On a Class of Polynomials with Integer Coefficients

Milan Janjić Department of Mathematics and Informatics University of Banja Luka Republic of Srpska, Bosnia and Herzegovina

                    agnus@blic.net

Abstract

We define a certain class of polynomials denoted by P_{n,m,p}(x), and give the combinatorial meaning of the coefficients. Chebyshev polynomials are special cases of P_{n,m,p}(x). We first show that P_{n,m,p}(x) can be expressed in terms of P_{n,0,p}(x). From this we derive that P_{n,2,2}(x) can be obtained in terms of trigonometric functions, from which we obtain some of its important properties. Some questions about orthogonality are also addressed. Furthermore, it is shown that P_{n,2,2}(x) fulfills the same threeterm recurrence as the Chebyshev polynomials. We also obtain some other recurrences for P_{n,m,p}(x) and its coefficients. Finally, we derive a formula for the coefficients of Chebyshev polynomials of the second kind.

1 Introduction

In the paper [1] I proved the following result:

Theorem 1. If a finite set X consists of n blocks of size p and an additional block of size m then, for n ≥ 0, k ≥ 0, the number f(n, k, m, p) of (n + k)-subsets of X intersecting each block of size p is

f(n, k, m, p) = ∑_{i=0}^{n}(−1)^{i}\binom{n}{i}\binom{np + m − ip}{n + k}.

I also proved the following relations for the function f [1]:

f(n, k, m, p) = ∑_{i=0}^{m} \binom{m}{i}f(n, k − i, 0, p),  (1)

f(n, k, m, p) = ∑_{i=0}^{t}(−1)^{i}\binom{t}{i}f(n, k + t, m + t − i, p),  (2)

f(n, k, m, p) = ∑_{i=1}^{p} \binom{p}{i}f(n − 1, k − i + 1, m, p),  (3)

f(n, k, m, p) = ∑_{i=0}^{n} ∑_{j=0}^{i} \binom{n}{i}\binom{i}{j}f(n − j, k − i + j, m, p − 1).  (4)

Furthermore, I showed that (−1)^{k}f(n, k, 0, 2) is the coefficient of the Chebyshev polynomial U_{n+k}(x) by x^{n−k}, and that (−1)^{k}f(n, k, 1, 2) is the coefficient of the Chebyshev polynomial T_{n+k−1}(x) by x^{n−k+1}.

Definition 2. We define the set of coefficients

{c(n, k, m, p) : n = m, m + 1, …; k = 0, 1, …, n}

such that

c(n, k, m, p) = (−1)^{\frac{n−k}{2}} f (\frac{n + k − 2m}{2}, \frac{n − k}{2} , m, p) ,

if n and k are of the same parity, and c(n, k, m, p) = 0 otherwise. The polynomials P_{n,m,p}(x) are defined to be

P_{n,m,p}(x) = ∑_{k=0}^{n} c(n, k, m, p)x^{k}.

Example 3. The Chebyshev polynomials are particular cases of P_{n,m,p}(x), obtained for m = 1, p = 2 and m = 0, p = 2, that is,

U_{n}(x) = P_{n,0,2}(x), T_{n}(x) = P_{n,1,2}(x).

The polynomial P_{n,2,2}(x) is the closest to the Chebyshev polynomials, and will be denoted simply by P_{n}(x).

In the next table we state the first few values of P_{n}(x).

x^{2}
2x^{3} − 2x
4x^{4} − 5x^{2} + 1
8x^{5} − 12x^{3} + 4x
16x^{6} − 28x^{4} + 13x − 1
32x^{7} − 64x^{5} + 38x^{3} − 6x.

Among coefficients of the above polynomials the following sequences from [2] appear: \underline{A024623}, \underline{A049611}, \underline{A055585}, \underline{A001844}, \underline{A035597}.

Triangles of coefficients for P_{n,m,2}(x), ( m = 2, 3, 4, 5, 6) are given in \underline{A136388}, \underline{A136389}, \underline{A136390}, \underline{A136397}, and \underline{A136398}, respectively.

