6.4 Working with Taylor Series

1. Working with Taylor Series

Title

Calculus II · Section 6.4

New series from old ones, then put them to work where antiderivatives run out

2. What this lesson gives you

Objectives

Section 6.3 built Taylor series one derivative at a time. That is slow, and for most functions it is hopeless. This lesson builds series by recycling the ones you already have, then uses them for two jobs nothing else in the course can do.

• Write the binomial series for any exponent, and say when it stops
• Build new series from Table 6.1 by substituting, multiplying, differentiating and integrating
• Use a series to settle a limit that resists other methods
• Solve a differential equation by assuming a power series and matching coefficients
• Evaluate a nonelementary integral as a series, with a guaranteed error

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 581-599 — learning objectives 6.4.1 to 6.4.5

In Section 6.3 you found Taylor series the honest way: differentiate, evaluate at the centre, divide by a factorial, repeat. That works for a handful of friendly functions, and then it drowns you. Try it on e to the minus x squared and by the fourth derivative you are pushing a polynomial of degree four around, with no pattern in sight.

This lesson is about working smarter. You will keep a short list of series you know by heart, and make every other series by bending one of them: substituting a new expression, multiplying by a power, differentiating or integrating term by term. The binomial series adds one more entry to that list, covering every power of one plus x.

Then you put series to work on two problems ordinary calculus cannot finish: differential equations with no formula solution, and integrals with no formula antiderivative. The payoff is a number you can trust, with an error bar you can prove.

3. Before anything new: the series you already own

Warm-up

Write these out before you reveal them. Every technique in this lesson starts from one of these four series, so if any of them is shaky, this is the moment to fix it rather than halfway through an integral.

Notice the pattern that helps you remember them. The exponential uses every power over its own factorial. Sine keeps only the odd powers, cosine only the even ones, and both alternate in sign. The geometric series is the simplest of all: every power, coefficient one.

Pay attention to where each one is valid. The exponential, sine and cosine series work for every real number. The geometric series only works when x lies strictly between minus one and one. That restriction will travel with every series you build from it, in a disguised form, and forgetting it is one of the traps in this lesson.

4. The binomial series

Section

Part 1

5. Whole-number powers: the binomial theorem

Concept

\[ (1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5 \]

For a whole-number exponent r, the coefficient of x to the n is a binomial coefficient, and the expansion stops after the x to the r term.

\[ \binom{r}{n} = \frac{r!}{n!\,(r-n)!}, \qquad (1+x)^r = \sum_{n=0}^{r}\binom{r}{n}x^n \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 581 — equations 6.6 and 6.7

You met this in algebra. Multiply out one plus x to the fifth and you get the fifth row of Pascal's triangle as coefficients: 1, 5, 10, 10, 5, 1. The triangle on the left builds each row from the one above, each entry the sum of the two entries over it.

The formula with factorials gives the same numbers without drawing the triangle. The coefficient of x to the n in one plus x to the r is r factorial over n factorial times r minus n factorial. For r equal to five and n equal to two that is 120 over 2 times 6, which is 10.

Now look at the right-hand panel. It shows what happens when the exponent is one half: the coefficients keep coming, forever. Factorials of fractions do not make sense, so the formula in the box cannot produce those numbers. The next slide finds a formula that can.

6. Any exponent: differentiate at zero

Concept

For an exponent r that is not a whole number, there is no finite expansion, so build the Maclaurin series from derivatives. The pattern shows itself after three.

\[ f(x) = (1+x)^r, \quad f(0) = 1 \]

\[ f'(x) = r(1+x)^{r-1}, \quad f'(0) = r \]

\[ f''(x) = r(r-1)(1+x)^{r-2}, \quad f''(0) = r(r-1) \]

\[ f^{(n)}(0) = r(r-1)(r-2)\cdots(r-n+1) \]

\[ \frac{f^{(n)}(0)}{n!} = \frac{r(r-1)\cdots(r-n+1)}{n!} = \binom{r}{n} \]

The last line defines the binomial coefficient for any real r: n falling factors on top, n factorial underneath.

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 582 — equation 6.8

Here you do the derivative route once, carefully, so that you never have to do it again for this family. Each derivative of one plus x to the r brings down the current exponent and lowers it by one. At x equal to zero the one plus x part becomes one, so only the string of brought-down numbers survives.

After n derivatives you have n factors: r, then r minus 1, then r minus 2, and so on down to r minus n plus 1. Divide by n factorial, as every Taylor coefficient requires, and you have the coefficient of x to the n.

The last line is really a definition. It takes the familiar binomial coefficient and rewrites it in a form that only needs r to be a number, not a whole number. When r is a whole number the two forms agree, because r factorial over r minus n factorial is exactly that product of n falling factors. When r is not a whole number, only the product form makes sense.

7. Reading the binomial series piece by piece

Notation

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 582-583 — definition of the binomial series, equation 6.9

Step through the notes one at a time. The whole series is controlled by a single number, the exponent r. Everything else is the same recipe every time: n falling factors on top, n factorial on the bottom, times x to the n.

The falling factors are where you will make mistakes, so practise them. For r equal to one half and n equal to three they are one half, minus one half, minus three halves: each one exactly one less than the one before. Count them; there must be exactly n.

The last note is the one to remember for exams. However strange r is, the radius of convergence is one. You need the absolute value of x below one, and what happens at the endpoints depends on r. That fact decides when you may use the series, and the trap later in this part shows what happens when you ignore it.

8. What happens when r is a whole number?

Prediction

\[ (1+x)^3: \quad \binom{3}{5} = \frac{3\cdot 2\cdot 1\cdot 0\cdot(-1)}{5!} \]

Commit to an answer before you reveal. Many people say undefined, because they expect something to go wrong when the product formula meets a whole number. Nothing goes wrong; something vanishes.

Write the five falling factors for r equal to three: 3, 2, 1, 0, minus 1. The zero is the fourth factor, and once it appears, it appears in every later coefficient too, since each coefficient includes all the factors of the one before. So the coefficients of x to the 4th, 5th, 6th and beyond are all zero.

That is how the general binomial series contains the old binomial theorem. For a whole-number exponent the infinite series quietly turns into a polynomial, and a polynomial is valid for every x, not just between minus one and one.

9. The radius is always one: the ratio test

Concept

Take the ratio of consecutive terms. Almost every factor cancels, leaving one new factor on top and one on the bottom.

\[ \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{r(r-1)\cdots(r-n)}{(n+1)!}\cdot\frac{n!}{r(r-1)\cdots(r-n+1)}\right||x| \]

\[ = \frac{|r-n|}{n+1}|x| \]

\[ \lim_{n\to\infty}\frac{|r-n|}{n+1}|x| = |x| \]

The limit is less than one exactly when the absolute value of x is less than one, whatever r is. The endpoints depend on r: both converge when r is at least 0; only x equal to 1 when r is between minus 1 and 0; neither when r is below minus 1.

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 582 — interval of convergence of the binomial series

This is the ratio test from Chapter 5, and it is worth seeing how much cancels. The coefficient of x to the n plus 1 has one more falling factor on top, r minus n, and one more factor in the factorial underneath, n plus 1. Everything else is shared and cancels.

So the ratio of consecutive terms is the absolute value of r minus n, over n plus 1, times the absolute value of x. As n grows, r becomes negligible next to n, and the fraction tends to one. The limit of the ratio is just the absolute value of x.

The ratio test says the series converges when that limit is below one and diverges when it is above one. So the radius is one for every exponent. The endpoints are delicate and the book only states the results: they depend on whether r is positive, between minus one and zero, or below minus one. You will rarely need them, but you should never use the series outside the interval from minus one to one.

10. Inside the radius and outside it

Picture it

Three partial sums of the square root's series against the root itself. Between minus 1 and 1 they pile onto the curve. Past 1, the extra terms make things worse: the degree 12 sum leaves the picture almost at once.

This picture is what a radius of convergence looks like. The solid curve is the square root of one plus x. The dashed curves are partial sums of its binomial series of degree 2, 5 and 12.

Inside the shaded band, between minus one and one, the higher the degree, the closer the fit: the degree 12 sum lies right on the curve. Now look past x equal to one. The degree 2 sum drifts away slowly, the degree 5 sum flies upward, and the degree 12 sum plunges off the bottom of the picture almost immediately.

