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Type | Label | Description |
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Statement | ||
Theorem | selberg2b 27401* | Convert eventual boundedness in selberg2 27400 to boundedness on any interval [π΄, +β). (We have to bound away from zero because the log terms diverge at zero.) (Contributed by Mario Carneiro, 25-May-2016.) |
β’ βπ β β+ βπ₯ β (1[,)+β)(absβ(((((Οβπ₯) Β· (logβπ₯)) + Ξ£π β (1...(ββπ₯))((Ξβπ) Β· (Οβ(π₯ / π)))) / π₯) β (2 Β· (logβπ₯)))) β€ π | ||
Theorem | chpdifbndlem1 27402* | Lemma for chpdifbnd 27404. (Contributed by Mario Carneiro, 25-May-2016.) |
β’ (π β π΄ β β+) & β’ (π β 1 β€ π΄) & β’ (π β π΅ β β+) & β’ (π β βπ§ β (1[,)+β)(absβ(((((Οβπ§) Β· (logβπ§)) + Ξ£π β (1...(ββπ§))((Ξβπ) Β· (Οβ(π§ / π)))) / π§) β (2 Β· (logβπ§)))) β€ π΅) & β’ πΆ = ((π΅ Β· (π΄ + 1)) + ((2 Β· π΄) Β· (logβπ΄))) & β’ (π β π β (1(,)+β)) & β’ (π β π β (π[,](π΄ Β· π))) β β’ (π β ((Οβπ) β (Οβπ)) β€ ((2 Β· (π β π)) + (πΆ Β· (π / (logβπ))))) | ||
Theorem | chpdifbndlem2 27403* | Lemma for chpdifbnd 27404. (Contributed by Mario Carneiro, 25-May-2016.) |
β’ (π β π΄ β β+) & β’ (π β 1 β€ π΄) & β’ (π β π΅ β β+) & β’ (π β βπ§ β (1[,)+β)(absβ(((((Οβπ§) Β· (logβπ§)) + Ξ£π β (1...(ββπ§))((Ξβπ) Β· (Οβ(π§ / π)))) / π§) β (2 Β· (logβπ§)))) β€ π΅) & β’ πΆ = ((π΅ Β· (π΄ + 1)) + ((2 Β· π΄) Β· (logβπ΄))) β β’ (π β βπ β β+ βπ₯ β (1(,)+β)βπ¦ β (π₯[,](π΄ Β· π₯))((Οβπ¦) β (Οβπ₯)) β€ ((2 Β· (π¦ β π₯)) + (π Β· (π₯ / (logβπ₯))))) | ||
Theorem | chpdifbnd 27404* | A bound on the difference of nearby Ο values. Theorem 10.5.2 of [Shapiro], p. 427. (Contributed by Mario Carneiro, 25-May-2016.) |
β’ ((π΄ β β β§ 1 β€ π΄) β βπ β β+ βπ₯ β (1(,)+β)βπ¦ β (π₯[,](π΄ Β· π₯))((Οβπ¦) β (Οβπ₯)) β€ ((2 Β· (π¦ β π₯)) + (π Β· (π₯ / (logβπ₯))))) | ||
Theorem | logdivbnd 27405* | A bound on a sum of logs, used in pntlemk 27455. This is not as precise as logdivsum 27382 in its asymptotic behavior, but it is valid for all π and does not require a limit value. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ (π β β β Ξ£π β (1...π)((logβπ) / π) β€ ((((logβπ) + 1)β2) / 2)) | ||
Theorem | selberg3lem1 27406* | Introduce a log weighting on the summands of Ξ£π Β· π β€ π₯, Ξ(π)Ξ(π), the core of selberg2 27400 (written here as Ξ£π β€ π₯, Ξ(π)Ο(π₯ / π)). Equation 10.4.21 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ (π β π΄ β β+) & β’ (π β βπ¦ β (1[,)+β)(absβ((Ξ£π β (1...(ββπ¦))((Ξβπ) Β· (logβπ)) β ((Οβπ¦) Β· (logβπ¦))) / π¦)) β€ π΄) β β’ (π β (π₯ β (1(,)+β) β¦ ((((2 / (logβπ₯)) Β· Ξ£π β (1...(ββπ₯))(((Ξβπ) Β· (Οβ(π₯ / π))) Β· (logβπ))) β Ξ£π β (1...(ββπ₯))((Ξβπ) Β· (Οβ(π₯ / π)))) / π₯)) β π(1)) | ||
Theorem | selberg3lem2 27407* | Lemma for selberg3 27408. Equation 10.4.21 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ (π₯ β (1(,)+β) β¦ ((((2 / (logβπ₯)) Β· Ξ£π β (1...(ββπ₯))(((Ξβπ) Β· (Οβ(π₯ / π))) Β· (logβπ))) β Ξ£π β (1...(ββπ₯))((Ξβπ) Β· (Οβ(π₯ / π)))) / π₯)) β π(1) | ||
Theorem | selberg3 27408* | Introduce a log weighting on the summands of Ξ£π Β· π β€ π₯, Ξ(π)Ξ(π), the core of selberg2 27400 (written here as Ξ£π β€ π₯, Ξ(π)Ο(π₯ / π)). Equation 10.6.7 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ (π₯ β (1(,)+β) β¦ (((((Οβπ₯) Β· (logβπ₯)) + ((2 / (logβπ₯)) Β· Ξ£π β (1...(ββπ₯))(((Ξβπ) Β· (Οβ(π₯ / π))) Β· (logβπ)))) / π₯) β (2 Β· (logβπ₯)))) β π(1) | ||
Theorem | selberg4lem1 27409* | Lemma for selberg4 27410. Equation 10.4.20 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ (π β π΄ β β+) & β’ (π β βπ¦ β (1[,)+β)(absβ((Ξ£π β (1...(ββπ¦))((Ξβπ) Β· ((logβπ) + (Οβ(π¦ / π)))) / π¦) β (2 Β· (logβπ¦)))) β€ π΄) β β’ (π β (π₯ β (1(,)+β) β¦ ((Ξ£π β (1...(ββπ₯))((Ξβπ) Β· Ξ£π β (1...(ββ(π₯ / π)))((Ξβπ) Β· ((logβπ) + (Οβ((π₯ / π) / π))))) / (π₯ Β· (logβπ₯))) β (logβπ₯))) β π(1)) | ||
Theorem | selberg4 27410* | The Selberg symmetry formula for products of three primes, instead of two. The sum here can also be written in the symmetric form Ξ£πππ β€ π₯, Ξ(π)Ξ(π)Ξ(π); we eliminate one of the nested sums by using the definition of Ο(π₯) = Ξ£π β€ π₯, Ξ(π). This statement can thus equivalently be written Ο(π₯)logβ2(π₯) = 2Ξ£πππ β€ π₯, Ξ(π)Ξ(π)Ξ(π) + π(π₯logπ₯). Equation 10.4.23 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ (π₯ β (1(,)+β) β¦ ((((Οβπ₯) Β· (logβπ₯)) β ((2 / (logβπ₯)) Β· Ξ£π β (1...(ββπ₯))((Ξβπ) Β· Ξ£π β (1...(ββ(π₯ / π)))((Ξβπ) Β· (Οβ((π₯ / π) / π)))))) / π₯)) β π(1) | ||
Theorem | pntrval 27411* | Define the residual of the second Chebyshev function. The goal is to have π (π₯) β π(π₯), or π (π₯) / π₯ βπ 0. (Contributed by Mario Carneiro, 8-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ (π΄ β β+ β (π βπ΄) = ((Οβπ΄) β π΄)) | ||
