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RE: a^n*b^n != (a*b)^n

• To: mathgroup@smc.vnet.net
• Subject: [mg12321] RE: [mg12267] a^n*b^n != (a*b)^n
• From: Ersek_Ted%PAX1A@mr.nawcad.navy.mil
• Date: Thu, 7 May 1998 18:52:07 -0400

mmichael@idirect.com  wrote:
|
|which as Mathematica claims does NOT equal to each other!!!
  (* discussion deleted  *)
|
|So generally that would mean: a^n*b^n != (a*b)^n |
|I tried to go and search for the basic proof of this equality.
Obviously |enough I couldn't find any :(
|For a, b being real and n being positive integer the equality is
|obvious. But for other cases - I don't know how to approach it. |
|While playing around with different examples I noticed that the above
|equality upholds for all the cases except when we have a and b being
|negative REAL numbers and n being p/q with q=2k |
|ANYTHING at all will be greatly appriciated, as I am completely
stuck!!! |

I am not a math professor, but an engineer with fairly proficient math 
skills.  I have spent a good bit of time thinking about this.  The 
conditions below do not include every possibility, but they do include
most  cases of practical significance.

((x)^a)^b==x^(a b)
If ( Positive[x]&&(Im[a]==0)&&(Im[b]==0)  ) Or
If ( (Im[x]==0)&&OddQ[a]&&IntegerQ[b] ) Or
If( (Im[x]==0)&&EvenQ[a]&&EvenQ[a*b] ) Or
If( (Im[b]==0)&&(-1<a<1)  )  (*  Even if x is complex!  *) Or
If( (Im[x]==0)&&IntegerQ[b]&&Not[EvenQ[a]]&&(Round[a/2]-a/2!=0.0) )

An example demonstrating the last case:
  Not[EvenQ[4.0]]->True,  (Round[4.0/2]-4.0/2)!=0.0->False
  And  (((-3)^4.0)^-2)!=(-3)^(-8.0),
  Not[EvenQ[4.1]]->True,  (Round[4.1/2]-4.1/2)!=0.0->True
  And  (((-3)^4.1)^-2)==(-3)^(-8.2)     
  Try it, and see that it works!        

On the other hand:
((x)^a)^b==-(x^(a b))
If ( (x<0)&&EvenQ[a]&&OddQ[a b] )

It would be neat to have conditions that are so complete that we could
say ((x)^a)^b==x^(a b)   If and only If one or more of the following
are true  ...........

_______________________________________ The way the ((x^a)^b) is
computed for limited cases is as follows:

Let (x) be real or complex
Let (a) be real
Let (b) be real

Recall:   E^(I*theta)=Cos[theta]+I*Sin[theta]
 (* that's I=Sqrt[-1],  it doesn't look right in ASC.  *)

Now let   mag=Abs[x];  ang=Arg[x];
Then  x=mag*E^(ang*I)
(x^a) evaluates to (mag^a)*E^(ang+a) Now you have to get (ang+a) between
-Pi, and +Pi. So the above evaluates to (mag^a)*E^(Mod[ang+a, 2*Pi]-Pi)
Then you do (x^a)^b and get,
(x^a)^b = (mag^a^b)*E^(Mod[ang+a,2*Pi]-Pi+b)


Ted Ersek