Results 1 to 2 of 2

Math Help - Deduce from Mobius inversion formula

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    33

    Deduce from Mobius inversion formula

    I saw this question in few book but no detail, hope someone can show the working. Use the Mobius inversion formula to prove

    1. \sum_{d\,\mid \,n}\mu(d)\tau(\displaystyle{\frac{n}{d}})=1

    2. \sum_{d\,\mid \,n}\mu(d)\sigma(\displaystyle{\frac{n}{d}})=n

    Thank you for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Both follow directly from Möbius inversion formula.

    Möbius inversion formula

    Given an arithmetic fuction f(n)

    Define: F(n)=\sum_{d|n}{f(d)}

    Then we have f(n)=\sum_{d|n}{\mu(d)\cdot{F\left(\frac{n}{d}\rig  ht)}}


     \tau(n)=\sum_{d|n}{1} (number of positive divisors) and  \sigma(n)=\sum_{d|n}{d}
    Last edited by PaulRS; April 15th 2008 at 05:01 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to apply Mobius inversion formula
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 2nd 2012, 09:16 AM
  2. Mobius Inversion problem
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 31st 2010, 07:36 PM
  3. Mobius Inversion Formula
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: February 26th 2010, 08:46 AM
  4. Mobius Inversion Formula
    Posted in the Number Theory Forum
    Replies: 9
    Last Post: December 10th 2009, 12:35 PM
  5. Mobius Inversion
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 26th 2008, 04:42 PM

Search Tags


/mathhelpforum @mathhelpforum