Theorem | pntrf 27412 | Functionality of the residual. Lemma for pnt 27463. (Contributed by Mario Carneiro, 8-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ π :β+βΆβ | ||
Theorem | pntrmax 27413* | There is a bound on the residual valid for all π₯. (Contributed by Mario Carneiro, 9-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ βπ β β+ βπ₯ β β+ (absβ((π βπ₯) / π₯)) β€ π | ||
Theorem | pntrsumo1 27414* | A bound on a sum over π . Equation 10.1.16 of [Shapiro], p. 403. (Contributed by Mario Carneiro, 25-May-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ (π₯ β β β¦ Ξ£π β (1...(ββπ₯))((π βπ) / (π Β· (π + 1)))) β π(1) | ||
Theorem | pntrsumbnd 27415* | A bound on a sum over π . Equation 10.1.16 of [Shapiro], p. 403. (Contributed by Mario Carneiro, 25-May-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ βπ β β+ βπ β β€ (absβΞ£π β (1...π)((π βπ) / (π Β· (π + 1)))) β€ π | ||
Theorem | pntrsumbnd2 27416* | A bound on a sum over π . Equation 10.1.16 of [Shapiro], p. 403. (Contributed by Mario Carneiro, 14-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ βπ β β+ βπ β β βπ β β€ (absβΞ£π β (π...π)((π βπ) / (π Β· (π + 1)))) β€ π | ||
Theorem | selbergr 27417* | Selberg's symmetry formula, using the residual of the second Chebyshev function. Equation 10.6.2 of [Shapiro], p. 428. (Contributed by Mario Carneiro, 16-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ (π₯ β β+ β¦ ((((π βπ₯) Β· (logβπ₯)) + Ξ£π β (1...(ββπ₯))((Ξβπ) Β· (π β(π₯ / π)))) / π₯)) β π(1) | ||
Theorem | selberg3r 27418* | Selberg's symmetry formula, using the residual of the second Chebyshev function. Equation 10.6.8 of [Shapiro], p. 429. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ (π₯ β (1(,)+β) β¦ ((((π βπ₯) Β· (logβπ₯)) + ((2 / (logβπ₯)) Β· Ξ£π β (1...(ββπ₯))(((Ξβπ) Β· (π β(π₯ / π))) Β· (logβπ)))) / π₯)) β π(1) | ||
Theorem | selberg4r 27419* | Selberg's symmetry formula, using the residual of the second Chebyshev function. Equation 10.6.11 of [Shapiro], p. 430. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ (π₯ β (1(,)+β) β¦ ((((π βπ₯) Β· (logβπ₯)) β ((2 / (logβπ₯)) Β· Ξ£π β (1...(ββπ₯))((Ξβπ) Β· Ξ£π β (1...(ββ(π₯ / π)))((Ξβπ) Β· (π β((π₯ / π) / π)))))) / π₯)) β π(1) | ||
Theorem | selberg34r 27420* | The sum of selberg3r 27418 and selberg4r 27419. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ (π₯ β (1(,)+β) β¦ ((((π βπ₯) Β· (logβπ₯)) β (Ξ£π β (1...(ββπ₯))((π β(π₯ / π)) Β· (Ξ£π β {π¦ β β β£ π¦ β₯ π} ((Ξβπ) Β· (Ξβ(π / π))) β ((Ξβπ) Β· (logβπ)))) / (logβπ₯))) / π₯)) β π(1) | ||
Theorem | pntsval 27421* | Define the "Selberg function", whose asymptotic behavior is the content of selberg 27397. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) β β’ (π΄ β β β (πβπ΄) = Ξ£π β (1...(ββπ΄))((Ξβπ) Β· ((logβπ) + (Οβ(π΄ / π))))) | ||
Theorem | pntsf 27422* | Functionality of the Selberg function. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) β β’ π:ββΆβ | ||
Theorem | selbergs 27423* | Selberg's symmetry formula, using the definition of the Selberg function. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) β β’ (π₯ β β+ β¦ (((πβπ₯) / π₯) β (2 Β· (logβπ₯)))) β π(1) | ||
Theorem | selbergsb 27424* | Selberg's symmetry formula, using the definition of the Selberg function. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) β β’ βπ β β+ βπ₯ β (1[,)+β)(absβ(((πβπ₯) / π₯) β (2 Β· (logβπ₯)))) β€ π | ||
Theorem | pntsval2 27425* | The Selberg function can be expressed using the convolution product of the von Mangoldt function with itself. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) β β’ (π΄ β β β (πβπ΄) = Ξ£π β (1...(ββπ΄))(((Ξβπ) Β· (logβπ)) + Ξ£π β {π¦ β β β£ π¦ β₯ π} ((Ξβπ) Β· (Ξβ(π / π))))) | ||
Theorem | pntrlog2bndlem1 27426* | The sum of selberg3r 27418 and selberg4r 27419. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) & β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ (π₯ β (1(,)+β) β¦ ((((absβ(π βπ₯)) Β· (logβπ₯)) β (Ξ£π β (1...(ββπ₯))((absβ(π β(π₯ / π))) Β· ((πβπ) β (πβ(π β 1)))) / (logβπ₯))) / π₯)) β β€π(1) | ||
Theorem | pntrlog2bndlem2 27427* | Lemma for pntrlog2bnd 27433. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) & β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β βπ¦ β β+ (Οβπ¦) β€ (π΄ Β· π¦)) β β’ (π β (π₯ β (1(,)+β) β¦ (Ξ£π β (1...(ββπ₯))(π Β· (absβ((π β(π₯ / (π + 1))) β (π β(π₯ / π))))) / (π₯ Β· (logβπ₯)))) β π(1)) | ||
Theorem | pntrlog2bndlem3 27428* | Lemma for pntrlog2bnd 27433. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) & β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β βπ¦ β (1[,)+β)(absβ(((πβπ¦) / π¦) β (2 Β· (logβπ¦)))) β€ π΄) β β’ (π β (π₯ β (1(,)+β) β¦ (Ξ£π β (1...(ββπ₯))(((absβ(π β(π₯ / π))) β (absβ(π β(π₯ / (π + 1))))) Β· ((πβπ) β (2 Β· (π Β· (logβπ))))) / (π₯ Β· (logβπ₯)))) β π(1)) | ||
