# Strange function

1. Aug 5, 2010

### zetafunction

how this function $$f(x)= 0^{-x}$$

should be understood? , if x is NEGATIVE, we find no problems, since 0 raised to any power is 0

but how about x being a positive real number ? , or x being a PURE COMPLEX or complex number ?

could we consider a 'regularization' to this f(x) so $$f(x)_{reg}=0$$ is always 0

2. Aug 5, 2010

### Office_Shredder

Staff Emeritus
I would imagine the same way you would understand $$\sqrt{x}$$. It's assumed that the domain is restricted only to numbers that make sense.

If you have some context where you think you really need to plug 1 into the function, you should post the full setup here

3. Aug 5, 2010

### Petr Mugver

In the real case, the function

$$f(x,y)=y^x=e^{x\log y}$$

is defined only for $$y>0$$

In the complex case, put

$$y=\rho e^{i\alpha}\qquad\textrm{and}\qquad x=a+ib$$

then

$$y^x=(\rho e^{i\alpha})^{(a+ib)}$$

and, after some calculations, you find

$$y^x=Re^{i\beta}$$

with

$$R=\rho^ae^{-\alpha b}\qquad\textrm{and}\qquad\beta=b\log\rho+\alpha a$$

So in the complex case you must have $$\rho\neq 0$$, which means $$y\neq 0$$.

I hope I didn't meke calculation errors...try it yourself!

Last edited: Aug 5, 2010