Journal of Inequalities and Applications

, 2012:167

Some properties of Chebyshev polynomials

Authors

    • Department of MathematicsSookmyung Women’s University
Open AccessResearch

DOI: 10.1186/1029-242X-2012-167

Cite this article as:
Kim, S. J Inequal Appl (2012) 2012: 167. doi:10.1186/1029-242X-2012-167
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Abstract

In this paper we obtain some new bounds for Chebyshev polynomials and their analogues. They lead to the results about zero distributions of certain sums of Chebyshev polynomials and their analogues. Also we get an interesting property about the integrals of certain sums of Chebyshev polynomials.

Keywords

Chebyshev polynomialsboundssumszeros

1 Introduction

Let T n ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq1_HTML.gif and U n ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq2_HTML.gif be the Chebyshev polynomials of the first kind and of the second kind, respectively. These polynomials satisfy the recurrence relations
T 0 ( x ) = 1 , T 1 ( x ) = x , T n + 1 ( x ) = 2 x T n ( x ) T n 1 ( x ) ( n 1 ) U 0 ( x ) = 1 , U 1 ( x ) = 2 x , U n + 1 ( x ) = 2 x U n ( x ) U n 1 ( x ) ( n 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ1_HTML.gif
(1)

Chebyshev polynomials are of great importance in many areas of mathematics, particularly approximation theory. Many papers and books [3, 4] have been written about these polynomials. Chebyshev polynomials defined on [ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq3_HTML.gif are well understood, but the polynomials of complex arguments are less so. Reported here are several bounds for Chebyshev polynomials defined on https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq4_HTML.gif including zero distributions of certain sums of Chebyshev polynomials. Moreover, we will introduce certain analogues of Chebyshev polynomials and study their properties. Also we get an interesting property about the integrals of certain sums of Chebyshev polynomials.

Other generalized Chebyshev polynomials (known as Shabat polynomials) have been introduced in [5] and they are studied in the theory of graphs on surfaces and curves over number fields. For a survey in this area, see [6].

For n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq5_HTML.gif and 2 ϵ > 1 + ϵ 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq6_HTML.gif, i.e., 0 ϵ < 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq7_HTML.gif, we let
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equa_HTML.gif
In detail,
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equb_HTML.gif
and it is easy to show that for an odd integer n,
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equc_HTML.gif

Since T n , 0 ( z ) = T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq8_HTML.gif, we may apply results about T n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq9_HTML.gif to those about T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif. But U n , 1 ( z ) = U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq11_HTML.gif and ϵ was a nonnegative real number less than 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq12_HTML.gif, and so properties of U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif will be investigated separately from those of U n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq14_HTML.gif.

2 New results

In this section we list some new results related to the bounds of Chebyshev polynomials T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif, U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif and their analogues T n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq9_HTML.gif, U n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq14_HTML.gif defined on https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq4_HTML.gif including zero distributions of certain sums of Chebyshev polynomials and their analogues. And we will get an interesting property about the integrals of certain sums of Chebyshev polynomials. We first begin with properties about bounds of T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif, U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif, T n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq9_HTML.gif and U n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq14_HTML.gif. We may compute that for n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq15_HTML.gif,
T n , ϵ ( z ) = { ( 1 + ϵ ) T n ( z ) if n is odd , ( 1 + ϵ ) T n ( z ) ϵ if n is even and 4 n , ( 1 + ϵ ) T n ( z ) + ϵ if n is even and 4 n , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equd_HTML.gif
and
U n , ϵ ( z ) = { ( 2 ϵ ) U n ( z ) if n is odd , ( 2 ϵ ) U n ( z ) 1 + ϵ if n is even and 4 n , ( 2 ϵ ) U n ( z ) + 1 ϵ if n is even and 4 n . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Eque_HTML.gif

Proposition 1Suppose thatzis a complex number satisfying | z | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq16_HTML.gif. Then for n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq5_HTML.gif,

| U n ( z ) | | U n 1 ( z ) | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equf_HTML.gif
and
| U n ( z ) | | T n ( z ) | 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ2_HTML.gif
(2)
Also
| U n , ϵ ( z ) | | U n 1 , ϵ ( z ) | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ3_HTML.gif
(3)
and
| U n , ϵ ( z ) | | T n , ϵ ( z ) | > { 1 + ϵ if n is odd , ϵ if n is even . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ4_HTML.gif
(4)

Proposition 1 will be used in the proofs of Theorems 4 and 6.

