6.3 Taylor and Maclaurin Series

1. Taylor and Maclaurin Series

Title

Calculus II · Section 6.3

Any smooth function, rebuilt from its derivatives at one point

2. What this lesson gives you

Objectives

Sections 6.1 and 6.2 found power series only for functions that look like a geometric series. This lesson finds the power series of almost any function you can differentiate, and tells you how far to trust it.

• Build a Taylor polynomial of any degree by matching derivatives at a point, and explain why each coefficient carries a factorial
• State Taylor's theorem with remainder and say what its unknown point c costs you
• Bound the error of an approximation, and turn a required accuracy into a degree
• Tell apart a Taylor series that converges from one that converges to the function
• Derive and use the standard expansions of the exponential, sine and cosine

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 561 — learning objectives 6.3.1 to 6.3.3

So far every power series you have met came from the geometric series: substitute, differentiate, integrate, and hope the function can be bent into the shape one over one minus something. That covers a surprising amount, but it cannot reach e to the x, or sine, or the cube root.

This lesson removes the restriction. If you can differentiate a function again and again at one point, you can write down its power series there, with no cleverness at all. The price is a new question: the series you build might not actually add up to the function, and even when it does, you usually stop after a few terms. So half of the lesson is about the error.

By the end you should be able to build any Taylor polynomial, say how accurate it is with a guaranteed number, choose a degree to hit a target accuracy, and explain why the exponential, sine and cosine equal their series everywhere.

3. Before anything new: two things a power series can do

Warm-up

Write your answer before you reveal it. The geometric series is the one power series you know completely: its sum, its interval, and the fact that it really equals one over one minus x there.

The second line is the important habit. Inside its interval of convergence you may differentiate a power series term by term, and the result is a new power series with the same radius. Each differentiation knocks every exponent down by one and multiplies each term by its old exponent.

Keep your eye on what happens to the constant term when you differentiate: it disappears, and the old linear coefficient becomes the new constant. That single observation, repeated, is how you will read off every coefficient of a power series from derivatives in the next few slides.

4. Where the coefficients come from

Section

Part 1

5. Suppose f already has a power series

Concept

Assume, for now, that f equals a power series centred at a. Forget convergence for a moment and ask what the coefficients would have to be.

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

\[ f(a) = c_0 + c_1 \cdot 0 + c_2 \cdot 0 + \cdots = c_0 \]

So the constant term is forced: it must be the value of f at the centre.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 561 — equation 6.4

This is a reverse-engineering argument. You do not yet know whether f has a power series; you pretend that it does and ask what the coefficients would be forced to be. If the answer is unique, you will have found the only possible candidate.

The trick is to evaluate at the centre. Every term except the first contains a factor of x minus a, and at x equal to a that factor is zero. The whole infinite tail collapses and only the constant term survives.

So the constant term has no choice: it must be f of a. This is exactly what you would want from an approximation anyway. Whatever else a good approximation of f near a does, it should at least give the right value at a.

6. Differentiate, then look at the centre again

Concept

\[ f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + 4c_4(x-a)^3 + \cdots \]

\[ f'(a) = c_1 \]

\[ f''(x) = 2c_2 + 3\cdot 2\,c_3(x-a) + 4\cdot 3\,c_4(x-a)^2 + \cdots \]

\[ f''(a) = 2c_2 \;\Longrightarrow\; c_2 = \frac{f''(a)}{2} \]

\[ f'''(a) = 3\cdot 2\,c_3 \;\Longrightarrow\; c_3 = \frac{f'''(a)}{3\cdot 2} \]

Each differentiation lowers every power by one and brings the old exponent down as a multiplier. After n differentiations the nth coefficient has collected every whole number from n down to 1.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 561-562 — the derivation of the coefficients

Now repeat the trick with derivatives. Differentiate once and the old linear coefficient becomes the constant term, so evaluating at the centre picks it out: the first coefficient must equal the slope of f at a.

Differentiate again and something new appears. The x minus a squared term becomes two times c two, so matching the second derivative forces c two to be half of f double prime of a. The third derivative produces three times two times c three, so c three is the third derivative divided by six.

Look at the multipliers that pile up: two, then three times two, then four times three times two. Each differentiation brings the current exponent down as a factor. By the time the nth coefficient has become a constant, it has been multiplied by n, n minus 1, and so on down to 1. That product is n factorial, and it is why every Taylor coefficient is divided by a factorial.

7. Why the factorial: differentiating a fourth power

Picture it

\[ \frac{d^4}{dx^4}(x-a)^4 = 4\cdot 3\cdot 2\cdot 1 = 4! \]

The term c four times x minus a to the fourth contributes nothing to the value or the first three derivatives at the centre, and exactly four factorial times c four to the fourth derivative. That factorial is what the coefficient must divide out.

The figure shows one term, x to the fourth, and its four derivatives. Each curve is steeper than the one before, and every one of them passes through the origin, except the last, which is the constant 24.

That is the whole story of the factorial. At the centre, the fourth-power term contributes nothing to the value, the slope, the second derivative or the third derivative; it only shows up in the fourth derivative, and there it shows up multiplied by four factorial.

So if you want the series to have the same fourth derivative as f at the centre, you must choose its coefficient to be the fourth derivative divided by 24. Leave out the factorial and your polynomial would have a fourth derivative 24 times too large. The same reasoning applies to every power, which is why the pattern is so rigid.

8. The Taylor series of f at a

Concept

\[ c_n = \frac{f^{(n)}(a)}{n!} \]

\[ \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \]

Taylor series (6.5) — If f has derivatives of all orders at a, this is the Taylor series for f at a. When the centre a is zero it is called the Maclaurin series.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 562 — definition and equation 6.5

Here is the result of the reverse engineering, written as a single formula. The nth coefficient is the nth derivative at the centre, divided by n factorial, and nothing else is possible.

Notice what the definition asks for: derivatives of all orders at a single point. Everything the series knows about f, it learns at a. It is remarkable that information at one point can ever reconstruct a function over a whole interval, and part four of this lesson will show it sometimes cannot.

The name Maclaurin series is used when the centre is zero. There is no new mathematics in it; it is simply the most common centre, because the powers of x are the simplest powers to work with. You will meet Maclaurin series for the exponential, sine and cosine in a few slides.

9. Reading the Taylor series piece by piece

Notation

Step through the four notes in order. The first is the heart of it: the derivative is evaluated at a, so it is a number, not a function. Students sometimes leave the derivative as a function of x inside the series; then it is not a power series at all.

The factorial note connects back to the figure you just saw. Its only job is to cancel the multipliers that repeated differentiation produces, so that the series has exactly the right derivatives at the centre.

The last two notes are about reading the series as an approximation. Near a, the powers of x minus a are small, and higher powers are smaller still, so the early terms carry most of the weight. And the n equal to zero term is just f of a, since the zeroth derivative is the function itself and zero factorial is one.

10. Theorem 6.6: there is only one candidate

Concept

Uniqueness of Taylor series (Theorem 6.6) — If f has a power series at a that converges to f on some open interval containing a, then that power series is the Taylor series for f at a.

\[ f(x) = \sum c_n(x-a)^n \text{ near } a \;\Longrightarrow\; c_n = \frac{f^{(n)}(a)}{n!} \]

So the series from Section 6.2, however it was found, must agree with the one built from derivatives. What the theorem does not promise is that the Taylor series converges to f at all; that question waits until Part 4.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 562 — Theorem 6.6

This theorem is short but it saves you a lot of work. It says that if a function has any power series representation at a, obtained by any method at all, then that series is the Taylor series. There cannot be two different power series for the same function about the same centre.

Here is why it matters. In Section 6.2 you found the series of arctangent or of the logarithm by integrating a geometric series. Uniqueness says those series are the Taylor series, so you can read derivatives off them without ever differentiating: the coefficient of x to the n times n factorial is the nth derivative at zero.

Be careful about what the theorem does not say. It assumes the function already has a power series converging to it. It does not promise that the Taylor series you build from derivatives will converge to f. That second question is harder, and it is settled by the remainder in part four.

11. Trap: forgetting the factorial

Trap

12. Taylor polynomials

Section

Part 2

13. Stop after n terms: the Taylor polynomial

Concept

\[ p_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \]

nth Taylor polynomial — The nth partial sum of the Taylor series. It needs only n derivatives at a, and it matches f and its first n derivatives at a. At a equal to zero it is the nth Maclaurin polynomial.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 563 — definition of Taylor polynomials

In practice you never add infinitely many terms. You stop, and the partial sum you stop at is a polynomial: the nth Taylor polynomial. It is the polynomial of degree at most n that agrees with f in value and in its first n derivatives at the centre.

