FZERO cannot continue because user-supplied function_handle, A and B must be floating-point scalars.

My code is:

p=0.51:0.001:1.013; 
fun1 = @(x) x.^2.*exp(-x)./(1-exp(-10)); 
fun2 = @(x,u) fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]).^2.*exp(-x)./(1-exp(-10)); 
fun3 = @(u) integral(@(x) fun1(x),0,0.5./(20-p)^0.5./(2.*u))+integral(@(x) fun2(x,u),0.5./(20-p)^0.5./(2.*u),10,'arrayvalued', true); 
fun11 = @(x) x.*exp(-x)./(1-exp(-10)); 
fun21 = @(x,u) fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]).*exp(-x)./(1-exp(-10));
fun31 = @(u) integral(@(x) fun11(x),0,0.5./(20-p)^0.5./(2.*u))+integral(@(x) fun21(x,u),0.5./(20-p)^0.5./(2.*u),10,'arrayvalued', true);
fun4 = @(x,u) (20-x+fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10])-p)^0.5.*exp(-x)./(1-exp(-10)); 
fun44= @(u) integral(@(x) (20-p).^0.5.*exp(-x)./(1-exp(-10)),0,0.5./(20-p)^0.5./(2.*u))+integral(@(x) fun4(x,u),0.5./(20-p)^0.5./(2.*u),10,'arrayvalued', true);
fun51 =@(p) p - feval(fun31,fzero(@(u) fun3(u)-0.9973,[0.02,1])); 
fun52 =@(p) feval(fun44,fzero(@(u) fun3(u)-0.9973,[0.02,1])) - integral(@(x) (20-x).^0.5.*exp(-x)./(1-exp(-10)),0,10);
plot3(p,fun51(p),fun52(p))
format long

Here is the error I get:

FZERO cannot continue because user-supplied function_handle ==> @(u)fun3(u)-0.9973 failed with the error below.

One argument must be a square matrix and the other must be a scalar. Use POWER (.^) for elementwise power.

I am sorry that my function is a little bit messy. But I am worried about that you may miss some information if I simplify it. Any help is appreciated, thanks!

[Walter Roberson]

Your code is confusing the parameter named p with the p you set to the numeric vector.

Answer 1

[Torsten]

Make a loop over the elements of p.

"fzero" cannot determine zeros of vector-valued functions (like in fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]) in fun2).

[James_111]

I put a loop in my code as follows:

for p=0.51:0.02:1.01;
fun1 = @(x) x.^2.*exp(-x)./(1-exp(-10)); 
fun2 = @(x,u) fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]).^2.*exp(-x)./(1-exp(-10)); 
fun3 = @(u) integral(@(x) fun1(x),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun2(x,u),0.5./(20-p)^0.5./(2.*u),10,'arrayvalued', true); 
fun11 = @(x) x.*exp(-x)./(1-exp(-10)); 
fun21 = @(x,u) fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]).*exp(-x)./(1-exp(-10));
fun31 = @(u) integral(@(x) fun11(x),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun21(x,u),0.5./(20-p).^0.5./(2.*u),10,'arrayvalued', true);
fun4 = @(x,u) (20-x+fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10])-p).^0.5.*exp(-x)./(1-exp(-10)); 
fun44= @(u) integral(@(x) (20-p).^0.5.*exp(-x)./(1-exp(-10)),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun4(x,u),0.5./(20-p).^0.5./(2.*u),10,'arrayvalued', true);
plot3(p,p - feval(fun31,fzero(@(u) fun3(u)-0.9973,[0.02,1])),feval(fun44,fzero(@(u) fun3(u)-0.9973,[0.02,1])) - integral(@(x) (20-x).^0.5.*exp(-x)./(1-exp(-10)),0,10),'bo')
format long
end

The good news is the error disappears. But just single point shows in the figure. Any one could help for this code?