The lesson is counter-intuitive but important. Outside the radius, more terms do not help; they make things worse, because the individual terms grow instead of shrinking. The function itself is perfectly well behaved there. The failure belongs to the series, not to the square root.

11. Example 6.17(a): the binomial series for a square root

Worked example

Find the binomial series for the square root of one plus x.

\[ \sqrt{1+x} = (1+x)^{1/2} \]

\[ r = \tfrac12 \]

\[ \binom{1/2}{0} = 1, \qquad \binom{1/2}{1} = \tfrac12 \]

\[ \binom{1/2}{2} = \frac{(1/2)(-1/2)}{2!} = -\frac18 \]

\[ \binom{1/2}{3} = \frac{(1/2)(-1/2)(-3/2)}{3!} = \frac{3/8}{6} = \frac{1}{16} \]

\[ \binom{1/2}{n} = \frac{(-1)^{n+1}}{n!}\cdot\frac{1\cdot3\cdot5\cdots(2n-3)}{2^n} \]

\[ \sqrt{1+x} = 1 + \frac12x - \frac18x^2 + \frac1{16}x^3 - \cdots \]

\[ \frac{(-1)^4}{3!}\cdot\frac{1\cdot3}{2^3} = \frac{3}{48} = \frac{1}{16} \;\checkmark \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 583 — Example 6.17a

Start by rewriting the root as a power, so that you can read off r equal to one half. Then work out the coefficients one at a time, writing every falling factor. It is tempting to skip straight to a pattern, but the pattern only becomes visible after you have done three or four by hand.

Look at what happens from the x squared term onward. Each new factor, one half minus k, is a negative number with an odd numerator: minus one half, minus three halves, minus five halves. Pulling out the signs gives the alternating sign; the numerators build the product of odd numbers; the halves build the power of two in the denominator.

The check at the end matters. Whenever you write a general term, substitute a value of n you already computed directly and make sure the two agree. Here the general formula at n equal to three gives one sixteenth, exactly the coefficient you found by hand, so the pattern is right.

12. Example 6.17(b): estimating the square root of 1.5

Worked example

Use the third-order Maclaurin polynomial to estimate the square root of 1.5, and bound the error with Taylor's theorem.

\[ p_3(x) = 1 + \tfrac12x - \tfrac18x^2 + \tfrac1{16}x^3 \]

\[ p_3(0.5) = 1 + 0.25 - 0.03125 + 0.0078125 \]

\[ p_3(0.5) = 1.2265625 \approx 1.2266 \]

\[ f^{(4)}(x) = -\frac{15}{16}(1+x)^{-7/2} \]

\[ |f^{(4)}(c)| \le \frac{15}{16} \]

\[ |R_3(0.5)| \le \frac{15/16}{4!}(0.5)^4 = \frac{15}{6144} \approx 0.00244 \]

\[ |1.2265625 - 1.2247449| = 0.0018 < 0.00244 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 583-584 — Example 6.17b and Figure 6.10

The estimate itself is just substitution: four terms of the series at x equal to one half, added up, give 1.2266. The more interesting part is how sure you can be of it.

Taylor's theorem says the error of a degree-three polynomial is the fourth derivative at some unknown point c, over 4 factorial, times the distance to the centre to the 4th power. You do not know c, so you bound the fourth derivative over the whole interval from 0 to one half. Because it is a negative power of one plus x, it is largest in size where one plus x is smallest, at x equal to 0.

The figure recreates the book's Figure 6.10. Near the centre the cubic and the root are indistinguishable; they only separate far out to the right. The final check compares with a calculator: the real error, 0.0018, is below the guaranteed bound, 0.00244, exactly as the theorem promised. A bound that the true error exceeded would mean a mistake somewhere.

13. Checkpoint 6.16: one over one plus x, squared

Worked example

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

\[ r = -2 \]

\[ \binom{-2}{n} = \frac{(-2)(-3)(-4)\cdots(-n-1)}{n!} \]

\[ \binom{-2}{n} = (-1)^n\frac{2\cdot3\cdots(n+1)}{n!} = (-1)^n\frac{(n+1)!}{n!} \]

\[ \binom{-2}{n} = (-1)^n(n+1) \]

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

\[ 1 - 0.2 + 0.03 - 0.004 + 0.0005 - 0.00006 = 0.82644 \approx \frac{1}{1.21} = 0.826446 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 584 — Checkpoint 6.16

A reciprocal square is the power minus two, so r is minus two. The falling factors now start at minus two and go down: minus two, minus three, minus four, and so on. Every one of them is negative, which is why a factor of minus one to the n comes out.

What is left on top is 2 times 3 times 4 up to n plus 1, which is n plus 1 factorial. Dividing by n factorial cancels everything but the last factor, so the coefficient is simply minus one to the n times n plus 1. The coefficients are 1, minus 2, 3, minus 4: easy to remember once you have seen them.

The check uses x equal to 0.1, well inside the radius. Six terms land within about six millionths of the exact value, one over 1.21. You can also see this series another way: it is the negative of the derivative of the alternating geometric series, since the derivative of one over one plus x is minus one over one plus x squared.

14. Trap: using the binomial series outside its radius

Trap

15. Terminates or runs forever?

Sorting

Everything turns on one question: is r a whole number that is zero or positive? If it is, one of the falling factors is eventually r minus r, which is zero, and the series stops and becomes a polynomial. If it is not, no factor is ever zero and the series goes on forever.

The item most people misfile is r equal to 2.5. It is positive, so it feels like the binomial theorem should apply, but the factors 2.5, 1.5, 0.5, minus 0.5 skip straight past zero without landing on it. Negative whole numbers such as minus one also run forever: the factors minus one, minus two, minus three move away from zero, never towards it.

The two buckets differ in more than length. A polynomial is valid for every x. An infinite binomial series is valid only for x between minus one and one, possibly with an endpoint. So sorting the exponent also tells you where you are allowed to use the result.

16. Slide r and the degree

Tweak it

Start with r equal to 2 and raise N. Once N passes 2, the curve stops changing: every extra coefficient is zero, and you are looking at the exact parabola one plus x squared, correct for every x on the screen.

Now set r to one half or minus one and raise N slowly. Watch the region between minus one and one: the curve settles down and stops moving as N grows, which is what convergence looks like. Then watch the region past x equal to one: as N grows the curve swings more and more violently, which is what divergence looks like.

Try r equal to minus one especially. The partial sums are 1 minus x plus x squared and so on, the alternating geometric series, and you can see them flip between too high and too low on the right as N changes from odd to even.

17. New series from known ones

Section

Part 2

18. Table 6.1: the series to know by heart

Concept

The book prints the logarithm's sum from n equal to 0, which would divide by zero, and gives arctan's interval as minus 1 less than x at most 1; the series in fact converges at both ends. Both are corrected here.

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 585 — Table 6.1

This table is your working vocabulary for the rest of the chapter. Every new series in this lesson is one of these seven, transformed. It is worth memorising the series column and the interval column together, because the interval is what stops you misusing a series.

Two small corrections to the printed table. The logarithm's series has n in the denominator, so it must start at n equal to 1, not 0; starting at 0 would divide by zero. And the arctangent series converges at both endpoints, minus one and one, which is how the famous series for pi over four is obtained at x equal to one.

Group them in your head. The exponential, sine and cosine are the factorial family: they converge everywhere. The geometric series, the logarithm, the arctangent and the binomial series have plain numbers in their denominators and radius one. Denominators tell you a lot about convergence: factorials grow fast enough to beat any power of x; plain numbers do not.

19. Match each function to its series

Matching

Three of these series begin with x minus something, which is exactly why this matching is worth practising. The fingerprint is the denominator.

The logarithm uses every power of x over plain n: x, then x squared over 2, then x cubed over 3. The arctangent uses only odd powers over plain odd numbers: x, x cubed over 3, x to the 5th over 5. Sine uses odd powers too, but over factorials: x cubed over 3 factorial, which is 6, not 3.

A quick way to sort sine from arctangent when you meet them is to look at the x cubed coefficient: minus one sixth for sine, minus one third for arctangent. The geometric and square-root series start with 1 because their functions equal 1 at x equal to zero.

20. Four moves that make new series

Concept

Section 6.2 showed that power series can be combined like polynomials. Each move below turns a table series into a new one with no derivatives computed.

