Single Variable Calculus, Chapter 4, 4.8, Section 4.8, Problem 16

Use Newton's Method to approximate the positive root of $2 \cos x = x^4$ correct to six decimal places.

Rewrite the equation in Standard Form

$2 \cos x - x^4 = 0$

Therefore, let $f(x) = 2 \cos x - x^4$. Then,


$
\begin{equation}
\begin{aligned}

f'(x) =& 2 \frac{d}{dx} (\cos x) - \frac{d}{dx} (x^4)
\\
\\
f'(x) =& 2 (- \sin x) - 4x^3
\\
\\
f'(x) =& -2 \sin x - 4x^3

\end{aligned}
\end{equation}
$


Using Approximation Formula


$
\begin{equation}
\begin{aligned}

x_{n + 1} =& x_n - \frac{f(x_n)}{f'(x_n)}
\\
\\
x_{n + 1} =& x_n - \frac{2 \cos x_n - x_n^4}{-2 \sin x_n 4x^3_n}
\\
\\
x_{n + 1} =& x_n + \frac{2 \cos x_n - x^4_n}{2 \sin x_n + 4x^3_n}


\end{aligned}
\end{equation}
$









To find the initial approximation $x_1$, we graph $y = 2 \cos x$ and $y = x^4$.

Based from the graph, it appears that they intersect at a point in $x$-coordinate where they are very close to -1 and 1. We have $x_1 = 1$ and $x_1 = -1$ as the initial approximation.

So we get


$
\begin{equation}
\begin{aligned}

x_2 =& x_1 + \frac{2 \cos x_1 - x_1^4}{2 \sin x_1 + 4x_1^3}
&&
\\
\\
x_2 =& 1 + \frac{2 \cos(1) - (1)^4}{2 \sin(1) + 4(1)^3}
&& x_2 = -1 + \frac{2 \cos (-1) - (-1)^4}{2 \sin (-1) + 4(-1)}
\\
\\
x_2 \approx & 1.014184
&& x_2 \approx - 1.014184
\\
\\
x_3 =& 1.014184 + \frac{f(1.014184)}{f'(1.014184)}
&& x_3 = - 1.014184 + \frac{f(-1.014184)}{f'(-1.014184)}
\\
\\
x_3 \approx & 1.013958
&& x_3 \approx - 1.013958
\\
\\
x_4 =& 1.013958 + \frac{f(1.013958)}{f'(1.013958)}
&& x_4 = - 1.013958 + \frac{f(-1.013958)}{f'(-1.013958)}
\\
\\
x_4 \approx & 1.013958
&& x_4 \approx -1.013958

\end{aligned}
\end{equation}
$


Since $x_3$ and $x_4$ agree to six decimal places, therefore $x \approx 1.013958$ and $x \approx -1.013958$