Remarks For a complex number z with | z | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq16_HTML.gif, we may follow the procedure of the proof of (2) to obtain
| T n ( z ) | | T n 1 ( z ) | 0 ( n 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equg_HTML.gif
and
| U n ( z ) | n + 1 ( n 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equh_HTML.gif
that is best possible since | U n ( 1 ) | = n + 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq17_HTML.gif. There seem to be larger lower bounds than 1 for | U n ( z ) | | T n ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq18_HTML.gif in (2). First, we observe that
min | z | 1 ( | U n ( z ) | | T n ( z ) | ) n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equi_HTML.gif
because | U n ( 1 ) | | T n ( 1 ) | = ( n + 1 ) 1 = n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq19_HTML.gif. Deciding in some general situations exactly where the minimum occurs seems to be extremely difficult. For example, machine calculation suggests that for n = 4 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq20_HTML.gif, | U 4 ( z ) | | T 4 ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq21_HTML.gif takes its minimum 3.91735… in | z | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq22_HTML.gif at four modulus 1 roots ± 0.9953 ± i 0.0964 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq23_HTML.gif of the polynomial
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equj_HTML.gif
But we may conjecture that, by numerical computations, the value
min | z | 1 ( | U n ( z ) | | T n ( z ) | ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equk_HTML.gif
occurs in | z | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq24_HTML.gif and lies between n 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq25_HTML.gif and n, where n 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq25_HTML.gif can be replaced by something larger. We now ask naturally what the minimum is for | z | = t > 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq26_HTML.gif. If one simply looks at the case z = t https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq27_HTML.gif, it seems that | U n ( z ) | | T n ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq18_HTML.gif is close to its minimum at z = t https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq27_HTML.gif. But
| U n ( t ) | | T n ( t ) | = U n ( t ) T n ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equl_HTML.gif
is the coefficient of x n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq28_HTML.gif in the power series expansion of t / ( 1 2 t x + t 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq29_HTML.gif. In fact,
t x 1 2 t x + x 2 = 1 1 2 t x + x 2 1 t x 1 2 t x + x 2 = n = 0 U n ( t ) x n n = 0 T n ( t ) x n = n = 0 ( U n ( t ) T n ( t ) ) x n . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equm_HTML.gif

For | z | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq16_HTML.gif, | U n ( z ) | | T n ( z ) | + 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq30_HTML.gif by (2). In the following proposition, we obtain an upper bound for arbitrary z = cos θ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq31_HTML.gif, https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq32_HTML.gif.

Proposition 2Let z = cos θ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq31_HTML.gif, where θ = α + i β https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq33_HTML.gifand β 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq34_HTML.gif. Then, for n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq5_HTML.gif,
| U n ( z ) | ( 1 + coth β coth n β ) | T n ( z ) | . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ5_HTML.gif
(5)
Remarks With the same notations with Proposition 2, it follows from
| z | 2 = | cos θ | 2 = 1 2 ( cos 2 α + cosh 2 β ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equn_HTML.gif
that, for | z | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq35_HTML.gif large, | β | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq36_HTML.gif is large. But coth β coth n β > 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq37_HTML.gif and
lim β ± coth β coth n β = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equo_HTML.gif
These imply that the upper bound
( 1 + coth β coth n β ) | T n ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equp_HTML.gif

is greater than 2 | T n ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq38_HTML.gif, but for | z | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq35_HTML.gif large, it is close to 2 | T n ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq38_HTML.gif. Also by machine computations (e.g., n = 4 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq39_HTML.gif and z = 100 + i 100 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq40_HTML.gif), we may check that the inequality (5) is sharp.

