Erschienen in:

Open Access 01.12.2012 | Research

Some properties of Chebyshev polynomials

verfasst von: Seon-Hong Kim

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2012

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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.

Competing interests

The author declares that they have no competing interests.

1 Introduction

Let

T_{n} (x)

and

U_{n} (x)

be the Chebyshev polynomials of the first kind and of the second kind, respectively. These polynomials satisfy the recurrence relations

\begin{aligned} T_{0} (x) = 1, T_{1} (x) = x, T_{n + 1} (x) = 2 x T_{n} (x) - T_{n - 1} (x) (n \geq 1) \\ U_{0} (x) = 1, U_{1} (x) = 2 x, U_{n + 1} (x) = 2 x U_{n} (x) - U_{n - 1} (x) (n \geq 1) . \end{aligned}

(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]

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 \geq 1

and

2 - ϵ > 1 + ϵ \geq 1

, i.e.,

0 \leq ϵ < 1 / 2

, 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)

, we may apply results about

T_{n, ϵ} (z)

to those about

T_{n} (z)

. But

U_{n, 1} (z) = U_{n} (z)

and ϵ was a nonnegative real number less than

1 / 2

, and so properties of

U_{n} (z)

will be investigated separately from those of

U_{n, ϵ} (z)

2 New results

In this section we list some new results related to the bounds of Chebyshev polynomials

T_{n} (z)

U_{n} (z)

and their analogues

T_{n, ϵ} (z)

U_{n, ϵ} (z)

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

T_{n} (z)

U_{n} (z)

T_{n, ϵ} (z)

and

U_{n, ϵ} (z)

. We may compute that for

n \geq 1

T_{n, ϵ} (z) = {\begin{matrix} (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, \end{matrix}

and

U_{n, ϵ} (z) = {\begin{matrix} (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 . \end{matrix}

Proposition 1 Suppose that z is a complex number satisfying

| z | \geq 1

. Then for

n \geq 1

| U_{n} (z) | - | U_{n - 1} (z) | \geq 1

and

| U_{n} (z) | - | T_{n} (z) | \geq 1 .

(2)

Also

| U_{n, ϵ} (z) | - | U_{n - 1, ϵ} (z) | \geq 1

(3)

and

| U_{n, ϵ} (z) | - | T_{n, ϵ} (z) | > {\begin{matrix} 1 + ϵ & if n is odd, \\ ϵ & if n is even . \end{matrix}

(4)

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

Remarks For a complex number z with

| z | \geq 1

, we may follow the procedure of the proof of (2) to obtain

| T_{n} (z) | - | T_{n - 1} (z) | \geq 0 (n \geq 1)

and

| U_{n} (z) | \geq n + 1 (n \geq 0)

that is best possible since

| U_{n} (1) | = n + 1

. There seem to be larger lower bounds than 1 for

| U_{n} (z) | - | T_{n} (z) |

in (2). First, we observe that

min_{| z | \geq 1} (| U_{n} (z) | - | T_{n} (z) |) \leq n

because

| U_{n} (1) | - | T_{n} (1) | = (n + 1) - 1 = n

. Deciding in some general situations exactly where the minimum occurs seems to be extremely difficult. For example, machine calculation suggests that for

n = 4

| U_{4} (z) | - | T_{4} (z) |

takes its minimum 3.91735… in

| z | \geq 1

at four modulus 1 roots

\pm 0.9953 \dots \pm i 0.0964 \dots

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 | \geq 1} (| U_{n} (z) | - | T_{n} (z) |)

occurs in

| z | = 1

and lies between

n - 1 / 2

and n, where

n - 1 / 2

can be replaced by something larger. We now ask naturally what the minimum is for

| z | = t > 1

. If one simply looks at the case

z = t

, it seems that

| U_{n} (z) | - | T_{n} (z) |

is close to its minimum at

z = t

. But

| U_{n} (t) | - | T_{n} (t) | = U_{n} (t) - T_{n} (t)

is the coefficient of

x^{n - 1}

in the power series expansion of

t / (1 - 2 t x + t^{2})

