Mixing
A series whose convergence is equivalent to the Riemann hypothesis
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It was claimed here that the convergence of the series$$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^3 n}tag1$$(where $Lambda$ is the Von Mangoldt function) is equivalent to the Riemann hypothesis. Is this true? That post provided a link to the Wikipedia article about the Von Mangoldt function, which does not mention this. Also, this page about the Von Mangoldt function in the context of the Riemann hypothesis makes no mention to that.
If it is true that the convergence of the series $(1)$ is equivalent to the Riemann hypothesis, then I would like to have a reference for that.
sequences-and-series reference-request riemann-hypothesis
asked Jan 21 at 14:13
José Carlos SantosJosé Carlos Santos
165k22132235
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add a comment |
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It was claimed here that the convergence of the series$$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^3 n}tag1$$(where $Lambda$ is the Von Mangoldt function) is equivalent to the Riemann hypothesis. Is this true? That post provided a link to the Wikipedia article about the Von Mangoldt function, which does not mention this. Also, this page about the Von Mangoldt function in the context of the Riemann hypothesis makes no mention to that.
If it is true that the convergence of the series $(1)$ is equivalent to the Riemann hypothesis, then I would like to have a reference for that.
sequences-and-series reference-request riemann-hypothesis
asked Jan 21 at 14:13
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
It was claimed here that the convergence of the series$$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^3 n}tag1$$(where $Lambda$ is the Von Mangoldt function) is equivalent to the Riemann hypothesis. Is this true? That post provided a link to the Wikipedia article about the Von Mangoldt function, which does not mention this. Also, this page about the Von Mangoldt function in the context of the Riemann hypothesis makes no mention to that.
If it is true that the convergence of the series $(1)$ is equivalent to the Riemann hypothesis, then I would like to have a reference for that.
sequences-and-series reference-request riemann-hypothesis
asked Jan 21 at 14:13
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
It was claimed here that the convergence of the series$$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^3 n}tag1$$(where $Lambda$ is the Von Mangoldt function) is equivalent to the Riemann hypothesis. Is this true? That post provided a link to the Wikipedia article about the Von Mangoldt function, which does not mention this. Also, this page about the Von Mangoldt function in the context of the Riemann hypothesis makes no mention to that.
If it is true that the convergence of the series $(1)$ is equivalent to the Riemann hypothesis, then I would like to have a reference for that.
sequences-and-series reference-request riemann-hypothesis
sequences-and-series reference-request riemann-hypothesis
asked Jan 21 at 14:13
José Carlos SantosJosé Carlos Santos
165k22132235
asked Jan 21 at 14:13
José Carlos SantosJosé Carlos Santos
165k22132235
asked Jan 21 at 14:13
José Carlos SantosJosé Carlos Santos
165k22132235
asked Jan 21 at 14:13
José Carlos SantosJosé Carlos Santos
165k22132235
asked Jan 21 at 14:13
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
add a comment |
add a comment |
2 Answers
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Claymath official description of the Riemann hypothesis claims the RH is true iff $pi(x) = Li(x)+O(x^{1/2}log x)$ so $psi(x) = x+O(x^{1/2}log^2 x)$ and I have been quite sloppy in that it implies the RH is true iff $$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{color{red}{3+epsilon}} n} < inftytag{1}$$
as with partial summation $$sum_{n le x} (Lambda(n)-1) frac{1}{n^{1/2}log^{a} n}=frac{psi(x)-x}{x^{1/2}log^a x}+sum_{n le x-1} (psi(n)-n)O(frac{1}{n^{3/2}log^a n})\ = O(log^{2-a}(x))+sum_{n le x} O(frac{1}{nlog^{a-2} n})$$
The point is to show an effective explicit formula (p.28) $$psi(x) =sum_{n le x} Lambda(n)= x - sum_{|Im(rho)| le T} frac{x^{rho}}{rho}+O(frac{xlog^2 x}{T})=x - sum_{kle K} 2Re(frac{x^{rho_k}}{rho_k})+O(frac{xlog^2 x}{K/log K})$$
where $K = N(T)$ and the density of zeros gives $K sim C T log T,T sim c K/log K$,$Im(rho_k) sim c k/log k$.
In this form, under the RH, with $K = x^{1/2}$ it yields
$$psi(x) =x +O(x^{1/2}log^{2+delta})$$
Plotting those things indicates the series may converge very slowly with $epsilon = 0$ and it is quite certain (under the RH) it converges with $epsilon = 2$.
answered Jan 21 at 18:04
reunsreuns
21.1k21250
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1
$begingroup$
Then I suggest that you delete this answer of yours, since it is not correct.
