Question

HELP!! Mobius inversion stuff
http://prntscr.com/4hqi84

Answer 1

Use induction and the definition of \(\mu\).

Answer 2

\[\mu = \begin{cases} 1 & if~ n=1\\0 & \text{n is not square free} \\(-1)^r& \text{otherwise} \end{cases}\]

Answer 3

That is not the definition of \(\mu\) as I know it. :-/

Answer 4

it is across :)
see above it says Mobius inversion

Answer 5

So, \(\mu\) is not the Mobius function?

Answer 6

it is Mobius function, let me take a snapshot and attach... looks I have messed up the meaning ;-;

Answer 7

here is the definition of \(\large \mu\) function from my textbook
http://prntscr.com/4hqmak

Answer 8

it is -.-
but it name mubius \(\mu\) function ,
what the function you know across ?

Answer 9

ok so , we need to prove one of them =0

Answer 10

I think the same symbol is used in different branches... when I searched i got multiple confusing definitions earlier

Answer 11

for
n
n+1
n+2
n+3

at least of of them should have \(p^2\) divisor .

Answer 12

mu is used every where , and there is several functions with mobius name
but there is only one function name mobius mu function :o

Answer 13

can we use this

there will be exactly TWO even numbers in a set of 4 consecitive integers ?

Answer 14

so the product of any four consecutive numbers is always NOT square free... this should conclude the proof ?

Answer 15

exactly :)
one of them would divid 4 :o
bhahaha cool

Answer 16

done , it is a proof :)

Answer 17

that was easy :) part b looks scary though >.<

Answer 18

its multiplicative function

Answer 19

Alright, since its multiplicative we can write

\[\large \sum \limits_{k=1}^n\mu(k!) = \sum \limits_{k=1}^n\left(\mu(1)*\mu(2)*\cdots \mu(k)\right) \]

Answer 20

next what

Answer 21

\(\mu(k!)=0\)
for k>=4

Answer 22

Oh yes got it :)

Answer 23

got it all ?

Answer 24

so we only need to show that
\(\mu (1!) +\mu (2!)+\mu (3!)=1\)

Answer 25

\[\large \begin{align} \\ \sum \limits_{k=1}^n\mu(k!) &= \sum \limits_{k=1}^n\left(\mu(1)*\mu(2)*\cdots \mu(k)\right) \\~\\ &= \sum \limits_{k=1}^3\left(\mu(1)*\mu(2)*\cdots \mu(k)\right) + 0 \end{align}\]

Answer 26

nice :)