Maclaurin Series Calculator

How to Use This Calculator

1. Select a standard function (eˣ, sin, cos, ln(1+x), 1/(1-x), arctan, √(1+x)).
2. Set the number of terms and x value.
3. The partial sum, true value, and error are shown instantly.
4. Use Custom (numerical) tab for any custom function via numerical derivatives.
5. Use Convergence tab to compare partial sums at n=2, 4, 6, 8.

Formula

Example

Frequently Asked Questions

• What is a Maclaurin series? ▼
  A Maclaurin series is a special case of a Taylor series centered at x = 0. It represents a smooth function as an infinite polynomial: f(x) = Σₙ₌₀^∞ f⁽ⁿ⁾(0)/n! · xⁿ = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + … Each coefficient is determined by the nth derivative of f evaluated at 0. For example, for f(x) = eˣ: every derivative is eˣ, so f⁽ⁿ⁾(0) = 1, giving eˣ = 1 + x + x²/2! + x³/3! + … Maclaurin series enable efficient computation of transcendental functions like sin, cos, and exp.
• What is the Maclaurin series for eˣ? ▼
  The Maclaurin series for eˣ is eˣ = Σₙ₌₀^∞ xⁿ/n! = 1 + x + x²/2 + x³/6 + x⁴/24 + … This series converges for all real (and complex) x, making it one of the most powerful series in mathematics. For x = 1: e ≈ 1 + 1 + 0.5 + 0.1667 + 0.0417 + … ≈ 2.71828. The series converges rapidly for |x| ≤ 1 — 10 terms give over 7 correct digits. For large x, more terms are needed before convergence becomes apparent.
• What is the Maclaurin series for sin(x) and cos(x)? ▼
  The Maclaurin series for sin(x) contains only odd powers: sin(x) = x − x³/6 + x⁵/120 − x⁷/5040 + … = Σₙ₌₀^∞ (−1)ⁿ x^(2n+1)/(2n+1)!. For cos(x), only even powers: cos(x) = 1 − x²/2 + x⁴/24 − x⁶/720 + … = Σₙ₌₀^∞ (−1)ⁿ x^(2n)/(2n)!. Both converge for all x. Example: sin(0.5) ≈ 0.5 − 0.5³/6 + 0.5⁵/120 ≈ 0.5 − 0.02083 + 0.00026 ≈ 0.47943. Actual sin(0.5) ≈ 0.47943. ✓
• What is the alternating series error bound? ▼
  For a convergent alternating series that satisfies the alternating series test (terms decrease in magnitude toward zero), the error from stopping after n terms is at most the absolute value of the (n+1)th term: |error| ≤ |aₙ₊₁|. This is the alternating series estimation theorem. Example: sin(0.1) approximated by the first two terms: 0.1 − 0.1³/6 = 0.1 − 0.0001667 = 0.0998333. The next term is 0.1⁵/120 ≈ 0.0000000833, so the error is less than 8.33 × 10⁻⁸. This gives a rigorous error bound without needing the exact answer.
• What is the radius of convergence for ln(1+x)? ▼
  The Maclaurin series for ln(1+x) is Σₙ₌₁^∞ (−1)ⁿ⁺¹ xⁿ/n = x − x²/2 + x³/3 − x⁴/4 + … The radius of convergence is R = 1, determined by the ratio test: the series converges for −1 < x ≤ 1 (it converges at x = 1 by the alternating series test, giving ln(2) ≈ 1 − 1/2 + 1/3 − 1/4 + …, and diverges at x = −1 which would be ln(0) = −∞). For |x| > 1, the series diverges. This means the Maclaurin series for ln(1+x) cannot approximate ln(3) directly — a different expansion centered elsewhere is needed.

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