![International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470
@ IJTSRD | Unique Paper ID – IJTSRD26662 | Volume – 3 | Issue – 5 | July - August 2019 Page 1342
formula for all the solutions of the recurrence relation.
Solving Linear Homogeneous Recurrence Relations of
Degree Two:
Theorem 1: Let c 1 and c 2 be real numbers. Suppose that r 2
– c 1 r – c 2 = 0 has two distinct roots r 1 and r 2. Then the
sequence {a n } is a solution to the recurrence relation a n = c
1 a n −1 + c 2 a n −2 if and only if for n = 0, 1, 2,…, where α 1
and α 2 are constants.
Example: What is the solution to the recurrence relation a n
= a n −1 + 2 a n −2 with a 0 = 2 and a 1 = 7 ? Solution: The
characteristic equation is r 2 − r − 2 = 0. Its roots are r = 2
and r = −1. Therefore, {a n } is a solution to the recurrence
relation if and only if a n = α 1 2 n + α 2 ( −1 ) n, for some
constants α 1 and α 2. To find the constants α 1 and α 2, note
that a 0 = 2 = α 1 + α 2 and a 1 = 7 = α 1 2 + α 2 ( −1 ). Solving
these equations, we find that α 1 = 3 and α 2 = −1. Hence, the
solution is the sequence {a n } with a n = 3∙2 n − ( −1 ) n.
To find an explicit formula for the Fibonacci numbers: The
sequence of Fibonacci numbers satisfies the recurrence
relation f n = f n −1 + f n −2 with the initial conditions: f 0 = 0
and f 1 = 1. Solution: The roots of the characteristic equation
r 2 – r – 1 = 0. For some constants α 1 and α 2: Using the
initial conditions f 0 = 0 and f 1 = 1, we have Solving, we
obtain.
Theorem 2 : Let c 1 and c 2 be real numbers with c 2 ≠ 0.
Suppose that r 2 – c 1 r – c 2 = 0 has one repeated root r 0.
Then the sequence {a n } is a solution to the recurrence
relation a n = c 1 a n −1 + c 2 a n −2 if and only if for n =
0,1,2,…, where α 1 and α 2 are constants.
Example: What is the solution to the recurrence relation a n
= 6a n −1 − 9 a n −2 with a 0 = 1 and a 1 = 6? Solution: The
characteristic equation is r 2 − 6 r + 9 = 0. The only root is r =
3. Therefore, {a n } is a solution to the recurrence relation if
and only if a n = α 1 3 n + α 2 n( 3 ) n where α 1 and α 2 are
constants. To find the constants α 1 and α 2, note that a 0 = 1
= α 1 and a 1 = 6 = α 1 ∙ 3 + α 2 ∙3. Solving, we find that α 1 =
1 and α 2 = 1. Hence, a n = 3 n + n 3 n.
Solving Linear Homogeneous Recurrence Relations of
Arbitrary Degree This theorem can be used to solve linear
homogeneous recurrence relations with constant
coefficients of any degree when the characteristic equation
has distinct roots. Theorem 3 : Let c 1, c 2,…, c k be real
numbers. Suppose that the characteristic equation r k – c 1 r
k −1 –⋯ – c k = 0 has k distinct roots r 1, r 2, …, r k. Then a
sequence {a n } is a solution of the recurrence relation a n =c
1 a n −1 + c 2 a n −2 + ….. + c k a n − k if and only if for n = 0,
1, 2, …, where α 1, α 2,…, α k are constants.
4. Method and Solution
Un = C1Un-1+ C2Un-2+ ……..+ Cd Un-d
Where C1, C2, ……. , Cd are number and Cd ≠0.
Pn = Un+ U' and q n =β U n
Pn = (C1 Un-1 +C2 Un-2 + ……. + Cd Un-d) + ( Cn U'n +C2U'n-2 + …..+
Cd U'n-d)
= C1 (Un-1 + U'n-1) + C2 (Un-2 + U'n-2 ) +…….+ Cd ( Un-d + U'n-d)
= C1 Pn-1 + C2 Pn-2 + …… + Cd P n-d
So, P n is a solution of the recurrence.
qn = β Un
= β ( C1 Un-1 + C2 Un-2 + ….. + Cd Un-d)
= C1 (βUn-1) + C2 (β Un-2) + …… + Cd (β Un-d)
= C1 qn-1 + C2 qn-2 + …. + Cd q n-d
So, q n is a solution of recurrence.
Example: What is the solution of recurrence relation.
