Proper Way To Compute An Upper Bound












4












$begingroup$


I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,



the authors used the average value $(log x)^c$, $c$ constant, of the number of divisors function $tau(d)=sum_{d|n}1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x (log x)^{2c},$$



where $2 delta <1/2$.



The questions are these:




  1. Is the main result invalid? The upper bound should be
    $$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
    m equiv a bmod q}} mu(m)right | ll x ^{1+2delta}.$$
    This is the best unconditional upper bound, under any known result, including Proposition 3.


  2. It is true that the proper upper bound $tau(d)^2 ll x^{2epsilon}$, $epsilon >0$, is not required here?


  3. Can we use this as a precedent to prove other upper bounds in mathematics?







nt.number-theory analytic-number-theory







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,



    the authors used the average value $(log x)^c$, $c$ constant, of the number of divisors function $tau(d)=sum_{d|n}1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
    $$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
    m equiv a bmod q}} mu(m)right | ll x (log x)^{2c},$$



    where $2 delta <1/2$.



    The questions are these:




    1. Is the main result invalid? The upper bound should be
      $$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
      m equiv a bmod q}} mu(m)right | ll x ^{1+2delta}.$$
      This is the best unconditional upper bound, under any known result, including Proposition 3.


    2. It is true that the proper upper bound $tau(d)^2 ll x^{2epsilon}$, $epsilon >0$, is not required here?


    3. Can we use this as a precedent to prove other upper bounds in mathematics?







    nt.number-theory analytic-number-theory







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,



      the authors used the average value $(log x)^c$, $c$ constant, of the number of divisors function $tau(d)=sum_{d|n}1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
      $$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
      m equiv a bmod q}} mu(m)right | ll x (log x)^{2c},$$



      where $2 delta <1/2$.



      The questions are these:




      1. Is the main result invalid? The upper bound should be
        $$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
        m equiv a bmod q}} mu(m)right | ll x ^{1+2delta}.$$
        This is the best unconditional upper bound, under any known result, including Proposition 3.


      2. It is true that the proper upper bound $tau(d)^2 ll x^{2epsilon}$, $epsilon >0$, is not required here?


      3. Can we use this as a precedent to prove other upper bounds in mathematics?







      nt.number-theory analytic-number-theory







      share|cite|improve this question











      $endgroup$




      I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,



      the authors used the average value $(log x)^c$, $c$ constant, of the number of divisors function $tau(d)=sum_{d|n}1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
      $$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
      m equiv a bmod q}} mu(m)right | ll x (log x)^{2c},$$



      where $2 delta <1/2$.



      The questions are these:




      1. Is the main result invalid? The upper bound should be
        $$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
        m equiv a bmod q}} mu(m)right | ll x ^{1+2delta}.$$
        This is the best unconditional upper bound, under any known result, including Proposition 3.


      2. It is true that the proper upper bound $tau(d)^2 ll x^{2epsilon}$, $epsilon >0$, is not required here?


      3. Can we use this as a precedent to prove other upper bounds in mathematics?







      nt.number-theory analytic-number-theory




      nt.number-theory analytic-number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 1:54









      GH from MO

      58.6k5146223




      58.6k5146223










      asked Dec 30 '18 at 19:37









      r. t.r. t.

      211




      211






















          1 Answer
          1






          active

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          6












          $begingroup$

          Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
          begin{align*}
          sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
          m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
          &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
          &ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
          end{align*}
          And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think the confusion in the paper is that they have an expression of the form $(A)^{1/2}(B)^{1/2}$ and refer therein to a bound of $x^{1/2}(log x)^c$ for "the first term in parenthesis" when they really mean this for $(A)^{1/2}$ rather than $(A)$, not to mention a possible discrepancy in $c$-values from one usage to the next, and the strange usage of $tau(x)^2$ in speaking of the average, when I would say $tau(q)^2$. OTOH, I don't think their "by crude estimates" (as you codify) need be more explicit here.
            $endgroup$
            – literature-searcher
            Dec 31 '18 at 9:14








          • 1




            $begingroup$
            My philosophy of writing is that it's better for authors to do the work once than to require each reader to do that work individually. In this case, it would take three more display-equation lines at most to write the argument and save readers the trouble; I think that's well worth it.
            $endgroup$
            – Greg Martin
            Dec 31 '18 at 18:00











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          1 Answer
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          active

          oldest

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          6












          $begingroup$

          Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
          begin{align*}
          sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
          m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
          &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
          &ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
          end{align*}
          And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think the confusion in the paper is that they have an expression of the form $(A)^{1/2}(B)^{1/2}$ and refer therein to a bound of $x^{1/2}(log x)^c$ for "the first term in parenthesis" when they really mean this for $(A)^{1/2}$ rather than $(A)$, not to mention a possible discrepancy in $c$-values from one usage to the next, and the strange usage of $tau(x)^2$ in speaking of the average, when I would say $tau(q)^2$. OTOH, I don't think their "by crude estimates" (as you codify) need be more explicit here.
            $endgroup$
            – literature-searcher
            Dec 31 '18 at 9:14








          • 1




            $begingroup$
            My philosophy of writing is that it's better for authors to do the work once than to require each reader to do that work individually. In this case, it would take three more display-equation lines at most to write the argument and save readers the trouble; I think that's well worth it.
            $endgroup$
            – Greg Martin
            Dec 31 '18 at 18:00
















