Question

HELP! Mobius inversion formula
http://prntscr.com/4hqt9l

Answer 1

using the hint, I should begin with\[\large
\begin{align}\\
\sum \limits_{d|n} \Lambda(d) =
\end{align}
\]

Answer 2

Since all divisors vanish except single prime powers :

\[\large
\begin{align}\\
\sum \limits_{d|n} \Lambda(d) &= \sum \limits_{p^i|n} \Lambda(p^i)
\end{align}\]

Answer 3

will try some other time , gn

Answer 4

gn

Answer 5

\[\large
\begin{align}\\
\sum \limits_{d|n} \Lambda(d) &= \sum \limits_{p^i|n} \Lambda(p^i) \\~\\
&=\log(p_1) + \log(p_1) + \cdots + \log (p_r)~~\\&\color{gray}{\text{(gives prime factorization of n)}}~\\~\\
&=\log(p_1^{e_1}p_2^{e_2}\cdots p_r^{e^r})\\~\\
&= \log (n)
\end{align}\]

Answer 6

apealing to mobius inversion formula
\[\large F(n) = \sum \limits_{d|n}f(d) \implies f(n) = \sum \limits_{d|n}\mu(d)F(n/d)\]

gives

\[\large \log (n) = \sum \limits_{d|n}\Lambda (d) \implies \Lambda(n) = \sum \limits_{d|n}\mu(d)\log(n/d)\]

Answer 7

first part is awesome ,
i dint get second part :o

Answer 8

i have just applied the mobius inversion formula

Answer 9

but we wanna
u(n/d) u(d) right ?

Answer 10

u(n/d) log d

Answer 11

hmmm , not sure still brain freazing >.<

Answer 12

?

Answer 13

from the first part we have this :
\[\large \log (n) =\sum \limits_{d|n} \Lambda(d) \]

Answer 14

apply mobius inversion formula, what do u get ?

Answer 15

\sum u(d)log(n/d)

Answer 16

yes, so we're done right ?

Answer 17

how ;_;

Answer 18

http://prntscr.com/4i1hj3

Answer 19

d is just a dummy variable :

as "d runs through n",
"n/d runs through n" too

Answer 20

and negative sighn ?

Answer 21

so

\[\large \begin{array}\\
\log (n) = \sum \limits_{d|n}\Lambda (d) \implies \Lambda(n) &= \sum \limits_{d|n}\mu(d)\log(n/d) \\~\\ &= \sum \limits_{d|n}\mu(n/d)\log(d) \\~\\ \end{array}\]

Answer 22

fine with this ?

Answer 23

yeah

Answer 24

lets work the second equality

Answer 25

\[\large \begin{array}\\
\log (n) = \sum \limits_{d|n}\Lambda (d) \implies \Lambda(n) &= \sum \limits_{d|n}\mu(d)\log(n/d) \\~\\ &= \\~\\\end{array}\]

Answer 26

expand log(n/d) = log(n) - log(d)

Answer 27

\[\large \begin{array}\\
\log (n) = \sum \limits_{d|n}\Lambda (d) \implies \Lambda(n) &= \sum \limits_{d|n}\mu(d)\log(n/d) \\~\\ &= \sum \limits_{d|n}\mu(d)\left[\log(n) - \log(d) \right] \\~\\\end{array}\]

Answer 28

and remember this result :

\[\large \begin{array}\\

\sum \limits_{d|n}\mu(d) = 0 \\~\\

\end{array}\]

Answer 29

oh :o

Answer 30

\[
\large \begin{array}\\
\log (n) = \sum \limits_{d|n}\Lambda (d) \implies \Lambda(n) &= \sum \limits_{d|n}\mu(d)\log(n/d) \\~\\
&= \sum \limits_{d|n}\mu(d)\left[\log(n) - \log(d) \right] \\~\\
&= \sum \limits_{d|n}\mu(d) \log(n) - \sum \limits_{d|n}\mu(d) \log(d) \\~\\
&= \log(n)\sum \limits_{d|n}\mu(d) - \sum \limits_{d|n}\mu(d) \log(d) \\~\\
&= 0 - \sum \limits_{d|n}\mu(d) \log(d) \\~\\

\end{array}

\]

Answer 31

ohhh !!

Answer 32

im not dude

Answer 33

ermm ok -.-
sry

Answer 34

i dnt even know what does that mean, really sry