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

Answer:

$x \approx 2.278863$

Step-by-step 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:

$x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)}$

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:

$x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)}$

$x_{1+1} = x_1 - \frac{h(x_1)}{h'(x_1)}$

$x_{2} = 2.2 - \frac{2.2 -3\sin(2.2)}{1 -3\cos(2.2)} = 2.28153641$

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

$x_{3} = 2.28153641 - \frac{2.28153641 -3\sin(2.28153641)}{1 -3\cos(2.28153641)} = 2.2788654$

$x_{4} = 2.2788654 - \frac{2.2788654 -3\sin(2.2788654)}{1 -3\cos(2.2788654)} = 2.2788627$

$x_{5} = 2.2788627 - \frac{2.2788627-3\sin(2.2788627)}{1 -3\cos(2.2788627)} = 2.2788627$

Hence:

$x \approx 2.278863$