Piltz Divisor Problem












2












$begingroup$


Let $tau_k(n)$ count the number of ways of representing $n$ as the product of $k$ natural numbers. It is known that:



$$D_k(x) = sum_{n leq x} tau_k(n) = xP_k(log x) + O(x ^{1 - frac{1}{k-1}}(log x)^{k-2}), ; forall k geq 2$$



Where $P_k$ is a polynomial of degree $k-1$ with leading coefficient $frac{1}{(k-1)!}$



I am asked to prove this. We may assume the base case as it is a well known result.



The assuming the result for all $l leq k$:



$$D_k(x) = sum_{mn leq x} tau_{k-1}(n) = sum_{nleq x}lfloor{frac{x}{n}}rfloortau_{k-1}(n)$$



$$= sum_{n leq x}(frac{x}{n} + O(1))tau_{k-1}(n) = xsum_{n leq x} frac{tau_{k-1}(n)}{n} + O(sum_{nleq x}tau_{k-1}(n))$$



Using Abel's Summation formula, we may see that:



$$sum_{n leq x} frac{tau_{k-1}(n)}{n} = P_k(log x) + O(x^{frac{-1}{k-1}}(log x)^{k-3})$$



So:



$$D_k(x) = xP_k(log x) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3}) + O(sum_{n leq x}tau_{k-1}(n))$$



Using the induction hypothesis:



$$O(sum_{n leq x}tau_{k-1}(n)) = O(xP_{k-1}(log x) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3}))$$



$$= O(x(log x)^{k-2}) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3})) = O(x(log x)^{k-2})$$



Thus we get:



$$D_k(x) = xP_k(log x) + O(x^{1- frac{1}{k-1}}(log x)^{k-3}) + O(x(log x)^{k-2})$$



$$= xP_k(log x)+O(x(log x)^{k-2})$$



Howwever, this error term is too large. How can I go about reducing it?






nt.number-theory analytic-number-theory divisors divisors-multiples







share|cite|improve this question











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  • 1




    $begingroup$
    If you use log instead of log, your formulas will be more readable.
    $endgroup$
    – Greg Martin
    Feb 2 at 21:49










  • $begingroup$
    This question is not of research level (so it would be more suitable to ask at math.stackexchange.com), but Greg Martin gave an excellent answer. In general, consulting the textbooks before asking a question might be useful. BTW I used Dirichlet's hyperbola method for a recent MO question here: mathoverflow.net/questions/321839/…
    $endgroup$
    – GH from MO
    Feb 2 at 23:50


















2












$begingroup$


Let $tau_k(n)$ count the number of ways of representing $n$ as the product of $k$ natural numbers. It is known that:



$$D_k(x) = sum_{n leq x} tau_k(n) = xP_k(log x) + O(x ^{1 - frac{1}{k-1}}(log x)^{k-2}), ; forall k geq 2$$



Where $P_k$ is a polynomial of degree $k-1$ with leading coefficient $frac{1}{(k-1)!}$



I am asked to prove this. We may assume the base case as it is a well known result.



The assuming the result for all $l leq k$:



$$D_k(x) = sum_{mn leq x} tau_{k-1}(n) = sum_{nleq x}lfloor{frac{x}{n}}rfloortau_{k-1}(n)$$



$$= sum_{n leq x}(frac{x}{n} + O(1))tau_{k-1}(n) = xsum_{n leq x} frac{tau_{k-1}(n)}{n} + O(sum_{nleq x}tau_{k-1}(n))$$



Using Abel's Summation formula, we may see that:



$$sum_{n leq x} frac{tau_{k-1}(n)}{n} = P_k(log x) + O(x^{frac{-1}{k-1}}(log x)^{k-3})$$



So:



$$D_k(x) = xP_k(log x) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3}) + O(sum_{n leq x}tau_{k-1}(n))$$



Using the induction hypothesis:



$$O(sum_{n leq x}tau_{k-1}(n)) = O(xP_{k-1}(log x) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3}))$$



$$= O(x(log x)^{k-2}) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3})) = O(x(log x)^{k-2})$$



Thus we get:



$$D_k(x) = xP_k(log x) + O(x^{1- frac{1}{k-1}}(log x)^{k-3}) + O(x(log x)^{k-2})$$



$$= xP_k(log x)+O(x(log x)^{k-2})$$



Howwever, this error term is too large. How can I go about reducing it?






