Finding Taylor Polynomial Approximations of Functions - AP Calculus BC

SECTION 1

Historical Context & Motivation

Long before electronic calculators existed, mathematicians and scientists needed practical ways to evaluate transcendental functions such as sin x, eˣ, and ln x. The key insight—that smooth functions can be represented as infinite sums of power terms—emerged gradually over several centuries and remains one of the most powerful ideas in all of analysis. Taylor polynomial approximations grew out of this tradition, providing finite truncations of infinite series that approximate a function near a chosen center point with remarkable accuracy. Understanding this history illuminates why the technique is central not only to calculus but also to numerical computation, physics, and engineering.

1671 • James Gregory's Series

Scottish mathematician James Gregory discovered series expansions for arctan x and several trigonometric functions, anticipating later general methods by decades.

1715 • Brook Taylor's Methodus Incrementorum

Brook Taylor published the general formula for expanding a function as a power series about a point, giving his name to the resulting polynomials and series.

1742 • Colin Maclaurin's Treatise of Fluxions

Maclaurin systematized the special case centered at a = 0, now called a Maclaurin series, and used it extensively in his rigorous defense of Newtonian calculus.

1821 • Cauchy's Convergence Criteria

Augustin-Louis Cauchy formalized convergence and the remainder term, placing Taylor's result on a rigorous analytical foundation that resolved earlier concerns about infinite series.

The central question these mathematicians addressed is elegantly simple: given a function f that is smooth (infinitely differentiable) near a point a, how well can a polynomial—built entirely from the derivatives of f at that single point—reproduce the behavior of f nearby? This question leads directly to the construction of Taylor polynomials and to deep investigations of how quickly, and under what conditions, the approximation converges to the original function.

SECTION 2

Core Principles & Definitions

A Taylor polynomial is constructed so that it matches a given function and as many of its derivatives as possible at a single point called the center of the expansion. The fundamental philosophy is that local derivative information encodes the shape of a function—its value, slope, concavity, rate of change of concavity, and so on—and a polynomial that preserves all of these features at one point will closely mimic the function in a neighborhood around it.

Center of Expansion (a)

The point about which the polynomial is built. When a = 0, the result is called a Maclaurin polynomial. The approximation is most accurate near this point.

Degree n of the Polynomial

The highest power of (x − a) retained. Increasing n generally improves accuracy and widens the interval over which the polynomial closely approximates f.

Derivative Matching

The nth-degree Taylor polynomial Pₙ(x) satisfies Pₙ⁽ᵏ⁾(a) = f⁽ᵏ⁾(a) for every k from 0 to n, ensuring the polynomial's first n derivatives equal those of f at the center.

Factorial Scaling

Each term is divided by k! to compensate for the chain of derivatives: differentiating (x − a)ᵏ k times yields k!, so dividing by k! normalizes the coefficient back to f⁽ᵏ⁾(a).

Remainder / Error

The difference f(x) − Pₙ(x) is the remainder Rₙ(x). The Lagrange remainder form provides an explicit bound on this error, crucial for justifying the approximation's quality.

✦

KEY TAKEAWAY

Think of a Taylor polynomial as a GPS recalculation anchored at a single known location. At that exact spot you know your position, heading, rate of turning, and so on. The polynomial uses all of that local information to predict where the road goes next. The more derivatives you include (the more 'sensors' you consult), the further ahead your prediction remains reliable—but stray too far from the anchor and the prediction drifts.

SECTION 3

Visual Explanation — Successive Polynomial Approximations

The solid curve is f(x) = sin x. The dashed pink line shows P₁(x) = x, which matches the function's value and slope at the origin but diverges quickly. The violet dashed curve is P₃, which also captures concavity, and the cyan curve is P₅, which tracks sin x closely over a much wider interval. Notice how each successive polynomial 'hugs' the true curve for a longer stretch away from the center a = 0.

The diagram above illustrates the central visual intuition of Taylor polynomials. At the center point a = 0, every polynomial in the family passes through the origin with the same slope as sin x, namely 1. The first-degree polynomial P₁(x) = x is simply the tangent line, which is a reasonable approximation only in a tiny neighborhood. By incorporating the third derivative, P₃(x) = x − x³/6 captures the concavity change and tracks the sine curve through roughly (−π, π). Adding the fifth-degree term yields P₅, which remains faithful to sin x over an even larger interval. This pattern generalizes: each additional term encodes one more piece of curvature information, extending the range of accurate approximation.