The table gives a geometric reading of the first few. Degree zero matches only the height, so it is a flat line. Degree one also matches the slope: it is the tangent line you used for linear approximation in Calculus One. Degree two also matches the curvature, so it bends the right way.

Each extra degree matches one more derivative, and each extra matched derivative makes the polynomial cling to the function over a slightly wider stretch around the centre. That is the picture to keep in mind for everything that follows.

14. Example 6.11: the logarithm at x = 1

Worked example

Find p0 through p3 for the natural logarithm at a equal to 1.

\[ f = \ln x, \quad f' = x^{-1}, \quad f'' = -x^{-2}, \quad f''' = 2x^{-3} \]

\[ f(1) = 0, \quad f'(1) = 1, \quad f''(1) = -1, \quad f'''(1) = 2 \]

\[ c_0 = 0, \quad c_1 = 1, \quad c_2 = -\tfrac12, \quad c_3 = \tfrac{2}{6} = \tfrac13 \]

\[ p_0 = 0, \quad p_1 = x - 1, \quad p_2 = (x-1) - \tfrac12(x-1)^2 \]

\[ p_3(x) = (x-1) - \tfrac12(x-1)^2 + \tfrac13(x-1)^3 \]

\[ p_3(1.5) = 0.5 - 0.125 + 0.041667 = 0.416667 \approx 0.405465 = \ln 1.5 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 563-564 — Example 6.11 and Figure 6.5

Follow the rhythm of the solution, because every Taylor polynomial problem has it: differentiate, evaluate at the centre, divide by the factorials, assemble. Here the logarithm has a clean pattern of derivatives because after the first one they are all powers of x.

The centre is 1 because the logarithm is not defined at zero, so it has no Maclaurin series. At 1 every power of x becomes 1, which makes the evaluation trivial. The third derivative is 2, and dividing by three factorial gives one third: the factorial cancels most of the growth in the derivatives.

In the figure, the flat line, the tangent line, the parabola and the cubic all pass through the point where the logarithm crosses the axis. Each one hugs the curve more closely near 1. The check at 1.5 shows the improvement in numbers: the quadratic is off by 0.03, the cubic by 0.011.

15. Checkpoint 6.10: one over x squared at x = 1

Worked example

\[ f(x) = \frac{1}{x^2} = x^{-2}, \quad a = 1 \]

\[ f' = -2x^{-3}, \quad f'' = 6x^{-4}, \quad f''' = -24x^{-5} \]

\[ f(1) = 1, \quad f'(1) = -2, \quad f''(1) = 6, \quad f'''(1) = -24 \]

\[ c_0 = 1, \quad c_1 = -2, \quad c_2 = \tfrac{6}{2} = 3, \quad c_3 = \tfrac{-24}{6} = -4 \]

\[ p_0 = 1, \quad p_1 = 1 - 2(x-1), \quad p_2 = p_1 + 3(x-1)^2 \]

\[ p_3(x) = 1 - 2(x-1) + 3(x-1)^2 - 4(x-1)^3 \]

\[ p_3(1.1) = 1 - 0.2 + 0.03 - 0.004 = 0.826 \approx 0.826446 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 564 — Checkpoint 6.10

Try this one yourself before revealing each line. The derivatives are all powers of x again, with the sign flipping every time and the size growing like a factorial: two, six, twenty-four.

The division is where the tidy pattern appears. Two over one factorial is two, six over two factorial is three, twenty-four over three factorial is four. The coefficients are just one, two, three, four with alternating signs, which is what you would get by differentiating the geometric series, as uniqueness predicts.

The check at 1.1 compares the cubic with the true value of one over 1.21. They agree to about four decimal places. Choosing a point close to the centre for a check is sensible: that is where a Taylor polynomial should be good, so a large disagreement there would reveal an arithmetic slip.

16. Does adding terms always help?

Prediction

\[ \ln 3 = 1.0986, \qquad p_1(3) = 2, \quad p_2(3) = 0, \quad p_3(3) = 2.6667 \]

Make a genuine prediction before you look at the answer. The natural expectation is that more terms always mean a better approximation, and the values at the top of the slide are there to test it.

They do not close in. The linear polynomial says 2, the quadratic says 0, the cubic says 2.67, and the next one would say minus 1.33. Each new term is bigger than the last, because at x equal to 3 the powers of x minus 1 are powers of 2, and they outgrow the one over n in the coefficients.

The lesson is that Taylor polynomials are local. They are built from information at the centre, and they are guaranteed to be good only near it. How near is controlled by the radius of convergence, which for the logarithm at 1 is exactly 1. Hold on to this example; it comes back in part five.

17. Example 6.12(a): the exponential's Maclaurin polynomials

Worked example

Find p0 through p3 and the nth Maclaurin polynomial of e to the x.

\[ f^{(n)}(x) = e^x \quad \text{for every } n \]

\[ f(0) = f'(0) = f''(0) = \cdots = f^{(n)}(0) = 1 \]

\[ c_n = \frac{1}{n!} \]

\[ p_0 = 1, \quad p_1 = 1 + x, \quad p_2 = 1 + x + \frac{x^2}{2}, \quad p_3 = p_2 + \frac{x^3}{6} \]

\[ p_n(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} = \sum_{k=0}^{n}\frac{x^k}{k!} \]

\[ p_3(1) = 1 + 1 + 0.5 + 0.16667 = 2.66667 \approx 2.71828 = e \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 564-565 — Example 6.12a and Figure 6.6

This is the easiest Taylor computation there is, and one of the most important. Every derivative of e to the x is e to the x, so every derivative at zero is one, and the nth coefficient is simply one over n factorial.

Write the general polynomial in sigma notation as well as expanded. The sigma form makes the pattern impossible to forget and is the form you will need when you take a limit as n grows in part four.

In the figure the four polynomials crowd together near zero, all passing through the point zero, one. To the right they fall behind the exponential, which outgrows every polynomial. To the left the exponential flattens toward zero while the polynomials, dominated by their highest power, dive or climb. At x equal to 1 the cubic gives 2.667 against e, 2.718; adding more terms closes that gap very quickly, as you will see.

18. Example 6.12(b): the sine's Maclaurin polynomials

Worked example

Find the Maclaurin polynomials of sin x and the general pattern.

\[ \sin x \to \cos x \to -\sin x \to -\cos x \to \sin x \]

\[ f(0) = 0, \quad f'(0) = 1, \quad f''(0) = 0, \quad f'''(0) = -1, \quad f^{(4)}(0) = 0 \]

\[ f^{(2m)}(0) = 0, \qquad f^{(2m+1)}(0) = (-1)^m \]

\[ p_1 = p_2 = x, \quad p_3 = p_4 = x - \frac{x^3}{3!}, \quad p_5 = p_6 = x - \frac{x^3}{3!} + \frac{x^5}{5!} \]

\[ p_{2m+1}(x) = p_{2m+2}(x) = \sum_{k=0}^{m}\frac{(-1)^k x^{2k+1}}{(2k+1)!} \]

\[ p_5(1) = 1 - \frac16 + \frac{1}{120} = 0.841667 \approx 0.841471 = \sin 1 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 565-566 — Example 6.12b and Figure 6.7

The derivatives of sine cycle with period four: sine, cosine, minus sine, minus cosine, and back to sine. At zero, the sines vanish and the cosines are plus or minus one. So half of the coefficients are zero, and the other half alternate in sign.

The even-numbered derivatives are the ones that vanish, which means only odd powers of x appear. That is no accident: sine is an odd function, and an odd function can only be built from odd powers. It also means consecutive polynomials come in equal pairs, p1 equals p2, p3 equals p4, and so on, because each even step adds a zero term.

In the figure, each odd degree adds another alternating term and follows the sine wave a little further before breaking away. The check at 1 shows how good even the quintic is: three nonzero terms give four correct decimal places. That efficiency is why calculators can evaluate sines with a handful of multiplications.

