2016/Promo/PJC/Q4

1. Evaluate the function at $x = 0$.

$\mathrm{f}(x) = 1 + \cos 2x$
$\mathrm{f}(0) = 1 + \cos(0) = 2$

2. Find the first and second derivatives and evaluate them at $x = 0$.

$\mathrm{f}'(x) = -2\sin 2x$
$\mathrm{f}'(0) = -2\sin(0) = 0$

$\mathrm{f}''(x) = -4\cos 2x$
$\mathrm{f}''(0) = -4\cos(0) = -4$

3. Find the third and fourth derivatives and evaluate them at $x = 0$.

$\mathrm{f}^{(3)}(x) = 8\sin 2x$
$\mathrm{f}^{(3)}(0) = 8\sin(0) = 0$

$\mathrm{f}^{(4)}(x) = 16\cos 2x$
$\mathrm{f}^{(4)}(0) = 16\cos(0) = 16$

4. Write down the Maclaurin series using the derived values.

The Maclaurin series is given by:
$\mathrm{f}(x) = \mathrm{f}(0) + x\mathrm{f}'(0) + \frac{x^2}{2!}\mathrm{f}''(0) + \frac{x^3}{3!}\mathrm{f}^{(3)}(0) + \frac{x^4}{4!}\mathrm{f}^{(4)}(0) + \dots$

Substituting the evaluated values:
$\mathrm{f}(x) = 2 + x(0) + \frac{x^2}{2}(-4) + \frac{x^3}{6}(0) + \frac{x^4}{24}(16) + \dots$
$\mathrm{f}(x) = 2 - 2x^2 + \frac{2}{3}x^4$

1. State or compute the Maclaurin series for $1 + \cos 2x$ from part (i)

From part (i), the derivatives of $\mathrm{f}(x) = 1 + \cos 2x$ evaluated at $x=0$ are:
$\mathrm{f}(0) = 2, \quad \mathrm{f}''(0) = -4, \quad \mathrm{f}^{(4)}(0) = 16$
(with odd derivatives being $0$).
The Maclaurin series for $\mathrm{f}(x)$ up to the $x^4$ term is:
$\mathrm{f}(x) \approx 2 - \frac{4}{2!}x^2 + \frac{16}{4!}x^4 = 2 - 2x^2 + \frac{2}{3}x^4$

2. Use the double angle identity to deduce the series for $\cos^2 x$

Using the double angle identity $\cos 2x = 2\cos^2 x - 1$, we can relate $\cos^2 x$ to $\mathrm{f}(x)$:
$2\cos^2 x = 1 + \cos 2x = \mathrm{f}(x)$
$\cos^2 x = \frac{1}{2}\mathrm{f}(x)$
Substituting the series from part (i):
$\cos^2 x \approx \frac{1}{2} \left( 2 - 2x^2 + \frac{2}{3}x^4 \right)$
$\cos^2 x \approx 1 - x^2 + \frac{1}{3}x^4$

1. State the standard Maclaurin series expansion for $\cos x$ from the List of Formulae up to the $x^4$ term.

From MF26,
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
$\cos x \approx 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4$

2. Square the series expansion and collect terms up to $x^4$ to verify the result.

$\cos^2 x = \left( 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + \dots \right)^2$
$= 1\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4\right) - \frac{1}{2}x^2\left(1 - \frac{1}{2}x^2\right) + \frac{1}{24}x^4(1) + \dots$
$= 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 - \frac{1}{2}x^2 + \frac{1}{4}x^4 + \frac{1}{24}x^4 + \dots$
$= 1 - x^2 + \left( \frac{1}{24} + \frac{1}{4} + \frac{1}{24} \right)x^4$
$= 1 - x^2 + \frac{8}{24}x^4$
$= 1 - x^2 + \frac{1}{3}x^4$
This matches the series deduced in part (ii), verifying its correctness.