6.1 Power Series and Functions

1. Power Series and Functions

Title

Calculus II · Section 6.1

Infinite polynomials, and the interval where they make sense

2. What this lesson gives you

Objectives

Chapter 5 was about series of numbers. Now every term carries a variable, so a series becomes a function of that variable, defined wherever it converges. This lesson answers the two questions that come first: where does such a series converge, and which familiar functions can be written this way?

• Recognise a power series and read off its centre and coefficients
• State the three possible shapes of the convergence set (Theorem 6.1), and why no others occur
• Find the radius of convergence with the ratio test
• Test both endpoints separately and report the full interval of convergence
• Read the geometric series backwards to represent a function as a power series

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 532-540 — learning objectives 6.1.1 to 6.1.3

Everything in Chapter 5 was about series of numbers: a fixed list of terms with one verdict, converges or diverges. In this chapter each term carries a variable, so a single series becomes a different series of numbers for every value of x. Where it converges, it defines a function.

That raises two questions, and this lesson answers both. First, for which x does such a series converge? The answer always has a remarkably tidy shape, an interval centred on the series, and the ratio test you already know finds most of it. Second, which functions can be written this way? The geometric series gives you the first family for free.

Keep your Chapter 5 tools close. The endpoints of every interval will need them, one at a time.

3. Before anything new: two facts from Chapter 5

Warm-up

Write both answers down before you reveal them. Everything in this lesson rests on them, and if either is shaky it is worth ten minutes with Sections 5.2 and 5.6 before going on.

The geometric series converges exactly when the ratio is between minus one and one, and then its sum is the first term divided by one minus the ratio. Notice that the condition is about the size of the ratio. That will become the size of x.

The ratio test looks at the limit of the size of each term divided by the one before it. Below one means absolute convergence, above one means divergence, and exactly one means the test has nothing to say. That last case is not a footnote here: it is precisely what happens at the ends of every interval you will compute today.

4. Power series and their centre

Section

Part 1

5. A polynomial that never stops

Concept

A power series is a series whose nth term is a constant times the nth power of x. Its partial sums are ordinary polynomials of higher and higher degree.

\[ \sum_{n=0}^{\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots \]

power series centred at 0 (6.1) — A series of this form, where x is a variable and the coefficients are constants. It defines a function of x wherever it converges.

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 532 — definition (6.1)

A power series looks like a polynomial whose degree never stops growing. Each coefficient is a fixed number; the only thing that varies is x. If you stop after the term of degree N, what you have is an honest polynomial, and it is called the Nth partial sum.

Look at the figure. The four solid curves are the partial sums of one particular power series, the one with coefficients one over n factorial. The first is a straight line, the next a parabola, then a cubic, then a quartic. Each one agrees with the dashed curve near zero, and each extra term extends the stretch where the agreement is good.

The dashed curve is what the whole infinite series adds up to. You will meet it properly in Section 6.3, where it turns out to be the exponential function. For now the point is the picture: an infinite polynomial can be a perfectly ordinary curve.

6. Why build functions out of powers

Intuition

Polynomials are the easiest functions there are: only adding and multiplying, trivial to differentiate and integrate, easy to evaluate by hand or by machine.

\[ S_N(x) = c_0 + c_1 x + \cdots + c_N x^N \quad \text{(a polynomial)} \]

If a hard function equals an infinite polynomial, you can approximate it by stopping early, and later in this chapter you will differentiate and integrate it term by term. Some functions can be written in no other way.

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 532, 537 — chapter introduction and 'Representing Functions as Power Series'

It is fair to ask why anyone would want to write a function as an infinite sum. The answer is that polynomials are the friendliest functions in mathematics. You can evaluate them with nothing but multiplication and addition, which is all a computer chip really does. You can differentiate and integrate them term by term without thinking.

So if a difficult function equals an infinite polynomial, you get two things. You can approximate it by cutting the series off, which is how calculators produce values of sine and the exponential. And you can do calculus on it one term at a time, which the next two sections turn into a machine for producing new series.

The book also points out a third use: some functions, including solutions of certain differential equations, can only be written as power series. The series is the definition.

7. Moving the centre to x = a

Concept

Replace x by x minus a and the series is built around a instead of around zero.

\[ \sum_{n=0}^{\infty} c_n (x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots \]

By convention the zeroth power is 1 even when its base is zero, so the first term is always the constant c-nought.

\[ x^0 = 1 \text{ at } x = 0, \qquad (x-a)^0 = 1 \text{ at } x = a \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 532 — definition (6.2)

A power series does not have to be built around zero. Replace every x by x minus a and the whole construction slides over so that it is built around a. The number a is the centre, and the powers now measure how far x is from a.

In the figure, the blue series has every coefficient equal to one and centre zero; it matches the function one over one minus x near zero. The orange series has the same coefficients but centre 3, and it matches a different function near 3. Each dashed partial sum only behaves on a band around its own centre, and the band moved when the centre did.

The convention about the zeroth power matters only at the centre itself, where x minus a is zero. Mathematicians agree that the zeroth power is one there too, so the first term of every power series is simply the constant c-nought. Without that agreement the formula would be undefined at the one point where it is easiest to evaluate.

8. Reading a power series piece by piece

Notation

Step through each annotation and connect it to the formula. The coefficients are the only part that differs from one power series to another, which is why every question about convergence ends up being a question about how the coefficients grow.

The centre is the part people misread most often. The definition subtracts it, so whatever is being subtracted from x is the centre. When you see x plus 5, rewrite it as x minus negative 5 before reading anything off.

Notice also that the index usually starts at zero. Some series in this lesson start at one instead, because their coefficient would involve dividing by zero when n is zero. That changes nothing about where the series converges; dropping or adding finitely many terms never does.

9. Every power series converges at its centre

Concept

Put x equal to a. Every power after the zeroth has a factor of zero.

\[ \sum_{n=0}^{\infty} c_n (a-a)^n = c_0 + c_1 \cdot 0 + c_2 \cdot 0 + \cdots \]

\[ = c_0 \]

The partial sums are all equal to c-nought, so they certainly converge. However wild the coefficients, the convergence set is never empty: it always contains the centre.

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 533 — Convergence of a Power Series

This is the one convergence fact you get for free. At x equal to a, every term after the first contains a factor of zero, so every partial sum equals the constant c-nought. A sequence that never changes certainly converges.

It sounds trivial, and in a sense it is, but it has a real consequence. The set of points where a power series converges can never be empty. Even the most badly behaved power series, like the one with factorial coefficients you will meet in a few slides, converges at its centre.

It also gives you a quick sanity check on any interval you compute later. If your interval of convergence does not contain the centre, something has gone wrong, most likely a sign in the centre.

