Recurrence relationships allow us to accurately determine logarithms, square roots, factorials, several sequences (like Fibonacci), roots of equations and more. This paper aims to describe a recurrence relationship and present some applications. It presents the concept of reciprocal matrix and allowed the approximation of non perfect square roots of positive integers and their representation in continuous fraction. In addition, there was a correspondence between this study and the so-called Fibonacci polynomials. As it is a new job, more studies are needed to expand its concepts and applications

Cruz, Edson Costa

da Silva, Hamilton Brito

de Matos, Fernando Cardoso

dos Santos, Matheus de Albuquerque Coelho

English

Brazilian Journals

Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100561  Study and Application of a Recurrence Relationship  Estudo e Aplicação de uma Relação de Recorrência  DOI:10.34117/bjdv6n12-520  Recebimento dos originais:  21/11/2020 Aceitação para publicação:  21/12/2020  Hamilton Brito da Silva Mestre em Biologia Ambiental. Instituto Federal de Educação, Ciência e Tecnologia do Pará Rodovia Mário Covas, nº 200, Jiboia Branca, Ananindeua, Pará, Brasil. E-mail: hamilton.britomath@gmail.com  Matheus de Albuquerque Coelho dos Santos Licenciado em Matemática. Instituto Federal de Educação, Ciência e Tecnologia do Pará Avenida José Bonifácio, 1977, Guamá, Belém, Pará, Brasil. E-mail: profmatheus01@gmail.com   Fernando Cardoso de Matos Doutor em Educação em Ciências e Matemática. Instituto Federal de Educação, Ciência e Tecnologia do Pará Travessa Mauriti, 1000, Pedreira, Belém, Pará, Brasil. E-mail: matos2001@gmail.com  Edson Costa Cruz Mestre em Geofísica. Instituto Federal de Educação, Ciência e Tecnologia do Pará Travessa Vileta, 1130, Pedreira, Belém, Pará, Brasil. E-mail: edson.cruz@ifpa.edu.br    ABSTRACT Recurrence relationships allow us to accurately determine logarithms, square roots, factorials, several sequences (like Fibonacci), roots of equations and more. This paper aims to describe a recurrence relationship and present some applications. It presents the concept of reciprocal matrix and allowed the approximation of non perfect square roots of positive integers and their representation in continuous fraction. In addition, there was a correspondence between this study and the so-called Fibonacci polynomials. As it is a new job, more studies are needed to expand its concepts and applications.  