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QveST [7]
2 years ago
7

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
Mathematics
1 answer:
MAXImum [283]2 years ago
4 0

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

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