Verified answer

a) F(pq) = Σ(d|pq) Λ(d)

............= Λ(1) + Λ(p) + Λ(q) + Λ(pq)

............= 0 + log p + log q + 0

............= log(pq), via product rule of logs.

b) This is similar to part a, but with more factors (note that Λ = 0 for numbers with more than one prime factor).

(Answer: log(p^3 q^5).)

c) Writing n = (p1)^(a1) * ... * (pk)^(ak) as the prime factorization of n,

F(n) = Λ(1) + Σ(j=1 to k) [Λ(pj) + Λ((pj)^2) + ... + Λ((pj)^(aj))] + 0, since Λ(d) = 0 if

d has more than one prime in its prime factorization

.......= Λ(1) + Σ(j=1 to k) [Λ(pj) + Λ((pj)^2) + ... + Λ((pj)^(aj))]

.......= 0 + Σ(j=1 to k) [log(pj) + log(pj) + ... + log(pj)]

.......= Σ(j=1 to k) aj * log(pj)

.......= Σ(j=1 to k) log ((pj)^(aj)), via power rule for logs

.......= log ((p1)^(a1) * ... * (pk)^(ak)), via product rule for logs

.......= log n.

(d) Λ(n) = Σ(d|n) μ(n/d) log d, via part (c) and Mobius Inversion Formula

............= Σ(d|n) μ(d) log(n/d), by reversing the order of summation

............= Σ(d|n) μ(d) (log n - log d)

............= log n * Σ(d|n) μ(d) - Σ(d|n) μ(d) log d

............= log n * 0 - Σ(d|n) μ(d) log d, fact about μ

............= - Σ(d|n) μ(d) log d.

I hope this helps!

You probably mean Λ(n)= log p if n=p^k for prime p and k > 0 [i.e. if n is a prime or a power of a prime] And I assume you mean that F(n) is the sum of Λ(d) for all divisors d of n.

a) If n=pq, what are its divisors? Clearly, 1, p, q and n itself -- and nothing else.

So F(n) = Λ(1) + Λ(p) + Λ(q) + Λ(n) = 0 + logp + logq + 0 = log(pq) = logn

b) In this case, the divisors of n are 1, p, p², p³, q, qp, qp², qp³, q², q²p, etc. up to q⁵p³.

But of al those numbers, only for p, p², p³, q, q², q³, q⁴ and q⁵ is Λ different from 0.

So Λ(n) = Λ(p) + Λ(ρ²) +... + Λ(q⁴) + Λ(q⁵) = logp + logp + logp + logq + logq + logq + logq + logq =

= 3logp + 5logq = log(p³) + log(q⁵) = log(p³q⁵) = logn

c) In fact, F(n)=logn for all n. For, let n = p^a * q^b * r^c * s^d *... where p, q, r, s... are the different prime divisors of n and a, b, c... their exponents in the prime factorization of n. Then the only divisors of n that are powers of primes are p, p², p³,... p^a; q, q², q³, ... q^b; r, r², r³,... r^c; etc. The Λ of all other divisors of n is 0, and the Λ of the divisors just listed is logp, logp, logp... (a times), logq, logq, logq... (b times), logr, logr, logr... (c times) etc. So the sum of all Λ's is just alogp + blogq + clogr +... = log(p^a)+log(q^b)+log(r^c)+... = log(p^a*q^b*r^c*...) = logn.

d) Möbius' inversion formula states just this, that if F(n) is the sum of Λ(d) over all divisors d of n, then Λ(n) is the sum of μ(n/d)F(d) over all divisors d of n for ANY arithmetic functions Λ and F. Apply this to aour case, where F(n)=logn, to get Λ(n) = Σ μ(n/d)log d. But as d runs over all divisors of n, n/d also runs over all divisors of n; so this sum can also be written Σμ(d)log(n/d) = Σμ(d)(logn - logd) = (Σμ(d))logn - Σμ(d)logd = -Σμ(d)logd, since Σμ(d) taken over all divisors of n is 0 (unless n=1, in which case logn=0 !).

This last assertion is an immediate consequence of the fact that every non-empty finite set has just as many subsets with odd numbers of elements as subsets with even numbers of elements, which is in turn a consequence of the binomial expansion of (1-1)^n = 0

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