6.2 Properties of Power Series

1. Properties of Power Series

Title

Calculus II · Section 6.2

One geometric series, and everything you can build from it

2. What this lesson gives you

Objectives

Section 6.1 gave you one power series you can trust completely, the geometric series. This lesson shows how to manufacture new ones from it by algebra and by calculus, and how to keep track of where each new series is valid.

• Add, subtract and scale power series, and find the interval of the result
• Build new series by multiplying by a power of x and by substituting for the variable
• Multiply two power series with the Cauchy product
• Differentiate and integrate a power series term by term, and retest the endpoints every time
• Explain why a function has only one power series at a given centre

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 544 — learning objectives 6.2.1 to 6.2.4

In Section 6.1 you met exactly one power series whose sum you know in closed form: the geometric series, one over one minus x. That sounds like a thin starting point, but this whole lesson is about how far one series can take you.

The idea is to treat a power series the way you treat a polynomial. You can add two of them, multiply by x, substitute something for x, multiply two together, and even differentiate and integrate. Each of those moves turns the geometric series into a series for a new function: rational functions, the logarithm, the arctangent.

Every move has a price, and the price is always paid in the interval of convergence. So alongside each operation you will learn what it does to the radius, and the one rule that never changes: the endpoints must be tested again by hand.

3. Before anything new: the one series you know

Warm-up

Write your answer before revealing. You should be able to produce the geometric series and its interval without thinking, because it is the seed for everything today.

The condition is that the size of x is less than 1. At x equal to 1 the series is one plus one plus one forever, and at minus 1 it bounces between one and zero. In both cases the terms do not shrink to zero, so the divergence test kills them. The interval is open at both ends.

Hold on to two facts from this: the formula one over one minus x makes sense for almost every x, but the series only for x strictly between minus 1 and 1. That gap between the function and its series is what every interval calculation today is really about.

4. Combining power series

Section

Part 1

5. A power series is a function you can do algebra on

Concept

On its interval, a power series is a function, and its partial sums are polynomials. Polynomials can be added, scaled and multiplied, so it is natural to hope power series can too.

\[ \frac{1}{1-x} \approx S_N(x) = 1 + x + \cdots + x^N \quad (|x| < 1) \]

The catch is visible in the picture: every statement about the series holds only where the series converges. Each operation in this lesson comes with a rule for the new interval.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 544 — introduction

Look at the figure. The solid curve is one over one minus x. The dashed curves are partial sums, which are just polynomials. Inside the green band they sit on top of the curve, and the more terms you take, the more of the band they cover.

Outside the band the story is completely different. To the left of minus 1 the partial sums swing up and down more wildly with each new term, and to the right of 1 they shoot upward even where the curve is negative. There, no number of terms helps.

That picture is the reason for every rule in this lesson. Inside the interval a power series behaves like a polynomial that never ends, so the usual algebra works. Outside it there is no function to do algebra on. So each time you build a new series, the first question is where it lives.

6. Theorem 6.2: combining power series

Concept

Suppose the two series below converge to f and g on a common interval I.

\[ \sum_{n=0}^{\infty} c_n x^n = f(x), \qquad \sum_{n=0}^{\infty} d_n x^n = g(x) \qquad (x \in I) \]

\[ \text{(i) } \sum_{n=0}^{\infty} (c_n \pm d_n)x^n = f(x) \pm g(x) \text{ on } I \]

\[ \text{(ii) } \sum_{n=0}^{\infty} b x^m c_n x^n = b x^m f(x) \text{ on } I \]

\[ \text{(iii) } \sum_{n=0}^{\infty} c_n (b x^m)^n = f(b x^m) \text{ wherever } b x^m \in I \]

Here m is a whole number, zero or more, and b is any real number. The same statements hold for series centred at any a.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 544 — Theorem 6.2

Theorem 6.2 bundles three operations. Part one says you may add or subtract two series term by term, as long as you stay where both converge. Part two says you may multiply every term by a constant times a power of x. Part three says you may substitute a constant times a power of x for the variable.

Read part three carefully, because it is the one you will use most. The new series converges to f of b x to the m, and it does so wherever b x to the m lands inside the old interval. The condition is on the substituted expression, and you have to solve it for x.

The book states everything for series centred at zero to keep the notation light. Nothing changes for a centre a: replace x by x minus a throughout.

7. Why adding is allowed: partial sums add

Concept

The proof of part (i) is two lines about partial sums. Fix x in I and name the partial sums of the two series.

\[ S_N(x) = \sum_{n=0}^{N} c_n x^n \to f(x), \qquad T_N(x) = \sum_{n=0}^{N} d_n x^n \to g(x) \]

A finite sum can be regrouped freely, so the partial sum of the combined series splits in two.

\[ \sum_{n=0}^{N} (c_n x^n + d_n x^n) = S_N(x) + T_N(x) \]

\[ \lim_{N\to\infty}\left(S_N(x) + T_N(x)\right) = f(x) + g(x) \]

The limit of a sum is the sum of the limits, which is exactly why both series must converge at the same x.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 545 — proof of Theorem 6.2(i)

This proof is short, and it is worth seeing why it works, because it tells you exactly what the hypothesis is for. A series is the limit of its partial sums. The partial sums of the combined series are finite sums, and a finite sum can be rearranged in any order you like, so each partial sum splits into a partial sum of the first series plus a partial sum of the second.

Now take the limit. The limit of a sum of two sequences is the sum of their limits, but only if both limits exist. That is the whole reason the theorem asks for a common interval: at a point where one of the two series diverges, one of the two limits does not exist and the argument has nothing to stand on.

Subtraction and multiplying by a constant go through in exactly the same way.

8. Example 6.4(a): the interval of a sum

Worked example

One series converges on the interval from minus 1 to 1, another on the interval from minus 2 to 2. Where does their sum converge?

\[ \sum a_n x^n \text{ on } (-1, 1), \qquad \sum b_n x^n \text{ on } (-2, 2) \]

\[ (-1, 1) \cap (-2, 2) = (-1, 1) \]

\[ \sum (a_n x^n + b_n x^n) \text{ converges on } (-1, 1) \]

\[ \sum a_n (1.5)^n = \sum (a_n + b_n)(1.5)^n - \sum b_n (1.5)^n \;\Rightarrow\; \text{contradiction} \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 545 — Example 6.4a

The first series converges on the interval from minus 1 to 1, the second on the wider interval from minus 2 to 2. The sum needs both at the same point, so it is guaranteed on their overlap, which is the smaller interval. The number lines in the figure make that overlap visible.

The check step shows that nothing is lost by being cautious here. Pick a point such as 1.5, where the first series diverges and the second converges. If the sum converged there, you could subtract the convergent second series from it and conclude that the first series converges too. It does not, so the sum cannot converge at 1.5.

Keep the structure of that argument: a convergent series plus a divergent one is always divergent. That settles every point between 1 and 2.

9. A convergent series plus a divergent one

Prediction

\[ \sum a_n x^n \text{ on } (-1, 1), \qquad \sum b_n x^n \text{ on } (-2, 2), \qquad x = 1.5 \]

Commit to one option before revealing. This is the argument from the check step of the previous example, turned into a question so that you meet it on your own.

