Determinants of two kinds of matrices whose elements involve sine functions

Michał Różański

doi:10.1515/math-2019-0096

Article Open Access

Determinants of two kinds of matrices whose elements involve sine functions

Michał Różański

Published/Copyright: November 13, 2019

Published by

Become an author with De Gruyter Brill

Submit Manuscript Author Information Explore this Subject

$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

The presented paper is strictly connected, among others, with the paper On the sum of some alternating series, Comp. Math. Appl. (2011), written by Wituła and Słota. A problem concerning the form of determinants formulated in the cited paper is solved here. Next, the obtained result is adapted to solve some system of linear equations and the description of the sum of alternating series.

Keywords: determinant; sine matrix; alternating series; Fourier series

MSC 2010: 11C20; 15A06; 42A05; 40A05

1 The main result

The presented paper is strictly connected with paper [1], and also [2, 3]. In paper [1] the following two matrices were considered

Δr:=sin⁡2ijπrνr×νr

and

Δr,n:=Δr,nijνr×νr,withΔr,nij:=sin⁡2ijπr,j≠n,π2r(r−2i),j=n,

where

νr:=r−12

for every r ∈ {3, 4, …} and n ∈ {1, 2, …, ν_r}. Also, the formula for determinant Δ_r (see Theorem 1 below) was given, but verified only for prime r < 1051, using Mathematica software. In this paper we prove that formula is valid for all natural r ≥ 3. Moreover, we also give the compact formula for the determinant of Δ_r,n. The formulas are given in the theorem below.

Theorem 1

For all natural r ≥ 3 and n ∈ {1, 2, …, ν_r} we have

detΔr=(−1)νr(νr−1)/2r4νr/2 (1)

and

detΔr,n=πrdetΔrcot⁡nπr. (2)

Remark 1

Let us also notice that in formula (1) the multiplier (–1)^{ν_r(ν_r–1)/2} vanishes if we reverse the order of columns (or rows) of Δ_r.

Remark 2

The hypothesis on the formula of detΔ_r were sent by the authors of [1] to The American - Mathematical Monthly, Section: Problems and Solutions. However, it was not published since it was considered insufficient and neither authors nor editors had an idea how to complete it.

Remark 3

The determinant related to our determinant, although having some other structure, is discussed in papers [4, 5].

Example 1

We give the explicit form of formulas (1) and (2) for r = 7 (ν₇ = 3, whence n ∈ {1, 2, 3}). Note that if we expand the determinants we get some nontrivial trigonometric equalities.

detΔ7=sin⁡2π7sin⁡4π7sin⁡6π7sin⁡4π7sin⁡8π7sin⁡12π7sin⁡6π7sin⁡12π7sin⁡18π7=−778detΔ7,1=5π14sin⁡4π7sin⁡6π73π14sin⁡8π7sin⁡12π7π14sin⁡12π7sin⁡18π7=−78πcot⁡π7detΔ7,2=sin⁡2π75π14sin⁡6π7sin⁡4π73π14sin⁡12π7sin⁡6π7π14sin⁡18π7=−78πcot⁡2π7detΔ7,3=sin⁡2π7sin⁡4π75π14sin⁡4π7sin⁡8π73π14sin⁡6π7sin⁡12π7π14=−78πcot⁡3π7

The paper is organized as follows. In Section 2 we prove four auxiliary lemmas which we use in Section 3 to prove Theorem 1. In section 4 we discuss two applications of the main result.

2 Auxiliary facts

The following lemmas are necessary for proving the main results. We will use the well-known trigonometric identities (see [6, 7, 8, 9])

∏k=1n−1sin⁡kπn=n2n−1, (3)

∏k=1n−1cos⁡kπn=sin⁡nπ22n−1=0,n∈2N,(−1)(n−1)/22n−1,n∈2N+1, (4)

where n ≥ 2 and

∑k=1nsin⁡kθ=12cot⁡θ2−cosn+12θ2sin⁡θ2=sin⁡nθ2sin⁡(n+1)θ2sin⁡θ2, (5)

∑k=1ncos⁡kθ=−12+sinn+12θ2sin⁡θ2=sin⁡nθ2cos⁡(n+1)θ2sin⁡θ2, (6)

where n ≥ 1 and θ ∉ 2πℤ.

