Remarks About the Square Equation. Functional Square Root of a Logarithm

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M A T H E M A T I C A L E C O N O M I C S No. 12(19) 2016

REMARKS ABOUT THE SQUARE EQUATION. FUNCTIONAL SQUARE ROOT OF A LOGARITHM

Arkadiusz Maciuk, Antoni Smoluk

Abstract: The study shows that the functional equation f (f (x)) = ln(1 + x) has a unique result in a semigroup of power series with the intercept equal to 0 and the function composition as an operation. This function is continuous, as per the work of Paulsen [2016]. This solution introduces into statistics the law of the one-and-a-half logarithm. Sometimes the natural growth processes do not yield to the law of the logarithm, and a double logarithm weakens the growth too much. The law of the one-and-a-half logarithm proposed in this study might be the solution. Keywords: functional square root, fractional iteration, power series, logarithm, semigroup. JEL Classification: C13, C65. DOI: 10.15611/me.2016.12.04.

1. Introduction

If the natural logarithm of random variable has normal distribution, then one considers it a logarithmic law – with lognormal distribution. If the function f being function composition to itself is a natural logarithm, that is f (f (x)) = ln(1 + x), then the function ( ) = (ln( )) applied to a random variable will give a new distribution law, which – similarly as the law of logarithm – is called the law of the one𝑔𝑔-and𝑥𝑥 -a-𝑓𝑓half logarithm.𝑥𝑥 A square root can be defined in any semigroup G. A semigroup is a non- empty set with one combined operation and a neutral element. The general form of a quadratic equation has the form of = , where . Solving such an equation2 is equivalent to answering the question when the element is a square𝑥𝑥 in𝑎𝑎 semigroup G. When the semigroup G is a set𝑎𝑎 ∈of𝐺𝐺 functions with the function composition as an operation and identity function as a neutral𝑎𝑎 element, the solution of the quadratic equation ( ) = o ( ) = ( ) is called a functional square root of function . 2 𝑓𝑓 𝑓𝑓 𝑥𝑥 𝑓𝑓 𝑓𝑓 𝑥𝑥 𝑔𝑔 𝑥𝑥 𝑔𝑔 Arkadiusz Maciuk, Antoni Smoluk Department of Mathematics and Cybernetics, Wrocław University of Economics [email protected] 40 Arkadiusz Maciuk, Antoni Smoluk

2. Examples

If is a semigroup of natural numbers = {0,1, … } with addition, then every even number = 2 is a square, as the quadratic equation is satisfied when 𝐺𝐺= , + = . A square in this caseℕ represents a simple parity of natural numbers. If,𝑛𝑛 however,𝑘𝑘 is a semigroup of natural numbers with the operation𝑥𝑥 of𝑘𝑘 multiplication,𝑘𝑘 𝑘𝑘 𝑛𝑛 then the quadratic equation boils down to the question: which natural numbersℕ are real squares. In this case, the answer is a little more difficult. The equation = has a solution if and only if in every prime number in the decomposition2 of the number into prime factors occurs an even number of times. Zero𝑥𝑥 is a square𝑛𝑛 because 0 = 0 × 0. One is a square because 1 = 1 × 1, two is not a square, neither 𝑛𝑛is three, but four is because 4 = 2 × 2. The number 36 is because 36 = 2 × 3 , and so on. If is a semigroup of rational numbers with the operation2 of2 addition, each rational number is a square, because = + , thus the number𝐺𝐺 ℚ 1 1 = is the solution. If is a semigroup of with the operation of mul- 𝑎𝑎 ∈ ℚ 𝑎𝑎 2 𝑎𝑎 2 𝑎𝑎 tiplication,1 then the answer to the question of which rational numbers are 𝑥𝑥 2 𝑎𝑎 𝐺𝐺 ℚ squares is more complex. The irreducible fraction , where , is a square if and only if both the numerator and denominator𝑚𝑚 are squares of 𝑛𝑛 𝑚𝑚 𝑛𝑛 ∈ ℕ natural numbers. The fraction is a square, however is not. Of course, numbers smaller than zero are not4 squares, but 0 is a square.12 If the semigroup 25 25 is a set of real numbers with addition, then similarly as in semigroup each number is a square. If this semigroup is considered with the operation 𝐺𝐺of multiplication, then anyℝ non-negative number is a square becauseℚ = , and negative numbers are not squares. In semigroup = {0,1,2} of residues modulo 3 with addition, each element𝑎𝑎 is a square because 3 0𝑎𝑎 = 0√𝑎𝑎+√0𝑎𝑎, 1 = 2 + 2, 2 = 1 + 1. The same set with the operation𝑍𝑍 of multiplication has other squares: 0 = 0 × 0, 1 = 1 × 1 = 2 × 2, but 2 is not a square. If the semigroup is a set of real polynomials Pol (F) with the operation of superposition, then in the semigroup squares are polynomials of degree zero, each monomial 𝐺𝐺of the form ( ) = + , where 0 is a square. The prerequisite for a polynomial to be a square, is for its degree to be a square in a semigroup of natural numbers𝑓𝑓 with𝑥𝑥 multiplication.𝑎𝑎𝑎𝑎 𝑏𝑏 𝑎𝑎 ≥

