Explicit Full-Tail Resource Bounds versus Finite Benchmarks
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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-20261013-advance-20261019-r06-asymptotic-resource-bound-foundations · 2026-10-13 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=1*n^0, g(n)=1*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(1, 0, 1, 0, Yes, Yes, Yes, [1, 1], [1/1, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 0, 1, 0, Yes, Yes, Yes, [1, 1], [1/1, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
2. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=2*n^1, g(n)=2*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(2, 1, 2, 3, Yes, No, No, [2, 1], does-not-exist)
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (2, 1, 2, 3, Yes, No, No, [2, 1], does-not-exist).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
3. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=3*n^2, g(n)=3*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(3, 2, 3, 0, No, Yes, No, does-not-exist, [1/3, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, 2, 3, 0, No, Yes, No, does-not-exist, [1/3, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
4. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=4*n^3, g(n)=4*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(4, 3, 4, 3, Yes, Yes, Yes, [4, 1], [1/4, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, 3, 4, 3, Yes, Yes, Yes, [4, 1], [1/4, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
5. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=5*n^4, g(n)=5*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(5, 4, 5, 0, No, Yes, No, does-not-exist, [1/5, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (5, 4, 5, 0, No, Yes, No, does-not-exist, [1/5, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
6. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=6*n^5, g(n)=6*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(6, 5, 6, 3, No, Yes, No, does-not-exist, [1/6, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (6, 5, 6, 3, No, Yes, No, does-not-exist, [1/6, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
7. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=7*n^0, g(n)=7*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(7, 0, 7, 0, Yes, Yes, Yes, [7, 1], [1/7, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (7, 0, 7, 0, Yes, Yes, Yes, [7, 1], [1/7, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
8. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=8*n^1, g(n)=1*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(8, 1, 1, 3, Yes, No, No, [8, 1], does-not-exist)
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (8, 1, 1, 3, Yes, No, No, [8, 1], does-not-exist).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
9. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=9*n^2, g(n)=2*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(9, 2, 2, 0, No, Yes, No, does-not-exist, [1/2, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (9, 2, 2, 0, No, Yes, No, does-not-exist, [1/2, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
10. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=10*n^3, g(n)=3*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(10, 3, 3, 3, Yes, Yes, Yes, [10, 1], [1/3, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (10, 3, 3, 3, Yes, Yes, Yes, [10, 1], [1/3, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
11. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=11*n^4, g(n)=4*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(11, 4, 4, 0, No, Yes, No, does-not-exist, [1/4, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (11, 4, 4, 0, No, Yes, No, does-not-exist, [1/4, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
12. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=12*n^5, g(n)=5*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(12, 5, 5, 3, No, Yes, No, does-not-exist, [1/5, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (12, 5, 5, 3, No, Yes, No, does-not-exist, [1/5, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
13. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=13*n^0, g(n)=6*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(13, 0, 6, 0, Yes, Yes, Yes, [13, 1], [1/6, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (13, 0, 6, 0, Yes, Yes, Yes, [13, 1], [1/6, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
14. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=14*n^1, g(n)=7*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(14, 1, 7, 3, Yes, No, No, [14, 1], does-not-exist)
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (14, 1, 7, 3, Yes, No, No, [14, 1], does-not-exist).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
15. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=15*n^2, g(n)=1*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(15, 2, 1, 0, No, Yes, No, does-not-exist, [1/1, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (15, 2, 1, 0, No, Yes, No, does-not-exist, [1/1, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
16. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=16*n^3, g(n)=2*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(16, 3, 2, 3, Yes, Yes, Yes, [16, 1], [1/2, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (16, 3, 2, 3, Yes, Yes, Yes, [16, 1], [1/2, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
17. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=17*n^4, g(n)=3*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(17, 4, 3, 0, No, Yes, No, does-not-exist, [1/3, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (17, 4, 3, 0, No, Yes, No, does-not-exist, [1/3, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
18. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=18*n^5, g(n)=4*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(18, 5, 4, 3, No, Yes, No, does-not-exist, [1/4, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (18, 5, 4, 3, No, Yes, No, does-not-exist, [1/4, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
19. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=19*n^0, g(n)=5*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(19, 0, 5, 0, Yes, Yes, Yes, [19, 1], [1/5, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (19, 0, 5, 0, Yes, Yes, Yes, [19, 1], [1/5, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
20. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=20*n^1, g(n)=6*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(20, 1, 6, 3, Yes, No, No, [20, 1], does-not-exist)
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (20, 1, 6, 3, Yes, No, No, [20, 1], does-not-exist).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
21. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=21*n^2, g(n)=7*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(21, 2, 7, 0, No, Yes, No, does-not-exist, [1/7, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (21, 2, 7, 0, No, Yes, No, does-not-exist, [1/7, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
22. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=22*n^3, g(n)=1*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(22, 3, 1, 3, Yes, Yes, Yes, [22, 1], [1/1, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (22, 3, 1, 3, Yes, Yes, Yes, [22, 1], [1/1, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
23. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=23*n^4, g(n)=2*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(23, 4, 2, 0, No, Yes, No, does-not-exist, [1/2, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (23, 4, 2, 0, No, Yes, No, does-not-exist, [1/2, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
24. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=24*n^5, g(n)=3*n^3, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(24, 5, 3, 3, No, Yes, No, does-not-exist, [1/3, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (24, 5, 3, 3, No, Yes, No, does-not-exist, [1/3, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
25. Positive-Coefficient Monomial Tail Classes and Witnesses
Fix a cost model and explicit nonnegative functions f(n)=25*n^0, g(n)=4*n^0, with positive integer coefficients. Give (a, p, b, q, f∈O(g)?, f∈Ω(g)?, f∈Θ(g)?, prescribed O witness [a,1] or does-not-exist, prescribed Ω witness [1/b,1] or does-not-exist). Witnesses apply to the full tail n≥1, not samples.
View answer and explanation
(25, 0, 4, 0, Yes, Yes, Yes, [25, 1], [1/4, 1])
Use the declared monomial family and compare exponents; retain the prescribed witnesses rather than substituting tighter constants or reducing 1/b. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (25, 0, 4, 0, Yes, Yes, Yes, [25, 1], [1/4, 1]).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-monomial-class · candidate
26. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[0, 1, 0], g=[1, 1, 2, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([0, 1, 0], [1, 1, 2, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0], [1, 1, 2, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
27. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[1, 2, 1], g=[1, 2, 0, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([1, 2, 1], [1, 2, 0, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1], [1, 2, 0, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
28. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[2, 3, 0], g=[1, 3, 1, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([2, 3, 0], [1, 3, 1, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 3, 0], [1, 3, 1, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
29. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[3, 1, 1], g=[1, 4, 2, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([3, 1, 1], [1, 4, 2, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([3, 1, 1], [1, 4, 2, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
30. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[4, 2, 0], g=[1, 5, 0, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([4, 2, 0], [1, 5, 0, 0], 1, 1, Yes, Yes, Yes)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([4, 2, 0], [1, 5, 0, 0], 1, 1, Yes, Yes, Yes).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
31. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[5, 3, 1], g=[1, 6, 1, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([5, 3, 1], [1, 6, 1, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3, 1], [1, 6, 1, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
32. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[6, 1, 0], g=[1, 7, 2, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([6, 1, 0], [1, 7, 2, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([6, 1, 0], [1, 7, 2, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
33. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[7, 2, 1], g=[1, 8, 0, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([7, 2, 1], [1, 8, 0, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([7, 2, 1], [1, 8, 0, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
34. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[8, 3, 0], g=[1, 9, 1, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([8, 3, 0], [1, 9, 1, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([8, 3, 0], [1, 9, 1, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
35. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[9, 1, 1], g=[1, 10, 2, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([9, 1, 1], [1, 10, 2, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([9, 1, 1], [1, 10, 2, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
36. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[10, 2, 0], g=[1, 11, 0, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([10, 2, 0], [1, 11, 0, 0], 1, 1, Yes, Yes, Yes)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([10, 2, 0], [1, 11, 0, 0], 1, 1, Yes, Yes, Yes).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
37. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[11, 3, 1], g=[1, 12, 1, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([11, 3, 1], [1, 12, 1, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([11, 3, 1], [1, 12, 1, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
38. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[12, 1, 0], g=[1, 13, 2, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([12, 1, 0], [1, 13, 2, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([12, 1, 0], [1, 13, 2, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
39. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[13, 2, 1], g=[1, 14, 0, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([13, 2, 1], [1, 14, 0, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([13, 2, 1], [1, 14, 0, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
40. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[14, 3, 0], g=[1, 15, 1, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([14, 3, 0], [1, 15, 1, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([14, 3, 0], [1, 15, 1, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
41. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[15, 1, 1], g=[1, 16, 2, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([15, 1, 1], [1, 16, 2, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([15, 1, 1], [1, 16, 2, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
42. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[16, 2, 0], g=[1, 17, 0, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([16, 2, 0], [1, 17, 0, 0], 1, 1, Yes, Yes, Yes)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([16, 2, 0], [1, 17, 0, 0], 1, 1, Yes, Yes, Yes).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
43. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[17, 3, 1], g=[1, 18, 1, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([17, 3, 1], [1, 18, 1, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([17, 3, 1], [1, 18, 1, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
44. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[18, 1, 0], g=[1, 19, 2, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([18, 1, 0], [1, 19, 2, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([18, 1, 0], [1, 19, 2, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
45. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[19, 2, 1], g=[1, 20, 0, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([19, 2, 1], [1, 20, 0, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([19, 2, 1], [1, 20, 0, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
46. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[20, 3, 0], g=[1, 21, 1, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([20, 3, 0], [1, 21, 1, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([20, 3, 0], [1, 21, 1, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
47. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[21, 1, 1], g=[1, 22, 2, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([21, 1, 1], [1, 22, 2, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([21, 1, 1], [1, 22, 2, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
48. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[22, 2, 0], g=[1, 23, 0, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([22, 2, 0], [1, 23, 0, 0], 1, 1, Yes, Yes, Yes)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([22, 2, 0], [1, 23, 0, 0], 1, 1, Yes, Yes, Yes).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
49. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[23, 3, 1], g=[1, 24, 1, 1]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([23, 3, 1], [1, 24, 1, 1], 2, 3, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([23, 3, 1], [1, 24, 1, 1], 2, 3, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
50. Nonnegative Polynomial Degrees Including Both Zero Polynomials
Fix the cost model. Coefficients in ascending powers are f=[24, 1, 0], g=[1, 25, 2, 0]; all are nonnegative. Give (original f coefficients, original g coefficients, degree f, degree g, O?, Ω?, Θ?). Use degree -1 for the zero polynomial; retain original vectors including trailing zeros.
