Bisection Method Code MATLAB

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Emmanuel Pardo-Cerezo
回答： Prosun 2024-9-24
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
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Aristi Christoforou
Aristi Christoforou 2021-4-14
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
Uttsa 2024-7-3
Whats the use of "eps" can you elaborate?

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回答（6 个）

David Hill
David Hill 2019-10-4
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);
  1 个评论
Justin Vaughn
Justin Vaughn 2022-10-10
Thank you for this because I was not sure of how to easily send a functino into my method's function. yours helped tremendously!

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SHUBHAM GHADOJE
SHUBHAM GHADOJE 2021-5-29
编辑：Walter Roberson 2024-7-12
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
Prathamesh Purkar
编辑：Walter Roberson 2021-12-3
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
narendran
narendran 2022-7-2
5cosx + 4.5572 -cos30cosx-ssin30sinx
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Walter Roberson
Walter Roberson 2022-7-2
Ran in:
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
Walter Roberson 2024-7-3
Note by the way that cos(30) is cos of 30 radians. It seems unlikely that is what is desired.

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Aman Pratap Singh
Aman Pratap Singh 2021-12-3
编辑：Walter Roberson 2021-12-3
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);
  1 个评论
Walter Roberson
Walter Roberson 2021-12-3
Ran in:
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'

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Prosun
Prosun 2024-9-24
% Clearing Screen
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+b)/2;
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;
else
a =c;
end
c = (a+b)/2;
fc = eval(subs(y,x,c));
end
fprintf('\nRoot is: %f\n', c);
end

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