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On the Dirichlet Eta Function Azkour M* Ministry of National Education, Vocational Training, Higher Education and Scientific Research, Morocco mp Co uta & tio d n e a li Azkour, J Appl Computat Math 2018, 7:4 l p M p a Journal of A t h DOI: 10.4172/2168-9679.1000420 f e o m l a a n t r ISSN: 2168-9679i c u s o J Applied & Computational Mathematics Commentary Open Access On the Dirichlet Eta Function Azkour M* Ministry of National Education, Vocational Training, Higher Education and Scientific Research, Morocco Abstract This paper gives a proof of the following result: if η(ρ)=0 and R(ρ)>0, then: (ρ ) limk = 2. k →∞ ηρk ( ) ∞ k R η = n−1 −s η ss= η is the Dirichlet eta function defined for (s)>0 by (ss) ∑n ( ) with χn(s)=(−1) n and kn( ) ∑ ( ) . n=1 n=1 Keywords: Dirichlet eta function; Riemann zeta function; Complex fs( ) "∀< 0R(ρ ) < 1: lim≠ 0, ∞ variable s→ρ fs(1− ) Introduction that is to say for every complex number ρ with 0<R(ρ)<1, we have: η (s) We know that the Riemann zeta function ζ is the analytic function lim≠∞ 0, of the complex variable s, defined in the half-plane R(s) >1 by Sarnak s→ρ η (1− s) [1] this means that: “if η(ρ)=0 and 0<R(ρ)<1, then η(1−ρ)=0”. − −−ss1 ζ (sn) =∑ =∏−(1 p) (1) So, for R(s) >0, we have: n−1 −s −s where the series ∑n is absolutely convergent for R(s)>1 and the η (ss) = ∑nn( ) with( sn) =( − 1) product ∏(1-p-s)-1 extends over all the prime numbers p P={2,3,5,7,…} n≥1 and as shown by Riemann, ζ can be continued analytically to \{1} as let’s denote: a meromorphic function and has a simple pole at s=1∊ with residue 1 k η = ≥ [2]. We also know that ζ is defined for any complex number s≠1 and kn(s) ∑ ( s) for k 1, = having R(s)>0 by n 1 n (ρ ) 1−s −1 if η(ρ)=0 and lim = ∞ then, ζ(s) =(1−2 ) η(s) (2) k →∞ ηρk ( ) where η is the Dirichlet eta function which is defined in the half-plane k+1(ρ ) R(s)>0 by [3] lim = ∞ k →∞ ηρk+1( ) n−1 −s η (sn) =( −1) (3) ρ ∑ k+1(ρ ) k and knowing that lim =−=−lim( 1)1 we have so, noticing that s : (1−21−s)−1≠0, we deduce that [4]: kk→∞ kk(ρ ) →∞ +1 “if ζ(s)=0 and∀ R∈(s)>0, then η(s)=0”. k+11(ρρ) k+ ( ) k+1(ρρ) k( ) k( ρ) −1 On the other hand, we know that ζ(s) is related to ζ(1 − s) by the lim = lim =lim = = = ∞ , kk→∞ ηρk+1 →∞ ηk+1( ρ) k→∞ ηρk( ) k+1( ρ) 1 Riemann functional equation [5] ( ) + −1 k(ρ) kk( ρρ) ( ) ∞ ∧(ss) =∧−(1 ) (4) this implies = ∞ ,and this result is absurd, so if η(ρ)=0, then where Λ(s)=π−s/2Γ(s/2)ζ(s) and Γ is the Euler gamma function. So, by the 1 two in eqns. (2) and (4), we obtain: k (ρ ) lim≠∞ , η (1− s) fs( ) k →∞ ηρk ( ) "∀< 0R(s) < 1: = ; fs( ) =π −ss/2 Γ( s/ 2)( 1 − 2) " η (s) fs(1− ) And we have Γ(s) does not vanish for any s in and has an infinity of simple poles with residue (−1)n/n! at s= −n where n=0,1,2,3,... etc., *Corresponding author: Azkour M, Ministry of National Education, Vocational Training, Higher Education and Scientific Research, Morocco, Tel: +21671766462; So E-mail: [email protected] fs( ) Received September 18, 2018; Accepted October 03, 2018; Published October lim=⇒= 0ρ 3,5,7,9... 