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Maclaurin
Series

Power series centred at zero, the universal language for approximating, integrating, and understanding every standard function in calculus.

Definition & Formula eˣ · sin x · cos x ln(1+x) · arctan x Binomial Series Error Estimation Limits via Series
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§ 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.

Why centre at zero? Most functions in calculus are defined and well-behaved at x = 0, making it the natural base point. Series centred at 0 are also the easiest to work with algebraically — substituting x = 0 into the series immediately gives f(0) = c₀ with no extra arithmetic.

§ 02The Maclaurin Formula

For any infinitely differentiable function f, the Maclaurin series is:

Maclaurin Series — Definition
f(x) = Σn=0 f(n)(0)/n! · xⁿ = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ···

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 of Degree N (partial sum)
PN(x) = Σn=0N f(n)(0)/n! · xⁿ = f(0) + f'(0)x + ··· + f(N)(0)xᴺ/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.

Maclaurin Series for eˣ
eˣ = Σn=0 xⁿ/n! = 1 + x + x²/2! + x³/3! + x⁴/4! + ··· Valid for all x ∈ ℝ  (R = ∞)

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.

Maclaurin Series for sin x
sin x = Σn=0 (−1)ⁿ x2n+1/(2n+1)! = x − x³/3! + x⁵/5! − x⁷/7! + ··· Valid for all x ∈ ℝ  (R = ∞)

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.

Maclaurin Series for cos x
cos x = Σn=0 (−1)ⁿ x2n/(2n)! = 1 − x²/2! + x⁴/4! − x⁶/6! + ··· Valid for all x ∈ ℝ  (R = ∞)

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.

Maclaurin Series for ln(1 + x)
ln(1+x) = Σn=1 (−1)n+1 xⁿ/n = x − x²/2 + x³/3 − x⁴/4 + ··· Valid for −1 < x ≤ 1  (R = 1)

arctan x — Inverse Tangent

Most efficiently derived by integrating 1/(1+x²) = Σ(−1)ⁿx²ⁿ term by term (see Power Series §08).

Maclaurin Series for arctan x
arctan x = Σn=0 (−1)ⁿ x2n+1/(2n+1) = x − x³/3 + x⁵/5 − x⁷/7 + ··· Valid for −1 ≤ x ≤ 1  (R = 1)
Euler's formula from the Maclaurin series Substitute ix into the series for eˣ (where i = √−1): eix = 1 + ix − x²/2! − ix³/3! + x⁴/4! + ··· Separate real and imaginary parts to recover exactly the series for cos x and i·sin x. This gives Euler's formula: eix = cos x + i sin x, and at x = π, the famous identity e + 1 = 0.

§ 04Complete Reference Table

All standard Maclaurin series in one place. These should be memorised, or at minimum, the top four.

FunctionMaclaurin SeriesFirst TermsR
Σ xⁿ/n!1 + x + x²/2 + x³/6 + ···
sin xΣ (−1)ⁿx²ⁿ⁺¹/(2n+1)!x − x³/6 + x⁵/120 − ···
cos xΣ (−1)ⁿx²ⁿ/(2n)!1 − x²/2 + x⁴/24 − ···
ln(1+x)Σ (−1)ⁿ⁺¹xⁿ/nx − x²/2 + x³/3 − ···1
arctan xΣ (−1)ⁿx²ⁿ⁺¹/(2n+1)x − x³/3 + x⁵/5 − ···1
1/(1−x)Σ xⁿ1 + x + x² + x³ + ···1
1/(1+x)Σ (−1)ⁿxⁿ1 − x + x² − x³ + ···1
sinh xΣ x²ⁿ⁺¹/(2n+1)!x + x³/6 + x⁵/120 + ···
cosh xΣ x²ⁿ/(2n)!1 + x²/2 + x⁴/24 + ···
(1+x)ᵏΣ C(k,n)xⁿ1 + kx + k(k−1)x²/2 + ···1 (k∉ℕ)
arcsin xΣ (2n)!x²ⁿ⁺¹/(4ⁿ(n!)²(2n+1))x + x³/6 + 3x⁵/40 + ···1
e−x²Σ (−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).

