Polynomials Associated with Reciprocation

Clark Kimberling Department of Mathematics University of Evansville 1800 Lincoln Avenue Evansville, IN 47722 USA

                      ck6@evansville.edu

Abstract Polynomials are defined recursively in various ways associated with reciprocation; e.g., S_{n+1}(x)/T_{n+1}(x) = S_{n}(x)/T_{n}(x) ± T_{n}(x)/S_{n}(x). Under certain conditions, the zeros of S_{n} interlace those of T_{n}. Identities for S_{n}, T_{n}, and related polynomials are derived, as well as recurrence relations and infinite sums involving roots of polynomials.

1 Introduction

A well-known problem [2] starts with the recurrence

x_{n+1} = x_{n} + 1/x_{n},  (1)

given that x_{0} = 1. The sequence (1, 2, 5, 29, …) thus determined is indexed in Sloane’s Online Encyclopedia of Integer Sequences [4] as \underline{A073833}. It is natural to ask what happens if the initial value 1 is replaced by an indeterminate x. The purpose of this paper is to respond to that question and related questions. For example, what if the recurrence is replaced by

x_{n+1} = x_{n} − 1/x_{n},  (2)

as in \underline{A127814}? The recurrence (2) leads to polynomials which are interesting because of the distribution of their zeros, as portended by the following definition: suppose V = (v_{1}, v_{2}, …, v_{m−1}) and W = (w_{1}, w_{2}, …, w_{m}) are lists of numbers satisfying

w_{1} < v_{1} < w_{2} < v_{2} < ⋯ < v_{m−1} < w_{m};

then V interlaces W.

Throughout this paper, except where otherwise stipulated, the letter c denotes an arbitrary nonzero complex number.

2 The recurrence S_{n+1}/T_{n+1} = (1/c)(S_{n}/T_{n} − T_{n}/S_{n})

Define polynomials S_{1} = S_{1}(x) = x, T_{1} = T_{1}(x) = 1, and

S_{n+1} = S_{n}^{2} − T_{n}^{2}, T_{n+1} = cS_{n}T_{n},  (3)

so that

S_{n+1}/T_{n+1} = (1/c)(S_{n}/T_{n} − T_{n}/S_{n}),  (4)

and for n ≥ 2,

T_{n} = c^{n−1}S_{1}S_{2} ⋯ S_{n−1},
S_{n+1} = (S_{n} − c^{n−1}S_{1}S_{2} ⋯ S_{n−1})(S_{n} + c^{n−1}S_{1}S_{2} ⋯ S_{n−1}).  (5)

If c = 1, then (4) is the recurrence (2) with x_{n} = S_{n}/T_{n}. For c > 0 and n ≥ 1, let Z_{n} be the list of zeros of S_{n} in increasing order, so that

Z_{1} = (0), Z_{2} = (−1, 1),

and Z_{3} is the ordered list consisting of the numbers (±c ± \sqrt{c^{2} + 4})/2; e.g., if c = 1, then Z_{3} = (−φ, −1/φ, 1/φ, φ), where φ = (1 + \sqrt{5})/2, the golden ratio.

If V and W are ordered lists, we shall employ the set-union symbol ∪ for the ordered union of the merged list formed by the numbers in V and W, thus: V ∪ W. Note that S_{n}

has degree 2^{n−1}, that T_{n} has degree 2^{n−1} − 1, and that ⋃_{k=1}^{n} Z_{k} lists the zeros of T_{n}.

Theorem 1. Suppose that c is a nonzero real number and n ≥ 1. Then ∪^{n}_{k=1}Z_{k} interlaces Z_{n+1}.

