On the possible applications of some theorems concerning the Number Theory to the various mathematical aspects and sectors of String Theory I
Michele Nardelli1,2
1Dipartimento di Scienze della Terra
Università degli Studi di Napoli Federico II, Largo S. Marcellino, 10 80138 Napoli, Italy
2
Dipartimento di Matematica ed Applicazioni “R. Caccioppoli”
Università degli Studi di Napoli “Federico II” – Polo delle Scienze e delle Tecnologie Monte S. Angelo, Via Cintia (Fuorigrotta), 80126 Napoli, Italy
Abstract
The aim of this paper is that of show the further and possible connections between the p-adic and adelic strings and Lagrangians with Riemann zeta function with some problems, equations and theorems in Number Theory.
In the Section 1, we have described some equations and theorems concerning the quadrature- and mean-convergence in the Lagrange interpolation. In the Section 2, we have described some equations and theorems concerning the difference sets of sequences of integers. In the Section 3, we have showed some equations and theorems regarding some problems of a statistical group theory (symmetric groups) and in the Section 4, we have showed some equations and theorems concerning the measure of the non-monotonicity of the Euler Phi function and the related Riemann zeta function.
In the Section 5, we have showed some equations concerning the p-adic and adelic strings, the zeta strings and the Lagrangians for adelic strings
In conclusion, in the Section 6, we have described the mathematical connections concerning the various sections previously analyzed. Indeed, in the Section 1, 2 and 3, where are described also various theorems on the prime numbers, we have obtained some mathematical connections with the Ramanujan’s modular equations, thence with the modes corresponding to the physical vibrations of the bosonic and supersymmetric strings and also with p-adic and adelic strings. Principally, in the Section 3, where is frequently used the Hardy-Ramanujan stronger asymptotic formula and are described some theorems concerning the prime numbers. With regard the Section 4, we have obtained some mathematical connections between some equations concerning the Euler Phi function, the related Riemann zeta function and the zeta strings and field Lagrangians for p-adic sector of adelic string (Section 5). Furthermore, in the Sections 1, 2, 3 and 4, we have described also various mathematical expressions regarding some frequency connected with the exponents of the Aurea ratio, i.e. with the exponents of the number
2 1 5+
=
Φ . We consider important remember that the number 7 of the various exponents is related to the compactified dimensions of the M-theory.
1. On some equations and theorems concerning the quadrature- and mean-convergence in the Lagrange interpolation. [1]
Let ( ) ( ) ( ) ( ) ( ) ( ) ≡ n n n n x x x x x x B 2 1 2 2 2 1 1 1 : (1.1)
be an aggregate of points, where for every n
1≥x1( )n >x2( )n >...>xn( )n ≥−1. (1.2)
Let f
( )
x be defined in the interval[
−1,+1]
(hence also the value of the Aurea ratio 0.618033987). We define the nth Lagrange-parabola of f( )
x with respect to B , as the polynomial of degree1
−
≤n , which takes at the points x1( ) ( )n ,x2n,...,xn( )n the values f
( )
x1( )n , f( )
x2( )n ,..., f( )
xn( )n . We denote this polynomial by Ln( )
f and we sometimes omit to indicate its dependence upon x and B . It is known, that( )
∑
( )
( ) ( )( )
∑
( ) ( )
= = ≡ = n i n i i i n i n i n f f x l x f x l x L 1 1 . (1.3)The functions li
( )
x called the fundamental functions of the interpolation, are polynomials of degree 1 − n and if( )
( )
∏
(
)
= − = ≡ n i i n x x x x 1 ω ω (1.4) then( )
( )
( )(
i i)
i x x x x x l − = ' ω ω . (1.5)It is known that if
ψ
( )
x is a polynomial of the m degree, then thLm+k
( ) ( )
ψ
≡ψ
x k =1,2,... (with k also prime number) (1.6) Whenψ
( )
x ≡1, we obtain from (1.6) and (1.3)
∑
( )
= ≡ n i i x l 1 1. (1.7)
∫
[
( )
( )
]
− ∞ → − = 1 1 2 0 lim f x Ln f dx n . (1.8) THEOREM I.Let p
( )
x be a function such thatp
( )
x ≥M >0 −1≤x≤+1, (1.9)∫
( )
−
1
1p x dx (1.10) exists.
It is known that there is an infinite sequence of polynomials
ω
0( ) ( )
x,ω
1 x ,... where the degree of( )
x nω
is n with( ) ( ) ( )
∫
−1 ≠ 1ω
n xω
m x p x dx 0 if n=m,∫
−( ) ( ) ( )
= 1 1ω
n xω
m x p x dx 0 if n≠m; coefficient of n x in( )
x =1 nω
.As known
ω
n( )
x has in[
−1,+1]
(hence also the value of the Aurea ratio 0.618033987) n different real roots. Then our relation (1.8) is true for any matrix formed of these roots. Or more generally, THEOREM Ia.Let
ω
n( )
x be the above polynomials, A and n B constants such that the equation n Rn( )
x ≡xn +...≡ωn( )
x +Anωn−1( )
x +Bnωn−2( )
x =0 (1.11)may have in
[
−1,+1]
(hence also the value of the Aurea ratio 0.618033987) n different real roots and Bn ≤0; then (1.8) holds also for the matrices formed by these roots.We prove Theorem Ia by proving the relation
∫
[
( )
( )
]
( )
− ∞ → − = 1 1 2 0 lim f x Ln f p x dx n , (1.12)which will be shown to be a consequence of p
( )
x ≥M and of the existence of∫
( )
− 1 1p x dx. From (1.12) it follows by (1.9) that
∫
[
( )
( )
]
∫
[
( )
( )
]
( )
− − ≤ − − ≤ 1 1 1 1 2 2 1 0 f x L f p x dx M dx f L x f n n , (1.13)and this by (1.12) establishes Theorem Ia. COROLLARY OF THEOREM Ia.
