����������� ������ 50 (1998), 105–110

UDK 517.58

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AN APPLICATION FOR THE CHEBYSHEV POLYNOMIALS

-Dur - de Cvijović and Jacek Klinowski

Abstract. Two sequences of polynomials for which all zeros, regardless of degree n, can be given by the following “simple formulae”

Γ_{n,m}(ξ) = cot ( \frac{(ξ + m)π}{n} ) and Δ_{n,m}(ξ) = tan ( \frac{(ξ + m)π}{n} ) (0 < ξ < 1)

(n = 1, 2,…; m = 0, 1,…, n − 1 and m ≠ (n − 1)/2 when ξ = 1/2 and n is odd in the case of Δ_{n,m}) are obtained from the linear combination of the Chebyshev polynomials of the first and second kind.

1. Introduction

In our recent work [1] on the Apostol formula for the Riemann zeta function [2] the problem has arisen of the closed-form evaluation of the following finite cotangent sum

S_{n}(q; ξ) = ∑_{p=0}^{q−1} cot^{n} (\frac{(ξ + p)π}{n} )  (0 < ξ < 1)  (1)

for any positive integer n. It appears that this sum is known only when n = 1, 2, 4 ([3], Section 4.4.7 and [4], Sections 29.1 and 30.1). In this note we shall establish a simple way of its summation. Namely, S_{n}(q; ξ) can be evaluated for any n by using the Newton identities (the Newton power sum formulae), since there exist the sequence of polynomials of which the numbers

cot (\frac{(ξ + p)π}{n} ) p = 0, 1,…, q − 1

are zeros. The polynomials in question are constructed by making use of the Chebyshev polynomials of the first and the second kind. A similar procedure leads to the analogous polynomials of which tan[(ξ + p)π/q] are zeros.

AMS Subject Classification: Primary 33 C 45; Secondary 33 C 90

Keywords and phrases: Chebyshev polynomials, polynomials, zeros of polynomials

2. Statement of results

Throughout the text T_{n}(x) and U_{n}(x) are, respectively, the Chebyshev polynomials of the first and second kind defined on [−1, 1] by [5, p. 776]

T_{n}(x) = cos(n arccos x) = \frac{n}{2} ∑_{k=0}^{[n/2]}(−1)^{k} \frac{(n − k − 1)!}{k! (n − 2k)!}(2x)^{n−2k}  (2a)

and

U_{n}(x) = \frac{sin((n + 1) arccos x)}{\sqrt{1 − x^{2}}} = ∑_{k=0}^{[n/2]}(−1)^{k} \frac{(n − k)!}{k! (n − 2k)!}(2x)^{n−2k}.  (2b)

Further, [x] denotes the largest integer not exceeding x while the principal values of arccot θ and of arctan θ are defined for any real θ by 0 < arccot θ < π and by −π/2 < arctan θ < π/2. Our results are as follows.

Theorem. Assume that n is a positive integer and that ξ is a real, 0 < ξ < 1. Let T_{n}(x) and U_{n}(x) be the Chebyshev polynomials defined in (2a,b). Consider

C_{n}(x; ξ) = (1 + x^{2})^{n/2}[T_{n}(\frac{x}{\sqrt{1 + x^{2}}} ) − \frac{cot(πξ)}{\sqrt{1 + x^{2}}} U_{n−1}(\frac{x}{\sqrt{1 + x^{2}}} )]  (3)

and

K_{n}(x; ξ) = (1 + x^{2})^{n/2}[T_{n}(\frac{1}{\sqrt{1 + x^{2}}} ) − x\frac{cot(πξ)}{\sqrt{1 + x^{2}}} U_{n−1}(\frac{1}{\sqrt{1 + x^{2}}} )]. (4)

Then we have:

(1) C_{n}(x; ξ) and K_{n}(x; ξ) are polynomials (with a parameter ξ) in a variable x, defined for any real x, such that:

(a) the polynomials C_{n}(x; ξ) are monic of degree exactly n, given by

C_{n}(x; ξ) = ∑_{k=0}^{n} \binom{n}{k}c_{k}(ξ)x^{n−k};  (5a)