SECTION 4

Mathematical Framework

The construction of a Taylor polynomial rests on a beautifully simple requirement: match f and its first n derivatives at the center a. This condition uniquely determines the coefficients of the polynomial, yielding the general formula presented below.

TAYLOR POLYNOMIAL OF DEGREE n

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

Equivalently in summation form: Pₙ(x) = Σ (k = 0 to n) f⁽ᵏ⁾(a)/k! · (x − a)ᵏ. Here f⁽ᵏ⁾(a) denotes the kth derivative of f evaluated at the center a, and k! = k × (k−1) × ⋯ × 1 is the factorial.

MACLAURIN POLYNOMIAL (SPECIAL CASE a = 0)

Pₙ(x) = f(0) + f′(0)x + f″(0)/2! · x² + f‴(0)/3! · x³ + ⋯ + f⁽ⁿ⁾(0)/n! · xⁿ

When the center is a = 0, every (x − a)ᵏ term simplifies to xᵏ, producing the Maclaurin polynomial. This is the form tested most frequently on the AP exam.

LAGRANGE REMAINDER (ERROR BOUND)

Rₙ(x) = f⁽ⁿ⁺¹⁾(c) / (n + 1)! · (x − a)ⁿ⁺¹ for some c between a and x

This remainder equals f(x) − Pₙ(x). In practice, to bound |Rₙ(x)| you replace |f⁽ⁿ⁺¹⁾(c)| by the maximum of |f⁽ⁿ⁺¹⁾| on the interval between a and x, since the exact value of c is unknown.

The coefficient structure f⁽ᵏ⁾(a)/k! deserves careful attention. The factorial in the denominator arises because differentiating (x − a)ᵏ exactly k times produces k! as a leading constant. Dividing by k! cancels this factor, so that when Pₙ is differentiated k times and evaluated at a, the result is precisely f⁽ᵏ⁾(a). This derivative-matching property is both the definition and the justification of the Taylor polynomial: it is the unique polynomial of degree at most n whose value and first n derivatives at a agree with those of f.

💡

AP Exam Tip

On free-response questions, the College Board expects you to show each derivative evaluation explicitly. Write out f(a), f′(a), f″(a), … in a table before substituting into the general formula. This organized approach earns full credit and reduces computational errors.

SECTION 5

Essential Taylor / Maclaurin Series You Must Know

The AP Calculus BC exam expects fluency with several standard Maclaurin series. Rather than deriving each one from scratch during the exam, you should memorize these expansions and be able to adapt them through substitution, differentiation, or integration to generate new series. The table below catalogs the most important ones, together with their intervals of convergence.

Function	Maclaurin Series	Interval of Convergence
eˣ	1 + x + x²/2! + x³/3! + ⋯ = Σ xᵏ/k!	(−∞, ∞)
sin x	x − x³/3! + x⁵/5! − ⋯ = Σ (−1)ᵏ x²ᵏ⁺¹/(2k+1)!	(−∞, ∞)
cos x	1 − x²/2! + x⁴/4! − ⋯ = Σ (−1)ᵏ x²ᵏ/(2k)!	(−∞, ∞)
1/(1 − x)	1 + x + x² + x³ + ⋯ = Σ xᵏ	(−1, 1)
ln(1 + x)	x − x²/2 + x³/3 − ⋯ = Σ (−1)ᵏ⁺¹ xᵏ/k	(−1, 1]

Standard Maclaurin series required for AP Calculus BC

Starting from two memorized 'parent' series—eˣ and 1/(1 − x)—you can generate many related series through substitution, differentiation, and integration. This strategy is heavily tested on the AP exam and eliminates the need to recompute derivatives from scratch.

