Solving Recurrences

```Homogeneous Linear
Recurrences
To solve such recurrences we must first know how
to solve an easier type of recurrence relation:
DEF: A linear recurrence relation is said to be
homogeneous if it is a linear combination of
the previous terms of the recurrence without
an additional function of n or a constant.
an = 2an-1 is homogeneous
an = 2an-1 + 2n-3 - an-3 is not.
bk = 2bk-1 - bk-2 + 4bk-3 + 2 is not.
Homogeneous Linear
Recurrences
To solve such recurrences we must first know how
to solve an easier type of recurrence relation:
DEF: A linear recurrence relation is said to be
homogeneous if it is a linear combination of
the previous terms of the recurrence without
an additional function of n or a constant.
an = 2an-1 is homogeneous
an = 2an-1 + 2n-3 - an-3 is not.
bk = 2bk-1 - bk-2 + 4bk-3 + 2 is not.
First Order Homogenous Recurrences
Quiz tomorrow from Solving Recurrence Relations
b0  2
bk  3bk 1k  0
Characteristic equation
D3
General Solution:
bk  A(root 1) k
bk  A(3) k
Using initial conditions:
b0  A(3) 0
2 A
Closed Form Solution:
bk  2(3) k
2
6
18
54
162
486
1458
4374
0
1
2
3
4
5
6
7
2
6
18
54
162
486
1458
4374
Solving Homogenous Recurrences
Finding a closed form solution to a recurrence
relation.
Multiply both sides by
a0  1
r n  2
a1  8
r  n 2 r n  r  n 2 r n1  2r  n 2 r n2  0
an  an 1  2an 2n  2
r n 2 n  r n 2 n1  2r n 2 n2  0
an  an 1  2an  2
Let
an  r n
r n  r n 1  2r n  2
r n  r n 1  2r n  2  0
r 2  r  2  0 Characteristic equation
r 2  2r  r  2  0
r (r  2)  (r  2)  0
(r  2)(r  1)  0
r2
roots
r  1
Example continued…
The closed form solution is given by:
Let
an  A(root1) n  B(root2) n
an  A(2) n  B(1) n
be the solution.
1
Where constants A and B are to
determined by employing initial
conditions. a0  1, a1  8
Plug in n = 0 in equation 1 to obtain:
a0  A(2)0  B(1)0 1  A  B
Plug in n = 1 in equation 1 to obtain:
a1  A(2)1  B(1)1  8  2 A  B
Adding the two equation, we get
A B 1
2A  B  8
3A  9
A  3, B  2
Example continued…
By substituting A = 3 and B = -2 in equation
1, we get the final solution in the closed
form:
an  3(2)  2(1)
n
Verification
an  3(2) n  2(1) n
Test Data
a0  3(2) 0  2(1) 0
a0  1
a0  3  2  1
a1  3(2)  2(1)
1
1
a1  3(2)  2(1)  6  2  8
a2  3(2)  2(1)  3 * 4  2 *1
2
2
 12  2  10
a3  3(2) 3  2(1) 3  24  2  26
a4  3(2) 4  2(1) 4  48  2  46
a5  3(2) 5  2(1) 5  96  2  98
a1  8
a2  10
a3  26
a4  46
a5  98
n
Example 2
b0  2
b1  2
bk  7bk 1  10bk 2k  2
D 2  7 D  10
D 2  7 D  10  0
( D  2)(D  5)  0
D  2, D  5
The closed form
solution is given by:
bk  A(root1) k  B(root2) k
bk  A(2) k  B(5) k
Using initial conditions:
bk  A(2) k  B(5) k
b0  A(2) 0  B(5) 0
2  A B
b1  A(2)1  B(5)1
2  2 A  5B
2  A B
2  2 A  5B
2
8
B
,A
3
3
8
2
bk  (2) k  (5) k
3
3
Solution of Fibonacci Recurrence
a0  1
a1  1
an  an1  an2n  2
r 2= r + 1
r 2 - r - 1 = 0 Characteristic equation
to obtain
r1 = (1+5)/2
r2 = (1-5)/2
General solution:
an = A [(1+5)/2]n +B [(1-5)/2]n
Fibonacci Recurrence
Use initial conditions a0 = 0, a1 = 1 to find A,B and obtain
specific solution.
