Taylor and Maclaurin Series

Exponential: eˣ

R = ∞ (converges for all x)

All derivatives of eˣ are eˣ, so every coefficient is e⁰/n! = 1/n!. Convergent everywhere.

Sine: sin x

R = ∞ (converges for all x)

Only odd powers. Signs alternate. The pattern arises because even derivatives of sin x at 0 are 0.

Cosine: cos x

R = ∞ (converges for all x)

Only even powers. Signs alternate. Note that d/dx [sin x] = cos x matches — term-by-term differentiation confirms the series.

Geometric: 1/(1−x)

R = 1 (converges for |x| < 1; diverges at both endpoints)

The geometric series. Substitute x → −x to get 1/(1+x), or x → x² to get 1/(1−x²), and so on.

Natural Log: ln(1+x)

R = 1 (converges for −1 < x ≤ 1)

Derived by integrating 1/(1+x) = Σ (−1)ⁿxⁿ. Converges at x = 1 (alternating harmonic) but diverges at x = −1 (harmonic series).

Technique 1: Substitution

Replace the variable with any expression. The radius of convergence scales accordingly.

Technique 2: Term-by-Term Differentiation

Differentiate both sides. Valid inside the open interval of convergence. The radius R stays the same; re-check endpoints.

Technique 3: Term-by-Term Integration

Integrate both sides from 0 to x (or from a to x). Useful for deriving series for ln, arctan, and other antiderivatives.

Approximating Function Values

Calculators and computer systems use Taylor polynomials internally to evaluate eˣ, sin x, ln x, and every transcendental function. For hand computation, truncate the series to the precision you need.

Evaluating Limits via Series

Replace functions with their Taylor series and cancel. Often faster than repeated applications of L'Hôpital's rule.

Integrating Functions Without Antiderivatives

Some functions have no elementary antiderivative. Taylor series allow integration to any desired accuracy.

Example 1 — Find the Maclaurin series for f(x) = x·eˣ

Example 2 — Find the interval of convergence for Σ (−1)ⁿ xⁿ / (n · 3ⁿ)

Example 3 — Maclaurin series for arctan x via integration

Example 4 — How many terms of cos x are needed to approximate cos(0.5) within 10⁻⁶?

Example 5 — Taylor series for ln x centered at a = 1

What is a Taylor series and why is it useful?

A Taylor series represents a smooth function f(x) as an infinite polynomial centered at a point a: f(x) = f(a) + f'(a)(x−a) + f''(a)/2!(x−a)² + f'''(a)/3!(x−a)³ + … = Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ. Taylor series are useful because polynomials are easy to differentiate, integrate, and evaluate numerically. They allow us to approximate transcendental functions like sin x, eˣ, and ln x to any desired precision, and they simplify calculating limits by replacing complex expressions with their polynomial equivalents.

What is the difference between a Taylor series and a Maclaurin series?

A Maclaurin series is simply a Taylor series centered at a = 0. The general Taylor series centered at a is: Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ. When a = 0, this becomes: Σ f⁽ⁿ⁾(0)/n! · xⁿ, which is the Maclaurin series. The Maclaurin series is used when working near x = 0 and gives simpler expressions. Taylor series centered at other values of a are used when you need accuracy near x = a instead of near the origin.

What is the radius of convergence and how do you find it?

The radius of convergence R is the number such that a power series Σ cₙ(x−a)ⁿ converges absolutely for |x−a| < R and diverges for |x−a| > R. Find R using the Ratio Test: compute L = lim |cₙ₊₁/cₙ| as n→∞, then R = 1/L. If L = 0, then R = ∞ (converges everywhere). If L = ∞, then R = 0. The interval of convergence is (a−R, a+R); you must check the endpoints separately by substituting them into the series and testing convergence. Example: for Σ xⁿ/n!, R = ∞ since |cₙ₊₁/cₙ| = 1/(n+1) → 0.

What are the most important Maclaurin series to memorize?

