Problem of The Day

[BigBeachBanana]

Those are the only ones I found. How can you tell if there's more, or not?

The reason I introduced the absolute value right off the bat is that, without it, I'm only assuming x>0, which cut off the other branch of the solution, i.e. x<0. I didn't consider that it would introduce the extraneous solution(s). However, from the check, in this case, it didn't. Is there a better way to solve it?

You're correct that without the absolute value, you're implying $x>0$ only, but you can do without it. You're going to have to consider 3 cases.
1) For $x>0$:
$$\log x\times(x-1)=0 \implies x=1$$2) For $x<0$, multiply both sides by -1:
$$-x^x=-x\\ (-x)^x=-x\\ x\log (-x)=\log(-x)\\ x\log (-x)-\log(-x)=0\\ \log(-x)\times(x-1)=0 \implies x=\pm 1$$3) For $x=0$, we get $0^0$, which can be undefined or 1. In either case, it's not a solution.

 

[Dr.Peterson]

You're correct that without the absolute value, you're implying $x>0$ only, but you can do without it. You're going to have to consider 3 cases.
1) For $x>0$:
$$\log x\times(x-1)=0 \implies x=1$$2) For $x<0$, multiply both sides by -1:
$$-x^x=-x\\ (-x)^x=-x\\ x\log (-x)=\log(-x)\\ x\log (-x)-\log(-x)=0\\ \log(-x)\times(x-1)=0 \implies x=\pm 1$$3) For $x=0$, we get $0^0$, which can be undefined or 1. In either case, it's not a solution.

Yes, this sort of case-by-case thinking is what I recommended.

But you have to be even more careful; it isn't always true that $-x^x=(-x)^x$. (Consider, for example, $x=-2$ or $x=-\frac{2}{3}$.)

My attempt:
$$x^x=x\\ x\log |x| = \log |x|$$...

I asked for a justification of this step, because you can't just replace some variables in an equation with their absolute values. You probably have something like this in mind: $$x^x=x\implies |x^x|=|x|\implies |x|^x=|x|\implies x\log |x| = \log |x|$$
The one-directional implication is the reason this could produce extraneous solutions, just as changing $x=-3$ to $|x|=|-3|$ would produce two solutions rather than one.

 

[Cubist]

The graph of |z^z - z| is interesting. Obviously where this surface dives down to 0 then z^z = z. The white dots are where the height of the surface is 0 at the two known solutions z=-1 and z=1. There's an interesting hump with a ridge between them!



A zoomed out view reveals extra solutions with non-zero imaginary components (I capped the height to 15 otherwise it shoots WAY up)...



Looks like a focaccia! At the bottom of the "finger holes", symmetrical about the real line, the surface seems to dive down to exactly zero. Here are the approximate coordinates (2.86295413 ± 3.22327384i), (3.72731487 ± 5.31804384i), (4.43317075 ± 7.193797i), (5.05673006 ± 8.94746967i), and I assume there'll be lots more of these if the graph is zoomed out more.

 

[Steven G]

Sorry, but I have a problem (actually a big problem) when you say that (−x)^x=−x follows from -x^x=-x
This was also pointed out by Dr Peterson.

 

[topsquark]

Okay. A better, much better thought out, proof that x = -1, 1.

$x = x^x$

$ln(x) = x ln(x)$

Let $x = r e^{i \theta }$

Then we have the system
$\begin{cases} r ln(r) ~ cos( \theta ) - \theta r ~ sin( \theta ) = ln(r) \\ r ln(r) ~ sin( \theta ) + \theta r ~ cos( \theta ) = \theta \end{cases}$

This can be rewritten in matrix form as
$\left ( \begin{matrix} cos( \theta ) - \dfrac{1}{r} & sin( \theta ) \\ sin( \theta ) & cos( \theta ) - \dfrac{1}{r} \end{matrix} \right ) \left ( \begin{matrix} ln(r) \\ \theta \end{matrix} \right ) = \left ( \begin{matrix} 0 \\ 0 \end{matrix} \right )$

So either ln(r) = 0 and $\theta = 0$ 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 $\theta = 0, ~ \pi$.) So we get $(r , \theta ) = (1, 0), ~ (1, \pi )$.

Now, the thing is that ln is a multi-valued function. The matrix actually is
$\begin{cases} r ln(r) ~ cos( \theta ) - \theta r ~ sin( \theta ) - 2 \pi n ~ sin( \theta )= ln(r) \\ r ln(r) ~ sin( \theta ) + \theta r ~ cos( \theta ) + 2 \pi n ~ cos( \theta )= \theta + 2 \pi m\end{cases}$
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

 

[topsquark]

Edit: Sorry, I mis-spoke. The determinant only gives $(r, \theta ) = (1, 0)$ as well. The original equations, given r = 1, give the $(r, \theta ) = (1, \pi )$ solution.

-Dan

 

[NobodyAnyone]

Why is math so hard????????

 

[topsquark]

Why is math so hard????????

Actually I had fun doing it.

-Dan

 

[Dr.Peterson]

Why is math so hard????????

In this case, it's because we like challenges, so we make it harder by enlarging the scope of the problem (like allowing complex numbers), or trying to use a different method (like logs). In fact, this whole thread is about posing an intentionally challenging problem for which the obvious answer is "wrong" (in the sense of missing a solution), so that others can have fun discussing it.

People get good at math by challenging themselves (and sometimes others).