How can I plot this function using Brent's method?
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Boran Kolcu
le 23 Nov 2020
Commenté : John D'Errico
le 24 Nov 2020
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/424623/image.png)
in the interval (0:0004; 0:0012).
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Manoj Kumar Koduru
le 23 Nov 2020
f=@(u) u*(1+0.7166/cos(25*sqrt(u)))-1.6901e-2; %Equation
a=-10;
b=+10;
err=0.001;
%Testing root is bracketed between [a b]
if f(a)*f(b) >=0
opts=struct('WindowsStyle','model','Interpreter','tex');
F=errordlg('The Root is out of the Brackets,increase a and b values'....
,'Roots Are Not Bracketed',opts); %Message Box
end
%Swapping a and b Contents
if abs(f(a)) < abs(f(b))
L=a; a=b; b=L;
end
c=a;
MFlag=1;
%Main Loop
delta =err; i=0;
while abs(b-a) >=err
i=i+1;
if f(a) ~=f(c)&&f(b) ~=f(c)
s=a*f(b)*f(c)/(f(a)-f(b))*(f(a)-f(c))+....
b*f(a)*f(c)/((f(b)*f(a))*(f(b)-f(c)))+...
c*f(a)*f(b)/((f(c)-f(a))*(f(c)-f(b))); %Inverse Quadratic Interpolation
else
s=b-f(b)*(b-a)/(f(b)-f(a)); %Secant method
end
if s<=(3*a+b)/4 || s>=b ||....
(MFlag==1 && abs(s-b) >= abs(b-c)/2) ||....
(MFlag==0 && abs(s-b) >= abs(c-d)/2) ||....
(MFlag==1 && abs(b-c) < abs(delta)) ||....
(MFlag==0 && abs(c-d) >= abs(delta))
s=(a+b)/2; %Bisection Method
MFlag=1;
else
MFlag=0;
end
%Calculate f(s)
d=c; c=b;
if f(a)*f(s) <0
b=s;
else
a=s;
end
%Swapping a and b contents
if abs(f(a)) < abs(f(b))
if abs(f(a)) < abs(f(b))
L=a; a=b; b=L;
end
end
ss(i,1)=s;
dd(i,1)=d;
ii(i,1)=i;
end
plot(1:i,ss,'Linewidth',2);
grid on
title('Brent Method')
xlabel('Number of Itterations')
ylabel('Root Contents')
2 commentaires
John D'Errico
le 24 Nov 2020
Please don't do obvious homework problems for students. They learn nothing from you, except to then post every homework question here.
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