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Power Series

Calculus 2 · Chapter 8: Power Series & Taylor SeriesSection 8.1 · lesson 39 of 5411 min · free

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In this lesson

Infinite polynomials, and where they are valid.

  • Read a power series as a geometric series in x
  • Apply the three-case convergence theorem
  • Find the radius with the Ratio Test
  • Test each endpoint separately - always

Transcript

Everything said in the video, section by section (1,576 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 8.1

Power Series

A structural engineer modeling the sway of a suspension cable needs the value of sine theta for a small angle. A calculator returns a decimal instantly. But the calculator itself does not know the sine function. Internally it evaluates a polynomial. A very long one, whose terms follow a predictable pattern. Polynomials are the only functions a processor can evaluate directly — because they require nothing but multiplication and addition. Every other function — sine, cosine, the exponential, the logarithm — is computed by replacing it with a polynomial that has infinitely many terms. Such an object is called a power series. But that substitution raises an immediate and unavoidable question. An infinite sum need not add up to anything finite. So which inputs are safe? That's what this section determines.

The Object

A Series Whose Terms Contain a Variable

A power series centered at a is a series whose terms are constant multiples of increasing powers of x minus a. And here's the crucial difference from Chapter Seven. Because x is a variable, a power series is not a single number the way those numerical series were. It's a machine. Feed it a value of x, and it produces a numerical series — which either converges or diverges. Let's find every value of x for which the sum of x to the n converges. Write out the terms. One, plus x, plus x squared, plus x cubed, and so on. Each term is the previous term multiplied by x. That is precisely a geometric series. First term one, common ratio r equals x. And from Chapter Seven, a geometric series converges exactly when the absolute value of the ratio is less than one. So this converges when absolute x is less than one. That is, minus one less than x less than one. And for those inputs, the geometric sum formula gives the value directly. One over one minus x. But the endpoints must be examined on their own — because the geometric criterion excludes them. At x equals one, the series becomes one plus one plus one plus one. Partial sums one, two, three, four. It diverges. At x equals minus one, it becomes one minus one plus one minus one. Partial sums oscillate between one and zero. It also diverges. Both failures are confirmed by the Divergence Test, since x to the n does not approach zero at either endpoint. So the conclusion. The series converges precisely on minus one to one, and there it equals one over one minus x. Now here's a point worth pausing on. Outside that interval, the expression one over one minus x is still a perfectly good number. At x equals three it equals minus one half. But the series is not equal to it. Because the series does not converge there. A power series represents a function only on the interval where it converges.

Figure 8.1.1

Adding Terms Helps Only Inside the Interval

The picture makes that concrete. The dark curve is the true function, one over one minus x. The lighter curves are partial sums. Through x squared, through x to the fourth, through x to the eighth. Inside the shaded band — minus one to one — watch what happens as you add terms. Each partial sum hugs the curve more closely. Adding terms pulls them toward the function. Now look outside the dashed lines at x equals minus one and x equals one. There, adding terms drives the partial sums away from the curve. Not toward it. That's what divergence looks like graphically. More effort makes the approximation worse, not better.

The Theorem

Exactly Three Shapes

That example produced a symmetric interval around the center. And that symmetry is not an accident. A theorem guarantees that the convergence set of any power series takes one of exactly three shapes. And it always spreads out symmetrically from the center. Case one. The series converges only at x equals a. The center, and nothing else. Case two. The series converges for every real number x. Case three. There's a positive number R such that the series converges absolutely whenever the distance from the center is less than R, and diverges whenever it exceeds R. That number R is called the radius of convergence. In case one set R to zero. In case two set R to infinity. Now — the theorem says nothing about the two points where the distance equals exactly R. And that final sentence carries real weight. At the endpoints, the theorem is silent. And the silence is genuine. All four combinations of endpoint behavior actually occur. Both included. Both excluded. Or one of each. So the interval of convergence — the complete set of x-values for which the series converges — requires you to test the endpoints separately. Determining R is a routine computation. Determining the endpoints requires separate testing, every single time.

