How to prove $sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1}$?












1














Prove that $$sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1},$$
where $mu$ is Möbius function, $phi$ is Euler's totient function, and $q$ is a positive integer.



I can get
begin{align}
sum_{dmid q} frac{mu(d)log d}{d}& = sum_{dmid q}frac{mu(d)}{d}sum_{pmid d}log p \
& = sum_{pmid q} log p sum_{substack{dmid q \ pmid d}} frac{mu(d)}{d}
= sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d},
end{align}
Let $d=pr$, then $mu(d)=mu(p)mu(r)=-mu(r)$,
$$ sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d}= - sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}.$$
But I don't know why
$$- sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}=-frac{phi(q)}{q} sum_{pmid q} frac{log p}{p-1}?$$



Can you help me?






elementary-number-theory arithmetic-functions







share|cite|improve this question
























  • Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
    – mathworker21
    Nov 12 at 5:40






  • 1




    @mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
    – reuns
    Nov 12 at 6:15












  • Just to make sure, does the sum over $p$ only involve prime values of $p$?
    – Batominovski
    Nov 12 at 10:28
















1














Prove that $$sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1},$$
where $mu$ is Möbius function, $phi$ is Euler's totient function, and $q$ is a positive integer.



I can get
begin{align}
sum_{dmid q} frac{mu(d)log d}{d}& = sum_{dmid q}frac{mu(d)}{d}sum_{pmid d}log p \
& = sum_{pmid q} log p sum_{substack{dmid q \ pmid d}} frac{mu(d)}{d}
= sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d},
end{align}
Let $d=pr$, then $mu(d)=mu(p)mu(r)=-mu(r)$,
$$ sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d}= - sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}.$$
But I don't know why
$$- sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}=-frac{phi(q)}{q} sum_{pmid q} frac{log p}{p-1}?$$



Can you help me?






elementary-number-theory arithmetic-functions







share|cite|improve this question
























  • Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
    – mathworker21
    Nov 12 at 5:40






  • 1




    @mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
    – reuns
    Nov 12 at 6:15












  • Just to make sure, does the sum over $p$ only involve prime values of $p$?
    – Batominovski
    Nov 12 at 10:28














1












1








1







Prove that $$sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1},$$
where $mu$ is Möbius function, $phi$ is Euler's totient function, and $q$ is a positive integer.



I can get
begin{align}
sum_{dmid q} frac{mu(d)log d}{d}& = sum_{dmid q}frac{mu(d)}{d}sum_{pmid d}log p \
& = sum_{pmid q} log p sum_{substack{dmid q \ pmid d}} frac{mu(d)}{d}
= sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d},
end{align}
Let $d=pr$, then $mu(d)=mu(p)mu(r)=-mu(r)$,
$$ sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d}= - sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}.$$
But I don't know why
$$- sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}=-frac{phi(q)}{q} sum_{pmid q} frac{log p}{p-1}?$$



Can you help me?






elementary-number-theory arithmetic-functions







share|cite|improve this question















Prove that $$sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1},$$
where $mu$ is Möbius function, $phi$ is Euler's totient function, and $q$ is a positive integer.



I can get
begin{align}
sum_{dmid q} frac{mu(d)log d}{d}& = sum_{dmid q}frac{mu(d)}{d}sum_{pmid d}log p \
& = sum_{pmid q} log p sum_{substack{dmid q \ pmid d}} frac{mu(d)}{d}
= sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d},
end{align}
Let $d=pr$, then $mu(d)=mu(p)mu(r)=-mu(r)$,
$$ sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d}= - sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}.$$
But I don't know why
$$- sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}=-frac{phi(q)}{q} sum_{pmid q} frac{log p}{p-1}?$$



Can you help me?






elementary-number-theory arithmetic-functions




elementary-number-theory arithmetic-functions






share|cite|improve this question















share|cite|improve this question













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edited Nov 12 at 5:45

























asked Nov 12 at 5:23









arithmetic1

456




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  • Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
    – mathworker21
    Nov 12 at 5:40






  • 1




    @mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
    – reuns
    Nov 12 at 6:15












  • Just to make sure, does the sum over $p$ only involve prime values of $p$?
    – Batominovski
    Nov 12 at 10:28


















  • Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
    – mathworker21
    Nov 12 at 5:40






  • 1




    @mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
    – reuns
    Nov 12 at 6:15












  • Just to make sure, does the sum over $p$ only involve prime values of $p$?
    – Batominovski
    Nov 12 at 10:28
















Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
– mathworker21
Nov 12 at 5:40




Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
– mathworker21
Nov 12 at 5:40




1




1




@mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
– reuns
Nov 12 at 6:15






@mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
– reuns
Nov 12 at 6:15














Just to make sure, does the sum over $p$ only involve prime values of $p$?
– Batominovski
Nov 12 at 10:28




Just to make sure, does the sum over $p$ only involve prime values of $p$?
– Batominovski
Nov 12 at 10:28










