Selberg Identity : Number Theory #Selberg #Identity #: #Number #Theory

Selberg Identity :

    For n ≥ 1 , we have   ⴷ(n) logn + 𝜮 ⴷ(d) ⴷ(n/d) = 𝜮 μ(d) log2 (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)    ……………

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    |

 u’’(n)  =   ( ⴷ’ * u ) (n) + (( ⴷ *   ⴷ)) * u ) (n)                                                                            

By multiplying on both sides with μ = u’ we get

         (u’’ * μ)(n)  = ⴷ’(n) + (ⴷ *   ⴷ)(n)

⇒     𝜮 μ(d) u’’(n/d) = ⴷ(n) logn + 𝜮 ⴷ(d) ⴷ(n/d) ………….

 Now

 u’’(n/d) = ( u’(n/d) )’

             = ( u(n/d)log (n/d) )’

             = ( u(n/d) log (n/d) ) log(n/d)

             = u(n/d) log2(n/d)

             = log2(n/d)         | since u(n) = 1 for all n|

∴ from , we have

     𝜮 μ(d) u’’(n/d) = ⴷ(n) logn + 𝜮 ⴷ(d) ⴷ(n/d)

⇒ 𝜮 μ(d) log2(n/d) = ⴷ(n) logn + 𝜮 ⴷ(d) ⴷ(n/d)  

                      *** Hence The Proof ***      

  * Bracket Function 


 * Wilson's Theorem  




















































#Selberg #Identity #: #Number #Theory

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