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Some Remarks on the Self-Exponential Function: Minimum Value, Inverse Function, and Indefinite Integral

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Some Remarks on the Self-Exponential Function: Minimum Value, Inverse Function, and Indefinite Integral Hindawi Publishing Corporation International Journal of Analysis Volume 2014, Article ID 195329, 7 pages http://dx.doi.org/10.1155/2014/195329 Research Article Some Remarks on the Self-Exponential Function: Minimum Value, Inverse Function, and Indefinite Integral J. L. González-Santander and G. Martín Departamento de Ciencias Experimentales y Matematicas,´ Universidad CatolicadeValencia“SanVicenteM´ artir”,´ C/Guillem de Castro 96, 46001 Valencia, Spain Correspondence should be addressed to J. L. Gonzalez-Santander;´ [email protected] Received 9 September 2014; Accepted 10 November 2014; Published 23 November 2014 Academic Editor: Shamsul Qamar Copyright © 2014 J. L. Gonzalez-Santander´ and G. Mart´ın. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Considering the function as a real function of real variable, what is its minimum value? Surprisingly, the minimum value is − reached for a negative value of . Furthermore, considering the function () = , ∈R and >0, two different expressions −1 in closed form for the inverse function can be obtained. Also, two different series expansions for the indefinite integral of −1 and are derived. The latter does not seem to be found in the literature. 1. Introduction complex except for negative integers. However, [4]saysthat, for <0, () is only defined if can be written as −/, Let us consider the following real function of real variable, where and arepositiveintegersand is odd. We will use :R → R : thisfactlateroninordertoanswerthefirstquestion. =() = = ( ) , This paper is organized such that each section is devoted exp log (1) to each of the questions raised above. and let us pose the following questions. 2. Minimum Value (1) What is the minimum value of ()? The usual way to answer the first question is just to solve the (2) Can its inverse function be expressed in closed form? equation () =;thatis, 0 (3) Is its indefinite integral known? exp ( log )(log +1) =0, (3) The function () is termed as self-exponential function so in [1,Section26:14]andcoupled exponential function in [2, 1 Equation 01.20], using in the latter the notation () = = . (4) cxt(). Probably, the most well-known property of () is just its great growth rate. In fact, the rate of increase of () Since as →∞is greater than the exponential function or the 1 −1 ( )=1− >0, factorial function [3, Chapter I. section 5] (5) ≻!≻ . (2) then (4) is a local minimum. Moreover, since there are no more local extrema and is a smooth function, (4) is the Regarding the domain of (),in[1,Section26:14],() global minimum; thus, is only defined as a real function for positive values of ,and 1 () =( ) =−1/ ≈ 0.692201. [2, p. 10] states that, for arguments less than zero, () is min (6) 2 International Journal of Analysis Nonetheless, this reasoning fails, because it does not take Notice that, despite the fact we have considered the log() into account negative values of . Therefore, we need to define function as a multivalued function in (10), () is a single- firstthedomainof() for negative values of .Despitethe valued function in (13), because we are considering () as a fact that this is essentially done in [4], in order to answer the real function. Figure 1 shows the plot of (). According to first question, we provide the following derivation. (13),for<0,theplotof() lies on the following curves: () =±|| =± ( (−)), 2.1. Domain of (). Let us consider first the case =0, ± exp log (15) where the function () does not exist. However, applying L’ H opital’sˆ rule, the following limit is finite: with a numerable infinite number of points. Notice that + and − signs in (15) occur for even and odd positive integers in lim =1. (13),respectively. →0+ (7) Nevertheless, the right derivative of () at =0is 2.2. Minimum Value of (). In order to calculate the infinite: minimum value of (),for<0,letussolvetheequation () = 0 ± ;thatis, lim () = lim (log +1)=−∞. →0+ →0+ (8) ± exp ( log (−))[log (−) +1] =0. (16) For =0̸ ,wemayrewrite() by using the signum function sgn() as Thus, 1 () = exp ( log (sgn () ||)) =− . (17) (9) = || exp ( log (sgn )) . Notice that Now, for =0̸ and ∈Z,wehave −1 1+−1 ( )=∓ . (18) 0, >0 ± ( ) = { log sgn + 2 (−1) , <0 (10) So, +() has a maximum and −() has a minimum in −1/ ∉ =(2 −) 1 (−) , (17),whichagreeswithFigure 1.However, Dom , so we have to get the best rational approximation −/ to −1/ −/ ∈ where () is the Heaviside function. Therefore, by applying in such a way that Dom .Moreover,since the minimum lies on the −() curve, and must be both Euler’s formula = cos +sin ,weobtain odd positive integers. In order to do so, let us consider the () sequence = || ( (2 −) 1 (−) ) exp ⌊10 0⌋+⌊10 0⌋ (mod 2) +1 ( )=− , 0 10 −1 || , >0 (19) ={ || [cos ( (2 −) 1 ) +sin ( (2 −) 1 )] , <0. 