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