Proof. First, suppose that c > 0. Clearly Z_{1} interlaces Z_{2}. It will be helpful to denote the zeros of S_{n} in increasing order by r_{n,i} where i = 1, 2, …, 2^{n−1}. To complete a two-part first induction step, we shall show that Z_{1} ∪ Z_{2} interlaces Z_{3}. The function f_{2} = S_{2}/S_{1} is continuous and rises strictly from −∞ to ∞ on each of the intervals (−∞, r_{11}) and (r_{11}, ∞). Therefore there exist unique numbers r_{31}, r_{32} such that

r_{31} < r_{21} < r_{32} < r_{11}, f_{2}(r_{31}) = −c, f_{2}(r_{32}) = c  (6)

and r_{33}, r_{34} such that

r_{11} < r_{33} < r_{22} < r_{34}, f_{2}(r_{33}) = −c, f_{2}(r_{34}) = c.  (7)

Thus, r_{31}, r_{33} are the zeros of S_{2} + cS_{1}, and r_{32}, r_{34} are the zeros of S_{2} − cS_{1}. Now S_{3} = (S_{2} + cS_{1})(S_{2} − cS_{1}), so that (6) and (7) imply that Z_{1} ∪ Z_{2} interlaces Z_{3}. As a general

induction hypothesis, assume for n ≥ 3 that ∪^{n−2}_{k=1}Z_{k} interlaces Z_{n−1} and that ∪^{n−1}_{k=1}Z_{k} interlaces Z_{n}. Write

∪^{n−2}_{k=1}Z_{k} = (ρ_{11}, ρ_{12}, …, ρ_{1m}), where m = 2^{n−2} − 1,
∪^{n−1}_{k=1}Z_{k} = (ρ_{21}, ρ_{11}, ρ_{22}, ρ_{12}, …, ρ_{2m}, ρ_{1m,}ρ_{2,m+1}).

The function f_{n} = S_{n}/(S_{1}S_{2} ⋯ S_{n−1}) is continuous and strictly increasing on each of the intervals

(−∞, ρ_{11}), (ρ_{11}, ρ_{12}), …, (ρ_{1,m−1}, ρ_{1m}),  (8)

with infinite limits (as in the argument above for f_{2}.) Let ρ_{31}, ρ_{32} be the unique numbers satisfying

ρ_{31} < ρ_{21} < ρ_{32} < ρ_{11}, f_{n}(ρ_{31}) = −c, f_{2}(ρ_{32}) = c.

Likewise applying the intermediate value theorem to the remaining intervals in (8), we conclude that ∪^{n}_{k=1}Z_{k} interlaces Z_{n+1}.

Now suppose that c < 0. The recurrences (3) show that the only exponents of c that occur in the polynomials S_{n} are even. Therefore, these polynomials and their zeros are identical to those already considered.

Theorem 2. If n ≥ 1 and r ∈ Z_{n}, then (cr ± \sqrt{c^{2}r^{2} + 4})/2 ∈ Z_{n+1}.

Proof. The proposition clearly holds for n = 1. The zeros of S_{2} are −1 and 1; let r_{1} be either of these, and let r_{2} be any number satisfying r_{1} = (1/c)(r_{2} − 1/r_{2}), so that

r_{1} = \frac{r_{2}^{2} − 1}{cr_{2}} = \frac{S_{2}(r_{2})}{T_{2}(r_{2})}.

Taking r_{1} = −1 gives S_{2}(r_{2}) + T_{2}(r_{2}) = 0 and taking r_{1} = 1 gives S_{2}(r_{2}) − T_{2}(r_{2}) = 0. Consequently,

S_{3}(r_{2}) = (S_{2}(r_{2}) + T_{2}(r_{2}))(S_{2}(r_{2}) − T_{2}(r_{2})),

which implies that the 4 numbers r_{2} are the zeros of S_{3}. Continuing with arbitrary zeros r_{1}, r_{2} of S_{2}, let r_{3} be any number satisfying r_{2} = (1/c)(r_{3} − 1/r_{3}), so that r_{2} = S_{2}(r_{3})/T_{2}(r_{3}). Then

r_{1} = \frac{1}{c}(r_{2} − \frac{1}{r_{2}} ) = \frac{1}{c}(\frac{S_{2}(r_{3})}{cr_{3}} − \frac{cr_{3}}{S_{2}(r_{3})})
= \frac{S_{2}^{2}(r_{3}) − c^{2}r_{3}^{2}}{cS_{2}(r_{3})T_{2}(r_{3})} = \frac{S_{3}(r_{3})}{T_{3}(r_{3})}.