Note that identities (3)–(6) were used in [10] to describe the Cartan matrix in complex simple Lie algebra.

Lemma 1

Let

Dm:=sin⁡jxim×m,

where x_i ∈ ℝ for 1 ≤ i ≤ m. Then

detDm=2m(m−1)/2∏k=1msin⁡xk∏1≤i<j≤m(cos⁡xj−cos⁡xi).

Proof

The proof can be found in [2, 3]. It is based on transforming D_m to Vandermonde matrix by means of elementary operations.□

Remark 4

Notice that matrix Δ_r is a special case of D_m for m = ν_r and x_i = 2iπ/r.

Lemma 2

For each natural number n ≥ 3 the equality

∏i=1νncos⁡iπn∏j=1νn∏k=12j−1sin⁡kπn=12νn2n4νn/2

is valid, where νn:=n−12.

Proof

By substituting first j = ν_n – i + 1 and next k = n – l we get

∏j=1νn∏k=12j−1sin⁡kπn=∏i=1νn∏k=12νn−2i+1sin⁡kπn=∏i=1νn∏l=n−2νn+2i−1n−1sin⁡(n−l)πn=∏i=1νn∏l=n−2νn+2i−1n−1sin⁡lπn. (7)

Let us assume first that n is even. Then n – 2ν_n = 2 and from (7) we obtain the equality

∏j=1νn∏k=12j−1sin⁡kπn=∏j=1νn∏k=2i+1n−1sin⁡kπn.

Substituting i=n2−m yields

∏i=1νncos⁡iπn=∏m=1νncos⁡(n2−m)πn=∏m=1νnsin⁡mπn.

Hence, by the successive algebraic operations we have

∏i=1νncos⁡iπn∏j=1νn∏k=12j−1sin⁡kπn2=∏i=1νncos⁡iπn∏i=1νnsin⁡iπn∏j=1νn∏k=12j−1sin⁡kπn∏j=1νn∏k=2j+1n−1sin⁡kπn=12νn∏i=1νnsin⁡2iπn∏j=1νn∏k=12j−1sin⁡kπn∏j=1νn∏k=2j+1n−1sin⁡kπn=12∏k=1n−1sin⁡kπnνn,

and from (3) we finally get

∏i=1νncos⁡iπn∏j=1νn∏k=12j−1sin⁡kπn2=n2nνn=122νn2n4νn.

Now, if we assume that n is odd, then n – 2ν_n = 1 and (7) yields

∏j=1νn∏k=12j−1sin⁡kπn=∏j=1νn∏k=2in−1sin⁡kπn.

Moreover, by substituting i = n – m we obtain

∏i=1νncos⁡iπn=∏m=νn+1n−1cos⁡(n−m)πn=(−1)(n−1)/2∏m=νn+1n−1cos⁡mπn,

whence

∏i=1νncos⁡iπn∏j=1νn∏k=12j−1sin⁡kπn2=(−1)(n−1)/2∏i=1νncos⁡iπn∏i=νn+1n−1cos⁡iπn∏j=1νn∏k=12j−1sin⁡kπn∏j=1νn∏k=2jn−1sin⁡kπn=(−1)(n−1)/2∏i=1n−1cos⁡iπn∏j=1νn∏k=1n−1sin⁡kπn=(−1)(n−1)/2∏i=1n−1cos⁡iπn∏k=1n−1sin⁡kπnνn.

