converging criteria problem in for loop in bisection method of finding roots.

problem: finding roots of the polynomial x3-x-11 by using bisection method.

Code:

close all
clear all
y = @(x) (x.^3)-x-11;
fplot(y);
grid on
xlo = input('Please enter the initial lower limit of x');
xuo = input('Please enter the initial upper limit of x');
x = (xlo+xuo)./2; % 1st approximation for finding root
for i = 1:1:1000
    f = y(x); % function value for approximation of x
    if f<0
        xlo = x;
        xuo = xuo;
    elseif f>0
        xlo = xlo;
        xuo = x;
    end
    x(i) = (xlo+xuo)./2;
    if abs(x-x(i))<=0.0001
        break
    else
        x = x(i);
    end
end
disp(i);
disp(x);

Comments: the converging criteria if abs(x-x(i))<=0.0001

break

else

x = x(i); stucks. why i can't find out. if i use == sign instead of <= then whole loop run and i get the desires root. but when i use <= this loop stucks. Please can anyone help me out.

Thanks in advance.

Answer 1

[Sam Chak]

Hi @MD

You can change the for-loop to a while-loop and x(i) to xnew. Also, insert the initial count. The new search interval criteria remains unchanged.

y = @(x) (x.^3) - x - 11;
fplot(y, [1 3])
grid on, xlabel x, ylabel y

xlo   = 2.0;
xuo   = 2.5;
x     = (xlo + xuo)/2;  % 1st approximation for finding root

count = 0;              % initial iteration count

while true              % can also use count < 1000 to break the loop
    f = y(x);           % function value for approximation of x
    if f < 0
        xlo = x;
        xuo = xuo;
    elseif f > 0
        xlo = xlo;
        xuo = x;
    end
    xnew = (xlo + xuo)/2;
    
    count = count + 1;  % increase count by 1
    
    if abs(x - xnew) <= 0.0001
        break
    else
        x = xnew;
    end
end
disp(count);            % change i to display count

    12

disp(x);

    2.3737

[MD]

Thank you Very much for your answer. I have also solved the problem by sligth modification. Please check this out:

close all
clear all
% Finding normal polynomial roots of single variable by using bisection method.
y = @(x) (x.^3)-x-11;
fplot(y);
grid on

xlo = 2.0;
xuo = 2.5;
for i = 1:1:1000
    x(i) = (xlo+xuo)./2; % 1st approximation for finding root
    f(i) = y(x(i)); % function value for approximation of x
    if f(i)<0
        xlo = x(i);
        xuo = xuo;
    elseif f(i)>0
        xlo = xlo;
        xuo = x(i);
    end
    x(i+1) = (xlo+xuo)./2;
    if abs(x(i+1)-x(i))<=0.00001
        break
    else
        x(i) = x(i+1);
    end
end
disp(i);

    15

disp(x(:,length(x)));

    2.3736

Wish you all the best.

[Sam Chak]

Yup @MD, your updated code for the bisection root-finding method works for the specific problem.

I also tested two other problems below, and one works, but the other doesn't.

close all
clear all
% Finding normal polynomial roots of single variable by using bisection method.

%% Test 1
% y = @(x) (x + 0.5).^2;  % <-- this one works

%% Test 2
y = @(x) (x - 0.5).^2;  % <-- this one does not

fplot(y, [-1 1]);
grid on

xlo = -1.0;
xuo =  1.0;
for i = 1:1:1000
    x(i) = (xlo+xuo)./2; % 1st approximation for finding root
    f(i) = y(x(i)); % function value for approximation of x
    if f(i)<0
        xlo = x(i);
        xuo = xuo;
    elseif f(i)>0
        xlo = xlo;
        xuo = x(i);
    end
    x(i+1) = (xlo+xuo)./2;
    if abs(x(i+1)-x(i))<=0.00001
        break
    else
        x(i) = x(i+1);
    end
end
disp(i);

    17

disp(x(:,length(x)));

   -1.0000

Answer 2

[Walter Roberson]

x = (xlo+xuo)./2; % 1st approximation for finding root

x is a scalar at that point -- so x(1) is the only defined element.

for i = 1:1:1000

for loop. i starts at 1, then next iteration will be 2, and so on.

        x(i) = (xlo+xuo)./2;

before that, x is a scalar. But here you store into x(i), which on the second iteration would be x(2), third iteration would be x(3) and so on. So afterwards x will be a vector with x(1) having the old value, then a bunch of zeros, and then x(i) having the new value.

            if abs(x-x(i))<=0.0001

For i > 1 then x has become a vector. You take the vector minus the most recent result, x(i) . You get a vector of resullts. You abs() that getting a vector of results. You compare the vector of results to 0.0001 . But when you ask if to process a non-scalar value, if considers the test to pass only if all of the values being tested are non-zero (that is, all of the values must be true). If even one entry in the vector did not satisfy abs(x-x(i))<=0.0001 then the if fails.

                    else
        x = x(i);

and when the if fails, you replace the vector x (of length i) with the scalar x(i) . So the next iteration, x is back to being a scalar again...

[MD]

Thank you for your academic brief but to the point explanation. My best wishes. Here i solved by slight modification:

close all

clear all

% Finding normal polynomial roots of single variable by using bisection

% method.

y = @(x) (x.^3)-x-11;

fplot(y);

grid on

xlo = input('Please enter the initial lower limit of x');

xuo = input('Please enter the initial upper limit of x');

for i = 1:1:1000

x(i) = (xlo+xuo)./2; % 1st approximation for finding root

f(i) = y(x(i)); % function value for approximation of x

if f(i)<0

xlo = x(i);

xuo = xuo;

elseif f(i)>0

xlo = xlo;

xuo = x(i);

end

x(i+1) = (xlo+xuo)./2;

if abs(x(i+1)-x(i))<=0.00001

break

else

x(i) = x(i+1);

end

end

disp(i);

disp(x(:,length(x)));

Answer 3

[MD]

close all

clear all

% Finding normal polynomial roots of single variable by using bisection

% method.

y = @(x) (x.^3)-x-11;

fplot(y);

grid on

xlo = input('Please enter the initial lower limit of x');

xuo = input('Please enter the initial upper limit of x');

for i = 1:1:1000

x(i) = (xlo+xuo)./2; % 1st approximation for finding root

f(i) = y(x(i)); % function value for approximation of x

if f(i)<0

xlo = x(i);

xuo = xuo;

elseif f(i)>0

xlo = xlo;

xuo = x(i);

end

x(i+1) = (xlo+xuo)./2;

if abs(x(i+1)-x(i))<=0.00001

break

else

x(i) = x(i+1);

end

end

disp(i);

disp(x(:,length(x)));