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On the zeros of some combinatorial polynomials
June 26, 2013
Toufik Mansour
Department of Mathematics, University of Haifa, 31905 Haifa, Israel
tmansour@univ.haifa.ac.il
Mark Shattuck
Department of Mathematics, University of Tennessee, Knoxville, TN 37996
shattuck@math.utk.edu
Abstract
Let an(q) denote the distribution on the set of involutions of size nfor the statistic which
records the number of fixed points. We show for a range of qvalues that the polynomial
Pn
i=0 ai(q)xialways has the smallest possible number of real zeros, that is, none when the
degree is even and one when the degree is odd. On the way, we show that the sequence
an(q) is log-convex for all q≥1. Our proof in the case q= 1 is elementary, while the
proof for the general case makes use of programming to estimate the zeros of some related
analytic functions in q. Furthermore, the sequence of real zeros obtained from the odd
case is shown to be monotonically convergent. We also consider the polynomial Pn
i=0 dixi,
where didenotes the number of derangements of an i-element set, and show that the same
holds for it. Our results extend recent ones concerning Fibonacci and Tribonacci coefficient
polynomials.
Keywords: zeros of polynomials, linear recurrences, involution, derangement.
2010 Mathematics Subject Classification: 11C08, 13B25.
1 Introduction
Garth, Mills, and Mitchell [1] considered the Fibonacci coefficient polynomial defined by pn(x) =
F1xn+F2xn−1+···+Fnx+Fn+1 and showed that it has no real zeros if nis even and exactly
one real zero if nis odd. This result was later extended by M´aty´as [4, 5] to polynomials whose
coefficients are given by more general second order recurrences (having constant coefficients).
M´aty´as and Szalay [6] showed that the same also holds true for the Tribonacci coefficient poly-
nomials qn(x) = T2xn+T3xn−1+···+Tn+1x+Tn+2 , a result which has been extended to
k-Fibonacci [2] and generalized Tribonacci [3] coefficient polynomials.
One might wonder for what other combinatorial sequences vidoes this result hold for the
polynomial f(x) = Pn
i=0 vixi. Here, we consider this question in two cases when the sequence
vicounts (i) derangements of size i(i.e., permutations all of whose cycles have length two or
more), or (ii) involutions of size i(i.e., permutations all of whose cycles have length one or two).
Let dndenote the sequence defined by
dn= (n−1)(dn−1+dn−2), n ≥2,
1
with d0= 1 and d1= 0. Note that dncounts the number of derangements of an n-element set.
See, for example, A000166 in [7]. In the next section, we show that the polynomial
d(x) =
n
X
i=0
dixi
always has the smallest possible number of real zeros and that the sequence of zeros in the odd
case is monotonically convergent.
Let an(q) denote the sequence of polynomials defined by
an(q) = qan−1(q) + (n−1)an−2(q), n ≥2,
with a0(q) = 1 and a1(q) = q. Note that an(q) gives the distribution for the statistic on the set
of involutions of size nwhich records the number of fixed points; see, e.g., A099174 in OEIS [7].
The special cases q= 1 and q= 2 of an(q) correspond, respectively, to sequences A000085 and
A005425 in [7] and occur in several settings in enumerative combinatorics.
In this paper, we will show that the polynomial a(x;q) defined by
a(x;q) =
n
X
i=0
ai(q)xi
has one real zero when the degree is odd and no real zeros when the degree is even for all qin
the interval [1, λ], where λ≈6.276. Furthermore, we show that the sequence of real zeros in the
odd case is increasing and convergent to zero. We first prove the case q= 1 and then extend the
arguments to obtain the result for all qin [1, λ], wherein we use programming (such as Maple)
to estimate real zeros of some analytic functions in qwhich are needed in our proof. We remark
that [1, λ] seems to be the largest interval in qfor which the technique presented here applies.
We note that the sequences under consideration in the current paper are given by second order
linear recurrences, but with variable instead of constant coefficients. This will require a different
technique than that used in earlier papers on Fibonacci coefficient polynomials and their rela-
tives. Instead of multiplying f(x) = Pn
i=0 vixiby a characteristic polynomial to transform it
into another polynomial whose coefficients are mostly zero (as was done in [1] and in subsequent
papers), we first apply a linear operator of a differential nature to f. This yields a first order
differential equation for fwhich can then be used to express it in a more convenient integral
form. In the case of vi=ai(q), which is considered in the third section, we will also need some
algebraic properties of these polynomials in order to complete our argument using an integral
form of a(x;q) (see Lemma 3.2 below).