Suppose the combined series converged at 1.5. The second series converges there too, and the difference of two convergent series is convergent. But that difference is the first series, which diverges at 1.5. So the assumption was impossible and the sum diverges.

The one loophole, and the subject of the next slide, is when both pieces diverge. Two divergent series can cancel each other exactly. A convergent one and a divergent one never can.

10. Can the sum converge on a bigger interval?

Counterexample

Try to build an example before revealing. The hint is in the previous slide: for the sum to survive outside the common interval, both series must diverge there, in a way that cancels.

The example takes the geometric series and adds a second series whose coefficients are one over 2 to the n minus 1. On its own that second series has radius 1, because of the minus 1 in each coefficient. Added to the geometric series, the ones cancel, and what is left is the series of x over 2 to the n, which has radius 2.

So read Theorem 6.2 as a guarantee, not as a formula. The sum is always valid on the common interval. It may be valid on more, and if you care about the exact interval of a combination, you find it directly from the combined coefficients.

11. Trap: keeping the larger interval

Trap

12. Multiplying by x and substituting

Section

Part 2

13. Multiplying by a power of x shifts the powers

Concept

Part (ii) multiplies every term by the same b times x to the m. The coefficients stay; every power moves up by m.

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

\[ \sum_{n=0}^{\infty} x^{n+3} = x^3 + x^4 + x^5 + \cdots \quad (|x| < 1) \]

At any single x, multiplying by x cubed is multiplying by a fixed number, and a constant multiple of a convergent series converges. So the interval does not change.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 544 — Theorem 6.2(ii)

Part two of the theorem is the gentlest operation. Multiplying every term by x cubed leaves the coefficients alone and pushes every power up by three, so the series for x cubed over one minus x starts at x cubed instead of at 1.

Why does the interval not change? Fix a value of x. Then x cubed is just a number, and multiplying every term of a series by the same number neither creates nor destroys convergence, unless the number is zero. So wherever the geometric series converges, so does the new one, and wherever it diverges, so does the new one.

In practice this move is almost always combined with a substitution, as in the next few examples, to tidy a numerator such as 3x into the series.

14. Substitution moves the condition too

Concept

Part (iii) replaces x by b times x to the m everywhere, including inside the condition.

\[ \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \quad \text{for } |u| < 1 \]

\[ u = -x^2: \quad \frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n} \quad \text{for } |-x^2| < 1 \]

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

Here the condition happened to come back unchanged. With u equal to 3x it would not: the condition becomes three times the size of x less than 1.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 544 — Theorem 6.2(iii)

Substitution is the workhorse of this section. You already know one over one minus u as a series in u, valid when the size of u is less than 1. Put any expression in x in place of u and you get a series in x, as long as the expression obeys the same condition.

Here u is minus x squared. The series becomes an alternating series in the even powers of x, and the condition becomes the size of x squared less than 1, which works out to the size of x less than 1. The interval happened to survive unchanged.

Do not let that lull you. The condition is always about u, and you always have to solve it for x. With u equal to 3x, the same work gives the size of x less than one third, a much smaller interval.

15. Reading Theorem 6.2(iii) piece by piece

Notation

Step through the annotations in order. The coefficients are the ones you already know, and nothing happens to them. What changes is the thing they multiply: every copy of the old variable is replaced by b x to the m.

The function side changes in exactly the same way, so the new series represents the composite function f of b x to the m. That is why the method works for any known series, not only the geometric one: substitute into both sides at once.

The last annotation is where marks are lost. The condition belongs to the substituted expression, not to x, and solving it for x is a separate step. Writing down the old interval unchanged is the single most common error with this theorem.

16. Substitution rescales the interval

Picture it

Each function blows up where its denominator is zero, and each series stops at the same distance from the centre. Replacing x by 3x reaches that point three times sooner.

Each of the three curves is the same function, one over one minus u, with a different expression for u. Each one blows up where its denominator is zero: x over 2 equal to 1 at x equal to 2, x equal to 1, and 3x equal to 1 at x equal to one third.

The coloured bars under the axis are the intervals of the corresponding series. Each bar reaches from the centre out to the distance of its blow-up point, in both directions. So replacing x by 3x makes the interval three times narrower, and replacing x by x over 2 makes it twice as wide.

Notice that the bars are symmetric even though the functions are not. On the left, nothing goes wrong with any of the three functions, yet the series stop anyway. A power series converges on a symmetric interval around its centre, and the nearest trouble spot in either direction sets the radius.

17. Example 6.4(b): the interval after substituting 3x

Worked example

The series of aₙ xⁿ converges on the interval from minus 1 to 1. Find the interval of the new series below.

\[ \sum_{n=0}^{\infty} a_n 3^n x^n \]

\[ a_n 3^n x^n = a_n (3x)^n \]

\[ \sum a_n (3x)^n \text{ converges when } -1 < 3x < 1 \]

\[ -\frac13 < x < \frac13 \]

\[ \sum (3x)^n = \frac{1}{1-3x} \quad \text{for } |x| < \frac13 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 545-546 — Example 6.4b

The given series is written with 3 to the n and x to the n as separate factors, and the first step is to see them as one power, 3x to the n. Once you see that, the new series is the old one with 3x in place of x.

Part three of the theorem then says the new series converges when 3x lies in the old interval, from minus 1 to 1. Solving a double inequality is just dividing all three parts by 3, which gives the interval from minus one third to one third.

The check uses the simplest possible choice of coefficients, all equal to 1. Then the new series is geometric with ratio 3x, and you know from scratch that it converges exactly when the size of 3x is less than 1. The general answer and the special case agree.

18. Checkpoint 6.4: substituting x over 2

Prediction

\[ \sum a_n x^n \text{ on } (-1, 1) \quad\Longrightarrow\quad \sum a_n \left(\frac{x}{2}\right)^n \text{ on } ? \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 546 — Checkpoint 6.4

Pick an answer before revealing. The instinct is to do something to the endpoints of the old interval, and the question is which thing.

The old series needs its variable between minus 1 and 1. The variable is now x over 2, so x over 2 must lie between minus 1 and 1, and multiplying through by 2 puts x between minus 2 and 2. Dividing x by 2 makes the variable move more slowly, so it takes a larger x to reach the edge.

If you chose the interval from minus one half to one half, you divided the old endpoints by 2 instead of solving the condition. Writing the inequality out every time avoids that error completely.

19. Example 6.5(a): three x over one plus x squared

Worked example

Use the series for one over one minus x to build a series for the function below, and find its interval.

\[ f(x) = \frac{3x}{1+x^2} \]

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

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

\[ f(x) = \sum_{n=0}^{\infty} 3(-1)^n x^{2n+1} = 3x - 3x^3 + 3x^5 - \cdots \]

\[ |-x^2| < 1 \iff |x| < 1 \quad\Rightarrow\quad (-1, 1) \]

\[ f\left(\tfrac12\right) = \frac{1.5}{1.25} = 1.2, \qquad \frac{1.5}{1 + 1/4} = 1.2 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 546-547 — Example 6.5a

The strategy for every example of this kind is the same: bend the function into the shape of a constant over one minus something. Here you pull out 3x and rewrite one plus x squared as one minus minus x squared. After that, the rest is mechanical.

