Doubt in declaring a function

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Pankaj
Pankaj el 28 de En. de 2016
Comentada: Walter Roberson el 28 de En. de 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

Respuesta aceptada

Walter Roberson
Walter Roberson el 28 de En. de 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 comentarios
Pankaj
Pankaj el 28 de En. de 2016
Thank you Walter, you cleared my doubt.
Walter Roberson
Walter Roberson el 28 de En. de 2016
Please Accept the answer to indicate you are finished with the Question.

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