Un = 2 Un-1 +Un-2 – 2 Un-3, U0 =3, U1=6, U2=0
The Characteristic equation is
xn- 2 xn-1 - xn-2+ 2xn-3 =0
x3-2x2-x+2 =0
x1=-1, x2=2, x3=1
The general solution is Un= β1x1
n+ β2x2
n+β3x3
n
Un= β1(–1)n+ β22n+β31n
U0=3 β1+β2+β3=3
U1=6 –β1+2β2+β3=6
U2=0 β1+4β2+β3=0
β1=–2, β2= –1, β3=6
The solution is Un= (–2) (–1)n+(–1)2n+6
5. Conclusion
A simple and efficient method for solving linear
homogeneous recursive relations has been introduced,
which mainly involves synthetic divisions. This method has
the advantage that it does not need to use generating
function techniques or solve a system of linear equations to
determine the unknowncoefficients ofthesolution.A variety
of techniques is available for finding explicit formulas for
special classes of recursivelydefinedsequences.Themethod
explained here works for the Fibonacci and other similarly
defined sequences.
Reference
[1] https://math.stackexchange.com/questions/246221/
what-are-linear-homogeneous-and-non-homoegenous-
recurrence-relations
[2] https://mathinsight.org/definition/recurrence_relatio
n
[3] Mircea I. Cîrnu.”Linear recurrence relations with the
coefficients in progression”; Annales Mathematicae et
Informaticae; 42 (2013) pp. 119–127;
https://www.researchgate.net/publication/26042484
9
[4] Yiu-Kwong Man. “Solving Linear Homogeneous
Recurrence Relation via the Inverse of Vander monde
Matrix”; Proceedings of the International Multi
Conference of Engineers andComputerScientists2018
Vol I IMECS 2018, March 14-16, 2018, Hong Kong
[5] A. G. Shannon, R. P. Loh,R. S. Melham,A. F. Horadam“A
Search for Solutions of a Functional Equation”
https://link.springer.com/chapter/10.1007/978-94-
009-0223-7_36
[6] https://www.usna.edu/Users/cs/crabbe/2004-
01/SI262/recur/recur.pdf
-
-](https://image.slidesharecdn.com/254impactoflinearhomogeneousrecurrentrelationanalysis-190912055116/85/Impact-of-Linear-Homogeneous-Recurrent-Relation-Analysis-2-320.jpg)
Impact of Linear Homogeneous Recurrent Relation Analysis
- 1. International Journal of Trend in Scientific Research and Development (IJTSRD) Volume 3 Issue 5, August 2019 Available Online: www.ijtsrd.com e-ISSN: 2456 – 6470 @ IJTSRD | Unique Paper ID – IJTSRD26662 | Volume – 3 | Issue – 5 | July - August 2019 Page 1341 Impact of Linear Homogeneous Recurrent Relation Analysis Thidar Hlaing University of computer Studies (Hpa-An), Kayin, Myanmar How to cite this paper: Thidar Hlaing "Impact of Linear Homogeneous Recurrent Relation Analysis" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456- 6470, Volume-3 | Issue-5, August 2019,pp.1341-1342, https://doi.org/10.31142/ijtsrd26662 Copyright © 2019 by author(s) and International Journal ofTrend inScientific Research and Development Journal. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (CC BY 4.0) (http://creativecommons.org/licenses/by /4.0) ABSTRACT A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence. A recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms are given; each further term of the sequence or array is defined as a function of the preceding terms. 1. INTRODUCTION A recurrence relation is an equation that defines a sequencebasedona rulethat gives the next term as a function of the previous term(s). The simplestformofa recurrence relation is the case where the next term depends only on the immediately previous term. If wedenotethe nnthtermin thesequence by xnxn, such a recurrence relation is of the form xn+1=f(xn)xn+1=f(xn) for some function ff. One such example is xn+1=2−xn/2.xn+1=2−xn/2. A recurrence relation can also be higher order, where the term xn+1xn+1 could depend not only on the previous term xnxn but also on earlier terms such as xn−1xn−1, xn−2xn−2, etc. A second order recurrence relation depends just on xnxn and xn−1xn−1 and is of the form xn+1=f(xn,xn−1)xn+1=f(xn,xn−1)for some function ff with two inputs. For example, the recurrence relation xn+1=xn+xn−1xn+1=xn+xn−1 cangeneratethe Fibonaccinumbers.To generate sequence basd on a recurrence relation, one must start with some initial values. For a first order recursion xn+1=f(xn)xn+1=f(xn), one just needs to start with an initial value x0x0 and can generate