Need help in MATLAB

[OfficerKenny]

Ok so I have to do this problem for a class



I am confused because aren't you supposed to know the tolerance before you are able to do the iterations?  If you are able to calculate tolerance how does one do so?

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[Conan]

a) is telling you to do 3 iterations
b) is telling you to figure out how many iterations it takes to reach a value with the tolerance of 10^-5, given the starting bracket.
i presume tolerance here means the distance of the current value from the minima.

[OfficerKenny]

a) is telling you to do 3 iterations
b) is telling you to figure out how many iterations it takes to reach a value with the tolerance of 10^-5, given the starting bracket.
i presume tolerance here means the distance of the current value from the minima.

Yeah I was confused by what tolerance meant as well, I assume that tolerance is the number of digits the midpoint has.  Like in this script here(This is from a different problem):

Code: [Select]

clear all;
xL = 0;
xR = 1.5;
f = @(x) (sin(x) - cos(3*x));
alpha = 10^-6;
maxloops = 25;
for i = 1:maxloops
xM = (xL+xR)/2; %xM = x*
if f(xM)*f(xR)<0
 xL=xM;
elseif f(xM)*f(xL)<0
 xR=xM;
end
if abs(f(xM))<alpha
 break
end
end
xM;
a = xM;
disp("The value of x* is");
disp(a);
disp("The value of f(x*) is");
disp(f(xM));
disp("The number of iterations attempted was");
disp(maxloops);
disp("The maximum number of iterations performed was")
disp(i);
disp("A grpah of the funtion with xL=0 to xR=1.5 is now being
displayed");
Problem3Plot;
I set it so that the script stops once it reaches a specific alpha value, the tolerance I think is the same thing as whats happening with alpha.  At least I hope so because I already submitted the assignment.