Remarks about the square equation… 41

3. Generalizations

Theoretically, the semigroup of words ( ) in the alphabet is im- portant. The product of the word = and = is a word = . The empty word denoted by 0 is a neutral element:𝑊𝑊 𝑋𝑋 0 = 0 = . The𝑋𝑋 word is a square if and only if = 𝑆𝑆. The𝑥𝑥𝑥𝑥 prerequisite𝑇𝑇 𝑎𝑎𝑎𝑎 of being a square𝑆𝑆𝑆𝑆 𝑥𝑥𝑥𝑥is 𝑥𝑥𝑥𝑥the appearance of each letter in the word an even number𝑆𝑆 of𝑆𝑆 times.𝑆𝑆 The word 𝑆𝑆 = satisfies a necessary𝑆𝑆 𝑇𝑇𝑇𝑇condition but it is not a square, and the word = satisfies a necessary condition and is a square, = , where𝑇𝑇 𝑎𝑎 𝑎𝑎𝑎𝑎=𝑎𝑎 . The equation = is easily solved; there is always one solution𝑆𝑆 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 or𝑏𝑏 the𝑏𝑏 equation is contradictory.2 𝑆𝑆 𝑇𝑇𝑇𝑇 A 𝑇𝑇quadratic𝑏𝑏𝑏𝑏𝑏𝑏 equation can be𝑇𝑇 generalized.𝑆𝑆 Let , and . One looks for element that in semigroup satisfies the condition = . A special case of this equation is the equation 𝑚𝑚= 𝑛𝑛. ∈Theℕ solution𝑎𝑎 ∈𝑚𝑚 𝐺𝐺of this𝑛𝑛 equation in some cases,𝑥𝑥 as in the above𝐺𝐺 examples,2 is almost trivial.𝑥𝑥 If 𝑎𝑎the semigroup is a set of functions defined in the domain𝑥𝑥 𝑎𝑎 transforming into itself and the operation in this group is function composition, then the solution to a quadratic𝐺𝐺 equation becomes a difficult problem. If𝑋𝑋 the set is a 𝑋𝑋set of permutations, this equation = does not have solutions when is an odd permutation, i.e. when its character2 is number 1; setting any 𝐺𝐺permutation with each other always gives𝑥𝑥 an 𝜎𝜎even permutation. The answer 𝜎𝜎to the question: for which permutations this equation has− a solution, is not obvious. What is existence in mathematics? This is a declaration that a certain object exists, i.e. the axiom of existence. Generally, we can say that in the set there is an object that satisfies the predicate , if the given below logical sen- tence is true: 𝔸𝔸 ( ) (𝜑𝜑 ). An absolute axiom of existence is considered to be true if its adoption does not lead to a conflict of theory.∃ 𝑎𝑎 ∈ Theorems𝔸𝔸 ∧ 𝜑𝜑 𝑎𝑎 of existence, as the one above, are conditional in nature: if the axioms are true, then the resulting theorems are also true. In a semigroup of real functions of one variable with the submission as the operation, the equation ( ) = ( ) where the function ( ) is given, can have an infinite number of2 solutions, and may not have a solution at all. For example, there might be𝑓𝑓 an𝑥𝑥 indefinite𝑔𝑔 𝑥𝑥 number of solutions𝑔𝑔 𝑥𝑥to equation ( ) = , i.e. involution. Involution is the identity function, function (2 ) = , hyperbole ( ) = for 0, and 0 for = 0 but also func- tions𝑓𝑓 𝑥𝑥 like 𝑥𝑥function ( ) = + 1 −for1 (0,1], ( ) = 1, for (1,2] 𝑓𝑓 𝑥𝑥 −𝑥𝑥 𝑓𝑓 𝑥𝑥 𝑥𝑥 𝑥𝑥 ≠ 𝑥𝑥 𝑓𝑓 𝑥𝑥 𝑥𝑥 𝑥𝑥 ∈ 𝑓𝑓 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 ∈ 42 Arkadiusz Maciuk, Antoni Smoluk

and ( ) = the remaining arguments. In contrast, the solution of equation ( ) = does not exist [Kuczma 1969]. Bödewart [Bödewart 1944; by Zhang2 𝑓𝑓 𝑥𝑥1997]−1𝑥𝑥 showed that if function : = [ , ] is continuous and strictly𝑓𝑓 𝑥𝑥 increasing,𝑥𝑥 then for any integer > 1 and , ( , ) with < equation ( ) = ( ) has a continuous𝑔𝑔 and𝐼𝐼 strictly𝑎𝑎 𝑏𝑏 increasing→ 𝐼𝐼 solution on I satisfying𝑛𝑛 ( ) ( ) < ( ) ( 𝑛𝑛). It is easy𝐴𝐴 to𝐵𝐵 show∈ 𝑎𝑎 that𝑏𝑏 this solution𝐴𝐴 𝐵𝐵 is unambiguous𝑓𝑓 𝑥𝑥 [Iga].𝑔𝑔 𝑥𝑥 Hence equations such as ( ) = or ( ) = ln𝑓𝑓 ( ) have a clear𝑔𝑔 unambiguous𝑎𝑎 ≤ 𝑓𝑓 𝐴𝐴 solution,𝑓𝑓 𝐵𝐵 ≤ which𝑔𝑔 𝑏𝑏 is a continuous2 and𝑥𝑥 strictly2 increasing function. In 1950 for the first time a method𝑓𝑓 𝑥𝑥 for solving𝑒𝑒 𝑓𝑓 the𝑥𝑥 equation𝑥𝑥 ( ) = was proposed [Kneser 1950], and to this day this equation has lived2 to see𝑥𝑥 several different methods of solution [Paulsen 2016]. 𝑓𝑓 𝑥𝑥 𝑒𝑒 4. Breakdown method

