Solving a homogeneous recurrence relation [duplicate]
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This question already has an answer here:
showing a genereral function from a generating function and recursive function
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How to solve $$a_n=4a_{n-1}-4a_{n-2}$$
$a_0=6$ and $a_1=8$
I found first few terms as $$6,8,8,0,... $$
But I don't know how to proceed.
recurrence-relations
edited Nov 13 at 6:22
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asked Nov 13 at 6:17
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marked as duplicate by Martin R, Community♦ Nov 13 at 6:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
0
down vote
favorite
This question already has an answer here:
showing a genereral function from a generating function and recursive function
2 answers
How to solve $$a_n=4a_{n-1}-4a_{n-2}$$
$a_0=6$ and $a_1=8$
I found first few terms as $$6,8,8,0,... $$
But I don't know how to proceed.
recurrence-relations
edited Nov 13 at 6:22
idea
2,09721024
asked Nov 13 at 6:17
DAVID
228
marked as duplicate by Martin R, Community♦ Nov 13 at 6:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Another one: math.stackexchange.com/q/1544784/42969.
– Martin R
Nov 13 at 6:21
1
Many many many questions about homogeneous linear recurrences have been asked and answered here before. Maybe do a little search for something like yours.
– Gerry Myerson
Nov 13 at 6:21
Yes...I should have looked that earlier. Thanks anyway
– DAVID
Nov 13 at 6:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
showing a genereral function from a generating function and recursive function
2 answers
How to solve $$a_n=4a_{n-1}-4a_{n-2}$$
$a_0=6$ and $a_1=8$
I found first few terms as $$6,8,8,0,... $$
But I don't know how to proceed.
recurrence-relations
edited Nov 13 at 6:22
idea
2,09721024
asked Nov 13 at 6:17
DAVID
228
This question already has an answer here:
showing a genereral function from a generating function and recursive function
2 answers
How to solve $$a_n=4a_{n-1}-4a_{n-2}$$
$a_0=6$ and $a_1=8$
I found first few terms as $$6,8,8,0,... $$
But I don't know how to proceed.
This question already has an answer here:
showing a genereral function from a generating function and recursive function
2 answers
recurrence-relations
recurrence-relations
edited Nov 13 at 6:22
idea
2,09721024
asked Nov 13 at 6:17
DAVID
228
edited Nov 13 at 6:22
idea
2,09721024
asked Nov 13 at 6:17
DAVID
228
edited Nov 13 at 6:22
idea
2,09721024
edited Nov 13 at 6:22
idea
2,09721024
edited Nov 13 at 6:22
idea
2,09721024
2,09721024
asked Nov 13 at 6:17
DAVID
228
asked Nov 13 at 6:17
DAVID
228
asked Nov 13 at 6:17
DAVID
228
228
marked as duplicate by Martin R, Community♦ Nov 13 at 6:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Community♦ Nov 13 at 6:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Another one: math.stackexchange.com/q/1544784/42969.
– Martin R
Nov 13 at 6:21
1
Many many many questions about homogeneous linear recurrences have been asked and answered here before. Maybe do a little search for something like yours.
– Gerry Myerson
Nov 13 at 6:21
Yes...I should have looked that earlier. Thanks anyway
– DAVID
Nov 13 at 6:28
add a comment |
1
Another one: math.stackexchange.com/q/1544784/42969.
– Martin R
Nov 13 at 6:21
1
Many many many questions about homogeneous linear recurrences have been asked and answered here before. Maybe do a little search for something like yours.
– Gerry Myerson
Nov 13 at 6:21
Yes...I should have looked that earlier. Thanks anyway
– DAVID
Nov 13 at 6:28
1
1
Another one: math.stackexchange.com/q/1544784/42969.
– Martin R
Nov 13 at 6:21
Another one: math.stackexchange.com/q/1544784/42969.
– Martin R
Nov 13 at 6:21
1
1
Many many many questions about homogeneous linear recurrences have been asked and answered here before. Maybe do a little search for something like yours.
– Gerry Myerson
Nov 13 at 6:21
Many many many questions about homogeneous linear recurrences have been asked and answered here before. Maybe do a little search for something like yours.
– Gerry Myerson
Nov 13 at 6:21
Yes...I should have looked that earlier. Thanks anyway
– DAVID
Nov 13 at 6:28
Yes...I should have looked that earlier. Thanks anyway
– DAVID
Nov 13 at 6:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Make its characteristic equation: $$x^2-4x+4=0$$
Solving which gives repeated root as $x=2$
So, its solution will be like: $$a_n=(A+Bn)2^n$$
Just substitute for initial conditions and you get $A=6$ and $B=-2$
So, $$a_n=(6-2n)cdot2^n$$
answered Nov 13 at 6:25
idea
2,09721024
I already got my answer. Thanks omega...
– DAVID
Nov 13 at 6:29
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Make its characteristic equation: $$x^2-4x+4=0$$
Solving which gives repeated root as $x=2$
So, its solution will be like: $$a_n=(A+Bn)2^n$$
Just substitute for initial conditions and you get $A=6$ and $B=-2$
So, $$a_n=(6-2n)cdot2^n$$
answered Nov 13 at 6:25
idea
2,09721024
I already got my answer. Thanks omega...
– DAVID
Nov 13 at 6:29
add a comment |
up vote
1
down vote
accepted
Make its characteristic equation: $$x^2-4x+4=0$$
Solving which gives repeated root as $x=2$
So, its solution will be like: $$a_n=(A+Bn)2^n$$
Just substitute for initial conditions and you get $A=6$ and $B=-2$
So, $$a_n=(6-2n)cdot2^n$$
answered Nov 13 at 6:25
idea
2,09721024
I already got my answer. Thanks omega...
– DAVID
Nov 13 at 6:29
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Make its characteristic equation: $$x^2-4x+4=0$$
Solving which gives repeated root as $x=2$
So, its solution will be like: $$a_n=(A+Bn)2^n$$
Just substitute for initial conditions and you get $A=6$ and $B=-2$
So, $$a_n=(6-2n)cdot2^n$$
answered Nov 13 at 6:25
idea
2,09721024
Make its characteristic equation: $$x^2-4x+4=0$$
Solving which gives repeated root as $x=2$
So, its solution will be like: $$a_n=(A+Bn)2^n$$
Just substitute for initial conditions and you get $A=6$ and $B=-2$
So, $$a_n=(6-2n)cdot2^n$$
answered Nov 13 at 6:25
idea
2,09721024
answered Nov 13 at 6:25
idea
2,09721024
answered Nov 13 at 6:25
idea
2,09721024
answered Nov 13 at 6:25
idea
2,09721024
2,09721024
I already got my answer. Thanks omega...
– DAVID
Nov 13 at 6:29
add a comment |
I already got my answer. Thanks omega...
– DAVID
Nov 13 at 6:29
I already got my answer. Thanks omega...
– DAVID
Nov 13 at 6:29
I already got my answer. Thanks omega...
– DAVID
Nov 13 at 6:29
add a comment |
1
Another one: math.stackexchange.com/q/1544784/42969.
– Martin R
Nov 13 at 6:21
1
Many many many questions about homogeneous linear recurrences have been asked and answered here before. Maybe do a little search for something like yours.
– Gerry Myerson
Nov 13 at 6:21
Yes...I should have looked that earlier. Thanks anyway
– DAVID
Nov 13 at 6:28