0 ∈ R,∈N, (11) where ⌊⌋ denotes the floor function.Noticethatthenumer- Since () is a real function, the complex part of (11) has ator and the denominator of (19) are both odd, so () = to be zero. For >0, () is never complex, and for <0 −/ irreducible, with , being odd positive integers. the complex part of () is zero when Therefore, (0) is a sequence of rational numbers for which ( ( )) () ( ) 0 lies on the curve − .Also, 0 is a monotonic ( (2 −) 1 ) =0→=− <0, ,∈Z+. decreasing sequence that satisfies sin 2 − 1 (12) lim (0)=−0 . →∞ (20) Therefore, substituting (12) into (11),thefunction() is ∈Q given by Defining now the rational number, , −1 ,>0 { =min { ( )} . (21) () = ∈N {(−1) || ,=− <0, ,∈Z+ 2 − 1 { () (13) Then, the minimum value of is min () =() ≈ −1.44467, and its domain is ∈Dom (22) =R+ ∪{=− ,,∈Z+}. Dom 2 − 1 (14) which is different from (6), as aforementioned. International Journal of Analysis 3 4 1 3 2 1 1 2 3 f(x) x −3 −2 −112 −1 x −1 W(x) −2 −2 Figure 1: Plot of as a real function of real variable. −3 3. Inverse Function About the second question, a closed form expression for −4 the inverse function does not seem to be found in the W0(x) usual literature (see [2, Chapter 2]). However, by using W (x) the Lambert function [5], () is very easily inverted. −1 The Lambert function is defined as the inverse func- Figure 2: Branches of the Lambert function for real arguments. tion of and it is implemented in MATHEMATICA by the commands ProductLog or LambertW.TheLambert function is a multivalued function that presents, for real 3.5 arguments, two branches: 0() (principal branch) and −1(). Figure 2 shows the plot of both branches. 3.0 Let us consider now on a little more general function than 2.5 (1), but, for simplicity, only for positive arguments; that is, 2.0 − () == ,∈R, >0. (x) (23) f 1.5 () Figure 3 shows the plot of for different values of . 1.0 It is easy to prove that 0.5 (1) =1, 0.5 1.0 1.5 2.0 lim () =1, →0+ (24) x 1 () =0←→= ,=0,̸ =−2 =1 =−1 =2 =0 +∞, > 0, { − Figure 3: Plot of for different values of . lim () = 0, = 0, (25) →0+ { {−∞, < 0, which agrees with Figure 3. and thus Solving (23),wehave −1 log () = exp ( (− )) (28) −log = log (26) − log / =log log . = . (29) (−log /) Applying now the Lambert function and taking into account (26),weobtain According to Figure 3, notice that, depending on the −1() log value of , sometimes is a double-valued function, so (− )= log we have two real values of the Lambert function in (29), () () (27) that is, 0 and −1 .Inthislattercase,wehaveused log the notation −1,0().Also,fromFigure 3,wecanseethat =− , −1 () is a single-valued function for >1when <0and 4 International Journal of Analysis for ∈ (0, 1) when >0. Therefore, taking into account (24) but, from (34) and (40), we arrive at (33): and Figure 3, we can consider the following cases. 1 () = = log , (i) Consider <0, (41) ℎ (1/) 0 (log ) − / { log , ≤≤1, { which, according to (37),convergesifandonlyif {−1,0 (− log /) −1 () = { (30) − 1 1/ { − log / ≤ ≤ , { , >1. (42) {0 (− log /) that is, the interval given in (35). −1 (ii) Consider >0, In [2,Equation02.03], () is termed as coupled root function. Since the latter reference is unaware of the closed − log / { ,1≤≤, expression (33), it performs numerically the following limit { (− /) −1 { −1,0 log [2, Equation 02.07]: () = { (31) { − / { log , 0<<1, −1 () =0, {0 (− log /) lim (43) →∞ log where we have defined −1 in order to show that () goes to infinity at a lower rate 1 / than logarithms. In fact, (43) is quite easily proved applying = ( )= . (32) (33) and the property By setting =−1,in(30),weobtaintheinversefunction lim =∞→ lim 0 () =∞, of (1) →∞ →∞ (44) { log ,−1/ ≤≤1, so that { {−1,0 (log ) −1 −1 () = () 1 { (33) lim = lim =0. (45) { →∞ →∞ ( ) { log ,>1. log 0 log {0 (log ) 4. Indefinite Integral Curiously, (33) is just the closed expression given in [6] for the following power tower: −1 4.1. Indefinite Integral of . For <0, according to (30),let (1/)⋅⋅⋅ () =(1/) , (34) us calculate −1 − log / which converges if and only if ∫ () = ∫ , >1. (46) 1 1 0 (− log /) −1/ ≤≤. (35) Performing the change of variables =−log /,wehave In order to see this, consider the power tower − log / − −1 ⋅⋅⋅ ∫ () =−∫ , (47) ℎ () = , (36) 1 0 0 () − which converges if and only if [7, 8] and expanding the exponential function ,in(47),we − 1/ obtain ≤≤ . (37) +1 ∞ (−) − log / +1 Taking in (36) the logarithm of both sides and plugging −1 ∫ () = ∑ ∫ . (48) back the ℎ() function definition, we obtain 1 =0 ! 0 0 () ⋅⋅⋅ log ℎ= log =ℎlog . (38) Performing now the change of variables 0() = (i.e., = − ), we get Performing now the change of variables ℎ= ,weget ∞ (−)+1 =− . −1 (+2) log (39) ∫ () = ∑ ∫ (+1) , (49) 1 =0 ! 0 Now, by using the principal branch of Lambert function, we can solve, for ℎ(), wherewehaveset −0 (− log ) log ℎ () = , (40) :=0 (− ).
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