Okay. A better, much better thought out, proof that x = -1, 1.
[imath]x = x^x[/imath]
[imath]ln(x) = x ln(x)[/imath]
Let [imath]x = r e^{i \theta }[/imath]
Then we have the system
[imath]\begin{cases} r ln(r) ~ cos( \theta ) - \theta r ~ sin( \theta ) = ln(r) \\ r ln(r) ~ sin( \theta ) + \theta r ~ cos( \theta ) = \theta \end{cases}[/imath]
This can be rewritten in matrix form as
[imath]\left ( \begin{matrix} cos( \theta ) - \dfrac{1}{r} & sin( \theta ) \\ sin( \theta ) & cos( \theta ) - \dfrac{1}{r} \end{matrix} \right ) \left ( \begin{matrix} ln(r) \\ \theta \end{matrix} \right ) = \left ( \begin{matrix} 0 \\ 0 \end{matrix} \right )[/imath]
So either ln(r) = 0 and [imath]\theta = 0[/imath] or the determinant of the operator is 0. After a bit of work we find that both conditions predict r = 1. (The determinant condition actually gives [imath]\theta = 0, ~ \pi[/imath].) So we get [imath](r , \theta ) = (1, 0), ~ (1, \pi )[/imath].
Now, the thing is that ln is a multi-valued function. The matrix actually is
[imath]\begin{cases} r ln(r) ~ cos( \theta ) - \theta r ~ sin( \theta ) - 2 \pi n ~ sin( \theta )= ln(r) \\ r ln(r) ~ sin( \theta ) + \theta r ~ cos( \theta ) + 2 \pi n ~ cos( \theta )= \theta + 2 \pi m\end{cases}[/imath]
where n and m are independent integers.
So far I can't solve the system. But I have a graph below for n = 1 and m = 2 and it predicts new solutions (where the red and blue curves intersect.) I have not been able to verify them yet. The green line is r = 1.
-Dan
