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An odd root of a negative real is a preimage, as the evaluator takes it #2313

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Rafael-SOWNet merged 1 commit into from
Nov 1, 2026
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An odd root of a negative real is a preimage, as the evaluator takes it #2313

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10 changes: 7 additions & 3 deletions BREAKING-CHANGES.md
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Expand Up @@ -141,8 +141,11 @@ the square root the solver takes after solving for `x^2`, where an equation has
for `0 < p < 1` its values lie in the sector `-p pi < arg <= p pi`, with `0`, and for `-1 < p < 0` in
`p pi <= arg < -p pi`, without `0`. `x^p = a` was solved to `a^(1/p)` with no condition, which is no root where `a` is
outside the sector: `sqrt(x) = a` was `{ a ^ 2 }`, and at `a = -1` that is `1`, whose square root is `1` and not `-1`. The root now carries
the sector, written with `arg`, as `abs`'s inversion carries its own condition. A number on the other side is decided
as before, and `|p| > 1` needs no condition, since every value is then a power. What the equation is provided is known
the sector, written with `arg`, as `abs`'s inversion carries its own condition. A negative real to a rational power
whose denominator is odd and below ten is the real root to the numerator, as the evaluator gives it, `(-8)^(1/3)` being
`-2`: with an odd numerator the negative reals are values as well, so `x^(1/3) = -2` is still `{ -8 }`, and with an even
one a positive value has a negative preimage too. A number on the other side is decided as before, and `|p| > 1` needs
no condition, since every value is then a power. What the equation is provided is known
while it is solved, as a provided expression is differentiated: a value known to be real, or declared real, has the
half-line for its sector, one known to lie on it needs no condition, and one known to lie off it has no root. A conjunct of a condition that
another conjunct already forces is not stated, so `a > -5 and a >= 0` is `a >= 0`, and a condition stated twice is
Expand All @@ -153,7 +156,8 @@ carried it once combined: a second derivative of `1/sqrt(x^2 + 13)` stated `not
| Input | Was (2.6.0) | Now |
|---|---|---|
| `"sqrt(x) = a".ToEntity().Solve("x")` | `{ a ^ 2 }` | `{ a ^ 2 provided a = 0 or -1/2 * pi < arg(a) and arg(a) <= 1/2 * pi }` |
| `"x^(1/3) = a".ToEntity().Solve("x")` | `{ a ^ 3 }` | `{ a ^ 3 provided a = 0 or -1/3 * pi < arg(a) and arg(a) <= 1/3 * pi }` |
| `"x^(1/3) = a".ToEntity().Solve("x")` | `{ a ^ 3 }` | `{ a ^ 3 provided a = 0 or -1/3 * pi < arg(a) and arg(a) <= 1/3 * pi or arg(a) = pi }` |
| `"x^(2/3) = 4".ToEntity().Solve("x")` | `{ 8 }` | `{ 8, -8 }` |
| `"x^(-1/2) = a".ToEntity().Solve("x")` | `{ a ^ (-2) }` | `{ a ^ (-2) provided not a = 0 and -1/2 * pi <= arg(a) and arg(a) < 1/2 * pi }` |
| `"(sqrt(x) = a) provided a in RR".ToEntity().Solve("x")` | `{ a ^ 2 provided a in RR }` | `{ a ^ 2 provided a in RR and a >= 0 }` |
| `"(sqrt(x) = a) provided a > 0".ToEntity().Solve("x")` | `{ a ^ 2 provided a > 0 }` | `{ a ^ 2 provided a > 0 }` |
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47 changes: 40 additions & 7 deletions Sources/AngouriMath/Functions/Continuous/Solvers/EquationSolver/InvertNode.Classes.cs
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Expand Up @@ -116,9 +116,16 @@ partial record Powf
return Enumerable.Empty<Entity>();
return Together(Number.GetAllRootsOf1(pow.Abs()).Select(root => Base.Invert(root * MathS.Pow(value, 1 / Exponent), x)));
}
else
return Base.Invert(MathS.Pow(value, 1 / Exponent), x)
?.Select(root => root.Provided(InTheRange(value)));
var principal = Base.Invert(MathS.Pow(value, 1 / Exponent), x)
?.Select(root => root.Provided(InTheRange(value)));
// With an even numerator a negative base has a positive power, (-8)^(2/3) = 4,
// so a positive value has the negative of its principal preimage for a
// preimage as well: x^(2/3) = 4 is 8 and -8.
if (principal is { } && TakesTheRealRootOfANegative(out var oddNumerator) && !oddNumerator)
return Base.Invert(-MathS.Pow(value, 1 / Exponent), x) is { } negative
? principal.Concat(negative.Select(root => root.Provided(new Greaterf(value, Integer.Zero))))
: null;
return principal;
}
// a ^ x = value => x = log(a, value)
// TODO: determine how we should return periodic roots as
Expand All @@ -138,12 +145,21 @@ private Entity InTheRange(Entity value)
if (Exponent.Evaled is not Real { EDecimal.IsFinite: true } p
|| p.EDecimal.Abs().CompareTo(PeterO.Numbers.EDecimal.One) >= 0)
return Boolean.True;
// A negative base with an odd numerator has a negative power, (-8)^(1/3) = -2, so
// the negative reals are in the range as well, and x^(1/3) = -2 is x = -8, as it
// was before the sector was stated.
var reachesTheNegatives = TakesTheRealRootOfANegative(out var oddNumerator) && oddNumerator;
// Where the value is known to be real -- by its codomain, or by what the equation
// it came from was provided -- arg(value) is 0 or pi, so the sector is the half-line
// value >= 0 for p > 0 and value > 0 for p < 0, and nothing is stated where that is
// known as well: sqrt(x) = a provided a > 0 is a ^ 2 provided a > 0.
Entity halfLine = p.IsPositive ? new GreaterOrEqualf(value, Integer.Zero) : new Greaterf(value, Integer.Zero);
Entity outside = p.IsPositive ? new Lessf(value, Integer.Zero) : new LessOrEqualf(value, Integer.Zero);
// known as well: sqrt(x) = a provided a > 0 is a ^ 2 provided a > 0. With the
// negatives it is every real, or every real but 0.
Entity halfLine = reachesTheNegatives
? (p.IsPositive ? Boolean.True : !value.EqualTo(Integer.Zero))
: p.IsPositive ? new GreaterOrEqualf(value, Integer.Zero) : new Greaterf(value, Integer.Zero);
Entity outside = reachesTheNegatives
? (p.IsPositive ? Boolean.False : value.EqualTo(Integer.Zero))
: p.IsPositive ? new Lessf(value, Integer.Zero) : new LessOrEqualf(value, Integer.Zero);
if (TreeAnalyzer.IsKnown(halfLine))
return Boolean.True;
if (TreeAnalyzer.IsKnown(outside))
Expand All @@ -153,9 +169,26 @@ private Entity InTheRange(Entity value)
var bound = (Exponent * MathS.pi).InnerSimplified;
var opposite = ((-Exponent).InnerSimplified * MathS.pi).InnerSimplified;
var angle = MathS.Arg(value);
return p.IsPositive
Entity sector = p.IsPositive
? value.EqualTo(Integer.Zero) | (new Lessf(opposite, angle) & new LessOrEqualf(angle, bound))
: !value.EqualTo(Integer.Zero) & new LessOrEqualf(bound, angle) & new Lessf(angle, opposite);
return reachesTheNegatives ? sector | angle.EqualTo(MathS.pi) : sector;
}

