Ramanujan stated this radical in his lost notebook:
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$
I don't have any idea on how to prove this.
Any help appreciated.
Thanks.
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Sign up to join this communityRamanujan stated this radical in his lost notebook:
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$
I don't have any idea on how to prove this.
Any help appreciated.
Thanks.
The correct period has length 4, namely (+,+,+,-)
$$x_1=\small+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}=2.7472\dots$$
The other roots of the quartic in $x$ are given by the patterns $\small(+,+,-,+),\; (+,-,+,+),\; (-,+,+,+)$, respectively
$$x_2=\small+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\cdots}}}}}}} = \frac{2-\sqrt 5 +\sqrt{15+6\sqrt 5}}{2}=2.5473\dots$$
$$x_3=\small+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\cdots}}}}}}} = \frac{2+\sqrt 5 -\sqrt{15-6\sqrt 5}}{2}=1.4888\dots$$
$$x_4=\small\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\cdots}}}}}}} = \frac{2-\sqrt 5 -\sqrt{15+6\sqrt 5}}{2}=-2.7833\dots$$
This immediately implies that the four roots obey the system, $$\begin{aligned} x_1^2 &= x_2+5\\ x_2^2 &= x_3+5\\ x_3^2 &= x_4+5\\ x_4^2 &= x_1+5\\ \end{aligned}$$ also studied by Ramanujan. (See this related post.) More generally, using any of the $2^4=16$ possible periods,
$$x = \pm\sqrt{a\pm \sqrt{a\pm \sqrt{a\pm \sqrt{a\pm\dots}}}}$$
will be the absolute value of a root of the 16th deg eqn,
$$x = (((x^2 - a)^2 - a)^2 - a)^2 - a\tag{1}$$
In his Notebooks IV (p.42-43), Ramanujan stated that (1) was a product of 4 quartic polynomials, one of which is the reducible,
$$(x^2-x-a)(x^2+x-a+1)=0\tag{2}$$
and the other three had coefficients in the cubic,
$$y^3+3y = 4(1+ay)\tag{3}$$
Using Mathematica to factor (1), we find that it is indeed a product of (2) and a 12th deg eqn with coefficients in a. After some manipulation, the 12 roots are,
$$x_n = -\frac{y-z}{4}\pm\frac{1}{2}\sqrt{\frac{(y-2)(y+z)z}{2y}}\tag{4}$$
where,
$$z =\pm\sqrt{y^2+4}\tag{5}$$
Since there are 4 sign changes and (3) gives 3 choices for $y$, this yields the 12 roots.
Note: For $a=5$ (as well as $a=2$), the cubic factors over $\mathbb{Q}$, hence no cubic irrationalities are involved, and one of the $x_n$ will give the value of the appropriate infinite nested radical.
P.S. Interestingly, for period length $n> 4$, not all the roots of the deg $2^n$ equation will be expressible as finite radical expressions for general $a$. The exception is $a=2$ where the solution involves roots of unity as discussed in this post.
If @Cocopops is correct, in that the +,- signs go like +,+,-,+,+,+,-,+,+,+, ... and the aperiodicity is just at the beginning, this is far less impressive.
Then if
$$x= \sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\dots}}}}}}} $$ then $$ y = \sqrt{5+x} = \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\dots}}}}}}}}, $$ so the pattern for $y$ is +,+,+,-,+,+,+,-,+,+,+, ... and we can say $$ (((y^2-5)^2-5)^2-5)^2-5 = -y. $$ Numerically we should be able to find a root. However finding the analytic expression still seems hard.
I'd like to suggest that we pose this as a dual question, what if the signs DO follow +,+,-,+,+,+,-,+,+,+,+,-, ...
