Need help with bisection of this function, any advise or help would be great! thanks in advance to all!

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ved 2015-11-3

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https://ww2.mathworks.cn/matlabcentral/answers/252581-need-help-with-bisection-of-this-function-any-advise-or-help-would-be-great-thanks-in-advance-to-a

编辑： Geoff Hayes 2015-11-3

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