Ramanujan’s sum

Jordan Bell

April 7, 2014

1 Definition

Let $q$ and $l$ be positive integers. Define

c_{q}(l)=\sum_{\stackrel{{\scriptstyle 1\leq j\leq q}}{{\gcd(h,q)=1}}}e^{-2\piihl%/q}=\sum_{\stackrel{{\scriptstyle 1\leq h\leq q}}{{\gcd(h,q)=1}}}e^{2\pi ihl/q%}=\sum_{\stackrel{{\scriptstyle 1\leq h\leq q}}{{\gcd(h,q)=1}}}\cos\frac{2\pihl%}{q}.

$c_{q}(l)$ is called Ramanujan’s sum.

2 Fourier transform on $\mathbb{Z}/q$ and the principal Dirichlet character modulo $q$

For $F:\mathbb{Z}/q\to\mathbb{C}$ , the Fourier transform $\widehat{F}:\mathbb{Z}/q\to\mathbb{C}$ of $F$ is defined by

\widehat{F}(k)=\frac{1}{q}\sum_{j\in\mathbb{Z}/q}F(j)e^{-2\pi ijk/q},\qquad k%\in\mathbb{Z}/q.

Define $\chi:\mathbb{Z}/q\to\mathbb{C}$ by $\chi(j)=1$ if $\gcd(j,q)=1$ and $\chi(j)=0$ if $\gcd(j,q)>1$ . $\chi$ is called the principal Dirichlet character modulo $q$ . The Fourier transform of $\chi$ is

\widehat{\chi}(k)=\frac{1}{q}\sum_{j\in\mathbb{Z}/q}\chi(j)e^{-2\pi ijk/q}=%\frac{1}{q}\sum_{\stackrel{{\scriptstyle 1\leq j\leq q}}{{\gcd(j,q)=1}}}e^{-2%\pi ijk/q}.

Therefore we can write Ramanujan’s sum $c_{q}(l)$ as $c_{q}(l)=q\cdot\widehat{\chi}(l)$ , thus $c_{q}=q\cdot\widehat{\chi}$ .

The above gives us an expression for $c_{q}(l)$ as a multiple of the Fourier transform of the principal Dirichlet character modulo $q$ . $c_{q}:\mathbb{Z}/q\to\mathbb{C}$ , and we can write the Fourier transform of $c_{q}$ as

	$\displaystyle\widehat{c_{q}}(k)$	$\displaystyle=$	$\displaystyle\frac{1}{q}\sum_{j\in\mathbb{Z}/q}c_{q}(j)e^{-2\pi ijk/q}$
		$\displaystyle=$	$\displaystyle\sum_{j\in\mathbb{Z}/q}\widehat{\chi}(j)e^{-2\pi ijk/q}.$

3 Dirichlet series

Here I am following Titchmarsh in §1.5 of his The theory of the Riemann zeta-function, second ed. Let $\mu$ be the Möbius function. The Möbius inversion formula states that if

g(q)=\sum_{d|q}f(d)

then

f(q)=\sum_{d|q}\mu\left(\frac{q}{d}\right)g(d).

( $\sum_{d|q}$ is a sum over the positive divisors of $q$ .)

Define

\eta_{q}(k)=\sum_{j\in\mathbb{Z}/q}e^{-2\pi ijk/q}.

We have (this is not supposed to be obvious)

\eta_{q}(k)=\sum_{d|q}c_{d}(k).

Therefore by the Möbius inversion formula we have

c_{q}(k)=\sum_{d|q}\mu\left(\frac{q}{d}\right)\eta_{d}(k).

(Hence $|c_{q}(k)|\leq\sum_{d|k}d=\sigma_{1}(k)$ , where $\sigma_{a}(k)=\sum_{d|k}d^{a}$ .)

If $q|k$ then $\eta_{q}(k)=q$ , and if $q\not|k$ then $\eta_{q}(k)=0$ . (To show the second statement: multiply the sum by $e^{-2\pi ik/q}$ , and check that this product is equal to the original sum. Since we multplied the sum by a number that is not $1$ , the sum must be equal to $0$ .) Thus we can express the Möbius function using Ramanujan’s sum as $\mu(q)=c_{q}(1)$ .

Because $\eta_{d}(k)=d$ if $k|d$ and $\eta_{d}(k)=0$ if $k\not|d$ , we have

c_{q}(k)=\sum_{d|q,d|k}\mu\left(\frac{q}{d}\right)d=\sum_{dr=q,d|k}\mu(r)d.

\frac{c_{q}(k)}{q^{s}}=\sum_{dr=q,d|k}\frac{1}{q^{s}}\mu(r)d=\sum_{dr=q,d|k}%\frac{1}{d^{s}r^{s}}\mu(r)d=\sum_{dr=q,d|k}\frac{1}{r^{s}}\mu(r)d^{1-s}.

Therefore

\sum_{q=1}^{\infty}\frac{c_{q}(k)}{q^{s}}=\sum_{q=1}^{\infty}\sum_{dr=q,d|k}%\frac{1}{r^{s}}\mu(r)d^{1-s}=\sum_{r=1}^{\infty}\sum_{d|k}\frac{1}{r^{s}}\mu(r%)d^{1-s}=\sum_{r=1}^{\infty}\frac{1}{r^{s}}\mu(r)\sum_{d|k}d^{1-s}.

Then

\sum_{q=1}^{\infty}\frac{c_{q}(k)}{q^{s}}=\sigma_{1-s}(k)\sum_{r=1}^{\infty}%\frac{1}{r^{s}}\mu(r)=\sigma_{1-s}(k)\frac{1}{\zeta(s)};

here we used that

\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}.

On the other hand, if rather than sum over $q$ we sum over $k$ , then we obtain

$\displaystyle\sum_{k=1}^{\infty}\frac{c_{q}(k)}{k^{s}}$	$\displaystyle=$	$\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^{s}}\sum_{d\|q,d\|k}\mu\left(\frac{q}%{d}\right)d$
	$\displaystyle=$	$\displaystyle\sum_{d\|q}\sum_{m=1}^{\infty}\frac{1}{(md)^{s}}\cdot\mu\left(%\frac{q}{d}\right)d$
	$\displaystyle=$	$\displaystyle\sum_{d\|q}\sum_{m=1}^{\infty}\frac{1}{m^{s}}\cdot\frac{1}{d^{s}}%\cdot\mu\left(\frac{q}{d}\right)d$
	$\displaystyle=$	$\displaystyle\sum_{m=1}^{\infty}\frac{1}{m^{s}}\sum_{d\|q}\frac{1}{d^{s}}\mu%\left(\frac{q}{d}\right)d$
	$\displaystyle=$	$\displaystyle\zeta(s)\cdot\sum_{d\|q}\mu\left(\frac{q}{d}\right)d^{1-s}.$

Ramanujan’s sum

1 Definition

2 Fourier transform on ℤ/q and the principal Dirichlet character modulo q

3 Dirichlet series

2 Fourier transform on $\mathbb{Z}/q$ and the principal Dirichlet character modulo $q$