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Why don’t mathematicians work on ‘difference-asymptotic’ of prime counting function?
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Mathematicians back in 19th century tried to find a function that satisfies $$lim_xtoinftyfracpi(x)f(x)=1$$ and $f(x)$ turns out to be $fracxln x$, or any function asymptotic to it(like $textLi(x)$). They proved it rigorously and now it is known as the Prime Number Theorem.
However, I don’t see much work on finding a function that satisfies $$lim_xtoinfty(pi(x)-g(x))=0$$ As far as I know, $$lim_xtoinfty(pi(x)-fracxln x)=infty$$ so $fracxln x$ cannot be a candidate of $g(x)$.
Moreover, if such function is discovered, it will be very useful in the sense that estimation of number of primes below some large $N$ can become more and more precise as $N$ grows. This would surely be more powerful than PNT.
Why only little work has done by mathematicians to figure out $g(x)$?
prime-numbers
asked Jul 24 at 0:42
Szeto
4,1281521
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up vote
4
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Mathematicians back in 19th century tried to find a function that satisfies $$lim_xtoinftyfracpi(x)f(x)=1$$ and $f(x)$ turns out to be $fracxln x$, or any function asymptotic to it(like $textLi(x)$). They proved it rigorously and now it is known as the Prime Number Theorem.
However, I don’t see much work on finding a function that satisfies $$lim_xtoinfty(pi(x)-g(x))=0$$ As far as I know, $$lim_xtoinfty(pi(x)-fracxln x)=infty$$ so $fracxln x$ cannot be a candidate of $g(x)$.
Moreover, if such function is discovered, it will be very useful in the sense that estimation of number of primes below some large $N$ can become more and more precise as $N$ grows. This would surely be more powerful than PNT.
Why only little work has done by mathematicians to figure out $g(x)$?
prime-numbers
asked Jul 24 at 0:42
Szeto
4,1281521
7
Well, even the Riemann hypothesis implies only $pi(x) - textli(x) = O(sqrtxlog x).$ Since $textli(x)$ is continuous and $pi(x)$ is highly discontinuous, studying the difference isn't very profitable. It seems to be to be almost equivalent to asking, "what numbers are prime?"
– Dzoooks
Jul 24 at 0:52
But, we can contrast the study of $pi(x)$ with the partition function: en.wikipedia.org/wiki/… , which has (an exact!) $g(x)$ in the form of an infinite series.
– Dzoooks
Jul 24 at 0:57
4
Such a function would essentially encode all information about sufficiently large prime numbers. For example, you could tell whether $x$ was prime by looking at $g(x)$ and $g(x-1)$.
– Jair Taylor
Jul 24 at 1:19
1
@Jair: So basically, it'd have to be $pi(x)$ up to some lower order error term that vanishes in the limit. That's actually pretty boring, unless you've got a closed form of some kind.
– Kevin
Jul 24 at 5:11
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Mathematicians back in 19th century tried to find a function that satisfies $$lim_xtoinftyfracpi(x)f(x)=1$$ and $f(x)$ turns out to be $fracxln x$, or any function asymptotic to it(like $textLi(x)$). They proved it rigorously and now it is known as the Prime Number Theorem.
However, I don’t see much work on finding a function that satisfies $$lim_xtoinfty(pi(x)-g(x))=0$$ As far as I know, $$lim_xtoinfty(pi(x)-fracxln x)=infty$$ so $fracxln x$ cannot be a candidate of $g(x)$.
Moreover, if such function is discovered, it will be very useful in the sense that estimation of number of primes below some large $N$ can become more and more precise as $N$ grows. This would surely be more powerful than PNT.
Why only little work has done by mathematicians to figure out $g(x)$?
prime-numbers
asked Jul 24 at 0:42
Szeto
4,1281521
Mathematicians back in 19th century tried to find a function that satisfies $$lim_xtoinftyfracpi(x)f(x)=1$$ and $f(x)$ turns out to be $fracxln x$, or any function asymptotic to it(like $textLi(x)$). They proved it rigorously and now it is known as the Prime Number Theorem.
However, I don’t see much work on finding a function that satisfies $$lim_xtoinfty(pi(x)-g(x))=0$$ As far as I know, $$lim_xtoinfty(pi(x)-fracxln x)=infty$$ so $fracxln x$ cannot be a candidate of $g(x)$.
