Question

Task 2 (25 points + 4 points for commenting): Write computer code to perform the Fixed.Point method and use your code to find the root of the following equation using an initial guess of 3 and a stopping criterion of 0.001%; f(x) e -4x For Fixed-Point, you do not have to code Matlab to take the derivatives of the function and check the g'(x). You can do that step by hand, and then show me your hand cal ulations to find the g(x) that works. Then, simply code in that g(x) Compare the Fixed Point method to the Modified Secant method using the same stopping criterion and the same initial guess. Use a perturbation factor of 0.1. For Fixed-Point: a) Make a plot of the ratio of the current iteration's approximate error to the previous iteration's approximate error versus the iteration number, Use "semilogy()" to make the y-axis logarithmic and the x-axis linear. Make your graph look nice. Explain your results. For both methods: b) c) d) How many iterations did your code take to find the root? What is your final guess of the root? Which method was faster and why?

Answer 1

The code given below is in the function file FixedPoint.m

The code given below is in the function file function [x, iter, Ea] = FixedPoint(g, x0, tol) %FIXEDPOINT Finds the fixed point of the function g(x) in the appropriate %interval. % Input - anonymous function handle -g, x0 - initial guess, tol - Error % tolerance. % Output - x - approximate fixed point, iter - no.of iterations, Ea - approximate % error. % Initialize iteration counter iter = 0: % Set maximum no. of iterations as 100 maxiter = 100: % Create array of errors Ea = zeros(1, maxiter): % First iteration x = g(x0): iter = iter + 1: Ea(iter) = abs((x - x0)/x)*100: while Ea(iter) > = tol && iter < = maxiter x0 = x: x = g(x0): iter = iter + 1: Ea(iter) = abs((x - x0)/x)*100: end Ea = Ea(1: iter): end \r\n function [x, iter, Ea] = FixedPoint(g, x0, tol) %FIXEDPOINT Finds the fixed point of the function g(x) in the appropriate %interval. % Input - anonymous function handle -g, x0 - initial guess, tol-Error % tolerance. % Output - x - approximate fixed point, iter-no.of iterations, Ea-approximate % error. % Initialize iteration counter iter = 0: % Set maximum no. of iterations as 100

function [x,iter,Ea] = FixedPoint(g,x0,tol)
%FIXEDPOINT Finds the fixed point of the function g(x) in the appropriate
%interval.
% Input - anonymous function handle -g, x0 - initial guess,tol-Error
% tolerance.
% Output - x - approximate fixed point,iter-no.of iterations,Ea-approximate
% error.
% Initialize iteration counter
iter = 0;
% Set maximum no. of iterations as 100
maxiter = 100;
% Create array of errors
Ea = zeros(1,maxiter);
% First iteration
x = g(x0);
iter = iter+1;
Ea(iter) = abs((x-x0)/x)*100;
while Ea(iter) >= tol && iter <= maxiter
x0 = x;
x = g(x0);
iter = iter+1;
Ea(iter) = abs((x-x0)/x)*100;
end
Ea = Ea(1:iter);
end

The code given below is in the function file ModifiedSecant.m

function [x,iter,Ea] = ModifiedSecant(f,delta,x0,tol)
%MODIFIEDSECANT Calculates the approximate root of f(x) using modified
%secant method.
% Input- f- anonymous function handle, delta- perturbation factor,
% x0 - initial guess, tol - Error tolerance
% Output- x - approximate root, iter - no.of iterations, Ea - approx. Error

% Initialize iteration counter
iter = 0;
% Set maximum number of iterations.
maxiter = 100;
% First iteration
x = x0 - delta*x0*f(x0)/(f(x0+delta*x0)-f(x0));
iter = iter+1;
Ea(iter) = abs((x-x0)/x)*100;

while Ea(iter) >= tol && iter <= maxiter
x0 = x;
x = x0 - delta*x0*f(x0)/(f(x0+delta*x0)-f(x0));
iter = iter+1;
Ea(iter) = abs((x-x0)/x)*100;
end
Ea = Ea(1:iter);

end

For the fixed point iteration to find the root of

defined the function g(x) as

We note that and the derivative is

which has an upper bound in the interval (2,4) as

because 1/x is a decreasing function. Thus the convergence criteria for fixed point iteration are met and we can apply fixed point iteration method to find the fixed point

(a) The result of fixed point method is given below

The ratio of current error to previous error seems to be tending to a limit that is less than 0.5 . This would indicate that the rate of convergence is linear, to say the least.

(b),(c),(d)

The comparison of results between fixed point and modified secant are shown below.

As can be seen fixed point iteration takes 14 iterations where as modified secant takes only 9 iterations. The final guess of the root would be approximately 2.1533. The reason modified secant method is faster is it has closer to quadratic rate of convergence whereas fixed point here as we saw has a linear rate of convergence.