Keywords: Recurrence relationship, Fibonacci Sequence, Reciprocal Matrix, Continued Fraction.   RESUMO Relações de recorrência permitem determinar com precisão logaritmos, raízes quadradas, fatoriais, diversas sequências (a Example da Fibonacci), raízes de equações e outros mais. O principal objetivo deste trabalho é descrever uma relação de recorrência e apresentar algumas aplicações. Ele apresenta o conceito de matriz recíproca e permitiu a aproximação de raízes quadradas não exatas de números inteiros positivos e sua representação em fração contínua. Além disso, houve a correspondência entre Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100562  este estudo e os chamados polinômios de Fibonacci. Por ser um trabalho novo, são necessários mais estudos para ampliar seus conceitos e aplicações.   Palavras-chave: Relação de recorrência, Sequência de Fibonacci, Matriz Recíproca, Fração Contínua.   1 INTRODUCTION A recurrence relationship is one where the first or first terms are given and the rest are obtained according to the previous terms. There are several examples of recurrence relationships such as the sequence of odd numbers, Fibonacci sequence and many others (ROSA, 2017; PEREIRA, 2014). Among the several problems that can be solved using the recurrence relation we have the approximation of non perfect square root of positive integers through the Pell’s equation (TEKCAN, 2007; KAPLAN & WILLIAMS, 1986). This paper aims to study a new recurrence relation that, like the Pell’s equation, allowed us to calculate with non perfect square root approximations of positive integers as well as their continuous fraction, Fibonacci polynomials, among other things.  2 RESULTS Consider that 𝑎𝑘,0 = 𝑏 , 𝑎0,𝑖 = 0 and the recurrence relationship below: 𝑎𝑘,𝑖 =ℎ𝑎𝑘,𝑖−1 + 𝑎𝑘−1,𝑖 (𝑰)  Condition of existence: ℎ ≥ −𝑏24 (ℎ and  𝑏 ∈ ℝ∗) We can organize the terms 𝑎𝑘,𝑖  in matrix form 𝐻𝑘×𝑖+1 not necessarily square: 𝐻𝑘,𝑖+1 =[    𝑎1,0𝑎2,0𝑎3,0𝑎1,1𝑎2,1𝑎3,1𝑎1,2𝑎2,2𝑎3,2⋯𝑎1,𝑖𝑎2,𝑖𝑎3,𝑖⋮ ⋱ ⋮𝑎𝑘,0 𝑎𝑘,1 𝑎𝑘,2⋯ 𝑎𝑘,𝑖]    =[    𝑏𝑏𝑏𝑎1,1𝑎2,1𝑎3,1𝑎1,2𝑎2,2𝑎3,2⋯𝑎1,𝑖𝑎2,𝑖𝑎3,𝑖⋮ ⋱ ⋮𝑏 𝑎𝑘,1 𝑎𝑘,2⋯ 𝑎𝑘,𝑖]      Example 1: We will use 𝑏 = 1 and ℎ = 2 until the term 𝑎3,2. 