Moreover, if such function is discovered, it will be very useful in the sense that estimation of number of primes below some large $N$ can become more and more precise as $N$ grows. This would surely be more powerful than PNT.
Why only little work has done by mathematicians to figure out $g(x)$?
prime-numbers
asked Jul 24 at 0:42
Szeto
4,1281521
asked Jul 24 at 0:42
Szeto
4,1281521
asked Jul 24 at 0:42
Szeto
4,1281521
asked Jul 24 at 0:42
Szeto
4,1281521
4,1281521
7
Well, even the Riemann hypothesis implies only $pi(x) - textli(x) = O(sqrtxlog x).$ Since $textli(x)$ is continuous and $pi(x)$ is highly discontinuous, studying the difference isn't very profitable. It seems to be to be almost equivalent to asking, "what numbers are prime?"
– Dzoooks
Jul 24 at 0:52
But, we can contrast the study of $pi(x)$ with the partition function: en.wikipedia.org/wiki/… , which has (an exact!) $g(x)$ in the form of an infinite series.
– Dzoooks
Jul 24 at 0:57
4
Such a function would essentially encode all information about sufficiently large prime numbers. For example, you could tell whether $x$ was prime by looking at $g(x)$ and $g(x-1)$.
– Jair Taylor
Jul 24 at 1:19
1
@Jair: So basically, it'd have to be $pi(x)$ up to some lower order error term that vanishes in the limit. That's actually pretty boring, unless you've got a closed form of some kind.
– Kevin
Jul 24 at 5:11
add a comment |Â
7
Well, even the Riemann hypothesis implies only $pi(x) - textli(x) = O(sqrtxlog x).$ Since $textli(x)$ is continuous and $pi(x)$ is highly discontinuous, studying the difference isn't very profitable. It seems to be to be almost equivalent to asking, "what numbers are prime?"
– Dzoooks
Jul 24 at 0:52
But, we can contrast the study of $pi(x)$ with the partition function: en.wikipedia.org/wiki/… , which has (an exact!) $g(x)$ in the form of an infinite series.
– Dzoooks
Jul 24 at 0:57
4
Such a function would essentially encode all information about sufficiently large prime numbers. For example, you could tell whether $x$ was prime by looking at $g(x)$ and $g(x-1)$.
– Jair Taylor
Jul 24 at 1:19
1
@Jair: So basically, it'd have to be $pi(x)$ up to some lower order error term that vanishes in the limit. That's actually pretty boring, unless you've got a closed form of some kind.
– Kevin
Jul 24 at 5:11
7
7
Well, even the Riemann hypothesis implies only $pi(x) - textli(x) = O(sqrtxlog x).$ Since $textli(x)$ is continuous and $pi(x)$ is highly discontinuous, studying the difference isn't very profitable. It seems to be to be almost equivalent to asking, "what numbers are prime?"
– Dzoooks
Jul 24 at 0:52
Well, even the Riemann hypothesis implies only $pi(x) - textli(x) = O(sqrtxlog x).$ Since $textli(x)$ is continuous and $pi(x)$ is highly discontinuous, studying the difference isn't very profitable. It seems to be to be almost equivalent to asking, "what numbers are prime?"
– Dzoooks
Jul 24 at 0:52
But, we can contrast the study of $pi(x)$ with the partition function: en.wikipedia.org/wiki/… , which has (an exact!) $g(x)$ in the form of an infinite series.
– Dzoooks
Jul 24 at 0:57
But, we can contrast the study of $pi(x)$ with the partition function: en.wikipedia.org/wiki/… , which has (an exact!) $g(x)$ in the form of an infinite series.
– Dzoooks
Jul 24 at 0:57
4
4
Such a function would essentially encode all information about sufficiently large prime numbers. For example, you could tell whether $x$ was prime by looking at $g(x)$ and $g(x-1)$.
– Jair Taylor
Jul 24 at 1:19
Such a function would essentially encode all information about sufficiently large prime numbers. For example, you could tell whether $x$ was prime by looking at $g(x)$ and $g(x-1)$.
– Jair Taylor
Jul 24 at 1:19
1
1
@Jair: So basically, it'd have to be $pi(x)$ up to some lower order error term that vanishes in the limit. That's actually pretty boring, unless you've got a closed form of some kind.
