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. 푥 푦 1 ln(푦 + 퐼)푦 Note that 푦 → 0 푎푠 푥 → ∞ (ln 푎)(퐼푎푥 − 퐼) 1 1 푘 푘 푥 ∴ lim = 푙푛푎 ∴ lim (퐼 + ) = lim [(퐼 + 푦)푦] 푥→0 푥푙푛푎 + 푙푛퐼 1 푥→∞ 푥 푦→0 lim 푙푛(푦 + 퐼)푦 (using corollary 4.1.1 ). 푦→0 1 1 푘 = 푙푛푎 . = [lim(퐼 + 푦)푦] = (퐼푒)푘 = 퐼푘푒푘 = 퐼푒푘 1 푦→0 푙푛 lim(푦 + 퐼)푦 푦→0 1 Corollary (4.1.3) = 푙푛푎 using corollary (4.1.1) 1 1 푙푛(퐼푒) 푥 푘 lim(퐼 + )푥 = (퐼푒)푘 = √퐼푒 , 푙푛푎 푙푛푎 푥→0 푘 = = where 푘 ≠ 1 & 푘 > 0. 푙푛 퐼 + 푙푛푒 푙푛퐼 + 1

Proof Corollary (4.2.1)

The immediate substitution of the value of 푥 in the 퐼푎푘푥 − 퐼 푘 푙푛푎 ∞ lim = above limit gives indeterminate form 퐼 , 푥→0 푙푛퐼 푥 + 1 + 푙푛퐼 푥 1 0 1 푙푛푎푘 i.e. lim(퐼 + )푥 = lim(퐼 + )0 = 퐼∞ 푥→0 푘 푥→0 푘 Proof 푦 So we need to treat this value as follow:- Put 푦 = 푘푥 → 푥 = 1 1 푘 푥 1 푥 푘 푘 푥 푘 푘 푦 → 0 푎푠 푥 → 0 lim(퐼 + )푥 = lim [(퐼 + )푥] = [lim(퐼 + )푥] 푥→0 푘 푥→0 푘 푥→0 푘 퐼푎푘푥−퐼 퐼푎푦−퐼 퐼푎푦−퐼 푥 1 1 lim 푙푛퐼 = lim 푦 푙푛퐼 = 푘. lim 푙푛퐼 put 푦 = → 푥 = 푘푦 → = 푥→0 푥+ 푦→0 + 푦→0 푦+ 푘 푥 푘푦 푙푛푎푘 푘 푘 푙푛푎 푙푛푎 As 푥 → 0 , 푦 → 0 using Th. (4.2) 1 1 푙푛푎 1 푥 푥 푘 = 푘. ( ) lim (퐼 + ) = lim [(퐼 + 푦)푦] 1 + 푙푛퐼 푥→0 푘 푦→0 1 Corollary (4.2.2) 1 푘 푥 = [lim(퐼 + 푦)푦] 퐼푒 − 퐼 1 푦→0 lim = 푥→0 푥 + 푙푛퐼 1 + 푙푛퐼 Using corollary (4.1.1) Proof 퐼 푘 푥 = (퐼푒)푘 = √퐼푒 Let 푦 = 퐼푒 − 퐼 , 푦 → 0 푎푠 푥 → 0 푦 + 퐼 = 퐼푒푥 → ln(푦 + 퐼) = 푙푛퐼 + 푥 푙푛푒 Theorem (4.2) 푥 = ln(푦 + 퐼) − 푙푛퐼 푥 (푙푛푎)[퐼푎푥−퐼] 푙푛푎 퐼푒 − 퐼 푦 lim = ∴ = 푥→0 푥푙푛푎+푙푛퐼 1+푙푛퐼 푥 + 푙푛퐼 ln(푦 + 퐼) − 푙푛퐼 + 푙푛퐼 Where 푎 > 0, 푎 ≠ 1 푥 푥 = (푙푛푎)[퐼푎 −퐼] 퐼푎 −퐼 1 Note that lim = lim 푙푛퐼 푥→ 0 푥푙푛푎 1 +푙푛퐼 푥→0 푥+ ln(푦 + 퐼) 푙푛푎 푦 Proof 1 = Let 푦 = 퐼푎푥 − 퐼 → 푦 + 퐼 = 퐼푎푥 → ln(푦 + 퐼) = 1 ln(푦 + 퐼)푦 ln 퐼 + ln 푎푥 퐼푒푥 − 퐼 1 → ln(푦 + 퐼) = ln 퐼 + 푥푙푛푎 → ∴ lim = lim 푥→0 푥 + 푙푛퐼 푦→0 1 ln(푦 + 퐼) − 푙푛퐼 ln(푦 + 퐼)푦 푥 = 푙푛푎 1 (ln 푎)(퐼푎푥 − 퐼) 퐼푎푥 − 퐼) = 1 = 푦 푙푛퐼 ln lim(푦 + 퐼) 푥푙푛푎 + 푙푛퐼 푥 + 푦→0 ( 푙푛푎 푦 using corollary (4.1.1) = 1 1 1 ln(푦 + 퐼) − 푙푛퐼 푙푛퐼 + = = 푙푛푎 푙푛푎 ln (퐼푒) 푙푛퐼 + 푙푛푒 푙푛퐼 + 1

Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains Neutrosophic Sets and Systems, Vol. 14, 2016 11

