← Back to course

Taylor & Maclaurin Series

You’ll be able to

Approximating a function with a polynomial

A Taylor polynomial approximates a function f near a center x = a by matching f’s value and its first n derivatives at a. The result is a polynomial that hugs the curve near a, getting better as you add terms. A Maclaurin series is simply a Taylor series centered at a = 0. Each coefficient uses a derivative of f divided by a factorial, so the polynomial "borrows" the local behavior of every derivative.

Taylor series centered at a
f(x) = Σ (n=0 to ∞) [ fⁿ(a) / n! ] (x − a)ⁿ = f(a) + f′(a)(x−a) + [f″(a)/2!](x−a)² + …
The nth coefficient is fⁿ(a)/n!. Setting a = 0 gives the Maclaurin series.

The three Maclaurin series to memorize

Three Maclaurin series appear constantly and should be automatic: eˣ = Σ xⁿ/n! = 1 + x + x²/2! + x³/3! + …; sin x = x − x³/3! + x⁵/5! − … (odd powers, alternating); and cos x = 1 − x²/2! + x⁴/4! − … (even powers, alternating). Knowing these cold lets you differentiate, integrate, and substitute to generate a huge family of related series without recomputing derivatives.

Building new series by substitution

The efficient way to find many series is to substitute into a known one rather than take derivatives. For e^(x²), replace x with x² in the eˣ series: e^(x²) = Σ (x²)ⁿ/n! = 1 + x² + x⁴/2! + …. Likewise the geometric series 1/(1 − x) = Σxⁿ (for |x| < 1) spawns 1/(1 + x²), arctan, and ln(1 + x) by substitution and term-by-term integration. This reuse is faster and far less error-prone than the derivative formula.

Worked example

Find the third-degree Maclaurin polynomial for f(x) = eˣ, and use it to approximate e^{0.1}.

  1. 1.All derivatives of eˣ are eˣ, and at x = 0 each equals 1, so every coefficient is 1/n!.
  2. 2.The degree-3 Maclaurin polynomial is P₃(x) = 1 + x + x²/2! + x³/3! = 1 + x + x²/2 + x³/6.
  3. 3.Substitute x = 0.1: 1 + 0.1 + (0.01)/2 + (0.001)/6.
  4. 4.Evaluate: 1 + 0.1 + 0.005 + 0.000167 ≈ 1.105167.
Answer: P₃(x) = 1 + x + x²/2 + x³/6, giving e^{0.1} ≈ 1.10517 — very close to the true value 1.105170918.
Tip

Do not grind out derivatives when a substitution will do. Need the series for x·sin x? Multiply the sin x series by x. Need e^(−x)? Replace x with −x. Reusing the memorized series is the intended, faster route.

Checkpoint

What is the Maclaurin series for eˣ?

Checkpoint

The Maclaurin series for f begins f(x) = 3 + 2x + 5x² + …. What is f″(0), the second derivative at 0?

On the exam

Coefficient and derivative are linked by a factorial: the coefficient of (x − a)ⁿ is fⁿ(a)/n!, so fⁿ(a) = n! × (that coefficient). AP loves asking for a high derivative at the center — read off the coefficient and multiply by n!.

Answer the 2 checkpoints as you read.

Sign in to save your progress