Need help with bisection of this function, any advise or help would be great! thanks in advance to all!
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this is the code I used for the first part of the question and is correct:
h=0.1;
t=36;
N=(t/h);
t=zeros(1,N+1);
p=zeros(1,N+1);
t(1)=0;
p(1)=87;
for n=1:N
t(n+1)=t(n)+h;
u=sin((2*pi*t)/12);
H=15*[nthroot(u,15)+1]
p1(n+1)=p(n)+h*[(0.016*p(n)*(100-p(n)))-H(n)];
p(n+1)=p(n)+(h/2)*[[(0.016*p(n)*(100-p(n)))-H(n)]+[(0.016*p(n+1)*(100-p(n+1)))-H(n+1)]];
end
What I need to find now is the largest value of H for which the solution for P(t) does not be 0 at t = 36.
I know some of the code :
H=@(t)15*[nthroot(u,15)+1];
upper-limit=100;
lower-limit=1;
mid-point=(upper-limit+lower-limit)/2;
while lower-limit-upper-limit>0
if (?)*(?)<0
upper-limit=mid-point;
else
lower-limit=mid-point;
end
end
if anyone knows how to fill this parts of my code it would be a huge help!! just confused about how to account for H changing and P at the sametime!!
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