Power Series Convergence

Section 9.2 Power Series Convergence

At the end of Section 9.1 Polynomial Approximations of Functions, we saw that these polynomial approximations can be thought of as partial sums of some larger infinite series. These infinite series are begging us to think about different notions of convergence, and at the end of the section, we saw that the polynomials, as the degree increases off to infinity, converge to match the function they are approximating, but this might be dependent on some interval of \(x\)-values. A domain, in a sense.

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In this section, we’ll investigate what it means for a power series, these infinite series of power functions, to converge. Let’s define a power series, and then we can think about convergence from there.

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Definition 9.2.1. Power Series.

A power series centered at \(x=a\) is an infinite series in the form:

\begin{equation*} \sum_{k=0}^\infty c_k(x-a)^k = c_0 + c_1(x-a)+c_2(x-a)^2+... \end{equation*}

where \(\{c_k\}_{k=0}^\infty\) is a sequence of real numbered coefficients.

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We have a good idea of how we can build these sequences of coefficients in order for the power series we construct to converge to specific functions that we are interested in.

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Before we state this formally, let’s write down what we mean when we talk about convergence of power series.

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One last thing: we have a kind of closure, so far, in our series. If we add up an infinite amount of numbers, then the infinite series might converge. If it does, it converges to a number (since sums of numbers are numbers). In a power series, though, we are adding up an infinite number of functions of \(x\text{.}\) If this series converges, it will converge to a function of \(x\) (since sums of functions are functions). So every power series is really a function.

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Subsection Interval of Convergence

Activity 9.2.1. Polynomial Division.

We’re going to do some fiddling with polynomials, and hopefully use this as a bridge to connect how we think of polynomials and power series with how we think about our traditional infinite series and the notions of convergence that we’ve already built.

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(a)

We’re going to factor some polynomials, but we might end up using some division. First, we’ll confirm some factors that we already know.

\begin{equation*} x^2-1 = (x-1)(x+1) \end{equation*}

We’ll confirm this by using division.


\(x-1\)	\(x^2\)	\(\)	\(-1\)

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(b)

Now let’s factor \(x^3-1\text{.}\) If the factors for this polynomial isn’t as familiar, it might be helpful to know that \((x-1)\) is also a factor of \(x^3-1\text{.}\)


\(x-1\)	\(x^3\)	\(\)	\(\)	\(-1\)

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(c)

Let’s try another one. Complete the following division.


\(x-1\)	\(x^4\)	\(\)	\(\)	\(\)	\(-1\)

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(d)

Can you generalize this? Find the formula for \(\dfrac{x^n-1}{x-1}\) for some positive integer \(n\text{.}\)


\(x-1\)	\(x^n\)	\(\)	\(\)	\(\)	\(-1\)

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(e)

Now that we have good evidence that

\begin{equation*} \sum_{k=0}^\infty x^k = \frac{x^n-1}{x-1}\text{,} \end{equation*}

We can apply a limit as \(n\to\infty\text{.}\)

\begin{align*} \sum_{k=0}^\infty x^k \amp =\lim_{n\to\infty} \sum_{k=0}^n x^k \\ \amp = \lim_{n\to\infty} \frac{x^n-1}{x-1} \end{align*}

For what values of \(x\) will this limit exist?

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(f)

Revisit Theorem 8.6.6 Geometric Series Convergence Criteria. Is there any difference for what we’ve just done compared to this result that we already know?

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In this activity, we see that we can rethink about our Geometric Series family of series as a power series! Then, instead of saying that we have some requirements on the “ratio” for the geometric series to converge, we can say that the power series \(g(x)=\displaystyle\sum_{k=0}^\infty x^k\) converges for \(x\)-values in the interval \((-1,1)\text{.}\)

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We noticed in Activity 9.1.4 How Do These Polynomials Converge? that the polynomials built to approximate the natural log function does seem to converge to the function as \(n\to\infty\text{,}\) but only for specific \(x\)-values. Hopefully we have some nice ideas as to why that happened: there is a vertical asymptote, and so maybe the “distance” from the center that this polynomial could approximate \(\ln(x)\) at is limited!