          6












          $begingroup$

          Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
          begin{align*}
          sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
          m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
          &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
          &ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
          end{align*}
          And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think the confusion in the paper is that they have an expression of the form $(A)^{1/2}(B)^{1/2}$ and refer therein to a bound of $x^{1/2}(log x)^c$ for "the first term in parenthesis" when they really mean this for $(A)^{1/2}$ rather than $(A)$, not to mention a possible discrepancy in $c$-values from one usage to the next, and the strange usage of $tau(x)^2$ in speaking of the average, when I would say $tau(q)^2$. OTOH, I don't think their "by crude estimates" (as you codify) need be more explicit here.
            $endgroup$
            – literature-searcher
            Dec 31 '18 at 9:14








          • 1




            $begingroup$
            My philosophy of writing is that it's better for authors to do the work once than to require each reader to do that work individually. In this case, it would take three more display-equation lines at most to write the argument and save readers the trouble; I think that's well worth it.
            $endgroup$
            – Greg Martin
            Dec 31 '18 at 18:00














          6












          6








          6





          $begingroup$

          Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
          begin{align*}
          sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
          m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
          &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
          &ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
          end{align*}
          And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
          $$






          share|cite|improve this answer









          $endgroup$



          Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
          begin{align*}
          sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
          m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
          &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
          &ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
          end{align*}
          And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 21:19









          Greg MartinGreg Martin

          8,78813560




          8,78813560












          • $begingroup$
            I think the confusion in the paper is that they have an expression of the form $(A)^{1/2}(B)^{1/2}$ and refer therein to a bound of $x^{1/2}(log x)^c$ for "the first term in parenthesis" when they really mean this for $(A)^{1/2}$ rather than $(A)$, not to mention a possible discrepancy in $c$-values from one usage to the next, and the strange usage of $tau(x)^2$ in speaking of the average, when I would say $tau(q)^2$. OTOH, I don't think their "by crude estimates" (as you codify) need be more explicit here.
            $endgroup$
            – literature-searcher
            Dec 31 '18 at 9:14








          • 1




            $begingroup$
            My philosophy of writing is that it's better for authors to do the work once than to require each reader to do that work individually. In this case, it would take three more display-equation lines at most to write the argument and save readers the trouble; I think that's well worth it.
            $endgroup$
            – Greg Martin
            Dec 31 '18 at 18:00


















          • $begingroup$
            I think the confusion in the paper is that they have an expression of the form $(A)^{1/2}(B)^{1/2}$ and refer therein to a bound of $x^{1/2}(log x)^c$ for "the first term in parenthesis" when they really mean this for $(A)^{1/2}$ rather than $(A)$, not to mention a possible discrepancy in $c$-values from one usage to the next, and the strange usage of $tau(x)^2$ in speaking of the average, when I would say $tau(q)^2$. OTOH, I don't think their "by crude estimates" (as you codify) need be more explicit here.
            $endgroup$
            – literature-searcher
            Dec 31 '18 at 9:14








          • 1




            $begingroup$
            My philosophy of writing is that it's better for authors to do the work once than to require each reader to do that work individually. In this case, it would take three more display-equation lines at most to write the argument and save readers the trouble; I think that's well worth it.
            $endgroup$
            – Greg Martin
            Dec 31 '18 at 18:00
















          $begingroup$
          I think the confusion in the paper is that they have an expression of the form $(A)^{1/2}(B)^{1/2}$ and refer therein to a bound of $x^{1/2}(log x)^c$ for "the first term in parenthesis" when they really mean this for $(A)^{1/2}$ rather than $(A)$, not to mention a possible discrepancy in $c$-values from one usage to the next, and the strange usage of $tau(x)^2$ in speaking of the average, when I would say $tau(q)^2$. OTOH, I don't think their "by crude estimates" (as you codify) need be more explicit here.
          $endgroup$
          – literature-searcher
          Dec 31 '18 at 9:14






          $begingroup$
          I think the confusion in the paper is that they have an expression of the form $(A)^{1/2}(B)^{1/2}$ and refer therein to a bound of $x^{1/2}(log x)^c$ for "the first term in parenthesis" when they really mean this for $(A)^{1/2}$ rather than $(A)$, not to mention a possible discrepancy in $c$-values from one usage to the next, and the strange usage of $tau(x)^2$ in speaking of the average, when I would say $tau(q)^2$. OTOH, I don't think their "by crude estimates" (as you codify) need be more explicit here.
          $endgroup$
          – literature-searcher
          Dec 31 '18 at 9:14






          1




          1




          $begingroup$
          My philosophy of writing is that it's better for authors to do the work once than to require each reader to do that work individually. In this case, it would take three more display-equation lines at most to write the argument and save readers the trouble; I think that's well worth it.
          $endgroup$
          – Greg Martin
          Dec 31 '18 at 18:00




          $begingroup$
          My philosophy of writing is that it's better for authors to do the work once than to require each reader to do that work individually. In this case, it would take three more display-equation lines at most to write the argument and save readers the trouble; I think that's well worth it.
          $endgroup$
          – Greg Martin
          Dec 31 '18 at 18:00


















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