![]() |
Metamath
Proof Explorer Theorem List (p. 276 of 484) | < Previous Next > |
Bad symbols? Try the
GIF version. |
||
Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List |
Color key: | ![]() (1-30784) |
![]() (30785-32307) |
![]() (32308-48350) |
Type | Label | Description |
---|---|---|
Statement | ||
Theorem | vmalogdivsum2 27501* | The sum Ξ£π β€ π₯, Ξ(π)log(π₯ / π) / π is asymptotic to logβ2(π₯) / 2 + π(logπ₯). Exercise 9.1.7 of [Shapiro], p. 336. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ (π₯ β (1(,)+β) β¦ ((Ξ£π β (1...(ββπ₯))(((Ξβπ) / π) Β· (logβ(π₯ / π))) / (logβπ₯)) β ((logβπ₯) / 2))) β π(1) | ||
Theorem | vmalogdivsum 27502* | The sum Ξ£π β€ π₯, Ξ(π)logπ / π is asymptotic to logβ2(π₯) / 2 + π(logπ₯). Exercise 9.1.7 of [Shapiro], p. 336. (Contributed by Mario Carneiro, 30-May-2016.) |
β’ (π₯ β (1(,)+β) β¦ ((Ξ£π β (1...(ββπ₯))(((Ξβπ) / π) Β· (logβπ)) / (logβπ₯)) β ((logβπ₯) / 2))) β π(1) | ||
Theorem | 2vmadivsumlem 27503* | Lemma for 2vmadivsum 27504. (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) / π) β (logβπ¦))) β€ π΄) β β’ (π β (π₯ β (1(,)+β) β¦ ((Ξ£π β (1...(ββπ₯))(((Ξβπ) / π) Β· Ξ£π β (1...(ββ(π₯ / π)))((Ξβπ) / π)) / (logβπ₯)) β ((logβπ₯) / 2))) β π(1)) | ||
Theorem | 2vmadivsum 27504* | The sum Ξ£ππ β€ π₯, Ξ(π)Ξ(π) / ππ is asymptotic to logβ2(π₯) / 2 + π(logπ₯). (Contributed by Mario Carneiro, 30-May-2016.) |
β’ (π₯ β (1(,)+β) β¦ ((Ξ£π β (1...(ββπ₯))(((Ξβπ) / π) Β· Ξ£π β (1...(ββ(π₯ / π)))((Ξβπ) / π)) / (logβπ₯)) β ((logβπ₯) / 2))) β π(1) | ||
Theorem | logsqvma 27505* | A formula for logβ2(π) in terms of the primes. Equation 10.4.6 of [Shapiro], p. 418. (Contributed by Mario Carneiro, 13-May-2016.) |
β’ (π β β β Ξ£π β {π₯ β β β£ π₯ β₯ π} (Ξ£π’ β {π₯ β β β£ π₯ β₯ π} ((Ξβπ’) Β· (Ξβ(π / π’))) + ((Ξβπ) Β· (logβπ))) = ((logβπ)β2)) | ||
Theorem | logsqvma2 27506* | The MΓΆbius inverse of logsqvma 27505. Equation 10.4.8 of [Shapiro], p. 418. (Contributed by Mario Carneiro, 13-May-2016.) |
β’ (π β β β Ξ£π β {π₯ β β β£ π₯ β₯ π} ((ΞΌβπ) Β· ((logβ(π / π))β2)) = (Ξ£π β {π₯ β β β£ π₯ β₯ π} ((Ξβπ) Β· (Ξβ(π / π))) + ((Ξβπ) Β· (logβπ)))) | ||
Theorem | log2sumbnd 27507* | Bound on the difference between Ξ£π β€ π΄, logβ2(π) and the equivalent integral. (Contributed by Mario Carneiro, 20-May-2016.) |
β’ ((π΄ β β+ β§ 1 β€ π΄) β (absβ(Ξ£π β (1...(ββπ΄))((logβπ)β2) β (π΄ Β· (((logβπ΄)β2) + (2 β (2 Β· (logβπ΄))))))) β€ (((logβπ΄)β2) + 2)) | ||
Theorem | selberglem1 27508* | Lemma for selberg 27511. Estimation of the asymptotic part of selberglem3 27510. (Contributed by Mario Carneiro, 20-May-2016.) |
β’ π = ((((logβ(π₯ / π))β2) + (2 β (2 Β· (logβ(π₯ / π))))) / π) β β’ (π₯ β β+ β¦ (Ξ£π β (1...(ββπ₯))((ΞΌβπ) Β· π) β (2 Β· (logβπ₯)))) β π(1) | ||
Theorem | selberglem2 27509* | Lemma for selberg 27511. (Contributed by Mario Carneiro, 23-May-2016.) |
β’ π = ((((logβ(π₯ / π))β2) + (2 β (2 Β· (logβ(π₯ / π))))) / π) β β’ (π₯ β β+ β¦ ((Ξ£π β (1...(ββπ₯))Ξ£π β (1...(ββ(π₯ / π)))((ΞΌβπ) Β· ((logβπ)β2)) / π₯) β (2 Β· (logβπ₯)))) β π(1) | ||
Theorem | selberglem3 27510* | Lemma for selberg 27511. Estimation of the left-hand side of logsqvma2 27506. (Contributed by Mario Carneiro, 23-May-2016.) |
β’ (π₯ β β+ β¦ ((Ξ£π β (1...(ββπ₯))Ξ£π β {π¦ β β β£ π¦ β₯ π} ((ΞΌβπ) Β· ((logβ(π / π))β2)) / π₯) β (2 Β· (logβπ₯)))) β π(1) | ||
Theorem | selberg 27511* | Selberg's symmetry formula. The statement has many forms, and this one is equivalent to the statement that Ξ£π β€ π₯, Ξ(π)logπ + Ξ£π Β· π β€ π₯, Ξ(π)Ξ(π) = 2π₯logπ₯ + π(π₯). Equation 10.4.10 of [Shapiro], p. 419. (Contributed by Mario Carneiro, 23-May-2016.) |
β’ (π₯ β β+ β¦ ((Ξ£π β (1...(ββπ₯))((Ξβπ) Β· ((logβπ) + (Οβ(π₯ / π)))) / π₯) β (2 Β· (logβπ₯)))) β π(1) | ||
Theorem | selbergb 27512* | Convert eventual boundedness in selberg 27511 to boundedness on [1, +β). (We have to bound away from zero because the log terms diverge at zero.) (Contributed by Mario Carneiro, 30-May-2016.) |
β’ βπ β β+ βπ₯ β (1[,)+β)(absβ((Ξ£π β (1...(ββπ₯))((Ξβπ) Β· ((logβπ) + (Οβ(π₯ / π)))) / π₯) β (2 Β· (logβπ₯)))) β€ π | ||
