How can I plot this function using Brent's method?

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Boran Kolcu
Boran Kolcu am 23 Nov. 2020
Kommentiert: John D'Errico am 24 Nov. 2020
in the interval (0:0004; 0:0012).

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Manoj Kumar Koduru
Manoj Kumar Koduru am 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 Kommentare
Boran Kolcu
Boran Kolcu am 23 Nov. 2020
Thank you so much!
John D'Errico
John D'Errico am 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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