※ 引述《Sfly (topos)》之銘言：
: ※ 引述《JohnMash (Paul)》之銘言：
: : 數論不等式
: : 試證
: : Σ_{n<R} μ(n)^2 τ_k(n) / φ(n) << ( ㏒ R)^k
: : 其中 n,k 為自然數 R為正實數 ㏒ 自然對數
: : μ(n) Moebius function
: : τ_k(n) the number of ways of writing n as a product of k natural numbers
: : φ(n) Euler totient function
: Since τ_k(p) / φ(p) = k/(p-1), so
: Σ_{n<R} μ(n)^2 τ_k(n) / φ(n)
: <= Π(1+k/(p-1)) << 1/Π(1-k/p) << 1/Π(1-1/p)^k << (log R)^k
: p<R p<R p<R
: The last part is followed from the Mertens' theorem.
: The case with the restriction gcd(n,W)=1 is similar.
我把它寫詳細一點
Since τ_k(p) / φ(p) = k/(p-1), so
Σ_{n<R} μ(n)^2 τ_k(n) / φ(n) <= Π_{p<R} (1+k/(p-1))
(1+k/(p-1)) < (1 + 1/(p-1))^k (二項式)
Π_{p<R} (1+k/(p-1)) < Π_{p<R} (1 + 1/(p-1))^k
令 p1,p2,...,pl<R
Π_{p<R} (1 + 1/(p-1))^k = Y^k
Y=(1 + 1/(p1-1)) (1 + 1/(p2-1)) .... (1 + 1/(pl-1))
又 (1 + 1/(p1-1)) =2
(1 + 1/(p2-1)) <= (1 + 1/p1)
(1 + 1/(p3-1)) <= (1 + 1/p2)
....
(1 + 1/(pl-1)) <= (1 + 1/p_{l-1})
Y=(1 + 1/(p1-1)) (1 + 1/(p2-1)) .... (1 + 1/(pl-1)) <= 2 (1 + 1/p1) (1 + 1/p2)
.... (1 + 1/p_{l-1})
又 1+ 1/p < e^(1/p)
Y<< e^( Σ_{p<R} 1/p)
又 Σ_{p<R} 1/p << ㏒ ㏒ R
Y << ㏒R
Σ_{n<R} μ(n)^2 τ_k(n) / φ(n) <= Y^k << ( ㏒R)^k
--
Sent from my Android
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 123.194.229.234
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1422816265.A.51A.html