Sum of Möbius function over Complementary Divisors of Square-Divisors

While considering the summatory function of the Liouville Function, we found that

which takes the value of 1 when n is a perfect square and 0 otherwise.  The following question was given as a challenge and we present its solution in this article.

Question
Find

Solution:-
From
     ${\displaystyle \sum _{d|n}\lambda \left(d\right)=1_{\text{PerfectSquare}}\left(n\right)}$
using the Möbius Inversion formula, we have
     ${\displaystyle \sum _{d|n}\mu \left(\frac{n}{d}\right)1_{\text{PerfectSquare}}\left(d\right)=\lambda \left(n\right)}$
The summands will be zero except when ${\displaystyle d=D^2}$.  Hence we get
     ${\displaystyle \sum _{D^2|n}\mu \left(\frac{n}{D^2}\right)=\lambda \left(n\right)}$
[End]

Möbius Inversion with Liouville Function

This article concerns Möbius Inversion with the Liouville Function.  For ${\displaystyle n=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}}$, the number of distinct prime factors of n is

Question
(i)  Prove that

(ii) Hence show that

Solution:-
(i) Note that ${\displaystyle \mu \left(d\right)=0}$ for divisors d containing any prime power of index 2 or higher.  We only need to consider divisors of the form ${\displaystyle d=p_{i_1}{p}_{i_2}...p_{i_h}}$, where ${\displaystyle h=0}$ is interpreted as ${\displaystyle d=1}$.
    ${\displaystyle \sum _{d|n}\mu \left(d\right)\lambda \left(d\right)}$
= ${\displaystyle \sum _{h=0}^k\sum _{p_{i_1},p_{i_2},...,p_{i_h}}\mu (p_{i_1}{p}_{i_2}...p_{i_h})\lambda (p_{i_1}{p}_{i_2}...p_{i_h})}$
= ${\displaystyle \sum _{h=0}^k\sum _{p_{i_1},p_{i_2},...,p_{i_h}}{(-1)}^h{(-1)}^{1+1+...+1}}$   (where 1+1+...+1 has h copies of 1)
= ${\displaystyle \sum _{h=0}^k\sum _{p_{i_1},p_{i_2},...,p_{i_h}}{(-1)}^{2h}}$ = ${\displaystyle \sum _{h=0}^k\sum _{p_{i_1},p_{i_2},...,p_{i_h}}1}$ = ${\displaystyle \sum _{h=0}^k\left(\begin{array}{c}k\\ h\end{array}\right)}$
= ${\displaystyle 2^k}$  = ${\displaystyle 2^{\omega \left(n\right)}}$
(ii)  Recall that ${\displaystyle \lambda \left(\frac{n}{d}\right)=\frac{\lambda \left(n\right)}{\lambda \left(d\right)}=\lambda \left(n\right)\lambda \left(d\right)}$
     From the result in part (i) we multiply both sides by ${\displaystyle \lambda \left(n\right)}$ to get
                  ${\displaystyle \sum _{d|n}\mu \left(d\right)\lambda \left(n\right)\lambda \left(d\right)=\lambda \left(n\right)2^{\omega \left(n\right)}}$
                  ${\displaystyle \sum _{d|n}\mu \left(d\right)\lambda \left(\frac{n}{d}\right)=\lambda \left(n\right)2^{\omega \left(n\right)}}$
    Treating this as the "Möbius Inversed" formula, the "original" formula is
                  ${\displaystyle \sum _{d|n}\lambda \left(d\right)2^{\omega \left(d\right)}=\lambda \left(n\right)}$
Hence
     ${\displaystyle \sum _{d|n}\lambda \left(\frac{n}{d}\right)2^{\omega \left(d\right)}}$ =   ${\displaystyle \sum _{d|n}\lambda \left(n\right)\lambda \left(d\right)2^{\omega \left(d\right)}}$ =   ${\displaystyle \lambda \left(n\right)\cdot \sum _{d|n}\lambda \left(d\right)2^{\omega \left(d\right)}}$
=   ${\displaystyle \lambda \left(n\right)\cdot \lambda \left(n\right)}$ =   ${\displaystyle \lambda {\left(n\right)}^2}$ =   ${\displaystyle 1}$

[End]

Liouville Function, Summatory of

We introduced the Liouville function here.  The summatory function of ${\displaystyle f\left(n\right)}$ is defined as

In this post, we examine the summatory function of the Liouville function.


Question
Prove that

Solution:-
   Note that for ${\displaystyle n=p^a}$ where p is a prime number,
   ${\displaystyle \sum _{d|n}\lambda \left(d\right)}$  =   ${\displaystyle \sum _{j=0}^a\lambda \left(p^j\right)}$  =   ${\displaystyle \sum _{j=0}^a{(-1)}^j}$
=   ${\displaystyle \frac{1\cdot (1-{(-1)}^{a+1})}{1-(-1)}}$  =   ${\displaystyle \frac{1-{(-1)}^{a+1}}{2}}$
=   ${\displaystyle 1_{\text{even}}\left(a\right)}$  =   ${\displaystyle \{\begin{array}{ll}1& \text{if}a\text{is even}\\ 0& \text{if}a\text{is odd}\end{array}}$       ("Even indicator function")