Taking r_{1} = −1 gives S_{3}(r_{3}) + T_{3}(r_{3}) = 0 and taking r_{1} = 1 gives S_{3}(r_{2}) − T_{3}(r_{3}) = 0, so that the 8 numbers r_{3} are the zeros of S_{4}. This inductive procedure shows that the zeros ρ of S_{n+1} arise from the zeros r of S_{n} by the rule r = (1/c)(ρ − 1/ρ), and solving this for ρ finishes the proof.

Theorem 3. If n ≥ 1, then

S_{n+1}(x) = (cx)^{2^{n−1}}S_{n}(\frac{x}{c} − \frac{1}{cx}).  (9)

More generally, if k = 2, 3, …, n + 1, then

S_{n+1} = T_{k}^{2^{n−k+1}}S_{n−k+2}(\frac{S_{k}}{T_{k}}).  (10)

Proof. First, we prove (9). The assertion clearly holds for n = 1. Assume that n > 1 and let m = 2^{n−1}. Denote the zeros of S_{n} by r_{k} for k = 1, 2, …, m. By Theorem 2,

S_{n+1}(x) = m ; ∏_{h=1}(x − \frac{cr_{h} − \sqrt{c^{2}r_{h}^{2} + 4}}{2})(x − \frac{cr_{h} + \sqrt{c^{2}r_{h}^{2} + 4}}{2})

= m ; ∏_{h=1}(x^{2} − cr_{h}x − 1)

= m ; ∏_{h=1}x(x − \frac{1}{x} − cr_{h})
= (cx)^{m}S_{n}(\frac{x}{c} − \frac{1}{cx}).

In case k = 2, equation (10) is essentially a restatement of (9), just proved. For k > 2, equations (10) will now be proved by induction on k. Assume for arbitrary k satisfying 2 ≤ k ≤ n that

S_{n+1} = T_{k−1}^{2^{n−k+2}}S_{n−k+3}(\frac{S_{k−1}}{T_{k−1}} ).  (11)

Substitute S_{k}/T_{k} for x in (9) with n replaced by n − k + 3 :

S_{n−k+3}(\frac{S_{k−1}}{T_{k−1}} ) = (\frac{cS_{k−1}}{T_{k−1}} )^{2^{n−k+1}}S_{n−k+2}(\frac{S_{k}}{T_{k}}),

so that by (11),

S_{n+1} = T_{k−1}^{2^{n−k+2}}(\frac{cS_{k−1}}{T_{k−1}} )^{2^{n−k+1}}S_{n−k+2}(\frac{S_{k}}{T_{k}})
= (cT_{k−1}S_{k−1})^{2^{n−k+1}}S_{n−k+2}(\frac{S_{k}}{T_{k}})
= T_{k}^{2^{n−k+1}}S_{n−k+2}(\frac{S_{k}}{T_{k}}).

Using Theorems 1-3, it is easy to establish the following properties of the polynomials and zeros:

1. S_{n}(−x) = S_{n}(x) for n ≥ 2, and if r ∈ Z_{n} then −r ∈ Z_{n} for n ≥ 1.

2. x^{2^{n−1}}S_{n}(1/x) = S_{n}(x) for n ≥ 3, and if r ∈ Z_{n} then 1/r ∈ Z_{n} for n ≥ 2.