Theorem | pntrlog2bndlem4 27429* | Lemma for pntrlog2bnd 27433. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) & β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ π = (π β β β¦ if(π β β+, (π Β· (logβπ)), 0)) β β’ (π₯ β (1(,)+β) β¦ ((((absβ(π βπ₯)) Β· (logβπ₯)) β ((2 / (logβπ₯)) Β· Ξ£π β (1...(ββπ₯))((absβ(π β(π₯ / π))) Β· ((πβπ) β (πβ(π β 1)))))) / π₯)) β β€π(1) | ||
Theorem | pntrlog2bndlem5 27430* | Lemma for pntrlog2bnd 27433. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) & β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ π = (π β β β¦ if(π β β+, (π Β· (logβπ)), 0)) & β’ (π β π΅ β β+) & β’ (π β βπ¦ β β+ (absβ((π βπ¦) / π¦)) β€ π΅) β β’ (π β (π₯ β (1(,)+β) β¦ ((((absβ(π βπ₯)) Β· (logβπ₯)) β ((2 / (logβπ₯)) Β· Ξ£π β (1...(ββπ₯))((absβ(π β(π₯ / π))) Β· (logβπ)))) / π₯)) β β€π(1)) | ||
Theorem | pntrlog2bndlem6a 27431* | Lemma for pntrlog2bndlem6 27432. (Contributed by Mario Carneiro, 7-Jun-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) & β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ π = (π β β β¦ if(π β β+, (π Β· (logβπ)), 0)) & β’ (π β π΅ β β+) & β’ (π β βπ¦ β β+ (absβ((π βπ¦) / π¦)) β€ π΅) & β’ (π β π΄ β β) & β’ (π β 1 β€ π΄) β β’ ((π β§ π₯ β (1(,)+β)) β (1...(ββπ₯)) = ((1...(ββ(π₯ / π΄))) βͺ (((ββ(π₯ / π΄)) + 1)...(ββπ₯)))) | ||
Theorem | pntrlog2bndlem6 27432* | Lemma for pntrlog2bnd 27433. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.) |
β’ π = (π β β β¦ Ξ£π β (1...(ββπ))((Ξβπ) Β· ((logβπ) + (Οβ(π / π))))) & β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ π = (π β β β¦ if(π β β+, (π Β· (logβπ)), 0)) & β’ (π β π΅ β β+) & β’ (π β βπ¦ β β+ (absβ((π βπ¦) / π¦)) β€ π΅) & β’ (π β π΄ β β) & β’ (π β 1 β€ π΄) β β’ (π β (π₯ β (1(,)+β) β¦ ((((absβ(π βπ₯)) Β· (logβπ₯)) β ((2 / (logβπ₯)) Β· Ξ£π β (1...(ββ(π₯ / π΄)))((absβ(π β(π₯ / π))) Β· (logβπ)))) / π₯)) β β€π(1)) | ||
Theorem | pntrlog2bnd 27433* | A bound on π (π₯)logβ2(π₯). Equation 10.6.15 of [Shapiro], p. 431. (Contributed by Mario Carneiro, 1-Jun-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ ((π΄ β β β§ 1 β€ π΄) β βπ β β+ βπ₯ β (1(,)+β)((((absβ(π βπ₯)) Β· (logβπ₯)) β ((2 / (logβπ₯)) Β· Ξ£π β (1...(ββ(π₯ / π΄)))((absβ(π β(π₯ / π))) Β· (logβπ)))) / π₯) β€ π) | ||
Theorem | pntpbnd1a 27434* | Lemma for pntpbnd 27437. (Contributed by Mario Carneiro, 11-Apr-2016.) Replace reference to OLD theorem. (Revised by Wolf Lammen, 8-Sep-2020.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β πΈ β (0(,)1)) & β’ π = (expβ(2 / πΈ)) & β’ (π β π β (π(,)+β)) & β’ (π β π β β) & β’ (π β (π < π β§ π β€ (πΎ Β· π))) & β’ (π β (absβ(π βπ)) β€ (absβ((π β(π + 1)) β (π βπ)))) β β’ (π β (absβ((π βπ) / π)) β€ πΈ) | ||
Theorem | pntpbnd1 27435* | Lemma for pntpbnd 27437. (Contributed by Mario Carneiro, 11-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β πΈ β (0(,)1)) & β’ π = (expβ(2 / πΈ)) & β’ (π β π β (π(,)+β)) & β’ (π β π΄ β β+) & β’ (π β βπ β β βπ β β€ (absβΞ£π¦ β (π...π)((π βπ¦) / (π¦ Β· (π¦ + 1)))) β€ π΄) & β’ πΆ = (π΄ + 2) & β’ (π β πΎ β ((expβ(πΆ / πΈ))[,)+β)) & β’ (π β Β¬ βπ¦ β β ((π < π¦ β§ π¦ β€ (πΎ Β· π)) β§ (absβ((π βπ¦) / π¦)) β€ πΈ)) β β’ (π β Ξ£π β (((ββπ) + 1)...(ββ(πΎ Β· π)))(absβ((π βπ) / (π Β· (π + 1)))) β€ π΄) | ||
Theorem | pntpbnd2 27436* | Lemma for pntpbnd 27437. (Contributed by Mario Carneiro, 11-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β πΈ β (0(,)1)) & β’ π = (expβ(2 / πΈ)) & β’ (π β π β (π(,)+β)) & β’ (π β π΄ β β+) & β’ (π β βπ β β βπ β β€ (absβΞ£π¦ β (π...π)((π βπ¦) / (π¦ Β· (π¦ + 1)))) β€ π΄) & β’ πΆ = (π΄ + 2) & β’ (π β πΎ β ((expβ(πΆ / πΈ))[,)+β)) & β’ (π β Β¬ βπ¦ β β ((π < π¦ β§ π¦ β€ (πΎ Β· π)) β§ (absβ((π βπ¦) / π¦)) β€ πΈ)) β β’ Β¬ π | ||
Theorem | pntpbnd 27437* | Lemma for pnt 27463. Establish smallness of π at a point. Lemma 10.6.1 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 10-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ βπ β β+ βπ β (0(,)1)βπ₯ β β+ βπ β ((expβ(π / π))[,)+β)βπ¦ β (π₯(,)+β)βπ β β ((π¦ < π β§ π β€ (π Β· π¦)) β§ (absβ((π βπ) / π)) β€ π) | ||
Theorem | pntibndlem1 27438 | Lemma for pntibnd 27442. (Contributed by Mario Carneiro, 10-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ πΏ = ((1 / 4) / (π΄ + 3)) β β’ (π β πΏ β (0(,)1)) | ||
Theorem | pntibndlem2a 27439* | Lemma for pntibndlem2 27440. (Contributed by Mario Carneiro, 7-Jun-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ πΏ = ((1 / 4) / (π΄ + 3)) & β’ (π β βπ₯ β β+ (absβ((π βπ₯) / π₯)) β€ π΄) & β’ (π β π΅ β β+) & β’ πΎ = (expβ(π΅ / (πΈ / 2))) & β’ πΆ = ((2 Β· π΅) + (logβ2)) & β’ (π β πΈ β (0(,)1)) & β’ (π β π β β+) & β’ (π β π β β) β β’ ((π β§ π’ β (π[,]((1 + (πΏ Β· πΈ)) Β· π))) β (π’ β β β§ π β€ π’ β§ π’ β€ ((1 + (πΏ Β· πΈ)) Β· π))) | ||
