Some polynomials associated with regular polygons

Acta Univ. Sapientiae, Mathematica,

Some polynomials associated with regular polygons

Seppo Mustonen

Department of Mathematics and Statistics

FI-00014 University of Helsinki, Finland email:[email protected]

Pentti Haukkanen

School of Information Sciences FI-33014 University of Tampere,

Finland

email:[email protected]

Jorma Merikoski

School of Information Sciences FI-33014 University of Tampere, Finland

email:[email protected]

Abstract. LetGn be a regularn-gon with unit circumradius, andm= bⁿ₂c,µ=bⁿ⁻¹₂ c. Let the edges and diagonals ofGn been1<· · ·< enm. We compute the coefficients of the polynomial

(x−e²_n1)· · ·(x−e²_nµ).

They appear to form a well-known integer sequence, and we study certain related sequences, too. We also compute the coefficients of the polynomial

(x−s²_n1)· · ·(x−s²_nm),

where

s_ni=cot(2i−1)π

2n , i=1, . . . , m.

We interpretsn1 as the sum of all individual edges and diagonals ofGn

divided by n. We also discuss the interpretation of sn2, . . . , snm, and present a conjecture on expressingsn1, . . . , snmusingen1, . . . , enm.

2010 Mathematics Subject Classification:11B83, 11C08, 15B36, 51M20 Key words and phrases:polynomial, regular polygon, eigenvalue

178

10.1515/ausm-2015-0005

1 Introduction

Throughout, G_n is a regular n-gon with unit circumradius, and m=

jn 2 k

, µ=

n−1 2

.

Long time ago Kepler observed [2] that the squares of the edge and diagonals ofG₇ are the zeros of the polynomialx³−7x²+14x−7. This raises a general question: Are the squares of (the lengths of) the edge and diagonals of G_n, excluding the diameter, the zeros of a monic polynomial of degree µ with integer coefficients?

Yes, they are. This follows from Savio’s and Suruyanarayan’s [6] results, which, however, do not give the polynomial explicitly. We will do it in Sec- tion2. A natural further question concerns the edge and diagonals themselves, instead of their squares. They are not zeros of a polynomial described above, but we will in Section 3 see that the squared sum of all individual edges and diagonals is the largest zero of a monic polynomial of degree m with inte- ger coefficients. We will study geometric interpretation of the square roots of the other zeros in Section 4. In Section 5, we will present a conjecture on expressing these square roots as simple linear combinations of the edge and di- agonals. We will in Section6notify that the coefficients of the first-mentioned polynomial form an OEIS [4] sequence, and also study OEIS sequences corre- sponding to certain related polynomials. Finally, we will complete our paper with conclusions and further questions in Section 7.

2 Squared chords

Let (the lengths of) the edge and diagonals of G_n be e_n1 < · · · < e_nm. Call them (the lengths of) thechords. Then

e_nk=2sinkπ

n , k=1, . . . , m.

Our problem is to find the coefficients a_mkand b_mk of the polynomials A_m(x) = (x−e²_n+2,1)· · ·(x−e²_n+2,m) =

x^m+a_m,m−1x^m−1+· · ·+a_m1x+a_m0, (1) wheren is even, and

B_m(x) = (x−e²_n1)· · ·(x−e²_nm) =x^m+b_m,m−1x^m−1+· · ·+b_m1x+b_m0, (2)

wherenis odd. We solve it in two theorems. Mustonen [3] found them exper- imentally and sketched their proofs.

Let tridiag_m(x, y)denote the symmetric tridiagonal m×mmatrix with all main diagonal entriesxand first super- and subdiagonal entriesy. Form≥2, define

A_m =tridiag_m(2, 1) and

B_m is as A_m but the (m, m) entry equals 3.

Also define A1= (2) andB1= (3). Denote by spec the (multi)set of eigenval- ues.

Lemma 1 For all m≥1, specA_m =

4sin² kπ n+2

k=1, . . . , m

={e²_n+2,1, . . . , e²_n+2,m}, (3) specB_m=

4sin²kπ n

k=1, . . . , m

={e²_n1, . . . , e²_nm}.

Proof.See [1,5,6].

