MATEMATIQKI VESNIK

58 (2006), 13–17

UDK 517.521

originalni nauqni rad research paper

ON HADAMARD TYPE POLYNOMIAL CONVOLUTIONS WITH REGULARLY VARYING SEQUENCES

Slavko Simić

Abstract. For a sequence of polynomials P_{n}(x) := ∑_{m⩽n} p_{m}x^{m}, n ⩾ 1, we give a necessary and sufficient condition for the asymptotic equivalence

P_{n}^{(α)}(x) := ∑_{m⩽n} c_{m}p_{m}x^{m} ∼ c_{n}P_{n}(x) (n → ∞),

to hold for each x ⩾ A and an arbitrary regularly varying sequence {c_{n}} of index α ∈ R.

Introduction

A sequence {p_{n}}_{n⩾1} of non-negative numbers generates a sequence of polynomials {P_{n}(x)}_{n⩾1} defined by P_{n}(x) := ∑_{m⩽n} p_{m}x^{m}.

A sequence {c_{n}}_{n⩾1} of positive numbers is regularly varying with index α ∈ R if it can be represented in the form c_{n} = n^{α}ℓ_{n}, where {ℓ_{n}} is a slowly varying sequence, i.e. satisfying ℓ_{[λn]} ∼ ℓ_{n} (n → ∞) for each λ > 0 ([1], [2]).

Some examples of slowly varying sequences are:

log^{a}(n + 1), a ∈ R; log^{b}(log(n + 1)), b ∈ R; exp(log^{c}(n + 1)), 0 < c < 1.

Our task here is to investigate asymptotic behavior of Hadamard-type convolutions P_{n}^{(α)}(x) := ∑_{m⩽n} c_{m}p_{m}x^{m} as n → ∞ (cf. [2]).

In [2] we introduced an operator T f(x) in the following way.

Definition. Let f ∈ C^{∞}[0, ∞). Then

T f(x) := \frac{xf′(x)}{f(x)}.

Under a more general framework, we obtained asymptotic behavior of P_{n}^{(α)}(x) supposing

T (T P_{n}(x)) < M,  (I)

where M does not depend on n or x.

AMS Subject Classification: 26 A 12

Keywords and phrases: Regular variation, polynomials, asymptotic behavior.

In this paper we find a necessary and sufficient condition for the given asymptotics avoiding the somewhat ambiguous condition (I).

Results

Theorem. Let A be a positive number. Then the asymptotic relation

P_{n}^{(α)}(x) ∼ n^{α}ℓ_{n}P_{n}(x)  (n → ∞),  (1)

holds for each x ⩾ A, α ∈ R, and an arbitrary slowly varying sequence {ℓ_{n}}, if and only if

T P_{n}(A) := \frac{AP′_{n}(A)}{P_{n}(A)} ∼ n (n → ∞).  (2)

Proof. Denote Q_{n}(x) := ∑_{m⩽n} P_{m}(x). We can see that the condition (2) is necessary if we put in (1): α = 1, ℓ_{n} = 1, x = A. That it is also sufficient can be proved using the following lemmas.

Lemma 1. Under the condition (2), for each real α we have

(i) n^{α}Q_{n}(A) → ∞; (ii) ∑_{m⩽n} m^{α}p_{m}A^{m} ∼ n^{α}P_{n}(A) (n → ∞).

Lemma 2. We have sup_{m⩽n}(mℓ_{m}) ∼ nℓ_{n}; inf_{m⩽n}(ℓ_{m}/m) ∼ ℓ_{n}/n (n → ∞).

Lemma 3. The function x ↦ \frac{xP′_{n}(x)}{P_{n}(x)} is non-decreasing for x > 0.

Lemma 4. (Stoltz’s lemma) If ∑_{m⩽n} b_{m} → ∞ and a_{n}/b_{n} → s as n → ∞, then

∑_{m⩽n} a_{m} / ∑_{m⩽n} b_{m} → s (n → ∞).