It is natural to ask about the bounds on the unit circle.

Proposition 3For | z | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq41_HTML.gif, we have
1 n + 1 | U n ( z ) | | T n ( z ) | < | U n ( z ) | . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ6_HTML.gif
(6)
Remarks The right inequality in (6) will be shown in Section 3 by using (2). So obtaining a better lower bound than 1 in (2) can improve this inequality. The left inequality in (6) is best possible in the sense that
1 n + 1 | U n ( ± 1 ) | = | T n ( ± 1 ) | = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equq_HTML.gif
For | z | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq41_HTML.gif, it is easy to see
| T n , ϵ ( z ) | < | U n , ϵ ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equr_HTML.gif
by (4), and it seems to be true that
1 n + 1 | U n , ϵ ( z ) | | T n , ϵ ( z ) | . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ7_HTML.gif
(7)

The proof of (6) will be given in Section 3 by using a well-known identity U n ( z ) = T n ( z ) + z U n 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq42_HTML.gif. But U n , ϵ ( z ) = T n , ϵ ( z ) + z U n 1 , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq43_HTML.gif does not hold. So we cannot use this to prove (7) if it is true.

All zeros of the polynomial T n ( z ) + U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq44_HTML.gif lie in ( 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq45_HTML.gif. More generally the convex combination of T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif and U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif has all its zeros in ( 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq45_HTML.gif. This will be proved in Proposition 5 below. So one might ask: where are the zeros of polynomials like T n ( z ) + z k U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq46_HTML.gif or U n ( z ) + z k T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq47_HTML.gif around | z | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq48_HTML.gif? The next theorem answers this for T n ( z ) + z k U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq46_HTML.gif.

Theorem 4Let P 1 ( z ) : = T n ( z ) + z k U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq49_HTML.giffor positive integersnandk. Then P 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq50_HTML.gifhas all its zeros in | z | < 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq51_HTML.gif. Furthermore, forkeven, P 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq50_HTML.gifhas at leastnreal zeros, and forkodd, P 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq50_HTML.gifhas at least n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq52_HTML.gifreal zeros.

Remarks Let P 2 ( z ) : = U n ( z ) + z k T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq53_HTML.gif for positive integers n and k. We can use the same method as in the proof of Theorem 4 to show that P 2 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq54_HTML.gif has at least n real zeros in ( 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq55_HTML.gif. Furthermore, for k even, there is no real zero outside [ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq56_HTML.gif, and for k odd, there is one more real zero on z < 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq57_HTML.gif.

Proposition 5The polynomial
( 1 λ ) T n ( z ) + λ U n ( z ) ( 0 λ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equs_HTML.gif

has all zeros in ( 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq55_HTML.gif.

Using analogues of T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif and U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif, we consider analogues of P 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq50_HTML.gif and investigate their zero distributions. Define
P 1 , ϵ ( z ) : = T n , ϵ ( z ) + z k U n , ϵ ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equt_HTML.gif

Theorem 6 P 1 , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq58_HTML.gifhas all its zeros in | z | < 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq59_HTML.gif.