𝑎1,1 =2𝑎1,0 + 𝑎0,1=21 + 0= 2 𝑎2,1 =2𝑎2,0 + 𝑎1,1=21 + 2=23 𝑎3,1 =2𝑎3,0 + 𝑎2,1=21 +23=65 𝑎1,2 =2𝑎1,1 + 𝑎0,2=22 + 0= 1 Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100563  𝑎2,2 =2𝑎2,1 + 𝑎1,2=223+ 1=65 𝑎3,2 =2𝑎3,1 + 𝑎2,2=265+65=56 𝐻3,3 =[    1 2 11236516556]      Example 2: Now if we do 𝑏 = 1 and ℎ = 1 until the term 𝑎5,4  we will have the following matrix:  We can observe the following property in the matrix 𝐻𝑘×𝑖+1. P1: The numbers 𝑎𝑘,1 (𝑘 → ∞) converge to: lim𝑘→∞𝑎𝑘,1 =√𝑏2 + 4ℎ − 𝑏2 (𝑰𝑰) Example 3: If 𝑏 = 1 and ℎ = 2, so: lim𝑘→∞𝑎𝑘,1 =√12 + 4.2 − 12= 1   2.1 Reciprocal Matrix If 𝑏 = 1 and ℎ = 1, the matrix 𝐻𝑘×𝑖+1 is denominated by reciprocal matrix because each term in the matrix not belonging to the first row and first column is equal to the reciprocal of the sum between the number on the left and the number immediately above of the term (and vice versa). As an example, in the reciprocal matrix 𝐻5×5 seen in example 2 the term2027=135+34=134+35. Based in (II), we have: lim𝑘→∞𝑎𝑘,1 = lim𝑘→∞11 + 𝑎𝑘−1,1=√5− 12=1𝜙, 𝜙 = 𝑔𝑜𝑙𝑑𝑒𝑛 𝑟𝑎𝑡𝑖𝑜.  Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100564  In fact, in matrix𝐻5×5 seen previously, we see that the sequence 𝑎𝑘,1 ∈ {1,12,23,35,58, … } is the ratio between a Fibonacci number (𝐹𝑘)  and its successor Fibonacci (𝐹𝑘+1), that is: lim𝑘→∞𝑎𝑘,1 =𝐹𝑘𝐹𝑘+1=1𝜙 Similarly, we have the limit: lim𝑘→∞𝑎𝑘,2 =√4𝜙2 + 1 − 12𝜙  2.2.1 Properties of Reciprocal Matrix P2: If the reciprocal matrix 𝐻𝑘×𝑖+1 is square, there will be a palindrome sequence in inverted L.  Example 4: In the matrix 𝐻5×5 in the example 2 we have:  P3: The product of the terms of 𝑎1,1 until the term  𝑎𝑘,1 is equal to: 𝑃 =∏𝑎𝑝,1𝑘𝑝=1=1𝐹𝑘+1 (𝑰𝑰𝑰) The demonstration of (𝑰𝑰𝑰) is easy because it is enough to note that each term 𝑎𝑘,1 =𝐹𝑘𝐹𝑘+1. Therefore: 𝑃 = 𝑎1,1. 𝑎2,1. 𝑎3,1. 𝑎4,1…𝑎𝑘,1 =𝐹1𝐹2.𝐹2𝐹3.𝐹3𝐹4.𝐹4𝐹5…𝐹𝑘𝐹𝑘+1=𝐹1𝐹𝑘+1=1𝐹𝑘+1 P4: Each term 𝑎𝑘,2 of reciprocal matrix is given by: 𝑎𝑘,2 =𝐹𝑘+1. 𝑞(𝑘 + 1)𝑞(𝑘 + 2) (𝑰𝑽) Where 𝑞(1) = 0, 𝑞(2) = 1, 𝑞(3) = 1 and (4) = 3  . The remaining values of 𝑞(𝑘)  are recursively found by the formula: 𝑞(𝑘 + 2) = 𝐹𝑘. [𝑞(𝑘 + 1) + 𝑞(𝑘). 𝐹𝑘+1](𝑽) Example 5: The term 𝑎4,2 =2027 can also be calculated as follows:  Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100565  𝑎4,2 =𝐹5. 𝑞(5)𝑞(6)→ 𝑞(5) = 𝐹3. [𝑞(4) + 𝑞(3).𝐹4] = 2. (3 + 1.3) = 12 𝑎4,2 =𝐹5. 𝑞(5)𝑞(6)=𝐹5. 𝑞(5)𝐹4. [𝑞(5) + 𝑞(4).𝐹5] =5.123. (12 + 3.5)=2027 Proof of (𝑰𝑽). Each term 𝑎𝑘,2 is constructed as follows: Columns  𝒂𝒌,𝟎 𝒂𝒌,𝟏 𝒂𝒌,𝟐 Row k-1  1 𝐹𝑘−1𝐹𝑘 𝒂𝒌−𝟏,𝟐 Row k  1 𝐹𝑘𝐹𝑘+1 𝒂𝒌,𝟐 𝑎𝑘,2 =1𝐹𝑘𝐹𝑘+1+ 𝑎𝑘−1,2=𝐹𝑘+1𝐹𝑘 + 𝐹𝑘+1. 𝑎𝑘−1,2 (𝑽𝑰) Let’s use the mathematical induction in (IV). For 𝑘 = 1 we have: 𝑎𝑘,2 =𝐹𝑘+1. 𝑞(𝑘 + 1)𝑞(𝑘 + 2)→ 𝑎1,2 =𝐹2. 𝑞(2)𝑞(3)=1.11= 1 (𝑇𝑟𝑢𝑒) If (IV) is valid for any 𝑘(𝑘 > 1), so it is valid for every 𝑘 + 1. Therefore: 𝑎𝑘+1,2 =𝐹𝑘+2. 𝑞(𝑘 + 2)𝑞(𝑘 + 3) (𝑽𝑰𝑰) Of (𝑽𝑰) we have: 𝑎𝑘+1,2 =𝐹𝑘+2𝐹𝑘+1 + 𝐹𝑘+2. 