Question

HELP!
http://prntscr.com/4g2pkf

Answer 1

@Adjax

Answer 2

\[\large \begin{align}\\

\sigma_s(n) &= \sum \limits_{d|n}d^s\\~\\
\sigma_0(n) &= \sum \limits_{d|n}d^0 = \sum \limits_{d|n}1 = \tau\\~\\
\sigma_1(n) &= \sum \limits_{d|n}d^1 = \sum \limits_{d|n}d = \sigma \\~\\


\end{align}\]

Answer 3

need help on part b

Answer 4

on what you are coonfused in part B?

Answer 5

the answer to the question is given in the hint

Answer 6

since n^s involves the multiplication of n with itself s times..therefore it's multiplicative(excuse me I haven't involved the rigors in this answer)

Answer 7

that may work :) we need to show
\[\large \sigma_s(ab) = \sigma_s(a) * \sigma_s(b)\]

Answer 8

isn't when ab is entered (it actually involves multiplication)./.so why you are dbl multiplications

Answer 9

Need to show
\[\large \sum \limits_{d|ab}d^s = \left(\sum \limits_{d|a}d^s\right)*\left( \sum \limits_{d|b}d^s\right)\]

Answer 10

hey I remebered :log ab = log a + log b(ok I know this function is not related to this EQN.(

But isn't that a and b is evaluated evaluated independently and the result is .....

Answer 11

Exactly a and b can be treated separately if we assume a and b are coprime !
the divisors of a*b involve the numbers that contain all the divisors of both a and b, so we can separate them :


\[\large \begin{align} \\

\sum \limits_{d|ab}d^s &= \sum \limits_{d|ab}(d_1d_2)^s \\~\\
&= \sum \limits_{d_1| a, ~d_2|b}(d_1^sd_2^s) \\~\\
&= \left(\sum \limits_{d_1| a}d_1^s\right) * \left( \sum \limits_{d_2| b}d_2^s \right)\\~\\

\end{align}\]

Answer 12

thanks! you have no idea how much you have helped me :)

Answer 13

hey look at this also:

http://www.willamette.edu/~cstarr/math356/stillwellHW/12-NT-functionsHWsn.pdf

Pg no.1 ..Question no.5

@rational

Answer 14

nice, thank you so much :)

Answer 15

write a novel titled :'Game of Coprimes:Songs of clashes of Maths on the brain'

Answer 16

lol part c looks trivial :

\[\large \begin{align}\\

\sigma_s(n) &= \sum \limits_{d|p^k}d^s\\~\\
&= (1)^s + (p)^s + (p^2)^s + \cdots + (p^k)^s \\~\\
&= 1 + p^s + p^{2s} + \cdots + p^{ks} \\~\\
& = \dfrac{(p^s)^{k+1}-1}{p^s - 1} \\~\\
\end{align}\]

Answer 17

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

Answer 18

@BSwan try last question when u get a chance, it looks tough compared to previous questions >.<

Answer 19

you solved the rest :O
alone , without me :O

Answer 20

@Adjax what sort of novel is it ?

Answer 21

im still working on last question

Answer 22

ok part a is ovc , right ?

Answer 23

whats ovc ?

Answer 24

ovc= oviouse
part b we need to prove that
\(\large \sigma_s= \) (something multiplicative )\(^n\)

Answer 25

its not obvious to me,
i forgot my brain in office today. can u show me the proof for part a -.-

Answer 26

awww , is there anyway we could take ur brain back :D

Answer 27

ugh for part a we consider \(n = p^k\)


\[\large \begin{align}\\

\sum \limits_{d|p^k} \sigma(d)

\end{align}\]

Answer 28

ok

\(\large \sigma_0 =\sum d^0=\sum 1 = \tau\)
\(\large \sigma_1 =\sum d^1=\sum d = \sigma \)

Answer 29

that question was done long back :/

Answer 30

im working on #23 :
http://prntscr.com/4g2prz

Answer 31

wait im looking for my brain ha ha ha
sry

Answer 32

yes we consider n=p^k

Answer 33

but wait

Answer 34

\(\large \sum_{d|p^k}\sigma (d)=\sum_{d|p^k}\sum_{d|p^k} d =\sum_{d|p^k}\sum_{d|p^k} \frac{n}{d} \)

Answer 35

we should use a different symbol for inside sum :

\[\large \large \sum_{d|p^k}\sigma (d)=\sum_{d|p^k}\sum_{c|d} c\]

Answer 36

oh yeah

Answer 37

give up , gn
i might try ltr :-|

Answer 38

only i thought of prove
\sum_{c|(d=p^r)} \frac{n}{c}= \frac{n}{d} \tau (d)

Answer 39

\(\large \sum_{c|(d=p^r)} \frac{n}{c}= \frac{n}{d} \tau (d)\)

Answer 40

im thinking of expanding out the left side summation :

\[ \large \sum_{d|p^k}\sigma (d)= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) + \cdots + \sigma(p^k) \]

Answer 41

does that look okay ?

Answer 42

and do this ?

p-1/p-1 + p^2-1/p-1 + ...+ p^k-1/p-1 ?

ok then ?

Answer 43

im not ure if its useful

Answer 44

its scares me to think out of the sum :o
i try to be in multible thingy as long as i can
hhmm
ok what do you think >.<

Answer 45

lets see :)

\[\large \begin{align} \\

\sum_{d|p^k}\sigma (d) &= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) + \cdots + \sigma(p^k) \\~\\

&= \dfrac{p^{0+1}-1}{p-1} + \dfrac{p^{1+1}-1}{p-1} + \cdots + \dfrac{p^{k+1}-1}{p-1} \\~\\
&= \dfrac{1}{p-1}\left( (p + p^2 + \cdots + p^{k+1}) - (k+1)\right) \\~\\

\end{align} \]

Answer 46

\[\large \begin{align} \\

&= \dfrac{1}{p-1}\left(\dfrac{p(p^{k+1}-1)}{p-1}- (k+1)\right) \\~\\

\end{align} \]

Answer 47

feels it wont result in anything useful :/

Answer 48

lets simplify right side as well

Answer 49

:D

Answer 50

RHS :

\[ \begin{align} \\

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

Answer 51

hmming

Answer 52

http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence

Answer 53

there should be another more meaningful way

Answer 54

i feel we're just doing the donkey work without trying to look for cool multiplicative/sum properties

Answer 55

i guess .
i have no clue though

Answer 56

wil try tomoro or post it in MSE

Answer 57

ok

Answer 58

post it there

Answer 59

https://math.stackexchange.com/questions/909131/for-any-positive-integer-n-show-that-sumdn-sigmad-sum-dnn-d-ta

Answer 60

gn then .

Answer 61

gn

Answer 62

:o