19. Example 6.12(c): the cosine's Maclaurin polynomials

Worked example

Repeat for cos x.

\[ \cos x \to -\sin x \to -\cos x \to \sin x \to \cos x \]

\[ f(0) = 1, \quad f'(0) = 0, \quad f''(0) = -1, \quad f'''(0) = 0, \quad f^{(4)}(0) = 1 \]

\[ f^{(2m)}(0) = (-1)^m, \qquad f^{(2m+1)}(0) = 0 \]

\[ p_0 = p_1 = 1, \quad p_2 = p_3 = 1 - \frac{x^2}{2!}, \quad p_4 = p_5 = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \]

\[ p_{2m}(x) = p_{2m+1}(x) = \sum_{k=0}^{m}\frac{(-1)^k x^{2k}}{(2k)!} \]

\[ \frac{d}{dx}\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) = -x + \frac{x^3}{6} = -\left(x - \frac{x^3}{6}\right) \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 566-567 — Example 6.12c and Figure 6.8; the book's 'fourth derivative is sin x' should read cos x

This is the same cycle of four derivatives, shifted by one place. Now the odd derivatives vanish at zero, so only even powers appear, again matching the symmetry: cosine is an even function.

The book says the fourth derivative is sin x in this example; it is cos x, which is why the pattern repeats. Small misprints like this are common, and the best defence is to recompute every derivative yourself rather than trust a printed line.

The check on this slide is a different kind of check. Instead of plugging in a number, it uses a relationship you already know: the derivative of cosine is minus sine. Differentiating the cosine's quartic gives exactly minus the sine's cubic. When two Taylor computations agree with a known identity like that, you can be confident both are right.

20. Slide the degree: the sine's polynomials

Tweak it

Move the slider slowly from 1 upward and watch two things: the stretch around the origin where the curve looks like a sine wave, and the ends of the window where it shoots off to plus or minus infinity.

Every polynomial eventually leaves, because a polynomial of odd degree must run off to infinity and the sine stays between minus one and one. But the point where it leaves moves further out as n grows. By degree 21 the curve tracks the wave from about minus 8 to 8, five humps of it, before it breaks away near the edges.

Notice also that moving from an odd degree to the next even degree does nothing. That is the zero coefficient of every even power, seen directly. The fact that the good stretch keeps widening, without limit, is the visual version of the statement that the sine's Maclaurin series converges to sine for every x.

21. Odd powers only, or even powers only?

Sorting

The rule behind this sort is symmetry. An odd function satisfies f of minus x equals minus f of x, and its even-numbered derivatives are also odd functions, which must be zero at zero. So only odd powers survive in its Maclaurin series. An even function is the mirror image: its odd derivatives vanish at zero, leaving only even powers.

Two items test whether you apply the rule to the whole function rather than a piece of it. The function x cos x is a product of an odd and an even function, which is odd. The function sine of x squared contains a sine, but the inside is even, and the whole thing is even.

This check costs nothing and catches many errors. If your series for an even function has an x cubed term, something has gone wrong before you even look at the numbers.

22. Checkpoint 6.11: one over one plus x

Worked example

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

\[ f' = -(1+x)^{-2}, \quad f'' = 2(1+x)^{-3}, \quad f''' = -6(1+x)^{-4} \]

\[ f^{(n)}(0) = (-1)^n n! \]

\[ c_n = \frac{(-1)^n n!}{n!} = (-1)^n \]

\[ p_0 = 1, \quad p_1 = 1 - x, \quad p_2 = 1 - x + x^2, \quad p_3 = 1 - x + x^2 - x^3 \]

\[ p_n(x) = \sum_{k=0}^{n}(-1)^k x^k \]

\[ \frac{1}{1-(-x)} = \sum(-x)^k, \qquad p_3(0.1) = 0.909 \approx 0.909091 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 567 — Checkpoint 6.11

Work this one before revealing it. The derivatives follow the pattern you saw for one over x squared: each brings down the current exponent, so the signs alternate and the sizes are factorials.

The division step is the satisfying one: the nth derivative at zero is plus or minus n factorial, and dividing by n factorial leaves just plus or minus one. The polynomial is one minus x plus x squared minus x cubed and so on.

The check uses uniqueness. You already know a power series for this function: the geometric series with minus x in place of x. Theorem 6.6 says the two must agree coefficient by coefficient, and they do. The numerical check at 0.1 confirms the arithmetic. This is a good example of two very different routes arriving at the same series.

23. Match each function to its third Maclaurin polynomial

Matching

Before matching term by term, look at the constant term, which is just the function's value at zero. The exponential and one over one plus x both equal 1 there; sine and the logarithm both equal 0. That splits the four polynomials into two pairs straight away.

Then look for factorials. Only the exponential's polynomial divides by 2 and by 6. The sine's polynomial has no x squared term, because sine is odd.

The logarithm is the interesting one. Its coefficients are one over n, not one over n factorial. Its derivatives at zero grow like factorials themselves, so most of the factorial cancels. That leftover one over n is why the logarithm's series converges only on a finite interval, while the exponential's, with a full factorial in the denominator, converges everywhere.

24. Taylor's theorem with remainder

Section

Part 3

25. The remainder: what the polynomial leaves out

Concept

\[ R_n(x) = f(x) - p_n(x) \]

nth remainder — The exact error made by using the nth Taylor polynomial in place of f at the point x.

Two questions depend on it. For one fixed n: how big is the error? And as n grows: does the error go to zero, so that the series really equals f?

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 567-568 — definition of the remainder

The remainder is defined in the simplest possible way: the function minus the polynomial. It is the exact error of the approximation at the point x, and for most functions you cannot compute it directly, because computing it would require knowing f of x exactly.

Two different questions both reduce to the remainder. In practice you fix a degree, use the polynomial, and want to know how far off you might be. In theory you want to know whether the Taylor series adds up to f, which happens exactly when the remainder shrinks to zero as n grows.

Taylor's theorem answers both at once, by giving a formula for the remainder that can be bounded even though it cannot be evaluated.

26. Degree zero: the remainder is the Mean Value Theorem

Concept

\[ R_0(x) = f(x) - p_0(x) = f(x) - f(a) \]

\[ f(x) - f(a) = f'(c)(x - a) \]

\[ R_0(x) = f'(c)(x-a) \]

In the picture, with f the square root, a equal to 1 and x equal to 4, the secant slope is one third, and c is 2.25, where one over twice the root of c equals one third.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 568 — the case n = 0

Start with the simplest case to see where the formula comes from. The zeroth Taylor polynomial is the constant f of a, so the remainder is how much f has changed between a and x.

The Mean Value Theorem, from Calculus One, says that change equals the slope at some intermediate point c times the distance. In the figure, the dashed secant from the centre to the point at 4 has slope one third, and the tangent at c equal to 2.25 is parallel to it. The red segment, the error of the flat approximation, is exactly that slope times the run of 3.

Notice what kind of information you get. You do not know c without solving an equation, but you do know it lies between a and x. That is the pattern of Taylor's theorem: an exact formula with one unknown point, which you then bound.

27. Taylor's theorem with remainder

Concept

Theorem 6.7 — Let f be differentiable n plus 1 times on an interval I containing a. For each x in I there is a number c strictly between a and x with the formula below. If the (n plus 1)st derivative is bounded by M on I, the error bound follows.

\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \]

\[ |f^{(n+1)}| \le M \text{ on } I \;\Longrightarrow\; |R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1} \]

The remainder looks exactly like the next term of the series, except that the derivative is taken at an unknown point c instead of at the centre.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 568 — Theorem 6.7

Read the remainder formula side by side with the Taylor series. It looks exactly like the first term you did not include, the one of degree n plus 1, except that the derivative is evaluated at an unknown point c instead of at the centre.

That substitution is the price of exactness. With the derivative at c, the formula is not an approximation of the error; it is the error. But since you cannot find c, you replace the derivative by the largest size it could have anywhere between a and x. That largest size is M, and it turns the formula into an inequality you can compute.

The hypotheses matter: f must have n plus 1 derivatives on an interval containing both the centre and the point x. You need one derivative more than the degree of the polynomial, because the error is controlled by the first derivative the polynomial does not match.

28. Reading the remainder formula piece by piece

Notation

Go through the notes in order and connect each one to a calculation you will do. The n plus 1 in the derivative is the most common place to slip: for a polynomial of degree two you need the third derivative, not the second.

The note about c is the conceptual heart of the theorem. The point exists, and the theorem guarantees it lies strictly between a and x, but it depends on x and on n in a way no formula captures. You never solve for it. You only ask how large the derivative could possibly be on the interval where c must live.

