basic question on creating a function and plotting

Hello everyone

i have the expression to calculate: Ke=K0*e*a*(a-1)/(1+e-a*e)

K0, e are known values and a is my variable so i create the function Ke(a).

But using the command: Ke1=inline('K0*e*a*(a-1)/(1+e-a*e)') does not apply the given values of K0 and e, that is, it returns the function Ke1(K0,e,a), instead of Ke1(a)

I am using MATLAB 2009b

Thanks in advance

Answer 1

[Salaheddin Hosseinzadeh]

Hi Vaggelis,

How about using anonymous function?

f = @(k0,e,a) k0.*e.*a.*(a-1)./(1+e-a.*e)

then you can easily evaluate f by making a range for a, lets say -10 < a < 10 and keep k0 and e constant, lets assume k0 is 1 and e is 2 so you can use f as f(1,2,a)

a = -10:.1:10;
plot(a,f(1,2,a));

hope this helps. Good luck!

[vaggelis vaggelakis]

That is a good idea of course! But i am interesting in using different functions that involve K0 and e and it wouldn't be smart to evaluate them in the parenthesis each time i want to change their values.(for a different problem K0 and e may need to change)

[Salaheddin Hosseinzadeh]

I also considered that you wanna change their value in the future so that I suggested this solution.

Although I made a mistake in the code, defining the anonymous function. I missed a @ sign

If you define your function as

f = @(k,e,a)

in fact you're making them all independent variables, so f is a function k e and a and you may change their value anytime. you may change k e a and get different answeres.

[Salaheddin Hosseinzadeh]

It would be a good practice to use inline, when the equation is to be given by the user.

In your case your equation is fixed! You probably don't need to use inline or anonymous function, I don't know what your intention is but I guess you can just write the code free of any complexity and getting involve function definition and get your satisfactory response, don't confuse your self!

Answer 2

[Amir]

Please try this code and compare the results:

disp('%%%%% Run1  %%%%%%');
syms K0 e a; 
F=K0*e*a*(a-1)/(1+e-a*e);
Ke1=matlabFunction(F)

disp('%%%%% Run2  %%%%%%');
K0=2; e=2.71;
syms a; 
F=K0*e*a*(a-1)/(1+e-a*e);
Ke1=matlabFunction(F)

Results are as below:

%%%%% Run1  %%%%%%

Ke1 = 

      @(K0,a,e)(K0.*a.*e.*(a-1.0))./(e-a.*e+1.0)

%%%%% Run2  %%%%%%

Ke1 = 

      @(a)(a.*(a-1.0).*(-2.71e2./5.0e1))./(a.*(2.71e2./1.0e2)-3.71e2./1.0e2)