Filomat 30:4 (2016), 1097–1102
DOI 10.2298/FIL1604097A
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
On the Constant Term of The Minimal Polynomial of cos
2π
n
over Q
Chandrashekar Adigaa , Ismail Naci Cangulb , H.N. Ramaswamyc
a Department
of Studies in Mathematics, University of Mysore, Manasagangothri, Mysore - 570006, India.
of Mathematics, Faculty of Arts and Science, Uludag University, Gorukle Campus, Bursa 16059, Turkey.
c Department of Studies in Mathematics, University of Mysore, Manasagangothri, Mysore - 570006, India.
b Department
Abstract. The algebraic numbers cos (2π/n) and 2 cos (π/n) play an important role in the theory of discrete
groups and has many applications because of their relation with Chebycheff polynomials. There are some
partial results in literature for the minimal polynomial of the latter number over rationals until 2012 when a
complete solution was given in [5]. In this paper we determine the constant term of the minimal polynomial
) over Q by a new method.
of cos( 2π
n
1. Introduction and Preliminaries
It is a well known result that the nth cyclotomic polynomial Φn (x) is a monic irreducible polynomial of
degree φ(n) with integer coefficients where φ(n) is the number of integers between 1 and n that are relatively
prime to n. Its roots are the primitive nth roots of unity. Thus
Φn (x) =
n
Y
(x − ζkn ),
k=1
(k,n)=1
where ζn = exp
2πi
n
2π
= cos( 2π
n ) + i sin( n ). Also
xn − 1 =
Y
Φd (x),
d|n
and this identity can be used to compute Φn (x). For example,
Φ1 (x) = x − 1, Φ2 (x) = x + 1, Φ3 (x) = x2 + x + 1, Φ4 (x) = x2 + 1.
2π
In 1933, D. H. Lehmer [3] gave a method for constructing the minimal polynomials of cos( 2π
n ) and sin( n )
using Φn (x). W. Watkins and J. Zeitlin [4] gave another method for computing the minimal polynomial
Ψn (x) of cos( 2π
n ) using Chebychev polynomials Ts (x) which are defined by,
Ts (cos θ) = cos(sθ),
2010 Mathematics Subject Classification. Primary 12E05, 33B10
Keywords. Cyclotomic polynomial, constant term.
Received: 13 August 2015; Accepted: 17 September 2015
Communicated by Gradimir Milovanović and Yilmaz Simsek
The second author is supported by the research fund of Uludag University, project nos: 2003-23, 2003-87, 2015-17, 2015-23.
Email addresses: c− [email protected] (Chandrashekar Adiga), [email protected] (Ismail Naci Cangul), [email protected] (H.N.
Ramaswamy)
C. Adiga et al. / Filomat 30:4 (2016), 1097–1102
for positive integers s and all real θ. We note that
(
de1 Ψn (x) =
1
φ(n)/2
1098
if n = 1, 2,
if n ≥ 3.
In fact they proved the following:
Theorem 1.1. ([4]) Let ψn (x) be the minimal polynomial of cos(2π/n) and let Ts (x) denote the sth Chebychev
polynomial.
a) If n = 2s + 1 is odd, then
Y
Ts+1 (x) − Ts (x) = 2s
ψd (x),
d|n
and
b) if n = 2s is even, then
Ts+1 (x) − Ts−1 (x) = 2s
Y
ψd (x).
d|n
Using the above result, I. N. Cangul [1] obtained the formula for Ψn (x) in terms of Chebychev polynomials. Recently M. Demirci and I. N. Cangul [2] have determined the constant term of Ψn (x) by making use
of the behavior of trigonometric functions sin nx and cos nx. We give here a direct simple proof making use
of Theorem 1.1.
2. Constant term of Ψn (x)
First we find the constant term of the polynomial Ts+1 (x) − Ts (x).
We have
Ts (cos θ) = cos(sθ) = Re((cos θ + i sin θ)s )
 s