all remaining terms using the recurrence relation. For a second orderrecursionxn+1=f(xn,xn−1)xn+1=f(xn,xn−1),oneneeds to begin with two values x0x0 and x1x1. Higher order recurrence relations require correspondingly more initial values. A linear homogeneous recurrence of order kk is expressed this way: A0an+A1an−1+A2an−2+⋯+Akan−k=0 2. Objectives The aim of this paper is to solve the linear recurrence relation xn+1 = a0 xn + a1 xn−1 + · ·· + an−1 x1 + an x0 , n = 0, 1, 2, . . . , when its constant coefficients are in arithmetic, respective geometric progression. Rather surprising, when the coefficients are in arithmetic progression, the solution is a sequence of certain generalized Fibonacci numbers, but not of usual Fibonacci numbers, while if they are in geometric progression the solution is again a geometric progression, with different ratio. Recurrence relations con bedividedinto two: Linear and Non-Linear. Linear homogeneous recurrence relations with constant coefficients; Solving linear homogeneous recurrence relations with constant coefficients; Solving linear homogeneous recurrence relations with constantcoefficientsofdegreetwoanddegree three; Generating functions; Using generating functions to solve recurrence relations. Some recurrence relations are solvable using algebraic techniques, but they’re often tricky, they require some math many of you don’t know, and it still won’t work for many relations. 3. Methodology Linear Homogeneous Recurrence Relations Definition: A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n −1 + c 2 a n −2 + ….. + c k a n − k, where c 1, c 2, ….,c k are real numbers, and c k ≠ 0 it is linear because the right- hand side is a sum of the previous terms of the sequence each multiplied by a function ofn.itis homogeneous because no terms occur that are not multiples of the a j s. Each coefficient is a constant. The degree is k because a n is expressed in terms of the previous k terms of the sequence. By strong induction, a sequence satisfying such arecurrence relation is uniquely determined by the recurrence relation and the k initial conditions a 0 = C 0, a 0 = C 1, …, a k −1 = C k −1. Examples of Linear Homogeneous Recurrence Relations P n = (1.11) P n-1 linear homogeneous recurrence relationof degree one f n = f n-1 + f n-2 linear homogeneous recurrence relation of degree two not linear H n = 2 H n −1 + 1 not homogeneous B n = n B n −1 coef icients are not constants. SolvingLinearHomogeneous RecurrenceRelations The basic approach is to look for solutions of the form a n = r n, where r is a constant. Note that a n = r n is a solution to the recurrence relation a n = c 1 a n −1 + c 2 a n −2 + ⋯ + c k a n − k if and only if r n = c 1 r n −1 + c 2 r n −2 + ⋯ + c k r n − k. Algebraic manipulationyieldsthecharacteristicequation: rk − c 1 r k −1 − c 2 r k −2 − ⋯ − c k −1 r − c k = 0 The sequence {a n } with a n = r n is a solution if and only if r is a solution to the characteristic equation. The solutions to the characteristic equation are called the characteristic roots of the recurrence relation. The roots areusedtogiveanexplicit IJTSRD26662
- 2. International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470 @ IJTSRD | Unique Paper ID – IJTSRD26662 | Volume – 3 | Issue – 5 | July - August 2019 Page 1342 formula for all the solutions of the recurrence relation. Solving Linear Homogeneous Recurrence Relations of Degree Two: Theorem 1: Let c 1 and c 2 be real numbers. Suppose that r 2 – c 1 r – c 2 = 0 has two distinct roots r 1 and r 2. Then the sequence {a n } is a solution to the recurrence relation a n = c 1 a n −1 + c 2 a n −2 if and only if for n = 0, 1, 2,…, where α 1 and α 2 are constants. Example: What is the solution to the recurrence relation a n = a n −1 + 2 a n −2 with a 0 = 2 and a 1 = 7 ? Solution: The characteristic equation is r 2 − r − 2 = 0. Its roots are r = 2 and r = −1. Therefore, {a n } is a solution to the recurrence relation if and only if a n = α 1 2 n + α 2 ( −1 ) n, for some constants α 1 and α 2. To find the constants α 1 and α 2, note that a 0 = 2 = α 1 + α 2 and a 1 = 7 = α 1 2 + α 2 ( −1 ). Solving these