Substitute minus x squared into the geometric series to get the alternating even powers, then multiply by 3x, which raises every power by one. The final series has only odd powers, which makes sense because the function is odd.

Now look at the figure, because it shows something surprising. The function three x over one plus x squared is perfectly smooth for every x, yet the partial sums leave it just past 1 and minus 1. The series is limited by the condition on the size of x squared, not by any visible misbehaviour of the function. The check at one half confirms the series does converge to the function inside the interval.

20. Match each substitution to its radius

Matching

Match each function to its radius before checking. Write each one as one over one minus u, name u, and solve the size of u less than 1 for x.

The first and second are quick: 4x gives a quarter, and minus x squared gives 1. The third is the one to watch: the size of x cubed over 8 less than 1 means the size of x cubed less than 8, so the size of x is less than 2. Taking a cube root is part of solving the condition.

The fourth needs a square root: 2 x squared less than 1 means x squared less than one half, so the size of x is less than one over root 2. None of these radii can be read directly off a coefficient.

21. Example 6.5(b): partial fractions first

Worked example

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

\[ \frac{1}{(x-1)(x-3)} = \frac{-1/2}{x-1} + \frac{1/2}{x-3} \]

\[ = \frac{1/2}{1-x} - \frac{1/2}{3-x} \]

\[ \frac{1/2}{3-x} = \frac{1/6}{1 - x/3} \]

\[ \frac{1/2}{1-x} = \sum \tfrac12 x^n \;(|x| < 1), \quad \frac{1/6}{1-x/3} = \sum \tfrac16 \left(\tfrac{x}{3}\right)^n \;(|x| < 3) \]

\[ f(x) = \sum_{n=0}^{\infty}\left(\frac12 - \frac{1}{6 \cdot 3^n}\right)x^n, \quad (-1, 1) \]

\[ c_0 = \frac12 - \frac16 = \frac13 = f(0), \qquad f\left(\tfrac12\right) = \frac{1}{(-0.5)(-2.5)} = 0.8 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 547 — Example 6.5b (the solution's first line prints the factor as 1 minus x; it should be x minus 1)

This function has two factors in the denominator, so it is not one geometric series but two. Partial fractions split it into pieces that each look like a constant over one minus something.

The cover-up rule gives the constants quickly: cover the factor x minus 1 and put x equal to 1 into what remains, and you get one over minus 2. The second piece works the same way. Then flip each denominator so it starts with a positive constant, and factor 3 out of the second so it becomes one minus x over 3. Each piece now expands, with its own interval.

The figure shows why the final interval is the smaller one. The first piece breaks at 1, the second only at 3, and the difference needs both. The check compares the constant term with f at zero, which is the fastest way to catch a missing factor. The book's solution misprints the factor as one minus x in its first line; the example itself is about x minus 1.

22. Checkpoint 6.5: your turn with partial fractions

Socratic

\[ f(x) = \frac{1}{(1-x)(x-2)} \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 547 — Checkpoint 6.5

Work this one fully on paper before revealing: partial fractions, two expansions, and the interval. It is the same pattern as the example, with one sign trap.

The partial fractions come out with both constants equal to minus 1. The piece minus one over x minus 2 is the same as plus one over 2 minus x, and factoring 2 out of that denominator gives one half over one minus x over 2. That piece is valid out to 2, while the one over one minus x piece stops at 1.

So the combined coefficient is minus 1 plus one over 2 to the n plus 1, and the interval is from minus 1 to 1. The constant term, minus one half, matches f at zero, which is one over minus 2: the check that takes five seconds.

23. Find the error: the condition carried across unchanged

Error analysis

\[ \frac{1}{3-x} = \sum_{n=0}^{\infty} \frac{x^n}{3^{n+1}} \quad \text{for } |x| < 3 \]

Look at the claimed expansion and try to find what is wrong before revealing the annotations. There are two separate errors, and a single substitution of x equal to zero exposes one of them.

The first error is algebraic. To get the geometric shape you must factor 3 out of the denominator, and that factor has to go somewhere: it becomes a one third in front of the whole series. Dropping it makes every coefficient three times too large.

The second error is the interval. The substituted variable is x over 3, so the condition is the size of x over 3 less than 1, which allows x all the way out to 3. Copying the parent's interval throws away most of the valid range. The corrected line at the bottom fixes both.

24. Reading a series backwards

Section

Part 3

25. Example 6.6: which function is this series?

Worked example

Find the function represented by the series below and its interval of convergence.

\[ \sum_{n=0}^{\infty} 2^n x^n \]

\[ \sum_{n=0}^{\infty} 2^n x^n = \sum_{n=0}^{\infty} (2x)^n \]

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

\[ |2x| < 1 \iff -\frac12 < x < \frac12 \]

\[ \frac{1}{1 - 2(1/4)} = 2 = 1 + \tfrac12 + \tfrac14 + \cdots \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 548 — Example 6.6

This example runs the machinery backwards: instead of turning a function into a series, you are handed a series and asked which function it is. The skill is recognising a disguised geometric series.

Two to the n times x to the n is 2x to the n, so the series is geometric with first term 1 and ratio 2x. Its sum is one over one minus 2x, and it converges exactly when the ratio has size less than 1, which gives the interval from minus one half to one half.

The check picks a point inside the interval where the series is one you can add up in your head. At one quarter the ratio is one half, and one plus a half plus a quarter and so on is 2, exactly what the formula gives.

26. Checkpoint 6.6: one over 3 to the n

Fill the middle

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

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 548 — Checkpoint 6.6

Fill in both blanks before checking. The worked line above the blanks already rewrote the series as a geometric series in x over 3, so the only question is what that means for the interval and for one value.

The ratio is x over 3, and the series converges when that ratio has size less than 1, so the size of x must be less than 3. At x equal to 1 the formula three over three minus x gives three halves.

If you want to be sure, add the series directly at x equal to 1: one plus a third plus a ninth and so on. A geometric series with ratio one third sums to one over two thirds, which is one and a half again.

27. A stream of payments is a power series

Concept

Money paid k years from now is worth less today, because money in hand earns interest. At rate r, a payment C in year k is worth its present value.

\[ P_k (1+r)^k = C \quad\Longrightarrow\quad P_k = \frac{C}{(1+r)^k} \]

A stream of payments is worth the sum of those present values, which is a series in the variable one over one plus r.

\[ P = \frac{C}{1+r} + \frac{C}{(1+r)^2} + \frac{C}{(1+r)^3} + \cdots \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 548-549 — present value, Example 6.7

The book's application of series is to money, and it rests on one idea: a dollar today is worth more than a dollar next year, because today's dollar can be invested and earn interest. The present value of a future payment is the amount you would need today to grow into it.

If the interest rate is r, a sum grows by a factor of one plus r each year. So a payment C in year k needs only C divided by one plus r to the k today. That is the present value formula on the slide.

A stream of payments is worth the sum of the present values of its payments. Written out, that sum is a geometric series in the variable one over one plus r, which is why this belongs in a lesson on power series.

