Resting with ‘Nested Radicals’

Standard

One night after hectic schedule at college, I was standing in front of black board with a chalk in my hand (to relax my mind!).

A thought of nested roots came to my mind and I wrote:
\sqrt{2 +\sqrt{2 +\sqrt{2 + \ldots}}} = 2
As I had `learnt’ from some olympiad book in class 8, the trick to solve such problems.
Let, \sqrt{2 +\sqrt{2 +\sqrt{2+ \ldots}}} = x
As we have infinite number of terms on left hand side, if we keep one of the two’s aside, we can rewrite it as:\\
\sqrt{2 + x} = x
\Rightarrow 2 + x = x^2
Solving this quadratic we will get x = 2, -1
Clearly x\geq 0, thus x = 2.
Now thrilled by this I proceeded to evaluate nested roots for other arithmetic operations in similar fashion:
\sqrt{2 -\sqrt{2 -\sqrt{2 - \ldots}}} = 1
\sqrt{2\sqrt{2\sqrt{2\ldots}}} = 2
\sqrt{\frac{2}{\sqrt{\frac{2}{\sqrt{\frac{2}{\vdots}}}}}} = \sqrt[3]{2}
Now I knew that no one knows the answer for:
1-1+1-1+1-1\ldots = ??
so, inspired by this I wrote:
\sqrt{1-\sqrt{1 +\sqrt{1-\ldots}}} = ??
But then one of my friend suggested:

Since we are dealing with infinite terms we can say

\sqrt{1-\sqrt{1 +x}} = x
\Rightarrow 1-\sqrt{1 +x} = x^2
\Rightarrow x^4 - 2x^2 - x = 0
\Rightarrow x(x^3 - 2x - 1) = 0
\Rightarrow x(x+1)(x^2-x-1) = 0
\Rightarrow x = 0, -1, \frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}
But I can’t say which one out of  0\quad \& \quad \frac{1+\sqrt{5}}{2} is solution.
I know that flaw in this argument is similar to that in saying:
1-1+1-1+1-1\ldots = 0, by pairing +1\quad \& \quad -1.
But I found the idea itself flawless and tried to apply it to my starting expression:
\sqrt{2 +\sqrt{2 + x}} = x
\Rightarrow 2 + \sqrt{2 + x} = x^2
\Rightarrow x^4 - 4x^2 - x +2 = 0
\Rightarrow (x-2)(x+1)(x^2 + x -1) = 0
\Rightarrow x = 2, - 1, \frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}
Unlike last case I can decide between 2\quad \& \quad \frac{-1+\sqrt{5}}{2}  [Thanks to Sagar Shrivastava]

Since, \sqrt{2 +\sqrt{2 +\sqrt{2+ \ldots}}} > \sqrt{2} > \frac{-1+\sqrt{5}}{2}

Thus, 2 is only answer.

Thus if I take out more and more 2’s out of square root, I will get more and more values of x and among them we will get only one appropriate value of x.

But still I can’t figure out the correct value of\sqrt{1-\sqrt{1 +\sqrt{1-\ldots}}} since I don’t have much experience with ‘series’ at this stage, but will surely investigate this question in detail by end of this semester.
If you have some ideas regarding this please do share.

Also these nested roots remind me of Ramanujam’s famous problem:
Find the value of : \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}}  (See this article by B. Sury)

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2 responses »

  1. Pingback: Nested Radical Sequences? | Gaurish4Math

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