Proof of Lemma 1. By partial summation we get ∑_{m⩽n} mp_{m}A^{m} = (n + 1)P_{n}(A) − Q_{n}(A). Hence, the condition (2) is equivalent to

nP_{n}(A)/Q_{n}(A) → ∞ (n → ∞)  (3)

Therefore, for n > n_{0} and fixed α ∈ R, we deduce

\frac{nP_{n}(A)}{Q_{n}(A)} > |α| + 1; \frac{Q_{n}(A) − Q_{n−1}(A)}{Q_{n}(A)} > \frac{|α| + 1}{n};

\frac{Q_{n−1}(A)}{Q_{n}(A)} < 1 − \frac{|α| + 1}{n} < exp(−\frac{|α| + 1}{n}).

Hence

Q_{n}(A) ≫ exp((|α| + 1) ∑_{m⩽n} 1/m) ≫ exp((|α| + 1) log n),

i.e. n^{α}Q_{n}(A) ≫ n^{α+|α|+1} and the part (i) is proved.

Denoting Δr_{n} := r_{n+1} − r_{n}, by (3) we get

\frac{P_{n}(A)Δn^{α}}{Δ(n^{α−1}Q_{n−1}(A))} = \frac{P_{n}(A)Δn^{α}}{n^{α−1}P_{n}(A) + Q_{n}(A)Δn^{α−1}} → α (n → ∞).

Now, applying part (i), Stoltz’s lemma and (3), we obtain

S_{n}(A) := ∑_{m⩽n} P_{m}(A)Δm^{α} ∼ αn^{α−1}Q_{n}(A) = o(n^{α}P_{n}(A)) (n → ∞).

Therefore, by partial summation we get

∑_{m⩽n} m^{α}p_{m}A^{m} = (n+1)^{α}P_{n}(A)−S_{n}(A) = (n+1)^{α}P_{n}(A)+o(n^{α}P_{n}(A)) (n → ∞),

and the part (ii) of Lemma 1 is also proved.

Lemma 2. is proved in [1, p. 23].

Proof of Lemma 3. Indeed, for x > 0 by Cauchy’s inequality, we get

x\frac{d}{dx} (\frac{xP′_{n}(x)}{P_{n}(x)}) = \frac{∑_{m⩽n} m^{2}p_{m}x^{m}}{∑_{m⩽n} p_{m}x^{m}} − (\frac{∑_{m⩽n} mp_{m}x^{m}}{∑_{m⩽n} p_{m}x^{m}} )^{2} ⩾ 0.

Hence T P_{n}(x) is monotone non-decreasing for x > 0.

Stoltz’s lemma is a classical one and is proved, for example, in [3, p. 30]. Now we can give the proof of the Theorem at the point x = A. By Lemmas 1

and 2, as n → ∞, we get

P_{n}^{α}(A) = ∑_{m⩽n} m^{α}ℓ_{m}p_{m}A^{m} ⩽ sup_{m≤n}(mℓ_{m}) ∑_{m⩽n} m^{α−1}p_{m}A^{m} ∼ n^{α}ℓ_{n}P_{n}(A),

and

∑_{m⩽n} m^{α}ℓ_{m}p_{m}A^{m} ⩾ inf_{m⩽n}(ℓ_{m}/m) ∑_{m⩽n} m^{α+1}p_{m}A^{m} ∼ n^{α}ℓ_{n}P_{n}(A).

Hence

1 ⩽ lim inf_{n}(P_{n}^{(α)}(A)/n^{α}ℓ_{n}P_{n}(A)) ⩽ lim sup_{n}(P_{n}^{(α)}(A)/n^{α}ℓ_{n}P_{n}(A)) ⩽ 1,

and the proof is done.

For x > A, by Lemma 3, we obtain

n ∼ AP′_{n}(A)/P_{n}(A) ⩽ xP′_{n}(x)/P_{n}(x) ⩽ n.

