Self-Similar Expressions

The self-similar move

Start with the tower of twos:

The expression is infinite, so there is no last radical to work back from and no obvious place to begin. Self-similarity hands us a foothold. Everything under the outermost radical, the entire $2 + \sqrt{2 + \sqrt{2 + \cdots}}$, is just $2$ added to another copy of the original tower. Name the tower $x$, and that inner copy is $x$ as well.

Both sides of $x = \sqrt{2 + x}$ are nonnegative, so squaring them is safe. The move is the same one from the previous lesson: raise both sides to a power to free the variable from under the root. Squaring gives

so the candidates are $x = 2$ and $x = -1$. This is exactly the trap the previous lesson set out: squaring is an even power, so it can introduce a candidate the original expression never had. That risk is why you always check. The check here is instant. A principal square root is never negative and every piece of the tower is positive, so $x$ cannot be $-1$. Discard the impostor and the value is

We will confirm at the end of the lesson that this tower really does settle on a number. Once that is confirmed, the $2$ is trustworthy and not the kind of phantom the caveat warns about.

Towers of every size

The move never changes. Only the number under the root does.

Continued fractions and the golden ratio

A continued fraction stacks divisions the way the tower stacked radicals, and it is self-similar in the very same way. Consider

Look under the first division bar. Below the leading $1 +$ sits $1$ divided by another copy of the identical endless fraction, which is again $x$. So the whole tail below the first $1 +$ is $\tfrac{1}{x}$, and the entire expression reads

Multiply through by $x$ to clear the fraction:

This does not factor over the integers, so use the quadratic formula:

Every term of the fraction is positive, so $x$ is positive, which rejects the negative root $\tfrac{1 - \sqrt{5}}{2} \approx -0.62$. The value is

the golden ratio, written $\varphi$. Notice that the equation we solved, $\varphi = 1 + \tfrac{1}{\varphi}$, is nothing more than the self-similarity restated: the golden ratio is precisely the number that equals one plus its own reciprocal.

Repeating decimals, seen fresh

You met repeating decimals in the geometric-series lesson, where each was summed as an infinite series. Self-similarity reaches the same fractions faster, because a repeating decimal already carries a shifted copy of itself. Take

Multiplying by $10$ slides every digit one place to the left:

The digits after the decimal point are untouched by the shift, so what remains is a perfect copy of the original $x$. That is the self-similarity, and it turns the endless decimal into a one-line equation:

When the repeating block is two digits long, shift by two places instead, which means multiplying by $100$.

The infinite geometric series, re-derived

The same move re-derives a formula you already trust. An infinite geometric series with first term $1$ and ratio $r$ is

Factor $r$ out of every term after the first. What remains inside the parentheses is $1 + r + r^2 + \cdots$, a perfect copy of the whole series $S$:

That is the self-similar equation. Solve it for $S$:

This is precisely the infinite-sum formula $\tfrac{a_1}{1 - r}$ from the geometric-series lesson, here with $a_1 = 1$, and the multiply-and-subtract shortcut you saw there was this same self-similarity wearing another costume. For a series that starts at $a_1$ instead of $1$, the identical factoring gives $S = a_1 + rS$ and hence $S = \tfrac{a_1}{1 - r}$.

Look back at what that derivation never checked. It began by writing $S =$, which quietly assumed that the series adds up to a number in the first place. Grant that one assumption and every step after it is airtight. The next section takes that assumption away and shows that the assumption, not the algebra, is the whole game.

When the self-similar equation lies

Every derivation so far ended with a sensible number, which makes it tempting to trust the move blindly. Do not. Run the identical steps on

the geometric series with ratio $2$. Factor $2$ out of every term after the first, and the copy of $S$ appears just as before:

Solve:

A sum of positive numbers has come out negative. Something is badly wrong, yet not a single line of the algebra is false. The factoring, the rearranging, and the solving are all correct. The mistake sits in the very first symbol. Writing $S =$ assumed there was a number to call $S$, and for this series there is not. Its running totals are $1, 3, 7, 15, 31, \ldots$, marching past every bound and never approaching anything. Because that first assumption was empty, every conclusion drawn from it, $S = -1$ included, is worthless.

A subtler case makes the same point without any runaway growth. Grandi’s series

has terms that never grow, so it looks tame. The self-similar move peels off the leading $1$ and finds a copy of the series in what remains, with every sign flipped:

A sum of whole numbers has come out as a half, a value no partial sum could ever equal. Worse, bracketing the terms differently suggests different answers. Grouping them in adjacent pairs gives

while shifting the brackets over by one gives

Three defensible manipulations, three different totals: $\tfrac{1}{2}$, $0$, and $1$. The running totals themselves bounce $1, 0, 1, 0, 1, \ldots$ forever and never settle on anything. When rearranging an expression changes its value, that is the signature of an expression with no single value to change.

Here is the lesson inside the lesson. The self-similar equation tells you the value the expression must have if it has one at all. It cannot tell you whether it has one. That second question, whether the expression settles on a number, has to be answered separately. Only once the answer is yes may you trust the value the equation hands you.

For a geometric series the answer is the condition from the geometric-series lesson: the running totals close in on a number exactly when $|r| < 1$. The reason is that only then does the tail shrink toward zero. With $r = 2$ that condition fails, which is why $S = -1$ was nonsense; with $r = \tfrac{1}{2}$ it holds, which is why $S = \tfrac{1}{1 - 1/2} = 2$ is trustworthy. For the nested radical there is no ratio to check, but a short argument settles it. Compute the tower one layer at a time and watch the partial values climb:

Each value is larger than the one before, because every new layer piles a little more under the roots. Yet none of them can reach $2$: whenever a value is below $2$, the next one is $\sqrt{2 + (\text{something below } 2)} < \sqrt{4} = 2$. That is why once the tower is below $2$ it stays below $2$. A list of numbers that keeps rising while it is held under a fixed ceiling has less and less room to move. That shrinking room makes it believable that the list closes in on a single value, which the equation then pins at exactly $2$. Turning that intuition into an airtight proof is a task for later study, but the picture is reliable here. The continued fractions behave the same way, their partial values closing in from above and below, so the numbers those self-similar equations produced are the honest ones.

Common mistakes