sage: foo=sin(x) > 0 ; foo
sin(x) > 0
sage: bar=cos(x)>0 ; bar
cos(x) > 0
I do not understand this :
sage: foo and bar
sin(x) > 0
sage: foo or bar
cos(x) > 0
As long as x has no value, these expressions can't be evaluated or simplified. Could some kind soul enlighten my confused mind ?
From my testing, it appears that foo is evaluating to False. Hence, in the expression "foo and bar", the foo is evaluating as False, so it skips evaluation of bar. As such, it is returning the first operand, foo, which is sin(x) > 0.
In the second expression, "foo or bar", the foo evaluates as False. With an "or" operator, the second operand must be evaluated. As such, it's returning the second operand, bar, cos(x) > 0.
You can see similar behavior of the "and" and "or" operators with the following:
sage: 0.0 and sin(x)
0.000000000000000
sage: 0.0 or sin(x)
sin(x)
Edit: Sorry, this probably didn't answer the original question. I can't tell you why foo is being evaluated as False, given the unknown value of x. However, I can reasonably assume that foo is being evaluated as False. If you plug in a value for x, like foo(2.0), then it correctly evaluates as True.
Okay. As jaydfox points out, there are three things :
and evaluates "lazily" from left to right, and stops at the first argument evaluated to False.
Anything that can't be proven True evaluates to False.
and returns the first False argument unevaluated.
The last one baffles me. it means that I can't write if sin(x>0 and cos(x>0): doSomething(): I would doSomething() if x was unbound....
I need to write if (sin(x)>0 and cos(x)>0) is True: doSomething().
Asked: 2020-07-15 23:38:13 +0100
Seen: 320 times
Last updated: Jul 16 '20
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