Question

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

approximation to the root of the equation x3+3x+sin(x)=5. What is the result?

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 - (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 - (f(1))/(f'(1)) = 1 + (1 - \sin(1))/(6 + \cos(1)) \approx 1.024238790`

`x_3 = x_2 - (f(x_2))/(f'(x_2)) \approx 1.024009549`

`x_4 = x_3 - (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.