Let $f(x)$ be defined for all rational $x$ in $0leq xleq 1$












1












$begingroup$


Let $f(x)$ be defined for all rational $x$ in $0leq xleq 1$
$$F(n)=sum_{k=1}^n fbigg(frac knbigg), quad F^* (n)=sum_{k=1\(k,n)=1}^nfbigg(frac knbigg).$$
Prove that $$F*=mu * F$$
where $*$ denoted the Dirichlet Multiplication and $mu$ is the Mobius Function.



I have knowledge about Elementary Number Theory, and now I'm studying Analytic Number Theory by Apostol, which I feel very hard to self-learn, and I've seen the proof of $phi(n)=sum_{dvert n} mu(d)n/d$ which uses a very elegant way, and I think most of people cannot have that idea in mind.

By the way, I have tried when $n=6$, it shows begin{align*}mu* F(6)=sum_{dvert 6}mubigg(frac 6dbigg)sum_{k=1}^d fbigg(frac kdbigg)end{align*}
And I just merely expanding everythings out, using the definitions of them, found it equals to $f(1/6)+f(5/6)$, it is true, but I cannot find any approach to prove it. Please help.






number-theory analytic-number-theory dirichlet-convolution







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I do not see a difference between $F$ and $F*$.
    $endgroup$
    – Mindlack
    Jan 9 at 9:06










  • $begingroup$
    @Mindlack Thanks for mentioning, I have added the detail.
    $endgroup$
    – kelvin hong 方
    Jan 9 at 10:51






  • 1




    $begingroup$
    Have you seen Möbius inversion formula?
    $endgroup$
    – Bruno Andrades
    Jan 9 at 11:02










  • $begingroup$
    @BrunoAndrades Yes, oh thank you! I didn't realize I can use that formula! I think I get it, thank you!
    $endgroup$
    – kelvin hong 方
    Jan 9 at 11:26
















1












$begingroup$


Let $f(x)$ be defined for all rational $x$ in $0leq xleq 1$
$$F(n)=sum_{k=1}^n fbigg(frac knbigg), quad F^* (n)=sum_{k=1\(k,n)=1}^nfbigg(frac knbigg).$$
Prove that $$F*=mu * F$$
where $*$ denoted the Dirichlet Multiplication and $mu$ is the Mobius Function.



I have knowledge about Elementary Number Theory, and now I'm studying Analytic Number Theory by Apostol, which I feel very hard to self-learn, and I've seen the proof of $phi(n)=sum_{dvert n} mu(d)n/d$ which uses a very elegant way, and I think most of people cannot have that idea in mind.

By the way, I have tried when $n=6$, it shows begin{align*}mu* F(6)=sum_{dvert 6}mubigg(frac 6dbigg)sum_{k=1}^d fbigg(frac kdbigg)end{align*}
And I just merely expanding everythings out, using the definitions of them, found it equals to $f(1/6)+f(5/6)$, it is true, but I cannot find any approach to prove it. Please help.






number-theory analytic-number-theory dirichlet-convolution







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I do not see a difference between $F$ and $F*$.
    $endgroup$
    – Mindlack
    Jan 9 at 9:06










  • $begingroup$
    @Mindlack Thanks for mentioning, I have added the detail.
    $endgroup$
    – kelvin hong 方
    Jan 9 at 10:51






  • 1




    $begingroup$
    Have you seen Möbius inversion formula?
    $endgroup$
    – Bruno Andrades
    Jan 9 at 11:02










  • $begingroup$
    @BrunoAndrades Yes, oh thank you! I didn't realize I can use that formula! I think I get it, thank you!
    $endgroup$
    – kelvin hong 方
    Jan 9 at 11:26














1












1








1





$begingroup$


Let $f(x)$ be defined for all rational $x$ in $0leq xleq 1$
$$F(n)=sum_{k=1}^n fbigg(frac knbigg), quad F^* (n)=sum_{k=1\(k,n)=1}^nfbigg(frac knbigg).$$
Prove that $$F*=mu * F$$
where $*$ denoted the Dirichlet Multiplication and $mu$ is the Mobius Function.



I have knowledge about Elementary Number Theory, and now I'm studying Analytic Number Theory by Apostol, which I feel very hard to self-learn, and I've seen the proof of $phi(n)=sum_{dvert n} mu(d)n/d$ which uses a very elegant way, and I think most of people cannot have that idea in mind.

