RE: a^n*b^n != (a*b)^n
- To: mathgroup@smc.vnet.net
- Subject: [mg12329] RE: [mg12267] a^n*b^n != (a*b)^n
- From: Ersek_Ted%PAX1A@mr.nawcad.navy.mil
- Date: Thu, 7 May 1998 18:52:16 -0400
I made a careless algebra mistake in my first transmission. I believe
the following is correct.
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)
(******* (ang*a) not (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
----------
|From: Ersek Ted
To: mathgroup@smc.vnet.net
|To: 'MATHGROUP@WOLFRAM.COM@INTERNET@PAXMB1' |Subject: [mg12329] RE: [mg12267]
a^n*b^n != (a*b)^n |Date: Wednesday, May 06, 1998 8:29AM |
|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
|