Hence xP′_{n}(x)/P_{n}(x) ∼ n (n → ∞) and we can apply the previous proof replacing A by x.

Comment. As the referee notes, the condition (2) is certainly less opaque then the former condition (I), but it still is opaque in that one has to do a calculation and some asymptotic approximations to decide if a candidate sequence satisfies it.

There is also a problem to determine the least possible A such that (2) holds. For instance, if p_{n} = a^{n} for some a > 0 then (2) holds for A > 1/a and fails

for A ⩽ 1/a.

Also, if p_{n} = 1/n! then (2) never holds; but for p_{n} = n! an easy calculation shows that (2) is valid for all A > 0.

Therefore we shall establish two simple criteria which can help to decide if a given sequence {p_{n}} satisfies (2) or not.

Proposition 1. If A lies inside the interval of convergence of ∑ p_{n}x^{n} then the condition (2) fails.

Proof. We have, as n → ∞, ∑_{m⩽n} p_{m}A^{m} → P (A), and consequently,

∑_{m⩽n} mp_{m}A^{m} → AP′(A).

Hence T P_{n}(A) → 0 (n → ∞).

But the divergence of ∑ p_{n}A^{n} does not imply that (2) is true. This can be seen from the following example.

Let p_{m} = 1 if m is in the factorial form and p_{m} = 0 otherwise. Then

P_{(n+1)!−1}(A) = A^{n!} + A^{(n−1)!} + · · · .

For A > 1, we have P_{(n+1)!−1}(A) ∼ A^{n!}, and

AP′_{(n+1)!−1}(A) = n!A^{n!} + (n − 1)!A^{(n−1)!} + ⋯ ∼ n!A^{n!} (n → ∞).

Therefore

T P_{(n+1)!−1}(A) ∼ \frac{n!}{(n + 1)! − 1} → 0 (n → ∞).

Proposition 2. If, for some A > 0,

lim_{n→∞} n (1 − \frac{1}{A} \frac{p_{n}}{p_{n+1}} ) = +∞,  (4)

then (2) holds.

Proof. Note that the condition (4) implies just a finite number of p_{n} = 0. Also, by Raabe’s convergence criteria, ∑ p_{n}A^{n} diverges.

Now, the condition (4) is equivalent to

1 + (n − 1) (1 − \frac{1}{A} \frac{p_{n−1}}{p_{n}} ) → +∞,

i.e.

(np_{n}A^{n} − (n − 1)p_{n−1}A^{n−1})/p_{n}A^{n} → +∞.

Applying Lemma 4, we get ∑_{m⩽n} p_{m}A^{m}/np_{n}A^{n} → 0 (n → ∞). It follows that

\frac{np_{n}A^{n}}{np_{n}A^{n} + ∑_{m⩽n−1} p_{m}A^{m}} → 1 (n → ∞),

i.e.

\frac{np_{n}A^{n}}{n ∑_{m⩽n} p_{m}A^{m} − (n − 1) ∑_{m⩽n−1} p_{m}A^{m}} → 1.

Applying Lemma 4 again, we obtain the condition (2).

Now it is not difficult to verify the above examples using Propositions 1 and 2.

REFERENCES

[1] N. H. Bingham, C. M. Goldie and J. L. Teugels, Regular Variation, Cambridge Univ. Press, 1987.

[2] S. Simić, Orthogonal polynomials and regularly varying sequences, Publ. Inst. Math. Belgrade (N.S) 70(84) (2001).

[3] G. Polya, G. Szegö, Aufgaben und Lehrsätze aus der Analysis, Springer-Verlag, Berlin, 1964.

(received 20.10.2003, in revised form 26.10.2005) Mathematical Institute, Kneza Mihaila 35/I, 11000 Belgrade, Serbia

E-mail: ssimic@turing.mi.sanu.ac.yu