COMPUTING DEDEKIND SUMS USING THE
EUCLIDEAN ALGORITHM
Anita Buckley
University of Ljubljana, Department of Mathematics
1000 Ljubljana, Jadranska 19, Slovenia
[email protected](Anita Buckley)
Abstract
This paper presents an algorithm for computing Dedekind sums σi 1r (a1 , . . . , an ) by using the
Euclidean algorithm and algebraic properties of σi . In Theorem 2.2 an explicit relation between
a polynomial α with rational coefficients, which is obtained by the Euclidean algorithm, and
σ0 , σ1 , . . . , σr−1 will be determined. The main advantage of the algorithm is that by knowing α
all σi can be calculated simultaneously. When a1 , . . . , an are relatively prime to r, the algorithm
can be put in a particularly simple and compact form. Several examples are explicitly computed
by using Mathematica.
Definition 0.1 Fix positive integers r and a1 , . . . , an . The sums σi are defined by
X
1
εi
σi =
,
a1 ) · · · (1 − εan )
r
(1
−
ε
ε∈µr
(1)
εai 6=1 ∀i=1,...,n
where ε runs over rth roots of unity. It is obvious that σi = σr+i . Therefore we only need to
consider σi for i = 0, 1, . . . , r − 1 and call it the ith Dedekind sum. If we want to stress that
not all a1 , . . . , an are relatively prime to r, then σi is called the ith generalised Dedekind sum.
When a1 , . . . , an are relatively prime to r the above sum is taken over all rth roots of unity. Let
n
Y
1 − tr
A=
(1 − tai )
and B =
.
(2)
1−t
i=1
Then by the Euclidean algorithm there exist polynomials α, β ∈ Q(t) such that
B
.
(3)
1 = αA + β
hcf (A, B)
Qn
1−tr
r−1
ai
.
This way obtained α is called the Inverse of
i=1 (1 − t ) modulo 1−t = 1 + t + · · · + t
Keywords: Dedekind sums, Euclidean algorithm, computing.
Presenting Author’s Biography
Anita Buckley. Her main research area is algebraic geometry. Currently
a teaching assistant for mathematics at the University of Ljubljana. Completed PhD in Mathematics at the University of Warwick, England in 2003.
Graduated in theoretical mathematics at the University of Ljubljana, Slovenia in 1998.
1 Introduction
There is an abundant amount of literature on Dedekind
sums, appearing in many areas of mathematics such
as analytic number theory [5], topology [7], combinatorial geometry [10], algorithmic complexity [8]
and singularity theory [11, 4]. Dedekind sums were
among many others studied also by Berndt, Carlitz,
Grosswald, Knuth, Rademacher and Zagier. In a recent
book [2] on this topic, the authors Beck and Robins
connect the Dedekind sums to the ’coin exchange
problem’ and also to the number of lattice points inside
a certain polyhedra. From here the discrete volume and