2 Answers
2






active

oldest

votes


















1














Let me write $n$ instead of $q$.
We have
begin{align}
sum_{d|n}frac{mu(d)log(d)}d
&=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
&=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
&=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
end{align}
Write $n=p^em$ with $pnmid m$.
Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
begin{align}
sum_{p|d|n}mu(d)frac nd
&=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
&=sum_{dmid m}mu(pd)frac{p^em}{pd}\
&=-sum_{dmid m}mu(d)frac{p^em}{pd}\
&=-p^{e-1}varphi(m)\
&=-frac{varphi(n)}{p-1}
end{align}






share|cite|improve this answer























  • Thank you very much!
    – arithmetic1
    Nov 28 at 8:28





















0














I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2



Using Dirichlet convolution
begin{eqnarray*}
- frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
&=& - frac{1}{n} left( Lambda ast varphi right) (n) \
&=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
&=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
&=& sum_{d mid n} frac{mu(d) log d}{d}.
end{eqnarray*}






share|cite|improve this answer





















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    2 Answers
    2






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    oldest

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    2 Answers
    2






    active

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    active

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    1














    Let me write $n$ instead of $q$.
    We have
    begin{align}
    sum_{d|n}frac{mu(d)log(d)}d
    &=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
    &=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
    &=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
    end{align}
    Write $n=p^em$ with $pnmid m$.
    Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
    begin{align}
    sum_{p|d|n}mu(d)frac nd
    &=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
    &=sum_{dmid m}mu(pd)frac{p^em}{pd}\
    &=-sum_{dmid m}mu(d)frac{p^em}{pd}\
    &=-p^{e-1}varphi(m)\
    &=-frac{varphi(n)}{p-1}
    end{align}






    share|cite|improve this answer























    • Thank you very much!
      – arithmetic1
      Nov 28 at 8:28


















    1














    Let me write $n$ instead of $q$.
    We have
    begin{align}
    sum_{d|n}frac{mu(d)log(d)}d
    &=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
    &=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
    &=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
    end{align}
    Write $n=p^em$ with $pnmid m$.
    Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
    begin{align}
    sum_{p|d|n}mu(d)frac nd
    &=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
    &=sum_{dmid m}mu(pd)frac{p^em}{pd}\
    &=-sum_{dmid m}mu(d)frac{p^em}{pd}\
    &=-p^{e-1}varphi(m)\
    &=-frac{varphi(n)}{p-1}
    end{align}






    share|cite|improve this answer























    • Thank you very much!
      – arithmetic1
      Nov 28 at 8:28
















    1












    1








    1






    Let me write $n$ instead of $q$.
    We have
    begin{align}
    sum_{d|n}frac{mu(d)log(d)}d
    &=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
    &=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
    &=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
    end{align}
    Write $n=p^em$ with $pnmid m$.
    Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
    begin{align}
    sum_{p|d|n}mu(d)frac nd
    &=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
    &=sum_{dmid m}mu(pd)frac{p^em}{pd}\
    &=-sum_{dmid m}mu(d)frac{p^em}{pd}\
    &=-p^{e-1}varphi(m)\
    &=-frac{varphi(n)}{p-1}
    end{align}






    share|cite|improve this answer














    Let me write $n$ instead of $q$.
    We have
    begin{align}
    sum_{d|n}frac{mu(d)log(d)}d
    &=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
    &=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
    &=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
    end{align}
    Write $n=p^em$ with $pnmid m$.
    Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
    begin{align}
    sum_{p|d|n}mu(d)frac nd
    &=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
    &=sum_{dmid m}mu(pd)frac{p^em}{pd}\
    &=-sum_{dmid m}mu(d)frac{p^em}{pd}\
    &=-p^{e-1}varphi(m)\
    &=-frac{varphi(n)}{p-1}
    end{align}







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 12 at 9:24

























    answered Nov 12 at 9:06









    Fabio Lucchini

    7,82311326




    7,82311326












    • Thank you very much!
      – arithmetic1
      Nov 28 at 8:28




















    • Thank you very much!
      – arithmetic1
      Nov 28 at 8:28


















    Thank you very much!
    – arithmetic1
    Nov 28 at 8:28






    Thank you very much!
    – arithmetic1
    Nov 28 at 8:28













    0














    I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2



    Using Dirichlet convolution
    begin{eqnarray*}
    - frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
    &=& - frac{1}{n} left( Lambda ast varphi right) (n) \
    &=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
    &=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
    &=& sum_{d mid n} frac{mu(d) log d}{d}.
    end{eqnarray*}






    share|cite|improve this answer


























      0














      I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2



      Using Dirichlet convolution
      begin{eqnarray*}
      - frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
      &=& - frac{1}{n} left( Lambda ast varphi right) (n) \
      &=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
      &=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
      &=& sum_{d mid n} frac{mu(d) log d}{d}.
      end{eqnarray*}






      share|cite|improve this answer
























        0












        0








        0






        I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2



        Using Dirichlet convolution
        begin{eqnarray*}
        - frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
        &=& - frac{1}{n} left( Lambda ast varphi right) (n) \
        &=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
        &=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
        &=& sum_{d mid n} frac{mu(d) log d}{d}.
        end{eqnarray*}






        share|cite|improve this answer












        I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2



        Using Dirichlet convolution
        begin{eqnarray*}
        - frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
        &=& - frac{1}{n} left( Lambda ast varphi right) (n) \
        &=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
        &=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
        &=& sum_{d mid n} frac{mu(d) log d}{d}.
        end{eqnarray*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 8:26









        arithmetic1

        456




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