For all bounded and R integrable f
( )
x we have for the matrices given in Theorem Ia lim 1( )
( )
01 − =
∫
− f x Ln f dx (1.14) and a fortiori there is quadrature convergence for these matrices.THEOREM II. If the sequence
( )
∑∫
( )
= − = n k k x dx l n C 1 1 1 2is unbounded as n→∞, there exists a continuous f6
( )
x such that for our matrix∫
[
( )
( )
]
− ∞ → − =+∞ 1 1 2 6 6 m li f x Ln f dx n .We have to prove (1.12) for the fundamental points given by the roots of the Rn
( )
x polynomials of (1.11). First we prove that
∫
( ) ( )
− ≥
1
1li x p xdx 0 i=1,2,...,n (with nalso prime number) (1.15)
and
∑∫
( ) ( )
∫
( )
= − − ≤ n i i x p x dx p x dx l 1 1 1 1 1 2 . (1.16)Consider the expression
∫
[
( )
( )
]
( )
− − 1 1 2 dx x p x l x li i . But li( )
x −li( )
x =Rn( ) ( )
x F x 2, where F
( )
x is a polynomial of degree(
n−2)
, in which the coefficient of the highest term is evidently 1/ω
'n( )
xi 2. Thus if( )
( )
( )
( )
2 2 0 0 x ... n x / 'n xi c xF =
ω
+ +ω
−ω
, by the orthogonality of the ωn( )
x ’s we have[
( )
( )
]
( )
( )
( ) ( )
∫
− − =∫
− − ≤ 1 1 1 1 2 2 2 2 0 ' x x p x dx B dx x p x l x l n i n n i i ω ω i.e.∫
( ) ( )
∫
( ) ( )
− ≤ − 1 1 1 1 2 dx x p x l dx x p x li i (1.17)which immediately establishes (1.15); by summation for i=1,2,...,n (with n also prime number) we obtain (1.16) in consequence of (1.7).
Let now Ω4 be an aggregate in
[
−1,+1]
(hence also the value of the Aurea ratio 0.618033987) formed of closed non-overlapping intervals. We prove that
( ) ( ) ( )
( ){( ) ( )
( ){ ( ){∑
∑ ∫
∑ ∫
Ω Ω Ω − − ≤ 4 4 4 1 1 2 1 1 2 n i x n i x n k x i i i k i x l x p x dx l x p x dx l . (1.18)
( )
( )
∫
( ) ( ) ( )
− + + + + ≡ − ≥ − 1 1 11 0 1 1i k Iik i k li x lk x p x dx if i≠k. (1.19) For by (1.5)( ) ( )
∫
−(
−)(
( )
−) ( ) ( )
= 1 1 ' ' 1 dx x p x R x x x x x R x R x R I n k i n k n i n ik . As i≠k,( )
(
)(
)
d( )
x d( )
x( )
x x x x x x R n n n i k n 2 3 3 0 0 +...+ − − + − = − − ω ω ω .Hence considering the definition of Rn
( )
x we have( ) ( )
∫
− −( ) ( )
= 1 1 2 2 ' ' x R x x p x dx R B I n k n i n n ik ωwhich proves (1.19), as Bn ≤0 and sign R'n
( ) ( ) ( )
xi R'n xk = −1i+k. Thus we have
( ) ( ) ( )
( ){( ) ( )
( )
( ) ( ) ( )
( ){ ( ){ ( ){ ( ){∑
∑ ∫
∑ ∫
∑ ∑
∫
Ω Ω Ω Ω Ω − − + − = − − = 4 4 4 4 4 1 1 1 1 2 1 1 1 n i x n i x xin xkn n k x i k i k i i k k i x l x p x dx l x p x dx l x l x p x dx l( ) ( )
( ) ( )
( ){( )
( ) ( )
( ){ ( ){∑ ∫
∫ ∑
∑ ∫
Ω Ω Ω − − ≤ − − − = 4 4 4 1 1 1 1 2 2 1 1 2 2 1 2 n i x xin xin dx x p x l dx x p x l dx x p x l i i i i ; thus (1.18) is proved.If Ω4 denotes the whole of the interval
[
−1,+1]
(hence also the value of the Aurea ratio 0.618033987), in consequence of (1.16) we have∑
∑ ∫
( ) ( ) ( )
∫
( )
= = − − ≤ n i n k k i x l x p x dx p x dx l 1 1 1 1 1 1 2 . (1.20)Let now f
( )
x be continuous,ϕ
( )
x the polynomial of degree n−1 that gives the best approximation to it in Tschebischeff’s sense for the interval[
−1,+1]
. Writef
( ) ( ) ( )
x −ϕ
x =∆ x (1.21)( ) ( )
1 1 max − ≤ − = n x f x ϕ x E (1.22)∫
[
( )
( )
]
( )
− − = 1 1 2 dx x p f L x f In n . Then by (1.6) we have∫
[
( )
( )
]
( )
∫
( ) ( )
∫
( ) ( )
− ∆ − ∆ ≤ − ∆ + − ∆ ≡ + = 1 2 1 2 1 2 '' ' 2 2 n n n n n x L p x dx x p x dx L p x dx I I I . (1.23)We have evidently
∫
( )
− − ≤ 1 1 2 1 2 ' E p x dx In n . (1.24) Further by (1.3)∑∑
( ) ( ) ( ) ( ) ( )
∫
∑
∑ ∫
( ) ( ) ( )
∫
( )
= = − − − − = = − ≤ ≤ ∆ ∆ = n i n n k k i n n i n k k i k i n x x l xl x p x dx E l xl x p x dx E p x dx I 1 1 1 2 1 1 1 1 2 1 1 1 1 1 2 4 2 '' (1.25) by (1.20); from (1.23), (1.24) and (1.25) we have
∫
( )
− − ≤ 1 1 2 1 6E p x dx In n (1.26)which by Weierstrass’ theorem establishes (1.12) for any continuous f
( )
x . LEMMA.Let B1 be a matrix satisfying the following expression
∫
( ) ( )
−
1
1li x p x dx, and Ω5 be a set of a finite
number of non-overlapping intervals in
[
−1,+1]
(thence also the value of the Aurea ratio 0.618033987) ; then for n>n0 we have
( ) ( )
( )
( ){∑ ∫
∫
Ω − < Ω 5 5 1 1 2 n i x i i x p x dx p x dx l . (1.27)We easily obtain this result if we consider the function
ψ
( )
x having the value 1 for points of Ω5and 0 elsewhere.