28. Example 6.7(a): twenty payments of 1.5 million

Worked example

A lottery pays 1.5 million dollars at the end of each year for 20 years. At 5 percent interest, what is that worth today?

\[ P_1 = \frac{1.5}{1.05} \approx 1.429 \]

\[ P_2 = \frac{1.5}{1.05^2} \approx 1.361 \]

\[ P = \sum_{k=1}^{20} \frac{1.5}{1.05^k} \]

\[ P = \frac{1.5}{1.05}\cdot\frac{1 - 1.05^{-20}}{1 - 1.05^{-1}} = \frac{1.5\left(1 - 1.05^{-20}\right)}{0.05} \]

\[ P \approx 18.693 \text{ million} < 30 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 548-549 — Example 6.7a

The figure shows the idea directly. Each dashed line is a payment of 1.5 million, but each bar is what that payment is worth today, and the bars shrink by a factor of 1.05 every year. The first payment is worth about 1.429 million today and the last only about 0.565 million.

The total is a finite geometric sum with twenty terms. The book adds it by computer; the finite geometric formula does it in one line, and both give 18.693 million.

The check makes a useful comparison: twenty payments of 1.5 million sound like 30 million, but they are worth well under that, and even under the 20 million lump sum offered as the first option. Waiting costs money.

29. Example 6.7(b) to (e): payments that never stop

Worked example

Now C dollars a year forever, at rate r. Find the present value and settle the lottery question.

\[ P = \sum_{n=0}^{\infty} \frac{C}{(1+r)^{n+1}} = \frac{C}{1+r}\sum_{n=0}^{\infty}\left(\frac{1}{1+r}\right)^n \]

\[ \sum_{n=0}^{\infty}\left(\frac{1}{1+r}\right)^n = \frac{1}{1 - \frac{1}{1+r}} = \frac{1+r}{r} \]

\[ P = \frac{C}{1+r}\cdot\frac{1+r}{r} = \frac{C}{r} \]

\[ P = \frac{1}{0.05} = 20 \text{ million} \]

\[ 20 = 20 > 18.693, \qquad \sum_{k=1}^{100}\frac{1}{1.05^k} \approx 19.848 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 549-550 — Example 6.7b-e

An annuity that pays forever sounds as if it should be worth an infinite amount. It is not, because the far-future payments are worth almost nothing today. The present values form a geometric series with ratio one over one plus r, which is less than 1 whenever the rate is positive.

Summing that series and simplifying gives a beautifully short answer: the present value of C a year forever is C divided by r. At 5 percent, a million a year forever is worth exactly 20 million today, the same as the lump sum.

The figure compares the two payment streams year by year. The 1.5 million stream rushes up and stops at 18.693. The perpetual stream climbs more slowly but keeps going, and after a hundred years it has already reached 19.848, closing in on 20. The lump sum and the perpetuity tie, and both beat the twenty-year option.

30. Exercise 77: what rate funds a pension forever?

Real world

\[ P = \frac{C}{r} \quad\Longrightarrow\quad r = \frac{C}{P} \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 558 — Exercises 76 and 77

Answer both parts before revealing. The perpetuity formula, present value equals payment over rate, can be solved for any one of its three quantities, and that makes it a real planning tool.

Solving for the rate gives payment divided by present value: 50,000 over a million, which is 5 percent. At 3 percent the same million supports only 30,000 dollars a year.

This is exactly how a university endowment or a family trust thinks about spending. If it spends no more than the interest each year, the principal is never touched and the payments can continue indefinitely. The algebra behind that rule is the sum of a geometric series.

31. Multiplying two series

Section

Part 4

32. Multiply like polynomials, collect by degree

Concept

Multiply every term of the first series by every term of the second, then gather the products that carry the same power of x.

\[ (c_0 + c_1 x + c_2 x^2 + \cdots)(d_0 + d_1 x + d_2 x^2 + \cdots) \]

\[ = c_0 d_0 + (c_1 d_0 + c_0 d_1)x + (c_2 d_0 + c_1 d_1 + c_0 d_2)x^2 + \cdots \]

The power of x in a product is the sum of the two powers, so the coefficient of x to the n collects every pair whose indices add up to n.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 550 — multiplying power series

Multiplying two power series is the same as multiplying two long polynomials: every term of the first meets every term of the second. The only question is bookkeeping, because infinitely many products land on each power of x if you are careless about organising them.

The grid is the organiser. Put the terms of the first series down the side and the terms of the second across the top. The cell in row i and column j holds the product of c i and d j, and it multiplies x to the power i plus j. So every cell on one anti-diagonal carries the same power of x, and each colour in the figure is one coefficient of the product.

Count the cells on each diagonal: one for the constant term, two for x, three for x squared. Each coefficient of the product is a finite sum, which is what makes the method practical.

33. Theorem 6.3: the Cauchy product

Concept

Suppose the two series converge to f and g on a common interval I. Build the new coefficients from the diagonals.

\[ e_n = c_0 d_n + c_1 d_{n-1} + \cdots + c_n d_0 = \sum_{k=0}^{n} c_k d_{n-k} \]

\[ \left(\sum_{n=0}^{\infty} c_n x^n\right)\left(\sum_{n=0}^{\infty} d_n x^n\right) = \sum_{n=0}^{\infty} e_n x^n = f(x)g(x) \text{ on } I \]

Cauchy product — The series whose nth coefficient is the sum of the products of coefficients with indices adding to n. The book omits the proof; it needs absolute convergence, which power series have inside their interval.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 551 — Theorem 6.3

Theorem 6.3 turns the grid into a formula. The nth coefficient of the product, e n, is the sum of c k times d of n minus k, for k from 0 to n: the n plus 1 cells on the nth diagonal.

The conclusion has two parts. The product series converges on the common interval, and there it converges to the product of the two functions. As with adding, the guarantee is on the common interval, the smaller of the two.

The book leaves out the proof. The reason it is harder than the proof for adding is that rearranging infinitely many terms into diagonals can change the sum of a series in general. Power series are safe because inside their interval they converge absolutely, and absolutely convergent series can be rearranged freely.

34. Reading the Cauchy coefficient

Notation

Walk through the annotations one at a time. The formula looks abstract, but every piece of it comes from one fact: x to the k times x to the n minus k is x to the n.

So the coefficient of x to the n collects exactly the pairs whose indices add up to n, and there are n plus 1 of them. That is a finite computation. No limits and no convergence questions arise in finding any single coefficient.

The last annotation names the mistake that the trap on a later slide is about. It is very tempting to guess that the nth coefficient of a product is the product of the nth coefficients. That would multiply x to the n by x to the n, which is x to the 2n: not even the right power.

