Infinite exponentiation

We can make sense of an infinite exponential, f[x] = x^x^x^..., by defining recursively:

\(f[1, x_]\  = \ x; 
f[n_, x_]\  := \ x^f[n - 1, x]\)

f[3, x]

The infinite exponential function f[x] can then be defined to be the limit of f[n,x] as n goes to infinity. This limit will not exist for every x. For example, f[1] = 1, but f[n,2] would quickly go to infinity as n increases. To see which values might work, we'll first plot and compare a few of the functions f[n,x].

\(p\  = 
      Plot[{f[3, x], \ f[4, x], \ f[5, x], 

          		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f[6, x], \ f[7, x], \ 
          f[8, x]}, 
		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {x, 0, 1.8}, \ 
        PlotRange -> {0, 10}]\)

There is some unusual behavior near 0 and the graphs split apart around x = 1.4, but the different functions f[n,x] agree closely between these numbers. The key to understanding f[x] is to realize that if this limit exists then the recursive definttion
   f[n,x] = x^f[n-1,x]
implies
    f[x] = x^f[x]
Let y = f[x]. Then y = x^y and we can solve: x = y^(1/y). The function y^(1/y) is the inverse of f[x], and vice-versa, at least on intervals containing 1 where these functions are 1-1. We can easily check for such intervals for the function y^(1/y).

Plot[y^(1/y),  {y, 0, 5}]

D[y^(1/y), y]

The derivative confirms that y^(1/y) is strictly increasing for 1 - Log[y] >= 0,
i.e., 0 < y <= E (this interval contains 1) and strictly decreasing for y > E. Therefore the desired inverse f[x] exists for 0 < x < E^(1/E).

N[E^(1/E)]

The plots confirm that the graphs of f[n,x] split apart near x = 1.4:

\(Plot[{f[6, x], \ f[7, x], \ f[8, x]}, \ {x, 1.3, 1.5}, 

      PlotRange -> {0, 3}]\)

The graphs of f[n,x] seem to split apart near 0. When n is odd, f[n,x] goes to 0 and when n is even f[n,x] goes to 1. However, as n increases, these functions do approach a common limit: the "even" functions shrink down and the "odd" functions creep up.

\(Plot[{f[5, x], f[6, x], f[7, x], f[8, x], 
		f[9, x], \ f[10, x], \ 
        f[11, x], \ f[12, x]}, 
	\ {x, 0, 0.5}, PlotRange -> {0, 1}]\)

We can obtain the graph of the inverse of y^(1/y) by plotting it parametrically.

\(q = \ 
      ParametricPlot[{y^\((1/y)\), y}, {y, 0, E}, 

        PlotStyle -> {RGBColor[1, 0, 0]}, PlotRange -> {0, 3}]\)

Let's compare this to the plots of f[n,x].

Show[p, q]

This analysis proves that f[x] exists for 0 < x < E^(1/E). More work would have to be done to prove that the limit of f[n,x] as n goes to infinity does not exist when
x > E^(1/E). Just because an inverse to y^(1/y) cannot be defined for y > E does not mean that the limit does not exist. The graphs indicate, however, that f[n,x] goes to infinity as n goes to infinity if x > E^(1/E).