ψ
( )
x is evidently bounded and R integrable, so that according to Fejér’s theorem∫
( ) ( )
∫
( ) ( )
− − ∞ → = 1 1 1 1 lim Ln p x dx x p x dx n ψ ψ . (1.28)But by the definition of
ψ
( )
x we may write
( ) ( )
( ) ( )
( ){∫
−∑ ∫
− Ω = 1 1 1 1 5 n i x i i n p xdx l x p x dx Lψ
, (1.29) further∫
( ) ( )
∫
( )
− = Ω 1 1ψ x p x dx 5 p x dx. (1.30)The expression (1.27) is an evident consequence of (1.28), (1.29) and (1.30). Now we consider the matrix B defined as in Theorem Ia. In consequence of (1.15) the Lemma is applicable; we obtain from (1.17) and (1.27)
( ) ( )
( ) ( )
( )
( ){ ( ){∑ ∫
∑ ∫
∫
Ω Ω − ≤ − < Ω 5 5 5 1 1 1 1 2 2 n i x xin i i i i x p x dx l x p x dx p x dx l , (1.31)and finally from (1.18)
( ) ( ) ( )
( ){( )
( ){∑
∑ ∫
∫
Ω Ω Ω − < 5 5 5 4 1 1 n i x xkn i k k i x l x p x dx p x dx l . (1.32)Let now f
( )
x be any bounded and R integrable function. Then in virtue of the Riemann integrability, to anyε
we can find a finite aggregate of non-overlapping open intervals of total length ≤ε
such that if we exclude these intervals, the oscillation of the function is ≤ε at any point of the remaining aggregate Ω6. We now define f7( )
x as follows: (i) in Ω6 let f7( ) ( )
x ≡ f x . (ii) if we denote the excluded intervals by(
p1,q1) (
... pv,qv)
, (v finite), the function f7( )
x is represented in(
pi,qi)
by the straight line connecting the point(
pi, f( )
pi)
and(
qi, f( )
qi)
. Thus we define f7( )
xfor the whole of
[
−1,+1]
(hence also the value of the Aurea ratio 0.618033987), and its oscillation is at any point ≤ε. But then f7( )
x may be uniformly approximated by a polynomialϕ
( )
x to withinε
2 . Let the degree of
ϕ
( )
x be m=m( )
ε
. Then we have
∫
[
( )
( )
]
( )
∫
[
( )
( )
]
( )
∫
[
(
)
]
( )
− − ≤ − − + − − − − ≤ ≡ 1 1 1 1 1 1 2 7 7 2 7 7 2 2 2 f x L f p xdx f f L f f p x dx dx x p f L x f In n n n∫
[
( )
( )
]
( )
∫
[
]
( )
∫
(
) ( )
− − + − − + − − ≡ + + ≤ 1 1 1 1 1 1 2 7 2 7 2 7 7 4 4 ' '' ' '' 2 f x Ln f p x dx f f p x dx Ln f f p x dx J n J n J n, (1.33) say. As the degree of approximation to f7( )
x is 2 , we have by (1.26) for ε n>m( )
ε
∫
( )
− ≤ 1 1 2 24 ' p x dx Jnε
. (1.34)Further as f
( )
x − f7( )
x differs from 0 only upon intervals, of which the total length is ≤ε and as( )
x f( )
x f( )
x M f x 2 max 2 1 7 ≤ ≡ − ≤ , we have∑∫
( )
= ≤ ν 1 2 16 '' i q p n i i dx x p M J . (1.35)For J'''n we may evidently write
∑∑
(
( )
( )
) ( )
(
( )
) ( ) ( ) ( )
∫
= = − − − = n i n k k i k k i i n f x f x f x f x l x l x p x dx J 1 1 1 1 7 7 '' ' .In consequence of the definition of f7
( )
x the terms of this sum differ from 0 only when xi and xk lie in intervals(
pl,ql)
and(
pµ,qµ)
respectively. Hence
( ) ( ) ( )
( )
( ) { ( ) {∑ ∑
∫
∑∫
= − ≤ ≤ l q l p i x xk p q i i i k i q p k i n M l xl x p x dx M p x dx J , , 1 2 1 1 2 '' ' 16 4 µ µ ν , with l,µ =1,2,...,ν (1.36)by (1.32). As the total length of the range of integration is ≤ε, it is evident by (1.33), (1.34), (1.35) and (1.36), that In →0 as n→∞. Hence the result. Let us write
( )
∑∫
( )
= − = n i i x dx l n S 1 1 1 2 ,and suppose this to be unbounded as n→∞. We shall prove that we can find a continuous function
( )
x f with∫
[
( )
( )
]
− ∞ → − =+∞ 1 1 2 sup lim f x Ln f dx n .By hypothesis there exists an infinite sequence n1<n2 <... with S
( ) ( )
n1 <S n2 <...→∞. For thesake of simplicity of notation we denote by m the mth element of this sequence nm.
Let the mth fundamental points be 1≥ξ1( )m >ξ2( )m >...>ξm( )m ≥−1. We regard them as abscissas and to any ξi( )m we adjoin an ordinate εi, where ε1,ε2,...,εm have arbitrarily the values + 1 or – 1. Thus we have m points; we connect them obtaining a continuous function ψε
( )
x with
ψ
ε( )
x ≤1 for −1≤x≤+1 (thence also x=0.618033987, i.e. the Aurea ratio) (1.37) and∫
( )
∑∑
∫
( ) ( )
− = = − = 1 1 1 1 1 1 2 m m k i m dx l x l x dx L µ ν µ ν εε
ε
ψ
. (1.38)By variation of the
ε
’s we obtain 2 different m ψε( )
x functions. For these functions we have by forming the sums of (1.38)
∑∫
( )
∑∫
( )
( )
− = = − = ε ψε ν ν 1 1 1 1 1 2 2 2 1 m m m L dx l x dx S m , (1.39)hence we may choose
ε
’s, so that for the corresponding ψε( )
x which we simply denote byψ
( )
x , we have∫
( )
( )
− ≥ 1 1 2 m S dx Lmψ
. (1.40)According to Weierstrass,
ψ
( )
x may be approximated by a polynomial fm( )
x of degreeµ
( )
m so that( )
2 3 ≤ xfm −1≤x≤+1 (hence also the value of the Aurea ratio 0.618033987) (1.41)
and
∫
( )
( )
− ≥ 1 1 2 2 1 m S dx f Lm m . (1.42)Now we select a partial sequence , ,...