From (3) and (4) we have

∏i=1νncos⁡iπn∏j=1νn∏k=12j−1sin⁡kπn2=12n−1n2n−1νn=122νn2n4νn,

which finally proves Lemma 2.□

Lemma 3

For each natural number n ≥ 3 and each i, j ∈ {1, 2, …, ν_n} we have

∑k=1νnsin⁡2ikπnsin⁡2kjπn=nδij4, (8)

where δ_ij denotes the Kronecker delta.

Proof

From the formula for the product of sines, we get

∑k=1νnsin⁡2ikπnsin⁡2kjπn=12∑k=1νncos⁡2(i−j)kπn−12∑k=1νncos⁡2(i+j)kπn. (9)

Assume first that i ≠ j. We have to consider three cases: when n is odd, when n is even and i + j is odd and when both n and i + j are even. We will investigate the second case, the other ones should be discussed analogically. Namely, let n be even and i + j odd. From (6) and (9) we obtain

∑k=1νnsin⁡2ikπnsin⁡2kjπn=sin⁡(n−2)(i−j)π2ncos⁡(i−j)π22sin⁡(i−j)πn−sin⁡(n−2)(i+j)π2ncos⁡(i+j)π22sin⁡(i+j)πn=0.

Now, assume that i = j. Then equality (9) takes the form

∑k=1νnsin⁡2ikπnsin⁡2kjπn=νn2−12∑k=1νncos⁡4ikπn. (10)

There are only two cases to consider: n is either odd or even. If n is odd, then from (6) and (10) we have

∑k=1νnsin⁡2ikπnsin⁡2kjπn=n−14+14−sin⁡2iπ4sin⁡2iπn=n4,

which finishes the proof.□

Remark 5

Identity (8) in less general form can be found in [9], in the chapter devoted to the finite sums.

Lemma 4

For each natural number n ≥ 3 and j ∈ {1, 2, …, ν_n} it is true that

∑k=1νn(n−2k)sin⁡2jkπn=n2cot⁡jπn.

Proof

Let us notice that, by introducing the auxiliary variable x, we can write the left-hand side of the examined equality in the following way

∑k=1νn(n−2k)sin⁡2ikπn=n∑k=1νnsin⁡2ikπn+niπ∑k=1νncos⁡2ikπxn′|x=1.

From (5) and (6) we get

∑k=1νn(n−2k)sin⁡2ikπn=n2cot⁡iπn−ncos⁡αiπ2sin⁡iπn+niπ−12+sin⁡αiπx2sin⁡iπxn′|x=1, (11)

where α := (2ν_n + 1)/n. Since

sin⁡αiπx2sin⁡iπxn′=αiπcos⁡αiπx2sin⁡iπxn−iπcos⁡iπxnsin⁡αiπx2nsin2⁡iπxn,

we obtain

niπ−12+sin⁡αiπx2sin⁡iπxn′|x=1=αncos⁡αiπ2sin⁡iπn−cos⁡iπnsin⁡αiπ2sin2⁡iπn.

Now, if n is even, then we have

niπ−12+sin⁡αiπx2sin⁡iπxn′|x=1=ncos⁡αiπ2sin⁡iπn−cos⁡αiπ2sin⁡iπn−cos⁡iπnsin⁡αiπ2sin2⁡iπn=ncos⁡αiπ2sin⁡iπn−sinαiπ+iπn2sin2⁡iπn=ncos⁡αiπ2sin⁡iπn−sin⁡iπ2sin2⁡iπn=ncos⁡αiπ2sin⁡iπn.

Similarly, if n is odd, then we get

niπ−12+sin⁡αiπx2sin⁡iπxn′|x=1=ncos⁡αiπ2sin⁡iπn−cos⁡iπnsin⁡iπ2sin2⁡iπn=ncos⁡αiπ2sin⁡iπn,

which, on the grounds of ((11), gives the desired identity.□

3 Proof of main result

Proof of Theorem 1

formula (1). From Lemma 1 we have

detΔr=2νr(νr−1)/2∏k=1νrsin⁡2kπr∏1≤i<j≤νrcos⁡2jπr−cos⁡2iπr=2νr(νr+1)/2∏k=1νrcos⁡kπr∏k=1νrsin⁡kπr∏1≤i<j≤νrcos⁡2jπr−cos⁡2iπr.

