OK, another maths question -
This is the double sum I have:
sum_(n = 1 to infinity) n^-s sum_(n = 1 to infinity) lamda (n) mu (n) n^-s
Where lamda (n) is the Liouville function and mu (n) is the Mobius function
lamda (n) defined as lamda (1) = 1 and if n = p1^a1 . p2^a2. p3^a3... then lamda (n) = (-1)^(a1 +a2 + a3...)
and mu (n) defined as mu(1) = 1 and mu(n) = (-1)^k if a1, a2, a3.... = 1 and mu(n) = 0 otherwise. (Where n = p1^a1 . p2^a2. p3^a3...)
Anyway, this is then rearranged to give:
sum_(n = 1 to infinity) [sum (d, the divisors of n) lamda (d) mu(d) ] n^-s
This function is said to be completely multiplicative, so as to allow rearrangement, but I can't follow why one would lead to the other. Any ideas on this - or any good online resources - I've searched but not found anything so far.
Many thanks! :)
This is the double sum I have:
sum_(n = 1 to infinity) n^-s sum_(n = 1 to infinity) lamda (n) mu (n) n^-s
Where lamda (n) is the Liouville function and mu (n) is the Mobius function
lamda (n) defined as lamda (1) = 1 and if n = p1^a1 . p2^a2. p3^a3... then lamda (n) = (-1)^(a1 +a2 + a3...)
and mu (n) defined as mu(1) = 1 and mu(n) = (-1)^k if a1, a2, a3.... = 1 and mu(n) = 0 otherwise. (Where n = p1^a1 . p2^a2. p3^a3...)
Anyway, this is then rearranged to give:
sum_(n = 1 to infinity) [sum (d, the divisors of n) lamda (d) mu(d) ] n^-s
This function is said to be completely multiplicative, so as to allow rearrangement, but I can't follow why one would lead to the other. Any ideas on this - or any good online resources - I've searched but not found anything so far.
Many thanks! :)
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