anonymous function.problem

Dear all,

I met a problem, xmax = fminbnd(@(x)(f(x)), 0, 5) this line is ok, however, xmax = fminbnd(@(x)(-f(x)), 0, 5) get an error, if I use inline , xmax = fminbnd(inline(-f(x)), 0, 5), it is also ok, I don't know what is the problem. Could anyone give me the answer.

Thanks!

matlab2018a win7

clearvars
syms t x
f = @(x)int(t*exp(-t^2),t, 0, x);
xmin = fminbnd(f, 0, 5)
fxmin = double( f(xmin) )

xmax = fminbnd(@(x)(f(x)), 0, 5)          %ok
xmax = fminbnd(inline(-f(x)), 0, 5)       %ok   
xmax = fminbnd(@(x)(-f(x)), 0, 5)         %error
xmax = fminbnd(@(x)(f(x)*(-1)), 0, 5)     %error 

[Eyal Ben-Hur]

Maybe try to define g(x) as:

g = @(x)(-f(x))

and than write:

xmax = fminbnd(@(x)(g(x)), 0, 5)

[liu]

Thanks for your comment, it still do not work , get the same error .

[TADA]

I think the problem is that the second anonymous functin wraps f in another function handle while inline unravels it:

inline(-f(x))

ans =

     Inline function:
     ans(x) = exp(-x.^2).*(1.0./2.0)-1.0./2.0

     
@(x) -f(x)

ans =

  function_handle with value:

    @(x)-f(x)

Answer 1

[madhan ravi]

clearvars
syms t x
f = int(t*exp(-t^2),t, 0, x);
f=matlabFunction(f)               %learn about matlabFunction
xmin = fminbnd(@(x)f(x), 0, 5)
fxmin = double( f(xmin) )
xmax = fminbnd(@(x)-f(x), 0, 5)     
 

[liu]

Thanks, it work!

[madhan ravi]

Anytime :)

Answer 2

[Walter Roberson]

You can only fminbnd numeric routines. int returns symbolic not numeric . you should be using integral()

[Christopher Creutzig]

Not for this particular input. Just don't use int inside an anonymous function; Madhan Ravi's answer shows how to do it. (And that way also creates an integral call for integrals not found in closed form.)

[liu]

ok!

f = @(x)integral(@(t)t.*exp(-t.^2), 0, x);
xmax = fminbnd(@(x)(-f(x)), 0, 5)        

[Walter Roberson]

Christopher, your reply implies that it would be okay to fminbnd a symbolic function, something like

syms t x
f(x) = int(t*exp(-t^2),t, 0, x);
xmin = fminbnd(f, 0, 5)
g(x) = -f(x)
xmax = fminbnd(g, 0, 5)

since, after all, you are not putting the int inside an anonymous function (a symbolic function is not an anonymous function.)

The xmin calculation with f succeeds in giving a numeric solution. The xmax calculation with g fails.

The fminbnd code includes an expression of the form value + value.*(variable == other_variable) . The symbolic calculation of the int() gives an exact expression that includes exp() terms. When you get further towards the maximum, the variable == other_variable becomes undecideable to symbolic numeric precision, so the symbolic toolbox leaves it in == form instead of making a decision. That leads to symbolic evaluation errors a small number of statements later. With the f formula, the values do not happen to get driven large enough to make the == undecideable to symbolic numeric precision and so fminbnd manages to emit something.

This is not a matter of whether the int is in an anonymous function: this is due to trying to fminbnd a non-numeric expression.

You can have int() inside an anonymous function for fminbnd, provided the anonymous function returns a numeric value:

 fminbnd(@(x) double(-f(x)), 0, 5)