Understanding the error term in the Siegel‒Walfisz theorem
$begingroup$
I'd like to inquire as to the nature of the error term in the Siegel‒Walfisz theorem. I understand that it takes the form
$$
Oleft( x expleft( -C sqrt{ln x} right) right).
$$
Yet this function is not one of those that one studied in one's first term calculus course and knows everything about. (For instance, a version of the prime number theorem gives an error term of $x/(ln x)^2$, which is a function that seems more familiar.)
My question is thus: How fast does the error term in the Siegel‒Walfisz theorem grow? How does it compare to better known error terms? (I'd be particularly interested in the error term $x/(ln x)^2$.)
For instance, writing
$$
x expleft( -C sqrt{ln x} right) = frac{x}{x^{C/sqrt{ln x}}},
$$
we find that it eventually supersedes $x^alpha$ for any $alpha < 1$.
calculus asymptotics analytic-number-theory
edited Jan 11 at 15:32
quid♦
36.9k95093
asked Jan 11 at 15:17
AlgebraicsAnonymousAlgebraicsAnonymous
1,067112
$endgroup$
add a comment |
$begingroup$
I'd like to inquire as to the nature of the error term in the Siegel‒Walfisz theorem. I understand that it takes the form
$$
Oleft( x expleft( -C sqrt{ln x} right) right).
$$
Yet this function is not one of those that one studied in one's first term calculus course and knows everything about. (For instance, a version of the prime number theorem gives an error term of $x/(ln x)^2$, which is a function that seems more familiar.)
My question is thus: How fast does the error term in the Siegel‒Walfisz theorem grow? How does it compare to better known error terms? (I'd be particularly interested in the error term $x/(ln x)^2$.)
For instance, writing
$$
x expleft( -C sqrt{ln x} right) = frac{x}{x^{C/sqrt{ln x}}},
$$
we find that it eventually supersedes $x^alpha$ for any $alpha < 1$.
calculus asymptotics analytic-number-theory
edited Jan 11 at 15:32
quid♦
36.9k95093
asked Jan 11 at 15:17
AlgebraicsAnonymousAlgebraicsAnonymous
1,067112
$endgroup$
add a comment |
$begingroup$
I'd like to inquire as to the nature of the error term in the Siegel‒Walfisz theorem. I understand that it takes the form
$$
Oleft( x expleft( -C sqrt{ln x} right) right).
$$
Yet this function is not one of those that one studied in one's first term calculus course and knows everything about. (For instance, a version of the prime number theorem gives an error term of $x/(ln x)^2$, which is a function that seems more familiar.)
My question is thus: How fast does the error term in the Siegel‒Walfisz theorem grow? How does it compare to better known error terms? (I'd be particularly interested in the error term $x/(ln x)^2$.)
For instance, writing
$$
x expleft( -C sqrt{ln x} right) = frac{x}{x^{C/sqrt{ln x}}},
$$
we find that it eventually supersedes $x^alpha$ for any $alpha < 1$.
calculus asymptotics analytic-number-theory
edited Jan 11 at 15:32
quid♦
36.9k95093
asked Jan 11 at 15:17
AlgebraicsAnonymousAlgebraicsAnonymous
1,067112
$endgroup$
I'd like to inquire as to the nature of the error term in the Siegel‒Walfisz theorem. I understand that it takes the form
$$
Oleft( x expleft( -C sqrt{ln x} right) right).
$$
Yet this function is not one of those that one studied in one's first term calculus course and knows everything about. (For instance, a version of the prime number theorem gives an error term of $x/(ln x)^2$, which is a function that seems more familiar.)
My question is thus: How fast does the error term in the Siegel‒Walfisz theorem grow? How does it compare to better known error terms? (I'd be particularly interested in the error term $x/(ln x)^2$.)
For instance, writing
$$
x expleft( -C sqrt{ln x} right) = frac{x}{x^{C/sqrt{ln x}}},
$$
we find that it eventually supersedes $x^alpha$ for any $alpha < 1$.
calculus asymptotics analytic-number-theory
calculus asymptotics analytic-number-theory
edited Jan 11 at 15:32
quid♦
36.9k95093
asked Jan 11 at 15:17
AlgebraicsAnonymousAlgebraicsAnonymous
1,067112
edited Jan 11 at 15:32
quid♦
36.9k95093
asked Jan 11 at 15:17
AlgebraicsAnonymousAlgebraicsAnonymous
1,067112
edited Jan 11 at 15:32
quid♦
36.9k95093
edited Jan 11 at 15:32
quid♦
36.9k95093
edited Jan 11 at 15:32
quid♦
36.9k95093
36.9k95093
asked Jan 11 at 15:17
AlgebraicsAnonymousAlgebraicsAnonymous
1,067112
asked Jan 11 at 15:17
AlgebraicsAnonymousAlgebraicsAnonymous
1,067112
asked Jan 11 at 15:17
AlgebraicsAnonymousAlgebraicsAnonymous
1,067112
1,067112
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
To compare to $frac{x}{(ln x)^2}$ it is helpful to write $(ln x)^2$ as an exponential. Namely $(ln x)^2 = exp( ln ln x)^2 = exp(2 ln ln x) $.
