Finding solutions to $ x^x = 2x$

Question

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.

Answer 1

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

Answer 2

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 : $$x\simeq 0.346323$$

Answer 3

(previous approach deleted,for archiving matters see edit-history)

Here is a list of 10 new solutions, internal precision 200 dec digits:

         x                                      |x^x-2x |
    ---------------------------------------  ----------------------  
 0     2.000000000000000                      0.0                     known solution
 1     3.116777334842701+3.115627209097009*I  8.249384166600113E-72   new solutions ...
 2     3.944366616159206+5.234957142265235*I  1.834796992244078E-88
 3     4.632829407543712+7.121955754816833*I  2.116951288747908E-98
 4     5.245455860616367+8.882459562041511*I  2.042549956851917E-105
 5     5.808250013483836+10.55849284332458*I  8.488793906217181E-111
 6     6.334925551361967+12.17222101970428*I  3.863116441982432E-115
 7     6.833781384222634+13.73716605764375*I  9.156097713919543E-119
 8     7.310302197236129+15.26235645509978*I  7.227764889511877E-122
 9     7.768342845867481+16.75420564534052*I  1.423061553268047E-124
10     8.210738840969926+18.21748045660466*I  5.736096368440588E-127

A q&d Pari/GP-routine:

 lg2 = log(2) ; \\ put it into a constant to avaoid recomputations
 for(c=0,10, 
       x=1+I; \\ just any complex initialization
       for(k=1,250,
               x = (lg2+ c*2*Pi*I)/log(x)+1
          );
       print( x, "  ", abs(x^x - 2*x));
      );      

A picture showing the first few complex zeros of $f(x)=x^x - 2x$. The green lines mark the zeros of the real part, the red lines that of the imaginary part, and where they cross (light color, and a marker) we have a complex zero. The image suggests, that we have one "linear" list of zeros as indicated in the numerical list above, approximating to some curve growing to linear shape.
Note, there are the two known zeros on the real axis ($x=2$,$ x \approx 0.34$)