I've been tasked with coming up with exam questions for a high school math contest to be hosted at my university. I offer the following equation, $$\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}=2$$ and ask for the solution for $x$.

Here's what I attempted so far. The first utilizes some pattern recognition, but it gives me two solutions (only one of which is correct). $$\begin{align*} \sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}&=2\\ \sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}&=4-x\\ \sqrt{x+\sqrt{x-\cdots}}&=x-(4-x)^2\\ 2&=x-(4-x)^2&\text{(from line 1)}\\ (x-6)(x-3)&=0 \end{align*}$$ $x=6$ is the extraneous solution. Where did I go wrong, and how can I fix this?

I know there's a closed form for non-alternating nested radicals $\sqrt{n+\sqrt{n+\cdots}}$ and $\sqrt{n-\sqrt{n-\cdots}}$, but I can't seem to find anything on alternating signs.

  • $\begingroup$ Nothing went wrong anywhere. Your implications lead from the top to the bottom, the proof should be in reverse. It's just unfinished. $\endgroup$ – user2345215 Jan 20 '15 at 19:05
  • $\begingroup$ But with $x=6$, I find the LHS to converge to $2.79129\ldots$. How would one approach the equation so as to only get a solution of $x=3$? $\endgroup$ – user170231 Jan 20 '15 at 19:09
  • $\begingroup$ Just a sanity check: Do you know what $\sqrt{x+\sqrt{x-\ldots}}$ actually means? How is it defined? $\endgroup$ – user2345215 Jan 20 '15 at 19:12
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    $\begingroup$ You need $x<4$ for the positive square root to be valid. $\endgroup$ – Joffan Jan 20 '15 at 19:12
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    $\begingroup$ In the second line, ......$=4-x$. The LHS is a square root which is positive. So $x<4$. After you take the square $(4-x)^2$, you introduced an extra root, which should be rejected at the end. $\endgroup$ – velut luna Jan 20 '15 at 19:13

The problem with that kind of expression with nested radicals is that the rigorous meaning of the "$\ldots$" at the end is not always clear (and I presume that's what some people were trying to point out in the comments).

In this case, I think you are considering the sequence given by $x_1 = \sqrt{x}$ and $x_{n+1} = \sqrt{x + \sqrt{x - x_n}}$ (for some fixed $x \geq 0$) and asking for values of $x$ for which $\lim_{n \to \infty} x_n = 2$.

The first thing to be worried about here is whether it's safe to take $\sqrt{x - x_n}$, that is, whether $x_n \leq x$. For $0 < x < 1$ everything goes south as $x_1 = \sqrt{x} > x$. For $x > 1$, $x_1 = \sqrt{x}$ is ok, but $x_2 = \sqrt{x + \sqrt{x - \sqrt{x}}}$ can be a problem if $x$ is small (try $x = 1.1$); to avoid these issues, let us assume $x \geq 2$, when clearly $x_n \leq \sqrt{x + \sqrt{x}} < x$.

Secondly, does $\lim_{n \to \infty} x_n$ always exist if $x \geq 2$? Yes it does, because the sequence is clearly increasing (why?) and bounded above by $\sqrt{x + \sqrt{x}}$ (why?).

Since the limit exists, we can call it $L = L(x)$: it satisfies the inequality $\sqrt{x} \leq L \leq \sqrt{x + \sqrt{x}}$ and the equation $L = \sqrt{x + \sqrt{x - L}}$, so that $L^4 - 2L^2x + L + x^2 - x = 0$.

Which values of $x$ could possibly have $L(x) = 2$? Well, they must satisfy $x^2 - 9x + 18 = 0$, so that $x = 3$ or $x = 6$. On the other hand, since $x$ also must satisfy $\sqrt{x} \leq 2$ (or $x \leq 4$), the solution $x = 6$ can be discarded.

What about $x = 3$, can it be proclaimed "the winner"? Not quite yet, we must verify that $L(3) = 2$: for that, we check that there are no other viable candidates for $L(3)$. To see this, notice that $L(3)$ must be a root of $L^4 - 6L^2 + L + 6 = (L+1)(L-2)(L^2 + L - 3)$. Since $2$ is the only root satisfying $\sqrt{3} \leq 2 \leq \sqrt{3 + \sqrt{3}}$, we can safely say $L(3) = 2$, and $x = 3$ is the only solution to the original problem.

