Finding Taylor Polynomial Approximations of Functions: Study Guide (Calculus BC 10.11)

BC-only material. Taylor polynomials are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

Prerequisite	Topic	What you use it for here
Higher-order derivatives	3.6	Finding f″, f‴, f⁽⁴⁾, … at the centre
Local linearity (tangent line approximation)	4.6	The tangent line is the degree-1 case of everything on this page
Factorials	—	Each coefficient divides by n! (with 0! = 1)
Alternating series error bound	10.10	Used again in Topic 10.12 to judge Taylor estimates

If any of these is shaky, see the Calculus BC roadmap or the previous topic, the alternating series error bound.

Notation on this page: f⁽ⁿ⁾(a) is the nth derivative of f evaluated at x = a, with f⁽⁰⁾ = f. Pₙ(x) is the Taylor polynomial of degree n (at most n).

From a tangent line to a better polynomial

In Topic 4.6 you approximated f near x = a with the tangent line

L(x) = f(a) + f′(a)(x − a)

The line has the same value and the same slope as f at x = a. Close to a, that is good. But a line cannot bend, so the error grows quickly once the graph curves.

The idea of a Taylor polynomial is simple: match more derivatives. A quadratic can also match f″(a), so it bends the same way as f at a. A cubic can match f‴(a) too, and so on. Each extra matching derivative makes the polynomial hug the graph over a wider stretch near a.

The coefficient formula, and where n! comes from

Write a polynomial centred at x = a:

P(x) = c₀ + c₁(x − a) + c₂(x − a)² + c₃(x − a)³ + … + cₙ(x − a)ⁿ

We want P⁽ᵏ⁾(a) = f⁽ᵏ⁾(a) for k = 0, 1, 2, …, n.

• At x = a, every term except c₀ is zero. So P(a) = c₀, and we need c₀ = f(a).
• Differentiate once: P′(x) = c₁ + 2c₂(x − a) + 3c₃(x − a)² + … At x = a, P′(a) = c₁. So c₁ = f′(a).
• Differentiate again: P″(x) = 2c₂ + 3 · 2 c₃(x − a) + … So P″(a) = 2c₂, and c₂ = f″(a)/2.
• Once more: P‴(a) = 3 · 2 · 1 · c₃ = 6c₃, so c₃ = f‴(a)/6 = f‴(a)/3!.

The pattern: differentiating (x − a)ᵏ exactly k times gives k · (k − 1) · … · 1 = k!, and every lower power has already vanished, while every higher power still contains a factor (x − a) that is zero at a. So

cₖ = f⁽ᵏ⁾(a) / k!

and the nth-degree Taylor polynomial for f about x = a is

Pₙ(x) = f(a) + f′(a)(x − a) + f″(a)/2! · (x − a)² + f‴(a)/3! · (x − a)³ + … + f⁽ⁿ⁾(a)/n! · (x − a)ⁿ

When a = 0 it is often called a Maclaurin polynomial. Notice that P₁(x) is exactly the tangent line.

Reading backwards. The formula also works in reverse. If you are given a Taylor polynomial about x = a, then

f⁽ᵏ⁾(a) = k! × (coefficient of (x − a)ᵏ)

For example, if the third-degree Taylor polynomial for f about x = −2 is 5 + 3(x + 2) − 2(x + 2)² + ½(x + 2)³, then f(−2) = 5, f′(−2) = 3, f″(−2) = 2! × (−2) = −4 and f‴(−2) = 3! × ½ = 3. Note that x + 2 = x − (−2), so the centre is −2, not 2.

Seeing the approximation improve

For f(x) = cos x about x = 0, the derivatives at 0 cycle through 1, 0, −1, 0, 1, 0, −1, … So the odd-power coefficients are all zero and

• P₂(x) = 1 − x²/2
• P₄(x) = 1 − x²/2 + x⁴/24
• P₆(x) = 1 − x²/2 + x⁴/24 − x⁶/720

The table shows the same story in numbers (values to 4 decimal places):

x	P₂(x)	P₄(x)	P₆(x)	cos x
0.5	0.875	0.8776	0.8776	0.8776
1	0.5	0.5417	0.5403	0.5403
2	−1	−0.3333	−0.4222	−0.4161
3	−3.5	−0.125	−1.1375	−0.9900

Two lessons:

1. Near the centre, low degrees already work well. At x = 0.5, P₄ matches cos x to 4 decimal places.
2. Far from the centre, you need higher degrees, and even P₆ is poor at x = 3.

“In many cases” the polynomials approach f as the degree increases, but not always. For f(x) = 1/(1 + x) about 0, the polynomials are 1 − x + x² − x³ + … At x = 0.5 their values 1, 0.5, 0.75, 0.625, 0.6875, … close in on f(0.5) = 2/3. At x = 2 they give 1, −1, 3, −5, 11, … and never settle near f(2) = 1/3. Topic 10.13 explains exactly where this happens (the interval of convergence).

Worked example 1: ln x about x = 2

Question. Find the third-degree Taylor polynomial for f(x) = ln x about x = 2. Use it to approximate ln 2.2, giving your answer in terms of ln 2 and as a decimal.

