# Sigma function help

• Apr 20th 2008, 09:36 PM
kleenex
Sigma function help
Let $\displaystyle \sigma_k(n) = \sum_{d|n}d^k$ (Note: $\displaystyle \sigma_1(n) = \sigma(n)$) the sum of the $\displaystyle k$th powers of divisors of n.

Is $\displaystyle \sigma_3(2) = 1^3 + 2^3 = 9$?

Also I need so help on the following:
• Give a formula for $\displaystyle \sigma_k(p)$ and $\displaystyle \sigma_k(p^a)$ where p is prime, a is positive integer.
My answer is $\displaystyle \sigma_k(p) = 1 + p^k$ and $\displaystyle \sigma_k(p^a) = 1+ \sum_{i=1}^a p^i$, right?
• Show the function $\displaystyle \sigma_k$ is multiplicative, but not completely multiplicative.
• Find a formula for $\displaystyle \sigma_k(n)$, where n has prime power factorization $\displaystyle n = p_1^{a_1}p_2^{a_2}...p_m^{a_m}$.
• Apr 20th 2008, 11:08 PM
Moo
Hello,

Quote:

Originally Posted by kleenex
Let $\displaystyle \sigma_k(n) = \sum_{d|n}d^k$ (Note: $\displaystyle \sigma_1(n) = \sigma(n)$) the sum of the $\displaystyle k$th powers of divisors of n.

Is $\displaystyle \sigma_3(2) = 1^3 + 2^3 = 9$?

I'd say yes

Quote:

Also I need so help on the following:[*]Give a formula for $\displaystyle \sigma_k(p)$ and $\displaystyle \sigma_k(p^a)$ where p is prime, a is positive integer.
My answer is $\displaystyle \sigma_k(p) = 1 + p^k$
Ok for this one

Quote:

and $\displaystyle \sigma_k(p^a) = 1+ \sum_{i=1}^a p^i$, right?
I don't agree...

$\displaystyle \sigma_k(p^a)=1^k+p^k+p^{2k}+\dots+p^{ak}=1+\sum_{ i=1}^a p^{ki}$

Plus, you can see that it is a geometrical sum :)
• Apr 21st 2008, 04:58 PM
ThePerfectHacker
This function is weakly multiplicative because if $\displaystyle F(n) = \sum_{d|n}f(d)$ and $\displaystyle f$ is weakly multiplicative then $\displaystyle F$ is weakly multiplicative. With that you can give a formula if you now the prime factorization. Note that $\displaystyle \sigma_k(p_1^{a_1} ... p_m^{a_m}) = \sigma_k(p_1^{a_1}) ... \sigma_k(p_m^{a_m})$ and now use what Moo used above.