How can I evaluate a function that is a function of other functions but all relate to the same parameter such as x

I have a bunch of functions that are related to each other through one parameter which is in my case altitude. I am having trouble understanding how MATlab is interpreting what I am writing.

func1=@(x) Some_Constant*x^2
  func2=Some_Other_Constant*func1(x)
  func3=Yet_Another_Constant*func2*func1

I would like to plot func3 vs x, how is Matlab interpreting what I wrote above and what is the right way to write the above code? Is there a good help page to answer my question? I have not been able to find one thus far.

Answer 1

[dpb]

Just write it like any other Matlab expression...

f1=@(x) C1*x.^2;
f2=@(x) C2*f1(x);
f3=@(x) C3*f1(x).*f2(x);

This way, of course, the values of the constants are embedded in the function definitions and are insensitive to subsequent changes in those values; if need to redefine C then define the functions as functions of (x,C) and pass the constants.

[Reid Castner]

I tried to write it like any other expression, however matlab wasn't understanding what I was trying to evaluate. Once I put my functions in parenthisis as you can see below, my expressions worked. I still don't understand why I need to put parenthisis around it, and if you have an explanation for that I would be glad to hear it. What is the difference between rho_h(h) and (rho_h(h))

rho=[.413,.467,.526,.590,.660,.736,.819,.909,1.007,1.112,1.225];
height=[10000:-1000:0];
p=polyfit(height,rho,2);
rho_h=@(h) p(1)*h^2+p(2)*h+p(3);
CL_H=@(h) 2*Weight/(rho_h(h))/V^2/S_w;
Drag_h=@(h) (.5*(rho_h(h))*V^2*S_w)*((f/S_w)+((CL_H(h))^2/pi/e/AR))*Cdi;

[Stephen23]

Nothing. There is no difference. Note that dpb's code does not use parentheses around the function calls. When I remove the parentheses from around the two rho_h calls in your code:

CL_H=@(h) 2*Weight/rho_h(h)/V^2/S_w;
Drag_h=@(h) (0.5*rho_h(h)*V^2*S_w)*((f/S_w)+((CL_H(h))^2/pi/e/AR))*Cdi;

it gives exactly the same answer (using the value 1 for all of the undefined variables):

>> Drag_h(1)
ans =
        1.132

[dpb]

NB: the real difference in the functions as you've written them vis a vis the way I did is the "dot" operator on the exponent. If you try to pass a vector h yours will fail for that reason. Perhaps you've misdiagnosed the cause of an error, maybe? We've not seen the error you received so can only guess, of course...