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Topic: Polynomial (not quite) division
Replies: 13   Last Post: Jan 16, 2014 11:07 AM

 Messages: [ Previous | Next ]
 Ken.Pledger@vuw.ac.nz Posts: 1,412 Registered: 12/3/04
Re: Polynomial (not quite) division
Posted: Jan 15, 2014 3:22 PM

hockleymitchell@gmail.com wrote:

> ....
> (1 + x^-2)(x^3 + 1) = x^3 + x + 1 + x^-2
>
> Now if I were to use the standard polynomial division algorithm (I appreciate
> that x^-2 + 1 is not a polynomial) to work out
>
> (x^3 + x + 1 + x^-2) / (x^3 + 1)
>
> then I would obtain a whole part of 1 and a remainder of x + x^-2. And this
> is clearly not right....

Allowing negative powers actually gives you the quotient 1 + x^-2
and remainder 0.

But a simple way to deal with it is to use genuine polynomials.
Your x^3 + x + 1 + x^-2 = x^-2.(x^5 + x^3 + x^2 + 1).
Ordinary division of x^5 + x^3 + x^2 + 1 by x^3 + 1 gives
x^5 + x^3 + x^2 + 1 = (x^3 + 1)(x^2 + 1) + 0
and then you can just reintroduce the factor x^-2.

Ken Pledger.

Date Subject Author
1/15/14 hockleymitchell@gmail.com
1/15/14 fom
1/15/14 hockleymitchell@gmail.com
1/15/14 Martin Shobe
1/15/14 fom
1/15/14 Ken.Pledger@vuw.ac.nz
1/16/14 William Elliot
1/16/14 hockleymitchell@gmail.com
1/16/14 Leon Aigret
1/16/14 Martin Shobe
1/16/14 Pubkeybreaker
1/16/14 Martin Shobe
1/16/14 Pubkeybreaker
1/16/14 Martin Shobe