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Re: Algebraic reduction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg108428] Re: Algebraic reduction
  • From: "Alexander Elkins" <alexander_elkins at hotmail.com>
  • Date: Wed, 17 Mar 2010 04:40:48 -0500 (EST)
  • References: <hnnk72$ckb$1@smc.vnet.net>

"Artur" <grafix at csl.pl> wrote in message news:hnnk72$ckb$1 at smc.vnet.net...
> Dear Mathematica Gurus.
> Who know how force simplification of polynomials (x^n - 1)/(x - 1)] by
> Mathematica
> Table[FullSimplify[(x^n - 1)/(x - 1)], {n, 1, 10}]
> Above eliminate denominator only in first 5 n
> Any idea ?
> Best wishes
> Artur
>

Here is a general solution to what you appear to be seeking to obtain that
does not use FullSimplify:

In[1]:=
Column[Table[PolynomialQuotient[#1,#2,#3]+PolynomialRemainder[#1,#2,#3]/#2,{
n,1,10}]&[x^n-1,x-1,x]]
Out[1]= 1
1+x
1+x+x^2
1+x+x^2+x^3
1+x+x^2+x^3+x^4
1+x+x^2+x^3+x^4+x^5
1+x+x^2+x^3+x^4+x^5+x^6
1+x+x^2+x^3+x^4+x^5+x^6+x^7
1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8
1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9

Of course for x^n-1 with n > 1 && Element[n, Integers] the
PolynomialRemainder[x^n-1,x-1,x] is always zero so you can also simplify
your particular example just using Factor:

In[2]:= Column[Table[Factor[x^n-1]/(x-1),{n,1,10}]]
Out[2]= 1
1+x
1+x+x^2
(1+x) (1+x^2)
1+x+x^2+x^3+x^4
(1+x) (1-x+x^2) (1+x+x^2)
1+x+x^2+x^3+x^4+x^5+x^6
(1+x) (1+x^2) (1+x^4)
(1+x+x^2) (1+x^3+x^6)
(1+x) (1-x+x^2-x^3+x^4) (1+x+x^2+x^3+x^4)

Perhaps you will also notice that the factors are the same as the cyclotomic
polynomials of all of the factors on n, so that  for x^n-1 with n > 1 &&
Element[n, Integers] the following is true:

In[3]:=
Table[Factor[x^n-1],{n,1,10}]==Table[Times@@Cyclotomic[Times@@@Tuples[Union[
#1^Range[0,#2]]&@@@FactorInteger[n]],x],{n,1,10}]
Out[3]= True

Union is used only to handle the special case of 1^Range[0,1] which produces
{{1, 1}} instead of the desired {{1}}

Note that x - 1 == Cyclotomic[1, x].
Hope this helps...






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