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An equation through abs is answered in decimals, with two non-real entries that are not roots #2139
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solve::x::x^2/abs(1-x^2) = x
→ { 0, 0.618…, 1.618…, 1.618… + 1.7E-9i, 1.618… + 9.9E-8i }
The answer is { 0, (sqrt(5) - 1)/2, (sqrt(5) + 1)/2 }, exactly:
x^2 = x * abs(1 - x^2)makesxa non-negative real.- The two non-real entries are approximations of
1.618…and satisfy nothing. - The case split on the sign of
1 - x^2gives each root exactly:x^2/(1-x^2) = xgives{ 0, (-1 ± sqrt(5))/2 }, andx^2/(x^2-1) = xgives the other.
Found while reading Modenov's Problems in Geometry for #2131.
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Happypig375 commentedon Oct 27, 2026
While these approximation related quirks are fixable early, remember to simplify these changes away for v3.
Rafael-SOWNet commentedon Oct 27, 2026
Will do. So that they can be found again, #2169 asks every such fix to link v3's downcasting decision (#1019) in its comment and in its test. A search for that comment then lists everything v3 has to take back out. The fix for this issue will be the first to carry the link.
A fix for an approximation quirk before v3 links the decision that re…
Rafael-SOWNet commentedon Oct 27, 2026
#2178 answers this exactly over the reals: with
Codomainset toReal,x^2/abs(1-x^2) = xis{ 0, (-1 + sqrt(5)) / 2, (1 + sqrt(5)) / 2 }. It also fixesabs(x^2 - 4) = 3x, which had a non-root4.0000000000000062 + 1.6e-7 ibeside{ 1, 4 }.The complex codomain, the default and what this issue shows, needs a decision. The wrong entries come from Newton's method run in the complex plane, which assumes a holomorphic function, and
|1 - x^2|is not one. A case split on the sign does not carry over either. Over the complex numbers|u| = ±uonly whereuis real, and roots need not make it real:x^2 + 1 = |x|has roots at±0.618 i, where|x|is0.618andxis not±0.618. So split-and-check finds roots and cannot say it found them all.Two ways:
xis inside a non-holomorphic function (abs,sgn,conj,re,im,arg) and also outside it. That is what|x - 2| = |x - 3|already gets (An equation between moduli is answered with points Newton's method finds on a curve #1573), for the same reason. It costs the equations Newton happened to answer right.I recommend 1, since not answering is a legitimate answer and a near-copy of a root is not one. Which do you prefer?
Over the reals, a modulus of a polynomial is split at the polynomial'…