2 Reduction to the case m = 0.

We shall first prove an analog of the formula (1) for polynomials. Theorem 4. The following equation is fulfilled:

P_{n,m,p}(x) = ∑_{i=0}^{m}(−1)^{i}\binom{m}{i}x^{m−i}P_{n−m−i,0,p}(x).

Proof. The following equation holds:

P_{n,m,p}(x) = ∑_{k=0}^{n}(−1)^{\frac{n−k}{2}} f (\frac{n + k − 2m}{2}, \frac{n − k}{2} , m, p) x^{k}.

Using (1) one obtains

P_{n,m,p}(x) = ∑_{k=0}^{n} ∑_{i=0}^{m}(−1)^{\frac{n−k}{2}} \binom{m}{i}f (r, s, 0, p) x^{k}.

where

r = \frac{n + k − 2m}{2}, s = \frac{n − k}{2} − i.

Changing the order of summation yields

P_{n,m,p}(x) = ∑_{i=0}^{m} \binom{m}{i}x^{m−i} ∑_{k=0}^{n}(−1)^{\frac{n−k}{2}} f (r, s, 0, p) x^{k−m+i}.

Terms in the sum on the right side of the preceding equation produce nonzero coefficients only in the case 0 ≤ s ≤ r, that is,

m − i ≤ k ≤ n − 2i.

It follows that

P_{n,m,p}(x) = ∑_{i=0}^{m}(−1)^{i}\binom{m}{i}x^{m−i} ∑_{k=m−i}^{n−2i}(−1)^{\frac{n−k}{2} −i}f (r, s, 0, p) x^{k−m+i}.

Denoting k − m + i = j we obtain

P_{n,m,p}(x) = ∑_{i=0}^{m}(−1)^{i}\binom{m}{i}x^{m−i} ∑_{j=0}^{n−m−i} c(n − m − i, j, 0, p)x^{j},

which means that

P_{n,m,p}(x) = ∑_{i=0}^{m}(−1)^{i}\binom{m}{i}x^{m−i}P_{n−m−i,0,p}(x).

According the the preceding theorem we can express P_{n}(x) in terms of Chebyshev polynomials of the second kind. Namely, for m = 2, n ≥ 4 we have

P_{n}(x) = x^{2}U_{n−2}(x) − 2xU_{n−3}(x) + U_{n−4}(x).  (5)

This allow us to express P_{n}(x) in terms of trigonometric functions.

Theorem 5. For each n ≥ 3 we have

P_{n}(cos θ) = − sin θ sin(n − 1)θ.  (6)

Proof. According to (5) and well-known property of Chebyshev polynomials we obtain

sin θP_{n}(cos θ) = cos^{2} θ sin(n − 1)θ − 2 cos θ sin(n − 2)θ + sin(n − 3)θ.

From the identity

2 cos θ sin(n − 2)θ = sin(n − 1)θ + sin(n − 3)θ

follows

sin θP_{n}(cos θ) = cos^{2} θ sin(n − 1)θ − sin(n − 1)θ = − sin^{2} θ sin(n − 1)θ.

Dividing by sin θ ≠ 0 we prove the theorem.

Note that this proof is valid for n ≥ 4. The case n = 3 can be checked directly.

In the following theorem we prove that P_{n}(x) have the same important property concerning zeroes as Chebyshev polynomials do.

Theorem 6. For n ≥ 3, the polynomial P_{n}(x) has all simple zeroes lying in the segment [−1, 1].

Proof. Since

U_{n}(1) = n + 1, U_{n}(−1) = (−1)^{n}(n + 1)

the equation (5) implies

P_{n}(1) = U_{n−2}(1) − 2U_{n−3}(1) + U_{n−4}(1) = n − 1 − 2(n − 2) + n − 3 = 0,

and

P_{n}(−1) = U_{n−2}(−1) + 2U_{n−3}(−1) + U_{n−4}(−1) =
= (−1)^{n−2}(n − 1) + 2(−1)^{n−3}(n − 2) + (−1)^{n−4}(n − 3) = 0.

Thus, x = −1 and x = 1 are zeroes of P_{n}(x). The remaining n − 2 zeroes are obtained from the equation

sin(n − 1)θ = 0,

and they are

x_{k} = cos \frac{kπ}{n − 1}, (k = 1, 2, …, n − 2).