The flowchart above highlights a crucial exam strategy. Rather than computing all derivatives of e⁻ˣ² from scratch (which becomes algebraically painful after the third derivative), you can simply substitute −x² for x in the known series for eˣ. Similarly, integrating the geometric series 1/(1 − x) = Σ xᵏ term by term yields −ln(1 − x) = Σ xᵏ⁺¹/(k + 1), which after rearrangement gives the standard series for ln(1 + x). These series manipulation techniques—substitution, term-by-term differentiation, and term-by-term integration—are among the most frequently assessed skills on the BC exam.

SECTION 6

Worked Example — Taylor Polynomial for ln x Centered at a = 1

Find the fourth-degree Taylor polynomial P₄(x) for f(x) = ln x centered at a = 1, and use it to approximate ln(1.2).

Constructing P₄(x) for ln x about a = 1

1
Step 1 — Compute Successive Derivatives

Starting from f(x) = ln x, compute the first four derivatives: f′(x) = 1/x = x⁻¹, f″(x) = −x⁻², f‴(x) = 2x⁻³, and f⁽⁴⁾(x) = −6x⁻⁴. Notice the pattern: f⁽ᵏ⁾(x) = (−1)ᵏ⁺¹ (k − 1)! x⁻ᵏ for k ≥ 1.
2
Step 2 — Evaluate Each Derivative at a = 1

Substitute x = 1 into each expression: f(1) = ln 1 = 0, f′(1) = 1, f″(1) = −1, f‴(1) = 2, and f⁽⁴⁾(1) = −6.

Derivative values: 0, 1, −1, 2, −6
3
Step 3 — Form the Coefficients f⁽ᵏ⁾(1)/k!

Divide each derivative value by the corresponding factorial: f(1)/0! = 0/1 = 0, f′(1)/1! = 1/1 = 1, f″(1)/2! = −1/2, f‴(1)/3! = 2/6 = 1/3, and f⁽⁴⁾(1)/4! = −6/24 = −1/4.

Coefficients: 0, 1, −1/2, 1/3, −1/4
4
Step 4 — Assemble the Polynomial

Multiply each coefficient by (x − 1)ᵏ and sum: P₄(x) = 0 + 1·(x − 1) + (−1/2)(x − 1)² + (1/3)(x − 1)³ + (−1/4)(x − 1)⁴.

P₄(x) = (x − 1) − (x − 1)²/2 + (x − 1)³/3 − (x − 1)⁴/4
5
Step 5 — Approximate ln(1.2)

Substitute x = 1.2, so (x − 1) = 0.2: P₄(1.2) = 0.2 − (0.04)/2 + (0.008)/3 − (0.0016)/4 = 0.2 − 0.02 + 0.002667 − 0.0004 = 0.182267. The true value is ln 1.2 ≈ 0.182322, so the error is approximately 0.000055.

P₄(1.2) ≈ 0.18227, with error ≈ 5.5 × 10⁻⁵

ℹ

Pattern Recognition

Notice that the coefficients follow the pattern (−1)ᵏ⁺¹/k for k ≥ 1. This matches the known Maclaurin series for ln(1 + u) with u = x − 1, confirming consistency. On the AP exam, recognizing such patterns can save significant time.

SECTION 7

Convergence, Strengths & Limitations

While Taylor polynomials are extraordinarily useful, understanding their limitations is just as important as knowing how to construct them. The approximation quality depends on the degree of the polynomial, the distance from the center, and the behavior of the function's higher-order derivatives. The table below contrasts the strengths and limitations of the method.

Strengths	Limitations
Converts transcendental functions into polynomials, which are easy to evaluate, differentiate, and integrate.	Accuracy degrades as \|x − a\| increases; the polynomial may diverge wildly outside the interval of convergence.
The Lagrange remainder provides a rigorous, quantitative error bound.	Computing the bound requires knowledge of \|f⁽ⁿ⁺¹⁾\| on the interval, which can be difficult for complex functions.
Series for standard functions can be adapted through substitution, differentiation, and integration.	Not all infinitely differentiable functions equal their Taylor series (e.g., f(x) = e⁻¹/ˣ² at x = 0 has all derivatives zero, so its Maclaurin series is identically 0, yet f is not 0 for x ≠ 0).
Higher-degree polynomials systematically improve approximation quality near the center.	Higher derivatives become increasingly laborious to compute for most functions.

