Question

HELP!
http://prntscr.com/4fxv0h

Answer 1

here u can use sum notation:P

Answer 2

no clue yet
say, \(\gcd(a,b) = 1\) and \(f(n) = 2^{\omega(n)}\)
guess i need to show :

\[\large 2^{\omega (ab)} = 2^{\omega (a)} . 2^{\omega (b)}\]

Answer 3

hmm one problemo
w(a) is not multiplicative right ?

Answer 4

how ever ,

w(n)= \(\sum_{p|n} p_i\)

Answer 5

\(\large \omega(6) = 2\)
\(\large \omega(5) = 1\)


\(\large \omega(6.5) = 3 \color{Red}{\ne} 2 = \omega(6) . \omega(5) \)

Answer 6

so \(\large \omega \) is not multiplicative

Answer 7

i know its not :o
i already said that :P
but sum notation omefrom def of w
so ,
\(\Huge 2^{\sum_{p|n} p_i} =\prod_{p|n} 2^{p_i}\)

Answer 8

thats it :)

Answer 9

w shouldn't be multiplicative, it should be additive so you can use the power properties to prove 2^w(n) is multiplicative. \[2^{w(m*n)}=2^{w(m)+w(n)}=2^{w(m)}.2^{w(n)}\]

Answer 10

nice :) i think that should work perfectly!
what do u think @BSwan

Answer 11

yeah yeah thats what i did xD

Answer 12

you did it like a prof so i couldn't understand sooner :P

Answer 13

\[\large \omega(n) = \sum_{p|n} 1\]

right ?

Answer 14

we need to count number of primes,
not add the primes

Answer 15

oh yeah :O hehe made a typo

Answer 16

\[\huge 2^{\sum \limits_{p|(ab)} 1} =2^{\sum \limits_{p|a} 1} . 2^{\sum \limits_{p|b} 1} \]

Answer 17

not sure how to represent this using PI function

Answer 18

does that mean below is true for any base t ?

\(\large t^{\omega(n)}\) is always multiplicative for all natural numbers \(t\) ?

Answer 19

nothing special about \(2\) right ?

Answer 20

hmmm i think so , since all exp functions are multiplicative
hmmm

Answer 21

that makes sense
next part

Answer 22

the special thing about 2 is that it gives the formula in second part :P
so i iguess we'll gonna use it

w(d) is number of primes fators of d right ?

Answer 23

Ohk.. yes

Answer 24

hmm

Answer 25

Since \(\large 2^{\omega(d)}\) is multiplicative, \(\large \sum \limits_{d|n} 2^{\omega(d)}\) is also multiplicative from `Theorem 6.4`;
\(\large \tau \) is multiplicative too.

So it suffices to prove it for a prime power : \(\large n = p^{k}\) :

\[\large \begin{align}\\

\sum \limits_{d|p^k}2^{\omega(d)} &= 2^0 + 2^1 + 2^1 + \cdots \text{k times} \\
&= 1 + 2k \\~\\
&=\tau\left((p^{k})^2\right)\\~\\
\end{align}\]

Answer 26

3 more questions to go before touching mobius inversion formula xD

Answer 27

what are the question ?
ill read them from my book
quick copy paste before i go

Answer 28

@rational QUICCKKKKKKKKKKKKKK

Answer 29

one question at a time :)
next question :
http://prntscr.com/4g2a68

Answer 30

no i want work online >.<
my eyes hurts , just give me the three to check them from my book :O

Answer 31

ok i saw that before >.<
what next ?

Answer 32

#22 :
http://prntscr.com/4g2pkf

Answer 33

#23 :
http://prntscr.com/4g2prz

Answer 34

@BSwan

Answer 35

got them , gn

Answer 36

gn :) il try these in the morning

Answer 37

hehe im reading this
http://assets.openstudy.com/updates/attachments/5300444ee4b024fe26e1a733-ganeshie8-1392541446906-ccc.png



from this question
http://openstudy.com/study#/updates/5300444ee4b024fe26e1a733

Answer 38

did you solve anyone of then @ganeshie8

Answer 39

wil try in the evening..

Answer 40

have u solved any ?

Answer 41

trying

Answer 42

#21 :

\[\large \begin{align}\\

\sum \limits_{d|p^{k}}\tau(d)^3 &= (0+1)^3 + (1+1)^3 + (2+1)^3 \cdots + (k+1)^3 \\
& = \left(\dfrac{(k+1)(k+2)}{2}\right)^2\\~\\
& = \left(1+2+3+\cdots +(k+1)\right)^2\\~\\
& = \left(\sum \limits_{d|p^{k}}\tau(d)\right)^2\\~\\
\end{align}\]

Answer 43

OH my i was stuck in the series >.<
thats cool