Taylor & Maclaurin Expansions — Working with Taylor and Maclaurin Series Expansions - Calculus 2

SECTION 1

Historical Context & Motivation

The desire to approximate complicated functions with simpler expressions stretches back centuries. Long before modern analysis formalized the notion of convergence, mathematicians in India, Europe, and beyond recognized that certain functions—trigonometric, exponential, logarithmic—could be represented as infinite sums of polynomial terms. The central insight was both elegant and practical: if a function is sufficiently smooth, its behavior near a chosen point is entirely determined by its derivatives at that point. This idea crystallized into what we now call Taylor series and, as a special case centered at zero, Maclaurin series.

c. 1400 • Kerala School of Mathematics

Madhava of Sangamagrama derives infinite series for sin x, cos x, and arctan x—anticipating European results by roughly 250 years. These expansions are transmitted through the works of his successors in southern India.

1715 • Brook Taylor's Methodus Incrementorum Directa

Taylor publishes his treatise containing the general formula for expanding a function about an arbitrary point a, establishing the series that bears his name.

1742 • Colin Maclaurin's Treatise of Fluxions

Maclaurin popularizes the special case of Taylor's formula centered at a = 0, providing systematic applications in geometry and mechanics that became standard across the curriculum.

1821 • Cauchy Formalizes Convergence

Augustin-Louis Cauchy provides rigorous definitions of convergence, remainder terms, and analytic functions, giving Taylor and Maclaurin series the theoretical foundation they needed.

The fundamental question these mathematicians addressed remains the guiding problem of this lesson: given a smooth function f(x), how can we construct a polynomial (potentially of infinite degree) that faithfully reproduces f near a specific point, and under what conditions does this polynomial converge to the original function? Answering this question gives us one of the most powerful tools in all of analysis—the ability to replace transcendental functions with sums we can compute, differentiate, and integrate term by term.

SECTION 2

Core Principles & Definitions

At the heart of Taylor and Maclaurin expansions lies a single, unifying idea: a function that is infinitely differentiable at a point can be reconstructed—at least locally—from the complete set of its derivatives at that point. Each successive derivative contributes a higher-order correction term, refining the approximation like adding finer details to a sketch. The following principles form the conceptual scaffolding for everything that follows.

Local Polynomial Approximation

Near a center point a, a smooth function f(x) can be approximated by a polynomial whose coefficients are determined by the derivatives f⁽ⁿ⁾(a). More derivatives yield a better local fit.

Taylor Series vs. Maclaurin Series

A Taylor series is centered at an arbitrary point a; a Maclaurin series is the special case where a = 0. Both share identical structure—the only difference is the choice of center.

Factorial Normalization

Each term is divided by n! because differentiating xⁿ repeatedly produces a factor of n!. The factorial in the denominator cancels this, ensuring the n-th coefficient equals the n-th derivative evaluated at a.

Radius of Convergence

A Taylor series converges within an interval centered at a, whose half-width R (the radius of convergence) may be finite or infinite. Outside this interval, the series diverges; at the endpoints, behavior must be checked individually.

Remainder and Error Bound

Truncating the series at degree n yields the n-th Taylor polynomial Tₙ(x). The remainder Rₙ(x) = f(x) − Tₙ(x) quantifies the approximation error and can be bounded using Taylor's theorem with remainder.

✦

KEY TAKEAWAY

Think of a Taylor series like a GPS recalculating your position. The zeroth-order term tells you where you are (the function value); the first derivative tells you the direction you're heading; the second derivative tells you how your path is curving. Each higher derivative is a finer correction—like adjusting for terrain, wind, or altitude—that collectively reconstruct the entire route. The more terms you keep, the longer the stretch of road your model faithfully predicts.

SECTION 3

Visual Explanation — Successive Polynomial Approximations

The diagram below illustrates how successive Taylor polynomials approximate f(x) = eˣ centered at a = 0 (i.e., Maclaurin polynomials). Each additional term captures more curvature information, extending the region over which the polynomial closely tracks the original function.

