16
$\begingroup$

In attempt to deepen my understanding of Dedekind sums, I've proven the following identity $$ \sum_{i = 0}^{t} \sum_{j = 0}^{b(t-i)} \left \lbrace c \left( t - i - \frac{j}{b} \right) \right \rbrace = \frac{t(t+1)}{4}(b - \gcd(b,c)), $$ where $b,c,t \in \mathbb{N}$ and $\lbrace \cdot \rbrace$ denotes the fractional part function.

I'd like to compute the following more general double sums $$ \sum_{i = 0}^{a t} \sum_{j = 0}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace \qquad \text{and} \qquad \sum_{i = 1}^{a t} \sum_{j = 1}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace \qquad a, b, c, t \in \mathbb{N}, $$ where $\lfloor \cdot \rfloor$ denotes the floor function. The sums clearly vanish if $a$ and $b$ are divisors of $c$ (by the vanishing of the fractional part) and (I expect) grow as $O(t^{2})$ otherwise. Are such sums in the literature, even for some special cases of pairwise coprime coefficients $a, b$ and $c$? Any help and/or hints are certainly appreciated!

Harder Problem(s): If the above is trivial, then suppose $a, b, c$ and $t$ are real and compute the following sums $$ \sum_{i = 0}^{\lfloor a t \rfloor} \left \lbrace b \left( t - \frac{i}{a} \right) \right \rbrace \qquad \text{and} \qquad \sum_{i = 0}^{\lfloor a t \rfloor} \sum_{j = 0}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace $$ as well as $$ \sum_{i = 1}^{\lfloor a t \rfloor} \left \lbrace b \left( t - \frac{i}{a} \right) \right \rbrace \qquad \text{and} \qquad \sum_{i = 1}^{\lfloor a t \rfloor} \sum_{j = 1}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace. $$

$\endgroup$
  • 1
    $\begingroup$ I want to look at this, but I want to make sure I'm looking at the right function. Calling your double sum $f(a,b,c,t)$, can you confirm that the values of $f(2,3,5,t)$ for $t=1,\dots,9$ are $7/3,43/6,29/2,73/3,110/3,103/2,413/6,266/3,111$? $\endgroup$ – Gerry Myerson Jul 7 '11 at 5:20
  • $\begingroup$ @Gerry: Yes. The next one in the sequence is $\frac{815}{6}$. $\endgroup$ – user02138 Jul 7 '11 at 5:36
4
$\begingroup$

Let $a, b$ and $c$ be positive integers with no common factor, i.e., $\gcd(a,b,c) = 1$. Define $a^{\prime} = \gcd(b,c)$, $b^{\prime} = \gcd(c,a)$, $c^{\prime} = \gcd(a,b)$ and $d = a^{\prime} b^{\prime} c^{\prime}$.

The following identity holds: \begin{align} \sum_{i = 0}^{at} \sum_{j = 0}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace = \tfrac{ab -d }{4} t^{2} + \gamma t \end{align} where $t \in \mathbb{Z}_{\geq 0}$ and \begin{align} \gamma & = a^{\prime} \ \mathfrak{s}(\tfrac{bc}{d},\tfrac{aa^{\prime}}{d}) + b^{\prime} \ \mathfrak{s}(\tfrac{ca}{d},\tfrac{bb^{\prime}}{d}) + c^{\prime} \ \mathfrak{s}(\tfrac{ab}{d}, \tfrac{cc^{\prime}}{d}) + \tfrac{1}{4}(a - a^{\prime} + b - b^{\prime} + c^{\prime} ) - \tfrac{a^{2} b^{2} + c^{2} (c^{\prime})^{2} + d^{2}}{12 a b c}. \end{align} Here, $\mathfrak{s}(p,q) = \tfrac{1}{4q} \sum_{k = 1}^{q-1} \cot(\tfrac{\pi k}{q}) \cot( \tfrac{\pi k p}{q} )$ is the standard Dedekind sum. This more general identity simplifies to the identity above (in the post) as well as $$ \sum_{i = 0}^{at} \sum_{j = 0}^{\lfloor t - i/a \rfloor} \left \lbrace c \left( t- \frac{i}{a} - j \right) \right \rbrace = \frac{t(t+1)}{4} (a - \gcd(a,c)), $$ after an application of the Dedekind Sum Reciprocity Law and taking $b = 1$.

$\endgroup$
  • 1
    $\begingroup$ Incredible work good sir /madam thankyou for sharing this with everyone $\endgroup$ – Adam Aug 22 '18 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.