X

s−k
s
k
s−k

= Re 
(k )cos θ i
sin θ
k=0







=





s−1
2
s−1
2
s−3
(s1 ) cos θ(−1) (1 − cos2 θ) + (s3 )cos3 θ(−1) 2 (1 − cos2 θ)
· · · + (ss ) coss θ,
s
s
s−2
s−2
(s0 )(−1) 2 (1 − cos2 θ) 2 + (s2 ) cos2 θ(−1) 2 (1 − cos2 θ) 2 +
· · · + (ss ) coss θ,
s−3
2
+
if s is odd,
if s is even.
Hence,
 P s+1
s−2k+1
s−2k+1

2

 k=1
(−1) 2 (s2k−1 )x2k−1 (1 − x2 ) 2 , if s is odd,
Ts (x) = 
P
s−2k

 s/2 (−1) s−2k
2 (s )x2k (1 − x2 ) 2 ,
if s is even.
2k
k=0
Thus, we have


0,



1,
Ts (0) = 


 −1,
if s is odd,
if s ≡ o (mod 4),
if s ≡ 2 (mod 4),
(2.1)
(2.2)
and so
(
Ts+1 (0) − Ts (0) =
−1, if s ≡ 0, 1 (mod 4),
1, if s ≡ 2, 3 (mod 4).
(2.3)
We find the constant term of the minimal polynomial of cos( 2π
n ) by considering the following cases:
C. Adiga et al. / Filomat 30:4 (2016), 1097–1102
1.
2.
3.
4.
5.
1099
n odd.
n = 2m, m is odd.
n = 2m .
n = 22 m, where m is odd.
n = 2β m, where β ≥ 3 and m is odd.
h φ(n) i
Theorem 2.1. If n is odd, then the constant term in 2
Proof. For n = 1, we have,
2
When n = 3,
h φ(n) i
2
h φ(n) i
2
2
2
ψn (x) is ±1.
1
ψn (x) = 2[ 2 ] ψ1 (x) = (x − 1).
ψn (x) = 2[
φ(3)
2 ]
ψ3 (x) = 2x + 1.
If n is an odd integer of the form n = 2s + 1, then the result is true for s = 0 and s = 1. Suppose the
result is true for all odd integers less than n = 2s + 1. We shall prove the result for n = 2s + 1. Then
P
φ(d)
s = n−1
d|n,d>n 2 and so by Theorem 1.1,
2 =
Y
Ts+1 (x) − Ts (x) = 2s
ψd (x)
d|n
= ψ1 (x) · 2s
Y
ψd (x)
d|n
d>1
= ψ1 (x)
Y
2[
φ(d)
2 ]
ψd (x).
d|n
d odd
d≥3
Hence,
Ts+1 (x) − Ts (x) = 1(x) 2[
where
φ(n)
2 ]
ψn (x),
Y
1(x) = ψ1 (x)
2[
(2.4)
φ(d)
2 ]
ψd (x).
d|n
d odd
d≥3, d,n
Now the constant term of Ts+1 (x) − Ts (x) is ±1 by (2.3). As n is odd, all its divisors d are also odd and hence
the constant term of 1(x), by induction hypothesis, is also ±1. Hence by (2.4), it follows that the constant
term of 2[
φ(n)
2 ]
ψn (x) is also ±1.
φ(n)
Theorem 2.2. If n = 2m , then the constant term in 2[ 2