the normal (or continuous) volume of polyhedra and
polytopes can be computed.
Recent research is focused on finding properties and
relations among Dedekind sums that would simplify
their computation [1]. The existing methods for
evaluating such sums are factorisation, multisection
and third most powerful method of partial fractions [6].
The main objective of this paper is to describe an algorithm for computing Dedekind sums using the Euclidean algorithm. The Sequential arithmetic time and
Parallel arithmetic complexity of the Euclidean algorithm are well understood [3]. In practice the fast
Fourier transform (FFT) algorithm is used.
In this section we will relate the Inverse function to
Dedekind sums. More precisely, we will find an equality connecting the polynomials
and
r−1
X
i
i ξi t an arbitrary
B
hcf(A,B) . Then
P
X
σr−i ti .
(4)
i=0
Remark 2.1 Running the Euclidean algorithm
on a computer gives α supported on monomials
1, t, . . . , tr−2−deg hcf(A,B) . Note also that α with such
support is unique.
We begin with two obvious equalities among Dedekind
sums:
ξi σi = 0,
P
i
since
i ξi ε = 0 for all ε ∈ µr such that
ai
ε 6= 1 ∀i = 1, . . . , n. This follows directly from
the definitions of B and σi . In particular, if applied
Pr−1
on B then i=0 σi = 0 always holds.
We proceed by giving an alternative description of the
Inverse function α. For all ε ∈ µr define
Aε =
1
(1−εa1 )···(1−εan ) ,
0,
if εai =
6 1 ∀i = 1, . . . , n
if εai = 1 for some i.
The expression Aε can be written as a polynomial in ε
with coefficients in Q. Moreover, in the proof of Theorem 2.2 a polynomial P (t) will be constructed such
that
P (ε) = Aε for all ε ∈ µr .
Observe that whenever εai 6= 1 for all i = 1, . . . , n also
For example, when r is prime one can simply substitute
ε with t in the polynomial expression of Aε to obtain α.
Theorem 2.2 Fix
Qn positive integers r andr a1 , . . . , an .
The Inverse of i=1 (1 − tai ) modulo 1−t
1−t denoted by
α equals
Pr−1
i=0
σr−i ti
modulo the polynomial
hcf (
r−1
Qn 1+t+···+t
ai ),1+···+tr−1 .
)
i=1 (1−t
P ROOF From now on fix a generator ε ∈ µr and denote the polynomial
r−1
X
=
σi
Aεj tj
j=1
1
r
εi
ε∈µr , εai 6=1 (1−εa1 )···(1−εan )
=
ε−i
ε∈µr , εai 6=1 (1−ε−a1 )···(1−ε−an )
=
P
P
P
(−1)n r1 ε∈µr , εai 6=1
εa1 +···+an −i
(1−εa1 )···(1−εan )
(−1)n σa1 +···+an −i
and
(5)
i
(i)
1
r
multiple of the
α(ε) = Aε = P (ε) holds by (3).
2 The main theorem
α
(ii) Denote by
polynomial
=
by p(t). Then
p(0) = 0,
p(1) = rσ0 ,
..
.
p(εi ) = rσi ,
..
.
p(εr−1 ) = rσr−1 .
(6)
The above formulae can be considered as a linear system of r equations in Aεj and variable ε
 0 