2 Derangement polynomials
We consider in this section the real zeros of the polynomial d(x) = Pn
i=0 dixi. We first will need
the following integral representation of d(x).
Lemma 2.1. Let −1< x0<0be fixed. Then we have
d(x) = Rx
x0
v(t)
t(1+t)e1/tdt +c
|x|e1/x ,−1< x < 0,(1)
where v(t) = −(n+ 1)dntn+2 −dn+1tn+1 + 1 and c=|x0|d(x0)e1/x0.
Proof. First observe that
(1 −x2)d(x)−x2d′(x)−x3d′(x) = −(n+ 1)dnxn+2 −dn+1xn+1 + 1.(2)
2
To see this, note that the coefficient of xion the left-hand side of (2) is zero if 2 ≤i≤n, by
the recurrence for dn, with the coefficient of xn+1 given by −ndn−ndn−1=−dn+1. Rewriting
(2) as
d′(x) + x−1
x2d(x) = −v(x)
x2(1 + x),−1< x < 0,
and solving this first-order linear differential equation by the usual method, gives (1).
We now prove the main result of this section.
Theorem 2.2. If n≥2, then the polynomial Pn
i=0 dixihas exactly one real zero if nis odd
and no real zeros if nis even.
Proof. Clearly, the polynomial d(x) = Pn
i=0 dixihas no positive zeros. If x≤ −1 and n≥5 is
odd, then
d(x) =
n
X
i=0
dixi= 1 +
n−1
2
X
i=1
(d2i+1x2i+1 +d2ix2i)
= 2x3+x2+ 1 +
n−1
2
X
i=2
(d2i+1x2i+1 +d2ix2i)<0,
being the sum of non-positive terms at least one of which is negative, since d2i+1x2i+1 <−d2ix2i
if i≥2 and 2x3+x2+ 1 ≤0. If n= 3, then d(x) = 2x3+x2+ 1, which has a single real zero
at x=−1. If x≤ −1 and nis even, then
d(x) =
n
X
i=0
dixi= 1 +
n
2
X
i=1
(d2i−1x2i−1+d2ix2i)>0.
So we consider the zeros of d(x) for −1< x < 0. To do so, we write d(x) = j(x)
|x|e1/x , where j(x) is
the numerator in (1). Note that the polynomial v(t) = −(n+ 1)dntn+2 −dn+1tn+1 +1 has a zero
at t=−1, by the identity dn+1 = (n+ 1)dn+ (−1)n+1 (see [8, p. 67]), whence the integrand is
bounded on the interval (−1,0).
Now assume n≥5 is odd. By Descartes’ rule of signs, the polynomial v(x) has two negative
zeros. Since v(x) achieves its minimum value over (−∞,0) at x=−dn+1
(n+2)dn, with
−dn+1
(n+ 2)dn
=−(n+ 1)dn+ (−1)n+1
(n+ 2)dn
>−1,
we see that the other negative zero of v(x), which we’ll denote by r, occurs in the interval
(−1,0). Note that v(x)<0 if −1< x < r and v(x)>0 if r < x < 0.
By the continuity of d(x), as d(0) >0, we may choose x0, with r < x0<0, such that d(x)>0
on [x0,0]. Note that
j′(x) = v(x)
x(1 + x)e1/x,
which is positive for −1< x < r and negative for r < x < 0. Since d(x0)>0, it follows that j(x)
is positive on the interval [r, x0), and thus so is d(x). Therefore, d(x)>0 on [r, 0). Furthermore,
d(−1) <0 since n≥5 is odd, which implies j(−1) <0, and thus j(x) has a single zero on the
interval (−1, r) since it is increasing there with j(r)>0. It follows that d(x) has a single zero
on the interval (−1, r) and thus on (−1,0). Furthermore, this zero is seen to have multiplicity
one. (See graphs of j(x) and v(x) below when n= 3,5,7).
3
Now assume nis even. Then v(x) has single negative zero at x=−1, with v(x)>0 if
−1< x < 0. Then j′(x)<0 if −1< x < 0, with j(x0)>0, where x0is chosen as in the odd
case above. Thus, j(x)>0 for all x∈(−1, x0], which implies d(x)>0 for all xin this interval.
