This document discusses linear recurrence relations with constant coefficients. It covers homogeneous solutions, particular solutions, and the total solution. It also discusses solving recurrence relations using generating functions. Key points:
- Homogeneous solutions are found by solving the characteristic equation.
- Particular solutions are found for homogeneous and non-homogeneous equations.
- The total solution is the sum of the homogeneous and particular solutions.
- Generating functions can also be used to solve recurrence relations.
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Linear Recurrence Relations Guide
1. TOPIC ON LINEAR RECURRENCE RELATIONS
WITH CONSTANT COEFFICIENTS:-
HOMOGEOUS SOLUTIONS,PARTICULAR
SOLUTIONS, TOTAL SOLUTIONS AND
SOLVING RECURRENCE RELATION USING
GENERATING FUNCTIONS
2. Introduction to Linear recurrence relations:- order and
degree of recurrence relations.
Linear recurrence relations with constant coefficients
Homogenous solutions
Paticular solutions for homogenous difference
equations
Particular solutions for non-homogenous difference
equations
Total solution and example
Generating functions
3. A recurrence relation is a functional relation between the
independent variable x, dependent variable f(x) and the
differences of various order of f (x). A recurrence relation is also
called a difference equation, and we will use these two terms
interchangeably.
Example1: The equation f (x + 3h) + 3f (x + 2h) + 6f (x + h) + 9f
(x) = 0 is a recurrence relation.
It can also be written as
ar+3 + 3ar+2 + 6ar+1 + 9ar = 0 yk+3 + 3yk+2 + 6yk+1 + 9yk = 0
Example2: The Fibonacci sequence is defined by the recurrence
relation ar = ar-2 + ar-1, r≥2,with the initial conditions a0=1 and a1=1.
4. The order of the recurrence relation or difference
equation is defined to be the difference between the
highest and lowest subscripts of f(x) or ar=yk.
Example1: The equation 13ar+20ar-1=0 is a first order
recurrence relation.
Example2: The equation 8f (x) + 4f (x + 1) + 8f (x+2) =
k (x)
5. The degree of a difference equation is defined to be the
highest power of f (x) or ar=yk
Example1: The equation y3
k+3+2y2
k+2+2yk+1=0 has the degree
3, as the highest power of yk is 3.
Example2: The equation a4
r+3a3
r-1+6a2
r-2+4ar-3 =0 has the
degree 4, as the highest power of ar is 4.
Example3: The equation yk+3 +2yk+2 +4yk+1+2yk= k(x) has the
degree 1, because the highest power of yk is 1 and its order is
3.
Example4: The equation f (x+2h) - 4f(x+h) +2f(x) = 0 has the
degree1 and its order is 2.
6. A Recurrence Relations is called linear if its degree is one.
The general form of linear recurrence relation with constant
coefficient is
C0 yn+r+C1 yn+r-1+C2 yn+r-2+⋯+Cr yn=R (n)
Where C0,C1,C2......Cn are constant and R (n) is same function of
independent variable n.
A solution of a recurrence relation in any function which satisfies
the given equation.
The equation is said to be linear non-homogeneous difference
equation if R (n) ≠ 0.
Example: Solve the difference equation
yK+4+4yK+3+8yK+2+8yK+1+4yK=0.
Solution: The characteristics equation is s4+4s3+8s2+8s+4=0
(s2+2s+2) (s2+2s+2)=0 s = -1±i,-1±i
Therefore, the homogeneous solution of the equation is given by
yK=(C1+C2 K)(-1+i)K+(C3 +C4 K)(-1-i)K
7. A linear homogeneous difference equation with constant coefficients is
given by
C0 yn+C1 yn-1+C2 yn-2+⋯......+Cr yn-r=0 ....... equation (i)
Where C0,C1,C2.....Cn are constants.
The solution of the equation (i) is of the form , where ∝1 is the
characteristics root and A is constant.
Substitute the values of A ∝K for yn in equation (1), we have
C0 A∝K+C1 A∝K-1+C2 A∝K-2+⋯....+Cr A∝K-r=0.......equation (ii)
After simplifying equation (ii), we have
C0 ∝r+C1 ∝r-1+C2 ∝r-2+⋯Cr=0..........equation (iii)
The equation (iii) is called the characteristics equation of the difference
equation.
If ∝1 is one of the roots of the characteristics equation, then is a
homogeneous solution to the difference equation.
To find the solution of the linear homogeneous difference equations, we
have the three cases that are discussed as follows:
8. Case1 If the characteristics equation has repeated real roots.
If ∝1=∝2, then (A1+A2 K) is also a solution.
If ∝1=∝2=∝3 then (A1+A2 K+A3 K2) is also a solution.
Similarly, if root ∝1 is repeated n times, then.
(A1+A2 K+A3 K2+......+An Kn-1)
The solution to the homogeneous equation.
