Proper Way To Compute An Upper Bound

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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 $tau(d)=sum_n1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
$$sum_q leq x^2deltatau(q)^2 left | sum_substackm leq x+2\
m equiv a bmod q mu(m)right | ll x (log x)^2c,$$



where $2 delta <1/2$.



The questions are these:



  1. Is the main result invalid? The upper bound should be
    $$sum_q leq x^2deltatau(q)^2 left | sum_substackm leq x+2\
    m equiv a bmod q mu(m)right | ll x ^1+2delta.$$

    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?










share|cite|improve this question




























    4















    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 $tau(d)=sum_n1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
    $$sum_q leq x^2deltatau(q)^2 left | sum_substackm leq x+2\
    m equiv a bmod q mu(m)right | ll x (log x)^2c,$$



    where $2 delta <1/2$.



    The questions are these:



    1. Is the main result invalid? The upper bound should be
      $$sum_q leq x^2deltatau(q)^2 left | sum_substackm leq x+2\
      m equiv a bmod q mu(m)right | ll x ^1+2delta.$$

      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?










    share|cite|improve this question


























      4












      4








      4








      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 $tau(d)=sum_n1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
      $$sum_q leq x^2deltatau(q)^2 left | sum_substackm leq x+2\
      m equiv a bmod q mu(m)right | ll x (log x)^2c,$$



      where $2 delta <1/2$.



      The questions are these:



      1. Is the main result invalid? The upper bound should be
        $$sum_q leq x^2deltatau(q)^2 left | sum_substackm leq x+2\
        m equiv a bmod q mu(m)right | ll x ^1+2delta.$$

        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?










      share|cite|improve this question
















      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 $tau(d)=sum_n1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
      $$sum_q leq x^2deltatau(q)^2 left | sum_substackm leq x+2\
      m equiv a bmod q mu(m)right | ll x (log x)^2c,$$



      where $2 delta <1/2$.



      The questions are these:



      1. Is the main result invalid? The upper bound should be
        $$sum_q leq x^2deltatau(q)^2 left | sum_substackm leq x+2\
        m equiv a bmod q mu(m)right | ll x ^1+2delta.$$

        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






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      edited Dec 31 '18 at 1:54









      GH from MO

      58.1k5144220




      58.1k5144220










      asked Dec 30 '18 at 19:37









      r. t.r. t.

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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
          beginalign*
          sum_q leq x^2deltatau(q)^2 bigg | sum_substackm leq x+2\
          m equiv a bmod q mu(m)bigg | &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q |mu(m)| \
          &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q 1 \
          &ll sum_q leq x^2deltatau(q)^2 frac xq = x sum_q leq x^2delta fractau(q)^2q.
          endalign*

          And this remaining sum is indeed $ll_delta (log x)^2c$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_q leq y fractau(q)^2q sim frac(log y)^44pi^2.
          $$






          share|cite|improve this answer























          • 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.

            – literature-searcher
            Dec 31 '18 at 9:14







          • 1





            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.

            – Greg Martin
            Dec 31 '18 at 18:00










          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6














          Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
          beginalign*
          sum_q leq x^2deltatau(q)^2 bigg | sum_substackm leq x+2\
          m equiv a bmod q mu(m)bigg | &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q |mu(m)| \
          &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q 1 \
          &ll sum_q leq x^2deltatau(q)^2 frac xq = x sum_q leq x^2delta fractau(q)^2q.
          endalign*

          And this remaining sum is indeed $ll_delta (log x)^2c$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_q leq y fractau(q)^2q sim frac(log y)^44pi^2.
          $$






          share|cite|improve this answer























          • 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.

            – literature-searcher
            Dec 31 '18 at 9:14







          • 1





            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.

            – Greg Martin
            Dec 31 '18 at 18:00















          6














          Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
          beginalign*
          sum_q leq x^2deltatau(q)^2 bigg | sum_substackm leq x+2\
          m equiv a bmod q mu(m)bigg | &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q |mu(m)| \
          &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q 1 \
          &ll sum_q leq x^2deltatau(q)^2 frac xq = x sum_q leq x^2delta fractau(q)^2q.
          endalign*

          And this remaining sum is indeed $ll_delta (log x)^2c$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_q leq y fractau(q)^2q sim frac(log y)^44pi^2.
          $$






          share|cite|improve this answer























          • 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.

            – literature-searcher
            Dec 31 '18 at 9:14







          • 1





            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.

            – Greg Martin
            Dec 31 '18 at 18:00













          6












          6








          6







          Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
          beginalign*
          sum_q leq x^2deltatau(q)^2 bigg | sum_substackm leq x+2\
          m equiv a bmod q mu(m)bigg | &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q |mu(m)| \
          &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q 1 \
          &ll sum_q leq x^2deltatau(q)^2 frac xq = x sum_q leq x^2delta fractau(q)^2q.
          endalign*

          And this remaining sum is indeed $ll_delta (log x)^2c$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_q leq y fractau(q)^2q sim frac(log y)^44pi^2.
          $$






          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
          beginalign*
          sum_q leq x^2deltatau(q)^2 bigg | sum_substackm leq x+2\
          m equiv a bmod q mu(m)bigg | &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q |mu(m)| \
          &le sum_q leq x^2deltatau(q)^2 sum_substackm leq x+2\ m equiv a bmod q 1 \
          &ll sum_q leq x^2deltatau(q)^2 frac xq = x sum_q leq x^2delta fractau(q)^2q.
          endalign*

          And this remaining sum is indeed $ll_delta (log x)^2c$ for some constant $c$; indeed, it's not hard to show that
          $$
          sum_q leq y fractau(q)^2q sim frac(log y)^44pi^2.
          $$







          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

          8,32813559




          8,32813559












          • 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.

            – literature-searcher
            Dec 31 '18 at 9:14







          • 1





            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.

            – Greg Martin
            Dec 31 '18 at 18:00

















          • 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.

            – literature-searcher
            Dec 31 '18 at 9:14







          • 1





            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.

            – Greg Martin
            Dec 31 '18 at 18:00
















          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.

          – literature-searcher
          Dec 31 '18 at 9:14






          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.

          – literature-searcher
          Dec 31 '18 at 9:14





          1




          1





          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.

          – Greg Martin
          Dec 31 '18 at 18:00





          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.

          – Greg Martin
          Dec 31 '18 at 18:00

















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