Each colored curve represents a Taylor polynomial of increasing degree approximating eˣ about a = 0. The pink dashed line (T₁) is the tangent line at the origin. The violet dashed curve (T₂) adds the quadratic term. The solid cyan curve (T₄) uses terms through x⁴/4! and hugs eˣ over a wider interval. Notice how all polynomials agree at the yellow center point a = 0.

Observe a critical pattern in the diagram: near x = 0, all of the Taylor polynomials are nearly indistinguishable from eˣ. As we move farther from the center, the lower-degree polynomials diverge from the true function, while higher-degree polynomials maintain accuracy over a broader domain. This visual captures the essence of local approximation—Taylor polynomials are most accurate near their center, and adding more terms extends the interval of faithful reproduction. For eˣ, the Maclaurin series has an infinite radius of convergence, meaning the series converges to eˣ for every real number x, given enough terms.

SECTION 4

Mathematical Framework

We now formalize the constructions previewed in the preceding sections. The two cornerstone formulas—the general Taylor series and its Maclaurin specialization—are accompanied by the remainder theorem that governs approximation error.

TAYLOR SERIES

f(x) = Σ (n=0 to ∞) [ f⁽ⁿ⁾(a) / n! ] · (x − a)ⁿ

f⁽ⁿ⁾(a) denotes the n-th derivative of f evaluated at the center a. The term (x − a)ⁿ measures displacement from the center, and n! normalizes the coefficient so that c_n = f⁽ⁿ⁾(a)/n!.

MACLAURIN SERIES (a = 0)

f(x) = Σ (n=0 to ∞) [ f⁽ⁿ⁾(0) / n! ] · xⁿ = f(0) + f′(0)x + f″(0)x²/2! + f‴(0)x³/3! + ⋯

This is simply the Taylor series with a = 0. Every textbook Maclaurin expansion—for eˣ, sin x, cos x, ln(1 + x), (1 + x)^α—follows directly from this formula.

TAYLOR REMAINDER (LAGRANGE FORM)

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

The remainder Rₙ(x) = f(x) − Tₙ(x) equals the next-order term evaluated at some c in the interval (a, x). If |Rₙ(x)| → 0 as n → ∞, the Taylor series converges to f(x).

RADIUS OF CONVERGENCE (RATIO TEST)

R = lim (n→∞) | cₙ / cₙ₊₁ | or equivalently R = 1 / lim sup (n→∞) |cₙ|^(1/n)

The series converges absolutely for |x − a| < R and diverges for |x − a| > R. At |x − a| = R (the boundary), convergence must be tested separately using comparison, alternating series, or other tests.

⚠

Common Pitfall

A function can be infinitely differentiable at a and still have a Taylor series that does not converge to the function. The classic counterexample is f(x) = e^−1/x² (with f(0) = 0), whose Maclaurin series is identically zero because every derivative at 0 vanishes, yet the function is nonzero for x ≠ 0. A function whose Taylor series equals the function on an interval is called analytic on that interval.

SECTION 5

Essential Maclaurin Series & Their Properties

A handful of Maclaurin series appear so frequently across mathematics, physics, and engineering that they should be committed to memory. The table below catalogs the most important expansions, their radii of convergence, and the general-term patterns that make them recognizable at a glance.

Function	Maclaurin Series	Radius R
eˣ	Σ xⁿ/n! = 1 + x + x²/2! + x³/3! + ⋯	∞
sin x	Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1)! = x − x³/3! + x⁵/5! − ⋯	∞
cos x	Σ (−1)ⁿ x²ⁿ/(2n)! = 1 − x²/2! + x⁴/4! − ⋯	∞
ln(1 + x)	Σ (−1)ⁿ⁺¹ xⁿ/n = x − x²/2 + x³/3 − ⋯	1
1/(1 − x)	Σ xⁿ = 1 + x + x² + x³ + ⋯	1
arctan x	Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1) = x − x³/3 + x⁵/5 − ⋯	1
(1 + x)ᵅ	Σ C(α,n) xⁿ, where C(α,n) = α(α−1)⋯(α−n+1)/n!	1 (for non-integer α)

Essential Maclaurin Series to Memorize

Horizontal bars represent convergence intervals on the real line. Series with R = ∞ extend in both directions (arrows). For finite R, open circles indicate divergence at the endpoint; filled circles indicate convergence. Note that ln(1 + x) converges at x = 1 (giving ln 2) but diverges at x = −1.