2 1 m
m f
f of sequence f1
( ) ( )
x,f2 x ,... and define a sequence of constants c1,c2,... in the following way. Let fm( )
x f1( )
x1 = and c1=1. Suppose mr−1, that is fmr−1
( )
xand cr−1, already defined, then we define
( )
( )
=∑
− − = ≤ − 1 1 1 1 1 max 1 , 4 min r r m k m k x r r x l c c (1.43)and mr as the least integer satisfying the following conditions: mr ≥
µ
( )
mr−1 +1 (1.44)∫
( )
∫
( )
− − − > 1 1 1 1 2 2 2 4 2 8 r m m r m m r L f dx c L f dx c r r r r ; (1.45)these 2 conditions can evidently be satisfied in consequence of (1.42) and limm→∞S
( )
m =∞. We now form with these cr and f( )
xr m the function
( )
∑
( )
∞ = = 1 r m rf x c x f r . (1.46) By (1.43) cr r 4 1 ≤ (1.47)and in consequence of (1.47) and (1.41) it is evident that the infinite series for f
( )
x uniformly converges in[
−1,+1]
i.e. f( )
x is continuous. Now we consider L( )
fr
m for a fixed value ρ of r . According to (1.44)
( )
∑
( )
∑
( )
− = ∞ = + = 1 1 ρ ρ ρ ρ r r m m r m r m f c f r x c L f r L ; hence∫
[
( )
]
∫
( )
∑
( )
∑
( )
− − ∞ + = ∞ = − + = − = 1 1 1 1 2 1 2 dx x f c f L c f L c dx f f L I r r m r m m r m m m m r r ρ ρ ρ ρ ρ ρ ρ ρ . (1.48) But in consequence of (1.47)( )
... 2 4 1 4 1 1 2 3 2 = + + + ≤∑
∞ =ρ r m rf x c r , (1.49)
∑
( )
∑
∑
( )( )
∞ + = ∞ + = = = + + ≤ ⋅ ≤ 1 1 1 2 ... 4 1 1 2 3 2 3 ρ ρ ν ν ρ ρ ρ r r m m r m m rL f c l x c r . (1.50) From (1.48), (1.49) and (1.50)∫
[
( )
]
− − = 1 1 2 4 dx f L c Im m mθ
ρ ρ ρ ρ (1.51)with
θ
≤1 (also 0.618033987, i.e. the value of the Aurea ratio). Further∫
( )
∫
( )
∫
( )
∫
( )
− − − − − − > − − > 1 1 1 1 1 1 2 1 1 1 2 2 2 2 2 16 2 8 16 8c L f dx c L f dx c L f dx dx f L c Im m m m m m m m m ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ , (1.52) and by (1.45)Imρ >4ρ −16. ρ =1,2,3,... (and also the prime numbers) (1.53)
With regard the eqs. (1.34) and (1.52), we note that can be related with the Aurea ratio by the numbers 8, 16 and 24. Indeed, we have:
( )
∫
− ≤ 1 1 2 24 ' p x dx Jnε
;( )
( )
( )
( )
∫
− −∫
− − >∫
− − ∫
− − > 1 1 1 1 1 1 2 1 1 1 2 2 2 2 2 16 2 8 16 8c L f dx c L f dx c L f dx dx f L c Im m m m m m m m m ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ;( )
Φ35/7 +( )
Φ −7/7+( )
Φ −21/7 +( )
Φ −42/7 =12; 12⋅2=24; 16 3 4 12⋅ = ; 8 3 2 12⋅ = . (1.54) In the expression (1.54), 1,6180339887... 2 1 5+ = =Φ , and the number 7 of the various exponents
is related to the compactified dimensions of the M-theory. Furthermore, we note that 8 and 24 are related with the “modes” that correspond to the physical vibrations of the bosonic strings and superstrings.
