Download Unit 2 * Combinatorics Recurrence Relation

Unit 2 – Combinatorics
Recurrence Relation
LINEAR NON HOMOGENEOUS RECURRENCE RELATIONS WITH
CONSTANT COEFFICIENTS
A recurrence relation of the form
an= C1an-1 + C2an-2 + ..........+Ckan-k +F(n)
..............(A)
Where C1,C2,.....,Cn are real numbers and F(n) is function not identically zero depending
only on n , is called a non homogeneous recurrence relation with constant coefficient
a n = C1an-1 + C2an-2 + ..........+Ckan-k +F(n)
................(B)
Is called associated homogenous recurrence relation.
PROCEDURE TO SOLVE NON HOMOGENEOUS RECURRENCE RELATION
The solution of non homogeneous recurrence relations is the sum of two solutions.
1. Solutions of associated homogeneous recurrence relations (RHS = 0)
2. Particular solution depending on the RHS of the given recurrence relation.
Step 1:
(a) If the RHS of the recurrence relation is a  a n  ....  ar nr then substitute
0
c0  c1 n  c2n2 ....  cr nr in place of an
place of an 1
1
c0  c1 (n 1)  c2 (n 1)2 ....  cr (n 1)r in
...... in the LHS of the given recurrence relation.
(b) If the RHS is an then we have
Case 1: If the base a of RHS is the characteristic root ,then the solution is of the cna n .
Therefore substitute cna n in place of an , c(n  1)a n1 in place of an 1 etc.
Case 2: If the base a of RHS is not a root, then solution if of the form ca n . Therefore
substitute ca n in place of an , ca n 1 in place of an 1 etc.
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MA204-DISCRETE MATHEMATICS
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Unit 2 – Combinatorics
Recurrence Relation
Step 2 : At the end of step -1, we get a polynomial in “n” with coefficients C0,C1,.....,Cn on
LHS. Now equating LHS, RHS and compare the coefficients find the constants
C0,C1,.....,Cn.
PROBLEMS
1. Solve an  3an1  2n an = 3an-1 + 2n with a1  3
Solution:
Given the non homogeneous recurrence relation is
an  3an1  2n  0
Its associated homogeneous equation is
an  3an1  0
Its characteristic equation is
r 3  0 , r  3
Now ,the solution of associated homogeneous equation is a ( h ) n   3n
To find particular solution
Since F(n)=2n is a polynomial of degree one , then the solution is of the form
an  cn  d (say) where c and d are constants.
Now the equation an  3an1  2n becomes
 cn  d  3an1  2n becomes
 cn  d  3(c(n  1)  d )  2n
 cn  d  3cn  3c  3d  2n
 2cn  2n  3c  2d  0
 (2c  2)n  (2d  3c)  0
 (2c  2)  0, (2d  3c)  0
Solving we get  c  1, d 
3
2
SNS COLLEGE OF TECHNOLOGY (AUTONOMOUS)
MA204-DISCRETE MATHEMATICS
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Unit 2 – Combinatorics
Recurrence Relation
 cn  d is a solution iff c  1, d 
 an( p )  n 
 General
3
2
3
is a particular solution
2
solution an  an( h )  an( p )
an  1 3n  n 
3
2
.............(A)
Given a1  3 . Put n=1 in (A) , we get
a1  1 31  1 
3  31 
31 
3
2
5
2
11
11
 1 
sub in (A), we get
2
6
 General
solution an  n     3n
2 6
3
11
2. Solve S (k )  5S (k  1)  6S (k  2)  2 with S (0)  1, S (1)  1
Solution :
Given the non homogeneous recurrence relation is
an  5an1  6an2  2  0
Its associated homogeneous equation is
an  5an1  6an2  0
Its characteristic equation is
r 2  5r  6  0,
r  2,3
The general solution is
 Sn( h )  1 (2)n   2 (3) n
SNS COLLEGE OF TECHNOLOGY (AUTONOMOUS)
MA204-DISCRETE MATHEMATICS
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Unit 2 – Combinatorics
Recurrence Relation
Recurrence Relations
Definition
A recurrence relation is an equation that recursively defines a sequence where the
next term is a function of the previous terms (Expressing Fn as some combination
of Fi with i<n).
Example:
Fibonacci series Fn=Fn-1+Fn-2, Tower of Hanoi – Fn=2Fn-1+1
Linear Recurrence Relations
A linear recurrence equation of degree k or order k is a recurrence equation which is
in the format xn=A1xn−1+A2xn−2+A3xn−3+…Akxn−k( An is a constant
and Ak≠0) on a sequence of numbers as a first-degree polynomial.
These are some examples of linear recurrence equations –
Recurrence
Initial
relations
values
Fn = Fn-1 + Fn-
a1 = a2 =
Fibonacci
2
1
number
Fn = Fn-1 + Fn-
a1 = 1,
Lucas
2
a2 = 3
Number
Fn = Fn-2 + Fn-
a1 = a2 =
Padovan
3
a3 = 1
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MA204-DISCRETE MATHEMATICS
Solutions
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Unit 2 – Combinatorics
Recurrence Relation
sequence
Fn = 2Fn-1 +
a1 = 0,
Fn-2
a2 = 1
Pell number
How to solve linear recurrence relation:
Suppose, a two ordered linear recurrence relation is Fn=AFn−1+BFn−2 where A
and B are real numbers.
The characteristic equation for the above recurrence relation is −
x2−Ax−B=0
Three cases may occur while finding the roots −
Case 1 − If this equation factors as (x−x1)(x−x2)=0 and it produces two distinct real
roots x1 and x2, then Fn=a(x1)n+b(x2)n is the solution. [Here, a and b are constants]
Case 2 − If this equation factors as (x−x1)2=0 and it produces single real root x1,
then Fn= (a+bn)(x1)n is the solution.
PROBLEM
1. Solve the recurrence relation Fn=5Fn-1−6Fn-2 where F0=1 and F1=4
Solution
The characteristic equation of the recurrence relation is −
x2−5x+6=0,
So, (x−3)(x−2)=0
Hence, the roots are −
x1=3 and x2=2
The roots are real and distinct. So, this is in the form of case 1
Hence, the solution is −
Fn=a(x1)n+b(x2)n Here, Fn=a3n+b2n (As x1=3 and x2=2)
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MA204-DISCRETE MATHEMATICS
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Unit 2 – Combinatorics
Recurrence Relation
Therefore,
1=F0=a30+b20=a+b
4=F1=a31+b21=3a+2b
Solving these two equations, we get a=2 and b=−1
Hence, the final solution is −
Fn=2.3n+(−1).2n
=2.(3n−2n)
2. Solve the recurrence relation Fn=10Fn-1−25Fn-2 where F0=3 and F1=17
Solution
The characteristic equation of the recurrence relation is −
X2−10x−25=0
So (x−5)2=0
Hence, there is single real root x1=5
As there is single real valued root, this is in the form of case 2
Hence, the solution is
Fn =(a+bn)(x1)n
3=F0= (a+b0)(5)0
17=F1= (a+b1)(5)1
=5a+5b
2
5
Solving these two equations, we get a=3 and b  .
2
Hence, the final solution is Fn   3  n  5n

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MA204-DISCRETE MATHEMATICS
5 
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