The factorial and the power pull in opposite directions. The power of the distance grows if the distance is more than one; the factorial always grows, and eventually faster than any power. Which one wins decides whether the error shrinks, and you will see in part four that for the exponential, sine and cosine the factorial always wins.

29. Why the theorem is true: build a function for Rolle

Concept

Fix x. Define a function of t that subtracts the Taylor polynomial centred at t, plus a correction term chosen so that it vanishes at t equal to a.

\[ g(t) = f(x) - \sum_{k=0}^{n}\frac{f^{(k)}(t)}{k!}(x-t)^k - R_n(x)\frac{(x-t)^{n+1}}{(x-a)^{n+1}} \]

\[ g(a) = f(x) - p_n(x) - R_n(x) = 0 \]

\[ g(x) = f(x) - f(x) - 0 = 0 \]

So g is zero at both ends, and Rolle's theorem hands you a c between a and x where g prime is zero.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 569 — proof of Theorem 6.7

This proof is clever, and it is worth following once. The idea is to build a function g of a new variable t that is zero at both ends of the interval, so that Rolle's theorem can find a point where its derivative is zero.

Read the definition of g as follows: for each t, subtract from f of x the Taylor polynomial centred at t, evaluated at x, and then subtract a correction proportional to x minus t to the power n plus 1. The correction is scaled so that at t equal to a it equals the full remainder.

Check the two ends yourself. At t equal to a, the sum is the ordinary Taylor polynomial, so g is f of x minus the polynomial minus the remainder, which is zero by definition. At t equal to x, every power of x minus t vanishes and only f of x minus f of x is left. Both ends are zero, so Rolle applies.

30. Differentiate g: the sum telescopes

Concept

\[ \frac{d}{dt}\left[\frac{f^{(k)}(t)}{k!}(x-t)^k\right] = \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1} \]

\[ g'(t) = -\frac{f^{(n+1)}(t)}{n!}(x-t)^n + (n+1)R_n(x)\frac{(x-t)^n}{(x-a)^{n+1}} \]

\[ \frac{f^{(n+1)}(c)}{n!} = (n+1)\frac{R_n(x)}{(x-a)^{n+1}} \]

\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 569 — proof of Theorem 6.7, continued

Now differentiate g with respect to t. Each term of the Taylor sum is a product of a derivative of f at t and a power of x minus t, so the product rule gives two pieces, one with the next derivative and one with the power lowered by one.

Line the pieces up and they cancel in pairs: the first piece of each term cancels the second piece of the term after it. This is exactly the telescoping you saw with series in Chapter 5. Only the very last piece survives, together with the derivative of the correction term.

Rolle gives a c where this derivative is zero. Setting it to zero, cancelling the common factor x minus c to the n, and dividing by n plus 1 gives the remainder formula, with n factorial times n plus 1 becoming n plus 1 factorial. You do not need to reproduce this proof on an exam, but it shows that the unknown c comes straight from Rolle's theorem.

31. The bound is a ceiling over the true error

Picture it

\[ |\sin x - p_3(x)| \le \frac{|x|^5}{5!} \]

Because the sine's p3 equals its p4, you may use the remainder of degree four, with the fifth derivative bounded by 1. The drawn error never crosses the ceiling.

The figure puts the theorem to a test. The solid curve is the actual error of the cubic approximation to sine, which you can compute here only because you know sine exactly. The dashed curve is the Taylor bound, x to the fifth over 120.

Why the fifth power, when the polynomial has degree three? Because the sine's cubic is also its polynomial of degree four: the x to the fourth coefficient is zero. So you are entitled to use the remainder after degree four, which involves the fifth derivative and five factorial, and gives a much sharper bound near zero. The book uses the same trick in Example 6.14.

The error never crosses the bound, which is what a bound is for. At x equal to 3 they are 1.64 and 2.03, not far apart; near zero both are tiny. The bound is a worst case, and for a well-behaved function like sine the worst case is not far from the truth.

32. Example 6.13(a, b): the cube root near 8

Worked example

Find p1 and p2 for the cube root of x at 8, and use them to estimate the cube root of 11.

\[ f = x^{1/3}, \quad f' = \frac{1}{3}x^{-2/3}, \quad f'' = -\frac{2}{9}x^{-5/3} \]

\[ f(8) = 2, \quad f'(8) = \frac{1}{12}, \quad f''(8) = -\frac{2}{288} = -\frac{1}{144} \]

\[ p_1 = 2 + \frac{x-8}{12}, \qquad p_2 = 2 + \frac{x-8}{12} - \frac{(x-8)^2}{288} \]

\[ p_1(11) = 2 + \frac{3}{12} = 2.25, \qquad p_2(11) = 2.25 - \frac{9}{288} = 2.21875 \]

\[ |2.25 - 2.22398| = 0.02602, \qquad |2.21875 - 2.22398| = 0.00523 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 570-571 — Example 6.13a, b and Figure 6.9

The centre 8 is chosen because it is a perfect cube near 11, so the cube root and its derivatives are easy to evaluate there. Eight to the two thirds is 4 and eight to the five thirds is 32; those two facts make the arithmetic clean.

Watch the second coefficient carefully. The second derivative at 8 is minus one over 144, and dividing by two factorial gives minus one over 288. Forgetting that division is the most common error in this example.

The figure shows both polynomials touching the cube root at 8. The tangent line lies above the curve, because the cube root bends downward; the parabola bends downward too, and follows the curve much further. The check against the true cube root of 11 shows the quadratic is about five times more accurate than the linear estimate. In real problems you will not know the true value, which is exactly why you need the error bound on the next slides.

33. Before bounding the error

Step zero

\[ |R_1(11)| \le \frac{M}{2!}|11 - 8|^2 \]

Answer this before you touch the formula, because most wrong error bounds go wrong here, in the choice of M. M is not the derivative at the centre, and it is not the derivative at the point x. It is the largest size of the derivative anywhere between them.

Why the whole interval? Because the unknown point c could be anywhere in it. If you take a value smaller than the largest possible, you are betting that c happens to avoid the worst spot, and your bound is no longer guaranteed.

For the cube root, the second derivative is a constant times x to the minus five thirds. A negative power of x gets smaller as x grows, so its size is largest at the left end of the interval, the centre 8. That is where you evaluate it.

34. Example 6.13(c): bounding both errors

Worked example

\[ |R_1(11)| \le \frac{1/144}{2!}(11-8)^2 = \frac{9}{288} = 0.03125 \]

\[ f'''(x) = \frac{10}{27}x^{-8/3}, \quad f'''(8) = \frac{10}{27\cdot 256} \approx 0.0014468 \]

\[ |R_2(11)| \le \frac{0.0014468}{3!}(11-8)^3 \approx 0.0065104 \]

\[ 0.02602 \le 0.03125, \qquad 0.00523 \le 0.00651 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 571-572 — Example 6.13c

The linear bound uses the second derivative at 8, which is one over 144 in size, divided by two factorial and multiplied by the distance squared, 9. The result, 0.03125, says the linear estimate 2.25 is within about three hundredths of the true cube root.

For the quadratic you need one more derivative. The third derivative is ten over twenty-seven times x to the minus eight thirds, positive and decreasing, so again it is largest at 8. Divide by three factorial and multiply by the distance cubed, 27.

The book prints 0.0011468 in that line where it means 0.0014468; its final bound of 0.0065104 is correct, so this is a typographical slip, not a wrong answer. The verification compares each bound with the actual error computed from the true cube root. Both actual errors sit under their bounds, and the quadratic's bound is about five times smaller, just as its actual error was.

35. Checkpoint 6.12: estimating the square root of 6

Worked example

\[ f = x^{1/2}, \quad f' = \tfrac12 x^{-1/2}, \quad f'' = -\tfrac14 x^{-3/2}, \quad f''' = \tfrac38 x^{-5/2} \]

\[ f(4) = 2, \quad f'(4) = \tfrac14, \quad f''(4) = -\tfrac{1}{32} \]

\[ p_1(6) = 2 + \tfrac{2}{4} = 2.5, \qquad p_2(6) = 2.5 - \tfrac{1}{64}\cdot 4 = 2.4375 \]

\[ |R_1(6)| \le \frac{1/32}{2!}\cdot 2^2 = 0.0625 \]

\[ |R_2(6)| \le \frac{3/256}{3!}\cdot 2^3 = 0.015625 \]

\[ |2.5 - 2.44949| = 0.0505 \le 0.0625, \qquad |2.4375 - 2.44949| = 0.0120 \le 0.0156 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 572 — Checkpoint 6.12

This is Example 6.13 again, with the square root centred at 4. Try each line yourself first. The only care needed is with the half-integer powers: four to the one half is 2, four to the three halves is 8, four to the five halves is 32.

