Proof of Mobius inversion formula - the other direction












3












$begingroup$


The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$



We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.



I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem.






elementary-number-theory mobius-inversion







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  • 1




    $begingroup$
    How about just doing it? Put the second into the first, and hack away....
    $endgroup$
    – Lord Shark the Unknown
    Jan 27 at 10:41










  • $begingroup$
    @LordSharktheUnknown not sure how I would do that, sorry
    $endgroup$
    – windircurse
    Jan 27 at 10:54






  • 2




    $begingroup$
    It's kind of hard to avoid using convolutions when there's one in the problem's statement.
    $endgroup$
    – jmerry
    Jan 27 at 10:57
















3












$begingroup$


The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$



We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.



I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem.






elementary-number-theory mobius-inversion







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    How about just doing it? Put the second into the first, and hack away....
    $endgroup$
    – Lord Shark the Unknown
    Jan 27 at 10:41










  • $begingroup$
    @LordSharktheUnknown not sure how I would do that, sorry
    $endgroup$
    – windircurse
    Jan 27 at 10:54






  • 2




    $begingroup$
    It's kind of hard to avoid using convolutions when there's one in the problem's statement.
    $endgroup$
    – jmerry
    Jan 27 at 10:57














3












3








3


1



$begingroup$


The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$



We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.



I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem.






elementary-number-theory mobius-inversion







share|cite|improve this question









$endgroup$




The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$



We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.



I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem.






elementary-number-theory mobius-inversion




elementary-number-theory mobius-inversion






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 27 at 10:39









windircursewindircurse

1,154820




1,154820








  • 1




    $begingroup$
    How about just doing it? Put the second into the first, and hack away....
    $endgroup$
    – Lord Shark the Unknown
    Jan 27 at 10:41










  • $begingroup$
    @LordSharktheUnknown not sure how I would do that, sorry
    $endgroup$
    – windircurse
    Jan 27 at 10:54






  • 2




    $begingroup$
    It's kind of hard to avoid using convolutions when there's one in the problem's statement.
    $endgroup$
    – jmerry
    Jan 27 at 10:57














  • 1




    $begingroup$
    How about just doing it? Put the second into the first, and hack away....
    $endgroup$
    – Lord Shark the Unknown
    Jan 27 at 10:41










  • $begingroup$
    @LordSharktheUnknown not sure how I would do that, sorry
    $endgroup$
    – windircurse
    Jan 27 at 10:54






  • 2




    $begingroup$
    It's kind of hard to avoid using convolutions when there's one in the problem's statement.
    $endgroup$
    – jmerry
    Jan 27 at 10:57








1




1




$begingroup$
How about just doing it? Put the second into the first, and hack away....
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:41




$begingroup$
How about just doing it? Put the second into the first, and hack away....
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:41












$begingroup$
@LordSharktheUnknown not sure how I would do that, sorry
$endgroup$
– windircurse
Jan 27 at 10:54




$begingroup$
@LordSharktheUnknown not sure how I would do that, sorry
$endgroup$
– windircurse
Jan 27 at 10:54




2




2




$begingroup$
It's kind of hard to avoid using convolutions when there's one in the problem's statement.
$endgroup$
– jmerry
Jan 27 at 10:57




$begingroup$
It's kind of hard to avoid using convolutions when there's one in the problem's statement.
$endgroup$
– jmerry
Jan 27 at 10:57










1 Answer
1






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oldest

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1












$begingroup$

We assume
begin{align*}
g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
end{align*}
and show the validity of
begin{align*}
f(m)=sum_{d|m}g(d)tag{2}
end{align*}



It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.




We start with the right-hand side of (2) and obtain
begin{align*}
color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
&=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
&=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
&=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
&=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
&=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
&=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
&,,color{blue}{=f(n)}tag{5}
end{align*}



and the claim (2) follows.




Comment:




  • In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.


  • In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.


  • In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.



Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.






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    1 Answer
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    active

    oldest

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    1












    $begingroup$

    We assume
    begin{align*}
    g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
    end{align*}
    and show the validity of
    begin{align*}
    f(m)=sum_{d|m}g(d)tag{2}
    end{align*}



    It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.




    We start with the right-hand side of (2) and obtain
    begin{align*}
    color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
    &=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
    &=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
    &=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
    &=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
    &=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
    &=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
    &,,color{blue}{=f(n)}tag{5}
    end{align*}



    and the claim (2) follows.




    Comment:




    • In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.


    • In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.


    • In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.



    Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      We assume
      begin{align*}
      g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
      end{align*}
      and show the validity of
      begin{align*}
      f(m)=sum_{d|m}g(d)tag{2}
      end{align*}



      It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.




      We start with the right-hand side of (2) and obtain
      begin{align*}
      color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
      &=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
      &=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
      &=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
      &=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
      &=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
      &=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
      &,,color{blue}{=f(n)}tag{5}
      end{align*}



      and the claim (2) follows.




      Comment:




      • In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.


      • In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.


      • In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.



      Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        We assume
        begin{align*}
        g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
        end{align*}
        and show the validity of
        begin{align*}
        f(m)=sum_{d|m}g(d)tag{2}
        end{align*}



        It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.




        We start with the right-hand side of (2) and obtain
        begin{align*}
        color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
        &=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
        &=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
        &=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
        &=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
        &=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
        &=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
        &,,color{blue}{=f(n)}tag{5}
        end{align*}



        and the claim (2) follows.




        Comment:




        • In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.


        • In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.


        • In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.



        Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.






        share|cite|improve this answer











        $endgroup$



        We assume
        begin{align*}
        g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
        end{align*}
        and show the validity of
        begin{align*}
        f(m)=sum_{d|m}g(d)tag{2}
        end{align*}



        It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.




        We start with the right-hand side of (2) and obtain
        begin{align*}
        color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
        &=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
        &=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
        &=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
        &=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
        &=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
        &=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
        &,,color{blue}{=f(n)}tag{5}
        end{align*}



        and the claim (2) follows.




        Comment:




        • In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.


        • In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.


        • In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.



        Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 28 at 17:23

























        answered Jan 27 at 23:17









        Markus ScheuerMarkus Scheuer

        63k460150




        63k460150






























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