A negative whole power is inverted to its roots, not to none (#2154)

Inverting x^n = v for a whole n multiplies an n-th root of v by each
root of unity of order n, which GetAllRootsOf1 lists by counting up to
n. For a negative n it counted nothing, so the inversion found no roots,
and an equation that reached one inside another was answered with the
empty set, a claim that it has none: 1/(1 + x^(-2)) = 1/2 came back { }
where its roots are 1 and -1, and 1/(1 + x^(-2)) = y came back { } for
every y. x^(-2) = 4 on its own was answered correctly by another route.

x^n = v for a negative n has the roots of x^|n| = 1/v, and an n-th root
of v times a root of unity of order |n| is each of them, so the roots of
unity are now counted to |n|. A negative power is never zero, so
x^(-2) = 0 is still answered with none, now on purpose; x^0 is not
inverted at all.

Measured on 2.6.0 and on this branch, in BREAKING-CHANGES.md.


Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>

Author: Rafael-SOWNet

BREAKING-CHANGES.md

@@ -42,6 +42,21 @@ A not-equal written `<>` is read as `not (a = b)`, and printed that way.
 
 ## Unreleased — since 2.6.0
 
+### An equation through a negative whole power is not answered with no roots
+
+**Silent: an empty answer becomes the roots.** Inverting `x^n = v` for a negative whole `n` multiplied by the roots
+of unity of order `n`, and there are none of order `-2`, so the inversion found no roots, and an equation that reached
+one inside another was answered with the empty set, a claim that it has no solution. `x^n = v` has the roots of
+`x^|n| = 1/v`, and they are found now. `x^(-2) = 4` on its own was already answered by another route
+([#2144](https://github.com/asc-community/AngouriMath/issues/2144)).
+
+| Input | Was (2.6.0) | Now |
+|---|---|---|
+| `"1/(1 + x^(-2)) = 1/2".Solve("x")` | `{ }` | `{ 1, -1 }` |
+| `"2/(1 + x^(-2)) = 1".Solve("x")` | `{ }` | `{ 1, -1 }` |
+| `"1/(2 + p^(-3)) = 1/3".Solve("p")` | `{ }` | `{ 1, -1/2 + i * 1/2 * sqrt(3), -1/2 + i * -1/2 * sqrt(3) }` |
+| `"1/(1 + x^(-2)) = y".Solve("x")` | `{ }` | `{ ((-1) / (-y) - 1) ^ (-1/2), -((-1) / (-y) - 1) ^ (-1/2) }` |
+
 ### A system with a product equation is solved branch by branch
 
 **Silent: solutions that were missing are answered.** `EquationSystem.Solve` eliminated `x * y = 0` for `y` as `0`,

Sources/.editorconfig

@@ -273,6 +273,9 @@ file_header_template=\nCopyright (c) 2019-2026 Angouri.\nAngouriMath is licensed
 [Tests/UnitTests/Convenience/DocCommentPlacementTest.cs]
 file_header_template=\nCopyright (c) 2019-2026 Angouri.\nAngouriMath is licensed under MIT.\nDetails: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.\nWebsite: https://am.angouri.org.\n
 
+[Tests/UnitTests/Algebra/SolveTest/NegativePowerInversionTest.cs]
+file_header_template=\nCopyright (c) 2019-2026 Angouri.\nAngouriMath is licensed under MIT.\nDetails: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.\nWebsite: https://am.angouri.org.\n
+
 [Tests/UnitTests/Common/LambdaArrowSyntaxTest.cs]
 file_header_template=\nCopyright (c) 2019-2026 Angouri.\nAngouriMath is licensed under MIT.\nDetails: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.\nWebsite: https://am.angouri.org.\n
 

Sources/AngouriMath/Functions/Continuous/Solvers/EquationSolver/InvertNode.Classes.cs

