## Monday, April 13, 2015

### 4=2 right?

Long time ago I was intrigued by a particular equation that I found in my Real Analysis book. It was not the traditional equation of a complicated polynomial expression equal to a number, rather it involved the idea of having a never ending expression:

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

I think the first challenge is to understand what the infinite exponents really mean. Actually, understanding that is the ket to find the solution of the equation.

It is surprising that such a weird looking expression has a very nice solution: $x=\sqrt{2}$.

As I said before, the mystery disappears when we actually understand what the equations really mean. In order to define an infinite exponential, we must use the language of limits.

First, lets define the sequence $a_n$ as follows:

$$a_1=x\,,$$
$$a_n=x^{a_{n-1}}\,.$$

Then, the equation really means to have

$$\lim_{n\to\infty}a_n=2\,.$$

Notice that defining the recurrence equation in the opposite order leads just to $a_n=x^{x^n}$.

Hence, if we know that the limit of the sequence is 2, we should have then

$$2=\lim_{n\to\infty}a_n=\lim_{n\to\infty}x^{a_{n-1}}=x^2\,,$$

which gives the desired solution.

We can notice that there was noting special about 2, thus we are able to reproduce this argument for any initial value, that is, if

$$x^{x^{x^{\dots}}}=r\,,$$

then this will have solution $$x=r^{1/r}$$.

Now, an interesting fact happens when we play a bit with this reasoning and consider the cases $r=2$ and $r=4$. The solution for $r=2$ as discussed above, is $x=\sqrt{2}$, and the solution for $r=4$ is $x=4^{1/4}=\sqrt{2}$... it seems we just proved that $2=4$.

If you pay close attention to this argument, we didn't divide by zero, like most of the mundane "proofs" that $1=0$ that you can find on the internet. We didn't take any wrong step, everything is well defined, everything exists, or does it?

We can pose this question in a slightly different manner, we have the equation

$$g(x)=r\,,$$

and we solved it by finding the inverse of $g$ to be

$$x=f(r)=r^{1/r}\,.$$

Going back to our basics in set theory (or precalculus), we know that a function has an inverse where it is bijective, so $f^{-1}=g$ or $g^{-1}=f$ only when these functions are bijective. Here is the graph of $f(r)$,

when $r$ goes to infinity, the graph keeps decreasing, with horizontal asymptote 1.

I graphed $f(r)$ since it is easier to look at this rather than plotting $g(x)$ (the computer would have hated me), but I remember that in my precalculus years, they told me that the graph of an inverse is really the reflection with respect to the line $y=x$, so we do have the graph of $g(x)$

Therefore we can see that the function $g(x)$ is actually a function up to the tip of the "nose", which is the maximum of $f(r)$. Using the first derivative test on $f(r)$ we have that this happens for $r=e$ and $x=e^{1/e}$. This means that $g(x)=r$ does not have any solutions for \$r

Hence out false proof has a more elegant reason for failing: injectivity.