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Say which infinity a quotient with a vanishing divisor tends to - #700
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The descent puts each part's own limit in place of the part, and for this shape that throws away the only thing that decides the answer: cos(x) / sin(x) at 0 becomes 1 / 0, which says nothing about which side the divisor vanishes from and so comes back NaN -- the claim that the limit does not exist, where on the right it is +oo and on the left -oo. Only 1 / x and the other quotients of polynomials escaped it, because those the solvers read outright once x has been moved out to infinity, and it is that accident which made the gap easy to miss. The side the divisor vanishes from is read off its first derivative that does not vanish with it. Where g(a) = 0 and the first non-vanishing derivative there is the k-th, g(x) has the sign of g_k(a) * (x - a)^k near a, which is the sign of g_k(a) on the right and that times (-1)^k on the left. That covers the first order, where sin(x) and e^x - 1 and x - a live, and the second, where 1 - cos(x) does -- and being even, that one looks the same from both sides, so 1 / (1 - cos(x)) now has a two-sided limit of +oo as well. Nothing is claimed unless a derivative comes out finite and non-zero at the point. An expression that is not differentiable there, or whose derivative diverges as sqrt(x)'s does, is left alone rather than guessed at: reading the sign wrongly would answer +oo where the truth is -oo, which is worse than not answering. Four derivatives is as far as it looks. Reached only where the substitution above answered nothing or NaN, so it costs nothing on any path that already worked, and whatever that answered is kept if this finds nothing better. 17 of the 35 new tests fail without it. Corpus unchanged at 101/117 with 0 wrong; suite 4044 passed, 0 failed.
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Fork CI, matrix run 30935777770 — success on windows-latest, ubuntu-latest and macos-latest, The branch For context on where this sits: I keep a corpus of 49 one-sided limits worked out by hand. Master answers 16 of them correctly, three of the misses being wrong answers rather than merely unanswered ones. #697 takes it to 42, #699 and this one together take it to 49. The three PRs are independent — |
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Codecov Report❌ Patch coverage is
Additional details and impacted files@@ Coverage Diff @@
## master #700 +/- ##
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- Coverage 80.99% 80.41% -0.59%
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Files 155 156 +1
Lines 13687 12944 -743
Branches 1957 2132 +175
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- Hits 11086 10409 -677
+ Misses 1990 1923 -67
- Partials 611 612 +1 ☔ View full report in Codecov by Harness. 🚀 New features to boost your workflow:
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3 participants
limitright(cos(x) / sin(x), x, 0)answers NaN. It is +oo. From the left it is -oo, and that is NaN too.The descent puts each part's own limit in place of the part, so the quotient becomes
1 / 0— which says nothing about which side the divisor vanishes from, and comes back NaN. NaN is not "I could not tell"; it is the claim that the limit does not exist.1 / xescapes this, which is why the gap is easy to miss: once x is moved out to infinity that one is a quotient of polynomials and the solvers read it outright. Anything whose divisor is not a polynomial —sin(x),tan(x),e^x - 1,ln(x)at 1 — fell through.How the side is read
Off the divisor's first derivative that does not vanish with it. Where
g(a) = 0and the first non-vanishing derivative there is the k-th,g(x)has the sign ofg_k(a) * (x - a)^knear a — so the sign ofg_k(a)on the right, and that times(-1)^kon the left.That covers the first order (
sin(x),e^x - 1,x - a) and the second (1 - cos(x),sin(x) - 1at pi/2). An even order looks the same from both sides, so1 / (1 - cos(x))gains a two-sided limit of +oo as well.Nothing is claimed unless a derivative comes out finite and non-zero at the point. An expression that is not differentiable there, or whose derivative diverges as
sqrt(x)'s does, is left alone rather than guessed at — reading the sign wrongly would answer +oo where the truth is -oo, which is worse than not answering at all. Four derivatives is as far as it looks.What it answers
1 / sin(x)at 0cos(x) / sin(x)at 01 / tan(x)at 0-2 / (e^x - 1)at 02 / ln(x)at 11 / x^3at 0cos(x) / (x - 1)^3at 11 / (1 - cos(x))at 03 / (sin(x) - 1)at pi/217 of the 35 new tests fail without the change.
The cases where the two sides genuinely disagree keep saying so:
1 / sin(x)and1 / x^3at 0 still have no two-sided limit, and there are tests pinning that, because turning a correct NaN into a wrong infinity is the failure mode this change could have had.Cost
Reached only where the substitution above answered nothing or NaN, so it costs nothing on any path that already worked, and whatever that answered is kept if this finds nothing better.
Measurements
Failed: 0, Passed: 4044, Skipped: 14, Total: 4058.Independent of my other open PRs;
git merge-treereports no conflict with any of them. It does compose with #697 — with both,sin(x) / (1 - cos(x))at 0 answers +oo on the right and -oo on the left, since l'Hopital's rule turns it intocos(x) / sin(x)and this reads that. On this branch alone it is still NaN.Fork CI to follow in a comment.