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A Neutrosophic Binomial Factorial Theorem with Their Refrains Neutrosophic Sets and Systems, Vol. 14, 2016 7 University of New Mexico A Neutrosophic Binomial Factorial Theorem with their Refrains Huda E. Khalid1 Florentin Smarandache2 Ahmed K. Essa3 1 University of Telafer, Head of Math. Depart., College of Basic Education, Telafer, Mosul, Iraq. E-mail: [email protected] 2 University of New Mexico, 705 Gurley Ave., Gallup, NM 87301, USA. E-mail: [email protected] 3 University of Telafer, Math. Depart., College of Basic Education, Telafer, Mosul, Iraq. E-mail: [email protected] Abstract. The Neutrosophic Precalculus and the Two other important theorems were proven with their Neutrosophic Calculus can be developed in many corollaries, and numerical examples as well. As a ways, depending on the types of indeterminacy one conjecture, we use ten (indeterminate) forms in has and on the method used to deal with such neutrosophic calculus taking an important role in indeterminacy. This article is innovative since the limits. To serve article's aim, some important form of neutrosophic binomial factorial theorem was questions had been answered. constructed in addition to its refrains. Keyword: Neutrosophic Calculus, Binomial Factorial Theorem, Neutrosophic Limits, Indeterminate forms in Neutrosophic Logic, Indeterminate forms in Classical Logic. 1 Introduction (Important questions) 2. Let 3 Q 1 What are the types of indeterminacy? √퐼 = 푥 + 푦 퐼 3 2 2 2 3 3 There exist two types of indeterminacy 0 + 퐼 = 푥 + 3푥 푦 퐼 + 3푥푦 퐼 + 푦 퐼 3 2 2 3 a. Literal indeterminacy (I). 0 + 퐼 = 푥 + (3푥 푦 + 3푥푦 + 푦 )퐼 3 As example: 푥 = 0, 푦 = 1 → √퐼 = 퐼. (5) 2 + 3퐼 (1) In general, b. Numerical indeterminacy. 2푘+1√퐼 = 퐼, (6) As example: where 푘 ∈ 푧+ = {1,2,3, … }. 푥(0.6,0.3,0.4) ∈ 퐴, (2) Basic Notes meaning that the indeterminacy membership = 0.3. 1. A component I to the zero power is Other examples for the indeterminacy com- undefined value, (i.e. 퐼0 is undefined), 퐼 ponent can be seen in functions: 푓(0) = 7 표푟 9 or since 퐼0 = 퐼1+(−1) = 퐼1 ∗ 퐼−1 = which is 2 퐼 푓(0 표푟 1) = 5 or 푓(푥) = [0.2, 0.3] 푥 … etc. impossible case (avoid to divide by 퐼). 2. The value of 퐼 to the negative power is Q 2 What is the values of 퐼 to the rational power? undefined value (i.e. 퐼−푛 , 푛 > 0 is 1. Let undefined). √퐼 = 푥 + 푦 퐼 Q 3 What are the indeterminacy forms in neutros- 0 + 퐼 = 푥2 + (2푥푦 + 푦2)퐼 ophic calculus? 푥 = 0, 푦 = ±1. (3) In classical calculus, the indeterminate forms In general, are [4]: 2푘 0 ∞ √퐼 = ±퐼 (4) , , 0 ∙ ∞ , ∞0, 00, 1∞, ∞ − ∞. (7) 0 ∞ where 푘 ∈ 푧+ = {1,2,3, … }. Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains 8 Neutrosophic Sets and Systems, Vol. 14, 2016 The form 0 to the power 퐼 (i.e. 0퐼 ) is an [2.1, 2.5] [2.1, 2.5] ∴ lim ln 푦 = lim = indeterminate form in Neutrosophic calculus; it is 푥→0 푥→0 1 1 tempting to argue that an indeterminate form of ln 푥 ln 0 [2.1, 2.5] [2.1, 2.5] type 0퐼 has zero value since "zero to any power is = = 퐼 1 −0 zero". However, this is fallacious, since 0 is not a −∞ power of number, but rather a statement about 2.1 2.5 = [ , ] = (−∞, −∞) limits. −0 −0 = −∞ Q 4 What about the form 1퐼? Hence 푦 = 푒−∞ = 0 The base "one" pushes the form 1퐼 to one OR it can be solved briefly by while the power 퐼 pushes the form 1퐼 to I, so 1퐼 is 푦 = 푥[2.1,2.5] = [02.1, 02.5] = [0,0] = 0. an indeterminate form in neutrosophic calculus. Example (3.2) Indeed, the form 푎퐼, 푎 ∈ 푅 is always an lim [3.5,5.9]푥[1,2] = [3.5,5.9] [9,11][1,2] = indeterminate form. 푥→[9,11] Q 5 What is the value of 푎퐼 , 푤 ℎ푒푟푒 푎 ∈ 푅? [3.5,5.9] [91, 112] = [(3.5)(9), (5.9)(121)] = 푥 퐼 Let 푦1 = 2 , 푥 ∈ 푅 , 푦2 = 2 ; it is obvious that [31.5,713.9]. lim 2푥 = ∞ , lim 2푥 = 0 , lim 2푥 = 1; while 푥→∞ 푥→−∞ 푥→0 Example (3.3) we cannot determine if 2퐼 → ∞ 표푟 0 표푟 1, lim [3.5,5.9] 푥[1,2] = [3.5,5.9] ∞[1,2] 푥→∞ 퐼 therefore we can say that 푦2 = 2 indeterminate = [3.5,5.9] [∞1, ∞2] form in Neutrosophic calculus. The same for 푎퐼, = [3.5 ∙ (∞) ,5.9 ∙ (∞)] where 푎 ∈ 푅 [2]. = (∞, ∞) = ∞. 2 Indeterminate forms in Neutrosophic Example (3.4) Logic Find the following limit using more than one [4,5]∙푥+1−1 It is obvious that there are seven types technique lim √ . 푥 of indeterminate forms in classical calculus [ ], 푥→0 4 Solution: 0 ∞ , , 0. ∞, 00, ∞0, 1∞, ∞ − ∞. The above limit will be solved firstly by using the 0 ∞ L'Hôpital's rule and secondly by using the As a conjecture, we can say that there are ten rationalizing technique. forms of the indeterminate forms in Neutrosophic Using L'Hôpital's rule calculus 1 −1 퐼 ∞ lim ([4, 5] ∙ 푥 + 1) ⁄2 [4,5] 퐼0 , 0퐼, , 퐼 ∙ ∞, , ∞퐼, 퐼∞, 퐼퐼, 푥→0 2 0 퐼 [4,5] 푎퐼(푎 ∈ 푅), ∞ ± 푎 ∙ 퐼 . = lim 푥→0 2√([4, 5] ∙ 푥 + 1) Note that: [4,5] 4 5 = = [ , ] = [2,2.5] 퐼 1 2 2 2 = 퐼 ∙ = 퐼 ∙ ∞ = ∞ ∙ 퐼. 0 0 Rationalizing technique [3] √[4,5] ∙ 푥 + 1 − 1 √[4,5] ∙ 0 + 1 − 1 3 Various Examples lim = Numerical examples on neutrosophic limits 푥→0 푥 0 would be necessary to demonstrate the aims of this √[4 ∙ 0, 5 ∙ 0] + 1 − 1 √[0, 0] + 1 − 1 = = work. 0 0 √0 + 1 − 1 0 Example (3.1) [1], [3] = = The neutrosophic (numerical indeterminate) values 0 0 = undefined. can be seen in the following function: Find lim 푓(푥), where 푓(푥) = 푥[2.1,2.5]. Multiply with the conjugate