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