Several powerful techniques extend these base expansions to broader families of functions. Substitution allows you to replace x with a more complex expression—for instance, substituting −x² into the series for eˣ immediately yields the expansion for e^−x². Term-by-term differentiation and integration are valid inside the interval of convergence: differentiating the series for sin x term by term produces the series for cos x, and integrating the geometric series 1/(1 + t) yields the series for ln(1 + x). Multiplication and division of series (Cauchy products) generate expansions for products and quotients like eˣ sin x or tan x, though the algebra grows quickly.

SECTION 6

Worked Example — Finding a Maclaurin Series and Bounding Error

We will find the Maclaurin series for f(x) = ln(1 + x), then use the fourth-degree Maclaurin polynomial T₄(x) to approximate ln(1.5) and bound the error via the alternating series estimation theorem.

Maclaurin Expansion of ln(1 + x) and Error Bound for ln(1.5)

1
Step 1 — Compute Successive Derivatives at a = 0

Begin with f(x) = ln(1 + x). We compute: f(0) = ln 1 = 0, f′(x) = 1/(1 + x) so f′(0) = 1, f″(x) = −1/(1 + x)² so f″(0) = −1, f‴(x) = 2/(1 + x)³ so f‴(0) = 2, f⁽⁴⁾(x) = −6/(1 + x)⁴ so f⁽⁴⁾(0) = −6. In general, f⁽ⁿ⁾(0) = (−1)ⁿ⁺¹(n − 1)! for n ≥ 1.
2
Step 2 — Assemble the Maclaurin Series

Substituting into the Maclaurin formula f(x) = Σ f⁽ⁿ⁾(0)/n! · xⁿ, the n-th coefficient for n ≥ 1 is (−1)ⁿ⁺¹(n − 1)!/n! = (−1)ⁿ⁺¹/n. Therefore:

ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + ⋯ = Σ (n=1 to ∞) (−1)ⁿ⁺¹ xⁿ/n, for −1 < x ≤ 1
3
Step 3 — Form the Fourth-Degree Polynomial T₄(x)

Truncating after the x⁴ term gives T₄(x) = x − x²/2 + x³/3 − x⁴/4. To approximate ln(1.5), set x = 0.5 (since ln(1.5) = ln(1 + 0.5)).
4
Step 4 — Evaluate T₄(0.5)

T₄(0.5) = 0.5 − (0.5)²/2 + (0.5)³/3 − (0.5)⁴/4 = 0.5 − 0.125 + 0.04167 − 0.01563.

T₄(0.5) ≈ 0.40104
5
Step 5 — Bound the Error Using the Alternating Series Estimation Theorem

Since the series for ln(1 + x) is alternating (for x > 0), the error |ln(1.5) − T₄(0.5)| is bounded by the absolute value of the first omitted term: |a₅| = (0.5)⁵/5 = 0.03125/5 = 0.00625. The true value is ln(1.5) ≈ 0.40546, confirming that our approximation 0.40104 has error ≈ 0.00442, safely within the bound 0.00625.

|Error| ≤ 0.00625

SECTION 7

Strengths, Limitations, and When to Use Taylor Series

Aspect	Strengths	Limitations
Approximation Power	Can approximate any analytic function to arbitrary precision by taking enough terms. Error bounds are computable.	The approximation degrades as x moves away from the center a. Outside the radius of convergence, the series diverges entirely.
Computational Utility	Replaces transcendental functions with polynomials—easy to differentiate, integrate, multiply, and evaluate on calculators/computers.	Slow convergence for some functions near the boundary of convergence (e.g., ln(2) via the Maclaurin series for ln(1 + x) converges very slowly).
Theoretical Insight	Reveals hidden connections—for instance, Euler's formula eⁱˣ = cos x + i sin x emerges naturally from comparing Taylor coefficients.	Not all infinitely differentiable functions are analytic—smooth non-analytic functions exist where the Taylor series converges to the wrong value.
Generality	Works for any center a, allowing expansions to be tailored to the region of interest.	Requires all derivatives at a to exist and be computable—not always feasible for piecewise or implicitly defined functions.

