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Math Help - Sigma function help

  1. #1
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    Sigma function help

    Let \sigma_k(n) = \sum_{d|n}d^k (Note: \sigma_1(n) =  \sigma(n)) the sum of the kth powers of divisors of n.

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

    Also I need so help on the following:
    • Give a formula for \sigma_k(p) and \sigma_k(p^a) where p is prime, a is positive integer.
    My answer is \sigma_k(p) = 1 + p^k and \sigma_k(p^a) = 1+ \sum_{i=1}^a p^i, right?
    • Show the function \sigma_k is multiplicative, but not completely multiplicative.
    • Find a formula for \sigma_k(n), where n has prime power factorization n = p_1^{a_1}p_2^{a_2}...p_m^{a_m}.
    Thank you for your help.
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  2. #2
    Moo
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    Hello,

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

    Is \sigma_3(2) = 1^3 + 2^3 = 9 ?
    I'd say yes

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

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

    \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
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  3. #3
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    This function is weakly multiplicative because if F(n) = \sum_{d|n}f(d) and f is weakly multiplicative then F is weakly multiplicative. With that you can give a formula if you now the prime factorization. Note that \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.
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