Strengths and limitations of Taylor polynomial approximations

✦

KEY TAKEAWAY

A Taylor polynomial is like a local weather forecast: brilliant for the next few hours (near the center), solid for a day or two (moderate |x − a|), and increasingly unreliable a week out (far from the center). More data points—analogous to higher-degree terms—extend the forecast window, but no polynomial can capture global behavior in every case. On the AP exam, always state the interval of convergence and use the Lagrange remainder to justify the accuracy of your approximation.

SECTION 8

Connection to Full Taylor Series & Advanced Theory

A Taylor polynomial of degree n is a finite object—a partial sum of the Taylor series, which is the infinite sum obtained by letting n → ∞. The transition from polynomial to series raises deep questions about convergence: does the infinite sum actually converge to f(x)? If so, for which values of x? These questions connect Taylor polynomials to the broader theory of power series and radius of convergence, topics you will study in the subsequent units of BC Calculus.

Feature	Taylor Polynomial Pₙ(x)	Taylor Series (n → ∞)
Number of terms	Finite (n + 1 terms)	Infinite
Equality with f	Approximation: f(x) ≈ Pₙ(x)	Exact (within radius of convergence): f(x) = Σ...
Error analysis	Lagrange remainder provides explicit bound	Error is zero when series converges to f
Practical computation	Always computable in finitely many steps	Requires convergence tests (Ratio, Root, etc.)
AP exam context	Construct and evaluate specific polynomials; bound error	Determine radius/interval of convergence; represent functions as series

Taylor Polynomial vs. Taylor Series

Looking ahead, the Taylor polynomial framework naturally extends to topics such as the ratio test for determining the radius of convergence, alternating series error estimation (which is often simpler than the Lagrange bound for alternating series), and the study of power series representations that enable integration of functions like e⁻ˣ² whose antiderivatives cannot be expressed in closed form. Mastering the polynomial construction in this lesson is the essential prerequisite for all of these advanced topics.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

The second-degree Taylor polynomial P₂(x) for a function f centered at a = 3 satisfies which of the following conditions?

A. P₂(3) = f(3), P₂′(3) = f′(3), but P₂″(3) ≠ f″(3)
B. P₂(3) = f(3), P₂′(3) = f′(3), and P₂″(3) = f″(3)
C. P₂(x) = f(x) for all x in the interval of convergence
D. P₂(3) = f(3) and P₂(x) has the same concavity as f for all x
E. P₂(3) = f(3) and P₂′(3) = f′(3), but the second derivative condition depends on the function

Show answer (B)

The correct answer is B. By definition, the nth-degree Taylor polynomial Pₙ satisfies Pₙ⁽ᵏ⁾(a) = f⁽ᵏ⁾(a) for k = 0, 1, …, n. For n = 2, this means P₂, P₂′, and P₂″ all agree with f and its derivatives at the center a = 3. Choice A is wrong because it claims P₂″(3) ≠ f″(3), which contradicts the defining property. Choice C confuses a finite polynomial approximation with exact equality, which holds only for the infinite series (and only within the interval of convergence). Choice D overstates the result—matching concavity at one point does not guarantee matching concavity everywhere. Choice E incorrectly suggests the second-derivative matching is conditional; for degree 2, it is guaranteed.

Question 2: BASIC CALCULATION

What is the third-degree Maclaurin polynomial for f(x) = eˣ?

A. 1 + x + x² + x³
B. 1 + x + x²/2 + x³/3
C. 1 + x + x²/2 + x³/6
D. x + x²/2 + x³/6
E. 1 + x + 2x² + 6x³

Show answer (C)

The correct answer is C. Since every derivative of eˣ is eˣ, we have f⁽ᵏ⁾(0) = e⁰ = 1 for all k. The Maclaurin polynomial is P₃(x) = 1/0! + x/1! + x²/2! + x³/3! = 1 + x + x²/2 + x³/6. Choice A omits the factorial denominators, dividing each term by 1 instead. Choice B uses k in the denominator rather than k!, confusing the Taylor series for eˣ with the series for −ln(1 − x). Choice D omits the constant term f(0)/0! = 1. Choice E multiplies by k! rather than dividing, producing the opposite error.