Interactive · Maclaurin Polynomial Approximation

PN(x) vs. true function

Function
Degree N
3
PN(1)
True f(1)

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.

Example — e−x² by substitution
eˣ = Σ xⁿ/n!  ⟹  e−x² = Σ (−x²)ⁿ/n! = Σ (−1)ⁿx2n/n! = 1 − x² + x⁴/2! − x⁶/3! + ···

Multiplication by a polynomial

Example — x·sin x by multiplication
x·sin x = x·(x − x³/6 + x⁵/120 − ···) = x² − x⁴/6 + x⁶/120 − ···

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.

Example — tan x = sin x / cos x (first three terms)
tan x = (x − x³/6 + x⁵/120 − ···)/(1 − x²/2 + x⁴/24 − ···) = x + x³/3 + 2x⁵/15 + ···

Composition

Example — esin x (first three terms)
Let u = sin x = x − x³/6 + ··· eᵘ = 1 + u + u²/2 + ··· = 1 + (x − x³/6) + (x − ···)²/2 + ··· = 1 + x + x²/2 − x³/3 + ···

§ 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.

Classic Limit via Series
limx→0 (sin x)/x = lim (x − x³/6 + ···)/x = lim (1 − x²/6 + ···) = 1
Another Classic
limx→0 (1 − cos x)/x² = lim (x²/2 − x⁴/24 + ···)/x² = lim (1/2 − x²/24 + ···) = 1/2

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.

When to use series vs. L'Hôpital's Rule Use series when: (1) the limit involves a composition like sin(x²) or ecos x − 1, (2) L'Hôpital's Rule would need to be applied more than twice, or (3) you need the limit of a ratio where both numerator and denominator have the same leading-order term. In these cases, series expansion is faster and less error-prone.

§ 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):

Lagrange Remainder
f(x) = PN(x) + RN(x) |RN(x)| ≤ M · |x|N+1 / (N+1)! where M = max |f(N+1)(t)| for t between 0 and x

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.

The series must converge at x to use the remainder bound Taylor's Remainder Theorem applies only within the interval of convergence. Outside the radius R, the partial sums diverge and the "error" grows without bound — the series is useless there regardless of N.
✦   ✦   ✦

§ 09Worked Examples

Twelve worked examples: deriving series, manipulating them, evaluating limits, computing integrals, and bounding errors.