Both error bounds take M at the centre 4, because every derivative here is a negative power of x and so shrinks as x grows. The quadratic's bound, 0.0156, is four times smaller than the linear one, 0.0625.

The figure turns the two answers into error bars. Each estimate is a dot with a guaranteed interval around it, and the true value of the square root of 6, the vertical line, lands inside both. That is the right way to think about an approximation: not as a number, but as a number with a certified margin.

36. Trap: taking M at the wrong end

Trap

37. Example 6.14(a): sine of pi over 18

Worked example

Use the fifth Maclaurin polynomial of sine to approximate the sine of 10 degrees, and bound the error.

\[ p_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}, \qquad x = \frac{\pi}{18} \approx 0.174533 \]

\[ p_5\left(\tfrac{\pi}{18}\right) = 0.174533 - 0.000886 + 0.000001 \approx 0.173648 \]

\[ \sin x - p_5(x) = R_6(x) = \frac{f^{(7)}(c)}{7!}x^7 \]

\[ |f^{(7)}(c)| = |{-\cos c}| \le 1 \]

\[ |R_6| \le \frac{1}{7!}\left(\frac{\pi}{18}\right)^7 \approx 9.79\times 10^{-10} \]

\[ p_5 - \sin\tfrac{\pi}{18} \approx 9.78\times 10^{-10} \le 9.79\times 10^{-10} \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 572-573 — Example 6.14a

Pi over 18 is ten degrees in radians, about 0.1745. At such a small input the terms of the sine's series shrink extremely fast: the cube term is under a thousandth and the fifth-power term is about one millionth.

The key move is to use the remainder after degree six rather than after degree five. Because the sixth-degree coefficient of sine is zero, the fifth and sixth Maclaurin polynomials are the same polynomial, so both remainders describe the same error. The later one has a seven factorial in the denominator and a seventh power of the input, which makes it far smaller.

The seventh derivative is minus cosine, never bigger than one, so M is one. The bound comes out below one billionth. The check compares with the true sine and finds an actual error of 9.78 times ten to the minus ten, a hair inside the bound of 9.79. That closeness is not a coincidence: for small inputs the bound is nearly the size of the first omitted term.

38. Example 6.14(b): where is p5 good to four decimals?

Worked example

For which x does the fifth Maclaurin polynomial approximate sin x to within 0.0001?

\[ |R_6(x)| \le \frac{|x|^7}{7!} \]

\[ \frac{|x|^7}{5040} \le 0.0001 \iff |x|^7 \le 0.504 \]

\[ |x| \le 0.504^{1/7} \approx 0.9068 \]

\[ \frac{0.906^7}{5040} \approx 0.0000994, \qquad \frac{0.907^7}{5040} \approx 0.0001002 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 573 — Example 6.14b

Now the question is turned around. Instead of fixing x and asking for the error, you fix the error you can tolerate and ask which x qualify. The same bound does the work, as a function of x.

Solving is ordinary algebra: multiply through by 5040 and take a seventh root. The answer is about 0.9068, which in degrees is roughly 52 degrees each side of zero.

The figure shows why the answer is so generous. A seventh power is extremely flat near zero, so the bound stays under a ten-thousandth for a long way before shooting up. The book writes less than 0.907; computing both sides shows that 0.907 itself is just over the tolerance and 0.906 is just under, so the safe statement is absolute x at most 0.9068. Checking the boundary like this is a good habit whenever you round.

39. Checkpoint 6.13: cosine of pi over 12

Worked example

\[ p_4(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}, \qquad x = \frac{\pi}{12} \approx 0.261799 \]

\[ p_4\left(\tfrac{\pi}{12}\right) = 1 - 0.03426946 + 0.00019573 = 0.96592627 \]

\[ |R_5(x)| \le \frac{|x|^6}{6!} = \frac{(0.261799)^6}{720} \]

\[ |R_5| \le 4.47\times 10^{-7} \]

\[ 0.9659263 - 0.9659258 \approx 4.47\times 10^{-7} \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 573 — Checkpoint 6.13

Pi over 12 is fifteen degrees. Evaluate the fourth Maclaurin polynomial term by term: the x squared term is about 0.034 and the fourth-power term about 0.0002.

For the error, use the same trick as in Example 6.14. The fifth-degree coefficient of cosine is zero, so the fourth and fifth polynomials coincide, and you may use the remainder after degree five. Its sixth derivative is minus cosine, bounded by one, giving a sixth power over 720.

The bound is 4.47 times ten to the minus seven, and the actual error, computed from the true cosine, is 4.47 times ten to the minus seven as well, to three digits. For small inputs the error is almost exactly the first omitted term, so the bound is nearly tight. Seven correct decimal places from three terms is a remarkable return.

40. How many terms for e to six decimals?

Estimation

\[ e = \sum_{k=0}^{n}\frac{1}{k!} + R_n(1), \qquad |R_n(1)| \le \frac{e}{(n+1)!} < \frac{3}{(n+1)!} \]

\[ n = 9: \quad \frac{3}{10!} \approx 8.3\times 10^{-7} < 10^{-6} \]

Guess before you compute. You might expect to need a lot of terms for six decimal places, because six decimals is a very demanding accuracy.

The remainder bound at x equal to 1 is e over n plus 1 factorial, since every derivative is the exponential and its largest value between 0 and 1 is e itself. You do not want e in a bound meant to compute e, so replace it by 3, which is safely larger.

Now just tabulate. At n equal to 8 the bound is about eight millionths, not good enough; at n equal to 9 it is under one millionth. Ten terms suffice, and the true error is even smaller, about three ten-millionths. The factorial in the denominator grows so fast that each extra term buys roughly another decimal place. Compare that with the p-series of Chapter 5, which could need thousands of terms for three decimals.

41. Exercise 132: a degree for sine on the whole of minus pi to pi

Worked example

Find the smallest n for which the Taylor bound guarantees an error of at most one thousandth for sin x on the interval from minus pi to pi.

\[ M = 1 \]

\[ |R_n(x)| \le \frac{\pi^{n+1}}{(n+1)!} \]

\[ n = 10: 0.00737, \quad n = 11: 0.00193, \quad n = 12: 0.00047 \]

\[ n = 12 \]

\[ |p_{12}(\pi) - \sin\pi| \approx 0.00045 \le 0.00047 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 578 — Exercise 132

This is a design problem: you want one polynomial that is good to three decimals everywhere on a whole interval, not at one point. The Taylor bound is largest where the distance from the centre is largest, so you plan for the worst point, pi.

With M equal to 1, the bound is pi to the n plus 1 over n plus 1 factorial. Early on, the power of pi wins and the bound is huge; from about n equal to 3 onward the factorial takes over. Tabulating shows the bound first drops under a thousandth at n equal to 12.

The check evaluates the actual polynomial at the worst point. Since sine's twelfth coefficient is zero, the degree-twelve polynomial is the degree-eleven one, and at pi it misses sin pi, which is zero, by 0.00045: inside the guarantee, and close to it, which shows the bound is not wasteful here.

42. Find the error: solving for c

Error analysis

Read each note and decide whether that part of the argument is valid. The first line is fine: Taylor's theorem really does express the remainder exactly with the second derivative at some unknown c.

The error is in pretending to know c. Picking the midpoint is a guess dressed up as a calculation, and the resulting number, 0.0235, is smaller than the true error of 0.026. So the answer is wrong, not just unjustified.

There is also a logical point here worth absorbing. If you could find c, you could compute the exact error, and then you would know the cube root of 11 exactly, which is what you were approximating in the first place. The theorem is useful precisely because it lets you avoid finding c, by bounding over every possible c at once.

