An observation on Non-Trivial Zeros of Riemann Zeta Function.
$begingroup$
I observed this property in month of July this year but unable to design a mathematical proof or mathematical way to state my observation. I need help to state this property.
We know that for certain values of 'b' in s = 1/2 + ib , ζ(s) = 0.
When I observed most of the values of 'b' here
I found that most of the primes are related to these values of 'b' in there square forms like:
Property: [b] = p^2 {where 'p' is a prime number and [b] is the Box Function}
example: b = 841.0363...
so, [841.0363...] = 29^2
below 200 there are only 6 primes which are not following the this above property.
I am searching a formula or a program to find out all those primes which follow above property but till now I didn't got any solution. I am also not good in programming please help me in this problem.
number-theory prime-numbers riemann-zeta zeta-functions
asked Dec 17 '18 at 13:50
DynamoDynamo
104517
$endgroup$
add a comment |
$begingroup$
I observed this property in month of July this year but unable to design a mathematical proof or mathematical way to state my observation. I need help to state this property.
We know that for certain values of 'b' in s = 1/2 + ib , ζ(s) = 0.
When I observed most of the values of 'b' here
I found that most of the primes are related to these values of 'b' in there square forms like:
Property: [b] = p^2 {where 'p' is a prime number and [b] is the Box Function}
example: b = 841.0363...
so, [841.0363...] = 29^2
below 200 there are only 6 primes which are not following the this above property.
I am searching a formula or a program to find out all those primes which follow above property but till now I didn't got any solution. I am also not good in programming please help me in this problem.
number-theory prime-numbers riemann-zeta zeta-functions
asked Dec 17 '18 at 13:50
DynamoDynamo
104517
$endgroup$
add a comment |
$begingroup$
I observed this property in month of July this year but unable to design a mathematical proof or mathematical way to state my observation. I need help to state this property.
We know that for certain values of 'b' in s = 1/2 + ib , ζ(s) = 0.
When I observed most of the values of 'b' here
I found that most of the primes are related to these values of 'b' in there square forms like:
Property: [b] = p^2 {where 'p' is a prime number and [b] is the Box Function}
example: b = 841.0363...
so, [841.0363...] = 29^2
below 200 there are only 6 primes which are not following the this above property.
I am searching a formula or a program to find out all those primes which follow above property but till now I didn't got any solution. I am also not good in programming please help me in this problem.
number-theory prime-numbers riemann-zeta zeta-functions
asked Dec 17 '18 at 13:50
DynamoDynamo
104517
$endgroup$
I observed this property in month of July this year but unable to design a mathematical proof or mathematical way to state my observation. I need help to state this property.
We know that for certain values of 'b' in s = 1/2 + ib , ζ(s) = 0.
When I observed most of the values of 'b' here
I found that most of the primes are related to these values of 'b' in there square forms like:
Property: [b] = p^2 {where 'p' is a prime number and [b] is the Box Function}
example: b = 841.0363...
so, [841.0363...] = 29^2
below 200 there are only 6 primes which are not following the this above property.
I am searching a formula or a program to find out all those primes which follow above property but till now I didn't got any solution. I am also not good in programming please help me in this problem.
number-theory prime-numbers riemann-zeta zeta-functions
number-theory prime-numbers riemann-zeta zeta-functions
asked Dec 17 '18 at 13:50
DynamoDynamo
104517
asked Dec 17 '18 at 13:50
DynamoDynamo
104517
asked Dec 17 '18 at 13:50
DynamoDynamo
104517
asked Dec 17 '18 at 13:50
DynamoDynamo
104517
asked Dec 17 '18 at 13:50
DynamoDynamo
104517
104517
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The $n$th zeta zero has $b(n) sim 2pi n/W(n/e)$, where $W$ is the Lambert function. The Lambert function grows as roughly the logarithm of its argument, so we see that the zeta zeroes get denser as $n$ increases, and at $n approx 10^4$, their imaginary parts should start hitting every integer. This turns out to be the case, as the last square of a prime it misses is $103^2 = 10609$.
