### Maclaurin Series

#### Definition

The Maclaurin series is a special case of the Taylor series, centred at $c=0$. The Maclaurin series of a function $f(x)$ is

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

The prime notation denotes the derivative of $f$ with respect to $x$.

Maclaurin series expansions have many applications, including evaluating definite integrals, finding the limit of a function or approximating the value of an expression.

#### Worked Example

###### Example 1

Use a Maclaurin series expansion to evaluate $\displaystyle \lim_{x\to0}\left[\frac{\cos{x}-1}{x^2}\right]$.

###### Solution

Note that this limit cannot be evaluated by substituting in $x=0$, as then the denominator would equal zero and the function would be undefined.

The limit can be found by expressing $\cos{x}$ as a Maclaurin series expansion. Only the first few terms are necessary; since the limit is to be evaluated as $x$ tends to zero, the higher terms in the expansion will tend to zero very rapidly and can therefore be suppressed.

Recall that the derivative of $\cos{x}$ is $-\sin{x}$, and that the derivative of $\sin{x}$ is $\cos{x}$. The Maclaurin series expansion of $\cos{x}$ is therefore given by

\begin{align} \cos{x} &= f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{\prime\prime\prime}(0)}{3!}x^3+\frac{f^{iv}(0)}{4!}x^3+ \ldots \\ &= \cos{0}-(\sin{0})x-\frac{\cos{0} }{2!}x^2+\frac{\sin{0} }{3!}x^3 + \frac{\cos{0} }{4!}x^4+\ldots \\ &= 1 - 0\cdot x - \frac{1}{2}x^2 + 0\cdot x^3 + \frac{1}{4}x^4 + \ldots \\ &= 1 - \frac{1}{2}x^2 + \frac{1}{4}x^4 + \ldots \end{align}

The limit therefore becomes

\begin{align} \lim_{x\to0}\left[\frac{\cos{x}-1}{x^2}\right] &= \lim_{x\to0}\left[\frac{(1 - \frac{1}{2}x^2 + \frac{1}{4}x^4 + \ldots) - 1}{x^2}\right], \\ &= \lim_{x\to0}\left[\frac{ - \frac{1}{2}x^2 + \frac{1}{4}x^4 + \ldots}{x^2}\right] \\ &= \lim_{x\to0}\left[-\frac{1}{2}+\frac{1}{4}x^2+\ldots\right], \\ \end{align} As $x$ tends to $0$ the terms involving powers of $x$ vanish, leaving us with

$\lim_{x\to0}\left[\frac{\cos{x}-1}{x}\right]=-\frac{1}{2}.$

#### Video Examples

##### Maclaurin Series Expansions
###### Example 1

Prof. Robin Johnson finds the first three terms in the Maclaurin expansion of $\sin{(2x)}$.

###### Example 2

Prof. Robin Johnson finds the first three terms in the Maclaurin expansion of $\ln{\Bigl(1-\dfrac{1}{2}x\Bigl)}$.

###### Example 3

Prof. Robin Johnson finds the first three terms in the Maclaurin expansion of $\dfrac{1}{2-3x}$.

###### Example 4

Prof. Robin Johnson finds the first three terms in the Maclaurin expansion of $\sqrt{3+2x}$.

###### Example 5

Prof Robin Johnson finds the first three terms in the Maclaurin expansion of $\dfrac{\sqrt{1-x}}{1+x}$.

###### Example 6

Prof. Robin Johnson finds the first four terms of the Maclaurin expansion of $(2-3x)^{^1/_2}$.

##### Using Maclaurin Series to find a Limit
###### Example 1

Prof. Robin Johnson uses the Maclaurin series to find \begin{align}\lim_{x \to 0} \left[\dfrac{e^{12\large{x}}-(1+2x)^{-1}}{\cos{(3x)}-(1-x^2)^{-2}}\right]\end{align}.

#### Workbook

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

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