2016/Promo/PJC/Q4

1. Evaluate $f(0)$.

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

2. Find the first and second derivatives of $f(x)$ and evaluate them at $x=0$.

$f'(x) = -2\sin 2x \implies f'(0) = -2\sin(0) = 0$
$f''(x) = -4\cos 2x \implies f''(0) = -4\cos(0) = -4$

3. Find the third and fourth derivatives of $f(x)$ and evaluate them at $x=0$.

$f^{(3)}(x) = 8\sin 2x \implies f^{(3)}(0) = 8\sin(0) = 0$
$f^{(4)}(x) = 16\cos 2x \implies f^{(4)}(0) = 16\cos(0) = 16$

4. Construct the Maclaurin series up to the term in $x^4$ using the standard formula.

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \dots$
$f(x) = 2 + (0)x + \frac{-4}{2}x^2 + \frac{0}{6}x^3 + \frac{16}{24}x^4 + \dots$
$f(x) = 2 - 2x^2 + \frac{2}{3}x^4 + \dots$

1. Work out the Maclaurin series for $f(x) = 1 + \cos 2x$ from part (i) since it is required for the deduction.

Evaluating the derivatives at $x=0$:
$f(0)=2, \quad f'(0)=0, \quad f''(0)=-4, \quad f^{(3)}(0)=0, \quad f^{(4)}(0)=16$

The Maclaurin series for $f(x)$ is:
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \dots$
$f(x) \approx 2 - \frac{4}{2}x^2 + \frac{16}{24}x^4$
$f(x) \approx 2 - 2x^2 + \frac{2}{3}x^4$

2. Relate $\cos^2 x$ to $f(x)$ using the double angle formula.

Using the double angle identity $\cos 2x = 2\cos^2 x - 1$, we have:
$1 + \cos 2x = 2\cos^2 x$
$f(x) = 2\cos^2 x$
$\cos^2 x = \frac{1}{2} f(x)$

3. Deduce the Maclaurin series for $\cos^2 x$.

Substitute the series for $f(x)$ into the relationship:
$\cos^2 x \approx \frac{1}{2} \left( 2 - 2x^2 + \frac{2}{3}x^4 \right)$
$\cos^2 x \approx 1 - x^2 + \frac{1}{3}x^4$

1. State the standard series for $\cos x$ from the formula list.

From the List of Formulae (MF26),
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
$= 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$

2. Square the standard series to derive the expansion for $\cos^2 x$ up to the term in $x^4$.

$\cos^2 x = \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots \right)^2$
Expanding and keeping only terms up to $x^4$:
$= 1^2 + 2(1)\left(-\frac{x^2}{2} + \frac{x^4}{24}\right) + \left(-\frac{x^2}{2}\right)^2 + \dots$
$= 1 - x^2 + \frac{x^4}{12} + \frac{x^4}{4} + \dots$
$= 1 - x^2 + \left(\frac{1}{12} + \frac{3}{12}\right)x^4 + \dots$
$= 1 - x^2 + \frac{1}{3}x^4 + \dots$
This matches the series deduced in part (ii), verifying its correctness.