# Question 1da1c

Jan 10, 2018

$f \left(x\right) = \sin \left(\frac{\pi}{5}\right) + \cos \left(\frac{\pi}{5}\right) x - \sin \left(\frac{\pi}{5}\right) {x}^{2} / 2 - \cos \left(\frac{\pi}{5}\right) {x}^{3} / 6 + \ldots$

#### Explanation:

The Maclaurin series is given by:

f(x) = sum_(n=0)^oof^n(0)/(n!)x^n

where ${f}^{n} \left(0\right)$ denotes the $n$th derivative of $f \left(x\right)$ evaluated at 0.

So as you have stated the list of the derivatives are as follows:

$f \left(x\right) = \sin \left(x + \frac{\pi}{5}\right)$
$f ' \left(x\right) = \cos \left(x + \frac{\pi}{5}\right)$
$f ' ' \left(x\right) = - \sin \left(x + \frac{\pi}{5}\right)$
$f ' ' ' \left(x\right) = - \cos \left(x + \frac{\pi}{5}\right)$

Evaluating these at $x = 0$ gives us the following list (it turns out these have the following exact values:

$f \left(0\right) = \sin \left(\frac{\pi}{5}\right) = \frac{1}{4} \sqrt{10 - 2 \sqrt{5}}$

$f ' \left(0\right) = \cos \left(\frac{\pi}{5}\right) = \frac{1}{4} \sqrt{1 + \sqrt{5}}$

$\to f ' ' \left(0\right) = - \sin \left(\frac{\pi}{5}\right) = - \frac{1}{4} \sqrt{10 - 2 \sqrt{5}}$

$\to f ' ' ' \left(0\right) = - \cos \left(\frac{\pi}{5}\right) = - \frac{1}{4} \sqrt{1 + \sqrt{5}}$

This will allow us to find the first 4 terms of the Maclaurin series (if you wish to find more terms and therefore have a more accurate series simply find further derivatives).

So from the definition of the Maclaurin series above:

f(x) = sum_(n=0)^oof^n(0)/(n!)x^n

=(f(0))/(0!)x^0+(f'(0))/(1!)x^1+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+...#

$= f \left(0\right) + f ' \left(0\right) x + \frac{f ' ' \left(0\right)}{2} {x}^{2} + \frac{f ' ' ' \left(0\right)}{6} {x}^{3} + \ldots$

It is now a simple matter of substituting the values worked out from the list above in place of ${f}^{n} \left(x\right)$ like so:

$f \left(x\right) = \sin \left(\frac{\pi}{5}\right) + \cos \left(\frac{\pi}{5}\right) x - \sin \left(\frac{\pi}{5}\right) {x}^{2} / 2 - \cos \left(\frac{\pi}{5}\right) {x}^{3} / 6 + \ldots$

You could stop here or continue and replace these trig terms with their exact values like so:

$= \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} + \frac{1}{4} \sqrt{1 + \sqrt{5}} \cdot x - \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} \cdot {x}^{2} / 2 - \frac{1}{4} \sqrt{1 + \sqrt{5}} \cdot {x}^{3} / 6 + \ldots$

$= \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} + \frac{1}{4} \sqrt{1 + \sqrt{5}} \cdot x - \frac{1}{8} \sqrt{10 - 2 \sqrt{5}} \cdot {x}^{2} - \frac{1}{24} \sqrt{1 + \sqrt{5}} \cdot {x}^{3} + \ldots$

Of course, as stated above, a Maclaurin series generally carries on to infinity. If you wish to find further terms you must calculate further derivatives of $f \left(x\right)$ until you you have reached your goal.