Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Dedekind sum $s(p,q)$ can be both positive and negative. What are the known lower/upper bounds in terms of p,q? (I would prefer something that grows not faster than q)

share|improve this question
add comment

2 Answers

up vote 6 down vote accepted

For a fixed $q$, the maximum is $$s(1,q)=-{1\over4}+{1\over6q}+{q\over12}$$ and the minimum is $s(q-1,q)=-s(1,q)$.

share|improve this answer
Wow, that's great. And 1/12 is especially delightful. Thanks! Can you also give a reference, what to cite? –  Dmitry Kerner May 4 '11 at 12:55
This follows from a finite-Fourier-series version of Cauchy-Schwartz. The earliest reference I'm aware of is H. Rademacher, Zur Theorie der Dedekindschen Summen, Math. Z. 63 (1956) 445-463. With some work, one can obtain even better bounds (see, e.g., <a href="front.math.ucdavis.edu/math.NT/0305421">my paper with S. Robins and S. Zacks</a>). –  matthias beck May 4 '11 at 15:26
$s(p,q)=q^{-2}\sum af(a)$ plus terms not relevant here, where $f(a)$ is the permutation of $1,2,\dots,q-1$ induced by multiplication by $p$ and reduction mod $q$. The maximum over all permutations $f$ (not just those arising from multiplication) of $\sum af(a)$ is attained when $f(a)=a$ for all $a$, and this corresponds to multiplication by 1; the minimum, when $f$ applied to $1,2,\dots,q-1$ gives $q-1,\dots,2,1$, and this corresponds to multiplication by $q-1$. –  Gerry Myerson May 5 '11 at 0:20
Thanks a lot for both comments! –  Dmitry Kerner May 5 '11 at 11:26
add comment



(most of the relevant stuff is due to our own @Gerry Myerson)

share|improve this answer
Sorry, I don't see any bound there, only some discussion on properties of the function. I'm completely ignorant in NT, I didn't mean some recent state of art result. There certainly must be some classical rough bound... –  Dmitry Kerner May 4 '11 at 0:04
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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