35. Example 6.8: one over (1 minus x)(1 minus x squared)

Worked example

Multiply the two geometric series below to build a series for their product on the interval from minus 1 to 1.

\[ \frac{1}{1-x} = 1 + x + x^2 + \cdots, \qquad \frac{1}{1-x^2} = 1 + x^2 + x^4 + \cdots \]

\[ (1 + x^2 + x^4 + \cdots) + (x + x^3 + x^5 + \cdots) + (x^2 + x^4 + \cdots) + \cdots \]

\[ = 1 + x + (1+1)x^2 + (1+1)x^3 + (1+1+1)x^4 + \cdots \]

\[ = 1 + x + 2x^2 + 2x^3 + 3x^4 + 3x^5 + \cdots \]

\[ e_n = \sum_{k=0}^{n} d_{n-k} = \left\lfloor \tfrac{n}{2} \right\rfloor + 1 \]

\[ \sum_{n=0}^{\infty} e_n \left(\tfrac12\right)^n \approx 2.6667 = \frac{1}{(1/2)(3/4)} = \frac83 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 551-552 — Example 6.8

The first series has every coefficient equal to 1. The second has 1 at the even powers and 0 at the odd ones. Multiplying them out row by row, as the book does, gives the rows on the slide, and collecting equal powers is a matter of counting how many rows contain each power.

The Cauchy formula confirms the pattern and explains it. Because every c is 1, the nth coefficient simply adds up d of n minus k for k from 0 to n, which counts the even numbers from 0 to n. That count goes 1, 1, 2, 2, 3, 3, rising by one every second step.

The figure shows the partial sums approaching the product function across the interval. The check at one half compares the series, summed numerically, with the function value eight thirds. Both factors converge on the interval from minus 1 to 1, so the product does too.

36. Before multiplying: which pairs make x to the fourth?

Step zero

\[ \left(\sum_{n=0}^{\infty} c_n x^n\right)\left(\sum_{n=0}^{\infty} d_n x^n\right) \]

Write down your list before revealing. The aim is to make the pattern automatic, so that you never multiply matching coefficients by mistake.

The pairs are c 0 with d 4, c 1 with d 3, and so on down to c 4 with d 0. Five pairs, and in each one the two indices add up to 4. Those are the five cells on one diagonal of the grid.

In general the coefficient of x to the n has n plus 1 products in it. If you ever write a coefficient with only one product in it, stop and check: only the constant term is that simple.

37. Checkpoint 6.7: squaring the geometric series

Worked example

Multiply the geometric series by itself to build a series for one over one minus x, squared.

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

\[ c_k = d_k = 1 \quad \text{for all } k \]

\[ e_n = \sum_{k=0}^{n} 1 \cdot 1 = n + 1 \]

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

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

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 552 — Checkpoint 6.7

This is the cleanest possible Cauchy product. Every coefficient of both factors is 1, so every product on every diagonal is 1, and the nth diagonal simply counts its own cells: n plus 1 of them.

So the square of the geometric series is one plus 2x plus 3x squared and so on, a series for one over one minus x, squared. The check at one half confirms it numerically: the series adds up to 4, and one over one half squared is 4.

Remember this answer. In a few slides you will reach the same series by a completely different route, differentiating the geometric series. The fact that the two routes agree is not luck, and the last part of the lesson explains why.

38. Trap: multiplying matching coefficients

Trap

39. Differentiating and integrating term by term

Section

Part 5

40. Calculus one term at a time

Concept

A polynomial is differentiated and integrated one term at a time. Theorem 6.4 says a power series can be too, inside its interval.

\[ f(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots \]

\[ f'(x) = c_1 + 2c_2 x + 3c_3 x^2 + \cdots \]

\[ \int f(x)\,dx = C + c_0 x + c_1 \frac{x^2}{2} + c_2 \frac{x^3}{3} + \cdots \]

This is not obvious: an infinite sum of functions can fail to be differentiable even when every term is. Power series are special, and the theorem is what licenses the move.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 552 — term-by-term differentiation and integration

For a polynomial, differentiating term by term is just the sum rule applied a finite number of times. For an infinite series it is a genuine theorem, because a limit of functions can behave very differently from the functions themselves. There are infinite sums of perfectly smooth functions whose total has corners.

Theorem 6.4 says that power series never do that inside their interval. You may differentiate and integrate them one term at a time, exactly as if they were polynomials, and the results converge to the derivative and the integral of the function.

This is what makes power series so useful. The geometric series gives you a whole family of functions just by differentiating and integrating it, and later in the course the same fact lets you solve differential equations with series.

41. Reading Theorem 6.4 piece by piece

Notation

\[ \int f(x)\,dx = C + \sum_{n=0}^{\infty} c_n \frac{(x-a)^{n+1}}{n+1}, \quad |x-a| < R \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 553 — Theorem 6.4

Go through the notes in order. The first is bookkeeping: the constant term differentiates to zero, so the derivative series starts at n equal to 1. When you reindex so that it starts at zero again, the coefficients shift by one place.

The second note explains why the radius survives. Differentiating multiplies the nth term by n, a factor that grows far too slowly to overturn the geometric growth or decay of x to the n. The ratio test sees n over n plus 1, which tends to 1, so the radius it computes does not change.

The third and fourth notes are the heart of it. You get the same radius, on the open interval, and nothing else. Whether either endpoint belongs to the new series is a new question, and the integral version at the bottom of the slide carries exactly the same warning.

42. Differentiate the partial sum, track the derivative

Picture it

On the left a partial sum hugs one over one minus x. Differentiate each of its terms and the result, on the right, hugs the derivative just as closely.

The left panel is the function one over one minus x together with a partial sum, the polynomial one plus x up to x to the fifth. They are close near zero and drift apart towards the edges, where more terms would be needed.

The right panel differentiates both. The derivative of the function is one over one minus x, squared, and the derivative of the polynomial is found term by term. The two stay close in the same region as before.

That is Theorem 6.4 in a picture: the derivative of the approximation approximates the derivative. As the number of terms grows, both panels tighten together, on the same interval.

43. Example 6.9(a): one over (1 minus x) squared by differentiating

Worked example

Differentiate the geometric series to find a series for g, and test the endpoints.

\[ g(x) = \frac{1}{(1-x)^2} = \frac{d}{dx}\left(\frac{1}{1-x}\right) \]

\[ \frac{d}{dx}(1 + x + x^2 + x^3 + \cdots) = 0 + 1 + 2x + 3x^2 + \cdots \]

\[ g(x) = \sum_{n=0}^{\infty}(n+1)x^n, \quad |x| < 1 \]

\[ \sum (n+1) \text{ diverges} \]

\[ \sum (-1)^n (n+1) \text{ diverges} \;\Rightarrow\; (-1, 1) \]

\[ \left(\sum x^n\right)^2 = \sum (n+1)x^n \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 553-554 — Example 6.9a

The derivative of one over one minus x is one over one minus x, squared, so differentiating the geometric series term by term gives a series for the new function. Each x to the n becomes n times x to the n minus 1, and reindexing turns that into n plus 1 times x to the n.

The radius is 1, guaranteed by the theorem. The endpoints need their own test. At both ends the terms n plus 1 grow in size, so they certainly do not tend to zero, and the divergence test settles each end in a line. The interval is open.

The check compares this result with Checkpoint 6.7, where squaring the geometric series gave exactly the same coefficients. Two different operations, one function, one series.

