Question

Suppose \(g(n) = \sum\limits_{i=1}^n f(i)\)

then \(f(n) = ?\)

Answer 1

@ikram002p @imqwerty

Answer 2

g(n) is multiplicative or f(n) ?

Answer 3

g(n) is just a partial sum

g or f need not be multiplicative

Answer 4

Suppose \(g(n) = \sum\limits_{i=1}^n f(i)\)

then \(f(n) = ?\)

Answer 5

basically we want to express \(f\) as a function of \(g\)

Answer 6

in other words, we want the \function, \(f\), back

Answer 7

ok
g(n)=f(1)+f(2)+...+f(n)
g(n)-f(n)=f(1)+...+f(n-1)
f(n)=g(n-1)-g(n)

Answer 8

Perfect!

Answer 9

wait, there seems to be a sign error... check again

Answer 10

:o this is how we do it :) thanks!! :D

Answer 11

wait no i have made some mistake
f(n)=g(n)-g(n-1)

:P

Answer 12

Yes :) its a trivial problem...
but I think it is a very good problem to enter the beautiful world of mobius inversion and stuff!

Answer 13

its recursive formula ok keep going

Answer 14

In light of above result, try finding below :

Suppose \(g(n) = \sum\limits_{d\mid n} f(d)\)

then \(f(n) = ?\)

Answer 15

then i need a shower lol

Answer 16

its not going to be trivial

Answer 17

you're euler if you're able to figure out \(f(n)\) on your own!

Answer 18

shower may not help that much..

Answer 19

here is the complete theorem :

Answer 20

well i start to think of it as this.
if \( n=p_1p_2...p_k\)
\(g(n)=\sum_{d|n} f(n)= f(p_1)f(p_2)...f(p_k)\)

Answer 21

Theorem :

If \(g(n) = \sum\limits_{d\mid n} f(d)\) then \(f(n)\) is given by

\[f(n) = \sum\limits_{d\mid n}\mu(d)g(n/d) \]

Answer 22

oh i was so close to that :P

Answer 23

i ws thinking like how we solved the previous problem
if we do that we get this-\[g(n)-\left[ f(d_1)+f(d_2)+f(d_3).. \right]=f(n)\]
so we need some g(n) which must equal \(f(d_1)+f(d_2)+f(d_3)..\) so as to express f(n) in terms of g(n)

Answer 24

here is an interesting result based on that theorem :

we know that \(\tau(n) = \sum\limits_{d\mid n}1\)

applying previous theorem gives \(1 = \sum\limits_{d\mid n} \mu(d)\tau(n/d)\)

Answer 25

impressive! ok brb

Answer 26

that is a nice start !

Answer 27

it took me several weeks to wrap my head around this theorem... I would be amazed if somebody gets this on the first day itself !

Answer 28

ikram, f and g need not be multiplicative...

Answer 29

lol m getting dizzy now xD i'll be back after i finish my home wrk :)

Answer 30

ermm i know, that property i used earlier which seems i'm the only one who use it xD

Answer 31

we had a lot of number theory for one day haha
maybe lets get back on this tomorrow

Answer 32

sure!

Answer 33

ermmm lol i hope it won't take me that much.