[Walter Roberson]

format long
pvals = 0.51:0.02:1.01;
nump = length(pvals);
y = zeros(1, nump);
z = zeros(1, nump);
for pidx = 1 : nump
    p = pvals(pidx);
    fun1 = @(x) x.^2.*exp(-x)./(1-exp(-10)); 
    fun2 = @(x,u) fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]).^2.*exp(-x)./(1-exp(-10)); 
    fun3 = @(u) integral(@(x) fun1(x),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun2(x,u),0.5./(20-p)^0.5./(2.*u),10,'arrayvalued', true); 
    fun11 = @(x) x.*exp(-x)./(1-exp(-10)); 
    fun21 = @(x,u) fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]).*exp(-x)./(1-exp(-10));
    fun31 = @(u) integral(@(x) fun11(x),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun21(x,u),0.5./(20-p).^0.5./(2.*u),10,'arrayvalued', true);
    fun4 = @(x,u) (20-x+fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10])-p).^0.5.*exp(-x)./(1-exp(-10)); 
    fun44= @(u) integral(@(x) (20-p).^0.5.*exp(-x)./(1-exp(-10)),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun4(x,u),0.5./(20-p).^0.5./(2.*u),10,'arrayvalued', true);
    y(pidx) = p - feval(fun31,fzero(@(u) fun3(u)-0.9973,[0.02,1]));
    z(pidx) = feval(fun44,fzero(@(u) fun3(u)-0.9973,[0.02,1])) - integral(@(x) (20-x).^0.5.*exp(-x)./(1-exp(-10)),0,10);
end
plot3(pvals, y, z, 'bo')

[James_111]

Thank you so much! How can I accept your this comment?

[Walter Roberson]

Just accept Torsten's answer, since he provided the key about looping over p.

Answer 2

[madhan ravi]

Put .^ instead of ^

[James_111]

Thanks. I change it, now the error becomes:

FZERO cannot continue because user-supplied function_handle ==> @(u)fun3(u)-0.9973 failed with the error below.

 A and B must be floating-point scalars.

Error in CaseB>@(p)p-feval(fun31,fzero(@(u)fun3(u)-0.9973,[0.02,1]))

Error in CaseB (line 29)
plot3(p,fun51(p),fun52(p))

[James_111]

Now the code is:

p=0.51:0.001:1.013; 
fun1 = @(x) x.^2.*exp(-x)./(1-exp(-10)); 
fun2 = @(x,u) fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]).^2.*exp(-x)./(1-exp(-10)); 
fun3 = @(u) integral(@(x) fun1(x),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun2(x,u),0.5./(20-p)^0.5./(2.*u),10,'arrayvalued', true); 
fun11 = @(x) x.*exp(-x)./(1-exp(-10)); 
fun21 = @(x,u) fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10]).*exp(-x)./(1-exp(-10));
fun31 = @(u) integral(@(x) fun11(x),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun21(x,u),0.5./(20-p).^0.5./(2.*u),10,'arrayvalued', true);
fun4 = @(x,u) (20-x+fzero(@(y) 2.*u.*y.*(20-p-x+y).^0.5-0.5,[0,10])-p).^0.5.*exp(-x)./(1-exp(-10)); 
fun44= @(u) integral(@(x) (20-p).^0.5.*exp(-x)./(1-exp(-10)),0,0.5./(20-p).^0.5./(2.*u))+integral(@(x) fun4(x,u),0.5./(20-p).^0.5./(2.*u),10,'arrayvalued', true);
fun51 =@(p) p - feval(fun31,fzero(@(u) fun3(u)-0.9973,[0.02,1])); 
fun52 =@(p) feval(fun44,fzero(@(u) fun3(u)-0.9973,[0.02,1])) - integral(@(x) (20-x).^0.5.*exp(-x)./(1-exp(-10)),0,10);
plot3(p,fun51(p),fun52(p))
format long

Error is:

FZERO cannot continue because user-supplied function_handle ==> @(u)fun3(u)-0.9973 failed with the error below.
 A and B must be floating-point scalars.

Any help is really appreciated, thanks!

Answer 3

[Andrea]

AUXILIOOO COMO RESUELVO ESTO?

Error using fzero>localFirstFcnEvalError FZERO cannot continue because user-supplied function_handle ==> @(t)((R*t)/(Vmol-b))-((psi*((R^2*Tc^2*(((1+(0.48+(1.574*w)-(0.176*w^2))*(1-sqrt(t/Tc))))^2))/Pc))/((Vmol+(epsilon*b))*(V+(sigma*b))))-P failed with the error below. Error: File: V.m Line: 49 Column: 9 Using 'V' as both the name of a variable and the name of a script is not supported.

[Torsten]

Using 'V' as both the name of a variable and the name of a script is not supported.

El mensaje de error es obvio. Cambie el nombre de la variable "V" en su código o asigne a su secuencia de comandos un nombre diferente.

Y dado que el identificador de su función se llamará con vectores para t, debe usar la multiplicación y división por componentes (.* y ./) en lugar de la multiplicación normal en los casos en que sea necesario.