From the formula for the difference of cosines we get

∏k=1νrsin⁡kπr∏1≤i<j≤νrcos⁡2jπr−cos⁡2iπr=(−2)νr(νr−1)/2∏k=1νrsin⁡kπr∏1≤i<j≤νrsin⁡(i+j)πr∏1≤i<j≤νrsin⁡(j−i)πr. (12)

Next, changing the indices gives

∏1≤i<j≤νrsin⁡(i+j)πr=∏j=2νr∏i=1j−1sin⁡(i+j)πr=∏j=2νr∏k=j+12j−1sin⁡kπr

and

∏1≤i<j≤νrsin⁡(j−i)πr=∏j=2νr∏i=1j−1sin⁡(j−i)πr=∏j=2νr∏k=1j−1sin⁡kπr.

Hence (12) can be written in the following way

∏k=1νrsin⁡kπr∏1≤i<j≤νrcos⁡2jπr−cos⁡2iπr=(−2)νr(νr−1)/2∏j=1νr∏k=12j−1sin⁡kπr.

It gives us

detΔr=(−1)νr(νr−1)/22νr2∏k=1νrcos⁡kπr∏j=1νr∏k=12j−1sin⁡kπr.

From Lemma 2 we obtain (1).□

Proof of Theorem 1

formula (2). Firstly, we show that

Δr−1=4rΔr.

For this purpose let us notice that

4rΔr⋅Δr=4rsin⁡2ijπrνr×νr⋅sin⁡2ijπrνr×νr=4r∑k=1νrsin⁡2ikπrsin⁡2kjπrνr×νr=δijνr×νr,

where the last equality results from Lemma 3.

Next, we can write the matrix Δr−1 in the form

Δr−1=(−1)i+jMji(Δr)detΔrνr×νr,

where M_ij(A) is the (i, j) minor of a matrix A. By comparing it with the values of elements of the matrix Δr−1 we get

Mij(Δr)=(−1)i+j4rdetΔrsin⁡2ijπr.

In order to compute the determinant of Δ_r,n let us notice that the following equality holds

Min(Δr,n)=Min(Δr)=(−1)i+n4rdetΔrsin⁡2inπr.

Hence, by applying the cofactor expansion along the n-th column we have

detΔr,n=∑i=1νr(−1)i+nΔr,ninMin(Δr,n)=2πr2detΔr∑i=1νr(r−2i)sin⁡2inπr.

At the end, by Lemma 4, the above formula takes the form given in the theorem.□

4 Applications of main result

In this section we present two applications of Theorem 1. The first is algebraic, the second – analytic.

Let us consider the following system of linear equations

xr,1sin⁡2πr+xr,2sin⁡4πr+…+xr,νrsin⁡2νrπr=π2r(r−2)xr,1sin⁡4πr+xr,2sin⁡6πr+…+xr,νrsin⁡4νrπr=π2r(r−4)…xr,1sin⁡2νrπr+xr,2sin⁡4νrπr+…+xr,νrsin⁡2νr2πr=π2r(r−2νr) (13)

where νr:=r−12 and r ≥ 3 is natural. Then from Cramer’s rule and Theorem 1 we get

xr,n=detΔr,ndetΔr=πrdetΔrcot⁡nπrdetΔr=πrcot⁡nπr.
Let us consider the sum of the alternating series of the following form

Sr,n=∑k=0∞1kr+n−1(k+1)r−n,

where r ≥ 2 is natural and n ∈ {1, 2, …, ν_r}.

