Bisection Method Code MATLAB

Problem 4 Find an approximation to (sqrt 3) correct to within 10−4 using the Bisection method (Hint: Consider f(x) = x 2 − 3.) (Use your computer code)

I have no idea how to write this code. he gave us this template but is not working. If you run the program it prints a table but it keeps running. for some reason the program doesnt stop.

tol = 1.e-10; 
a = 1.0;  
b = 2.0; 
nmax = 100; 
% Initialization  
itcount = 0; 
error = 1.0; 
% Graph of the function 
xval = linspace(a,b,100); 
for i=1:100 
    fval(i) = func(xval(i)); 
end 
plot(xval,fval); 
grid on;  
hold on; 
% iteration begins here 
while (itcount <= nmax && error >= tol) 
itcount = itcount + 1; 
% Generate and save iteratres 
x = a + (b-a)/2; 
z(itcount) = x; 
fa = func(a);  
fb = func(b); 
fx = func(x);  
error = abs(fx); 
%  error = abs(x - xold); 
if (error < tol) 
x_final = x; 
else 
if (fa*fx < 0) 
% root is between a and x 
b = x; 
else 
% root is between x and b 
a = x; 
end 
end 
plot(z(1:itcount),zeros(itcount,1),'r+'); 
pause(5)  
end 
if (itcount < nmax); 
val = func(x); 
fprintf(1,'Converged solution after %5d iterations',itcount); 
fprintf(1,' is %15.7e, %e \n',x_final, val); 
else fprintf(1,'Not converged after %5d iterations',nmax); 
end 
function val = func(x) 
%val = x^3 + 4 * x^2 - 10; 
val = x^3 - x - 3; 
%val = sin(x);  
end 

[tala saffarini]

can u help me??

[Aristi Christoforou]

function[x]=bisect(m)
a=1;
b=3;
k=0;
while b-a>eps*b
    x=(a+b)/2
    if x^2>m
        b=x
    else
        a=x
    end
    k=k+1
end

[Uttsa]

Whats the use of "eps" can you elaborate?

Answer 1

[David Hill]

function c = bisectionMethod(f,a,b,error)%f=@(x)x^2-3; a=1; b=2; (ensure change of sign between a and b) error=1e-4
c=(a+b)/2;
while abs(f(c))>error
    if f(c)<0&&f(a)<0
        a=c;
    else
        b=c;
    end
    c=(a+b)/2;
end

Not much to the bisection method, you just keep half-splitting until you get the root to the accuracy you desire. Enter function above after setting the function.

f=@(x)x^2-3;
root=bisectionMethod(f,1,2);

[Justin Vaughn]

Thank you for this because I was not sure of how to easily send a functino into my method's function. yours helped tremendously!

Answer 2

[SHUBHAM GHADOJE]

function c = bisectionMethod(f,j,k,error)
%f=@(x)x^2-3;
%j=1;
%k=2;
%(ensure change of sign between a and b)
%error=1e-4
c=(j+k)/2;
while abs(f(c))>error
    if f(c)<0&&f(a)<0
        j=c;
    else
        k=c;
    end
    c=(j+k)/2;
end

Answer 3

[Prathamesh Purkar]

tol = 1.e-10;
a = 1.0;
b = 2.0;
nmax = 100;
% Initialization
itcount = 0;
error = 1.0;
% Graph of the function
xval = linspace(a,b,100);
for i=1:100 
    fval(i) = func(xval(i));
end 
plot(xval,fval);
grid on;
hold on; 
% iteration begins here
while (itcount <= nmax && error >= tol)
    itcount = itcount + 1;
    % Generate and save iteratres
    x = a + (b-a)/2;
    z(itcount) = x;
    fa = func(a);
    fb = func(b);
    fx = func(x);
    error = abs(fx);
    % error = abs(x - xold);
    if (error < tol)
        x_final = x;
    else
        if (fa*fx < 0)
            % root is between a and x
            b = x;
        else
            % root is between x and b
            a = x;
        end
    end
    plot(z(1:itcount),zeros(itcount,1),'r+');
    pause(5)
end
if (itcount < nmax);
    val = func(x);
    fprintf(1,'Converged solution after %5d iterations',itcount);
    fprintf(1,' is %15.7e, %e \n',x_final, val);
else
    fprintf(1,'Not converged after %5d iterations',nmax);
end

function val = func(x)
%val = x^3 -x + 1;
val = x^3 -x + 1;
%val = sin(x);
end

Answer 4

[narendran]

5cosx + 4.5572 -cos30cosx-ssin30sinx

[Image Analyst]

@narendran it doesn't look like this is an answer to the original question. I think you posted this in the wrong place.

[Walter Roberson]

syms x

y = 5*cos(x) + 4.5572 - cos(30)*cos(x)-sin(30)*sin(x)

y = 

fplot(y, [-20 20]); yline(0)

vpasolve(y,x)

ans = 

[Walter Roberson]

Note by the way that cos(30) is cos of 30 radians. It seems unlikely that is what is desired.

Answer 5

[Aman Pratap Singh]

f = @(x)('x^3-2x-5');
a = 2;
b = 3;
eps = 0.001;
m = (a+b)/2; 
fprintf('\nThe value of, after bisection method, m is %f\n', m);
while abs(b-a)>eps 
    if (f(a)*f(m))<0
        b=m;
    else
        a=m;
    end
    m = (a+b)/2;
end
fprintf('\nThe value of, after bisection method, m is %f\n', m);

[Walter Roberson]

f = @(x)('x^3-2x-5');

That means that f will become a function handle that, given any input, will return the character vector ['x', '^', '3', '-', '2', 'x', '-', '5'] which is unlikely to be what you want to have happen.

f(0)

ans = 'x^3-2x-5'

f(1)

ans = 'x^3-2x-5'

f(rand(1,20))

ans = 'x^3-2x-5'

Answer 6

[albertson]

f(x) = x3-2x-5 (xl = 0, xu = 4, x0 = 0, εs = 0.0001%)

Answer 7

[AMAN DEV]

bisection method

equation (x^3-0.165(x^2)+3.993*10^-4

[Walter Roberson]

This does not appear to be a solution to the question?