Case2: If the characteristics equation has one imaginary root.
If α+iβ is the root of the characteristics equation, then α-iβ is also
the root, where α and β are real.
Thus, (α+iβ)K and (α-iβ)K are solutions of the equations. This
implies
(α+iβ)K A1+(α-iβ)K A2
Is also a solution to the characteristics equation, where A1 and
A2 are constants which are to be determined.
9. Case3: If the characteristics equation has repeated imaginary
roots.
When the characteristics equation has repeated imaginary
roots,
(C1+C2 k) (α+iβ)K +(C3+C4 K)(α-iβ)K
Is the solution to the homogeneous equation
Example1: Solve the difference equation ar-3ar-1+2ar-2 =0
Solution: The characteristics equation is given by
s2-3s-1+2s-2 => s2-3s+2=0 or (s-1)(s-2)=0
⇒ s = 1, 2
Therefore, the homogeneous solution of the equation is
given by
ar=C 1 +C2.2r or c C 1 (1)K+C2(2)K
10. Example2: Solve the difference equation 9yK+2-6yK+1+yK=0.
Solution: The characteristics equation is
9s2-6s+1=0 or (3s-1)2=0
⇒ s =1/3,1/3
Therefore, the homogeneous solution of the equation is given by
yK=(C1+C2 k)(1/3)k
Example3: Solve the difference equation yK-yK-1-yK-2=0.
Solution: The characteristics equation is s2-s-1=0
s=1+√5
2
Therefore, the homogeneous solution of the equation is
.
C1 (1+√5) and C2(1-√5 )
2 2
11. (A) Homogeneous Linear Difference Equations and
Particular Solution:
We can find the particular solution of the difference equation when the
equation is of homogeneous linear type by putting the values of the
initial conditions in the homogeneous solutions.
Example1: Solve the difference equation 2ar-5ar-1+2ar-2=0 and find
particular solutions such that a0=0 and a1=1.
Solution: The characteristics equation is 2s2-5s+2=0
(2s-1)(s-2)=0
s = and 2.
Therefore, the homogeneous solution of the equation is given by
ar(h)= C1+C2 .2r..........equation (i)
Putting r=0 and r=1 in equation (i), we get
a0=C1+C2=0...........equation (a)
a1= C1+2C2=1...........equation.(b)
Solving eq (a) and (b), we have
C1=-2/3and C2=2/3
Hence, the particular solution is
Ar(p) =-2/3(1/2)r+2/3.2r
12. Example2: Solve the difference equation ar-4ar-1+4ar-2=0
and find particular solutions such that a0=0 and a1=6.
Solution: The characteristics equation is
s2-4s+4=0 or (s-2)2=0 s = 2, 2
Therefore, the homogeneous solution of the equation is
given by
ar(n)=(C1+C2 r).2r.............. equation (i)
Putting r = 0 and r = 1 in equation (i), we get
a0=(C1+0).20 = 1 ∴C1=1
a1=(C1+C2).2=6 ∴C1+C2=3⇒C2=2
Hence, the particular solution is
ar(P)=(1+2r).2r.
13. . Theorems that give us a method for solving non-
homogeneous recurrences are listed below:-
Theorem 1:- If f(n) is a solution to the associated
homogeneous recurrence, and g(n) is a solution to the non-
homogeneous recurrence, then f(n) + g(n) is also a solution
to the non-homogeneous recurrence.
Theorem 2:- Suppose the non-homogeneous term is of the
form p(n)sn, where p(n) is a polynomial of degree r and s is a
constant. If s is not a root of the characteristic equation, then
there is a particular solution of the form q(n)sn , where q is a
polynomial of degree at most r. If s is a root of the
characteristic equation of multiplicity t, then there is a
particular solution of the form nt q(n)sn, where q is a
polynomial of degree at most r.
14. Theorem 3:- If the non-homogeneous term is of
the form p 1(n) s1
n +p 2(n)s2
n+ · · · + p m(n)sm
n,
then there is a particular solution of the form f
1(n) + f2(n) + · · · + f m(n) where, for each i, f
i(n) is a particular solution to the non
homogeneous recurence a n= α 1a n-1 + α 2a n-2 + · · ·
+ α kan-k + pi(n)si
n
Procedure for solving non-homogeneous recurrences. 1. Write
down the associated homogeneous recurrence and find its general
solution.
2. Find a particular solution the non-homogeneous recurrence.
This may involve solving several simpler non-homogeneous
recurrences (using this same procedure).
3. Add all of the above solutions together to obtain the general
solution to the non-homogeneous recurrence.
4. Use the initial conditions to get a system of k equations in k
unknowns, then solve it to obtain the solution you want.
