Taylor Series

Approximating a function near a chosen center with polynomials built from its derivatives.

Taylor polynomials match a function near the center point

T₅(x) = x - x³/3! + x⁵/5!

At x = 1, error is about 1.96e-4.

degree 5

Definition

A Taylor Series expands a function as an infinite polynomial centred at a point $a$ :

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

When $a = 0$ , this is called a Maclaurin series.

The key insight: if you know all the derivatives of $f$ at one point, you know $f$ everywhere (in its radius of convergence).

Essential Maclaurin series to memorise: $e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \qquad (\text{all } x)$

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

$\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \qquad (\text{all } x)$

Maclaurin series for eˣ

For $f(x) = e^x$ , every derivative $f^{(n)}(x) = e^x$ , so $f^{(n)}(0) = 1$ for all $n$ :

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$

At $x = 1$ : $e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \cdots \approx 2.71828\ldots$

Try it

Write the first four non-zero terms of the Maclaurin series for $\cos x$ . Use it to approximate $\cos(0.1)$ .

Solution

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

At $x = 0.1$ : $\cos(0.1) \approx 1 - \frac{0.01}{2} + \frac{0.0001}{24} = 1 - 0.005 + 0.00000417 \approx 0.99500$

The exact value is $\cos(0.1) \approx 0.99500$ — excellent accuracy with just three terms.

Related concepts

Needs first

Derivatives Series & Convergence

Limits

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