Understanding the error term in the Siegel‒Walfisz theorem












1












$begingroup$


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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 11 at 15:32









      quid♦

      36.9k95093




      36.9k95093










      asked Jan 11 at 15:17









      AlgebraicsAnonymousAlgebraicsAnonymous

      1,067112




      1,067112






















          1 Answer
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          1












          $begingroup$

          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 $beta <1/2$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          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 $beta <1/2$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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 $beta <1/2$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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 $beta <1/2$).






          share|cite|improve this answer











          $endgroup$



          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 $beta <1/2$).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 17:33

























          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


















          • $begingroup$
            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


















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