By the way, I have tried when $n=6$, it shows begin{align*}mu* F(6)=sum_{dvert 6}mubigg(frac 6dbigg)sum_{k=1}^d fbigg(frac kdbigg)end{align*}
And I just merely expanding everythings out, using the definitions of them, found it equals to $f(1/6)+f(5/6)$, it is true, but I cannot find any approach to prove it. Please help.






number-theory analytic-number-theory dirichlet-convolution







share|cite|improve this question











$endgroup$




Let $f(x)$ be defined for all rational $x$ in $0leq xleq 1$
$$F(n)=sum_{k=1}^n fbigg(frac knbigg), quad F^* (n)=sum_{k=1\(k,n)=1}^nfbigg(frac knbigg).$$
Prove that $$F*=mu * F$$
where $*$ denoted the Dirichlet Multiplication and $mu$ is the Mobius Function.



I have knowledge about Elementary Number Theory, and now I'm studying Analytic Number Theory by Apostol, which I feel very hard to self-learn, and I've seen the proof of $phi(n)=sum_{dvert n} mu(d)n/d$ which uses a very elegant way, and I think most of people cannot have that idea in mind.

By the way, I have tried when $n=6$, it shows begin{align*}mu* F(6)=sum_{dvert 6}mubigg(frac 6dbigg)sum_{k=1}^d fbigg(frac kdbigg)end{align*}
And I just merely expanding everythings out, using the definitions of them, found it equals to $f(1/6)+f(5/6)$, it is true, but I cannot find any approach to prove it. Please help.






number-theory analytic-number-theory dirichlet-convolution




number-theory analytic-number-theory dirichlet-convolution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 10:50







kelvin hong 方

















asked Jan 9 at 8:53









kelvin hong 方kelvin hong 方

3508




3508








  • 3




    $begingroup$
    I do not see a difference between $F$ and $F*$.
    $endgroup$
    – Mindlack
    Jan 9 at 9:06










  • $begingroup$
    @Mindlack Thanks for mentioning, I have added the detail.
    $endgroup$
    – kelvin hong 方
    Jan 9 at 10:51






  • 1




    $begingroup$
    Have you seen Möbius inversion formula?
    $endgroup$
    – Bruno Andrades
    Jan 9 at 11:02










  • $begingroup$
    @BrunoAndrades Yes, oh thank you! I didn't realize I can use that formula! I think I get it, thank you!
    $endgroup$
    – kelvin hong 方
    Jan 9 at 11:26














  • 3




    $begingroup$
    I do not see a difference between $F$ and $F*$.
    $endgroup$
    – Mindlack
    Jan 9 at 9:06










  • $begingroup$
    @Mindlack Thanks for mentioning, I have added the detail.
    $endgroup$
    – kelvin hong 方
    Jan 9 at 10:51






  • 1




    $begingroup$
    Have you seen Möbius inversion formula?
    $endgroup$
    – Bruno Andrades
    Jan 9 at 11:02










  • $begingroup$
    @BrunoAndrades Yes, oh thank you! I didn't realize I can use that formula! I think I get it, thank you!
    $endgroup$
    – kelvin hong 方
    Jan 9 at 11:26








3




3




$begingroup$
I do not see a difference between $F$ and $F*$.
$endgroup$
– Mindlack
Jan 9 at 9:06




$begingroup$
I do not see a difference between $F$ and $F*$.
$endgroup$
– Mindlack
Jan 9 at 9:06












$begingroup$
@Mindlack Thanks for mentioning, I have added the detail.
$endgroup$
– kelvin hong 方
Jan 9 at 10:51




$begingroup$
@Mindlack Thanks for mentioning, I have added the detail.
$endgroup$
– kelvin hong 方
Jan 9 at 10:51




1




1




$begingroup$
Have you seen Möbius inversion formula?
$endgroup$
– Bruno Andrades
Jan 9 at 11:02




$begingroup$
Have you seen Möbius inversion formula?
$endgroup$
– Bruno Andrades
Jan 9 at 11:02












$begingroup$
@BrunoAndrades Yes, oh thank you! I didn't realize I can use that formula! I think I get it, thank you!
$endgroup$
– kelvin hong 方
Jan 9 at 11:26




$begingroup$
@BrunoAndrades Yes, oh thank you! I didn't realize I can use that formula! I think I get it, thank you!
$endgroup$
– kelvin hong 方
Jan 9 at 11:26










