Question

Use Newton’s method to approximate the indicated root of the equation correct to six decimal places. The positive root of 3 sin x = x

Answer 1

`x \approx 2.278863`

Explanation:

Required

The positive root of
`3\sin(x) = x`

Equate to 0

`0 = x -3\sin(x)`

So, we have our function to be:

`h(x) = x -3\sin(x)`

Differentiate the above function:

`h'(x) = 1 -3\cos(x)`

Using Newton's method of approximation, we have:

`x_(n+1) = x_n - (h(x_n))/(h'(x_n))`

Plot the graph of
`h(x) = x -3\sin(x)` to get
`x_1` --- see attachment for graph

From the attached graph, the first value of x is at 2.2; so:

`x_1 = 2.2`

So, we have:

`x_(n+1) = x_n - (h(x_n))/(h'(x_n))`

`x_(1+1) = x_1 - (h(x_1))/(h'(x_1))`

`x_(2) = 2.2 - (2.2 -3\sin(2.2))/(1 -3\cos(2.2)) = 2.28153641`

The process will be repeated until the digit in the 6th decimal place remains unchanged

`x_(3) = 2.28153641 - (2.28153641 -3\sin(2.28153641))/(1 -3\cos(2.28153641)) = 2.2788654`

`x_(4) = 2.2788654 - (2.2788654 -3\sin(2.2788654))/(1 -3\cos(2.2788654)) = 2.2788627`

`x_(5) = 2.2788627 - (2.2788627-3\sin(2.2788627))/(1 -3\cos(2.2788627)) = 2.2788627`

Hence:

`x \approx 2.278863`