Our goal is to solve the equation ( ) = ln ( ) which we will further refer to as the Kopociński equation, from the author of the question of whether the natural logarithm in analytic functions’𝑓𝑓�𝑓𝑓 𝑥𝑥 semigroup� 𝑥𝑥 is a square and what its analytical range is. The result is the well-known fact that the functional square root of function 1 is an analytic function in the neighborhood of zero [see Paulsen 2016].𝑥𝑥 Hence the functional square root of function ln (1 + ), as an inverse function𝑒𝑒 − of this function, is also an analytic function in the neighborhood of zero. To find its analytical sequence, we can use one of several𝑥𝑥 methods. Examination of the function ( ) = ln (1 + ) instead of the function ln ( ) is beneficial due to the fact that in its point 0 this function takes the value 0. The function develops into the𝑔𝑔 power𝑥𝑥 series 𝑥𝑥 𝑥𝑥

( ) = ∞ , 𝑛𝑛 𝑛𝑛 ( )( ) 𝑔𝑔 𝑥𝑥 � 𝑎𝑎 𝑥𝑥 where = . Because, ( )(0𝑛𝑛)=0= 0 and ( )(0) = ( 1) !, so 𝑛𝑛 ! 𝑔𝑔 0 ( ) 0 𝑛𝑛+1 𝑛𝑛 ( ) = 𝑛𝑛 hence ( ) = + , for | | < 1. 𝑎𝑎 𝑛𝑛 𝑛𝑛 𝑔𝑔 𝑔𝑔 − 𝑛𝑛 One looks∞ for a− 1function𝑛𝑛+1 ( ) = + + 1 analytic2 1 3 in the neighborhood 𝑔𝑔 𝑥𝑥 ∑𝑛𝑛=0 𝑛𝑛+1 𝑥𝑥 𝑔𝑔 𝑥𝑥 𝑥𝑥 − 2 𝑥𝑥 3 𝑥𝑥 − ⋯ 𝑥𝑥 of zero and satisfying the equation ( ) = + . The 𝑓𝑓 𝑥𝑥 𝑏𝑏0 𝑏𝑏1𝑥𝑥 ⋯ composition of real analytic functions is a real analytic1 2 function1 3 to solve 2 3 a quadratic equation to calculate the coefficients𝑓𝑓�𝑓𝑓 𝑥𝑥 � 𝑥𝑥 − 𝑥𝑥 of the𝑥𝑥 function− ⋯ .

𝑏𝑏𝑛𝑛 ∈ ℝ 𝑔𝑔 Remarks about the square equation… 43

Symbol indicates a semigroup of power series with operation of function composition. This semigroup includes analytic functions at zero, in particular function 𝕂𝕂( ) = ln (1 + ). If the power series satisfies a quadratic equation in semi- Lemma.𝑔𝑔 𝑥𝑥 𝑥𝑥 group , then coefficient = 0. = o Evidence.𝕂𝕂 Le 𝑏𝑏0, where meets the quadratic equation in semigroup . Of course (0) = 0, because (0) = 0. It follows that (0) = ( ) = 0. Zero is ℎthus 𝑓𝑓a fixed𝑓𝑓 point𝑓𝑓 function ; also a fixed point of function 𝕂𝕂for ( ) = ℎ ( ) = (0) = 𝑔𝑔 . This function has exactlyℎ one 0 0 fixed𝑓𝑓 𝑏𝑏 point, hence = 0 because here is theℎ same𝑏𝑏 as , which completes 0 0 0 the proof.ℎ ℎ 𝑏𝑏 𝑓𝑓�𝑓𝑓 𝑏𝑏 � 𝑓𝑓 𝑏𝑏 𝑔𝑔 0 We have 𝑏𝑏 ℎ 𝑔𝑔 ( ) = + ( + + + ) +

2 0 1 0 1 2 ( 1) (𝑓𝑓�𝑓𝑓+𝑥𝑥 � +𝑏𝑏 𝑏𝑏+ 𝑏𝑏 ) +𝑏𝑏 𝑥𝑥 =𝑏𝑏 ∞𝑥𝑥 ⋯ + 1𝑛𝑛 2 2 − 𝑛𝑛+1 The lemma𝑏𝑏2 𝑏𝑏 shows0 𝑏𝑏1 that𝑥𝑥 𝑏𝑏2𝑥𝑥 ⋯ ⋯ � 𝑥𝑥 𝑛𝑛=0 𝑛𝑛 ( ) = ( + + ) + ( + + ) + 2 2 2 1 1 2 2 11 2 1 ℎ 𝑥𝑥 ( 𝑏𝑏 +𝑏𝑏 𝑥𝑥 +𝑏𝑏 𝑥𝑥 ) +⋯ =𝑏𝑏 𝑏𝑏 𝑥𝑥 𝑏𝑏+ 𝑥𝑥 +⋯ , 2 3 2 3 2 3 3 1 2 so = 1 𝑏𝑏because𝑏𝑏 𝑥𝑥 𝑏𝑏(0𝑥𝑥) = 1⋯. Hence⋯ 𝑥𝑥=−1or𝑥𝑥 = 𝑥𝑥1. If⋯ = 1 then ( ) 2 ′ ( ) 1= + + + + + 1 + 1= +1 + ; 𝑏𝑏 2 𝑓𝑓 2 𝑏𝑏 2 𝑏𝑏 − 1 2 𝑏𝑏 1 3 we obtain equality2 + =2 , that2 is = . If the above equality had ℎ 𝑥𝑥 𝑥𝑥 𝑏𝑏 𝑥𝑥 ⋯ 𝑏𝑏 𝑥𝑥 𝑏𝑏 𝑥𝑥 ⋯ ⋯ 𝑥𝑥 − 2 𝑥𝑥 3 𝑥𝑥 ⋯ the value of 1 then ( ) = 1 + 1( + + ) + = 𝑏𝑏2 𝑏𝑏2 − 2 𝑏𝑏 2 − 4 𝑏𝑏1 + , that is 0 = 2 + = , which is2 contradictory.2 − ℎ 𝑥𝑥 𝑥𝑥 − 𝑏𝑏2𝑥𝑥 − ⋯ 𝑏𝑏2 −𝑥𝑥 𝑏𝑏2𝑥𝑥 ⋯ ⋯ Other1 factors2 1 are3 calculated recursively. 1 2 3 2 2 2 𝑥𝑥 − The𝑥𝑥 expression𝑥𝑥 − ⋯ ( + −+𝑏𝑏 +𝑏𝑏 −+ ) is a polynomial wherein the value of the -th power2 of the variable𝑛𝑛 , for𝑛𝑛 , depends on 1 2 𝑛𝑛 the number of possible 𝑏𝑏breakdowns𝑥𝑥 𝑏𝑏 𝑥𝑥 of⋯ the𝑏𝑏 number𝑥𝑥 ⋯ into numbers. The breakdown of a of a natural𝑘𝑘 number 1 is a finite 𝑥𝑥sequence𝑘𝑘 ≥ of𝑛𝑛 natural numbers different from zero summing up to . Number 1 𝑘𝑘has one𝑛𝑛 breakdown (1), number 2 has two ( 1,1) and (𝑘𝑘2≥). Number 3 has four breakdowns (1,1,1), (1,2), (2,1) and (3), and so on. 𝑘𝑘 44 Arkadiusz Maciuk, Antoni Smoluk

