devotion & mathematics

Selberg Identity :     For n ≥ 1 , we have    ⴷ(n) logn + 𝜮 ⴷ(d) ⴷ(n/d) = 𝜮 μ(d) log 2 (n/d). Proof :      Suppose n ≥ 1 .   We know that log n = ( ⴷ * u ) (n)   ⇒ u(n)logn = ( ⴷ * u ) (n) | since u(n) = 1)   ⇒ u’(n) = ( ⴷ * u ) (n)    …………… ① By differentiating on both sides , we get   u’’(n) = ( ⴷ * u )’ (n)              = ( ⴷ’ * u ) (n) + ( ⴷ * u’ ) (n)             =  ( ⴷ’ * u ) (n) + ( ⴷ *  ( ⴷ * u ) (n)    | Since from    ① |  u’’(n)  =   ( ⴷ’ * u ) (n) + ( ( ⴷ *    ⴷ)) * u ) (n)                                     ...

    Bracket Function :                   The function I : R → Z defined by I(x) = n where n ≤ x < n+1 is called the bracket function or the step function or the integral part function .   Notation :                    Integral part of x in R is denoted by I(x) or [x]. Definition :                     If x 𝞊 R , x - [x] is called the Fractional part of x.    Note :         [x] ≤ x < [x] + 1  or x-1 < [x] ≤ x.       For every x 𝞊 R, x ≥ [x] ,  hence the fractional part of x is always non-negative       x 𝞊 Z ⇔ x=[x].   For example :         * For 14/3 = 4.6666... , [ 14/3] = 4.        * For 3/4, [3/4] = 0   because 0 < 3/4 < 1.        * For π = 22/7 , we...

Perfect Number :     A number n is called a perfect number if the sum of all divisors of n > 1 , is equal to 2n.      For example :             Let n = 28.                  The divisors of 28 are 1,2,4,7,14,28.                 Also the sum of these divisors = 1+2+4+7+14+28                                                                   = 56                                                                  = 2 x 28                             ...

   Wilson’s Theorem :   Statement :   If p is a prime then (p-1)! + 1 = 0 ( mod p ) Proof :          Suppose p is a prime number..     For p=2, we have (p-1)! + 1 = 2 and 2 ≡ 0 ( mod 2 )    i.e. the statement is true for p=2.   Let p > 2 and a 𝟄 Z such that 1 ≤ a ≤ p-1.   Since p is prime , we have (a,p) = 1 and hence the linear congruence ax ≡ 1 ( mod p) has    unique solution, say x 0 .   Let a’ 𝟄 x 0 — and   1 ≤ a’ ≤ p-1 where x 0 — is the congruence class of x 0 . Clearly aa’ ≡ 1 ( mod p ) a’ = a ð   a 2 ≡ 1 ( mod p ) ð   p / a 2 -1 ð   p/(a-1) or p/ (a+1)   for p/(a-1) and a > 0 =>   a+1 = p ð   a = p-1   ∴ a’ = a ð   either a = 1 or a= p-1 similarly if a’ ≠ a then a 𝟄 { 2,3,…,p-2}. ∴ the distinct a , a’ belong to the set { 2,3,…,p-2} containing (p-3) elements. Th...

Fermat's Theorem : Statement :                       If  p is a prime and (a,p) = 1 then a p-1 ≡ 1 (mod p). Proof :                Suppose p is a prime number and (a,p) = 1.         Since (a,p) = 1 then the numbers a,2a,3a,…,(p-1) are divided by p and the remainders are 1,2,3 …,                 (p-1) ; not necessarily in this order.          Let a ≡ r 1 ( mod p), 2a ≡ r 2 ( mod p) , … , (p-1)a ≡ r p-1 ( mod p)         Since r 1 , r 2 , … r p-1 are remainders obtained when a,2a,…,(p-1)a are divided by p.         ∴ r 1. r 2 . … .r p-1 = 1 . 2 .3 …. . (p-1).         Multiplying the above congruent relations : a . 2a. … . (p-1) ≡ r 1 r 2 …r p-1 ( mod p ) ð   {1.2…....