Therefore, d(x)>0 if −1< x < 0, which implies d(x) has no real zeros when nis even.
–0.1
–0.05
0
0.05
0.1
0.15
0.2
–1 –0.8 –0.6 –0.4 –0.2
x
–0.05
–0.04
–0.03
–0.02
–0.01
0
0.01
0.02
0.03
–0.6 –0.5 –0.4 –0.3 –0.2 –0.1
x
–0.05
–0.04
–0.03
–0.02
–0.01
0
0.01
0.02
0.03
–0.4 –0.3 –0.2 –0.1
x
Graph of j3(x) Graph of j5(x) Graph of j7(x)
–1
–0.8
–0.6
–0.4
–0.2
0
0.2
0.4
0.6
0.8
1
–1.2 –1 –0.8 –0.6 –0.4 –0.2
x
–2
–1
0
1
2
–1.2 –1 –0.8 –0.6 –0.4 –0.2
x
–14
–12
–10
–8
–6
–4
–2
0
2
4
–1.2 –1 –0.8 –0.6 –0.4 –0.2
x
Graph of v3(x)/(1 + x2) Graph of v5(x)/(1 + 10x2) Graph of v7(x)/(1 + 60x2)
3 Involution polynomials
We will need the log-convexity of the polynomials an(q).
Lemma 3.1. If q≥1, then
an−1(q)an+1(q)> a2
n(q), n ≥3.(3)
Proof. Here, we will denote an(q) by an. To prove (3), we will show by induction the following
pair of inequalities: n+ 1
na2
n> an−1an+1 > a2
n, n ≥3.(4)
The n= 3 case of (4) is easily verified. We first prove the right-hand inequality in (4) with n
replaced by n+ 1. If n≥3, then anan+2 > a2
n+1 if and only if
an(qan+1 + (n+ 1)an)> a2
n+1,
i.e.,
(n+ 1)a2
n> an+1(an+1 −qan) = nan−1an+1,
which holds by the induction hypothesis.
For the left-hand inequality in (4), first note that n+2
n+1 a2
n+1 > anan+2 if and only if
n+ 2
n+ 1a2
n+1 > an(qan+1 + (n+ 1)an),
i.e.,
(n+ 2)an+1(qan+nan−1)>(n+ 1)an(qan+1 + (n+ 1)an),
4
which reduces to
n(n+ 2)an−1an+1 >(n+ 1)2a2
n−qanan+1 .(5)
To show (5), note that n(n+ 2)a2
n>(n+ 1)2a2
n−qanan+1 since qanan+1 > a2
n. Thus, by the
induction hypothesis, we have
n(n+ 2)an−1an+1 > n(n+ 2)a2
n>(n+ 1)2a2
n−qanan+1,
which gives (5) and completes the proof.
We will also need the following integral representation of a(x;q).
Lemma 3.2. Let x0<0and qbe fixed. Then for all x < 0, we have
a(x;q) = Rx
x0
u(t;q)
t2e−q/t+1/2t2dt +c
|x|e−q/x+1/2x2,(6)
where u(t;q) = −(n+ 1)an(q)tn+2 −an+1(q)tn+1 + 1 and c=|x0|a(x0;q)e−q/x0+1/2x2
0.
Proof. First observe that
(1 −qx −x2)a(x;q)−x3a′(x;q) = −(n+ 1)an(q)xn+2 −an+1(q)xn+1 + 1.(7)
Note that the coefficient of xion the left-hand side of (7) is zero if 2 ≤i≤nsince ai(q) =
qai−1(q) + (i−1)ai−2(q). One may rewrite (7) as
a′(x;q) + x2+qx −1
x3a(x;q) = −u(x;q)
x3,
where u(x;q) is as given. Solving this first-order linear differential equation using the standard
technique gives (6).
Remark 3.3. Note that substituting x= 0 in formula (6) produces the indeterminate ∞
∞.
However, the formula is still valid in this case, since
lim
x→0−
a(x;q) = lim
x→0− Rx
x0
u(t;q)
t2e−1/t+1/2t2dt +c
|x|e−1/x+1/2x2!
= lim
x→0− u(x;q)
x2e−1/x+1/2x2
−x1
x+1
x2−1
x3e−1/x+1/2x2!
