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Estimate about $sum_{nleq x}mu(n)log frac xn$
$begingroup$
I'm finding the estimate of $$sum_{nleq x}mu(n)log frac xn$$
There is a formula saying that $$sum_{nleq x}f(n)Gbigg(frac xnbigg)=sum_{nleq x}g(n)Fbigg(frac xnbigg)$$
where $F(x)=sum_{nleq x}f(n)$ and $G(x)=sum_{nleq x}g(n)$. So I want to express $log x$ in the form $$log x=sum_{nleq x}g(n)$$
for some function $g$. but what I really know is $$log n=sum_{d|n}mu(d)logdfrac nd$$
not the same as what I want. Any suggestion? I also appreciate different way to do this.
asymptotics analytic-number-theory
edited Jan 29 at 20:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 4:04
kelvin hong 方kelvin hong 方
83719
$endgroup$
add a comment |
$begingroup$
I'm finding the estimate of $$sum_{nleq x}mu(n)log frac xn$$
There is a formula saying that $$sum_{nleq x}f(n)Gbigg(frac xnbigg)=sum_{nleq x}g(n)Fbigg(frac xnbigg)$$
where $F(x)=sum_{nleq x}f(n)$ and $G(x)=sum_{nleq x}g(n)$. So I want to express $log x$ in the form $$log x=sum_{nleq x}g(n)$$
for some function $g$. but what I really know is $$log n=sum_{d|n}mu(d)logdfrac nd$$
not the same as what I want. Any suggestion? I also appreciate different way to do this.
asymptotics analytic-number-theory
edited Jan 29 at 20:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 4:04
kelvin hong 方kelvin hong 方
83719
$endgroup$
1
$begingroup$
What is $mu(n)$?
$endgroup$
– Stefan Lafon
Jan 29 at 4:24
$begingroup$
@StefanLafon It is the Möbius Function.
$endgroup$
– kelvin hong 方
Jan 29 at 4:27
2
$begingroup$
$int_1^infty (sum_{n le x} mu(n)(log x -log n)) x^{-s-1}dx = frac{1}{s^2zeta(s)}$ so assuming the zeros are simple and after a lot of estimates to show it converges, there is the explicit formula $sum_{n le x} mu(n)(log x -log n) = sum Res(frac{1}{s^2zeta(s)} x^s) = sum_rho frac{x^rho}{rho^2 zeta'(rho)}-frac{zeta'(0)}{zeta(0)^2}+frac{log x}{zeta(0)} +sum_{k=1}^infty frac{x^{-2k}}{(-2k)^2 zeta'(-2k)}$. The RH is true iff it is $O(x^{1/2+epsilon})$.
$endgroup$
– reuns
Jan 29 at 5:24
add a comment |
$begingroup$
I'm finding the estimate of $$sum_{nleq x}mu(n)log frac xn$$
There is a formula saying that $$sum_{nleq x}f(n)Gbigg(frac xnbigg)=sum_{nleq x}g(n)Fbigg(frac xnbigg)$$
where $F(x)=sum_{nleq x}f(n)$ and $G(x)=sum_{nleq x}g(n)$. So I want to express $log x$ in the form $$log x=sum_{nleq x}g(n)$$
for some function $g$. but what I really know is $$log n=sum_{d|n}mu(d)logdfrac nd$$
not the same as what I want. Any suggestion? I also appreciate different way to do this.
asymptotics analytic-number-theory
edited Jan 29 at 20:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 4:04
kelvin hong 方kelvin hong 方
83719
$endgroup$
I'm finding the estimate of $$sum_{nleq x}mu(n)log frac xn$$
There is a formula saying that $$sum_{nleq x}f(n)Gbigg(frac xnbigg)=sum_{nleq x}g(n)Fbigg(frac xnbigg)$$
where $F(x)=sum_{nleq x}f(n)$ and $G(x)=sum_{nleq x}g(n)$. So I want to express $log x$ in the form $$log x=sum_{nleq x}g(n)$$
for some function $g$. but what I really know is $$log n=sum_{d|n}mu(d)logdfrac nd$$
not the same as what I want. Any suggestion? I also appreciate different way to do this.
asymptotics analytic-number-theory
asymptotics analytic-number-theory
edited Jan 29 at 20:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 4:04
kelvin hong 方kelvin hong 方
83719
edited Jan 29 at 20:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 4:04
kelvin hong 方kelvin hong 方
83719
edited Jan 29 at 20:07
Martin Sleziak
44.9k10122277
edited Jan 29 at 20:07
Martin Sleziak
44.9k10122277
edited Jan 29 at 20:07
Martin Sleziak
44.9k10122277
44.9k10122277
asked Jan 29 at 4:04
kelvin hong 方kelvin hong 方
83719
asked Jan 29 at 4:04
kelvin hong 方kelvin hong 方
83719
asked Jan 29 at 4:04
kelvin hong 方kelvin hong 方
83719
83719
1
$begingroup$
What is $mu(n)$?
$endgroup$
– Stefan Lafon
Jan 29 at 4:24
$begingroup$
@StefanLafon It is the Möbius Function.