44. Example 6.9(b): summing (n + 1) over 4 to the n

Worked example

Use the series just found to evaluate a numerical series.

\[ \sum_{n=0}^{\infty} \frac{n+1}{4^n} \]

\[ \sum_{n=0}^{\infty} \frac{n+1}{4^n} = \sum_{n=0}^{\infty} (n+1)\left(\tfrac14\right)^n \]

\[ = \frac{1}{(1 - 1/4)^2} = \frac{1}{(3/4)^2} \]

\[ = \frac{16}{9} \approx 1.777778 \]

\[ S_5 \approx 1.775391, \quad S_{10} \approx 1.777774 \to 1.777778 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 554 — Example 6.9b

Here the new series earns its keep: it evaluates a numerical series you could not easily add by hand. The trick is to recognise the numerical series as the power series evaluated at a particular x.

One over 4 to the n is one quarter to the n, so the series is the derivative series at x equal to one quarter. One quarter lies inside the interval, so the formula applies, and one over three quarters squared is sixteen ninths.

The check adds the terms directly: six terms already give 1.7754 and eleven give 1.77777, homing in on 1.77778. Many numerical series in textbooks and exams are built this way, so when you see n or n plus 1 multiplying a geometric term, think of a derivative.

45. Checkpoint 6.8: differentiate once more

Socratic

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

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 554 — Checkpoint 6.8

Differentiate before revealing. Every term of one plus 2x plus 3x squared and so on goes down by one power and picks up a factor: 2x becomes 2, 3x squared becomes 6x, 4x cubed becomes 12x squared.

The general term is n times n plus 1 times x to the n minus 1, which reindexes to n plus 1 times n plus 2 times x to the n. The function is the derivative of one over one minus x, squared, which is two over one minus x, cubed.

The radius is still 1, and a numerical check at 0.3 matches the function to six decimal places. Repeated differentiation of the geometric series gives a series for every power of one over one minus x.

46. Exercise 91: the sum of n over 2 to the n

Estimation

\[ \sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2} \quad\Rightarrow\quad \sum_{n=1}^{\infty}\frac{n}{2^n} = \frac{1/2}{1/4} = 2 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 559 — Exercise 91

Make an estimate before revealing. The figure is designed to make that hard: the terms first rise, from one half to one half again, then shrink, and it is not obvious where the partial sums will stop.

The derivative route settles it. Differentiating the geometric series gives the sum of n x to the n minus 1, and multiplying by x, which is part two of Theorem 6.2, gives the sum of n x to the n. That equals x over one minus x, squared, and at one half it is exactly 2.

The dots confirm it: the partial sums level off at the dashed line. Multiplying by x after differentiating is a move you will use often, to fix up the power before evaluating.

47. Integrating: the constant comes from the centre

Concept

Integrating term by term produces an antiderivative plus an unknown constant. To pin it down, evaluate both sides at the centre, where every power of x vanishes.

\[ F(x) = C + c_0 x + c_1 \frac{x^2}{2} + \cdots \quad\Longrightarrow\quad F(0) = C \]

So if the function you want is known at zero, the constant is that value. For the logarithm and the arctangent, both are zero at zero, so C is zero.

\[ \ln(1 + 0) = 0, \qquad \tan^{-1}(0) = 0 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 555 — Example 6.10

Integrating a series term by term gives an antiderivative, and every antiderivative comes with an unknown constant. You cannot skip it, because a power series with the wrong constant term represents a different function.

The centre is where the constant is easy to find. At x equal to zero every power of x vanishes, so the integrated series collapses to its constant C. Whatever value the function you want takes at zero, that is C.

For the two examples coming up, the natural logarithm of one plus x and the arctangent, the value at zero is zero, so C is zero. That is not always so, and you should always write the step down.

48. Example 6.10(a): the series for ln(1 + x)

Worked example

Integrate the series of the derivative to find a series for f and its interval.

\[ f(x) = \ln(1+x), \qquad f'(x) = \frac{1}{1+x} \]

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

\[ \int(1 - x + x^2 - \cdots)\,dx = C + x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \]

\[ \ln 1 = 0 = C \]

\[ \ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}, \quad |x| < 1 \]

\[ x = 1: \; 1 - \tfrac12 + \tfrac13 - \cdots \text{ conv.}, \quad x = -1: \; -\sum \tfrac1n \text{ div.} \]

\[ 0.405465 - 0.401042 = 0.0044 < \frac{(1/2)^5}{5} = 0.00625 \quad (-1, 1] \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 555 — Example 6.10a

The logarithm is not a rational function, so no substitution into the geometric series will produce it directly. But its derivative, one over one plus x, is geometric, with u equal to minus x. So expand the derivative, then integrate back.

Integrating one term at a time turns each power of x into the next power divided by its new exponent. Setting x equal to zero fixes the constant at zero. The radius is 1 by the theorem, and now the endpoints: at 1 you get the alternating harmonic series, which converges; at minus 1 you get the negative of the harmonic series, which diverges.

The figure shows both ends. On the right the partial sums reach the curve at ln 2; on the left they fall away as the curve does. The check uses the alternating series error bound at one half. One subtlety from the book: the theorem only guarantees the series equals the logarithm inside the interval; that it equals ln 2 at the endpoint needs Abel's theorem, which is beyond this course.

49. Slide N: how many terms of the logarithm series?

Tweak it

Start with N at 4 and increase it slowly. Watch three regions: the middle of the interval, the neighbourhood of x equal to 1, and the region beyond 1.

In the middle, a handful of terms is already indistinguishable from the logarithm. Near 1 the convergence is slow, because the terms there are one over n, which shrink only gently. Near minus 1 the partial sums dive, following the logarithm down toward its vertical asymptote.

Beyond 1 something opposite happens: adding terms makes things worse. Each term there is a power of a number larger than 1 divided by n, and those terms grow, so the partial sums swing further away with every step. That is what outside the interval of convergence looks like.

50. Example 6.10(b): the series for the arctangent

Worked example

\[ f(x) = \tan^{-1} x, \qquad f'(x) = \frac{1}{1+x^2} \]

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

\[ \int(1 - x^2 + x^4 - \cdots)\,dx = C + x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \]

\[ C = \tan^{-1}(0) = 0 \]

\[ \tan^{-1} x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}, \quad |x| < 1 \]

\[ x = \pm 1: \; \pm\left(1 - \tfrac13 + \tfrac15 - \cdots\right) \text{ conv.} \;\Rightarrow\; [-1, 1] \]

\[ 0.463648 - 0.463467 = 0.000181 < \frac{(1/2)^9}{9} \approx 0.000217 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 555-556 — Example 6.10b

This is the same recipe as the logarithm, with a different substitution. The derivative of the arctangent is one over one plus x squared, which is geometric with u equal to minus x squared. Expand, integrate term by term, and fix the constant with the arctangent of zero, which is zero.

The result has only odd powers, divided by the odd numbers, with alternating signs. The radius is 1. At both endpoints the terms alternate in sign and shrink to zero, so the alternating series test gives convergence at both, and the interval is closed.

Look at the figure: the arctangent is defined and smooth for every x, levelling off toward plus and minus pi over 2, yet the series gives up beyond 1. At the endpoints the partial sums still meet the curve, at plus and minus pi over 4. The check at one half shows four terms already agree to about two ten-thousandths.