We shall now state an immediate consequence of (6) which shows that values of P_{n}(x), (x ∈ [−1, 1]) lie inside the unit circle.

Corollary 7. For n ≥ 3 and x ∈ [−1, 1] we have

P_{n}(x)^{2} + x^{2} ≤ 1.

Example 8. Dividing P_{n}(x) by 2^{n−2} we obtain a polynomial with the leading coefficient 1. Thus, its sup norm on [−1, 1] is ≤ \frac{1}{2^{n−2}} , which means that \frac{1}{2^{n−2}} P_{n}(x) has at most 2 times greater sup norm, comparing with the sup norm of T_{n}(x), which is minimal.

Taking the derivative in the equation (6) we obtain the following equation for extreme points of P_{n}(x) :

(n − 1) tan θ + tan(n − 1)θ = 0.

The values θ = 0, and θ = π obviously satisfied this equation, which implies that endpoints x = −1 and x = 1 are extreme points. The remaining extreme points of P_{3}(x) are x = arctan \sqrt{2} and x = − arctan \sqrt{2}.

3 Orthogonality

In this section we investigate the set {P_{n}(x) : n = 2, 3, 4, …} concerning to the problem of orthogonality, with respect to some standard Jacobi’s weights.

The first result is for the weight \frac{1}{\sqrt{1−x^{2}}} of Chebyshev polynomials of the first kind.

So, when n = m = 2, ∫_{−1}^{1} \frac{[P_{n}(x)]^{2}}{\sqrt{1−x^{2}}} dx = \frac{3}{8}π.

Theorem 9. The following equation holds:

∫_{−1}^{1} \frac{P_{n}(x)P_{m}(x)}{\sqrt{1 − x^{2}}}dx = \begin{cases}\frac{π}{4}, & if m = n > 2; \\ −\frac{π}{8}, & if |n − m| = 2; \\ 0, & otherwise. \end{cases}

Proof. Puting x = cos θ implies

I = ∫_{−1}^{1} \frac{P_{n}(x)P_{m}(x)}{\sqrt{1 − x^{2}}}dx = ∫_{0}^{π} P_{n}(cos θ)P_{m}(cos θ)dθ.

Using (6) we obtain

I = ∫_{0}^{π} sin^{2} θ sin(n − 1)θ sin(m − 1)θdθ.

Transforming the integrating function we obtain

sin^{2} θ sin(n − 1)θ sin(m − 1)θ = \frac{1}{4} cos(n − m)θ − \frac{1}{4} cos(n + m − 2)θ−
−\frac{1}{8} cos(n − m − 2)θ − \frac{1}{8} cos(n − m + 2)θ + \frac{1}{8} cos(n + m − 4)θ + \frac{1}{8} cos(n + m)θ.

Taking into account that m, n ≥ 3 we conclude that integrals of the terms on the right side of the preceding equation are zero if n ≠ m and |n − m| ≠ 2. If n = m we obtain I = \frac{π}{4}, and I = −\frac{π}{8}, if |n − m| = 2.

Corollary 10. Each subset of the set {P_{n}(x) : n ≥ 3}, not containing polynomials P_{k}(x) and P_{m}(x) such that |k − m| = 2 and k ≠ m is orthogonal.

The next result concerns the weight \sqrt{1 − x^{2}} of Chebyshev polynomials of the second kind. The result is similar to the result of the preceding theorem.

First of all we have

∫_{−1}^{1} \sqrt{1 − x^{2}}P_{n}(x)P_{m}(x)dx = \begin{cases}\frac{5π}{32}, & if (n, m) = (3, 3); \\ −\frac{π}{32}, & if (n, m) = (2, 4) or (4, 2). \end{cases}

Theorem 11. For m, n such that (m, n) ≠ (2, 4) and (4, 2) we have

∫_{−1}^{1} \sqrt{1 − x^{2}}P_{n}(x)P_{m}(x)dx = \begin{cases}\frac{3π}{16}, & if m = n > 3; \\ −\frac{π}{8}, & if |n − m| = 2; \\ \frac{π}{32}, & if |n − m| = 4; \\ 0, & otherwise. \end{cases}

Proof. In this case we have

∫_{−1}^{1} \sqrt{1 − x^{2}}P_{n}(x)P_{m}(x)dx = ∫_{0}^{π} sin^{2} θP_{n}(cos θ)P_{m}(cos θ)dθ.