The five essential Maclaurin series are: (1) eˣ = 1 + x + x²/2! + x³/3! + … = Σ xⁿ/n!, R = ∞. (2) sin x = x − x³/3! + x⁵/5! − x⁷/7! + … = Σ (−1)ⁿx^(2n+1)/(2n+1)!, R = ∞. (3) cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + … = Σ (−1)ⁿx^(2n)/(2n)!, R = ∞. (4) 1/(1−x) = 1 + x + x² + x³ + … = Σ xⁿ, R = 1. (5) ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + … = Σ (−1)ⁿ⁺¹xⁿ/n, R = 1. These five series can be manipulated by substitution, differentiation, and integration to derive hundreds of other series.

How do you find the Taylor series of a function using derivatives?

To find the Taylor series of f(x) centered at a: Step 1 — compute f(a), f'(a), f''(a), f'''(a), and so on. Step 2 — write the general term: cₙ = f⁽ⁿ⁾(a) / n!. Step 3 — assemble the series: f(x) = Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ. Example: for f(x) = eˣ at a = 0, all derivatives equal eˣ, so f⁽ⁿ⁾(0) = 1 for all n. Thus eˣ = Σ xⁿ/n! = 1 + x + x²/2! + x³/3! + … . Recognize patterns: for sin x, odd-degree terms alternate in sign; for cos x, even-degree terms alternate.

How do you manipulate known series instead of computing derivatives from scratch?

Three powerful techniques avoid repeated differentiation: (1) Substitution — replace x with another expression. For example, since eˣ = Σ xⁿ/n!, substituting x → −x² gives e^(−x²) = Σ (−1)ⁿx^(2n)/n!. (2) Differentiation — differentiate both sides of a known series term by term. Since d/dx [1/(1−x)] = 1/(1−x)², differentiating Σ xⁿ gives Σ n·xⁿ⁻¹ = 1/(1−x)². (3) Integration — integrate both sides term by term. Since 1/(1+x) = Σ (−1)ⁿxⁿ, integrating gives ln(1+x) = Σ (−1)ⁿxⁿ⁺¹/(n+1). These techniques preserve the radius of convergence (integration/differentiation may require re-checking endpoints).

What is Taylor's Remainder Theorem and how do you use it?

Taylor's Remainder Theorem (Lagrange form) states that if you use the first n+1 terms of the Taylor series, the error Rₙ(x) satisfies |Rₙ(x)| ≤ M/(n+1)! · |x−a|ⁿ⁺¹, where M is a bound on |f⁽ⁿ⁺¹⁾| on the interval between a and x. To use it: Step 1 — identify the degree n of your approximation. Step 2 — find M by bounding |f⁽ⁿ⁺¹⁾| on the relevant interval. Step 3 — compute M/(n+1)! · |x−a|ⁿ⁺¹. If this is less than your desired error tolerance, your approximation is sufficiently accurate. For alternating series, the simpler alternating series estimation theorem often applies: the error is at most the absolute value of the first omitted term.

How do you use Taylor series to evaluate limits?

Taylor series simplify limits that produce indeterminate forms. Replace functions with their series expansions and cancel leading terms. Example: lim(x→0) (sin x − x)/x³. Replace sin x = x − x³/6 + x⁵/120 − … , so sin x − x = −x³/6 + higher-order terms. Then (sin x − x)/x³ = (−x³/6 + …)/x³ = −1/6 + higher-order terms → −1/6. This is faster than applying L'Hôpital's rule three times. General strategy: expand to just enough terms to see which ones survive the cancellation, then take the limit directly.

What is the interval of convergence and how do you determine it completely?

The interval of convergence is all x-values for which a power series converges. To find it completely: Step 1 — use the Ratio Test to find the radius of convergence R. The series converges absolutely on (a−R, a+R). Step 2 — test the endpoints x = a−R and x = a+R separately. Substitute each into the series and apply a convergence test (usually the Alternating Series Test or p-series comparison). Step 3 — write the final interval using [ for a convergent endpoint and ( for a divergent endpoint. Example: ln(1+x) = Σ (−1)ⁿ⁺¹xⁿ/n has R = 1. At x = 1: alternating harmonic series — converges. At x = −1: harmonic series — diverges. Interval of convergence: (−1, 1].