Doubt in declaring a function

4 vues (au cours des 30 derniers jours)
Pankaj
Pankaj le 28 Jan 2016
Commenté : Walter Roberson le 28 Jan 2016
I have a doubt regarding declaration of a function, kindly conside the following code
fun = @GVF
sqrt_estimt = fminsearch(@(theta)ODE_fit(fun,t,y,Q,theta(1),theta(2), iniVal), theta0);
..
function err = ODE_fit(fun, exp_t, exp_y, Q, theta(1), theta(2), iniVal)
% exp_y = Experimental observation at time exp_t
[t,y] = ode45(@(t,X)fun(t,X,n, S0, Q, theta(1),theta(2)),exp_t,iniVal);
err = sum(sum((y-exp_y).^2)); % compute error between experimental y and fitted y
end
..
function dy = GVF(x, y, Q, T, g)
% T = theta(1)
% g = theta(2);
A = y*T;
V = Q/A;
P = T+2*y;
R = A/P;
Sf = (.14*V)^2/(R^(4/3));
Fr = V/sqrt(g*y);
dy = (Sf)/(1-Fr^2);
end
..
The above does not work, but the following does: Is there a way to make the above way function? Thanks
sqrt_estimt = fminsearch(@(theta)ODE_fit(fun, t, y, Q, theta, iniVal), theta0);
..
function err = ODE_fit(fun, exp_t, exp_y, Q, theta, iniVal)
% exp_y = Experimental observation at time exp_t
[t,y] = ode45(@(t,X)fun(t,X,n, S0, Q, theta(1),theta(2)),exp_t,iniVal);
err = sum(sum((y-exp_y).^2)); % compute error between experimental y and fitted y
end
..
function dy = GVF(x, y, Q, T, g)
T = theta(1)
g = theta(2);
A = y*T;
V = Q/A;
P = T+2*y;
R = A/P;
Sf = (.14*V)^2/(R^(4/3));
Fr = V/sqrt(g*y);
dy = (Sf)/(1-Fr^2);
end

Réponse acceptée

Walter Roberson
Walter Roberson le 28 Jan 2016
No, there is not. You cannot name an element of a matrix in a function header. You can use two different variables though.
function err = ODE_fit(fun, exp_t, exp_y, Q, theta1, theta2, iniVal)
% exp_y = Experimental observation at time exp_t
[t,y] = ode45(@(t,X)fun(t,X,n, S0, Q, theta1,theta2),exp_t,iniVal);
err = sum(sum((y-exp_y).^2)); % compute error between experimental y and fitted y
end
  2 commentaires
Pankaj
Pankaj le 28 Jan 2016
Thank you Walter, you cleared my doubt.
Walter Roberson
Walter Roberson le 28 Jan 2016
Please Accept the answer to indicate you are finished with the Question.

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