On Thu, Feb 21, 2019 at 1:39 PM Kai Torben Ohlhus <
address@hidden> wrote:
Dear Sir/Madam,
Greetings,
I am using Octave to solve and equation and to find a root using bisection method. The method works for one value, but in fact I want to calculate for a vector of values.
I wonder if this method of fining root or other methods support entering vector array of values?
Regards
Best,
Kai
> Dear Kai,
>
> Thank you for the email,
>
> I have not used those provided in the link yet. I will type my example and uploaded as Octave file.
>
> X=85:-1:50;
> Y=5:1:40;
> Xr=X*pi/180;
> Yr=Y*pi/180;
> N=(Xr.*cosd(Y))-(Yr.*cosd(X));
> D=(cosd(Y)-cosd(X));
> C=N./D;
>
> for j= length(C);
> y= @(F) (F*pi/180)+cotd(F)-C(j);
> end
>
> FL=30;
> FR=70;
>
> for i= 1:1000
> FM=(FL+FR)/2;
> if y(FL)*y(FM) < 0
> FR=FM;
> else
> FL=FM;
> end
> if abs (y(FL)) < 1.0E-10;
> break
> end
> end
>
> fprintf('Number of iterations: %d\n', i);
> fprintf('phi_note: %f\n', FL);
>
> Regards
> Peshawa Salih
>
Dear Peshawa Salih,
Please use bottom-posting [1] for this mailing-list and keep the mailing-list in the CC, as many people are having better ideas than I have ;-)
Your example is not clear to me. Your problem seems to be: Find a real scalar `F`, such that `y(F) = 0`, where `y` : R --> R^36. But in your code despite the for loop over 36 elements, you get a function `y` of scalar input and output and only the last C(36) is taken for `y`.
And why do you write the bisection method yourself?
Best,
Kai
Bisection2.m