Corollary (4.2.3) 5 Numerical Examples 퐼푒푘푥 − 퐼 푘 lim = Example (5.1) 푥→0 푙푛퐼 4푥 푥 + 1 + 푙푛퐼 퐼5 −퐼 Evaluate the limit lim 푙푛퐼 푘 푥→0 푥+ Proof 푙푛54 푦 Solution let 푦 = 푘푥 → 푥 = 푘 퐼54푥−퐼 4푙푛5 lim 푙푛퐼 = (using corollary 4. 2.1) 푦 → 0 푎푠 푥 → 0 푥→0 푥+ 1+푙푛퐼 푙푛54 퐼푒푘푥−퐼 퐼푒푦−퐼 퐼푒푦−퐼 lim 푙푛퐼 = lim 푦 푙푛퐼 = 푘. lim using 푥→0 푥+ 푦→0 + 푦→0 푦+푙푛퐼 Example (5.2) 푘 푘 푘 퐼푒4푥−퐼 Corollary (4.2.2) to get Evaluate the limit lim 푥→0 퐼32푥−퐼 1 푘 = 푘. ( ) = Solution 1 + 푙푛퐼 1 + 푙푛퐼 푙푛3[퐼푒4푥 − 퐼] 푙푛퐼 ∗ (푥 + ) 푙푛퐼 4 Theorem (4.3) 퐼푒4푥 − 퐼 (푥 + ) ln (퐼 + 푘푥) lim = lim 4 푥→0 퐼32푥 − 퐼 푥→0 푙푛3[퐼32푥 − 퐼] 푙푛퐼 lim = 푘(1 + 푙푛퐼) ∗ (푥 + ) 푥→0 푥 푙푛퐼 푙푛32 (푥 + ) Proof 푙푛32 ln (퐼 + 푘푥) ln(퐼 + 푘푥) − 푙푛퐼 + 푙푛퐼 푙푛3[퐼푒4푥 − 퐼] lim = lim lim 푥→0 푥→0 푙푛퐼 푙푛퐼 푥 푥 푥→0 (푥 + ) lim (푥 + ) Let 푦 = ln(퐼 + 푘푥) − 푙푛퐼 → 푦 + 푙푛퐼 = ln(퐼 + 4 푥→0 4 = 2푥 ∗ ) 푙푛3[퐼3 − 퐼] 푙푛퐼 푘푥 lim lim (푥 + 2) 푦 푙푛퐼 푦 푥→0 푙푛퐼 푥→0 푙푛3 푒 푒 − 퐼 퐼 푒 − 퐼 (푥 + 2) 푒푦+푙푛퐼 = 퐼 + 푘푥 → 푥 = = 푙푛3 푘 푘 (using corollary (4.2.3) on numerator & corollary 푦 → 0 푎푠 푥 → 0 (4.2.1) on denominator ) ln(퐼 + 푘푥) − 푙푛퐼 + 푙푛퐼 lim 4 푙푛퐼 푥→0 푥 = 1 + 푙푛퐼 ∗ 4 = 1. 푦 + 푙푛퐼 2푙푛3 푙푛퐼 = lim 푦 2 푦→0 퐼 푒 − 퐼 1 + 푙푛퐼 푙푛3 푘 푘 푘 5 Conclusion lim 퐼 푒푦−퐼 = 푦 푦→0 lim(퐼 푒 −퐼) 푦+푙푛퐼 푦→0 푦+푙푛퐼 In this article, we introduced for the first time using corollary (4.2.2) to get the result a new version of binomial factorial theorem 푘 containing the literal indeterminacy (I). This = = 푘(1 + 푙푛퐼) 1 theorem enhances three corollaries. As a 1 + 푙푛퐼 conjecture for indeterminate forms in classical Theorem (4.4) calculus, ten of new indeterminate forms in Neutrosophic calculus had been constructed. Prove that, for any two real numbers 푎, 푏 Finally, various examples had been solved. 퐼a푥−퐼 lim = 1 , where 푎, 푏 > 0 & 푎, 푏 ≠ 1 푥→0 퐼 푥−퐼 Proof b References The direct substitution of the value 푥 in the above limit conclude that 0 ,so we need to treat it as [1] F. Smarandache. Neutrosophic Precalculus and 0 Neutrosophic Calculus. EuropaNova Brussels, follow: 2015. 푥 푙푛a[퐼a − 퐼] 푥푙푛a + 푙푛퐼 [2] F. Smarandache. Introduction to Neutrosophic 퐼a푥 − 퐼 ∗ lim = lim 푥푙푛a + 푙푛퐼 푙푛a Statistics. Sitech and Education Publisher, Craiova, 푥→0 퐼b푥 − 퐼 푥→0 푙푛b[퐼b푥 − 퐼] 푥푙푛b + 푙푛퐼 2014. ∗ 푥푙푛b + 푙푛퐼 푙푛b [3] H. E. Khalid & A. K. Essa. Neutrosophic Pre- 푙푛a[퐼a푥 − 퐼] calculus and Neutrosophic Calculus. Arabic lim lim( 푥푙푛a + 푙푛퐼) 푥푙푛a + 푙푛퐼 푙푛b version of the book. Pons asbl 5, Quai du Batelage, = 푥→푥 ∗ 푥→0 ∗ 푙푛b[퐼b푥 − 퐼] lim ( 푥푙푛b + 푙푛퐼) 푙푛a Brussells, Belgium, European Union 2016. lim 푥→푥 푥푙푛b + 푙푛퐼 푥→0 [4] H. Anton, I. Bivens & S. Davis, Calculus, 7th (using Th.(4.2) twice (first in numerator second in Edition, John Wiley & Sons, Inc. 2002. denominator )) 푙푛a 푙푛퐼 푙푛b = 1+푙푛퐼 ∗ ∗ = 1. 푙푛b 푙푛퐼 푙푛a Received: November 7, 2016. Accepted: November 14, 2016 1+푙푛퐼

Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains