Minimal Polynomials of Algebraic Cosine Values, II

Pinthira Tangsupphathawat Department of Mathematics

Phranakhon Rajabhat University

Bangkok 10220 Thailand

 t.pinthira@hotmail.com

Takao Komatsu

School of Mathematics and Statistics Wuhan University Wuhan 430072 China

   komatsu@whu.edu.cn

Vichian Laohakosol Department of Mathematics

Faculty of Science Kasetsart University Bangkok 10900 Thailand

    fscivil@ku.ac.th

Abstract

We derive alternative proofs of two recent results due to Gürtas, and a new recursion related to the minimal polynomials of algebraic cosine values. We determine the minimal polynomials of algebraic values of the sine function evaluated at rational multiples of π. We also correct a formula from our earlier work.

1 Introduction

In 1933, Lehmer proved the following result [3]: let n ∈ ℕ, n > 2, k ∈ {1, 2, …, n} with gcd(k, n) = 1. Then the value 2 cos(2kπ/n) is an algebraic integer of degree φ(n)/2 (sequence \underline{A023022} in the OEIS [7]) whose minimal polynomial is ψ_{n}(x) ∈ ℤ[x], where

ψ_{n}(x + x^{−1}) = x^{−φ(n)/2}Φ_{n}(x),  (1)

where Φ_{n} is the nth cyclotomic polynomial and φ denotes Euler’s totient function (sequence \underline{A000010} in the OEIS).

The first and the third authors [8] derived some reduction formulae, based on the work of Surowski and McCombs, [6], that can be used to completely determine explicit forms of the polynomials ψ_{n}(x). Almost simultaneously, Gürtas [2] considered the same problem, but used a different approach based on Ramanujan’s trigonometric sum and its connection with the power sum function to derive a recursive relation among the coefficients of ψ_{n}, as well as certain other properties. The present work has three objectives.

1. To amend the formula in part IV of Theorem 5 of our earlier work [8].

2. To give alternative proofs of the last two results (Theorems 3.4 and 3.8) in [2] based on the results and ideas in [8].

3. To derive more useful recursive relations among the coefficients of ψ_{n} and to determine minimal polynomials of the sine values at rational multiples of π.

Since there is no universal agreement regarding the notation used in most of the works in this area, especially those appearing in [8] and [2], throughout this work we stick to the one employed in [8]. Throughout, let

X = x + x^{−1}, X_{s} = x^{s} + x^{−s} (s ∈ ℕ).

2 A correction

The formula stated in part IV of Theorem 5 in [8] is not correct as it was derived from the erroneous usage of (1) at n = 2. We now establish its correct version.

Proposition 1. For e ∈ ℕ, e ≥ 2, the minimal polynomial of 2 cos(2kπ/2^{e}), where k ∈ {1, 2, …, 2^{e}}, gcd(k, 2) = 1, is

ψ_{2^{e}}(x) = ∑_{k=0}^{⌊2^{e−3}⌋}(−1)^{k} (\binom{2^{e−2} − k}{k} + \binom{2^{e−2} − k − 1}{k − 1}) x^{2^{e−2}−2k}.

Proof. Following the approach in the proof of Theorem 5 part IV in [8], using (1) and the well-known relation (see e.g. [8, Lemma 3 B])

Φ_{mq^{e}}(x) = Φ_{mq}(x^{q^{e−1}}),

we have

ψ_{2^{e}}(X) = ψ_{2^{e}} (x + x^{−1}) = x^{−φ(2^{e})/2}Φ_{2^{e}}(x) = x^{−2^{e−2}}Φ_{2}(x^{2^{e−1}}).

Substituting Φ_{2}(T) = 1 + T [4, Example 2.46, p. 65], we get

ψ_{2^{e}}(X) = x^{−2^{e−2}} (1 + x^{2^{e−1}}) = x^{−2^{e−2}} + x^{2^{e−2}} = X_{2^{e−2}}

= ∑_{k=0}^{⌊2^{e−2}/2⌋}(−1)^{k} (\binom{2^{e−2} − k}{k} + \binom{2^{e−2} − k − 1}{k − 1}) X^{2^{e−2}−2k},

where the last equality comes from [8, Lemma 4]. Since the polynomial

ψ_{2^{e}}(X) − ∑_{k=0}^{⌊2^{e−2}/2⌋}(−1)^{k} (\binom{2^{e−2} − k}{k} + \binom{2^{e−2} − k − 1}{k − 1}) X^{2^{e−2}−2k}

vanishes for infinitely many real (or complex) values of X = x + 1/x, it is identically zero, i.e., all its coefficients are zero, and the result follows.