2. On some equations and theorems concerning the difference sets of sequences of integers. [2] A set of integers u1<u2 <... will be called an A -set if its difference set does not contain the square of a positive integer. Let A
( )
x denote the greatest number of integers that can be selected fromx ,..., 2 ,
1 to form an A -set and let us write
( ) ( )
x x A x a = . (2.1)Let N be a large integer and let us write M =
[ ]
N . Let( )
( )
( )
[ ]
∑
∑
= = = = N z M z z e z e T 1 1 2 2α α α . (2.2)Let u1,u2,...,uA( )N be a maximal A -set selected from 1,2,...,N and let
( )
( )
( )∑
= =AN x x u e F 1α
α
. (2.3)We are going to investigate the integral
=
∫
( ) ( )
1 0 2α
α
α
T d F E . (2.4) Obviously,( ) ( ) ( )
( )
(
)
( )
( ) ( )∫
∫ ∑
∑
∑
∑
= = = = + − = = − = − = 1 0 1 0 1 1 1 , , 2 0 2 0 1 N A y N A x M z xyz x y z x u y u d z e u e u e d T F F Eα
α
α
α
α
α
α
α
(2.5) since u1,u2,...,uA( )N is an A -set. LEMMA 1If a, are integers such that b a≤b, and
β
is an arbitrary real number (also a prime number) then
( )
+ − ≤∑
=β
β
2 1 , 1 min b a k e b a k . (2.6)(For
β
=0, the right hand side is defined by a b =a 0 , min .) LEMMA 2
Let N,p,q be integers and
α
a real number (also prime number) such that( )
p,q =1, N≥3, (2.7) 1≤q≤N1/2/logN (2.8) and 12 q q p < −α
. (2.9) Then( )
2 / 1 21 < q N Tα
. (2.10) LEMMA 3Let N,p,q be integers and
α
,β
real numbers (also prime numbers) such that N≥9, (2.11)( )
p,q =1, (2.12) 1≤q≤ N , (2.13)α
= +β
q p (2.14) and N q N N 2 1 log < ≤β
. (2.15) Then( )
2 / 1 log 30 <β
α
q N T . (2.16) LEMMA 4Let N,p,q be integers, R,Q,
α
real numbers (also prime numbers) such that N ≥1,( )
p,q =1, Q q R≤ ≤ ≤ 1 , (2.17) N ≤Q≤N (2.18) and qQ q p < 1 −α
. (2.19) Then( )
(
)
1/2 2 / 1 log 14 7 Q N R N T + <α
. (2.20) LEMMA 5For any positive integer x, 1≤a
( )
x ≤1x . (2.21)
LEMMA 6
For any positive integers x and y , we have
A
(
x+y) ( ) ( )
≤A x +A y , (2.22) A( )
xy ≤xA( )
y , (2.23) a( ) ( )
xy ≤a y , (2.24)( )
a( )
y x y x a + ≤ 1 . (2.25) LEMMA 7Let q ,,t N be positive integers, p integer,
α
,β
real numbers (also prime numbers) such thatβ
α
− = q p . (2.26). Let( ) ( )
( )
=∑
∑
= = N j q s j e q sp e q t a F 1 1 2 1 2β
α
, (2.27) so that if( )
p,q =1 then( ) ( ) ( )
∑
= = 1 1 j j e t aF
α
α
for q=1, F1( )
α
=0 for q>1 (where( )
p,q =1). (2.28) ThenLEMMA 8
Let Nt, be positive integers, R,Q real numbers (also prime numbers) such that
8 e N≥ , (2.30) t /N, (2.31) 1≤R≤N1/2/logN , (2.32) N N Q N log 2 1/2< ≤ . (2.33) Then 2
( )
3/2 <( ) ( ) ( )
(
−)
3/2 + 2( ) (
)
1/2 −1/2 + log 12600 log log 1260a t at a N N N a t tN N Q N t a +120(
a( ) ( )
t −a N)
2{
7N3/2R3/2logN +20N2(
logN)
1/2Q−1/2R}
+26000a2( )
t t2{
1/2(
)
3 11/2 2(
)
1/2 5/2 3}
( )
{
3/2 1/2 1/2(
)
1/2}
log 2 140 log 2 log 3N− N R + N N Q− R + a t N R− + NQ N . (2.34) Let us write( ) ( ) ( )
∑
= = N j j e t a G 1α
α
. Then =∫
( ) ( )
=∫
( ) ( )
+∫
(
( )
−( )
)
( )
1 0 1 0 1 0 2 2 2 2α
α
α
α
α
α
α
α
α
α
T d G T d F G T d F E where E=0 by (2.5). Hence∫
( ) ( )
=−∫
(
( )
−( )
)
( )
1 0 1 0 2 2 2α
α
α
α
α
α
α
T d F G T d G . (2.35) Here( ) ( )
( ) ( ) ( ) (
)
( )
[ ]
∫
∫
∑
∑
∑
= − = = = = 1 0 1 0 1 2 1 1 2α
α
α
α
α
α
α
T d at e j at e k e z d G N z N k N j( )
∫ ∑
(
(
)
)
( )
∑
≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤− + = = + − = 1 0 1 , 1 1 , 1 0 2 2 2 2 1 N z N k j N z N k j z k j t a d z k j e t aα
α
. (2.36) If 1 2 1≤z2 ≤ N − , z>0, (2.37) 1 2 1≤ j≤ N + (2.38)then the numbers j , k = j+z2, z satisfy the conditions
j−k+z2 =0, 1≤ j, k≤N, 1≤z≤ N since k j z N N =N − + + ≤ + = 1 2 1 2 2 .
By (2.30), the number of the positive integers z satisfying (2.37) is at least 4 4 2 1 2 2 4 2 1 2 N N N N N N N N = − ≥ − ≥ = − ≥ −
while (2.38) holds for
2 1 2 N N + >
integers j . Thus (2.36) yields that
∫
( ) ( )
( )
∑
( )
( )
≤ ≤ ≤ ≤− + = = ⋅ ⋅ > = 1 0 1 , 1 0 2 / 3 2 2 2 2 2 8 1 2 4 1 N z N k j z k j N t a N N t a t a d T Gα