3. Suppose that c > 0, and let r_{n} denote the greatest zero of S_{n}. Then (r_{n}) is a strictly increasing sequence, and

lim_{n→∞}r_{n} = \{ \begin{matrix}(1 − c)^{−1/2}, & if 0 < c < 1; \\ ∞, & if c ≥ 1. \end{matrix}

An outline of a proof follows. Of course, r_{n} = (cr_{n−1}+\sqrt{c^{2}r_{n−1}^{2} + 4})/2, so that (r_{n}) is strictly increasing. If 0 < c < 1, then, as is easily proved, r_{n} < (1 − c)^{−1/2} for all n, so that a limit r exists; since r = (cr + \sqrt{c^{2}r^{2} + 4})/2, we have r = (1 − c)^{−1/2}. On the other hand, supposing c ≥ 1, if (r_{n}) were bounded above, then the previous argument would give r = (1 − c)^{−1/2}, but this is not a real number. Therefore r_{n} → ∞.

4. If n ≥ 3, then

S_{n} = S_{n−1}^{2} + c^{2}S_{n−1}S_{n−2}^{2} − c^{2}S_{n−2}^{4}.  (12)

To prove this recurrence, from (5) we obtain both

c^{2n−4}(S_{1}S_{2} ⋯ S_{n−3})^{2}S_{n−2}^{2} = S_{n−1}^{2} − S_{n}

and

c^{2n−6}(S_{1}S_{2} ⋯ S_{n−3})^{2} = S_{n−2}^{2} − S_{n−1},

and (12) follows by eliminating (S_{1}S_{2} ⋯ S_{n−3})^{2}.

In the case c = 1, the recurrence (12) is used (e.g., [1]) to define the Gorškov-Wirsing polynomials, for which the initial polynomials are 2x − 1 and 5x^{2} − 5x + 1 rather than x and x^{2} − 1.

3 The polynomials V_{n} = S_{n} − c^{n−1}S_{1}S_{2} ⋯ S_{n−1}

Equation (5) shows that n ≥ 2, the polynomial S_{n+1} factors. We use one of factors to define a sequence of polynomials

V_{n} = V_{n}(x) = S_{n} − c^{n−1}S_{1}S_{2} ⋯ S_{n−1},  (13)

so that

S_{n}(x) = V_{n−1}(x)V_{n−1}(−x),  (14)
2S_{n}(x) = V_{n}(x) + V_{n}(−x).  (15)

Suppose that n ≥ 3. Substitute n − 1 for n in (13) and then multiply both sides of the result by cS_{n−1} to obtain

cV_{n−1}S_{n−1} = cS_{n−1}^{2} − c^{n−1}S_{1}S_{2} ⋯ S_{n−1},

so that by (13),

V_{n} = S_{n} − cS_{n−1}^{2} + cV_{n−1}S_{n−1},

and by (14), and (15) with n − 1 substituted for n,

V_{n}(x) = V_{n−1}(x)V_{n−1}(−x) − c(\frac{V_{n−1}(x) + V_{n−1}(−x)}{2})^{2}

+cV_{n−1}(x)\frac{V_{n−1}(x) + V_{n−1}(−x)}{2}.

Thus, we have the following recurrence for the polynomials V_{n} :

V_{n} = V_{n−1}(x)V_{n−1}(−x) + \frac{c}{4}[V_{n−1}^{2}(x) − V_{n−1}^{2}(−x)],

for n ≥ 3. Another recurrence for these polynomials stems directly from (9) and (14):

V_{n+1}(x) = (cx)^{2^{n−1}}V_{n}(\frac{x}{c} − \frac{1}{cx}),

for n ≥ 2.

For the remainder of this section, assume that c > 0. For n ≥ 2, let Z_{n}^{+} denote the ordered list of zeros of V_{n} Since Z_{n}^{+} ⊂ Z_{n}, we need only observe that in the rule given by Theorem 2 for forming zeros, the half of the numbers in Z_{n} that descend from r = 1 in Z_{1}^{+} are the numbers that comprise Z_{n}^{+}. That is, if n ≥ 2and r ∈ Z_{n}^{+} then

(cr ± \sqrt{c^{2}r^{2} + 4})/2 ∈ Z_{n+1}^{+}.