nt.number-theory analytic-number-theory divisors divisors-multiples







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If you use log instead of log, your formulas will be more readable.
    $endgroup$
    – Greg Martin
    Feb 2 at 21:49










  • $begingroup$
    This question is not of research level (so it would be more suitable to ask at math.stackexchange.com), but Greg Martin gave an excellent answer. In general, consulting the textbooks before asking a question might be useful. BTW I used Dirichlet's hyperbola method for a recent MO question here: mathoverflow.net/questions/321839/…
    $endgroup$
    – GH from MO
    Feb 2 at 23:50
















2












2








2





$begingroup$


Let $tau_k(n)$ count the number of ways of representing $n$ as the product of $k$ natural numbers. It is known that:



$$D_k(x) = sum_{n leq x} tau_k(n) = xP_k(log x) + O(x ^{1 - frac{1}{k-1}}(log x)^{k-2}), ; forall k geq 2$$



Where $P_k$ is a polynomial of degree $k-1$ with leading coefficient $frac{1}{(k-1)!}$



I am asked to prove this. We may assume the base case as it is a well known result.



The assuming the result for all $l leq k$:



$$D_k(x) = sum_{mn leq x} tau_{k-1}(n) = sum_{nleq x}lfloor{frac{x}{n}}rfloortau_{k-1}(n)$$



$$= sum_{n leq x}(frac{x}{n} + O(1))tau_{k-1}(n) = xsum_{n leq x} frac{tau_{k-1}(n)}{n} + O(sum_{nleq x}tau_{k-1}(n))$$



Using Abel's Summation formula, we may see that:



$$sum_{n leq x} frac{tau_{k-1}(n)}{n} = P_k(log x) + O(x^{frac{-1}{k-1}}(log x)^{k-3})$$



So:



$$D_k(x) = xP_k(log x) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3}) + O(sum_{n leq x}tau_{k-1}(n))$$



Using the induction hypothesis:



$$O(sum_{n leq x}tau_{k-1}(n)) = O(xP_{k-1}(log x) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3}))$$



$$= O(x(log x)^{k-2}) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3})) = O(x(log x)^{k-2})$$



Thus we get:



$$D_k(x) = xP_k(log x) + O(x^{1- frac{1}{k-1}}(log x)^{k-3}) + O(x(log x)^{k-2})$$



$$= xP_k(log x)+O(x(log x)^{k-2})$$



Howwever, this error term is too large. How can I go about reducing it?






nt.number-theory analytic-number-theory divisors divisors-multiples







share|cite|improve this question











$endgroup$




Let $tau_k(n)$ count the number of ways of representing $n$ as the product of $k$ natural numbers. It is known that:



$$D_k(x) = sum_{n leq x} tau_k(n) = xP_k(log x) + O(x ^{1 - frac{1}{k-1}}(log x)^{k-2}), ; forall k geq 2$$



Where $P_k$ is a polynomial of degree $k-1$ with leading coefficient $frac{1}{(k-1)!}$



I am asked to prove this. We may assume the base case as it is a well known result.



The assuming the result for all $l leq k$:



$$D_k(x) = sum_{mn leq x} tau_{k-1}(n) = sum_{nleq x}lfloor{frac{x}{n}}rfloortau_{k-1}(n)$$



$$= sum_{n leq x}(frac{x}{n} + O(1))tau_{k-1}(n) = xsum_{n leq x} frac{tau_{k-1}(n)}{n} + O(sum_{nleq x}tau_{k-1}(n))$$



Using Abel's Summation formula, we may see that:



$$sum_{n leq x} frac{tau_{k-1}(n)}{n} = P_k(log x) + O(x^{frac{-1}{k-1}}(log x)^{k-3})$$



So:



$$D_k(x) = xP_k(log x) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3}) + O(sum_{n leq x}tau_{k-1}(n))$$



Using the induction hypothesis:



$$O(sum_{n leq x}tau_{k-1}(n)) = O(xP_{k-1}(log x) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3}))$$



$$= O(x(log x)^{k-2}) + O(x^{1 - frac{1}{k-1}}(log x)^{k-3})) = O(x(log x)^{k-2})$$



Thus we get:



$$D_k(x) = xP_k(log x) + O(x^{1- frac{1}{k-1}}(log x)^{k-3}) + O(x(log x)^{k-2})$$



$$= xP_k(log x)+O(x(log x)^{k-2})$$



Howwever, this error term is too large. How can I go about reducing it?