(b) the polynomials K_{n}(x; ξ) are of degree n − 1 when ξ = 1/2 and n is odd and of degree n otherwise, and they are given by

K_{n}(x; ξ) = ∑_{k=0}^{n} \binom{n}{k}c_{k}(ξ)x^{k}.  (5b)

Here, in (a) and (b), c_{k}(ξ) has the same meaning:

c_{2r}(ξ) = (−1)^{r}, c_{2r+1}(ξ) = (−1)^{r+1} cot(πξ)  (0 < ξ < 1, r ∈ N_{0}). (5c)

(2) For any fixed n, all zeros of C_{n}(x; ξ) and K_{n}(x, ξ) are respectively given by:

(a) Γ_{n,m}(ξ) = cot (\frac{(ξ + m)π}{n} ), m = 0, 1,…, n − 1;

(b) Δ_{n,m}(ξ) = tan (\frac{(ξ + m)π}{n} ), m = 0, 1,…, n − 1, m ≠ (n − 1)/2 when

ξ = 1/2 and n is odd.

3. Proof of the Theorem

(1a) and (2a). In order to prove these two parts of the theorem, we derive the following formulae for any real x

C_{n}(x; ξ) = (1 + x^{2})^{n/2}[cos(n arccot x) − cot(πξ) sin(n arccot x)]  (6a)
= csc^{n}(arccot x)[cos(n arccot x) − cot(πξ) sin(n arccot x)]  (6b)
= ∑_{k=0}^{[n/2]}(−1)^{k}\binom{n}{2k}x^{n−2k} − cot(πξ) ∑_{k=0}^{[(n−1)/2]}(−1)^{k}\binom{n}{2k + 1}x^{n−2k−1}(6c)

on the assumption that arccot x takes its principal value.

First, in view of (3), the formula in (6a) readily follows since it is easy to show, starting from the definitions of the Chebyshev polynomials in (2a,b) and using the relationship cos θ = cot θ(1 + cot^{2} θ)^{−1/2} that the following holds

cos(n arccot x) = T_{n}(\frac{x}{\sqrt{1 + x^{2}}} ), sin(n arccot x) = \frac{1}{\sqrt{1 + x^{2}}} U_{n−1}(\frac{x}{\sqrt{1 + x^{2}}} )

for any real x. Moreover, the formula in (6b) follows at once from (6a), considering that for any real θ and integer r we have csc θ = (1 + cot^{2} θ)^{1/2} when θ ≠ rπ (r ∈ Z).

Next, from (2a,b) we also have

(1 + x^{2})^{\frac{n}{2}} T_{n}(\frac{x}{\sqrt{1 + x^{2}}} ) = \frac{n}{2} ∑_{k=0}^{[n/2]}(−1)^{k} \frac{(n − k − 1)!}{k! (n − 2k)!}(2x)^{n−2k}(1 + x^{2})^{k}, (7a)

(1 + x^{2})^{\frac{n−1}{2}} U_{n−1}(\frac{x}{\sqrt{1 + x^{2}}} ) = ∑_{k=0}^{[(n−1)/2]}(−1)^{k} \frac{(n − k − 1)!}{k! (n − 1 − 2k)!}(2x)^{n−2k−1}(1 + x^{2})^{k}, (7b)

and it is thus clear that C_{n}(x; ξ) (see (3)) are polynomials in x of degree n which can be expressed in terms of the finite sums in (7a,b).

However, our polynomials can be represented in much simpler way as it is suggested in (6c). Indeed, note that the real and imaginary parts of z = (x + i)^{n} are

Re z = (1 + x^{2})^{n/2} cos(n arccot x) and Im z = (1 + x^{2})^{n/2} sin(n arccot x).

On the other hand, upon setting z^{∗} = (x − i)^{n} and making use of the binomial theorem, we arrive at

Re z = \frac{z + z^{∗}}{2} = ∑_{k=0}^{[n/2]}(−1)^{k}\binom{n}{2k}x^{n−2k},

Im z = \frac{z − z^{∗}}{2i} = ∑_{k=0}^{[(n−1)/2]}(−1)^{k}\binom{n}{2k + 1}x^{n−2k−1}

so that the formula in (6c) and the proposed formula in (5a), as its immediate consequence, follow without difficulty. It is obvious from (5a) that C_{n}(x; ξ) are monic polynomials of degree exactly n.

