MathGroup Archive 1997

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

Search the Archive

Fwd: Why doesn't this simplify further?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg5997] Fwd: [mg5964] Why doesn't this simplify further?
  • From: BobHanlon at aol.com
  • Date: Sat, 8 Feb 1997 22:38:24 -0500
  • Sender: owner-wri-mathgroup at wolfram.com

test = {(1/4)^n 4^(n+3), (1/4)^n 2^(2n), 
	(1/x)^n x^(n+3), (1/x)^n (x/2)^(2n), (1/x)^n (x)^(2n)}

  1 n  3 + n   1 n  2 n   1 n  3 + n   1 2 n  1 n  2 n
{(-)  4     , (-)  2   , (-)  x     , (-)    (-)  x   , 
  4            4          x            2      x
 
   1 n  2 n
  (-)  x   }
   x

test // PowerExpand

  1 n  3 + n   1 n  2 n   3   1 2 n  n   n
{(-)  4     , (-)  2   , x , (-)    x , x }
  4            4              2

The differences are due to the different internal representations:

test // FullForm

List[Times[Power[Rational[1, 4], n], 
 
   Power[4, Plus[3, n]]], 
 
  Times[Power[Rational[1, 4], n], Power[2, Times[2, n]]], 
 
  Times[Power[Power[x, -1], n], Power[x, Plus[3, n]]], 
 
  Times[Power[Rational[1, 2], Times[2, n]], 
 
   Power[Power[x, -1], n], Power[x, Times[2, n]]], 
 
  Times[Power[Power[x, -1], n], Power[x, Times[2, n]]]]

Reciprocals of integers are represented as Rational rather than Power.

revPowerExpand[expr_] := PowerExpand[expr /. 
	{k_^(m_Integer x_) -> (k^m)^x,
	 Rational[1, x_Integer]^pwr_ -> x^(-pwr)}];

test // revPowerExpand

         3   1 n  n   n
{64, 1, x , (-)  x , x }
             4

---------------------
Forwarded message:
From:	hucka at eecs.umich.edu (Michael Hucka)
To: mathgroup at smc.vnet.net
To:	mathgroup at smc.vnet.net

I encountered the following in mma 2.2:

In[14]:= PowerExpand[ (1/4)^n 2^(2 n)]

          1 n  2 n
Out[14]= (-)  2
          4

Now, shouldn't the above be equal to 1, assuming n is real?  Or am I missing
something really obvious here?  I don't understand why PowerExpand doesn't
simplify this further.

-- 
Mike Hucka   hucka at umich.edu   http://www.eecs.umich.edu/~hucka
     University
 PhD to be, computational models of human visual processing (AI Lab)     of
   UNIX systems administrator & programmer/analyst (EECS DCO)
        Michigan




  • Prev by Date: Re: pattern test and default value
  • Next by Date: Re: pattern test and default value
  • Previous by thread: special ColorFunction
  • Next by thread: Why doesn't this simplify further?