Proper Way To Compute An Upper Bound












2














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 ca.classical-analysis-and-odes cv.complex-variables







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    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 ca.classical-analysis-and-odes cv.complex-variables







    share|cite|improve this question









    New contributor




    r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      2












      2








      2







      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 ca.classical-analysis-and-odes cv.complex-variables







      share|cite|improve this question









      New contributor




      r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      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 ca.classical-analysis-and-odes cv.complex-variables




      nt.number-theory ca.classical-analysis-and-odes cv.complex-variables






      share|cite|improve this question









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      r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question









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      edited 3 hours ago









      Martin Sleziak

      2,91032028




      2,91032028






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      asked 3 hours ago









      r. t.

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          1 Answer
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          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





















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            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


























              3














              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
























                3












                3








                3






                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












                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 2 hours ago









                Greg Martin

                8,08313458




                8,08313458






















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