Advanced Higher Maths
Maclaurin Series

Method 1: If you have memorised the standard power series for \(\raise 0.3pt{e^x}\) (as above) you can just substitute \(3x\) for \(\raise 0.3pt{x}\) and obtain the answer almost immediately. This method receives full credit in exams.

$$ \begin{eqnarray} f(x) &=& e^{3x}\\[6pt] &=& 1+3x+\small\frac{(3x)^2}{2!}\normalsize+\small\frac{(3x)^3}{3!}\normalsize+\small\,\cdots \\[6pt] &=& 1+3x+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{27}{6}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] &=& 1+3x+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] \end{eqnarray} $$

Method 2: If you haven't memorised the power series for \(\raise 0.3pt{e^x}\small,\) you can use Maclaurin expansion from scratch:

$$ \begin{matrix} f(x)=e^{3x} \:&\: f(0)=1 \\[5pt] f'(x)=3e^{3x} \:&\: f'(0)=3 \\[5pt] f''(x)=9e^{3x} \:&\: f''(0)=9 \\[5pt] f'''(x)=27e^{3x} \:&\: f'''(0)=27 \\ \end{matrix} $$

Then we substitute into the general Maclaurin series:

$$ \begin{eqnarray} f(x)\!&=&\!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\frac{f'''(0)}{3!}\normalsize x^3\!\small+\normalsize\!\small\,\small\cdots \\[6pt] &=& 1+3x+\small\frac{9}{2!}\normalsize\tiny\,\normalsize x^2+\small\frac{27}{3!}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] &=& 1+3x+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \end{eqnarray} $$

Method 1: If you have memorised the standard power series for \(\text{sin}\,x\) you can just substitute \(4x\) for \(x\small.\)

$$ \begin{eqnarray} f(x) &=& \text{sin}\,4x\\[6pt] &=& 4x-\small\frac{(4x)^3}{3!}\normalsize+\small\,\cdots \\[6pt] &=& 4x-\small\frac{64}{3!}\normalsize x^3+\small\,\cdots \\[6pt] &=& 4x-\small\frac{32}{3}\normalsize x^3+\small\,\cdots \\[6pt] \end{eqnarray} $$

Method 2: If you haven't memorised the power series for \(\text{sin}\,x\small,\) use Maclaurin expansion from scratch.

$$ \begin{matrix} f(x)=\text{sin}\,4x \:&\: f(0)=0 \\[5pt] f'(x)=4\,\text{cos}\,4x \:&\: f'(0)=4 \\[5pt] f''(x)=-16\,\text{sin}\,4x \:&\: f''(0)=0 \\[5pt] f'''(x)=-64\,\text{cos}\,4x \:&\: f'''(0)=-64 \\ \end{matrix} $$

Then substitute into the general Maclaurin series:

$$ \begin{eqnarray} f(x)\!&=&\!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\frac{f'''(0)}{3!}\normalsize x^3\!\small+\normalsize\!\small\,\small\cdots \\[6pt] &=& 0+4x+0-\small\frac{64}{3!}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] &=& 4x-\small\frac{32}{3}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \end{eqnarray} $$

In the first two examples, we obtained the following:

$$ \begin{eqnarray} e^{3x} &=& 1+3x+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3+\small\,\cdots\\[6pt] \text{sin}\,4x &=& 4x-\small\frac{32}{3}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] \end{eqnarray} $$

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 \(\raise 0.3pt{x}\) than \(3\small.\)