We therefore need to calculate the integral

∫_{0}^{π} sin^{4} θ sin(n − 1)θ sin(m − 1)θdθ.

We have

sin^{4} θ sin(n − 1)θ sin(m − 1)θ = \frac{3}{16} cos(n − m)θ − \frac{3}{16} cos(n + m − 2)θ+

+\frac{1}{32} cos(n − m − 4)θ + \frac{1}{32} cos(n − m + 4)θ − \frac{1}{32} cos(n + m − 6)θ − \frac{1}{32} cos(n + m + 2)θ−

−\frac{1}{8} cos(n − m − 2)θ − \frac{1}{8} cos(n − m + 2)θ + \frac{1}{8} cos(n + m − 4)θ + \frac{1}{8} cos(n + m)θ.

The integral of each term on the right side with m ≠ n, |m − n| ≠ 2, |n − m| ≠ 4 is zero. For these particular values we easily obtain the desired result.

Taking, for instance, the weight (1 − x^{2})^{\frac{3}{2}} in a similar way one obtains

∫_{−1}^{1}(1 − x^{2})^{\frac{3}{2}} P_{n}(x)P_{m}(x)dx = \begin{cases}\frac{7π}{64}, & if (m, n) = (3, 3); \\ \frac{21π}{128}, & if (m, n) = (4, 4); \\ −\frac{π}{128}, & if (m, n) = (4, 2) or (2, 4); \\ −\frac{7π}{64}, & if (m, n) = (3, 5) or (5, 3). \end{cases}

For m, n such that (m, n) ≠ (2, 4), (4, 2), (3, 5), and (5, 3) we have

∫_{−1}^{1}(1 − x^{2})^{\frac{3}{2}} P_{n}(x)P_{m}(x)dx = \begin{cases}\frac{5π}{32}, & if m = n > 4; \\ −\frac{15π}{128}, & if |n − m| = 2; \\ \frac{3π}{64}, & if |n − m| = 4; \\ −\frac{π}{128}, & if |n − m| = 6; \\ 0, & otherwise. \end{cases}

Considering the weight 1 leads to the following result: Theorem 12. If m and n are of different parity then

∫_{−1}^{1} P_{n}(x)P_{m}(x)dx = 0.

Proof. In this case we need to calculate the integral

∫_{0}^{π} sin^{3} θ sin(n − 1)θ sin(m − 1)θdθ.

We have

sin^{3} θ sin(n − 1)θ sin(m − 1)θ = −\frac{1}{16} sin(n − m + 3)θ + \frac{1}{16} sin(n − m − 3)θ+

+\frac{1}{16} sin(n + m + 1)θ − \frac{1}{16} sin(n + m − 5)θ + \frac{3}{16} sin(n − m + 1)θ−

−\frac{3}{16} sin(n − m − 1)θ − \frac{3}{16} sin(n + m − 1)θ + \frac{1}{16} sin(n + m − 3)θ.

Since m and n are of different parity each function on the right is of the form sin 2kθ, which implies that its integral equals zero.

4 Some recurrence relations

In this section we prove some recurrence relation for P_{n,m,p}(x) as well as some recurrence relations for their coefficients. Theorem 13. For each integer t ≥ 0 we have

P_{n,m,p}(x) = ∑_{i=0}^{t}(−1)^{t−i}\binom{t}{i}x^{i}P_{n+2t−i,m+t−i,p}(x).

Proof. Translating (2) into the equation for coefficients we obtain

c(n, k, m, p) = ∑_{i=0}^{t}(−1)^{i+t}\binom{t}{i}c(n + 2t − i, k − i, m + t − i, p).

Multiplying by x^{k} yields

c(n, k, m, p)x^{k} = ∑_{i=0}^{t}(−1)^{i+t}\binom{t}{i}x^{i}c(n + 2t − i, k − i, m + t − i, p)x^{k−i},

which easily implies the claim of the theorem.