0=a0 = A [(1+5)/2]0 +B [(1-5)/2]0 = A +B
That is A +B = 0 --- (1)
1=a1 = A [(1+5)/2]1 +B [(1-5)/2]1 = A(1+5)/2 +B (1-5)/2
= (A+B )/2 + (A-B )5/2
That is (A+B )/2 + (A-B )5/2 = 1 or
(A+B ) + (A-B )5 = 2 -- (2)
First equation give B = -A. Plug into 2nd:
A = 1/5, B = -1/5
n
1 1 5 
an 

5 
2
1 1 5 
 





2
5



n
Case where the roots are identical
Finding a closed form solution to a recurrence
relation when the roots are identical.
t0  1
t1  3
tk  6tk 1  9tk 2k  2
D  6D  9
2
D 2  6D  9  0
( D  3) 2  0
D3
Test Data
The closed form
solution is given by:
t k  A( root 1) k  Bk ( root 2) k
t k  A(3) k  Bk (3) k
t0  1
t1  3
t2  9
t3  27
t 4  81
t5  243
t6  729
Where constants A and B are to determined by
employing initial conditions.
t k  A(3) k  Bk (3) k
t0  1
I
t1  3
Plug in k = 0 in equation 1 to obtain:
Plug in k= 1 in equation 1 to obtain:
t0  A(3) 0  B * 0 * (3)0  1  A
A  1.
t1  A(3)1  B *1* (3)1  3  3 A  3B
A B 1
A 1
B0
tk  (3)
k
A variation of the previous Example
Finding a closed form solution to a recurrence
relation when the roots are identical.
t0  0
t1  3
tk  6tk 1  9tk 2k  2
D  6D  9
2
D 2  6D  9  0
( D  3) 2  0
D3
Test Data
The closed form
solution is given by:
t k  A( root 1) k  Bk ( root 2) k
t k  A(3) k  Bk (3) k
t0  0
t1  3
t 2  18
t3  81
t 4  324
Where constants A and B are to determined by
employing initial conditions.
t k  A(3) k  Bk (3) k
t0  0
t0  0
t1  3
I
t 2  18
t1  3
t3  81
t 4  324
Plug in k = 0 in equation 1 to obtain:
Plug in k= 1 in equation 1 to obtain:
t0  A(3) 0  B * 0 * (3) 0  0  A
A  0.
t1  A(3)1  B *1* (3)1  3  3 A  3B
A B 1
A0
B 1
tk  k (3)
k
Relationship between Recurrence Relation and
Sequences
Finding sequences that satisfy a recurrence
relation.
bk  7bk 1 10bk 2k  2
Characteristic equation
D 2  7 D  10
b0, b1, b2, ….
D  7 D  10  0
( D  2)(D  5)  0
D  2, D  5
2
1, 2, 4, 8,
The given recurrence is
satisfied by the following
sequences:
Or
…
1, 5, 25, 125,
1, (2)1, (2)2, (2)3 , …
1, (5)1, (5)2, (5)3 , …
…
Verification
bk  7bk 1 10bk 2k  2
1, 2, 4, 8,
…
b2  7b1  10b0
b2  7 * 2  10*1  4
b3  7b2  10b1
 7 * 4  10* 2  28  20  8
1, 5, 25, 125,
…
b2  7b1  10b0
b2  7 * 5  10*1  25
b3  7b2  10b1
 7 * 25  10* 5  175 50  125
Practice Problem
a0  2
a1  3
a k  3a k 1  2a k 2 k  2
List first 5 terms of sequence generated by the recurrence.
Solve the recurrence in the close form.
Step 1: Write characteristic equation.
Step 2: Find roots of the characteristic equation.
Step 3: Write the general solution.
ak  A(root1) k  B(root2) k
Step 4: Find constants A and B using boundary conditions.
Step 5: Substitute constants in the general solution.
Solving Recurrences using
a0  2
a1  3
a k  3a k 1  2a k 2 k  2
2
3
5
9
17
33
65
129
257
513
2
3
=3*A2-2*A1
=3*A3-2*A2
=3*A4-2*A3
=3*A5-2*A4
=3*A6-2*A5
=3*A7-2*A6
=3*A8-2*A7
=3*A9-2*A8
3rd order Homogenous Recurrences
a 0  2,
Where A, B, and C are
constants to be
determined by using
boundary conditions.
Let r = 0
a1  1,
a2  1
ar  6ar 1  11ar 2  6ar 3
Characteristic equation
a0  A(1) 0  B(2) 0  C (3) 0
D 3  6D 2  11D  6
2  A B C
A B C  2
D  6D  11D  6  0
3
2
Roots D = 1, D = 2, and D = 3
General Solution:
ar  A(1)  B(2)  C(3)
r
r
r
(I)
Let r = 1
a1  A(1)1  B(2)1  C (3)1
1  A  2 B  3C
A  2 B  3C  1
(II)
3rd order Homogenous Recurrences
Let r = 2
The solution is not complete.
a 2  A(1) 2  B(2) 2  C (3) 2
1  A  4 B  9C
A  4 B  9C  1
(III)
We have to solve (I), (II), and
(III) simultaneously for A, B,
and C.
A BC  2
A  2 B  3C  1
A  4 B  9C  1
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