Fzero with defined scalar parameters

2 vues (au cours des 30 derniers jours)
Francesco
Francesco le 16 Déc 2014
Hello, I am approaching to MATLAB by myself, but I have a problem: I'm writing a script, that I will simplify as follow
k=0.1805;
n=4;
r_t=1.78e-3;
I=151;
d_r=0.06;
d=0.0009/d_r;
delt=10;
f=0.907; % Parameter that I want to set and change when needed
r_c=r_t*sqrt(f);
q_r=0.0072*((1+2/n)/((1+(4/n))^(3/10)))*(I^(7/5))/(r_t*sqrt(f))^(12/5)
y=k*delt/((q_r)*d_r*r_t*sqrt(f))
x=fzero('( exp(-(x^2)))/(x*sqrt(pi))+erf(x) -1-0.1805*10/((0.0072*((1+2/4)/((1+(4/4))^(3/10)))*(151^(7/5))/(1.78e-3*sqrt(0.907))^(12/5)*(0.06)*(1.78e-3)*sqrt(0.907)))',3)
Now the problem arises when I compute x: it is computed as the zero of exp(-(x^2)))/(x*sqrt(pi))+erf(x) -y , but I can't introduce "y" in the string of fzero, so I've inserted all the numerical expression. This is a big deal, because everytime I want to change "f" I have to find it in the string and change its value, while I would like to substitute it only once, when I define it. If I want to change any other parameter, the problem is still the same... I've read about fmin, function_handle etc etc, but I was not able to understand those tools, because by now I know very little about MATLAB. Could you kindly help me to solve this particular problem? Please, I beg you also to be detailed, because nothing is so obvious to me, now. Thanks a lot!

Réponse acceptée

Matt J
Matt J le 16 Déc 2014
Modifié(e) : Matt J le 16 Déc 2014
You should not be using strings to specify the function whose root you are trying to find. It's a deprecated syntax. You should be using anonymous functions
y=1;
fun=@(x) exp(-(x^2))/(x*sqrt(pi))+erf(x) -y;
x=fzero(fun,x0)
For other options, see also Passing Extra Parameters.

Plus de réponses (2)

Thorsten
Thorsten le 16 Déc 2014
If FUN is parameterized, you can use anonymous functions to capture the problem-dependent parameters. Suppose you want to solve the equation given in the function myfun, which is parameterized by its second argument c. Here myfun is an M-file function such as
function f = myfun(x,c)
f = cos(c*x);
To solve the equation for a specific value of c, first assign the value to c. Then create a one-argument anonymous function that captures that value of c and calls myfun with two arguments. Finally, pass this anonymous function to FZERO:
c = 2; % define parameter first
x = fzero(@(x) myfun(x,c),0.1)
In your case use y instead of c and define myfun as exp(-(x^2)))/(x*sqrt(pi))+erf(x)

Francesco
Francesco le 16 Déc 2014
Thank you both, you help me to save a lot of time!

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