## Doubt in declaring a function

### Pankaj (view profile)

on 28 Jan 2016
Latest activity Commented on by Walter Roberson

on 28 Jan 2016

### Walter Roberson (view profile)

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

### Tags ### Walter Roberson (view profile)

on 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

Pankaj

### Pankaj (view profile)

on 28 Jan 2016
Thank you Walter, you cleared my doubt.
Walter Roberson

### Walter Roberson (view profile)

on 28 Jan 2016
Please Accept the answer to indicate you are finished with the Question.