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

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Liming Fang
Liming Fang 2020-4-26
评论： Walter Roberson 2020-4-26
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...

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采纳的回答

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
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Liming Fang
Liming Fang 2020-4-26
编辑：Walter Roberson 2020-4-26
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
Walter Roberson 2020-4-26
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

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Walter Roberson
Walter Roberson 2020-4-26
syms x y
f = symfun(piecewise(x-1<=y+1, x-1, y+1),[x y])

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