What is a Taylor series?

A Taylor series expresses a function as an infinite sum of power terms built from its derivatives at a single point. Truncating it gives a polynomial that approximates the function near that point, getting better as you include more terms.

What is the difference between a Taylor and a Maclaurin series?

A Maclaurin series is simply a Taylor series centred at x = 0. The Taylor series is the general case centred at any point x = a.

What is the radius of convergence?

It is the distance from the centre within which the series actually converges to the function. For example, the series for 1/(1−x) only converges for |x| < 1, so its radius of convergence is 1; outside that the polynomial diverges even though the function is still defined.

Why are Taylor series useful?

They let computers evaluate sin, cos, eˣ and logs; they linearise hard problems in physics and engineering; and they underpin numerical methods, error analysis and much of applied mathematics.

Taylor Series Explained (Interactive) — Polynomials That Become Functions

Q: Why are Taylor series useful?

They let computers evaluate sin, cos, eˣ and logs; they linearise hard problems in physics and engineering; and they underpin numerical methods, error analysis and much of applied mathematics.

How does a calculator find $\sin(1.2)$ or $e^{0.5}$ ? It doesn’t “know” them — it adds up a polynomial. A Taylor series rebuilds almost any smooth function as an infinite sum of powers of $x$ , using only the function’s derivatives at a single point. Add terms one at a time and watch a plain polynomial mould itself into the curve.

The big idea

Suppose you want to copy a function near $x=0$ using a polynomial. Match the value there and you get a horizontal line at the right height. Also match the slope and you get the tangent line — better. Match the curvature (second derivative) too and a parabola hugs the curve more closely. Keep matching higher derivatives and the polynomial clings to the function over a wider and wider range.

A Taylor series is the polynomial that agrees with a function’s value and all its derivatives at one point.

Add terms, watch it fit

Slide the number of terms up and watch the gold approximation snap onto the navy curve. For $\sin x$ , $\cos x$ and $e^x$ the fit just keeps improving everywhere. For $1/(1-x)$ , something different happens — keep an eye on what occurs past $x=1$ .

InteractiveTaylor polynomials converging

Loading interactive…

More terms → a closer fit. The 1/(1−x) example diverges outside |x| < 1, however many terms you add.

Text description ↓

An interactive plot comparing a function (navy) with its Maclaurin polynomial (gold) as the number of terms increases. Maclaurin series used:

sin x = x − x³/3! + x⁵/5! − … (converges for all x)
cos x = 1 − x²/2! + x⁴/4! − … (converges for all x)
eˣ = 1 + x + x²/2! + x³/3! + … (converges for all x)
1/(1−x) = 1 + x + x² + x³ + … (converges only for |x| < 1)

The Taylor formula

Centred at a point $a$ , the Taylor series of $f$ is

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

Each coefficient is a derivative divided by a factorial. The $n!$ is what makes it work: differentiating $(x-a)^n$ exactly $n$ times produces $n!$ , which the coefficient cancels, so the $n$ -th derivative of the series matches $f^{(n)}(a)$ perfectly.

When it works — and when it fails

A truncated Taylor series is only an approximation, and it’s only trustworthy within the radius of convergence: the distance from the centre inside which the infinite sum actually equals the function. Outside it, adding terms makes things worse, not better.

The geometric series $1+x+x^2+\cdots=\frac{1}{1-x}$ is the classic warning. It converges only for $|x|<1$ . At $x=1$ the function blows up, and the polynomial can never follow it there — which is exactly what you saw diverging in the interactive.

Worked examples

1Worked example — deriving the series for eˣ

Every derivative of $e^x$ is $e^x$ , and $e^0=1$ , so every coefficient $f^{(n)}(0)=1$ :

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

2Worked example — approximating sin(0.2)

Using $\sin x \approx x - \tfrac{x^3}{6}$ with $x=0.2$ :

0.2 - \frac{0.2^3}{6} = 0.2 - 0.001333\ldots \approx 0.19867.

The true value is $0.198669\ldots$ — two terms already give five-figure accuracy near 0.

Your turn

Write down the first three non-zero terms of the Maclaurin series for $\cos x$ .

Show the answer ↓

$\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$ , i.e. $1 - \tfrac{x^2}{2} + \tfrac{x^4}{24}$ .

The big idea

A Taylor series is the polynomial that agrees with a function’s value and all its derivatives at one point.

Add terms, watch it fit

InteractiveTaylor polynomials converging

Loading interactive…

More terms → a closer fit. The 1/(1−x) example diverges outside |x| < 1, however many terms you add.

Text description ↓

An interactive plot comparing a function (navy) with its Maclaurin polynomial (gold) as the number of terms increases. Maclaurin series used:

sin x = x − x³/3! + x⁵/5! − … (converges for all x)
cos x = 1 − x²/2! + x⁴/4! − … (converges for all x)
eˣ = 1 + x + x²/2! + x³/3! + … (converges for all x)
1/(1−x) = 1 + x + x² + x³ + … (converges only for |x| < 1)

The Taylor formula

Centred at a point $a$ , the Taylor series of $f$ is

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

When it works — and when it fails

Worked examples

1Worked example — deriving the series for eˣ

Every derivative of $e^x$ is $e^x$ , and $e^0=1$ , so every coefficient $f^{(n)}(0)=1$ :

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

2Worked example — approximating sin(0.2)

Using $\sin x \approx x - \tfrac{x^3}{6}$ with $x=0.2$ :

0.2 - \frac{0.2^3}{6} = 0.2 - 0.001333\ldots \approx 0.19867.

The true value is $0.198669\ldots$ — two terms already give five-figure accuracy near 0.

Your turn

Write down the first three non-zero terms of the Maclaurin series for $\cos x$ .

Show the answer ↓

$\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$ , i.e. $1 - \tfrac{x^2}{2} + \tfrac{x^4}{24}$ .

Taylor Series, Visualised

The big idea

Add terms, watch it fit

The Taylor formula

When it works — and when it fails

Worked examples

Frequently asked questions

Keep exploring

Want a tutor to walk you through it?

Taylor Series, Visualised

The big idea

Add terms, watch it fit

The Taylor formula

When it works — and when it fails

Worked examples

Frequently asked questions

Keep exploring

Want a tutor to walk you through it?