Finally, we get an interesting property about the integrals of sums of Chebyshev polynomials. Observe that
0 2 π T n ( e i θ ) d θ = 0 2 π U n ( e i θ ) d θ = T n ( 0 ) 2 π = { ( 1 ) n / 2 if n is even , 0 if n is odd https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equu_HTML.gif
and
0 2 π | T n ( e i θ ) | 2 d θ ( or 0 2 π | U n ( e i θ ) | 2 d θ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equv_HTML.gif
equals
2 π sum of the squares of all coefficients of T n ( z ) ( or U n ( z ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equw_HTML.gif
For example, from T 4 ( z ) = 8 z 4 8 z 2 + 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq60_HTML.gif and U 4 ( z ) = 16 z 6 80 z 4 + 24 z 2 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq61_HTML.gif we can calculate
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equx_HTML.gif

and we see that these two integrals are different. But for P 1 ( z ) = T n ( z ) + z k U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq62_HTML.gif and P 2 ( z ) = U n ( z ) + z k T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq63_HTML.gif, the integrals have the same value.

Proposition 7For z = e i θ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq64_HTML.gif,
0 2 π | P 1 ( e i θ ) | 2 d θ = 0 2 π | P 2 ( e i θ ) | 2 d θ . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equy_HTML.gif
Remark It seems to be true that for k large, z = e i θ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq64_HTML.gif,
0 2 π | P 1 ( e i θ ) | d θ 0 2 π | P 2 ( e i θ ) | d θ , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equz_HTML.gif
but
lim k 0 2 π | P 1 ( e i θ ) | d θ = lim k 0 2 π | P 2 ( e i θ ) | d θ . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equaa_HTML.gif

These remain open problems.

3 Proofs

Proof of Proposition 1 Suppose that z is a complex number satisfying | z | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq16_HTML.gif. Using (1), for n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq5_HTML.gif, we have
| U n + 1 ( z ) | 2 | U n ( z ) | | U n 1 ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equab_HTML.gif
and
| U n + 1 ( z ) | | U n ( z ) | | U n ( z ) | | U n 1 ( z ) | . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equac_HTML.gif
Then by recurrence,
| U n ( z ) | | U n 1 ( z ) | | U 1 ( z ) | | U 0 ( z ) | = | 2 z | 1 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ8_HTML.gif
(8)
By (8) and the identity
T n ( z ) = 1 2 ( U n ( z ) U n 2 ( z ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equad_HTML.gif
we have
| T n ( z ) | 1 2 | U n ( z ) | + 1 2 | U n 2 ( z ) | 1 2 | U n ( z ) | + 1 2 ( | U n ( z ) | 2 ) = | U n ( z ) | 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ9_HTML.gif
(9)
Next we prove the results about U n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq65_HTML.gif and T n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq66_HTML.gif. For n odd and 4 n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq67_HTML.gif, it follows from the definition of U n , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq65_HTML.gif and (8) that
| U n , ϵ ( z ) | | U n 1 , ϵ ( z ) | = | ( 2 ϵ ) U n ( z ) | | ( 2 ϵ ) U n 1 ( z ) ( 1 ϵ ) | | ( 2 ϵ ) U n ( z ) | | ( 2 ϵ ) U n 1 ( z ) | ( 1 ϵ ) = ( 2 ϵ ) ( | U n ( z ) | | U n 1 ( z ) | ) ( 1 ϵ ) 2 ϵ 1 + ϵ = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equae_HTML.gif
This inequality
| U n , ϵ ( z ) | | U n 1 , ϵ ( z ) | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equaf_HTML.gif
for other three cases (i.e., n odd and 4 n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq68_HTML.gif, n even and 4 n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq67_HTML.gif, n even and 4 n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq68_HTML.gif) can be proved in the same way. Finally, for n odd, by 2 ϵ > 1 + ϵ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq69_HTML.gif and (9), we have
| U n , ϵ ( z ) | | T n , ϵ ( z ) | = ( 2 ϵ ) | U n ( z ) | ( 1 + ϵ ) | T n ( z ) | > ( 1 + ϵ ) ( | U n ( z ) | | T n ( z ) | ) 1 + ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equag_HTML.gif
In the same way, we can check that for n even,
| U n , ϵ ( z ) | | T n , ϵ ( z ) | > ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equah_HTML.gif

