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Finding the general formula for a sequence:

1

1

$begingroup$

I have a sequence:

$a_0 = 0; a_1 = 4; a_2 = 9; a_n = 4a_n-1 - 5a_n-2 + 2a_n-3$

I want to find the general formula.

sequences-and-series combinatorics discrete-mathematics recurrence-relations characteristic-functions

share|cite|improve this question

edited Apr 4 at 11:31

J. Lastin

asked Apr 1 at 0:38

J. LastinJ. Lastin

12412

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with the standard methods of solving such recursions (such as characteristic polynomials)? If you are, this is a standard example. If you are not, this is a good introduction.
  $endgroup$
  – lulu
  Apr 1 at 0:41
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I believe your computations for the sequence start to go wrong at $84$.
  $endgroup$
  – amd
  Apr 1 at 1:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I agree with @amd; I got 81 instead of 84
  $endgroup$
  – J. W. Tanner
  Apr 1 at 1:36
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I have a sequence:

$a_0 = 0; a_1 = 4; a_2 = 9; a_n = 4a_n-1 - 5a_n-2 + 2a_n-3$

I want to find the general formula.

sequences-and-series combinatorics discrete-mathematics recurrence-relations characteristic-functions

share|cite|improve this question

edited Apr 4 at 11:31

J. Lastin

asked Apr 1 at 0:38

J. LastinJ. Lastin

12412

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with the standard methods of solving such recursions (such as characteristic polynomials)? If you are, this is a standard example. If you are not, this is a good introduction.
  $endgroup$
  – lulu
  Apr 1 at 0:41
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I believe your computations for the sequence start to go wrong at $84$.
  $endgroup$
  – amd
  Apr 1 at 1:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I agree with @amd; I got 81 instead of 84
  $endgroup$
  – J. W. Tanner
  Apr 1 at 1:36
  

  
  
  
  

add a comment | 

$begingroup$

I have a sequence:

$a_0 = 0; a_1 = 4; a_2 = 9; a_n = 4a_n-1 - 5a_n-2 + 2a_n-3$

I want to find the general formula.

sequences-and-series combinatorics discrete-mathematics recurrence-relations characteristic-functions

share|cite|improve this question

edited Apr 4 at 11:31

J. Lastin

asked Apr 1 at 0:38

J. LastinJ. Lastin

12412

$endgroup$

share|cite|improve this question

edited Apr 4 at 11:31

J. Lastin

asked Apr 1 at 0:38

J. LastinJ. Lastin

12412

share|cite|improve this question

edited Apr 4 at 11:31

J. Lastin

asked Apr 1 at 0:38

J. LastinJ. Lastin

12412

asked Apr 1 at 0:38

J. LastinJ. Lastin

12412

asked Apr 1 at 0:38

J. LastinJ. Lastin

12412

3 Answers
3

active

oldest

votes

2

$begingroup$

These are called linear difference equations. Solve the characteristic equation and use the initial values to find the appropriate constants. After, working it out, the solution to your problem is $2^n+3n-1$.

share|cite|improve this answer

answered Apr 1 at 6:37

Geethu JosephGeethu Joseph

3358

$endgroup$

add a comment | 

1

$begingroup$

HINT: Note that for all $ninBbbN$ you have
$$beginpmatrixa_n+3\ a_n+2\ a_n+1endpmatrix
=beginpmatrix4&-5&2\1&0&0\0&1&0endpmatrix
beginpmatrixa_n+2\ a_n+1\ a_nendpmatrix.$$

share|cite|improve this answer

answered Apr 1 at 2:26

ServaesServaes

30.6k342101

$endgroup$

add a comment | 

0

$begingroup$

$lambda^n = 4lambda^n-1 - 5lambda^n-2 + 2lambda^n-3$

$lambda^3 - 4lambda^2 + 5lambda - 2 = (lambda - 2)(lambda -1)^2$

$lambda_1 = 2; lambda_2,3 = 1$

$a_n = alpha(lambda_1)^n + beta(lambda_2)^n + gammacdot n(lambda_3)^n$

$a_0 = 0 = alpha + beta$

$a_1 = 4 = 2alpha +beta +gamma$

$a_2 = 9 = 4alpha +beta +2gamma$

$alpha = 1; beta = -1; gamma = 3$

$a_n = 2^n +3n - 1$

share|cite|improve this answer

answered Apr 4 at 11:32

J. LastinJ. Lastin

12412

$endgroup$

add a comment | 

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2

$begingroup$

These are called linear difference equations. Solve the characteristic equation and use the initial values to find the appropriate constants. After, working it out, the solution to your problem is $2^n+3n-1$.

share|cite|improve this answer

answered Apr 1 at 6:37

Geethu JosephGeethu Joseph

3358

$endgroup$

add a comment | 

2

$begingroup$

These are called linear difference equations. Solve the characteristic equation and use the initial values to find the appropriate constants. After, working it out, the solution to your problem is $2^n+3n-1$.