10. Three power series from the book, written out

Worked example

Identify the centre and the coefficients, and write the first few terms.

\[ \text{(a) }\sum_{n=0}^{\infty}\frac{x^n}{n!} \qquad \text{(b) }\sum_{n=0}^{\infty} n!\,x^n \qquad \text{(c) }\sum_{n=0}^{\infty}\frac{(x-2)^n}{(n+1)3^n} \]

\[ a = 0, \quad c_n = \frac{1}{n!}: \quad 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

\[ a = 0, \quad c_n = n!: \quad 1 + x + 2!\,x^2 + 3!\,x^3 + \cdots \]

\[ a = 2, \quad c_n = \frac{1}{(n+1)3^n} \]

\[ 1 + \frac{x-2}{2\cdot 3} + \frac{(x-2)^2}{3\cdot 3^2} + \frac{(x-2)^3}{4\cdot 3^3} + \cdots \]

\[ \left.\sum_{n=0}^{\infty}\frac{(x-2)^n}{(n+1)3^n}\right|_{x=2} = c_0 = \frac{1}{1\cdot 3^0} = 1 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 532-533 — the three examples after the definition

These are the three series the book introduces straight after the definition, and they will reappear as Example 6.1. Get used to reading a series before doing anything with it: find the centre, then find the coefficient by removing the power of x minus a.

In the first two series nothing is subtracted from x, so the centre is zero, and the coefficients are one over n factorial and n factorial. They look like near relatives, yet you will see that one converges everywhere and the other only at zero.

The third series is centred at 2. Its coefficient combines a factor n plus one with a power of 3, and writing out the first few terms by putting n equal to 0, 1, 2 and 3 into it is the best way to see what the notation means. The check confirms the free fact from the previous slide: at the centre, only the first term survives, and it equals one.

11. Match each series to its centre

Matching

For each series, ask which value of x makes the base equal to zero. That value is the centre. Do this before revealing the answer, because two of the four are designed to catch you.

The series in x plus 4 has centre negative 4, since the definition subtracts the centre. The series in 2x minus 1 is centred at one half: factor out the 2, and the base becomes 2 times the quantity x minus one half. The extra power of 2 simply joins the coefficient.

The series in 3 minus x is centred at 3, not at negative 3. Three minus x is the negative of x minus 3, so its nth power is negative one to the n times the nth power of x minus 3, and the alternating sign is again absorbed into the coefficient. In every case the rule is the same: the centre is where the base vanishes.

12. Trap: reading the centre off a plus sign

Trap

13. Where a power series converges

Section

Part 2

14. One series, a different verdict at every x

Concept

Fix a value of x and a power series becomes an ordinary series of numbers, with its own verdict. The geometric series shows all the behaviours at once.

\[ \sum_{n=0}^{\infty} x^n: \quad x = \tfrac12 \to 2, \quad x = -\tfrac12 \to \tfrac23, \quad x = 1.05 \to \text{diverges} \]

So the real question about a power series is not whether it converges, but for which x. The answer is a set of real numbers, and the rest of this part finds its shape.

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 532 — the geometric series as a power series

This is the change of viewpoint the whole lesson turns on. A power series is not one series of numbers; it is a family of them, one for each value of x. Choose a value, and you are back in Chapter 5 with an ordinary series to test.

The figure shows the running totals of the geometric series at four values of x. At one half they settle quickly at 2. At negative one half they settle at two thirds, approaching from alternate sides. At 0.9 they converge too, but slowly: after thirty terms they are still well short of 10. At 1.05 the running total runs off the top of the chart.

So the useful question is not does it converge, but for which x does it converge. The answer is a set of real numbers. The next few slides show that this set can only take a very restricted shape.

15. What shape can the convergence set take?

Prediction

\[ \left\{ x : \sum_{n=0}^{\infty} c_n x^n \text{ converges} \right\} \]

Commit to an answer before revealing. You have one example to reason from, the geometric series, whose convergence set is the open interval from minus one to one. Ask yourself whether that shape is an accident of one example or something forced.

It is forced. The key fact on the next slide shows that convergence at any point drags along every point that is closer to the centre. So the set cannot have holes and cannot be lopsided, except possibly at its two ends.

The options that fail are worth a look. The set is not always the interval from minus one to one: other series have other radii, including zero and infinity. And two separate intervals can never happen, because convergence at a far point would fill in the gap between them.

16. The key fact: convergence spreads inward

Concept

Suppose the series converges at some nonzero d. Then its terms tend to zero, so from some N on they are at most 1 in size.

\[ c_n d^n \to 0 \;\Longrightarrow\; |c_n d^n| \le 1 \text{ for } n \ge N \]

Now take any x closer to 0 than d is, and compare term by term.

\[ |c_n x^n| = |c_n d^n|\left|\frac{x}{d}\right|^n \le \left|\frac{x}{d}\right|^n \]

\[ |x| < |d| \;\Longrightarrow\; \left|\frac{x}{d}\right| < 1 \;\Longrightarrow\; \sum \left|\frac{x}{d}\right|^n \text{ converges} \]

By the comparison test the series converges absolutely at x. One point of convergence brings the whole open interval of radius |d| with it.

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 533-534 — proof of Theorem 6.1

This is the engine of the proof of Theorem 6.1, and it is short enough to follow line by line. Start from a single point d where the series converges. Its terms must tend to zero, by the divergence test, so from some point on they are at most one in size.

Now take any x that is closer to the centre than d is. Write each term at x as the term at d times the ratio x over d raised to the nth power. The first factor is at most one, so the term at x is at most the nth power of the size of x over d. Those powers form a geometric series with ratio less than one, which converges.

By the comparison test from Section 5.4, the series at x converges absolutely. One point of convergence has produced a whole interval of absolute convergence. Notice that the argument only used the distance from x to the centre, never its sign, which is why the interval is symmetric.

17. Convergence spreads inward, divergence spreads outward

Picture it

Read the same fact backwards: if the series diverged at some x star, it could not converge anywhere farther out, because convergence there would drag x star along. So divergence spreads outward, convergence inward, and they meet at a single distance from the centre.

The figure puts the key fact and its mirror image on one line. The yellow dot at 3 is a point of convergence, so everything strictly between minus 3 and 3 is forced to converge, shown in green.

The red dot at minus 4 is a point of divergence. If the series converged anywhere farther than 4 from the centre, the key fact would force it to converge at minus 4 as well, which it does not. So divergence pushes outward, on both sides, beyond distance 4.

What remains undecided is the stretch between distance 3 and distance 4, marked with question marks. As you learn about more points, that stretch narrows, and in the limit it closes down on a single distance from the centre. That distance is the radius of convergence.

18. Theorem 6.1: exactly three possibilities

Concept

• (i) converges at x = a only, and diverges for every other x
• (ii) converges for every real x
• (iii) there is an R > 0 with convergence for |x − a| < R and divergence for |x − a| > R

The proof takes the set of points of convergence. If it is neither the single centre nor everything, divergence somewhere bounds it, and its least upper bound is the R of case (iii).

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 533-535 — Theorem 6.1 and Figure 6.2

The three number lines are the book's Figure 6.2. Every power series you will ever meet is exactly one of these. In the first, only the centre converges and the radius is zero. In the second, every real number converges and the radius is infinite. In the third, there is a finite positive radius.

The book's proof collects all the points where the series converges into one set. If that set is just the centre, you are in case one; if it is everything, case two. Otherwise there is a point of divergence, so by the previous slide the set is bounded, and its least upper bound is the number R. The fact that every bounded set of reals has a least upper bound belongs to real analysis, so the book quotes it rather than proving it.

Case three is the common one, and its two question marks are where your work will concentrate.

19. Reading Theorem 6.1(iii) piece by piece

Notation

Read each part of case three against the number line. The expression for the size of x minus a is the distance from x to the centre. The theorem only ever talks about that distance, which is why left and right of the centre are treated identically.