For n ≥ 3, no number in Z_{n}^{+} is rational, so that V_{n} is irreducible over the rational integers. Let r_{n} denote the greatest zero of S_{n}, and also of V_{n}. Then

r_{n+1} = (cr_{n} + \sqrt{c^{2}r_{n}^{2} + 4})/2,

and from this recurrence easily follows

r_{n+1} − 1/r_{n+1} = cr_{n},  (16)

of which the left-hand side is the distance from the least positive zero of S_{n+1} to the greatest.

4 The case c = 1

We turn now to the case that c = 1; that is, the immediate generalization of (2) to the case that the initial value is S_{1}(x) = x. The first four polynomials S_{n} and V_{n} are as shown here:

n S_{n}  V_{n}
1 x
2 x^{2} − 1  x − 1
3 x^{4} − 3x^{2} + 1  x^{2} − x − 1
4 x^{8} − 7x^{6} + 13x^{4} − 7x^{2} + 1 x^{4} − x^{3} − 3x^{2} + x + 1

Arrays of coefficients for S_{n} are indexed [4] as \underline{A147985} and \underline{A147990}, and for T_{n} as \underline{A147986}. The polynomials V_{n} are related by the equation V_{n}(x) = U_{n}(−x) to polynomials U_{n} presented at \underline{A147989}.

As mentioned in Section 2, the greatest zero r_{n} of S_{n} grows without bound as n → ∞. In order to discuss r_{n} in some detail, define

z(x) = (x + \sqrt{x^{2} + 4})/2,

so that r_{1} = 1 and r_{n} = z(r_{n−1}) for n ≥ 2. The sequence (r_{n}) has some interesting properties arising from (16). For example, if x is the positive number satisfying 1 + 1/x = x, then x = r_{2} = (1 + \sqrt{5})/2, and inductively, if x is the positive number satisfying

\frac{1}{r_{1}} + \frac{1}{r_{2}} + ⋯ + \frac{1}{r_{n}} + \frac{1}{x} = x,  (17)

then x = r_{n+1}. Equation (17) shows how the numbers r_{n} arise naturally without reference to polynomials. Since

r_{n+1} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + ⋯ + \frac{1}{r_{n}} + \frac{1}{r_{n+1}} ,

we have ∑^{∞} ; k=1 \frac{1}{r_{k}} = ∞.

Theorem 4. If n ≥ 1, and r_{n} is the greatest zero of S_{n}, then

\sqrt{2n} − 1 < r_{n+1} < \sqrt{2n + 1}.

Proof. Taking c = 1 in (16) and squaring give

\frac{1}{r_{n+1}^{2}} = r_{n}^{2} − r_{n+1}^{2} + 2,

whence

∑^{n+1} ; k=2 \frac{1}{r_{k}^{2}} = r_{1}^{2} − r_{n+1}^{2} + 2n,

r_{n+1}^{2} = 1 + 2n − ∑^{n+1} ; k=2 \frac{1}{r_{k}^{2}}
< 1 + 2n,

so that r_{n+1} < \sqrt{2n + 1}.

We turn next to an inductive proof that

\sqrt{2n} − 1 < r_{n+1}  (18)

for all n. This is true for n = 1, and we assume it true for arbitrary n and wish to prove that r_{n+2} > \sqrt{2n + 2} − 1. We begin with the easily proved inequality

4n^{2} + 2n + 1 + 4n\sqrt{2n} + 4n + 2\sqrt{2n} < (2n + 2\sqrt{2n} + 1)(2n + 2).

Taking the square root of both sides,

2n + \sqrt{2n} + 1 < (\sqrt{2n} + 1)(\sqrt{2n + 2}),

so that

2n < (\sqrt{2n} + 1)(\sqrt{2n + 2} − 1).

Expanding and adding appropriate terms to both sides,

2n − 2\sqrt{2n + 2} + 5 > 4(2n + 2) + 2n + 1 − 4\sqrt{2n}\sqrt{2n + 2} − 4\sqrt{2n + 2} + 2\sqrt{2n}.

Taking the square root of both sides,

\sqrt{2n − 2\sqrt{2n} + 5} > 2\sqrt{2n + 2} − \sqrt{2n} − 1.

