Why we need to have $S(x)=O(x)$?












2












$begingroup$


Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:




Given two real-valued functions $S(x)$ and $T(x)$ such that
$$T(x)=sum_{nle x}S(frac xn),xge1.$$
If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
$$S(x)sim cx~text{as}~xtoinfty$$ implies
$$T(x)sim cxlog x~text{as}~x rightarrow infty.$$




This is one of my answers:



$$lim_{xtoinfty}frac{T(x)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}S(frac xn)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}(frac{cx}n+o(x))}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}frac 1n}{log x}=lim_{xtoinfty}frac{log x+O(1)}{log x}=1$$



As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!






number-theory analytic-number-theory







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:




    Given two real-valued functions $S(x)$ and $T(x)$ such that
    $$T(x)=sum_{nle x}S(frac xn),xge1.$$
    If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
    $$S(x)sim cx~text{as}~xtoinfty$$ implies
    $$T(x)sim cxlog x~text{as}~x rightarrow infty.$$




    This is one of my answers:



    $$lim_{xtoinfty}frac{T(x)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}S(frac xn)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}(frac{cx}n+o(x))}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}frac 1n}{log x}=lim_{xtoinfty}frac{log x+O(1)}{log x}=1$$



    As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!






    number-theory analytic-number-theory







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:




      Given two real-valued functions $S(x)$ and $T(x)$ such that
      $$T(x)=sum_{nle x}S(frac xn),xge1.$$
      If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
      $$S(x)sim cx~text{as}~xtoinfty$$ implies
      $$T(x)sim cxlog x~text{as}~x rightarrow infty.$$




      This is one of my answers:



      $$lim_{xtoinfty}frac{T(x)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}S(frac xn)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}(frac{cx}n+o(x))}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}frac 1n}{log x}=lim_{xtoinfty}frac{log x+O(1)}{log x}=1$$



      As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!






      number-theory analytic-number-theory







      share|cite|improve this question











      $endgroup$




      Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:




      Given two real-valued functions $S(x)$ and $T(x)$ such that
      $$T(x)=sum_{nle x}S(frac xn),xge1.$$
      If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
      $$S(x)sim cx~text{as}~xtoinfty$$ implies
      $$T(x)sim cxlog x~text{as}~x rightarrow infty.$$




      This is one of my answers:



      $$lim_{xtoinfty}frac{T(x)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}S(frac xn)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}(frac{cx}n+o(x))}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}frac 1n}{log x}=lim_{xtoinfty}frac{log x+O(1)}{log x}=1$$



      As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!






      number-theory analytic-number-theory




      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 Jan 8 at 12:33

























      asked Jan 8 at 11:16







      user604594





























          1 Answer
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          $begingroup$

          The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.



          On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
            $endgroup$
            – Peter Humphries
            Jan 9 at 14:50












          • $begingroup$
            @Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
            $endgroup$
            – W-t-P
            Jan 9 at 15:55










          • $begingroup$
            sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
            $endgroup$
            – Peter Humphries
            Jan 9 at 15:57












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          $begingroup$

          The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.



          On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
            $endgroup$
            – Peter Humphries
            Jan 9 at 14:50












          • $begingroup$
            @Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
            $endgroup$
            – W-t-P
            Jan 9 at 15:55










          • $begingroup$
            sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
            $endgroup$
            – Peter Humphries
            Jan 9 at 15:57
















          1












          $begingroup$

          The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.



          On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
            $endgroup$
            – Peter Humphries
            Jan 9 at 14:50












          • $begingroup$
            @Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
            $endgroup$
            – W-t-P
            Jan 9 at 15:55










          • $begingroup$
            sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
            $endgroup$
            – Peter Humphries
            Jan 9 at 15:57














          1












          1








          1





          $begingroup$

          The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.



          On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.






          share|cite|improve this answer









          $endgroup$



          The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.



          On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 15:47









          W-t-PW-t-P

          1,749612




          1,749612












          • $begingroup$
            Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
            $endgroup$
            – Peter Humphries
            Jan 9 at 14:50












          • $begingroup$
            @Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
            $endgroup$
            – W-t-P
            Jan 9 at 15:55










          • $begingroup$
            sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
            $endgroup$
            – Peter Humphries
            Jan 9 at 15:57


















          • $begingroup$
            Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
            $endgroup$
            – Peter Humphries
            Jan 9 at 14:50












          • $begingroup$
            @Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
            $endgroup$
            – W-t-P
            Jan 9 at 15:55










          • $begingroup$
            sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
            $endgroup$
            – Peter Humphries
            Jan 9 at 15:57
















          $begingroup$
          Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
          $endgroup$
          – Peter Humphries
          Jan 9 at 14:50






          $begingroup$
          Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
          $endgroup$
          – Peter Humphries
          Jan 9 at 14:50














          $begingroup$
          @Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
          $endgroup$
          – W-t-P
          Jan 9 at 15:55




          $begingroup$
          @Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
          $endgroup$
          – W-t-P
          Jan 9 at 15:55












          $begingroup$
          sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
          $endgroup$
          – Peter Humphries
          Jan 9 at 15:57




          $begingroup$
          sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
          $endgroup$
          – Peter Humphries
          Jan 9 at 15:57


















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