Exploring Prime Numbers and Modular Functions III: On the Exponential of Prime Number via Dedekind Eta Function. Edigles Guedes

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1 Explorig Prime Numbers ad Modular Fuctios III: O the Expoetial of Prime Number via Dedekid Eta Fuctio Edigles Guedes edigles.guedes@gmail.com December 7, 203 Every way of a ma seems right to himself, but the Lord is the tester of hearts. Proverbs 2:2 ABSTRACT. The mai objective this paper is to develop asymptotic formulas for the expoetial of prime umber, usig Dedekid eta fuctio.. INTRODUCTION As cosequece of the prime umber theorem, I get the asymptotic formula for the th prime umber, deoted by : () ~ l. that (2) M. Pervouchie, i Mémories de la Société physic-mathématique de Kasa, [, page 848] deduced l + l l l + 24 l l. O the other had, Erest Cesáro, i Sur ue formule empirique de M. Pervouchie [, page 849], I ecouter the formula (3) l l 2 l + l l + (l l )2 l l + l 2(l ) 2, i moder otatio, (4) l l 2 = l + l l + (l l )2 l l + l 2(l ) 2 + O ( (l ) 2). O the other had, the Dedekid eta fuctio was itroduced by Richard Dedekid, i 877, ad is defied i the half-plae H = {τ: I(τ) > 0} by the equatio [2, page 47] (5) iτ 2 η(τ) = e ( e2ikτ ). k= The ifiite product has the form ( x ), where x = e 2iτ. If τ H, the x <, so the product coverges absolutely ad is ozero. I [2, page 48], I have () 2 l y = k e 2ky k e2k y 2 (y y ). Hece, k= k=

2 2 (7) l y = 2 k ad (8) y = for y > 0. k= l y 2 e2k y k k= 2 k e 2ky + (y y ) k= e2k y I this paper, I prove, amog other thigs, that ( l 2 (l ) l l + ( + l + 3 l l (l ) 2 2(l ) 2) [ (l LEMMA. I have 2. THEOREM k e 2ky + y, k= + l k k= l ) 2 k k= e2k l ] 2(l ) 2. e2k l 2 2 p e e 2( 2 2) + (2 2 + l ) l, where deotes the th prime umber defied by : N R +. Proof. Multiplyig (3) by 2(l ) 2, I obtai 2 (l ) 2 2(l ) 3 (l ) 2 l l 2(l ) 2 l (l l 2) (l l ) 2 + l l, thereupo, lettig e, I fid 2 2 p e e l 2 2 (l 2) (l ) 2 + l, 2 2 p e 2( 2 2) + (2 2 + l ) l. e COROLLARY. I have 2 2 p e e 2( 2 2) + (2 2 + l ) l, where deotes the th prime umber defied by : N R + ad deotes the floor fuctio. Proof. I ote that e e, arisig from the defiitio of the floor fuctio. THEOREM. For sufficietly large, the 2 l l [ l 24 k 2k 4 k k= k= ] [ ( 2 2 l (l ) 2)] e2k l

3 3 + [ (l l ) k 2k 2 k k= where deotes the th prime umber. k= Proof. I put (7) ad (8) ito Lemma, as followig 2 2 p e 2 l e [ 24 k k= e2k + (2 2 + l ) [2 k ] { [ + l + 3 l l (l ) 2 2(l ) 2]} e2k l 2(l ) 2, k= 4 k e 2k ] (2 2) k= e2k 2 k e 2k k= Dividig both members of above equatio by l + ( )]. 2 2 p e e (2 2 + l ) 2 l [ 24 k e2k 4 k e 2k k= + [2 k k= e2k k= 2 k e 2k k= I retur to replace by l i the previous equatio ] ( l ) + ( )] l. 2(l ) 2 (9) [2(l ) 2 l + l l ] 2 l l [ 24 k e2k 4 l k 2k k= + [2 k k= e2k l k= 2 k 2k k= 2(l ) 2 l + l l. l ] [ (l ) 2 l 2 2(l ) 2 l + l l ] + (l l )] Multiplyig both members of (9) by [2(l )2 +2 l +l l ], I ecouter 2(l ) 2 2 l l [ 24 k e2k 4 l k 2k + [2 k k= cosequetly, e2k l k= 2 k 2k k= k= l ] [ ( 2 2 l (l ) 2)] + (l )] { [ + l l + 3 l l (l ) 2 2(l ) 2]} 2(l ) 2,

4 4 2 l l [ l 24 k 2k 4 k k= + [ (l l ) k 2k 2 k k= This completes the Theorem. COROLLARY 2. For sufficietly large, the ( l 2 (l ) k= k= e2k l 2(l ) 2. l l + ( + l + 3 l l (l ) 2 2(l ) 2) [ (l where deotes the th prime umber. Proof. I ote that, for N >, ] [ ( 2 2 l (l ) 2)] e2k l ] { [ + l + 3 l l (l ) 2 2(l ) 2]} + l () k 2k 0. From Theorem ad (), I obtai ( l 2 (l ) k= l l + ( + l + 3 l l (l ) 2 2(l ) 2) [ (l This completes the proof. k k= l ) 2 k k= + l k k= l ) 2 k k= e2k l ] 2(l ) 2, e2k l e2k l ] 2(l ) 2. e2k l CONJECTURE: I observe that, umerically, is better approximate if I add the term so ( l 2 (l ) l l + ( + l + 3 l l (l ) 2 2(l ) 2) [ (l + l k k= l ) 2 k l + 24 l l. REFERENCES k= e2k l ] 2(l ) 2 e2k l 5 +, 2 l 24 l l

5 5 [] Cesàro, Erest, Sur ue formule empirique de M. Pervouchie, Comptes redus hebdomadaires des séaces de l Academia des scieces 9: , (894). (Frech) [2] Apostol, Tom M., Modular fuctios ad Dirichlet series i umber theory, Spriger Verlag, καιδιζlκαιs