- What Is a Maclaurin Series?
- The Maclaurin Formula
- Deriving the Standard Series
- Complete Reference Table
- Interactive Approximation Diagram
- Manipulating Series: Substitution, Products, Composition
- Using Maclaurin Series to Evaluate Limits
- Error Bounds: Taylor's Remainder Theorem
- Worked Examples
- Common Mistakes
- 10-Question Quiz
- Continue in Series & Sequences
§ 01What Is a Maclaurin Series?
A Maclaurin series is a Taylor series centred at zero: an infinite polynomial whose coefficients are determined entirely by the derivatives of a function at the single point x = 0.
Colin Maclaurin (1698–1746) popularised this special case of Brook Taylor's more general result. While the Taylor series can be centred at any point a, the Maclaurin series is centred at a = 0, making it the simplest and most natural choice for the standard functions of calculus.
The key idea is this: if a function f is infinitely differentiable at 0, then it can be written as an infinite polynomial whose n-th coefficient is f⁽ⁿ⁾(0)/n!. The resulting series converges to f(x) on some interval around zero: and for functions like eˣ, sin x, and cos x, that interval turns out to be the entire real line.
§ 02The Maclaurin Formula
For any infinitely differentiable function f, the Maclaurin series is:
Each coefficient cn = f⁽ⁿ⁾(0)/n! is the n-th derivative of f evaluated at 0, divided by n factorial. The formula is derived by assuming f has a power series representation Σ cₙxⁿ and then differentiating both sides repeatedly, setting x = 0 each time to isolate the coefficients.
Deriving the coefficient formula
Suppose f(x) = c₀ + c₁x + c₂x² + c₃x³ + ··· Then:
- At x = 0: f(0) = c₀ ⟹ c₀ = f(0)
- Differentiate once, set x = 0: f'(0) = c₁ ⟹ c₁ = f'(0)
- Differentiate twice, set x = 0: f''(0) = 2c₂ ⟹ c₂ = f''(0)/2!
- In general: f⁽ⁿ⁾(0) = n! · cₙ ⟹ cₙ = f⁽ⁿ⁾(0)/n!
The Maclaurin polynomial PN(x) is the best polynomial approximation to f near x = 0 in the sense that all its derivatives up to order N match those of f at 0. As N → ∞, PN(x) → f(x) within the radius of convergence.
§ 03Deriving the Standard Series
The five essential Maclaurin series are derived by computing successive derivatives at 0 and applying the formula. It is worth doing these derivations at least once: the pattern of derivatives reveals deep structure.
eˣ — Exponential
All derivatives of eˣ equal eˣ, so f⁽ⁿ⁾(0) = e⁰ = 1 for all n.
sin x — Sine
Derivatives of sin x cycle with period 4: sin x, cos x, −sin x, −cos x, ... At x = 0: 0, 1, 0, −1, 0, 1, 0, −1, ... Only odd powers survive.
cos x — Cosine
Derivatives of cos x cycle identically but shifted: cos x, −sin x, −cos x, sin x, ... At x = 0: 1, 0, −1, 0, 1, 0, ... Only even powers survive.
ln(1 + x) — Natural Logarithm
f(x) = ln(1+x): f(0) = 0. f'(x) = 1/(1+x), f'(0) = 1. f''(x) = −1/(1+x)², f''(0) = −1. In general f⁽ⁿ⁾(0) = (−1)ⁿ⁺¹(n−1)! for n ≥ 1.
arctan x — Inverse Tangent
Most efficiently derived by integrating 1/(1+x²) = Σ(−1)ⁿx²ⁿ term by term (see Power Series §08).
§ 04Complete Reference Table
All standard Maclaurin series in one place. These should be memorised, or at minimum, the top four.