− 1,



 + 1,



0,



 ± 2,
]
ψn (x) is defined as
if m = 0,
if m = 1,
if m = 2,
if m ≥ 3.
Proof. It is easy to check the result for m = 0, 1 and 2.
Suppose n = 2m , m ≥ 3. By induction we show that constant term in 2[
by Theorem 1.1,
Y
Ts+1 (x) − Ts−1 (x) = 2s
ψd (x).
d|n
φ(n)
2 ]
ψn (x) is ±2. If n = 2s, then again
C. Adiga et al. / Filomat 30:4 (2016), 1097–1102
Hence for n = 2m , m ≥ 3,
T2m−1 +1 (x) − T2m−1 −1 (x) = 22
m−1
Y
1100
ψd (x)
d|2m
= 22 ψ1 (x) ψ2 (x) ψ4 (x)
Y
2[
φ(d)
2 ]
ψd (x),
d|2m
d>4
since 2m−1 − 2 =
Hence,
1
2
P
d|2m , d>4
φ(d), we have, 22
m−1
= 22 .2
P
d|2m , d>4
φ(d)
2
.
T2m−1 +1 (x) − T2m−1 −1 (x) = 22 (x − 1)(x + 1)x
Y
2[
φ(d)
2 ]
ψd (x).
d|2m
d>4
Note that constant term in LHS is 0 by (2.2). Also from (2.1) the coefficient of x in Ts+1 (x) − Ts−1 (x) is 2s when
s ≡ 0 (mod 4). Hence coefficient of" x in T2m−1 +1 (x) − T#2m−1 −1 (x) = 2m
φ(2m )
φ(d)
Q
[ 2 ]
ψd (x) 2[ 2 ] ψ2m (x).
= constant term in 22 (x − 1)(x + 1)
d|2m−1 2
d>4
Now by using the induction
of x in T2m−1 +1 (x) − T2m−1 −1 (x) =
hypothesis,
it follows that, the coefficient
φ(2m )
φ(2m )
2
m−3
[ 2 ]
]
[
2 (±2) × constant term in 2
ψ2m (x) . Thus constant term in 2 2 ψ2m (x) = ±2.
Theorem 2.3. If n = 2s and s is odd, then the constant term in 2[
φ(n)
2 ]
ψn (x) is ±1.
Proof. For s = 1 and 3 the proof is clear.
Suppose the result is true for all m < n and m ≡ 2 (mod 4). We shall prove the result for n. If n = 2s, we have,
Y
Ts+1 (x) − Ts−1 (x) = 2s
ψd (x)
d|n
= 2 ψ1 (x) ψ2 (x) 2s−1
Y
ψd (x).
d|n
d>2
Further
if n ≡ 2 (mod 4), then, the divisors d of n are either odd or of the form d ≡ 2 (mod 4) and so
1 P
[φ(d)]
= 12 (n − φ(1) − φ(2)) = s − 1. Hence
d|n
2
d>2
Ts+1 (x) − Ts−1 (x) = 2 ψ1 (x) ψ2 (x)
Y
2
φ(d)
2
ψd (x)
d|n
d>2
= 2 h(x) 2[
φ(n)
2 ]
ψn (x), where
h(x) = ψ1 (x) ψ2 (x)
Y
2
φ(d)
2
ψd (x).
d|n
n>d>2
By induction hypothesis the constant term of 2 h(x) is ±2. Hence, using (2.1), we get the desired result.
Theorem 2.4. Let n = 22 .pα , where p is an odd prime and α ≥ 1. Then the constant term in 2[
φ(n)
2 ]
Proof. We prove the theorem by induction on α. By Theorem 1.1, we have,
Y
Y φ(d)
α
T2pα +1 (x) − T2pα −1 (x) = 22p
ψd (x) = 22 (x − 1)(x + 1)x
2[ 2 ] ψd (x).
d|22 pα
d|22 pα
d,1,2,4
ψn (x) is ±p.
C. Adiga et al. / Filomat 30:4 (2016), 1097–1102
1101
By (2.1), the coefficient of x in LHS is −22 pα which is equal to the constant term in
Y φ(d)
22 (x − 1)(x + 1)
2[ 2 ] ψd (x).
d|22 pα
d,1,2,4
Now
Y
22 (x − 1)(x + 1)
2[
φ(d)
2 ]
Y
ψd (x) = 22 (x − 1)(x + 1)
d|22 pα
d,1,2,4
2[
φ(d)
2 ]
ψd (x)×
d=p,p2 ,...,pα