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In general, we can notice that this isn’t new: we have families of infinite series that have specific values of variables for which they converge.

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A big thing to notice is that these power series, by definition, include exponentials in them (since \(x^k\) is a power function of \(x\) but is also an exponential function of \(k\)). This means that they’re great candidates to use the Ratio Test. Since we have a variable \(x\) (and we don’t know if this variable is taking on positive or negative values), we’ll need to test these series for absolute convergence.

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Activity 9.2.2. Some Power Series and their Convergence.

Let’s consider a couple of power series and apply some convergence tests to them in order for us to find out how it might converge.

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(a)

Consider the power series:

\begin{equation*} \sum_{k=1}^\infty \frac{(x-1)^k}{k^2}\text{.} \end{equation*}

In order for us to apply the Ratio Test, we’ll actually need to consider the positive-term version:

\begin{equation*} \sum_{k=1}^\infty \frac{|x-1|^k}{k^2}\text{.} \end{equation*}

Apply the Ratio Test. What do you get in the limit of the ratio between terms?

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(b)

What kind of result from the Ratio Test guarantees convergence for the series? What are the \(x\)-values that guarantee convergence?

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(c)

The Ratio Test is inconclusive when the limit is equal to 1. What \(x\)-values does this happen at? Consider the power series evaluated at each of these \(x\)-values. Do these series converge or diverge?

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(d)

Consider the power series:

\begin{equation*} \sum_{k=1}^\infty \frac{(x-1)^k}{\sqrt{k}}\text{.} \end{equation*}

Find the interval of \(x\)-values for which this series converges and test the end points of the interval in the same way as earlier. If it differs, explain why.

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Definition 9.2.2. Interval of Convergence.

For a power series \(\displaystyle\sum_{k=0}^\infty c_k(x-a)^k\) centered at \(x=a\text{,}\) the interval of \(x\)-values for which the power series converges is called the Interval of Convergence. The distance from the center to endpoints of the interval is called the Radius of Convergence

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So this is how we’ll think about convergence! When we think about power series and their convergence, we’re specifically thinking about convergence for specific inputs. These series are really families of infinite series, and we can try to explain their convergence criteria.

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And for a power series, there is always some \(x-\)value for which if converges. For the power series

\begin{equation*} f(x)=\sum_{k=0}^\infty c_k(x-a)^k \end{equation*}

centered at \(x=a\text{,}\) then as long as \(\{c_k\}\) is a sequence of real numbers, then the series converges at \(x=a\text{.}\) We get \(f(a)=c_0\text{,}\) the constant term. This should match with how we thought about these series originally! They came from polynomial approximations of our functions, where of course the polynomials needed to match the function value at the center. We were thinking about tangent lines, tangent quadratics, tangent cubics, etc. They need to be tangent, and so they “converge” at that single center \(x\)-value at least.

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Example 9.2.3.

For the power series \(f(x)=\displaystyle\sum_{k=0}^\infty \frac{3^k(x^k)}{2k+1}\text{,}\) find the interval of convergence. Along the way, it will likely be helpful to identify the center and the radius of convergence.

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Solution.

Let’s apply the Ratio Test. We’ll technically be applying this to \(\displaystyle\sum_{k=0}^\infty \frac{3^k|x|^k}{2k+1}\text{.}\)

\begin{align*} \lim_{k\to\infty} \frac{3^{k+1}|x|^{k+1}}{2(k+1)+1}\cdot \frac{2k+1}{3^k|x|^k}\amp =\lim_{k\to\infty} \frac{3|x|(2k+1)}{2k+3}\\ \amp = 3|x| \end{align*}

For us to conclude that this series converges, we need the limit from the ratio test to be less than 1.

\begin{equation*} 3|x|\lt 1 \to |x|\lt \frac{1}{3} \end{equation*}

This series is centered at 0 with a radius of convergence of \(\frac{1}{3}\text{.}\) So we know that this series converges for \(-\dfrac{1}{3}\lt x \lt \dfrac{1}{3}\text{.}\)

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Since the Ratio Test is inconclusive when \(x=-\dfrac{1}{3}\) and \(x=\dfrac{1}{3}\text{,}\) we’ll test those individually.