43. Does the series equal f?

Section

Part 4

44. Example 6.15: the Taylor series of one over x at 1

Worked example

\[ f = x^{-1}, \quad f' = -x^{-2}, \quad f'' = 2x^{-3}, \quad f''' = -3!\,x^{-4} \]

\[ f^{(n)}(1) = (-1)^n n! \]

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

\[ \frac{|a_{n+1}|}{|a_n|} = |x-1| < 1 \iff 0 < x < 2 \]

\[ x = 2: \sum(-1)^n, \qquad x = 0: \sum 1 \quad \text{both diverge} \]

\[ \sum_{n=0}^{10}(-0.5)^n = 0.666992 \approx \frac{1}{1.5}, \qquad \text{interval } (0, 2) \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 573-574 — Example 6.15

This example moves from polynomials to the full series. The derivatives of one over x at 1 are plus or minus n factorial, so after dividing by n factorial each coefficient is plus or minus one.

With the series in hand, the next question is where it converges. The ratio test gives the absolute value of x minus 1 less than one, an open interval from 0 to 2. Then you must test the endpoints separately, as in Section 6.1: at both, the terms have size one and do not go to zero, so both endpoint series diverge.

The figure shows what the interval means visually. Inside it, the degree-ten polynomial lies almost on top of one over x. Outside it, to the right of 2, both polynomials swing away violently, and higher degree makes it worse. The check at 1.5, well inside, shows eleven terms agreeing with two thirds to three decimals.

45. Checkpoint 6.14: one over x at 2

Worked example

\[ f^{(n)}(x) = (-1)^n n!\,x^{-(n+1)} \;\Longrightarrow\; f^{(n)}(2) = \frac{(-1)^n n!}{2^{n+1}} \]

\[ \frac1x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(x-2)^n \]

\[ \frac{|a_{n+1}|}{|a_n|} = \frac{|x-2|}{2} < 1 \iff 0 < x < 4 \]

\[ x = 4: \sum\frac{(-1)^n}{2}, \qquad x = 0: \sum\frac12 \quad \text{both diverge} \]

\[ \sum_{n=0}^{20}\frac{(-1)^n}{2^{n+1}} = 0.3333335 \approx \frac13, \qquad \text{interval } (0, 4) \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 574 — Checkpoint 6.14

Reuse the derivatives you just found; only the evaluation point changes. At 2, each derivative carries a power of 2 in the denominator, one more than the order of the derivative.

The series is the same alternating pattern, now in powers of x minus 2 divided by powers of 2. The ratio test gives the distance from 2, halved, less than one, so the radius is 2 and the open interval runs from 0 to 4.

Notice the pattern across the two examples: centred at 1 the radius was 1, centred at 2 it is 2. In both cases the interval stops exactly at zero, where one over x blows up. A Taylor series cannot reach past the nearest point where the function misbehaves. The check at 3 shows twenty-one terms agreeing with one third to about seven decimals.

46. Converging is not the same as converging to f

Intuition

The ratio test tells you where the Taylor series adds up to some number. It never looks at f, so it cannot tell you that the number is f of x.

\[ \lim_{n\to\infty} p_n(x) \text{ exists} \quad\text{vs}\quad \lim_{n\to\infty} p_n(x) = f(x) \]

For one over x at 1 you can settle the second question by algebra, because one over x is a geometric series in one minus x. For most functions there is no such trick, and you need the remainder.

\[ \frac1x = \frac{1}{1-(1-x)} = \sum_{n=0}^{\infty}(1-x)^n, \quad |1-x| < 1 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 574 — the discussion after Example 6.15

Slow down on this distinction, because it is the most subtle idea in the lesson. The ratio test answers the question of whether the Taylor series adds up to something. It does not know what f is, so it cannot possibly tell you that the something is f of x.

For one over x the gap can be closed by algebra: one over x can be rewritten as a geometric series in one minus x, which you know converges to exactly one over x on the same interval. So in that example the series converges to the function it came from.

For most functions, e to the x or sine for instance, there is no geometric series in disguise. You need a general tool that compares the series with f directly, and the remainder is exactly that comparison.

47. Theorem 6.8: the remainder decides

Concept

Convergence of Taylor series (Theorem 6.8) — Suppose f has derivatives of all orders on an interval I containing a. Then the Taylor series converges to f of x for every x in I if and only if the remainder tends to zero for every x in I.

\[ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \iff \lim_{n\to\infty}R_n(x) = 0 \]

\[ p_n(x) = f(x) - R_n(x) \to f(x) \iff R_n(x) \to 0 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 574-575 — Theorem 6.8

This theorem is almost a definition once you see it. The partial sums of the Taylor series are the Taylor polynomials, and each polynomial differs from f by the remainder. So the polynomials approach f of x exactly when the remainder approaches zero.

In practice you almost never compute the remainder exactly. You use the bound from Taylor's theorem: M over n plus 1 factorial times the distance to the power n plus 1. If you can show that bound goes to zero, the squeeze theorem forces the remainder to zero, and the series equals f.

One detail matters for the examples. M may depend on the interval, but it must not grow with n too fast. For the exponential, sine and cosine it does not grow with n at all, which is what makes the next three examples work.

48. The factorial beats any power

Concept

\[ \lim_{n\to\infty}\frac{|x|^{n+1}}{(n+1)!} = 0 \quad \text{for every real } x \]

This is the fact that makes the remainder vanish for the exponential, sine and cosine. It follows from Section 5.3's divergence test: the series of these terms converges by the ratio test, so its terms must go to zero.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 576 — the limit used in Example 6.16

Look at the bars: five to the n over n factorial. For the first few n, multiplying by five each step beats multiplying by n, so the bars grow, peaking at about 26 at n equal to 4 and 5. After that, each step multiplies by five over n, which is less than one, and the bars collapse. By n equal to 15 the value is two hundredths.

The same thing happens for any fixed x, however large. Once n is bigger than x, each step multiplies by less than one, and soon by very much less than one. The factorial always wins in the end.

The book proves this with a neat shortcut: the series of these quantities converges for every x by the ratio test, and the terms of a convergent series must tend to zero by the divergence test of Section 5.3. You can use this limit freely from now on.

49. Example 6.16(a): the exponential equals its series everywhere

Worked example

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

\[ \frac{|x|^{n+1}}{(n+1)!}\cdot\frac{n!}{|x|^n} = \frac{|x|}{n+1} \to 0 < 1 \]

\[ \text{interval of convergence } (-\infty, \infty) \]

\[ |R_n(x)| \le \frac{e^b}{(n+1)!}|x|^{n+1} \]

\[ 0 \le |R_n(x)| \le e^b\cdot\frac{|x|^{n+1}}{(n+1)!} \to 0 \]

\[ \sum_{n=0}^{10}\frac{2^n}{n!} = 7.388995 \approx 7.389056 = e^2 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 575-576 — Example 6.16a and Exercise 161

There are two separate jobs here, and the example does them in order. First the ratio test: the ratio of consecutive terms is x over n plus 1, which goes to zero for every x, so the series converges everywhere.

Second, the remainder. Pick any bound b on the size of x. Every derivative is the exponential, which is increasing, so on that interval its largest value is e to the b. That number is fixed; it does not change as n grows. The remainder bound is that fixed number times the quantity you just saw tends to zero, so the squeeze theorem sends the remainder to zero.

The numerical check at x equal to 2 uses Exercise 161: eleven terms of two to the n over n factorial give 7.388995, and e squared is 7.389056. Both jobs were needed. The first says the series settles; only the second says it settles on e to the x.

50. Example 6.16(b): the sine equals its series everywhere

Worked example

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

\[ \frac{|x|^{2n+3}}{(2n+3)!}\cdot\frac{(2n+1)!}{|x|^{2n+1}} = \frac{x^2}{(2n+3)(2n+2)} \to 0 \]

\[ |f^{(n+1)}(c)| \le 1 \;\Longrightarrow\; |R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!} \]

\[ |R_n(x)| \to 0 \quad \text{for every } x \]

\[ p_{15}(3) = 0.1411197 \approx 0.1411200 = \sin 3 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 576 — Example 6.16b

Follow the same two steps. The ratio test for the sine's series compares terms two powers apart, so the ratio has x squared on top and two consecutive whole numbers below. That tends to zero for every x.

For the remainder, sine is even easier than the exponential. Every derivative of sine is one of sine, cosine, minus sine or minus cosine, and all four are between minus one and one everywhere. So M is one on any interval, and the bound is the power of the distance over the factorial, which you know tends to zero.

The check at 3 is a strong one. Three is far from the centre, the polynomials were visibly wrong there on the slider for small degrees, and yet the degree-fifteen polynomial matches sin 3 to six decimal places. The series really does converge to sine, not just near zero but everywhere.

