#5223 The sign of x has two normal forms, x/abs(x) and abs(x)/x, and their difference does not simplify

[David Scherfgen]

Written by Claude.

The simplifier knows that abs(x)^2 is x^2: it reduces (x/abs(x))^2 and (abs(x)/x)^2 to 1, abs(x)^3 to x^2*abs(x), x^2/abs(x)^3 to 1/abs(x), abs(x)/x^2 to 1/abs(x) and x/abs(x)^2 to 1/x, so at most one power of abs(x) survives in a product and the even part is moved to x. But it has no single form for the sign of x: x*abs(x)^(-1) stays x/abs(x) and abs(x)*x^(-1) stays abs(x)/x, and a sum of the two is not recognized as 0 by the simplifier or by expand, only by ratsimp and radcan which happen to produce abs(x)^2 - x^2 on the way.

(%i1) display2d : false$
(%i2) [x/abs(x), abs(x)/x, (x/abs(x))*(abs(x)/x), (x/abs(x))^2, (abs(x)/x)^2];
(%o2) [x/abs(x),abs(x)/x,1,1,1]
(%i3) [abs(x)^3, x^2/abs(x)^3, abs(x)/x^2, x/abs(x)^2, (x/abs(x))^3];
(%o3) [x^2*abs(x),1/abs(x),1/abs(x),1/x,x/abs(x)]
(%i4) x/abs(x) - abs(x)/x;
(%o4) x/abs(x)-abs(x)/x
(%i5) expand((1 + x/abs(x))*(1 - abs(x)/x));
(%o5) x/abs(x)-abs(x)/x
(%i6) expand((1 + x/abs(x))*(1 - x/abs(x)));
(%o6) 0
(%i7) [ratsimp(x/abs(x) - abs(x)/x), radcan(x/abs(x) - abs(x)/x), is(equal(x/abs(x), abs(x)/x))];
(%o7) [0,0,true]
(%i8) [abs(x^3)^(1/3), abs(-x^3)^(-1/3), abs(x)^(2/3)];
(%o8) [x^(2/3)*abs(x)^(1/3),1/(x^(2/3)*abs(x)^(1/3)),x^(2/3)]

(%o5) and (%o6) are the same product written with the two forms of the sign; only the second collapses. (%o8) shows the same gap for a root: abs(x^3)^(1/3) is abs(x), but abs(x^3) is first normalized to x^2*abs(x) and the root then distributed over it, while abs(x)^(2/3) alone does become x^(2/3).

Why it matters. With domain : real the principal branch of a power of a negative quantity, as needed next to gamma_incomplete, is naturally written as a phase alpha + beta*x/abs(x) with constant alpha and beta, and such phases from different sources, an antiderivative and its derivative, or two terms of gamma_expand, have to cancel under expand. They do so only if every source writes the sign the same way; the work on the domain : real branch had to arrange that by hand (multiplying the sign into the phase by the parity of the power of x that comes with it, and avoiding abs(z)^s for the modulus).

Proposal. Pick one canonical form for the sign and have simptimes produce it. Since the simplifier already keeps at most one power of abs(x) and moves the even part to x, the natural rule is the one it applies for positive exponents extended to negative ones: abs(x)^(-1)*x and abs(x)*x^(-1) both to x/abs(x), and in general x^p*abs(x)^q with p + q even to x^(p+q) and with p + q odd to x^(p+q-1)*abs(x) or x^(p+q+1)/abs(x), whichever keeps the exponent of abs(x) in {-1, 1}. Then (%o4) and (%o5) are 0 by simplification. For a fractional power, (x^(2*m)*abs(x))^(1/(2*m+1)) could be recognized as abs(x), but that is a smaller matter.

[Stavros Macrakis]

The two forms are equal for reals:

abs(-1)/-1 => -1
(-1)/abs(-1) => -1
signum(-1) => -1

but not equal in general. The simplest case:

abs(%i)/%i => -%i
%i/abs(%i) => %i
signum(%i) => %i

Presumably "the sum of the two" should be the difference of the two?

[David Scherfgen]

Yes to both. "Sum" should be "difference", as the reproducer shows. And the forms are equal for a real x only; the request is for the case the simplifier already treats as real: the clauses in timesin that turn abs(x)/x^2 into 1/abs(x) and x^2/abs(x)^3 into 1/abs(x) test csign of the argument and leave a complex z alone, and the missing case is the same rule for the exponent 1, in the same clauses, under the same test. With declare(z, complex), abs(z)/z and z/abs(z) would stay distinct, as they should.

[David Scherfgen]

First part of this is fixed by commit [139e38].
The fractional powers remain to be done.