Equivalently,

\sqrt{2n} − 1 + \sqrt{2n − 2\sqrt{2n} + 5} > 2\sqrt{2n + 2} − 2,

so that by the induction hypothesis (18),

r_{n+1} + \sqrt{2n − 2\sqrt{2n} + 5} > 2\sqrt{2n + 2} − 2.  (19)

The inequality (18), after squaring, adding 4, and taking square roots, gives

\sqrt{r_{n+1}^{2} + 4} > \sqrt{2n − 2\sqrt{2n} + 5.}

In view of (19), therefore,

r_{n+2} = \frac{r_{n+1} + \sqrt{r_{n+1}^{2} + 4}}{2} > \sqrt{2n + 2} − 1.

Theorem 4 implies that lim_{n→∞}r_{n}/\sqrt{n} = \sqrt{2} and that

\frac{1}{2n + 1} < \frac{1}{r_{n+1}^{2}} < \frac{1}{(\sqrt{2n} − 1)^{2}} .

Consequently,

∑^{∞} ; n=1 \frac{1}{r_{n}^{2}} = ∞, ∑^{∞} ; n=1 \frac{1}{nr_{n}} < ∞, and ∑^{∞} ; n=1 \frac{1}{2^{n−1}r_{n}} < ∞.

The second and third sums are approximately 2.26383447 and 1.518737247. For more digits of the latter, see \underline{A154310}.

5 The case c = 2

It is easy to prove that there is exactly one choice of c > 0 in (3) for which the resulting polynomial S_{n}(x) + iT_{n}(x) has the form

(x + a + bi)^{2^{n}}

for some real a and b and all n ≥ 2. The unique values are c = 2 and (a, b) = (0, 1). In this case, the first three polynomials S_{n}, T_{n}, V_{n} are as shown here:

n S_{n}  T_{n}  V_{n}
1 x  1
2 x^{2} − 1 2x x^{2} − 2x − 1
3 x^{4} − 6x^{2} + 1 4x^{3} − 4x x^{4} − 4x^{3} − 6x^{2} + 4x + 1

Arrays of coefficients for S_{n} and T_{n} are included as subarrays [4] of arrays closely associated with Pascal’s triangle. Specifically, for S_{n} see \underline{A096754}, \underline{A135670}, and \underline{A141665}; for T_{n}, see \underline{A095704} and \underline{A135685}. In the same way, coefficients for V_{n} can be read from A108086, modified in accord with the identity V_{n}(x) = U_{n}(−x).

The fact that the zeros of T_{n} interlace those of S_{n} is an example of Theorem 1. However, in this case, one can also appeal to the Hermite-Biehler theorem: if S and T are nonconstant polynomials with real coefficients, then the polynomials S and T have interlacing zeros if and only if all the zeros of the polynomial S + iT lie either in the upper half-plane or the lower half-plane. For a discussion of this theorem and related matters, see Rahman and Schmeisser [3, pp. 196–209].

6 The case c = 2i

Suppose that S_{n} is a square for some n, and write S_{n}(x) = H_{n}^{2}(x). Then

S_{n}(\frac{x}{c} − \frac{1}{cx}) = H_{n}^{2}(\frac{x}{c} − \frac{1}{cx}).

By (9),

S_{n+1}(x) = (cx)^{2^{n−1}}H_{n}^{2}(\frac{x}{c} − \frac{1}{cx}),  (20)

which implies that S_{n+1} is a square. It is easy to show that the only nonzero choices of c for which S_{3} is a square are ±2i. Equation (20) gives the recurrence

H_{n+1}(x) = (2ix)^{2^{n−2}}H_{n}(\frac{i}{2x} − \frac{ix}{2}),

which implies

|H_{n+1}(e^{iθ})| = 2^{deg H_{n}} |H_{n}(sin θ)|

for all real θ. Another recurrence, which follows readily from (12), is

H_{n} = 2H_{n−2}^{4} − H_{n−1}^{2}.