51. Checkpoint 6.15: the cosine's series

Worked example

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

\[ \frac{|x|^{2n+2}}{(2n+2)!}\cdot\frac{(2n)!}{|x|^{2n}} = \frac{x^2}{(2n+2)(2n+1)} \to 0 \]

\[ |R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!} \to 0 \]

\[ \frac{d}{dx}\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 576 — Checkpoint 6.15

This is the same argument as for sine, and you should be able to write it yourself. The ratio of consecutive terms has x squared over two consecutive whole numbers, which tends to zero, so the cosine's series converges everywhere.

Every derivative of cosine is bounded by one in size, so the remainder is at most the power of the distance over the factorial, and it tends to zero. The series converges to cosine for every real x.

The check uses term-by-term differentiation, the tool from the warm-up. Differentiate the sine's series and each odd power becomes an even power with the factorial reduced by one: x cubed over three factorial becomes x squared over two factorial. The result is exactly the cosine's series, as it must be because the derivative of sine is cosine.

52. A series that converges, but not to f

Counterexample

Look at the graph first. The function is positive everywhere except at zero, but near zero it is extraordinarily flat: at x equal to one half it is under two hundredths, and at x equal to one tenth it is about ten to the minus forty-four.

That flatness is so extreme that every derivative at zero is exactly zero. Proving it takes l'Hospital's rule repeatedly, and you may take it on trust here. So every Maclaurin coefficient is zero, and the Maclaurin series is the zero series.

The zero series converges for every x, as easily as a series can. But it converges to zero, and the function is not zero anywhere except the origin. Its remainder is the function itself, which does not shrink as n grows. This is the example to quote whenever someone assumes that a convergent Taylor series must equal its function.

53. Two questions, two tools

Comparison

Fill in the blanks, keeping the two questions strictly apart. Where a series converges is answered by the ratio test, together with separate endpoint checks. Whether it converges to the function is answered only by the remainder.

The rows for the exponential show the two tools agreeing: the ratio test says everywhere, and the remainder bound confirms the sum is e to the x everywhere. The row for one over x reminds you that the ratio test gives an open interval and the endpoints must be tested on their own.

The last two rows are the counterexample. The ratio test is satisfied everywhere, trivially, since every coefficient is zero. The remainder, however, is the whole function, so the series equals f only at the single point zero. If you only ever ran the ratio test, you would never notice anything was wrong.

54. The standard expansions

Section

Part 5

55. Five expansions worth knowing by heart

Concept

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

The first three have bounded derivatives on any interval, so the factorial always wins and they hold for every x. The last two come from the geometric series, and inherit its radius of one.

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, pp. 575-576 — Example 6.16 and Checkpoint 6.15; the last two rows from §6.1-6.2

These five series come up so often that deriving them every time would waste your effort. Learn them the way you learned the derivatives of sine and cosine, and learn the interval with each one, because the interval is part of the fact.

Look for the family resemblances. The exponential has every power over its factorial. Sine keeps the odd ones and cosine the even ones, both with alternating signs; together they split the exponential's terms between them. The geometric series has no factorials at all, and the logarithm has one over n, which is what the geometric series becomes when you integrate it.

The right-hand column tells you where each one is safe. The first three have infinite radius because their derivatives are bounded and the factorial always wins. The last two are tied to the geometric series and stop at distance one from the centre. The logarithm's series happens to converge at x equal to 1 as well, giving the alternating harmonic series for the logarithm of two.

56. Deriving the series for ln(1 + x)

Worked example

\[ f = \ln(1+x), \quad f' = (1+x)^{-1}, \quad f'' = -(1+x)^{-2}, \quad f''' = 2(1+x)^{-3} \]

\[ f(0) = 0, \qquad f^{(n)}(0) = (-1)^{n+1}(n-1)! \quad (n \ge 1) \]

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

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

\[ p_{10}(0.5) = 0.405435 \approx 0.405465 = \ln 1.5 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 579 — Exercise 154, in the form centred at 0

This derivation shows why the logarithm behaves so differently from the exponential. After the first derivative you are differentiating one over one plus x, which you did in Checkpoint 6.11, so the derivatives at zero are factorials with alternating signs, one step behind.

Divide by n factorial and almost everything cancels, leaving one over n. That is the crucial difference. The exponential's coefficients have a full factorial in the denominator; the logarithm's have only n. With so little in the denominator, the powers of x win as soon as x is bigger than one, and the series diverges.

The check at one half uses ten terms and misses the true logarithm of 1.5 by three hundred-thousandths. Because the series alternates with shrinking terms, the error is smaller than the first omitted term, and the check confirms that too. You could also have reached this series by integrating the geometric series, as in Section 6.2; uniqueness says the two routes must agree, and they do.

57. A finite radius, drawn

Picture it

The logarithm keeps going smoothly past x equal to 1, but its series cannot follow. The polynomials of degree 5 and 10 agree with it on the left of the dashed line and fly apart on the right.

The logarithm of one plus x is a perfectly smooth function for every x bigger than minus one. Nothing special happens to it at x equal to 1. So why should its series stop there?

Because a power series converges on an interval symmetric about its centre. The logarithm blows up at minus one, distance one to the left of the centre, so the radius can be at most one, and the series fails at the same distance to the right as well, even though the function is fine there.

In the figure, the degree-five and degree-ten polynomials lie almost exactly on the logarithm to the left of the dashed line. To the right of it they separate, one upward and one downward, and higher degrees separate faster. The radius is a property of the series, not only of the function you can see.

58. Push the degree past the radius

Edge cases

Use the slider to feel the difference between inside and outside the radius. Fix your eye on x equal to 0.5 first and raise n: the curve there settles and stops moving, at the logarithm of 1.5, about 0.405.

Now watch x equal to 1.5. As n rises, the value there flips up and down with a growing swing, because the terms 1.5 to the n over n grow without bound. No choice of n gives a good approximation there.

Right at x equal to 1 the behaviour is in between: the values wobble but slowly settle on the logarithm of 2, because the alternating harmonic series converges, though very slowly. This is exactly the endpoint behaviour you tested by hand in Section 6.1, now visible on the screen.

59. Trap: using a series outside its radius

Trap

60. Exercise 162: which function is this?

Prediction

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

\[ S_{10} = \sum_{n=0}^{10}\frac{(-1)^n(2\pi)^{2n}}{(2n)!} = 1.000301 \approx \cos 2\pi = 1 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 579 — Exercise 162

Recognising a series as a known function evaluated at a point is a skill you will use a lot in the next section. Look for the fingerprints: the signs, which powers appear, and which factorial sits in the denominator.

Here the signs alternate, only even powers appear, and the denominator is the factorial of the same even number. That is the cosine's series, evaluated at 2 pi. Since the cosine's series converges to cosine everywhere, the sum is cos 2 pi, which is 1.

The partial sum is instructive. Eleven terms give 1.000301, still wrong in the fourth decimal place, even though eleven terms gave six-decimal accuracy near zero. Two pi is far from the centre, so the early terms are huge before the factorial takes control. Convergence everywhere does not mean fast convergence everywhere.

61. Exercise 170: a limit from a series

Worked example

\[ \lim_{x\to 0}\frac{\cos x - 1}{x^2} \]

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

\[ \frac{\cos x - 1}{x^2} = -\frac{1}{2} + \frac{x^2}{24} - \frac{x^4}{720} + \cdots \]

\[ \lim_{x\to 0}\frac{\cos x - 1}{x^2} = -\frac12 \]

\[ \frac{\cos 0.01 - 1}{0.0001} = -0.4999958 \approx -0.5 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 580 — Exercise 170

Limits of the form zero over zero were a job for l'Hospital's rule in Calculus One. Series give you another method that is often faster and shows you more.

Replace the cosine by its series. The constant 1 cancels, leaving a series that starts with minus x squared over two. Divide by x squared and every exponent drops by two, so the series now starts with the constant minus one half, followed by terms that all contain x.

As x goes to zero every term with an x vanishes, and the limit is minus one half. The series tells you more than the limit: it says the quotient is about minus one half plus x squared over 24 for small x. The numerical check at 0.1 and 0.01 shows the quotient creeping toward minus one half from above, just as that extra positive term predicts.