Y
Y




φ(d)
φ(d)


 [ φ(222pα ) ]
[ 2 ]
[ 2 ]
2
α
2
ψ
(x)
ψ
(x)
ψ
(x)
.
2
2




d
d
2 p







d=2p,2p2 ,...,2pα


d=22 p,22 p2 ,...,22 pα−1
Now by Theorem 2.1, the constant term in
Y
2[
φ(d)
2 ]
ψd (x)
d=p,p2 ,...,pα
is ±1. From Theorem 2.3, the constant term in
Y
2[
φ(d)
2 ]
ψd (x)
d=2p,2p2 ,...,2pα
is ±1. By induction hypothesis, the constant term in
Y
2[
φ(d)
2 ]
ψd (x)
d=22 p,22 p2 ,...,22 pα−1
is (±p)α−1 . Thus,
−22 pα = −22 (±1) (±1) (±p)α−1 × constant term in 2[
φ(22 pα )
]
2
ψ22 pα (x)
which implies the result.
Remark 2.5. Now suppose n = 22 m where m = p1 α1 . . . pk αk with p1 , . . . , pk distinct odd primes. Then
αk Y
α1
T2pα1 ... pαk +1 (x) − T2pα1 ... pαk −1 (x) = 22p1 ... pk
ψd (x).
1
k
1
k
d|n
= 22 (x − 1)(x + 1)x
Y
2[
φ(d)
2 ]
ψd (x).
d|n
d,1,2,4
As above, the coefficient of x in LHS is
−22 pα1 1 . . . pαk k ,
which is equal to the constant term in
22 (x − 1)(x + 1)
Y
2[
φ(d)
2 ]
ψd (x).
α
α
d|22 p1 1 ...pk k
d,1,2,4
Now the constant term in
Y
2[
φ(d)
2 ]
ψd (x)
α
d=22 pi ,22 pi 2 ,...,22 pi i
is equal to (±pi )αi by Theorem 2.4, for i = 1, 2, . . . , k. Hence the constant term in 2[
φ(n)
2 ]
ψn (x) must be ±1.
C. Adiga et al. / Filomat 30:4 (2016), 1097–1102
1102
Theorem 2.6. Suppose n = 2β pα , where p is an odd prime with α ≥ 1 and β ≥ 3. Then the constant term in
2[
φ(n)
2 ]
ψn (d) is ±1.
Proof. We have, by Theorem 1.1,
n
T2β−1 pα +1 (x) − T2β−1 pα −1 (x) = 2 2
Y
Y
ψd (x) = 22 (x − 1) (x + 1) x ×
2[
φ(d)
2 ]
ψd (x)
β α
d|n
d|2 p
d,1,2,4
Now by (2.1), the coefficient of x in LHS is 2β pα . Now the theorem follows from the fact that the constant
terms in the products:
Y
Y
2[
d=23 ,...,2β
φ(d)
2 ]
ψd (x) and
2[
φ(d)
2 ]
ψd (x)
d=22 p,22 p2 ,...,22 pα
are respectively (±2)β−2 and (±p)α by Theorem 2.2 and Theorem 2.4.
Remark 2.7. Finally we suppose n = 2β m where β ≥ 3 and m an odd integer. Then the constant term in 2[
is ±1. We omit the proof as it is similar to the case n = 22 m, m odd, given in Remark 2.5.
φ(n)
2 ]
ψn (x)
References
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[3] M. Demirci and I. N. Cangul, The constant term of the minimal polynomial of cos(2π/n) over Q, Fixed point theory and Applications,
2013:77 (2013) 1-8.
[4] D. H. Lehmer, A note on trigonometric algebraic numbers, Am. Math. Mon., 40, (1933) 165-166.
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