Theorem 1 In(1),

a_mk= (−1)^m−k

m+1+k 2k+1

. (4)

Proof.Denoting

P_m(x) =x^m+

m−1X

k=0

(−1)^m−k

m+1+k 2k+1

x^k, our claim is that

P_m(x) =A_m(x) (5)

for all m≥1. Expanding det(xI_m−A_m) along the last row, we have A_m+1(x) = (x−2)A_m(x) −A_m−1(x)

for all m≥2. Since

P₁(x) =x−2=A₁(x)

and

P₂(x) =x²−4x+3=A₂(x), the claim (5) follows by showing that

P_m+1(x) = (x−2)P_m(x) −P_m−1(x) (6) for all m ≥ 2. Mustonen [3] did it by using Mathematica. We will do the

computations algebraically in the appendix.

The formula (4) yields a_mm = 1, consistently with the coefficient of x^m in (1). It also allows to define a₀₀=1. The polynomial

A˜_m+1(x) = (x−4)A_m(x) =x^m+1+α_m+1,mx^m+· · ·+α_m+1,1x+α_m+1,0 (7) hase²_n+2,m+1=4 as the additional zero. By (4),

α_m+1,k= (−1)^m−k+1

m+k 2k−1

+4

m+1+k 2k+1

. (8)

(We define ⁿ_k

=0 ifk < 0.) Theorem 2 In(2),

b_mk= (−1)^m−k2m+1 m−k

m+k 2k+1

= (−1)^m−k

m+1+k 2k+1

+

m+k 2k+1

. (9) Proof. The second equation follows from trivial computation. To show the first, denote

Q_m(x) =x^m+

m−1X

k=0

(−1)^m−k2m+1 m−k

m+k 2k+1

x^k and claim that

Q_m(x) =B_m(x) (10)

for all m≥1. Expanding det(xI_m−B_m), we have B_m+1(x) = (x−3)A_m(x) −A_m−1(x) for all m≥2. Since

Q₁(x) =x−3=B₁(x)

and

Q₂(x) =x²−5x+5=B₂(x), the claim (10) follows by showing that

Q_m+1(x) = (x−3)P_m(x) −P_m−1(x) (11) for all m≥2. Mustonen [3] did also this by using Mathematica, and we will do the computations algebraically in the appendix.

Fork=m, the first expression in (9) is undefined but the second is defined.

(We define ⁿ_k

=0 ifn < k.) It givesb_mm=1, the coefficient ofx^m in (2). It also allows to define b₀₀=1.

Corollary 1 The sum of all individual squared chords of G_n is n². Their product is nⁿ.

Proof.By Theorems 1and 2 (or by [7, Eqs. (20) and (24)]), we obtain e²_2m,1+· · ·+e²_2m,m−1 = −a_m−1,m−2=2(m−1),

e²_2m+1,1+· · ·+e²_2m+1,m = −b_m,m−1=2m+1, and

e²_2m,1· · ·e²_2m,m−1= (−1)^ma_m−1,0=m, e²_2m+1,1· · ·e²_2m+1,m = (−1)^mb_m0=2m+1.

Denoting by Σ_n the sum and by Π_n the product of all individual squared chords ofG_n, we therefore have

Σ_2m=2m·2(m−1) +m·4= (2m)², Σ_2m+1= (2m+1)(2m+1) = (2m+1)², and

Π_2m=m^2m4^m= (2m)^2m, Π_2m+1= (2m+1)^2m+1.

3 Sum of chords

The sum of all individual chords of G_n is S_n=ns_n, where

s_n=e_n1+· · ·+e_n,m−1+ ¹₂e_nm =e_n1+· · ·+e_n,m−1+1 ifnis even, and

s_n=e_n1+· · ·+e_nm

if n is odd, is the sum of different (lengths of) chords but the diameter is halved.