3 Alternative proofs of three results

In this section, based upon our work in [8], we give new proofs of the following two results due to Gürtas [2].

Theorem 2. ([2, Theorem 3.4]) Let q ∈ ℕ be odd ≥ 3, and let d = φ(q)/2 = φ(2q)/2. If

ψ_{q}(2x) = 2^{d} ∑_{i=0}^{d}(−1)^{i}e_{i}x^{d−i}

and

ψ_{2q}(2x) = 2^{d} ∑_{i=0}^{d}(−1)^{i}e′_{i}x^{d−i},

then e_{i} = (−1)^{i}e′_{i} (i = 0, 1, …, d).

Theorem 3. ([2, Theorem 3.8]) Let n > 4. If n is divisible by 4, then ψ_{n}(2x) is a polynomial consisting of even powers of x only.

To prove Theorem 2, we need a lemma.

Lemma 4. Let q ∈ ℕ be odd ≥ 3, and let d = φ(q)/2. Then

ψ_{2q}(x) = (−1)^{d} ψ_{q}(−x).

Proof. Using (1) and [4, Problem 2.57 (d)], we have

ψ_{2q}(X) = ψ_{2q} (x + x^{−1}) = x^{−d}Φ_{2q}(x) = x^{−d}Φ_{q}(−x)
= x^{−d}(−x)^{d}ψ_{q} (−x − x^{−1}) = (−1)^{d}ψ_{q}(−X).

Since the polynomial (in X) expression ψ_{2q}(X) − (−1)^{d} ψ_{q}(−X) vanishes for infinitely many real (or complex) values of X = x + 1/x, it must vanish identically, i.e., all its coefficients are identically zero, yielding ψ_{2q}(x) = (−1)^{d} ψ_{q}(−x).

Proof of Theorem 2. Using Lemma 4, we get

2^{d} ∑_{i=0}^{d}(−1)^{i}e′_{i}(2x)^{d−i} = ψ_{2q}(2x) = (−1)^{d} ψ_{q}(−2x) = 2^{d} ∑_{i=0}^{d}(−1)^{i}e_{i}(−2x)^{d−i},

and the result follows at once from equating coefficients. □

To prove Theorem 3, we need another lemma.

Lemma 5. Let e ∈ ℕ, let p be a prime, and let q ∈ ℕ, q > 1 with gcd(q, p) = 1. Then

ψ_{p^{e}q}(X) = \frac{ψ_{q}(X_{p^{e}})}{ψ_{q}(X_{p^{e−1}})}.

Proof. Using (1) and the well-known relation (see e.g. [8, Lemma 3 B]) Φ_{mq^{e}}(x) = Φ_{mq}(x^{q^{e−1}}), we get

ψ_{p^{e}q}(X) = ψ_{p^{e}q} (x + x^{−1}) = x^{−φ(p^{e}q)/2}Φ_{p^{e}q}(x) = x^{−φ(p^{e}q)/2}Φ_{pq}(x^{p^{e}−1})

= x^{−φ(p^{e}q)/2} \frac{Φ_{q} ((x^{p^{e−1}})^{p})}{Φ_{q} (x^{p^{e−1}})} = x^{−φ(p^{e}q)/2} \frac{(x^{p^{e}})^{φ(q)/2}ψ_{q}(x^{p^{e}} + x^{−p^{e}})}{(x^{p^{e}−1})^{φ(q)/2}ψ_{q}(x^{p^{e}−1} + x^{−p^{e−1}})}

= \frac{ψ_{q} (X_{p^{e}})}{ψ_{q}(X_{p^{e−1}})}.

Proof of Theorem 3. Let n = 2^{e}q, where e ≥ 2 and q is an odd positive integer chosen such that n = 2^{e}q > 4. From Lemma 5, and [8, Lemma 4], i.e.,

X_{s} = ∑_{k=0}^{⌊s/2⌋}(−1)^{k} (\binom{s − k}{k} + \binom{s − k − 1}{k − 1}) X^{s−2k},

with the convention that \binom{n}{r} = 0 for negative r, we have

ψ_{n}(2X) = ψ_{2^{e}q}(2X) = \frac{ψ_{q} (∑_{k=0}^{2^{e}/2}(−1)^{k} (\binom{2^{e}−k}{k} + \binom{2^{e}−k−1}{k−1}) (2X)^{2^{e}−2k})}{ψ_{q} (∑_{k=0}^{2^{e−1}/2}(−1)^{k} (\binom{2^{e−1}−k}{k} + \binom{2^{e−1}−k−1}{k−1}) (2X)^{2^{e−1}−2k})}.