equations, we find that α 1 = 3 and α 2 = −1. Hence, the solution is the sequence {a n } with a n = 3∙2 n − ( −1 ) n. To find an explicit formula for the Fibonacci numbers: The sequence of Fibonacci numbers satisfies the recurrence relation f n = f n −1 + f n −2 with the initial conditions: f 0 = 0 and f 1 = 1. Solution: The roots of the characteristic equation r 2 – r – 1 = 0. For some constants α 1 and α 2: Using the initial conditions f 0 = 0 and f 1 = 1, we have Solving, we obtain. Theorem 2 : Let c 1 and c 2 be real numbers with c 2 ≠ 0. Suppose that r 2 – c 1 r – c 2 = 0 has one repeated root r 0. Then the sequence {a n } is a solution to the recurrence relation a n = c 1 a n −1 + c 2 a n −2 if and only if for n = 0,1,2,…, where α 1 and α 2 are constants. Example: What is the solution to the recurrence relation a n = 6a n −1 − 9 a n −2 with a 0 = 1 and a 1 = 6? Solution: The characteristic equation is r 2 − 6 r + 9 = 0. The only root is r = 3. Therefore, {a n } is a solution to the recurrence relation if and only if a n = α 1 3 n + α 2 n( 3 ) n where α 1 and α 2 are constants. To find the constants α 1 and α 2, note that a 0 = 1 = α 1 and a 1 = 6 = α 1 ∙ 3 + α 2 ∙3. Solving, we find that α 1 = 1 and α 2 = 1. Hence, a n = 3 n + n 3 n. Solving Linear Homogeneous Recurrence Relations of Arbitrary Degree This theorem can be used to solve linear homogeneous recurrence relations with constant coefficients of any degree when the characteristic equation has distinct roots. Theorem 3 : Let c 1, c 2,…, c k be real numbers. Suppose that the characteristic equation r k – c 1 r k −1 –⋯ – c k = 0 has k distinct roots r 1, r 2, …, r k. Then a sequence {a n } is a solution of the recurrence relation a n =c 1 a n −1 + c 2 a n −2 + ….. + c k a n − k if and only if for n = 0, 1, 2, …, where α 1, α 2,…, α k are constants. 4. Method and Solution Un = C1Un-1+ C2Un-2+ ……..+ Cd Un-d Where C1, C2, ……. , Cd are number and Cd ≠0. Pn = Un+ U' and q n =β U n Pn = (C1 Un-1 +C2 Un-2 + ……. + Cd Un-d) + ( Cn U'n +C2U'n-2 + …..+ Cd U'n-d) = C1 (Un-1 + U'n-1) + C2 (Un-2 + U'n-2 ) +…….+ Cd ( Un-d + U'n-d) = C1 Pn-1 + C2 Pn-2 + …… + Cd P n-d So, P n is a solution of the recurrence. qn = β Un = β ( C1 Un-1 + C2 Un-2 + ….. + Cd Un-d) = C1 (βUn-1) + C2 (β Un-2) + …… + Cd (β Un-d) = C1 qn-1 + C2 qn-2 + …. + Cd q n-d So, q n is a solution of recurrence. Example: What is the solution of recurrence relation. Un = 2 Un-1 +Un-2 – 2 Un-3, U0 =3, U1=6, U2=0 The Characteristic equation is xn- 2 xn-1 - xn-2+ 2xn-3 =0 x3-2x2-x+2 =0 x1=-1, x2=2, x3=1 The general solution is Un= β1x1 n+ β2x2 n+β3x3 n Un= β1(–1)n+ β22n+β31n U0=3 β1+β2+β3=3 U1=6 –β1+2β2+β3=6 U2=0 β1+4β2+β3=0 β1=–2, β2= –1, β3=6 The solution is Un= (–2) (–1)n+(–1)2n+6 5. Conclusion A simple and efficient method for solving linear homogeneous recursive relations has been introduced, which mainly involves synthetic divisions. This method has the advantage that it does not need to use generating function techniques or solve a system of linear equations to determine the unknowncoefficients ofthesolution.A variety of techniques is available for finding explicit formulas for special classes of recursivelydefinedsequences.Themethod explained here works for the Fibonacci and other similarly defined sequences. Reference [1] https://math.stackexchange.com/questions/246221/ what-are-linear-homogeneous-and-non-homoegenous- recurrence-relations [2] https://mathinsight.org/definition/recurrence_relatio n [3] Mircea I. Cîrnu.”Linear recurrence relations with the coefficients in progression”; Annales Mathematicae et Informaticae; 42 (2013) pp. 119–127; https://www.researchgate.net/publication/26042484 9 [4] Yiu-Kwong Man. “Solving Linear Homogeneous Recurrence Relation via the Inverse of Vander monde Matrix”; Proceedings of the International Multi Conference of Engineers andComputerScientists2018 Vol I IMECS 2018, March 14-16, 2018, Hong Kong [5] A. G. Shannon, R. P. Loh,R. S. Melham,A. F. Horadam“A Search for Solutions of a Functional Equation” https://link.springer.com/chapter/10.1007/978-94- 009-0223-7_36 [6] https://www.usna.edu/Users/cs/crabbe/2004- 01/SI262/recur/recur.pdf - -