Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

When is x^y = != E^(y*Log[x])

• To: mathgroup at smc.vnet.net
• Subject: [mg66236] When is x^y = != E^(y*Log[x])
• From: ted.ersek at tqci.net
• Date: Fri, 5 May 2006 05:02:22 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

The Mathematica documentation for Power says:
For complex numbers (x^y) gives the principal value of  ( E^(y*Log[x] ).
This is consistent with reference books.

I wanted to see where in the extended complex plane this identity applies.
Also when it doesn't apply how do we determine (x^y).
So consider the following:


In[1]:=
  x=0; y=-3;
  {x^y, E^(y*Log[x])}

Out[3]=
  {ComplexInfinity, Infinity}


The documentation wasn't wrong in the above example because 0, -3 are not
complex numbers.
However, I have seen books that imply the identity above works for any (x,y).
Well I can see the above identity doesn't apply when x=0 and y is negative.
The above identity doesn't apply in the following cases either.

In[4]:=
  x=-2;  y=(2+I)*Infinity;
  {x^y, E^(y*Log[x])}

Out[6]=
  {Indeterminate, 0}



In[7]:=
  x=5-6*I;  y=-Infinity*I;
  {x^y, E^(y*Log[x])}

Out[9]=
  {Indeterminate, 0}



In[10]:=
  x=y=Infinity;
  {x^y, E^(y*Log[x])}

Out[11]=
  {ComplexInfinity, Infinity}


Then you might say the above identity doesn't apply when Abs[y]==Infinity,
but the above identity does apply in the next example.

In[12]:=
  x=(1+2*I)*Infinity;  y=(-1+2*I)*Infinity;
  {x^y, E^(y*Log[x])}

Out[14]=
  {0, 0}


Could somebody provide conditions on (x,y) that are necessary and
sufficient for
E^(y*Log[x])  to return the same thing as (x^y).

-------------
  Ted Ersek

• Follow-Ups:
  • Re: When is x^y = != E^(y*Log[x])
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>