Advanced Higher Maths
Maclaurin Series

Method 1: If you have memorised the standard power series for $e^x$ (as above) you can just substitute $3x$ for $x$ and obtain the answer almost immediately. This method receives full credit in SQA marking schemes.

$$\begin{array}{rcl}f(x)& =& e^{3x}\\ & =& 1+3x+\frac{{3x}^2}{2!}+\frac{{3x}^3}{3!}+\cdots \\ & =& 1+3x+\frac{9}{2!}{x}^2+\frac{27}{3!}{x}^3+\cdots \\ & =& 1+3x+\frac{9}{2}{x}^2+\frac{9}{2}{x}^3+\cdots \end{array}$$

Method 2: If you haven't memorised the power series for $e^x,$ you can use Maclaurin expansion from scratch:

$$\begin{array}{cc}f(x)=e^{3x}& f(0)=1\\ f^{\prime }(x)=3e^{3x}& f^{\prime }(0)=3\\ f^{″}(x)=9e^{3x}& f^{″}(0)=9\\ f^{‴}(x)=27e^{3x}& f^{‴}(0)=27\end{array}$$

Then we substitute into the general Maclaurin series:

$$\begin{array}{rcl}f(x)& =& f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2!}{x}^2+\frac{f^{‴}(0)}{3!}{x}^3+\cdots \\ & =& 1+3x+\frac{9}{2!}{x}^2+\frac{27}{3!}{x}^3+\cdots \\ & =& 1+3x+\frac{9}{2}{x}^2+\frac{9}{2}{x}^3+\cdots \end{array}$$

Method 1: If you have memorised the standard power series for $sinx$ you can just substitute $4x$ for $x.$

$$\begin{array}{rcl}f(x)& =& sin4x\\ & =& 4x-\frac{(4x{)}^3}{3!}+\cdots \\ & =& 4x-\frac{64}{3!}{x}^3+\cdots \\ & =& 4x-\frac{32}{3}{x}^3+\cdots \end{array}$$

Method 2: If you haven't memorised the power series for $sinx,$ use Maclaurin expansion from scratch.

$$\begin{array}{cc}f(x)=sin4x& f(0)=0\\ f^{\prime }(x)=4cos4x& f^{\prime }(0)=4\\ f^{″}(x)=-16sin4x& f^{″}(0)=0\\ f^{‴}(x)=-64cos4x& f^{‴}(0)=-64\end{array}$$

Then substitute into the general Maclaurin series:

$$\begin{array}{rcl}f(x)& =& f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2!}{x}^2+\frac{f^{‴}(0)}{3!}{x}^3+\cdots \\ & =& 0+4x+0-\frac{64}{3!}{x}^3+\cdots \\ & =& 4x-\frac{32}{3}{x}^3+\cdots \end{array}$$

In the first two examples, we obtained the following:

$$\begin{array}{rcl}{e}^{3x}& =& 1+3x+\frac{9}{2}{x}^2+\frac{9}{2}{x}^3+\cdots \\ sin4x& =& 4x-\frac{32}{3}{x}^3+\cdots \end{array}$$

To obtain the Maclaurin expansion for the product of these two functions, we simply multiply their respective expansions, ignoring anything that will multiply to a higher power of $x$ than $3.$

$$\begin{array}{rcl}& & e^{3x}sin4x\\ & =& \left(1+3x+\frac{9}{2}{x}^2+\frac{9}{2}{x}^3+\cdots \right)\left(4x-\frac{32}{3}{x}^3+\cdots \right)\\ & =& 1.4x+3x.4x+\frac{9}{2}{x}^2.4x-1.\frac{32}{3}{x}^3+\cdots \\ & =& 4x+12x^2+18x^3-\frac{32}{3}{x}^3+\cdots \\ & =& 4x+12x^2+\frac{54}{3}{x}^3-\frac{32}{3}{x}^3+\cdots \\ & =& 4x+12x^2+\frac{22}{3}{x}^3+\cdots \end{array}$$

The beauty of this solution is that we don't have to go to the bother of differentiating $e^{3x}sin4x,$ finding Maclaurin series for each expression within the derivative and then combining them.

Instead, we just differentiate the Maclaurin series for $e^{3x}sin4x$ term-by-term. Simple!

$$\begin{array}{c}{e}^{3x}sin4x=4x+12x^2+\frac{22}{3}{x}^3+\cdots \\ \frac{d}{dx}\left(e^{3x}sin4x\right)=4+24x+22x^2+\cdots \end{array}$$

$\frac{d}{dx}(tanx)=se{c}^{2}x$ so we can integrate term-by-term to obtain the Maclaurin expansion for $tanx.$

$$\begin{array}{rcl}\tan x& =& \int \left(1+x^2+\frac{2}{3}{x}^4+\cdots \right)dx\\ & =& x+\frac{1}{3}{x}^3+\frac{2}{15}{x}^5+\cdots \end{array}$$

Finally we substitute substitute $2x$ for $x,$ and simplify:

$$\begin{array}{rcl}\tan 2x& =& 2x+\frac{1}{3}(2x{)}^3+\frac{2}{15}(2x{)}^5+\cdots \\ & =& 2x+\frac{1}{3}(8x^3)+\frac{2}{15}(32x^5)+\cdots \\ & =& 2x+\frac{8}{3}{x}^3+\frac{64}{15}{x}^5+\cdots \end{array}$$

This question combines Maclaurin series and the quotient rule for differentiation.

$$\begin{array}{rcl}f(x)& =& f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2!}{x}^2+\cdots \\ & =& 1+\left(\frac{1}{1+1^4}\right)x+\frac{f^{″}(0)}{2!}{x}^2+\cdots \end{array}$$

Before we go any further, we need to differentiate to obtain $f^{″}(x).$

$$\begin{array}{rcl}{f}^{″}(x)& =& \frac{\{1+(x+1{)}^4\}.1-(x+1)\{4(x+1{)}^3\}}{\{1+(x+1{)}^4{\}}^2}\end{array}$$

There is no need to simplify the expression above, because its only purpose is to evaluate $f^{″}(0).$

$$\begin{array}{rcl}{f}^{″}(0)& =& \frac{1+1^4-(1)(4(1^3))}{(1+1^4{)}^2}\\ & =& \frac{1+1-4}{2^2}\\ & =& \frac{-2}{4}\\ & =& -\frac{1}{2}\end{array}$$

Putting this together, we obtain:

$$\begin{array}{rcl}f(x)& =& 1+\left(\frac{1}{1+1^4}\right)x-\frac{1}{2}\left(\frac{x^2}{2!}\right)+\cdots \\ & =& 1+\frac{1}{2}x-\frac{1}{4}{x}^2+\cdots \end{array}$$