How to multiply function handles?

Hello,

I would like to calculate and to plot the following equation:

f(t,x) = (beta_hat*s(t,x)*[I(t,x)]^(beta_hat-1))*exp(-[I(t,x)]^beta_hat)

where:

      s(t,x): step-function(330 for x=[0;2000])
                           (350 for x=[2000;3000])
                           (390 for x=[3000;4000])

     I(t,x): Integral(exp(-B_hat/x(u))/C_hat)) from x=0 to t
     beta_hat, C_hat and B_hat are known parameters

My MATLAB-script looks like that:

beta_hat = 4.2915822 
B_hat = 1861.6186657 
C_hat = 58.9848692  

%%Step-function x(t)

syms t 
y(t) = (exp(-B_hat/((heaviside(t)-heaviside(t-2000))*(330)+(heaviside(t-2000)-heaviside(t-3000))*(350)+...
       (heaviside(t-3000)-heaviside(t-4000))*(390))))/C_hat;  
fnum=matlabFunction(y);
Inum=@(x)integral(fnum,0,x);

f = @(x)(beta_hat*fnum(x).*Inum(x).^(beta_hat-1)).*exp(-(Inum(x).^beta_hat))
ezplot(f,[0,14000])

But if I plot my function f the figure isn´t right. There should be a smooth curve, similar to the bell-shaped curve.

Does anybody recognize my fault?

Answer 1

[Walter Roberson]

What value are you assuming that heaviside(0) has?

As you are working with the Symbolic toolbox anyhow, have you considered doing a symbolic integration with int() instead of changing to a numeric function and using integral() ?

[Max]

Hello Walter,

heaviside(t=0) should be 330 and heaviside(t=4000) should be 390. But I didn´t know how to realize that step-function in MATLAB which I could use for calculation the Integral.

I´ve also tried that: I=int(y); f_new = (beta_hat*y.*(I)^(beta_hat-1)).*exp(-((I).^beta_hat)) ezplot(f_new) But,I couldn´t plot that expression.

[Max]

Hi Walter,

thank you for your answer. I found found the mistake. It wasn´t a problem in MATLAB, but in a mathematical equation.

[Walter Roberson]

heaviside(x) is defined to be 0 for x < 0,, and is defined to be 1 for x > 0. You cannot have heaviside(t=4000) be 390 .

The value of heaviside(x) for x = 0 is not precisely defined. In some systems, heaviside(0) is undefined. In some systems, heaviside(0) is 1/2 . In some systems, heaviside(0) is 0. In some systems, heaviside(0) is 1.

As of R2015a, you can use the new sympref() to control the value that is used for heaviside(0)

Remember, no matter what value you use, heaviside is discontinuous, so you should avoid using numeric integration with it, unless you split your integral up into disjoint ranges at each boundary, integrate each separately, and total the results.

Answer 2

[Tiago Araujo]

I need to do a integral with a function by combining two functions, something like this:

syms q r
x(q,r) =  q   + 2*r;
y(q,r) =  q*9 + r;

G = @(q,r) x+y

integral2(G,1,2,1,2)

It is only a example, my real problem is much more complex. How can i do that?

[Steven Lord]

The only function handle you have in your code is G.

But if x and y were defined as function handles, you can't multiply function handles. What you can do is multiply the values obtained by evaluating the function handles.

x = @(q, r) q + 2*r;
y = @(q, r) 9*q + r;
G = @(q, r) x(q, r) .* y(q, r); % let's multiply rather than add
integral2(G, 1, 2, 1, 2)

ans = 68.4167

[Tiago Araujo]

Please, look at my real problem:

% Coordenadas no espaço X,Y,Z
P1 = [1 1 1];
P2 = [1 1 1.4];
P3 = [1.2 1 1];
P4 = [1.2 1 1.4];
P5 = [1 1.3 1];
P6 = [1 1.3 1.4];
P7 = [1.2 1.3 1];
P8 = [1.2 1.3 1.4];

% Transformação de Coordenadas
C1 = (1/8)*(1+q)*(1-r)*(1-t);
C2 = (1/8)*(1+q)*(1+r)*(1-t);
C3 = (1/8)*(1-q)*(1+r)*(1-t);
C4 = (1/8)*(1-q)*(1-r)*(1-t);
C5 = (1/8)*(1+q)*(1-r)*(1+t);
C6 = (1/8)*(1+q)*(1+r)*(1+t);
C7 = (1/8)*(1-q)*(1+r)*(1+t);
C8 = (1/8)*(1-q)*(1-r)*(1+t);

% Coordenadas em função de q,r e t

xi = C1*P1(:,1)+C2*P2(:,1)+C3*P3(:,1)+C4*P4(:,1)+...
    C5*P5(:,1)+C6*P6(:,1)+C7*P7(:,1)+C8*P8(:,1);

yi = C1*P1(:,2)+C2*P2(:,2)+C3*P3(:,2)+C4*P4(:,2)+...
    C5*P5(:,2)+C6*P6(:,2)+C7*P7(:,2)+C8*P8(:,2);

zi = C1*P1(:,3)+C2*P2(:,3)+C3*P3(:,3)+C4*P4(:,3)+...
    C5*P5(:,3)+C6*P6(:,3)+C7*P7(:,3)+C8*P8(:,3);

% Jacobiano

J =  [diff(xi,q) diff(yi,q) diff(zi,q)
      diff(xi,r) diff(yi,r) diff(zi,r)
      diff(xi,t) diff(yi,t) diff(zi,t)];

% Função G(q,r,t)

F = @(q,r,t) xi(q,r,t) .* yi(q,r,t)./zi(q,r,t)
G = det(J)*F
    
% Solução analítica

integral3(G,-1, 1, -1, 1, -1, 1)

It doesnt work.

[Walter Roberson]

You have F as a function handle. You cannot multiply a function handle by anything.

You are also mixing symbolic expressions and function handles.

Stick with symbolic all the way through to creating G. Then use matlabFunction. Be sure to use the 'vars' option to control the order of variables.

[Tiago Araujo]

You're great! Thanks, matlabFunction solved my problem! I didnt know this function.

[James Morrison]

Thank you @Steven Lord! worked perfectly :)