1. Derivatives. f(x) = ln x, f′(x) = 1/x, f″(x) = −1/x², f‴(x) = 2/x³.
2. Evaluate at the centre x = 2. f(2) = ln 2, f′(2) = 1/2, f″(2) = −1/4, f‴(2) = 2/8 = 1/4.
3. Divide by the factorials.
   • c₀ = ln 2
   • c₁ = (1/2)/1! = 1/2
   • c₂ = (−1/4)/2! = −1/8
   • c₃ = (1/4)/3! = 1/24
4. Write the polynomial. P₃(x) = ln 2 + ½(x − 2) − ⅛(x − 2)² + (1/24)(x − 2)³
5. Approximate at x = 2.2, so x − 2 = 0.2.
   • ½(0.2) = 0.1
   • −⅛(0.04) = −0.005
   • (1/24)(0.008) = 1/3000 ≈ 0.000333
   • P₃(2.2) = ln 2 + 0.1 − 0.005 + 0.000333 = ln 2 + 143/1500 ≈ 0.788481, using ln 2 ≈ 0.693147.

Checks.

• Matching: P₃(2) = ln 2 and P₃′(2) = ½, as required.
• Calculator comparison (where allowed): ln 2.2 ≈ 0.788457, so the estimate is off by about 0.000023. The tangent line alone gives ln 2 + 0.1 ≈ 0.793147, off by about 0.0047. The extra terms improved the estimate by a factor of about 200.
• Common slip: forgetting the factorials gives ln 2 + 0.1 − 0.01 + 0.002 ≈ 0.785147, which is off by about 0.0033, far worse than the correct P₃.

Worked example 2: a polynomial from a table (no formula for f)

On the exam you often do not get a formula. You get derivative values, and you build the polynomial from them.

Context. The depth of water in a fictional reservoir is D(t) metres, where t is hours after midnight. D has derivatives of all orders. At t = 2 the following values are known.

t	D(t) (m)	D′(t) (m/h)	D″(t) (m/h²)	D‴(t) (m/h³)
2	12	−0.8	0.3	−0.06

(a) Write the third-degree Taylor polynomial for D about t = 2. (b) Use it to estimate the depth at t = 2.5. (c) What is the coefficient of (t − 2)² telling you?

1. (a) Coefficients. c₀ = 12; c₁ = −0.8; c₂ = 0.3/2 = 0.15; c₃ = −0.06/6 = −0.01. P₃(t) = 12 − 0.8(t − 2) + 0.15(t − 2)² − 0.01(t − 2)³
2. (b) Evaluate at t = 2.5 (so t − 2 = 0.5):
   • 12
   • −0.8(0.5) = −0.4
   • 0.15(0.25) = 0.0375
   • −0.01(0.125) = −0.00125
   • P₃(2.5) = 11.63625, so D(2.5) ≈ 11.64 m.
3. (c) Interpret. The coefficient 0.15 is D″(2)/2. Since D″(2) > 0, the depth is falling (D′ < 0) but its rate of fall is slowing at t = 2. The quadratic term bends the estimate up a little compared with the tangent line, which gives 11.6 m.

Checks. Size: the depth starts at 12 m and falls at about 0.8 m/h, so after half an hour roughly 11.6 m is expected; the small corrections (+0.0375, −0.00125) shrink fast, as they should close to the centre. Units: each term is in metres, because D⁽ᵏ⁾ is in m/hᵏ and (t − 2)ᵏ is in hᵏ.

Degree versus number of terms

A question may ask for “the third-degree Taylor polynomial” or for “the first three nonzero terms”. These can differ.

• For sin x about 0, the derivatives at 0 are 0, 1, 0, −1, 0, … So P₃(x) = x − x³/6, which has two nonzero terms. P₄ is the same polynomial, because the x⁴ coefficient is 0.
• The first three nonzero terms are x − x³/6 + x⁵/120, which is P₅.

Always check which one is asked for.

Common misconceptions

• Forgetting the factorial. The coefficient of (x − a)³ is f‴(a)/3! = f‴(a)/6, not f‴(a) and not f‴(a)/3.
• Using x instead of (x − a). About x = 2 the powers are (x − 2)ᵏ, not xᵏ. Only a Maclaurin polynomial (a = 0) uses plain powers of x.
• Evaluating derivatives at the wrong point. The coefficients use f⁽ᵏ⁾(a), the values at the centre, never at the point you are estimating.
• Sign of the centre. (x + 3) means the centre is x = −3.
• Confusing the coefficient with the derivative when reading backwards. If the coefficient of (x − a)² is 4, then f″(a) = 2! × 4 = 8, not 4.
• Thinking a Taylor polynomial is good everywhere. It is built to fit at the centre. Far away it can be badly wrong, as P₂(3) = −3.5 for cos x shows.
• Mixing up degree and number of terms (see the section above).
• Using degrees for trig functions. The derivative rules for sin and cos need radians.

Where this leads

An estimate is only useful if you know how far off it might be. The next topic, Topic 10.12, the Lagrange error bound, gives a guaranteed limit on |f(x) − Pₙ(x)|. After that, Topic 10.13 finds where the infinite version (a power series) converges, and Topic 10.14 writes full Taylor and Maclaurin series. See the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.