of the numerator: 푥→0 Solution: Let 푦 = 푥[2.1,2.5] → ln 푦 = [2.1, 2.5] ln 푥 Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains Neutrosophic Sets and Systems, Vol. 14, 2016 9 √[4, 5]푥 + 1 − 1 √[4, 5]푥 + 1 + 1 Again, Solving by using L'Hôpital's rule lim ∙ 푥→0 푥 √[4, 5]푥 + 1 + 1 푥2 + 3푥 − [1, 2]푥 − [3, 6] lim 2 푥→−3 푥 + 3 (√[4, 5]푥 + 1) − (1)2 2 푥 + 3 − [1, 2] = lim = lim 푥→0 푥 (√[4, 5]푥 + 1 + 1) 푥→−3 1 2 (−3) + 3 − [1, 2] [4, 5] ∙ 푥 + 1 − 1 = lim = lim 푥→−3 1 푥→0 푥 ∙ (√[4, 5]푥 + 1 + 1) = −6 + 3 − [1, 2] [4, 5] ∙ 푥 = −3 − [1, 2] = lim 푥→0 푥 ∙ (√[4, 5]푥 + 1 + 1) = [−3 − 1, −3 − 2] = [−5, −4] [4, 5] = lim 푥→0 (√[4, 5]푥 + 1 + 1) The above two methods are identical in results. [4, 5] [4, 5] = = 4 New Theorems in Neutrosophic Limits √1 + 1 (√[4, 5] ∙ 0 + 1 + 1) Theorem (4.1) (Binomial Factorial ) [4, 5] 4 5 1 푥 = = [ , ] = [2, 2.5]. lim (퐼 + ) = 퐼푒 ; I is the literal indeterminacy, 2 2 2 푥→∞ 푥 e = 2.7182828 Identical results. Proof Example (3.5) 푥 0 1 1 푥 1 푥 1 Find the value of the following neutrosophic limit (퐼 + ) = ( ) 퐼푋 ( ) + ( ) 퐼푋−1 ( ) 푥 0 푥 1 푥 푥2+3푥−[1,2]푥−[3,6] lim using more than one 푥 1 2 푥 1 3 푥→−3 푥+3 + ( ) 퐼푋−2 ( ) + ( ) 퐼푋−3 ( ) technique . 2 푥 3 푥 푥 1 4 Analytical technique [1], [3] + ( ) 퐼푋−4 ( ) + ⋯ 푥2+3푥−[1,2]푥−[3,6] 4 푥 lim 1 퐼 1 푥→−3 푥+3 = 퐼 + 푥. 퐼. + (1 − ) By substituting 푥= -3 , 푥 2! 푥 2 퐼 1 2 퐼 1 2 (−3) + 3 ∙ (−3) − [1, 2] ∙ (−3) − [3, 6] + (1 − ) (1 − ) + (1 − ) (1 − ) lim 3! 푥 푥 4! 푥 푥 푥→−3 −3 + 3 3 9 − 9 − [1 ∙ (−3), 2 ∙ (−3)] − [3, 6] (1 − ) + ⋯ = 푥 0 1 It is clear that → 0 푎푠 푥 → ∞ 0 − [−6, −3] − [3, 6] [3, 6] − [3,6] 푥 = = 1 퐼 퐼 퐼 0 0 ∴ lim(퐼 − )푥 = 퐼 + 퐼 + + + + ⋯ = 퐼 + [ ] [ ] 푥→∞ 푥 2! 3! 4! 3 − 6, 6 − 3 −3, 3 푛 = = , ∑∞ 퐼 0 0 푛=1 푛! 0 which has undefined operation , since 0 ∈ 1 푥 ∞ 1 0 ∴ lim(퐼 + ) = 퐼푒, where e = 1 + ∑ 푛=1 , I is the 푥→∞ 푥 푛! [−3, 3]. Then we factor out the numerator, and literal indeterminacy . simplify: 푥2 + 3푥 − [1, 2]푥 − [3, 6] lim = Corollary (4.1.1) 푥→−3 푥 + 3 1 (푥 − [1, 2]) ∙ (푥 + 3) lim(퐼 + 푥)푥 = 퐼푒 lim = lim (푥 − [1,2]) 푥→0 푥→−3 (푥 + 3) 푥→−3 Proof:- 1 = −3 − [1,2] = [−3, −3] − [1,2] Put 푦 = = −([3,3] + [1,2]) = [−5, −4]. 푥 It is obvious that 푦 → ∞ , as 푥 → 0 1 1 ∴ lim(퐼 + 푥)푥 = lim (퐼 + )푦 = 퐼푒 푥→0 푦→∞ 푦 ( using Th. 4.1 ) Corollary (4.1.2) 푘 lim (퐼 + )푥 = 퐼푒푘 , where k > 0 & 푘 ≠ 0 , I is the 푥→∞ 푥 literal indeterminacy. Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains 10 Neutrosophic Sets and Systems, Vol. 14, 2016 Proof 푦 1 = 푙푛푎. = 푙푛푎. 푥 푘 ( ) 1 푘 푘 ln 푦 + 퐼 ln(푦 + 퐼) lim (퐼 + )푥 = lim [(퐼 + )푘] 푦 푥→∞ 푥 푥→∞ 푥 푘 푘 1 Put 푦 = → 푥푦 = 푘 → 푥 = = 푙푛푎.
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