nt.number-theory analytic-number-theory divisors divisors-multiples




nt.number-theory analytic-number-theory divisors divisors-multiples






share|cite|improve this question















share|cite|improve this question













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edited Feb 2 at 23:48









GH from MO

58.6k5146223




58.6k5146223










asked Feb 2 at 21:36









user366818user366818

1163




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  • 1




    $begingroup$
    If you use log instead of log, your formulas will be more readable.
    $endgroup$
    – Greg Martin
    Feb 2 at 21:49










  • $begingroup$
    This question is not of research level (so it would be more suitable to ask at math.stackexchange.com), but Greg Martin gave an excellent answer. In general, consulting the textbooks before asking a question might be useful. BTW I used Dirichlet's hyperbola method for a recent MO question here: mathoverflow.net/questions/321839/…
    $endgroup$
    – GH from MO
    Feb 2 at 23:50
















  • 1




    $begingroup$
    If you use log instead of log, your formulas will be more readable.
    $endgroup$
    – Greg Martin
    Feb 2 at 21:49










  • $begingroup$
    This question is not of research level (so it would be more suitable to ask at math.stackexchange.com), but Greg Martin gave an excellent answer. In general, consulting the textbooks before asking a question might be useful. BTW I used Dirichlet's hyperbola method for a recent MO question here: mathoverflow.net/questions/321839/…
    $endgroup$
    – GH from MO
    Feb 2 at 23:50










1




1




$begingroup$
If you use log instead of log, your formulas will be more readable.
$endgroup$
– Greg Martin
Feb 2 at 21:49




$begingroup$
If you use log instead of log, your formulas will be more readable.
$endgroup$
– Greg Martin
Feb 2 at 21:49












$begingroup$
This question is not of research level (so it would be more suitable to ask at math.stackexchange.com), but Greg Martin gave an excellent answer. In general, consulting the textbooks before asking a question might be useful. BTW I used Dirichlet's hyperbola method for a recent MO question here: mathoverflow.net/questions/321839/…
$endgroup$
– GH from MO
Feb 2 at 23:50






$begingroup$
This question is not of research level (so it would be more suitable to ask at math.stackexchange.com), but Greg Martin gave an excellent answer. In general, consulting the textbooks before asking a question might be useful. BTW I used Dirichlet's hyperbola method for a recent MO question here: mathoverflow.net/questions/321839/…
$endgroup$
– GH from MO
Feb 2 at 23:50












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$begingroup$

You've discovered some of the primary motivation for the invention of the Dirichlet hyperbola method. Since $tau_k = tau_{k-1}*1$ (which you are already using), you can take advantage of the extra parameter ($U$ and $V$ in the linked document, instead of just $x$) to get a better error term—just as in the proof of the base case $k=2$.






share|cite|improve this answer









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    7












    $begingroup$

    You've discovered some of the primary motivation for the invention of the Dirichlet hyperbola method. Since $tau_k = tau_{k-1}*1$ (which you are already using), you can take advantage of the extra parameter ($U$ and $V$ in the linked document, instead of just $x$) to get a better error term—just as in the proof of the base case $k=2$.






    share|cite|improve this answer









    $endgroup$


















      7












      $begingroup$

      You've discovered some of the primary motivation for the invention of the Dirichlet hyperbola method. Since $tau_k = tau_{k-1}*1$ (which you are already using), you can take advantage of the extra parameter ($U$ and $V$ in the linked document, instead of just $x$) to get a better error term—just as in the proof of the base case $k=2$.






      share|cite|improve this answer









      $endgroup$
















        7












        7








        7





        $begingroup$

        You've discovered some of the primary motivation for the invention of the Dirichlet hyperbola method. Since $tau_k = tau_{k-1}*1$ (which you are already using), you can take advantage of the extra parameter ($U$ and $V$ in the linked document, instead of just $x$) to get a better error term—just as in the proof of the base case $k=2$.






        share|cite|improve this answer









        $endgroup$



        You've discovered some of the primary motivation for the invention of the Dirichlet hyperbola method. Since $tau_k = tau_{k-1}*1$ (which you are already using), you can take advantage of the extra parameter ($U$ and $V$ in the linked document, instead of just $x$) to get a better error term—just as in the proof of the base case $k=2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 2 at 21:53









        Greg MartinGreg Martin

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