n=0}^\infty x^n$ is a power series centered at $a=0$ where $c_n = 1$ for all $n$.” But this just gives a name to the kind of mathematical object that the infinite sum is, it doesn’t really tell us what it means.
• Let’s look at the formula $\sum_{n=0}^\infty x^n$ again. We could substitute different numbers for the variable $x$. For example, in Warmup exercise 1, we substituted three different values for $x$: $\frac{9}{10}, \frac{11}{10}$, and $-1$. We could have chosen other values for $x$. We could have chosen any real number for $x$ and asked if the corresponding series converges or diverges. This might seem like a simple fact that won’t actually help us understand what $\sum_{n=0}^\infty x^n$ means, but actually, this is an excellent start!

The power series represents a family, where each member of the family is a series which depends on $x$. Let’s update our notation to reflect this dependence on $x$ and give our power series a name:

\[P(x) = \sum_{n=0}^\infty x^n\]

This is very suggestive notation! It should make you think of…a function! With this notation $P(x)$, we saw in Warmup exercise 1 that:

1. the series $P(\frac{9}{10})$ converges,
2. the series $P(\frac{11}{10})$ diverges, and
3. the series $P(-1)$ diverges.

Remember that if a series converges, it converges to a number (even if we don’t specify what that number is). Let’s update what we just said with this in mind:

1. the number $P(\frac{9}{10})$ exists,
2. the number $P(\frac{11}{10})$ does not exist, and
3. the number $P(-1)$ does not exist.

We could test whether $P(x)$ exists for other values of $x$ too, but we can already see that $P(x)$ exists for some values of $x$ and not for others. This should make you think of…the domain of the function!

Recall that a number $b$ is in the domain of a function $f(x)$ if $f(b)$ exists (a.k.a. is a number); a number $b$ is not in the domain of a function $f(x)$ if $f(b)$ does not exist (a.k.a. is not a number). (If you need a refresher on domains of functions, see Lesson 2 of the MAT 1375 course hub here.)

The upshot is that the power series $P(x)$, when thought of a function, has a domain which consists of all the $x$-values for which the series $P(x)$ converges. For now, let’s rename the domain of $P(x)$ as its “set of convergence” (we’ll rename it again later in the lesson).

So we saw in Warmup exercise 1 that:

1. $x=\frac{9}{10}$ is in the set of convergence for $P(x)$,
2. $x=\frac{11}{10}$ is not in the set of convergence for $P(x)$
3. $x=-1$ is not in the set of convergence for $P(x)$

We’re just saying the same thing again and again in a different way. Again, we could test whether other values of $x$ are in the set of convergence for $P(x)$. Let’s ask that question: what is the complete set of convergence for $P(x)$?

Again, lucky for us, $P(x) = \sum_{n=0}^\infty x^n$ is a geometric series for every possible value of $x$ that we could plug in. So let’s treat $P(x)$ as a geometric series itself, but with a varying common ratio $r$! From this perspective the common ratio for $P(x)$ is $r=x$. Knowing what we know about geometric series, we can now say conclusively:

• if $|x| < 1$, then the series $ \sum_{n=0}^\infty x^n$ converges, and
• if $|x| \geq 1$, then the series $ \sum_{n=0}^\infty x^n$ diverges.

Finally, we’ll say that the set of convergence for the $ \sum_{n=0}^\infty x^n$ is the open interval $(-1,1)$.

So what does $\sum_{n=0}^\infty x^n$ mean? Well we know that on the interval $(-1,1)$ the function $P(x)$ is defined. What is this function?

Again, we’ll use what we learned about geometric in Lesson 14 (link here): when $|r| < 1$, the geometric series $\sum_{n=1}^\infty ar^{n-1} = \sum_{n=0}^\infty ar^n$ converges to $\frac{a}{1-r}$. For us, $a=1$ and $r=x$. So on $(-1,1)$ $\sum_{n=0}^\infty x^n$ converges to the function $\frac{1}{1-x}$. All of this together is what we mean when we say $\sum_{n=0}^\infty x^n$!

We can see what $\sum_{n=0}^\infty x^n$ means graphically too and much more quickly. Figure 1 below shows graphs of the first several partial sums of the series $\sum_{n=0}^\infty x^n$. The partial sums are just polynomials! The figure below is cluttered but click on it (or here) to navigate to the interactive Desmos graph. By default, only the first few partial sums are displayed. Click on the grey circles on the left of the screen to turn on the other partial sums one by one. The black dotted curve is the graph of the function $f(x) = \frac{1}{1-x}$. As you can see, as the degree of the partial sum increases, the closer the polynomial gets to this function…but this is only true for the set of convergence $(-1,1)$. Outside this set, the partial sums diverge and the graphs are getting further away from the graph of $f(x) = \frac{1}{1-x}$.

Again, we’re seeing that the power series $\sum_{n=0}^\infty x^n$ converges to $f(x) = \frac{1}{1-x}$ on the interval $(-1,1)$ and diverges outside this interval.

