Solution to the Equation $x^x = 0$.

Question

If I solve the Equation $x^x = 0$, then I get $x^x=0$:

x ln⁡(x)= -∞ or ln⁡x $e^{\ln⁡(x)}$ =-∞

and if we take the Lambert W function on both sides we get

ln⁡(x)=∞,

And I put it on Wolfram Alpha then it shows undefined? Is $x^x$ never equal to zero,or is it in a complex domain?.

Answer 1

Whenever you ask: "What are the solutions to this equation?" one crucial component is to specify over what admissible solution space you are looking for solutions. This is because equations can have a wastely different number of solutions depending on what kind o "x" we allow.

Take $x(x^2+1)(x^2-2)e^{x} = 0$ for instance. Over the integers, this has a single solution: $x=0$. If we allow real numbers, then also $\pm \sqrt{2}$ are solutions. If we allow complex numbers, then $x=\pm i$ are solutions as well. If we allow the extended real number line, $x=-\infty$ is a solution. If we allow Quaternions, then there are infinitely many solutions. If you allow $x$ to be an element of some matrix space, you can get all sorts of solution sets.

And if you are unsure of what the solution space should/could be, then it is also important to mention that, since quite often possible to extend the space somehow to get solutions.

Now, in the case at hand we have

$$0 \overset{!}{=} x^x \overset{\text{def}}{=} e^{x\log x} $$

Now over the reals, there is no solution. Over the extended reals, $e^x = 0$ is achieved for $x=-\infty$. However $x\log x=-\infty$ has so solution over the extended reals. If we allow ourselves to extend further into the complex domain, we have $\log(z) = \log(re^{i\phi}) = \log r + i\phi$ for any complex number $z$.

Taking $z=-r = re^{i\pi}$ we have $z\log z = -r(\log r + i\pi) \to -\infty$ as $r\to\infty$, so one could say that $x=-\infty +0i = \infty \hat z$, where $\hat z=e^{i\pi}\in S^1$ is a solution in a sort of "extended complex plane".

Note that this is a different kind of extended complex plane than the Riemann Sphere, which you'll find if you google for this term. The Riemann sphere is obtained by gluing all points at infinity together into a single point, whereas here, we attach a circle of so-called directed infinities $\{\infty\cdot \hat e_z \mid |\hat e_z|=1 \} = \infty \cdot S^1$ to the complex plane.

We can verify numerically that $x=-\infty$ this is a solution in python:

import numpy as np

for x in np.linspace(-1+0j, -100+0j):
    print( x**x )

Converges against $0$ very quickly. Note that we have to add the $0j$ to make sure python treats it as a complex number, otherwise you will get errors!

Answer 2

Any positive number raised to a positive power gives a positive number. So, $x^x$ is always positive in $\mathbb R^+$.

For, negative $x$, that is, in $\mathbb R^-$, the function $x^x$ cannot be defined. This is because $(-2)^{-2}$ doesn't make any sense (in $\mathbb R$).

At $x=0$, the function $x^x$ is simply not defined as $0^0$ is undefined.

That is why $x^x=0$ has no solution in reals.

Now, let us consider $f:\mathbb C\to \mathbb C$ defined as $f(z)=z^z$. If we consider $z=a+ib$, then the modulus of $z^z$ is $(a^2+b^2)^{\frac a 2}.e^{-b\arg (z)}$, which can never be $0$. This proves that $f(z)=0$ has no solutions.

In case you don't understand, how I got the modulus, check complex exponentiation.