MathGroup Archive 2000

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

Search the Archive

Re: Question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg25266] Re: [mg25258] Question
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Sun, 17 Sep 2000 17:33:35 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Actually, I forgot to mention that this problem is almost  trivial, provided
you do not insist on  Mathematica doing all the work. We of course assume
that m is a rational number. This is how it goes:

In[1]:=
Factor[x^3 + 8]

Out[1]=
                    2
(2 + x) (4 - 2 x + x )

In[2]:=
% /. x -> m^(1/3)

Out[2]=
      1/3          1/3    2/3
(2 + m   ) (4 - 2 m    + m   )


So the answer will be

Out[25]=
       1/3    2/3
4 - 2 m    + m
-----------------
      8 + m

Mathematica can also verify this

In[27]:=
FullSimplify[(4 - 2*m^(1/3) + m^(2/3))/(m + 8) ==
   1/(m^(1/3) + 2), Element[m, Reals]]

Out[27]=
True

 Now I am getting worried if I have not inadvertently done your homework for
you ...



on 00.9.17 6:17 PM, Andrzej Kozlowski at andrzej at tuins.ac.jp wrote:

> on 00.9.17 5:47 PM, Steven Spear at smitsky at mindspring.com wrote:
> 
>> Hi, can anyone be so kind as to tell me how this problem:
>> 
>> 1/cbrt(m) +2    (One over the Cube Root of "m" plus Two)
>> 
>> ...can be solved in Mathematica by rationalizing the denominator? Thank you.
>> Steve
>> 
>> 
>> 
>> 
> 
> As long as your m is a number the following will work for roots of degree less
> than 5:
> 
> RationalizeDenominator1[expr_] :=
> 
> FullSimplify[expr, ComplexityFunction ->
> (
> Count[#, _?
> (MatchQ[Denominator[#], Power[_, _Rational] _. + _.] &),
> {0, Infinity}
> ] + If[FreeQ[#, Root], 0, 1] &
> )
> ]
> 
> For example:
> 
> In[3]:=
> RationalizeDenominator1[1/(3^(1/3) + 2)]
> 
> Out[3]=
> 1          1/3    2/3
> -- (4 - 2 3    + 3   )
> 11
> 
> or
> 
> In[4]:=
> RationalizeDenominator1[1/(7^(1/4) - 3)]
> 
> Out[4]=
> 1
> -- (-27 - 3 Sqrt[7] - Sqrt[2 (63 + 44 Sqrt[7])])
> 74
> 
> 
> It can also be one for expressions involving roots of degree higher than 5 but
> it is considerably more complicated. In fact already once explained how to do
> this in this list (in 1999).

-- 
Andrzej Kozlowski
Toyama International University
JAPAN

http://platon.c.u-tokyo.ac.jp/andrzej/
http://sigma.tuins.ac.jp/



  • Prev by Date: Re: Use of Miscellaneous`RealOnly` Package
  • Next by Date: Re: Newbie question
  • Previous by thread: Re: Question
  • Next by thread: powersum with a function with variable count of parameters