share|cite|improve this answer

answered Apr 1 at 6:37

Geethu JosephGeethu Joseph

3358

$endgroup$

add a comment | 

$begingroup$

These are called linear difference equations. Solve the characteristic equation and use the initial values to find the appropriate constants. After, working it out, the solution to your problem is $2^n+3n-1$.

share|cite|improve this answer

answered Apr 1 at 6:37

Geethu JosephGeethu Joseph

3358

$endgroup$

share|cite|improve this answer

answered Apr 1 at 6:37

Geethu JosephGeethu Joseph

3358

answered Apr 1 at 6:37

Geethu JosephGeethu Joseph

3358

answered Apr 1 at 6:37

Geethu JosephGeethu Joseph

3358

1

$begingroup$

HINT: Note that for all $ninBbbN$ you have
$$beginpmatrixa_n+3\ a_n+2\ a_n+1endpmatrix
=beginpmatrix4&-5&2\1&0&0\0&1&0endpmatrix
beginpmatrixa_n+2\ a_n+1\ a_nendpmatrix.$$

share|cite|improve this answer

answered Apr 1 at 2:26

ServaesServaes

30.6k342101

$endgroup$

add a comment | 

1

$begingroup$

HINT: Note that for all $ninBbbN$ you have
$$beginpmatrixa_n+3\ a_n+2\ a_n+1endpmatrix
=beginpmatrix4&-5&2\1&0&0\0&1&0endpmatrix
beginpmatrixa_n+2\ a_n+1\ a_nendpmatrix.$$

share|cite|improve this answer

answered Apr 1 at 2:26

ServaesServaes

30.6k342101

$endgroup$

add a comment | 

$begingroup$

HINT: Note that for all $ninBbbN$ you have
$$beginpmatrixa_n+3\ a_n+2\ a_n+1endpmatrix
=beginpmatrix4&-5&2\1&0&0\0&1&0endpmatrix
beginpmatrixa_n+2\ a_n+1\ a_nendpmatrix.$$

share|cite|improve this answer

answered Apr 1 at 2:26

ServaesServaes

30.6k342101

$endgroup$

share|cite|improve this answer

answered Apr 1 at 2:26

ServaesServaes

30.6k342101

answered Apr 1 at 2:26

ServaesServaes

30.6k342101

answered Apr 1 at 2:26

ServaesServaes

30.6k342101

0

$begingroup$

$lambda^n = 4lambda^n-1 - 5lambda^n-2 + 2lambda^n-3$

$lambda^3 - 4lambda^2 + 5lambda - 2 = (lambda - 2)(lambda -1)^2$

$lambda_1 = 2; lambda_2,3 = 1$

$a_n = alpha(lambda_1)^n + beta(lambda_2)^n + gammacdot n(lambda_3)^n$

$a_0 = 0 = alpha + beta$

$a_1 = 4 = 2alpha +beta +gamma$

$a_2 = 9 = 4alpha +beta +2gamma$

$alpha = 1; beta = -1; gamma = 3$

$a_n = 2^n +3n - 1$

share|cite|improve this answer

answered Apr 4 at 11:32

J. LastinJ. Lastin

12412

$endgroup$

add a comment | 

0

$begingroup$

$lambda^n = 4lambda^n-1 - 5lambda^n-2 + 2lambda^n-3$

$lambda^3 - 4lambda^2 + 5lambda - 2 = (lambda - 2)(lambda -1)^2$

$lambda_1 = 2; lambda_2,3 = 1$

$a_n = alpha(lambda_1)^n + beta(lambda_2)^n + gammacdot n(lambda_3)^n$

$a_0 = 0 = alpha + beta$

$a_1 = 4 = 2alpha +beta +gamma$

$a_2 = 9 = 4alpha +beta +2gamma$

$alpha = 1; beta = -1; gamma = 3$

$a_n = 2^n +3n - 1$

share|cite|improve this answer

answered Apr 4 at 11:32

J. LastinJ. Lastin

12412

$endgroup$

add a comment | 

$begingroup$

$lambda^n = 4lambda^n-1 - 5lambda^n-2 + 2lambda^n-3$

$lambda^3 - 4lambda^2 + 5lambda - 2 = (lambda - 2)(lambda -1)^2$

$lambda_1 = 2; lambda_2,3 = 1$

$a_n = alpha(lambda_1)^n + beta(lambda_2)^n + gammacdot n(lambda_3)^n$

$a_0 = 0 = alpha + beta$

$a_1 = 4 = 2alpha +beta +gamma$

$a_2 = 9 = 4alpha +beta +2gamma$

$alpha = 1; beta = -1; gamma = 3$

$a_n = 2^n +3n - 1$

share|cite|improve this answer

answered Apr 4 at 11:32

J. LastinJ. Lastin

12412

$endgroup$

share|cite|improve this answer

answered Apr 4 at 11:32

J. LastinJ. Lastin

12412

answered Apr 4 at 11:32

J. LastinJ. Lastin

12412

answered Apr 4 at 11:32

J. LastinJ. Lastin

12412