 □

Proof of Proposition 2 Using the identity U n ( z ) = T n ( z ) + z U n 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq42_HTML.gif, for z = cos θ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq31_HTML.gif we have
| U n ( z ) T n ( z ) | = | 1 + z U n 1 ( z ) T n ( z ) | = | 1 + cos θ tan n θ sin θ | = | 1 + cot θ tan n θ | 1 + | cot θ tan n θ | . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equai_HTML.gif
If we set θ = α + i β https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq70_HTML.gif, https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq71_HTML.gif, then
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equaj_HTML.gif
Since z ¯ = cos θ ¯ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq72_HTML.gif, it suffices to consider the case β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq73_HTML.gif. The above implies
| tan θ | 2 = sin 2 α cos 2 α + sinh 2 β cosh 2 β ( cos 2 α + sinh 2 β ) 2 sinh 2 β cosh 2 β sinh 4 β . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equak_HTML.gif
So
| tan n θ | coth n β . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ10_HTML.gif
(10)
Also
| sin θ | 2 = sin 2 α cosh 2 β + cos 2 α sinh 2 β ( sin 2 α + cos 2 α ) sinh 2 β https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equal_HTML.gif
and
| cos θ | 2 = cos 2 α cosh 2 β + sin 2 α sinh 2 β ( sin 2 α + cos 2 α ) cosh 2 β , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equam_HTML.gif
and so
| cot θ | coth β . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equ11_HTML.gif
(11)
Now we see with (10) and (11) that
| cot θ tan n θ | coth β coth n β . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equan_HTML.gif

 □

Proof of Proposition 3 Suppose that z is a complex number satisfying | z | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq24_HTML.gif. First, it follows from (2) that
| T n ( z ) U n ( z ) | < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equao_HTML.gif
Using the identity U n ( z ) = T n ( z ) + z U n 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq42_HTML.gif, we have
| T n ( z ) U n ( z ) | = | 1 z U n 1 ( z ) U n ( z ) | 1 | U n 1 ( z ) U n ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equap_HTML.gif
and so it is enough to show that
| U n 1 ( z ) U n ( z ) | 1 1 n + 1 = n n + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equaq_HTML.gif
We use induction on n. For n = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq74_HTML.gif, | U 0 ( z ) U 1 ( z ) | = | 1 2 z | = 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq75_HTML.gif. Assume the result holds for k. Then
| U k + 1 ( z ) U k ( z ) | = | 2 z U k 1 ( z ) U k ( z ) | 2 | U k 1 ( z ) U k ( z ) | 2 k k + 1 = k + 2 k + 1 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equar_HTML.gif
and
| U k ( z ) U k + 1 ( z ) | k + 1 k + 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equas_HTML.gif