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

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 585 — combining Table 6.1 with the properties of Section 6.2

These four moves come from Section 6.2, where you learned that power series can be added, multiplied by powers of x, differentiated and integrated term by term inside their interval of convergence. Here they become a recipe for producing new series without touching a derivative.

Substitution is the most common. Whatever sits where the variable was, say minus x squared in the exponent, replaces u everywhere in the series. Multiplying by a power of x shifts every exponent up. Differentiating and integrating act on each term with the power rule.

Keep an eye on the last column. Substitution changes the interval, because the old condition was on u and must be rewritten in x. The other moves keep the radius the same, although integration or differentiation can change what happens at an endpoint.

21. Why recycling beats differentiating

Intuition

Try the derivative route on e to the minus x squared. Each derivative is a longer polynomial times the same exponential.

\[ f'(x) = -2x\,e^{-x^2}, \qquad f''(x) = (4x^2 - 2)\,e^{-x^2} \]

\[ f'''(x) = (-8x^3 + 12x)\,e^{-x^2} \]

\[ f^{(4)}(x) = (16x^4 - 48x^2 + 12)\,e^{-x^2}, \qquad \frac{f^{(4)}(0)}{4!} = \frac{12}{24} = \frac12 \]

Four derivatives to earn one coefficient, and no general pattern in sight. The substitution on the previous slide gave the same one half, the coefficient of x to the 4th over 2 factorial, and every other coefficient too, in one line.

This slide is here to make you appreciate the shortcut. Differentiating e to the minus x squared four times is legal, and each derivative is a polynomial times the same exponential, but the polynomials grow and their coefficients show no pattern you could turn into a general term.

After all that work you get one coefficient: the fourth derivative at zero is 12, and 12 over 24 is one half. The substitution method gives the coefficient of x to the 4th directly as one over 2 factorial, which is the same one half, and it gives every other coefficient too, in one line.

So the rule of thumb is: before you differentiate anything, ask whether the function is a table function wearing a disguise. Most of the time it is.

22. Example 6.18(a): cos of the square root of x

Worked example

Find the Maclaurin series of cos of the square root of x using Table 6.1.

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

\[ (\sqrt x)^{2n} = x^n \]

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

\[ \cos\sqrt x = 1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} + \cdots \]

\[ \text{valid for } x \ge 0 \]

\[ 1 - \tfrac12 + \tfrac1{24} - \tfrac1{720} = 0.540278 \approx \cos 1 = 0.540302 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 585-586 — Example 6.18a

Write the cosine series in a placeholder variable u first. That habit makes substitution mechanical: every u becomes the root of x, and nothing else changes. Even powers of a square root are ordinary powers, so u to the 2n becomes x to the n.

Notice two things that did not change. The sign pattern and the factorial, 2n factorial, are exactly as they were; substitution never touches the index. And the new series has every power of x, not only even ones, which is what makes it look unfamiliar.

The figure shows why the domain matters. The series makes sense for negative x, but the function does not, since you cannot take the square root of a negative number in the real numbers. So the book restricts it to x at least zero. The check at x equal to 1 compares four terms with cos 1, and they agree to four decimal places.

23. Example 6.18(b): sinh x from two exponentials

Worked example

\[ \sinh x = \frac{e^x - e^{-x}}{2} \]

\[ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}, \qquad e^{-x} = \sum_{n=0}^{\infty}\frac{(-x)^n}{n!} \]

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

\[ n \text{ even}: 0, \qquad n \text{ odd}: \frac{2x^n}{n!} \]

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

\[ 1 + \tfrac16 + \tfrac1{120} = 1.175000 \approx \sinh 1 = 1.175201 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 586 — Example 6.18b

The hyperbolic sine is half the difference between e to the x and e to the minus x. Both have series you know, the second by substituting minus x, and both converge everywhere, so you may subtract them term by term.

Look at the bracket, one minus minus one to the n. For even n it is 1 minus 1, zero; for odd n it is 1 plus 1, two. So every even-power term cancels and every odd-power term doubles. Halving leaves the odd powers over their factorials: sine's series without the alternating signs.

The figure makes the cancellation visible: the two dashed curves are half of e to the x and minus half of e to the minus x, and their sum is the solid S-shaped curve through the origin. The dotted cubic, the first two terms of the series, hugs it across the middle of the picture. Three terms at x equal to 1 give 1.175000 against the true 1.175201.

24. Checkpoint 6.17: sine of x squared

Worked example

\[ \text{Find the Maclaurin series for } \sin(x^2) \]

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

\[ (x^2)^{2n+1} = x^{4n+2} \]

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

\[ \sin(x^2) = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \cdots \]

\[ 1 - \tfrac16 + \tfrac1{120} - \tfrac1{5040} = 0.841468 \approx \sin 1 = 0.841471 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 586 — Checkpoint 6.17

Same method, different table entry. Write sine in the placeholder u, then replace every u by x squared. The only algebra is the exponent: x squared to the power 2n plus 1 is x to the 4n plus 2, because a power of a power multiplies the exponents.

So the powers now jump by four: x squared, x to the 6th, x to the 10th. The factorials are unchanged, 1, 3 factorial, 5 factorial, because the index did not change. It is a common slip to write 6 factorial under x to the 6th; resist it.

The series holds for every x, because sine's series does and x squared is always a real number. The check at x equal to 1 is a nice one: x squared is also 1 there, so the new series at 1 must equal sine's series at 1, and it does, 0.841468 against 0.841471.

25. Find the error: a substitution applied halfway

Error analysis

\[ \text{correct: } e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!} \]

Read the wrong line carefully. The numerator is right: minus x squared to the n is minus one to the n times x to the 2n. The error is in the denominator. Someone saw x to the 2n and wrote 2n factorial underneath, as if the factorial should match the power.

Substitution replaces the variable inside the series; it never touches the index or the coefficients. The exponential series has n factorial, so after substituting it still has n factorial.

The third note gives you a check you can always run. At x equal to 1 the wrong series gives 1 minus one half plus one twenty-fourth, which is the cosine series, adding up to cos 1, about 0.540. But e to the minus 1 is about 0.368. Evaluating at one convenient point catches most substitution errors in seconds.

26. Example 6.19: differentiating to reach one over a square root

Worked example

Use the series for the square root of one plus x to find the series for one over that root.

\[ \frac{d}{dx}\sqrt{1+x} = \frac{1}{2\sqrt{1+x}} \]

\[ \frac{1}{\sqrt{1+x}} = 2\,\frac{d}{dx}\sqrt{1+x} \]

\[ \frac{d}{dx}\left(1 + \tfrac12x - \tfrac18x^2 + \tfrac1{16}x^3 - \tfrac{5}{128}x^4\right) \]

\[ = \tfrac12 - \tfrac14x + \tfrac3{16}x^2 - \tfrac5{32}x^3 \]

\[ \frac{1}{\sqrt{1+x}} = 1 - \tfrac12x + \tfrac38x^2 - \tfrac5{16}x^3 + \cdots \]

\[ \frac{1}{\sqrt{1+x}} = 1 + \sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\cdot\frac{1\cdot3\cdot5\cdots(2n-1)}{2^n}x^n \]

\[ \binom{-1/2}{2} = \frac{(-1/2)(-3/2)}{2!} = \frac38 \;\checkmark \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 586-587 — Example 6.19

The two functions are related by differentiation: the derivative of the square root of one plus x is one over twice that root. So the reciprocal root is twice the derivative of the root, and you can get its series by differentiating the series you already have.

Differentiate each term with the power rule: the constant disappears, one half x becomes one half, minus one eighth x squared becomes minus one quarter x, and so on. Then double every coefficient. Each step is simple, which is the whole point; the book notes that doing this directly from the binomial definition is harder.

The general term now has odd numbers running up to 2n minus 1 instead of 2n minus 3, because differentiating shifted every term down by one place. The check is independent: compute the coefficient of x squared straight from the binomial formula with r equal to minus one half, and it matches the three eighths you found by differentiating.