@@ -103,7 +103,19 @@ partial record Powf
                 if (Base.ContainsNode(x))
                 {
                     if (Exponent is Integer { EInteger: var pow })
-                        return Together(Number.GetAllRootsOf1(pow).Select(root => Base.Invert(root * MathS.Pow(value, 1 / Exponent), x)));
+                    {
+                        // x^n = v has |n| roots for a negative n as well, those of x^|n| = 1/v:
+                        // each n-th root of v times a root of unity of order |n|, whose n-th
+                        // power is 1. GetAllRootsOf1 counts up to its argument, so given -2 it
+                        // gave no roots at all, and 1/(1 + x^(-2)) = 1/2 was answered { }. A
+                        // negative power is never zero, so x^(-2) = 0 is the one that has none.
+                        // https://github.com/asc-community/AngouriMath/issues/2144
+                        if (pow.IsZero)
+                            return null;
+                        if (pow.Sign < 0 && value.Evaled is Complex { IsZero: true })
+                            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);
                 }  

Sources/Tests/UnitTests/Algebra/SolveTest/NegativePowerInversionTest.cs

@@ -0,0 +1,63 @@
+//
+// Copyright (c) 2019-2026 Angouri.
+// AngouriMath is licensed under MIT.
+// Details: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.
+// Website: https://am.angouri.org.
+//
+
+using AngouriMath;
+using AngouriMath.Extensions;
+using System.Linq;
+using Xunit;
+using static AngouriMath.Entity;
+using static AngouriMath.Entity.Set;
+
+namespace AngouriMath.Tests.Algebra.SolveTest
+{
+    /// <summary>
+    /// Inverting a negative whole power found no roots: the roots of unity it multiplies by were
+    /// counted up to the exponent, and up to -2 is none. So an equation that reaches one inside
+    /// another, as <c>1/(1 + x^(-2)) = 1/2</c> does, was answered with the empty set, a claim that
+    /// it has no roots, where it has 1 and -1.
+    /// <a href="https://github.com/asc-community/AngouriMath/issues/2144">#2144</a>
+    /// </summary>
+    [Trait("Area", "Algebra")]
+    public sealed class NegativePowerInversionTest
+    {
+        private static Set Solve(string statement, string x = "x") => statement.ToEntity().Solve(x);
+
+        private static Entity IsIn(string value, Set set) => value.ToEntity().In(set).Simplify();
+
+        [Theory]
+        [InlineData("1/(1 + x^(-2)) = 1/2", "1", "-1", "2")]
+        [InlineData("1/(2 + x^(-3)) = 1/3", "1", "-1/2 + 1/2 * sqrt(3) * i", "-1")]
+        [InlineData("2/(1 + x^(-2)) = 1", "1", "-1", "0")]
+        [InlineData("1/(3 + (x + 1)^(-2)) = 1/4", "0", "-2", "1")]
+        public void EveryRootIsFound(string equation, string root, string otherRoot, string nonRoot)
+        {
+            var solved = Solve(equation);
+            Assert.Equal(Entity.Boolean.True, IsIn(root, solved));
+            Assert.Equal(Entity.Boolean.True, IsIn(otherRoot, solved));
+            Assert.Equal(Entity.Boolean.False, IsIn(nonRoot, solved));
+        }
+
+        /// <summary>
+        /// With a symbol on the right the answer is the two roots in terms of it, which at
+        /// y = 1/2 are the roots above.
+        /// </summary>
+        [Fact]
+        public void TheRootsInASymbolAreTheRootsAtEachValue()
+        {
+            var solved = Assert.IsType<FiniteSet>(Solve("1/(1 + x^(-2)) = y"));
+            Assert.Equal(2, solved.Count);
+            var atAHalf = solved.Select(root => root.Substitute("y", "1/2".ToEntity()).Simplify()).ToList();
+            Assert.Contains(atAHalf, root => root == 1);
+            Assert.Contains(atAHalf, root => root == -1);
+        }
+
+        /// <summary>A negative power is never zero, so this one does have none.</summary>
+        [Fact]
+        public void ANegativePowerIsNeverZero()
+            => Assert.Equal(Set.Empty, Solve("1/(1 + x^(-2)) = 1"));
+    }
+}