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When \(x=-\dfrac{1}{3}\)

\begin{align*} \amp \sum_{k=0}^\infty \frac{3^k\left(-\frac{1}{3}\right)^k}{2k+1} \\ \amp = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \end{align*}

We can apply the Alternating Series Test!

\begin{align*} \amp \lim_{k\to\infty} \frac{1}{2k+1}\\ \amp = 0 \end{align*}

This series converges!

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When \(x=\dfrac{1}{3}\)

\begin{align*} \amp \sum_{k=0}^\infty \frac{3^k\left(\frac{1}{3}\right)^k}{2k+1} \\ \amp = \sum_{k=0}^\infty \frac{1}{2k+1} \end{align*}

We can apply the Rational Comparison Theorem! Since the difference in degrees is 1, we know that this series could be compared to the Harmonic series, and so it diverges.

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So the interval of convergence is \(\left[-\dfrac{1}{3},\dfrac{1}{3}\right)\text{.}\)

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Subsection Operations on Power Series

Theorem 9.2.4. Operations on Power Series.

For two power series \(\displaystyle\sum_{k=0}^\infty c_kx^k\) and \(\displaystyle\sum_{k=0}^\infty d_kx^k\) that converge to \(f(x)\) and \(g(x)\) (respectively) on the interval of convergence \(I\text{,}\) we can consider the following operations to combine power series.

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Sum: \(\displaystyle \sum_{k=0}^\infty (c_k+d_k)x^k\) converges to \(f(x)+g(x)\) on \(I\text{.}\)
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Difference: \(\displaystyle \sum_{k=0}^\infty (c_k-d_k)x^k\) converges to \(f(x)-g(x)\) on \(I\text{.}\)
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Product: If \(bx^n\) is a power function, then \(\displaystyle \sum_{k=0}^\infty b(c_k)x^{k+n}\) converges to \((bx^m)f(x)\) on \(I\text{.}\)
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Composition: If \(bx^n\) is a power function, then \(\displaystyle \sum_{k=0}^\infty c_k(b^k)(x^{nk})\) converges to \(f(bx^n)\) when \(bx^n\) is in \(I\text{.}\)
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We can do something similar with some of our calculus operations: differentiation and integration.

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Theorem 9.2.5. Differentiating and Integrating Power Series.

If \(\displaystyle\sum_{k=0}^\infty c_k(x-a)^k\) converges to \(f(x)\) on an interval of convergence with a radius \(R\gt0\text{,}\) then:

\(\displaystyle\sum_{k=0}^\infty kc_k(x-a)^{k-1}\) converges to \(f'(x)\text{.}\)
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\(\displaystyle\sum_{k=0}^\infty \frac{c_k(x-a)^{k+1}}{k+1}\) converges to \(F(x)\text{,}\) an antiderivative of \(f(x)\text{.}\)
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Both of these converge on an interval of convergence centered at \(x=a\) with radius \(R\text{.}\)

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Note 9.2.6.

We’re being weird about naming the interval of convergence for these. The issue is that when we differentiate, we might lose closed endpoints of an interval. Similarly, when we antidifferentiate, we could add endpoints to an open interval.

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We can see this in some of the examples that follow, but the intervals of convergence are going to be identical except for possibly at the endpoints.

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Let’s finish this here by revisiting the power series from Example 9.2.3.

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Example 9.2.7.

Let’s reconsider the power series \(\displaystyle\sum_{k=0}^\infty \frac{3^k(x^k)}{2k+1}\text{.}\) We know, from Example 9.2.3, that this power series converges to \(f(x)\) for \(x\)-values in the interval \(\left[-\dfrac{1}{3},\dfrac{1}{3}\right)\text{.}\)

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(a)

Find a power series that converges to \(x^7f(x)\text{.}\) What is the interval of convergence?