. In fact,

\begin{array}{rcl} \frac{t x}{1 - 2 t x + x^{2}} & = & \frac{1}{1 - 2 t x + x^{2}} - \frac{1 - t x}{1 - 2 t x + x^{2}} \\ = & \sum_{n = 0}^{\infty} U_{n} (t) x^{n} - \sum_{n = 0}^{\infty} T_{n} (t) x^{n} = \sum_{n = 0}^{\infty} (U_{n} (t) - T_{n} (t)) x^{n} . \end{array}

For

| z | \geq 1

| U_{n} (z) | \geq | T_{n} (z) | + 1

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_IEq32_HTML.gif

Proposition 2 Let

z = cos θ

, where

θ = α + i β

and

β \neq 0

. Then, for

n \geq 1

| U_{n} (z) | \leq (1 + coth β coth n β) | T_{n} (z) | .

(5)

Remarks With the same notations with Proposition 2, it follows from

| z |^{2} = | cos θ |^{2} = \frac{1}{2} (cos 2 α + cosh 2 β)

that, for

| z |

large,

| β |

is large. But

coth β coth n β > 1

and

lim_{β \to \pm \infty} coth β coth n β = 1 .

These imply that the upper bound

(1 + coth β coth n β) | T_{n} (z) |

is greater than

2 | T_{n} (z) |

, but for

| z |

large, it is close to

2 | T_{n} (z) |

. Also by machine computations (e.g.,

n = 4

and

z = 100 + i 100

), we may check that the inequality (5) is sharp.

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

Proposition 3 For

| z | = 1

, we have

\frac{1}{n + 1} | U_{n} (z) | \leq | T_{n} (z) | < | U_{n} (z) | .

(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

\frac{1}{n + 1} | U_{n} (\pm 1) | = | T_{n} (\pm 1) | = 1 .

For

| z | = 1

, it is easy to see

| T_{n, ϵ} (z) | < | U_{n, ϵ} (z) |

by (4), and it seems to be true that

\frac{1}{n + 1} | U_{n, ϵ} (z) | \leq | T_{n, ϵ} (z) | .

(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)

. But

U_{n, ϵ} (z) = T_{n, ϵ} (z) + z U_{n - 1, ϵ} (z)

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)

lie in

(- 1, 1)

. More generally the convex combination of

T_{n} (z)

and

U_{n} (z)

has all its zeros in

(- 1, 1)

. 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)

U_{n} (z) + z^{k} T_{n} (z)

around

| z | = 1

? The next theorem answers this for

T_{n} (z) + z^{k} U_{n} (z)

Theorem 4 Let

P_{1} (z) : = T_{n} (z) + z^{k} U_{n} (z)

for positive integers n and k. Then

P_{1} (z)

has all its zeros in

| z | < 1

. Furthermore, for k even,

P_{1} (z)

has at least n real zeros, and for k odd,

P_{1} (z)

has at least

n - 1

real zeros.

Remarks Let

P_{2} (z) : = U_{n} (z) + z^{k} T_{n} (z)

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)

has at least n real zeros in

(- 1, 1)

. Furthermore, for k even, there is no real zero outside

[- 1, 1]

, and for k odd, there is one more real zero on

z < - 1

Proposition 5 The polynomial

(1 - λ) T_{n} (z) + λ U_{n} (z) (0 \leq λ \leq 1)

has all zeros in

(- 1, 1)

Using analogues of

T_{n} (z)

and

U_{n} (z)

, we consider analogues of

P_{1} (z)

and investigate their zero distributions. Define

P_{1, ϵ} (z) : = T_{n, ϵ} (z) + z^{k} U_{n, ϵ} (z) .

