Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier...
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Transcript of Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier...
![Page 1: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/1.jpg)
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 homogeneousan = 2an-1 + 2n-3 - an-3 is not.bk = 2bk-1 - bk-2 + 4bk-3 + 2 is not.
![Page 2: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/2.jpg)
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 homogeneousan = 2an-1 + 2n-3 - an-3 is not.bk = 2bk-1 - bk-2 + 4bk-3 + 2 is not.
![Page 3: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/3.jpg)
First Order Homogenous Recurrences
032
1
0
kbbb
kk
3DCharacteristic equation
kk
kk
Ab
rootAb
)3(
)1(
Using initial conditions:
kkb )3(2
General Solution:
AAb
2)3( 0
0
Closed Form Solution:
2 0 26 1 618 2 1854 3 54162 4 162486 5 4861458 6 14584374 7 4374
Quiz tomorrow from Solving Recurrence Relations
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Solving Homogenous Recurrences
Finding a closed form solution to a recurrence relation.
2281
21
1
0
naaaaa
nnn
12
0)1)(2(0)2()2(
0222
rr
rrrrrrrr
02
2
2
21
21
21
nnn
nnn
nn
nnn
rrr
rrr
ra
Letaaa
Multiply both sides by2 nr
02
0222122
22122
nnnnnn
nnnnnn
rrr
rrrrrr
022 rr Characteristic equation
roots
![Page 5: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/5.jpg)
Example continued…The closed form solution is given by:
nnn
nnn
BAa
rootBrootAa
)1()2(
)2()1(
Where constants A and B are to determined by employing initial conditions. 8,1 10 aa
Plug in n = 0 in equation 1 to obtain:
Plug in n = 1 in equation 1 to obtain:
BABAa 1)1()2( 000
BABAa 28)1()2( 111
821
BABA
Adding the two equation, we get 2,393
BA
A
1
Let be the solution.
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Example continued…By substituting A = 3 and B = -2 in equation 1, we get the final solution in the closed form:
nnna )1(2)2(3
Verification
98296)1(2)2(3
46248)1(2)2(3
26224)1(2)2(3
102121*24*3)1(2)2(3
826)1(2)2(3)1(2)2(3
123)1(2)2(3
)1(2)2(3
555
444
333
222
1
111
0
000
a
a
a
a
aa
aa
a nnn
9846261081
5
4
3
2
1
0
aaaaaa
Test Data
![Page 7: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/7.jpg)
Example 2
210722
21
1
0
kbbbbb
kkk
5,20)5)(2(0107
1072
2
DDDDDD
DD
The closed form solution is given by:
kkk
kkk
BAb
rootBrootAb
)5()2(
)2()1(
Using initial conditions:
BABAb
BABAb
BAb kkk
522)5()2(
2)5()2(
)5()2(
111
000
38,
32522
2
AB
BABA
kkkb )5(
32)2(
38
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Solution of Fibonacci Recurrence
Solve the quadratic equation:r 2
= r + 1r 2
- r - 1 = 0 to obtain r1 = (1+5)/2 r2 = (1-5)/2General solution:
an = A [(1+5)/2]n +B [(1-5)/2]n
211
21
1
0
naaaaa
nnn
Characteristic equation
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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 +BThat 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 )5/2That 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 = -1/5Final answer:
nn
na
251
51
251
51
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Case where the roots are identical
Finding a closed form solution to a recurrence relation when the roots are identical.
29631
21
1
0
kttttt
kkk
30)3(
096
96
2
2
2
DD
DD
DD
Test DataThe closed form solution is given by:
kkk
kkk
BkAt
rootBkrootAt
)3()3(
)2()1(
7292438127931
6
5
4
3
2
1
0
ttttttt
![Page 11: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/11.jpg)
Where constants A and B are to determined by employing initial conditions.
I
Plug in k = 0 in equation 1 to obtain:
Plug in k= 1 in equation 1 to obtain:
.11)3(*0*)3( 00
0
AABAt
1333)3(*1*)3( 11
1
BABABAt
01
BA
31
)3()3(
1
0
tt
BkAt kkk
kkt )3(
![Page 12: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/12.jpg)
A variation of the previous Example
Finding a closed form solution to a recurrence relation when the roots are identical.
29630
21
1
0
kttttt
kkk
30)3(
096
96
2
2
2
DD
DD
DD
Test DataThe closed form solution is given by:
kkk
kkk
BkAt
rootBkrootAt
)3()3(
)2()1(
324811830
4
3
2
1
0
ttttt
![Page 13: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/13.jpg)
Where constants A and B are to determined by employing initial conditions.
I
Plug in k = 0 in equation 1 to obtain:
Plug in k= 1 in equation 1 to obtain:
.00)3(*0*)3( 00
0
AABAt
1333)3(*1*)3( 11
1
BABABAt
10
BA
30
)3()3(
1
0
tt
BkAt kkk
kk kt )3(
324811830
4
3
2
1
0
ttttt
![Page 14: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/14.jpg)
Relationship between Recurrence Relation and Sequences
Finding sequences that satisfy a recurrence relation.
2107 21 kbbb kkk
5,20)5)(2(0107
1072
2
DDDDDD
DD
Characteristic equation
The given recurrence is satisfied by the following sequences:
b0, b1, b2, ….
1, (2)1, (2)2, (2)3 , …
1, (5)1, (5)2, (5)3 , … 1, 2, 4, 8, …
1, 5, 25, 125, …
Or
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Verification2107 21 kbbb kkk
1, 5, 25, 125, …
820282*104*7107
41*102*7107
123
2
012
bbbb
bbb
1, 2, 4, 8, …
125501755*1025*7107
251*105*7107
123
2
012
bbbb
bbb
![Page 16: Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type…](https://reader037.fdocuments.in/reader037/viewer/2022090107/5a4d1c127f8b9ab0599f7f00/html5/thumbnails/16.jpg)
Practice Problem
22332
21
1
0
kaaaaa
kkk
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.
Step 4: Find constants A and B using boundary conditions.Step 5: Substitute constants in the general solution.
kkk rootBrootAa )2()1(
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Solving Recurrences using Spreadsheet
2359173365129257513
22332
21
1
0
kaaaaa
kkk
23=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
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3rd order Homogenous Recurrences
321 6116 rrrr aaaa
1,1,2
2
1
0
aaa
Characteristic equation
6116 23 DDD
06116 23 DDD
Roots D = 1, D = 2, and D = 3
General Solution:rrr
r CBAa )3()2()1(
Where A, B, and C are constants to be determined by using boundary conditions.Let r = 0
22
)3()2()1( 0000
CBACBA
CBAa
Let r = 1
132321
)3()2()1( 1111
CBACBA
CBAa
(I)
(II)
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3rd order Homogenous Recurrences
Let r = 2
194941
)3()2()1( 2222
CBACBA
CBAa
We have to solve (I), (II), and (III) simultaneously for A, B, and C.
(III)
194 CBA
132 CBA
2 CBA
The solution is not complete.