Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Ken.Pledger@vuw.ac.nz

Posts: 1,374
Registered: 12/3/04
Re: Polynomial (not quite) division
Posted: Jan 15, 2014 3:22 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

In article <102d1a32-f67d-49cd-bd17-da8f75477404@googlegroups.com>,
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.



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.