We can also estimate the probability of each square of a prime not having a zeta zero near it. The density of zeta zeroes is $W(n/e)/(2pi)$, so we expect each square of a prime to not be near a zeta zero with probability $1 - W(p^2/e)/(2pi)$. Summing this over all primes where this value is positive, we get an expected count of $7.7$, which is reasonably close to the actual value of 6.
answered Dec 17 '18 at 15:43
eyeballfrogeyeballfrog
6,228629
$endgroup$
$begingroup$
Thanks for your answer but are you sure that after n=10^4 their imaginary part start hitting every integer. If yes then please tell me where I can see its proof.
$endgroup$
– Dynamo
Dec 17 '18 at 16:02
$begingroup$
@AdarshKumar My statement is probably a little too strong, but it is known that there is some $n$ after which $[b(n)]$ hits every integer. It is also known how to count the number of zeros with imaginary part less than $b$, with formula given here. Keeping only the dominant terms and inverting gives my estimate in the answer. It is possible there are integers missed that are significantly larger than $10^4$, but prime squares have very low density at large $n$, making them unlikely to be counterexamples anyways.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:09
$begingroup$
so it means that after some n which you say that will surely exist after which b(n) hits every integer than it is also possible all primes whose square is greater than n also exist . So my property is true till infinite primes. ??
$endgroup$
– Dynamo
Dec 17 '18 at 17:15
$begingroup$
Yes, this property will hold for all sufficiently large primes. Though since this is true because it holds for all sufficiently large integers, of which primes are a subset, I'm not sure how much this tells us about primes.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:18
$begingroup$
ok thanks a lot to giving your time to solve my problem.
$endgroup$
– Dynamo
Dec 17 '18 at 17:21
|
show 1 more comment
Your Answer
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1 Answer
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active
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1 Answer
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oldest
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active
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votes
$begingroup$
The $n$th zeta zero has $b(n) sim 2pi n/W(n/e)$, where $W$ is the Lambert function. The Lambert function grows as roughly the logarithm of its argument, so we see that the zeta zeroes get denser as $n$ increases, and at $n approx 10^4$, their imaginary parts should start hitting every integer. This turns out to be the case, as the last square of a prime it misses is $103^2 = 10609$.
We can also estimate the probability of each square of a prime not having a zeta zero near it. The density of zeta zeroes is $W(n/e)/(2pi)$, so we expect each square of a prime to not be near a zeta zero with probability $1 - W(p^2/e)/(2pi)$. Summing this over all primes where this value is positive, we get an expected count of $7.7$, which is reasonably close to the actual value of 6.
answered Dec 17 '18 at 15:43
eyeballfrogeyeballfrog
6,228629
$endgroup$
$begingroup$
Thanks for your answer but are you sure that after n=10^4 their imaginary part start hitting every integer. If yes then please tell me where I can see its proof.
$endgroup$
– Dynamo
Dec 17 '18 at 16:02
$begingroup$
@AdarshKumar My statement is probably a little too strong, but it is known that there is some $n$ after which $[b(n)]$ hits every integer. It is also known how to count the number of zeros with imaginary part less than $b$, with formula given here. Keeping only the dominant terms and inverting gives my estimate in the answer. It is possible there are integers missed that are significantly larger than $10^4$, but prime squares have very low density at large $n$, making them unlikely to be counterexamples anyways.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:09
$begingroup$
so it means that after some n which you say that will surely exist after which b(n) hits every integer than it is also possible all primes whose square is greater than n also exist . So my property is true till infinite primes. ??
$endgroup$
– Dynamo
Dec 17 '18 at 17:15
$begingroup$
Yes, this property will hold for all sufficiently large primes. Though since this is true because it holds for all sufficiently large integers, of which primes are a subset, I'm not sure how much this tells us about primes.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:18
$begingroup$
ok thanks a lot to giving your time to solve my problem.
$endgroup$
– Dynamo
Dec 17 '18 at 17:21
|
show 1 more comment
$begingroup$
The $n$th zeta zero has $b(n) sim 2pi n/W(n/e)$, where $W$ is the Lambert function. The Lambert function grows as roughly the logarithm of its argument, so we see that the zeta zeroes get denser as $n$ increases, and at $n approx 10^4$, their imaginary parts should start hitting every integer. This turns out to be the case, as the last square of a prime it misses is $103^2 = 10609$.
