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I'm reading the book Concrete Mathematics and on page 27(chapter sums and recurrences) there is a text I simply cannot understand:

This trick is a special case of a general technique that can reduce
virtually any recurrence of the form

$a_nT_n = b_nT_{n-1} + c_n$

To a sum. The idea is to multiply both sides by a summation factor,
$S_n$:

$S_na_nT_n = S_nb_nT_{n-1} + S_nc_n$

This factor $S_n$ is cleverly chosen to make

$S_nb_n = S_{n-1}a_{n-1}$ //THIS IS MY FIRST QUESTION, How did they
get this result from the previous equation?

Then if we write $S_n=S_na_nT_n$ we have a sum recurrence

$S_n = S_{n-1} + S_nC_n$ //HOW DID THEY GET THIS?

summation recurrence-relations

share|cite|improve this question

asked Mar 12 at 15:09

EduardoEduardo

1125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps is isn't derived but rather defined so that is the case. So maybe define $S_0=1$ and for $n>0$ define $S_n = S_{n-1}a_{n-1}/b_n$?
  $endgroup$
  – MPW
  Mar 12 at 15:13
  

  
  
  
  

add a comment | 

0

1

$begingroup$

I'm reading the book Concrete Mathematics and on page 27(chapter sums and recurrences) there is a text I simply cannot understand:

This trick is a special case of a general technique that can reduce
virtually any recurrence of the form

$a_nT_n = b_nT_{n-1} + c_n$

To a sum. The idea is to multiply both sides by a summation factor,
$S_n$:

$S_na_nT_n = S_nb_nT_{n-1} + S_nc_n$

This factor $S_n$ is cleverly chosen to make

$S_nb_n = S_{n-1}a_{n-1}$ //THIS IS MY FIRST QUESTION, How did they
get this result from the previous equation?

Then if we write $S_n=S_na_nT_n$ we have a sum recurrence

$S_n = S_{n-1} + S_nC_n$ //HOW DID THEY GET THIS?

summation recurrence-relations

share|cite|improve this question

asked Mar 12 at 15:09

EduardoEduardo

1125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps is isn't derived but rather defined so that is the case. So maybe define $S_0=1$ and for $n>0$ define $S_n = S_{n-1}a_{n-1}/b_n$?
  $endgroup$
  – MPW
  Mar 12 at 15:13
  

  
  
  
  

add a comment | 

$begingroup$

I'm reading the book Concrete Mathematics and on page 27(chapter sums and recurrences) there is a text I simply cannot understand:

This trick is a special case of a general technique that can reduce
virtually any recurrence of the form

$a_nT_n = b_nT_{n-1} + c_n$

To a sum. The idea is to multiply both sides by a summation factor,
$S_n$:

$S_na_nT_n = S_nb_nT_{n-1} + S_nc_n$

This factor $S_n$ is cleverly chosen to make

$S_nb_n = S_{n-1}a_{n-1}$ //THIS IS MY FIRST QUESTION, How did they
get this result from the previous equation?

Then if we write $S_n=S_na_nT_n$ we have a sum recurrence

$S_n = S_{n-1} + S_nC_n$ //HOW DID THEY GET THIS?

summation recurrence-relations

share|cite|improve this question

asked Mar 12 at 15:09

EduardoEduardo

1125

$endgroup$

share|cite|improve this question

asked Mar 12 at 15:09

EduardoEduardo

1125

share|cite|improve this question

asked Mar 12 at 15:09

EduardoEduardo

1125

asked Mar 12 at 15:09

EduardoEduardo

1125

asked Mar 12 at 15:09

EduardoEduardo

1125