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

Question

#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 # (6dp)

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)
.
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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.