Interval of convergence

In our motivating example, we saw that on the set of convergence, the power series $P(x)$ converges to a function $f(x)$; outside that set, the series diverges. For a general power series $P(x) = \sum_{n=0}^\infty c_n (x-a)^n$, we want to know the set of convergence. We typically won’t need to know what function the power series $P(x)$ converges to until Lesson 20 (link here).

Easy convergence?

Question: There is always one really easy value of $x$ where the power series $P(x) = \sum_{n=0}^\infty c_n (x-a)^n$ converges. Take a second now to guess what it is.

The set of convergence is an interval

We see that every power series converges at its center. That is, the center is always in the set of convergence. It turns out that the set of convergence is always an interval…and the center is the center of that interval! We’ll now rename the “set of convergence” of a power series as the interval of convergence.

We’ll see why this set is actually an interval through some upcoming examples which use the ratio test, which we saw in Lesson 18 (link here). In our motivating example above, the interval was open because it excluded both endpoints. In general the interval of convergence may include or exclude either endpoint, so the interval might be open, closed, or half open/half closed.

More examples

Video 1 below quickly introduces us to some of the concepts above and shows us two examples. Don’t forget: the ratio test is inconclusive when the limit $\rho = 1$. This happens at the endpoints of the interval of convergence, so to determine whether an endpoint is included or excluded from the interval, you’ll have to test it separately. (It is a common mistake to forget to check convergence at interval endpoints.)

Video 2 below shows two more examples where we need to find the interval of convergence for a power series. Notice that the power series in Video 1 above were centered at 0 but the power series in Video 2 are not centered at 0.

Series representations of functions – introduction

We saw that the power series $\sum_{n=0}^\infty x^n$ converges to the function $f(x)=\frac{1}{1-x}$ on the interval of convergence. We’ll see a lot more about this kind of series representation of functions when we see Taylor series in Lesson 20 (link here).

For now we can easily build on the motivating example by replacing $x$ with other functions of $x$ to say, for example:

• The power series $\sum_{n=0}^\infty x^{2n}$ converges to the function $f(x)=\frac{1}{1-x^2}$.
• The power series $\sum_{n=0}^\infty (-1)^n x^n$ converges to the function $f(x)=\frac{1}{1-(-1 \cdot x)} = \frac{1}{1+x}$.

We can can also integrate or differentiate series representations of functions term by term to get series representations of new functions. For example:

• Let $f(x) = \ln(1+x)$. Notice that its derivative $f'(x) = \frac{1}{1+x}$. We saw above that on the interval of convergence, $f'(x)=\frac{1}{1+x}$ is represented by the series $P'(x)= \sum_{n=0}^\infty (-1)^n x^n$. We want a series representation of $f(x) = \ln(1+x)$, not of $f'(x)=\frac{1}{1+x}$, so we can integrate $P'(x) = \sum_{n=0}^\infty (-1)^n x^n$ term by term: $f(x) = \ln(1+x)$ is represented by the series $P(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} +C$ for some constant $C$. To determine $C$ we can choose any $x$ value in the interval of convergence; let’s choose $x=0$. Then $f(0) = \ln(1+0) = \ln(1) = 0$. And the series $P(0) = \sum_{n=0}^\infty (-1)^n \frac{0^{n+1}}{n+1} +C = C + 0 + 0+0 + \cdots = C$. Since we want $P(x)$ to represent $f(x)$ on the interval of convergence, we must have that $P(0) = f(0)$, so $C =0$. Therefore, the series representation of $f(x) = \ln(1+x)$ is $P(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}$. Recall the interactive Desmos graph (link here) from Lesson 11 on Taylor Polynomials (link here). We’ll come back to this example once again in Lesson 20 (link here).
• Now let $f(x) = \frac{2x}{1-x^2}$. Notice that $f(x)$ is the derivative $F'(x)$ of $F(x) = \ln(1-x^2)$. If we can find a series representation of $F(x)$, then we can differentiate it term by term to get a series representation of $f(x)$. We just saw that the series $\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}$ represents $\ln(1+x)$. We can replace $x$ by $-x^2$ to see that the series $P(x) = \sum_{n=0}^\infty (-1)^n \frac{(-x^2)^{n+1}}{n+1}$ represents $F(x) = \ln(1-x^2)$. When we simplify it, we see that $P(x) = – \sum_{n=0}^\infty \frac{x^{2(n+1)}}{n+1}$. Now we differentiate term by term to see that $P'(x) = – \sum_{n=0}^\infty \frac{2(n+1)x^{2(n+1)-1}}{n+1}$ which we simplify to $-2 \sum_{n=0}^\infty x^{2n+1}$. In conclusion, the function $f(x) = \frac{2x}{1-x^2}$ is represented on the interval of convergence by the power series $P'(x) = -2 \sum_{n=0}^\infty x^{2n+1}$.

Video 3 below shows another series representation of a function. It starts with our motivating example as well.