On a sum of modified Bessel functions

Mediterranean Journal of Mathematics manuscript No. (will be inserted by the editor)

On a sum of modified Bessel functions ´ ad Baricz · Tibor K. Pog´ Arp´ any

Received: date / Accepted: date

Abstract In this paper we consider a sum of modified Bessel functions of the first kind of which particular case is used in the study of Kanter’s sharp modified Bessel function bound for concentrations of some sums of independent symmetric random vectors. We present some monotonicity and convexity properties for that sum of modified Bessel functions of the first kind, as well as some Tur´ an type inequalities, lower and upper bounds. Moreover, we point out an error in Kanter’s paper [8]. Keywords Modified Bessel functions · Concentration bounds · Functional inequalities Mathematics Subject Classification (2000) 39B62 · 33C10 · 33C15

1 Introduction Special functions like modified Bessel functions of the first and second kind, Iν and Kν , are frequently used in probability theory and statistics, see for example the papers of Fotopoulos and Venkata [7], Kanter [8], Marchand and Perron [9, 10], Robert [15], Yuan and Kalbfleisch [21], to mention a few. In [8] Kanter deduced a sharp bound for the probability that a sum of independent symmetric random vectors lies in a symmetric convex set, by improving in one dimension an inequality first proved by Kolmogorov. For a very recently deduced concentration inequality for sums of independent isotropic random vectors we refer to the paper of Cranston and Molchanov [6]. In deducing ´ Baricz A. Department of Economics, Babe¸s-Bolyai University, Cluj-Napoca 400591, Romania E-mail: [email protected] T.K. Pog´ any Faculty of Maritime Studies, University of Rijeka, Rijeka 51000, Croatia E-mail: [email protected]

´ ad Baricz, Tibor K. Pog´ Arp´ any

2

the above mentioned result in [8] the function Φ : (0, ∞) → (0, ∞), defined by Φ(x) = e−x [I0 (x) + I1 (x)] , plays an important role. Some properties of Φ were deduced in [8, Lemma 4.4] and [8, Lemma 4.5]. Recently, Mattner and Roos [11], by using other properties of the function Φ, shortened the proof of [8, Theorem 4.1], in which Kanter deduced his concentration bound for sums of independent random vectors. The properties proved in [11, Lemma 1.4] complement the study of Φ from [8]. Our aim in this paper is twofold: in one hand to generalize some of the results on Φ from [8, 11] to the function Φν : (0, ∞) → (0, ∞), defined by Φν (x) = e−x x−ν [Iν (x) + Iν+1 (x)] , and on the other hand to point out a gap in the proof of [8, Lemma 4.4], which in turn implies that Kanter’s proof of [8, Theorem 4.1] is not complete. To achieve our goal, we present some monotonicity and convexity properties, lower and upper bounds, and Tur´ an type inequalities for the function Φν . In the study of the Tur´ an type inequalities for the function Φν one of the key tools is the Neumann integral formula concerning the product of two modified Bessel functions with different parameters. 2 Monotonicity and convexity properties of the function Φν Before we present the main results of this paper we recall some definitions, which will be used in the sequel. A function f : (0, ∞) → R is said to be completely monotonic if f has derivatives of all orders and satisfies (−1)m f (m) (x) ≥ 0 for all x > 0 and m ∈ {0, 1, 2, . . . }. A function g : (0, ∞) → (0, ∞) is said to be logarithmically convex, or simply log-convex, if its natural logarithm ln g is convex, that is, for all x, y > 0 and λ ∈ [0, 1] we have λ

1−λ

g(λx + (1 − λ)y) ≤ [g(x)] [g(y)]

.

A similar characterization of log-concave functions also holds. We also note that every completely monotonic function is log-convex, see [20, p. 167]. Now, by definition, a function h : (0, ∞) → (0, ∞) is said to be geometrically (or multiplicatively) convex if it is convex with respect to the geometric mean, that is, if for all x, y > 0 and all λ ∈ [0, 1] the inequality h(xλ y 1−λ ) ≤ [h(x)]λ [h(y)]1−λ holds. The function h is called geometrically concave if the above inequality is reversed. Observe that, actually the geometrical convexity of a function h means that the function ln h is a convex function of ln x in the usual sense. We also note that the differentiable function g is log-convex (log-concave) if and only if x 7→ g 0 (x)/g(x) is increasing (decreasing), while the differentiable function h is geometrically convex (concave) if and only if the function x 7→