Taylor Series: Strengths vs. Limitations

✦

KEY TAKEAWAY

Taylor series are the Swiss Army knife of analysis, but like any tool, they have a sweet spot. They excel when you need a polynomial stand-in for a complicated function near a specific point—computing definite integrals with no closed-form antiderivative, evaluating limits of indeterminate forms, or solving differential equations with series methods. However, when you need uniform accuracy over a large interval, techniques like Padé approximants, Fourier series, or Chebyshev polynomials may outperform truncated Taylor series.

SECTION 8

Connections to Advanced Theory

Taylor and Maclaurin series are not merely computational shortcuts—they are foundational structures that recur throughout higher mathematics. The ideas introduced here ramify into several advanced domains, and understanding these connections deepens your appreciation of why these series are central to the mathematical canon.

This Lesson's Concept	Advanced Extension
Maclaurin series of eˣ, sin x, cos x	Euler's formula eⁱˣ = cos x + i sin x unifies these series in the complex plane, providing the gateway to complex analysis and Fourier theory.
Taylor polynomial approximation Tₙ(x)	The theory of jets in differential geometry generalizes Taylor polynomials to functions on manifolds, encoding local behavior at higher orders.
Radius of convergence	In complex analysis, R equals the distance from the center to the nearest singularity in ℂ, explaining why 1/(1 + x²) has R = 1 despite being smooth on all of ℝ.
Lagrange remainder Rₙ(x)	Asymptotic analysis and the big-O notation formalize remainder estimates, forming the basis for computational complexity and numerical methods.
Term-by-term differentiation / integration	Power series solutions of ODEs (Frobenius method, Airy functions, Bessel functions) exploit this property to solve equations with no elementary closed-form solutions.

From Taylor Series to Advanced Mathematics

One particularly striking connection deserves emphasis. The function f(x) = 1/(1 + x²) is perfectly smooth for all real x, yet its Maclaurin series 1 − x² + x⁴ − x⁶ + ⋯ converges only for |x| < 1. From a purely real-variable standpoint this seems mysterious—where is the obstruction? The answer lies in complex analysis: the function has poles at x = ±i, which are distance 1 from the origin in the complex plane. The radius of convergence always equals the distance to the nearest complex singularity, a profound fact that only becomes visible when the domain is extended to ℂ. This illustrates how Taylor series naturally bridge real and complex analysis, and why a deeper study of series leads inevitably into the complex plane.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the Maclaurin series for cos x contains only even powers of x, while the Maclaurin series for sin x contains only odd powers. Relate your answer to properties of the functions themselves.

Show answer

The function cos x is an even function, meaning cos(−x) = cos x, so its Taylor series about x = 0 must also be even—only even-powered terms survive. Similarly, sin x is an odd function (sin(−x) = −sin x), so its series must be odd, consisting exclusively of odd powers. Formally, the n-th derivative of cos x at 0 vanishes whenever n is odd (since differentiating cos gives ±sin, which equals 0 at x = 0), and the n-th derivative of sin x at 0 vanishes whenever n is even. This parity constraint manifests as alternating zero and nonzero coefficients in the Maclaurin formula.

Question 2: BASIC CALCULATION

Find the Maclaurin series for f(x) = e^−x² up to and including the x⁶ term.

Show answer

Start from the known series eᵘ = 1 + u + u²/2! + u³/3! + ⋯ and substitute u = −x². This gives

e^{−x²}

= 1 + (−x²) + (−x²)²/2! + (−x²)³/3! + ⋯ = 1 − x² + x⁴/2 − x⁶/6 + ⋯. Through the x⁶ term, the result is

e^{−x²}

≈ 1 − x² + x⁴/2 − x⁶/6. Notice how substitution is far more efficient than computing six derivatives of

e^{−x²}

from scratch.

Question 3: INTERMEDIATE

Use a known Maclaurin series to evaluate the integral ∫₀^(0.5) (sin x)/x dx to four decimal places, justifying how many terms are needed.