Theorem 6

P_{1, ϵ} (z)

has all its zeros in

| z | < 1

Finally, we get an interesting property about the integrals of sums of Chebyshev polynomials. Observe that

\int_{0}^{2 π} T_{n} (e^{i θ}) d θ = \int_{0}^{2 π} U_{n} (e^{i θ}) d θ = T_{n} (0) \cdot 2 π = {\begin{matrix} {(- 1)}^{n / 2} & if n is even, \\ 0 & if n is odd \end{matrix}

and

\int_{0}^{2 π} | T_{n} (e^{i θ}) |^{2} d θ (or \int_{0}^{2 π} | U_{n} (e^{i θ}) |^{2} d θ)

equals

2 π \cdot sum of the squares of all coefficients of T_{n} (z) (or U_{n} (z)) .

For example, from

T_{4} (z) = 8 z^{4} - 8 z^{2} + 1

and

U_{4} (z) = 16 z^{6} - 80 z^{4} + 24 z^{2} - 1

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)

and

P_{2} (z) = U_{n} (z) + z^{k} T_{n} (z)

, the integrals have the same value.

Proposition 7 For

z = e^{i θ}

\int_{0}^{2 π} | P_{1} (e^{i θ}) |^{2} d θ = \int_{0}^{2 π} | P_{2} (e^{i θ}) |^{2} d θ .

Remark It seems to be true that for k large,

z = e^{i θ}

\int_{0}^{2 π} | P_{1} (e^{i θ}) | d θ \approx \int_{0}^{2 π} | P_{2} (e^{i θ}) | d θ,

but

lim_{k \to \infty} \int_{0}^{2 π} | P_{1} (e^{i θ}) | d θ = lim_{k \to \infty} \int_{0}^{2 π} | P_{2} (e^{i θ}) | d θ .

These remain open problems.

3 Proofs

Proof of Proposition 1 Suppose that z is a complex number satisfying

| z | \geq 1

. Using (1), for

n \geq 1

, we have

| U_{n + 1} (z) | \geq 2 | U_{n} (z) | - | U_{n - 1} (z) |

and

| U_{n + 1} (z) | - | U_{n} (z) | \geq | U_{n} (z) | - | U_{n - 1} (z) | .

Then by recurrence,

| U_{n} (z) | - | U_{n - 1} (z) | \geq | U_{1} (z) | - | U_{0} (z) | = | 2 z | - 1 \geq 1 .

(8)

By (8) and the identity

T_{n} (z) = \frac{1}{2} (U_{n} (z) - U_{n - 2} (z)),

we have

\begin{array}{rcl} | T_{n} (z) | & \leq & \frac{1}{2} | U_{n} (z) | + \frac{1}{2} | U_{n - 2} (z) | \\ \leq & \frac{1}{2} | U_{n} (z) | + \frac{1}{2} (| U_{n} (z) | - 2) \\ = & | U_{n} (z) | - 1 . \end{array}

(9)

Next we prove the results about

U_{n, ϵ} (z)

and

T_{n, ϵ} (z)

. For n odd and

4 ∣ n - 1

, it follows from the definition of

U_{n, ϵ} (z)

and (8) that

\begin{array}{rcl} | U_{n, ϵ} (z) | - | U_{n - 1, ϵ} (z) | & = & | (2 - ϵ) U_{n} (z) | - | (2 - ϵ) U_{n - 1} (z) - (1 - ϵ) | \\ \geq & | (2 - ϵ) U_{n} (z) | - | (2 - ϵ) U_{n - 1} (z) | - (1 - ϵ) \\ = & (2 - ϵ) (| U_{n} (z) | - | U_{n - 1} (z) |) - (1 - ϵ) \\ \geq & 2 - ϵ - 1 + ϵ = 1 . \end{array}

This inequality

| U_{n, ϵ} (z) | - | U_{n - 1, ϵ} (z) | \geq 1

for other three cases (i.e., n odd and

4 ∤ n - 1

, n even and

4 ∣ n - 1

, n even and

4 ∤ n - 1

) can be proved in the same way. Finally, for n odd, by

2 - ϵ > 1 + ϵ

and (9), we have

\begin{array}{rcl} | 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) |) \geq 1 + ϵ . \end{array}

In the same way, we can check that for n even,

| U_{n, ϵ} (z) | - | T_{n, ϵ} (z) | > ϵ .