𝑎𝑘,2(𝑽𝑰𝑰𝑰) Like 𝑎𝑘,2 =𝐹𝑘+1.𝑞(𝑘+1)𝑞(𝑘+2), so: 𝑎𝑘+1,2 =𝐹𝑘+2𝐹𝑘+1 + 𝐹𝑘+2. 𝑎𝑘,2=𝐹𝑘+2𝐹𝑘+1 + 𝐹𝑘+2.𝐹𝑘+1.𝑞(𝑘+1)𝑞(𝑘+2) 𝑎𝑘+1,2 =𝐹𝑘+2. 𝑞(𝑘 + 2)𝐹𝑘+1. [𝑞(𝑘 + 2) + 𝐹𝑘+2. 𝑞(𝑘 + 1)]=𝐹𝑘+2. 𝑞(𝑘 + 2)𝐹𝑘+1. [𝑞(𝑘 + 2) + 𝑞(𝑘 + 1). 𝐹𝑘+2] 𝑎𝑘+1,2 =𝐹𝑘+2. 𝑞(𝑘 + 2)𝑞(𝑘 + 3) (𝑿)  (𝑞. 𝑒. 𝑑)  Proof of  (𝑽). We have already proved to be valid (IV). From dividing numerator and denominator of (IV) by 𝑞(𝑘 + 1), we have: 𝑎𝑘,2 =𝐹𝑘+1. 𝑞(𝑘 + 1)𝑞(𝑘 + 2)=𝐹𝑘+1𝑞(𝑘+2)𝑞(𝑘+1)=𝐹𝑘+1𝐹𝑘.[𝑞(𝑘+1)+𝑞(𝑘).𝐹𝑘+1]𝑞(𝑘+1)=𝐹𝑘+1𝐹𝑘 +𝑞(𝑘)𝑞(𝑘+1). 𝐹𝑘. 𝐹𝑘+1 (𝑿𝑰) Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100566  Comparing (XI) with (VI), we have: 𝑎𝑘,2 =𝐹𝑘+1𝐹𝑘 +𝑞(𝑘)𝑞(𝑘+1). 𝐹𝑘. 𝐹𝑘+1=𝐹𝑘+1𝐹𝑘 + 𝐹𝑘+1. 𝑎𝑘−1,2 (𝑿𝑰𝑰) From the above equation it follows that:  𝐹𝑘 +𝑞(𝑘)𝑞(𝑘 + 1). 𝐹𝑘. 𝐹𝑘+1 = 𝐹𝑘 + 𝐹𝑘+1. 𝑎𝑘−1,2  𝑞(𝑘)𝑞(𝑘 + 1). 𝐹𝑘. 𝐹𝑘+1 = 𝐹𝑘+1. 𝑎𝑘−1,2 𝑎𝑘−1,2 =𝐹𝑘. 𝑞(𝑘)𝑞(𝑘 + 1)→ 𝑎𝑘,2 =𝐹𝑘+1. 𝑞(𝑘 + 1)𝑞(𝑘 + 2)  The above result (equation (IV)) has already been proven to be valid and therefore proves that (V) is also true.  P5: The product of terms of 𝑎1,2 until 𝑎𝑘,2 is given by: 𝑃 =∏𝑎𝑝,2𝑘𝑝=1=∏ 𝐹𝑛+1𝑘𝑛=1𝐹𝑘. [𝑞(𝑘 + 1) + 𝑞(𝑘). 𝐹𝑘+1] (𝑿𝑰𝑽) Proof of (𝑿𝑰𝑽). 𝑃 =∏𝑎𝑝,2𝑘𝑝=1= 𝑎1,2. 𝑎2,2. 𝑎3,2. 𝑎4,2…𝑎𝑘,2 𝑃 =𝐹2. 𝑞(2)𝑞(3).𝐹3. 𝑞(3)𝑞(4).𝐹4. 𝑞(4)𝑞(5).𝐹5. 𝑞(5)𝑞(6)…𝐹𝑘+1. 𝑞(𝑘 + 1)𝑞(𝑘 + 2) 𝑃 =𝐹2. 𝑞(2)𝑞(3).𝐹3. 𝑞(3)𝑞(4).𝐹4. 𝑞(4)𝑞(5).𝐹5. 𝑞(5)𝑞(6)…𝐹𝑘+1. 𝑞(𝑘 + 1)𝑞(𝑘 + 2)=𝐹2. 𝐹3. 𝐹4. 𝐹5…𝐹𝑘+1. 𝑞(2)𝑞(𝑘 + 2) 𝑃 =𝐹2. 𝐹3. 𝐹4. 𝐹5…𝐹𝑘+1. 1𝑞(𝑘 + 2)=∏ 𝐹𝑛+1𝑘𝑛=1𝐹𝑘. [𝑞(𝑘 + 1) + 𝑞(𝑘). 𝐹𝑘+1]  Example 6: The product of terms of 𝑎1,2 until 𝑎5,2 is: 𝑃 =∏ 𝐹𝑛+15𝑛=1𝐹5. [𝑞(6) + 𝑞(5). 𝐹6]=𝐹2. 𝐹3. 𝐹4. 𝐹5. 𝐹6𝐹5. (81 + 12.8)=1.2.3.5.85. (81 + 96)=1659 Really, 𝑃 = 1.23.34.2027.216295=1659  3 APPLICATIONS OF RECURRENCE RELATIONSHIP 3.1 Determine of approximation of non perfect square roots of positive integers To approximate non perfect square roots of positive integers we can use the term 𝑎𝑘,1 with (𝑰𝑰) in a matrix 𝐻𝑘×2.  Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100567  Example 7: Let’s use the term 𝑎5,1 of a matrix 𝐻5×2 to approximate √2.  Solution: We know that lim𝑘→∞𝑎𝑘,1 =√𝑏2+4ℎ−𝑏2. We must choose satisfactory numbers 𝑏 and ℎ to make √2 appear in the expression. Let’s use 𝑏 = 2 and ℎ = 1. Doing this we have: 𝑎𝑘,1 ≈√𝑏2 + 4ℎ − 𝑏2→ 𝑎5,1 ≈√22 + 4.1 − 22= √2 − 1 So, let's calculate the term 𝑎5,1which will be a good approximation for √2 − 1. 𝑎1,0 = 2 , 𝑎0,1 = 0  and 𝑎𝑘,1 =1𝑎𝑘,0+𝑎𝑘−1,1 𝑎1,1 =1𝑎1,0 + 𝑎0,1=12 + 0=12 𝑎2,1 =1𝑎2,0 + 𝑎1,1=12 +12=25 𝑎3,1 =1𝑎3,0 + 𝑎2,1=12 +25=512 𝑎4,1 =1𝑎4,0 + 𝑎3,1=12 +512=1229 𝑎5,1 =1𝑎5,0 + 𝑎4,1=12 +1229=2970 The value 𝑎5,1 =2970 is a good approximation to √2 − 1. Therefore, we can do: √2 − 1 ≈2970→  √2 ≈9970≈ 1,4142. Example 8: Let’s use the term 𝑎5,1 of a matrix 𝐻5×2 to approximate √0,3.  