$$ \begin{eqnarray} && e^{3x}\tiny\,\normalsize\text{sin}\,4x\\[6pt] &=& \small\left(\!\normalsize1\!\small+\normalsize\!3x\!\small+\normalsize\!\small\frac{9}{2}\normalsize x^2\!\small+\normalsize\!\small\frac{9}{2}\normalsize x^3\!\small+\normalsize\!\small\cdots\small\!\right)\left(\!\normalsize4x\small-\normalsize\!\small\frac{32}{3}\normalsize x^3\!\small+\normalsize\!\small\cdots\small\!\right)\normalsize \\[6pt] &=& 1.4x+3x.4x+\small\frac{9}{2}\normalsize x^2.4x-1.\small\frac{32}{3}\normalsize x^3+\normalsize\!\small\,\cdots \\[6pt] &=& 4x+12x^2+18x^3-\small\frac{32}{3}\normalsize x^3+\normalsize\!\small\,\cdots \\[6pt] &=& 4x+12x^2+\small\frac{54}{3}\normalsize x^3-\small\frac{32}{3}\normalsize x^3+\normalsize\!\small\,\cdots \\[6pt] &=& 4x+12x^2+\small\frac{22}{3}\normalsize x^3+\normalsize\!\small\,\cdots \end{eqnarray} $$

The beauty of this solution is that we don't have to go to the bother of differentiating \(\raise 0.3pt{e^{3x}\,\text{sin}\,4x\small,}\) finding Maclaurin series for each expression within the derivative and then combining them.

Instead, we just differentiate the Maclaurin series for \(\raise 0.3pt{e^{3x}\,\text{sin}\,4x}\) term-by-term. Simple!

$$ \begin{gather} e^{3x}\,\text{sin}\,4x=4x+12x^2+\small\frac{22}{3}\normalsize x^3+\normalsize\!\small\,\cdots\\[6pt] \small\frac{d}{dx}\left(\normalsize e^{3x}\,\text{sin}\,4x\small\right)\normalsize=4+24x+22x^2+\small\,\cdots \\[6pt] \end{gather} $$

\(\large\frac{d}{dx}\normalsize(\text{tan}\,x)=\text{sec}^{2}\,x\) so we can integrate term-by-term to obtain the Maclaurin expansion for \(\text{tan}\,x\small.\)

$$ \begin{eqnarray} \tan\,x&=& \!\!\int\!\small\left(\normalsize 1\small+\normalsize x^2+\small\frac{2}{3}\normalsize x^4\small+\cdots\normalsize\small\right)\normalsize\,dx \\[6pt] &=& x+\small\frac{1}{3}\normalsize x^3+\small\frac{2}{15}\normalsize x^5+\small\ \cdots\\[6pt] \end{eqnarray} $$

Finally we substitute substitute \(\raise 0.2pt{2x}\) for \(\raise 0.3pt{x}\small,\) and simplify:

$$ \begin{eqnarray} \tan\,2x &=& 2x+\small\frac{1}{3}\normalsize (2x)^3+\small\frac{2}{15}\normalsize (2x)^5+\small\ \cdots\\[6pt] &=& 2x+\small\frac{1}{3}\normalsize (8x^3)+\small\frac{2}{15}\normalsize (32x^5)+\small\ \cdots\\[6pt] &=& 2x+\small\frac{8}{3}\normalsize x^3+\small\frac{64}{15}\normalsize x^5+\small\ \cdots\\[6pt] \end{eqnarray} $$

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

$$ \begin{eqnarray} f(x)\!&=&\!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\,\small\cdots \\[6pt] &=& 1+\small\left(\!\frac{1}{1\!+\!1^4}\!\right)\normalsize x+\frac{f''(0)}{2!}\normalsize x^2\!\normalsize+\small\,\cdots \end{eqnarray} $$

Before we go any further, we need to differentiate to obtain \(f''(x)\small.\)

$$ \begin{eqnarray} f''(x)\!&=& \small\frac{\{1\!+\!(x\!+\!1)^4\}.1-(x\!+\!1)\{4(x\!+\!1)^3\}}{\{1\!+\!(x\!+\!1)^4\}^2}\\[6pt] \end{eqnarray} $$

There is no need to simplify the expression above, because its only purpose is to evaluate \(f''(0)\small.\)

$$ \begin{eqnarray} f''(0)\!&=& \small\frac{1+1^4-(1)\left(4(1^3)\right)}{(1+1^4)^2}\\[6pt] &=& \small\frac{1+1-4}{2^2}\\[6pt] &=& \small\frac{-2}{4}\\[6pt] &=& -\!\small\frac{1}{2}\\[6pt] \end{eqnarray} $$

