Proving by Möbius' inversion formula if $frac nphi(n)=sumlimits_dmid nf(d)$ then $f(n)=fracmu^2(n)phi(n)$ The 2019 Stack Overflow Developer Survey Results Are InProve that $ fracnphi(n) = sumlimits_d mid n fracmu^2(d)phi(d) $Looking for help understanding the Möbius Inversion FormulaProve that if $n>10$ then $sum_dmid nphi(phi(d))<frac35n$Prove by strong induction that if $sum_dmid n f(d) =F(n)$, and $F(n)$ is multiplative, then so is $f(n)$How to show the existence of a primitive root modulo $p$?Divisor sum or möbius inversion of the Jacobi Symbol?How do I prove the following equality using Mobius inversion.Proving the following using Mobius Inversion.What is the explicit formula for $Phi(x)=sumlimits_n=1^xphi(n)$?Does $sumlimits_n=1^infty lnleft(fracp_np_n - 1right)$ converge?
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Proving by Möbius' inversion formula if $frac nphi(n)=sumlimits_dmid nf(d)$ then $f(n)=fracmu^2(n)phi(n)$
The 2019 Stack Overflow Developer Survey Results Are InProve that $ fracnphi(n) = sumlimits_d mid n fracmu^2(d)phi(d) $Looking for help understanding the Möbius Inversion FormulaProve that if $n>10$ then $sum_dmid nphi(phi(d))<frac35n$Prove by strong induction that if $sum_dmid n f(d) =F(n)$, and $F(n)$ is multiplative, then so is $f(n)$How to show the existence of a primitive root modulo $p$?Divisor sum or möbius inversion of the Jacobi Symbol?How do I prove the following equality using Mobius inversion.Proving the following using Mobius Inversion.What is the explicit formula for $Phi(x)=sumlimits_n=1^xphi(n)$?Does $sumlimits_n=1^infty lnleft(fracp_np_n - 1right)$ converge?
$begingroup$
Prove by Möbius' inversion formula if $dfrac nphi(n)=sumlimits_dmid nf(d)$ then $f(n)=dfracmu^2(n)phi(n)$.
Notice $phi$ is the Euler totient function and $mu$ is the Mobius function.
Recall that Möbius' inversion formula states if $g(n)=sumlimits_dmid nf(d)$, then $f(n)=sumlimits_dmid nmu(d)gleft(fracndright)$. In this question, $g(n)=dfracnphi(n)$, then $$f(n)=sum_dmid nmu(d)g(fracnd)=sum_dmid nmu(d)fracfracndphi(fracnd).$$
From this I got trouble in simply the last equation.
I have read a similar question about prove $dfracnphi(n)=sumlimits_dmid ndfracmu^2(d)phi(d)$. But they were using the fact that both sides are multiplicative.
number-theory
edited Mar 30 at 16:20
Saad
20.4k92452
asked Mar 30 at 16:07
Jaqen ChouJaqen Chou
487110
$endgroup$
add a comment |
$begingroup$
Prove by Möbius' inversion formula if $dfrac nphi(n)=sumlimits_dmid nf(d)$ then $f(n)=dfracmu^2(n)phi(n)$.
Notice $phi$ is the Euler totient function and $mu$ is the Mobius function.
Recall that Möbius' inversion formula states if $g(n)=sumlimits_dmid nf(d)$, then $f(n)=sumlimits_dmid nmu(d)gleft(fracndright)$. In this question, $g(n)=dfracnphi(n)$, then $$f(n)=sum_dmid nmu(d)g(fracnd)=sum_dmid nmu(d)fracfracndphi(fracnd).$$
From this I got trouble in simply the last equation.
I have read a similar question about prove $dfracnphi(n)=sumlimits_dmid ndfracmu^2(d)phi(d)$. But they were using the fact that both sides are multiplicative.
number-theory
edited Mar 30 at 16:20
Saad
20.4k92452
asked Mar 30 at 16:07
Jaqen ChouJaqen Chou
487110
$endgroup$
add a comment |
$begingroup$
Prove by Möbius' inversion formula if $dfrac nphi(n)=sumlimits_dmid nf(d)$ then $f(n)=dfracmu^2(n)phi(n)$.
Notice $phi$ is the Euler totient function and $mu$ is the Mobius function.
Recall that Möbius' inversion formula states if $g(n)=sumlimits_dmid nf(d)$, then $f(n)=sumlimits_dmid nmu(d)gleft(fracndright)$. In this question, $g(n)=dfracnphi(n)$, then $$f(n)=sum_dmid nmu(d)g(fracnd)=sum_dmid nmu(d)fracfracndphi(fracnd).$$
From this I got trouble in simply the last equation.
I have read a similar question about prove $dfracnphi(n)=sumlimits_dmid ndfracmu^2(d)phi(d)$. But they were using the fact that both sides are multiplicative.
number-theory
edited Mar 30 at 16:20
Saad
20.4k92452
asked Mar 30 at 16:07
Jaqen ChouJaqen Chou
487110
$endgroup$
Prove by Möbius' inversion formula if $dfrac nphi(n)=sumlimits_dmid nf(d)$ then $f(n)=dfracmu^2(n)phi(n)$.
Notice $phi$ is the Euler totient function and $mu$ is the Mobius function.
Recall that Möbius' inversion formula states if $g(n)=sumlimits_dmid nf(d)$, then $f(n)=sumlimits_dmid nmu(d)gleft(fracndright)$. In this question, $g(n)=dfracnphi(n)$, then $$f(n)=sum_dmid nmu(d)g(fracnd)=sum_dmid nmu(d)fracfracndphi(fracnd).$$
From this I got trouble in simply the last equation.
I have read a similar question about prove $dfracnphi(n)=sumlimits_dmid ndfracmu^2(d)phi(d)$. But they were using the fact that both sides are multiplicative.
number-theory
number-theory
edited Mar 30 at 16:20
Saad
20.4k92452
asked Mar 30 at 16:07
Jaqen ChouJaqen Chou
487110
edited Mar 30 at 16:20
Saad
20.4k92452
asked Mar 30 at 16:07
Jaqen ChouJaqen Chou
487110
edited Mar 30 at 16:20
Saad
20.4k92452
edited Mar 30 at 16:20
Saad
20.4k92452
edited Mar 30 at 16:20
Saad
20.4k92452
20.4k92452
asked Mar 30 at 16:07
Jaqen ChouJaqen Chou
487110
asked Mar 30 at 16:07
Jaqen ChouJaqen Chou
487110
asked Mar 30 at 16:07
Jaqen ChouJaqen Chou
487110
487110
add a comment |
add a comment |
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