/// <summary>
/// Whether a negative real base has a real power, as the evaluator takes it: a rational
/// power whose denominator is odd and below ten is the real root of a negative real to
/// the numerator, as Number.Pow's search for a good root finds it, and any other is the
/// principal power. (-8)^(1/3) is -2, (-8)^(2/3) is 4, and (-1)^(1/11) is not real.
/// </summary>
private bool TakesTheRealRootOfANegative(out bool oddNumerator)
{
oddNumerator = false;
if (Exponent is not Rational { ERational: var ratio } || ratio.Denominator.IsEven
|| ratio.Denominator.CompareTo(PeterO.Numbers.EInteger.FromInt32(10)) >= 0)
return false;
oddNumerator = !ratio.Numerator.IsEven;
return true;
}
}

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17 changes: 16 additions & 1 deletion Sources/Tests/UnitTests/Algebra/SolveTest/FractionalPowerRangeTest.cs
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Expand Up @@ -19,6 +19,12 @@ namespace AngouriMath.Tests.Algebra
/// solution. <c>sqrt(x) = a</c> was answered <c>{ a ^ 2 }</c>, which is wrong at
/// <c>a = -1</c>: <c>sqrt((-1)^2)</c> is <c>1</c>.
/// <a href="https://github.com/asc-community/AngouriMath/issues/212">#212</a>
/// <para/>
/// A negative real to a rational power with an odd denominator below ten is the real root to
/// the numerator, which is what the evaluator gives: <c>(-8)^(1/3)</c> is <c>-2</c>. So with
/// an odd numerator the negative reals are values too, and <c>x^(1/3) = -2</c> is
/// <c>x = -8</c>, which the sector alone answered <c>{ }</c>; and with an even one a positive
/// value has a negative preimage as well, <c>(-8)^(2/3) = 4</c>.
/// </summary>
[Trait("Area", "Algebra")]
public sealed class FractionalPowerRangeTest
Expand All @@ -40,7 +46,8 @@ private static Entity OnlyRoot(string equation)
[InlineData("sqrt(x) = a", "-i", false)]
[InlineData("sqrt(x) = a", "-1", false)]
[InlineData("x^(1/3) = a", "1 + i", true)]
[InlineData("x^(1/3) = a", "-1", false)]
[InlineData("x^(1/3) = a", "-1", true)]
[InlineData("x^(1/3) = a", "-i", false)]
[InlineData("x^(-1/2) = a", "-i", true)]
[InlineData("x^(-1/2) = a", "i", false)]
[InlineData("x^(-1/2) = a", "0", false)]
Expand All @@ -63,6 +70,14 @@ public void TheRootHoldsWhereItsConditionDoes(string equation, string value, boo
[InlineData("sqrt(x) = -1", "{ }")]
[InlineData("sqrt(x) = i", "{ -1 }")]
[InlineData("sqrt(x) = -i", "{ }")]
[InlineData("x^(1/3) = -2", "{ -8 }")]
[InlineData("x^(1/5) = -2", "{ -32 }")]
[InlineData("x^(-1/3) = -1/2", "{ -8 }")]
[InlineData("x^(3/5) = -8", "{ -32 }")]
[InlineData("x^(2/3) = 4", "{ 8, -8 }")]
[InlineData("x^(4/3) = 16", "{ 8, -8 }")]
[InlineData("x^(2/3) = -4", "{ }")]
[InlineData("x^(1/11) = -1", "{ }")]
public void ANumberIsDecided(string equation, string expected)
=> Assert.Equal(expected.ToEntity(), equation.ToEntity().Solve("x"));

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