Self-Similar Expressions: Additional Practice (Set 2)

Squaring merges $x = \sqrt{2+x}$ with $x = -\sqrt{2+x}$. Test $x = -1$ in the negative branch.

$-\sqrt{2 + (-1)} = -\sqrt{1} = -1 \ \checkmark$

So $-1$ solves $x = -\sqrt{2 + x}$, not the original, which is why it is extraneous.

Let $x = 0.\overline{81}$. Then $100x = 81 + x$, so $99x = 81$.

$x = \frac{81}{99} = \frac{9}{11}$

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

Multiply through by $x$ and move all terms to one side.

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

The positive root is $\varphi = \tfrac{1 + \sqrt{5}}{2}$.

The first three settle on $\tfrac{2}{3}$, $3$, and $2$. The last has ratio $2$, so its terms grow.

$2, \ 6, \ 14, \ 30, \ \ldots \ \text{grow without bound}$

The running totals never settle, so it has no finite value.

The equation only reports the value an expression must have if it has one, so you must confirm existence.

$\text{partial values close in} \ \Longrightarrow \ \text{a value exists}$

For a geometric series that condition is $|r| < 1$; for a tower it is partial values that rise but stay bounded.

Name it $x$. Then $x = \sqrt{7x}$, so

$x^2 - 7x = 0 \quad\Longrightarrow\quad x(x - 7) = 0.$

The tower exceeds $1$, so $x = 7$. In general $\sqrt{k\sqrt{k\cdots}} = k$.

Writing $S =$ assumes the series settles on a number, which happens only when the terms shrink.

$|r| < 1 \quad\Longrightarrow\quad S = \frac{1}{1 - r}$

If $|r| \ge 1$ there is no $S$ and the equation solves to a false value.

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, $x = 7$.

A decimal expansion pins down a definite point, so the value always exists.

$10^6 x = 142857 + x \quad\Longrightarrow\quad 999999 x = 142857$

Since the value is guaranteed to exist, the self-similar equation reports it correctly, giving an exact fraction.

Set $x = \sqrt{20 + x}$, then square and factor.

$x^2 - x - 20 = 0 \quad\Longrightarrow\quad (x - 5)(x + 4) = 0$

The value is the nonnegative candidate, $x = 5$.

Self-similarity means the whole expression reappears inside itself, so the inner copy can be written as $x$.

$x = \sqrt{2 + x}, \qquad x = 1 + \frac{1}{x}, \qquad S = 1 + rS$

Each is a finite equation obtained by naming the inner copy.

Substitute $r = \tfrac{1}{2}$, which satisfies $|r| < 1$.

$S = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$

So $1 + \tfrac{1}{2} + \tfrac{1}{4} + \cdots = 2$.