2 The Maclaurin series

Consider a function $f\left(x\right)$ which can be differentiated at $x=0$ as often as we please. For example $e^x,cosx,sinx$ would fit into this category but $\left|x\right|$ would not.

Let us assume that $f\left(x\right)$ can be represented by a power series in $x$ :

$f\left(x\right)=b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4+\cdots ={\sum }_{p=0}^{\infty }{b}_p{x}^p$

where $b_0,b_1,b_2,\dots$ are constants to be determined.

If we substitute $x=0$ then, clearly $f\left(0\right)=b_0$

The other constants can be determined by further differentiating and, on each differentiation, substituting $x=0$ . For example, differentiating once:

$f^{\prime }\left(x\right)=0+b_1+2b_{2}x+3b_3{x}^2+4b_4{x}^3+\cdots$

so, putting $x=0$ , we have $f^{\prime }\left(0\right)=b_1$ .

Continuing to differentiate:

$f^{″}\left(x\right)=0+2b_2+3\left(2\right)b_{3}x+4\left(3\right)b_4{x}^2+\cdots$

so

$f^{″}\left(0\right)=2b_2\text{or}{b}_2=\frac{1}{2}{f}^{″}\left(0\right)$

Further:

$f^{‴}\left(x\right)=3\left(2\right)b_3+4\left(3\right)\left(2\right)b_{4}x+\cdots$ so $f^{‴}\left(0\right)=3\left(2\right)b_3\text{implying}{b}_3=\frac{1}{3\left(2\right)}{f}^{‴}\left(0\right)$

Continuing in this way we easily find that (remembering that $0!=1$ )

$b_n=\frac{1}{n!}{f}^{\left(n\right)}\left(0\right)n=0,1,2,\dots$

where $f^{\left(n\right)}\left(0\right)$ means the value of the $n^{th}$ derivative at $x=0$ and $f^{\left(0\right)}\left(0\right)$ means $f\left(0\right)$ .

Bringing all these results together we have:

Example 4

Find the Maclaurin expansion of   $cosx$ .

Solution

Here $f\left(x\right)=cosx$ and, differentiating a number of times:

$f\left(x\right)=cosx,f^{\prime }\left(x\right)=-sinx,f^{″}\left(x\right)=-cosx,f^{‴}\left(x\right)=sinx\text{etc.}$

Evaluating each of these at $x=0$ :

$f\left(0\right)=1,f^{\prime }\left(0\right)=0,f^{″}\left(0\right)=-1,f^{‴}\left(0\right)=0\text{etc.}$

Substituting into $f\left(x\right)=f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{″}\left(0\right)+\frac{x^3}{3!}{f}^{‴}\left(0\right)+\cdots$ , gives:

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

The reader should confirm (by finding the radius of convergence) that this series is convergent for all values of $x$ . The geometrical approximation to $cosx$ by the first few terms of its Maclaurin series are shown in Figure 6.

Figure 6 :

Task!

Find the Maclaurin expansion of $ln\left(1+x\right)$ .

(Note that we cannot find a Maclaurin expansion of the function $lnx$ since $lnx$ does not exist at $x=0$ and so cannot be differentiated at $x=0$ .)

Find the first four derivatives of $f\left(x\right)=ln\left(1+x\right)$ :

Answer

Now obtain $f\left(0\right),f^{\prime }\left(0\right),f^{″}\left(0\right),f^{‴}\left(0\right)$ :

Answer

Hence, obtain the Maclaurin expansion of $ln\left(1+x\right)$ :

Answer

Now obtain the radius of convergence and consider the situation at the boundary values:

Answer

Figure 7 :

Note that when $x=1$ $ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\cdots$ so the alternating harmonic series converges to $ln2\simeq 0.693$ , as stated in Section 16.2, page 17.

The Maclaurin expansion of a product of two functions: $f\left(x\right)g\left(x\right)$ is obtained by multiplying together the Maclaurin expansions of $f\left(x\right)$ and of $g\left(x\right)$ and collecting like terms together. The product series will have a radius of convergence equal to the smaller of the two separate radii of convergence.

Example 5

Find the Maclaurin expansion of  $e^{x}ln\left(1+x\right)$ .

Solution

Here, instead of finding the derivatives of $f\left(x\right)=e^{x}ln\left(1+x\right)$ , we can more simply multiply together the Maclaurin expansions for ${\text{e}}^x$ and $ln\left(1+x\right)$ which we already know:

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

and

$ln\left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots -1<x\le 1$

The resulting power series will only be convergent if $-1<x\le 1$ . Multiplying:

(You must take care not to miss relevant terms when carrying through the multiplication.)

Task!

Find the Maclaurin expansion of ${cos}^{2}x$ up to powers of $x^4$ . Hence write down the expansion of ${sin}^{2}x$ to powers of $x^6$ .

First, write down the expansion of $cosx$ :

Answer

Now, by multiplication, find the expansion of ${cos}^{2}x$ :

Answer

Now obtain the expansion of ${sin}^{2}x$ using a suitable trigonometric identity:

Answer

Find the Maclaurin expansion of $tanhx$ up to powers of $x^5$ .

Solution

The first two derivatives of $f\left(x\right)=tanhx$ are

$f^{\prime }\left(x\right)={\text{sech}}^{2}x{f}^{″}\left(x\right)=-2{\text{sech}}^{2}xtanhx{f}^{‴}\left(x\right)=4{\text{sech}}^{2}x{tanh}^{2}x-2{\text{sech}}^{4}x\cdots$

giving $f\left(0\right)=0,f^{\prime }\left(0\right)-1,f^{″}\left(0\right)=0f^{‴}\left(0\right)=-2\cdots$

This leads directly to the Maclaurin expansion as $tanhx=1-\frac{1}{3}{x}^3+\frac{2}{15}{x}^5\cdots$

Example 7

The relationship between the wavelength, $L$ , the wave period, $T$ , and the water depth, $d$ , for a surface wave in water is given by: $L=\frac{g{T}^2}{2\pi }tanh\left(\frac{2\pi d}{L}\right)$

In a particular case the wave period was 10 s and the water depth was 6.1 m. Taking the acceleration due to gravity, $g$ , as $9.81\text{m}{\text{s}}^{-2}$ determine the wave length.

[Hint: Use the series expansion for $tanhx$ developed in Example 6.]

Solution

Substituting for the wave period, water depth and $g$ we get

$L=\frac{9.81\times 10^2}{2\pi }tanh\left(\frac{2\pi \times 6.1}{L}\right)=\frac{490.5}{\pi }tanh\left(\frac{12.2\pi }{L}\right)$

The series expansion of $tanhx$ is given by $tanhx=x-\frac{x^3}{3}+\frac{2x^5}{15}+\cdots$

Using the series expansion of $tanhx$ we can approximate the equation as

$L=\frac{490.5}{\pi }\left\{\left(\frac{12.2\pi }{L}\right)-\frac{1}{3}{\left(\frac{12.2\pi }{L}\right)}^3+\cdots \right\}$

Multiplying through by $\pi L^3$ the equation becomes

$\pi L^4=490.5\times 12.2\pi L^2-\frac{490.5}{3}\times {\left(12.2\pi \right)}^3$

This equation can be rewritten as $L^4-5984.1L^2+2930198=0$

Solving this as a quadratic in $L^2$ we get $L=74$ m.

Using Newton-Raphson iteration this can be further refined to give a wave length of 73.9 m.