α
α
. (2.39)Now, we have to give an upper estimate for the right hand side of (2.35). For q=1,2,...,
[ ]
Q and 1 ,..., 1 , 0 − = q p , let + − = qQ q p pQ q p Ipq 1 , 1 ,and let S denote the set of those real numbers (also prime numbers)
α
for whichQ Q 1 1 1 < ≤ − −
α
holds andα
∉Ip,q for 1≤q≤R, 0≤ p≤q−1,( )
p,q =1. (2.40) Then∫
(
( )
( )
)
( )
∫
(
( )
( )
)
( )
∫
( )
( ) ( )
+ − − − + − ≤ − = − Q Q Q Q d T G F d T G F d T G F / 1 / 1 2 2 / 1 1 / 1 2 2 1 0 2 2α
α
α
α
α
α
α
α
α
α
α
α
( )
( ) ( )
( )
( ) ( )
( ) [ ]∑ ∑ ∫
∫
= =− ≤ ≤ + + = − + − + R q q p q p Ipq S E E E d T G F d T G F 2 1 , 1 1 3 2 1 2 2 2 2 ,α
α
α
α
α
α
α
α
. (2.41)Let us estimate the term E1 at first. For any complex numbers u, v, we have
u2 −v2 = uu−vv =
(
u−v)
u +v(
u −v)
≤ u−vu +vu−v = u−v(
u +v)
== u−v
(
(
u−v)
+v + v)
≤ u−v(
u−v +2v)
= u−v2+2u−vv. (2.42)Furthermore, applying Lemma 7 with p=0 q, =1,
α
=β
, we have F1( ) ( )
α
=Gα
there, thus weobtain that
F
( ) ( )
α
−Gα
≤H(
t, N,1,α
)
. (2.43) The expressions (2.42) and (2.43) yield that
∫
( )
( ) ( )
∫
( ) ( ) ( )
∫
( ) ( ) ( ) ( )
+ − + − + − ≤ − + − ≤ − = Q Q Q Q Q Q d T G G F d T G F d T G F E / 1 / 1 / 1 / 1 / 1 / 1 2 2 2 1α
α
α
α
α
α
α
α
2α
α
α
α
α
∫
(
) ( )
∫
(
) ( ) ( )
+ − + − + = + ≤ Q Q Q Q E E d T G N t H d T N t H / 1 / 1 / 1 / 1 1 1 2 '' 2 ' , 1 , , 2 , 1 , ,α
α
α
α
α
α
α
. (2.44)Furthermore, for
α
≤logN/N, we use the trivial inequality
( )
∑
( )
∑
= = ≤ = ≤ = M z M z N M z e T 1 2 / 1 1 2 1α
α
, (2.45)while for logN/N <
α
≤1/Q(<1/2 N , by (2.33)), we apply Lemma 3. In this way, we obtain that∫
(
( ) ( )
)
( )
( )
≤ + ⋅ ⋅ ⋅ ⋅ + + − < N d N N t a N N t t a N N a t a E / 1 2 / 1 '' 1 1 1 2 αα
π
∫
{
(
( ) ( )
)
( ) (
)
}
( )
≤ < + ⋅ ⋅ + + − + N N N d N t a N t t a N N a t a / log / 1 2 / 1 2 1 1 2 αα
α
α
π
∫
{
(
( ) ( )
)
( ) (
)
}
( )
≤ < < ⋅ ⋅ + + − + Q N N d N t a N t t a N N a t a / 1 / log 2 / 1 log 30 2 1 1 2 αα
α
α
α
π
{
( ) ( ) ( )
(
)
5/2 2( )
3/2}
( ) ( ) ( )
(
)
3/2 2 1 5 2 2 N N a t a t a N t t a N N a t a t a N − + ⋅ ⋅ + − <( )
( ) ( ) ( )
(
) (
)
∫
∫
≤ < < ≤ + − + ⋅ ⋅ ⋅ ⋅ + N N N N N Q d N N N a t a t a N t t a N N d / log / 1 log / 1/ 2 / 3 2 / 1 2 / 3 2 1 log 15 5 log 2 1 α αα
α
α
α
( )
(
)
∫
≤ < ⋅ + Q N N d N N t t a / 1 / log 2 / 1 2 / 1 2 1 log 5 30 αα
α
. (2.46) Here∫
≤ < = N N N N d / log / 1 log log 2 1 αα
α
,∫
∫
≤ < +∞ − = ⋅ = < Q N N N N N N N N d d / 1 / log log / 2 / 1 2 / 1 2 / 3 2 / 3 log 4 log 2 2 1 2 1 αα
α
α
α
∫
∫
≤ < = ⋅ = < Q N N Q Q Q d d / 1 / log / 1 0 2 / 1 2 / 1 2 / 1 2 / 1 4 1 2 2 1 2 1 αα
α
α
α
.Thus with respect to (2.30), (2.33) and a
( ) ( )
t ≥a N (by (2.24) and (2.31)),E1''<2a
( ) ( ) ( )
t(
a t −a N)
N3/2+20a2( )
t tN1/2 +a( ) ( ) ( )
t(
a t −a N)
N3/2loglogN+10a2( )
t tN1/2logN+ +60a( ) ( ) ( )
t(
at −a N)
N3/2+600a2( ) (
t tN logN)
1/2Q−1/2 <63a( ) ( ) ( )
t(
a t −a N)
N3/2loglogN +( )
( ) ( ) ( )
(
)
3/2 2( ) (
)
1/2 1/2 2 / 1 2 / 1 2 log 630 log log 63 log 20 1 log 30 ≤ − + − + + a t at a N N N a t tN N Q N Q N N tN t aNow we are going to estimate E1'+E2. If 2≤q≤Q, 1≤ p≤q−1 then
α
∈Ip,q implies thatq q q qQ q 2 1 2 1 1 1 1− ≥ − = ≥
α
.Thus for 2≤q≤Q, 1≤ p≤q−1 and
( )
p,q =1, Lemmas 1 and 7 (where F1( )
α
=0 in this case) and the trivial inequality (2.45) yield that
∫
( )
−( ) ( )
≤∫
( ) ( )
+∫
( ) ( )
≤∫
( ) ( )
+ q p pq pq pq I I I I d T F d T G d T F d T G F , , , , 2 2 2 2 2α
α
α
α
α
α
α
α
α
α
α
α
α
+∫
( ) ( )
=∫
( ) ( )
+ ⋅( )
≤ q p pq I I N q t a qQ d T F d N q t a , , 2 / 1 2 2 2 2 / 1 2 2 2 2 1 2 2 1α
α
α
α
∫
(
)
( )
+ − + + ≤ qQ qQ N t a d q p T q N t H / 1 / 1 2 / 1 2 2 , , ,β
β
β
. Hence(
) ( )
(
)
( )
( ) [ ]∫
∑ ∑
∫
+ − = =− ≤ ≤ + − ≤ + + + ≤ + Q Q R q q p q p qQ qQ N t a d q p T q N t H d T N t H E E / 1 / 1 2 1 , 1 1 / 1 / 1 2 / 1 2 2 2 2 ' 1 , ,1,α
α
α
, , ,β
β
β
(
)
( )
( ) [ ]∑ ∑ ∫
