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I need help simplifying the radical $$y=\sqrt{x-\sqrt{x+\sqrt{x-...}}}$$ The above expression can be rewritten as $$y=\sqrt{x-\sqrt{x+y}}$$ Squaring on both sides, I get $$y^2=x-\sqrt{x+y}$$ Rearranging terms and squaring again yields $$x^2+y^4-2xy^2=x+y$$ At this point, deriving an expression for $y$, completely independent of $x$ does not seem possible. This is the only approach to solving radicals which I'm aware of. Any hints to simplify this expression further/simplify it with a different approach will be appreciated.

EDIT: Solving the above quartic expression for $y$ on Wolfram Alpha, I got 4 possible solutions

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    $\begingroup$ It is possible, but you have to solve a quartic, which is tedious (but doable in principle). $\endgroup$
    – Deepak
    Jul 28, 2020 at 4:45
  • $\begingroup$ @Deepak I'm not yet familiar with a method of solving quartic expressions. Could you provide a reference for solving quartics? $\endgroup$
    – Manan
    Jul 28, 2020 at 4:47
  • $\begingroup$ I believe the answer is $\frac{\sqrt{4x+3}-1}{2}$ $\endgroup$
    – Arjun
    Jul 28, 2020 at 4:52
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    $\begingroup$ Here is the link from where i learnt,youtube.com/watch?v=BO1T7ebJlO8 $\endgroup$
    – Arjun
    Jul 28, 2020 at 4:58
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    $\begingroup$ There is a tag that you could usefully add of "nested-radicals" $\endgroup$ Jul 28, 2020 at 9:31

3 Answers 3

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Perhaps it is more instructive to consider instead the following: let $$y = \sqrt{x - \sqrt{x + \sqrt{x - \sqrt{x + \cdots}}}}, \\ z = \sqrt{x + \sqrt{x - \sqrt{x + \sqrt{x - \cdots}}}},$$ so that if $y$ and $z$ exist, they satisfy the system $$y = \sqrt{x - z}, \\ z = \sqrt{x + y},$$ or $$y^2 = x - z, \\ z^2 = x + y.$$ Consequently $$0 = z^2 - y^2 - y - z = (z-y-1)(y+z).$$ It follows that either $z = -y$ or $z = 1 + y$. The first case is impossible for $x \in \mathbb R$ since by convention we take the positive square root, so both $y, z > 0$. In the second case, we can substitute back into the first equation to obtain $y^2 = x - (1+y)$, hence $$y = \frac{-1 + \sqrt{4x-3}}{2},$$ where again, we discard the negative root.

So far, what we have shown is that if such a nested radical for $y$ converges, it must converge to this value. It is not at all obvious from the above whether a given choice of $x$ results in a real-valued $y$, for any meaningful definition of $y$ must be as the limit of the sequence $$y = \lim_{n \to \infty} y_n, \\ y_n = \underbrace{\sqrt{x - \sqrt{x + \sqrt{x - \cdots \pm \sqrt{x}}}}}_{n \text{ radicals}},$$ and although the choice $x = 1$ appears at first glance permissible, we quickly run into problems; $y_3 = \sqrt{1 - \sqrt{1 + \sqrt{1}}} \ne \mathbb R$. In particular, we need $x$ to satisfy the relationship $$x \ge \sqrt{x + \sqrt{x}},$$ which leads to the cubic $x^3 - 2x^2 + x - 1$ with unique real root $$x = \frac{1}{3} \left(2+\sqrt[3]{\frac{25-3 \sqrt{69}}{2}}+\sqrt[3]{\frac{25+3 \sqrt{69}}{2}}\right) \approx 1.7548776662466927600\ldots.$$ However, any such $x$ meeting this condition will lead to a convergent sequence. The idea is to show that $|y_{n+2} - y| < |y_n - y|$ for all $n \ge 1$; then since $\lim y_n$ has at most one unique limiting value as established above, the result follows.

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Consider the final relation you have obtained as a quadratic equation in $x$,i.e: $$x^2-(2y^2+1)x+y^4-y=0$$ Solving the above gives you $$x=y^2+y+1 \text{ or } x=y^2-y$$ Individually solve these the quadratics in $y$ to obtain the four solutions you got from Wolfram Alpha.

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    $\begingroup$ Both answers are good, but this one is more natural +1 $\endgroup$
    – Arjun
    Jul 28, 2020 at 5:02
  • $\begingroup$ @Arjun Thanks a lot! $\endgroup$ Jul 28, 2020 at 5:03
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Note that :

$$(x-y^2)^2 = x+y \implies (x-y^2)^2 - y^2 = x+y-y^2 \implies (x-y^2-y)(x-y^2+y) = x-y^2+y \\ \implies \boxed{(x-y^2+y)(x-y^2-y-1) = 0}$$

So either one is correct.


Note : The problem is that one is still not sure when the radical above converges i.e. what is the set of all $x$ for which $\sqrt{x + \sqrt{x-\sqrt{x+...}}}$ forms a convergent sequence.

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  • $\begingroup$ I was solved to suppose an integral, where this radical was the integrand- the limit of integration was 1 to 3, so I suppose only positive expressions have to be assumed? $\endgroup$
    – Manan
    Jul 28, 2020 at 5:08
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    $\begingroup$ Yes, of course. You would run into square root problems if any intermediate $a_n$ became negative. Of course, if this is one of those IIT type questions where you just need an answer, then you need not bother and can go ahead with whatever expressions you get for $y$ in terms of $x$. $\endgroup$ Jul 28, 2020 at 5:09
  • $\begingroup$ No, this is actually a subjective problem where I need an explicit method to solve. $\endgroup$
    – Manan
    Jul 28, 2020 at 5:14
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    $\begingroup$ Ok, so my question is : do you need to prove explicit convergence of the radical in an interval? See, if you don't justify the convergence, you can do what you like, right? You might know all those proofs with sequences like $1+1+1+...$ in videos ; you don't want to land in a situation like that. $\endgroup$ Jul 28, 2020 at 5:17
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    $\begingroup$ Yes, I think you should do that, if the selection is enough then I think we can close the conversation. +1 to your question! $\endgroup$ Jul 28, 2020 at 5:22

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