 □

Proof of Theorem 4 All zeros of T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif and U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif are real and lie in ( 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq55_HTML.gif and for 1 k n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq76_HTML.gif,
T n ( cos ( 2 k 1 ) π 2 n ) = 0 , U n ( cos k π n + 1 ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equat_HTML.gif
For convenience, by removing ‘cos’ and the constant π, cos ( 2 k 1 ) π 2 n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq77_HTML.gif and cos k π n + 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq78_HTML.gif can be identified with the ascending chain of rational numbers ( 2 k 1 ) 2 n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq79_HTML.gif and k n + 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq80_HTML.gif, respectively. We may calculate that for n odd
{ 2 k 1 2 n < k n + 1 < 2 k + 1 2 n , 1 k < n + 1 2 , 2 k + 1 2 n = k + 1 n + 1 , k = n 1 2 , 2 k 1 2 n < k + 1 n + 1 < 2 k + 1 2 n , n 1 2 < k < n 1 , 2 k 1 2 n < k + 1 n + 1 , k = n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equau_HTML.gif
and for n even
{ 2 k 1 2 n < k n + 1 < 2 k + 1 2 n , 1 k < n 2 , 2 k 1 2 n < k n + 1 < k + 1 n + 1 < 2 k + 1 2 n , k = n 2 , 2 k 1 2 n < k + 1 n + 1 < 2 k + 1 2 n , n 2 < k < n 1 , 2 k 1 2 n < k + 1 n + 1 , k = n 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equav_HTML.gif
By using the above and denoting ■ a zero of T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif, □ a zero of z k U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq81_HTML.gif, we see that, for n even, all zeros between −1 and 1 listed in increasing order are of the form
, https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equaw_HTML.gif
where the center https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq82_HTML.gif is the 0 that is the zero of z k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq83_HTML.gif. For n odd, all zeros listed in increasing order are of the form
, https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equax_HTML.gif
where the center https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq84_HTML.gif means that all those three numbers □, ■, □ are cos π 2 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq85_HTML.gif and one □ comes from the zero of z k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq83_HTML.gif. Now consider sign changes using the above two chains in increasing order so that for n odd or even we may check that, if k is even, P 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq50_HTML.gif has at least n real zeros in ( 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq55_HTML.gif, and if k is odd, P 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq50_HTML.gif has at least n 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq52_HTML.gif real zeros in ( 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq55_HTML.gif. On the other hand, the zeros z of P 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq50_HTML.gif satisfy
| T n ( z ) U n ( z ) | = | z | k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equay_HTML.gif

If | z | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq22_HTML.gif, then | T n ( z ) | | U n ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq86_HTML.gif, which contradicts (2). Thus all zeros of P 1 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq50_HTML.gif lie in | z | < 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq87_HTML.gif. □

‘Bad pairs’ of polynomial zeros were defined in [2]. It is an easy consequence of Fell [1] that, if the all zeros of T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif and U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif form ‘good pairs’, their convex combination has all its zeros real.

Proof of Proposition 5 Following the proof of Theorem 4, we may see that for n even, all zeros T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif and U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif between −1 and 1 listed in increasing order are of the form
. https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equaz_HTML.gif
For n odd, all zeros listed in increasing order are of the form
, https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equba_HTML.gif

where the center https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq88_HTML.gif means that both numbers □, ■ are cos π 2 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq85_HTML.gif. Thus we can see that for n even, all zeros of T n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq10_HTML.gif and U n ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq13_HTML.gif form good pairs, and for n odd, all pairs from integral polynomials T n ( z ) / z https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq89_HTML.gif and U n ( z ) / z https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq90_HTML.gif are good. It follows that, by Fell [1], all zeros of the convex combination are real and in ( 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq55_HTML.gif. □

Proof of Theorem 6 The zeros z of P 1 , ϵ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq58_HTML.gif satisfy
| T n , ϵ ( z ) U n , ϵ ( z ) | = | z | k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equbb_HTML.gif

If | z | 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq22_HTML.gif, then | T n , ϵ ( z ) | | U n , ϵ ( z ) | https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq91_HTML.gif, which contradicts (4). □

Proof of Proposition 7 Using | z | 2 = z z ¯ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_IEq92_HTML.gif, we have
| P 1 ( z ) | 2 = ( T n ( e i θ ) + e i k θ U n ( e i θ ) ) ( T n ( e i θ ) + e i k θ U n ( e i θ ) ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equbc_HTML.gif
and
| P 2 ( z ) | 2 = ( U n ( e i θ ) + e i k θ T n ( e i θ ) ) ( U n ( e i θ ) + e i k θ T n ( e i θ ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equbd_HTML.gif
So we only need to show that
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Eqube_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-167/MediaObjects/13660_2011_Article_309_Equbf_HTML.gif

But this equality follows from just replacing the variable θ by −θ. □

Acknowledgements

The author wishes to thank Professor Kenneth B. Stolarsky who let the author know some questions in this paper. The author is grateful to the referee of this paper for useful comments and suggestions that led to further development of an earlier version. This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2010-0011010).

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© Kim; licensee Springer 2012

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