In the case t = 1, m = 1, p = 2 we obtain the following formula, expressing P_{n}(x) in terms of Chebyshev polynomials of the first kind:

P_{n}(x) = xT_{n−1}(x) − T_{n−2}(x).

From this we easily conclude that P_{n}(x) satisfies the same three-term recurrence as Chebyshev polynomials.

Corollary 14. The polynomials P_{n,m,2}(x) satisfy the following equation:

P_{n,m,2}(x) = 2xP_{n−1,m,2}(x) − P_{n−2,m,2}(x),

with initial conditions

P_{0,m,2}(x) = x^{m}, P_{1,m,2}(x) = 2x^{m+1} − mx^{m−1}.

Combining equations (1) and (4) we obtain

f(n, k, m, p) = ∑_{i=0}^{n} ∑_{j=0}^{i} ∑_{t=0}^{m} \binom{n}{i}\binom{m}{t}\binom{i}{j}f(n − j, k − i + j − t, 0, p − 1).

Translating this equation into an equation for the coefficients, we obtain

c(n, k, m, p) = ∑_{i=0}^{ñ} ∑_{j=0}^{i} ∑_{t=0}^{m} \binom{ñ}{i}\binom{m}{t}\binom{i}{j}(−1)^{i−j+t}c(n − m − i − t, k − m + i − 2j + t, 0, p − 1),

where ñ = \frac{n+k−2m}{2}.

Applying the preceding equation several times we obtain the following:

Corollary 15. The coefficients of P_{n,m,p}(x) can be expressed as functions of coefficients of Chebyshev polynomials of the second kind.

Converting (3) into the equation for coefficients we obtain

c(n, k, m, p) = ∑_{i=1}^{p}(−1)^{i−1}\binom{p}{i}c(n − i, k + i − 2, m, p).

This equation implies the following:

Corollary 16. The coefficients of P_{n,m,p}(x) can be expressed in terms of coefficients of the polynomials P_{n′,m,p}(x), where n′ < n.

We shall finish the paper with a formula for coefficients of Chebyshev polynomials of the second kind. Taking p = 2 in (4) we obtain

f(n, k, m, 2) = ∑_{i=0}^{n} ∑_{j=0}^{i} \binom{n}{i}\binom{i}{j}f(n − j, k − i + j, m, 1).

Since f(r, s, m, 1) = \binom{m}{s} we have

f(n, k, m, 2) = ∑_{i=0}^{n} ∑_{j=0}^{i} \binom{n}{i}\binom{i}{j}\binom{m}{k − i + j}.

For m = 0, in the sum on the right side of this equation only terms with k = i − j remains. We thus obtain

f(n, k, 0, 2) = ∑_{s=0}^{n−k} \binom{n}{s}\binom{n − s}{k}.

Accordingly, the following formula follows.

Corollary 17. For coefficients c(n, k) of Chebyshev polynomial U_{n}(x) hold

c(n, k) = (−1)^{\frac{n−k}{2}} ∑_{i=0}^{k} \binom{\frac{n+k}{2}}{i}\binom{\frac{n+k}{2} − i}{\frac{n−k}{2}},

if n i k are of the same parity and c(n, k) = 0 otherwise.

References

[1] M. Janjić, An enumerative function, http://arxiv.org/abs/0801.1976.

[2] N. J. Sloane, The Encyclopedia of Integer Sequences, published electronically at

      http://www.research.att.com/∼njas/sequences/

2000 Mathematics Subject Classification: Primary 05A10; Secondary 33C99. Keywords: Chebyshev polynomials, binomial coefficients, recurrence relations.

(Concerned with sequences \underline{A001844}, \underline{A024623}, \underline{A035597}, \underline{A049611}, \underline{A055585}, \underline{A136388}, \underline{A136389}, \underline{A136390}, \underline{A136397}, \underline{A136398}.)

Received April 12 2008; revised version received October 29 2008; December 7 2008. Published in Journal of Integer Sequences, December 11 2008.

Return to Journal of Integer Sequences home page.