Strictly inside the radius, the series converges, and absolutely, thanks to the key fact. Strictly outside, it diverges; in fact the terms do not even tend to zero there.

Exactly at distance R there are just two points, one on each side of the centre. The theorem deliberately says nothing about them. They can both converge, both diverge, or split, and the only way to find out is to substitute them into the series and test the resulting series of numbers.

20. The interval and the radius of convergence

Concept

interval of convergence — The set of all real x at which the power series converges.

radius of convergence — The number R in case (iii): the series converges within distance R of the centre and diverges beyond it.

\[ R = 0 \;\text{(only } x = a\text{)}, \qquad R = \infty \;\text{(all real } x\text{)} \]

The interval has length twice R. For the geometric series it is the open interval from minus 1 to 1, so the radius is 1.

\[ \sum_{n=0}^{\infty} x^n: \quad \text{interval } (-1, 1), \quad R = 1 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 534 — definition of interval and radius of convergence

Two names for two different things. The interval of convergence is a set: every x where the series converges, endpoints included or not. The radius is a single number: how far from the centre the convergence reaches.

The radius can be zero or infinite, and the conventions on the slide cover those cases. A radius of zero means the interval is just the centre. An infinite radius means the interval is the whole real line.

Two series can share a radius and still have different intervals, because the radius ignores the endpoints. You will see four such series side by side later in the lesson. When a question asks for the interval of convergence, a radius alone is never a complete answer.

21. Which endpoint can misbehave?

Counterexample

Write your example and your argument before revealing. A good way to build the example is to start from the alternating harmonic series, which converges, and the harmonic series, which diverges, and ask which power series turns into one at x equal to 1 and the other at x equal to minus 1.

Putting an extra factor of negative one to the n into the coefficients does it. At x equal to 1 the signs alternate and the series converges; at minus 1 the two sign factors cancel and you get the harmonic series, which diverges.

The second half is the key fact again. Convergence at 3 forces convergence at every point less than 3 from the centre, and minus 2 is such a point. Asymmetry is only possible at points exactly R from the centre, never inside.

22. Find the error: a convergence set that cannot exist

Error analysis

Before revealing the annotations, ask yourself which of the three shapes this claimed set could be. It is an interval, but it is not centred at zero, which is the centre of this series.

The contradiction comes straight from the key fact. If the series converges at 2, then it converges absolutely at every x of size less than 2, including all the negative numbers down to minus 2. The claimed set leaves all of those out.

An answer like this usually comes from solving an inequality with the absolute value dropped, so that only the positive half of the solution survives. Whenever you finish an interval of convergence, check that it is centred at the centre of the series. It is a ten-second check that catches a whole class of errors.

23. Finding the radius

Section

Part 3

24. The ratio test with a variable inside

Concept

Apply the ratio test to the absolute values of the terms. The powers of x minus a cancel down to a single factor.

\[ \rho = \lim_{n\to\infty}\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n}\right| \]

\[ \rho = |x-a| \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| = L\,|x-a| \]

\[ \rho < 1 \iff |x-a| < \frac1L \quad\Longrightarrow\quad R = \frac1L \]

If that coefficient limit L is zero, the ratio is below one for every x and the radius is infinite; if L is infinite, only the centre survives.

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 535 — 'we typically apply the ratio test'

Here is the method you will use for almost every power series in this chapter. Apply the ratio test to the absolute values of the terms, treating x as a fixed number. The powers of x minus a cancel down to a single factor, and what is left is that factor times the limit of the ratio of consecutive coefficients.

Call that coefficient limit L. The ratio test says the series converges absolutely when L times the distance from x to a is below one, which rearranges to the distance being below one over L. So the radius is one over L.

The two extreme cases fall out naturally. If L is zero, the product is zero for every x, which is below one, so the radius is infinite. If L is infinite, the product is infinite for every x except the centre, so the radius is zero. Example 6.1 shows all three cases in turn.

25. Example 6.1(a): factorials downstairs

Worked example

\[ \sum_{n=0}^{\infty}\frac{x^n}{n!} \]

\[ \rho = \lim_{n\to\infty}\left|\frac{x^{n+1}}{(n+1)!}\cdot\frac{n!}{x^n}\right| \]

\[ \rho = \lim_{n\to\infty}\left|x\right|\frac{n!}{(n+1)!} \]

\[ \frac{n!}{(n+1)!} = \frac{n!}{(n+1)\,n!} = \frac{1}{n+1} \]

\[ \rho = |x|\lim_{n\to\infty}\frac{1}{n+1} = 0 < 1 \]

\[ \text{interval } (-\infty, \infty), \quad R = \infty \]

\[ S_{20}(5) = \sum_{n=0}^{20}\frac{5^n}{n!} \approx 148.41315 \approx e^5 = 148.41316 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 535-536 — Example 6.1a

Follow the ratio line by line. After the powers cancel you are left with the size of x times the ratio of n factorial to n plus one factorial. The key identity is that n plus one factorial is n plus one times n factorial, so that ratio is simply one over n plus one.

For any fixed x, however large, dividing it by n plus one gives something that tends to zero. So the ratio test limit is zero, which is below one, for every real x. The radius is infinite.

The figure shows what this means at x equal to 5. The terms first grow, to about 26 at n equal to 4 and 5, because five is bigger than the early factorials. But once n passes 4 each ratio is below one and the factorial takes over for good. The partial sum with twenty-one terms is already within a hundred-thousandth of e to the fifth, a value you will be able to justify in Section 6.3.

26. Example 6.1(b): factorials upstairs

Worked example

\[ \sum_{n=0}^{\infty} n!\,x^n \]

\[ \rho = \lim_{n\to\infty}\left|\frac{(n+1)!\,x^{n+1}}{n!\,x^n}\right| \]

\[ \rho = \lim_{n\to\infty}(n+1)|x| \]

\[ \rho = |x|\lim_{n\to\infty}(n+1) = \infty > 1 \]

\[ \text{interval } \{0\}, \quad R = 0 \]

\[ 10!\,(0.1)^{10} \approx 0.00036, \quad 25!\,(0.1)^{25} \approx 1.55, \quad 30!\,(0.1)^{30} \approx 265 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 536 — Example 6.1b

Now the factorial is in the numerator, and the same cancellation leaves n plus one times the size of x. For any x other than zero this grows without bound, so the ratio test says diverge. Only the centre survives, and the radius is zero.

The figure makes the point at a very small x, one tenth. The vertical scale is logarithmic. The terms at first get smaller, because multiplying by one tenth beats the early factorials. But from n equal to 10 on, the ratio of consecutive terms is bigger than one, and after that the terms grow without limit: past one at n equal to 25, past a hundred million by 40. A smaller x only delays the moment.

The book's solution contains a misprint here: it says the series converges only for x not equal to zero, when it means only for x equal to zero, as the sentences around it make clear.