Part (2a) is easily established by putting x = Γ_{n,m}(ξ) in (6b) and showing that C_{n}(Γ_{m,n}(ξ); ξ) = 0 for any n and m = 0, 1,…, n − 1.

Observe first that in general if k is an integer, then arccot(cot x) = x − kπ, kπ < x < (k + 1)π. Thus arccot(Γ_{n,m}(ξ)) = (ξ + m)π/n, since 0 < (ξ + m)π/n < π (m = 0, 1,…, n − 1) provided that 0 < ξ < 1. Hence, on setting x = Γ_{n,m}(ξ) in (6b), we have after simplification

C_{n}(Γ_{n,m}(ξ); ξ) = (−1)^{m} csc^{n} (\frac{(ξ + m)π}{n} ) [cos(πξ) − cot(πξ) sin(πξ)] = 0.

Since C_{n}(x; ξ) is of degree n, it has exactly n zeros, and so the numbers Γ_{n,m}(ξ) (m = 0, 1,…, n − 1) are all zeros of C_{n}(x; ξ). In view of the properties of the cotangent functions, these zeros are real and simple.

(1b), (2b). First, it can be shown working along the same lines as in the proof of Part (1a) that for any real x we have

K_{n}(x; ξ) = (1 + x^{2})^{n/2}[cos(n arctan x) − cot(πξ) sin(n arctan x)]  (8a)
= sec^{n}(arctan x)[cos(n arctan x) − cot(πξ) sin(n arctan x)]  (8b)
= ∑_{k=0}^{[n/2]}(−1)^{k}\binom{n}{2k}x^{2k} − cot(πξ) ∑_{k=0}^{[(n−1)/2]}(−1)^{k}\binom{n}{2k + 1}x^{2k+1} (8c)

if arctan x takes its principal value. To do this, we need the relationship cos θ = (1 + cot^{2} θ)^{−1/2} and to note that the use of z = (xi + 1)^{n} leads to the desired result in (8c).

Next, on rewriting (8c) the proposed formula for K_{n}(x; ξ) in (5b) is obtained and it is trivial to verify that the polynomials K_{n}(x; ξ) are not monic; they are of degree exactly n − 1 when ξ = 1/2 and n is odd and of degree n otherwise. We also have

K_{n}(x; ξ) = x^{n}C_{n}(1/x; ξ).  (9)

To prove Part (2b) recall that, in general, if z_{0} is a non-vanishing zero of the polynomial P_{n}(z), then 1/z_{0} is a zero of the polynomial z^{n}P_{n}(1/z) (see for instance [6], p. 180). Hence, because of (9) the zeros of K_{n}(x; ξ) are the reciprocals of the zeros of C_{n}(x; ξ), i.e. the reciprocals of Γ_{n,m}(ξ). Clearly, the case when Γ_{n,m}(ξ) = 0 must be excluded.

This completes the proof of the Theorem.

4. Concluding remarks

An interesting application of the Chebyshev polynomials, which we have failed to find in the literature (for instance, in the standard text on these “classical orthogonal polynomilas” [7]), is described in this note: two sequences of polynomials, C_{n}(x; ξ) and K_{n}(x; ξ), are obtained, for which all zeros, regardless of the degree, can be given by a single “simple formula”. This porcedure, however, fails to give similar polynomials when applied to the sine, cosine and their reciprocals.

Recall that similarly all zeros τ_{m} and υ_{m} of the Chebyshev polynomilas T_{n}(x) and U_{n}(x) are also given by a “simple formula” (a single analytical expression) [5, Entry 22.16.4] as follows:

τ_{m} = cos \frac{(2m − 1)π}{2n} , υ_{m} = cos \frac{mπ}{n + 1} (m = 1, 2,…, n).