Since the polynomial (in X)

ψ_{2^{e}q}(2X)ψ_{q} ( ∑_{k=0}^{2^{e−1}/2}(−1)^{k} (\binom{2^{e−1} − k}{k} + \binom{2^{e−1} − k − 1}{k − 1}) (2X)^{2^{e−1}−2k})

−ψ_{q} (∑_{k=0}^{2^{e}/2}(−1)^{k} (\binom{2^{e} − k}{k} + \binom{2^{e} − k − 1}{k − 1}) (2X)^{2^{e}−2k})

vanishes for infinitely many real values of X = x + 1/x, it must vanish identically, and since ψ_{s}(X) is a monic polynomial, we identically have

ψ_{n}(2x)ψ_{q} ( ∑_{k=0}^{2^{e−1}/2}(−1)^{k} (\binom{2^{e−1} − k}{k} + \binom{2^{e−1} − k − 1}{k − 1}) (2x)^{2^{e−1}−2k})

= ψ_{q} (∑_{k=0}^{2^{e}/2}(−1)^{k} (\binom{2^{e} − k}{k} + \binom{2^{e} − k − 1}{k − 1}) (2x)^{2^{e}−2k}) .  (2)

Observe next that from e ≥ 2, the arguments of ψ_{q} in the second factor on the left and in the right-hand expression, namely,

∑_{k=0}^{2^{e}/2}(−1)^{k} (\binom{2^{e} − k}{k} + \binom{2^{e} − k − 1}{k − 1}) (2x)^{2^{e}−2k} and

∑_{k=0}^{2^{e−1}/2}(−1)^{k} (\binom{2^{e−1} − k}{k} + \binom{2^{e−1} − k − 1}{k − 1}) (2x)^{2^{e−1}−2k}

are polynomials with even exponents. Thus, in (2) the right-hand expression and the second polynomial factor on the left contain only even powers of x. This forces the polynomial ψ_{n}(2x) to contain only even powers of x. □

We end this section by presenting another derivation of ψ_{p} [6, Theorem 2.1] based on the following binomial identity which is identity (1.60) in [1, p. 8].

∑_{k=0}^{⌊n/2⌋}(−1)^{k}\binom{n − k}{k} (xy)^{k} (x + y)^{n−2k} = \frac{x^{n+1} − y^{n+1}}{x − y} .  (3)

Theorem 6. Let p = 2s + 1 be an odd prime. The minimal polynomial of 2 cos(2π/p) is

ψ_{p}(x) = ∑_{j=0}^{⌊s/2⌋}(−1)^{j}\binom{s − j}{j}x^{s−2j} − ∑_{j=1}^{⌊(s+1)/2⌋}(−1)^{j}\binom{s − j}{j − 1}x^{s−(2j−1)}.

Proof. Putting y = 1/x in (3), we get

∑_{k=0}^{⌊n/2⌋}(−1)^{k}\binom{n − k}{k} (x + \frac{1}{x})^{n−2k} = \frac{x^{n+1} − 1/x^{n+1}}{x − 1/x} .  (4)

Adopting the convention that \binom{m}{j} = 0 if j < 0, and using (4), we get

∑_{k=0}^{⌊n/2⌋}(−1)^{k}\binom{n − k − 1}{k − 1} (x + \frac{1}{x})^{n−2k} = ∑_{ℓ=0}^{⌊n/2⌋−1}(−1)^{ℓ+1}\binom{n − ℓ − 2}{ℓ} (x + \frac{1}{x})^{n−2ℓ−2}

= − \frac{x^{n−1} − 1/x^{n−1}}{x − 1/x} .  (5)

Adding (4) and (5), we get

∑_{k=0}^{⌊n/2⌋}(−1)^{k}\binom{n − k}{k} (x + \frac{1}{x})^{n−2k} + ∑_{k=0}^{⌊n/2⌋}(−1)^{k}\binom{n − k − 1}{k − 1} (x + \frac{1}{x})^{n−2k}