□

Proof of Proposition 2 Using the identity

U_{n} (z) = T_{n} (z) + z U_{n - 1} (z)

, for

z = cos θ

we have

\begin{array}{rcl} | \frac{U_{n} (z)}{T_{n} (z)} | & = & | 1 + z \frac{U_{n - 1} (z)}{T_{n} (z)} | = | 1 + cos θ \frac{tan n θ}{sin θ} | \\ = & | 1 + cot θ tan n θ | \leq 1 + | cot θ tan n θ | . \end{array}

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_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

\bar{z} = cos \bar{θ}

, it suffices to consider the case

β > 0

. The above implies

| tan θ |^{2} = \frac{{sin}^{2} α {cos}^{2} α + {sinh}^{2} β {cosh}^{2} β}{{({cos}^{2} α + {sinh}^{2} β)}^{2}} \leq \frac{{sinh}^{2} β {cosh}^{2} β}{{sinh}^{4} β} .

| tan n θ | \leq coth n β .

(10)

Also

\begin{array}{rcl} | sin θ |^{2} & = & {sin}^{2} α {cosh}^{2} β + {cos}^{2} α {sinh}^{2} β \\ \geq & ({sin}^{2} α + {cos}^{2} α) {sinh}^{2} β \end{array}

and

\begin{array}{rcl} | cos θ |^{2} & = & {cos}^{2} α {cosh}^{2} β + {sin}^{2} α {sinh}^{2} β \\ \leq & ({sin}^{2} α + {cos}^{2} α) {cosh}^{2} β, \end{array}

and so

| cot θ | \leq coth β .

(11)

Now we see with (10) and (11) that

| cot θ tan n θ | \leq coth β coth n β .

□

Proof of Proposition 3 Suppose that z is a complex number satisfying

| z | = 1

. First, it follows from (2) that

| \frac{T_{n} (z)}{U_{n} (z)} | < 1 .

Using the identity

U_{n} (z) = T_{n} (z) + z U_{n - 1} (z)

, we have

| \frac{T_{n} (z)}{U_{n} (z)} | = | 1 - z \frac{U_{n - 1} (z)}{U_{n} (z)} | \geq 1 - | \frac{U_{n - 1} (z)}{U_{n} (z)} |

and so it is enough to show that

| \frac{U_{n - 1} (z)}{U_{n} (z)} | \leq 1 - \frac{1}{n + 1} = \frac{n}{n + 1} .

We use induction on n. For

n = 1

| \frac{U_{0} (z)}{U_{1} (z)} | = | \frac{1}{2 z} | = \frac{1}{2}

. Assume the result holds for k. Then

| \frac{U_{k + 1} (z)}{U_{k} (z)} | = | 2 z - \frac{U_{k - 1} (z)}{U_{k} (z)} | \geq 2 - | \frac{U_{k - 1} (z)}{U_{k} (z)} | \geq 2 - \frac{k}{k + 1} = \frac{k + 2}{k + 1},

and

| \frac{U_{k} (z)}{U_{k + 1} (z)} | \leq \frac{k + 1}{k + 2} .

□

Proof of Theorem 4 All zeros of

T_{n} (z)

and

U_{n} (z)

are real and lie in

(- 1, 1)

and for

1 \leq k \leq n

T_{n} (cos \frac{(2 k - 1) π}{2 n}) = 0, U_{n} (cos \frac{k π}{n + 1}) = 0 .