= lim
x→0−1
−x2−x+ 1= 1 = a(0; q).
3.1 The case q= 1
In this subsection, we focus on the real zeros of the polynomials a(x;q) in the case q= 1. Here,
we will denote the q= 1 case of an(q) by an. We now show the following inequality for the
sequence an.
Lemma 3.4. We have
an+1
n<(n+ 1)nan
n−1, n ≥1.(8)
Proof. Before we prove (8), let us first make some preliminary observations. Let bn:= an
an−1.
Note that bn< bn+1 for all n≥3, by Lemma 3.1. The recurrence for anmay be written
equivalently as
bn+1 = 1 + n
bn
, n ≥1,
5
so that
bn<1 + n
bn
, n ≥3.
Then b2
n−bn−n < 0 implies
bn<1 + √1 + 4n
2<1
2+√n+ 1, n ≥3.(9)
Thus, by (9),
an=
n
Y
i=1
bi<
n
Y
i=1 1
2+√i+ 1, n ≥1.(10)
We now show (8). One may verify (8) directly for 1 ≤n≤14, so let us assume n≥15 for the
remainder of the proof. Note that (8) may be written equivalently as
anbn
n<(n+ 1)n.(11)
For (11), it is enough to show
1
2+√n+ 1nn
Y
i=1 1
2+√i+ 1<(n+ 1)n,(12)
by (9) and (10). Since the left-hand side of (12) may be written as
1
2+√n+ 1nn
Y
i=1 1
2+√i+ 11
21
2+√n+ 2 −i1
2
,
and since
1
2+√i+ 11
2+√n+ 2 −i≤ 1
2+rn+ 3
2!2
,1≤i≤n,
it follows that the left-hand side of (12) is at most
"1
2+√n+ 1 1
2+rn+ 3
2!#n
.
So to show (12), it suffices to show
1
2+√n+ 1 1
2+rn+ 3
2!< n + 1.(13)
Note that
(√2 + 1)√n+ 3 <2(√2−1)n+3√2
2−4, n ≥15,(14)
since it is true for n= 15 and hence for all n > 15 as
√2 + 1
√n+ 4 + √n+ 3 <2√2−2, n ≥15,
where the last inequality was obtained by replacing nwith n+ 1 on both sides of (14) and
subtracting the ncase.
By (14), we then have
1
2+√n+ 1 1
2+rn+ 3
2!=1
4+1
2 √n+ 1 + rn+ 3
2!+r(n+ 1)(n+ 3)
2
<1
4+√n+ 3
21 + 1
√2+n+ 2
√2< n + 1, n ≥15,
which gives (13) and completes the proof.
6
Let a(x) = a(x; 1). We now prove the main result in the case q= 1.
Theorem 3.5. The polynomial Pn
i=0 aixihas exactly one real zero if nis odd and no real zeros
if nis even.
Proof. Clearly, a(x) = Pn
i=0 aixihas no positive zeros as it has positive coefficients. So assume
x < 0. We show that a(x) has one negative zero when nis odd and that a(x)>0 for all xwhen
nis even. First assume nis odd. Let u(x) = u(x; 1), where u(x;q) is as defined in Lemma 3.2.
We first show u(x)>0 if x < 0, that is,
−(n+ 1)anxn+2 −an+1xn+1 + 1 >0, x < 0.(15)
Since u′(x) = −(n+ 1)xn((n+ 2)anx+an+1), we see that u(x) achieves a minimum value over
x < 0 at x=−an+1
(n+2)an. So to show (15) is to show u−an+1
(n+2)an>0 when nis odd, i.e.,
1>(n+ 1)an−an+1
(n+ 2)ann+2
+an+1 −an+1
(n+ 2)ann+1
,
which may be rewritten as
(n+ 2)n+2an+1
n> an+2
n+1, n odd.(16)
Note that (16) holds by Lemma 3.4, which establishes (15).
By Lemma 3.2, we have
a(x) = h(x)
|x|e−1/x+1/2x2, x < 0,(17)
where
h(x) = Zx
x0
u(t)
t2e−1/t+1/2t2dt +|x0|a(x0)e−1/x0+1/2x2
0.