15. Example 15 Solve an = 4a n-1−4a n-2+n. 2n +3n +4, n ≥ 2, a0 = 0, a1 =
1. The associated homogeneous recurrence is an = 4a n-1 − 4a n-2 . Its
characteristic equation is x2 = 4x − 4, or (x − 2)2 = 0. Thus, 2 is the
only root and it has multiplicity 2. The general solution to the
associated homogeneous recurrence is c 12n + c 2n2n .
To
find a particular solution to the non-homogeneous recurrence, we add
together particular solutions to the three “simpler” non-homogeneous
recurrences: • a n= 4a n-1 − 4a n-2 + n2n , (i)
a n= 4a n-1 − 4a n-2 + 3n , and (ii)
a n= 4a n-1− 4a n-2 + 4. (iii)
Let’s find a particular solution to
equation (iii) first. The non homogeneous term is 4 = 4 · 1 n , and
since 1 is not a root of the characteristic equation, Theorem 2 says
there is a particular solution of the form c1n . To determine c, plug
c1n=c into the non-homogeneous recurrence and get 16 c =
4c − 4c + 4 = 4. Thus, a particular solution to the recurrence under
consideration is a n = 4
16. . Now let’s do the same for equation (ii). The
non-homogeneous term is 3n = 1 · 3n . Since 3 is
not a root of the characteristic equation, Theorem
5 says there is a particular solution of the form
c3n . To determine c, plug c3n into the non-
homogeneous recurrence and get c3n = 4c3n-1−
4c3n-2 + 3n . After dividing by the common factor
of 3n-2 , we have 9c = 12c − 4c + 9, or c = 9.
Thus, a particular solution to the recurrence under
consideration is an = 9 · 3n .
17. Let’s find a particular solution to equation
(i). The non homogeneous term is n2n .
Since 2 is a root of the characteristic
equation with multiplicity 2, NHLRR 2 says
there is a particular solution of the form n2
(un + v)2n .
To determine u and v, plug this expression
into the non homogeneous recurrence and get
n2 (un+v)2n = 4(n−1)2 (u(n−1)+v)2n-1
−4(n−2) 2 (u(n−2)+v)2n-2 +n2n .
19. The first two equations tell us nothing. The
third equation implies u = 1/6. Substituting
this value into the last equation and solving
gives v = 1/2. Thus, a particular solution to
the given recurrence is n2 ( 1/6 n + 1/2 )2n
. By Theorem 6, combining the three
particular solutions just obtained gives a
particular solution to an = 4a n-1 − 4a n-2+
n2n+ 3n + 4. It is 4 + 9 · 3n + n2 ( 1/6 n
+ 1 /2 )2n .
20. The total solution or the general solution of a non-
homogeneous linear difference equation with constant
coefficients is the sum of the homogeneous solution
and a particular solution. If no initial conditions are
given, obtain n linear equations in n unknowns and
solve them, if possible to get total solutions.
If y(h) denotes the homogeneous solution of the
recurrence relation and y(p) indicates the particular
solution of the recurrence relation then, the total
solution or the general solution y of the recurrence
relation is given by
y =y(h)+y(p).
21. in the previous example, to find total solution we have
to combine the homogenous solution and particular
solution and then solve them using initial conditions
as shown:-
The general solution is c 12n + c 2n2n + 4 + 9 · 3n + n2 (
1/6 n + 1/2 )2n . To determine the constants, use the
initial conditions. a0 = 0 ⇒ c 1 + 4 + 9 = 0
a1 = 1 ⇒ 2c 1+ 2c 2 + 4 + 27 + (1/6 + 1/2)2 = 1 The
first equation says c1 = −13. Plugging this into the
second equation gives c2 = −8/3. Thus, the solution to
the recurrence is an = (−13)2n + (−8/3)n2n + 4 + 9 · 3n +
n2 ( 1 6 n + 1 2 )2n .
22. Generating function is a method to solve the
recurrence relations.
Let us consider, the sequence a0, a1, a2....ar of
real numbers. For some interval of real numbers
containing zero values at t is given, the function
G(t) is defined by the series
G(t)= a0,
a1t+a2 t2+⋯+ar tr+............equation (i)
This function G(t) is called the generating
function of the sequence ar.
23. Example: Solve the recurrence relation ar+2-3ar+1+2ar=0
By the method of generating functions with the initial
conditions a0=2 and a1=3.
Solution: Let us assume that
Multiply equation (i) by tr and summing from r = 0 to ∞, we
have
(a2+a3 t+a4 t2+⋯)-3(a1+a2 t+a3 t2+⋯)+2(a0+a1 t+a2 t2+⋯)=0
[∴ G(t)=a0+a1 t+a2 t2+⋯]
+2G(t)=0............equation (ii)
24. Put t=1 on both sides of equation (iii) to find A. Hence
-1=- A ∴ A = 1
Put t= 1/2 on both sides of equation (iii) to find B. Hence
1/2 =1/2 B ∴ B = 1
Thus G (t) = .Hence,ar=1+2r.