Solution: A good possibility is to use 𝑏 =  1 and ℎ = −740. 𝑎𝑘,1 ≈√𝑏2 + 4ℎ − 𝑏2→ 𝑎3,1 ≈√0,3 − 12 𝑎1,0 = 1 , 𝑎0,1 = 0  and 𝑎𝑘,1 =−740(𝑎𝑘,0+𝑎𝑘−1,1) 𝑎1,1 =−740(𝑎1,0 + 𝑎0,1)=−740 𝑎2,1 =−740(𝑎2,0 + 𝑎1,1)=−740 (1 −740)= −733 𝑎3,1 =−740(𝑎3,0 + 𝑎2,1)=−740 (1 −733)= −2311040 Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100568  𝑎4,1 =−740(𝑎4,0 + 𝑎3,1)=−740 (1 −2311040)= −182809 𝑎5,1 =−740(𝑎5,0 + 𝑎4,1)=−740(1 −182809)= −556325080 Assim: 𝑎5,1 ≈√0,3 − 12≈ −556325080→ √0,3 ≈687712540≈ 0,5484.  3.2 Continued Fraction of non perfect square root of positive integers  With the non perfect square root approximations of a positive integer as in the previous topic, it is easy to determine the continuous fractions. Example 9: Using the example 7, let’s determine the continued fractions of  √2 remembering that in this example 𝑎𝑘,1 ≈ √2 − 1. Solution: 𝑎1,1 =12= [0; 2] 𝑎2,1 =25=12 +12= [0; 2,2] 𝑎3,1 =512=12 +25=12 +12+12= [0; 2,2,2] 𝑎4,1 =1229=12 +512=12 +12+12+12= [0; 2,2,2,2] 𝑎5,1 =2970=12 +1229=12 +12+12+12+12= [0;2,2,2,2,2] If we continue this indefinitely, we will always obtain as a continuous fraction [0; 2̅]. Therefore: √2 − 1 = [0; 2̅] → √2 = 1 + [0; 2̅] = [1; 2̅] Example 10: Let’s calculate the continued fractions of √5. Solution: We have already seen that in the reciprocal matrix lim𝑘→∞𝑎𝑘,1 =√5−12 which allows us to state that √5 ≈ 2. 𝑎𝑘,1 + 1. So just calculate the terms 𝑎𝑘,1 and observe the behavior of a possible period, remembering that 𝑏 = ℎ = 1. √5 ≈ 2. 𝑎1,1 + 1 = 2.1 + 1 = 3 = [3] Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100569  √5 ≈ 2. 𝑎2,1 + 1 = 2.12+ 1 = 2 = [2] √5 ≈ 2. 𝑎3,1 + 1 = 2.23+ 1 =73= [2; 3] √5 ≈ 2. 𝑎4,1 + 1 = 2.35+ 1 =115= [2; 5] √5 ≈ 2. 𝑎5,1 + 1 = 2.58+ 1 =188= [2; 4] In doing so, we find the next continuous fractions to be [2; 4,3],  [2;4,5] , [2;4,4] , [2; 4,4,3], [2; 4,4,5] , [2; 4,4,4], [2; 4,4,4, 3], [2; 4,4,4, 5] , [2; 4,4,4, 4], [2; 4,4,4, 4,3], [2; 4,4,4, 4,5] , [2; 4,4,4, 4,4]. Following these steps definitely, we find as a continuous fraction √5 = [2; 4̅].  Based on this, it’s possible to obtains the following expressions ∀𝑘 ∈ ℕ∗  where [𝑎0; 𝑎1, 𝑎2, … , 𝑎𝑛] represent continuous fractions and 𝐹𝑘is the Fibonacci number. 2.𝐹3𝑘𝐹3𝑘+1= [1; 4,4,… ,4⏟    𝑘−1 𝑣𝑒𝑧𝑒𝑠, 3] (𝑿𝑽) 2.𝐹3𝑘+1𝐹3𝑘+2= [1; 4,4,… ,4⏟    𝑘−1 𝑣𝑒𝑧𝑒𝑠, 5] (𝑿𝑽𝑰) 2.