Theorem 3 For all n≥3,

s_n=cot π 2n. Proof.We have [7, Eq. (21)]

n−1X

k=1

sinkπ

n =cot π

2n. (12)

Ifn is even, this implies s_n=

m−1X

k=1

2sinkπ n +1

2 ·2=

m−1X

k=1

sinkπ n +1+

2m−1X

k=m+1

sinkπ n =

2m−1X

k=1

sinkπ

n =cot π 2n. Ifn is odd, then

s_n= Xm k=1

2sinkπ n =

Xm k=1

sinkπ n +

X2m k=m+1

sinkπ n =

X2m k=1

sinkπ

n =cot π 2n.

Iss_na zero of a monic polynomial of degreemwith integer coefficients? Yes fors₄=cot^π₈ =1+√

2; it is a zero ofx²−2x−1. On the other hand, it is easy to see that s₅ = cot₁₀^π = p

5+2√

5 is not a zero of such a polynomial. But

s²₅=5+2√

5 is a zero ofx²−10x+5, and the other zero is5−2√

5=cot^{2 3π}₁₀. Also s²₄ =3+2√

2has this property: it is a zero of x²−6x+1, and the other zero is3−2√

2=cot^{2 3π}₈ . Generally, denoting

s_ni =cot(2i−1)π

2n , i=1, . . . , m,

this motivates us to study for even nthe coefficients of the polynomial U_m(x) = (x−s²_n1)· · ·(x−s²_nm) =x^m+u_m,m−1x^m−1+· · ·+u_m1x+u_m0,(13) and for oddn those of

V_m(x) = (x−s²_n1)· · ·(x−s²_nm) =x^m+v_m,m−1x^m−1+· · ·+v_m1x+v_m0. (14) We will see that they all are integers. The largest zero is s²_n=s²_n1.

Mustonen [3] found the following theorem experimentally and also presented its proof. Yaglom and Yaglom [9, Eqs. (7) and (8)] formulated (16) differently.

Theorem 4 In(13),

u_mk= (−1)^k n

2k

. (15)

In(14),

v_mk= (−1)^k n

2k+1

. (16)

Proof.We have [10]

cotnt= P_m

k=0(−1)^k _2kⁿ

cot^n−2kt P_m

k=0(−1)^k _2k+1ⁿ

cot^n−2k−1t. (17)

Denote

t_i = (2i−1)π

2n , i=1, . . . , m.

Since cotnt_i =0, (17) yields Xm

k=0

(−1)^k n

2k

cot^n−2kt_i =0. (18)

First assume neven. The polynomial U˜_m(x) =

Xm k=0

(−1)^m−k n

2k

x^k

is monic and has degree m. For alli=1, . . . , m, U˜_m(s²_ni) =

Xm k=0

(−1)^m−k 2m

2k

s^2k_ni = Xm

l=0

(−1)^l

2m 2m−2l

s^2m−2l_ni

= Xm

l=0

(−1)^l 2m

2l

s^2m−2l_ni = Xm

l=0

(−1)^l n

2l

cot^n−2lt_i =0 by (18). Hence

U˜_m(x) = (x−s²_n1)· · ·(x−s²_nm) =U_m(x), and (15) follows.

Second, assume nodd. The polynomial V˜_m(x) =

Xm

k=0

(−1)^m−k n

2k+1

x^k

is monic and has degree m. For alli=1, . . . , m, V˜_m(s²_ni) =

Xm

k=0

(−1)^m−k

2m+1 2k+1

s^2k_ni=

Xm

l=0

(−1)^l

2m+1 2m−2l+1

s^2m−2l_ni = s⁻¹_ni

Xm

l=0

(−1)^l

2m+1 2m−2l+1

s^2m+1−2l_ni =s⁻¹_ni Xm

l=0

(−1)^l

2m+1 2l

s^2m+1−2l_ni

=s⁻¹_ni Xm

l=0

(−1)^l n

2l

cot^n−2lt_i =0, again by (18). Hence

V˜_m(x) = (x−s²_n1)· · ·(x−s²_nm) =V_m(x),

and (16) follows.

Corollary 2 The number s²_n is the largest zero of the polynomial x^m+u_m,m−1nx^m−1+· · ·+u_m1n^m−1x+u_m0n^m if n is even, and that of

x^m+v_m,m−1nx^m−1+· · ·+v_m1n^m−1x+v_m0n^m if n is odd.