Let γ ∈ (0, π). Applying the theory of Fourier series one can prove that the expansion of the 2π-periodic function

f(x)=1,π−γ<|x|≤π,12,|x|=π−γ,0,|x|<π−γ,

into the cosine series is given by the formula

f(x)=γπ−2π∑n=1∞sin⁡((π−γ)n)ncos⁡(nx).

In the special case, by substituting x = π – γ we get

π2−γ=∑n=1∞1nsin⁡(2nγ),

whence by taking γ=kπr, we obtain the identity

π2r(r−2k)=∑n=1∞1nsin⁡2knπr. (14)

Let us notice that we have

∑n=1νrSr,nsin⁡2knπr=∑n=1⌊r/2⌋Sr,nsin⁡2knπr,

since 2kr/2πr=0 when r is even. Hence

∑n=1νrSr,nsin⁡2knπr=∑n=1⌊r/2⌋sin⁡2knπr∑j=0∞1jr+n−1(j+1)r−n=∑n=1⌊r/2⌋∑j=0∞sin⁡2k(jr+n)πrjr+n+sin⁡2k((j+1)r−n)πr(j+1)r−n=∑n=1∞1nsin⁡2knπr=(14)π2r(r−2k).

Therefore S_r,n are the solutions to the system ((13), so we have

Sr,n=πrcot⁡nπr, (15)

for natural r ≥ 2 and n ∈ {1, 2, …, ν_r}.

Remark 6

Formula ((15) can be also derived in a purely analytical way, from the residue theory, which will be omitted here.

Remark 7

In [11], series of similar type as above are considered. But the Author focused on increasing the speed of their convergence, and not on the form of sum of these series.

References

[1] Wituła R., Słota D., On the sum of some alternating series, Comput. Math. Appl., 2011, 62(6), 2658–2664, 10.1016/j.camwa.2011.08.008.Search in Google Scholar

[2] Bagby R., Battaile M., The determinant of a matrix of sines: problem 10849, Amer. Math. Monthly, 2002, 109, 81–82, 10.2307/2695784.Search in Google Scholar

[3] Proskuryakov I.V., Problems in Linear Algebra, 1978, Moscow: Mir.Search in Google Scholar

[4] Da Fonseca C.M., Kowalenko V., On a finite sum with powers of cosines, Appl. Anal. Discrete Math., 2013, 7, 354–377, 10.2298/AADM130822018D.Search in Google Scholar

[5] Merca M., Problem 98*, Eur. Math. Soc. Newsl., 2012, 83, 58.Search in Google Scholar

[6] Beebee J., Some trigonometric identities related to exact covers, Proc. Amer. Math. Soc., 1991, 112, 329–338, 10.2307/2048723.Search in Google Scholar

[7] Knapp M.P., Sines and cosines of angles in arithmetic progression, Math. Mag., 2009, 82, 371–372, 10.4169/002557009X478436.Search in Google Scholar

[8] Olver F.W.J., Lozier D.W., Boisvert R.F., Clark C.W., NIST Handbook Mathematical Functions, 2010, Cambridge: Cambridge University Press.Search in Google Scholar

[9] Prudnikov A.P., Brychkov Y.A., Marichev O.I., Integrals and Series in Elementary Functions, vol. 1, 1986, New York: Gordon and Breach Science Publishers.Search in Google Scholar

[10] Damianou P.A., A beautiful sine formula, Amer. Math. Monthly, 2014, 121, 120–135, 10.4169/amer.math.monthly.121.02.120.Search in Google Scholar

[11] Amore P., Convergence acceleration of series through a variational approach, J. Math. Anal. Appl., 2006, 323, 63–77, 10.1016/j.jmaa.2005.09.091.Search in Google Scholar

Received: 2019-01-14

Accepted: 2019-06-19

Published Online: 2019-11-13

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Articles in the same Issue

https://doi.org/10.1515/math-2019-0096

Keywords for this article

determinant; sine matrix; alternating series; Fourier series

Creative Commons

BY 4.0