27. Example 6.1(c): the ratio part

Worked example

\[ \sum_{n=0}^{\infty}\frac{(x-2)^n}{(n+1)3^n} \]

\[ \rho = \lim_{n\to\infty}\left|\frac{(x-2)^{n+1}}{(n+2)3^{n+1}}\cdot\frac{(n+1)3^n}{(x-2)^n}\right| \]

\[ \rho = \lim_{n\to\infty}\left|\frac{(x-2)(n+1)}{3(n+2)}\right| \]

\[ \rho = \frac{|x-2|}{3} \]

\[ |x-2| < 3 \iff -3 < x - 2 < 3 \iff -1 < x < 5 \]

\[ R = 3, \quad \text{converges absolutely on } (-1, 5) \]

\[ x = 4: \rho = \tfrac23 < 1, \qquad x = 6: \rho = \tfrac43,\; \frac{(4/3)^{30}}{31} \approx 181 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 536 — Example 6.1c

This is the typical case, with a finite radius and a centre that is not zero. Form the ratio carefully: the next term has n plus 2 and a power of 3 one higher. The powers of x minus 2 and of 3 cancel down to single factors, and the ratio of n plus one to n plus 2 tends to one.

What remains is the distance from x to 2, divided by 3. Setting that below one gives a distance below 3, so the radius is 3 and the open interval runs from minus 1 to 5, centred at 2 as it should be.

The check samples a point on each side of the boundary. At x equal to 4 the ratio is two thirds, comfortably below one. At x equal to 6 it is four thirds, and the terms there visibly grow, to about 181 by n equal to 30. The ratio test has done its job. What it cannot do is decide x equal to minus 1 and x equal to 5, which is the next part of the lesson.

28. Radius zero, finite, or infinite?

Sorting

Sort each series by thinking about the ratio of consecutive coefficients, not by computing everything. If that ratio grows without bound, only the centre survives. If it tends to zero, every x works. If it settles at a positive number, the radius is the reciprocal of that number.

Factorials in the numerator, and the even faster-growing n to the n, give a coefficient ratio that grows without bound, so their radius is zero. Factorials in the denominator win against any fixed power, including ten to the n, so those radii are infinite.

The two in the middle are the ones worth a second look. For x to the n over n the coefficient ratio tends to one, so the radius is one. For x to the n over 2 to the n the ratio is one half, so the radius is its reciprocal, 2. It is easy to write one half there by mistake.

29. Stretch the radius

Tweak it

\[ \rho = \frac{|x|}{b} < 1 \iff |x| < b \quad\Longrightarrow\quad R = b \]

The curve is a partial sum with twenty-one terms of a geometric series in x over b. The ratio of consecutive terms is x over b, so the series converges exactly when x is within b of zero. The radius is b.

Start with b at 2 and move the slider slowly. Inside the band from minus b to b the curve is calm, close to the function it represents. Outside that band the twentieth power takes over and the curve shoots off the screen. As you raise b, the calm stretch widens in step with it.

This is the reciprocal rule from the last slides made visible. The coefficients here are one over b to the n, so their ratio is one over b, and the radius is the reciprocal of that, b itself. Coefficients that shrink faster give a wider interval.

30. The endpoints

Section

Part 4

31. At the edge, the ratio test goes silent

Concept

At the two points exactly R from the centre, the ratio comes out as exactly one, and the ratio test gives no verdict. In Example 6.1(c) those points are minus 1 and 5.

\[ \rho = \frac{|x-2|}{3} = 1 \iff x = -1 \text{ or } x = 5 \]

At each endpoint the power series is just a series of numbers. Substitute, and use whichever Chapter 5 test fits.

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 536 — 'The ratio test is inconclusive if rho = 1'

At exactly distance R from the centre, the ratio test limit equals one, the one value where the test gives no verdict. This is not bad luck in one example; it happens at the ends of every finite interval, because R was defined as the distance where the limit crosses one.

So at each endpoint, the power series becomes an ordinary series of numbers, and you test it with whatever Chapter 5 tool fits. The figure shows the two endpoints of Example 6.1(c). On the left, at minus 1, the running totals zig-zag in toward ln 2. On the right, at 5, they keep climbing, tracking a logarithm.

Two points at the same distance from the centre, the same ratio limit, and opposite verdicts. That is why every endpoint is its own small problem.

32. Example 6.1(c): testing both endpoints

Worked example

The ratio test gave convergence on the open interval from minus 1 to 5. Settle the two ends.

\[ \frac{(-3)^n}{(n+1)3^n} = \frac{(-1)^n 3^n}{(n+1)3^n} = \frac{(-1)^n}{n+1} \]

\[ \sum_{n=0}^{\infty}\frac{(-1)^n}{n+1} = 1 - \frac12 + \frac13 - \frac14 + \cdots \text{ converges} \]

\[ \frac{3^n}{(n+1)3^n} = \frac{1}{n+1} \]

\[ \sum_{n=0}^{\infty}\frac{1}{n+1} = 1 + \frac12 + \frac13 + \cdots = \infty \]

\[ \text{interval } [-1, 5), \quad R = 3 \]

\[ \sum_{n=0}^{200000}\frac{(-1)^n}{n+1} \approx 0.693150 \approx \ln 2, \qquad \sum_{n=0}^{1000}\frac{1}{n+1} \approx 7.49 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 536-537 — Example 6.1c

Substitute each endpoint and simplify before choosing a test. At x equal to minus 1, the base x minus 2 becomes minus 3, and its nth power is negative one to the n times 3 to the n. The powers of 3 cancel with the denominator, leaving the alternating harmonic series. Its terms shrink to zero in size and decrease, so the alternating series test gives convergence.

At x equal to 5 the base becomes plus 3, the powers of 3 cancel again, and you get the harmonic series, which diverges. So minus 1 is in and 5 is out.

The number line records the answer with the usual conventions: a filled dot and a square bracket for an included endpoint, a hollow dot and a round bracket for an excluded one. The numerical check backs this up: two hundred thousand terms at minus 1 agree with ln 2 to about five decimal places, while a thousand terms at 5 give only 7.49 and are still climbing.

33. Checkpoint 6.1: the sum of x to the n over n

Worked example

\[ \sum_{n=1}^{\infty}\frac{x^n}{n} \]

\[ \rho = \lim_{n\to\infty}\left|\frac{x^{n+1}}{n+1}\cdot\frac{n}{x^n}\right| = |x|\lim_{n\to\infty}\frac{n}{n+1} \]

\[ \rho = |x| < 1 \iff -1 < x < 1, \quad R = 1 \]

\[ \sum_{n=1}^{\infty}\frac{1}{n} \text{ diverges} \]

\[ \sum_{n=1}^{\infty}\frac{(-1)^n}{n} = -1 + \frac12 - \frac13 + \cdots \text{ converges} \]

\[ \text{interval } [-1, 1), \quad R = 1 \]

\[ S_{10}(-1) \approx -0.6456, \quad S_{200000}(-1) \approx -0.693145 \approx -\ln 2 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 537 — Checkpoint 6.1

Try this one yourself before stepping through it; it is the book's checkpoint for Example 6.1. The ratio of consecutive terms is the size of x times n over n plus one, which tends to the size of x. So the radius is one, centred at zero.

At x equal to 1 the series is the harmonic series, which diverges. At x equal to minus 1 the signs alternate, and the alternating series test gives convergence. So the interval is closed on the left and open on the right.

The check uses a fact you will prove in Section 6.2: the sum at minus 1 is minus ln 2. The partial sums close in on it, slowly, as alternating harmonic sums always do: ten terms are only good to about one decimal place, and two hundred thousand terms to about five.