Each of these breakdowns corresponds to -th element permutation with number repetitions whose sum is . For example, for = 4 and = 8 there are five possibilities: (1 1 1 5), (1 1 2 4), (1 1 3 3𝑛𝑛), (1 2 2 3) and (2 2 2 2). The breakdown (1 1 1 5) occurs 4𝑘𝑘 times, breakdown (1 1 3𝑘𝑘 3) 6 times, (1 1 2 4) and (1 2 2 3) 12 times, and the breakdown (2 2 2 2) occurs once. Hence, after expanding the coefficient of is equal 4 + 12 + 6 + 12 ∞ +𝑖𝑖 4 . 8 3 𝑖𝑖=1 𝑖𝑖 1 5 2In order to determine2 2 �∑ the2 formula𝑏𝑏 𝑥𝑥 � 4 for the coefficient𝑥𝑥 at the -th power𝑏𝑏 𝑏𝑏 of 1 2 4 1 3 1 2 3 2 function𝑏𝑏 𝑏𝑏 𝑏𝑏 it is𝑏𝑏 sufficient𝑏𝑏 to𝑏𝑏 𝑏𝑏consider𝑏𝑏 𝑏𝑏 all the possible breakdowns of the number , whereas the coefficient determined by a one-element 𝑘𝑘breakdown is multiplied𝑓𝑓 by , the coefficient determined by a two-element breakdown is multiplied𝑘𝑘 by , and so on. For example, the number 2 can be represented as 1 a breakdown composed𝑏𝑏 of a single element 2 or in the form of (1,1) it is the 2 coefficient at 𝑏𝑏 and has the form of + . Breakdown 3 is respec- tively 3, (1,2), 2(1,1,1) which means that the coefficient2 at is equal to 1 2 2 1 ( ) + (2𝑥𝑥 ) + ( ) = (𝑏𝑏1𝑏𝑏+ 𝑏𝑏) +𝑏𝑏 2 ; 4 is 4, 3(1,3), (2,2), (1,1,2), (1,1,1,1), hence the3 coefficient at 2 is ( ) 2+ (2𝑥𝑥 + ) + 1 3 2 1 2 3 1 1 3 1 1 2 𝑏𝑏 (𝑏𝑏 ) 𝑏𝑏+ 𝑏𝑏( 𝑏𝑏) = 𝑏𝑏 𝑏𝑏(1 + 𝑏𝑏 )𝑏𝑏+ 2 𝑏𝑏4 + 𝑏𝑏 +𝑏𝑏 . And so2 on, 1 4 2 1 3 2 comparing2 the thus4 obtained polynomial3 of𝑥𝑥 coefficients𝑏𝑏 𝑏𝑏3 2 𝑏𝑏to the𝑏𝑏 correspond-𝑏𝑏 𝑏𝑏 3 1 2 4 1 1 4 1 1 2 3 2 1 2 3 ing𝑏𝑏 𝑏𝑏values𝑏𝑏 of𝑏𝑏 the𝑏𝑏 expansion𝑏𝑏 𝑏𝑏 of the𝑏𝑏 Maclaurin𝑏𝑏 𝑏𝑏 series𝑏𝑏 𝑏𝑏of the𝑏𝑏 function𝑏𝑏 𝑏𝑏 ln (1 + ) 𝑖𝑖 one obtains: 𝑏𝑏 1 𝑥𝑥 ( + ) = 2 2 1 1 2 1 2 𝑏𝑏 + (𝑏𝑏 𝑏𝑏+ )− = 3 2 3 1 2 1 1 3 1 + 2 𝑏𝑏 𝑏𝑏+ 3 𝑏𝑏 𝑏𝑏+ (𝑏𝑏 + ) = 4 3 2 4 2 1 2 3 1 2 3 1 1 4 1 2 + 3 𝑏𝑏 +𝑏𝑏 3𝑏𝑏 𝑏𝑏 +𝑏𝑏 2 𝑏𝑏 𝑏𝑏 +𝑏𝑏4 𝑏𝑏 𝑏𝑏 + (− + ) = . 5 2 2 2 2 3 5 2 3 1 2 3 1 3 1 2 4 1 2 4 1 1 5 This𝑏𝑏 system𝑏𝑏 is𝑏𝑏 solved𝑏𝑏 𝑏𝑏 recursively:𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 = 1, = , = ⋮ ( , ) = = ; 1 1 1 1 1 1 5 2 𝑏𝑏1 =𝑏𝑏2 − 4 𝑏𝑏+3 𝑏𝑏(1+𝑏𝑏,1 �,3 −)𝜙𝜙2=𝑏𝑏1 𝑏𝑏2 �+ 2 �3=− 8� ; 48 1 1 1 1 7 5 3 𝑏𝑏4 = 𝑏𝑏1+𝑏𝑏1 �− 4 𝜙𝜙( 3 ,𝑏𝑏1 𝑏𝑏, 2 𝑏𝑏, 3 )� =2 �− 4 48� = − 96 . 1 1 1 1 55 109 4 𝑏𝑏5 𝑏𝑏1+𝑏𝑏1 �5 − 𝜙𝜙4 𝑏𝑏1 𝑏𝑏2 𝑏𝑏3 𝑏𝑏4 � 2 �5 − 384� 3840 Remarks about the square equation… 45