– Kevin
Jul 24 at 5:11
@Jair: So basically, it'd have to be $pi(x)$ up to some lower order error term that vanishes in the limit. That's actually pretty boring, unless you've got a closed form of some kind.
– Kevin
Jul 24 at 5:11
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
In brief, because what you're suggesting is overwhelmingly more powerful than the multiplicative difference, to the point where none of the known techniques can even come close. It's not that this isn't studied; indeed, 'additive differences' on the PNT and related functions - but as noted in a comment, they're usually only as good as being able to say $pi(x)=f(x)+O(x^alpha)$ for some $alpha$ (typically with $alphagt 1/2$). Note that these sorts of asymptotics imply the 'multiplicative' equalities you mention (since then $pi(x)/f(x)=1+O^*(x^alpha-1)$), but the additive results you're requesting are even stronger than reducing $alpha$ to zero.
answered Jul 24 at 0:57
Steven Stadnicki
40.1k765119
add a comment |Â
up vote
4
down vote
The question you are suggesting, in its exact form, would basically be a lost fight. Should you be able to find such function $g(x)$, this would mean
$$forall epsilon > 0 exists x_0 in mathbbN forall x in mathbbN, x > x_0: |pi(x) - g(x)| < epsilon.$$
If you plug in $epsilon = 1/2$, this tells you that from some $x_0$ on the function value $g(x)$ differs from the $x$-th prime by less than a half, in other words, that you could just round it to the nearest integer to get the $x$-th prime. I doubt anyone nowadays hopes that a closed form like that could be found.
answered Jul 24 at 6:53
The Vee
2,135723
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
In brief, because what you're suggesting is overwhelmingly more powerful than the multiplicative difference, to the point where none of the known techniques can even come close. It's not that this isn't studied; indeed, 'additive differences' on the PNT and related functions - but as noted in a comment, they're usually only as good as being able to say $pi(x)=f(x)+O(x^alpha)$ for some $alpha$ (typically with $alphagt 1/2$). Note that these sorts of asymptotics imply the 'multiplicative' equalities you mention (since then $pi(x)/f(x)=1+O^*(x^alpha-1)$), but the additive results you're requesting are even stronger than reducing $alpha$ to zero.
answered Jul 24 at 0:57
Steven Stadnicki
40.1k765119
add a comment |Â
up vote
9
down vote
accepted
In brief, because what you're suggesting is overwhelmingly more powerful than the multiplicative difference, to the point where none of the known techniques can even come close. It's not that this isn't studied; indeed, 'additive differences' on the PNT and related functions - but as noted in a comment, they're usually only as good as being able to say $pi(x)=f(x)+O(x^alpha)$ for some $alpha$ (typically with $alphagt 1/2$). Note that these sorts of asymptotics imply the 'multiplicative' equalities you mention (since then $pi(x)/f(x)=1+O^*(x^alpha-1)$), but the additive results you're requesting are even stronger than reducing $alpha$ to zero.
answered Jul 24 at 0:57
Steven Stadnicki
40.1k765119
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
In brief, because what you're suggesting is overwhelmingly more powerful than the multiplicative difference, to the point where none of the known techniques can even come close. It's not that this isn't studied; indeed, 'additive differences' on the PNT and related functions - but as noted in a comment, they're usually only as good as being able to say $pi(x)=f(x)+O(x^alpha)$ for some $alpha$ (typically with $alphagt 1/2$). Note that these sorts of asymptotics imply the 'multiplicative' equalities you mention (since then $pi(x)/f(x)=1+O^*(x^alpha-1)$), but the additive results you're requesting are even stronger than reducing $alpha$ to zero.
answered Jul 24 at 0:57
Steven Stadnicki
40.1k765119
In brief, because what you're suggesting is overwhelmingly more powerful than the multiplicative difference, to the point where none of the known techniques can even come close. It's not that this isn't studied; indeed, 'additive differences' on the PNT and related functions - but as noted in a comment, they're usually only as good as being able to say $pi(x)=f(x)+O(x^alpha)$ for some $alpha$ (typically with $alphagt 1/2$). Note that these sorts of asymptotics imply the 'multiplicative' equalities you mention (since then $pi(x)/f(x)=1+O^*(x^alpha-1)$), but the additive results you're requesting are even stronger than reducing $alpha$ to zero.