On a sum of modified Bessel functions

3

xh0 (x)/h(x) is increasing (decreasing). See for example [3] for more details on geometrically convex (concave) functions and their relations with continuous univariate distributions. The following result is motivated by [8, Lemma 4.5] and [11, Lemma 1.4]. Part a of Theorem 1 generalizes the statement that Φ is completely monotonic, see [11, Lemma 1.4], while part d for a = 0 provides a generalization of [8, Lemma 4.5]. The right-hand side of (1) extends inequality [11, eq. (9)]. The sharpness of (1) is discussed below in Remark 1. Theorem 1 The following assertions are valid: Φν is completely monotonic on (0, ∞) for all ν ≥ − 12 ; Φν is log-convex and geometrically concave on (0, ∞) for all ν ≥ − 12 ; ν 7→ Φν (x) is decreasing on [0, ∞) for all x > 0; ν+ 21 x (x + Φν (x) is increasing on (0, ∞) for all ν ≥ − 12 and a ∈ 7→ a) ν 1 0, 2 + 4 ; e. the inequality r ν+ 12 ν + 21 2 1 ν+ 12 < Φν (x) < π · ν+ 12 (1) 1 ν 1 ν 2ν+ 2 Γ (ν + 1) x + 2 + x+ + 1

a. b. c. d.

2

is valid for all ν >

− 21

4

2

4

and x > 0;

Proof Observe that since r

r 2 2 cosh x and I 21 (x) = sinh x, πx πx q we have that Φ− 21 (x) = π2 . Thus, in what follows we assume that ν > − 12 . a. We proceed somewhat similar as in the case of complete monotonicity of the function x 7→ e−x x−ν Iν (x). About this function we know that it is completely monotonic on (0, ∞) for all ν ≥ − 12 . The case ν > − 12 was proved by N˚ asell [13] and the case ν = − 12 was pointed out in [2]. In [13] N˚ asell used the integral representation [18, p. 79] ν Z 1 1 ν− 12 −xt 1 2x Iν (x) = √ 1 − t2 e dt, ν > − , (2) 1 2 πΓ ν + 2 −1 I− 12 (x) =

to prove the above mentioned complete monotonicity. Now, if we change ν to ν + 1 in (2) and we use integration by parts, we obtain ν+1 Z 1 1 ν+ 12 −xt 2x Iν+1 (x) = √ e dt 1 − t2 3 πΓ ν + 2 −1 ν Z 1 1 1 ν+ 12 −xt 2 2x =√ 1 − t2 xe dt 3 πΓ ν + 2 −1 ν Z 1 1 ν− 21 −xt 2x = −√ 1 − t2 te dt, 1 πΓ ν + 2 −1


4

and this in view of (2) yields the Laplace transform representation Z 1 ν− 12 −x(1+t) √ 1 Ψν (x) = π2ν Γ ν + e dt, Φν (x) = (1 − t) 1 − t2 2 −1

(3)

which, by the easy half of Bernstein’s theorem (see [17]) is known to imply that Ψν is completely monotonic: Z 1 ν− 12 −x(1+t) (n) (1 − t)(1 + t)n 1 − t2 (−1)n [Ψν (x)] = e dt > 0, −1

− 12 ,

x > 0 and n ∈ {0, 1, . . . }. Thus, the function Φν is completely for all ν > monotonic on (0, ∞) for all ν > − 21 . b. The first part of the assertion follows from part a. Namely, it is known that every completely monotonic function is log-convex, see [20, p. 167]. Now, for convenience let us introduce the notation φν (x) = x−ν [Iν (x) + Iν+1 (x)] . Then by using the recurrence relations [18, p. 79] −ν 0 x Iν (x) = x−ν Iν+1 (x),

Iν+2 (x) − Iν (x) = −

2ν + 1 Iν+1 (x) x

we have φ0ν (x) − φν (x) = −(2ν + 1)x−(ν+1) Iν+1 (x) and consequently Φ0ν (x) φ0 (x) − φν (x) 2ν + 1 Iν+1 (x) = ν =− , Φν (x) φν (x) x Iν (x) + Iν+1 (x) that is, xΦ0ν (x) = −(2ν + 1) Φν (x)

1 Iν (x) Iν+1 (x)

.