( ) = 1 = ( , , … , ) Generally we have: , 𝑛𝑛 . These cal- 1 −1 culations show that: 𝑏𝑏1 𝑏𝑏𝑛𝑛+1 2 � 𝑛𝑛+1 − 𝜙𝜙𝑛𝑛 𝑏𝑏2 𝑏𝑏3 𝑏𝑏𝑛𝑛 � 5 5 109 497 127 11569 ( ) = + + + 42 483 964 38405 307206 134407 20643848 𝑥𝑥 3127𝑥𝑥 57 𝑥𝑥 1219255𝑥𝑥 𝑥𝑥 165677473𝑥𝑥 𝑥𝑥 𝑓𝑓 𝑥𝑥 𝑥𝑥 − + − − + − 928972809 59454259210 13079937024011 885730939𝑥𝑥 20163875141𝑥𝑥 𝑥𝑥 − + 112113745920012 4080940351488013 252312616027𝑥𝑥 9565074633871𝑥𝑥 − + 81618807029760014 4897128421785600015 691138954263097𝑥𝑥 5061676927076641𝑥𝑥 − + 548478383239987200016 6216088343386521600017 95993669516238563𝑥𝑥 𝑥𝑥 − 191810726024498380800018 1245671625068799013𝑥𝑥 +− 4165032907960536268800019 903291663542261320331𝑥𝑥 + 4081732249801325543424000020 0.25 + 0.104167 0.0520833 +𝑥𝑥 0.0283854 − ⋯ ≈ 2 0.01617843 + 0.00944944 0.005604095 𝑥𝑥 − 𝑥𝑥 + 0.0033667𝑥𝑥 −6 0.00205074𝑥𝑥 7 + 0.00126665𝑥𝑥 8 − 0.000790029𝑥𝑥9 + 0.000494099𝑥𝑥 10− 𝑥𝑥 11 0.000309135𝑥𝑥 −12 + 0.00019532𝑥𝑥 13 0.000126𝑥𝑥01 +− 0.0000814287𝑥𝑥 14 0.000050046𝑥𝑥15 16 +− 0.0000299078𝑥𝑥 17 0.0000221301𝑥𝑥 −18 . 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 Figure 1 shows a polynomial ( )19 which is the sum of the20 first twenty terms 𝑥𝑥 − 𝑥𝑥 of a power series f together with functions ln (1 + ) and function ( ) = 20 ln 1 + ( ) for the open interval𝑓𝑓 𝑥𝑥 ( 1,1). 3 20 𝑥𝑥 𝑓𝑓 𝑥𝑥 � 𝑓𝑓20 𝑥𝑥 � −

46 Arkadiusz Maciuk, Antoni Smoluk

Fig. 1. Selection of function graphs f(x)represented by a polynomial f20(x), function ln(1 + x) and function ln(1 + f20(x)) Source: own elaboration.

Fig. 1. Selection of function graphs Source: own elaboration. Remarks about the square equation… 47

5. Matrix method

Definition. Let represent a subsemigroup of a semigroup of the power series for which = 0 and = 1. 𝕂𝕂0 𝕂𝕂

Semigroup contains𝑏𝑏 all0 the power𝑏𝑏1 series of the form + + . 0 𝕂𝕂 2 If function is odd and satisfies2 a quadratic equation, then also Note. 𝑥𝑥 𝑎𝑎 𝑥𝑥 ⋯ function satisfies it too. If ( ) = then also ( ) = . This no- tice is a simple consequence𝑓𝑓 of the definition of parity. −𝑓𝑓 𝑓𝑓 𝑓𝑓 𝑔𝑔 �−𝑓𝑓 −𝑓𝑓 � 𝑔𝑔 Theorem on uniqueness. The Kopociński equation in semigroup has a uniquely determined solution. 𝕂𝕂0 Evidence. This theorem means that there is always one and only one solution. We shall give below an alternative method of calculating the coefficient of an analytic function with relation to the method referring to the par- tition described above. Both methods differ only in the way of deriving equations – the equations are the same. Consider the auxiliary matrix with an infinite number of elements … 𝐴𝐴 … 𝑎𝑎11 𝑎𝑎12 𝑎𝑎13 … , 𝑎𝑎21 𝑎𝑎22 𝑎𝑎 23 � � 𝑎𝑎31 𝑎𝑎32 𝑎𝑎33 where = 1, = for 2, = 0 for < , = ⋮ ⋮ ⋮ ⋱ ( , … , ) for 1 < < . The function is a unequivocally defined 1 1 1 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑖𝑖 𝑘𝑘 𝑛𝑛+1 polynomial;𝑎𝑎 its arguments𝑎𝑎 are𝑏𝑏 coefficients𝑛𝑛 ≥ calculated𝑎𝑎 before.𝑖𝑖 Enumerating𝑛𝑛 𝑎𝑎 the 𝑘𝑘 𝑛𝑛 1 𝑛𝑛 𝑘𝑘 𝑛𝑛 coefficients𝜙𝜙 𝑏𝑏 𝑏𝑏of these polynomials𝑘𝑘 𝑛𝑛 is not easy. 𝜙𝜙This –th column of the matrix is used to calculate the coefficient . In that column the coefficient occurs twice, and the remaining terms are well known and 𝑛𝑛depend on the coefficients 𝑛𝑛 𝑛𝑛 calculated previously. This column𝑏𝑏 is thus a sequence: 𝑏𝑏