answered Jul 24 at 0:57
Steven Stadnicki
40.1k765119
edited Jul 24 at 1:05
answered Jul 24 at 0:57
Steven Stadnicki
40.1k765119
answered Jul 24 at 0:57
Steven Stadnicki
40.1k765119
answered Jul 24 at 0:57
Steven Stadnicki
40.1k765119
40.1k765119
add a comment |Â
add a comment |Â
up vote
4
down vote
The question you are suggesting, in its exact form, would basically be a lost fight. Should you be able to find such function $g(x)$, this would mean
$$forall epsilon > 0 exists x_0 in mathbbN forall x in mathbbN, x > x_0: |pi(x) - g(x)| < epsilon.$$
If you plug in $epsilon = 1/2$, this tells you that from some $x_0$ on the function value $g(x)$ differs from the $x$-th prime by less than a half, in other words, that you could just round it to the nearest integer to get the $x$-th prime. I doubt anyone nowadays hopes that a closed form like that could be found.
answered Jul 24 at 6:53
The Vee
2,135723
add a comment |Â
up vote
4
down vote
The question you are suggesting, in its exact form, would basically be a lost fight. Should you be able to find such function $g(x)$, this would mean
$$forall epsilon > 0 exists x_0 in mathbbN forall x in mathbbN, x > x_0: |pi(x) - g(x)| < epsilon.$$
If you plug in $epsilon = 1/2$, this tells you that from some $x_0$ on the function value $g(x)$ differs from the $x$-th prime by less than a half, in other words, that you could just round it to the nearest integer to get the $x$-th prime. I doubt anyone nowadays hopes that a closed form like that could be found.
answered Jul 24 at 6:53
The Vee
2,135723
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The question you are suggesting, in its exact form, would basically be a lost fight. Should you be able to find such function $g(x)$, this would mean
$$forall epsilon > 0 exists x_0 in mathbbN forall x in mathbbN, x > x_0: |pi(x) - g(x)| < epsilon.$$
If you plug in $epsilon = 1/2$, this tells you that from some $x_0$ on the function value $g(x)$ differs from the $x$-th prime by less than a half, in other words, that you could just round it to the nearest integer to get the $x$-th prime. I doubt anyone nowadays hopes that a closed form like that could be found.
answered Jul 24 at 6:53
The Vee
2,135723
The question you are suggesting, in its exact form, would basically be a lost fight. Should you be able to find such function $g(x)$, this would mean
$$forall epsilon > 0 exists x_0 in mathbbN forall x in mathbbN, x > x_0: |pi(x) - g(x)| < epsilon.$$
If you plug in $epsilon = 1/2$, this tells you that from some $x_0$ on the function value $g(x)$ differs from the $x$-th prime by less than a half, in other words, that you could just round it to the nearest integer to get the $x$-th prime. I doubt anyone nowadays hopes that a closed form like that could be found.
answered Jul 24 at 6:53
The Vee
2,135723
answered Jul 24 at 6:53
The Vee
2,135723
answered Jul 24 at 6:53
The Vee
2,135723
answered Jul 24 at 6:53
The Vee
2,135723
2,135723
add a comment |Â
add a comment |Â
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7
Well, even the Riemann hypothesis implies only $pi(x) - textli(x) = O(sqrtxlog x).$ Since $textli(x)$ is continuous and $pi(x)$ is highly discontinuous, studying the difference isn't very profitable. It seems to be to be almost equivalent to asking, "what numbers are prime?"
– Dzoooks
Jul 24 at 0:52
But, we can contrast the study of $pi(x)$ with the partition function: en.wikipedia.org/wiki/… , which has (an exact!) $g(x)$ in the form of an infinite series.
– Dzoooks
Jul 24 at 0:57
4
Such a function would essentially encode all information about sufficiently large prime numbers. For example, you could tell whether $x$ was prime by looking at $g(x)$ and $g(x-1)$.
– Jair Taylor
Jul 24 at 1:19
1
@Jair: So basically, it'd have to be $pi(x)$ up to some lower order error term that vanishes in the limit. That's actually pretty boring, unless you've got a closed form of some kind.
– Kevin
Jul 24 at 5:11