infinite – Lucky's Notes

Couple weeks ago my math teacher gave this problem to the class:

$x^{x^{x^{\cdot}}} = 2$

Solve for x

Here the tower of x’s is supposedly infinite (although I have no way of showing that in Latex).

The solution is simple, and it involves a single trick. Notice that the tower is infinite; thus if you remove an x from it, the power doesn’t change. Thus the exponent of the left hand side is equal to the whole left hand side. As the left hand side is equal to the right hand side which is 2, by substitution you get $x^2 = 2$ , and $x = \sqrt{2}$ .

There are a couple of problems in this approach.

Consider the similar equation, $x^{x^{x^{\cdot}}} = 4$ . If we use the same logic, it follows that $x = \sqrt[4]{4}$ or $x = \sqrt{2}$ .

Clearly that is not possible, as the left hand side equals $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}}$ , which cannot both equal 2 and 4. Evaluating it on a computer confirms that 2 is the value of this expression.

Even weirder is the case where $x^{x^{x^{\cdot}}} = 3$ ; when $x = \sqrt[3]{3}$ , the value of the expression converges to 2.4780526802882967 for the tower, where algebra dictates it should be 3.

What is going on here?

Introducing the Product Log function

The Product log function (sometimes called the Lambert W function or Omega function is an interesting function. It is usually written as $W(x)$ , with this definition:

$y = xe^x \Longleftrightarrow W(y)=x$

Compare this with the natural log:

$y = e^x \Longleftrightarrow \ln(y) = x$

The Product log function allows you to solve some interesting problems that can’t be solved with normal logarithms:

Example 1

$\begin{array}{rcl} 3x &=& 3^x \\ \frac{3x}{3^x} &=& 1 \\ 3x \cdot e^{-x \ln 3} &=& 1 \\ x \cdot e^{-x \ln 3} &=& \frac{1}{3} \\ -x \ln 3 \cdot e^{-x \ln 3} &=& \frac{-\ln 3}{3} \\ W(\frac{-\ln 3}{3}) &=& -x \ln 3 \\ x &=& -\frac{W(\frac{-\ln 3}{3})}{\ln 3} \end{array}$

Example 2

$\begin{array}{rcl} x^x &=& 100 \\ e^{x \ln x} &=& 100 \\ x \ln x &=& \ln 100 \\ e^{\ln x} \cdot \ln x &=& \ln 100 \\ W(\ln 100) &=& \ln x \\ x &=& e^{W(\ln 100)} \end{array}$

So the pattern is to manipulate the equation into $z e^z = k$ , where $k$ does not contain the variable. From there, $W(k) = z$ , and we can solve it because the variable is in $z$ .

A closed form expression

$x^{x^{x^{\cdot}}} = -\frac{W(-\ln x)}{\ln x}$

Similar to the two examples, we start by solving $y = x^y$ for $y$ :

$\begin{array}{rcl} y &=& x^y \\ \frac{y}{x^y}&=&1 \\ y \cdot e^{-y \ln x} &=& 1 \\ -y \ln x \cdot e^{-y \ln x} &=& -\ln x \\ W(-\ln x) &=& -y \ln x \\ y &=& -\frac{W(-\ln x)}{\ln x} \end{array}$

For the infinite tower scenario, let $y = x^{x^{x^{\cdot}}}$ . Then, $y = x^y$ , since the tower is infinite already. From what we just did, $y = -\frac{W(-\ln x)}{\ln x}$ .

Let’s use this formula to compute $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}}$ :

$\begin{array}{rcl} \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}} &=& -\frac{W(-\ln \sqrt{2})}{\ln \sqrt{2}} \\ &=& \frac{\ln 2}{\ln (2^{\frac{1}{2}})} \\ &=& 2 \end{array}$

Why is it that $W(- \ln \sqrt{2}) = - \ln 2$ ? $-\ln \sqrt{2} = -\frac{1}{2} \ln 2$ , so $W(-\frac{1}{2} \ln 2) = -\ln 2$ because $-\ln 2 \cdot e^{-\ln 2} = -\frac{1}{2} \ln 2$ .

Similarly this rule applies to other numbers: $W(-\frac{1}{n} \ln n) = -\ln n$ .

Limitations of the W function

The plot of $y = W(x)$ looks like this:

Here, the function ‘stops’ at the point $(-\frac{1}{e}, -1)$ , because there is no value $y$ when $y e^y < -\frac{1}{e}$ .

So if $W(x)$ is only meaningful when $x \ge -\frac{1}{e}$ , then in the previous closed form expression $-\ln x \ge -\frac{1}{e}$ , which means $x \le e^{\frac{1}{e}}$ .

This is easily imagined: $5^{5^{5^{\cdot}}} = \infty$ and there is no solution to $y = 5^y$ . So the graph of $y = -\frac{W(-\ln x)}{\ln x}$ looks like this:

Surprisingly, the formula does not fail for $\sqrt[3]{3}$ , as $-\frac{W(-\ln \sqrt[3]{3})}{-\ln \sqrt[3]{3}}$ gives the correct answer of 2.478. Where I said previously that $W(-\frac{1}{n} \ln n) = -\ln n$ , this only applies when $n \le e$ ; otherwise the two diverge.

Recap

The expression $x^{x^{x^{\cdot}}}$ is only valid for $x \le e^{\frac{1}{e}}$ . The value of the expression can never exceed $e$ (unless of course it’s infinity).

There is also a lower bound for x: it doesn’t work unless $x > e^{-e}$ ; I didn’t explain the reasoning in this post.

Lucky's Notes

Notes on math, coding, and other stuff

Tag: infinite

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