34. Exercise 13: doubling the variable

Worked example

\[ \sum_{n=1}^{\infty}\frac{(2x)^n}{n} \]

\[ \rho = \lim_{n\to\infty}\left|\frac{(2x)^{n+1}}{n+1}\cdot\frac{n}{(2x)^n}\right| = 2|x|\lim_{n\to\infty}\frac{n}{n+1} = 2|x| \]

\[ 2|x| < 1 \iff |x| < \tfrac12, \quad R = \tfrac12 \]

\[ \sum_{n=1}^{\infty}\frac{1^n}{n} = \sum\frac1n \text{ diverges} \]

\[ \sum_{n=1}^{\infty}\frac{(-1)^n}{n} \text{ converges} \]

\[ \text{interval } \left[-\tfrac12, \tfrac12\right), \quad R = \tfrac12 \]

\[ u = 2x \in [-1, 1) \iff x \in \left[-\tfrac12, \tfrac12\right) \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 541 — Exercise 13

This exercise is the checkpoint series with x replaced by 2x. Doing it from scratch first, then checking by substitution, shows you how the two are related.

In the ratio, the factor that survives is 2x instead of x, so the limit is twice the size of x. Setting that below one gives a size of x below one half, so the radius has halved. At x equal to one half, 2x is 1 and you get the harmonic series, which diverges. At x equal to minus one half, 2x is minus 1 and you get the alternating harmonic series, which converges.

The check shows a shortcut you can use whenever a series is a known one evaluated at a multiple of x. If the series in u converges exactly for u in some interval, the series in 2x converges exactly when 2x is in that interval. Halving the ends gives the answer directly.

35. Squares downstairs: both endpoints close

Worked example

The series of Exercise 59, which the book asks you to plot.

\[ \sum_{n=1}^{\infty}\frac{x^n}{n^2} \]

\[ \rho = |x|\lim_{n\to\infty}\frac{n^2}{(n+1)^2} = |x| \]

\[ |x| < 1, \quad R = 1 \]

\[ \sum_{n=1}^{\infty}\frac{1}{n^2} \text{ converges} \]

\[ \sum_{n=1}^{\infty}\left|\frac{(-1)^n}{n^2}\right| = \sum \frac{1}{n^2} < \infty \]

\[ \text{interval } [-1, 1], \quad R = 1 \]

\[ \sum_{n=1}^{1000}\frac{1}{n^2} \approx 1.64393 < \frac{\pi^2}{6} \approx 1.64493, \quad \sum\frac{(-1)^n}{n^2} = -\frac{\pi^2}{12} \approx -0.82247 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 543 — Exercise 59

The ratio test here gives exactly the same radius as the last two examples, because n squared over n plus one squared still tends to one. The difference is entirely at the endpoints.

At x equal to 1 the series is a p-series with p equal to 2, which converges by the rule from Section 5.3. At x equal to minus 1 the terms alternate, but you do not even need the alternating series test: their sizes form the same convergent p-series, so the series converges absolutely. Both endpoints are included.

The check uses the exact values, pi squared over 6 and minus pi squared over 12. With positive terms, a partial sum must sit below the full sum, and a thousand terms give 1.64393, just below 1.64493. Compare this series with Checkpoint 6.1: the extra power of n in the denominator is what closed the right-hand end.

36. One radius, four different intervals

Picture it

Four series, one radius. The ratio test hands all four the same open interval; the four different answers come entirely from the endpoint tests.

These four series all have radius one, and the ratio test cannot tell them apart. The number lines show that their intervals are all different, and that every combination of open and closed ends occurs.

The geometric series is open at both ends because its terms at x equal to plus or minus one have size one and do not tend to zero. Dividing by n closes one end, whichever end gives the alternating signs. Dividing by n squared closes both, because the p-series with p equal to 2 converges absolutely.

Take this picture with you as a warning. The radius tells you where the answer is interesting; it does not tell you the answer at the two points where it is most interesting.

37. Complete the endpoint table

Comparison

Fill in every blank before you check. For each one, substitute the endpoint and name the series you get: a geometric series with ratio one or minus one, a harmonic series, an alternating harmonic series, or a p-series.

The row for the series with the extra factor of negative one to the n is the mirror image of the row above it. At x equal to 1 its signs alternate, so it converges; at minus 1 the two sign factors multiply to plus one and you get the harmonic series.

If a row gives you trouble, go back to the question of whether the terms at that endpoint even tend to zero. If they do not, as for the geometric series, the divergence test settles it in one line.

38. Trap: letting the ratio test decide the endpoints

Trap

39. Exercise 5: where must it converge?

Discrimination

\[ \sum_{n=0}^{\infty} a_n (x-3)^n \text{ converges at } x = 6 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 541 — Exercise 5

Start by finding the centre, which is 3, and the distance from the centre to the point of known convergence, which is 3 as well. The key fact then tells you the radius is at least 3, so every point strictly less than 3 away from the centre must converge.

Measure each candidate's distance from 3. The points 1, 2, 3 and 5.99 are clearly inside. The tiny positive number 0.000001 is inside too, by the smallest of margins: its distance is 2.999999.

The point 0 is the interesting one. Its distance from the centre is exactly 3, the same as 6. If the radius happens to be exactly 3, then 0 and 6 are the two endpoints, and endpoints can disagree, as you have seen several times. So convergence at 0 is not guaranteed.

40. Power series as functions

Section

Part 5

41. The geometric series, read backwards

Concept

The partial sums of the geometric series have a closed form. Multiply by one minus x and almost everything cancels.

\[ (1-x)(1 + x + \cdots + x^N) = 1 - x^{N+1} \]

\[ S_N(x) = \frac{1 - x^{N+1}}{1-x} \]

When x is between minus 1 and 1, the power in the numerator dies away, so the partial sums tend to one over one minus x.

\[ |x| < 1: \quad x^{N+1} \to 0 \;\Longrightarrow\; \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 537 — equation (6.3)

This is the fact that turns power series into a way of writing functions. Multiply the Nth partial sum of the geometric series by one minus x, and every term cancels except the first and the last. Dividing back gives a closed form for the partial sum.

Now let N grow. If x is between minus one and one, the power of x in the numerator shrinks to zero, and the partial sums tend to one over one minus x. That is the geometric series formula from Section 5.2, with the ratio called x.

The new idea is the direction you read it in. In Chapter 5 you started with the series and found its sum. Here you start with the function one over one minus x and say that, on the interval from minus one to one, this function is a power series. Every representation in this lesson comes from bending a function into this shape.

42. Reading the representation piece by piece

Notation

Look at each annotation in turn. The left side is a function you have known for years; it is defined everywhere except at x equal to 1. The right side is a power series with every coefficient equal to one.

The equals sign needs care. It asserts that the two sides have the same value, and that is only true where the right side has a value at all, which is on the interval from minus one to one. Outside, the left side is still perfectly defined but the right side diverges.

So the condition on the size of x is part of the statement, not a detail to drop. When you write a power series for a function, always write its interval beside it. A later slide shows the nonsense that follows when the condition is forgotten.