4 Interpreting s

, k = 1, . . . , b

c, n odd

The zeros of A_m(x) and B_m(x) describe the squared chords of G_2m+2 and G_2m+1, respectively, excluding the diameter. The largest zero ofU_m(x),s²_2m,1= s²_2m, and that of V_m(x), s²_2m+1,1 = s²_2m+1, describe the squared sum of chords but halving the diameter. In other words, the sum of all individual chords of G_n is divided by n and the result is squared.

What about the other zeros?

Let the vertices of G_n be P₀, . . . , P_n−1, where P_k = (cos^kπ_n,sin^kπ_n). Then e_nk= P₀P_k=2sin^kπ_n,k=1, . . . , m. Since P₀P_n−k=P₀P_k, we define e_n,n−k= e_nk,k=1, . . . , m.

Fix n and denote e_k = e_nk for brevity. Assume that 3k < n; i.e., k < ⁿ₃. Then the line segmentsP₀P_2kandP_kP_n−kintersect; letQ_kbe their intersection point and denotex_k=P₀Q_k. Because 4Q_kP₀P_k∼4Q_kP_2kP_n−k, we have

x_k

e_2k−x_k = e_k e_3k. Hence

x_k= e_ke_2k

e_k+e_3k = 2sin^kπ_n sin^2kπ_n sin^kπ_n +sin^3kπ_n = 2sin^kπ_n sin^2kπ_n

sin(^2kπ_n − ^kπ_n) +sin(^2kπ_n + ^kπ_n) = sin^kπ_n sin^2kπ_n

sin^2kπ_n cos^kπ_n =tankπ n . Ifn is odd, then

tankπ n =cot

π

2 − kπ 2m+1

=cot[2(m−k) +1]π

2n =s_n,m−k+1.

Thus s_n,m−k+1 = P₀Q_k,k = 1, . . . ,bⁿ⁻¹₃ c. In other words, the bⁿ⁻¹₃ c smallest zeros of V_m(x) are the squared line segments P₀Q_k,k=1, . . . ,bⁿ⁻¹₃ c. Musto- nen [3] found this experimentally. The largest zero is already interpreted, but the interpretation of the rest of zeros remains open. For some experimental observations, see [3]. Interpretation of the zeros ofU_m(x), except the largest, remains open, too.

5 Expressing s

, . . . , s

using e

, . . . , e

Mustonen’s [3] experiments make conjecture that, givenn, there are numbers λ⁽ⁱ⁾_nk∈{0,±1},i, k=1, . . . , m, such that

s_ni=λ⁽ⁱ⁾_n1e_n1+· · ·+λ⁽ⁱ⁾_n,m−1e_n,m−1+λ⁽ⁱ⁾_nme_nm⁰ , i=1, . . . , m, where

e_nm⁰ = ₁

2e_nm ifn is even, e_nm ifn is odd.

In other words, cot(2i−1)π

2n =2 h

λ⁽ⁱ⁾_n1sinπ

n+· · ·+λ⁽ⁱ⁾_n,m−1sin(m−1)π

n +θ_nλ⁽ⁱ⁾_nmsinmπ n

i , where

θ_n= ₁

2 ifn is even, 1 ifn is odd.

This is true by (12) when i= 1(s_n1 =s_n,λ⁽¹⁾_n1 =· · ·=λ⁽¹⁾_nm =1) but remains generally open.

For example, let n=15. Denoting s_k =s_15,k and e_k= e_15,k for brevity, we have [3, p. 17]

s₁= e₁+ e₂+ e₃+ e₄+ e₅+ e₆+ e₇

s₂= e₃+ e₆

s₃= e₅

s₄= e₁− e₂+ e₃− e₄+ e₅− e₆+ e₇

s₅= −e₃+ e₆

s₆= e₁− e₂+ e₃+ e₄− e₅+ e₆− e₇ s₇= e₁+ e₂− e₃− e₄+ e₅+ e₆− e₇.