Theorem | pntibndlem2 27440* | Lemma for pntibnd 27442. The main work, after eliminating all the quantifiers. (Contributed by Mario Carneiro, 10-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ πΏ = ((1 / 4) / (π΄ + 3)) & β’ (π β βπ₯ β β+ (absβ((π βπ₯) / π₯)) β€ π΄) & β’ (π β π΅ β β+) & β’ πΎ = (expβ(π΅ / (πΈ / 2))) & β’ πΆ = ((2 Β· π΅) + (logβ2)) & β’ (π β πΈ β (0(,)1)) & β’ (π β π β β+) & β’ (π β π β β) & β’ (π β π β β+) & β’ (π β βπ₯ β (1(,)+β)βπ¦ β (π₯[,](2 Β· π₯))((Οβπ¦) β (Οβπ₯)) β€ ((2 Β· (π¦ β π₯)) + (π Β· (π₯ / (logβπ₯))))) & β’ π = ((expβ(π / (πΈ / 4))) + π) & β’ (π β π β ((expβ(πΆ / πΈ))[,)+β)) & β’ (π β π β (π(,)+β)) & β’ (π β ((π < π β§ π β€ ((π / 2) Β· π)) β§ (absβ((π βπ) / π)) β€ (πΈ / 2))) β β’ (π β βπ§ β β+ ((π < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (π Β· π)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) | ||
Theorem | pntibndlem3 27441* | Lemma for pntibnd 27442. Package up pntibndlem2 27440 in quantifiers. (Contributed by Mario Carneiro, 10-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ πΏ = ((1 / 4) / (π΄ + 3)) & β’ (π β βπ₯ β β+ (absβ((π βπ₯) / π₯)) β€ π΄) & β’ (π β π΅ β β+) & β’ πΎ = (expβ(π΅ / (πΈ / 2))) & β’ πΆ = ((2 Β· π΅) + (logβ2)) & β’ (π β πΈ β (0(,)1)) & β’ (π β π β β+) & β’ (π β βπ β (πΎ[,)+β)βπ£ β (π(,)+β)βπ β β ((π£ < π β§ π β€ (π Β· π£)) β§ (absβ((π βπ) / π)) β€ (πΈ / 2))) β β’ (π β βπ₯ β β+ βπ β ((expβ(πΆ / πΈ))[,)+β)βπ¦ β (π₯(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (π Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) | ||
Theorem | pntibnd 27442* | Lemma for pnt 27463. Establish smallness of π on an interval. Lemma 10.6.2 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 10-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ βπ β β+ βπ β (0(,)1)βπ β (0(,)1)βπ₯ β β+ βπ β ((expβ(π / π))[,)+β)βπ¦ β (π₯(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (π Β· π)) Β· π§) < (π Β· π¦)) β§ βπ’ β (π§[,]((1 + (π Β· π)) Β· π§))(absβ((π βπ’) / π’)) β€ π) | ||
Theorem | pntlemd 27443 | Lemma for pnt 27463. Closure for the constants used in the proof. For comparison with Equation 10.6.27 of [Shapiro], p. 434, π΄ is C^*, π΅ is c1, πΏ is Ξ», π· is c2, and πΉ is c3. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) β β’ (π β (πΏ β β+ β§ π· β β+ β§ πΉ β β+)) | ||
Theorem | pntlemc 27444* | Lemma for pnt 27463. Closure for the constants used in the proof. For comparison with Equation 10.6.27 of [Shapiro], p. 434, π is Ξ±, πΈ is Ξ΅, and πΎ is K. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) β β’ (π β (πΈ β β+ β§ πΎ β β+ β§ (πΈ β (0(,)1) β§ 1 < πΎ β§ (π β πΈ) β β+))) | ||
Theorem | pntlema 27445* | Lemma for pnt 27463. Closure for the constants used in the proof. The mammoth expression π is a number large enough to satisfy all the lower bounds needed for π. For comparison with Equation 10.6.27 of [Shapiro], p. 434, π is x2, π is x1, πΆ is the big-O constant in Equation 10.6.29 of [Shapiro], p. 435, and π is the unnamed lower bound of "for sufficiently large x" in Equation 10.6.34 of [Shapiro], p. 436. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) β β’ (π β π β β+) | ||
Theorem | pntlemb 27446* | Lemma for pnt 27463. Unpack all the lower bounds contained in π, in the form they will be used. For comparison with Equation 10.6.27 of [Shapiro], p. 434, π is x. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) β β’ (π β (π β β+ β§ (1 < π β§ e β€ (ββπ) β§ (ββπ) β€ (π / π)) β§ ((4 / (πΏ Β· πΈ)) β€ (ββπ) β§ (((logβπ) / (logβπΎ)) + 2) β€ (((logβπ) / (logβπΎ)) / 4) β§ ((π Β· 3) + πΆ) β€ (((π β πΈ) Β· ((πΏ Β· (πΈβ2)) / (;32 Β· π΅))) Β· (logβπ))))) | ||
Theorem | pntlemg 27447* | Lemma for pnt 27463. Closure for the constants used in the proof. For comparison with Equation 10.6.27 of [Shapiro], p. 434, π is j^* and π is Δ΅. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) β β’ (π β (π β β β§ π β (β€β₯βπ) β§ (((logβπ) / (logβπΎ)) / 4) β€ (π β π))) | ||
Theorem | pntlemh 27448* | Lemma for pnt 27463. Bounds on the subintervals in the induction. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) β β’ ((π β§ π½ β (π...π)) β (π < (πΎβπ½) β§ (πΎβπ½) β€ (ββπ))) | ||
Theorem | pntlemn 27449* | Lemma for pnt 27463. The "naive" base bound, which we will slightly improve. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) β β’ ((π β§ (π½ β β β§ π½ β€ (π / π))) β 0 β€ (((π / π½) β (absβ((π β(π / π½)) / π))) Β· (logβπ½))) | ||