Hence for ${\displaystyle n=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}}$
   ${\displaystyle \sum _{d|n}\lambda \left(d\right)}$  =   ${\displaystyle \sum _{i_1,i_2,...,i_k}\lambda (p_1^{i_1}{p}_2^{i_2}...p_k^{i_k})}$ =   ${\displaystyle \sum _{i_1,i_2,...,i_k}\lambda \left(p_1^{i_1}\right)\lambda \left(p_2^{i_2}\right)...\lambda \left(p_k^{i_k}\right)}$
=   ${\displaystyle \sum _{i_1=0}^{a_1}\lambda \left(p_1^{i_1}\right)\sum _{i_2=0}^{a_2}\lambda \left(p_2^{i_2}\right)...\sum _{i_1=0}^{a_k}\lambda \left(p_k^{i_k}\right)}$
=   ${\displaystyle 1_{\text{even}}\left(a_1\right)1_{\text{even}}\left(a_2\right)...1_{\text{even}}\left(a_k\right)}$ 
=   ${\displaystyle \{\begin{array}{ll}1& \text{if}\forall i{a}_i\in 2\mathbb{Z}\\ 0& \text{otherwise}\end{array}}$      =   ${\displaystyle \{\begin{array}{ll}1& \text{if}n\text{is a perfect square}\\ 0& \text{otherwise}\end{array}}$

Challenge: Find ${\displaystyle \sum _{D^2|n}\mu \left(\frac{n}{D^2}\right)}$, where ${\displaystyle \mu \left(\right)}$ is the Möbius function.  Solution here.

Liouville Function - Introduction

The Liouville function, after Joseph Liouville, is defined for ${\displaystyle n=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}}$ as

interpreted so that ${\displaystyle \lambda \left(1\right)=1}$.  It is quite obvious that

This, coupled with the fact that the Liouville function is completely multiplicative (see below), we have

 

Question
Prove that the Liouville function is completely multiplicative i.e.

for all integers m and n (not necessarily coprime).
Solution:-
   Let ${\displaystyle p_1,p_2,...,p_k}$ be the list of primes appearing in m and/or n. Write ${\displaystyle m=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}}$ and ${\displaystyle n=p_1^{b_1}{p}_2^{b_2}...p_k^{b_k}}$, in which any of the non-negative integers ${\displaystyle a_1,a_2,...,a_k }$ and ${\displaystyle b_1,b_2,...,b_k }$ can be zero.  Then
                     ${\displaystyle mn=p_1^{a_1+b_1}{p}_2^{a_2+b_2}...p_k^{a_k+b_k}}$
Consequently
   ${\displaystyle \lambda \left(mn\right)={(-1)}^{(a_1+b_1)+(a_2+b_2)+...+(a_k+b_k)}}$
            ${\displaystyle ={(-1)}^{(a_1+a_2+...+a_k)+(b_1+b_2+...+b_k)}}$
            ${\displaystyle ={(-1)}^{a_1+a_2+...+a_k}{(-1)}^{b_1+b_2+...+b_k}}$
            ${\displaystyle =\lambda \left(m\right)\lambda \left(n\right)}$
[End]


More about the Liouville function here, and a result involving Möbius Inversion function here.

A Polar Double Integral

  ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\int }_0^{1+sin\theta }\theta r\text{d}r\text{d}\theta }$
= ${\displaystyle {\int }_0^{\frac{\pi }{2}}\theta {\left[\frac{1}{2}{r}^2\right]}_0^{1+sin\theta }\text{d}\theta }$
= ${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{1}{2}\theta {(1+sin\theta )}^2\text{d}\theta }$
= ${\displaystyle \frac{1}{2}{\int }_0^{\frac{\pi }{2}}\theta (1+2sin\theta +{sin}^2\theta )\text{d}\theta }$
= ${\displaystyle \frac{1}{2}{\int }_0^{\frac{\pi }{2}}\theta (1+2sin\theta +\frac{1-cos2\theta }{2})\text{d}\theta }$
= ${\displaystyle \frac{1}{2}{\int }_0^{\frac{\pi }{2}}\theta (\frac{3}{2}+2sin\theta -\frac{1}{2}cos2\theta )\text{d}\theta }$

= ${\displaystyle \frac{3}{4}{\int }_0^{\frac{\pi }{2}}\theta \text{d}\theta +{\int }_0^{\frac{\pi }{2}}\theta (sin\theta -\frac{1}{4}cos2\theta )\text{d}\theta }$
= ${\displaystyle \frac{3}{4}{\left[\frac{{\theta }^2}{2}\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\theta \frac{\text{d}}{\text{d}\theta }(cos\theta +\frac{1}{8}sin2\theta )\text{d}\theta }$
= ${\displaystyle \frac{3}{8}\cdot \frac{{\pi }^2}{4}-{\left[\theta (cos\theta +\frac{1}{8}sin2\theta )\right]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}cos\theta +\frac{1}{8}sin2\theta \text{d}\theta }$
= ${\displaystyle \frac{3{\pi }^2}{32}-0+{[sin\theta -\frac{1}{16}cos2\theta ]}_0^{\frac{\pi }{2}}}$
= ${\displaystyle \frac{3{\pi }^2}{32}+[(1+\frac{1}{16})-(0-\frac{1}{16})]}$
= ${\displaystyle \frac{3{\pi }^2}{32}+\frac{9}{8}}$