43. Two readings of one equation

Intuition

\[ \frac{1}{1-r} = 1 + r + r^2 + \cdots, \quad |r| < 1 \]

Read left to right, as in Chapter 5, it adds up a series: given the ratio, find the sum. Read right to left, it writes a function as a power series. The second reading is the new one.

Any time you can bend a function into the shape one over one minus something, the something plays the part of r, and the condition that its size is below 1 gives the interval for free.

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 538 — lead-in to Example 6.3

The same identity has two jobs. Reading left to right, as you did in Chapter 5, you have a series and you want its sum. Reading right to left, you have a function and you want a series for it. Both directions are the same equation.

The practical recipe comes from the second reading. Look at a function and try to write it as some constant times one over one minus something. Whatever sits in the something plays the role of r: the series is the constant times the powers of r, and the interval is wherever the size of r is less than one.

The next few examples are all this recipe. The only skill is the algebra that reveals the shape: turning a plus into a minus of a negative, factoring a constant out of the denominator, or rewriting around a new centre.

44. Example 6.2: partial sums against the function

Worked example

Compare f with its partial sums of degree 2, 4 and 6 on the interval from minus 1 to 1.

\[ f(x) = \frac{1}{1-x}, \qquad S_N(x) = \sum_{n=0}^{N} x^n \]

\[ S_2 = 1 + x + x^2, \quad S_4 = S_2 + x^3 + x^4, \quad S_6 = S_4 + x^5 + x^6 \]

\[ S_2 = 1.75, \quad S_4 = 1.9375, \quad S_6 = 1.984375, \quad f = 2 \]

\[ S_2 = 2.71, \quad S_4 = 4.0951, \quad S_6 = 5.2170, \quad f = 10 \]

\[ f - S_N = \frac{x^{N+1}}{1-x}: \quad \frac{(0.5)^7}{0.5} = 0.015625 = 2 - 1.984375 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 537-538 — Example 6.2 and Figure 6.3

The book asks for a sketch, and the figure is it. The three partial sums are polynomials of degree 2, 4 and 6. Near zero they are all practically on top of the function. Toward x equal to 1, the function climbs to infinity and every polynomial falls behind, the higher-degree ones less badly.

The numbers make the same point. At one half, the three partial sums are 1.75, 1.9375 and 1.984375, closing in on 2. At 0.9 they are 2.71, 4.10 and 5.22, while the function is 10: the series does converge there, but six terms are nowhere near enough.

The check comes from the closed form on the previous slide. The exact gap between the function and the Nth partial sum is x to the N plus one over one minus x. At one half with N equal to 6 that is exactly the difference you computed. The same formula explains the picture: the error is small when x is small, and blows up near 1.

45. Add terms and watch the fit

Tweak it

Start with N at 4 and raise it one step at a time. Inside the interval from minus one to one, look at a single point, such as x equal to one half. The curve's height there creeps up toward 2 and then stops moving. Near x equal to 0.9 it takes many more steps to approach 10.

Now watch outside the interval. To the right of 1, each extra term pushes the curve higher, while the function it is supposed to represent is negative there. To the left of minus 1, the curve flips between very high and very low values as N switches between even and odd.

This is the picture behind the words interval of convergence. Inside, adding terms always helps. Outside, adding terms always makes things worse, and no value of N will ever fix it.

46. Inside the interval they hug; outside they peel away

Picture it

The closed form explains the picture. When x is bigger than 1 in size, the power in the numerator grows without bound, so each extra term makes the partial sum worse, not better.

\[ |x| > 1: \quad |x|^{N+1} \to \infty \]

The figure freezes three positions of the slider: 5, 10 and 20 extra powers. Inside the shaded band the three curves sit on top of the function so closely that they are hard to tell apart, except near the ends.

Outside the band they separate from the function and from each other. On the right the function is negative, minus 5 at x equal to 1.2, while the partial sums there are 9.93, 32.15 and 225. On the left they swing up and down with growing amplitude.

The closed form of the partial sums explains why. When x is bigger than one in size, the power in the numerator grows without bound, so every extra term adds a larger error. The function continues past the ends of the band, but the series does not follow it.

47. How many terms at x = 0.99?

Estimation

\[ \left|\frac{1}{1-x} - S_N(x)\right| = \frac{x^{N+1}}{1-x} < 10^{-6} \quad \text{at } x = 0.99 \]

\[ (0.99)^{N+1} < 10^{-8} \iff N + 1 > \frac{\ln 10^{-8}}{\ln 0.99} \approx 1832.8 \]

Commit to a guess before revealing. Most people guess far too low, because near the centre a handful of terms is plenty.

The exact error formula makes this a short computation. The error is 0.99 to the N plus one, divided by one hundredth, and you need it below one millionth, so the power itself must be below one hundred-millionth. Taking natural logarithms of both sides, and remembering that the logarithm of 0.99 is negative so the inequality flips, gives N plus one at least about 1833.

So it takes more than eighteen hundred terms to pin down a value of 100 to six decimal places. Convergence near the edge of the interval is real, but slow. This is exactly why later sections care about choosing the centre well: a series centred near the point you need converges much faster there.

48. Checkpoint 6.2: one over one minus x squared

Worked example

\[ f(x) = \frac{1}{1-x^2}, \qquad S_N(x) = \sum_{n=0}^{N} x^{2n} \]

\[ \frac{1}{1-x^2} = \frac{1}{1-r}, \quad r = x^2 \]

\[ \frac{1}{1-x^2} = \sum_{n=0}^{\infty}(x^2)^n = 1 + x^2 + x^4 + x^6 + \cdots \]

\[ |x^2| < 1 \iff |x| < 1 \]

\[ S_2 = 1 + x^2 + x^4, \quad S_4 = S_2 + x^6 + x^8, \quad S_6 = S_4 + x^{10} + x^{12} \]

\[ S_2 = 1.3125, \quad S_4 \approx 1.332031, \quad S_6 \approx 1.333252 \to \tfrac43 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 538 — Checkpoint 6.2

This checkpoint is Example 6.2 with x squared in place of x. The function has the geometric shape with ratio x squared, so its series is the powers of x squared: only even powers of x appear.

The interval comes from the ratio's size being below one. The square of x is below one exactly when the size of x is below one, so the interval is again from minus one to one. This time the function blows up at both ends, and the figure shows the partial sums failing at both ends together.

The partial sums asked for, with N equal to 2, 4 and 6, are polynomials of degree 4, 8 and 12. Be careful with that: N counts terms of the series, not the degree. At one half they are 1.3125, 1.332 and 1.33325, closing in on four thirds. Every one of them is an even function, symmetric about the vertical axis, just like the function itself.

49. Example 6.3(a): one over one plus x cubed

Worked example

\[ f(x) = \frac{1}{1+x^3} \]

\[ \frac{1}{1+x^3} = \frac{1}{1-(-x^3)} \]

\[ \frac{1}{1-(-x^3)} = \sum_{n=0}^{\infty}(-x^3)^n \]

\[ \sum_{n=0}^{\infty}(-1)^n x^{3n} = 1 - x^3 + x^6 - x^9 + \cdots \]

\[ |-x^3| < 1 \iff |x|^3 < 1 \iff |x| < 1 \]

\[ 1 - \tfrac18 + \tfrac1{64} - \tfrac1{512} + \tfrac1{4096} \approx 0.888916 \approx \tfrac89 = 0.888889 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 538-539 — Example 6.3a

The move here is to see a plus as a minus of a negative. One plus x cubed is one minus the quantity negative x cubed, and now the geometric shape is visible, with ratio negative x cubed.

