Self-Similar Expressions: Additional Practice (Set 1)

Set $x = \sqrt{12 + x}$, then square and factor.

$x^2 - x - 12 = 0 \quad\Longrightarrow\quad (x - 4)(x + 3) = 0$

The value is the nonnegative candidate, $x = 4$.

Subtract $1$ from both sides of the self-similar identity.

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

So the reciprocal of the golden ratio is exactly one less than the golden ratio, about $0.618$.

Set $x = \sqrt{30 + x}$, then square and factor.

$x^2 - x - 30 = 0 \quad\Longrightarrow\quad (x - 6)(x + 5) = 0$

The value is the nonnegative candidate, $x = 6$.

The equation only reports the value an expression must have if it has one.

$1 + 2 + 4 + \cdots \ \Longrightarrow \ S = -1 \ \text{(false)}$

Only after confirming the expression settles on a number is the reported value trustworthy.

The repeating block $5$ is one digit long, so use a factor of $10$.

$10x = 5.\overline{5} = 5 + x$

Then $9x = 5$, so $x = \tfrac{5}{9}$.

The algebra is correct, but there is no number to name.

$1, \ 3, \ 7, \ 15, \ \ldots \ \text{grow without bound}$

Since the series settles on nothing, every consequence of writing $S =$ is worthless.

Let $x = 0.\overline{18}$. Then $100x = 18 + x$, so $99x = 18$.

$x = \frac{18}{99} = \frac{2}{11}$

Reducing by $9$ gives $\tfrac{2}{11}$.

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

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

The negative root is rejected because the fraction is positive. This is the golden ratio $\varphi$.

Subtract $\tfrac{1}{5}S$ and solve.

$\tfrac{4}{5}S = 1 \quad\Longrightarrow\quad S = \frac{5}{4}$

This matches $\tfrac{1}{1 - 1/5} = \tfrac{5}{4}$.

Under the outer radical sits $k$ times a copy of the whole tower, which is $x$.

$x = \sqrt{kx}$

Squaring gives $x^2 = kx$, so $x = k$ (rejecting $x = 0$).

A sum exists only if the running totals close in on one number.

$1, \ 0, \ 1, \ 0, \ \ldots \ \text{never settle}$

Since the partial sums bounce forever, the series has no sum, and $S = 1 - S$ giving $\tfrac{1}{2}$ is not valid.

Multiplying by $10$ shifts the digits one place, leaving the repeating tail unchanged.

$10x = 3.\overline{3} = 3 + x$

The tail is a copy of $x$, so $9x = 3$ and $x = \tfrac{1}{3}$.