1
1
1 ...
1
Aε 
ε
ε2 . . .
εr−1  
 1



2
4
2(r−1)   Aε2

ε
ε
... ε

 1

..

 ... ... ... ...

.
r−1
1 ε
... ...
ε
Aεr−1



= r


σ0
σ1
σ2
..
.
σr−1



.


Since the Vandermonde matrix is invertible, we can
solve the system and obtain
Aεk
=
=
Pr−1
σi εk(r−i)
i=0
Pr−1
i=0
σr−i εki
for k = 1,2, . . . ,r − 1.
3
The computing algorithm
The Euclidean algorithm and Theorem
2.2 will be used
to calculate σi r1 (a1 , . . . , an ) for i = 0, 1, . . . , r − 1.
We will split the description of our method into three
steps. As before denote
A=
n
Y
(1 − tai )
i=1
A
B
hcf (A, B)
and
A1 = Aε0 =
σi =
α(t) =
r−1
X
σr−i ti + γ(t)
(8)
Lemma 3.1 Equalities (i) and (ii) uniquely determine
γ.
σr−i = 0.
P ROOF Write
This proves that for all ε ∈ µr the polynomial
P (t) =
B
.
hcf (A, B)
i=0
i=0
r−1
X
to obtain α.
Step II) Recall that by Remark 2.1 thus obtained polynomial α(t) is spanned on 1, t, . . . , tr−2−deg hcf(A,B) .
By Theorem 2.2 there exists a unique polynomial
γ(t) ∈ Q(t) of degree hcf (A, B) such that
Moreover, by (ii)
r−1
X
1 − tr
.
1−t
Step I) Run the Euclidean algorithm on
i=0
r−1
X
B=
and
σr−i t
i
α(t) =
(7)
r−1
X
αi ti .
(9)
i=0
i=0
evaluated in ε equals Aε . Since we are working over algebraically closed C, our polynomial P (t) is uniquely
determined by its r values and support 1, t, . . . , tr−1 .
On the other hand recall that for all
We will separate cases for n odd or even.
ε ∈ µr such that εai 6= 1 ∀i = 1, . . . , n
the Inverse function α(ε) equals Aε = P (ε). This
means that
α(t) − P (t)
(t −
ε∈µr
εai 6=1 ∀i=1,...,n
σi = (−1)n σa1 +···+an −i .
ε)
1−tr
ai
i=1 (1−t ), 1−t
Qn
−σa1 +···+an −r+i = − αr−a1 +···+an +i − γ) .
From this
)
for somepolynomial γ with rational
coefficients of deQn
1−tr
ai
gree hcf
i=1 (1 − t ) , 1−t . In other words
1−tr
1−t
α(t) ≡ P (t) modulo
hcf
Q
n
i=1
(10)
Therefore αi − γ = σr−i equals
=
1−tr
1−t
(
When n is odd compare the coefficients at ti and
tr−a1 +···+an +i in (8), where
denotes the smallest
nonnegative residue mod r. Recall from (i) that
=
Q
γ(t)
γ(t) hcf
For the sake of simpler notation we will restrict the
proof to hcf (A, B) = 1 which holds if and only if
a1 , . . . , an are relatively prime to r. In this case γ is
a constant. The general case can be proved in the same
way by using equalities (i) and (ii) and comparing the
coefficients in (8).
r
(1 − tai ) , 1−t
1−t
.
γ=
1
αi + αr−a1 +···+an +i
2
(11)
for any and thus for all i = 0, . . . , r − 1.
When n is even we use equality (ii) to determine γ:
r−1
X
i=0
αi − rγ =
r−1
X
i=0
σi = 0
implies
γ=
r−1
1X
r
αi .
(12)
i=0
Step III) By (8) the ith coefficient of the above obtained
B
polynomial α − γ hcf(A,B)
is exactly σr−i .
4 Examples
In the following examples it will be shown how to use
Mathematica to compute the polynomials α and
γ. In this way all Dedekind sums σi are calculated
simultaneously.
In[1] :=
PolynomialExtendedGCD[
