ON SIMPLE POLYNOMIAL BOUNDS FOR THE EXPONENTIAL FUNCTION

Asia Pac. J. Math. 2022 9:6 ON SIMPLE POLYNOMIAL BOUNDS FOR THE EXPONENTIAL FUNCTION CHRISTOPHE CHESNEAU1 , YOGESH J. BAGUL2,∗ , RAMKRISHNA M. DHAIGUDE3 1 2 Université de Caen Normandie, LMNO, Campus II, Science 3, 14032, Caen, France Department of Mathematics, K. K. M. College, Manwath, Dist: Parbhani(M. S.)-431505, India 3 Department of Mathematics, Government Vidarbha Institute of Science and Humanities, Amravati(M.S.)-444604, India ∗ Corresponding author: yjbagul@gmail.com Received Jan. 3, 2022 Abstract. In this article, we offer a new polynomial or polynomial-exponential bounds for the exponential function. Its main interest is to be both simple and sharp, under some clear conditions on the parameters involved. Applications are given for a probability function and the Kummer beta function. 2010 Mathematics Subject Classification. 26D07, 33B10, 33B20. Key words and phrases. exponential inequality; productlog function; normal distribution; Kummer beta function. 1. Introduction The natural exponential functions are extremely important in many branches of science and mathematics. Sometimes we require the bounds of such a function on the interval [0, 1] for a specific purpose. One obvious upper bound is given in the following inequality: For any x ∈ [0, 1), (1) ex ≤ 1 . 1−x The inequality (1) is coarser and its refinement is given in [3]. For some other sharp bounds, we refer to [1,3] and the references therein. The bounds in the present literature are somewhat complex in nature; there is still a need for tractable and sharp bounds in all branches of DOI: 10.28924/APJM/9-6 ©2022 Asia Pacific Journal of Mathematics 1 Asia Pac. J. Math. 2022 9:6 2 of 7 mathematics. The main goal of this paper is to achieve useful and simpler polynomial or polynomial-exponential bounds for exponential functions than those available in the literature. 2. Main Result The following result presents the main finding of the study. Proposition 1. For any x ∈ [0, 1] and a ∈ R, we have   sign(a)eax ≤ sign(a) ax(1 − x) + x2 (ea − 1) + 1 , where sign(a) = −1 if a < 0, sign(0) = 0 and sign(a) = 1 for a > 0. Proof. We propose a proof based on the analysis of appropriate functions. To begin, let us consider the following function: (2) f (x; a) = eax − ax(1 − x) − x2 (ea − 1) − 1. Then, upon differentiation, we have f ′ (x; a) = aeax + 2(a − ea + 1)x − a. The equation f ′ (x; a) = 0 has two solutions only into [0, 1], which are x0 = 0 and x1 = [ya − W (ya eya )] /a, where ya = a2 /[2(a + 1 − ea )] and W (x) is the productlog function defined by W (x)eW (x) = x. Clearly, the nature of x0 and x1 are informative on the possible sign of f (x; a). Since f (x; a) is continuous with f (0; a) = 0 and f (1; a) = 0, it is enough to study the nature of the extremum x0 ; if x0 a local minimum, then x1 is a local maximum, and vice versa. First, let us notice that f ′′ (x; a) = a2 eax + 2(a − ea + 1), which implies that f ′′ (0; a) = a2 + 2(a − ea + 1) = 2φ(a), where φ(a) = 1 + a + a2 − ea . 