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I am interested in the behavior of the "extended square root", such as: $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}} ...$$ $$\sqrt{n\pm\sqrt{n\pm\sqrt{n\pm\sqrt{n\pm}}}} ...$$ I've noticed after running a python script to evaluate these, they usually converge to a value after 5 or so layers of the nested square root.

I have found that:

$$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n}}}} = 1 + \sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n}}}}$$

and

$$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+...}}}} * \sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-...}}}} = n$$

But more interesting to me is the behvaior that these show on the interval of 'n', where:

$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} = \sqrt{6-\sqrt{6-\sqrt{6-\sqrt{6}}}}$$ $$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}} = \sqrt{12-\sqrt{12-\sqrt{12-\sqrt{12}}}}$$ $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12}}}} = \sqrt{20-\sqrt{20-\sqrt{20-\sqrt{20}}}}$$

Seems an easy way to construct these ugly monster squares, you can make these equality expressions at an index i with the following: $$f(i)=i^2+i$$ $$f(i+1)=i^2-i$$

My question is, this sequence of equality follows $$f(i) = 2*i + f(i-1)$$ $$(0,2,6,12,20,30,42,56,...)$$

This smells like a log rule to me, and I'd love some help associating it with that if anyone has an explanation. Feel free to correct me on anything I've got wrong here. I can also provide the python scripts I've been using to run these numbers. :)

desmos graphs of calculated values and their lines

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  • $\begingroup$ Funny, laughed. $\endgroup$
    – Shean
    Commented Apr 6 at 19:02

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$$ x = \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\dots}}{}}}\implies x =\sqrt{n+x} $$ So $$ x^2-n-x=0\implies x=\frac{1\pm\sqrt{1+4n}}{2} $$ And if the nested square roots are negative we instead get: $$ x^2-n+x=0\implies x=\frac{-1\pm\sqrt{1+4n}}{2} $$ Hope this helps

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  • $\begingroup$ I used this substitution rule as well, but had some questions. So, by this second formula, shouldn't $$3=\frac{-1\pm\sqrt{13}}{2}$$ Doesn't seem true, based on evaluating the square root. Seems like there's a function on n when looking at the data, and it's related to the sequence I presented at the bottom. Unless I've been doing my calculations incorrectly, the "subtractive extended sqrt" of 12 is 3, not 48 as could be constructed by this formula. Let me know if I'm off on this please, thanks! $\endgroup$
    – Ivy Darling
    Commented Apr 6 at 18:00
  • $\begingroup$ This should be $1+4n$, shouldn't it? $\endgroup$
    – Command Master
    Commented Apr 6 at 18:09
  • $\begingroup$ Yes @CommandMaster thank you for pointing that out $\endgroup$
    – Masd
    Commented Apr 6 at 18:10
  • $\begingroup$ @IvyDarling if you put it in the formula: $$3=\frac{-1+\sqrt{1+12\times4}}{2}$$ $\endgroup$
    – Masd
    Commented Apr 6 at 18:11
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    $\begingroup$ Quadratic formula like, very pretty, thanks, I love it $\endgroup$
    – Ivy Darling
    Commented Apr 6 at 18:23

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