Theorem | pntlemq 27450* | Lemma for pntlemj 27452. (Contributed by Mario Carneiro, 7-Jun-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) & β’ (π β βπ¦ β (π(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (πΎ Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ π = (((ββ(π / (πΎβ(π½ + 1)))) + 1)...(ββ(π / (πΎβπ½)))) & β’ (π β π β β+) & β’ (π β (((πΎβπ½) < π β§ ((1 + (πΏ Β· πΈ)) Β· π) < (πΎ Β· (πΎβπ½))) β§ βπ’ β (π[,]((1 + (πΏ Β· πΈ)) Β· π))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ (π β π½ β (π..^π)) & β’ πΌ = (((ββ(π / ((1 + (πΏ Β· πΈ)) Β· π))) + 1)...(ββ(π / π))) β β’ (π β πΌ β π) | ||
Theorem | pntlemr 27451* | Lemma for pntlemj 27452. (Contributed by Mario Carneiro, 7-Jun-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) & β’ (π β βπ¦ β (π(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (πΎ Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ π = (((ββ(π / (πΎβ(π½ + 1)))) + 1)...(ββ(π / (πΎβπ½)))) & β’ (π β π β β+) & β’ (π β (((πΎβπ½) < π β§ ((1 + (πΏ Β· πΈ)) Β· π) < (πΎ Β· (πΎβπ½))) β§ βπ’ β (π[,]((1 + (πΏ Β· πΈ)) Β· π))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ (π β π½ β (π..^π)) & β’ πΌ = (((ββ(π / ((1 + (πΏ Β· πΈ)) Β· π))) + 1)...(ββ(π / π))) β β’ (π β ((π β πΈ) Β· (((πΏ Β· πΈ) / 8) Β· (logβπ))) β€ ((β―βπΌ) Β· ((π β πΈ) Β· ((logβ(π / π)) / (π / π))))) | ||
Theorem | pntlemj 27452* | Lemma for pnt 27463. The induction step. Using pntibnd 27442, we find an interval in πΎβπ½...πΎβ(π½ + 1) which is sufficiently large and has a much smaller value, π (π§) / π§ β€ πΈ (instead of our original bound π (π§) / π§ β€ π). (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) & β’ (π β βπ¦ β (π(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (πΎ Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ π = (((ββ(π / (πΎβ(π½ + 1)))) + 1)...(ββ(π / (πΎβπ½)))) & β’ (π β π β β+) & β’ (π β (((πΎβπ½) < π β§ ((1 + (πΏ Β· πΈ)) Β· π) < (πΎ Β· (πΎβπ½))) β§ βπ’ β (π[,]((1 + (πΏ Β· πΈ)) Β· π))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ (π β π½ β (π..^π)) & β’ πΌ = (((ββ(π / ((1 + (πΏ Β· πΈ)) Β· π))) + 1)...(ββ(π / π))) β β’ (π β ((π β πΈ) Β· (((πΏ Β· πΈ) / 8) Β· (logβπ))) β€ Ξ£π β π (((π / π) β (absβ((π β(π / π)) / π))) Β· (logβπ))) | ||
Theorem | pntlemi 27453* | Lemma for pnt 27463. Eliminate some assumptions from pntlemj 27452. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) & β’ (π β βπ¦ β (π(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (πΎ Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ π = (((ββ(π / (πΎβ(π½ + 1)))) + 1)...(ββ(π / (πΎβπ½)))) β β’ ((π β§ π½ β (π..^π)) β ((π β πΈ) Β· (((πΏ Β· πΈ) / 8) Β· (logβπ))) β€ Ξ£π β π (((π / π) β (absβ((π β(π / π)) / π))) Β· (logβπ))) | ||
Theorem | pntlemf 27454* | Lemma for pnt 27463. Add up the pieces in pntlemi 27453 to get an estimate slightly better than the naive lower bound 0. (Contributed by Mario Carneiro, 13-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) & β’ (π β βπ¦ β (π(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (πΎ Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) β β’ (π β ((π β πΈ) Β· (((πΏ Β· (πΈβ2)) / (;32 Β· π΅)) Β· ((logβπ)β2))) β€ Ξ£π β (1...(ββ(π / π)))(((π / π) β (absβ((π β(π / π)) / π))) Β· (logβπ))) | ||
Theorem | pntlemk 27455* | Lemma for pnt 27463. Evaluate the naive part of the estimate. (Contributed by Mario Carneiro, 14-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) & β’ (π β βπ¦ β (π(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (πΎ Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) β β’ (π β (2 Β· Ξ£π β (1...(ββ(π / π)))((π / π) Β· (logβπ))) β€ ((π Β· ((logβπ) + 3)) Β· (logβπ))) | ||
Theorem | pntlemo 27456* | Lemma for pnt 27463. Combine all the estimates to establish a smaller eventual bound on π (π) / π. (Contributed by Mario Carneiro, 14-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β π β (π[,)+β)) & β’ π = ((ββ((logβπ) / (logβπΎ))) + 1) & β’ π = (ββ(((logβπ) / (logβπΎ)) / 2)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) & β’ (π β βπ¦ β (π(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (πΎ Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ (π β βπ§ β (1(,)+β)((((absβ(π βπ§)) Β· (logβπ§)) β ((2 / (logβπ§)) Β· Ξ£π β (1...(ββ(π§ / π)))((absβ(π β(π§ / π))) Β· (logβπ)))) / π§) β€ πΆ) β β’ (π β (absβ((π βπ) / π)) β€ (π β (πΉ Β· (πβ3)))) | ||