(4)

+1

Since the function x 7→ Iν+1 (x)/Iν (x) is increasing on (0, ∞) for all ν ≥ − 12 (see [19, p. 201] or [21, p. 446]), it follows that x 7→ 1/ [Iν (x)/Iν+1 (x) + 1] is increasing on (0, ∞) for all ν ≥ − 21 . Now, by using (4) clearly the function x 7→ xΦ0ν (x)/Φν (x) is decreasing on (0, ∞) for all ν ≥ − 12 , as we required. c. By using the infinite series representation of the modified Bessel function of the first kind [18, p. 77] Iν (x) =

X n≥0

2n+ν 1 2x n!Γ (n + ν + 1)

we have φν (2x) =

X n≥0

an (ν)x2n +

X n≥0

bn (ν)x2n+1 ,

(5)


5

where an (ν) =

1 2ν n!Γ (n + ν + 1)

and

bn (ν) =

1 . 2ν n!Γ (n + ν + 2)

Thus, to prove that the function ν 7→ Φν (x) = e−x φν (x) is decreasing on [0, ∞) for all x > 0, it is enough to show that ν 7→ an (ν) and ν 7→ bn (ν) are decreasing on [0, ∞) for all n ∈ {0, 1, . . . }, that is, for all ν ≥ 0 and n ∈ {0, 1, . . . } we have ∂ log(an (ν)) 1 ∂an (ν) = = − log 2 − ψ(n + ν + 1) < 0, (6) ∂ν an (ν) ∂ν 1 ∂bn (ν) ∂ log(bn (ν)) = = − log 2 − ψ(n + ν + 2) < 0, ∂ν bn (ν) ∂ν

(7)

where ψ(x) = Γ 0 (x)/Γ (x) denotes the digamma function. Since ψ(x) > 0 for all x > x∗ , where x∗ ' 1.461632144 . . . is the abscissa of the minimum of the Γ function, the inequality (6) clearly holds for all ν ≥ 0 and n ∈ {1, 2, . . . }, while the inequality (7) clearly holds for all ν ≥ 0 and n ∈ {0, 1, . . . }. Thus, we just need to verify the inequality (6) when n = 0, that is, ψ(ν + 1) + log 2 > 0 for ν ≥ 0. When ν = 0 this is true, since ψ(1) ' −0.5772156649 . . . and log 2 = 0.6931471805. . .. Now, suppose that ν > 0. According to Batir [5, Lemma 1.7] the inequality ψ(ν + 1) > log ν + 21 is valid for all ν > 0. Thus ψ(ν + 1) + log 2 > log(2ν + 1) > 0 for all ν > 0, as we required. d. For x > 0 and ν ≥ 0 let us consider the inequality [16, eq. (20)] Iν (x) < xIν−1 (x)

ν−

1 2

1 q 2 , + x2 + ν − 12

which implies that q Iν (x) > Iν+1 (x)

x2 + ν + x

1 2 2

+

ν+ x

1 2

>1+

ν + 12 , x

where x > 0 and ν ≥ −1. Observe that this implies the inequality 1 1 Φ0ν (x) >− ν+ [log Φν (x)] = ν Φν (x) 2 x+ 2 +

ν+ 21 #0 ν 1 = − log x + + , 2 4 "

0

1 4

where ν ≥ − 12 and x > 0. Thus, the function x 7→ log

h

x+

ν 2

+

1 1 ν+ 2 4

Φν (x)

i

is increasing on (0, ∞) for all ν ≥ − 12 . This in turn implies that the function 1 x 7→ (x + a)ν+ 2 Φν (x) is increasing on (0, ∞) for all ν ≥ − 12 and a ∈ 0, ν2 + 41 as a product of two positive and increasing functions x 7→

x+a x + ν2 +

ν+ 21 1 4

ν+ 12 ν 1 and x 7→ x + + Φν (x). 2 4


6

e. It is known that e−x > 1 − x for x > 0 and thus

2−t t

ν+ 21

ν+ 12 t 1 2 < e− 2 (ν+ 2 ) t

for all ν > − 12 and t ∈ (0, 2]. By using this inequality together with (3) we obtain ν+ 12 Z 2 2−t 1 t2ν e−xt dt Φν (x) = √ ν t π2 Γ ν + 12 0 Z 2 ν+ 12 t 1 1 2 e− 2 (ν+ 2 ) t2ν e−xt dt − 21 and x > 0. Here in the last step we used the well-known formula Z ∞ Γ (α + 1) tα e−βt dt = , β α+1 0 which is valid for all α > −1 and β > 0. Now, for the left-hand side of (1) we use part d together with (5) for x = 0. Namely, for all x > 0 and ν > − 21 we have

