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Supposedly, the infinitely nested radical $$\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}\tag1\label{1}$$ converges to $$\frac {A-1}{6}+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt 3}\right)\tag2$$ where $A=\sqrt{4a-7}$. In fact, that's how Ramanujan arrived at the famous nested radical $$\sqrt{2-\sqrt{2+\sqrt{2+\ldots}}}=2\sin\left(\frac {\pi}{18}\right)\tag3$$ So I'm wondering how you would prove $\ref{1}$. My best try was to set $(1)$ equal to $x$ and substitute to get $$x=\sqrt{a-\sqrt{a+\sqrt{a+x}}}$$to get an octic polynomial. But I'm not too sure how the trigonometry made its way into the generalization.

And I also wonder if there is a way to generalize this even further to possible $$\sqrt{a-\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}}$$ which has period $4$ instead of $3$.

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Here's how I duplicated Ramanujan's result, with a lot of help from Wolfy.

Start with

$x=\sqrt{a-\sqrt{a+\sqrt{a+x}}} $.

Squaring, $x^2-a =-\sqrt{a+\sqrt{a+x}} $.

Squaring again, $(x^2-a)^2 =a+\sqrt{a+x} $ or $(x^2-a)^2-a =\sqrt{a+x} $.

A final squaring gives $((x^2-a)^2-a)^2-a =x $.

Setting $a=2$ gives $x =((x^2-2)^2-2)^2-2 $.

According to Wolfy, there are 8 real roots of this (including -1 and 2), though it identifies 3 as complex roots while giving them as real numbers. These are $-1, 2, -1.8794, 0.34730, 1.5321, -1.8019, -0.44504, 1.2470, $ though, for some reason, it calls the last 3 complex.

I have no idea how Wolfy got the roots. Each one of them is expressed with $\sqrt{3}$ and expressions involving it and $i$, some of them raised to the $\frac13$ and $\frac23$ powers.

It turns out the the fourth of these, 0.34730, is the desired root. This can be gotten by asking Wolfy for "root of $x=\sqrt{2-\sqrt{2+\sqrt{2+x}}}$" which gives a value slightly less than $0.35$.

Here is how I went from the exact expression for this root given by Wolfy to Ramanujan's expression:

$\begin{array}\\ x &=\dfrac{-(1-i \sqrt{3})}{2^{2/3} (-1+i \sqrt{3})^{1/3}}-\frac12 (1+i \sqrt{3}) (\frac12 (-1+i \sqrt{3}))^{1/3} \\ &=\dfrac{-1+i \sqrt{3}}{2^{2/3} (-1+i \sqrt{3})^{1/3}}-\frac12 (1+i \sqrt{3}) (\frac12 (-1+i \sqrt{3}))^{1/3} \\ &=2^{-2/3}(-1+i \sqrt{3})^{2/3}-\frac12 (1+i \sqrt{3}) (\frac12 (-1+i \sqrt{3}))^{1/3} \\ &=(\frac12(-1+i \sqrt{3}))^{2/3}-\frac12 (1+i \sqrt{3}) (\frac12 (-1+i \sqrt{3}))^{1/3} \\ &=(e^{2\pi i/3})^{2/3}-e^{\pi i/3} (e^{2\pi i/3})^{1/3} \\ &=e^{4\pi i/9}-e^{(\pi i)(1/3+2/9)} \\ &=e^{4\pi i/9}-e^{(\pi i)(5/9)} \qquad\text{(I was worried here)}\\ &=e^{4\pi i/9}-e^{(\pi i)(1-4/9)} \qquad\text{(but then I thought of this)}\\ &=e^{4\pi i/9}-e^{\pi i}e^{-4\pi i/9} \\ &=e^{4\pi i/9}+e^{-4\pi i/9} \\ &=2\cos(4\pi /9) \qquad\text{since }2\cos(z)=e^{iz}+e^{-iz} \\ &=2\sin(\pi/2-4\pi /9) \qquad\text{since } \cos(z) = \sin(\pi/2-z)\\ &=2\sin(\pi/18)\\ \end{array} $

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  • $\begingroup$ Where did you get $$\frac {-(1-i\sqrt{3})}{2^{2/3}(-1+i\sqrt{3})^{1/3}}-\frac {1}{2}(1+i\sqrt{3})(\frac {1}{2}(-1+i\sqrt{3}))^{1/3}$$ $\endgroup$ – Frank Oct 16 '16 at 19:02
  • $\begingroup$ Wolfy. The 4th root that it gives as the solutions to $x =((x^2-2)^2-2)^2-2$ in the exact form. $\endgroup$ – marty cohen Oct 16 '16 at 21:51
  • $\begingroup$ What if $a=8$? How would you simplify that? $\endgroup$ – Frank Oct 17 '16 at 2:21
  • $\begingroup$ I wouldn't but, entering "Select[Solve[-8 + (-8 + (-8 + x^2)^2)^2 == x, x], Element[x /. #1, Reals] & ]" gives 8 reals of which most are like the one above, only worse. Two simple one are $ (1\pm\sqrt(33))/2$. $\endgroup$ – marty cohen Oct 17 '16 at 2:31
  • $\begingroup$ More: The numeric values are, removing very small imaginary parts, x≈-3.34471, x≈-3.22668, x≈-2.37228, x≈-2.18479, x≈2.15761, x≈2.41147, x≈3.1871, x≈3.37228. $\endgroup$ – marty cohen Oct 17 '16 at 2:33

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