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

Question

$x^{3} - x = 2$ $x^3-x=2$

(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.)

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Answer 1 · 2016-11-07T21:13:32

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

We have:
$x^{3} - x = 2 \Rightarrow x^{3} - x - 2 = 0$ $x^3-x=2 => x^3-x-2 = 0$

Let $f (x) = x^{3} - x - 2$ $f(x) = x^3-x-2$ 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)
.
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(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);

enter image source here

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);

enter image source here

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.