Does the expression have a closed form? In general, what about radicals of the form $$ \sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a- \ldots}}}}}}}}}? $$
Based on the above results by piezas, I change them to the form of trigonometric as follows
$ x_1=\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+x_1}}}}=\frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}=2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{8 \pi }{15}\right)$
$ x_2=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+x_2}}}}=\frac{2-\sqrt 5 +\sqrt{15+6\sqrt 5}}{2}=2 \cos \left(\frac{4 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right)$
$ x_3=\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+x_3}}}}=\frac{2+\sqrt 5 -\sqrt{15-6\sqrt 5}}{2}=2 \cos \left(\frac{2 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{11 \pi }{15}\right)$
$ x_4=\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-x_4}}}}=\frac{-2+\sqrt 5 +\sqrt{15+6\sqrt 5}}{2}=2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{6 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right)$
then we have
$ \sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\left(2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{8 \pi }{15}\right)\right)}}}}=2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{8 \pi }{15}\right) $
$ \sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\left(2 \cos \left(\frac{4 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right)\right)}}}}=2 \cos \left(\frac{4 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right) $
$ \sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\left(2 \cos \left(\frac{2 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{11 \pi }{15}\right)\right)}}}}=2 \cos \left(\frac{2 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{11 \pi }{15}\right) $
$ \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\left(2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{6 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right)\right)}}}}=2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{6 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right) $
In addition, I also find the following examples
$ \sqrt{5-\left(2 \cos \left(\frac{\pi }{21}\right)+2 \cos \left(\frac{5 \pi }{21}\right)+2 \cos \left(\frac{17 \pi }{21}\right)\right)}=2 \cos \left(\frac{\pi }{21}\right)+2 \cos \left(\frac{5 \pi }{21}\right)+2 \cos \left(\frac{17 \pi }{21}\right) $
$ \sqrt{5+\left(2 \cos \left(\frac{2 \pi }{21}\right)+2 \cos \left(\frac{8 \pi }{21}\right)+2 \cos \left(\frac{10 \pi }{21}\right)\right)}=2 \cos \left(\frac{2 \pi }{21}\right)+2 \cos \left(\frac{8 \pi }{21}\right)+2 \cos \left(\frac{10 \pi }{21}\right) $
$ \sqrt{5+\sqrt{5-\left(2 \cos \left(\frac{3 \pi }{17}\right)+2 \cos \left(\frac{5 \pi }{17}\right)+2 \cos \left(\frac{7 \pi }{17}\right)+2 \cos \left(\frac{11 \pi }{17}\right)\right)}}=2 \cos \left(\frac{3 \pi }{17}\right)+2 \cos \left(\frac{5 \pi }{17}\right)+2 \cos \left(\frac{7 \pi }{17}\right)+2 \cos \left(\frac{11 \pi }{17}\right) $
