Consider the ##n##th Ramanujan sum, $$c_n(k) = \sum_{\substack{m = 1\\(m,n) = 1}}^n \exp\left\{2\pi i \frac{k m}{n}\right\}$$ Show that $$c_n(k) = \sum_{d\mid (k,n)} d\, \mu\left(\frac{n}{d}\right)$$
Can you define the terms please?
#3
Euge
Gold Member
MHB
POTW Director
2,072
245
The function ##\mu## is the Möbius function: it is defined by setting ##\mu(1) = 1## and for integers ##n > 1##, ##\mu(n) = (-1)^r## if ##n## is a product of ##r## distinct prime factors. Otherwise ##\mu(n) = 0##.
The notation ##(a,b)## means the gcd of ##a## and ##b##, and ##\sum_{d\mid (k,n)}## means the sum over all positive divisors ##d## of ##(k,n)##.
I looked at the case where ##k=1## first. Proved the Mobius inversion formula. Answered the question for ##k=1##. Then proved the general case.
Case ##k=1##.
Consider
\begin{align*}
c_n (1) = \sum_{m=1 , (m,n)=1}^n \exp \left\{ 2 \pi i \frac{m}{n} \right\}
\end{align*}
We wish to show:
\begin{align*}
c_n (1) = \sum_{d|(1,n)} d \; \mu \left( \frac{n}{d} \right) = \mu (n)
\end{align*}
The fractions ##\frac{l}{n}##, ##l=1,2, \dots, n## and the fractions defined by ##\frac{m}{d}##, ##(m,d)=1##, ##1 \leq m \leq d## for the set of divisors of ##n##, ##\{ d_1, d_2, \dots , d_s \}##, are the same set of fractions. As such:
\begin{align*}
\sum_{d|n} \sum_{m=1 , (m,d)=1}^d f \left( \frac{m}{d} \right) = \sum_{l=1}^n f \left( \frac{l}{n} \right)
\end{align*}
We use this, for ##n>1##,
\begin{align*}
\sum_{d|n} c_d (1) = \sum_{d|n} \sum_{m=1 , (m,d)=1}^d \exp \left\{ 2 \pi i \frac{m}{d} \right\} = \sum_{l=1}^n \exp \left\{ 2 \pi i \frac{l}{n} \right\} = \dfrac{\exp \left\{ 2 \pi i \right\} - 1}{\exp \left\{ 2 \pi i \frac{1}{n} \right\} - 1} = 0
\end{align*}
If ##n=1##:
\begin{align*}
\sum_{d|1} c_d (1) = \sum_{d|1} \sum_{m=1 , (m,d)=1}^d \exp \left\{ 2 \pi i \frac{m}{d} \right\} = 1
\end{align*}
Next consider the same sum for ##\mu (n)##. First take ##n>1##. We have ##n = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r}##. Note the divisors ##d## of ##n## such that ##\mu (d) \not= 0## are the product of members of subsets of: ##\{ p_1 , p_2 , \cdots , p_r \}## (the empty set corresponds to ##d=1##). As there are ##\dfrac{r!}{k! (r-k)!}## ways of choosing subsets of size ##k##, the following:
\begin{align*}
\sum_{k=0}^r \dfrac{r!}{k! (r-k)!} (-1)^k
\end{align*}
will tell us how many more subsets of even number there are than subsets of odd number (where ##k=0## term corresponds to the empty set). The above sum is equal to ##[1 + (-1)]^r = 0##, meaning that there are an equal number of subsets of even number and subsets and odd number. As such:
\begin{align*}
\sum_{d|n} \mu (d) = 0 \qquad (n>1) .
\end{align*}
For ##n=1##:
\begin{align*}
\sum_{d|1} \mu (1) = 1 .
\end{align*}
This second result can be used to prove the Mobius inversion formula:
\begin{align*}
\text{if} \qquad
g (n) = \sum_{d|n} f(d)
\qquad \text{then} \qquad
f (n) = \sum_{d|n} \mu(d) g \left( \frac{n}{d} \right)
\end{align*}
If the divisors of ##n## are ##\{ d_1 , d_2, \dots d_s \}##, define ##\tilde{d}_i = \frac{n}{d_i}## for each ##i##. The ##\{ \tilde{d}_1 , \tilde{d}_2, \dots \tilde{d}_s \}## are the ##s## original distinct divisors in a new order, and obviously, ##d_i = \frac{n}{\tilde{d}_i}##. So that
\begin{align*}
\sum_{d|n} \mu(d) g \left( \frac{n}{d} \right) = \sum_{i=1}^s \mu (d_i) g (\tilde{d}_i) = \sum_{i=1}^s \mu (\tilde{d}_i) g (d_i) = \sum_{d|n} \mu \left( \frac{n}{d} \right) g (d)
\end{align*}
So we have the alternative version of the Mobius inversion formula:
\begin{align*}
\text{if} \qquad
g (n) = \sum_{d|n} f(d)
\qquad \text{then} \qquad
f (n) = \sum_{d|n} \mu \left( \frac{n}{d} \right) g (d) .