$endgroup$
– kelvin hong 方
Jan 29 at 4:27
2
$begingroup$
$int_1^infty (sum_{n le x} mu(n)(log x -log n)) x^{-s-1}dx = frac{1}{s^2zeta(s)}$ so assuming the zeros are simple and after a lot of estimates to show it converges, there is the explicit formula $sum_{n le x} mu(n)(log x -log n) = sum Res(frac{1}{s^2zeta(s)} x^s) = sum_rho frac{x^rho}{rho^2 zeta'(rho)}-frac{zeta'(0)}{zeta(0)^2}+frac{log x}{zeta(0)} +sum_{k=1}^infty frac{x^{-2k}}{(-2k)^2 zeta'(-2k)}$. The RH is true iff it is $O(x^{1/2+epsilon})$.
$endgroup$
– reuns
Jan 29 at 5:24
add a comment |
1
$begingroup$
What is $mu(n)$?
$endgroup$
– Stefan Lafon
Jan 29 at 4:24
$begingroup$
@StefanLafon It is the Möbius Function.
$endgroup$
– kelvin hong 方
Jan 29 at 4:27
2
$begingroup$
$int_1^infty (sum_{n le x} mu(n)(log x -log n)) x^{-s-1}dx = frac{1}{s^2zeta(s)}$ so assuming the zeros are simple and after a lot of estimates to show it converges, there is the explicit formula $sum_{n le x} mu(n)(log x -log n) = sum Res(frac{1}{s^2zeta(s)} x^s) = sum_rho frac{x^rho}{rho^2 zeta'(rho)}-frac{zeta'(0)}{zeta(0)^2}+frac{log x}{zeta(0)} +sum_{k=1}^infty frac{x^{-2k}}{(-2k)^2 zeta'(-2k)}$. The RH is true iff it is $O(x^{1/2+epsilon})$.
$endgroup$
– reuns
Jan 29 at 5:24
1
1
$begingroup$
What is $mu(n)$?
$endgroup$
– Stefan Lafon
Jan 29 at 4:24
$begingroup$
What is $mu(n)$?
$endgroup$
– Stefan Lafon
Jan 29 at 4:24
$begingroup$
@StefanLafon It is the Möbius Function.
$endgroup$
– kelvin hong 方
Jan 29 at 4:27
$begingroup$
@StefanLafon It is the Möbius Function.
$endgroup$
– kelvin hong 方
Jan 29 at 4:27
2
2
$begingroup$
$int_1^infty (sum_{n le x} mu(n)(log x -log n)) x^{-s-1}dx = frac{1}{s^2zeta(s)}$ so assuming the zeros are simple and after a lot of estimates to show it converges, there is the explicit formula $sum_{n le x} mu(n)(log x -log n) = sum Res(frac{1}{s^2zeta(s)} x^s) = sum_rho frac{x^rho}{rho^2 zeta'(rho)}-frac{zeta'(0)}{zeta(0)^2}+frac{log x}{zeta(0)} +sum_{k=1}^infty frac{x^{-2k}}{(-2k)^2 zeta'(-2k)}$. The RH is true iff it is $O(x^{1/2+epsilon})$.
$endgroup$
– reuns
Jan 29 at 5:24
$begingroup$
$int_1^infty (sum_{n le x} mu(n)(log x -log n)) x^{-s-1}dx = frac{1}{s^2zeta(s)}$ so assuming the zeros are simple and after a lot of estimates to show it converges, there is the explicit formula $sum_{n le x} mu(n)(log x -log n) = sum Res(frac{1}{s^2zeta(s)} x^s) = sum_rho frac{x^rho}{rho^2 zeta'(rho)}-frac{zeta'(0)}{zeta(0)^2}+frac{log x}{zeta(0)} +sum_{k=1}^infty frac{x^{-2k}}{(-2k)^2 zeta'(-2k)}$. The RH is true iff it is $O(x^{1/2+epsilon})$.
$endgroup$
– reuns
Jan 29 at 5:24
add a comment |
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1
$begingroup$
What is $mu(n)$?
$endgroup$
– Stefan Lafon
Jan 29 at 4:24
$begingroup$
@StefanLafon It is the Möbius Function.
$endgroup$
– kelvin hong 方
Jan 29 at 4:27
2
$begingroup$
$int_1^infty (sum_{n le x} mu(n)(log x -log n)) x^{-s-1}dx = frac{1}{s^2zeta(s)}$ so assuming the zeros are simple and after a lot of estimates to show it converges, there is the explicit formula $sum_{n le x} mu(n)(log x -log n) = sum Res(frac{1}{s^2zeta(s)} x^s) = sum_rho frac{x^rho}{rho^2 zeta'(rho)}-frac{zeta'(0)}{zeta(0)^2}+frac{log x}{zeta(0)} +sum_{k=1}^infty frac{x^{-2k}}{(-2k)^2 zeta'(-2k)}$. The RH is true iff it is $O(x^{1/2+epsilon})$.
$endgroup$
– reuns
Jan 29 at 5:24