Ramanujan sum RamanujanSum 2002-01-20 23:37:25 2006-10-06 14:29:55 Definition Ramanujan trigonometric sum \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \PMlinkescapeword{fixed} \PMlinkescapeword{name} \PMlinkescapeword{equivalent} \PMlinkescapeword{theory} For positive integers $s$ and $n$, the complex number $$c_s(n)=\sum_{\substack{0\le k<n \\ (k,n)=1}}e^{2\pi iks/n}$$ is referred to as a \emph{Ramanujan sum}, or a \emph{Ramanujan trigonometric sum}. Since $e^{2\pi i}=1$, an equivalent definition is $$c_s(n)=\sum_{k\in r(n)}e^{2\pi iks/n}$$ where $r(n)$ is some reduced residue system mod $n$, meaning any subset of $\mathbb{Z}$ containing exactly one element of each invertible residue class mod $n$. Using a symmetry argument about roots of unity, one can show $$ \sum_{d|s}c_s(n)= \begin{cases} s & \textrm{if $s|n$} \\ 0 & \text{otherwise.} \end{cases} $$ Applying M\"obius inversion, we get $$c_s(n)=\sum_{\substack{d|n \\ d|s}}\mu(n/d)d =\sum_{d|(n,s)}\mu(n/d)d$$ which shows that $c_s(n)$ is a real number, and indeed an integer. In particular $c_s(1)=\mu(s)$. More generally, $$c_{st}(mn)=c_s(m)c_t(n)\text{ if }(m,t)=(n,s)=1\;.$$ Using the Chinese remainder theorem, it is not hard to show that for any fixed $n$, the function $s\mapsto c_s(n)$ is multiplicative: $$c_s(n)c_t(n)=c_{st}(n) \text{ if }(s,t)=1 \;.$$ If $m$ is invertible mod $n$, then the mapping $k\mapsto km$ is a permutation of the invertible residue classes mod $n$. Therefore $$c_s(mn)=c_s(n)\text{ if }(m,s)=1\;.$$ \textbf{Remarks: } Trigonometric sums often make convenient apparatus in number theory, since any function on a quotient ring of $\mathbb{Z}$ defines a periodic function on $\mathbb{Z}$ itself, and conversely. For another example, see Landsberg-Schaar relation. Some writers use different notation from ours, reversing the roles of $s$ and $n$ in the expression $c_s(n)$. The name ``Ramanujan sum'' was introduced by Hardy.