27. Checkpoint 6.18: one over one plus x, to the three halves

Worked example

\[ \text{Find the binomial series for } (1+x)^{-3/2} \]

\[ \frac{d}{dx}(1+x)^{-1/2} = -\tfrac12(1+x)^{-3/2} \]

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

\[ \frac{d}{dx}\left(1 - \tfrac12x + \tfrac38x^2 - \tfrac5{16}x^3 + \tfrac{35}{128}x^4\right) \]

\[ = -\tfrac12 + \tfrac34x - \tfrac{15}{16}x^2 + \tfrac{35}{32}x^3 \]

\[ (1+x)^{-3/2} = 1 - \tfrac32x + \tfrac{15}8x^2 - \tfrac{35}{16}x^3 + \cdots \]

\[ (1+x)^{-3/2} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\cdot\frac{1\cdot3\cdots(2n+1)}{2^n}x^n \]

\[ \binom{-3/2}{2} = \frac{(-3/2)(-5/2)}{2!} = \frac{15}{8} \;\checkmark \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 587 — Checkpoint 6.18

This is the same move one level further down. Differentiating the minus one half power gives minus one half times the minus three halves power, so the function you want is minus two times that derivative.

Start from the series you just built in Example 6.19. Differentiate term by term, then multiply by minus two. The minus two cancels the leading minus one half to give 1, and the remaining coefficients come out as minus three halves, fifteen eighths, minus thirty-five sixteenths.

Look at how the odd products have shifted again: now they run 1 times 3 times 5 up to 2n plus 1. Each differentiation pushes the product up by one odd number. The check, as before, computes one coefficient straight from the binomial formula with r equal to minus three halves: minus three halves times minus five halves over 2 is fifteen eighths.

28. Integrating: arctan from a geometric series

Concept

Exercise 211 runs the fourth move. Substitute minus t squared into the geometric series, then integrate from 0 to x.

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

\[ \arctan x = \int_0^x\frac{dt}{1+t^2} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1} \]

At x equal to one half, four terms give 0.463467 against the true 0.463648: an error of 0.00018, under the next term, 0.00022.

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 597 — Exercise 211

This is the fourth move, integration, and it gives one of the most useful series in the table. Substitute minus t squared into the geometric series to get the series for one over one plus t squared. The derivative of arctangent is exactly that function, so integrating from 0 to x gives the arctangent.

Integrate term by term: t to the 2n becomes x to the 2n plus 1 over 2n plus 1. No factorials anywhere, which is why this series converges slowly compared with sine's.

The figure shows the partial sums hugging arctangent between minus one and one and breaking away outside. Arctangent itself is smooth everywhere, so the limit on the radius comes from the geometric series you started with: one over one plus t squared misbehaves at the complex points plus and minus i, a distance one from the origin. The numerical check at one half shows the alternating error bound at work: the error is less than the first omitted term.

29. Complete the table of new series

Comparison

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 596 — Exercises 175 and 177 use the same substitutions

For each row, first identify the table series and the substitution, then rewrite the old condition on u as a condition on x. That last step is where the interest lies.

The exponential works for every u, so it works for every x. The geometric series needs u between minus one and one; with u equal to 3x that means x between minus one third and one third. The cosine works for every u, but u is the root of x, which needs x at least zero.

The binomial row is Exercise 175: a power of one plus something is always the binomial series, and u equal to x squared keeps the radius one. The logarithm row needs care at the endpoints: the logarithm's series converges for u from just above minus one up to one, so with u equal to 2x, x runs from just above minus one half up to one half, including one half.

30. A limit that a series settles in one line

Worked example

Use a series to evaluate a limit that would take L'Hôpital's rule three rounds.

\[ \lim_{x\to 0}\frac{\sin x - x}{x^3} \]

\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \]

\[ \sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - \cdots \]

\[ \frac{\sin x - x}{x^3} = -\frac16 + \frac{x^2}{120} - \cdots \]

\[ \lim_{x\to 0}\frac{\sin x - x}{x^3} = -\frac16 \]

\[ \frac{\sin 0.1 - 0.1}{0.001} = -0.166583 = -\tfrac16 + \tfrac{0.01}{120} \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 597 — Exercises 218 to 225 use the same expansions

This is the fourth objective of the lesson, and it is one of the most satisfying uses of series. The limit is zero over zero, and L'Hôpital's rule would need three rounds of differentiation, each checked for the indeterminate form. The series does it in one expansion.

Expand the sine. The x cancels the leading term exactly, which is why the original expression was zero over zero. What is left starts with minus x cubed over 6, and dividing by x cubed leaves minus one sixth plus terms that all carry a power of x. As x goes to zero, those terms vanish.

The figure shows more than the limit. The quotient is undefined at zero, marked by the hollow dot, but it approaches minus one sixth, and the dashed parabola, the next two terms of the series, describes how it leaves that value. The numerical check at x equal to 0.1 agrees with the two-term series to six decimal places.

31. Trap: the substituted series keeps the old interval

Trap

32. Solving differential equations with series

Section

Part 3

33. The method: assume a series, then match coefficients

Concept

Most differential equations have no solution you can write with the functions you know. So guess a power series with unknown coefficients and let the equation tell you what they are.

\[ y = \sum_{n=0}^{\infty}c_nx^n, \qquad y' = \sum_{n=1}^{\infty}nc_nx^{n-1} \]

The key fact is uniqueness: if two power series are equal on an interval, their coefficients are equal, power by power, because each coefficient is a derivative at the centre divided by a factorial.

\[ \sum a_nx^n = \sum b_nx^n \;\Longrightarrow\; a_n = b_n = \frac{f^{(n)}(0)}{n!} \text{ for every } n \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 587-588 — solving differential equations with power series; uniqueness of power series

Most differential equations you meet in science have no solution you can write as a formula. The power-series method sidesteps that: assume the solution is a power series with unknown coefficients, and let the equation decide what the coefficients must be.

Differentiating a power series is easy, so both sides of the equation become power series. Then you use uniqueness: if two power series are equal for every x near zero, they must have the same coefficient for every power. Each coefficient is forced to be a derivative at zero divided by a factorial, so there is no freedom for two different series to add up to the same function.

Matching the coefficient of each power gives an equation between the unknown coefficients, usually a recurrence that computes each one from earlier ones. The initial conditions fix the first one or two, and the recurrence does the rest.

34. Example 6.20: a growth equation solved by series

Worked example

\[ y' = y, \qquad y(0) = 3 \]

\[ y = c_0 + c_1x + c_2x^2 + c_3x^3 + \cdots \]

\[ y' = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \cdots \]

\[ c_1 = c_0, \quad 2c_2 = c_1, \quad 3c_3 = c_2, \quad (n+1)c_{n+1} = c_n \]

\[ c_0 = y(0) = 3 \]

\[ c_1 = 3, \quad c_2 = \frac32, \quad c_3 = \frac{3}{3\cdot2}, \quad c_n = \frac{3}{n!} \]

\[ y = 3\sum_{n=0}^{\infty}\frac{x^n}{n!} = 3e^x \]

\[ y' = 3e^x = y, \qquad y(0) = 3e^0 = 3 \;\checkmark \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 587-588 — Example 6.20

You already know the answer to this one, which makes it the right place to learn the method: you can watch the series rediscover the exponential. Assume a series, differentiate it, and set the two equal.

Matching powers gives the recurrence: n plus 1 times c n plus 1 equals c n. In words, each coefficient is the previous one divided by the next whole number. The initial condition is the easiest step to overlook: setting x equal to zero kills every term except the constant, so c zero is y of zero, which is 3.

Unwinding the recurrence gives 3 divided by n factorial, and you should recognise three times the exponential series. The final check substitutes 3 e to the x back into the equation and the initial condition. The power of the method is that the same steps work when the answer is not a function you already know.

35. Checkpoint 6.19: doubling the rate

Worked example

\[ y' = 2y, \qquad y(0) = 5 \]

\[ y' = c_1 + 2c_2x + 3c_3x^2 + \cdots \]

\[ (n+1)c_{n+1} = 2c_n \]

\[ c_0 = 5 \]

\[ c_1 = 10, \quad c_2 = \frac{2\cdot10}{2} = 10, \quad c_n = \frac{5\cdot 2^n}{n!} \]

\[ y = 5\sum_{n=0}^{\infty}\frac{(2x)^n}{n!} = 5e^{2x} \]

\[ y' = 10e^{2x} = 2y, \qquad y(0) = 5 \;\checkmark \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 588 — Checkpoint 6.19

The only change from Example 6.20 is the factor of two on the right, and it shows up in exactly one place: the recurrence becomes n plus 1 times c n plus 1 equals twice c n. Each step now multiplies by two as well as dividing by the next whole number.