For convenience, by removing ‘cos’ and the constant π,

cos \frac{(2 k - 1) π}{2 n}

and

cos \frac{k π}{n + 1}

can be identified with the ascending chain of rational numbers

\frac{(2 k - 1)}{2 n}

and

\frac{k}{n + 1}

, respectively. We may calculate that for n odd

{\begin{matrix} \frac{2 k - 1}{2 n} < \frac{k}{n + 1} < \frac{2 k + 1}{2 n}, & 1 \leq k < \frac{n + 1}{2}, \\ \frac{2 k + 1}{2 n} = \frac{k + 1}{n + 1}, & k = \frac{n - 1}{2}, \\ \frac{2 k - 1}{2 n} < \frac{k + 1}{n + 1} < \frac{2 k + 1}{2 n}, & \frac{n - 1}{2} < k < n - 1, \\ \frac{2 k - 1}{2 n} < \frac{k + 1}{n + 1}, & k = n - 1 \end{matrix}

and for n even

{\begin{matrix} \frac{2 k - 1}{2 n} < \frac{k}{n + 1} < \frac{2 k + 1}{2 n}, & 1 \leq k < \frac{n}{2}, \\ \frac{2 k - 1}{2 n} < \frac{k}{n + 1} < \frac{k + 1}{n + 1} < \frac{2 k + 1}{2 n}, & k = \frac{n}{2}, \\ \frac{2 k - 1}{2 n} < \frac{k + 1}{n + 1} < \frac{2 k + 1}{2 n}, & \frac{n}{2} < k < n - 1, \\ \frac{2 k - 1}{2 n} < \frac{k + 1}{n + 1}, & k = n - 1 . \end{matrix}

By using the above and denoting ■ a zero of

T_{n} (z)

, □ a zero of

z^{k} U_{n} (z)

, we see that, for n even, all zeros between −1 and 1 listed in increasing order are of the form

■ □ ■ □ \dots ■ □ ■ □ \underset{⏟}{□} □ ■ □ ■ \dots □ ■ □ ■,

where the center

\underset{⏟}{□}

is the 0 that is the zero of

z^{k}

. For n odd, all zeros listed in increasing order are of the form

■ □ ■ □ \dots ■ □ ■ □ \underset{⏟}{□ ■ □} □ ■ □ ■ \dots □ ■ □ ■,

where the center

\underset{⏟}{□ ■ □}

means that all those three numbers □, ■, □ are

cos \frac{π}{2} = 0

and one □ comes from the zero of

z^{k}

. 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)

has at least n real zeros in

(- 1, 1)

, and if k is odd,

P_{1} (z)

has at least

n - 1

real zeros in

(- 1, 1)

. On the other hand, the zeros z of

P_{1} (z)

satisfy

| \frac{T_{n} (z)}{U_{n} (z)} | = | z |^{k} .

| z | \geq 1

, then

| T_{n} (z) | \geq | U_{n} (z) |

, which contradicts (2). Thus all zeros of

P_{1} (z)

lie in

| z | < 1

. □

‘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)

and

U_{n} (z)

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)

and

U_{n} (z)

between −1 and 1 listed in increasing order are of the form

■ □ ■ □ \dots ■ □ □ ■ □ ■ \dots □ ■ □ ■ .

For n odd, all zeros listed in increasing order are of the form

■ □ ■ □ \dots ■ □ \underset{⏟}{□ ■} □ ■ □ ■ \dots □ ■ □ ■,

where the center

\underset{⏟}{□ ■}

means that both numbers □, ■ are

cos \frac{π}{2} = 0

. Thus we can see that for n even, all zeros of

T_{n} (z)

and

U_{n} (z)

form good pairs, and for n odd, all pairs from integral polynomials

T_{n} (z) / z

and

U_{n} (z) / z

are good. It follows that, by Fell [1], all zeros of the convex combination are real and in

(- 1, 1)

. □

Proof of Theorem 6 The zeros z of

P_{1, ϵ} (z)

satisfy

| \frac{T_{n, ϵ} (z)}{U_{n, ϵ} (z)} | = | z |^{k} .

| z | \geq 1

, then

| T_{n, ϵ} (z) | \geq | U_{n, ϵ} (z) |

, which contradicts (4). □

Proof of Proposition 7 Using

{| z |}^{2} = z \bar{z}

, 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}))

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})) .

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).

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Competing interests

The author declares that they have no competing interests.

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