Here’s a strange math problem that I encountered as an undergraduate:

What is the solution to the following equation?

[Note: The order of exponents here is such that the upper ones are taken first.  For example, you should read $2^{3^2}$ as $2^{(3^2)} = 512$ and not as $(2^3)^2 = 64$.]

As it happens, there’s a handy trick for solving this equation, and that’s to use both sides as an exponent for $x$.  This gives

From the first equation, though, the left hand side is just $2$.  So now we’re left with simply $x^2 = 2$, which means $x = \sqrt{2}$.

Not bad, right?  Apparently the conclusion is that

Where things get weird is when you try to solve an almost identical variant of this problem.  In particular, let’s try to solve:

We can do the same trick as before, using both sides of the equation as an exponent for $x$, and this gives

so that we’re left with $4 = x^4$.  The solution to this equation is, again, $x = \sqrt{2}$.

But now you should be worried, because apparently we have reached the conclusion that

So which is it?  What is the correct value of $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$?  Is it 2, or is it 4?

## Yes, but which one is the real answer?

Maybe in the world of purely abstract mathematics, it’s not a problem to have two different answers to a single straightforward mathematical operation.  But in the real world this is not a tolerable situation.

The reasoning above raised a straightforward question — what is $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$? — and provided two conflicting answers: $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = 2$ and $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = 4$.  Both of these equations are correct, but which one should you really believe?

Suppose that you don’t really believe either of those two equations (which, at this point, you probably shouldn’t), and you want to figure out for yourself what the value of $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$ is.  How would you do it?

One simple protocol that you could do with a calculator or a spreadsheet is this:

• Make an initial guess for what you think is the correct value of $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$.
• Take your guess and raise $\sqrt{2}$ to that power.
• Take the answer you get and raise $\sqrt{2}$ to that power.
• Repeat that last step a bunch of times.

Try this process out, and you will almost certainly get one of two answers:

If you initially guessed that $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$ was any number less than 4, then you will arrive at the conclusion that $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = 2$.

If your initial guess was something larger than 4, though, you will instead get to the conclusion that $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = \infty$.

The situation can be illustrated something like this:

Only if your initial guess was exactly 4, and if your calculator gave you the exact correct answer at every step, will you ever see the solution $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = 4$.  A single error in the 16th decimal place anywhere along the way will instead lead you to a final answer of either $2$ or $\infty$.

In this sense $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = 2$ is a much better answer than $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = 4$.  The latter is true only in a hypothetical world of perfect exactness, while the former is true even if your starting conditions are a little uncertain, or if your calculator makes mistakes along the way, or (most importantly) there’s some small additional factor that you haven’t taken into consideration.

## Scientists adjudicate between mathematical realities

For the most part, this has been a silly little exercise.  But it actually does illustrate something that is part of the job of a physicist, or anyone else who uses math as a tool.  Physicists spend a lot of time solving equations that describe (or are supposed to describe) the physical world.  But finding a solution to some equations is not the end of process.  We also have to check whether the solution we came up with is meaningful in the real world, which is full of inexactnesses.  For example, the equation that describe the forces acting on a pencil on my desktop will tell me that the pencil can be non-moving either when lying on its side or when balanced on its point.  But only one of those two situations really deserves to be called a “solution”.

So, as for me, if you ask me whether $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = 2$ or $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = 4$, I’ll go with 2.

Because, as Napoleon the pig understood, some equations are more equal than others.

1. August 3, 2014 6:43 am

Very interesting problem. I also prefer the answer “2”. Below I try to explain why.

You look for the solutions of a fix point problem T(y) = y, where the operator T(y) is defined as sqrt(2)^y. We know that the solution to the original problem (i.e. what is the value of the infinite sequence of sqrt(2) exponents) solves the fixed point problem. However, we don’t know, if it is a unique solution. Actually, you showed that it has at least two solutions: 2 and 4. It seems that although the fixed point property is a necessary condition, it is not sufficient to select the solution of the original problem.

We can express the solution of the original problem in a different way, as a limit of a finite sequence of sqrt(2) exponents, where the number of elements of the sequence goes to infinity. We can denote the expression, which limit we are calculating, as (T^n)(y), where (T^2)(y) = T(T(y)) , (T^3)(y) = T(T(T(y))) and so on, and we are taking a limit as n -> infinity. If the limit exists, it will be equal to the solution of the original problem. Above you showed that this limit is equal to 2 – it’s like solving the fixed point problem from the initial guess equal sqrt(2).

August 3, 2014 11:46 am

Yes, there is something nice about the fact that the initial guess sqrt(2), which is in some sense the most natural choice, gives the answer 2.

And you’re right to point out that in the technical language, both 2 and 4 are called “fixed points”. In this case 2 is a “stable fixed point” and 4 is an “unstable fixed point.”

2. January 5, 2015 6:19 pm

I first became aware of this paradox in high school, and spent many an afternoon figuring it out, armed with nothing but a slide rule. It is a favorite of mine, because it once taught me a great deal.

Two comments: First, in order to get from the original equation with infinite exponents to

x^2 = 2, all you have to do is notice that exponent on the LHS is the same as the whole LHS. It has one less “x” in it, but one less than infinity is still infinity. So the infinite exponent is exactly the same as the infinite LHS.

This implies that IF the equation has a solution, then it is legitimate to substitute the RHS for the exponent of the LHS, and you get X^2 = 2.

In fact, the equation DOES have a solution, and therefore the trick works, and the answer that you get (sqrt(2)) is a solution to the infinite equation.

When you try the same thing with the tree of exponents = 4, you substitute to get x^4=4, assuming again that the equation DOES have a solution. But in this case, the equation does not have a solution, and that is why your manipulation fails.

This leads to the question, for what values of x does the equation have a solution? If it does have a solution, clearly the solution is the x root of x.

If you plot the function f(x)=xth root of x, you find some nice surprises. When x=0, the function is undefined; but when x is very close to zero, the function is a small number raised to a high power, which approaches zero very smoothly. Then the curve rises slowly and reaches a maximum. You can do a little calculus to find the maximum, or you can find it with trial and error. The answer (as I remember from high school) is that the maximum occurs when x=e, and the maximum is, of course, the eth root of e=1.44.

After that value, the function comes gradually back down, and approaches 1 when
x->infinity. Because the function is doubled over, it does not have an inverse over its entire range. If you take just the portion to the left of the maximum, then the infinite tree of powers of x is indeed an inverse function; but I know of no way to express the inverse function for x to the right of the eth root of e.

January 5, 2015 6:44 pm

Thanks for the great comment. I hadn’t realized yet the connection to e.

• June 19, 2015 8:40 am

I programmed a computer to calculate this function in 1967, and won a first in Mathematics for it at the science fair. Forty years later, I ran into Steiner’s problem.