Theorem | pntleme 27457* | Lemma for pnt 27463. Package up pntlemo 27456 in quantifiers. (Contributed by Mario Carneiro, 14-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β (π β β+ β§ π < π)) & β’ (π β πΆ β β+) & β’ π = (((π + (4 / (πΏ Β· πΈ)))β2) + (((π Β· (πΎβ2))β4) + (expβ(((;32 Β· π΅) / ((π β πΈ) Β· (πΏ Β· (πΈβ2)))) Β· ((π Β· 3) + πΆ))))) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) & β’ (π β βπ β (πΎ[,)+β)βπ¦ β (π(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· πΈ)) Β· π§) < (π Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· πΈ)) Β· π§))(absβ((π βπ’) / π’)) β€ πΈ)) & β’ (π β βπ§ β (1(,)+β)((((absβ(π βπ§)) Β· (logβπ§)) β ((2 / (logβπ§)) Β· Ξ£π β (1...(ββ(π§ / π)))((absβ(π β(π§ / π))) Β· (logβπ)))) / π§) β€ πΆ) β β’ (π β βπ€ β β+ βπ£ β (π€[,)+β)(absβ((π βπ£) / π£)) β€ (π β (πΉ Β· (πβ3)))) | ||
Theorem | pntlem3 27458* | Lemma for pnt 27463. Equation 10.6.35 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 8-Apr-2016.) (Proof shortened by AV, 27-Sep-2020.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β βπ₯ β β+ (absβ((π βπ₯) / π₯)) β€ π΄) & β’ π = {π‘ β (0[,]π΄) β£ βπ¦ β β+ βπ§ β (π¦[,)+β)(absβ((π βπ§) / π§)) β€ π‘} & β’ (π β πΆ β β+) & β’ ((π β§ π’ β π) β (π’ β (πΆ Β· (π’β3))) β π) β β’ (π β (π₯ β β+ β¦ ((Οβπ₯) / π₯)) βπ 1) | ||
Theorem | pntlemp 27459* | Lemma for pnt 27463. Wrapping up more quantifiers. (Contributed by Mario Carneiro, 14-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β βπ₯ β β+ (absβ((π βπ₯) / π₯)) β€ π΄) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β βπ β (0(,)1)βπ₯ β β+ βπ β ((expβ(π΅ / π))[,)+β)βπ¦ β (π₯(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· π)) Β· π§) < (π Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· π)) Β· π§))(absβ((π βπ’) / π’)) β€ π)) & β’ (π β π β β+) & β’ (π β π β€ π΄) & β’ πΈ = (π / π·) & β’ πΎ = (expβ(π΅ / πΈ)) & β’ (π β (π β β+ β§ 1 β€ π)) & β’ (π β βπ§ β (π[,)+β)(absβ((π βπ§) / π§)) β€ π) β β’ (π β βπ€ β β+ βπ£ β (π€[,)+β)(absβ((π βπ£) / π£)) β€ (π β (πΉ Β· (πβ3)))) | ||
Theorem | pntleml 27460* | Lemma for pnt 27463. Equation 10.6.35 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 14-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) & β’ (π β π΄ β β+) & β’ (π β βπ₯ β β+ (absβ((π βπ₯) / π₯)) β€ π΄) & β’ (π β π΅ β β+) & β’ (π β πΏ β (0(,)1)) & β’ π· = (π΄ + 1) & β’ πΉ = ((1 β (1 / π·)) Β· ((πΏ / (;32 Β· π΅)) / (π·β2))) & β’ (π β βπ β (0(,)1)βπ₯ β β+ βπ β ((expβ(π΅ / π))[,)+β)βπ¦ β (π₯(,)+β)βπ§ β β+ ((π¦ < π§ β§ ((1 + (πΏ Β· π)) Β· π§) < (π Β· π¦)) β§ βπ’ β (π§[,]((1 + (πΏ Β· π)) Β· π§))(absβ((π βπ’) / π’)) β€ π)) β β’ (π β (π₯ β β+ β¦ ((Οβπ₯) / π₯)) βπ 1) | ||
Theorem | pnt3 27461 | The Prime Number Theorem, version 3: the second Chebyshev function tends asymptotically to π₯. (Contributed by Mario Carneiro, 1-Jun-2016.) |
β’ (π₯ β β+ β¦ ((Οβπ₯) / π₯)) βπ 1 | ||
Theorem | pnt2 27462 | The Prime Number Theorem, version 2: the first Chebyshev function tends asymptotically to π₯. (Contributed by Mario Carneiro, 1-Jun-2016.) |
β’ (π₯ β β+ β¦ ((ΞΈβπ₯) / π₯)) βπ 1 | ||
Theorem | pnt 27463 | The Prime Number Theorem: the number of prime numbers less than π₯ tends asymptotically to π₯ / log(π₯) as π₯ goes to infinity. This is Metamath 100 proof #5. (Contributed by Mario Carneiro, 1-Jun-2016.) |
β’ (π₯ β (1(,)+β) β¦ ((Οβπ₯) / (π₯ / (logβπ₯)))) βπ 1 | ||
Theorem | abvcxp 27464* | Raising an absolute value to a power less than one yields another absolute value. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π΄ = (AbsValβπ ) & β’ π΅ = (Baseβπ ) & β’ πΊ = (π₯ β π΅ β¦ ((πΉβπ₯)βππ)) β β’ ((πΉ β π΄ β§ π β (0(,]1)) β πΊ β π΄) | ||
Theorem | padicfval 27465* | Value of the p-adic absolute value. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) β β’ (π β β β (π½βπ) = (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) | ||
Theorem | padicval 27466* | Value of the p-adic absolute value. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) β β’ ((π β β β§ π β β) β ((π½βπ)βπ) = if(π = 0, 0, (πβ-(π pCnt π)))) | ||
Theorem | ostth2lem1 27467* | Lemma for ostth2 27486, although it is just a simple statement about exponentials which does not involve any specifics of ostth2 27486. If a power is upper bounded by a linear term, the exponent must be less than one. Or in big-O notation, π β π(π΄βπ) for any 1 < π΄. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ ((π β§ π β β) β (π΄βπ) β€ (π Β· π΅)) β β’ (π β π΄ β€ 1) | ||
Theorem | qrngbas 27468 | The base set of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π = (βfld βΎs β) β β’ β = (Baseβπ) | ||
Theorem | qdrng 27469 | The rationals form a division ring. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π = (βfld βΎs β) β β’ π β DivRing | ||
Theorem | qrng0 27470 | The zero element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π = (βfld βΎs β) β β’ 0 = (0gβπ) | ||
Theorem | qrng1 27471 | The unity element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π = (βfld βΎs β) β β’ 1 = (1rβπ) | ||