Starting from c zero equal to 5, the coefficients are 5 times 2 to the n over n factorial. Group the 2 to the n with the x to the n and you get powers of 2x over factorials: the exponential series evaluated at 2x.

The figure shows how the partial sums approach the answer. Every partial sum starts at the initial value 5 with slope 10, because those two coefficients are fixed first. Each extra term from the recurrence bends the polynomial further up, and seven terms are already close to 5 e to the 2x across the whole picture.

36. Before matching: line up the powers

Step zero

\[ 2c_2 + 6c_3x + 12c_4x^2 + \cdots = x\left(c_0 + c_1x + c_2x^2 + \cdots\right) \]

This slide is the single most important habit in the whole method. Before you equate coefficients, both sides must be written as plain sums of powers of x, with the powers lined up.

Multiply the x through the right-hand bracket. Now the right side starts at x to the 1: c zero x, c one x squared, and so on. It has no constant term at all. The left side does have one, 2 c two, and its partner on the right is zero. That is why c two is zero in Airy's equation.

The classic slip is to compare the first term on each side, 2 c two with c zero, just because they are both written first. They multiply different powers of x, so there is no reason for them to be equal. Always compare coefficients of the same power.

37. Airy's equation: a recurrence three steps long

Concept

Airy's equation models light near a caustic and a quantum particle in a linear field. Its solutions are not elementary. Assume a series and differentiate twice.

\[ y'' = xy \]

\[ y'' = 2\cdot1\,c_2 + 3\cdot2\,c_3x + 4\cdot3\,c_4x^2 + 5\cdot4\,c_5x^3 + \cdots \]

\[ xy = c_0x + c_1x^2 + c_2x^3 + \cdots \]

\[ 2c_2 = 0, \quad 3\cdot2\,c_3 = c_0, \quad 4\cdot3\,c_4 = c_1, \quad 5\cdot4\,c_5 = c_2 \]

\[ n(n-1)c_n = c_{n-3} \quad (n \ge 3) \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 588-589 — Airy's equation; the book first prints it as y′ − xy = 0, a misprint for y″ − xy = 0

Airy's equation is the simplest equation whose solutions are genuinely new functions. It appears in optics, near the bright edge of a rainbow, and in quantum mechanics. Note the book's misprint: it first writes y prime minus x y equals zero, but the example that follows, and the name, belong to the second-derivative equation.

Differentiate the series twice. Each term drops two powers and picks up two factors: 2 times 1 times c two, 3 times 2 times c three times x, and so on. Multiplying y by x lifts every power by one. So the coefficient of x to the m on the left is the coefficient m plus 2 on the right, shifted.

Matching gives the recurrence: n times n minus 1 times c n equals c n minus 3. It links coefficients three apart, which means the coefficients split into three separate chains: one starting from c zero, one from c one, and one from c two.

38. Example 6.21: Airy's equation solved

Worked example

\[ y'' - xy = 0, \qquad y(0) = a, \quad y'(0) = b \]

\[ c_2 = 0, \quad c_3 = \frac{c_0}{3\cdot2}, \quad c_4 = \frac{c_1}{4\cdot3} \]

\[ c_5 = \frac{c_2}{5\cdot4} = 0, \quad c_6 = \frac{c_0}{6\cdot5\cdot3\cdot2}, \quad c_7 = \frac{c_1}{7\cdot6\cdot4\cdot3} \]

\[ c_0 = a, \qquad c_1 = b \]

\[ y = a\left(1 + \frac{x^3}{6} + \frac{x^6}{180} + \cdots\right) + b\left(x + \frac{x^4}{12} + \frac{x^7}{504} + \cdots\right) \]

\[ y_1'' = x + \tfrac{x^4}{6}, \qquad xy_1 = x + \tfrac{x^4}{6} + \tfrac{x^7}{180} \]

They agree up to x to the 7th, and that leftover is cancelled by the next term of the series, whose second derivative is x to the 7th over 180.

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 589-590 — Example 6.21

Follow the three chains separately. The chain from c two dies at once, because c two is zero and every later member is a multiple of it. The chain from c zero produces c three, c six, c nine; the chain from c one produces c four, c seven, c ten.

The initial conditions pick out the two free coefficients. The value of y at zero is c zero, so c zero is a. The derivative at zero is the coefficient of x, so c one is b. Grouping everything multiplied by a, and everything multiplied by b, gives two independent solutions.

The check is a real substitution. Take the first three terms of the a-part, differentiate twice, and compare with x times the same terms. They agree up to x to the 4th, and the only mismatch, x to the 7th over 180, is exactly what the next term of the series cancels. That is how a truncated series solution behaves: correct up to the order you kept.

39. What the two Airy solutions look like

Picture it

Both curves were drawn by summing their power series with the recurrence, 180 terms each. The same few lines of algebra produce waves on the left and runaway growth on the right.

Here are the two solutions you just found, drawn by summing 180 terms of each series with the recurrence. No closed formula was used; the series is the definition of these functions.

Read the equation as a statement about curvature. The second derivative equals x times y. Where x is negative, the curvature has the opposite sign to y, so the curve always bends back toward the axis, and you get waves. Further left, the factor x is larger, so the bending is stronger and the waves get shorter.

Where x is positive, the curvature has the same sign as y, so any positive value bends further upward, and the solutions run away. One simple equation produces both behaviours. That is the payoff of the power-series method: it gives you access to functions that no combination of the familiar ones can express.

40. Checkpoint 6.20: a recurrence four steps long

Worked example

\[ y'' + x^2y = 0, \qquad y(0) = a, \quad y'(0) = b \]

\[ y'' = 2c_2 + 6c_3x + 12c_4x^2 + \cdots, \quad x^2y = c_0x^2 + c_1x^3 + \cdots \]

\[ 2c_2 = 0, \qquad 6c_3 = 0 \]

\[ n(n-1)c_n = -c_{n-4} \quad (n \ge 4) \]

\[ c_4 = -\frac{c_0}{12}, \quad c_5 = -\frac{c_1}{20}, \quad c_8 = \frac{c_0}{672}, \quad c_9 = \frac{c_1}{1440} \]

\[ y = a\left(1 - \frac{x^4}{12} + \frac{x^8}{672} - \cdots\right) + b\left(x - \frac{x^5}{20} + \frac{x^9}{1440} - \cdots\right) \]

\[ \left(1 - \tfrac{x^4}{12}\right)'' + x^2\left(1 - \tfrac{x^4}{12}\right) = -\tfrac{x^6}{12}, \quad \left(\tfrac{x^8}{672}\right)'' = \tfrac{x^6}{12} \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 590 — Checkpoint 6.20

Now the right side is x squared times y, which lifts every power by two. So the lowest two powers on the left, the constant and the x term, have no partners, and c two and c three are both zero.

From x squared on, the coefficients must cancel, which gives n times n minus 1 times c n equals minus c n minus 4. The chains now run four apart, and there are four of them, but the chains from c two and c three are all zeros. The minus sign makes the surviving terms alternate.

The figure shows both solutions oscillating on both sides of the origin, because x squared is never negative: the curvature always opposes y. The check substitutes the first two terms of the a-part into the equation and finds a leftover minus x to the 6th over 12, exactly cancelled by the second derivative of the next term, x to the 8th over 672.

41. Trap: matching coefficients without shifting the index

Trap

42. Put the power-series method in order

Ranking

Think about what each step needs before it can happen. You cannot differentiate until you have a series, and you cannot substitute until you have the derivatives. After substituting, the two sides are sums with different index shifts, so shifting has to come before equating, or you compare different powers.

Equating gives the recurrence. The initial conditions fix the first coefficients, and only then can you unwind the recurrence into actual numbers.

The final step says recognise the series if you can. Sometimes, as in Example 6.20, you recognise the exponential. Often, as in Airy's equation, you cannot, and the series itself is the answer. That is not a failure; it is a new function.

43. Integrals with no elementary antiderivative

Section

Part 4

44. What elementary means

Concept

elementary function — A function built from powers, exponentials, logarithms and trig functions (and their inverses) by finitely many sums, products, quotients and compositions.