2 Since ea > 1 + a for any a ∈ R∗ , we have φ′ (a) = 1 + a − ea < 0, which implies that φ(a) is decreasing for any a ∈ R. Let us now study the nature of the extremum x = 0 according to a > 0 and a < 0 via the second derivative test. Asia Pac. J. Math. 2022 9:6 3 of 7 • For a > 0, we have φ(a) < φ(0) = 0, implying that f ′′ (x0 ; a) = f ′′ (0; a) < 0. Thus, x0 is a local maximum point, implying that x1 is necessary a local minimum point: For any x ∈ [0, 1] and a > 0, we have f (x; a) ≤ min(f (0; a), f (1; a)) = 0, which implies the desired inequality. • For a < 0, we have φ(a) > φ(0) = 0, implying that f ′′ (x0 ; a) = f ′′ (0; a) > 0. Thus, x0 is a local minimum point, implying that x1 is necessary a local maximum point: For any x ∈ [0, 1] and a < 0, we have f (x; a) ≥ max(f (0; a), f (1; a)) = 0, which implies the desired inequality. This ends the proof.  Remark 1. For the case a > 0, some alternative proofs can be given. By the series expansion of the exponential function, since x ∈ [0, 1], we have e ax −1= +∞ X (ax)k k! k=1 = ax + +∞ X (ax)k k=2 k! ≤ ax + x 2 +∞ k X a k=2 k! = ax + x2 (ea − 1 − a) = ax(1 − x) + x2 (ea − 1). Remark 2. For the case a < 0 and x ∈ [0, 1], the proposed lower bound improves the famous inequality eax ≥ 1 + ax. Indeed, by using ea ≥ 1 + a, we have eax ≥ ax(1 − x) + x2 (ea − 1) + 1 = 1 + ax + x2 (ea − 1 − a) ≥ 1 + ax. Remark 3. Proposition 1 can be extended to any bounded interval of the form [0, c] for x, with c > 0. In this case, it is enough to replace x by x/c. That is, for any x ∈ [0, c], since x/c ∈ [0, 1], we have sign(a)eax/c ≤ sign(a)  1  ax(c − x) + x2 (ea − 1) + c2 , 2 c Remark 4. For x ∈ [0, 1] and a = x or a = −x, Proposition 1 yields simple polynomial bounds for 2 2 ex and e−x , respectively. More precisely, we have: 2 ex ≤ x2 (1 − x) + x2 (ex − 1) + 1 and 2 −x2 (1 − x) + x2 (e−x − 1) + 1 ≤ e−x . Asia Pac. J. Math. 2022 9:6 2 4 of 7 2 The functions ex and e−x are involved in a plethora of mathematical and physical quantities, more or less complex. Our bounds can be of interest for direct bounds of these quantities. See, for instance, [2]. A graphical illustration of Proposition 1 is given in Figure 1. It shows the curve of the a = −1 a = −1 a = −1 a = −1 3 4 5 6 f(x;a) 2e−04 4e−04 −6e−04 −8e−04 f(x;a) −4e−04 6e−04 −2e−04 0e+00 8e−04 function f (x; a) defined by (2) for some values of a > 0 and a < 0. 0.0 3 4 5 6 0e+00 −1e−03 a=1 a=1 a=1 a=1 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 x 0.6 0.8 1.0 x (a) (b) Figure 1. Curves of f (x; a) as defined by (2) for x ∈ [0, 1] and (a) some values of a > 0 and (b) some values of a < 0. In Figure 1, we clearly identify the extrema x0 and x1 as described in the proof of Proposition 1. Also, the sharpness of the obtained bounds can be observed; the maximum of magnitude being between 10−4 and 10−3 for the considered values of a. 3. Applications Two applications of Proposition 1 are examined in this section. 