1 Answer
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$begingroup$

Thanks to @BrunoAndrades, I should first prove $F=u * F^*$, where $u(n)=1$ for all positive integer $n$, which is quite easier:
begin{align*}F(n)&=sum_{k=1}^nfbigg(dfrac knbigg)
end{align*}
Among all fractions $k/n$ where $1leq kleq n$, after we convert it to the simplest fraction form, it is of the form $k'/n'$ where $(k',n')=1$, then since $n'|n$ and $fbigg(dfrac{k'}{n'}bigg)$ is contained in the summation of $F^*(n')$, hence we finally can conclude that $$F(n)=sum_{d|n}sum_{k=1\(k,d)=1}^dfbigg(dfrac kdbigg)=sum_{d|n}F^*(d)=u*F^*$$
So that Möbius Inversion Formula tell us $$F^*=mu * F.$$






share|cite|improve this answer









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    $begingroup$

    Thanks to @BrunoAndrades, I should first prove $F=u * F^*$, where $u(n)=1$ for all positive integer $n$, which is quite easier:
    begin{align*}F(n)&=sum_{k=1}^nfbigg(dfrac knbigg)
    end{align*}
    Among all fractions $k/n$ where $1leq kleq n$, after we convert it to the simplest fraction form, it is of the form $k'/n'$ where $(k',n')=1$, then since $n'|n$ and $fbigg(dfrac{k'}{n'}bigg)$ is contained in the summation of $F^*(n')$, hence we finally can conclude that $$F(n)=sum_{d|n}sum_{k=1\(k,d)=1}^dfbigg(dfrac kdbigg)=sum_{d|n}F^*(d)=u*F^*$$
    So that Möbius Inversion Formula tell us $$F^*=mu * F.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Thanks to @BrunoAndrades, I should first prove $F=u * F^*$, where $u(n)=1$ for all positive integer $n$, which is quite easier:
      begin{align*}F(n)&=sum_{k=1}^nfbigg(dfrac knbigg)
      end{align*}
      Among all fractions $k/n$ where $1leq kleq n$, after we convert it to the simplest fraction form, it is of the form $k'/n'$ where $(k',n')=1$, then since $n'|n$ and $fbigg(dfrac{k'}{n'}bigg)$ is contained in the summation of $F^*(n')$, hence we finally can conclude that $$F(n)=sum_{d|n}sum_{k=1\(k,d)=1}^dfbigg(dfrac kdbigg)=sum_{d|n}F^*(d)=u*F^*$$
      So that Möbius Inversion Formula tell us $$F^*=mu * F.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Thanks to @BrunoAndrades, I should first prove $F=u * F^*$, where $u(n)=1$ for all positive integer $n$, which is quite easier:
        begin{align*}F(n)&=sum_{k=1}^nfbigg(dfrac knbigg)
        end{align*}
        Among all fractions $k/n$ where $1leq kleq n$, after we convert it to the simplest fraction form, it is of the form $k'/n'$ where $(k',n')=1$, then since $n'|n$ and $fbigg(dfrac{k'}{n'}bigg)$ is contained in the summation of $F^*(n')$, hence we finally can conclude that $$F(n)=sum_{d|n}sum_{k=1\(k,d)=1}^dfbigg(dfrac kdbigg)=sum_{d|n}F^*(d)=u*F^*$$
        So that Möbius Inversion Formula tell us $$F^*=mu * F.$$






        share|cite|improve this answer









        $endgroup$



        Thanks to @BrunoAndrades, I should first prove $F=u * F^*$, where $u(n)=1$ for all positive integer $n$, which is quite easier:
        begin{align*}F(n)&=sum_{k=1}^nfbigg(dfrac knbigg)
        end{align*}
        Among all fractions $k/n$ where $1leq kleq n$, after we convert it to the simplest fraction form, it is of the form $k'/n'$ where $(k',n')=1$, then since $n'|n$ and $fbigg(dfrac{k'}{n'}bigg)$ is contained in the summation of $F^*(n')$, hence we finally can conclude that $$F(n)=sum_{d|n}sum_{k=1\(k,d)=1}^dfbigg(dfrac kdbigg)=sum_{d|n}F^*(d)=u*F^*$$
        So that Möbius Inversion Formula tell us $$F^*=mu * F.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 11:34









        kelvin hong 方kelvin hong 方

        3508




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