Putting this together, we obtain:

$$ \begin{eqnarray} f(x)\!&=& 1+\small\left(\!\frac{1}{1\!+\!1^4}\!\right)\normalsize x-\small\frac12\normalsize\left(\!\small\frac{x^2}{2!}\normalsize\!\right)\!\normalsize+\small\,\cdots \\[6pt] &=& 1+\small\frac12\,x-\small\frac14\,x^2\normalsize+\small\,\cdots \end{eqnarray} $$

(a) (i)  If you have memorised the standard power series for \(\raise 0.3pt{e^x}\) you can just substitute \(2x\) for \(\raise 0.3pt{x}\) and obtain the answer almost immediately. This method receives full credit.

$$ \begin{eqnarray} e^{2x} &=& 1+2x+\small\frac{(2x)^2}{2!}\normalsize+\small\frac{(2x)^3}{3!}\normalsize+\small\,\cdots \\[6pt] &=& 1+2x+\small\frac{4}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{8}{6}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] &=& 1+2x+2x^2+\small\frac{4}{3}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] \end{eqnarray} $$

Alternatively, if you haven't memorised the power series for \(\raise 0.3pt{e^x}\small,\) you can use Maclaurin expansion from scratch:

$$ \begin{matrix} f(x)=e^{2x} \:&\: f(0)=1 \\[5pt] f'(x)=2e^{2x} \:&\: f'(0)=2 \\[5pt] f''(x)=4e^{2x} \:&\: f''(0)=4 \\[5pt] f'''(x)=8e^{2x} \:&\: f'''(0)=8 \\ \end{matrix} $$

Then you would substitute into the general Maclaurin series:

$$ \begin{eqnarray} e^{2x}\!&=&\!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\frac{f'''(0)}{3!}\normalsize x^3\!\small+\normalsize\!\small\,\small\cdots \\[6pt] &=& 1+2x+\small\frac{4}{2!}\normalsize\tiny\,\normalsize x^2+\small\frac{8}{3!}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] &=& 1+2x+2x^2+\small\frac{4}{3}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \end{eqnarray} $$

(a) (ii)  If you have memorised the standard power series for \(\text{sin}\,x\) you can just substitute \(3x\) for \(x\small.\)

$$ \begin{eqnarray} \text{sin}\,3x &=& 3x-\small\frac{(3x)^3}{3!}\normalsize+\small\,\cdots \\[6pt] &=& 3x-\small\frac{27}{3!}\normalsize x^3+\small\,\cdots \\[6pt] &=& 3x-\small\frac{9}{2}\normalsize x^3+\small\,\cdots \\[6pt] \end{eqnarray} $$

Alternatively, if you haven't memorised the power series for \(\text{sin}\,x\small,\) you would need to use Maclaurin expansion from scratch.

$$ \begin{matrix} f(x)=\text{sin}\,3x \:&\: f(0)=0 \\[5pt] f'(x)=3\,\text{cos}\,3x \:&\: f'(0)=3 \\[5pt] f''(x)=-9\,\text{sin}\,3x \:&\: f''(0)=0 \\[5pt] f'''(x)=-27\,\text{cos}\,3x \:&\: f'''(0)=-27 \\ \end{matrix} $$

Then you would have to substitute into the general Maclaurin series:

$$ \begin{eqnarray} \text{sin}\,3x\! &=& \!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\frac{f'''(0)}{3!}\normalsize x^3\!\small+\normalsize\!\small\,\small\cdots \\[6pt] &=& 0+3x+0-\small\frac{27}{3!}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \\[6pt] &=& 3x-\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3+\small\,\cdots \end{eqnarray} $$

(b)  Making sense of this part of the question requires us to notice that the given expression is a composite function. Specifically:

Let \(f(x)=e^{2x}\) and let \(g(x)=\text{sin}\,3x\small.\)
Then \(f(g(x) = f(\text{sin}\,3x) = e^{\large{2\,\text{sin}\,3x}}\small.\)

So we can obtain the required power series by making use of this fact.

