finding root using false position method

Good evening\morning

I try to write a code that calculate the root of a nonlinear function using False Position Method, but I get an infinite loop. I use the same loop for the Bisection Method and it's work.

     clc
    x0 = input('enter the value of x0 = ');
     x1 = input('enter the value of x1 = ');

     tolerance=input('inter the tolerance =  ');

     f =@(x) sin(2*pi*x)+ exp(1.2*x) + x - 2.5;

   for i=0:inf

            x2= x1 - (f(x1)* (x1-x0)/(f(x1)-f(x0))) 

            c = f(x2)

            absolute_c= abs(c);
            if  absolute_c < tolerance
                 break 
            end

            if f(x0)*c <0
                          x1=x2;
                  continue
            else
                x0=x2;
                continue
            end
     end
i

[Samanta]

1.Use the False Position Method to find the root of the equation e^x−3x=0. Start with the initial guesses x_0=0 and x_1 =1

Answer 1

[Fangjun Jiang]

Do a plot to find out the curve. If you put the right initial value, it could solve the problem.

ezplot(f)

x0=-6
x1=6
Tolerance=0.001

It reached the end at i==590

A better approach is to check whether f(x0)*f(x1)<0 right after the input().

[Ayda]

thanxs

[Fangjun Jiang]

A better approach is to check whether f(x0)*f(x1)<0 right after the input().

Answer 2

[Mahesha MG]

http://numericalcomputation.blogspot.in/2012/11/matlab-false-position-method.html

Answer 3

[Polash Roy]

clc
clear all
close all
f=@(r) exp((-5e-3)*r)*cos((sqrt(2000-.01*r^2)*.05))-.01;
a=100;
b=550;
for i=1:10
    x0=a;
    x1=b;
    fprintf('\n Hence root lies between (%.4f,%.0f)',a,b)
    x2(i)=x0-(x1-x0)/(f(x1)-f(x0))*f(x0);
    if f(x2(i))*f(x0)>0
        b=x2(i);
    else
        a=x2(i);
    end
    fprintf('\n Therefore, x2=%.4f \n Here, f(x20=%.4f',x2(i),f(x2(i)))
    p=x2(i);
end
for i=1:10
    eror(i)=p-x2(i);
end
Answer=p
plot(eror)
grid on;
title('Plot of error')
xlabel('iterations')
ylabel('Error')

Answer 4

[Aman Pratap Singh]

clc

% Setting x as symbolic variable
syms x;

% Input Section
y = input('Enter non-linear equations: ');
a = input('Enter first guess: ');
b = input('Enter second guess: ');
e = input('Tolerable error: ');

% Finding Functional Value
fa = eval(subs(y,x,a));
fb = eval(subs(y,x,b));

% Implementing Bisection Method
if fa*fb > 0 
    disp('Given initial values do not bracket the root.');
else
    c = a - (a-b) * fa/(fa-fb);
    fc = eval(subs(y,x,c));
    fprintf('\n\na\t\t\tb\t\t\tc\t\t\tf(c)\n');
    while abs(fc)>e
        fprintf('%f\t%f\t%f\t%f\n',a,b,c,fc);
        if fa*fc< 0
            b =c;
            fb = eval(subs(y,x,b));
        else
            a =c;
            fa = eval(subs(y,x,a));
        end
        c = a - (a-b) * fa/(fa-fb);
        fc = eval(subs(y,x,c));
    end
    fprintf('\nRoot is: %f\n', c);
end

[Walter Roberson]

You should never eval() a symbolic expression. Symblic expressions are not character vectors containing MATLAB code. Sometimes they look like MATLAB code, but they contain parts that are not MATLAB, and some of the functions have a different parameter order than MATLAB uses.

If you have fully substituted for all numeric variables, then

fa = double(subs(y,x,a));
fb = double(subs(y,x,b));

Answer 5

[Samanta]

Use the False Position Method to find the root of the equation e^x−3x=0. Start with the initial guesses x_0=0 and x_1 =1

[Walter Roberson]

This solution does not explain how to use False Position Method, so it is not clear how it answers the question?