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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.

Shobhit
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    Great question. One from his personal goddess, no doubt. – Bennett Gardiner Oct 05 '13 at 04:22
  • Why is it called lost note book? – jimjim Oct 05 '13 at 05:29
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    I think the repetition is misunderstood. $\frac{2 + \sqrt{5} + \sqrt{15 - 6\sqrt{5}}}{2}$ is the value where the +,- signs go like +,+,-,+,+,+,-,+,+,+,-,+,+,+,-,+ i.e. periodic in +,+,-,+. Ramanujan published this problem in the Journal of the Indian Math. Society. – Cocopuffs Oct 05 '13 at 05:31
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    Oh. That's much more believable then, whilst still very interesting, do you mean he published it with proof? Do you know if that paper is available? – Bennett Gardiner Oct 05 '13 at 05:41
  • @BennettGardiner He published this as part of a series of problems and asked the readers to submit proofs. Of course, the answer is found as one of the roots of $x = (((x^2 - 5)^2 - 5)^2 - 5)^2 - 5$, and you can check numerically which. Maybe Ramanujan had a better way in mind. – Cocopuffs Oct 05 '13 at 05:53
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    It seems i gave a similar method for a slightly different period, (ie ++-), and took 8 negative hits, because the problem was presented as having an increasing number of signs between each + sign. The method i gave allowed for any period of signs, by repeating $a$ at the point of the first period, and solving for that. – wendy.krieger Oct 05 '13 at 07:16
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    See: Bruce C. Berndt , "Ramanujan's Notebooks IV" ,(pp. 42-45). The companion given

    has this signature $+ - - + + - - + + ...$

     – Alan Oct 05 '13 at 07:22
  • @wendy.krieger, your method was still incorrect surely? Given that the problem certainly couldn't have been for the period (+,+,-) I don't see how it could have been correct - can you explain further how you allow for "any period" of signs? – Bennett Gardiner Oct 05 '13 at 07:31
  • If you have any period of signs, say '++-+-', you could write eg $x = ++-+- x$, where the + and - preserve the radix, and the x substitutes the first repetition. So eg $x=\sqrt{5+\sqrt{5+\dots}}$. the repetition is after the first sign (ie + x), so you could write $x=\sqrt{5+x}$. If the period were two places, it might be $x=\sqrt{5-\sqrt{5+x}}$, etc. You can then solve these by squaring and rearranging the sides. – wendy.krieger Oct 05 '13 at 07:38
  • Also, i mistrad + + -. ie solved the wrong problem, but the method is correct for a period. But it was suggested at the time i made my answer, that it ought be increasing number of + signs, ie 2, 3, 4, 5, 6, ..., rather than 2, 3, 3, 3, ... @BennettGardiner – wendy.krieger Oct 05 '13 at 07:45
  • see above comment,also given is:$$\sqrt{5+\sqrt{5-\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5-\dots}}}}}}$$ $$= \frac{\sqrt{5} - 2 +\sqrt{13 - 4\sqrt{5}} +\sqrt{50 + 12\sqrt{5} - 2\sqrt{65 - 20\sqrt{5}}}}{4}$$ – Alan Oct 05 '13 at 08:03
  • That one is even more impressive, these all seem to end up as polynomials of degree $\ge 5$, so there is no formula for the solution, they must be special cases that can be solved in radicals?

    @wendy.krieger, I think I understand now. I apologise for the ridiculous number of down votes, which I think I started... although your solution could have been explained a little better it probably wasn't fair to be that harsh.

     – Bennett Gardiner Oct 05 '13 at 08:17
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    iPads are not TeX friendly if you ever used one. So some jumps are given simply for the complexity of the process. Many of the questions get answered before i can get to them (time zones), and even though i wrought hyperbolic geometry from raw principles, trig and calculus is largely beyond me. @BennettGardiner – wendy.krieger Oct 05 '13 at 08:29

3 Answers3

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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.

Community
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Tito Piezas III
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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 $$ \sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a- \ldots}}}}}}}}}? $$

Bennett Gardiner
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  • You could reduce $x=\sqrt{a+\sqrt{a- \dots}}$ to $x=\sqrt{a+\sqrt{a-x}}$. whence $x^2-a = \sqrt{a-x}$, thence $x^4-2ax^2+a^2=a-x$, and solve for $x^4-2ax^2+x-a^2-a = 0$. That's the method i was trying to say in my response. – wendy.krieger Oct 05 '13 at 07:50
  • @wendy.krieger that will be wrong too.:) – Shobhit Oct 05 '13 at 09:33
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    @BennetGardiner only if i have known that i was interpreting the +,- are like you have solved(if they are), i would have solved that too, still thanks. Also i was also thinking if there is any closed form for the +,- interpretations i made. – Shobhit Oct 05 '13 at 09:37
  • It would not be wrong for alternating signs, which is what i was trying to show. It's wrong for Bennet Gardiner's example, which would require replacing $x$ with $\sqrt{a+\sqrt{x}}$, in the surd. But the method is correct: it's just that i have difficulties spotting the period. OK @Shobhit – wendy.krieger Oct 05 '13 at 10:11
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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

Or refer to my website:

http://eslpower.org

http://eslpower.org/Notebook.htm

Chen Shuwen
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