Some properties of Chebyshev polynomials

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

Cite this article as:
Kim, SH. 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 polynomials bounds sums zeros 

1 Introduction

Let T n ( x ) Open image in new window and U n ( x ) Open image in new window 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 ) . Open image in new window
(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 ] Open image in new window are well understood, but the polynomials of complex arguments are less so. Reported here are several bounds for Chebyshev polynomials defined on Open image in new window 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 Open image in new window and 2 ϵ > 1 + ϵ 1 Open image in new window, i.e., 0 ϵ < 1 / 2 Open image in new window, we let
and it is easy to show that for an odd integer n,

Since T n , 0 ( z ) = T n ( z ) Open image in new window, we may apply results about T n , ϵ ( z ) Open image in new window to those about T n ( z ) Open image in new window. But U n , 1 ( z ) = U n ( z ) Open image in new window and ϵ was a nonnegative real number less than 1 / 2 Open image in new window, and so properties of U n ( z ) Open image in new window will be investigated separately from those of U n , ϵ ( z ) Open image in new window.

2 New results

In this section we list some new results related to the bounds of Chebyshev polynomials T n ( z ) Open image in new window, U n ( z ) Open image in new window and their analogues T n , ϵ ( z ) Open image in new window, U n , ϵ ( z ) Open image in new window defined on Open image in new window 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 ) Open image in new window, U n ( z ) Open image in new window, T n , ϵ ( z ) Open image in new window and U n , ϵ ( z ) Open image in new window. We may compute that for n 1 Open image in new window,
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 , Open image in new window
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 . Open image in new window

Proposition 1Suppose thatzis a complex number satisfying | z | 1 Open image in new window. Then for n 1 Open image in new window,

| U n ( z ) | | U n 1 ( z ) | 1 Open image in new window
and
| U n ( z ) | | T n ( z ) | 1 . Open image in new window
(2)
Also
| U n , ϵ ( z ) | | U n 1 , ϵ ( z ) | 1 Open image in new window
(3)
and
| U n , ϵ ( z ) | | T n , ϵ ( z ) | > { 1 + ϵ if n is odd , ϵ if n is even . Open image in new window
(4)

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

Remarks For a complex number z with | z | 1 Open image in new window, we may follow the procedure of the proof of (2) to obtain
| T n ( z ) | | T n 1 ( z ) | 0 ( n 1 ) Open image in new window
and
| U n ( z ) | n + 1 ( n 0 ) Open image in new window
that is best possible since | U n ( 1 ) | = n + 1 Open image in new window. There seem to be larger lower bounds than 1 for | U n ( z ) | | T n ( z ) | Open image in new window in (2). First, we observe that
min | z | 1 ( | U n ( z ) | | T n ( z ) | ) n Open image in new window
because | U n ( 1 ) | | T n ( 1 ) | = ( n + 1 ) 1 = n Open image in new window. Deciding in some general situations exactly where the minimum occurs seems to be extremely difficult. For example, machine calculation suggests that for n = 4 Open image in new window, | U 4 ( z ) | | T 4 ( z ) | Open image in new window takes its minimum 3.91735… in | z | 1 Open image in new window at four modulus 1 roots ± 0.9953 ± i 0.0964 Open image in new window of the polynomial
But we may conjecture that, by numerical computations, the value
min | z | 1 ( | U n ( z ) | | T n ( z ) | ) Open image in new window
occurs in | z | = 1 Open image in new window and lies between n 1 / 2 Open image in new window and n, where n 1 / 2 Open image in new window can be replaced by something larger. We now ask naturally what the minimum is for | z | = t > 1 Open image in new window. If one simply looks at the case z = t Open image in new window, it seems that | U n ( z ) | | T n ( z ) | Open image in new window is close to its minimum at z = t Open image in new window. But
| U n ( t ) | | T n ( t ) | = U n ( t ) T n ( t ) Open image in new window
is the coefficient of x n 1 Open image in new window in the power series expansion of t / ( 1 2 t x + t 2 ) Open image in new window. 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 . Open image in new window

For | z | 1 Open image in new window, | U n ( z ) | | T n ( z ) | + 1 Open image in new window by (2). In the following proposition, we obtain an upper bound for arbitrary z = cos θ Open image in new window, Open image in new window.