$ \sqrt{5-\sqrt{5+\left(2 \cos \left(\frac{2 \pi }{17}\right)+2 \cos \left(\frac{4 \pi }{17}\right)+2 \cos \left(\frac{8 \pi }{17}\right)+2 \cos \left(\frac{16 \pi }{17}\right)\right)}}=2 \cos \left(\frac{2 \pi }{17}\right)+2 \cos \left(\frac{4 \pi }{17}\right)+2 \cos \left(\frac{8 \pi }{17}\right)+2 \cos \left(\frac{16 \pi }{17}\right) $
$ \sqrt{3-\left(2 \cos \left(\frac{2 \pi }{13}\right)+2 \cos \left(\frac{6 \pi }{13}\right)+2 \cos \left(\frac{8 \pi }{13}\right)\right)}=2 \cos \left(\frac{2 \pi }{13}\right)+2 \cos \left(\frac{6 \pi }{13}\right)+2 \cos \left(\frac{8 \pi }{13}\right) $
$ \sqrt{3+\left(2 \cos \left(\frac{\pi }{13}\right)+2 \cos \left(\frac{3 \pi }{13}\right)+2 \cos \left(\frac{9 \pi }{13}\right)\right)}=2 \cos \left(\frac{\pi }{13}\right)+2 \cos \left(\frac{3 \pi }{13}\right)+2 \cos \left(\frac{9 \pi }{13}\right) $
$ \sqrt{6-\sqrt{6+\left(2 \cos \left(\frac{\pi }{21}\right)+2 \cos \left(\frac{5 \pi }{21}\right)+2 \cos \left(\frac{17 \pi }{21}\right)\right)}}=2 \cos \left(\frac{\pi }{21}\right)+2 \cos \left(\frac{5 \pi }{21}\right)+2 \cos \left(\frac{17 \pi }{21}\right) $
$ \sqrt{6+\sqrt{6-\left(2 \cos \left(\frac{2 \pi }{21}\right)+2 \cos \left(\frac{8 \pi }{21}\right)+2 \cos \left(\frac{10 \pi }{21}\right)\right)}}=2 \cos \left(\frac{2 \pi }{21}\right)+2 \cos \left(\frac{8 \pi }{21}\right)+2 \cos \left(\frac{10 \pi }{21}\right) $
$ \sqrt{8-\sqrt{8+\sqrt{8-\left(2 \cos \left(\frac{2 \pi }{9}\right)+2 \cos \left(\frac{3 \pi }{9}\right)+2 \cos \left(\frac{5 \pi }{9}\right)\right)}}}=2 \cos \left(\frac{2 \pi }{9}\right)+2 \cos \left(\frac{3 \pi }{9}\right)+2 \cos \left(\frac{5 \pi }{9}\right) $
$ \sqrt{8-\sqrt{8-\sqrt{8+\left(2 \cos \left(\frac{\pi }{9}\right)+2 \cos \left(\frac{2 \pi }{9}\right)+2 \cos \left(\frac{6 \pi }{9}\right)\right)}}}=2 \cos \left(\frac{\pi }{9}\right)+2 \cos \left(\frac{2 \pi }{9}\right)+2 \cos \left(\frac{6 \pi }{9}\right) $
$ \sqrt{8+\sqrt{8-\sqrt{8-\left(2 \cos \left(\frac{\pi }{9}\right)+2 \cos \left(\frac{3 \pi }{9}\right)+2 \cos \left(\frac{4 \pi }{9}\right)\right)}}}=2 \cos \left(\frac{\pi }{9}\right)+2 \cos \left(\frac{3 \pi }{9}\right)+2 \cos \left(\frac{4 \pi }{9}\right) $
$ \sqrt{11+\left(2 \cos \left(\frac{\pi }{5}\right)+2 \cos \left(\frac{\pi }{5}\right)+2 \cos \left(\frac{2 \pi }{5}\right)\right)}=2 \cos \left(\frac{\pi }{5}\right)+2 \cos \left(\frac{\pi }{5}\right)+2 \cos \left(\frac{2 \pi }{5}\right) $
$ \sqrt{11-\left(2 \cos \left(\frac{\pi }{5}\right)+2 \cos \left(\frac{2 \pi }{5}\right)+2 \cos \left(\frac{2 \pi }{5}\right)\right)}=2 \cos \left(\frac{\pi }{5}\right)+2 \cos \left(\frac{2 \pi }{5}\right)+2 \cos \left(\frac{2 \pi }{5}\right) $
In case of nested square roots of 2, we have proved that
For any rational number $\frac{n}{m}$, both $\cos \left(\frac{n}{m}\pi\right)$ and $\sin \left(\frac{n}{m}\pi\right)$ can be represented as cyclic infinite nested square roots of 2, of which the cyclic period is less than $(m−1)/2$.
for example,
$ 2 \cos \left(\frac{\pi }{13}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{13}\right)}}}}}} $
$ 2 \cos \left(\frac{355 \pi }{113}\right)=-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{355 \pi }{113}\right)}}}}}}}}}}}}}} $
please see the below link for more
https://math.stackexchange.com/a/4232525/954936
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