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Re: When is x^y = != E^(y*Log[x])
*To*: mathgroup at smc.vnet.net
*Subject*: [mg66264] Re: When is x^y = != E^(y*Log[x])
*From*: "Nagu" <thogiti at gmail.com>
*Date*: Sat, 6 May 2006 01:54:39 -0400 (EDT)
*References*: <e3f555$s5h$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Hi Ted,
x^y = exp(y log(x)) is the y-th power function where log denotes a
branch of logarithm.
x^y is a multifunction.
The general definition of y-th power function can be defined as
follows:
x^y = (exp(y ln|x| + i y arg(x)))*(exp(i 2 pi y))^k, where k is any
complex number.
If we x^y defined on any open set S, and is continuous on S, then it is
a branch of y-th power function. Clearly, S does not contain the point
zero as an epsilon-neighborhood of zero has many preimages.
The principal branch can be defined as follows:
x^y = exp(y ln|x| + i y arg(x)), for -pi< arg(x) < pi.
Also note that, when y is integer, then every branch of x^y is same as
a polynomial function; when y = 1/n, each branch coincides with the
inverse of x^n.
Hope this helps.
Nagu
ted.ersek at tqci.net wrote:
> 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
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