( , ( , . . , ), … , ( , . . , ), , 0,0, … ).

Hence 𝑛𝑛 2=𝑛𝑛 1 = 𝑛𝑛−,1 this means𝑛𝑛−1 𝑛𝑛 that1 to 𝑛𝑛calculate−1 𝑛𝑛 the coefficient 𝑏𝑏 𝜙𝜙 𝑏𝑏 𝑏𝑏 𝜙𝜙 𝑏𝑏 𝑏𝑏 𝑏𝑏 we have linear equations in the form of 2 = where the size is 1 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 determined 𝑎𝑎by coefficients𝑎𝑎 𝑏𝑏 calculated in advance and free terms. This means 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 that,𝑏𝑏 firstly, a solution exists, and secondly, that𝑏𝑏 there𝑐𝑐 is only one which𝑐𝑐 was to be proved. Of course, the Kopociński equation in semigroup has a solution for any analytic function, and not only for the function ln(1 + ). 𝕂𝕂0 𝑥𝑥 48 Arkadiusz Maciuk, Antoni Smoluk

There is a third possible way of calculating the coefficients which also proves the theorem; it stems from the formula for the derivative of function 𝑛𝑛 composition. Just remember that the power series -th derivative𝑏𝑏 is equal to ! . In the quadratic equation there naturally appears 𝑘𝑘the product of Cauchy’s 𝑘𝑘 sequences;𝑘𝑘 𝑏𝑏 if we multiply the power strings, their coefficient sequences are multiplied according to Cauchy’s product. So if = { }, = { } then the Cauchy product is equal 𝑎𝑎 𝑎𝑎𝑛𝑛 𝑏𝑏 𝑏𝑏𝑛𝑛

𝑛𝑛 .

� 𝑎𝑎𝑛𝑛−𝑖𝑖𝑏𝑏𝑖𝑖 If so, ( ) = , ( ) =𝑖𝑖=0 , then ( ) ( ) = , where ∞ 𝑛𝑛 ∞ 𝑛𝑛 ∞ 𝑛𝑛 𝑓𝑓 𝑥𝑥 ∑𝑛𝑛=0 𝑎𝑎𝑛𝑛𝑥𝑥 𝑔𝑔 𝑥𝑥 ∑𝑛𝑛=0 𝑏𝑏𝑛𝑛𝑥𝑥 𝑓𝑓 𝑥𝑥 𝑔𝑔 𝑥𝑥 ∑𝑛𝑛=0 𝑐𝑐𝑛𝑛𝑥𝑥 = . = ( , , , … ) 𝑛𝑛 Let 𝑐𝑐and𝑛𝑛 let∑𝑖𝑖 =0 𝑎𝑎𝑛𝑛mean−𝑖𝑖𝑏𝑏𝑖𝑖 the n-th row of the matrix shifted to the right by one place, that is = (0, , , … ). Of 0 1 2 𝑛𝑛 course 𝑏𝑏= 𝑏𝑏 𝑏𝑏, where𝑏𝑏 = 𝑤𝑤is the n-th power of Cachy’s series b. 𝑛𝑛 𝑛𝑛 1 𝑛𝑛 2 If, 𝐴𝐴 = ( , 𝑛𝑛, , … ), then (𝑛𝑛 ) = 𝑤𝑤 . Hence:𝑎𝑎 𝑎𝑎 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑤𝑤 𝑏𝑏 𝑏𝑏 𝑟𝑟𝑒𝑒 𝑏𝑏 𝑏𝑏𝑛𝑛 ∞ 𝑘𝑘 𝑛𝑛0 𝑛𝑛1 𝑛𝑛2 𝑘𝑘=𝑛𝑛 𝑛𝑛𝑛𝑛 𝑏𝑏 𝑏𝑏 𝑏𝑏 𝑏𝑏 ( ) = o 𝑔𝑔( )𝑥𝑥= ∞∑ 𝑏𝑏( )𝑥𝑥; 𝑘𝑘 𝑘𝑘 so we have an equationℎ 𝑥𝑥 =𝑔𝑔 𝑔𝑔 𝑥𝑥 �where𝑏𝑏 𝑔𝑔 𝑥𝑥 is the n-th coefficient of 𝑘𝑘=1 expansion of a function ln (1 + 𝑘𝑘) into series. So we have a recursive patterns 𝑘𝑘 ∑𝑖𝑖=1 𝑖𝑖 𝑖𝑖 𝑘𝑘 𝑛𝑛 in the form of: = ,𝑎𝑎 = 𝑏𝑏 𝑏𝑏 ,𝑎𝑎 where and = . Eventually = 𝑥𝑥. The𝑘𝑘 equation for the coefficient has the 1 𝑘𝑘 𝑘𝑘 𝑛𝑛+1 𝑘𝑘 𝑖𝑖=0 𝑛𝑛 𝑖𝑖 𝑘𝑘−𝑖𝑖 𝑛𝑛 𝑛𝑛+1 form of 2 =𝑏𝑏 𝑏𝑏 𝑏𝑏 . ∑ 𝑏𝑏 𝑏𝑏 𝑘𝑘 ∈ ℕ 𝑏𝑏 𝑛𝑛+1 𝑛𝑛 𝑘𝑘 𝑛𝑛 𝑛𝑛 𝑘𝑘 𝑘𝑘 𝑏𝑏 𝑎𝑎 𝑘𝑘−𝑏𝑏1 𝑏𝑏 𝑏𝑏 𝑏𝑏𝑘𝑘 𝑎𝑎𝑘𝑘 −6. ∑Abstraction𝑖𝑖=1 𝑎𝑎𝑖𝑖 𝑘𝑘 and identical objects

