Idea of Mobius Inversion

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  • 7/26/2019 Idea of Mobius Inversion

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    Mertens function

    Anant Saxena

    December 28, 2015

    1 Basic Definitions(n, r) = Remainder of

    n

    r

    We start with a discrete variable n (n can only be an integer). Hence,

    n= 1 + 1 +. . . ntimes

    Now, using the mobius inversion formula:

    1 =n

    r=1

    (r)n

    r

    Where,x is the floor function on xWe will call this series basic sum.

    2 Mertens function

    2.1 Series

    We start by considering:

    1

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    n= 1 + 1 +. . . ntimes

    n= (0 + 1) + . . . 2ntimes

    n= (0 + 0 + 1) + . . . 3ntimes

    ...

    n= (0 + 0 + 0 n1times

    +1) + . . . n2times

    (1)

    Multiplying the rth equation with (r)

    2.2 Example n =3

    When n= 3

    n= 1 + 1 + 1

    n= 0 + 1 + 0 + 1 + 0 + 1

    n= 0 + 0 + 1 + 0 + 0 + 1 + 0 + 0 + 1

    (2)

    Now we apply the following algorithm: we add another 0 appropriatelyto make it look as the the 1st nterms are repeating in each equation :

    n= 1 + 1 + 1

    n= 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1(3,2)

    n= 0 + 0 + 1 + 0 + 0 + 1 + 0 + 0 + 1

    (3)

    Now we add and 1s in each equation to make each equations initial termsto look like the first n terms

    n= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1(n1) n

    1

    n= 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0(n2)(n1)n

    2

    + 1

    (3,2)

    n= 0 + 0 + 1 + 0 + 0 + 1 + 0 + 0 + 1

    (4)

    2

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    Multiplying the rth equation with (r)

    (1)n= (1)(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1(n1)n

    1

    )

    (2)n= (2)(0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0(n2)n

    2

    + 1

    (3,2)

    )

    (3)n= (3)(0 + 0 + 1 + 0 + 0 + 1 + 0 + 0 + 1)

    (5)

    Adding all the above:

    (1)n= (1)(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1(n1) n

    1)(2)n= (2)(0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0(n2)

    n

    2

    + 1

    (3,2)

    )

    (3)n= (3)(0 + 0 + 1 + 0 + 0 + 1 + 0 + 0+)(n n)n

    3

    )

    (6)

    We notice the first n= 3 terms is the basic sum. Hence,

    (1)n= (1)(

    1 + 1 + 1 +

    1 + 1 + 1 +

    1 + 1 + 1 (n1)n

    1

    )

    (2)n= (2)(0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0(n2) n2+ 1(3,2)

    )

    (3)n= (3)(0 + 0 + 1 =1

    + 0 + 0 + 1 =1

    + 0 + 0 + 1) =1

    (n n)n

    3

    )

    (7)

    2.3 Generalization of Sum

    Adding the a generalization of the above equations together:

    n

    n

    r=1

    (r) = nn

    r=1

    (r)(n r) n

    r +n

    r=1

    (r)(n, r)

    = nn

    r=1

    (r) =n

    r=1

    (r)rn

    r

    +

    nr=1

    (r)(n, r)

    3

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    = nn

    r=1

    (r) =

    n

    r=1

    (r)r nr+

    n

    r=1

    (r)(n, r)

    = nn

    r=1

    (r) =n

    r=1

    (r)rn

    r

    +

    nr=1

    (r)(nn

    r

    r)

    IDEA FAIL :(

    4