Re: polynomial division
- To: mathgroup at smc.vnet.net
- Subject: [mg41805] Re: [mg41770] polynomial division
- From: Ken Levasseur <klevasseur at mac.com>
- Date: Fri, 6 Jun 2003 09:50:36 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Olivier:
Unless q(x) is a divisor of p(x), there will be a remainder r(x) and
p(x)/q(x) = m(x) + r(x)/q(x).
If q(x) = x^2+x+1 and p(x) = 1, then m(x) = 0 and r(x) = 1. In any case,
the remainder term can be expressed as an extended power series. To
approximate it you do something like this:
In[13]:=
Series[1/(x^2+x+1),{x,0,6}]
Out[13]=
3 4 6 7
1 - x + x - x + x + O[x]
If some power of x is a divisor of q(x), then the approximation is an
extended power series.
In[14]:=
Series[1/(x^5+x^2),{x,0,6}]
Out[14]=
-2 4 7
x - x + x + O[x]
Ken Levasseur
Math. Sciences
UMass Lowell
> From: oldodo2000 at yahoo.fr (oldodo2000)
To: mathgroup at smc.vnet.net
> Date: Wed, 04 Jun 2003 08:34:53 -0400 (EDT)
> To: mathgroup at smc.vnet.net
> Subject: [mg41805] [mg41770] polynomial division
>
> Hi,
>
> I need to divide a polynom p(x)/q(x).
> but PolynomialQuotient and PolynomialRemainder function are OK if
> degree of p(x)>degree of q(x).
>
> If we consider by example f(x)=x^2+x+1
> How can I find an approximation of the result of g(x)=1/(x^2+x+1)
>
> Thanks
>
> Olivier
>