Proposition 2Let z = cos θ Open image in new window, where θ = α + i β Open image in new windowand β 0 Open image in new window. Then, for n 1 Open image in new window,
| U n ( z ) | ( 1 + coth β coth n β ) | T n ( z ) | . Open image in new window
(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 ) Open image in new window
that, for | z | Open image in new window large, | β | Open image in new window is large. But coth β coth n β > 1 Open image in new window and
lim β ± coth β coth n β = 1 . Open image in new window
These imply that the upper bound
( 1 + coth β coth n β ) | T n ( z ) | Open image in new window

is greater than 2 | T n ( z ) | Open image in new window, but for | z | Open image in new window large, it is close to 2 | T n ( z ) | Open image in new window. Also by machine computations (e.g., n = 4 Open image in new window and z = 100 + i 100 Open image in new window), 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 Open image in new window, we have
1 n + 1 | U n ( z ) | | T n ( z ) | < | U n ( z ) | . Open image in new window
(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 . Open image in new window
For | z | = 1 Open image in new window, it is easy to see
| T n , ϵ ( z ) | < | U n , ϵ ( z ) | Open image in new window
by (4), and it seems to be true that
1 n + 1 | U n , ϵ ( z ) | | T n , ϵ ( z ) | . Open image in new window
(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 ) Open image in new window. But U n , ϵ ( z ) = T n , ϵ ( z ) + z U n 1 , ϵ ( z ) Open image in new window 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 ) Open image in new window lie in ( 1 , 1 ) Open image in new window. More generally the convex combination of T n ( z ) Open image in new window and U n ( z ) Open image in new window has all its zeros in ( 1 , 1 ) Open image in new window. 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 ) Open image in new window or U n ( z ) + z k T n ( z ) Open image in new window around | z | = 1 Open image in new window? The next theorem answers this for T n ( z ) + z k U n ( z ) Open image in new window.

Theorem 4Let P 1 ( z ) : = T n ( z ) + z k U n ( z ) Open image in new windowfor positive integersnandk. Then P 1 ( z ) Open image in new windowhas all its zeros in | z | < 1 Open image in new window. Furthermore, forkeven, P 1 ( z ) Open image in new windowhas at leastnreal zeros, and forkodd, P 1 ( z ) Open image in new windowhas at least n 1 Open image in new windowreal zeros.

Remarks Let P 2 ( z ) : = U n ( z ) + z k T n ( z ) Open image in new window 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 ) Open image in new window has at least n real zeros in ( 1 , 1 ) Open image in new window. Furthermore, for k even, there is no real zero outside [ 1 , 1 ] Open image in new window, and for k odd, there is one more real zero on z < 1 Open image in new window.

Proposition 5The polynomial
( 1 λ ) T n ( z ) + λ U n ( z ) ( 0 λ 1 ) Open image in new window

has all zeros in ( 1 , 1 ) Open image in new window.

Using analogues of T n ( z ) Open image in new window and U n ( z ) Open image in new window, we consider analogues of P 1 ( z ) Open image in new window and investigate their zero distributions. Define
P 1 , ϵ ( z ) : = T n , ϵ ( z ) + z k U n , ϵ ( z ) . Open image in new window

Theorem 6 P 1 , ϵ ( z ) Open image in new windowhas all its zeros in | z | < 1 Open image in new window.

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 Open image in new window
and
0 2 π | T n ( e i θ ) | 2 d θ ( or 0 2 π | U n ( e i θ ) | 2 d θ ) Open image in new window
equals
2 π sum of the squares of all coefficients of T n ( z ) ( or U n ( z ) ) . Open image in new window
For example, from T 4 ( z ) = 8 z 4 8 z 2 + 1 Open image in new window and U 4 ( z ) = 16 z 6 80 z 4 + 24 z 2 1 Open image in new window we can calculate

and we see that these two integrals are different. But for P 1 ( z ) = T n ( z ) + z k U n ( z ) Open image in new window and P 2 ( z ) = U n ( z ) + z k T n ( z ) Open image in new window, the integrals have the same value.

Proposition 7For z = e i θ Open image in new window,
0 2 π | P 1 ( e i θ ) | 2 d θ = 0 2 π | P 2 ( e i θ ) | 2 d θ . Open image in new window
Remark It seems to be true that for k large, z = e i θ Open image in new window,
0 2 π | P 1 ( e i θ ) | d θ 0 2 π | P 2 ( e i θ ) | d θ , Open image in new window
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 . Open image in new window

These remain open problems.

3 Proofs

Proof of Proposition 1 Suppose that z is a complex number satisfying | z | 1 Open image in new window. Using (1), for n 1 Open image in new window, we have
| U n + 1 ( z ) | 2 | U n ( z ) | | U n 1 ( z ) | Open image in new window
and
| U n + 1 ( z ) | | U n ( z ) | | U n ( z ) | | U n 1 ( z ) | . Open image in new window
Then by recurrence,
| U n ( z ) | | U n 1 ( z ) | | U 1 ( z ) | | U 0 ( z ) | = | 2 z | 1 1 . Open image in new window
(8)
By (8) and the identity
T n ( z ) = 1 2 ( U n ( z ) U n 2 ( z ) ) , Open image in new window
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 . Open image in new window
(9)
Next we prove the results about U n , ϵ ( z ) Open image in new window and T n , ϵ ( z ) Open image in new window. For n odd and 4 n 1 Open image in new window, it follows from the definition of U n , ϵ ( z ) Open image in new window 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 . Open image in new window
This inequality
| U n , ϵ ( z ) | | U n 1 , ϵ ( z ) | 1 Open image in new window
for other three cases (i.e., n odd and 4 n 1 Open image in new window, n even and 4 n 1 Open image in new window, n even and 4 n 1 Open image in new window) can be proved in the same way. Finally, for n odd, by 2 ϵ > 1 + ϵ Open image in new window 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 . Open image in new window
In the same way, we can check that for n even,
| U n , ϵ ( z ) | | T n , ϵ ( z ) | > ϵ . Open image in new window

