Question

Use Newton’s Method with initial approximation x1=1 to find x4, the third

Answer 1

Let $f(x) = x^3 + 3x + \sin(x) - 5$. Using Newton's method to approximate a solution to $f(x) = 0$, we consider the recurrence

$\begin{cases} x_1 = 1 \\ x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)} & \text{for } n \ge 1 \end{cases}$

Differentiating $f(x)$ gives

$f'(x) = 3x^2 + 3 + \cos(x)$

Then

$x_2 = 1 - \dfrac{f(1)}{f'(1)} = 1 + \dfrac{1 - \sin(1)}{6 + \cos(1)} \approx 1.024238790$

$x_3 = x_2 - \dfrac{f(x_2)}{f'(x_2)} \approx 1.024009549$

$x_4 = x_3 - \dfrac{f(x_3)}{f'(x_3)} \approx \boxed{1.024009528}$

which agrees numerically with the actual root of $f(x)$ up to at least 9 digits after the decimal point.