How to implement a function within another one ?

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Martin Matin
Martin Matin on 18 Mar 2015
Commented: Martin Matin on 20 Mar 2015
Hi, I'd like to use the bisection method to find the root of a function "Fetoile" The method works but I have to iterate it for some values of a parameter "i". The problem is, the bisection function does not use "i".
Here is the code that gives me the results for "Fetoile" :
function Fetoile = q4(v)
% datas
vitesse_rotation= [83.7758040957278, 119.587013890941, 155.398223686155, 191.209433481369, 227.020643276582, 262.831853071796, 298.643062867009, 334.454272662223, 370.265482457437, 406.07669225265, 441.887902047864, 477.699111843078, 513.510321638291, 549.321531433505, 585.132741228718, 620.943951023932, 656.755160819146, 692.566370614359, 728.377580409573, 764.188790204786]
couple=[206.41064280052, 243.796265372056, 256.741969803922, 258.849308707433, 255.11836152428, 247.801181783381, 238.039957653591, 226.445249315901, 213.320493034029, 198.699623003111, 181.889713594511, 163.718280099678, 145.175662563345, 126.558747418095, 108.009552378088, 89.6025609484197, 71.3773301900352, 53.3538548851947, 35.5406860579882, 17.9395273160688]
rho=1.2;
S=2.28;
Cx=0.31;
mveh=1360;
g=9.81;
f0=0.0136;
f2=4.*10.^(-7);
ntrans=0.92;
atrans=1.04;
btrans=0.0025;
R=0.31;
i=[9.62,5.87,3.91,2.90,2.31];
teta=0;
%end of datas
for m = 1:5
Cmax=spline(vitesse_rotation,couple,i(m).*(v)/R);
Faero = (1/2)*rho*S*Cx*(v).^2;
Frlt = mveh*g*cos(teta)*(f0+f2*(v).^2);
Fgrav = mveh*g*sin(teta);
Fmotmax = ntrans*(i(m)/R)*Cmax;
Fetoile = Faero + Frlt + Fgrav - Fmotmax;
tabFetoile(m) = Fetoile;
end %end of for loop
disp('The results are :');
disp(tabFetoile);
And here is the code for the bisection, I tried a loop on the values of "vmin" are values that I'm sure are negatives.
function root = bishoz(f,vmin,vmax)
resultat = [];
for m = 1:5
vmin=[6,9.5,13.5,16.5,19];
vmax = 150;
fvmin(m) = f(vmin(m));
fvmax = f(vmax);
if fvmax*fvmin(m) > 0
disp('Valeurs de même signe.');
else
root = (vmax + vmin(m))/2;
err = abs(f(root));
end
while err > 1e-7
if fvmax*f(root)<0
vmin(m) = root;
else
vmax = root;
end %end of if
root = (vmax + vmin(m))/2;
err = abs(f(root));
end %end of while loop
resultat(m)=root;
end %end of for loop
disp(resultat);

Accepted Answer

John
John on 18 Mar 2015
I read through your example and it looks like the language functionality you are looking for are nested functions. Please see this for more info http://www.mathworks.com/help/matlab/matlab_prog/nested-functions.html
  3 Comments
Martin Matin
Martin Matin on 20 Mar 2015
It is okay, I finally did by making a script which I looped 5 times and I used a function handle to call q4.

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More Answers (1)

Xia
Xia on 18 Mar 2015
Sorry I read your code twice but got lost. Would you please epitomize your question by a simple example? Maybe it'll be a little better for other guys to get into this question..
  1 Comment
Martin Matin
Martin Matin on 18 Mar 2015
I have a function that depends of "i" and "v" I have another function, bisecting the first one.
I want to make to make the bisection vary in terms of "i". But because the reach of "i" is only the first one, it doesn't work. It gives me 5 times the last value of "i".

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