1 ν x+ + 2 4

ν+ 21 Φν (x) > =

1 1

2ν+ 2

1 ν+ 2

ν+ 12

1 2

ν+ 12

1 1

2ν+ 2

ν+

Φν (0) 1 . 2ν Γ (ν + 1)

Remark 1 Let us focus on the inequality (1). Observe that the lower and upper bounds are sharp as x tends to infinity, while the lower bound in (1) is sharp as x tends to 0. Namely, by using the asymptotic formula [1, p. 377] ex 4ν 2 − 1 (4ν 2 − 1)(4ν 2 − 9) Iν (x) ∼ √ 1− + − . . . , 1!(8x) 2!(8x)2 2πx which holds for large values of x and for fixed ν > −1, one has r 2(ν 2 + (ν + 1)2 ) − 1 2 −ν− 12 1− + ... , Φν (x) ∼ x π 1!(8x)


7

as x → ∞. This means that Φν (x) → 0 as x → ∞ for each ν > − 12 . On the other hand, when ν > − 21 both bounds in (1) tend to 0 as x → ∞, and thus indeed the lower and upper bounds are sharp as x tends to infinity. Finally, we note that the sharpness of the upper bound in (1), denoted by Uν (x), can be seen also from the asymptotic relations ! r 2 3 ν + 12 ν + 12 ν − 12 2 −3 −ν− 21 +O x , 1− + Uν (x) = x π 2x 8x2

Φν (x) = x

−ν− 21

r

2 π

ν + 12 1− 2x

2 +

ν+

1 2 2

ν + 32 8x2

ν−

1 2

!

+O x

−3

.

The next result is motivated by the Turán type inequalities for modified Bessel functions of the first and second kind. These kind of inequalities are named after the Hungarian mathematician Paul Turán who proved a similar inequality for Legendre polynomials. For more details see the recent paper [4] and the references therein. Theorem 2 The function ν 7→√Φν (x) is log-concave on (−1, ∞) for x > 0, 1 ν π2 Γ ν + Φ (x) is completely monotonic while the function ν 7→ Ψν (x) = ν 2 and log-convex on − 21 , ∞ for x > 0. Moreover, the next Tur´ an type inequality is valid for all ν > 12 and x > 0 2

0 < [Φν (x)] − Φν−1 (x)Φν+1 (x) ≤

1 ν+

2

1 2

[Φν (x)] .

(8)

In addition, the left-hand side of (8) holds true for all ν > −1 and x > 0. Proof First we show that for each fixed b ∈ (0, 2] and each x > 0, the function ν 7→ Φν+b (x)/Φν (x) is decreasing, where ν ≥ −(b + 1)/2, ν > −1. For this, we consider Neumann’s formula [18, p. 441] Iα (x)Iβ (x) =

2 π

Z

π 2

Iα+β (2x cos θ) cos ((α − β)θ) dθ,

(9)

0

which holds for all α + β > −1. Using this we find that for 2ν + ε + b > −1 αν (x) = Iν (x)Iν+b+ε (x) − Iν+b (x)Iν+ε (x) Z π2 4 =− I2ν+b+ε (2x cos θ) sin(bθ) sin(εθ)dθ, π 0 βν (x) = Iν (x)Iν+b+ε+1 (x) − Iν+b (x)Iν+ε+1 (x) + Iν+1 (x)Iν+b+ε (x) − Iν+b+1 (x)Iν+ε (x) Z π2 8 =− I2ν+b+ε+1 (2x cos θ) sin(bθ) sin(εθ) cos θdθ, π 0