By continuity, we may choose −1< x0<0 sufficiently close to zero such that a(x0)>0 since
a(0) = 1, whence h(x0)>0. Note that
h′(x) = u(x)
x2e−1/x+1/2x2>0, x < 0,
by (15), with h(−1) ≤0 by (17) since
a(−1) =
n
X
i=0
ai(−1)i= (−an+an−1) + (−an−2+an−3) + ···+ (−a1+a0)≤0
for nodd (with equality only if n= 1). It follows that h(x), and thus a(x), has one negative
zero whenever nis odd, and it belongs to the interval [−1,0). Since h′(x)>0 for all x < 0 with
h(x) analytic on the interval (−∞,0), this zero necessarily has multiplicity one.
Now suppose nis even. We will show that a(x)>0 for all xby induction, the n= 2 case clear
since a(x) = 2x2+x+ 1 in that case. By Descartes’ rule of signs, the polynomial u(x) has one
negative zero when nis even, which we denote by r. We first show a(r)>0. To do so, note
initially that if x < −an−1
an, then
a(x) =
n
X
i=0
aixi=
n−2
X
i=0
aixi+xn−1(an−1+anx)>0,
since Pn−2
i=0 aixi>0, by the induction hypothesis, and since xn−1(an−1+anx)>0. So to show
a(r)>0, it suffices to show r < −an−1
an. Since u(0) = 1 and u′(x) has one sign change for x < 0
7
when nis even, and it is from positive to negative, we need only verify u−an−1
an>0, which
holds for neven if and only if
(n+ 1)anan+2
n−1−anan+1an+1
n−1< an+2
n,
or
(n+ 1)anan+2
n−1−an(an+nan−1)an+1
n−1< an+2
n.
The latter inequality reduces to
anan+2
n−1< an+2
n+a2
nan+1
n−1,
which is seen to hold as anan+2
n−1< a2
nan+1
n−1. Thus a(r)>0, as desired.
We now represent a(x) once again as in (17) above. This time, however, we choose x0=r.
Then h′(x)>0 for r < x < 0 and h′(x)<0 for x < r, which implies that the minimum value
of h(x) for x < 0 is achieved at x=r. Now a(r)>0 implies h(r)>0, whence h(x)>0 for all
x < 0. Thus, a(x)>0 for all xby (17), which completes the induction and the proof in the
even case.
3.2 A generalization
Here, we consider the problem of determining the nature of the real zeros of the polynomials
a(x;q) =
n
X
i=0
ai(q)xi.
We will show that a(x;q) has the smallest possible number of real zeros over a range of q-values
by generalizing the argument given above in the case q= 1. Let bn(q) := an(q)
an−1(q)if n≥1. At
times, we will suppress the qargument in either an(q) or bn(q) when the context in clear.
The next two lemmas concern properties of the rational functions bn(q).
Lemma 3.6. If n≥1, then bn(q)2> n for all q > 1.
Proof. We will show for all q > 1 that
√nan−1< an<(q+√n−1)an−1, n ≥2,(18)
by induction, the n= 2 case clear. If n≥2, then
an+1 =qan+nan−1< qan+n
√nan= (q+√n)an,
by the induction hypothesis, which gives the right-hand inequality in (18) in the n+ 1 case.
On the other hand, we have
an+1 =qan+nan−1> qan+n
q+√n−1an≥√n+ 1an, n ≥2,(19)
which gives the left-hand inequality in (18) in the n+ 1 case and completes the proof of (18).
Note that the second inequality in (19) is seen to hold since it is equivalent to
q2+n−pn2−1≥2q
√n+ 1 + √n−1, n ≥2,
which is true, since for n≥2, one has
q2> q > 2q
√3 + 1 ≥2q
√n+ 1 + √n−1.
8
Lemma 3.7. If n≥1, then b′
n(q)>0for all q > 1.
Proof. Let b′
n=b′
n(q) and q > 1. We will show 0 < b′
n≤1 for all n≥1 by induction,
the n= 1 and n= 2 cases clear. If n≥2, then differentiating both sides of the recurrence
bn+1bn=qbn+n, and rearranging, gives
b′
n(bn+1 −q) = bn(1 −b′
n+1),
i.e.,
nb′
n=b2
n(1 −b′
n+1),(20)
since n
bn=bn+1 −q. By the induction hypothesis, we have b′
n>0, which taken together with
(20), implies b′
n+1 ≤1. On the other hand, if b′
n+1 ≤0, then b′
n≤1 implies
b2
n=nb′
n
1−b′
n+1 ≤n,
which contradicts Lemma 3.6. Thus, 0 < b′
n+1 ≤1, which completes the induction and the
proof.