𝐹3𝑘+2𝐹3𝑘+3= [1; 4,4,… ,4⏟    𝑘 𝑣𝑒𝑧𝑒𝑠] (𝑿𝑽𝑰𝑰) 3.3 Fibonacci Polynomials Fibonacci polynomials (generalization of Fibonacci numbers) are defined by the second order linear recurrence relation: 𝐹𝑛(𝑥) = {0, 𝑖𝑓 𝑛 = 01, 𝑖𝑓 𝑛 = 1𝑥. 𝐹𝑛−1(𝑥) + 𝐹𝑛−2(𝑥), 𝑖𝑓 𝑛 ≥ 2=(𝑥 + √𝑥2 + 4 )𝑛− (𝑥 − √𝑥2 + 4 )𝑛2𝑛 . √𝑥2 + 4  We have 𝐹𝑛(1) = 𝐹𝑛. Example 11: Let’s determine the 6th 𝐹𝑛(𝑥). 𝐹1(𝑥) = 1 𝐹2(𝑥) = 𝑥 𝐹3(𝑥) = 𝑥2 + 1 𝐹4(𝑥) = 𝑥3 + 2𝑥 𝐹5(𝑥) = 𝑥4 + 3𝑥2 + 1 𝐹6(𝑥) = 𝑥5 + 4𝑥3 + 3𝑥 Fibonacci polynomials relate to the recurrence relationship of this work as follows: lim𝑛→∞𝐹𝑛(𝑥)𝐹𝑛+1(𝑥)=√𝑥2 + 4 − 𝑥2 Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100570  This is equivalent to finding lim𝑛→∞𝑎𝑛,1 , when we have 𝑏 = 𝑥  and ℎ = 1  in the recurrence relationship 𝑎𝑛,1 =1𝑎𝑛,0+𝑎𝑛−1,1, 𝑎𝑛,0 = 𝑏 = 𝑥 , 𝑎0,1 = 0 . So we have: lim𝑛→∞𝐹𝑛(1)𝐹𝑛+1(1)=√5 − 12=1𝜙→ 𝑎𝑢𝑡𝑜𝑚𝑎𝑡𝑖𝑐 𝑑𝑒𝑚𝑜 lim𝑛→∞𝐹𝑛(2)𝐹𝑛+1(2)= √2 − 1 4 FINAL CONSIDERATIONS The recurrence relationship studied in this work provided us with tools to determine approximation for non perfect square roots of positive integers as well as their representation by continuous fraction and relation to Fibonacci polynomials. The concept of reciprocal matrix proved to be innovative and with intriguing properties when it is closely related to the Fibonacci sequence and golden ratio. However, as it is a new tool, further studies can still be done to expand the fields of application of these approaches.   Brazilian Journal of Development       Braz. J. of  Develop., Curitiba, v. 6, n.12, p.100561-100571 dec. 2020.               ISSN 2525-8761   100571  REFERENCE  HENRIQUE, F. Resolução de Recorrências. São Paulo: [s.n.], (2019).  ISBN 9781796412352  KAPLAN, P., WILLIAMS, K.S. Pell’s Equations 𝒙𝟐 −𝒎𝒚𝟐 = −𝟏,−𝟒 and Continued Fractions, Jour. Number Theory.23 (1986), 169–182.  PEREIRA, M. V. Recorrências –Problemas e Aplicações. Dissertação de Mestrado, Universidade de Brasília. Brasília, 2014.  ROSA, M. A. A importância das Relações de Recorrência para melhoria do Ensino-Aprendizagem da Matemática Discreta. Dissertação de Mestrado, Universidade Estadual de Campinas. Campinas, 2017.  TEKCAN, A. The Pell Equation 𝒙𝟐 −𝑫𝒚𝟐 = ±𝟒. Applied Mathematical Sciences, Vol. 1, 2007, no. 8, 363 – 369. 

Study and Application of a Recurrence Relationship / Estudo e Aplicação de uma Relação de Recorrência

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Study and Application of a Recurrence Relationship / Estudo e Aplicação de uma Relação de Recorrência

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