(1 − t4 )(1 − t5 )2 (1 − t7 )(1 − t10 )(1 − t14 ),
PolynomialQuotient[1 − t33 , 1 − t, t]]
2
3
4
40 t
t
7t
Out[1] := {1, { 113
+ 3711t + 568
27 + 11
297 + 11
1135 t6
30 t7
15 t8
541 t9
47 t5
+ 11 + 297 + 11 + 11 + 297 + 4 t10
t12
15 t13
30 t14
1135 t15
47 t16
+4 t11 + 541
297 18 + 11 19 + 11 20 +
297 21 + 1123
17
t
t
37 t
40 t
113 t
+ 711
+ 568
+ 2911t
297 + 1125 + 1126 +
27
28
41 t
40 t
161 t27
1054 t24
+ 297 + 11 + 11 − 297 + 4011t
29
30
31
t
+ 4111t + 1054
+ 2911t ,
297
44
134 t
3 t2
53 t42
271 t43
86
− 27 − 297 + 11 +· · ·− 297 − 297 − 2911t }}.
In step II) γ is calculated using (4) since n = 6 is even.
Thus
Load the standard package
<< Algebra‘PolynomialExtendedGCD‘.
Example 4.1 We start with examples where a1 , . . . , an
are relatively prime to r.
1
(4, 5, 12) :
23
In step I) α and β are calculated by
γ=
31
1 2431
1 X
221
αi =
=
.
33 i=0
33 27
81
Step III) yields σi as the coefficient at t33−i in
In[2] :=
• σi
α − γ PolynomialQuotient[
1 − t33 , 1 − t, t]
118
809 t
566 t2
727 t3
1864 t4
81 5+ 891 +
8917 − 891 8− 891 9
6
t
t
t
t
t
+ 1376
+ 974
− 891
− 1216
− 808
891
891
891 13
89114
11
12
10
t
1216 t
t
− 891
+ 10381t + 10381t − 808
891 −
891
974 t15
1376 t16
1864 t17
727 t18
566 t19
+ 891 + 891 − 891 − 891 + 891
22
t20
118 t21
t23
t24
+ 809
− 22181t − 82891
+ 731
891 +
81
89128
26
27
25
809 t
2914 t
t
t
+ 809
+ 890
891 29+ 891 30−
891
891
890 t
731 t
82 t31
221 t32
+ 891 + 891 − 891 − 81 .
Out[2] :=
In[1] :=
PolynomialExtendedGCD[
(1 − t4 )(1 − t5 )(1 − t12 ),
PolynomialQuotient[1 − t23 , 1 − t, t]]
2
3
4
5
5
t
Out[1] := {1, {− 23
− 323t − 23
− 623t − 223t + 423t
5 t6
6 t7
13 t8
3 t10
11 t11
t12
t13
+ 23 − 23 − 23 + 23 − 23 − 323
− 323
15
16
17
19
20
14
t
t
t
t21
t
− 923
− 623
+ 723
− 1123t − 1023t − 423
,
+ 523
2
3
4
5
6
28
25 t
2t
5t
9t
9t
3t
8 t7
−
−
+
−
−
+
+
23 8 23 9 23 10 23
23
23
23
2314
12
t
t11
t13
t
+ 623t + 223t + 223
+ 223
− 2623t + 423
+ 423
11 t16
11 t17
t18
6 t19
4 t20
3 t15
− 23 + 23 + 23 − 23 − 23 − 23 }}.
In step II) γ is calculated using (3) since n = 3 is odd.
Thus
γ=
1
(4, 5, 5, 7, 10, 14) :
33
Step I) computes α by
• σi
1
1
3
α0 + α23−4+5+12 = (α0 + α2 ) = − .
2
2
23
In step III) we obtain σi as the coefficient at t23−i in
In[2] :=
α − γ PolynomialQuotient[
1 − t23 , 1 − t, t]
Out[2] :=
2
3
4
It is easy to verify that
σ0 = −σ21 , σ22 = −σ22 = 0, . . .
as expected.
5
6
t
2
+ 223t − 323t + 23
+ 723t + 823t
− 23
7
8
9
t10
t11
− 323t − 1023t + 323t + 623
− 823
14
15
16
17
t
t
t
t18
+ 823
− 623
− 323
+ 1023t + 323
21
t20
t22
t19
− 723
− t23 + 323
.
− 823
Example 4.2 In the following examples we consider two generalised Dedekind sums, where not all
a1 , . . . , an are relatively prime to r. In this case γ(t) is
a polynomial of degree hcf (A, B) which will be computed by generalising the proof of Lemma 3.1.
1
(1, 2, 3) :
12
In this case hcf (A, B) = (1 + t)(1 + t + t2 ). Thus in
step I) α is obtained by running the Euclidean algorithm
