How to find the maximum and minimum values of a piecewise function?

I have defined the following piecewise function:

s = piecewise(0<theta<50*pi/180, 0, 50*pi/180<theta<100*pi/180, curv_arm, 100*pi/180<theta<2*pi, curv_par);

where `curv_arm` and `curv_par` are functions that I wrote at the beginning. And I used the command fplot to display the plot of this function. Now I want to obtain the maximum and the minimum value of this function in the interval [0,pi] but I'm getting the following error:

"s" was previously used as a variable,
conflicting with its use here as the name of a function 
or command

I'm trying with the following line:

[x,fval] = fminbnd(@s,0,pi)

The plot is the following:

[Shivam Kaushik]

sigmoid_exact = @(x)(((-4<x) & (x<4))*1/(1+exp(-x)));

sigmoid_approx = @(x) (((0<x) & (x<4))*(1-0.5*((1-0.25*x)^(2)))+((-4<x) & (x<0))*(0.5*((1+0.25*x)^(2)))+(x>4)*1+(x<-4)*0);

error = @(x) (sigmoid_exact-sigmoid_approx) ;

how to obtain plot
piecewise non linear approximation of sigmoid function 
i have to minimze error

Answer 1

[Star Strider]

I’m not quite sure what you want to do. This will get you started:

curv_arm = 3;
curv_par = 5;
s = @(theta, curv_arm, curv_par) [0*((0<theta) & (theta<50*pi)) + curv_arm.*((50*pi/180<theta) & (theta<100*pi/180)) + curv_par.*((100*pi/180<theta) & (theta<2*pi))];

[x,fval] = fminbnd(@(theta)s(theta, curv_arm, curv_par),0,pi)

You cannot use a symbolic expression with fminbnd, and ‘matlabFunction’ will not create a function from your piecewise function. You can easily create your own function for numeric applications using essentially the same code, and ‘logical indexing’, as I did here.

Make any necessary changes to get the result you want.

[Edgar Dominguez Ugalde]

What I want to do is obtain the maximum and minimum values of the plot that I posted above in a certain range. I know that the minimum value is 0, but obviusly not the maximum. I think I forgot to define "s" as a function of theta and curv_arm and curv_par values are the following:

 curv_arm = L/2*(1-cos((pi*(theta-betha1))/betha1))
 curv_par = L*(1-(theta-(100*pi/180))/(betha2)+((1)/(2*pi))*(sin(2*pi*(theta-(100*pi/180))/(betha2))))

where L = 2.5, betha1 = 50*pi/180, betha2 = 260*pi/180

[Star Strider]

That makes things considerably easier.

This works:

L = 2.5;
betha1 = 50*pi/180;
betha2 = 260*pi/180;
curv_arm = @(theta) L/2*(1-cos((pi*(theta-betha1))/betha1));
curv_par = @(theta) L*(1-(theta-(100*pi/180))/(betha2)+((1)/(2*pi)).*(sin(2*pi*(theta-(100*pi/180))/(betha2))))
s = @(theta) [0*((0<theta) & (theta<50*pi)) + curv_arm(theta).*((50*pi/180<theta) & (theta<100*pi/180)) + curv_par(theta).*((100*pi/180<theta) & (theta<2*pi))];

[x,fval] = fminbnd(@(theta) -s(theta),0,pi)

x =
     1.7454

fval =
    -2.5000

Since fminbnd searches for the minimum, it is necessary to negate ‘s’ in the fminbnd call, essentially ‘flipping’ it upside-down. The result then is that the maximum is at 1.7454 radians, and the function maximum is +2.5000 (since you need to ‘negate the negation’ to get the correct value for the maximum).

Note that it is necessary to create ‘curv_arm’ and ‘curv_par’ as functions of ‘theta’ and then call them as functions in the piecewise function.

[Edgar Dominguez Ugalde]

Thank you! That was exacty what I was trying. Now I would like to ask you: if I want to differentiate the function "s" with respect to "theta" and then plot the result, how can I do it?

[Star Strider]

My pleasure!

Doing it symbolically is likely not an option because of the discontinuities, especially if you later want to evaluate it numerically. I would evaluate it over a suitably fine ‘theta’ vector and then use the gradient function to calculate the numerical derivative. The advantage to using gradient is that the result vector is the same length as the argument vector, unlike diff with a result that is one element shorter than the argument vector.

[Edgar Dominguez Ugalde]

OK, I'll try. I really appreciate your help.

[Star Strider]

As always, my pleasure.