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How to make polynomial division agree with Euclidean division when the variable $X$ takes numerical values?

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$begingroup$

Consider two polynomials $A(X), B(X)$ both $in mathbbZ[X]$ such that $deg (B) < deg(A)$. And let us perform a long polynomial division of $A(X)$ over $B(X)$,
$$A(X)=Q(X)B(X)+R(X)$$
$$Q(X)=leftlfloorfracA(X)B(X)rightrfloor quad mboxis the quotient quad mboxand $R(x)$ is the remainder$$
Now consider the variable $X$ taking positive integer value , say $X=n$ and perform the Euclidean division of $A(n)$ over $B(n)$, we get $A(n)=q(n)B(n)+r(n)$ where $q(n)$ and $r(n)$ are the usual quotient and remainder. What we can get is there is not always agreement between Long polynomial division and Euclidean divison when the variable takes numerical integer values. Namely
$$Q(n) not= q(n) quad mboxand quad R(n)not= r(n)$$
To fix ideas here is an example:
Fix a positive integer $N$ and consider
$$A(X)=X^2-1, quad B(X)=X+N$$
so that synthetic or polynomial division gives:
$$Q(X)=X-N quad mboxand quad R(X)=N^2-1$$
It is clear for this example that both divisions do not agree for $n<N^2-N-1$ we even have a negative numerical quotient $Q(n)$ for $1<n<N$
, while both $A(n)$ and $B(n)$ are positive integers.

I am interested in finding a general way (if it exist) which yield an algebraic expression of new quotient and remainder $Q^'(X)$ and $R^'(X)$ such that $A(X)=Q^'(X)B(X)+R^'(X)$ and $Q^'(n)=q(n), R^'(n)=r(n)$, restricted to the simplest case where $deg(A)=2$, $deg(B)=1$.

euclidean-algorithm

share|cite|improve this question

asked Mar 30 at 21:33

HassanBHassanB

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What would $Q(X)$ be if $B)X)$ were to equal $2X+N$?
  $endgroup$
  – Dilip Sarwate
  Mar 30 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DilipSarwate it will be $Q(X)=fracX2-fracN4$ and $R(X)=fracN^24-1$ but this is not the question....
  $endgroup$
  – HassanB
  Mar 30 at 21:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So your polynomials are in $mathbb Z(X)$ to begin with but it is not necessary for quotients and remainders to be members of $mathbb Z(X)$? Why not begin with $mathbb Q(X()$?
  $endgroup$
  – Dilip Sarwate
  Mar 30 at 22:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  looking for solution when $deg(A)=2$ and $B(X)=X+N$
  $endgroup$
  – HassanB
  Mar 31 at 0:14
  
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Consider two polynomials $A(X), B(X)$ both $in mathbbZ[X]$ such that $deg (B) < deg(A)$. And let us perform a long polynomial division of $A(X)$ over $B(X)$,
$$A(X)=Q(X)B(X)+R(X)$$
$$Q(X)=leftlfloorfracA(X)B(X)rightrfloor quad mboxis the quotient quad mboxand $R(x)$ is the remainder$$
Now consider the variable $X$ taking positive integer value , say $X=n$ and perform the Euclidean division of $A(n)$ over $B(n)$, we get $A(n)=q(n)B(n)+r(n)$ where $q(n)$ and $r(n)$ are the usual quotient and remainder. What we can get is there is not always agreement between Long polynomial division and Euclidean divison when the variable takes numerical integer values. Namely
$$Q(n) not= q(n) quad mboxand quad R(n)not= r(n)$$
To fix ideas here is an example:
Fix a positive integer $N$ and consider
$$A(X)=X^2-1, quad B(X)=X+N$$
so that synthetic or polynomial division gives:
$$Q(X)=X-N quad mboxand quad R(X)=N^2-1$$
It is clear for this example that both divisions do not agree for $n<N^2-N-1$ we even have a negative numerical quotient $Q(n)$ for $1<n<N$
, while both $A(n)$ and $B(n)$ are positive integers.

I am interested in finding a general way (if it exist) which yield an algebraic expression of new quotient and remainder $Q^'(X)$ and $R^'(X)$ such that $A(X)=Q^'(X)B(X)+R^'(X)$ and $Q^'(n)=q(n), R^'(n)=r(n)$, restricted to the simplest case where $deg(A)=2$, $deg(B)=1$.

euclidean-algorithm

share|cite|improve this question

asked Mar 30 at 21:33

HassanBHassanB

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What would $Q(X)$ be if $B)X)$ were to equal $2X+N$?
  $endgroup$
  – Dilip Sarwate
  Mar 30 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DilipSarwate it will be $Q(X)=fracX2-fracN4$ and $R(X)=fracN^24-1$ but this is not the question....
  $endgroup$
  – HassanB
  Mar 30 at 21:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So your polynomials are in $mathbb Z(X)$ to begin with but it is not necessary for quotients and remainders to be members of $mathbb Z(X)$? Why not begin with $mathbb Q(X()$?
  $endgroup$
  – Dilip Sarwate
  Mar 30 at 22:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  looking for solution when $deg(A)=2$ and $B(X)=X+N$
  $endgroup$
  – HassanB
  Mar 31 at 0:14
  
  

  
  
  
  

add a comment | 

$begingroup$

Consider two polynomials $A(X), B(X)$ both $in mathbbZ[X]$ such that $deg (B) < deg(A)$. And let us perform a long polynomial division of $A(X)$ over $B(X)$,
$$A(X)=Q(X)B(X)+R(X)$$
$$Q(X)=leftlfloorfracA(X)B(X)rightrfloor quad mboxis the quotient quad mboxand $R(x)$ is the remainder$$
Now consider the variable $X$ taking positive integer value , say $X=n$ and perform the Euclidean division of $A(n)$ over $B(n)$, we get $A(n)=q(n)B(n)+r(n)$ where $q(n)$ and $r(n)$ are the usual quotient and remainder. What we can get is there is not always agreement between Long polynomial division and Euclidean divison when the variable takes numerical integer values. Namely
$$Q(n) not= q(n) quad mboxand quad R(n)not= r(n)$$
To fix ideas here is an example:
Fix a positive integer $N$ and consider
$$A(X)=X^2-1, quad B(X)=X+N$$
so that synthetic or polynomial division gives:
$$Q(X)=X-N quad mboxand quad R(X)=N^2-1$$
It is clear for this example that both divisions do not agree for $n<N^2-N-1$ we even have a negative numerical quotient $Q(n)$ for $1<n<N$
, while both $A(n)$ and $B(n)$ are positive integers.

I am interested in finding a general way (if it exist) which yield an algebraic expression of new quotient and remainder $Q^'(X)$ and $R^'(X)$ such that $A(X)=Q^'(X)B(X)+R^'(X)$ and $Q^'(n)=q(n), R^'(n)=r(n)$, restricted to the simplest case where $deg(A)=2$, $deg(B)=1$.

euclidean-algorithm

share|cite|improve this question

asked Mar 30 at 21:33

HassanBHassanB

62

$endgroup$

share|cite|improve this question

asked Mar 30 at 21:33

HassanBHassanB

62

share|cite|improve this question

asked Mar 30 at 21:33

HassanBHassanB

62

asked Mar 30 at 21:33

HassanBHassanB

62

asked Mar 30 at 21:33

HassanBHassanB

62