 □

Proof of Proposition 2 Using the identity U n ( z ) = T n ( z ) + z U n 1 ( z ) Open image in new window, for z = cos θ Open image in new window 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 | . Open image in new window
Since z ¯ = cos θ ¯ Open image in new window, it suffices to consider the case β > 0 Open image in new window. 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 . Open image in new window
So
| tan n θ | coth n β . Open image in new window
(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 Open image in new window
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 , Open image in new window
and so
| cot θ | coth β . Open image in new window
(11)
Now we see with (10) and (11) that
| cot θ tan n θ | coth β coth n β . Open image in new window

 □

Proof of Proposition 3 Suppose that z is a complex number satisfying | z | = 1 Open image in new window. First, it follows from (2) that
| T n ( z ) U n ( z ) | < 1 . Open image in new window
Using the identity U n ( z ) = T n ( z ) + z U n 1 ( z ) Open image in new window, 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 ) | Open image in new window
and so it is enough to show that
| U n 1 ( z ) U n ( z ) | 1 1 n + 1 = n n + 1 . Open image in new window
We use induction on n. For n = 1 Open image in new window, | U 0 ( z ) U 1 ( z ) | = | 1 2 z | = 1 2 Open image in new window. 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 , Open image in new window
and
| U k ( z ) U k + 1 ( z ) | k + 1 k + 2 . Open image in new window

 □

Proof of Theorem 4 All zeros of T n ( z ) Open image in new window and U n ( z ) Open image in new window are real and lie in ( 1 , 1 ) Open image in new window and for 1 k n Open image in new window,
T n ( cos ( 2 k 1 ) π 2 n ) = 0 , U n ( cos k π n + 1 ) = 0 . Open image in new window
For convenience, by removing ‘cos’ and the constant π, cos ( 2 k 1 ) π 2 n Open image in new window and cos k π n + 1 Open image in new window can be identified with the ascending chain of rational numbers ( 2 k 1 ) 2 n Open image in new window and k n + 1 Open image in new window, 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 Open image in new window
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 . Open image in new window
By using the above and denoting ■ a zero of T n ( z ) Open image in new window, □ a zero of z k U n ( z ) Open image in new window, we see that, for n even, all zeros between −1 and 1 listed in increasing order are of the form
, Open image in new window
where the center Open image in new window is the 0 that is the zero of z k Open image in new window. For n odd, all zeros listed in increasing order are of the form
, Open image in new window
where the center Open image in new window means that all those three numbers □, ■, □ are cos π 2 = 0 Open image in new window and one □ comes from the zero of z k Open image in new window. 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 ) Open image in new window has at least n real zeros in ( 1 , 1 ) Open image in new window, and if k is odd, P 1 ( z ) Open image in new window has at least n 1 Open image in new window real zeros in ( 1 , 1 ) Open image in new window. On the other hand, the zeros z of P 1 ( z ) Open image in new window satisfy
| T n ( z ) U n ( z ) | = | z | k . Open image in new window

If | z | 1 Open image in new window, then | T n ( z ) | | U n ( z ) | Open image in new window, which contradicts (2). Thus all zeros of P 1 ( z ) Open image in new window lie in | z | < 1 Open image in new window. □

‘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 ) Open image in new window and U n ( z ) Open image in new window 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 ) Open image in new window and U n ( z ) Open image in new window between −1 and 1 listed in increasing order are of the form
. Open image in new window
For n odd, all zeros listed in increasing order are of the form
, Open image in new window

where the center Open image in new window means that both numbers □, ■ are cos π 2 = 0 Open image in new window. Thus we can see that for n even, all zeros of T n ( z ) Open image in new window and U n ( z ) Open image in new window form good pairs, and for n odd, all pairs from integral polynomials T n ( z ) / z Open image in new window and U n ( z ) / z Open image in new window are good. It follows that, by Fell [1], all zeros of the convex combination are real and in ( 1 , 1 ) Open image in new window. □

Proof of Theorem 6 The zeros z of P 1 , ϵ ( z ) Open image in new window satisfy
| T n , ϵ ( z ) U n , ϵ ( z ) | = | z | k . Open image in new window

If | z | 1 Open image in new window, then | T n , ϵ ( z ) | | U n , ϵ ( z ) | Open image in new window, which contradicts (4). □

Proof of Proposition 7 Using | z | 2 = z z ¯ Open image in new window, 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 ) ) Open image in new window
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 ) ) . Open image in new window
So we only need to show that

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).

Copyright information

© Kim; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Department of MathematicsSookmyung Women’s UniversitySeoulKorea

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