Several examples of C_{n}(x; ξ) and K_{n}(x; ξ), 0 < ξ < 1, are:
C_{1}(x; ξ) = x − cot(πξ),  K_{1}(x; ξ) = − cot(πξ)x + 1,
C_{2}(x; ξ) = x^{2} − 2 cot(πξ)x − 1,  K_{2}(x; ξ) = −x^{2} − 2 cot(πξ)x + 1,
C_{3}(x; ξ) = x^{3} − 3 cot(πξ)x^{2} − 3x + cot(πξ),
K_{3}(x; ξ) = cot(πξ)x^{3} − 3x^{2} − 3 cot(πξ)x + 1,
C_{4}(x; ξ) = x^{4} − 4 cot(πξ)x^{3} − 6x^{2} + 4 cot(πξ)x + 1,
K_{4}(x; ξ) = x^{4} + 4 cot(πξ)x^{3} − 6x^{2} − 4 cot(πξ)x + 1.

A straightforward consequence of the results is that the cotangent sum S_{n}(q; ξ) in (1) is in fact the power sum of the zeros of C_{q}(x; ξ) and this enables the closedform evaluation of S_{n}(q; ξ) for any n by using the Newton identities [8, p. 179], for instance. This gives for example

S_{3}(q; ξ) = q^{3}(cot^{3}(πξ) + cot(πξ)) − q cot(πξ),
S_{5}(q; ξ) = q^{5}(cot^{5}(πξ) + \frac{5}{3} cot^{3}(πξ) + \frac{2}{3} cot(πξ))
− q^{3}(\frac{5}{3} cot^{3}(πξ) + \frac{5}{3} cot^{3}(πξ)) + q cot(πξ).

Finally, note that the sums deduced here

∑_{k=0}^{[n/2]}(−1)^{k}\binom{n}{2k}x^{n−2k} = (1 + x^{2})^{n/2}T_{n}(\frac{x}{\sqrt{1 + x^{2}}} )

= (1 + x^{2})^{n/2} cos(n arccot x) = csc^{n}(arccot x) cos(n arccot x)

and

∑_{k=0}^{[(n−1)/2]}(−1)^{k}\binom{n}{2k + 1}x^{n−2k−1} = (1 + x^{2})^{(n−1)/2}U_{n−1}(\frac{x}{\sqrt{1 + x^{2}}} )

= (1 + x^{2})^{n/2} sin(n arccot x) = csc^{n}(arccot x) sin(n arccot x)

(see (3) in conjunction with (6)) are not listed in [3, section 10.28], the most extensive compilation of sums, and in [4, Section 4.2.3]. It is interesting that Ramanujan [9, Entry 21, p. 32] showed the polynomial nature of (1 + x^{2})^{n/2} sin(n arctan x) without deriving the summation formula

∑_{k=0}^{[(n−1)/2]}(−1)^{k}\binom{n}{2k + 1}x^{2k+1} = x(1 + x^{2})^{(n−1)/2}U_{n−1}(\frac{1}{\sqrt{1 + x^{2}}} )

(see (4) in conjunction with (8a) and (8c)).

REFERENCES

[1] -D. Cvijović and J. Klinowski, Finite cotangent sums and the Riemann zeta function, Proc. Royal Society, Section A (in press).

[2] T. M. Apostol, Another elementary proof of formula for ζ(2n), Am. Math. Monthly 80 (1973), 425–431.

[3] E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, N.J., 1975.

[4] A. P. Prudnikov, Yu. A. Brychkov and O. I. Marichev, Integrals and Series, Vol. 1, Gordon and Breach Science Publ., New York, 1986.

[5] M. Abramowitz and I. Stegun (eds), Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, Dover Publ., New York, 1972.

[6] E. J. Barbeau, Polynomials, Springer-Verlag, New York, 1989.

[7] T. J. Rivlin, Chebyshev Polynomials: From Approximation Theory to Algebra and Number Theory, 2nd ed., John Willey, Chichester, 1990.

[8] P. M. Cohen, Algebra, Vol. 1, John Wiley, Chichester, 1982.

[9] B. C. Brendt, Ramanujan’s Notebooks, Part IV, Springer, New York, 1994.

(received 13.01.1998.)

Department of Chemistry, University of Cambridge, Lensfield Road, Cambridge CB2 1EW, U.K. E-mail: dcl133@cam.ac.uk, jk18@cam.ac.uk