In math forums you can find various problems like the following.

\(\sqrt{12+\sqrt{12+\sqrt{12 + \ldots}}} = x\)

Usually with the remark „Solve this!“.

This is a very old kind of presenting and doing mathematics from the centuries before the invention of precise limits. Of course, the old ones were also great ones. They knew most of the time (but not always) what this limit exactly means. Their computations, however, were often based on assumptions which were checked only by experience, not by rigid reasoning. And sometimes they were simply wrong.

The problem above is „solved“ by observing that one step more yields the equation

\(\sqrt{12+x} = x.\)

As a next dubious step, this is squared and yields

\(12+x = x^2 \quad \Longleftrightarrow \quad x=-3 \text{ or } x=4\)

Since negative values do not make sense, the „solution“ is x=4.

Let us start with trying to precisely formulate the meaning of the left hand side of our problem. Clearly, it is impossible to follow the three dots „…“ infinitely. So, on a computer, we have to stop at some point. We face the strange fact, that we have to think from left to right, but compute from right to left. E.g.,

\(\sqrt{12+\sqrt{12 + \sqrt{12}}} = 3.9915\ldots\)

That, of course, is what the old ones did, and since the „solution“ presented above agrees with this computation, it should be correct.

The problem can easily be formalized as follows, Let us imagine the actual process of the computation, which goes from right to left, from the inner square root towards the outer square roots. Thus we apply the function

\(f(x) = \sqrt{12+x}\)

over and over again, starting from the value

\(x_0 = \sqrt{12}\)

Formally, for all n,

\(x_{n+1} = f(x_n)\).

If this converges and if f is continuous, then it must converge against a value x with

\(f(x) = x\)

This is called a fixed point of f. So much about the first semester theory of such iterations. We have also learned, that the above way to derive x=4 is correct under the assumption that the sequence converges.

If that assumption is not met the result can be completely wrong. Here is an example.

\(2 \cos(2 \cos(2 \cos(2 \ldots))) = x\)

Let us try our „solution“. We proceed as above and get

\(2 \cos(x) = x.\)

This does in fact have a unique solution

\(x \approx 1.0298\ldots\)

But the convergence is not going to happen, no matter, where you start.

>showlarge iterate("2*cos(x)",1,20)' 1 1.0806 0.94159 1.17701 0.767382 1.43946 0.261915 1.93179 -0.706411 1.52139 0.0987668 1.99025 -0.814529 1.37242 0.39415 1.84665 -0.54473 1.71053 -0.278568 1.9229 -0.689747

The behavior is in fact chaotic. The reason for this is that the derivative of

\(f(x) = 2 \cos(x)\)

in the solution is greater than 1.

So, you got to be careful in mathematics. And you got to get things nailed down to the level of exactness and rigid thinking. Even then, you got to know the trick on how to show the convergence of your sequence.

By the way, the sequence above is not completely chaotic. But the study of it might get difficult.

>x=iterate("2*cos(x)",1,10000); >plot2d(x,>points,style=".."):