62. Exercise 131: integrating e to the minus x squared

Worked example

The function e to the minus x squared has no elementary antiderivative, but its series integrates term by term.

\[ e^{-x^2} \approx 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots + \frac{x^{12}}{6!} \]

\[ \int_0^1 e^{-x^2}\,dx \approx \sum_{k=0}^{6}\frac{(-1)^k}{k!\,(2k+1)} \]

\[ 1 - 0.333333 + 0.1 - 0.023810 + 0.004630 - 0.000758 + 0.000107 = 0.746836 \]

\[ |0.746836 - 0.746824| = 0.000012 < \frac{1}{7!\cdot 15} \approx 0.000013 \]

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 578 — Exercise 131

The bell-shaped function e to the minus x squared is central to probability, and it has no antiderivative you can write with elementary functions. Series give you a way around that.

Substitute minus x squared into the exponential's polynomial. Each power of x doubles, and the signs alternate. Then integrate each power from 0 to 1, which is easy: x to the 2k integrates to one over 2k plus 1. Seven terms of a simple alternating sum approximate the integral.

The check compares with a numerical integral from Simpson's rule, 0.746824. The series answer is 0.746836, off by twelve millionths, and that gap is less than the first term left out, as it must be for an alternating series with shrinking terms. In the next section you will do this systematically.

63. Complete the reasoning about the factorial

Fill the middle

\[ |R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1} \]

Fill in both blanks, then read the whole sentence as an argument. It is the reason three of the five standard expansions converge to their functions everywhere.

The argument has two parts. First, the derivatives of the exponential on a fixed interval, and of the sine and cosine anywhere, are bounded by a number that does not depend on n. Second, the factorial in the denominator eventually grows faster than any fixed power of the distance, as the bar chart showed.

Together they force the remainder bound to zero, and Theorem 6.8 turns that into equality between the function and its series. If either part fails, for instance derivatives that grow like factorials as the logarithm's do, the argument breaks, and the radius may be finite.

64. The small-angle approximation, priced

Real world

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series §6.3, p. 578 — Exercise 137, sin x approximated by x

In physics you will be told to replace the sine of a small angle by the angle itself when solving the pendulum equation. It is usually presented as a convention. Taylor's theorem makes it a calculation with a guaranteed error.

The first-degree polynomial of sine is theta, and it equals the second-degree one, so you may use the remainder after degree two. The third derivative of sine is minus cosine, bounded by one, giving an error of at most theta cubed over six.

At 0.2 radians, about eleven and a half degrees, the bound is 0.00133, less than seven tenths of a percent of the sine itself. The figure shows the gap between the line and the curve, invisible at small angles and obvious by one radian. Now when someone says small angle, you can ask how small and answer with a number.

65. Putting it together

Section

Part 6

66. Pattern: from a function to a guaranteed approximation

Pattern

1. Differentiate until you see the pattern, then evaluate at the centre.
2. Divide the nth derivative by n factorial to get each coefficient; assemble the polynomial or the series.
3. For accuracy: bound the (n plus 1)st derivative by its largest size M between a and x; the error is at most M over (n plus 1) factorial times the distance to the power n plus 1.
4. For a tolerance: increase n until that bound drops below it.
5. For the whole series: use the ratio test for where it converges, and the remainder for whether it converges to f.

The diagram is the whole lesson as one process. The top row builds the approximation: derivatives, evaluated at the centre, divided by factorials, assembled into a polynomial. The bottom row prices it: bound the next derivative, compute the remainder bound, and decide.

The two exits at the bottom correspond to the two uses you have seen. For a fixed degree, the bound is a guaranteed error, and if you have a tolerance you raise the degree until the bound meets it. For the whole series, you ask whether the bound tends to zero as the degree grows, which is how you prove the series equals the function.

When you face a new problem, place yourself in the diagram first. Most exam questions start at one box and ask you to get to another.

67. Put an error bound in order

Ranking

Put the steps in order, then check your ordering against the reasoning in the explanation. The order is not arbitrary: each step needs the one before.

The most common failure is to skip the second step and simply plug the centre, or the point x, into the derivative. The maximum has to be taken over the whole stretch between them, because that is where the unknown c lives.

The last step is also easy to skip. The number you compute is a statement about where the true value lies, and saying so explicitly, the true value is within this distance of the estimate, is what makes the calculation useful to someone else. An error bound with no statement attached is just a number; with the statement it is a guarantee.

68. Check: the Taylor coefficient

Check

Answer before you reveal. The question gives you all four derivatives at the centre, but only one of them matters for the coefficient of the cube, and it has to be divided by the right factorial.

Each wrong option corresponds to a real slip: forgetting the factorial altogether, dividing by the degree instead of its factorial, or using the factorial of the wrong degree. If you chose one of them, go back to the fourth-power figure and the idea that n differentiations bring down n factorial.

A quick self-test for next time: the coefficient of the square would be 4 over two factorial, which is 2, and the constant term is just f of 2, which is 3. Every coefficient comes from the same recipe: the derivative of that order at the centre, divided by the factorial of that order. The derivatives of other orders play no part in it.

69. Check: bounding a remainder

Check

Work this out on paper before choosing. Two decisions matter: which derivative controls the error of a cubic, and how large that derivative can be on the interval.

A cubic's error is controlled by the fourth derivative, and for the exponential that is the exponential again, which is at most e to the one half on the interval from 0 to one half. Since that is less than 2, 2 is a safe M. The option with M equal to 1 is the classic mistake of evaluating only at the centre.

The actual error, about 0.0029, is under the correct bound, 0.0052. A bound bigger than the actual error is fine; a bound smaller than it would be wrong. Choosing M a little too large, as 2 is here, costs you only a slightly looser guarantee.

70. Check: converging to f

Check

This question separates the two ideas of part four. Read each option and ask whether the flat function, e to the minus one over x squared, breaks it.

It breaks three of them. Its Maclaurin series converges, everywhere, but not to the function; it has all its derivatives, but does not equal its series near zero; and the ratio test is perfectly happy with it. Only the statement about the remainder survives, because it is exactly Theorem 6.8.

If you picked the first option, you have merged the two questions of part four into one. Keep them apart: the ratio test says where the series settles, and only the remainder says what it settles on. A single counterexample is enough to sink a general claim, and the flat function is the one to reach for.

71. Explain the unknown point c

Explain it to yourself

Write your explanation before you read the model answer. It is the kind of question that shows whether you understand Taylor's theorem or only its formula.

The key idea is that the theorem trades an unknown number for a known interval. You do not know c, and finding it would be as hard as finding the exact error. But you do know that c lies between a and x, and bounding the derivative over that interval turns the exact formula into a bound you can compute.

Compare this with the Mean Value Theorem, which has the same kind of unknown point. You have used that theorem for years without ever finding its c, for exactly the same reason. It is enough to know where c lives.

72. Exit ticket

Exit ticket

\[ f(x) = e^{-x}, \quad a = 0 \]

Do this without looking back. It uses every step of the lesson once: derivatives at the centre, the polynomial, an evaluation, and an error bound.

The derivatives of e to the minus x alternate in sign, so the polynomial alternates too. For the bound, the fourth derivative is e to the minus x again, which on the interval from 0 to 0.2 is largest at 0, where it equals 1. That is why M is 1 here, not something larger: this derivative decreases.

The answer checks itself: the estimate 0.818667 and the true value 0.818731 differ by 0.000064, just inside the bound of 0.0000667. If your bound was smaller than the actual error, look again at which derivative and which M you used.

73. Recap

Recap

\[ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \quad\text{whenever}\quad R_n(x) \to 0 \]

Next, Section 6.4 puts these expansions to work: new series from old by substitution, multiplication and integration, and limits and integrals no other method can reach.

Stewart, Calculus: Early Transcendentals 8e, §11.10 Taylor and Maclaurin Series §11.10, pp. 759-773 — the same material in Stewart

The table is the lesson in five lines. Taylor coefficients come from matching derivatives, which is where the factorial comes from. Taylor polynomials are the partial sums and the practical approximations. Taylor's theorem writes the error exactly with an unknown point, and the bound turns that into a number you can use.

The last row is the conceptual one. A Taylor series equals its function exactly where the remainder goes to zero. Convergence alone is not enough, as the flat function showed.

In the next section you will stop differentiating. With the five standard expansions in hand, you can build new series by substituting, multiplying and integrating, and use them to evaluate limits and integrals that no other method reaches.