Example 01 — Derive the series for eˣ from scratch
Use the Maclaurin formula to derive the series for f(x) = eˣ.
  1. f(x) = eˣ. All derivatives: f⁽ⁿ⁾(x) = eˣ for all n.
  2. Evaluate at 0: f⁽ⁿ⁾(0) = e⁰ = 1 for all n ≥ 0.
  3. Coefficients: cₙ = f⁽ⁿ⁾(0)/n! = 1/n!.
  4. eˣ = Σ xⁿ/n! = 1 + x + x²/2! + x³/3! + ···
  5. Radius: Ratio Test gives lim|x/(n+1)| = 0, so R = ∞.
eˣ = Σn=0 xⁿ/n!,   R = ∞
Example 02 — Derive the series for sin x
Use the Maclaurin formula to find the series for sin x.
  1. Derivatives cycle: sin x, cos x, −sin x, −cos x, sin x, ...
  2. At x = 0: f(0)=0, f'(0)=1, f''(0)=0, f'''(0)=−1, f⁽⁴⁾(0)=0, f⁽⁵⁾(0)=1, ...
  3. Non-zero values occur at odd n: f⁽²ᵏ⁺¹⁾(0) = (−1)ᵏ.
  4. Coefficients c₂ₙ₊₁ = (−1)ⁿ/(2n+1)!, all even coefficients = 0.
  5. sin x = x − x³/3! + x⁵/5! − x⁷/7! + ···
sin x = Σ (−1)ⁿx²ⁿ⁺¹/(2n+1)!,   R = ∞
Example 03 — Series for e−x by substitution
Find the Maclaurin series for e−x.
  1. Start from eˣ = Σ xⁿ/n!. Substitute x → −x.
  2. e−x = Σ (−x)ⁿ/n! = Σ (−1)ⁿxⁿ/n!.
  3. = 1 − x + x²/2! − x³/3! + ···
  4. Same radius R = ∞.
e−x = Σ (−1)ⁿxⁿ/n!,   R = ∞
Example 04 — Series for cos(x²) by substitution
Find the Maclaurin series for cos(x²).
  1. cos u = Σ (−1)ⁿu²ⁿ/(2n)!. Substitute u = x².
  2. cos(x²) = Σ (−1)ⁿ(x²)²ⁿ/(2n)! = Σ (−1)ⁿx⁴ⁿ/(2n)!
  3. = 1 − x⁴/2! + x⁸/4! − x¹²/6! + ···
cos(x²) = Σ (−1)ⁿx4n/(2n)!,   R = ∞
Example 05 — Evaluate a limit using series
Evaluate limx→0 (eˣ − 1 − x)/x².
  1. Expand eˣ: eˣ = 1 + x + x²/2 + x³/6 + ···
  2. Numerator: eˣ − 1 − x = x²/2 + x³/6 + ···
  3. Divide by x²: (x²/2 + x³/6 + ···)/x² = 1/2 + x/6 + ···
  4. As x → 0: limit = 1/2.
lim = 1/2
Example 06 — Evaluate lim (1 − cos x)/x²
Compute limx→0 (1 − cos x)/x².
  1. cos x = 1 − x²/2 + x⁴/24 − ···
  2. 1 − cos x = x²/2 − x⁴/24 + ···
  3. (1 − cos x)/x² = 1/2 − x²/24 + ···
  4. As x → 0: limit = 1/2.
lim = 1/2
Example 07 — Integrate sin(x)/x using series
Express ∫ sin(x)/x dx as a power series.
  1. sin x = x − x³/6 + x⁵/120 − ···
  2. sin(x)/x = 1 − x²/6 + x⁴/120 − ··· (divide each term by x)
  3. This is valid for x ≠ 0; the function extends continuously to x = 0 with value 1 (since lim sin x/x = 1).
  4. Integrate: ∫ sin(x)/x dx = C + x − x³/18 + x⁵/600 − ···
  5. = C + Σn=0 (−1)ⁿ x2n+1 / ((2n+1)·(2n+1)!)
∫ sin(x)/x dx = Σ (−1)ⁿx2n+1/((2n+1)(2n+1)!) + C
Example 08 — Series for ln(1 − x)
Find the Maclaurin series for ln(1 − x).
  1. ln(1+u) = Σ (−1)ⁿ⁺¹uⁿ/n for |u| ≤ 1.
  2. Substitute u = −x: ln(1 − x) = Σ (−1)ⁿ⁺¹(−x)ⁿ/n = Σ (−1)²ⁿ⁺¹xⁿ/n = −Σ xⁿ/n.
  3. ln(1 − x) = −x − x²/2 − x³/3 − x⁴/4 − ···
  4. Valid for −1 ≤ x < 1 (note: the inequality flips at x = 1 since ln(0) = −∞).
ln(1−x) = −Σn=1 xⁿ/n,   −1 ≤ x < 1
Example 09 — Approximate e to 4 decimal places
How many terms of the series for eˣ are needed to compute e = e¹ to within 0.0001?
  1. 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]).
  2. We need e/(N+1)! < 0.0001. Since e < 3: need 3/(N+1)! < 0.0001, i.e. (N+1)! > 30000.
  3. 8! = 40320 > 30000, so N+1 = 8, N = 7.
  4. Check: P₇(1) = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720 + 1/5040 ≈ 2.71825.
  5. True e ≈ 2.71828: error ≈ 0.00003 < 0.0001. ✓
7 terms (n = 0 to 6) suffice; e ≈ 2.7183
Example 10 — Series for x·eˣ
Find the Maclaurin series for f(x) = x·eˣ.
  1. eˣ = Σ xⁿ/n!. Multiply by x:
  2. x·eˣ = x·Σ xⁿ/n! = Σ xn+1/n!.
  3. Re-index (let m = n+1): = Σm=1 xm/(m−1)!
  4. First terms: x + x²/1! + x³/2! + x⁴/3! + ···= x + x² + x³/2 + x⁴/6 + ···
x·eˣ = Σn=1 xⁿ/(n−1)!,   R = ∞
Example 11 — Euler's formula check
Show that substituting x = π into e = cos π + i sin π gives Euler's identity e + 1 = 0.
  1. e = cos π + i sin π (Euler's formula, derived from Maclaurin series of eˣ with x = iπ).
  2. cos π = −1, sin π = 0.
  3. e = −1 + i·0 = −1.
  4. Therefore e + 1 = −1 + 1 = 0. ✓
e + 1 = 0 — confirmed
Example 12 — Series for sin²x using double-angle identity
Find the first four non-zero terms of the Maclaurin series for sin²x.
  1. Use the identity sin²x = (1 − cos 2x)/2.
  2. cos 2x = 1 − (2x)²/2! + (2x)⁴/4! − (2x)⁶/6! + ···= 1 − 2x² + 2x⁴/3 − 4x⁶/45 + ···
  3. 1 − cos 2x = 2x² − 2x⁴/3 + 4x⁶/45 − ···
  4. sin²x = (1 − cos 2x)/2 = x² − x⁴/3 + 2x⁶/45 − ···
sin²x = x² − x⁴/3 + 2x⁶/45 − ···

