Expression for the inverse of Euler's totient function $phi^{-1}$












2












$begingroup$


I have to demonstrate that $$phi^{-1}(n)= prod_{p|n}(1-p)$$ where $phi(n)$ is the Euler's totient function.



I know that I can write $phi$ in terms of the Mobius function $mu$ as$$phi(n)= sum_{n|d} mu(d) cdot left(frac{n}{d}right)$$
and I tried to substitute this in the definition of Dirichlet's inverse $$f^{-1}(n)=- frac{1}{f(1)} sum_{d|n \d<n} f left(frac{n}{d}right)f^{-1}(d) $$ but I can't find a solution.



Can anyone give me a hint?






number-theory totient-function dirichlet-convolution







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I have to demonstrate that $$phi^{-1}(n)= prod_{p|n}(1-p)$$ where $phi(n)$ is the Euler's totient function.



    I know that I can write $phi$ in terms of the Mobius function $mu$ as$$phi(n)= sum_{n|d} mu(d) cdot left(frac{n}{d}right)$$
    and I tried to substitute this in the definition of Dirichlet's inverse $$f^{-1}(n)=- frac{1}{f(1)} sum_{d|n \d<n} f left(frac{n}{d}right)f^{-1}(d) $$ but I can't find a solution.



    Can anyone give me a hint?






    number-theory totient-function dirichlet-convolution







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      I have to demonstrate that $$phi^{-1}(n)= prod_{p|n}(1-p)$$ where $phi(n)$ is the Euler's totient function.



      I know that I can write $phi$ in terms of the Mobius function $mu$ as$$phi(n)= sum_{n|d} mu(d) cdot left(frac{n}{d}right)$$
      and I tried to substitute this in the definition of Dirichlet's inverse $$f^{-1}(n)=- frac{1}{f(1)} sum_{d|n \d<n} f left(frac{n}{d}right)f^{-1}(d) $$ but I can't find a solution.



      Can anyone give me a hint?






      number-theory totient-function dirichlet-convolution







      share|cite|improve this question









      $endgroup$




      I have to demonstrate that $$phi^{-1}(n)= prod_{p|n}(1-p)$$ where $phi(n)$ is the Euler's totient function.



      I know that I can write $phi$ in terms of the Mobius function $mu$ as$$phi(n)= sum_{n|d} mu(d) cdot left(frac{n}{d}right)$$
      and I tried to substitute this in the definition of Dirichlet's inverse $$f^{-1}(n)=- frac{1}{f(1)} sum_{d|n \d<n} f left(frac{n}{d}right)f^{-1}(d) $$ but I can't find a solution.



      Can anyone give me a hint?






      number-theory totient-function dirichlet-convolution




      number-theory totient-function dirichlet-convolution






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 14 at 16:25









      Phi_24Phi_24

      1938




      1938






















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          $begingroup$

          Since $phi(n)$ is multiplicative, i.e., satisfies $phi(nm) = phi(n)phi(m)$ for all $m,n$ with
          $gcd(n,m) = 1$,
          it follows from $phi(p^k)= p^{k-1}(p-1)$ that
          $$sum_{d | n} phi(d) = n, qquad phi(n) = sum_{d | n} mu(d) frac{n}{d}$$
          by the Moebius inversion formula. Thus
          $$phi^{-1}(n) = sum_{d | n} d mu(d)$$
          by the definition of the Dirichlet inverse. But we have, for every multiplicative arithmetic function $f$ that
          $$
          sum_{dmid n}mu(d)f(d)=prod_{pmid n}(1-f(p)).
          $$
          Just check this for prime powers and use the multiplicativity of $f$. Now take $f=id$ to obtain
          $$phi^{-1}(n)= prod_{p|n}(1-p).$$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            Since $phi(n)$ is multiplicative, i.e., satisfies $phi(nm) = phi(n)phi(m)$ for all $m,n$ with
            $gcd(n,m) = 1$,
            it follows from $phi(p^k)= p^{k-1}(p-1)$ that
            $$sum_{d | n} phi(d) = n, qquad phi(n) = sum_{d | n} mu(d) frac{n}{d}$$
            by the Moebius inversion formula. Thus
            $$phi^{-1}(n) = sum_{d | n} d mu(d)$$
            by the definition of the Dirichlet inverse. But we have, for every multiplicative arithmetic function $f$ that
            $$
            sum_{dmid n}mu(d)f(d)=prod_{pmid n}(1-f(p)).
            $$
            Just check this for prime powers and use the multiplicativity of $f$. Now take $f=id$ to obtain
            $$phi^{-1}(n)= prod_{p|n}(1-p).$$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Since $phi(n)$ is multiplicative, i.e., satisfies $phi(nm) = phi(n)phi(m)$ for all $m,n$ with
              $gcd(n,m) = 1$,
              it follows from $phi(p^k)= p^{k-1}(p-1)$ that
              $$sum_{d | n} phi(d) = n, qquad phi(n) = sum_{d | n} mu(d) frac{n}{d}$$
              by the Moebius inversion formula. Thus
              $$phi^{-1}(n) = sum_{d | n} d mu(d)$$
              by the definition of the Dirichlet inverse. But we have, for every multiplicative arithmetic function $f$ that
              $$
              sum_{dmid n}mu(d)f(d)=prod_{pmid n}(1-f(p)).
              $$
              Just check this for prime powers and use the multiplicativity of $f$. Now take $f=id$ to obtain
              $$phi^{-1}(n)= prod_{p|n}(1-p).$$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Since $phi(n)$ is multiplicative, i.e., satisfies $phi(nm) = phi(n)phi(m)$ for all $m,n$ with
                $gcd(n,m) = 1$,
                it follows from $phi(p^k)= p^{k-1}(p-1)$ that
                $$sum_{d | n} phi(d) = n, qquad phi(n) = sum_{d | n} mu(d) frac{n}{d}$$
                by the Moebius inversion formula. Thus
                $$phi^{-1}(n) = sum_{d | n} d mu(d)$$
                by the definition of the Dirichlet inverse. But we have, for every multiplicative arithmetic function $f$ that
                $$
                sum_{dmid n}mu(d)f(d)=prod_{pmid n}(1-f(p)).
                $$
                Just check this for prime powers and use the multiplicativity of $f$. Now take $f=id$ to obtain
                $$phi^{-1}(n)= prod_{p|n}(1-p).$$






                share|cite|improve this answer











                $endgroup$



                Since $phi(n)$ is multiplicative, i.e., satisfies $phi(nm) = phi(n)phi(m)$ for all $m,n$ with
                $gcd(n,m) = 1$,
                it follows from $phi(p^k)= p^{k-1}(p-1)$ that
                $$sum_{d | n} phi(d) = n, qquad phi(n) = sum_{d | n} mu(d) frac{n}{d}$$
                by the Moebius inversion formula. Thus
                $$phi^{-1}(n) = sum_{d | n} d mu(d)$$
                by the definition of the Dirichlet inverse. But we have, for every multiplicative arithmetic function $f$ that
                $$
                sum_{dmid n}mu(d)f(d)=prod_{pmid n}(1-f(p)).
                $$
                Just check this for prime powers and use the multiplicativity of $f$. Now take $f=id$ to obtain
                $$phi^{-1}(n)= prod_{p|n}(1-p).$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 14 at 19:58

























                answered Jan 14 at 19:47









                Dietrich BurdeDietrich Burde

                79.3k647103




                79.3k647103






























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