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The following does not answer the Question, but may shed some light on it. For convenience, name the two equations eq1 and eq2, respectively. eq1 = Abs[1 - (-1)^L Sqrt[1 - a^2] Cos[2 b]] == 1 - (-1)^L Sqrt[1 - a^2] Cos[2 b]; eq2 = Abs[1 - E^(I π L) Sqrt[1 - a^2] Cos[2 b]] == 1 - E^(I π L) Sqrt[1 - a^2] Cos[2 b]; The essence of the matter appears to be ...


3

Probably some internal weirdness with the ComplexityFunction, but: Simplify[Sqrt[1/(a + b c d e )] Sqrt[a + b c d e ]==1] // PowerExpand Simplify[1 == Sqrt[a + b*c*d*e] Sqrt[1/(a + b*c*d*e)], x4 > 0] // PowerExpand (* True True *)


1

Maybe obvious but if you want to test if $z$ represents an integer you can use: Assuming[z \[Element] Integers, Simplify[z \[Element] Integers]] This gives True


5

Basically IntegerQ is for determining whether the object inside it is an integer, not whether it represents an integer. Since z is a Symbol, we get False. As for your function, I'm assuming you're using some variant of If[EvenQ[z], ___]. As you said, this won't work because EvenQ will always return False on a symbol since z does not have head Integer. ...


3

According to the documentation (ref/IntegerQ): IntegerQ[expr] returns False unless expr is manifestly an integer (i.e. has head Integer). [emph added] Evidently the assumption does not make z an actual integer, that is, an expression whose head is Integer.


2

Trivially, in this case you can use Re[x] ^= x. However, you can't use upvalues with expressions like Sqrt[Pi^2] (which evaluates to Pi). You could use $Post and Refine, but it's sort of a hack. $Post = Refine[#, x \[Element] Reals] & Re[x] (* x *) Sqrt[x^2] (* Abs[x] *) Note that this system isn't very flexible. If you want to add another ...


1

Since the expression isn't guaranteed to have a unique value in general when the input variables can have multiple discrete values, we have to use an approach that can give you all possible results. This can be done with Reduce instead of Simplify. Here are two examples: expression /. List@ToRules[ Reduce[Join[{expression == a b + a b^2 + a^2 b}, ...



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