\end{align*}
Taking an example:
\begin{align*}
& \sum_{d|8} \mu(d) g \left( \frac{8}{d} \right)
\nonumber \\
& = \sum_{d|8} \mu(d) \sum_{d'|\frac{8}{d}} f \left( d' \right)
\nonumber \\
& = \mu(1) \sum_{d'|8} f \left( d' \right) + \mu(2) \sum_{d'|\frac{8}{2}} f \left( d' \right) + \mu(4) \sum_{d'|\frac{8}{4}} f \left( d' \right) + \mu(8) \sum_{d'|\frac{8}{8}} f \left( d' \right)
\nonumber \\
& = \mu(1) f \left( 1 \right) + \mu(2) f \left( 1 \right) + \mu(4) f \left( 1 \right) + \mu(4) f \left( 1 \right)
\nonumber \\
& + \mu(1) f \left( 2 \right) + \mu(2) f \left( 2 \right) + \mu(4) f \left( 2 \right) + \qquad 0
\nonumber \\
& + \mu(1) f \left( 4 \right) + \mu(2) f \left( 4 \right) + \qquad 0 \qquad + \quad \;\; 0
\nonumber \\
& + \mu(1) f \left( 8 \right) + \quad \;\;\; 0 \quad \;\;\; + \quad \;\;\; 0 \quad \;\;\; + \quad \;\;\; 0
\nonumber \\
& = f \left( 1 \right) \sum_{d|8} \mu(d) + f \left( 2 \right) \sum_{d|4} \mu(d) + f \left( 4 \right) \sum_{d|2} \mu(d) + f \left( 8 \right) \sum_{d|1} \mu(1)
\nonumber \\
& = \sum_{d'|8} f \left( d' \right) \sum_{d|\frac{8}{d'}} \mu(d)
\nonumber \\
& = f \left( 8 \right) .
\end{align*}
where we have used ##(*)##.
Let ##\{ d_1 , d_2, \dots d_s \}## be the divisors of ##n##. Define
\begin{align*}
\epsilon (d_i',d_j) =
\left\{
\begin{matrix}
1 & d_i' | \frac{n}{d_j} \\
0 & d_i' \nmid \frac{n}{d_j}
\end{matrix}
\right.
\end{align*}
Note ##d_i' | \frac{n}{d_j}## if and only if ##d_j | \frac{n}{d_i'}##. Thus
\begin{align*}
\epsilon (d_i',d_j) =
\left\{
\begin{matrix}
1 & d_j | \frac{n}{d_i'} \\
0 & d_j \nmid \frac{n}{d_i'}
\end{matrix}
\right.
\end{align*}
So
\begin{align*}
\sum_{d|n} \mu(d) g \left( \frac{n}{d} \right) & = \sum_{d|n} \mu(d) \sum_{d'|\frac{n}{d}} f \left( d' \right)
\nonumber \\
& = \sum_{j=1}^s \mu(d_j) \sum_{i=1}^s \epsilon (d_i',d_j) f \left( d_i' \right)
\nonumber \\
& = \sum_{i=1}^s f \left( d_i' \right) \sum_{j=1}^s \epsilon (d_i',d_j) \mu(d_j)
\nonumber \\
& = \sum_{d'|n} f \left( d' \right) \sum_{d|\frac{n}{d'}} \mu(d)
\nonumber \\
& = f \left( n \right)
\end{align*}
where we have used ##(*)##.
Write
\begin{align*}
f (n) = \sum_{m=1 , (m,n)=1}^n \exp \left\{ 2 \pi i \frac{m}{n} \right\}
\end{align*}
then
\begin{align*}
g (n) = \sum_{d|n} \sum_{m=1 , (m,n)=1}^d \exp \left\{ 2 \pi i \frac{m}{d} \right\}
= \left\{
\begin{matrix}
1 & n=1 \\
0 & n>1
\end{matrix}
\right.
\end{align*}
and then
\begin{align*}
f (n) = \sum_{d|n} \mu \left( \frac{n}{d} \right) g(d) = \mu (n)
\end{align*}
So finally,
\begin{align*}
\mu (n) = \sum_{m=1 , (m,n)=1}^n \exp \left\{ 2 \pi i \frac{m}{n} \right\}
\end{align*}Next, the general case, ##1 \leq k##:
Consider
\begin{align*}
c_n (k) = \sum_{m=1 , (m,n)=1}^n \exp \left\{ 2 \pi i \frac{km}{n} \right\}
\end{align*}
We wish to prove
\begin{align*}
c_n (k) = \sum_{d|(k,n)} d \; \mu \left( \frac{n}{d} \right)
\end{align*}
Write ##f(n) = c_n (k)##. When ##n \nmid k##,
\begin{align*}
g (n) = \sum_{d|n} \sum_{m=1 , (m,d)=1}^d \exp \left\{ 2 \pi i \frac{km}{d} \right\}
= \sum_{l=1}^n \exp \left\{ 2 \pi i \frac{kl}{n} \right\}
= \dfrac{\exp \left\{ 2 \pi i k \right\} - 1}{\exp \left\{ 2 \pi i \frac{k}{n} \right\} - 1} = 0
\end{align*}
When ##n | k##,
\begin{align*}
g (n) = \sum_{d|n} \sum_{m=1 , (m,d)=1}^d \exp \left\{ 2 \pi i \frac{km}{d} \right\}
= \sum_{l=1}^n \exp \left\{ 2 \pi i \frac{kl}{n} \right\}
= n .
\end{align*}
That is:
\begin{align*}
g (n) = \sum_{d|n} \sum_{m=1 , (m,d)=1}^d \exp \left\{ 2 \pi i \frac{km}{d} \right\} =
\left\{
\begin{matrix}
n & n | k \\
0 & n \nmid k
\end{matrix}
\right.
\end{align*}
So that
\begin{align*}
f (n) = \sum_{d|n} \mu \left( \frac{n}{d} \right) g (d) = \sum_{d|(k,n)} d \; \mu \left( \frac{n}{d} \right)
\end{align*}