Theorem | qrngneg 27472 | The additive inverse in the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π = (βfld βΎs β) β β’ (π β β β ((invgβπ)βπ) = -π) | ||
Theorem | qrngdiv 27473 | The division operation in the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.) |
β’ π = (βfld βΎs β) β β’ ((π β β β§ π β β β§ π β 0) β (π(/rβπ)π) = (π / π)) | ||
Theorem | qabvle 27474 | By using induction on π, we show a long-range inequality coming from the triangle inequality. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) β β’ ((πΉ β π΄ β§ π β β0) β (πΉβπ) β€ π) | ||
Theorem | qabvexp 27475 | Induct the product rule abvmul 20662 to find the absolute value of a power. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) β β’ ((πΉ β π΄ β§ π β β β§ π β β0) β (πΉβ(πβπ)) = ((πΉβπ)βπ)) | ||
Theorem | ostthlem1 27476* | Lemma for ostth 27488. If two absolute values agree on the positive integers greater than one, then they agree for all rational numbers and thus are equal as functions. (Contributed by Mario Carneiro, 9-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ (π β πΉ β π΄) & β’ (π β πΊ β π΄) & β’ ((π β§ π β (β€β₯β2)) β (πΉβπ) = (πΊβπ)) β β’ (π β πΉ = πΊ) | ||
Theorem | ostthlem2 27477* | Lemma for ostth 27488. Refine ostthlem1 27476 so that it is sufficient to only show equality on the primes. (Contributed by Mario Carneiro, 9-Sep-2014.) (Revised by Mario Carneiro, 20-Jun-2015.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ (π β πΉ β π΄) & β’ (π β πΊ β π΄) & β’ ((π β§ π β β) β (πΉβπ) = (πΊβπ)) β β’ (π β πΉ = πΊ) | ||
Theorem | qabsabv 27478 | The regular absolute value function on the rationals is in fact an absolute value under our definition. (Contributed by Mario Carneiro, 9-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) β β’ (abs βΎ β) β π΄ | ||
Theorem | padicabv 27479* | The p-adic absolute value (with arbitrary base) is an absolute value. (Contributed by Mario Carneiro, 9-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ πΉ = (π₯ β β β¦ if(π₯ = 0, 0, (πβ(π pCnt π₯)))) β β’ ((π β β β§ π β (0(,)1)) β πΉ β π΄) | ||
Theorem | padicabvf 27480* | The p-adic absolute value is an absolute value. (Contributed by Mario Carneiro, 9-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) β β’ π½:ββΆπ΄ | ||
Theorem | padicabvcxp 27481* | All positive powers of the p-adic absolute value are absolute values. (Contributed by Mario Carneiro, 9-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) β β’ ((π β β β§ π β β+) β (π¦ β β β¦ (((π½βπ)βπ¦)βππ )) β π΄) | ||
Theorem | ostth1 27482* | - Lemma for ostth 27488: trivial case. (Not that the proof is trivial, but that we are proving that the function is trivial.) If πΉ is equal to 1 on the primes, then by complete induction and the multiplicative property abvmul 20662 of the absolute value, πΉ is equal to 1 on all the integers, and ostthlem1 27476 extends this to the other rational numbers. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) & β’ πΎ = (π₯ β β β¦ if(π₯ = 0, 0, 1)) & β’ (π β πΉ β π΄) & β’ (π β βπ β β Β¬ 1 < (πΉβπ)) & β’ (π β βπ β β Β¬ (πΉβπ) < 1) β β’ (π β πΉ = πΎ) | ||
Theorem | ostth2lem2 27483* | Lemma for ostth2 27486. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) & β’ πΎ = (π₯ β β β¦ if(π₯ = 0, 0, 1)) & β’ (π β πΉ β π΄) & β’ (π β π β (β€β₯β2)) & β’ (π β 1 < (πΉβπ)) & β’ π = ((logβ(πΉβπ)) / (logβπ)) & β’ (π β π β (β€β₯β2)) & β’ π = ((logβ(πΉβπ)) / (logβπ)) & β’ π = if((πΉβπ) β€ 1, 1, (πΉβπ)) β β’ ((π β§ π β β0 β§ π β (0...((πβπ) β 1))) β (πΉβπ) β€ ((π Β· π) Β· (πβπ))) | ||
Theorem | ostth2lem3 27484* | Lemma for ostth2 27486. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) & β’ πΎ = (π₯ β β β¦ if(π₯ = 0, 0, 1)) & β’ (π β πΉ β π΄) & β’ (π β π β (β€β₯β2)) & β’ (π β 1 < (πΉβπ)) & β’ π = ((logβ(πΉβπ)) / (logβπ)) & β’ (π β π β (β€β₯β2)) & β’ π = ((logβ(πΉβπ)) / (logβπ)) & β’ π = if((πΉβπ) β€ 1, 1, (πΉβπ)) & β’ π = ((logβπ) / (logβπ)) β β’ ((π β§ π β β) β (((πΉβπ) / (πβππ))βπ) β€ (π Β· ((π Β· π) Β· (π + 1)))) | ||
Theorem | ostth2lem4 27485* | Lemma for ostth2 27486. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) & β’ πΎ = (π₯ β β β¦ if(π₯ = 0, 0, 1)) & β’ (π β πΉ β π΄) & β’ (π β π β (β€β₯β2)) & β’ (π β 1 < (πΉβπ)) & β’ π = ((logβ(πΉβπ)) / (logβπ)) & β’ (π β π β (β€β₯β2)) & β’ π = ((logβ(πΉβπ)) / (logβπ)) & β’ π = if((πΉβπ) β€ 1, 1, (πΉβπ)) & β’ π = ((logβπ) / (logβπ)) β β’ (π β (1 < (πΉβπ) β§ π β€ π)) | ||
Theorem | ostth2 27486* | - Lemma for ostth 27488: regular case. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) & β’ πΎ = (π₯ β β β¦ if(π₯ = 0, 0, 1)) & β’ (π β πΉ β π΄) & β’ (π β π β (β€β₯β2)) & β’ (π β 1 < (πΉβπ)) & β’ π = ((logβ(πΉβπ)) / (logβπ)) β β’ (π β βπ β (0(,]1)πΉ = (π¦ β β β¦ ((absβπ¦)βππ))) | ||
