Taylor series, explained

By David Mashiah · 01.11.2026

A Taylor series rebuilds a function from its derivatives at one point, as a polynomial. Cut it off after a few terms and you get an approximation that is very good near that point. It is how calculators evaluate sin and e^x, and the quickest way through many limits.

f (x) = n = 0 \sum \infty \frac{f ^{(n)} ( a )}{n !} (x - a)^{n}

e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots

sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots, cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots

ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots, \frac{1}{1 - x} = 1 + x + x^{2} + \dots (∣ x ∣ < 1)

R_{N} (x) = \frac{f ^{(N + 1)} ( ξ )}{( N + 1 )!} (x - a)^{N + 1}

The idea

Pick a point a. The Taylor polynomial of degree N is the polynomial whose value and first N derivatives at a match the function’s. Each extra term matches one more derivative, so the fit stays good further from a. Around a = 0 it is also called a Maclaurin series.

How accurate is it

The error after degree N is the Lagrange remainder: the next derivative at some point ξ between a and x, times (x − a)^(N+1)/(N+1)!. The factorial grows fast, which is why a handful of terms is often enough near a.

For alternating series like sin and cos, there is a simpler bound: the error is smaller than the first term you left out.

Limits without L’Hôpital

Replace each function by its first terms and most 0/0 limits fall apart. sin x / x = (x − x³/6 + …)/x → 1. (1 − cos x)/x² = (x²/2 − …)/x² → 1/2. One expansion often replaces several rounds of L’Hôpital.

Where it converges

Some series work everywhere: e^x, sin and cos converge for every x. Others stop at a radius. 1/(1 − x) and ln(1 + x) only converge for |x| < 1, and adding terms outside that range makes the approximation worse, not better. Try x = 2 in the calculator.

Worked example

Approximate e^0.5 to degree 3: 1 + 0.5 + 0.125 + 0.0208 = 1.6458. The true value is 1.6487, so the error is about 0.0029. These are the calculator’s default inputs.

For sin 1 to degree 5: 1 − 1/6 + 1/120 = 0.84167, against the true 0.84147. The error, about 0.0002, is below the first omitted term, 1/5040 ≈ 0.0002.

Taylor polynomial calculator

P_N(x) around a = 0, against the exact value

Function Degree N x

P_N(x)–

Exact value–

Error–

Common mistakes

Forgetting the factorials, or dividing by n instead of n!.
Using a series outside its interval of convergence, such as ln(1 + x) at x = 3.
Mixing up degree and number of terms. The degree-5 polynomial of sin x has only three nonzero terms.