From part (a) above:
\(f(x)=1+2x+2x^2+\small\frac{4}{3}\normalsize\tiny\,\normalsize x^3+\small\,\cdots\)
\(g(x)=3x-\frac{9}{2}x^3+\small\,\cdots\)

So, ignoring terms with powers higher than 3:

$$ \begin{eqnarray} f(g(x)) &=& f(3x-\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3)\\[6pt] &=&1+2(3x\!-\!\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3)+2{(3x\!-\!\small\frac{9}{2}\normalsize\small\,\normalsize x^3)}^2\\[6pt] &&\:\:+\small\frac{4}{3}\normalsize{(3x\!-\!\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3)}^3+\small\,\cdots\\[6pt] &\approx& 1+\!6x-\!9x^3+\!2(9x^{2})+\!\small\frac{4}{3}\normalsize(27x^{3})\\[6pt] &=& 1+6x-9x^3+18x^2+36x^3\\[6pt] &=& 1+6x+18x^2+27x^3\\[6pt] \end{eqnarray} $$

(a)  If you have memorised the standard power series for \(\text{cos}\,x\) you can just substitute \(3x\) for \(x\) and obtain the answer almost immediately. This method receives full credit.

$$ \begin{eqnarray} \text{cos}\,3x &=& 1-\small\frac{(3x)^2}{2!}\normalsize+\small\frac{(3x)^4}{4!}\normalsize-\small\,\cdots \\[6pt] &=& 1-\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{81}{24}\normalsize\tiny\,\normalsize x^4-\small\,\cdots \\[6pt] &=& 1-\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{27}{8}\normalsize\tiny\,\normalsize x^4-\small\,\cdots \\[6pt] \end{eqnarray} $$

Alternatively, if you haven't memorised the power series, you can use Maclaurin expansion from scratch:

$$ \begin{matrix} f(x)=\text{cos}\,3x \:&\: f(0)=1 \\[5pt] f'(x)=-3\,\text{sin}\,3x \:&\: f'(0)=0 \\[5pt] f''(x)=-9\,\text{cos}\,3x \:&\: f''(0)=-9 \\[5pt] f'''(x)=27\,\text{sin}\,3x \:&\: f'''(0)=0 \\[5pt] f^{\textsf{iv}}(x)=81\,\text{cos}\,3x \:&\: f^{\textsf{iv}}(0)=81 \\ \end{matrix} $$

Then you would substitute into the general Maclaurin series:

$$ \begin{eqnarray} \text{cos}\,3x\! &=& \!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\frac{f'''(0)}{3!}\normalsize x^3\!\small+\normalsize\!\small\frac{f^{\textsf{iv}}(0)}{4!}\,\normalsize x^4+\small\,\small\cdots \\[6pt] &=& 1+0+\small\frac{-9}{2!}\normalsize\tiny\,\normalsize x^2+0+\small\frac{81}{4!}\normalsize\tiny\,\normalsize x^4+\small\,\cdots \\[6pt] &=& 1-\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{27}{8}\normalsize\tiny\,\normalsize x^4\small\,\cdots \end{eqnarray} $$

(b)  We can make use of the simple fact that \(\text{cos}^{2}\,3x = (\text{cos}\,3x)(\text{cos}\,3x)\small.\)

So, truncating our power series after the term in \(x^4\small,\) we obtain the following approximation for \(\text{cos}^{2}\,3x\small.\)

$$ \begin{eqnarray} && \left(1-\small\frac{9}{2}\normalsize x^2+\small\frac{27}{8}\normalsize x^4\right)\left(1-\small\frac{9}{2}\normalsize x^2+\small\frac{27}{8}\normalsize x^4\right)\\[6pt] &=& 1-\small\frac{9}{2}\normalsize x^2+\small\frac{27}{8}\normalsize x^4-\small\frac{9}{2}\normalsize x^2+\small\frac{81}{4}\normalsize x^4+\small\frac{27}{8}\normalsize x^4\\[6pt] &=& 1-9x^2+\small\frac{216}{8}\tiny\,\normalsize x^4\\[6pt] &=& 1-9x^2+27x^4\\[6pt] \end{eqnarray} $$