Taylor Series Explained: Formula, Examples, and Applications

Taylor series allow us to represent complex functions as infinite sums of polynomial terms. This is one of the most powerful ideas in mathematics, enabling us to approximate functions like $e^x$ , $\sin x$ , and $\ln(1+x)$ to any desired accuracy.

This guide explains how Taylor and Maclaurin series work, how to compute their terms, and how to determine when a series converges.

The Taylor Series Formula

The Taylor series of a function $f(x)$ centred at $x = a$ is:

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

Each term uses the $n$ -th derivative of $f$ evaluated at $a$ , divided by $n!$ (n factorial), multiplied by $(x - a)^n$ .

Maclaurin Series

A Maclaurin series is simply a Taylor series centred at $a = 0$ :

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots

Worked Examples

Example 1: Maclaurin Series for e^x

Every derivative of $e^x$ is $e^x$ , and $e^0 = 1$ . So $f^{(n)}(0) = 1$ for all $n$ .

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

This series converges for all real values of $x$ .

Example 2: Maclaurin Series for sin(x)

The derivatives of $\sin x$ cycle: $\sin x, \cos x, -\sin x, -\cos x, \ldots$ Evaluating at $x = 0$ gives $0, 1, 0, -1, \ldots$

Only odd powers appear:

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

Example 3: Taylor Series for ln(x) about x = 1

We centre the series at $a = 1$ . The derivatives of $\ln x$ are:

$f(x) = \ln x$ , so $f(1) = 0$
$f'(x) = \frac{1}{x}$ , so $f'(1) = 1$
$f''(x) = -\frac{1}{x^2}$ , so $f''(1) = -1$
$f'''(x) = \frac{2}{x^3}$ , so $f'''(1) = 2$

Substituting into the Taylor formula:

\ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-1)^n}{n}

This series converges for $0 < x \leq 2$ .

Example 4: Approximating a Value

Use the Maclaurin series for $e^x$ to approximate $e^{0.5}$ using four terms.

e^{0.5} \approx 1 + 0.5 + \frac{(0.5)^2}{2} + \frac{(0.5)^3}{6} = 1 + 0.5 + 0.125 + 0.02083 = 1.6458

The true value is approximately 1.6487, so four terms give a very good approximation.

Convergence

Not every Taylor series converges for all $x$ . The radius of convergence $R$ tells you the interval where the series is valid. You can find it using the ratio test:

R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|

The series converges when $|x - a| < R$ and diverges when $|x - a| > R$ . At the boundary points, convergence must be checked individually.

Common Maclaurin Series

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ (converges for all $x$ )
$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ (converges for all $x$ )
$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$ (converges for all $x$ )
$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ (converges for $|x| < 1$ )

Try It Yourself

Explore Taylor and Maclaurin series expansions with our free Taylor Series Calculator. Enter any function and see its series expansion, term by term, with convergence information. Since Taylor series rely on repeated differentiation, our Derivative Calculator can help verify individual terms.