10, 2018 s→ρ fs(1− ) Citation: Azkour M (2018) On the Dirichlet Eta Function. J Appl Computat Math 7: And 420. doi: 10.4172/2168-9679.1000420 fs( ) lim =∞⇒−−−−= ...,8,6,4,2 ρ Copyright: © 2018 Azkour M. This is an open-access article distributed under the s→ρ fs(1− ) terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and Then source are credited. J Appl Computat Math, an open access journal Volume 7 • Issue 4• 1000420 ISSN: 2168-9679 Citation: Azkour M (2018) On the Dirichlet Eta Function. J Appl Computat Math 7: 420. doi: 10.4172/2168-9679.1000420 Page 2 of 2 let’s take: k (ρ ) if η(ρ)=0 and lim =λ ∈ \{0}, then m k →∞ ηρ k ( ) η = ηm(s)=ηk(s)+ηk,m(s) with km, (ss) ∑ n( ) nk= +1 kk++11(ρρ) ( ) +1(ρρ) ( ) ( ρ) −1 so, if η(ρ)=0, then: lim kk= lim = lim k= = λ, kk→∞ →∞ k→∞ ηρ+1( ) ηk+1( ρ) ηρkk( ) +1( ρ) 1 k + −1 ηρ( ) λ km, (ρ) ( ρρ) ( ) limηρm ( ) = 0and lim = −1 k kk mm→∞ →∞ ηρ( ) that is to say k and we have n≥0: λ=2, ρ �k+n(ρ ) n k if η(ρ)=0 and 0<R(ρ)<1, then ∀=( −1) k (ρ ) kn+ (ρρ) (1− ) so, using f , we have: lim�k =∃ 0,2, and lim k =∃=∃0,2,eg . lim eik n,α kk→∞ →∞ ( k→∞ ) ρ ηρkk( ) ηρ( ) mm kn+ (ρ ) n k 1+ =+− 11( ) =fkm,ρ ( ) ∑∑ρ + and nn=11k ( ) = kn �k (ρ ) and ηρ( ) 0,2, ∃ η(1− ρ) �( ρ) lim k= =0,1,γ ≠ 1, ∞ , ∃= lim kk m 1 limfxρ ( ) =−+−+ 1 1 1 1 ... +−( 1) = kk→∞ − ρ →∞ →∞ m, �(1 ) 0,2, ∃−ηρk ( ) �(1 ρ) m k k m/x→ 0 2 ηρk (1− ) that is to say ρ m �k ( ) ρ R lim =0, ∞∃ , m ∑kn+ ( ) for 0< (ρ)<1: →∞ ρ ηρ k �(1− ρ ) kn+ ( ) n=1 km, ( ) 1 k 1+ lim ∑ =+=+=1 lim 1 lim mm→∞ ρρρ→∞ m→∞ m/xx→→ 0 n=1 kkk( ) m/ 0 ( ) m/ x → 0 ( ) 2 kk(ρ) ηρ(1− ) if lim =0and lim ≠∞ , then kk→∞ →∞ kk(1− ρ) ηρ( ) then ηρkm, ( ) ηρkk( ) ηρ( ) 1 ηρ1− ρ 1+ lim =−=1 lim kk( ) ( ) 0 mk→∞ →∞ lim =0 = → ηρkk( ) ( ρ) k(ρ) 2 k →∞ m/x 0 ηρkk( ) (1−∃ ρ) 0,2, that is to say 1 ηρk (1− ) that is to say if η(ρ)=0 and <R(ρ ) <1and lim ≠∞ ηρ( ) 1 2 k →∞ ηρ( ) lim k = k k →∞ ρ 2 then k ( ) (ρ ) So lim k = 0 →∞ k ηρ1− k (ρ ) k ( ) lim= 2 . k →∞ ηρ( ) but, it is very likely that if η(ρ)=0, then k References (ρ ) limk = 2 ≠ 0 1. Sarnak P (2004) Problems of the Millennium: The Riemann Hypothesis. k →∞ ηρk ( ) Princeton University and Courant Institute of Math. Sciences. let f be the function of the variable x defined by 2. Borwein P, Choi S, Rooney B, Weirathmueller A (2007) The Riemann n,α Hypothesis: A Resource for the Afficionado and Virtuoso Alike. Springer, Berlin, α αα αHeidelberg NewYork, Hong Kong, London, Milan Paris, Tokyo. xxx n x fxn,α ( ) =1 − + − +... +−( 1) ++ + + 3. Jaiswal AK, Khandelwal A (2009) A text book of computer based numerical and x123 x x xn statistical techniques, p: 618 Then 4. Iyengar SRK, Jain RK (2009) Numerical Methods, New Age International Publishers, pp: 182-184. n 1 limfn,α ( x) =−+−+ 1 1 1 1 ... +−( 1) = (Grandi'series) n→∞ 5. Zaliznaik V (2008) Essentials of scientific computing (1stedn.), Numerical → 2 nx/0 Methods in Science and Engineering. J Appl Computat Math, an open access journal Volume 7 • Issue 4 • 1000420 ISSN: 2168-9679 .
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