The Computation

Finding R with the Ratio Test

The Ratio Test from Chapter Seven is the standard instrument for locating R. Because it responds to the size of the distance from the center, in exactly the way the theorem describes. Applied to a power series, the general term is c n times x minus a, to the n. So the ratio always factors into a part depending on x, and a part depending only on the coefficients. The absolute ratio is the absolute ratio of consecutive coefficients, times the absolute distance from the center. Taking the limit gives L of x equals the distance, times K — where K is the limit of the coefficient ratios. Now set L of x less than one and solve for the distance. That yields the radius immediately. If K is a positive finite number, then R equals one over K. Let's run one completely. The sum of x minus two to the n, over n times three to the n. The ratio works out to n over three times n plus one, times the distance. Which tends to the distance over three. Set that below one. The distance from two is less than three. So R equals three. And the open interval is minus one to five, obtained by adding two to each part. Now test the endpoints. Each produces a numerical series, handled by the Chapter Seven tests. At x equals five: five minus two to the n is three to the n. And the three to the n in the denominator cancels it exactly. What's left is the sum of one over n. The harmonic series. Which diverges. So x equals five is excluded. At x equals minus one: minus one minus two to the n is minus three to the n, which is minus one to the n, times three to the n. Again the three to the n cancels. What's left is the alternating harmonic series. Which converges. So x equals minus one is included. The interval of convergence is closed at minus one, open at five. With radius three. And notice — the two endpoints behave differently, even though they sit the same distance from the center. A reminder that endpoint behavior cannot be inferred from the radius.

Worked Example 3

Radius Zero and Radius Infinity

Two extreme cases are worth seeing, because they bracket everything else. First. Analyze the sum of n factorial, times x to the n. Here the coefficients are n factorial. Using n plus one factorial equals n plus one times n factorial, the factorials cancel cleanly. The absolute ratio is n plus one, times the absolute value of x. For any fixed nonzero x, that quantity grows without bound. So L is infinite, and the series diverges. Only x equals zero survives — where the series equals its constant term. So R equals zero, and the interval of convergence is the single point zero. Coefficients that grow rapidly shrink the interval. Second. Analyze the sum of x to the n over n factorial. Now the factorial sits in the denominator. So the same cancellation runs the other direction. The absolute ratio is absolute x, over n plus one. For any fixed x, the denominator grows without bound while the numerator stays constant. So L equals zero, less than one, for every real x. The series converges absolutely everywhere. R equals infinity, and the interval is the whole real line. And this particular series defines e to the x. Which is precisely why exponential values can be computed reliably for any input at all. The table collects four series spanning every case the theorem permits. And the key observation is this. The radius alone never determines the interval. The endpoints supply independent information, every time.

Key Takeaways

What to Carry Forward

Five takeaways. A power series is a function-valued object. Each x produces a numerical series, and the series represents a function only where it converges. The convergence set has one of exactly three shapes. A single point, the whole line, or a symmetric interval of radius R about the center. The Ratio Test locates R, because the ratio factors as the distance from the center times a coefficient limit K. Giving R equals one over K. The theorem is silent at the endpoints. So substitute each one and test the resulting numerical series with the Chapter Seven tools. All four combinations genuinely occur. And rapidly growing coefficients shrink the interval — n factorial x to the n gives radius zero. While rapidly shrinking coefficients expand it — x to the n over n factorial gives infinite radius, and defines e to the x. Next up is section eight point two — Representation of Functions as Power Series. Where we build new series from the geometric one.

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FAQ

Questions students ask us

About Power Series

What is a power series in calculus?
A power series centered at a is a series of the form \sum c_n(x-a)^n, whose terms are constant multiples of increasing powers of x-a. Because x is a variable, each input produces a different numerical series that either converges or diverges. For example, \sum x^n is geometric with ratio x, so it converges to \frac{1}{1-x} exactly when -1<x<1.
How do you find the radius of convergence of a power series using the Ratio Test?
Apply the Ratio Test to the general term c_n(x-a)^n, take the limit, and solve L(x)<1 for |x-a|. The limit factors as |x-a|\,K, where K=\lim|c_{n+1}/c_n|. If K is positive and finite, then R=1/K. For \sum \frac{(x-2)^n}{n\,3^n}, the limit is |x-2|/3, which gives R=3 and the open interval (-1,5).
What are the possible intervals of convergence for a power series?
Every power series converges on one of three kinds of sets. It converges only at the center x=a (R=0), or for every real x (R=\infty), or there is a positive finite R with absolute convergence for |x-a|<R and divergence for |x-a|>R. For instance, \sum n!\,x^n has R=0, while \sum x^n/n! has R=\infty and equals e^x.
Do you need to check the endpoints of the interval of convergence separately?
Yes. When 0<R<\infty, the radius does not tell you whether the endpoints x=a\pm R are included. Substitute each endpoint and test the resulting numerical series. For \sum \frac{(x-2)^n}{n\,3^n}, x=5 gives the divergent harmonic series, while x=-1 gives the convergent alternating harmonic series, so the interval is [-1,5). All four endpoint combinations occur.

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