Functions : and : are equivalent, if there are isomorphisms : and : such that, = , when 1 1 1 2 2 2 = 𝑓𝑓 𝑋𝑋. Equivalent→ 𝑌𝑌 𝑓𝑓 functions𝑋𝑋 → 𝑌𝑌 work the same way in isomorphic 1 1 2 2 1 2 2 1 2 1 collection.−1𝜑𝜑 There𝑋𝑋 → is𝑋𝑋 no difference𝜑𝜑 𝑌𝑌 whether→ 𝑌𝑌 we permute𝜑𝜑 letters∘ 𝑓𝑓 o𝑓𝑓r if∘ we𝜑𝜑 permute 1 2 1 numbers.𝑓𝑓 𝜑𝜑2 If∘ 𝑓𝑓we∘ talk𝜑𝜑 about continuous functions, the isomorphisms are home- omorphisms of topological spaces. Functions ln( ) and ln(1 + ) are equivalent. 𝑥𝑥 𝑥𝑥 Remarks about the square equation… 49

Let mean translation; ( ) = + 1; naturally, each translation is a square in the group of translations, for ( ) = , where ( ) = + . 𝜏𝜏 𝜏𝜏 𝑥𝑥 𝑥𝑥 We solved the Kopociński equation in the form of = ln which is1 2 equivalent to form = ln𝜎𝜎. 𝜎𝜎 Introducing𝜏𝜏 𝜎𝜎determination𝑥𝑥 𝑥𝑥 = we obtain −1 = ln−1. If one change𝑔𝑔 ∘ s𝑔𝑔 the operation∘ 𝜏𝜏 in a semigroup−1 we obtain∘ the𝜏𝜏 exact∘ 𝜏𝜏 ∘ equation𝑔𝑔 ∘ 𝜏𝜏 in the form = ln, where is 1 1 1 the𝑓𝑓 new𝑔𝑔 ∘ operation𝜏𝜏 defined𝑓𝑓 by∘ 𝜏𝜏the∘ 𝑓𝑓 formula = . In a semigroup 1 1 with the new operation with unity there is a translation𝑓𝑓 ∗ 𝑓𝑓 because ∗ = . 𝑓𝑓 ∗ 𝑔𝑔 𝑓𝑓 ∘ 𝜏𝜏 ∘ −𝑔𝑔1 −1 The law of one-and-a-half logarithm in the zero position,𝜏𝜏 that is,𝜏𝜏 after∗ 𝑔𝑔translation𝑔𝑔 means that on the random variable function ln is applied. While the law of the one-and-a-half logarithm in neutral position, that is without moving,𝜏𝜏 means that on the random variable function𝑔𝑔 ∘ ∘ 𝜏𝜏 ln applied. −1 𝜏𝜏 ∘ 𝑔𝑔 ∘ ∘ 𝜏𝜏 Hypothesis. Let mean a semigroup of power series with a positive radius of convergence; each series represents an analytic function, hence it 𝑎𝑎 is . If 𝕂𝕂 is the solution for the Kopociński equation, then the question arises whether g is an analytic function, so if . This is a dif- 𝑎𝑎 0 0 ficult𝕂𝕂 ⊂problem𝕂𝕂 𝑔𝑔in ∈the𝕂𝕂 field of analytic functions and power series. For if 𝑎𝑎 , then the equality o = seems paradoxical.𝑔𝑔 An∈ 𝕂𝕂 analytic function is obtained from the series composition of zero-convergence. The nature of 𝑎𝑎 coeffi𝑔𝑔 ∉ 𝕂𝕂cients of function𝑔𝑔 is𝑔𝑔 unknown.𝑓𝑓 They arise from the algebraic sum of very large numbers, so they can be themselves very large. This means that 𝑛𝑛 the radius of𝑏𝑏 convergence of𝑔𝑔 this series can be zero. This is the content of our assumptions. There is a full analogy to the equation = 1 that has no solution in a multiplicative semigroups of real numbers,2 and has two solutions in a multiplicative semigroups of complex numbers.𝑥𝑥 Mathematics− knows many examples of contradictory equations that after the expansion of permis- sible objects have obtained generalized solutions. We suppose that the natural logarithm without shift is not a square in the set of analytic functions with real coefficients. The argument in favor of this hypothesis is the following remark. If the areas of these features are the same, then there appears a contradiction. ln(1) = 0 and, ( ) = ln ( ), so (1) = , where > 0, because it cannot be 0. But ( ) = 0, so ( ) = (0), hence the contradiction. 𝑔𝑔�𝑔𝑔 𝑥𝑥 � 𝑥𝑥 𝑔𝑔 𝛼𝛼 𝛼𝛼 𝛼𝛼 ≤ 𝑔𝑔 𝛼𝛼 𝑔𝑔�𝑔𝑔 𝛼𝛼 � 𝑔𝑔 50 Arkadiusz Maciuk, Antoni Smoluk

