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
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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