x
−1
x
+1
=
x−2x
+1
x−1
In this form, the CAS returns the original equation. If you add domain constraints on x like this
x
−1
x
+1
=
x−2x
+1
x−1
|x = y
2
and y > 0
the CAS will return true. I believe that this effect occurs because the CAS does not always interpret the
square root function as exponentiation to the power of 1/2; see [2.10] CAS square root simplification
limitations.
If both expressions are fractions, you can try this function:
frctest(a,b)
func
expand(getnum(a))*expand(getdenom(b))=expand(getnum(b))*expand(getdenom(a))
Endfunc
This function cross-multiplies the expanded numerators and denominators of the equation
expressions. a and b are the expressions on the left- and right-hand sides of the equality. For the
example above,
frctest((√(x)-1)/(√(x)+1),(x-2*√ (x)+1)/(x-1))
returns true.
Alternatively, try solving the expression for a variable in the expression. For example,
solve((√(x)-1)/(√(x)+1)=(x-2*√ (x)+1)/(x-1),x)
returns true, indicating that the two expressions are equal.
(Credit for domain constraints method to Glenn E. Fisher; solve() method to Bhuvanesh Bhatt)
[2.3] Use when() for absolute value in integrals
AMS 2.03 and later support the when() function in integration, so it can be used for numeric solutions
of integrals. This is especially useful when the integrand includes expressions using the absolute
value. For example, if you try to evaluate this integral
¶
−2
3
x
2
− 1
dx
with this command
á(abs(x^2-1),x,-2,3)
the calculator is busy for several seconds, then returns the original expression. The integral can be
2 - 2
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