| Function | Maclaurin Series | First Terms | R |
|---|---|---|---|
| eˣ | Σ xⁿ/n! | 1 + x + x²/2 + x³/6 + ··· | ∞ |
| sin x | Σ (−1)ⁿx²ⁿ⁺¹/(2n+1)! | x − x³/6 + x⁵/120 − ··· | ∞ |
| cos x | Σ (−1)ⁿx²ⁿ/(2n)! | 1 − x²/2 + x⁴/24 − ··· | ∞ |
| ln(1+x) | Σ (−1)ⁿ⁺¹xⁿ/n | x − x²/2 + x³/3 − ··· | 1 |
| arctan x | Σ (−1)ⁿx²ⁿ⁺¹/(2n+1) | x − x³/3 + x⁵/5 − ··· | 1 |
| 1/(1−x) | Σ xⁿ | 1 + x + x² + x³ + ··· | 1 |
| 1/(1+x) | Σ (−1)ⁿxⁿ | 1 − x + x² − x³ + ··· | 1 |
| sinh x | Σ x²ⁿ⁺¹/(2n+1)! | x + x³/6 + x⁵/120 + ··· | ∞ |
| cosh x | Σ x²ⁿ/(2n)! | 1 + x²/2 + x⁴/24 + ··· | ∞ |
| (1+x)ᵏ | Σ C(k,n)xⁿ | 1 + kx + k(k−1)x²/2 + ··· | 1 (k∉ℕ) |
| arcsin x | Σ (2n)!x²ⁿ⁺¹/(4ⁿ(n!)²(2n+1)) | x + x³/6 + 3x⁵/40 + ··· | 1 |
| e−x² | Σ (−1)ⁿx²ⁿ/n! | 1 − x² + x⁴/2 − x⁶/6 + ··· | ∞ |
§ 05Interactive Approximation Diagram
Select a function and drag the slider to add more terms to the Maclaurin polynomial. Watch how the approximation improves near x = 0 and how it eventually covers a wider range. The gold curve is the true function; teal is the polynomial approximation PN(x).
PN(x) vs. true function
Notice that for eˣ, sin x, and cos x the polynomial approximation eventually fits everywhere: these have R = ∞. For ln(1+x), the approximation breaks down visibly for x > 1 or x ≤ −1 no matter how many terms you use, illustrating R = 1.
§ 06Manipulating Series — Substitution, Products, Composition
The real power of Maclaurin series lies not in deriving new ones from scratch, but in manipulating known series to reach new results in seconds.
Substitution
Replace x with any expression u(x) in a known series. The result converges wherever |u(x)| < R.
Multiplication by a polynomial
Division (long division of series)
To find series for f(x)/g(x), perform long division of the two power series, collecting coefficients of xⁿ. This works as long as g(0) ≠ 0.
Composition
§ 07Using Maclaurin Series to Evaluate Limits
Many indeterminate limits of the form 0/0 or ∞/∞ that resist L'Hôpital's Rule yield effortlessly to the Maclaurin series approach: expand numerator and denominator, cancel the leading terms, and read off the limit.
The series method is particularly useful when L'Hôpital's Rule requires many applications or when the limit involves compositions. For example, evaluating limx→0(ex − 1 − x)/x² is immediate from the series: (x + x²/2 + ··· − x)/x² = (x²/2 + ···)/x² → 1/2.
§ 08Error Bounds — Taylor's Remainder Theorem
When you truncate the Maclaurin series after N terms, the error is controlled by Taylor's Remainder Theorem (Lagrange form):
For alternating series that satisfy the Alternating Series Test, the simpler bound from the ASET applies: the error is at most the absolute value of the first omitted term. This is often easier to use in practice.
Example: bounding the error for cos x
Approximate cos(0.1) using P₄(x) = 1 − x²/2 + x⁴/24. The next term is |x⁶/6!| = (0.1)⁶/720 ≈ 1.4 × 10⁻¹⁰. The error is at most this: effectively zero for practical purposes.
§ 09Worked Examples
Twelve worked examples: deriving series, manipulating them, evaluating limits, computing integrals, and bounding errors.
- f(x) = eˣ. All derivatives: f⁽ⁿ⁾(x) = eˣ for all n.
- Evaluate at 0: f⁽ⁿ⁾(0) = e⁰ = 1 for all n ≥ 0.
- Coefficients: cₙ = f⁽ⁿ⁾(0)/n! = 1/n!.
- eˣ = Σ xⁿ/n! = 1 + x + x²/2! + x³/3! + ···
- Radius: Ratio Test gives lim|x/(n+1)| = 0, so R = ∞.
- Derivatives cycle: sin x, cos x, −sin x, −cos x, sin x, ...
- At x = 0: f(0)=0, f'(0)=1, f''(0)=0, f'''(0)=−1, f⁽⁴⁾(0)=0, f⁽⁵⁾(0)=1, ...
