Results 1 to 2 of 2

Math Help - Mobius Function for 2^X

  1. #1
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943

    Mobius Function for 2^X

    We have to find the Mobius function \mu for the set 2^X such that F(Y) = \sum_{Z\subset Y}f(Z) for any Y\subset X\Longleftrightarrow f(Y)=\sum_{Z\subset Y}\mu(Z)F(Y\backslash Z).

    I started doing this for a few basic sets. If X=\{a,b\}, then

    \mu(\emptyset) = 1
    \mu(\{a\}) = \mu(\{b\}) = -1
    \mu(\{a,b\}) = 1

    I know the \mu-function for integers is

    \mu(n)=\bigg\{\begin{array}{lr}0:&\mbox{n~is~not~s  quarefree}\\(-1)^{p(n)}:&\mbox{n~is~squarefree}\end{array}

    where p(n) is the number of prime factors of n.

    EDIT: I've had a recent breakthrough (I think) and I believe that \mu(Y)=(-1)^{|Y|}. So unless that's incorrect, I don't need any help.
    Last edited by redsoxfan325; April 11th 2010 at 07:47 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by redsoxfan325 View Post
    We have to find the Mobius function \mu for the set 2^X such that F(Y) = \sum_{Z\subset Y}f(Z) for any Y\subset X\Longleftrightarrow f(Y)=\sum_{Z\subset Y}\mu(Z)F(Y\backslash Z).

    I started doing this for a few basic sets. If X=\{a,b\}, then

    \mu(\emptyset) = 1
    \mu(\{a\}) = \mu(\{b\}) = -1
    \mu(\{a,b\}) = 1

    I know the \mu-function for integers is

    \mu(n)=\bigg\{\begin{array}{lr}0:&\mbox{n~is~not~s  quarefree}\\(-1)^{p(n)}:&\mbox{n~is~squarefree}\end{array}

    where p(n) is the number of prime factors of n.

    EDIT: I've had a recent breakthrough (I think) and I believe that \mu(Y)=(-1)^{|Y|}. So unless that's incorrect, I don't need any help.
    Should that be F(Y) = \sum_{Z\subset Y}f(Z) for any Y\subset X\Longleftrightarrow f(Y)=\sum_{Z\subset Y}\mu(Z, Y)F(Z), where \mu(Z, Y)= (-1)^{l(Z,Y)} and l(Z, Y) is the length of the interval of [Z,Y] in the order-theoretic term?

    In your example, I think if F(Y) is the cardinality of the set Y = {a, b}, then we can apply
    \mu (\emptyset, \{a,b\})= (-1)^2=1, \mu(\emptyset, \{a\}) = -1, \mu(Y, Y) = 1, etc.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mobius Map
    Posted in the Geometry Forum
    Replies: 1
    Last Post: December 11th 2010, 04:20 PM
  2. Mobius Function Proof
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 21st 2010, 11:31 AM
  3. Mobius transformation
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 27th 2009, 07:39 AM
  4. Mobius transformations
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 8th 2009, 11:06 AM
  5. Help on Mobius function
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: April 16th 2008, 03:54 AM

Search Tags


/mathhelpforum @mathhelpforum