We study the zero coefficients in general. If and only ifd=gcd(n, 2i−1)> 1, thenG_n ”inherits” the chord

s_ni=cot(2i−1)π 2n

fromG_d. Then the chords ofG_dare enough to expresss_ni, and the coefficients of the remaining chords are zero. Indeed, in our example,

s₂=s_15,2=cot3π

30 =cot π 10 =2

sinπ

5 +sin2π 5

, s₃=s_15,3 =cot5π

30 =cotπ

6 =2 sinπ 3, s₅=s_15,5 =cot9π

30 =cot3π 10 =2

−sinπ

5 +sin2π 5

, showing that s₃ is ”inherited” fromG₃, and s₂ and s₅ from G₅.

So we conjecture additionally that if and only if n is a prime or a power of 2, then eachλ⁽ⁱ⁾_nk∈{±1}. Mustonen [3] gives also other experimental results and conjectures about the structure of the three-dimensional array(λ⁽ⁱ⁾_nk), and presents an efficient algorithm to compute these numbers.

6 Connections with OEIS sequences

The (lexicographically ordered) sequence (a_mk) is A053122 in OEIS. Its first six terms area₀₀=1,a₁₀ = −2,a₁₁=1,a₂₀ =3,a₂₁= −4,a₂₂ =1.

The OEIS sequence A132460 consists of the numbers t_n0 =1, n=0, 1, 2, . . . ,

t_nk= (−1)^k(

n−k k

+

n−k−1 k−1

), n=2, 3, . . . , k=1, . . . , m.

The first six terms of its subsequence corresponding to odd values of n are t₁₀ = 1= b₀₀, t₃₀ = 1= b₁₁, t₃₁ = −3= b₁₀, t₅₀ =1 = b₂₂, t₅₁ = −5= b₂₁, t₅₂=5=b₂₀. In general,b_mk=t_2m+1,m−k.

Also the characteristic polynomials of certain other tridiagonal matrices have connections with OEIS sequences. We study two of them.

Let tridiag(a,b,c) denote the tridiagonal matrix with main diagonal, sub- diagonal and superdiagonal entries those of vectors a, b and c, respectively, and denote x^(k)=x, . . . , x,k copies. Form≥3, define

C_m=tridiag((2^(m)),((−1)^(m−2),−2),(−2,(−1)^(m−2)))

and

C₂=

2 −2

−2 2

, C₁= (2).

Form≥1, consider the polynomial

C_m(x) =det(xI_m−C_m) =x^m+c_m,m−1x^m−1+· · ·+c_m1x+c_m0 and define C₀(x) = 1, c₀₀ = c_mm = 1. The sequence A140882 consists of the numbers (−1)^mc_mk. Since C₀(x) = 1, C₁(x) = x−2, C₂(x) = x²− 4x, C₃(x) = x³−6x²+8x, its first ten terms are 1, 2,−1, 0,−4, 1, 0,−8, 6,−1, as listed in [4].

We have xA˜₁(x) =x²−4x= C₂(x) and xA˜₂(x) = x³−6x²+8x = C₃(x), and generally

C_m+1(x) =xA˜_m(x) (19) for allm≥1. This can be proved similarly to the proofs of Theorems1and2.

By (8), a formula for A140882 is then obtained. By (19), (7) and (3), specC_m=specA_m−2∪{0, 4}=

4sin² kπ 2m−2

k=0, . . . , m−1

form≥3.

Finally, the sequence A136672 motivates us to study the polynomial F_m+1(x) = (x−2)A_m(x) =x^m+1+f_m+1,mx^m+· · ·+f_m+1,1x+f_m+1,0 (20) and its connections with the matrix Dm, defined by

D_m=tridiag((2^(m)),((−1)^(m−2), 0),((−1)^(m−1))) ifm≥3, and

D₂=

2 −1 0 2

, D₁= (2).

By Theorem 1,

f_m+1,k= (−1)^m−k+1(

m+k 2k−1

+2

m+1+k 2k+1

). (21)

For m≥1, consider the polynomial

D_m(x) =det(xI_m−D_m) =x^m+d_m,m−1x^m−1+· · ·+d_m1x+d_m0

and define D₀(x) =1, d₀₀ = d_mm= 1. The sequence A136672 consists of the numbers(−1)^md_mk. We have D₀(x) =1,D₁(x) =x−2,D₂(x) =x²−4x+4, D₃(x) =x³−6x²+11x−6. So its first ten terms are1, 2,−1, 4,−4, 1, 6,−11, 6,−1, as listed in [4].

