Closed brianmickel closed 5 years ago
Hi @brianmickel,
in the second example, x^(9/3)
is the same as x^3
. However, x^3
and (x^9)^(1/3)
have different graphs (x^3
is defined and has negative values for x<0
, while (x^9)^(1/3)
is not defined).
In the first example, you are right I could say the result is 0
but only if I could also add the constraint when x>=0
. If x<0
then the subtraction of those two is not zero, but rather is undefined. Since Algebrite can't track that constraint when x>=0
, it can't give 0
because that would be incorrect when x<0
, hence the expression is left as is.
This is indeed a limitation of Algebrite, however I can't let that simplification happen without having a constraint system in place to specify for which values of x
that simplification holds, otherwise I get incorrect results in other places
Closing for the time being.
in:
simplify(x^(5/2) - sqrt(x^5))
out:x^(5/2)-(x^5)^(1/2)
expected out:0
in:
simplify(8*x^(9/3) - 8*(x^9)^(1/3))
out:8*(x^3-(x^9)^(1/3))
expected out:0