How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$

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.

• Great question. One from his personal goddess, no doubt. Oct 5, 2013 at 4:22
• 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. Oct 5, 2013 at 5:31
• 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? Oct 5, 2013 at 5:41
• 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. Oct 5, 2013 at 7:16
• See: Bruce C. Berndt , "Ramanujan's Notebooks IV" ,(pp. 42-45). The companion given has this signature $+ - - + + - - + + ...$
– Alan
Oct 5, 2013 at 7:22

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.

• I need more explanation. Where can I refer? Oct 3, 2020 at 17:45

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 $$\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a- \ldots}}}}}}}}}?$$

• 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. Oct 5, 2013 at 7:50
• @wendy.krieger that will be wrong too.:) Oct 5, 2013 at 9:33
• @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. Oct 5, 2013 at 9:37
• 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 Oct 5, 2013 at 10:11

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,

$$2 \cos \left(\frac{\pi }{13}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{13}\right)}}}}}}$$

$$2 \cos \left(\frac{355 \pi }{113}\right)=-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{355 \pi }{113}\right)}}}}}}}}}}}}}}$$