I have encountered some problems when using Simplify
with expressions containing a square root and isolated a following test case
Simplify[x - #[y], x == #[y]] & /@ {Sin, Sqrt, #^s &, #^(3/2) &}
with a result
{0, x - Sqrt[y], 0, x - y^(3/2)}
- Why is
x-Sqrt[y]
not simplified to0
as in the case ofx-Sin[y]
? - What is the logic behind simplifying
x-y^s
to0
, but keeping x-y^(3/2)? Why do all the above differences vanish when we replace
-
with==
? In such caseSimplify[x == #[y], x == #[y]] & /@ {Sin, Sqrt, #^s &, #^(3/2) &}
results in
{True, True, True, True}
Sqrt[y]
'sFullForm
is reallyPower[y, Rational[1/2]]
, just asy^(3/2)
'sFullform
isPower[y, Rational[3/2]]
, so it seems likely that the same cause (fractional powers?) is leading to the observed output in the two cases. $\endgroup$Simplify[x - y^(3/2) == 0, x == y^(3/2)] ==> True
butSimplify[PossibleZeroQ[x - y^(3/2)], x == y^(3/2)] ==> False
. $\endgroup$FullForm
ofSqrt
, it is no different than the casex == y^(3/2)
. But still, whyx == y^s
behaves in a different way? $\endgroup$