We will often denote the sum Pm
i=0 ai(q)ximore specifically by am(x;q). Since the even case
concerning the real zeros of am(x;q) will follow in much the same way as it did when q= 1, we
will concern ourselves primarily with the case when mis odd. Furthermore, since it is easily
seen that am(x;q)>0 if x > 0 and am(x;q)<0 if x < −1 for all odd mif q≥1, then we may
restrict our attention to the interval [−1,0].
Let us first require that am(x;q) be increasing on the interval (−1,0) since am(−1; q)≤0 and
am(0; q)>0 if q≥1. Write
d
dxam(x;q) =
m−1
X
i=0
(i+ 1)ai+1(q)xi=q+
n
X
i=1
ℓi,
where m= 2n+ 1, n≥0, and
ℓi= 2ia2i(q)x2i−11 + (2i+ 1)a2i+1(q)
2ia2i(q)x,1≤i≤n.
In order to ensure that d
dx am(x;q) be positive on the interval (−1,0), let us first find for each
ℓia uniform lower bound on (−1,0) as a function of q. Note that if ℓi<0, then making the
substitution x=−2ia2i(q)
(2i+1)a2i+1(q)y, where 0 ≤y≤(2i+1)a2i+1 (q)
2ia2i(q), implies
−ℓi=(2ia2i(q))2i
((2i+ 1)a2i+1(q))2i−1y2i−1(1 −y)
≤(2ia2i(q))2i
((2i+ 1)a2i+1(q))2i−12i−1
2i2i−11
2i=2i−1
2i+ 1 2i−1a2i(q)2i
a2i+1(q)2i−1.
Thus, regardless of whether ℓiis positive or negative, it is always the case that
ℓi≥ −2i−1
2i+ 12i−1a2i(q)2i
a2i+1(q)2i−1.
Therefore, for all n≥0 and q≥1, we have
d
dx a2n+1(x;q)≥fn(q),−1< x < 0,(21)
9
where
fn(q) := q−
n
X
i=1 2i−1
2i+ 12i−1a2i(q)2i
a2i+1(q)2i−1.
For each n, we seek the largest interval of the form [1, t] such that fn(q)≥0 for all q∈[1, t].
For such q, note that a2n+1(x;q) is increasing on the interval (−1,0), by (21). To this end, let
q∗
n:= max{t≥1 : fn(q)≥0 for all q∈[1, t]},
where we take q∗
n=∞if fn(q)≥0 for all q≥1. Note that q∗
nis a decreasing function of nsince
fn(q)> fn+1(q) for all positive q. By Maple, we have that q∗
nis finite if and only if n≥5. If
0≤j≤nand 1 ≤q≤q∗
n, then we are guaranteed that
d
dxa2j+1 (x;q)≥fj(q)≥fn(q)≥0,−1< x < 0,(22)
and thus a2j+1(x;q) has exactly one zero on the interval [−1,0) for all such jand qsince
a2j+1(−1; q)≤0 and a2j+1(0; q)>0. Note that we require q≥1 here and elsewhere since we
do not have a generalization of Lemma 3.1 for q < 1 and a meaningful one may not indeed be
possible (note also that the sequence an(q) need not be increasing for such q).
It will also be necessary that inequality (16) is suitably generalized. To this end, given n≥0,
let eqndenote the q≥1 such that
a2n+4(q)b2n+4 (q)2n+4 = (2n+ 5)2n+5.(23)
Note that, by Lemmas 3.7 and 3.4, the left-hand side of (23) is an increasing function of q
for each fixed nand is strictly less than the right-hand side when q= 1, whence eqnis uniquely
determined. Thus, if 1 ≤q≤eqn, then the left-hand side of (23) is less than or equal (2n+5)2n+5,
which provides a generalization of (16) for such q.
One could then attempt to verify the inequalities
a2j+4(q)b2j+4 (q)2j+4 ≤(2j+ 5)2j+5 , j ≥n, (24)
for all 1 ≤q≤eqn, which would then show (see below) that a2j+3(x;q) has exactly one real zero
for all such jand q. By Lemma 3.7, it would suffice to verify (24) in the case q=eqn.