Show answer

Since sin x = x − x³/3! + x⁵/5! − x⁷/7! + ⋯, dividing by x gives (sin x)/x = 1 − x²/3! + x⁴/5! − x⁶/7! + ⋯. Integrating term by term from 0 to 0.5: ∫₀^0.5 (sin x)/x dx = [x − x³/(3·3!) + x⁵/(5·5!) − x⁷/(7·7!) + ⋯] from 0 to 0.5. Evaluating: 0.5 − (0.125)/(18) + (0.03125)/(600) − (0.0078125)/(35280) + ⋯ = 0.5 − 0.006944 + 0.0000521 − 0.000000221 + ⋯ ≈ 0.4931. The series is alternating with decreasing terms, so the error after the x⁴ term is at most 0.0000521, and after the x⁶ term at most 2.21 × 10⁻⁷, which is well within four-decimal accuracy. Thus ∫₀^0.5 (sin x)/x dx ≈ 0.4931.

Question 4: APPLIED

In physics, the relativistic kinetic energy is K = mc²(γ − 1) where γ = (1 − v²/c²)^−1/2. Use the binomial series expansion of γ to show that K ≈ ½mv² for v ≪ c, and find the next correction term.

Show answer

Let u = v²/c². Then γ =

(1 − u)^{−1/2}

. Using the binomial series (1 + x)^α with x = −u and α = −1/2: γ = 1 + (1/2)u + (3/8)u² + ⋯ = 1 + v²/(2c²) + 3v⁴/(8c⁴) + ⋯. Therefore K = mc²(γ − 1) = mc²[v²/(2c²) + 3v⁴/(8c⁴) + ⋯] = ½mv² + (3/8)mv⁴/c² + ⋯. The leading term ½mv² is the classical kinetic energy, recovering Newtonian mechanics in the low-speed limit. The first relativistic correction is (3/8)mv⁴/c², which becomes significant only when v is an appreciable fraction of c.

Question 5: CRITICAL THINKING

Let f(x) = 1/(1 + x²). Its Maclaurin series is 1 − x² + x⁴ − x⁶ + ⋯ with radius of convergence R = 1, despite f being infinitely differentiable on all of ℝ. Explain why R = 1 using the complex-analytic viewpoint, and then find the Taylor series of f centered at a = 1 and determine its radius of convergence.

Show answer

From the complex-analytic viewpoint, f(z) = 1/(1 + z²) = 1/((z + i)(z − i)) has simple poles at z = ±i. The radius of convergence of a Taylor series centered at a equals the distance from a to the nearest singularity in ℂ. For a = 0, this distance is |0 − i| = 1, confirming R = 1. For the series centered at a = 1, the nearest singularity is either i or −i. The distances are |1 − i| = √2 and |1 − (−i)| = √2. Therefore the Taylor series about a = 1 has R = √2. To find the series, write f(x) = 1/(1 + x²) and compute f(1) = 1/2, f′(1) = −2x/(1+x²)²|_{x=1} = −1/2, and continue. Alternatively, use partial fractions and geometric series after rewriting in terms of (x − 1). The key insight is that the real-line smoothness of f is irrelevant—complex singularities constrain convergence, and shifting the center changes R because it changes the distance to the poles at ±i.

SUMMARY

Summary — Taylor & Maclaurin Series Expansions

Summary

A Taylor series represents a smooth function f(x) as the infinite sum Σ f⁽ⁿ⁾(a)/n! · (x − a)ⁿ, with the Maclaurin series being the special case a = 0. Each term encodes derivative information at the center: the constant term gives the function's value, the linear term its slope, the quadratic term its concavity, and so on. The series converges within a radius of convergence R determined by the ratio or root test (or, from the complex viewpoint, the distance to the nearest singularity in ℂ), and truncating at the n-th term produces the Taylor polynomial Tₙ(x) whose error is bounded by the Lagrange remainder or the alternating series estimation theorem.

Essential Maclaurin series—for eˣ, sin x, cos x, ln(1 + x), 1/(1 − x), arctan x, and the binomial series (1 + x)ᵅ—serve as building blocks from which more complex expansions are derived through substitution, term-by-term differentiation and integration, and series multiplication. These techniques transform Taylor series from a theoretical construct into a versatile computational tool for approximating integrals, evaluating limits, solving differential equations, and bridging real and complex analysis.