We can also estimate the probability of each square of a prime not having a zeta zero near it. The density of zeta zeroes is $W(n/e)/(2pi)$, so we expect each square of a prime to not be near a zeta zero with probability $1 - W(p^2/e)/(2pi)$. Summing this over all primes where this value is positive, we get an expected count of $7.7$, which is reasonably close to the actual value of 6.
answered Dec 17 '18 at 15:43
eyeballfrogeyeballfrog
6,228629
$endgroup$
$begingroup$
Thanks for your answer but are you sure that after n=10^4 their imaginary part start hitting every integer. If yes then please tell me where I can see its proof.
$endgroup$
– Dynamo
Dec 17 '18 at 16:02
$begingroup$
@AdarshKumar My statement is probably a little too strong, but it is known that there is some $n$ after which $[b(n)]$ hits every integer. It is also known how to count the number of zeros with imaginary part less than $b$, with formula given here. Keeping only the dominant terms and inverting gives my estimate in the answer. It is possible there are integers missed that are significantly larger than $10^4$, but prime squares have very low density at large $n$, making them unlikely to be counterexamples anyways.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:09
$begingroup$
so it means that after some n which you say that will surely exist after which b(n) hits every integer than it is also possible all primes whose square is greater than n also exist . So my property is true till infinite primes. ??
$endgroup$
– Dynamo
Dec 17 '18 at 17:15
$begingroup$
Yes, this property will hold for all sufficiently large primes. Though since this is true because it holds for all sufficiently large integers, of which primes are a subset, I'm not sure how much this tells us about primes.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:18
$begingroup$
ok thanks a lot to giving your time to solve my problem.
$endgroup$
– Dynamo
Dec 17 '18 at 17:21
|
show 1 more comment
$begingroup$
The $n$th zeta zero has $b(n) sim 2pi n/W(n/e)$, where $W$ is the Lambert function. The Lambert function grows as roughly the logarithm of its argument, so we see that the zeta zeroes get denser as $n$ increases, and at $n approx 10^4$, their imaginary parts should start hitting every integer. This turns out to be the case, as the last square of a prime it misses is $103^2 = 10609$.
We can also estimate the probability of each square of a prime not having a zeta zero near it. The density of zeta zeroes is $W(n/e)/(2pi)$, so we expect each square of a prime to not be near a zeta zero with probability $1 - W(p^2/e)/(2pi)$. Summing this over all primes where this value is positive, we get an expected count of $7.7$, which is reasonably close to the actual value of 6.
answered Dec 17 '18 at 15:43
eyeballfrogeyeballfrog
6,228629
$endgroup$
The $n$th zeta zero has $b(n) sim 2pi n/W(n/e)$, where $W$ is the Lambert function. The Lambert function grows as roughly the logarithm of its argument, so we see that the zeta zeroes get denser as $n$ increases, and at $n approx 10^4$, their imaginary parts should start hitting every integer. This turns out to be the case, as the last square of a prime it misses is $103^2 = 10609$.
We can also estimate the probability of each square of a prime not having a zeta zero near it. The density of zeta zeroes is $W(n/e)/(2pi)$, so we expect each square of a prime to not be near a zeta zero with probability $1 - W(p^2/e)/(2pi)$. Summing this over all primes where this value is positive, we get an expected count of $7.7$, which is reasonably close to the actual value of 6.
answered Dec 17 '18 at 15:43
eyeballfrogeyeballfrog
6,228629
answered Dec 17 '18 at 15:43
eyeballfrogeyeballfrog
6,228629
answered Dec 17 '18 at 15:43
eyeballfrogeyeballfrog
6,228629
answered Dec 17 '18 at 15:43
eyeballfrogeyeballfrog
6,228629
6,228629
$begingroup$
Thanks for your answer but are you sure that after n=10^4 their imaginary part start hitting every integer. If yes then please tell me where I can see its proof.
$endgroup$
– Dynamo
Dec 17 '18 at 16:02
$begingroup$
@AdarshKumar My statement is probably a little too strong, but it is known that there is some $n$ after which $[b(n)]$ hits every integer. It is also known how to count the number of zeros with imaginary part less than $b$, with formula given here. Keeping only the dominant terms and inverting gives my estimate in the answer. It is possible there are integers missed that are significantly larger than $10^4$, but prime squares have very low density at large $n$, making them unlikely to be counterexamples anyways.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:09
$begingroup$
so it means that after some n which you say that will surely exist after which b(n) hits every integer than it is also possible all primes whose square is greater than n also exist . So my property is true till infinite primes. ??