8

which are negative for all b ∈ (0, 2] and ε ∈ (0, 2]. Consequently, the expression Φν (x)Φν+b+ε (x)−Φν+b (x)Φν+ε (x) = e−2x x−(2ν+b+ε) [αν (x) + βν (x) + αν+1 (x)] is negative, that is, we obtain Φν+b+ε (x)/Φν+ε (x) < Φν+b (x)/Φν (x). Now, since ν 7→ Φν+b (x)/Φν (x) is decreasing, it follows that the function ν 7→ log[Φν+b (x)] − log[Φν (x)] is decreasing too. This implies that the function ν 7→ ∂ log[Φν (x)]/∂ν is decreasing on (−1, ∞). Now, since ν 7→ Φν (x) is logconcave, it follows that for all ν1 , ν2 > −1, x > 0 and α ∈ [0, 1] we have α

1−α

Φαν1 +(1−α)ν2 (x) ≥ [Φν1 (x)] [Φν2 (x)]

,

and choosing ν1 = ν − 1, ν2 = ν + 1 and α = 12 we arrive at left-hand side of the Tur´ an type inequality (8), but just for ν > 0. In what follows we show that this inequality is actually valid for all ν > −1. For this, observe that Φ ∆ν (x)

2

= [Φν (x)] −Φν−1 (x)Φν+1 (x) = e−2x x−2ν [∆ν (x) + ∆ν+1 (x) + Θν (x)] ,

where 2

∆ν (x) = [Iν (x)] − Iν−1 (x)Iν+1 (x) and Θν (x) = Iν (x)Iν+1 (x) − Iν−1 (x)Iν+2 (x). Since ∆ν (x) > 0 for all x > 0 and ν > −1 (see for example [2]), to prove the left-hand side of (8), we just need to show that the expression Θν (x) is positive. By using (9) we obtain Z π2 2 Θν (x) = I2ν+1 (2x cos θ) (cos θ − cos(3θ)) dθ π 0 Z π2 8 = I2ν+1 (2x cos θ)(cos θ) sin2 θ dθ > 0 π 0 for all ν > −1 and x > 0. Finally, since m Z 1 ∂ m Ψν (x) 1 2 ν− 12 = (1 − t)(1 − t ) (−1)m log e−x(1+t) dt > 0 2 ∂ν m 1 − t −1 for all ν > − 12 , x > 0 and m ∈ {0, 1, . . . }, indeedthe function ν 7→ Ψν (x) is completely monotonic and log-convex on − 21 , ∞ for x > 0. Consequently, for all ν1 , ν2 > − 12 , x > 0 and α ∈ [0, 1] we have α

1−α

Ψαν1 +(1−α)ν2 (x) ≤ [Ψν1 (x)] [Ψν2 (x)]

,

and choosing ν1 = ν − 1, ν2 = ν + 1 and α = 21 we arrive at the Turán type inequality 2 [Ψν (x)] − Ψν−1 (x)Ψν+1 (x) ≤ 0, which is equivalent to the right-hand side of (8).


9

3 Remarks on Kanter’s paper In this section we would like to point out an error in Kanter’s paper [8]. Namely, in the beginning of the proof of [8, Lemma 4.4] the author claimed that the inequality Z Z 1 π 1 π −2r(1−cos t) e (1 + cos t)dt ≥ (cos t)2r (1 + cos t)dt, (10) π 0 π 0 that is, r −2r Φ(2r) ≥ p(2r) = C2r 2 ,

(11)

is valid for r nonnegative integer. However, the proof of the inequality [8, p. 232] Φ(2r) Φ(2r + 2) ≤ (12) p(2r) p(2r + 2) is not correct, and hence the proof of (11) is not complete. Namely, there is a typographical error in the definition of p(2r) in [8, p. 232], and in the proof of the above inequality the author used the inequalities r 2r + 1 r + 1 Φ(2r + 2) Φ(2r + 2) Φ(2r) ≤ · ≤ , p(2r) 2r + 2 r p(2r + 2) p(2r + 2) but we can see that the second inequality is not true. In an effort to prove (12) we deduced part d of Theorem 1, however, this does not help us in the proof of (12). More precisely, by using part d of Theorem 1 for a = ν2 + 14 , we have that s s + 14 Φ(s + t) ≥ Φ(s) s + t + 41 for all s, t > 0, and consequently Φ(2r) 2r + 1 ≤ p(2r) 2r + 2

s

2r + 2 + 2r + 41

1 4

·

Φ(2r + 2) , p(2r + 2)