Expanding gives the powers of negative x cubed. Each power splits into an alternating sign and a power of x cubed, so the series runs one minus x cubed plus x to the sixth minus x to the ninth, stepping up by three each time. The interval is where the size of negative x cubed is below one, which is where the size of x is below one.

Now look at the figure, because it shows something surprising. The function blows up only at minus 1; to the right of 1 it is smooth and small. Yet the series diverges to the right of 1 as well. The interval of convergence is symmetric about the centre, so trouble at distance one on the left limits the series at distance one on the right. The check at one half confirms the representation to four decimal places with five terms.

50. Before solving the inequality

Step zero

\[ \frac{x^2}{4-x^2} = \frac{x^2}{4}\cdot\frac{1}{1-\left(\frac{x}{2}\right)^2} \]

Write your answer before revealing. The rewriting has already been done; the question is what inequality the ratio must satisfy and what it means for x.

The ratio is the square of x over 2. Its size is below one exactly when the size of x over 2 is below one, which means the size of x is below 2. Another way to see it: x squared over 4 below one means x squared below 4, and the square root of 4 is 2.

The tempting wrong answer, 4, comes from treating x squared over 4 as though it were x over 4. That confuses the square with the variable. When the ratio contains a power of x, always solve for x itself at the end, and remember that the square root undoes the square on both sides of the inequality.

51. Example 6.3(b): x squared over four minus x squared

Worked example

\[ f(x) = \frac{x^2}{4-x^2} \]

\[ \frac{x^2}{4-x^2} = \frac{x^2}{4\left(1 - \frac{x^2}{4}\right)} = \frac{x^2}{4}\cdot\frac{1}{1-\left(\frac{x}{2}\right)^2} \]

\[ \frac{x^2}{4-x^2} = \frac{x^2}{4}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{2n} \]

\[ \frac{x^2}{4}\cdot\frac{x^{2n}}{4^n} = \frac{x^{2n+2}}{4^{n+1}} \]

\[ \frac{x^2}{4-x^2} = \sum_{n=0}^{\infty}\frac{x^{2n+2}}{4^{n+1}} = \frac{x^2}{4} + \frac{x^4}{4^2} + \frac{x^6}{4^3} + \cdots \]

\[ |x| < 2: \quad \text{interval } (-2, 2) \]

\[ \sum_{n=0}^{\infty}\frac{1}{4^{n+1}} = \frac{1/4}{1 - 1/4} = \frac13 = \frac{1^2}{4-1^2} \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, pp. 539-540 — Example 6.3b

The function is not yet in geometric shape, because the denominator starts with 4, not 1. Factoring 4 out of the denominator fixes that, and leaves an extra factor of x squared over 4 in front.

Expanding the geometric part with ratio x over 2, squared, gives powers of x squared over 4. Multiplying each term by the factor in front combines the powers of x and of 4, and the series comes out as x to the 2n plus 2 over 4 to the n plus 1. Writing out the first terms, x squared over 4 plus x to the fourth over 16 and so on, is a good check that the indices are right.

The interval is from minus 2 to 2, open at both ends, because at x equal to plus or minus 2 every term equals one. The check at x equal to 1 is exact: the series becomes a geometric series with first term and ratio both a quarter, which sums to one third, the same as the function's value.

52. Checkpoint 6.3: x cubed over two minus x

Worked example

\[ f(x) = \frac{x^3}{2-x} \]

\[ \frac{x^3}{2-x} = \frac{x^3}{2\left(1-\frac{x}{2}\right)} = \frac{x^3}{2}\cdot\frac{1}{1-\frac{x}{2}} \]

\[ \frac{x^3}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n = \sum_{n=0}^{\infty}\frac{x^{n+3}}{2^{n+1}} \]

\[ \frac{x^3}{2} + \frac{x^4}{4} + \frac{x^5}{8} + \cdots \]

\[ \left|\frac{x}{2}\right| < 1 \iff |x| < 2: \quad \text{interval } (-2, 2) \]

\[ \sum_{n=0}^{\infty}\frac{1}{2^{n+1}} = 1 = \frac{1^3}{2-1}, \quad S_5 = 0.984375 \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 540 — Checkpoint 6.3

Try this before stepping through it; it combines both moves from Example 6.3(b). Factor 2 out of the denominator to get a leading 1, and the ratio is x over 2.

The factor x cubed over 2 in front shifts every power up by three and every power of 2 up by one, so the series is x to the n plus 3 over 2 to the n plus 1. Its first term is x cubed over 2, not a constant, which is fine: a power series may have some coefficients equal to zero.

The interval is where the size of x over 2 is below one, from minus 2 to 2, with both ends excluded because the terms there have size 4 and do not tend to zero. The check at x equal to 1 is exact: the series is one half plus one quarter plus one eighth and so on, which sums to 1, the value of the function. Six terms already give 0.984375.

53. Exercise 33: one over x, centred at 1

Worked example

Use the hint in the exercise to write one over x as a power series centred at 1.

\[ f(x) = \frac1x, \quad a = 1 \]

\[ \frac1x = \frac{1}{1-(1-x)} \]

\[ \frac{1}{1-(1-x)} = \sum_{n=0}^{\infty}(1-x)^n \]

\[ (1-x)^n = (-1)^n(x-1)^n \;\Longrightarrow\; \frac1x = \sum_{n=0}^{\infty}(-1)^n (x-1)^n \]

\[ |x-1| < 1 \iff 0 < x < 2: \quad \text{interval } (0, 2), \quad R = 1 \]

\[ \sum_{n=0}^{\infty}\left(-\tfrac12\right)^n = \frac{1}{1+\frac12} = \frac23 = \frac{1}{1.5} \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 542 — Exercise 33

One over x cannot have a power series centred at zero, because it is not even defined there. Centred at 1 it works, and the hint shows how: write x as 1 minus the quantity 1 minus x, so the function takes the geometric shape with ratio 1 minus x.

The series is then the powers of 1 minus x. To present it properly as a series centred at 1, rewrite each power in terms of x minus 1: one minus x is the negative of x minus 1, so the coefficients alternate in sign.

The interval is where the distance from x to 1 is below one, which is from 0 to 2. At 0 every term is 1, and at 2 the terms alternate between 1 and minus 1, so both ends diverge. The figure shows the partial sums hugging the curve on that band and peeling away beyond 2. The check at 1.5 is exact: a geometric series with ratio negative one half sums to two thirds.

54. The radius reaches to the nearest trouble

Concept

In every example so far, the radius is the distance from the centre to the nearest point where the function blows up.

The interval is symmetric, so trouble on one side limits the series on the other side too, even where the function itself is perfectly smooth.

\[ \frac{1}{1+x^3} \text{ is smooth at } x = 1.2, \text{ but } \sum(-x^3)^n \text{ diverges there} \]

Look down the table. In every example, the radius equals the distance from the centre to the nearest point where the function blows up. That is not a coincidence, and it gives you a quick way to predict a radius before computing it.