A more general assertion. In semigroup the generalized Kopociński equation has a unique solution. If the solution is not an analytic function, then 0 there is a solution in the set – a generalized𝕂𝕂 solution. The proof of this theorem – mutatis mutandis – is a repetition of the proof 𝕂𝕂0 of the theorem about the existence of the equation = . If the function satisfies the equation = , then also for any the equality = is true. This means𝑚𝑚 𝑛𝑛that in the semigroup𝑔𝑔 ∘ 𝑔𝑔 fraction𝑓𝑓 1/3 is equal𝑔𝑔 to𝑘𝑘 𝑘𝑘the fraction𝑘𝑘𝑘𝑘 7/21. So𝑔𝑔 it is 𝑓𝑓enough to solve equations𝑘𝑘 ∈of ℕform = 0 𝑔𝑔for numbers𝑓𝑓 and that are relatively prime. 𝕂𝕂 𝑚𝑚 𝑛𝑛 The aforementioned G semigroup of positive rational numbers𝑔𝑔 with the𝑓𝑓 operation of 𝑛𝑛multiplication𝑚𝑚 is in fact a commutative group – Abelian. This group is isomorphic to the group of M sequences of integers with a finite number of terms different from zero. The isomorphism identifying the two groups is formed from the extension of a function associating to the first number ‘n’ a sequence that at the n-th location has number one, and zeros other- wise. M is essentially a module of such sequences above the ring of integers. So the prime number 2 has the number zero and the assigned string (1,0,0, …). Number 3 has the number 1, so it is assigned to a string(0,1,0,0, … ); prime number 5 has number 2, so it corresponds to the string (0,0,1,0,0, … ), and so on. Prime numbers are, in fact, a base for module M. The number 4 is identi- fied with a string 2(1,0,0, … ) = (2,0,0, … ), number 6 is in this convention is a sum of a series of strings for 2 and 3, that is (1,1,0,0, … ). Rational number 15/8 is a string: 3(1,0,0, . . ) + (0,1,0,0, … ) + (0,0,1,0,0, . . ) = ( 3,1,1,0,0, … ) This isomorphism identifies strings of integers of the specified property with the rational −numbers. Multiplication of rational numbers is− adding these strings; division on the other hand is subtraction. The index of integer number is what the sum of terms within its string is called. If so, = 24 = 2 3 then its index is the number four – the sum of the exponents of primes occur-3 ring𝑛𝑛 in the factorization of this number. If the integer n is the𝑛𝑛 k-th power ∙of another integer its index is divisible by k. The index 125 is divisible by 3 because 125 = 5 . The concept 3of the index allows the hypothesis, which, if true, includes as a special case Fermat’s Last Theorem. Fermat’s Last Theorem says that if n is greater than or equal 3, there are natural numbers different from 0 satisfying the equation + = . In other words, if denotes the set of natural numbers 𝑛𝑛different𝑛𝑛 from𝑛𝑛 zero which are in the form , where 𝑛𝑛 , then if 3 𝑥𝑥and 𝑦𝑦, 𝑧𝑧 then + . 𝐹𝐹 ⊂ ℕ 𝑛𝑛 𝑘𝑘 𝑛𝑛 ∈ ℕ ≥ 𝑎𝑎 𝑏𝑏 ∈ 𝐹𝐹𝑛𝑛 𝑎𝑎 𝑏𝑏 ∉ 𝐹𝐹𝑛𝑛 Remarks about the square equation… 51

The multiplicative semigroup of natural numbers different from zero is isomorphic to the commutative semigroup∗ of words in an infinite alphabet, in which there are prime numbers. ℕIn this semigroup, the uniqueness of factorization of natural numbers into prime factors is an obvious fact. The defined above index of an integer is equal to the length of the word that represents this number. Thus, each prime number is an index of 1, the product of two primes has an index of 2, and so on. The number 0 does not have an index, and the number 1 has the index 0. The generalized hypothesis. If n is greater than equal to 3 and x and y are natural numbers different from zero, then the index of number + is not divisible by n. If this generalized hypothesis is true, then it naturally𝑛𝑛 follows𝑛𝑛 Fermat’s Theorem. 𝑥𝑥 𝑦𝑦

7. Concluding remarks

Mapping : described by the pattern ( ) = , as men- tioned above, covers the semigroup , that is, each element of the semigroup 0 0 is a square. At 𝜙𝜙the𝕂𝕂 same→ 𝕂𝕂 time we have to stress that semigroup𝜙𝜙 𝑔𝑔 𝑔𝑔 elements∘ 𝑓𝑓 may 0 have different radiuses of convergence𝕂𝕂 and the radius of composition convergence of such series is dependent on factors – and it may vary. The functions presented here can be looked at from a purely formal per- spective. These are new non-elementary features that enhance the mathematical arsenal. On this occasion, we would like to recall a flat pseudo-mathematical joke. – I found a new smooth function. –Yes? And what did you do with it? – I differentiated it. Secondly, speaking about the probability theory we can discuss not only the logarithmic-normal disintegration, but also about disin- tegrations such as semi-logarithmic-normal, half-logarithmic-normal and so on. The one-and-a-half logarithm is the function. Furthermore, the article is a pictorial representation of a method of3 indefinite coefficients very useful in solving function equations. 𝑔𝑔

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