Taylor series calculator

Enter f(x), the centre a, the degree and a point x. The coefficients come from repeated symbolic derivatives at a, and the polynomial is compared with the real function at x.

Function f(x)Use x, numbers, + - * / ^, parentheses, pi, e and sin, cos, tan, exp, ln, log, sqrt, abs with brackets.

Centre a0 gives the Maclaurin series.

Degree nA whole number from 0 to 10.

Evaluate at xWhere to compare the polynomial with f.

Taylor polynomial1 + x + 0.5*x^2 + 0.16666667*x^3 + 0.041666667*x^4

Polynomial value at x: 2.708333333
f(x): 2.718281828
Polynomial minus f(x): -0.009948
First term left out: 0.008333

With centre 0 this is the Maclaurin polynomial.

Runs in your browser: nothing you enter is sent anywhere.

How to use it

Type f(x) and the centre a.
Choose the degree n, from 0 to 10.
Enter a point x to compare the polynomial with the real value.

How it works

The polynomial is P(x) = Σ f⁽ᵏ⁾(a) / k! · (x − a)ᵏ for k = 0 to n. Terms with a zero coefficient are left out of the printed text. P(x) is evaluated with Horner’s rule and compared with f(x) computed directly.

Worked example

f(x) = eˣ around 0, degree 4: P(x) = 1 + x + 0.5x² + 0.16666667x³ + 0.041666667x⁴. At x = 1, P(1) = 2.7083333 while e = 2.7182818, so the polynomial is 0.0099485 too small. The first term left out is 1/120 ≈ 0.0083333.

Tips

Raise the degree or move the centre closer to x to improve accuracy.
Even functions like cos(x) have only even powers; odd ones like sin(x) only odd powers.

Frequently asked questions

What is the difference between Taylor and Maclaurin?

A Maclaurin series is a Taylor series centred at 0. Set the centre to 0 to get it.

How are the coefficients found?

The coefficient of (x − a)^k is the k-th derivative of f at a divided by k!. The tool differentiates f symbolically k times and evaluates it at a.

How good is the approximation?

It is best near the centre and gets worse as x moves away. The “first term left out” row is the next term of the series at your x, a rough size of the error, not a guaranteed bound.

Why does it fail for ln(x) at 0?

ln is not defined at 0, so its derivatives there are not numbers and no Taylor polynomial around 0 exists. Use a centre such as 1.

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All tools in this category: Math ·Updated November 3, 2026