MathGroup Archive 2001

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Factor[1+x^4]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg26889] Re: Factor[1+x^4]
  • From: paradaxiom at my-deja.com
  • Date: Fri, 26 Jan 2001 01:27:32 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

By default Factor[] factors over integers. If you want Factor[]
to factor into Sqrt[2] and I you have to give this to the
Extension option:

In[1]:= f = Factor[1 + z^4, Extension -> {I, Sqrt[2]}]

Out[1]= (1/4)*(Sqrt[2] - (1 + I)*z)*
              (Sqrt[2] - (1 - I)*z)*
              (Sqrt[2] + (1 - I)*z)*
              (Sqrt[2] + (1 + I)*z)

In[2]:= Expand[f]

Out[2]= 1 + z^4


Of course, in general you don't know what Extension to pick beforehand.
So one thing you can do is the following:

In[1]:= poly = 1 + z^4

Out[1]= 1 + z^4

The find the extension you want by looking at the roots of the
polynomial:

In[2]:= extension[poly_, z_] :=
    ComplexExpand[
       Roots[poly == 0, z] /. {_ == a_ -> a} /. {Or -> List}
       ]

In[3]:= extension[poly, z]

Out[3]= {(1 + I)/Sqrt[2],
       -((1 - I)/Sqrt[2]),
       -((1 + I)/Sqrt[2]),
         (1 - I)/Sqrt[2]}

In[4]:= Factor[poly, Extension -> extension[poly, z]]

Out[4]= (1/4)*(Sqrt[2] - (1 + I)*z)*
              (Sqrt[2] - (1 - I)*z)*
              (Sqrt[2] + (1 - I)*z)*
              (Sqrt[2] + (1 + I)*z)



I hope this helps a bit. Perhaps someone knows a better way to
go about this?

//Marten


Sent via Deja.com
http://www.deja.com/


  • Prev by Date: Re: Overriding Power
  • Next by Date: Re: Overriding Power
  • Previous by thread: Re: Factor[1+x^4]
  • Next by thread: Re: Factor[1+x^4]