Question 3: INTERMEDIATE

Let f(x) = cos x. Which expression gives the coefficient of (x − π/2)⁴ in the Taylor polynomial for f centered at a = π/2?

A. f'''(π/2) / 3! = sin(π/2) / 6 = 1/6
B. −sin(π/2) / 4! = −1/24
C. sin(π/2) / 4! = 1/24
D. cos(π/2) / 3! = 0
E. f⁽⁴⁾(π/2) / 4! = cos(π/2) / 24 = 0

Show answer (E)

The correct answer is E. The coefficient of (x − a)⁴ in the Taylor polynomial centered at a is f⁽⁴⁾(a)/4!. For f(x) = cos x, the derivatives cycle with period 4: f'(x) = −sin x, f''(x) = −cos x, f'''(x) = sin x, f⁽⁴⁾(x) = cos x. Evaluating at a = π/2 gives f⁽⁴⁾(π/2) = cos(π/2) = 0, so the coefficient is 0/24 = 0. Choice A incorrectly uses the third derivative f'''(π/2) = sin(π/2) = 1 but divides by 3! instead of using the fourth derivative and 4!, giving 1/6. Choice B incorrectly identifies the fourth derivative as −sin x, giving −sin(π/2)/4! = −1/24. Choice C incorrectly uses the second derivative f''(π/2) = −cos(π/2) = 0 but divides by 4! instead of 2!, arriving at 0 with the wrong reasoning. Choice D uses the correct fourth derivative f⁽⁴⁾(π/2) = cos(π/2) = 0 but divides by 3! instead of 4!, which happens to give 0 but applies the wrong factorial. Only E correctly identifies f⁽⁴⁾(x) = cos x and evaluates f⁽⁴⁾(π/2)/4! = cos(π/2)/24 = 0 with the proper formula.

Question 4: APPLIED

5 pts

Let f(x) = √x. (a) Find the third-degree Taylor polynomial P₃(x) for f centered at a = 4. (b) Use P₃ to approximate √4.1 and express your answer as a fraction. (c) Use the Lagrange error bound to find an upper bound on |f(4.1) − P₃(4.1)|. (d) Is your approximation in (b) an overestimate or underestimate of √4.1? Justify your answer using the sign of the first omitted term or concavity.

Show answer

Point 1 (Derivatives and evaluation, 1 pt): f(x) =

x^{1/2}

, f′(x) = (1/2)

x^{−1/2}

, f″(x) = (−1/4)

x^{−3/2}

, f‴(x) = (3/8)

x^{−5/2}

. At a = 4: f(4) = 2, f′(4) = 1/4, f″(4) = −1/32, f‴(4) = 3/256. Point 2 (Polynomial assembly, 1 pt): P₃(x) = 2 + (1/4)(x − 4) + (−1/32)/2! · (x − 4)² + (3/256)/3! · (x − 4)³ = 2 + (1/4)(x − 4) − (1/64)(x − 4)² + (1/512)(x − 4)³. Point 3 (Approximation as exact fraction, 1 pt): P₃(4.1) = 2 + (1/4)(0.1) − (1/64)(0.01) + (1/512)(0.001) = 2 + 1/40 − 1/6400 + 1/512000. Finding a common denominator of 512000: 2 = 1024000/512000, 1/40 = 12800/512000, 1/6400 = 80/512000, 1/512000 = 1/512000. So P₃(4.1) = (1024000 + 12800 − 80 + 1)/512000 = 1036721/512000 ≈ 2.024845703125. Point 4 (Lagrange error bound, 1 pt): f⁽⁴⁾(x) = (−15/16)

x^{−7/2}

. On [4, 4.1], |f⁽⁴⁾(x)| is maximized at x = 4, giving |f⁽⁴⁾(4)| = 15/(16 · 128) = 15/2048. The bound is |R₃(4.1)| ≤ (15/2048)/4! · (0.1)⁴ = (15/2048) · (1/24) · (1/10000) = 15/491,520,000 ≈ 3.05 × 10⁻⁸. Point 5 (Overestimate/underestimate with justification, 1 pt): The first omitted term in the expansion is f⁽⁴⁾(4)/4! · (x − 4)⁴ = (−15/2048)/24 · (0.1)⁴ < 0, since f⁽⁴⁾(4) < 0. Because the next correction would subtract a positive quantity, P₃(4.1) is an overestimate of √4.1. Equivalently, √x is concave down for x > 0, and tangent-line-based approximations of concave-down functions overestimate.

