[Date Index]
[Thread Index]
[Author Index]
Re: PolynomialMod
*To*: mathgroup at smc.vnet.net
*Subject*: [mg121357] Re: PolynomialMod
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 12 Sep 2011 04:21:03 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201109111128.HAA12187@smc.vnet.net> <52BDE91D-1ACE-471D-A992-DF8902ADBE7D@mimuw.edu.pl> <sig.0235ae8fc1.4E6CFABB.9040206@csl.pl> <6763942A-FD6A-4C87-AB48-51BF0CC71021@mimuw.edu.pl>
In fact, for univariate polynomials you can equally well use =
PolynomialRemaineder:
PolynomialRemainder[1+3 x+5 x^2+5 x^3+5 x^4+3 x^5+x^6,x^3-b x-c,x]
x^2 (b^2+5 b+3 c+5)+x (3 b^2+2 b c+5 b+5 c+3)+3 b c+c^2+5 c+1
Andrzej
On 11 Sep 2011, at 21:38, Andrzej Kozlowski wrote:
>
> On 11 Sep 2011, at 20:15, Artur wrote:
>
>> Mayby I was used wrong example
>>
>> Example 1.
>> f = 1 + 3 x + 5 x2 + 5 x3 + 5 x4 + 3 x5 + x6;
>>
>> If we have smaller order equation for example
>> x^5-x-1=0
>> that x^5=x+1
>> multiply both sides by x
>> x^6=x^2+x
>> if we substitute in f x^6->x^2+x and x^5->x+1 we have
>> f = 1 + 3 x + 5 x2 + 5 x3 + 5 x4
>> + 3 (x+1) +(x^2+x);
>> now after reduction f=4 + 7 x + 6 x^2 + 5 x^3 + 5 x^4
>>
>> That same result we will obtain uses
>>
>> PolynomialMod[1 + 3 x + 5 x^2 + 5 x^3 + 5 x^4 + 3 x^5 + x^6, x^5 - x =
- 1]
>>
>> of course f=0 is divisible by my p from previous example but we can =
do similar cycle of substitutions as in Example 1 but if f<>0 isn't =
divisible
>>
>> in my example let symbolically
>>
>> x^3-b x-c==0
>>
>> x^3=b x+c
>> x^4=b x^2+c x
>> x^5=b x^3+c x^2=b (b x+c)+c x^2=b^2 x+b c+c x^2
>> x^6=b^2 x^2+b c x+c x^3=b^2 x^2+b c x+c (b x+c)=b^2 x^2+2 b c =
x+c^2
>> and apply these equations to starting function f
>>
>> f=1 + 3 x + 5 x
>> 2
>> + 5 (b x+c) + 5 (b x^2+c x) + 3 (b^2 x+b c+c x^2) + (b^2 x^2+2 b c =
x+c^2)
>> after reduction we have finally result (Collect by x)
>>
>>
>> f=(1 + 5 c + 3 b c + c^2) + (3 + 5 b + 3 b^2 + 5 c + 2 b c) x + (5 =
+ 5 b + b^2 + 3 c) x^2
>>
>> generally g=0 and f is some function (not necessary zero)
>>
>> Artur Jasinski
>>
>
>
> I think what you want to do is to use PolynomialReduce and not =
PolynomialMod. In your case
>
> In[11]:= Last[
> PolynomialReduce[1 + 3*x + 5*x^2 + 5*x^3 + 5*x^4 + 3*x^5 + x^6,
> x^3 - b*x - c, x]]
>
> Out[11]= x^2*(b^2 + 5*b + 3*c + 5) +
> x*(3*b^2 + 2*b*c + 5*b + 5*c + 3) +
> 3*b*c + c^2 + 5*c + 1
>
>
>
> Andrzej Kozlowski
>
Prev by Date:
**Re: Column vectors should be interpreted as simple lists where**
Next by Date:
**Re: execution model: Function vs. delayed execution**
Previous by thread:
**Re: PolynomialMod**
Next by thread:
**Re: PolynomialMod**
| |