Elementary does not mean simple: a tangle of roots, exponentials and sines is still elementary. What matters is that the antiderivative of e to the minus x squared is provably not elementary, so none of the techniques of Chapter 3 can ever find it.

\[ \int e^{-x^2}\,dx = \;? \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 590 — evaluating nonelementary integrals

Elementary is a technical word here. It means built from powers, roots, exponentials, logarithms, trig functions and their inverses using finitely many arithmetic operations and compositions. A complicated formula can still be elementary.

The bell curve in the figure is e to the minus x squared, and the shaded area from 0 to 1 is a perfectly definite number. But it was proved in the nineteenth century that no elementary function has e to the minus x squared as its derivative. So substitution, parts, partial fractions and every other technique of Chapter 3 must fail on it, not because you are not clever enough, but because there is no answer of that form.

Series give you a way around. You cannot write the antiderivative as a finite formula, but you can write it as an infinite series, and evaluate that series to any accuracy you want.

45. Example 6.22(a): the integral of e to the minus x squared as a series

Worked example

Express the indefinite integral as an infinite series.

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

\[ e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots \]

\[ \int x^{2n}\,dx = \frac{x^{2n+1}}{2n+1} \]

\[ \int e^{-x^2}\,dx = C + \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)\,n!} \]

\[ = C + x - \frac{x^3}{3} + \frac{x^5}{5\cdot2!} - \frac{x^7}{7\cdot3!} + \cdots \]

\[ \frac{d}{dx}\frac{x^{2n+1}}{(2n+1)\,n!} = \frac{x^{2n}}{n!} \;\checkmark \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 590-591 — Example 6.22a

Two moves, both from Part 2. First substitute minus x squared into the exponential series, which gives even powers over factorials with alternating signs. Then integrate term by term.

Integrating a power series inside its interval of convergence is allowed, and here the interval is every real number, so there are no restrictions. Each term x to the 2n becomes x to the 2n plus 1 over 2n plus 1, and the factorial just comes along.

Look at the denominators that result: the new power times the old factorial, so 3, then 5 times 2, then 7 times 6. The check differentiates the general term back and recovers the integrand's general term, which is the right way to check any antiderivative, series or not.

46. Example 6.22(b): the area from 0 to 1 within 0.01

Worked example

\[ \int_0^1 e^{-x^2}\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)\,n!} \]

\[ = 1 - \frac13 + \frac1{10} - \frac1{42} + \frac1{216} - \cdots \]

\[ |\text{error}| < \text{first omitted term} \]

\[ 1 - \tfrac13 + \tfrac1{10} - \tfrac1{42} = 0.742857 \]

\[ |\text{error}| < \frac{1}{216} \approx 0.00463 < 0.01 \]

\[ |0.746824 - 0.742857| = 0.00397 < 0.00463 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 591 — Example 6.22b

Put x equal to 1 into the series. Every power of 1 is 1, and at the lower limit, 0, every term vanishes. What remains is a series of plain numbers: 1, minus one third, one tenth, minus one forty-second, one over 216, and so on.

The signs alternate and the terms shrink to zero, so the alternating series bound from Chapter 5 applies: stopping at any point, the error is smaller than the first term you left out. After four terms the next term is one over 216, less than half of a hundredth. So four terms, 0.742857, meet the target.

The figure shows why this bound is honest. Each dot is a partial sum, and each bar is as long as the next term in each direction. The partial sums zigzag around the true value, and every bar covers it. The check compares with Simpson's rule: the actual error, about 0.004, sits just under the guaranteed 0.0046.

47. Why an alternating series is easy to trust

Intuition

When the terms alternate in sign and shrink steadily to zero, each partial sum overshoots the total and the next one undershoots it. The total is always trapped between two consecutive partial sums.

\[ S_3 = 0.742857 < \int_0^1 e^{-x^2}\,dx < S_4 = 0.747487 \]

\[ |\text{error after } S_k| < |\text{term } k+1| \]

So the size of the first term you leave out is a guaranteed error bar. That is what made four terms enough in Example 6.22, and it is why nonelementary integrals whose series alternate are so pleasant to compute.

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 591 — the alternating series bound from §5.5, used in Example 6.22

This is the reason you can put a number on the accuracy so easily. When terms alternate in sign and shrink, each partial sum overshoots, and the next one undershoots, by less each time. The true value is squeezed between any two consecutive partial sums.

In Example 6.22 the third and fourth partial sums were 0.742857 and 0.747487, and the true area, 0.746824, sits between them. The gap between them is exactly the next term, which is why the next term is a guaranteed bound on the error.

Keep the conditions in mind: alternating signs, sizes decreasing, sizes tending to zero. When a series does not alternate, for instance the series for the integral of e to the plus x squared, you need Taylor's theorem instead, and the bound is harder to get.

48. How many terms for six decimals?

Estimation

\[ \int_0^1 e^{-x^2}\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)\,n!}, \qquad \text{error} < 10^{-6} \]

\[ \frac{1}{17\cdot 8!} \approx 1.46\times10^{-6}, \qquad \frac{1}{19\cdot 9!} \approx 1.45\times10^{-7} \]

Make a guess before you calculate. If your instinct says dozens or hundreds of terms, that comes from series such as the p-series of Chapter 5, which converge painfully slowly. This one is different.

The terms are one over 2n plus 1 times n factorial, and factorials grow ferociously. By n equal to 8 the term is about one and a half millionths, just too big to leave out. By n equal to 9 it is about fifteen hundred-millionths. So summing n from 0 through 8, nine terms, guarantees six decimal places.

Compare with the reciprocal squares, where four-decimal accuracy took ten thousand terms. Series whose coefficients have factorials in the denominator are the ones that make practical computation possible, and that is why the exponential, sine and cosine series are the workhorses of numerical calculation.

49. Checkpoint 6.21: integrating cos of root x

Worked example

\[ \int_0^1\cos\sqrt x\,dx \text{ to within } 0.01 \]

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

\[ \int\cos\sqrt x\,dx = C + \sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{(n+1)(2n)!} \]

\[ \int_0^1\cos\sqrt x\,dx = 1 - \frac14 + \frac1{72} - \frac1{2880} + \cdots \]

\[ 1 - \tfrac14 = 0.75, \quad |\text{error}| < 0.0139 \]

\[ 1 - \tfrac14 + \tfrac1{72} = 0.763889, \quad |\text{error}| < 0.00035 \]

\[ 2\int_0^1 u\cos u\,du = 2(\cos 1 + \sin 1 - 1) = 0.763547 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 591 — Checkpoint 6.21

Reuse the series from Example 6.18: cos of root x has every power of x over 2n factorial. Integrating raises each power by one and divides by the new power, so the denominators become n plus 1 times 2n factorial.

Evaluating from 0 to 1 gives 1, minus one quarter, one seventy-second, minus one over 2880. The signs alternate and the terms shrink, so use the alternating bound. Two terms would leave an error up to one seventy-second, about 0.014, which is not good enough for 0.01. Three terms leave an error below one over 2880, about 0.00035.

The check is a treat: this particular integral does have an elementary antiderivative. Substitute u equal to root x, so the integral becomes twice the integral of u cos u, and integrate by parts. The exact value 0.763547 differs from the three-term estimate by 0.00034, just inside the bound.

50. Slide the number of terms

Tweak it

Start at N equal to 3 and look at where the curve goes wrong. Near zero it is excellent, but past about x equal to 1.5 it bends away sharply, because the last term you kept dominates.

Now increase N. Each increase pushes the good region further out. The true curve rises and levels off near 0.886, half the square root of pi, and by N around 25 the series matches it all the way to x equal to 3.

The series converges for every x, because the factorials always win in the end, but notice how many terms the far region needs. At x equal to 3 the early terms are large, 3, then 9, then about 24, before the factorials pull them down. Converges everywhere does not mean converges fast everywhere.