3.1. Bound of a useful probability. Let X be a random variable with the standard normal 2 /2 distribution, i.e., with probability density function f (x) = 1/(2π)−1/2 e−x , x ∈ R. Then, the following proposition gives an evaluation of the probability that the event {0 ≤ X ≤ t} occurs with t ∈ [0, 1]. Proposition 2. Let X be a random variable with the standard normal distribution. For any t ∈ [0, 1], we have     1 3 3 1 4 t P (0 ≤ X ≤ t) ≥ √ − t + t + 8γ 3, +t , 2 8 2 2π where γ(a, x) = Rx 0 ta−1 e−t dt is the standard incomplete gamma function. Asia Pac. J. Math. 2022 9:6 5 of 7 Proof. By applying Proposition 1 with x ∈ [0, 1] and a = −x/2, we have e−x 2 /2 1 ≥ − x2 (1 − x) + x2 (e−x/2 − 1) + 1, 2 which implies that   Z t Z t Z Z 1 t 2 1 1 t 3 2 2 −x/2 P (0 ≤ X ≤ t) ≥ √ x dx + t xe dx − x dx + x dx + − 2 0 2 0 2π 0 0     t 3 3 1 4 1 +t . − t + t + 8γ 3, =√ 2 8 2 2π The proof of Proposition 2 ends.  In Proposition 2, it is intriguing to see how polynomial and gamma functions appear to bound a probability function. Figure 2 illustrates the lower bound in Proposition 2 by showing the curve of the function    x 1 3 3 1 4 F (x) = P (0 ≤ X ≤ x) − √ − x + x + 8γ 3, +x 2 8 2 2π 0.2 0.0 0.1 F(x) 0.3 0.4 for x ∈ [0, 1]. 0.0 0.2 0.4 0.6 0.8 1.0 x Figure 2. Curve for F (x) for x ∈ [0, 1]. In Figure 2, we see that the bound is very sharp for x ∈ [0, 0.6]. However, we do not claim that it is the “sharpest lower bound ever” for P (0 ≤ X ≤ t), but just an interesting application of our main result. Asia Pac. J. Math. 2022 9:6 6 of 7 Remark 5. Proposition 2 can be used to bound the cumulative distribution function of X; Since, for t ∈ [0, 1], P (X ≤ t) = 1/2 + P (0 ≤ X ≤ t), we have     1 3 3 1 4 1 t P (X ≤ t) ≥ √ − t + t + 8γ 3, +t − , 2 8 2 2 2π 3.2. Bound of the Kummer beta function. Proposition 1 can be used for approximation purposes. For instance, let us consider the Kummer beta function defined by (3) I(a, α, β) = Z 1 0 eax xα−1 (1 − x)β−1 dx, with a ∈ R, α > 0 and β > 0. This function has found numerous applications in probability and statistics. See [4–7] in this regard, and it remained complicated to evaluate with simple functions. Thanks to Proposition 1, the following result can be proved. Proposition 3. Let I(a, α, β) be the Kummer beta function as defined by (3). For any a ∈ R, we have sign(a)I(a, α, β) ≤ sign(a) [aB(α + 1, β + 1) + (ea − 1)B(α + 2, β) + B(α, β)] , where B(α, β) = R1 0 xα−1 (1 − x)β−1 dx is the standard beta function. Proof. The proof is a direct application of Proposition 1 and basic integral properties: • For a > 0, by applying Proposition 1, we get eax ≤ ax(1 − x) + x2 (ea − 1) + 1, which implies that, after some developments, I(a, α, β) ≤ aB(α + 1, β + 1) + (ea − 1)B(α + 2, β) + B(α, β). • For a < 0, by also applying Proposition 1, we get the reverse inequality: I(a, α, β) ≥ aB(α + 1, β + 1) + (ea − 1)B(α + 2, β) + B(α, β), The proof of Proposition 3 is complete.  Thus, thanks to Proposition 3, the standard beta function can be used to evaluate quite precisely the Kummer beta function. Other applications can be given; this study just opens a door for more in this direction. Asia Pac. J. Math. 2022 9:6 7 of 7 References [1] J. Bae, Optimal polynomial bounds for the exponential function, Math. Inequal. Appl. 3 (2013), 763–782. https://doi.org/10.7153/mia-16-58. [2] I.S. Gradshteyn, I.M. Ryzhik, Table of integrals, series, and products, Academic Press, New York, 2007. [3] S.-H. 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