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    $\begingroup$ +1, Finally someone who understands! (but maybe I shouldn't have been so lazy and should have written an answer myself) $\endgroup$ – user2345215 Jan 20 '15 at 20:18

You know that:

$$\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}=A = 2$$

and hence

$$A = \sqrt{x+\sqrt{x-A}} = 2$$

or equivalently

$$\sqrt{x+\sqrt{x-2}} = 2$$

Clearly, $\sqrt{x-2}$ is well defined when $$x \geq 2. ~~~(1)$$ Then, squaring both side, you get:

$$x+ \sqrt{x-2} = 4 \Rightarrow \sqrt{x-2} = 4-x ~~~(2).$$

Since $\sqrt{x-2} \geq 0$, then also $4-x \geq0$, and hence

$$x \leq 4. ~~~(3)$$

Joining conditions $(1)$ and $(3)$, one obtain the existence set for $x$:

$$x \geq 2 \wedge x \leq 4 \Rightarrow 2\leq x \leq 4. ~~~(4)$$

Going back to $(3)$, we can square both side and we get:

$$x-2 = (4-x)^2 \Rightarrow x-2=16+x^2-8x \Rightarrow x^2-9x+18=0 \Rightarrow $$ $$\Rightarrow (x-3)(x-6) = 0. ~~~(5)$$

The solution of $(5)$ are $x_1 = 3$ and $x_2 = 6$, but according to $(4)$, only $x_1 = 3$ is feasible.

  • $\begingroup$ why downvote? Please, don't hide behind the hood!!! $\endgroup$ – the_candyman Jan 20 '15 at 19:39
  • $\begingroup$ Because you didn't check that $3$ actually is a solution. You can't just assume it out of the blue that there's a solution. For example, $x^2+1=0$ has no real solution. And this is especially needed when working with limits. $\endgroup$ – user2345215 Jan 20 '15 at 20:11
  • $\begingroup$ Ok... when $x=3$, then: $$\sqrt{3+\sqrt{3-2}} = \sqrt{3+\sqrt{1}} = \sqrt{4} = 2$$. When $x=6$, then:$$\sqrt{6+\sqrt{6-2}} = \sqrt{6+\sqrt{4}} = \sqrt{8} \neq 2$$ $\endgroup$ – the_candyman Jan 20 '15 at 20:13
  • $\begingroup$ Not that, you have to check that $\sqrt{3+\sqrt{3-\ldots}} = 2$. You only know that if this holds, then $\sqrt{3+\sqrt{3-2}}=2$, but that's not enough! $\endgroup$ – user2345215 Jan 20 '15 at 20:14
  • $\begingroup$ Ok, indeed the way I solved the problem is like searching for the fixed point of the discrete time map $$y_{n+1} = \sqrt{x+\sqrt{x-y_n}}$$. That is: $$y = \sqrt{x + \sqrt{x-y}}.$$ We want that $y=2$ is the fixed point and this happens only when $x=3$. This point converges if $\left|\frac{\partial y_{n+1}}{\partial y_{n}}\right| < 1$ evaluated for $y_n = y = 2$ and $x=3$. One get that $$\left.\frac{\partial y_{n+1}}{\partial y_{n}}\right|_{x=3, y_n=2}=-0.125,$$ so $y=2$ is attractive when $x=3$. Hence, the limit converges to $2$ when $x=3$. $\endgroup$ – the_candyman Jan 20 '15 at 20:33

In the second line, ......$=4-x$. The LHS is a square root which is positive. So $x<4$. After you take the square $(4-x)^2$, you introduced an extra root, which should be rejected at the end.

  • $\begingroup$ Still not finished. What if $x=3$ can be rejected too by some arcane manipulations? $\endgroup$ – user2345215 Jan 20 '15 at 19:17
  • $\begingroup$ 3<4, no need to reject. $\endgroup$ – velut luna Jan 20 '15 at 19:19
  • $\begingroup$ You are missing my point. I didn't say that this rejects it, but what if you do more manipulations, subtract something, square, ... and it follows by other means that $x\ne 3$? $\endgroup$ – user2345215 Jan 20 '15 at 19:21
  • $\begingroup$ You are working from the beginning by A --> B --> C --> D .....If you've done nothing wrong mathematically, it's impossible that the correct solution be ruled out from your final answer!! $\endgroup$ – velut luna Jan 20 '15 at 19:34
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    $\begingroup$ user2345215, you are right $\endgroup$ – velut luna Jan 20 '15 at 20:09

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