= =− ≤ ≤ + − + + ≤ R q q p q p qQ qQ N R t a d q p T q N t H 1 1 , 1 0 / 1 / 1 2 / 1 2 2 2 , , ,β
β
β
. (2.48) To estimate +β
q pT , we use Lemmas 2 and 3 for
β
≤logN /N and logN/N <β
≤1/qQ, respectively. We obtain with respect to (2.30), (2.32) and (2.33) that (for q≤R,( )
p,q =1)(
)
{
(
( ) ( )
)
( )
(
)
}
∫
∫
+ − ≤ + + + − < + qQ qQ N N d q N N q t t a N N a t a d q p T q N t H / 1 / 1 log / 2 / 1 2 2 2 4 2 2 2 2 2 21 1 16 2 , , , ββ
β
π
β
β
β
∫
{
(
( ) ( )
)
( )
(
)
}
≤ < < + + − + qQ N N d q N N q t t a N N a t a / 1 / log 2 / 1 2 2 2 4 2 2 2 2 log 30 1 16 2 ββ
β
β
π
(
( ) ( )
)
( )
(
( ) ( )
)
2 2 2 / 1 2 2 2 2 4 2 2 2 2 60 21 log 1 16 2 log 2 a t a N N q N N N N q t t a N N a t a N N + − + + − <π
(
)
∫
∫
( )
≤ < < ≤ − − < ⋅ ⋅ + qQ N N N N qQ d q N N q t t a d q N / 1 / log log / 1/ 2 / 1 2 2 4 2 2 2 / 1 2 / 1 2 / 1 log 11 480 log β ββ
β
β
β
β
( ) ( )
(
)
2 3/2 1/2 1/2 2( )
2 7/2 2(
( ) ( )
)
2 2 120 21 log 11 log 32 log 84a t −a N N N⋅q + ⋅ N⋅N a t t q ⋅ ⋅ N⋅ + a t −a N N < − −(
)
−∫
− +( )
(
)
∫
=(
( ) ( )
−)
⋅ − + qQ qQ q N N N a t a d N N q t t a d q N / 1 0 / 1 0 2 / 1 2 / 3 2 2 / 3 2 / 1 2 2 / 7 2 2 2 / 1 2 / 1 2 / 1 log 84 log 10560 logβ
β
β
β
( )
2 1/2(
)
3 7/2(
( ) ( )
) (
2 2)
1/2 1/2( )
1/2 2( )
2 7/2 2 2 10560 2 log 120 log 7392a t t N N q + at −a N N N q ⋅ qQ + a t t q N + − − −(
)
1/2( )
5/2(
( ) ( )
)
2{
84 3/2log 1/2 240 2(
log)
1/2 1/2 1}
2( )
2 5 2 logN ⋅ qQ − = a t −a N N N⋅q− + N N Q− q− +a t t{
7392N−1/2(
logN)
3q7/2+4224N2(
logN)
1/2Q−5/2q}
since(
x+ y)
2 ≤2x2 +2y2for any real numbers (or also prime numbers) x, . Thus (2.48) yields with respect to (2.30) and y (2.32) that
{
(
( ) ( )
)
(
(
)
)
( )
[ ]∑∑
= − = − − − + + ⋅ − < + R q q p t t a q Q N N q N N N a t a E E 1 1 0 2 2 1 2 / 1 2 / 1 2 2 / 1 2 / 3 2 2 ' 1 84 log 240 log(
(
)
(
)
)}
( )
{
(
( ) ( )
)
[ ]∑
= − − + + < R − q N a t a N R t a q Q N N q N N 1 2 2 / 1 2 2 2 / 5 2 / 1 2 2 / 7 3 2 / 1 log 4224 log 7392(
)
(
3/2 ⋅ 1/2 + 2 1/2 −1/2)
+ 2( )
2(
−1/2(
)
3 9/2 + 2(
)
1/2 −5/2 2)}
+ log 4224 log 7392 log 240 log 84N N q N N Q a t t N N q N N Q q+a2
( )
t(
N1/2/logN)
2N1/2 ≤(
a( ) ( )
t −a N)
2{
84N3/2R3/2logN +240N2(
logN)
1/2Q−1/2R}
+ 2( )
2{
1/2(
)
3 11/2 2(
)
1/2 5/2 3}
2( )
3/2 64 1 log 4224 log 7392N N R N N Q R a t N t t a + + + − − . (2.49)Finally, in order to estimate E , we use Lemma 4. Namely, if 3 α∈S then there exist integers p, q such that 1≥q≥Q, 0≤ p≤q−1,
( )
p,q =1 and qQ q p < 1 −α
;by (2.40), q satisfies also R<q. Thus (2.17) and (2.19) in Lemma 4 hold. Hence, Lemma 4 yields that
( )
(
)
1/2 2 / 1 log 14 7 sup Q N R N T S + ≤ ∈α
α .Thus we obtain applying Parseval’s formula that
( )
2( ) ( )
2( )
( )
2( )
2{
1/2 1/2 1/2(
)
1/2}
3 F G T d supT F d G d 7N R 14Q logN E S S S S + < + ≤ − = − ∈∫
∫
∫
α
α
α
α
α
α
α
α
α
α( )
( )
={
+(
)
}
(
( )
+( )
)
= + −∫
F d∫
G d N1/2R 1/2 Q1/2 N 1/2 A N a2 t N 1 0 1 0 2 2 log 14 7α
α
α
α
(
)
{
+}
(
( )
+( )
)
≤{
+(
)
}
(
( )
+( )
)
= = N1/2R−1/2 Q1/2 N 1/2 a N N a2 t N N1/2R−1/2 Q1/2 N 1/2 a t N a t N log 14 7 log 14 7( )
{
3/2 1/2 1/2(
)
1/2}
log 28 14N R NQ N t a + = − (2.50)by Lemma 5 and since a
( ) ( )
N ≤a t by (2.24) in Lemma 6 and (2.31). Collecting the results (2.35), (2.39), (2.41), (2.44), (2.47), (2.49) and (2.50) together, we obtain that
( )
<∫
( ) ( )
≤ + + ≤(
+)
+ + = +(
+)
+ < 1 0 3 2 ' 1 '' 1 3 2 '' 1 ' 1 3 2 1 2 2 / 3 2 2 2 8 1 E E E E E E E E E E E d T G N t aα
α
α
( ) ( ) ( )
(
)
3/2 2( ) (
)
1/2 1/2(
( ) ( )
)
2 log 1260 log log 126a t a t −a N N N + a t tN N Q + a t −a N < −
{
84N3/2R3/2logN+240N2(
logN)
1/2Q−1/2R}
+a2( )
t t2{
7392N−1/2(
logN)
3R11/2 +4224N2(
)
1/2 5/2 3}
2( )
3/2( )
{
14 3/2 1/2 28 1/2(
log)
1/2}
64 1 logN Q− R + a t N +a t N R− + NQ N . (2.51) Subtracting 2( )
3/2 64 1 N ta and then multiplying by
9,142857 10 7 64 64 7 64 1 8 1 1 1 < = = = − − − , we obtain that
a2
( )
t N3/2 <1260a( ) ( ) ( )
t(
at −a N)
3/2loglogN +12600a2( ) (
