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I know that $x^x$ for all $x>0$

but what is negative values for that function which give a real number

for example $$f(-1)=(-1)^{-1}=-1\in R$$

I try to put sequence for that but i faild

is there any help

thanks for all

Ayman Hourieh
  • 39,603
mnsh
  • 5,875

3 Answers3

3

$x^x$ is well defined as a real function for $$(0,\infty) \cup \{ -\frac{m}{2n+1}| m, n \in {\mathbb N} \}$$

N. S.
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2

Set $y = -x$.

Thus, we have:

$x^x = (-y)^{-y}$ when $y \ge 0$

Or equivalently,

$$x^x = \frac{1}{(-y)^{y}}$$

Which can be simplified to:

$$x^x = \frac{1}{(-1)^{y} y^y}$$

Therefore, the domain of $x^x$ consists of both reals and complex numbers depending on the value of $(-1)^y$ or to be more precise depending on the value of $y$.

response
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For negative values of $x$, when $x$ is not an integer, you run into surly problems involving complex numbers. These entail a study of the complex log function and its branches.

Clearly, $x\mapsto x^x$ makes sense for positive $x$. It also makes sense for negative integer values of $x$. It is not defined at $0$.

ncmathsadist
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  • but when x is not negative integer like x=-e its give complex value – mnsh May 21 '13 at 14:30
  • Yes, because of a choice of a branch of the log. $(-e)^{-e} = (e^{1 + (2n + 1)i\pi})^{-e}.$ Now you must pick an $n$, for an infinity of values is possible from this expression. – ncmathsadist May 21 '13 at 14:39
  • @ncmathsadist $e^{\ln{x}}\neq{x}$ when $x<0$. Take $x=-1$, for example. – Arbuja Dec 15 '15 at 13:35