Thus, given any fixed n≥0 and 1 ≤q≤min{q∗
n,eqn}, we know that a2j+1 (x;q) has exactly one
real zero if 0 ≤j≤n, and so to show that am(x;q) always has the smallest possible number
of real zeros for qin the given range, it will suffice to verify (24) for all j≥n. So in order to
maximize the range in qfor which our technique will work, we choose nsuch that min{q∗
n,eqn}
is largest among all n≥0. This value, which we denote by λ, will yield the best possible qby
the present method.
Lemma 3.8. The maximum value of min{q∗
n,eqn}over n≥0occurs when n= 140 and is given
by λ=q∗
140 ≈6.2758547971.
Proof. Using Maple, we determine the approximate values of eqnfor 0 ≤n≤150 and find that
eqnis increasing over such n, with eq139 < q∗
139 and eq140 > q∗
140. Since q∗
nis a decreasing function
of nand since eq139 ≈6.2579383791 is less than q∗
140 ≈6.2758547971, it follows that the largest
value of min{q∗
n,eqn}is achieved when n= 140 and is given by q∗
140.
Lemma 3.9. If 0≤j≤140 and 1≤q≤λ, then the polynomial a2j+1(x;q)has exactly one
real zero, and it belongs to the interval [−1,0).
Proof. This follows from (22) and the subsequent discussion, along with Lemma 3.8.
We now show the inequalities in (24) in the case when n= 140.
10
Lemma 3.10. If j≥140 and 1≤q≤λ, then
a2j+4(q)b2j+4 (q)2j+4 <(2j+ 5)2j+5 .(25)
Proof. Using Maple, one may verify that the sequence eqjis increasing for 0 ≤j≤250. Since
λ < eqjif j≥140, it follows from Lemma 3.7 that (25) is satisfied when 140 ≤j≤250. So let
us assume j > 250. Similar to the proof of Lemma 3.4 above, we have for all n≥1,
bn(q)≤q
2+rn+q2
4, q ≥1,
and
an(q) =
n
Y
i=1
bi(q)≤
n
Y
i=1 q
2+ri+q2
4!
=
n
Y
i=1 q
2+ri+q2
4!1
2 q
2+rn+ 1 −i+q2
4!1
2
≤ q
2+rn+ 1
2+q2
4!n
, q ≥1.
Thus, to show (25), it is enough to show
q
2+r2j+ 4 + q2
4! q
2+r2j+ 5
2+q2
4!<2j+ 5, j > 250,(26)
where 1 ≤q≤λ. Since λ < 6.5 and the left-hand side of (26) is an increasing function of q, it
is enough to verify (26) in the case when q= 6.5, i.e.,
13
4+r2j+ 4 + 169
16 ! 13
4+r2j+ 5
2+169
16 !<2j+ 5, j > 250.(27)
Let pjdenote the left-hand side of (27). Note that
pj+1 −pj<6.5
q2j+ 6 + 169
16 +q2j+ 4 + 169
16
+3.25
q2j+7
2+169
16 +q2j+5
2+169
16
+4j+683
16
√22j+ 4 + 169
16
<2, j > 250,(28)
since the first inequality holds for all j > 0 and since the middle quantity is a decreasing function
of j(being the sum of three decreasing functions) which is less than 2 when j= 251. Then (27)
follows from (28) since it holds when j= 251, which completes the proof of (25).
We now can give a generalization of Theorem 3.5.
Theorem 3.11. If 1≤q≤λ, then the polynomial Pn
i=0 ai(q)xihas exactly one real zero if n
is odd and no real zeros if nis even.
Proof. Clearly, a(x;q) has no positive zeros or zeros less than −1, so we may restrict attention
to the interval [−1,0]. First assume n= 2j+ 1 is odd. By Lemma 3.9, we may assume j≥141.
11
Let u(x;q) = −(n+ 1)an(q)xn+2 −an+1(q)xn+1 + 1. To show that u(x;q)>0 if x < 0, we must
show that u−an+1(q)
(n+2)an(q);q>0, or, equivalently,
an+1(q)bn+1 (q)n+1 <(n+ 2)n+2, n ≥283 odd.(29)
Inequality (29) holds by Lemma 3.10, which implies u(x;q)>0 if x < 0. The remainder of the
proof in the odd case uses Lemma 3.2 and follows now in much the same way as it did when
q= 1.
