IB Analysis and Approaches HL · Differential calculus · Section 12 of 14

Taylor and Maclaurin expansions

3 worked examples · By Álvaro López

Key idea

A Taylor polynomial approximates a function near a point with an expression that uses only sums and products. If the point is $x = 0$ it is called Maclaurin. It is useful for estimating values without a calculator and for solving difficult limits. This is Higher Level content and appears in both Paper 1 and Paper 2.

The idea: copy the derivatives

To approximate $f$ near $x = a$ we look for a polynomial $P$ that agrees with $f$ at $a$ and also has the same derivatives there: $P (a) = f (a)$ , $P^{'} (a) = f^{'} (a)$ , $P^{''} (a) = f^{''} (a)$ , and so on up to order $n$ . If the function and the polynomial match this closely at $a$ (same value, same gradient, same curvature...), they also match at nearby points.

The difference between $f$ and the polynomial is the error or remainder. It can be bounded with Lagrange's formula: $R_{n} (x) = \frac{f ^{(n + 1)} ( c )}{( n + 1 )!} (x - a)^{n + 1}$ for some $c$ between $a$ and $x$ . The IB does not usually ask you for this formula, but it explains something important: the error is small when $x$ is close to $a$ and when $n$ is large.

Standard Maclaurin series you should know

These series are obtained by successive differentiation and are in the IB formula booklet.

Notice the patterns: $sin x$ has only odd powers (it is an odd function), $cos x$ only even powers (it is an even function), and the signs of both alternate. In the binomial series, with $p$ a positive integer, you obtain the binomial theorem and the series terminates.

Graph of sin x and its Maclaurin polynomials P1, P3 and P5: the higher the degree, the more closely they hug the curve near the origin

Limits with Maclaurin

If there is an indeterminate limit as $x \to 0$ , you can substitute each function by its expansion and simplify. The high-degree terms disappear and what remains is the quotient of the first relevant terms. It is an alternative to L'Hôpital, and often shorter when you would need to differentiate three times.

Worked examples

Example 1 Basic

Taylor expansion of a square root

Let $f (x) = x$ .

a) Find the Taylor polynomial of degree $2$ of $f$ at $x = 4$ .

b) Use it to approximate $4.2$ . Write the result as a fraction.

Solution

a) Calculate the derivatives at $a = 4$ :

$f (4) = 2$ ;
$f^{'} (x) = \frac{1}{2 x}$ , so $f^{'} (4) = \frac{1}{4}$ ;
$f^{''} (x) = - \frac{1}{4 x ^{3/2}}$ , so $f^{''} (4) = - \frac{1}{32}$ .

$P_{2} (x) = 2 + \frac{1}{4} (x - 4) + \frac{- 1/32}{2 !} (x - 4)^{2} = 2 + \frac{x - 4}{4} - \frac{( x - 4 ) ^{2}}{64} .$

b) With $x = 4.2$ we have $x - 4 = \frac{1}{5}$ : $P_{2} = 2 + \frac{1}{20} - \frac{1/25}{64} = 2 + \frac{1}{20} - \frac{1}{1600} = \frac{3200 + 80 - 1}{1600} = \frac{3279}{1600} = 2.049375.$ The real value is $4.2 \approx 2.049390$ : the error is a few millionths.

Example 2 Medium

Maclaurin of sin(3x) and a limit

a) Find the Maclaurin expansion of $sin (3 x)$ up to the term in $x^{5}$ .

b) Use the result to calculate $x \to 0 lim \frac{sin ( 3 x ) - 3 x}{x ^{3}}$ .

Solution

a) Start from $sin u = u - \frac{u ^{3}}{3 !} + \frac{u ^{5}}{5 !} - \dots$ and substitute $u = 3 x$ : $sin (3 x) = 3 x - \frac{27 x ^{3}}{6} + \frac{243 x ^{5}}{120} - \dots = 3 x - \frac{9}{2} x^{3} + \frac{81}{40} x^{5} - \dots$

b) The numerator is $sin (3 x) - 3 x = - \frac{9}{2} x^{3} + \frac{81}{40} x^{5} - \dots$ . Dividing by $x^{3}$ : $\frac{sin ( 3 x ) - 3 x}{x ^{3}} = - \frac{9}{2} + \frac{81}{40} x^{2} - \dots x \to 0 - \frac{9}{2} .$

Example 3 Hard

The product of two series

Let $f (x) = e^{x} cos x$ .

a) Find the Maclaurin expansion of $f$ up to the term in $x^{3}$ , multiplying the series of $e^{x}$ and $cos x$ .

b) Calculate $x \to 0 lim \frac{e ^{x} cos x - 1 - x}{x ^{3}}$ .

c) Deduce the value of $f^{'''} (0)$ .

Solution

a) Up to degree $3$ : $e^{x} \approx 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6}$ and $cos x \approx 1 - \frac{x ^{2}}{2}$ . Multiply and keep the terms up to $x^{3}$ : $(1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6}) (1 - \frac{x ^{2}}{2}) = 1 + x + (\frac{1}{2} - \frac{1}{2}) x^{2} + (\frac{1}{6} - \frac{1}{2}) x^{3} + \dots = 1 + x - \frac{x ^{3}}{3} + \dots$ Therefore $f (x) \approx 1 + x - \frac{x ^{3}}{3}$ (the term in $x^{2}$ cancels).

b) $e^{x} cos x - 1 - x = - \frac{x ^{3}}{3} + (terms of degree \geq 4)$ . Dividing by $x^{3}$ and letting $x \to 0$ , the limit is $- \frac{1}{3}$ .

c) The coefficient of $x^{3}$ in the series is $\frac{f ^{'''} ( 0 )}{3 !}$ . Then $\frac{f ^{'''} ( 0 )}{6} = - \frac{1}{3}$ , that is, $f^{'''} (0) = - 2$ .

In the exam: state the degree to which you expand and why (the denominator is $x^{3}$ , so you must reach $x^{3}$ ). If you leave lower-degree terms unsimplified or forget one, the limit comes out wrong. The series coefficients are $\frac{f ^{(k)} ( 0 )}{k !}$ : that relation lets you find derivatives without differentiating.

Common mistakes

Summary

Taylor at $a$ : $k = 0 \sum n \frac{f ^{(k)} ( a )}{k !} (x - a)^{k}$ . Maclaurin is the case $a = 0$ .
Series: $e^{x}$ , $sin x$ (odd powers), $cos x$ (even powers), $ln (1 + x)$ , $arctan x$ and $(1 + x)^{p}$ .
You may substitute ( $g (0) = 0$ ), add and multiply series, keeping the degree required.
For limits at $0$ , substitute the functions by their expansions and simplify.
The coefficient of $x^{n}$ is $\frac{f ^{(n)} ( 0 )}{n !}$ : it lets you obtain derivatives from the series.

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp