How to use function like `min` in symfun?(I got an error)

For example, i want to create a symbolic function: f(x,y) = min(x-1,y+1);

when f(x,y) = x+y, everything is ok, but when it comes to min, error occured.

syms x y
f = symfun(min([x-1 y+1])),[x y])  

I got error like this:

"Unable to convert expression into double array."

Help...

Answer 1

[Thiago Henrique Gomes Lobato]

You can't use min or max in symbolic variables in matlab. A work around can be this one:

syms x1 x2 real;
xlt = symfun(1/2*(x1+x2-abs(x1-x2)),[x1,x2]); %min
xgt = symfun(1/2*(x1+x2+abs(x1-x2)),[x1,x2]); %max

I took this answer for a comment here https://de.mathworks.com/matlabcentral/answers/102485-why-are-symbolic-variables-not-allowed-when-using-the-functions-max-min-in-the-symbolic-math-toolbox , which is also from a question very similar to yours

[Liming Fang]

wow, i will try it soon, thanks in advance. BTW, why min,max is banned?functions like sin is ok

[Walter Roberson]

min() and max() are not banned,

>> min(sym(5), sym(-7))
ans =
-7

However, min() and max() want to be able to resolve which value is larger as soon as they are called, so for example if you have

syms x; min(sym(5), x-7)

then min() is not willing to leave that as an unresolved min(5, x-7) to be worked out later: it wants to compute the answer right then, and when x-7 cannot be converted to numeric to figure it out right then, then they fail.

[Liming Fang]

ok i got it.

Does this mean that functions involved condition won't work?

And the answer below menthons "piecewise", i wonder whether it is used to solve functions involved condions.

[Walter Roberson]

It depends upon how the condition is constructed.

Functions you should avoid for symbolic expressions that include unresolved symbolic variables include min(), max(), and mod() . Most other functions are willing to postpone evaluation when they are passed symbolic variables, but those three do odd or undesirable things.

And the answer below menthons "piecewise", i wonder whether it is used to solve functions involved condions.

Yes, it is. You can even differentiate piecewise -- but the boundary conditions might not get the right value, as piecewise() expressions are often discontinuous on the boundaries and differentiating piecewise() does not return a special "undefined" result for the boundary.

However, solve() sometimes has difficulty with piecewise, especially something of the form piecewise(condition, expression, condition, expression...) == value . Sometimes you need to solve() against each of the expressions and then test to see whether the result meets the conditions

Answer 2

[Walter Roberson]

syms x y
f = symfun(piecewise(x-1<=y+1, x-1, y+1),[x y])  

[Liming Fang]

Thanks!