= \frac{(x^{2} − 1)(x^{n−1} + 1/x^{n+1})}{x − 1/x} = x^{n} + 1/x^{n} = X_{n}.  (6)

We claim that the sum of the X_{n}’s with odd and even indices are given, respectively, by

X_{1} + X_{3} + ⋯ + X_{2t+1} = ∑_{j=0}^{t}(−1)^{j}\binom{2t + 1 − j}{j}X^{2t+1−2j}  (7)

X_{2} + X_{4} + ⋯ + X_{2t} = ∑_{j=0}^{t−1}(−1)^{j}\binom{2t − j}{j}X^{2t−2j} − 2δ_{t},  (8)

where δ_{t} = 0 for even t, and δ_{t} = 1 for odd t. We begin with (7), which holds trivially for t = 0. Assume that it holds up to t − 1, i.e., assume

X_{1} + X_{3} + ⋯ + X_{2t−1} = ∑_{j=0}^{t−1}(−1)^{j}\binom{2t − 1 − j}{j}X^{2t−1−2j} = ∑_{k=1}^{t}(−1)^{k−1}\binom{2t − k}{k − 1}X^{2t+1−2k}. (9)

From (6), with n = 2t + 1, we get

X_{2t+1} = ∑_{k=0}^{t}(−1)^{k} (\binom{2t + 1 − k}{k} + \binom{2t − k}{k − 1}) X^{2t+1−2k}

= X^{2t+1} + ∑_{k=1}^{t}(−1)^{k} (\binom{2t + 1 − k}{k} + \binom{2t − k}{k − 1}) X^{2t+1−2k}.  (10)

The identity (7) follows from induction by adding (9) and (10). We proceed now to verify (8). When t = 1, the right-hand expression is equal to

X^{2} − 2 = (x + \frac{1}{x})^{2} − 2 = x^{2} + \frac{1}{x^{2}} = X_{2},

and we are done in this case. Assume that it holds up to t − 1, i.e., assume that

X_{2} + X_{4} + ⋯ + X_{2t−2} = ∑_{j=0}^{t−2}(−1)^{j}\binom{2t − 2 − j}{j}X^{2t−2−2j} − 2δ_{t−1}

= ∑_{k=1}^{t−1}(−1)^{k−1}\binom{2t − 1 − k}{k − 1}X^{2t−2k} − 2δ_{t−1}  (11)

From (6), with n = 2t, we get

X_{2t} = ∑_{k=0}^{t}(−1)^{k} (\binom{2t − k}{k} + \binom{2t − k − 1}{k − 1}) X^{2t−2k}

= X^{2t} + ∑_{k=1}^{t}(−1)^{k} (\binom{2t − k}{k} + \binom{2t − k − 1}{k − 1}) X^{2t−2k}.  (12)

Adding (11) and (12), we get

X_{2} + ⋯ + X_{2t−2} + X_{2t} = X^{2t} + ∑_{k=1}^{t−1}(−1)^{k}\binom{2t − k}{k}X^{2t−2k} + (−1)^{t}2 − 2δ_{t−1},

and using the definition of δ_{t}, the identity (8) follows by induction.

From (1) and the shape of the p^{th} cyclotomic polynomial [4, Chapter 2], we get

ψ_{p}(X) = ψ_{p} (x + x^{−1}) = x^{−s}Φ_{p}(x) = 1 + (x + x^{−1}) + ⋯ + (x^{s} + x^{−s}) = 1 + X_{1} + ⋯ + X_{s}.

For odd s = 2t + 1, using (7) and (8) we have

ψ_{p}(X) = (1 + X_{2} + X_{4} + ⋯ + X_{2t}) + (X_{1} + X_{3} + ⋯ + X_{2t+1})

= (1 + ∑_{j=0}^{t−1}(−1)^{j}\binom{2t − j}{j}X^{2t−2j} − 2δ_{t}) + (∑_{j=0}^{t}(−1)^{j}\binom{2t + 1 − j}{j}X^{2t+1−2j})

= ∑_{j=0}^{t}(−1)^{j}\binom{2t − j}{j}X^{2t−2j} + ∑_{j=0}^{t}(−1)^{j}\binom{2t + 1 − j}{j}X^{2t+1−2j}

= ∑_{k=1}^{t+1}(−1)^{k−1}\binom{2t − k + 1}{k − 1}X^{2t+2−2k} + ∑_{j=0}^{t}(−1)^{j}\binom{2t + 1 − j}{j}X^{2t+1−2j}.