The argument for the even case also follows in a similar manner as before and in fact applies to
all q≥1.
4 Convergence of zeros
In this section, we prove that the sequence of real zeros of a(x;q) and d(x) for nodd is mono-
tonically convergent.
Proposition 4.1. If 1≤q≤λ, then the sequence of real zeros of the polynomials a(x;q)for n
odd is monotonically increasing and converges to 0.
Proof. We treat the q= 1 case, as the proof for the general case is similar. We first show that
the sequence of zeros increases. Where needed, we will denote the sum Pn
i=0 aiximore explicitly
by an(x) if n≥0, where ai=ai(1). Let sdenote the real zero of a2m−1(x). Note that s=−1
when m= 1, so let us assume m≥2. The proof of Theorem 3.5 above shows that −1< s < 0
in this case. Observe that a2m−1(s) = 0 may be written as
1 + s+
m−1
X
i=1 a2i
a2i+1
+sa2i+1s2i= 0.(30)
Then we have a2m−2
a2m−1+s < 0, for if not, and a2m−2
a2m−1≥ −s, then a2i
a2i+1 ≥ −sfor all 1 ≤i≤m−1,
since the ratio an
an+1 is decreasing, by Lemma 3.1. But then the left-hand side of (30) would be
positive as 1 + s > 0 and Pm−1
i=1 a2i
a2i+1 +sa2i+1s2i≥0 (being a sum of non-negative terms).
Then a2m−2
a2m−1+s < 0 implies a2m
a2m+1 +s < 0, again since the ratio is decreasing. Therefore,
a2m+1(s) = a2m
a2m+1
+sa2m+1s2m+a2m−1(s)
=a2m
a2m+1
+sa2m+1s2m<0,
which implies that the zero of a2m+1 (x) is greater than that of a2m−1(x).
To show that the sequence of real zeros converges to zero, we fix y,−1< y < 0, and show that
a2m+1(y)<0 for msufficiently large. To do so, we first bound the ratio bn:= an
an−1from below.
Since bn+1 = 1 + n
bnand bn+1 ≤n+1
nbnif n≥3, by (4), we have
(n+ 1)b2
n−nbn−n2≥0,
which gives
bn≥n
2n+ 2 1 + √4n+ 5, n ≥1.(31)
From (31), we see that an−1
an→0 as n→ ∞ and that an=Qn
i=1 biis larger than Cnfor any
positive constant Cfor all nsufficiently large. This implies
a2m+1(y) =
m
X
i=0 a2i
a2i+1
+ya2i+1y2i<0,
for all msufficiently large, which completes the proof.
12
Modifying the above proof somewhat yields the same result for the polynomials d(x).
Proposition 4.2. The sequence of real zeros of the polynomials d(x)for n≥3odd is mono-
tonically increasing and converges to 0.
Proof. First note that
d2m−2d2m+1 =1
2m−1(d2m−1+ 1)((2m+ 1)d2m−1)
=1
2m−1((2m+ 1)d2m−1d2m+ (2m+ 1)d2m−d2m−1−1)
>2m+ 1
2m−1d2m−1d2m> d2m−1d2m,
whence d2m−2d2m+1 > d2m−1d2mfor all m≥1. The monotonicity of the real zeros of the d(x)
now follows as in the first part of the proof of Proposition 4.1 above.
Recall that dn∼n!
efor large n(see [8, p. 67]) and note that dn−1
dn→0 as n→ ∞. Using
Stirling’s formula for n!, the convergence of the real zeros now follows in a manner similar to
the second part of the proof of Proposition 4.1 above.
References
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Integer Seq. 10 (2007), Art. 07.6.8.
[2] T. Mansour and M. Shattuck, Polynomials whose coefficients are k-Fibonacci numbers,
Ann. Math. Inform. 40 (2012), 57–76.
[3] T. Mansour and M. Shattuck, Polynomials whose coefficients are generalized Tribonacci
numbers, Appl. Math. Comput. 219 (2013), 8366–8374.
[4] F. M´aty´as, On the generalization of the Fibonacci-coefficient polynomials, Ann. Math.
Inform. 34 (2007), 71–75.
[5] F. M´aty´as, Further generalizations of the Fibonacci-coefficient polynomials, Ann. Math.
Inform. 35 (2008), 123–128.
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38 (2011), 95–98.
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