Infinite nested radical and fractions

Question

By dealing with infinite radicals I'm pretty sure that we always choose the positive roots, as an example:

$\displaystyle\sqrt{1+\displaystyle\sqrt{1+...} }$

$\displaystyle\sqrt{1+x}=x$

$x^2=x+1$

This gives us two roots $φ$ and $\frac{-1}{φ}$ Since the other is negative so the infinite radical is just $φ$

The argument holds for infinite fractions, denominator must be $>0$

But what if it was a complex number, We know that there's not inequalities with complex numbers how we will deal with that?

But what about the radical and infinite fractions the gives us two values and both of them are either positive or $>0$

How do we know what we should choose in analytical way?

Example: $(x-1)(x-2)=0$

$x^2-3x+2=0$

We know $x>0$ then divide both sides by x

$x-3+\frac{2}{x}=0$

$x=3-\frac{2}{x}$

$x=3-\frac{2}{3-\frac{2}{x}}$

$x=3-\frac{2}{3-\frac{2}{...}}$

So I have convert it into infinite nested fractions, But how can I know that if this is 1 or 2?

The same question for radicals

Another example

$2^x=x^2$

For tetrating $√(2^{√(2{√(2^{..})}}$

Can we use calculus to know?

Answer 1

The thing is... it's both! If you substitute $x=1$, everything will collapse to give you a final answer of $x=3-2=1$. If you substitute $x=2$ instead, you'll end up with $x=3-\frac{2}{2}=3-1=2$.

As for why this happens, just think about it: you start with the equation $x^2-3x+2=0$ and you know that the roots are $1$ and $2$. You then proceed to try and construct a continued fraction representing these roots, but since the roots are integers, their "continued fraction" representations are just themselves.

The ambiguity in this case (which root does it evaluate to?) comes from the fact that the original equation has two roots. Continued fractions can only be said to unambiguously equal a number when that number is the only root of whatever equation the continued fraction is derived from (that, and the condition on the denominator that you already noted).

Long story short: your continued fraction equals both $1$ and $2$, in a sense. Which one it is equal to depends on which one you want it to be equal to.

Answer 2

I think this is a major problem of rookies of math (including me), because we do not understand the mechanism of such 'trick'. In general, it seems it is very reasonable for people to claim $x=\sqrt{1+\sqrt{1+\cdots}}=\sqrt{1+x}$. However, one important note here is that we assume the existence of this limit. But how do we actually prove it?

We shall define a sequence $\{x_n\}$, where $x_n=\underbrace{\sqrt{1+\sqrt{1+\cdots}}}_{n\text{ square roots}}$. Then we use some tricks to prove the sequence has a limit in $\mathbb{R}$. For example, in this case we use $\{x_n\}$ is monotonic increasing (trivial), and it is bounded above by $2$ (use induction). So by monotone convergence theorem, we have $\{x_n\}$ exists, this makes our step $$\text{Let } x=\sqrt{1+\sqrt{1+\cdots}}$$ to be legal.

Now in your case, we shall apply same idea. Let $\{y_n\}$ be the sequence where $y_n=\underbrace{3-2/(3-2/\cdots)}_{n\text{ times}}$. You shall see $y_1=\dfrac{8}{3},y_2=\dfrac{15}{7},\cdots$. So it is reasonable to guess $\{y_n\}$ is bounded below by some number, and is strictly decreasing.

Now how do we restrict out the ambiguously in solving the limit? The idea is to eliminate through setting a larger lower bound of the sequence. (This sounds so dumb, but actually, you can see $2=3-2/2$, so is there any chance the sequence can escape away below $2$, and starts converging to $1$?) That's why in usual, we will directly use $2$ as lower bound (induction proof also), and we actually do not need to consider this kind of uncertainty.