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

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 , and .

There are a couple of problems in this approach.

Consider the similar equation, . If we use the same logic, it follows that or .

Clearly that is not possible, as the left hand side equals , 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 ; when , 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 , with this definition:

Compare this with the natural log:

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

#### Example 1

#### Example 2

So the pattern is to manipulate the equation into , where does not contain the variable. From there, , and we can solve it because the variable is in .

### A closed form expression

Similar to the two examples, we start by solving for :

For the infinite tower scenario, let . Then, , since the tower is infinite already. From what we just did, .

Let’s use this formula to compute :

Why is it that ? , so because .

Similarly this rule applies to other numbers: .

### Limitations of the W function

The plot of looks like this:

Here, the function ‘stops’ at the point , because there is no value when .

So if is only meaningful when , then in the previous closed form expression , which means .

This is easily imagined: and there is no solution to . So the graph of looks like this:

Surprisingly, the formula does not fail for , as gives the correct answer of 2.478. Where I said previously that , this only applies when ; otherwise the two diverge.

### Recap

The expression is only valid for . The value of the expression can never exceed (unless of course it’s infinity).

There is also a lower bound for x: it doesn’t work unless ; I didn’t explain the reasoning in this post.

hi! nice post on this curious expression. I was wondering why x has to be greater than e^{-e} can you send me a reference for the proof, please? I can’t figure it out thanks!

LikeLike

Hi. Unfortunately I don’t have any proof for that statement, but basically what happens is it doesn’t converge because the result ‘oscillates’ between two different values if x is smaller than a certain amount.

LikeLike

So, why doesn’t the method your teacher used work? I see the contradiction, but is there a better explanation?

LikeLike

The maximum value of an infinite tower is e, so there is no x such that a tower of x’s evaluates to 3 or 4.

The original argument shows (correctly) that tower(x) = 4 implies x = 2. But it doesn’t show that tower(2) = 4.

LikeLike

One thing I’ve never figured out – if x^x^x….^x = 2 (= y), then if you raise both sides to the power x, you get x^x^x….^x^x (=y) = 2^x, because the left side is unaffected, as its an infinite series. Therefore, you have now shown that y = 2 and y = 2^x, implying x=1. Why doesn’t this work?

LikeLike

Because the tower of x’s by convention means the limit of the sequence u(0) = x, u(i+1) = x^u(i), which converges for x upto 1.44-ish. Your argument would apply to the alternative sequence u(0) = x, u(i+1) = u(i)^x, which does not converge for any x > 1.

I suggest you put x = sqrt(2) and calculate the first few terms of each of the two sequences – it will quickly become obvious.

LikeLike

Hi. Thanks for the reply. Yeah, I spotted that too, shortly after my post. It’s interesting how powers are not associative (i.e. x^(x^x) does not equal (x^x)^x), unlike all the other operators everyone is used to (multiplication, addition, etc). I’d incorrectly assumed that because there were no brackets in the original tetration equation, that it would, but I guess that shows the dangers of making assumptions. Still, a bit ambiguous to write the problem x^x^x^.. without brackets given associativity doesn’t hold. Anyway, thanks for your reply

LikeLike