Newton-Raphson method

I have the function f(x)=sen(x)-2x^2 with x0=1.1 and iterating until the approximate relative error (Ea) is less than the estimated error Es=0.02%. The exact value of the root is 0.4810 and I need to get the find the percent true relative error (Ev).

This is what i have done, but it doesn't run with the trigonometric function.

%This program is for you to run it, is problem 2 of HW1.
f=@(x) x.^3-(12.1*x.^2)+(45.86*x)-52.976 %f(x) of Problem 2
x0=input("Enter x0_");%Your initial "x"
EE=input("Enter EE__");%your Absolute relative approximated percentage error
vv=input("Enter vv_");%Is the true value, the real known root
nrhw1(f,x0,EE,vv)

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

[Walter Roberson]

f = @(x) sin(x)-2*x.^2

Answer 1

[Monisha Nalluru]

Relative error is used in Newton Raphson method inorder to find whther the root matches the given tolerance.

Relative error can bee found using the following formula relative_error = abs( f(x+1) -f(x)/f(x+1)) * 100

As an example

 function [r] = newton(x1)
 n = 1000;
 tol_error = 0.0002;
 r = x1;
 for t=1:n
    x_plus = x1 - (my_func(x1)/my_func_diff(x1));
    relative_error(t) = abs((x_plus - x1)/(x_plus))*100;  % calculating relative error
    disp(relative_error(t));
    fun_val(t) = my_func(x1);  
    if (relative_error(t) < tol_error)  % checking whether relative error is less than tolerance
        display(approx_error(t));
        r = x_plus;
        break;                 
    else
        x1 = x_plus;
    end
 end
 end
 
 function y = my_func_diff(x)
y = cos(x)-4*x;
end
function y = my_func(x)
y = sin(x)-2*x^2;
end

Hope this helps!