51. Normal probabilities are integrals of the bell curve

Concept

If scores are normally distributed with mean mu and standard deviation sigma, the chance that a score lands between a and b is an area under a bell curve.

\[ P(a < X < b) = \frac{1}{\sigma\sqrt{2\pi}}\int_a^b e^{-(x-\mu)^2/(2\sigma^2)}\,dx \]

\[ z = \frac{x-\mu}{\sigma} \quad\Longrightarrow\quad P = \frac{1}{\sqrt{2\pi}}\int_{(a-\mu)/\sigma}^{(b-\mu)/\sigma}e^{-z^2/2}\,dz \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 591-592 — equations 6.10 and 6.11, Figure 6.11

This is where e to the minus x squared earns its fame. Test scores, measurement errors and heights all tend to follow the normal distribution, and the probability that a value lands between a and b is the area under this bell curve between them.

The substitution to the z score is the standardisation you may know from statistics. It moves the mean to zero and scales the spread so that the standard deviation becomes one. Every normal probability question then becomes the same integral of e to the minus z squared over 2, just with different limits.

That integral has no elementary antiderivative, so there is no formula to plug the limits into. Statistics tables and calculators get their numbers from series, or from numerical methods, and the next two examples show how the series route works.

52. Example 6.23: scores between 100 and 200

Worked example

Scores are normal with mean 100 and standard deviation 50. Estimate the probability of a score between 100 and 200 with six terms of the series, and bound the error.

\[ P = \frac{1}{\sqrt{2\pi}}\int_0^2 e^{-z^2/2}\,dz \]

\[ e^{-z^2/2} = \sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{2^n\,n!} \]

\[ \int_0^2 e^{-z^2/2}\,dz = 2 - \frac86 + \frac{32}{40} - \frac{128}{336} + \frac{512}{3456} - \frac{2048}{42240} + \cdots \]

\[ 2 - 1.333333 + 0.8 - 0.380952 + 0.148148 - 0.048485 = 1.185378 \]

\[ P \approx 0.398942 \times 1.185378 = 0.4729 \]

\[ |\text{error}| < \frac{1}{\sqrt{2\pi}}\cdot\frac{2^{13}}{13\cdot2^6\cdot6!} \approx 0.00546 \]

\[ |0.4772 - 0.4729| = 0.0043 < 0.00546 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 592-593 — Example 6.23

Convert the scores to z scores first: 100 is the mean, so z is zero, and 200 is two standard deviations of 50 above it, so z is 2. Then substitute minus z squared over 2 into the exponential series and integrate term by term, exactly as in Example 6.22.

At z equal to 2 the terms start large, 2, then minus 1.33, then 0.8, because powers of 2 grow before the factorials take over. The series still alternates and the terms shrink, so after six terms the error is below the seventh term, divided by the square root of 2 pi: about 0.0055.

There is a misprint to know about. The book asks for six terms, prints the five-term value 0.4922, and then quotes the six-term error bound. The five-term sum has error bound 0.0193, and the true answer 0.4772 is not within 0.0055 of it. The six-term value 0.4729 is, which is the check: half of the famous 95 percent.

53. Checkpoint 6.22: scores between 100 and 150

Worked example

\[ P(100 < X < 150) = \frac{1}{\sqrt{2\pi}}\int_0^1 e^{-z^2/2}\,dz \]

\[ z: 0 \to 1 \]

\[ \int_0^1 e^{-z^2/2}\,dz = 1 - \frac16 + \frac1{40} - \frac1{336} + \frac1{3456} - \cdots \]

\[ 1 - 0.166667 + 0.025 - 0.002976 + 0.000289 = 0.855646 \]

\[ P \approx 0.398942 \times 0.855646 = 0.341354 \]

\[ |\text{error}| < \frac{0.398942}{42240} \approx 0.0000094 \]

\[ |0.341354 - 0.341345| = 0.0000089 < 0.0000094 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 593 — Checkpoint 6.22

Now the upper score is one standard deviation above the mean, so the limits are 0 and 1. Every power of 1 is 1, so the terms are just the denominators: 1, 6, 40, 336, 3456, and they shrink very fast.

Five terms give 0.855646, and dividing by the square root of 2 pi gives 0.341354. The next term is one over 42,240, which after the division is below one hundred-thousandth. So this estimate is good to about five decimal places.

The check uses a standard normal table, which gives 0.341345. The difference, 0.0000089, is just under the bound. When the true error is that close to the bound, it is a sign the bound is sharp: the first omitted term really is almost the whole error.

54. Where the 68 and 95 percent rules come from

Real world

\[ P(-1 < Z < 1) = 2 \times \frac{1}{\sqrt{2\pi}}\int_0^1 e^{-z^2/2}\,dz \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 593 — the Analysis after Example 6.23

You have now computed the two numbers behind the rule every statistics course quotes. The bell curve is symmetric about the mean, so the probability of landing within one standard deviation on either side is twice the area from the mean to one standard deviation.

Twice 0.3414 is 0.6827, the 68 percent. Twice 0.4772 is 0.9545, the 95 percent. Those numbers are not measured from data; they are areas under e to the minus z squared over 2, and you computed them with series.

That is the point of this whole part of the lesson. A question with no formula answer, how likely is a score within two standard deviations, gets a numerical answer with a guaranteed accuracy, using nothing more than the exponential series and the alternating series bound.

55. The period of a pendulum

Section

Part 5

56. A pendulum and its period

Picture it

The period is the time for one full swing, out and back. The exact formula is an elliptic integral, the kind that first appeared in computing the arc length of an ellipse, and it has no elementary antiderivative.

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 593-594 — Figure 6.12

The last application is the one that historically drove the study of these integrals. A pendulum of length L is released from an angle theta max and swings back and forth. Its period is the time for one full swing, out and back.

You may have met the formula 2 pi times the square root of L over g in physics. That formula comes from pretending the swing is very small, replacing sin theta by theta in the equation of motion. The exact period, for a swing of any size, is given by an elliptic integral, and it has no elementary antiderivative.

The right side of the figure previews where this is going. The small-angle formula is the first term of a series, and the correction factor one plus k squared over 4 is the second. For a 30 degree swing the correction is under two percent.

57. Reading the period integral piece by piece

Notation

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 593-594 — Example 6.24, the period of a pendulum

Step through the parts. The square root of L over g sets the time scale; everything else is a pure number that depends only on how far the pendulum swings. Longer pendulums swing more slowly, and on the Moon, with smaller g, every pendulum slows down.

The number k is the sine of half the largest angle. A small swing makes k small, and a swing all the way to horizontal makes k about 0.71. The integrand is one over the square root of one minus k squared sin squared theta, which is one plus u to the minus one half with u equal to minus k squared sin squared theta: a binomial series.

Since k is less than one and sine is at most one, the u inside is always less than one in size, so the binomial series converges for every theta in the integral. The variable of integration is a substitution variable, not the pendulum's actual angle, so do not try to read it off the picture.

58. Example 6.24: the period with one term, then two

Worked example

\[ \frac{1}{\sqrt{1-k^2\sin^2\theta}} = 1 + \frac12k^2\sin^2\theta + \frac{1\cdot3}{2!\,2^2}k^4\sin^4\theta + \cdots \]

\[ T \approx 4\sqrt{\tfrac{L}{g}}\cdot\frac{\pi}{2} = 2\pi\sqrt{\tfrac{L}{g}} \]

\[ \frac{\pi}{2}\left(\tfrac12k^2 + \tfrac38k^4 + \cdots\right) < \frac{\pi k^2}{2}\cdot\frac{1}{1-k^2} \]

\[ \int_0^{\pi/2}\sin^2\theta\,d\theta = \int_0^{\pi/2}\frac{1-\cos 2\theta}{2}\,d\theta = \frac{\pi}{4} \]

\[ T \approx 4\sqrt{\tfrac{L}{g}}\left(\frac{\pi}{2} + \frac{k^2}{2}\cdot\frac{\pi}{4}\right) = 2\pi\sqrt{\tfrac{L}{g}}\left(1 + \frac{k^2}{4}\right) \]

\[ k = 0: \quad 2\pi\sqrt{\tfrac{L}{g}}(1 + 0) = 2\pi\sqrt{\tfrac{L}{g}} \;\checkmark \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, pp. 594-595 — Example 6.24

Expand the integrand with the binomial series for r equal to minus one half, replacing x by minus k squared sin squared theta. The minus signs cancel, so every term is positive: 1, plus one half k squared sin squared theta, plus three eighths k to the 4th sin to the 4th theta, and so on.