So in the one you divide by $exp(C sqrt{ln x}) $ and in the other by $exp(2 ln ln x)$. Now $sqrt{ln x}$ grows much faster than $2 ln ln x$, and thus you error-term is smaller (as you divide by something larger).
More generally, the error term is better than $frac{x}{(ln x)^k}$ for any positive integer $k$.
So the short is, better than $frac{x}{(ln x)^k}$ for any $k ge 1$ yet worse than $x^{alpha}$ for $alpha lt 1$, as you said.
It is also worse than $xexp (-C (ln x)^{beta}) $ for $1/2 < beta <1$ (and better for $beta <1/2$).
answered Jan 11 at 15:31
quid♦quid
36.9k95093
$endgroup$
$begingroup$
Could you elaborate on what happens if in the very last line, the logarithm is replaced by a double logarithm?
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:28
$begingroup$
That'd be worse than $x / ln x $ as $(ln ln x)^b$ would be smaller than $ln ln x$ (at least when $b$ is less than $1$ which is what I meant to impose).
$endgroup$
– quid♦
Jan 11 at 17:31
$begingroup$
I think that's a $beta$ and not a $b$. But thank you very much indeed. Gladly, I just found my brain somewhere, so that it was not entirely necessary...
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:37
add a comment |
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1 Answer
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votes
$begingroup$
To compare to $frac{x}{(ln x)^2}$ it is helpful to write $(ln x)^2$ as an exponential. Namely $(ln x)^2 = exp( ln ln x)^2 = exp(2 ln ln x) $.
So in the one you divide by $exp(C sqrt{ln x}) $ and in the other by $exp(2 ln ln x)$. Now $sqrt{ln x}$ grows much faster than $2 ln ln x$, and thus you error-term is smaller (as you divide by something larger).
More generally, the error term is better than $frac{x}{(ln x)^k}$ for any positive integer $k$.
So the short is, better than $frac{x}{(ln x)^k}$ for any $k ge 1$ yet worse than $x^{alpha}$ for $alpha lt 1$, as you said.
It is also worse than $xexp (-C (ln x)^{beta}) $ for $1/2 < beta <1$ (and better for $beta <1/2$).
answered Jan 11 at 15:31
quid♦quid
36.9k95093
$endgroup$
$begingroup$
Could you elaborate on what happens if in the very last line, the logarithm is replaced by a double logarithm?
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:28
$begingroup$
That'd be worse than $x / ln x $ as $(ln ln x)^b$ would be smaller than $ln ln x$ (at least when $b$ is less than $1$ which is what I meant to impose).
$endgroup$
– quid♦
Jan 11 at 17:31
$begingroup$
I think that's a $beta$ and not a $b$. But thank you very much indeed. Gladly, I just found my brain somewhere, so that it was not entirely necessary...
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:37
add a comment |
$begingroup$
To compare to $frac{x}{(ln x)^2}$ it is helpful to write $(ln x)^2$ as an exponential. Namely $(ln x)^2 = exp( ln ln x)^2 = exp(2 ln ln x) $.
So in the one you divide by $exp(C sqrt{ln x}) $ and in the other by $exp(2 ln ln x)$. Now $sqrt{ln x}$ grows much faster than $2 ln ln x$, and thus you error-term is smaller (as you divide by something larger).
More generally, the error term is better than $frac{x}{(ln x)^k}$ for any positive integer $k$.
So the short is, better than $frac{x}{(ln x)^k}$ for any $k ge 1$ yet worse than $x^{alpha}$ for $alpha lt 1$, as you said.
It is also worse than $xexp (-C (ln x)^{beta}) $ for $1/2 < beta <1$ (and better for $beta <1/2$).
answered Jan 11 at 15:31
quid♦quid
36.9k95093
$endgroup$
$begingroup$
Could you elaborate on what happens if in the very last line, the logarithm is replaced by a double logarithm?
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:28
$begingroup$
That'd be worse than $x / ln x $ as $(ln ln x)^b$ would be smaller than $ln ln x$ (at least when $b$ is less than $1$ which is what I meant to impose).