For even s = 2t, using (7) and (8) we have

ψ_{p}(X) = (1 + X_{2} + X_{4} + ⋯ + X_{2t}) + (X_{1} + X_{3} + ⋯ + X_{2t−1})

= (1 + ∑_{j=0}^{t−1}(−1)^{j}\binom{2t − j}{j}X^{2t−2j} − 2δ_{t}) + (∑_{j=0}^{t−1}(−1)^{j}\binom{2t − 1 − j}{j}X^{2t−1−2j})

= ∑_{j=0}^{t}(−1)^{j}\binom{2t − j}{j}X^{2t−2j} + ∑_{k=1}^{t}(−1)^{k−1}\binom{2t − k}{k − 1}X^{2t+1−2k},

and the assertion of the theorem holds for ψ_{p}(X). As argued before, this relation holds for infinitely many values of X = x + 1/x yielding it to be an identity, and the desired result follows.

4 More properties

Lemmas 4 and 5 enable us to deduce the next interesting result.

Theorem 7. Let q ∈ ℕ be odd ≥ 3, d := φ(q)/2.

(a) We have ψ_{q}(−X)ψ_{q}(X) = (−1)^{d}ψ_{q}(X^{2} − 2);

(b) If ψ_{q}(x) := ∑_{i=0}^{d} a_{i}x^{i}, then for ℓ ∈ {0, 1, …, d} we have

∑_{\begin{matrix}0≤i,j≤d \\ i+j=2ℓ \end{matrix}}(−1)^{i}a_{i}a_{j} = (−1)^{d} ∑_{k=ℓ}^{d} \binom{k}{ℓ}a_{k}(−2)^{k−ℓ}

∑_{\begin{matrix}0≤i,j≤d \\ i+j is odd \end{matrix}}(−1)^{i}a_{i}a_{j} = 0.

Proof. (a) Taking p = 2, e = 1 in Lemma 5 and equating with the expression in Lemma 4, we get

\frac{ψ_{q}(X_{2})}{ψ_{q}(X)} = ψ_{2q}(X) = (−1)^{d}ψ_{q}(−X).

The result follows at once from X_{2} = x^{2} + x^{−2} = (x + x^{−1})^{2} − 2 = X^{2} − 2.

(b) From part (a), arguing as before we see that the polynomial

ψ_{q}(−X)ψ_{q}(X) − (−1)^{d}ψ_{q}(X^{2} − 2)

vanishes identically, and since each polynomial ψ_{q} is monic, we get ψ_{q}(−x)ψ_{q}(x) = (−1)^{d}ψ_{q}(x^{2}− 2), i.e.,

∑_{i=0}^{d} a_{i}(−x)^{i} ∑_{j=0}^{d} a_{j}x^{j} = (−1)^{d} ∑_{k=0}^{d} a_{k}(x^{2} − 2)^{k}.

The first and second assertions follow from equating the coefficients of the even (respectively, odd) powers of x on both sides. The second assertion can also be verified by noting that since i + j is odd, the integers i and j have different parities. Therefore, (−1)^{i}a_{i}a_{j} + (−1)^{j}a_{j}a_{i} = 0 for every such pair.

4.1 Minimal polynomials of sine values

Lehmer [3, Theorem 2, p. 166] also proved the result about the values of the sine function at rational multiples of π. We now use an analysis in [5, Theorem 3.9, pp. 37–39], which amends inaccuracies in the case n ≡ 4 (mod 8) of [3, Theorem 2, p. 166], to determine the minimal polynomial of sin(2kπ/n) \underline{A178182}.

Theorem 8. A. Let n ∈ ℕ, k ∈ {1, 2, …, n} with n > 2, n ≠ 4 and gcd(k, n) = 1.

(a) If n is odd, then 2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is ψ_{4n}(x) of degree φ(n).

(b) If n ≡ 2 (mod 4), then 2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is ψ_{2n}(x) of degree φ(n).

(c) If n ≡ 0 (mod 8), then 2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is ψ_{n}(x) of degree φ(n)/2.

(d) If n ≡ 4 (mod 8), n > 4 and k ≡ n/4 (mod 4), then 2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is ψ_{n/4}(x) of degree φ(n)/4.