Theorem | ostth3 27487* | - Lemma for ostth 27488: p-adic case. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) & β’ πΎ = (π₯ β β β¦ if(π₯ = 0, 0, 1)) & β’ (π β πΉ β π΄) & β’ (π β βπ β β Β¬ 1 < (πΉβπ)) & β’ (π β π β β) & β’ (π β (πΉβπ) < 1) & β’ π = -((logβ(πΉβπ)) / (logβπ)) & β’ π = if((πΉβπ) β€ (πΉβπ), (πΉβπ), (πΉβπ)) β β’ (π β βπ β β+ πΉ = (π¦ β β β¦ (((π½βπ)βπ¦)βππ))) | ||
Theorem | ostth 27488* | Ostrowski's theorem, which classifies all absolute values on β. Any such absolute value must either be the trivial absolute value πΎ, a constant exponent 0 < π β€ 1 times the regular absolute value, or a positive exponent times the p-adic absolute value. (Contributed by Mario Carneiro, 10-Sep-2014.) |
β’ π = (βfld βΎs β) & β’ π΄ = (AbsValβπ) & β’ π½ = (π β β β¦ (π₯ β β β¦ if(π₯ = 0, 0, (πβ-(π pCnt π₯))))) & β’ πΎ = (π₯ β β β¦ if(π₯ = 0, 0, 1)) β β’ (πΉ β π΄ β (πΉ = πΎ β¨ βπ β (0(,]1)πΉ = (π¦ β β β¦ ((absβπ¦)βππ)) β¨ βπ β β+ βπ β ran π½ πΉ = (π¦ β β β¦ ((πβπ¦)βππ)))) | ||
The surreal numbers can be represented in several equivalent ways. In [Alling], Norman Alling made this notion explicit by giving a set of axioms that all representations admit, then proving that there is an order and birthday preserving bijection between any systems that satisfy these axioms. In this section, we start with the definition of surreal numbers given in [Gonshor] and derive Alling's axioms. After deriving them we no longer refer to the explicit definition of surreals. In particular, we never take advantage of the fact that the empty set is a surreal number under our definition. | ||
Syntax | csur 27489 | Declare the class of all surreal numbers (see df-no 27492). |
class No | ||
Syntax | cslt 27490 | Declare the less-than relation over surreal numbers (see df-slt 27493). |
class <s | ||
Syntax | cbday 27491 | Declare the birthday function for surreal numbers (see df-bday 27494). |
class bday | ||
Definition | df-no 27492* |
Define the class of surreal numbers. The surreal numbers are a proper
class of numbers developed by John H. Conway and introduced by Donald
Knuth in 1975. They form a proper class into which all ordered fields
can be embedded. The approach we take to defining them was first
introduced by Hary Gonshor, and is based on the conception of a
"sign
expansion" of a surreal number. We define the surreals as
ordinal-indexed sequences of 1o and
2o, analagous to Gonshor's
( β ) and ( + ).
After introducing this definition, we will abstract away from it using axioms that Norman Alling developed in "Foundations of Analysis over Surreal Number Fields." This is done in an effort to be agnostic towards the exact implementation of surreals. (Contributed by Scott Fenton, 9-Jun-2011.) |
β’ No = {π β£ βπ β On π:πβΆ{1o, 2o}} | ||
Definition | df-slt 27493* | Next, we introduce surreal less-than, a comparison relation over the surreals by lexicographically ordering them. (Contributed by Scott Fenton, 9-Jun-2011.) |
β’ <s = {β¨π, πβ© β£ ((π β No β§ π β No ) β§ βπ₯ β On (βπ¦ β π₯ (πβπ¦) = (πβπ¦) β§ (πβπ₯){β¨1o, β β©, β¨1o, 2oβ©, β¨β , 2oβ©} (πβπ₯)))} | ||
Definition | df-bday 27494 | Finally, we introduce the birthday function. This function maps each surreal to an ordinal. In our implementation, this is the domain of the sign function. The important properties of this function are established later. (Contributed by Scott Fenton, 11-Jun-2011.) |
β’ bday = (π₯ β No β¦ dom π₯) | ||
Theorem | elno 27495* | Membership in the surreals. (Shortened proof on 2012-Apr-14, SF). (Contributed by Scott Fenton, 11-Jun-2011.) |
β’ (π΄ β No β βπ₯ β On π΄:π₯βΆ{1o, 2o}) | ||
Theorem | sltval 27496* | The value of the surreal less-than relation. (Contributed by Scott Fenton, 14-Jun-2011.) |
β’ ((π΄ β No β§ π΅ β No ) β (π΄ <s π΅ β βπ₯ β On (βπ¦ β π₯ (π΄βπ¦) = (π΅βπ¦) β§ (π΄βπ₯){β¨1o, β β©, β¨1o, 2oβ©, β¨β , 2oβ©} (π΅βπ₯)))) | ||
Theorem | bdayval 27497 | The value of the birthday function within the surreals. (Contributed by Scott Fenton, 14-Jun-2011.) |
β’ (π΄ β No β ( bday βπ΄) = dom π΄) | ||
Theorem | nofun 27498 | A surreal is a function. (Contributed by Scott Fenton, 16-Jun-2011.) |
β’ (π΄ β No β Fun π΄) | ||
Theorem | nodmon 27499 | The domain of a surreal is an ordinal. (Contributed by Scott Fenton, 16-Jun-2011.) |
β’ (π΄ β No β dom π΄ β On) | ||
Theorem | norn 27500 | The range of a surreal is a subset of the surreal signs. (Contributed by Scott Fenton, 16-Jun-2011.) |
β’ (π΄ β No β ran π΄ β {1o, 2o}) |
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