§ 10Common Mistakes

Mistake 1 — Using the series outside its radius of convergence The series for ln(1+x) is only valid for −1 < x ≤ 1. Plugging in x = 2 gives a divergent series — the result is meaningless, even if you compute many partial sums.
Mistake 2 — Wrong sign pattern for sin x and ln(1+x) sin x has (−1)ⁿ with odd powers x²ⁿ⁺¹; ln(1+x) has (−1)ⁿ⁺¹ with all powers xⁿ (starting n=1). Getting the sign factor or the power pattern wrong produces garbage coefficients.
Mistake 3 — Forgetting the factorial denominator cₙ = f⁽ⁿ⁾(0)/n!, not f⁽ⁿ⁾(0). The n! is always in the denominator. For eˣ, cₙ = 1/n!, not 1.
Mistake 4 — Computing limits without enough terms When evaluating lim (f(x) − g(x))/xⁿ, expand both f and g to at least n+1 terms. If you stop too early, the leading terms cancel and you get 0/0 instead of the true answer.
Mistake 5 — Confusing Maclaurin and Taylor series A Maclaurin series is centred at a = 0; a Taylor series centred at a ≠ 0 uses (x − a)ⁿ, not xⁿ. The coefficients are f⁽ⁿ⁾(a)/n!, not f⁽ⁿ⁾(0)/n!.

§ 1110-Question Quiz

Select one answer per question and click Check All Answers.

Score: 0 / 10
Q 01
The Maclaurin series coefficient cₙ is equal to:
Q 02
The Maclaurin series for eˣ starts:
Q 03
Which powers appear in the Maclaurin series for sin x?
Q 04
limx→0 (sin x)/x = ?
Q 05
The series for e−x² is obtained from eˣ by:
Q 06
The radius of convergence of the Maclaurin series for ln(1+x) is:
Q 07
Euler's formula e + 1 = 0 follows from substituting x = iπ into:
Q 08
limx→0 (1 − cos x)/x² = ?
Q 09
The first non-zero coefficient of the Maclaurin series for sin²x (using sin²x = (1−cos 2x)/2) is the coefficient of:
Q 10
The series for arctan x at x = 1 gives the formula:

§ 12Continue in Series & Sequences

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