however, this is not less or equal than Φ(2r + 2)/p(2r + 2). Summarizing, the proof of [8, Lemma 4.4] is not complete, which implies that the proof of [8, Lemma 4.3] is not complete, and then the proof of [8, Theorem 4.1] is not complete too. However, [8, Theorem 4.1] has been proved recently by Mattner and Roos [11] by using a different approach, and this in particular implies that the inequality (11) is valid for r ∈ {0, 1, 2, . . . }. More precisely, Kanter [8, p. 222] pointed out that if the convex set is {0} and the random variables are considered to be ±1 valued random variables, then in particular [8, Theorem 4.1] implies (11). Now, in what follows we present an analytic proof of the extension of (11) to r ≥ 0. Namely, we prove the following result. We mention here that this result was proved by Mattner and Roos [11, eq. (29)] in a more general setting by using a completely different approach.


10

Theorem 3 The extension of Kanter’s inequality (11) to real variable, that is, Γ (2r + 1) −2r Φ(2r) = e−2r [I0 (2r) + I1 (2r)] ≥ 2 2 , (13) Γ (r + 1) is valid for r ≥ 0. Proof Observe that to prove (13) it is enough to show that (10) is valid for all r ≥ 0. The case r = 0 is obvious, and thus in what follows we assume that r > 0. First we consider the integral Z π 1 (cos t)2r (1 + cos t)dt ϕ(r) = π 0 and we show that the right-hand side of (10) and (13) coincide. Substituting x = cos t we get Z Γ r + 12 1 2 1 x2r 1 1 √ dx = B r + , ϕ(r) = =√ . π 0 π 2 2 πΓ (r + 1) 1 − x2 Making use of the Legendre duplication formula, 1 1 Γ (2z) = π − 2 22z−1 Γ (z)Γ z + 2 for z = r + 12 , we conclude that ϕ(r) =

Γ (2r + 1) −2r 2 , Γ 2 (r + 1)

as we required. Now, let us consider the integral Z πh i 1 A(r) = e2r(cos t−1) − (cos t)2r (1 + cos t) dt. π 0 To prove (10) we show that A(r) is positive for r > 0. For this we shall use the Okamura’s variant of the second integral mean–value theorem [12, 14], which states that if f : [a, b] → R is a monotone function and g : [a, b] → R is integrable, then there exists a ξ ∈ [a, b] such that Z b Z ξ Z b f (t)g(t)dt = f (a+ ) g(t)dt + f (b− ) g(t)dt . (14) a

a

ξ

Choosing [a, b] = [0, π], t 7→ f (t) = 1 + cos t, which is monotone (decreasing) and t 7→ g(t) = e2r(cos t−1) −(cos t)2r , which is integrable, for a fixed 0 ≤ ξ ≤ π, by (14) we have Z Z i i 2 1 h 2r(t−1) dt 2 ξ h 2r(cos t−1) e − (cos t)2r dt = e − t2r √ . A(r) = π 0 π cos ξ 1 − t2


11

Let hr (t) denote the integrand of the last integral, that is, hr (t) =

e2r(t−1) − t2r √ . 1 − t2

Now, we consider two cases. The first case is when ξ ∈ 0, π2 . In this case the integrand hr (t) is positive for all r > 0, since et−1 ≥ 1 + (t − 1) = t > 0 for t ∈ [cos ξ,1]. Consequently, we have A(r) > 0 for r > 0. The second case is when ξ ∈ π2 , π , that is, −1 ≤ cos ξ ≤ 0. Observe that in this case we have Z 0 Z 1 2 A(r) = hr (t)dt + hr (t)dt = Aξ + B, π cos ξ 0 and since t 7→ e2r(t−1) is monotone increasing on R for r > 0, and t 7→ 1 t2r (1 − t2 )− 2 is an even function, the first integral’s modulus |Aξ | cannot overgrow B. Indeed, Z − cos ξ Z − cos ξ π |Aξ | = hr (−t) dt < hr (t)dt 2 0 0 Z − cos ξ Z 1 π ≤ hr (t) dt + hr (t)dt = B . 2 0 − cos ξ Hence A(r) is nonnegative for all r > 0. This completes the proof.

Acknowledgement ´ Baricz was supported by the János Bolyai Research ScholarThe research of A. ship of the Hungarian Academy of Sciences. The authors wish to acknowledge the referee’s comments and suggestions which enhanced this paper.

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