$endgroup$
– José Carlos Santos
Jan 21 at 18:07
add a comment |
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A good place to start is by reading Terry Tao's post "The Riemann hypothesis in various settings"
answered Jan 21 at 15:47
ConradConrad
67335
$endgroup$
$begingroup$
so (if we can't find a cancellation in the phases of the explicit formula) we really need $log^{3+epsilon} n$ instead of $log^3 n$
$endgroup$
– reuns
Jan 21 at 18:25
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Claymath official description of the Riemann hypothesis claims the RH is true iff $pi(x) = Li(x)+O(x^{1/2}log x)$ so $psi(x) = x+O(x^{1/2}log^2 x)$ and I have been quite sloppy in that it implies the RH is true iff $$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{color{red}{3+epsilon}} n} < inftytag{1}$$
as with partial summation $$sum_{n le x} (Lambda(n)-1) frac{1}{n^{1/2}log^{a} n}=frac{psi(x)-x}{x^{1/2}log^a x}+sum_{n le x-1} (psi(n)-n)O(frac{1}{n^{3/2}log^a n})\ = O(log^{2-a}(x))+sum_{n le x} O(frac{1}{nlog^{a-2} n})$$
The point is to show an effective explicit formula (p.28) $$psi(x) =sum_{n le x} Lambda(n)= x - sum_{|Im(rho)| le T} frac{x^{rho}}{rho}+O(frac{xlog^2 x}{T})=x - sum_{kle K} 2Re(frac{x^{rho_k}}{rho_k})+O(frac{xlog^2 x}{K/log K})$$
where $K = N(T)$ and the density of zeros gives $K sim C T log T,T sim c K/log K$,$Im(rho_k) sim c k/log k$.
In this form, under the RH, with $K = x^{1/2}$ it yields
$$psi(x) =x +O(x^{1/2}log^{2+delta})$$
Plotting those things indicates the series may converge very slowly with $epsilon = 0$ and it is quite certain (under the RH) it converges with $epsilon = 2$.
answered Jan 21 at 18:04
reunsreuns
21.1k21250
$endgroup$
1
$begingroup$
Then I suggest that you delete this answer of yours, since it is not correct.
$endgroup$
– José Carlos Santos
Jan 21 at 18:07
add a comment |
$begingroup$
Claymath official description of the Riemann hypothesis claims the RH is true iff $pi(x) = Li(x)+O(x^{1/2}log x)$ so $psi(x) = x+O(x^{1/2}log^2 x)$ and I have been quite sloppy in that it implies the RH is true iff $$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{color{red}{3+epsilon}} n} < inftytag{1}$$
as with partial summation $$sum_{n le x} (Lambda(n)-1) frac{1}{n^{1/2}log^{a} n}=frac{psi(x)-x}{x^{1/2}log^a x}+sum_{n le x-1} (psi(n)-n)O(frac{1}{n^{3/2}log^a n})\ = O(log^{2-a}(x))+sum_{n le x} O(frac{1}{nlog^{a-2} n})$$
The point is to show an effective explicit formula (p.28) $$psi(x) =sum_{n le x} Lambda(n)= x - sum_{|Im(rho)| le T} frac{x^{rho}}{rho}+O(frac{xlog^2 x}{T})=x - sum_{kle K} 2Re(frac{x^{rho_k}}{rho_k})+O(frac{xlog^2 x}{K/log K})$$
where $K = N(T)$ and the density of zeros gives $K sim C T log T,T sim c K/log K$,$Im(rho_k) sim c k/log k$.
In this form, under the RH, with $K = x^{1/2}$ it yields
$$psi(x) =x +O(x^{1/2}log^{2+delta})$$
Plotting those things indicates the series may converge very slowly with $epsilon = 0$ and it is quite certain (under the RH) it converges with $epsilon = 2$.
answered Jan 21 at 18:04
reunsreuns
21.1k21250
$endgroup$
1
$begingroup$
Then I suggest that you delete this answer of yours, since it is not correct.
$endgroup$
– José Carlos Santos
Jan 21 at 18:07
add a comment |
$begingroup$
Claymath official description of the Riemann hypothesis claims the RH is true iff $pi(x) = Li(x)+O(x^{1/2}log x)$ so $psi(x) = x+O(x^{1/2}log^2 x)$ and I have been quite sloppy in that it implies the RH is true iff $$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{color{red}{3+epsilon}} n} < inftytag{1}$$
as with partial summation $$sum_{n le x} (Lambda(n)-1) frac{1}{n^{1/2}log^{a} n}=frac{psi(x)-x}{x^{1/2}log^a x}+sum_{n le x-1} (psi(n)-n)O(frac{1}{n^{3/2}log^a n})\ = O(log^{2-a}(x))+sum_{n le x} O(frac{1}{nlog^{a-2} n})$$
The point is to show an effective explicit formula (p.28) $$psi(x) =sum_{n le x} Lambda(n)= x - sum_{|Im(rho)| le T} frac{x^{rho}}{rho}+O(frac{xlog^2 x}{T})=x - sum_{kle K} 2Re(frac{x^{rho_k}}{rho_k})+O(frac{xlog^2 x}{K/log K})$$
where $K = N(T)$ and the density of zeros gives $K sim C T log T,T sim c K/log K$,$Im(rho_k) sim c k/log k$.