The first four of these polynomials are as follows:

H_{3}(x) = x^{2} + 1

H_{4}(x) = x^{4} − 6x^{2} + 1

H_{5}(x) = x^{8} + 20x^{6} − 26x^{4} + 20x^{2} + 1

H_{6}(x) = x^{16} − 88x^{14} + 92x^{12} − 872x^{10} + 1990x^{8} − 872x^{6} + 92x^{4} − 88x^{2} + 1

Coefficients for the polynomials H_{7} and H_{8} are given at \underline{A154308}.

7 The recurrence P_{n+1}/Q_{n+1} = (1/c)(P_{n}/Q_{n} + Q_{n}/P_{n})

We return now to the recurrence (1), with initial value x_{0} = P_{1} = P_{1}(x) = x. Taking Q_{1} = Q_{1}(x) = 1 leads to sequences P_{n} and Q_{n} defined by

P_{n} = P_{n−1}^{2} + cQ_{n−1}^{2} and Q_{n} = cP_{n−1}Q_{n−1}.

The properties of these polynomials are analogous to those of the polynomials S_{n} and T_{n} already discussed. Indeed,

P_{n}(x) = S_{n}(ix)  (21)

for n ≥ 2, so that the zeros of P_{n} are ir, where r ranges through Z_{n}, and, if c > 0, we have interlaced lists of zeros on the imaginary axis. The recurrence (12) holds without change; that is, for n ≥ 3, we have

P_{n} = P_{n−1}^{2} + P_{n−1}P_{n−2}^{2} − P_{n−2}^{4}.

Putting x = ie^{iθ} in (9) and applying (21) lead to

|P_{n+1}(e^{iθ})| = |c|^{deg P_{n}} |P_{n}(\frac{2}{c} cos θ)|

for all real θ.

For c = 1, coefficient arrays are given for the polynomials P_{n} and Q_{n} are indexed as \underline{A147987} and \underline{A147988}, respectively.

8 Concluding remarks

The author is grateful to a referee for pointing out various properties associated with polynomials discussed in this paper — properties which may be worth further study. For example, the interlacing of zeros in Theorem 1 implies that for fixed n, the polynomials T_{n} and S_{n} are consecutive members of some sequence of orthogonal polynomials. A consequence of Theorem 2 is that there exists Euclidean straightedge-and-compass constructions for the zeros of T_{n} and S_{n}. The manner in which S_{n+1} arises from the argument of S_{n} in (9) is similar to the Joukowski transform. Indeed (9) can be written as

e^{−i2^{n−1}θ}S_{n+1}(e^{iθ}) = c^{2^{n−1}}S_{n}(\frac{2i}{c} sin θ),

so that, apart from a constant, the modulus of S_{n+1} on the unit circle is the modulus of S_{n} on a line segment. This has consequences for estimates, such as the derivative estimates of the Bernstein-Markov type.

References

[1] H. L. Montgomery, Ten Lectures on the Interface between Analytic Number Theory and Harmonic Analysis, CBMS Regional Conference Series in Mathematics, 84. Published for the American Mathematical Society, 1994, page 187.

[2] D. J. Newman, A Problem Seminar. Problem Books in Mathematics, Springer-Verlag, 1982. Problem 60.

[3] Q. Rahman and G. Schmeisser, Analytic Theory of Polynomials, Oxford University Press, 2002.

[4] N. J. A. Sloane, On-Line Encyclopedia of Integer Sequences,

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

2000 Mathematics Subject Classification: Primary 26C10; Secondary 26C15. Keywords: polynomial, interlacing zeros.

(Concerned with sequences \underline{A073833}, \underline{A095704}, \underline{A096754}, \underline{A108086}, \underline{A127814}, \underline{A135670}, \underline{A135685}, \underline{A141665}, \underline{A147985}, \underline{A147986}, \underline{A147987}, \underline{A147988}, \underline{A154308}, \underline{A154310}.)

Received January 11, 2009; revised version received March 18 2009. Published in Journal of Integer Sequences, March 20 2009.

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