on
In[1] :=
PolynomialExtendedGCD[
(1 − t)(1 − t2 )(1 − t3 ),
PolynomialQuotient[
1 − t12 , (1 − t)(1 + t)(1 + t + t2 ), t]]
• σi
2
4
5
6
7
7
Out[1] :={1, { 12
+ 512t − t4 + t4 + t2 − t3 + 1112t ,
19
7t
11 t2
4 t3
t4
11 t5
12 + 12 − 12 − 3 − 3 + 12 }}.
In step II) we determine γ = γ0 + γ1 t + γ2 t2 + γ3 t3 .
From (i) it follows
σ0 = −σ6
σ1 = −σ5
σ7 = −σ11
σ3 = −σ3 = 0
σ2 = −σ4
σ8 = −σ10
σ9 = −σ9 = 0.
Moreover, (ii) and
Since
11
X
1 − t12
= 1 − t + t 2 + t6 − t7 + t8
(1 − t)(1 + t)(1 + t + t2 )
σ12−i ti
i=0
equals
imply
α−γ
σ0 − σ1 + σ2 + σ6 − σ7 + σ8 = 0.
By comparing the coefficients in
11
X
σ12−i ti and α − γ
i=0
we get
7
σ0 = − 12
− γ0
1
+ γ0 − γ1
σ11 = − 12
σ10 = 41 + γ1 − γ2
5
− γ0 + γ2 − γ3
σ9 = − 12
5
σ8 = − 12 − γ1 + γ3 − γ4
5
+ γ0 − γ2 + γ4 − γ5
σ7 = 12
σ6 = −γ0 + γ1 − γ3 + γ5
σ5 = −γ1 + γ2 − γ4
σ4 = −γ2 + γ3 − γ5
σ3 = −γ3 + γ4
σ2 = −γ4 + γ5
σ1 = −γ5 .
1 − t12
(1 − t)(1 + t)(1 + t + t2 )
the above translates into the following system of equations
7
− γ0 = 31 + γ0
− 12
11
12
−γ0 + γ1 − γ2 = γ2 − γ3
+ γ0 − γ1 = γ3
−γ1 + γ2 − γ3 = 0
and
7
1
1
1
− −γ0 +γ3 −γ2 +γ3 − −γ0 − +γ3 + −γ2 +γ3 = 0.
12
3
2
4
Its solution is
Next apply (ii) on the following multiples of
In step III) we obtain σi as the coefficient at t12−i in
α − γ PolynomialQuotient[
1 − t12 , (1 − t)(1 + t)(1 + t + t2 ), t]
− 81 −
Out[2] :=
• σi
In this case
hcf (A, B) =
=
2
5t
24
Recall that by (i)
σi = σ1+2+3+4−i = σ10−i holds for i = 0, 1, . . . , 11.
11
t
11 t2
7 t3
γ=− + +
+
.
24 6
24
24
In[2] :=
4
5
B
hcf(A,B)
=
B
hcf(A,B) :
1−t12
(1−t)(1+t)(1+t2 )(1+t+t2 )
3
5
6
=1−t+t −t +t ,
B
hcf(A,B) (1
+ t + t 2 ) = 1 + t4 + t8 ,
6
t
t
− 12
+ 12
+ 524t + t8
7
8
10
t11
+ 724t + t6 − t6 − 724
.
=
B
2
hcf(A,B) (1 + t)(1 + t )
B
2
3
hcf(A,B) (1 + t + t + t )
3
6
9
= 1+t +t +t
1
(1, 2, 3, 4) :
12
and B.
Therefore
(1 + t)(1 + t2 )(1 + t + t2 )
σ0 − σ1 + σ3 − σ5 + σ6 = 0,
σ0 + σ4 + σ8 = 0,
σ0 + σ3 + σ6 + σ9 = 0,
P11
i=0 σi = 0.
(1 + t + t2 )(1 + t + t2 + t3 ).
Step I) calculates α by running the Euclidean algorithm
on
In[1] :=
PolynomialExtendedGCD[
(1 − t)(1 − t2 )(1 − t3 )(1 − t4 ),
PolynomialQuotient[
1 − t12 , (1 − t)(1 + t)(1 + t2 )(1 + t + t2 ), t]]
2
1 − t12
(1 − t)(1 + t)(1 + t2 )(1 + t + t2 )
3
4
5
7
t
Out[1] := {1, {− 12
− 12
+ t4 − 512t − 512t + 512t ,
3
4
5
13 t
t2
19
− 5 6t − 5 3t − t3
12 + 126 + 6 7
8
9
t
t
+ t6 + 56t + 512
− 512
}}.
In step II) we determine
γ = γ0 + γ1 t + γ2 t2 + γ3 t3 + γ4 t4 + γ5 t5 .
The system of the above equations has a unique solution
γ=−
37 11 t 5 t2
t3
t4
7 t5
−
−
+
+ +
.
72
24
36
18
8
72
Finally, in step III) we obtain σi as the coefficient at
t12−i in
In[2] :=
α−γPolynomialQuotient[
1 − t12 , (1 − t)(1 + t)(1 + t2 )(1 + t + t2 ), t]
Out[2] :=
2
3
4
5
t
5
− 536t − 572t − 772t − 36
+ 572t
− 72
7 t7
7 t8
5 t9
t10
t11
7 t6
.
+ 72 + 36 + 72 + 72 − 36 − 772
5 References
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