Question 5: CRITICAL THINKING

3 pts

A function g is infinitely differentiable, and the Maclaurin series for g converges to g(x) for all x. The known values are g(0) = 1, g′(0) = 0, g″(0) = −2, and for n ≥ 3, g⁽ⁿ⁾(0) = g⁽ⁿ⁻²⁾(0). (a) Find the first four nonzero terms of the Maclaurin series for g. (b) Identify the function g and justify your answer.

Show answer

Point 1 (Generating derivative values using the recurrence, 1 pt): Using g⁽ⁿ⁾(0) = g⁽ⁿ⁻²⁾(0) for n ≥ 3, along with g(0) = 1, g′(0) = 0, and g″(0) = −2: | n | g⁽ⁿ⁾(0) | Source | |---|---------|--------| | 0 | 1 | Given | | 1 | 0 | Given | | 2 | −2 | Given | | 3 | g⁽¹⁾(0) = 0 | Recurrence | | 4 | g⁽²⁾(0) = −2 | Recurrence | | 5 | g⁽³⁾(0) = 0 | Recurrence | | 6 | g⁽⁴⁾(0) = −2 | Recurrence | All odd-order derivatives at 0 equal 0. For even order, g⁽ⁿ⁾(0) = −2 for all n ≥ 2 even, and g(0) = 1. Point 2 (First four nonzero terms written correctly, 1 pt): Since all odd-order derivatives are zero, the Maclaurin series contains only even powers of x. The first four nonzero terms are: • n = 0: g(0)/0! · x⁰ = 1 • n = 2: g″(0)/2! · x² = (−2/2)x² = −x² • n = 4: g⁽⁴⁾(0)/4! · x⁴ = (−2/24)x⁴ = −x⁴/12 • n = 6: g⁽⁶⁾(0)/6! · x⁶ = (−2/720)x⁶ = −x⁶/360 Therefore: g(x) = 1 − x² − (1/12)x⁴ − (1/360)x⁶ − ⋯ Point 3 (Identifying g(x) = 3 − 2cosh x with justification, 1 pt): The full Maclaurin series can be written as: g(x) = 1 + ∑(k=1 to ∞) (−2)/(2k)! · x²ᵏ = 1 − 2 ∑(k=1 to ∞) x²ᵏ/(2k)! Recall that cosh x = ∑(k=0 to ∞) x²ᵏ/(2k)! = 1 + ∑(k=1 to ∞) x²ᵏ/(2k)!, so ∑(k=1 to ∞) x²ᵏ/(2k)! = cosh x − 1. Substituting: g(x) = 1 − 2(cosh x − 1) = 1 − 2 cosh x + 2 = 3 − 2 cosh x Verification: g(0) = 3 − 2 cosh(0) = 3 − 2(1) = 1 ✓. Since the Maclaurin series for cosh x converges for all x, the series for g also converges for all x ✓.

SUMMARY

Summary

A Taylor polynomial of degree n centered at a is the unique polynomial Pₙ(x) = Σ (k = 0 to n) f⁽ᵏ⁾(a)/k! · (x − a)ᵏ whose value and first n derivatives match those of f at the center. When a = 0 the result is called a Maclaurin polynomial. The factorial scaling f⁽ᵏ⁾(a)/k! ensures the derivative-matching condition is satisfied, and the Lagrange remainder provides a rigorous bound on the approximation error.

For the AP exam, memorize the standard Maclaurin series for eˣ, sin x, cos x, 1/(1 − x), and ln(1 + x), and practice the three key manipulation techniques—substitution, term-by-term differentiation, and term-by-term integration—to derive new series quickly. Higher-degree polynomials extend the interval of accurate approximation, but accuracy always degrades as you move farther from the center. Mastering Taylor polynomials prepares you for the full theory of power series and convergence that forms the capstone of AP Calculus BC.