Finding solutions to $x^x = 2x$

A friend claims it isn't possible to find a closed form for the smaller positive real solution of $x^x = 2x$. Numerically we have seen that $0.346...$ and $2$ are solutions, but are failing to do anything but approximate the first solution.

The first attempt to find a closed form was using the same trick as solving $$x^{x^{x^x...}} = 2$$ Where raising $x$ to the power of both sides gives: $$x^{x^{x^x...}} = x^2$$

$$\Rightarrow 2 = x^2$$ $$\Rightarrow x = \sqrt{2}$$

But without an infinite power tower we got nowhere.

We know $f(z) = z^z$ is analytic with inverse $f^{-1}(z) = e^{W(ln(z))}$ where $W(z)$ is the Lambert-W function, But $x = e^{W(ln(2x))}$ doesn't seem any better.

I am just wondering how to approach such a problem.

• Hint: Take the natural log of both sides and see what you get – ASKASK Jan 21 '15 at 3:54
• @ASKASK That hint doesn't lead anywhere special. $(x - 1)\log_2 x = 1$ – Axoren Jan 21 '15 at 4:07

I will give you an example problem:

An exponential tetration is where $x$ is exponentiated by itself $n$ times, as for example

$$_{ }^{ 4 }{ x }={ x }^{ {\displaystyle x }^{ {\displaystyle x }^{\displaystyle x } } }$$

As $n$ approaches $\infty$, then for some $0$, the function bifurcates at point $B$, splitting into the upper branch where $n$ is even and the lower branch where $n$ is odd, even though in both cases, $n$ approaches $\infty$ .

Point $A$ is the origin $(0,0)$ and point $C$ is $(0,1)$

Let $T$ be the area of $ABC$ as defined by the upper and lower branches, and the line $x=0$

How do you Solve this?

This problem is a little easier to solve if we looked at the function(s) in the other way. The curve as defined by the upper and lower branches, joining at point $B,$ can alternatively be described by the implicit equation

$${ x }^{ { \displaystyle x }^{\displaystyle y } }-y=0$$

where $x(y)$ arches smoothly from $(0,0)$ to $(1,0)$. Then any one of several possible numerical integration methods can be used, once we have an exact implicit equation. For example, given a valuie $0\le y\le 1$, numerically determine $x$ using the secant method, and use it as a data point for numerical integration.

The critical point $B$ is $$\left( \dfrac { 1 }{ { e }^{ e } } ,\dfrac { 1 }{ e } \right)$$

A more straightforward way to numerically integrate this is to use a loop to exponentially tetrate $0\le x\le { e }^{ -e }$ for some very large $2n$ and $2n+1$, and use the difference as a data point for numerical integration.

Thus, one can check the numerical results from both methods for confidence in final accuracy.

Note: The other critical point for the infinite exponential tetration is $$({ e }^{ \frac { 1 }{\displaystyle e } },e)$$

which is the limit of convergence, i.e., for $x>{ e }^{ \frac { 1 }{ e } }$ it does not converge

Today and as far as I know about special functions, there is no closed form for the non-trivial root of $x^x=2x$, even with the LambertW function. This would require a more advanced or more general special function which is not defined and referenced yet.

As already mentioned, in practice the root $x\simeq 0.3456323$ is obtained thanks to numerical calculus.

Another way, but not useful in practice because more complicated, is the inversion of function in terms of series : $x^x=2x$ is equivalent to $x^{x-1}=2$ and to $(x-1)\ln(x)=\ln(2)$

Consider the function $y=(x-1)\ln(x)$ . The serie expansion and inversion, which is a booring calculus, leads to : $$x=1-y^{1/2}+\frac{1}{4}y+\frac{1}{96}y^{3/2}-\frac{59}{92160}y^{5/2}-\frac{1}{92880}y^{3}-\frac{2783}{20643840}y^{7/2}-\frac{1}{24192}y^{4}-\frac{1060117}{118908518400}y^{9/2}-\frac{1}{4838400}y^{5}+...$$ In the present case $y=\ln(2)$ Puting it into the preceeding equation yields to : $$x\simeq 0.346323$$

I don't know, whether this counts as another solution, but I found this in the complex numbers for an initial $z=20-20I$ using Newton-Iteration in Pari/GP

{ l2 = log(2); z = 20  - 20*I
for(k=1,50,
err = (l2 + z - z*exp(z))/( 1 - (1+z)*exp(z)) ;
z=z-err ;
if(abs(err)<1e-80,break());
);
print(z) }


the values :

  z
\\ result: 0.00067963071061921624623 - 18.812728043232459702*I
y=exp(z)
\\ result: 1.0000013310516279147 + 0.036844586179231404316*I

[exp(z*y) ; 2*y]
\\ result:
[2.0000026621032558295 + 0.073689172358462808632*I]
[2.0000026621032558295 + 0.073689172358462808632*I]


We need $\exp(z \cdot y)$ here, simply $y^y = 2y$ does not work; we use here another branch of the log for the finding of $z$

I did not find any more roots for $z^z - 2z=0$ besides the two given one $r_1 \approx 0.346...$ and $r_2 = 2$, not even in the complex plane.

One can often find such roots when using the Newton-iteration: one starts at some complex value and iterates to approach a root and then find an arbitrary well approximation of that root; I used $f(z) = (z-1) \cdot \log(z) / \log(2) - 1$ and its derivative for the Newton-rootfinding procedure.
Here is a picture to which of the roots the Newton-algorithm converges when started at some complex value $z_0$ and iterated: it either finds $r_1$ or $r_2$ except of course where $z=0$ or $z=1$ which cannot be evaluated due to singularities. The square area shown is from the unit-square in the complex plane; $z_0$ from the blue area let me find root $r_2=2$ and $z_0$ from the green area let me find root $r_1$. No other roots were found in that area. Lighter color indicates, that the root is found quickly(few iterations) darker color indicate more iterations. The boundary between green and blue is likely fractal (but I don't have a good/reliable analysis) :