The reason is that the series cannot converge at a point where the function has no value, and by the key fact it cannot converge beyond that distance on either side. So the blow-up caps the radius, and trouble on one side limits the series on the other side too. That is why the series for one over one plus x cubed fails at 1.2, where the function is perfectly smooth.

One honest warning. The rule really concerns trouble in the complex numbers, which this course does not cover. One over one plus x squared has no real blow-up at all, yet its series has radius 1. So use the rule to predict and to check, but let the ratio test and the geometric condition have the final word.

55. Find the error: one plus two plus four equals minus one

Error analysis

Read the claim and decide where it goes wrong before revealing the annotations. Every individual step looks legal: a correct identity, then substitution.

The identity is only true when the size of x is below one, and x equal to 2 is far outside that interval. On the left, one over one minus 2 is a perfectly good number, minus 1. On the right there is no number at all: the partial sums are 1, 3, 7, 15 and so on, doubling and adding one each time, growing without bound. A sum of positive numbers could never equal minus 1 anyway.

This is the reason to write the interval next to every representation. The equals sign between a function and its power series is a statement about the interval of convergence and nowhere else.

56. Trap: more terms will stretch the interval

Trap

57. Rank the tenth partial sum's accuracy

Ranking

\[ \left|\frac{1}{1-x} - S_{10}(x)\right| = \frac{|x|^{11}}{|1-x|} \]

Use the exact error formula on the slide: the size of x to the eleventh power, divided by the size of one minus x. You do not need a calculator to get the order right, only to confirm it.

The biggest influence is the distance from the centre, because an eleventh power shrinks very fast for small x and grows toward one as x approaches the edge. So 0.1 is by far the best, and the two points at size 0.9 are the worst.

The subtle part is the pairs at the same distance. At minus one half the denominator is one and a half, while at plus one half it is one half, so the error on the negative side is three times smaller. The same happens at plus and minus 0.9. Near the blow-up at 1 the function is steep and hard to follow; on the far side it is gentle.

58. Echoes that add up

Real world

Write your answer before revealing. The first echo carries 0.6 of the energy, the second 0.6 of that, and so on, so the total is a geometric series with ratio 0.6. The sum is one over one minus 0.6, which is two and a half times the original sound. The running totals in the figure level off at exactly that height.

Now treat the reflection ratio r as a variable. The total is the power series in r with every coefficient one, and it is finite exactly on the interval of convergence you found at the start of Part 5. As r approaches 1 the total grows without bound.

This is the mathematics of audio feedback. A microphone picks up its own speaker, the signal goes around the loop, and each pass multiplies it by some gain. Below a gain of one the loop settles. At one or above the series diverges, and you hear it as a howl.

59. Putting it together

Section

Part 6

60. Pattern: finding an interval of convergence

Pattern

1. Identify the centre a: the number subtracted from x.
2. Ratio test on the absolute values: simplify to a constant times |x − a|, or to 0, or to infinity.
3. Solve ρ < 1 for the open interval; its half-width is R.
4. Substitute each endpoint and test the numerical series with Chapter 5 tools (p-series, alternating series, divergence test).
5. Write the interval with the right bracket at each end.
6. To represent a function, bend it into a/(1 − r); the interval is |r| < 1.

This is the procedure to follow every time, and the diagram shows its shape: one computation for the middle of the interval, then two separate problems for the ends.

Start by identifying the centre, because a sign error there shifts everything. Then run the ratio test with x held fixed and simplify until you have a constant times the distance from x to the centre. Setting that below one gives the radius and the open interval.

Then, and this is the step people skip, substitute each endpoint and name the series of numbers you get. The divergence test, the p-series rule, the alternating series test and the comparison test are the usual tools. Only then write the interval, with the correct bracket at each end. For a representation, the last item is a shortcut: once a function is in geometric shape, the interval comes straight from the ratio.

61. Check: reading off the radius

Check

The coefficients here are one over 5 to the n, so the ratio of consecutive terms is the distance from x to 1, divided by 5. Setting that below one gives a distance below 5, so the radius is 5.

The wrong options are each a real number from the problem in the wrong role. One fifth is the coefficient ratio, whose reciprocal is the radius. One is the centre. Six is the right-hand end of the interval, the centre plus the radius. Keep the three roles apart: centre, radius, ends.

62. Check: the endpoints

Check

The ratio test gives radius one about the centre 3, so the open interval is from 2 to 4. All four options agree on that; the question is entirely about the endpoints.

At x equal to 2 the base is minus 1, the terms are negative one to the n over n, and the alternating series test gives convergence. At x equal to 4 the base is 1 and you get the harmonic series, which diverges. So the square bracket belongs at 2 and the round one at 4. If you chose the reverse, check which endpoint makes the base negative.

63. Check: representing a function

Check

Bend the function into the geometric shape first. One plus 4 x squared is one minus the quantity negative 4 x squared, so the ratio is negative 4 x squared and the series is its powers.

The interval comes from the size of the ratio being below one: 4 x squared below one, so x squared below a quarter, so the size of x below one half. The most common error is to write the interval of the plain series in x squared, which is the size of x below one. The factor of 4 inside the ratio shrinks the interval.

64. Explain the silence at the endpoints

Explain it to yourself

Write your explanation before revealing. A good answer names the ratio test limit at the endpoints and says why the two ends can still differ.

The radius was defined as the distance at which the ratio test limit crosses one, so at that distance the limit is exactly one, the single case where the test has no verdict. That is true of every power series with a finite positive radius.

The two endpoints give terms of the same size, because the size is all the ratio test sees. What differs is the sign. On one side the powers of a negative number make the signs alternate, which can rescue convergence through the alternating series test. On the other side all the signs agree, and the same sizes may be too big to converge.

65. Exit ticket

Exit ticket

\[ \sum_{n=1}^{\infty}\frac{n\,x^n}{2^n} \]

OpenStax Calculus Volume 2, §6.1 Power Series and Functions §6.1, p. 541 — Exercise 15

This brings the whole procedure together on one series from the exercise set. Compute the ratio, find the radius, then test both ends.

The ratio is the size of x over 2 times n plus one over n, which tends to the size of x over 2. So the radius is 2. At x equal to 2 the powers of 2 cancel and the terms are just n, which do not tend to zero. At minus 2 the terms are negative one to the n times n, which do not tend to zero either. Both ends fail the divergence test, and the interval is open at both ends.

If you wanted to reach for the alternating series test at minus 2, notice that it does not apply: its terms must shrink to zero in size, and these grow.

66. Recap

Recap

\[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \quad |x| < 1 \]

Next, Section 6.2 differentiates and integrates power series term by term, which turns this one geometric series into series for logarithms and arctangents.

Stewart, Calculus: Early Transcendentals 8e, §11.8 Power Series §11.8, pp. 746-751 — the same material in Stewart

A power series is an infinite polynomial built around a centre. It always converges at the centre, and Theorem 6.1 says its convergence set is only ever the centre alone, the whole line, or an interval of finite radius around the centre.

To find the interval, use the ratio test to get the radius, then test each endpoint on its own with the Chapter 5 tools. The endpoints can go either way, independently, and a radius alone is never a complete answer.

The geometric series, read backwards, is your first source of representations: bend a function into the shape one over one minus r and the interval is where r has size below one. Section 6.2 differentiates and integrates these series term by term, which turns this one family into series for logarithms and arctangents.