View answer and explanation
([24, 1, 0], [1, 25, 2, 0], 1, 2, Yes, No, No)
Classify the declared polynomial family with the zero cases explicit; a finite benchmark is not the full-tail argument. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([24, 1, 0], [1, 25, 2, 0], 1, 2, Yes, No, No).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-polynomial-degree · candidate
51. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=0, b=1, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(0, 1, 1, 0, Yes, 0n+1≤1(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 1, 1, 0, Yes, 0n+1≤1(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
52. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=1, b=3, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(1, 3, 4, 0, Yes, 1n+3≤4(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 3, 4, 0, Yes, 1n+3≤4(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
53. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=2, b=5, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(2, 5, 7, 0, Yes, 2n+5≤7(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (2, 5, 7, 0, Yes, 2n+5≤7(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
54. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=3, b=7, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(3, 7, 10, 0, Yes, 3n+7≤10(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, 7, 10, 0, Yes, 3n+7≤10(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
55. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=4, b=9, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(4, 9, 13, 0, Yes, 4n+9≤13(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, 9, 13, 0, Yes, 4n+9≤13(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
56. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=5, b=11, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(5, 11, 16, 0, Yes, 5n+11≤16(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (5, 11, 16, 0, Yes, 5n+11≤16(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
57. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=6, b=13, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(6, 13, 19, 0, Yes, 6n+13≤19(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (6, 13, 19, 0, Yes, 6n+13≤19(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
58. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=7, b=15, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(7, 15, 22, 0, Yes, 7n+15≤22(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (7, 15, 22, 0, Yes, 7n+15≤22(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
59. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=8, b=17, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(8, 17, 25, 0, Yes, 8n+17≤25(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (8, 17, 25, 0, Yes, 8n+17≤25(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
60. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=9, b=19, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(9, 19, 28, 0, Yes, 9n+19≤28(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (9, 19, 28, 0, Yes, 9n+19≤28(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
61. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=10, b=21, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(10, 21, 31, 0, Yes, 10n+21≤31(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (10, 21, 31, 0, Yes, 10n+21≤31(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
62. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=11, b=23, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(11, 23, 34, 0, Yes, 11n+23≤34(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (11, 23, 34, 0, Yes, 11n+23≤34(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
63. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=12, b=25, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(12, 25, 37, 0, Yes, 12n+25≤37(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (12, 25, 37, 0, Yes, 12n+25≤37(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
64. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=13, b=27, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(13, 27, 40, 0, Yes, 13n+27≤40(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (13, 27, 40, 0, Yes, 13n+27≤40(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
65. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=14, b=29, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(14, 29, 43, 0, Yes, 14n+29≤43(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (14, 29, 43, 0, Yes, 14n+29≤43(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
66. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=15, b=31, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(15, 31, 46, 0, Yes, 15n+31≤46(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (15, 31, 46, 0, Yes, 15n+31≤46(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
67. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=16, b=33, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(16, 33, 49, 0, Yes, 16n+33≤49(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (16, 33, 49, 0, Yes, 16n+33≤49(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
68. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=17, b=35, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(17, 35, 52, 0, Yes, 17n+35≤52(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (17, 35, 52, 0, Yes, 17n+35≤52(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
69. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=18, b=37, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(18, 37, 55, 0, Yes, 18n+37≤55(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (18, 37, 55, 0, Yes, 18n+37≤55(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
70. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=19, b=39, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(19, 39, 58, 0, Yes, 19n+39≤58(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (19, 39, 58, 0, Yes, 19n+39≤58(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
71. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=20, b=41, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(20, 41, 61, 0, Yes, 20n+41≤61(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (20, 41, 61, 0, Yes, 20n+41≤61(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
72. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=21, b=43, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(21, 43, 64, 0, Yes, 21n+43≤64(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (21, 43, 64, 0, Yes, 21n+43≤64(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
73. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=22, b=45, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(22, 45, 67, 0, Yes, 22n+45≤67(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (22, 45, 67, 0, Yes, 22n+45≤67(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
74. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=23, b=47, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(23, 47, 70, 0, Yes, 23n+47≤70(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (23, 47, 70, 0, Yes, 23n+47≤70(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
75. An Explicit All-Input Linear Resource Witness
Use natural n≥0, nonnegative integers a=24, b=49, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).
