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