- Non-zero values occur at odd n: f⁽²ᵏ⁺¹⁾(0) = (−1)ᵏ.
- Coefficients c₂ₙ₊₁ = (−1)ⁿ/(2n+1)!, all even coefficients = 0.
- sin x = x − x³/3! + x⁵/5! − x⁷/7! + ···
- Start from eˣ = Σ xⁿ/n!. Substitute x → −x.
- e−x = Σ (−x)ⁿ/n! = Σ (−1)ⁿxⁿ/n!.
- = 1 − x + x²/2! − x³/3! + ···
- Same radius R = ∞.
- cos u = Σ (−1)ⁿu²ⁿ/(2n)!. Substitute u = x².
- cos(x²) = Σ (−1)ⁿ(x²)²ⁿ/(2n)! = Σ (−1)ⁿx⁴ⁿ/(2n)!
- = 1 − x⁴/2! + x⁸/4! − x¹²/6! + ···
- Expand eˣ: eˣ = 1 + x + x²/2 + x³/6 + ···
- Numerator: eˣ − 1 − x = x²/2 + x³/6 + ···
- Divide by x²: (x²/2 + x³/6 + ···)/x² = 1/2 + x/6 + ···
- As x → 0: limit = 1/2.
- cos x = 1 − x²/2 + x⁴/24 − ···
- 1 − cos x = x²/2 − x⁴/24 + ···
- (1 − cos x)/x² = 1/2 − x²/24 + ···
- As x → 0: limit = 1/2.
- sin x = x − x³/6 + x⁵/120 − ···
- sin(x)/x = 1 − x²/6 + x⁴/120 − ··· (divide each term by x)
- This is valid for x ≠ 0; the function extends continuously to x = 0 with value 1 (since lim sin x/x = 1).
- Integrate: ∫ sin(x)/x dx = C + x − x³/18 + x⁵/600 − ···
- = C + Σn=0∞ (−1)ⁿ x2n+1 / ((2n+1)·(2n+1)!)
- ln(1+u) = Σ (−1)ⁿ⁺¹uⁿ/n for |u| ≤ 1.
- Substitute u = −x: ln(1 − x) = Σ (−1)ⁿ⁺¹(−x)ⁿ/n = Σ (−1)²ⁿ⁺¹xⁿ/n = −Σ xⁿ/n.
- ln(1 − x) = −x − x²/2 − x³/3 − x⁴/4 − ···
- Valid for −1 ≤ x < 1 (note: the inequality flips at x = 1 since ln(0) = −∞).
- The remainder after N terms for eˣ at x = 1: |RN| ≤ e/(N+1)! (since |f⁽ᴺ⁺¹⁾(t)| ≤ e for all t ∈ [0,1]).
- We need e/(N+1)! < 0.0001. Since e < 3: need 3/(N+1)! < 0.0001, i.e. (N+1)! > 30000.
- 8! = 40320 > 30000, so N+1 = 8, N = 7.
- Check: P₇(1) = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720 + 1/5040 ≈ 2.71825.
- True e ≈ 2.71828: error ≈ 0.00003 < 0.0001. ✓
- eˣ = Σ xⁿ/n!. Multiply by x:
- x·eˣ = x·Σ xⁿ/n! = Σ xn+1/n!.
- Re-index (let m = n+1): = Σm=1∞ xm/(m−1)!
- First terms: x + x²/1! + x³/2! + x⁴/3! + ···= x + x² + x³/2 + x⁴/6 + ···
- eiπ = cos π + i sin π (Euler's formula, derived from Maclaurin series of eˣ with x = iπ).
- cos π = −1, sin π = 0.
- eiπ = −1 + i·0 = −1.
- Therefore eiπ + 1 = −1 + 1 = 0. ✓
- Use the identity sin²x = (1 − cos 2x)/2.
- cos 2x = 1 − (2x)²/2! + (2x)⁴/4! − (2x)⁶/6! + ···= 1 − 2x² + 2x⁴/3 − 4x⁶/45 + ···
- 1 − cos 2x = 2x² − 2x⁴/3 + 4x⁶/45 − ···
- sin²x = (1 − cos 2x)/2 = x² − x⁴/3 + 2x⁶/45 − ···
§ 10Common Mistakes
§ 1110-Question Quiz
Select one answer per question and click Check All Answers.