The goal of this part of assignment is to explore the convergence of the Newton,

Question

The goal of this part of assignment is to explore the convergence of the Newton, Secant, and Fixed Point methods to the solution of sin(x) = a for different values of a. a = 1/2. With a = a root of the equation is alpha = pi/6 = 0.52 Prove that the following fixed point iteration converges to the root. xn+1 = xn-sin(xn) + 1/2 Use xQ = 0.6 for the Newton's and fixed point methods, and x0 = 0.6, X1 = 0.5 for the Secant method. Discuss the order of convergence for each method as compared to the theoretical results.

Explanation / Answer

a) First, we show that pi/6 is a fixed point

pi/6 - sin(pi/6) + 1/2 = pi/6 - 1/2 + 1/2 = pi/6

The derivative of x - sin(x) + 1/2 = 1 - cos(x)

This is less than 1 (and greater than -1) near pi/6, so it converges.

At pi/6, this equals 1 - cos(pi/6) = 1 - sqrt(3)/2 =0.133974596215561

This gives the ratio of the error between steps.

Newton Method

For f(x) = sin(x) - 1/2, the derivative is cos(x), so the Newton step is

x - (sin(x) - 1/2)/cos(x)

Newton step

Fixed Point Method

Secant Method

We see, as they are supposed to, that the Newton step has quadratic convergence, (the ratio of

the error/the square of the previous error goes to a constant)

the fixed point method has linear convergence (the error ratio tends to a constant)

and for the secant method that we have superlinear convergence (the ratio of error steps goes to

0).

Step x-step f(x) f'(x) step x-step+1 Error Diff-step/(diff-step-1)^2 0 0.6 0.064642 0.825335615 -0.07832 0.5216773 -0.00192 -0.32917 1 0.521677349 -0.00166 0.866984518 0.00192 0.5235977 -1.1E-06 -0.28772 2 0.523597713 -9.2E-07 0.866025935 1.06E-06 0.5235988 -3.3E-13 -0.2886 3 0.523598776 -2.8E-13 0.866025404 3.26E-13 0.5235988

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