Theorem | selberg2lem 27513* | Lemma for selberg2 27514. Equation 10.4.12 of [Shapiro], p. 420. (Contributed by Mario Carneiro, 23-May-2016.) |
β’ (π₯ β β+ β¦ ((Ξ£π β (1...(ββπ₯))((Ξβπ) Β· (logβπ)) β ((Οβπ₯) Β· (logβπ₯))) / π₯)) β π(1) | ||
Theorem | selberg2 27514* | Selberg's symmetry formula, using the second Chebyshev function. Equation 10.4.14 of [Shapiro], p. 420. (Contributed by Mario Carneiro, 23-May-2016.) |
β’ (π₯ β β+ β¦ (((((Οβπ₯) Β· (logβπ₯)) + Ξ£π β (1...(ββπ₯))((Ξβπ) Β· (Οβ(π₯ / π)))) / π₯) β (2 Β· (logβπ₯)))) β π(1) | ||
Theorem | selberg2b 27515* | Convert eventual boundedness in selberg2 27514 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 27516* | Lemma for chpdifbnd 27518. (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 27517* | Lemma for chpdifbnd 27518. (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 27518* | 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 27519* | A bound on a sum of logs, used in pntlemk 27569. This is not as precise as logdivsum 27496 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 27520* | Introduce a log weighting on the summands of Ξ£π Β· π β€ π₯, Ξ(π)Ξ(π), the core of selberg2 27514 (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 27521* | Lemma for selberg3 27522. 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 27522* | Introduce a log weighting on the summands of Ξ£π Β· π β€ π₯, Ξ(π)Ξ(π), the core of selberg2 27514 (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 27523* | Lemma for selberg4 27524. 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 27524* | 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 27525* | 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 27526 | Functionality of the residual. Lemma for pnt 27577. (Contributed by Mario Carneiro, 8-Apr-2016.) |
β’ π = (π β β+ β¦ ((Οβπ) β π)) β β’ π :β+βΆβ | ||
Theorem | pntrmax 27527* | 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 27528* | 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 27529* | 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 27530* | 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 27531* | 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 27532* | 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 27533* | 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 27534* | The sum of selberg3r 27532 and selberg4r 27533. (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 27535* | Define the "Selberg function", whose asymptotic behavior is the content of selberg 27511. (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 27536* | 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 27537* | 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 27538* | 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 27539* | 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 27540* | The sum of selberg3r 27532 and selberg4r 27533. (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 27541* | Lemma for pntrlog2bnd 27547. 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 27542* | Lemma for pntrlog2bnd 27547. 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 27543* | Lemma for pntrlog2bnd 27547. 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 27544* | Lemma for pntrlog2bnd 27547. 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 27545* | Lemma for pntrlog2bndlem6 27546. (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 27546* | Lemma for pntrlog2bnd 27547. 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 27547* | 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 27548* | Lemma for pntpbnd 27551. (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 27549* | Lemma for pntpbnd 27551. (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 27550* | Lemma for pntpbnd 27551. (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 27551* | Lemma for pnt 27577. 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 27552 | Lemma for pntibnd 27556. (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 27553* | Lemma for pntibndlem2 27554. (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 27554* | Lemma for pntibnd 27556. 