(e) If n ≡ 4 (mod 8), n > 4 and k ≢ n/4 (mod 4), then 2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is ψ_{n/2}(x) of degree φ(n)/4.

B. If n = 1 or n = 2, then 2 sin(2π/n) = 0 is an algebraic integer whose minimal polynomial is x. If n = 4, then 2 sin(2π/n) = 2 is an algebraic integer whose minimal polynomial is x−2, while 2 sin(2·3π/n) = −2 is an algebraic integer whose minimal polynomial is ψ_{2}(x) = x+2.

Proof. A. Note that

2 sin (\frac{2kπ}{n} ) = 2 cos 2π (\frac{k}{n} − \frac{1}{4}) = 2 cos (\frac{2π(4k − n)}{4n}) .

If n is odd, then the fraction (4k − n)/4n is in reduced form and so 2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is the same as that of 2 cos (\frac{2π(4k−n)}{4n}), i.e., ψ_{4n}(x) with deg ψ_{4n} = φ(n).

If n ≡ 2 (mod 4), then \frac{4k−n}{4n} = \frac{2k−n/2}{2n}, where the last fraction is in reduced form. Thus,

2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is that of 2 cos (\frac{2π(2k−n/2)}{2n}), i.e., ψ_{2n}(x) with deg ψ_{2n} = φ(2n)/2 = φ(n).

If n ≡ 0 (mod 4), then there are two subcases.

• If n ≡ 0 (mod 8), then \frac{4k−n}{4n} = \frac{k−n/4}{n} is in reduced form because gcd(k, n) = 1. Thus,

2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is that of 2 cos (\frac{2π(k−n/4)}{n}), i.e., ψ_{n}(x) with deg ψ_{n} = φ(n)/2.

• If n ≡ 4 (mod 8), then the fraction (4k−n)/4n reduces to one with denominator n/4 in case k ≡ n/4 (mod 4) and denominator n/2 otherwise. In the former case, 2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is that of 2 cos (\frac{2π(k−n/4)/4}{n/4}), i.e., ψ_{n/4}(x) with degree φ(n/4)/2 = φ(n)/4. In the latter case, 2 sin(2kπ/n) is an algebraic integer whose minimal polynomial is that of 2 cos (\frac{2π(k−n/4)/2}{n/2}), i.e., ψ_{n/2}(x) with degree φ(n/2)/2 = φ(n)/4.

The assertions in part B are easily checked directly.

5 Acknowledgments

The first author is supported by Phranakhon Rajabhat University. The third author is supported by the Center for Advanced Studies in Industrial Technology and the Faculty of Science, Kasetsart University. The authors thank the referee for several corrections and suggestions.

References

[1] H. W. Gould, Combinatorial Identities: A Standardized Set of Tables Listing 500 Binomial Coefficient Summations, Morgantown Printing and Binding Co., Morgantown, 1972.

[2] Y. Z. Gürtas, The minimal polynomial of cos(2π/n), Commun. Korean Math. Soc. 31 (2016), 667–682.

[3] D. H. Lehmer, A note on trigonometric algebraic numbers, Amer. Math. Monthly 40 (1933), 165–166.

[4] R. Lidl and H. Niederreiter, Finite Fields, Encyclopedia of Mathematics and Its Applications, Volume 20, Addison-Wesley, 1983.

[5] I. Niven, Irrational Numbers, Carus Monographs, Vol. 11, The Mathematical Association of America, 1956.

[6] D. Surowski and P. McCombs, Homogeneous polynomials and the minimal polynomial of cos(2π/n), Missouri J. Math. Sci. 15 (2003), 4–14.

[7] N. J. A. Sloane, The Online Encyclopedia of Integer Sequences, 2018, available at

   https://oeis.org.

[8] P. Tangsupphathawat and V. Laohakosol, Minimal polynomials of algebraic cosine values at rational multiples of π, J. Integer Sequences 19 (2016), Article 16.2.8.

2010 Mathematics Classification: Primary 11R04; Secondary 33B10, 11J72. Keywords: cosine function, algebraic value, multiple of π.

(Concerned with sequences \underline{A000010}, \underline{A023022}, and \underline{A178182}.)

Received March 31 2017; revised versions received January 6 2018; April 20 2018; September 3 2018. Published in Journal of Integer Sequences, December 10 2018.

Return to Journal of Integer Sequences home page.