Since F₁(x) = x−2 = D₁(x), F₂(x) = x²−4x+4 = D₂(x), and F₃(x) = x³−6x²+11x−6=D₃(x), it seems that

D_m(x) =F_m(x) (22)

for all m≥1. This can be proved similarly to the previous proofs. By (21), a formula for A136672 follows. By (22), (20) and (3),

specD_m=specA_m−1∪{2}=

4sin² kπ 2m

k=1, . . . , m−1

∪{2} form≥2.

7 Conclusions and further questions

The squared chords of G_n, excluding the diameter, are the zeros of a monic polynomial of degree µ with integer coefficients. Including the diameter, the degree ism.

The squared sum of all individual chords is the largest zero of a monic polynomial of degree m with integer coefficients. An equivalent fact is that the squared sum of all different (lengths of) chords but the diameter is halved, is a zero of such a polynomial. The zeros of this polynomial seem to be linear combinations of the chords with all coefficients 0or ±1.

Lemma 1, stating thate²_n1, . . . , e²_nµ are the eigenvalues of a tridiagonal ma- trix with integer entries, follows from certain properties of the Chebychev polynomials. So squared chords have interesting connections with these top- ics. But what about s²_n1, . . . , s²_nm? Are also they the eigenvalues of such a tridiagonal matrix? This question remains open.

The coefficients of the polynomial (x−e²_n1)· · ·(x−e²_nµ) form an OEIS se- quence, and so do also those of certain related polynomials. What about the coefficients of(x−s²_n1)· · ·(x−s²_nm)? Do also they form such a sequence? This question remains open, too.

Appendix: Proofs of (6) and (11)

Proof of (6)

(x−2)P_m(x) −P_m−1(x)

= (x−2) Xm k=0

(−1)^m−k

m+1+k 2k+1

x^k−

m−1X

k=0

(−1)^m−1−k

m+k 2k+1

x^k

−x^m+1+

m−1X

k=0

(−1)^m−k

m+1+k 2k+1

x^k+1−2 Xm k=0

(−1)^m−k

m+1+k 2k+1

x^k

−

m−1X

k=0

(−1)^m−1−k

m+k 2k+1

x^k

=x^m+1+ Xm k=1

(−1)^m+1−k

m+k 2k−1

x^k+2

Xm k=0

(−1)^m+1−k

m+1+k 2k+1

x^k

−

m−1X

k=0

(−1)^m+1−k

m+k 2k+1

x^k

=x^m+1−

2m 2m−1

+2

2m+1 2m+1

x^m +

m−1X

k=1

(−1)^m+1−k

m+k 2k−1

+2

m+1+k 2k+1

−

m+k 2k+1

x^k + (−1)^m+1

2

m+1 1

− m

1

=x^m+1− (2m+2)x^m+

m−1X

k=1

(−1)^m+1−k

m+2+k 2k+1

x^k+ (−1)^m+1(m+2)

=

m+1X

k=0

(−1)^m+1−k

m+1+1+k 2k+1

x^k=P_m+1(x).

Proof of (11)

(x−3)P_m(x) −P_m−1(x)

=· · ·=x^m+1−

2m 2m−1

+3

2m+1 2m+1

x^m +

m−1X

k=1

(−1)^m+1−k

m+k 2k−1

+3

m+1+k 2k+1

−

m+k 2k+1

x^k + (−1)^m+1

3

m+1 1

− m

1

=x^m+1− (2m+3)x^m+

m−1X

k=1

(−1)^m+1−k 2m+3 m−k+1

m+1+k 2k+1

x^k + (−1)^m+1(2m+3)

=x^m+1+ Xm k=0

(−1)^m+1−k2(m+1) +1 m+1−k

m+1+k 2k+1

x^k=Q_m+1(x).

References

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http://mathworld.wolfram.com/Tangent.html

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[10] http://functions.wolfram.com/ElementaryFunctions/Cot/27/01/

0002/

Received: 4 August 2014