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

```

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