View answer and explanation
(24, 49, 73, 0, Yes, 24n+49≤73(n+1) for all n≥0)
The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (24, 49, 73, 0, Yes, 24n+49≤73(n+1) for all n≥0).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-linear-witness · candidate
76. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^0, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,0]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(0, 1, 0, [[0, 1, 0]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 1, 0, [[0, 1, 0]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
77. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^1, g(n)=n^3, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,1]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(1, 3, 1, [[0, 0, 0], [1, 1, 1]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 3, 1, [[0, 0, 0], [1, 1, 1]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
78. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^2, g(n)=n^5, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,2]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(2, 5, 2, [[0, 0, 0], [1, 1, 1], [2, 4, 32]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (2, 5, 2, [[0, 0, 0], [1, 1, 1], [2, 4, 32]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
79. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^3, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,3]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(3, 1, 3, [[0, 0, 0], [1, 1, 1], [2, 8, 2], [3, 27, 3]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, 1, 3, [[0, 0, 0], [1, 1, 1], [2, 8, 2], [3, 27, 3]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
80. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^4, g(n)=n^3, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,4]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(4, 3, 4, [[0, 0, 0], [1, 1, 1], [2, 16, 8], [3, 81, 27], [4, 256, 64]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, 3, 4, [[0, 0, 0], [1, 1, 1], [2, 16, 8], [3, 81, 27], [4, 256, 64]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
81. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^5, g(n)=n^5, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,5]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(5, 5, 5, [[0, 0, 0], [1, 1, 1], [2, 32, 32], [3, 243, 243], [4, 1024, 1024], [5, 3125, 3125]], Yes, [Yes, Yes, Yes], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (5, 5, 5, [[0, 0, 0], [1, 1, 1], [2, 32, 32], [3, 243, 243], [4, 1024, 1024], [5, 3125, 3125]], Yes, [Yes, Yes, Yes], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
82. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^0, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,6]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(0, 1, 6, [[0, 1, 0], [1, 1, 1], [2, 1, 2], [3, 1, 3], [4, 1, 4], [5, 1, 5], [6, 1, 6]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 1, 6, [[0, 1, 0], [1, 1, 1], [2, 1, 2], [3, 1, 3], [4, 1, 4], [5, 1, 5], [6, 1, 6]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
83. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^1, g(n)=n^3, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,0]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(1, 3, 0, [[0, 0, 0]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 3, 0, [[0, 0, 0]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
84. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^2, g(n)=n^5, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,1]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(2, 5, 1, [[0, 0, 0], [1, 1, 1]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (2, 5, 1, [[0, 0, 0], [1, 1, 1]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
85. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^3, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,2]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(3, 1, 2, [[0, 0, 0], [1, 1, 1], [2, 8, 2]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, 1, 2, [[0, 0, 0], [1, 1, 1], [2, 8, 2]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
86. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^4, g(n)=n^3, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,3]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(4, 3, 3, [[0, 0, 0], [1, 1, 1], [2, 16, 8], [3, 81, 27]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, 3, 3, [[0, 0, 0], [1, 1, 1], [2, 16, 8], [3, 81, 27]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
87. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^5, g(n)=n^5, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,4]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(5, 5, 4, [[0, 0, 0], [1, 1, 1], [2, 32, 32], [3, 243, 243], [4, 1024, 1024]], Yes, [Yes, Yes, Yes], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (5, 5, 4, [[0, 0, 0], [1, 1, 1], [2, 32, 32], [3, 243, 243], [4, 1024, 1024]], Yes, [Yes, Yes, Yes], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
88. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^0, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,5]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(0, 1, 5, [[0, 1, 0], [1, 1, 1], [2, 1, 2], [3, 1, 3], [4, 1, 4], [5, 1, 5]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 1, 5, [[0, 1, 0], [1, 1, 1], [2, 1, 2], [3, 1, 3], [4, 1, 4], [5, 1, 5]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
89. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^1, g(n)=n^3, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,6]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(1, 3, 6, [[0, 0, 0], [1, 1, 1], [2, 2, 8], [3, 3, 27], [4, 4, 64], [5, 5, 125], [6, 6, 216]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 3, 6, [[0, 0, 0], [1, 1, 1], [2, 2, 8], [3, 3, 27], [4, 4, 64], [5, 5, 125], [6, 6, 216]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
90. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^2, g(n)=n^5, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,0]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(2, 5, 0, [[0, 0, 0]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (2, 5, 0, [[0, 0, 0]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
91. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^3, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,1]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(3, 1, 1, [[0, 0, 0], [1, 1, 1]], Yes, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, 1, 1, [[0, 0, 0], [1, 1, 1]], Yes, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
92. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^4, g(n)=n^3, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,2]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(4, 3, 2, [[0, 0, 0], [1, 1, 1], [2, 16, 8]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, 3, 2, [[0, 0, 0], [1, 1, 1], [2, 16, 8]], No, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
93. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^5, g(n)=n^5, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,3]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(5, 5, 3, [[0, 0, 0], [1, 1, 1], [2, 32, 32], [3, 243, 243]], Yes, [Yes, Yes, Yes], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (5, 5, 3, [[0, 0, 0], [1, 1, 1], [2, 32, 32], [3, 243, 243]], Yes, [Yes, Yes, Yes], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
94. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^0, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,4]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(0, 1, 4, [[0, 1, 0], [1, 1, 1], [2, 1, 2], [3, 1, 3], [4, 1, 4]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 1, 4, [[0, 1, 0], [1, 1, 1], [2, 1, 2], [3, 1, 3], [4, 1, 4]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
95. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^1, g(n)=n^3, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,5]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(1, 3, 5, [[0, 0, 0], [1, 1, 1], [2, 2, 8], [3, 3, 27], [4, 4, 64], [5, 5, 125]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 3, 5, [[0, 0, 0], [1, 1, 1], [2, 2, 8], [3, 3, 27], [4, 4, 64], [5, 5, 125]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
96. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^2, g(n)=n^5, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,6]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(2, 5, 6, [[0, 0, 0], [1, 1, 1], [2, 4, 32], [3, 9, 243], [4, 16, 1024], [5, 25, 3125], [6, 36, 7776]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (2, 5, 6, [[0, 0, 0], [1, 1, 1], [2, 4, 32], [3, 9, 243], [4, 16, 1024], [5, 25, 3125], [6, 36, 7776]], Yes, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
97. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^3, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,0]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(3, 1, 0, [[0, 0, 0]], Yes, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, 1, 0, [[0, 0, 0]], Yes, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
98. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^4, g(n)=n^3, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,1]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(4, 3, 1, [[0, 0, 0], [1, 1, 1]], Yes, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, 3, 1, [[0, 0, 0], [1, 1, 1]], Yes, [No, Yes, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
99. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^5, g(n)=n^5, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,2]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(5, 5, 2, [[0, 0, 0], [1, 1, 1], [2, 32, 32]], Yes, [Yes, Yes, Yes], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (5, 5, 2, [[0, 0, 0], [1, 1, 1], [2, 32, 32]], Yes, [Yes, Yes, Yes], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
100. Finite Sample Results versus Independently Known Tail Classes
Fix f(n)=n^0, g(n)=n^1, with n natural and the algebraic convention 0^0=1. Sample the complete window [0,3]. Give (p, q, bound, every [n,f(n),g(n)] sample, all sampled f≤g?, independently derived [O?,Ω?,Θ?], do samples alone suffice?, fixed finite-benchmark-insufficiency label).
View answer and explanation
(0, 1, 3, [[0, 1, 0], [1, 1, 1], [2, 1, 2], [3, 1, 3]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof)
Use the explicit family for asymptotic classification. A short sample can agree despite the wrong eventual order, or fail at zero despite a valid eventual upper bound. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 1, 3, [[0, 1, 0], [1, 1, 1], [2, 1, 2], [3, 1, 3]], No, [Yes, No, No], No, finite-benchmark-insufficient-for-a-full-tail-proof).
Reviewed step-by-step derivation
Asymptotic resource bounds · qf-ext19-asymptotic-benchmark-insufficiency · candidate
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