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 27555* | Lemma for pntibnd 27556. Package up pntibndlem2 27554 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 27556* | Lemma for pnt 27577. 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 27557 | Lemma for pnt 27577. 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 27558* | Lemma for pnt 27577. 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 27559* | Lemma for pnt 27577. 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 27560* | Lemma for pnt 27577. 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 27561* | Lemma for pnt 27577. 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 27562* | Lemma for pnt 27577. 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 27563* | Lemma for pnt 27577. 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 27564* | Lemma for pntlemj 27566. (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 27565* | Lemma for pntlemj 27566. (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 27566* | Lemma for pnt 27577. The induction step. Using pntibnd 27556, 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 27567* | Lemma for pnt 27577. Eliminate some assumptions from pntlemj 27566. (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 27568* | Lemma for pnt 27577. Add up the pieces in pntlemi 27567 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 27569* | Lemma for pnt 27577. 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 27570* | Lemma for pnt 27577. 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 27571* | Lemma for pnt 27577. Package up pntlemo 27570 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 27572* | Lemma for pnt 27577. 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 27573* | Lemma for pnt 27577. 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 27574* | Lemma for pnt 27577. 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 27575 | 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 27576 | 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 27577 | 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 27578* | 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 27579* | 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 27580* | 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 27581* | Lemma for ostth2 27600, although it is just a simple statement about exponentials which does not involve any specifics of ostth2 27600. 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 27582 | 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 27583 | 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 27584 | 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 27585 | 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 27586 | 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 27587 | 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 27588 | 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 27589 | Induct the product rule abvmul 20713 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 27590* | Lemma for ostth 27602. 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 27591* | Lemma for ostth 27602. Refine ostthlem1 27590 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 27592 | 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 27593* | 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 27594* | 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 27595* | 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 27596* | - Lemma for ostth 27602: 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 20713 of the absolute value, πΉ is equal to 1 on all the integers, and ostthlem1 27590 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 27597* | Lemma for ostth2 27600. (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 27598* | Lemma for ostth2 27600. (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 27599* | Lemma for ostth2 27600. (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 27600* | - Lemma for ostth 27602: 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))) |
< Previous Next > |
Copyright terms: Public domain | < Previous Next > |