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Coding a function to find zeros using a variant of the bisection/secant method
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Lavorizia Vaughn
on 28 Sep 2021
Commented: Walter Roberson
on 29 Sep 2021
Hello, I have a programming assignment where I have to implement a matlab function that is a variant of the bisection and secant method. Please see attachment for exact details. I am having problems with the code. Will someone help me with the necessary fixes? My code is as follows: —————————————————————————————————————-
if true
% code
function p = bisection(f, a, b, tol)
w=1;
for i=1:100
p= a +(w*f(a)*(a-b)/f(b)-w*f(a));
fprintf(a,b,p,f(p));
if f(p)*f(b)>0
w=0.5;
else
w=1;
a=b;
end
b=p;
if abs(b-a)<tol
abs(f(p))<tol
break;
end
end.
0 Comments
Accepted Answer
Walter Roberson
on 28 Sep 2021
No, no error about needing to use () . However,
output = bisection(@(x) sin(x)+3*cos(x).^2, -pi/2, pi/2, 1e-5)
function p = bisection(f, a, b, tol)
w=1;
for i=1:100
p= a +(w*f(a)*(a-b)/f(b)-w*f(a));
fprintf(a,b,p,f(p));
if f(p)*f(b)>0
w=0.5;
else
w=1;
a=b;
end
b=p;
if abs(b-a)<tol
abs(f(p))<tol
break;
end
end
end
This is because the first parameter to fprintf must be one of:
The first parameter you are passing is a, which is numeric but is not a valid file identifier.
6 Comments
Walter Roberson
on 29 Sep 2021
fprintf('Lovorizia, the value for a was %g .\nAnd b was %g .\n, Then there was p, which was %g .\n And lastly, if you still care, f(p) was %g .\n', a, b, p, f(p));
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