51. Put the arctangent derivation in order

Ranking

Order the steps before checking. The logic runs in one direction: you need a geometric series before you can integrate anything, so the derivative comes first.

The two steps people most often misplace are the constant and the endpoints. The constant cannot be found until the integration has produced it. The endpoints come last, because the theorem you used to integrate only promised the open interval, so testing them is always the final job.

52. Complete the endpoint table

Comparison

Fill in every blank before checking. All four series have radius 1, and the table makes you notice that the radius is the one thing you never need to recompute after differentiating or integrating.

The endpoint columns are where the work is. The geometric series and its derivative diverge at both ends because their terms do not shrink. The logarithm series converges at 1, where it alternates, and diverges at minus 1, where it becomes the negative harmonic series. The arctangent series converges at both ends by the alternating series test.

Read the last column top to bottom and you see four different intervals with the same radius. That is the whole point of this part of the lesson in one table.

53. One radius, four different sets of endpoints

Picture it

All four series come from the same geometric series, and all four have radius 1. Differentiating lost nothing at the ends; integrating gained one endpoint for the logarithm and both for the arctangent.

The four number lines line up the results you have just derived. Every one of them stretches from minus 1 to 1, because every one came from the geometric series by operations that keep the radius.

The endpoint dots tell different stories. Differentiating made the terms bigger, by a factor of n, and the geometric series had no endpoints to lose. Integrating made the terms smaller, by a factor of n, and that was enough to rescue the endpoint at 1 for the logarithm and both endpoints for the arctangent.

There is a useful rule of thumb here: integration tends to gain endpoints and differentiation tends to lose them. But it is only a rule of thumb, which is why you test each end directly every time.

54. Trap: carrying the endpoints across

Trap

55. Checkpoint 6.9: integrating the logarithm series

Worked example

Integrate the series for ln of one plus x term by term to find its indefinite integral.

\[ \int \ln(1+x)\,dx = \int \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}\,dx \]

\[ \int \frac{(-1)^{n+1}x^n}{n}\,dx = \frac{(-1)^{n+1}x^{n+1}}{n(n+1)} \]

\[ \int \ln(1+x)\,dx = C + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n+1}}{n(n+1)} \]

\[ = C + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{12} - \cdots \]

\[ (1.5)\ln 1.5 - 0.5 \approx 0.108198 = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(0.5)^{n+1}}{n(n+1)} \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 556 — Checkpoint 6.9

Integrate term by term before looking at the steps. Each term x to the n over n becomes x to the n plus 1 over n times n plus 1, and the signs come along unchanged.

The first few terms are x squared over 2, minus x cubed over 6, plus x to the fourth over 12, and the radius is still 1. The constant C is whatever you need: it is an indefinite integral.

The check is a genuinely independent one. Integration by parts gives the antiderivative one plus x times the logarithm of one plus x, minus x, which is zero at zero. At one half it equals 0.108198, and so does the series. Two methods, one answer.

56. Exercise 115: computing pi with the arctangent series

Worked example

The arctangent of one over root 3 is pi over 6. How many terms give pi to within 0.001?

\[ \pi = 6\tan^{-1}\left(\tfrac{1}{\sqrt3}\right) = 6\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\left(\frac{1}{\sqrt3}\right)^{2k+1} \]

\[ \left(\frac{1}{\sqrt3}\right)^{2k+1} = \frac{1}{3^k\sqrt3} \quad\Rightarrow\quad \pi = 2\sqrt3\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)3^k} \]

\[ |\pi - 6S_N| < \frac{2\sqrt3}{(2N+3)3^{N+1}} \]

\[ N = 4: \; 0.001296, \qquad N = 5: \; 0.000366 < 0.001 \]

\[ |\pi - 3.1413088| \approx 0.000284 < 0.000366 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 560 — Exercise 115

This exercise shows why the arctangent series matters: it computes pi. The arctangent of one over root 3 is pi over 6, so six times the arctangent series at that point is a series for pi.

Simplifying the power is the key step. An odd power of one over root 3 is one over 3 to the k, times one over root 3, so each term is divided by a power of 3 as well as by an odd number. Because the series alternates with shrinking terms, the error after stopping is less than the next term, and that bound first drops below a thousandth at N equal to 5.

The figure compares this with the series at x equal to 1, which also gives pi. The fast series is settled after six terms; the slow one is still zigzagging after thirteen. The actual error with six terms is 0.000284, inside the bound, which is the check.

57. The slow series for pi

Estimation

\[ \frac{\pi}{4} = \tan^{-1} 1 = 1 - \frac13 + \frac15 - \frac17 + \cdots \]

\[ \frac{1}{2N+3} < 0.001 \iff 2N + 3 > 1000 \iff N \ge 499 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 560 — Exercise 114

Make a guess before revealing. This is the famous series one minus a third plus a fifth and so on, and it looks innocent enough.

The alternating series bound says the error is less than the next term, one over 2N plus 3. For that to be under a thousandth, 2N plus 3 must exceed 1000, so N must be at least 499: about five hundred terms, compared with six for the series at one over root 3.

The difference is where you evaluate the series. At x equal to 1 there is no small number raised to a power to shrink the terms, only the odd denominators. Evaluate a power series well inside its interval and it converges fast; evaluate it at the edge and it crawls.

58. Does the radius change?

Sorting

Sort each operation before checking, and for each one ask what happens to the condition on the size of x.

Multiplying by x cubed and termwise calculus keep the radius, for the reasons in Theorems 6.2 and 6.4. Substituting minus x squared also keeps it, but for a different reason: the size of x squared is less than 1 exactly when the size of x is less than 1, so the condition happens to come out the same.

Substituting 3x or x over 2 rescales the radius, to a third or to 2. The common thread is that you find the new radius by solving the new condition, never by guessing from the operation's name.

59. Only one series

Section

Part 6

60. Theorem 6.5: a power series is unique

Concept

Suppose two power series centred at a are equal for every x in an open interval around a. Then they are the same series, coefficient by coefficient.

\[ \sum_{n=0}^{\infty} c_n (x-a)^n = \sum_{n=0}^{\infty} d_n (x-a)^n \text{ near } a \quad\Longrightarrow\quad c_n = d_n \text{ for all } n \]

So whichever route you take, squaring, differentiating, substituting, you must arrive at the same coefficients. If two methods disagree, one of them has an error.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, pp. 556-557 — Theorem 6.5

You have now built several series by more than one route, and they always agreed. Theorem 6.5 says they had to. If two power series with the same centre give the same values on any open interval around that centre, then their coefficients are identical.

The bar charts show one instance. Squaring the geometric series with the Cauchy product and differentiating it term by term are completely different calculations, and they produce the same coefficients, 1, 2, 3, and so on.

The practical consequence is freedom. You may use whichever method is easiest for a given function: substitution, multiplication, differentiation or integration. The answer does not depend on the route. And if two routes ever disagree, you know for certain that one of them contains an error.

