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; zeros1 Introduction
Let
and
be the Chebyshev polynomials of the first kind and of the second kind, respectively.
These polynomials satisfy the recurrence relations
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
are well understood, but the polynomials of complex arguments are less so. Reported
here are several bounds for Chebyshev polynomials defined on
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].

In detail,

and it is easy to show that for an odd integer n,

Since
, we may apply results about
to those about
. But
and ϵ was a nonnegative real number less than
, and so properties of
will be investigated separately from those of
.
2 New results
In this section we list some new results related to the bounds of Chebyshev polynomials
,
and their analogues
,
defined on
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
,
,
and
. We may compute that for
,
and
Proposition 1Suppose thatzis a complex number satisfying
. Then for
,
and
Also
and
Proposition 1 will be used in the proofs of Theorems 4 and 6.
Remarks For a complex number z with
, we may follow the procedure of the proof of (2) to obtain
and
that is best possible since
. There seem to be larger lower bounds than 1 for
in (2). First, we observe that
because
. Deciding in some general situations exactly where the minimum occurs seems to be
extremely difficult. For example, machine calculation suggests that for
,
takes its minimum 3.91735… in
at four modulus 1 roots
of the polynomial

But we may conjecture that, by numerical computations, the value
occurs in
and lies between
and n, where
can be replaced by something larger. We now ask naturally what the minimum is for
. If one simply looks at the case
, it seems that
is close to its minimum at
. But
is the coefficient of
in the power series expansion of
. In fact,
For
,
by (2). In the following proposition, we obtain an upper bound for arbitrary
,
.
Proposition 2Let
, where
and
. Then, for
,
Remarks With the same notations with Proposition 2, it follows from
that, for
large,
is large. But
and
These imply that the upper bound
is greater than
, but for
large, it is close to
. Also by machine computations (e.g.,
and
), we may check that the inequality (5) is sharp.
It is natural to ask about the bounds on the unit circle.
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
by (4), and it seems to be true that
The proof of (6) will be given in Section 3 by using a well-known identity
. But
does not hold. So we cannot use this to prove (7) if it is true.
All zeros of the polynomial
lie in
. More generally the convex combination of
and
has all its zeros in
. This will be proved in Proposition 5 below. So one might ask: where are the zeros
of polynomials like
or
around
? The next theorem answers this for
.
Theorem 4Let
for positive integersnandk. Then
has all its zeros in
. Furthermore, forkeven,
has at leastnreal zeros, and forkodd,
has at least
real zeros.
Remarks Let
for positive integers n and k. We can use the same method as in the proof of Theorem 4 to show that
has at least n real zeros in
. Furthermore, for k even, there is no real zero outside
, and for k odd, there is one more real zero on
.
Proposition 5The polynomial
Using analogues of
and
, we consider analogues of
and investigate their zero distributions. Define
Theorem 6
has all its zeros in
.
Finally, we get an interesting property about the integrals of sums of Chebyshev polynomials. Observe that
and
equals
For example, from
and
we can calculate

and we see that these two integrals are different. But for
and
, the integrals have the same value.
Remark It seems to be true that for k large,
,
but
These remain open problems.
3 Proofs
Proof of Proposition 1 Suppose that z is a complex number satisfying
. Using (1), for
, we have
and
Then by recurrence,
By (8) and the identity
we have
Next we prove the results about
and
. For n odd and
, it follows from the definition of
and (8) that
This inequality
for other three cases (i.e., n odd and
, n even and
, n even and
) can be proved in the same way. Finally, for n odd, by
and (9), we have
In the same way, we can check that for n even,
□
Proof of Proposition 2 Using the identity
, for
we have

Since
, it suffices to consider the case
. The above implies
So
Also
and
and so
Now we see with (10) and (11) that
□
Proof of Proposition 3 Suppose that z is a complex number satisfying
. First, it follows from (2) that
and so it is enough to show that
We use induction on n. For
,
. Assume the result holds for k. Then
and
□
Proof of Theorem 4 All zeros of
and
are real and lie in
and for
,
For convenience, by removing ‘cos’ and the constant π,
and
can be identified with the ascending chain of rational numbers
and
, respectively. We may calculate that for n odd
and for n even
By using the above and denoting ■ a zero of
, □ a zero of
, we see that, for n even, all zeros between −1 and 1 listed in increasing order are of the form
where the center
is the 0 that is the zero of
. For n odd, all zeros listed in increasing order are of the form
where the center
means that all those three numbers □, ■, □ are
and one □ comes from the zero of
. 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,
has at least n real zeros in
, and if k is odd,
has at least
real zeros in
. On the other hand, the zeros z of
satisfy
If
, then
, which contradicts (2). Thus all zeros of
lie in
. □
‘Bad pairs’ of polynomial zeros were defined in [2]. It is an easy consequence of Fell [1] that, if the all zeros of
and
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
and
between −1 and 1 listed in increasing order are of the form
For n odd, all zeros listed in increasing order are of the form
where the center
means that both numbers □, ■ are
. Thus we can see that for n even, all zeros of
and
form good pairs, and for n odd, all pairs from integral polynomials
and
are good. It follows that, by Fell [1], all zeros of the convex combination are real and in
. □
Proof of Theorem 6 The zeros z of
satisfy
If
, then
, which contradicts (4). □
Proof of Proposition 7 Using
, we have
and
So we only need to show that

and

But this equality follows from just replacing the variable θ by −θ. □
Competing interests
The author declares that they have no competing interests.
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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