$endgroup$
– Dynamo
Dec 17 '18 at 17:15
$begingroup$
Yes, this property will hold for all sufficiently large primes. Though since this is true because it holds for all sufficiently large integers, of which primes are a subset, I'm not sure how much this tells us about primes.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:18
$begingroup$
ok thanks a lot to giving your time to solve my problem.
$endgroup$
– Dynamo
Dec 17 '18 at 17:21
|
show 1 more comment
$begingroup$
Thanks for your answer but are you sure that after n=10^4 their imaginary part start hitting every integer. If yes then please tell me where I can see its proof.
$endgroup$
– Dynamo
Dec 17 '18 at 16:02
$begingroup$
@AdarshKumar My statement is probably a little too strong, but it is known that there is some $n$ after which $[b(n)]$ hits every integer. It is also known how to count the number of zeros with imaginary part less than $b$, with formula given here. Keeping only the dominant terms and inverting gives my estimate in the answer. It is possible there are integers missed that are significantly larger than $10^4$, but prime squares have very low density at large $n$, making them unlikely to be counterexamples anyways.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:09
$begingroup$
so it means that after some n which you say that will surely exist after which b(n) hits every integer than it is also possible all primes whose square is greater than n also exist . So my property is true till infinite primes. ??
$endgroup$
– Dynamo
Dec 17 '18 at 17:15
$begingroup$
Yes, this property will hold for all sufficiently large primes. Though since this is true because it holds for all sufficiently large integers, of which primes are a subset, I'm not sure how much this tells us about primes.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:18
$begingroup$
ok thanks a lot to giving your time to solve my problem.
$endgroup$
– Dynamo
Dec 17 '18 at 17:21
$begingroup$
Thanks for your answer but are you sure that after n=10^4 their imaginary part start hitting every integer. If yes then please tell me where I can see its proof.
$endgroup$
– Dynamo
Dec 17 '18 at 16:02
$begingroup$
Thanks for your answer but are you sure that after n=10^4 their imaginary part start hitting every integer. If yes then please tell me where I can see its proof.
$endgroup$
– Dynamo
Dec 17 '18 at 16:02
$begingroup$
@AdarshKumar My statement is probably a little too strong, but it is known that there is some $n$ after which $[b(n)]$ hits every integer. It is also known how to count the number of zeros with imaginary part less than $b$, with formula given here. Keeping only the dominant terms and inverting gives my estimate in the answer. It is possible there are integers missed that are significantly larger than $10^4$, but prime squares have very low density at large $n$, making them unlikely to be counterexamples anyways.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:09
$begingroup$
@AdarshKumar My statement is probably a little too strong, but it is known that there is some $n$ after which $[b(n)]$ hits every integer. It is also known how to count the number of zeros with imaginary part less than $b$, with formula given here. Keeping only the dominant terms and inverting gives my estimate in the answer. It is possible there are integers missed that are significantly larger than $10^4$, but prime squares have very low density at large $n$, making them unlikely to be counterexamples anyways.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:09
$begingroup$
so it means that after some n which you say that will surely exist after which b(n) hits every integer than it is also possible all primes whose square is greater than n also exist . So my property is true till infinite primes. ??
$endgroup$
– Dynamo
Dec 17 '18 at 17:15
$begingroup$
so it means that after some n which you say that will surely exist after which b(n) hits every integer than it is also possible all primes whose square is greater than n also exist . So my property is true till infinite primes. ??
$endgroup$
– Dynamo
Dec 17 '18 at 17:15
$begingroup$
Yes, this property will hold for all sufficiently large primes. Though since this is true because it holds for all sufficiently large integers, of which primes are a subset, I'm not sure how much this tells us about primes.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:18
$begingroup$
Yes, this property will hold for all sufficiently large primes. Though since this is true because it holds for all sufficiently large integers, of which primes are a subset, I'm not sure how much this tells us about primes.
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:18
$begingroup$
ok thanks a lot to giving your time to solve my problem.
$endgroup$
– Dynamo
Dec 17 '18 at 17:21
$begingroup$
ok thanks a lot to giving your time to solve my problem.
$endgroup$
– Dynamo
Dec 17 '18 at 17:21
|
show 1 more comment
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