In this form, under the RH, with $K = x^{1/2}$ it yields
$$psi(x) =x +O(x^{1/2}log^{2+delta})$$
Plotting those things indicates the series may converge very slowly with $epsilon = 0$ and it is quite certain (under the RH) it converges with $epsilon = 2$.
answered Jan 21 at 18:04
reunsreuns
21.1k21250
$endgroup$
Claymath official description of the Riemann hypothesis claims the RH is true iff $pi(x) = Li(x)+O(x^{1/2}log x)$ so $psi(x) = x+O(x^{1/2}log^2 x)$ and I have been quite sloppy in that it implies the RH is true iff $$sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{color{red}{3+epsilon}} n} < inftytag{1}$$
as with partial summation $$sum_{n le x} (Lambda(n)-1) frac{1}{n^{1/2}log^{a} n}=frac{psi(x)-x}{x^{1/2}log^a x}+sum_{n le x-1} (psi(n)-n)O(frac{1}{n^{3/2}log^a n})\ = O(log^{2-a}(x))+sum_{n le x} O(frac{1}{nlog^{a-2} n})$$
The point is to show an effective explicit formula (p.28) $$psi(x) =sum_{n le x} Lambda(n)= x - sum_{|Im(rho)| le T} frac{x^{rho}}{rho}+O(frac{xlog^2 x}{T})=x - sum_{kle K} 2Re(frac{x^{rho_k}}{rho_k})+O(frac{xlog^2 x}{K/log K})$$
where $K = N(T)$ and the density of zeros gives $K sim C T log T,T sim c K/log K$,$Im(rho_k) sim c k/log k$.
In this form, under the RH, with $K = x^{1/2}$ it yields
$$psi(x) =x +O(x^{1/2}log^{2+delta})$$
Plotting those things indicates the series may converge very slowly with $epsilon = 0$ and it is quite certain (under the RH) it converges with $epsilon = 2$.
answered Jan 21 at 18:04
reunsreuns
21.1k21250
edited Jan 22 at 12:58
answered Jan 21 at 18:04
reunsreuns
21.1k21250
answered Jan 21 at 18:04
reunsreuns
21.1k21250
answered Jan 21 at 18:04
reunsreuns
21.1k21250
21.1k21250
1
$begingroup$
Then I suggest that you delete this answer of yours, since it is not correct.
$endgroup$
– José Carlos Santos
Jan 21 at 18:07
add a comment |
1
$begingroup$
Then I suggest that you delete this answer of yours, since it is not correct.
$endgroup$
– José Carlos Santos
Jan 21 at 18:07
1
1
$begingroup$
Then I suggest that you delete this answer of yours, since it is not correct.
$endgroup$
– José Carlos Santos
Jan 21 at 18:07
$begingroup$
Then I suggest that you delete this answer of yours, since it is not correct.
$endgroup$
– José Carlos Santos
Jan 21 at 18:07
add a comment |
$begingroup$
A good place to start is by reading Terry Tao's post "The Riemann hypothesis in various settings"
answered Jan 21 at 15:47
ConradConrad
67335
$endgroup$
$begingroup$
so (if we can't find a cancellation in the phases of the explicit formula) we really need $log^{3+epsilon} n$ instead of $log^3 n$
$endgroup$
– reuns
Jan 21 at 18:25
add a comment |
$begingroup$
A good place to start is by reading Terry Tao's post "The Riemann hypothesis in various settings"
answered Jan 21 at 15:47
ConradConrad
67335
$endgroup$
$begingroup$
so (if we can't find a cancellation in the phases of the explicit formula) we really need $log^{3+epsilon} n$ instead of $log^3 n$
$endgroup$
– reuns
Jan 21 at 18:25
add a comment |
$begingroup$
A good place to start is by reading Terry Tao's post "The Riemann hypothesis in various settings"
answered Jan 21 at 15:47
ConradConrad
67335
$endgroup$
A good place to start is by reading Terry Tao's post "The Riemann hypothesis in various settings"
answered Jan 21 at 15:47
ConradConrad
67335
answered Jan 21 at 15:47
ConradConrad
67335
answered Jan 21 at 15:47
ConradConrad
67335
answered Jan 21 at 15:47
ConradConrad
67335
67335
$begingroup$
so (if we can't find a cancellation in the phases of the explicit formula) we really need $log^{3+epsilon} n$ instead of $log^3 n$
$endgroup$
– reuns
Jan 21 at 18:25
add a comment |
$begingroup$
so (if we can't find a cancellation in the phases of the explicit formula) we really need $log^{3+epsilon} n$ instead of $log^3 n$
$endgroup$
– reuns
Jan 21 at 18:25
$begingroup$
so (if we can't find a cancellation in the phases of the explicit formula) we really need $log^{3+epsilon} n$ instead of $log^3 n$
$endgroup$
– reuns
Jan 21 at 18:25
$begingroup$
so (if we can't find a cancellation in the phases of the explicit formula) we really need $log^{3+epsilon} n$ instead of $log^3 n$
$endgroup$
– reuns
Jan 21 at 18:25
add a comment |
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