61. Why it is true: differentiate and set x equal to a

Concept

Call the common function f. Setting x equal to a kills every term but the first.

\[ f(a) = c_0 = d_0 \]

Differentiate both series term by term, which Theorem 6.4 allows, and set x equal to a again.

\[ f'(x) = c_1 + 2c_2(x-a) + \cdots \;\Rightarrow\; f'(a) = c_1 = d_1 \]

\[ f''(x) = 2c_2 + 3\cdot 2c_3(x-a) + \cdots \;\Rightarrow\; f''(a) = 2c_2 = 2d_2 \]

\[ f^{(n)}(a) = n!\,c_n = n!\,d_n \;\Rightarrow\; c_n = d_n \]

Every coefficient is fixed by a derivative of f at the centre, which is exactly the formula Section 6.3 will turn into Taylor series.

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 557 — proof of Theorem 6.5

The proof is a neat trick, and it is worth understanding because it leads directly into the next section. Setting x equal to the centre kills every term except the constant, so the two constant terms must both equal f of a.

Now differentiate both series, which Theorem 6.4 allows, and set x equal to a again. Again everything vanishes except the first term, which is now c 1 on one side and d 1 on the other. Differentiating twice leaves 2 c 2 and 2 d 2, and in general the nth derivative at a is n factorial times the nth coefficient.

So every coefficient is determined by the function: c n is the nth derivative at a divided by n factorial. Two series for the same function therefore have the same coefficients. Section 6.3 takes that formula and uses it to build series directly from derivatives.

62. Exercise 97: ln(1 minus x squared), two ways

Worked example

Find the series for ln of one minus x squared by two different routes and compare.

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

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

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

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

\[ |x| < 1; \quad x = \pm1: -\sum \tfrac1m \text{ diverges} \;\Rightarrow\; (-1, 1) \]

\[ -\sum_{m=1}^{\infty}\frac{(1/2)^{2m}}{m} \approx -0.287682 = \ln 0.75 \]

OpenStax Calculus Volume 2, §6.2 Properties of Power Series §6.2, p. 559 — Exercise 97

This example puts the uniqueness theorem to work. There are two natural routes to a series for the logarithm of one minus x squared, and you are going to take both.

The first route substitutes minus x squared into the logarithm series you derived in Example 6.10. The signs combine to make every term negative, leaving minus the sum of x to the 2n over n. The second route differentiates first, gets a geometric series after pulling out minus 2x, and integrates back, with the constant zero because the logarithm of 1 is zero.

After reindexing, the two answers are identical, just as Theorem 6.5 requires. The interval is open, because at both ends the series is the negative harmonic series. The numerical check at one half matches the logarithm of three quarters.

63. Putting it together

Section

Part 7

64. Pattern: building a series from a known one

Pattern

1. Find the geometric shape. Rewrite the target as a constant over one minus u, or as a product, derivative or antiderivative of such a thing.
2. Split products with partial fractions, or multiply two known series with the Cauchy formula.
3. Differentiate or integrate term by term when the target's derivative or antiderivative is geometric; fix C at the centre.
4. Solve for the radius from the substituted condition: the size of u less than 1, or the smaller of two intervals.
5. Test each endpoint separately. No operation hands you the ends.

This is the procedure for the whole section. The diagram shows the three ways in and the one way out. Whatever the target function, look for the geometric shape directly, or in its partial fractions, or in its derivative or antiderivative.

The list gives the order of work. Rewrite into a constant over one minus u. Split products or multiply known series. Use termwise calculus when the derivative or antiderivative is the geometric one, and fix the constant at the centre.

The last two steps are the ones that are most often skipped, and they are the ones that make an answer complete. Solve the condition for the radius; then test each endpoint on its own. No operation in this lesson ever hands you the endpoints for free.

65. Check: substituting x squared

Check

Work out the condition before choosing. The old series needs its variable to lie between minus 4 and 4, and the new variable is x squared.

Since x squared is never negative, the condition becomes x squared less than 4, which means the size of x is less than 2. The substitution shrank the interval, because squaring makes numbers bigger than 1 grow faster than they would on their own.

66. Check: a Cauchy coefficient

Check

Use the diagonal: the coefficient of x cubed is the sum of four products, c 0 d 3, c 1 d 2, c 2 d 1 and c 3 d 0. All the c's are 1 and the d's alternate in sign, so the four products are minus 1, plus 1, minus 1, plus 1.

They cancel to zero. That is a good sign, not a suspicious one: the product of the two functions is one over one minus x squared, whose series has no odd powers at all. Checking a coefficient against the product function is always worth the few seconds it takes.

67. Check: endpoints after differentiating

Check

Start with the radius, which Theorem 6.4 keeps at 1, and then test each end of the derivative series separately.

At x equal to 1 the derivative series is the harmonic series, which diverges. At minus 1 it is the alternating harmonic series, which converges. So the derivative series has lost exactly one of the original series' endpoints: differentiating multiplied the terms by n, and at 1 that was enough to tip the balance.

68. Explain the radius and the ends

Explain it to yourself

Write both sentences before revealing. If you can explain this, you understand the whole of Theorem 6.4, not just its statement.

Inside the radius, the powers of x shrink geometrically, and multiplying or dividing by n cannot overturn geometric decay. Outside, the powers grow geometrically, and no factor of n can tame that growth either. So the radius is unchanged.

At an endpoint, the powers of x have size exactly 1, so they neither shrink nor grow. The factor of n, or one over n, is then all that is left to decide convergence, and it can go either way. That is why the endpoints are always retested, and why integration tends to gain them and differentiation tends to lose them.

69. Exit ticket

Exit ticket

\[ f(x) = \ln(1 + 2x) \]

This combines two of the lesson's moves: a substitution into the logarithm series, followed by a full interval calculation. Do it on paper before revealing.

Replacing x by 2x in the series for the logarithm of one plus x gives the series with 2 to the n in each coefficient. The condition is the size of 2x less than 1, so the radius is one half.

At x equal to one half, 2x equals 1, and the series is the alternating harmonic series again, which converges. At minus one half it is the negative harmonic series, which diverges. So the interval runs from minus one half, excluded, to one half, included: the logarithm's pattern of endpoints, rescaled.

70. Recap

Recap

\[ \frac{1}{1-x} \;\to\; \frac{1}{(1-x)^2},\; \ln(1+x),\; \tan^{-1}x, \; \dots \]

Next, Section 6.3 reads the coefficients straight off the derivatives at the centre, the idea behind the uniqueness proof, and builds Taylor series for functions that are not geometric at all.

Stewart, Calculus: Early Transcendentals 8e, §11.9 Representations of Functions as Power Series §11.9, pp. 752-758 — the same material in Stewart

The table is the lesson in one place. Adding and multiplying series work on the common interval. Multiplying by a power of x changes nothing about convergence. Substitution rescales the interval, and you find the new one by solving the condition. Termwise differentiation and integration keep the radius exactly.

The last column is the one to remember: apart from multiplying by a power of x, every operation leaves the endpoints as an open question, and you settle them by testing each end directly.

Everything today started from the geometric series. The uniqueness proof showed that each coefficient is a derivative at the centre divided by a factorial. Section 6.3 turns that observation into Taylor series, which give power series for functions such as sine and the exponential that are not built from the geometric series at all.