Part (a) keeps only the 1. Its integral from 0 to pi over 2 is pi over 2, and four times that is 2 pi: the familiar small-angle formula drops out. The book then shows the part you dropped is at most pi k squared over 2 times one over one minus k squared, which is small when k is small.

Part (b) keeps the second term too. It needs one integral, sin squared from 0 to pi over 2, which the half-angle identity gives as pi over 4. Factoring out pi over 2 leaves the correction factor one plus k squared over 4. The check is the limiting case: with no swing, k is zero, and the improved formula must reduce to the small-angle one. It does.

59. Exercise 249: a 10 metre pendulum swung 30 degrees

Worked example

Length 10 metres, largest angle pi over 6. Take g equal to 9.8 metres per second squared.

\[ k = \sin\tfrac{\pi}{12} = 0.258819, \qquad k^2 = 0.066987 \]

\[ T_0 = 2\pi\sqrt{\tfrac{10}{9.8}} = 6.3470\text{ s} \]

\[ T \approx 6.3470\times(1 + 0.016747) = 6.4533\text{ s} \]

\[ T \approx 6.3470\left(1 + \tfrac{k^2}{4} + \tfrac{9k^4}{64}\right) = 6.4573\text{ s} \]

\[ |6.4575 - 6.4533| = 0.0042, \qquad |6.4575 - 6.4573| = 0.0002 \]

OpenStax Calculus Volume 2, §6.4 Working with Taylor Series §6.4, p. 599 — Exercises 249 and 251

Now put numbers in. The largest angle is pi over 6, 30 degrees, so k is the sine of 15 degrees, about 0.2588, and k squared over 4 is about 0.0167. The small-angle period of a 10 metre pendulum is about 6.347 seconds, and the correction stretches it to 6.453 seconds.

Exercise 251 adds the third term of the binomial series. Its integral, sin to the 4th from 0 to pi over 2, is 3 pi over 16, and after simplifying the correction gains nine sixty-fourths of k to the 4th. That gives 6.4573 seconds.

The figure compares all three estimates with the exact elliptic integral, computed numerically, across swings up to 160 degrees. At 30 degrees every estimate except the flat small-angle line agrees to three decimal places. At large swings each extra term helps, but near a full half-circle no finite number of terms is enough, because the integrand blows up.

60. How much longer does a 30 degree swing take?

Fill the middle

\[ \frac{T}{T_0} \approx 1 + \frac{k^2}{4} = 1 + \frac{0.066987}{4} = 1.016747 \]

Read the ratio off the formula above. The period divided by the small-angle period is one plus k squared over 4, which is 1.016747. The extra part, 0.016747, is the fraction by which the period grows.

As a percentage that is about 1.7 percent. On a 6.35 second swing it is roughly a tenth of a second per period, which adds up quickly: over a day, a pendulum clock swinging this wide would lose more than twenty minutes compared with one calibrated for tiny swings.

This is why real pendulum clocks are built to swing through small angles, where the small-angle formula is accurate and the period barely depends on how wide the swing is.

61. Putting it together

Section

Part 6

62. Pattern: choosing the route to a series

Pattern

1. Is it one plus u to a power? Binomial series, then substitute for u, and rewrite the radius condition in x.
2. Is it a table function of something else? Substitute. Only the variable changes, never the index or the factorials.
3. Is it a derivative or integral of one? Differentiate or integrate the table series term by term; the radius stays.
4. An equation for y rather than a formula? Assume a series, line up the powers, match coefficients, apply the initial conditions.
5. Need a number? Evaluate the series and bound the error, with the alternating series bound when the signs alternate, Taylor's theorem otherwise.

When you need a series, run down this list in order and stop at the first route that fits. The first four routes each take a line or two; the last one, computing derivatives at the centre, can take a page and may never reveal a pattern.

For each route, remember its hazard. Substitution changes the interval, so translate the condition into x. Substitution never changes the index or the factorials. Differentiation and integration keep the radius. When the question gives you an equation rather than a formula, the route is the power-series method, and its hazard is comparing different powers.

When you need a number rather than a formula, finish by bounding the error. If the series alternates with shrinking terms, the first omitted term is your bound. Otherwise use Taylor's theorem, as in Example 6.17.

63. Check: a binomial coefficient

Check

Write r equal to one third and the two falling factors, one third and minus two thirds. Their product is minus two ninths, and dividing by 2 factorial gives minus one ninth.

The three wrong answers are three different habits to watch for: dropping the sign of the second factor, forgetting the factorial, and confusing the coefficient of x with the coefficient of x squared. If you got it wrong, find which habit it was.

You can confirm the answer with the expansion the book prints in Exercises 194 to 198, which begins 1 plus x over 3 minus x squared over 9. Checking a coefficient against a printed expansion, or against a quick numerical test at a small x, takes seconds and catches most slips.

64. Check: a coefficient from a differential equation

Check

Match coefficients: each step multiplies by 3 and divides by the next whole number, starting from 2. So c one is 6, c two is 9, and c three is 9 as well, because 9 times 3 over 3 is 9.

The general formula is 2 times 3 to the n over n factorial, and the solution is 2 e to the 3x. Its x cubed coefficient is 2 times 27 over 6, which is 9.

The distractors come from dropping the factorial, which gives 54, or dropping the initial value, which gives 4.5. Both are common, and both are caught by checking against the closed form when you can recognise it.

65. Check: how many terms are enough

Check

The series alternates and its terms decrease to zero, so the error after stopping is less than the first term you omit. After one term that is one forty-second, far too big. After two terms it is one over 1320, about 0.00076, which is below 0.001.

So two terms suffice. Three also work, but the question asks for the fewest. The actual error of the two-term estimate is 0.00074, just under the bound.

These terms come from integrating the series for sine of x squared that you built in Checkpoint 6.17: each term x to the 4n plus 2 integrates to x to the 4n plus 3 over 4n plus 3.

66. Explain why matching coefficients is allowed

Explain it to yourself

Write your explanation before you reveal the model answer. The key word is uniqueness, but the explanation should say why power series are unique, not just that they are.

If a power series equals a function f near zero, then differentiating n times and setting x to zero kills every term except the nth, which shows the nth coefficient must be the nth derivative of f at zero over n factorial. Two series for the same function therefore have the same coefficients, because both are equal to those same derivatives.

Without uniqueness, the whole method collapses: two sums could be equal while their coefficients differed, and you could not read a recurrence off an equation. Every coefficient equation in Part 3 rests on this one fact.

67. Exit ticket

Exit ticket

\[ \int_0^{1/2}\frac{dx}{1+x^4} \]

This question uses almost every idea in the lesson, so name your moves as you go. The integrand is a geometric series with u equal to minus x to the 4th. One half is inside the radius, since one half to the 4th is well below one.

Integrate term by term from 0 to one half: each x to the 4n becomes one half to the 4n plus 1, over 4n plus 1. The terms alternate and shrink rapidly: 0.5, then 0.00625, then 0.000217, then about 0.0000094.

The alternating series bound says three terms leave an error below 0.0000094, well inside 0.0001. The estimate is 0.493967, and Simpson's rule confirms 0.493958. If you reached for derivatives or a trig substitution, look again at the pattern slide: the first route that fits is the cheapest.

68. Recap

Recap

\[ (1+x)^r = 1 + rx + \frac{r(r-1)}{2!}x^2 + \frac{r(r-1)(r-2)}{3!}x^3 + \cdots \]

Next, Chapter 7 turns to parametric equations and polar coordinates, where the same calculus is done along curves instead of graphs.

Stewart, Calculus: Early Transcendentals 8e, §11.11 Applications of Taylor Polynomials §11.11, pp. 774-783 — applications of Taylor polynomials in Stewart; the binomial series is in §11.10

You now have five tools. The binomial series covers every power of one plus x, with radius one, and stops only for whole-number exponents. The four moves, substitute, multiply, differentiate, integrate, turn the seven table series into almost any series you will need, with no derivatives computed.

Series settle limits by exposing the leading term, solve differential equations by turning them into recurrences for coefficients, and evaluate integrals that have no formula answer. In each case the analytic difficulty becomes arithmetic on coefficients.

Every answer that is a number should come with an error bound: the first omitted term for an alternating series, Taylor's theorem otherwise. That closes Chapter 6. Chapter 7 moves from functions written as series to curves written parametrically and in polar coordinates.