Problems with integral in fsolve

13 Ansichten (letzte 30 Tage)
Oeer
Oeer am 9 Apr. 2019
Kommentiert: Walter Roberson am 9 Apr. 2019
Hey
I have three functions with three unknown variables, but I can't find a solution for them. the features look like this:
functions.PNG
I've tried to do the following code:
m = 0.31;
r = 0.05;
delta = 0.02;
y_bar = 2;
w_1 = 5;
w_2_bar = 5;
sigma = 1;
b_bar = 3;
z_0=[0.5,3,4];
sol=fsolve(@(z)([
w_1*(1-z(1))-z(2)-((1+delta)/((1-m)*(1+r)))*int((b_bar+(1-z(1))*(1-m)*x+(1-m)*(1+r)*z(2))*(1/(2*pi*sigma^2))*exp(-((x-w_2_bar)^2)/(2*sigma^2)),x,0,z(3))
+((1+delta)/(1+r))*int(((1-z(1))*k+(1+r)*z(2))*(1/(2*pi*sigma^2))*exp(-((k-w_2_bar)^2)/(2*sigma^2)),k,z(3),inf);
z(1)*w_1+int(z(1)*h*(1/(2*pi*sigma^2))*exp(-((h-w_2_bar)^2)/(2*sigma^2)),h,0,z(3))
-int((b_bar-m*((1-z(1))*y+(1+r)*z(2)))*(1/(2*pi*sigma^2))*exp(-((y-w_2_bar)^2)/(2*sigma^2)),y,0,z(3));
z(3)-(1/(1-z(1))*y_bar+((1+r)/(1-z(1)))*z(2))]),z_0);
Hope there is one who can correct my code so it works, thank you very much if it is you. :)

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 9 Apr. 2019
Code attached.
It is not fast code. You could improve the performance by changing the symbolic integrations into numeric integrations.
Or, since you are using symbolic integration, you could process your function once with symbolic z variables, and simplify. vpasolve() does a good job on what is left, or you can matlabFunction() and fsolve()
  1 Kommentar
Walter Roberson
Walter Roberson am 9 Apr. 2019
Having the int() inside the fsolve() is slow, and the int() have closed form representations if you use symbolic z. The performance gains from executing once symbolically and using the result with vpasolve() or fsolve() is substantial.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Torsten
Torsten am 9 Apr. 2019
Bearbeitet: Torsten am 9 Apr. 2019
function main
z_0 = [0.5,3,4];
options = optimset('TolFun',1e-8,'TolX',1e-8);
sol = fminsearch(@fun,z_0,options)
end
function res = fun(z)
m = 0.31;
r = 0.05;
delta = 0.02;
y_bar = 2;
w_1 = 5;
w_2_bar = 5;
sigma = 1;
b_bar = 3;
re(1) = w_1*(1-z(1))-z(2)-((1+delta)/((1-m)*(1+r)))*integral(@(x)(b_bar+(1-z(1))*(1-m)*x+(1-m)*(1+r)*z(2))*(1/(2*pi*sigma^2))*exp(-((x-w_2_bar).^2)/(2*sigma^2)),0,z(3),'ArrayValued',1)...
+((1+delta)/(1+r))*integral(@(k)((1-z(1))*k+(1+r)*z(2))*(1/(2*pi*sigma^2)).*exp(-((k-w_2_bar).^2)/(2*sigma^2)),z(3),inf,'ArrayValued',1);
re(2) = z(1)*w_1+integral(@(h)z(1)*h*(1/(2*pi*sigma^2)).*exp(-((h-w_2_bar).^2)/(2*sigma^2)),0,z(3),'ArrayValued',1)...
-integral(@(y)(b_bar-m*((1-z(1))*y+(1+r)*z(2)))*(1/(2*pi*sigma^2)).*exp(-((y-w_2_bar).^2)/(2*sigma^2)),0,z(3),'ArrayValued',1);
re(3) = z(3)-(1/(1-z(1))*y_bar+((1+r)/(1-z(1)))*z(2));
res = norm(re)
end

Produkte


Version

R2018b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by