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Solution.

\begin{align*} f(x) \amp = \sum_{k=0}^\infty \frac{3^k(x^k)}{2k+1}\\ x^7f(x) \amp = x^7 \sum_{k=0}^\infty \frac{3^k(x^k)}{2k+1}\\ \amp = \sum_{k=0}^\infty \frac{3^k(x^k)(x^7)}{2k+1}\\ \amp = \sum_{k=0}^\infty \frac{3^kx^{k+7}}{2k+1} \end{align*}

The interval of convergence doesn’t change: \(\left[-\dfrac{1}{3},\dfrac{1}{3}\right)\text{.}\)

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(b)

Find a power series that converges to \(f(x^2)\text{.}\) What is the interval of convergence?

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Solution.

\begin{align*} f(x) \amp = \sum_{k=0}^\infty \frac{3^k(x^k)}{2k+1}\\ f(x^2) \amp = \sum_{k=0}^\infty \frac{3^k((x^2)^k)}{2k+1}\\ \amp = \sum_{k=0}^\infty \frac{3^k(x^{2k})(x^7)}{2k+1} \end{align*}

The interval of convergence is \(\left(-\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)\text{.}\)

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(c)

Find a power series that converges to \(f'(x)\text{.}\) What is the interval of convergence?

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Solution.

\begin{align*} f(x) \amp = \sum_{k=0}^\infty \frac{3^k(x^k)}{2k+1}\\ f'(x) \amp = \ddx{\sum_{k=0}^\infty \frac{3^k(x^k)}{2k+1}}\\ \amp = \sum_{k=0}^\infty \ddx{\frac{3^k(x^k)}{2k+1}}\\ \amp = \sum_{k=0}^\infty \frac{3^k\ddx{x^k}}{2k+1}\\ \amp = \sum_{k=0}^\infty \frac{k3^k(x^{k-1})}{2k+1} \end{align*}

Since \(k\) is in the coefficient, starting the index at \(k=0\) means that the “first” term is 0: we can re-index if we’d like, but it is not necessary.

\begin{equation*} \sum_{k=1}^\infty \frac{k3^k(x^{k-1})}{2k+1} \hspace{0.5cm} \text{or} \hspace{0.5cm} \sum_{k=0}^\infty \frac{(k+1)3^{k+1}(x^k)}{2k+3} \end{equation*}

In any of these cases, the interval of convergence is \(\left(-\dfrac{1}{3},\dfrac{1}{3}\right)\text{.}\)

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These operations aren’t that useful without a purpose. Similarly, these power series aren’t that interesting without knowing what functions they converge to.

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We’ll now start putting some things we’ve learned together.

We have ideas of power series representations for \(\sin(x)\text{,}\) \(\cos(x)\text{,}\) and \(e^x\text{.}\)
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We know the geometric series, which we can use to get a power series representation for \(\dfrac{1}{1-x}\text{.}\)
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We have a general way of building terms of these power series, using the polynomial approximations and the formula for building those terms.
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We have ways of combining, differentiating, and anti-differentiating power series.
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We’ll move forward and demonstrate some really great ways of constructing power series representations of some new functions. And then we’ll show some very fun uses of these power series representations to illustrate how friendly they are to work with.

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Practice Problems Practice Problems

1.

Describe the following terms in your own words, and explain how they relate to a power series.

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(a)

Center.

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(b)

Interval of Convergence.

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(c)

Radius of Convergence.

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2.

Does a power series ever diverge everywhere? Explain why not, using an example.

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3.

For a power series that converges for \(\left|\frac{x-3}{4}\right| \lt 1\) and diverges elsewhere, find the center of the power series, the radius of convergence, and interval of convergence.

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4.

For a power series that converges for \(\left|4(x+1)\right| \lt 1\) and diverges elsewhere, find the center of the power series, the radius of convergence, and interval of convergence.

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5.

For each of the following power series, determine the interval of convergence. Be specific on which convergence test you are applying and how you are finding the interval of convergence.

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