$endgroup$
– quid♦
Jan 11 at 17:31
$begingroup$
I think that's a $beta$ and not a $b$. But thank you very much indeed. Gladly, I just found my brain somewhere, so that it was not entirely necessary...
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:37
add a comment |
$begingroup$
To compare to $frac{x}{(ln x)^2}$ it is helpful to write $(ln x)^2$ as an exponential. Namely $(ln x)^2 = exp( ln ln x)^2 = exp(2 ln ln x) $.
So in the one you divide by $exp(C sqrt{ln x}) $ and in the other by $exp(2 ln ln x)$. Now $sqrt{ln x}$ grows much faster than $2 ln ln x$, and thus you error-term is smaller (as you divide by something larger).
More generally, the error term is better than $frac{x}{(ln x)^k}$ for any positive integer $k$.
So the short is, better than $frac{x}{(ln x)^k}$ for any $k ge 1$ yet worse than $x^{alpha}$ for $alpha lt 1$, as you said.
It is also worse than $xexp (-C (ln x)^{beta}) $ for $1/2 < beta <1$ (and better for $beta <1/2$).
answered Jan 11 at 15:31
quid♦quid
36.9k95093
$endgroup$
To compare to $frac{x}{(ln x)^2}$ it is helpful to write $(ln x)^2$ as an exponential. Namely $(ln x)^2 = exp( ln ln x)^2 = exp(2 ln ln x) $.
So in the one you divide by $exp(C sqrt{ln x}) $ and in the other by $exp(2 ln ln x)$. Now $sqrt{ln x}$ grows much faster than $2 ln ln x$, and thus you error-term is smaller (as you divide by something larger).
More generally, the error term is better than $frac{x}{(ln x)^k}$ for any positive integer $k$.
So the short is, better than $frac{x}{(ln x)^k}$ for any $k ge 1$ yet worse than $x^{alpha}$ for $alpha lt 1$, as you said.
It is also worse than $xexp (-C (ln x)^{beta}) $ for $1/2 < beta <1$ (and better for $beta <1/2$).
answered Jan 11 at 15:31
quid♦quid
36.9k95093
edited Jan 11 at 17:33
answered Jan 11 at 15:31
quid♦quid
36.9k95093
answered Jan 11 at 15:31
quid♦quid
36.9k95093
answered Jan 11 at 15:31
quid♦quid
36.9k95093
36.9k95093
$begingroup$
Could you elaborate on what happens if in the very last line, the logarithm is replaced by a double logarithm?
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:28
$begingroup$
That'd be worse than $x / ln x $ as $(ln ln x)^b$ would be smaller than $ln ln x$ (at least when $b$ is less than $1$ which is what I meant to impose).
$endgroup$
– quid♦
Jan 11 at 17:31
$begingroup$
I think that's a $beta$ and not a $b$. But thank you very much indeed. Gladly, I just found my brain somewhere, so that it was not entirely necessary...
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:37
add a comment |
$begingroup$
Could you elaborate on what happens if in the very last line, the logarithm is replaced by a double logarithm?
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:28
$begingroup$
That'd be worse than $x / ln x $ as $(ln ln x)^b$ would be smaller than $ln ln x$ (at least when $b$ is less than $1$ which is what I meant to impose).
$endgroup$
– quid♦
Jan 11 at 17:31
$begingroup$
I think that's a $beta$ and not a $b$. But thank you very much indeed. Gladly, I just found my brain somewhere, so that it was not entirely necessary...
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:37
$begingroup$
Could you elaborate on what happens if in the very last line, the logarithm is replaced by a double logarithm?
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:28
$begingroup$
Could you elaborate on what happens if in the very last line, the logarithm is replaced by a double logarithm?
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:28
$begingroup$
That'd be worse than $x / ln x $ as $(ln ln x)^b$ would be smaller than $ln ln x$ (at least when $b$ is less than $1$ which is what I meant to impose).
$endgroup$
– quid♦
Jan 11 at 17:31
$begingroup$
That'd be worse than $x / ln x $ as $(ln ln x)^b$ would be smaller than $ln ln x$ (at least when $b$ is less than $1$ which is what I meant to impose).
$endgroup$
– quid♦
Jan 11 at 17:31
$begingroup$
I think that's a $beta$ and not a $b$. But thank you very much indeed. Gladly, I just found my brain somewhere, so that it was not entirely necessary...
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:37
$begingroup$
I think that's a $beta$ and not a $b$. But thank you very much indeed. Gladly, I just found my brain somewhere, so that it was not entirely necessary...
$endgroup$
– AlgebraicsAnonymous
Jan 11 at 17:37
add a comment |
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