Taylor Series
Approximate a function by a polynomial built from its derivatives at one point. The Taylor (or Maclaurin) polynomial step by step, with exact coefficients, and a picture of how each extra term hugs the curve a little longer.
Why Taylor Series?
Polynomials are easy: they need only adding and multiplying. A Taylor polynomial copies a difficult function near one point by matching its value, its slope, its curvature and so on. Calculators use this idea to work out sin, cos and ex; physicists use it when they write sin θ ≈ θ for small angles.
Close to the centre the approximation is excellent; further away it drifts off, and more terms push that point further out. For some functions, like ex and sin x, the full series is exact everywhere; for others, like ln(1 + x), only within a radius of convergence.
The Approximation, Step by Step
PₙStep-by-Step Solution
Famous Series
Both are exact for every x when all terms are taken.
These only converge for −1 < x ≤ 1 and −1 < x < 1.
For some ξ between a and x (Lagrange form). Each extra term multiplies the error by roughly (x − a)/(n + 1).
Sources & Verification
- J. Stewart, Calculus: Early Transcendentals, 8th ed., Cengage, 2015 — Sec. 11.10 (Taylor and Maclaurin Series) and 11.11 (Applications of Taylor Polynomials).
Frequently Asked Questions
How do you find a Taylor polynomial?
Take the derivatives of f at the centre a, divide the k-th one by k! and multiply by (x − a)ᵏ: P(x) = f(a) + f′(a)(x − a) + f″(a)/2!·(x − a)² + … For eˣ at a = 0 every derivative is 1, so P₅(x) = 1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120.
What is the difference between a Taylor and a Maclaurin series?
A Maclaurin series is simply a Taylor series around a = 0. Around another centre the powers are (x − a)ᵏ instead of xᵏ; for √x around a = 4 that gives 2 + (x − 4)/4 − (x − 4)²/64 + (x − 4)³/512.
What is the Taylor series of sin x?
sin x = x − x³/3! + x⁵/5! − x⁷/7! + … (only odd powers, with alternating signs). cos x = 1 − x²/2! + x⁴/4! − … These series are exact for every x; with 7 terms P₇(2) = 0.907937 against sin 2 = 0.909297.
How accurate is a Taylor approximation?
Very accurate close to the centre, less further away. The error is about the size of the first term you left out: R = f⁽ⁿ⁺¹⁾(ξ)/(n + 1)!·(x − a)ⁿ⁺¹. For eˣ with 6 terms the error at x = 1 is 0.0016; each extra term divides it by roughly (n + 1).
Why is sin θ ≈ θ for small angles?
Because the Taylor series of sin θ starts with θ − θ³/6. For a small angle in radians θ³/6 is tiny: at θ = 0.1 the difference is less than 0.0002. That is the small-angle approximation used for pendulums and optics.
Where does a Taylor series converge?
Some series, like those of eˣ, sin x and cos x, converge for every x. Others only within a radius: 1/(1 − x) = 1 + x + x² + … and ln(1 + x) need −1 < x < 1 (ln(1 + x) also at x = 1). Outside that range adding more terms makes the approximation worse, not better.