Self-Similar Expressions: Free Response

Everything under the first radical is $72$ plus another copy of the exact same endless tower, so that inner copy is the same value $x$. Replacing it by $x$ turns the infinite expression into one finite equation.

$x = \sqrt{72 + x}$

Squaring removes the radical and leaves the $72$ untouched.

$x^2 = 72 + x \quad\Longrightarrow\quad x^2 - x - 72 = 0 \quad\Longrightarrow\quad (x - 9)(x + 8) = 0$

The two candidates are $x = 9$ and $x = -8$.

A principal square root is never negative, and every layer of this tower is built from a positive number added under more positive numbers, so the whole expression can only be nonnegative.

$x = \sqrt{72+x} \ \ge \ 0$

That rules out $x=-8$ on structural grounds alone, not because it fails some outside check: a negative number can never equal a principal square root in the first place. The genuine value is $x = 9$, and indeed $\sqrt{72+9}=\sqrt{81}=9$ confirms it.

Below the very first $4+$ sits $1$ divided by the rest of the continued fraction, and that rest is built exactly like the whole thing: the same $4$, the same endless nesting. It is a perfect copy of the entire expression, so it is the same value $x$, making the tail $\frac{1}{x}$.

$x = 4 + \frac{1}{x}$

That is Equation 1.

Equation 2, $x = 4 + \frac14$, treats the deeper copy of the fraction as though it were simply the number $4$ again rather than the full endless value $x$; it stops the recursion after one layer instead of letting it run forever.

$4 + \frac14 \ne 4 + \frac{1}{x} \quad \text{unless } x = 4$

Equation 3, $x = \frac{1}{4+x}$, puts the leading $4$ inside the fraction rather than outside it, which describes a continued fraction that never has a $4+$ sitting out front, not this one.

Multiply the correct equation by $x$ to clear the fraction.

$x^2 = 4x+1 \quad\Longrightarrow\quad x^2-4x-1=0$

The quadratic formula gives

$x = \frac{4\pm\sqrt{20}}{2} = 2\pm\sqrt5.$

Every term of the continued fraction is positive, so its value must be positive, which rules out $2-\sqrt5 \approx -0.24$ on structural grounds. The genuine value is $x=2+\sqrt5 \approx 4.24$.

The repeating block is the single digit $8$, so multiply by $10$ to slide the digits one place.

$10x = 8.8888\ldots = 8 + x$

Subtract $x$ from both sides and solve.

$9x = 8 \quad\Longrightarrow\quad x = \frac{8}{9}$

Every one of the self-similar expressions elsewhere in this lesson had to earn its value: a tower needed its partial values to climb toward a ceiling, and a series needed the ratio's absolute value below $1$. A decimal is different, because writing down digits after a decimal point already names one exact point on the number line before any algebra begins.

$x = 0.8888\ldots$

Because that existence is guaranteed from the start, multiplying by a power of ten and solving is safe every time, with no convergence check required.

Collect the $S$ terms and solve.

$S - 4S = 1 \quad\Longrightarrow\quad -3S = 1 \quad\Longrightarrow\quad S = -\frac13$

This is exactly the value Devon reports, so every step of the algebra is correct.

Add the terms one at a time.

$1, \quad 1+4=5, \quad 5+16=21, \quad 21+64=85$

Each new term is four times the one before it, so the running total grows faster with every step and never levels off or approaches a fixed ceiling.

Part A shows every algebraic step from $S=1+4S$ to $S=-\frac13$ is valid, so the flaw cannot be in the manipulation. Part B shows the partial sums climb without bound.

$1,\ 5,\ 21,\ 85,\ \ldots \ \text{grow without bound}$

Devon's error is committed before any algebra begins, at the moment of naming $S$. The guarded, correct condition is that the self-similar equation for a geometric series gives a trustworthy value only when the ratio's absolute value is below $1$; here the ratio is $4$, so the condition fails, and $-\frac13$ is not a real sum.

Everything under the first radical is $110$ plus another copy of the whole tower, so

$x = \sqrt{110+x}.$

Squaring gives $x^2-x-110=0$, which factors as $(x-11)(x+10)=0$, with candidates $x=11$ and $x=-10$. A principal square root built from positive numbers can never be negative, which rules out $x=-10$ on that structural ground alone. The genuine value is $x=11$, and $\sqrt{110+11}=\sqrt{121}=11$ confirms it.

Factor the ratio $5$ out of every term after the first, exposing a copy of the whole series.

$S = 1 + 5\left(1+5+25+\cdots\right) = 1+5S$

Solve for $S$.

$S-5S=1 \quad\Longrightarrow\quad -4S=1 \quad\Longrightarrow\quad S=-\frac14$

Add the terms one at a time.

$1, \quad 1+5=6, \quad 6+25=31, \quad 31+125=156$

Each total grows by more than the one before it, with no sign of leveling off. Since the running totals climb without any bound, the series has no genuine value, so the candidate found in part B is not the sum of anything real.

Part A supplies a case where the claim holds: the nested radical settles, its partial values climbing toward a ceiling, and $x=11$ is genuinely trustworthy. That alone does not make the claim true in general.

Parts B and C together supply the counterexample. The series' self-similar equation reports a candidate, but its partial sums climb without any bound, so the series has no value for that candidate to be. One infinite self-similar expression whose equation reports a number it does not actually settle on is enough to refute a claim about EVERY such expression.

$x=11\ \text{(settles)} \qquad S=-\tfrac14\ \text{(does not settle)}$

The honest, guarded form of the claim is that a self-similar equation gives the value an infinite expression must have if it settles on a number, never a guarantee that it does; that has to be checked separately, by the ratio for a series or by a bounded, climbing pattern for a tower.