t tN logN)
1/2Q−1/2 +120(
a( ) ( )
t −a N)
2{
7N3/2R3/2logN+20N2(
logN)
1/2Q−1/2R}
+26000a2( )
t t2{
3N−1/2(
logN)
3R11/2 +2N2(
logN)
1/2 Q−5/2R3}
+140a( )
t{
N3/2R−1/2+2NQ1/2(
logN)
1/2}
.3. On various equations and theorems regarding some problems of a statistical group theory (symmetric groups). [3]
Let S stand for the symmetric group with n n letters, P a generic element of it and O
( )
P its order. Then we haveTHEOREM 1
For almost all P ’s in S , i.e. with the exception of n o
( )
n! P ’s at most, O( )
P is divisible by all prime powers not exceeding
( )
− + n n n n n n log log log log log log log 3 1 log log logω
(3.1)if only
ω
( )
n +∞ arbitrarily slowly.Since the P ’s in a conjugacy class H of S have the same order, we may denote by n O
( )
H the common order of its elements and it is natural to ask the corresponding statistical theorem for( )
HO . The total number of conjugacy classes in S is, as well known, n p
( )
n , the number of partitions of n. Now, in this Section, we prove the following theorem:THEOREM 2
For almost all classes H , i.e. with exception of o
(
p( )
n)
classes, O( )
H is divisible by all prime powers not exceeding
( )
− + ⋅ n n n n n n log log log log 5 1 log 6 2π
ω
(3.2)if only
ω
( )
n +∞ arbitrarily slowly.This is again best possible in the following strong sense. THEOREM 3
If
ω
( )
n +∞ arbitrarily slowly, then almost no classes H (i.e. only o(
p( )
n)
of it) have the property that O( )
H is divisible by all primes not exceeding
( )
+ + ⋅ n n n n n n log log log log 5 1 log 6 2π
ω
(3.2b)Now we turn to the proof of Theorem 2. Let, for y>0,
( )
∏
∑
( )
∞ − ∞ = − − = − = 11 0 1 v n ny vy p ne e y f . (3.3)For this we have the classical functional equation
( )
+ − = y y y yf y f 6 24 exp 4 2 1π
2π
2π
(3.4)and hence for y→+0
( )
(
( )
)
+ = y y o y f 6 exp 2 1 1 2π
π
. (3.5)Let Y =Y
( )
n →∞ with n to be determined later and let q run through all prime powers with q≤Y( )
n . (3.6)Let further pq
( )
n be the number of all partitions of n with the property that no summand is divisible by q . Then we have for y>0
( )
( )
( )
∑
∞∏
= − − = − = 0 / 1 1 n q n ny ny q qy f y f e e n p . (3.7) Putting∑
( )
( )
≤ = Y q Y def q n h n p . we get( )
( )
( )
∑
∞∑
= ≤ − = 0 n q Y ny Y qy f y f e n h . (3.8)Using (3.5) we get for all q ’s in (3.8)
( )
( )
( )
− + = y q q o qy f y f 1 1 1 6 exp 1 1π
2 (3.9) if only qy→0. (3.10) Hence, if y and Y 1are sufficiently small, we have
∑
( )
∑
∞ = ≤ − − < − < 0 2 2 1 1 1 6 exp log 5 1 1 1 1 6 exp 2 n q Y ny Y y Y Y Y q y q e n hπ
π
. Putting n n Y y defλ
π
=
− = 1 1 6 , we get( )
( )
∑
( )
∞ = − − − − < ≤ = 0 1 6 exp log 5 m my Y ny Y n Y n Y n Y Y e m h e n h e n h λπ
and hence( )
− < − < n Y Y Y n Y n Y Y n hY 2 1 1 6 2 exp log 5 1 6 2 exp log 5π
π
. (3.11)
( )
⋅ ≈ n n n p 6 2 exp 3 4 1π
(3.12)which gives for all sufficiently large n,
( )
( )
− ⋅ < Y n n n p Y Y n hY 6 exp log 8 5π
. (3.13) Now choosing n n Y log 6 5 4⋅ ⋅ =π
, (3.14)the restriction (3.10) is satisfied and hence (3.13) gives
( )
( )
→0 n p n hY for n→∞. (3.15)Now, there is a one-to-one correspondence between the conjugacy classes H of S and partitions n n=m1n1+m2n2 +...+mknk 1≤n1<n2 <...<nk (3.16)
of n; moreover
O
( )
H =[
n1,n2,...,nk]
V . (3.17)Hence O
( )
H is divisible by a prime power q if and only if q is the divisor of some summand n j and hY( )
n is an upper bound for the number of conjugacy classes H of S whose order is not n divisible by some prime power q≤Y . Hence (3.15) means that for almost all classes H the quantity O( )
H is divisible by all prime powers not exceedingn n log 6 5 4⋅
π
⋅ . (3.18)Next we consider the divisibility of O
( )
H by the prime powers q satisfying n n q n n log 6 10 log 6 5 4 ⋅ ≤ ≤ ⋅ ⋅π
π
. (3.19)Taking into account the Euler-Legendre “Pentagonalsatz” according to which for Rez>0 the
relation