Question

Find the root of the equation. Give your answers correct to six decimal places?

`x^3-x=2`

(a) Use Newton's method with x1 = 1.

(b) Solve the equation using x1 = 0.6 as the initial approximation.

(c) Solve the equation using x1 = 0.58. (You definitely need a programmable calculator for this part.)

Answer 1

The solution is `x=1.521380` (6dp)

Explanation:

We have:
`x^3-x=2 => x^3-x-2 = 0`

Let `f(x) = x^3-x-2` then `f'(x)=3x^2-1` and we can use Newton's method using the iterative formula;

`x_(n+1) = x_n - f(x_n) / (f'(x_n))`
`:. x_(n+1) = x_n - (x_n^3-x_n-2) / (3x_n^2-1)`

(a) If we start with `x_0=1`, then we can tabulate the results as follows (in this case using Excel working to 8dp);

So we see that very rapidly the Newton-Rhapson method converges to the solution `x=1.521380` (6dp)
.

(b) If we start with `x_0=0.6=`, then we can tabulate the results as follows (in this case using Excel working to 8dp);

So again we see that the Newton-Rhapson method converges to the solution `x=1.521380` (6dp), but this time it takes a few more steps.

(c) If we start with `x_0=0.58=`, then we can tabulate the results as follows (in this case using Excel working to 8dp);

So again we see that the Newton-Rhapson method converges to the solution `x=1.521380` (6dp), but this time it takes many more steps.