Self-Similar Expressions: Practice

Set $x = \sqrt{42 + x}$, then square and factor.

$x^2 - x - 42 = 0 \quad\Longrightarrow\quad (x - 7)(x + 6) = 0$

The value is the nonnegative candidate, $x = 7$.

Self-similarity gives $x = 1 + \tfrac{1}{x}$, so $x^2 - x - 1 = 0$. Apply the quadratic formula and keep the positive root.

$x = \frac{1 + \sqrt{5}}{2} \approx 1.618$

The negative root is rejected because every term of the fraction is positive.

Multiply through by $x$.

$x^2 = x + 1 \quad\Longrightarrow\quad x^2 - x - 1 = 0$

This is the defining equation of the golden ratio.

Let $x = 0.\overline{12}$. The block is two digits, so $100x = 12.\overline{12} = 12 + x$.

$99x = 12 \quad\Longrightarrow\quad x = \frac{12}{99} = \frac{4}{33}$

Reducing $\tfrac{12}{99}$ by $3$ gives $\tfrac{4}{33}$.

Subtract $\tfrac{1}{2}S$ and solve.

$S - \tfrac{1}{2}S = 2 \quad\Longrightarrow\quad \tfrac{1}{2}S = 2 \quad\Longrightarrow\quad S = 4$

This matches $\tfrac{a_1}{1 - r} = \tfrac{2}{1 - 1/2} = 4$.

Self-similarity gives $x = 2 + \tfrac{1}{x}$, so $x^2 - 2x - 1 = 0$. The quadratic formula gives

$x = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}.$

The fraction is positive, so $x = 1 + \sqrt{2} \approx 2.414$.

The algebra is correct, so the error is in writing $S =$ at all. The running totals $1, 3, 7, 15, \ldots$ grow without bound.

$S = 1 + 2S \quad\Longrightarrow\quad -S = 1$

Since the series settles on no number, the value $-1$ is meaningless.

The continued fraction $\varphi = 1 + \tfrac{1}{1 + \tfrac{1}{\cdots}}$ contains a copy of itself below the first $1 +$.

$\varphi = 1 + \frac{1}{\varphi}$

So the golden ratio equals one plus its own reciprocal.

Subtract $\tfrac{1}{3}S$ and solve.

$S - \tfrac{1}{3}S = 1 \quad\Longrightarrow\quad \tfrac{2}{3}S = 1 \quad\Longrightarrow\quad S = \frac{3}{2}$

This matches $\tfrac{1}{1 - 1/3} = \tfrac{3}{2}$.

Let $x = 0.\overline{6}$. Then $10x = 6 + x$, so $9x = 6$.

$x = \frac{6}{9} = \frac{2}{3}$

So $0.\overline{6} = \tfrac{2}{3}$.

Here the whole fraction $x$ appears in the denominator, so $x = \tfrac{1}{1 + x}$, giving $x^2 + x - 1 = 0$.

$x = \frac{-1 + \sqrt{5}}{2} \approx 0.618$

The value is positive, so the negative root is rejected. This is $\tfrac{1}{\varphi}$.

The whole tower is a principal square root, so its value cannot be negative.

$x = \sqrt{2 + x} \ \ge \ 0$

The negative candidate solves the squared equation but not the original, so it is discarded.