Abstract

Many researchers have been working on recurrence relation which is an important topic not only in mathematics but also in physics, economics and various applications in computer science. There are many useful results on recurrence relation sequence but there main problem to find any term of recurrence relation sequence we need to find all previous terms of recurrence relation sequence. There were many important theorems obtained on recurrence relations. In this paper we have given special identity for generalized kth order recurrence relation. These identities are very useful for finding any term of any order of recurrence relation sequence. Authors define a special formula in this paper by this we can find direct any term of a recurrence relation sequence. In this recurrence relation sequence to find any terms we need to find all previous terms so this result is very important. There is important property of a relation between coefficients of recurrence relation terms and roots of a polynomial for second order relation but in this paper, we gave this same property of recurrence relation of all higher order recurrence relation. So finally, we can say that this theorem is valid all order of recurrence relation only condition that roots are distinct. So, we can say that this paper is generalization of property of a relation between coefficients of recurrence relation terms and roots of a polynomial. Theorem: - Let C<sub>1</sub> and C<sub>2</sub> are arbitrary real numbers and suppose the equation <img src=image/13422456_01.gif> (1) Has X<sub>1</sub> and X<sub>2</sub> are distinct roots. Then the sequence <img src=image/13422456_25.gif> is a solution of the recurrence relation <img src=image/13422456_02.gif> (2) <img src=image/13422456_03.gif>. For n= 0, 1, 2 …where β<sub>1</sub> and β<sub>2</sub> are arbitrary constants. Proof: - First suppose that <img src=image/13422456_25.gif> of type <img src=image/13422456_04.gif> we shall prove <img src=image/13422456_25.gif> is a solution of recurrence relation (2). Since X<sub>1</sub>, X<sub>2</sub> and X<sub>3</sub> are roots of equation (1) so all are satisfied equation (1) so we have<img src=image/13422456_05.gif>, <img src=image/13422456_06.gif>. Consider <img src=image/13422456_07.gif><img src=image/13422456_08.gif>. This implies <img src=image/13422456_09.gif>. So the sequence <img src=image/13422456_25.gif> is a solution of the recurrence relation. Now we will prove the second part of theorem. Let <img src=image/13422456_10.gif> is a sequence with three <img src=image/13422456_11.gif>. Let <img src=image/13422456_12.gif>. So <img src=image/13422456_13.gif> (3). <img src=image/13422456_14.gif> (4). Multiply by X<sub>1</sub> to (3) and subtracts from (4). We have <img src=image/13422456_15.gif> similarly we can find <img src=image/13422456_16.gif>. So we can say that values of β<sub>1</sub> and β<sub>2</sub> are defined as roots are distinct. So non- trivial values ofβ<sub>1</sub> and β<sub>2</sub> can find and we can say that result is valid. Example: Let <img src=image/13422456_25.gif> be any sequence such that <img src=image/13422456_17.gif> n≥3 and a<sub>0</sub>=0, a<sub>1</sub>=1, a<sub>2</sub>=2. Then find a<sub>10</sub> for above sequence. Solution: The polynomial of above sequence is <img src=image/13422456_18.gif>. Solving this equation we have roots are 1, 2, and 3 using above theorem we have <img src=image/13422456_19.gif> (7). Using a<sub>0</sub>=0, a<sub>1</sub>=1, a<sub>2</sub>=2 in (7) we have β<sub>1</sub>+β<sub>2</sub>+β<sub>3</sub>=0 (8). β<sub>1</sub>+2β<sub>2</sub>+3β<sub>2</sub>=1 (9).β<sub>1</sub>+4β<sub>2</sub>+9β<sub>3</sub>=2 (10) Solving (8), (9) and (10) we have <img src=image/13422456_20.gif>, <img src=image/13422456_21.gif>, <img src=image/13422456_22.gif>. This implies <img src=image/13422456_23.gif>. Now put n=10 we have a<sub>10</sub>=-27478. Recurrence relation is a very useful topic of mathematics, many problems of real life may be solved by recurrence relations, but in recurrence relation there is a major difficulty in the recurrence relation. If we want to find 100th term of sequence, then we need to find all previous 99 terms of given sequence, then we can get 100th term of sequence but above theorem is very useful if coefficients of recurrence relation of given sequence satisfies the condition of the above theorem, then we can apply above theorem and we can find direct any term of sequence without finding all previous terms.

Highlights

  • Theorem3.1: - Let c1, c2 and c3 are arbitrary real numbers and suppose the equationIn Number Theory there are many special types of sequences

  • Fibonacci Sequence and Luca Sequence are special type of sequences obtained from recurrence relation with given initial terms

  • Recurrence relations are used in mathematics as well as economics; physics and are very useful in solving real life problems. [4,5], we can calculate growth in economics by recurrence techniques

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Summary

Introduction

Theorem3.1: - Let c1, c2 and c3 are arbitrary real numbers and suppose the equation. In Number Theory there are many special types of sequences. Fibonacci Sequence and Luca Sequence are special type of sequences obtained from recurrence relation with given initial terms. Network marketing business is a special type of recurrence relation and many problems of Network marketing business can be solved by recurrence methods.[6,7] In recurrence relation for finding any term of sequence we need to find all previous terms of sequence but by using the theorem given in the paper we can find direct any term of sequence. In Number Theory there are many special types of Sequences Both Fibonacci Sequence and Luca Sequence are special type of recurrence relation with given initial terms. The sequence < an > is a solution of the recurrence relation an = c1an−1 + c2an−2 + c3an−3n ≥ 3 (3.2) If and only if an = β1x1n + β2x2n + β3x3n for n = 0, 1, 2, .... Non-trivial values of β1, β2and β3can find and we can say that result is valid

Fourth Order Recurrence Relation
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