ans = 
Does isAlways Make an Unwarranted Assumption that a Variable is real?
이전 댓글 표시
Define a sym variable
syms v
isAlways can't prove that v is real. Makes sense.
isAlways(in(v,'real'))
Now make an assumption
assume(v > 0); % (1)
Is that assumption a sufficient condition to imply that v is real?
isAlways(in(v,'real')) %(2)
Apparently it does.
But we don't see that v is real in the assumptions
assumptions(v)
as we would if stated explicitly
assumeAlso(v,'real');
assumptions(v)
Now define v as a complex number, which clears all of the assumptions
v = sym(1+1i);
assumptions(v)
Here, v satisfies assumption (1) because symbolic gt only compares the real parts of both sides (though the doc page does not state that explicitly)
isAlways(v > 0)
But satisfying assumption (1) in this case does not imply the truth of condition (2) (thankfully)
isAlways(in(v,'real'))
Seems like the correct way for the software to interpret (1) would be Re(v) > 0, in accordance with the de facto definition of symbolic gt, which would provide no information for evaluating (2).
댓글 수: 9
Matt J
2026년 1월 28일
In my opinion, the thing that is wrong is,
>> isAlways( sym(1+1i) > 0)
ans =
logical
1
I never really understood this convention even for numeric variables, but certainly not for symbolic variables.
Paul
2026년 1월 29일
Matt J
2026년 1월 29일
That's my opinion, yes.
Paul
2026년 1월 30일
the cyclist
2026년 1월 31일
편집: the cyclist
2026년 1월 31일
I guess I would have expected
1 + i > 0
to return NaN, since ordering is not defined on complex numbers.
Matt J
2026년 1월 31일
x >= x % returns false?
I can see the case for returning true iff the inequality is satisfied with equality.
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