Newton-Raphson method

Question

Alexa Fernanda Cantú García 2021-1-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/712223-newton-raphson-method

回答： Monisha Nalluru 2021-1-11

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)

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Walter Roberson 2021-1-8

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

Walter Roberson 2021-1-8

在 MATLAB Online 中打开

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

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Answer 1

Monisha Nalluru 2021-1-11

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/712223-newton-raphson-method#answer_595123

在 MATLAB Online 中打开

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