Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.


share|cite|improve this question
Great question. One from his personal goddess, no doubt. – Bennett Gardiner Oct 5 '13 at 4:22
Why is it called lost note book? – Arjang Oct 5 '13 at 5:29
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 5 '13 at 5:31
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 5 '13 at 7:16
See: Bruce C. Berndt , "Ramanujan's Notebooks IV" ,(pp. 42-45). The companion given has this signature $+ - - + + - - + + ...$ – Alan Oct 5 '13 at 7:22

2 Answers 2

up vote 11 down vote accepted

As pointed out by Cocopuffs (and Alan), the correct period has length 4, namely +,+,-,+. More generally, using any of the $2^4=16$ possible periods,

$$x = \sqrt{a\pm \sqrt{a\pm \sqrt{a\pm \sqrt{a\pm\dots}}}}$$

will be the absolute value of a root of the 16th deg eqn,

$$x = (((x^2 - a)^2 - a)^2 - a)^2 - a\tag{1}$$

Ramanujan (Notebooks IV, p.42-43) stated that (1) was a product of 4 quartic polynomials, one of which is the reducible,


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,

$$x_n = -\frac{y-z}{4}\pm\frac{1}{2}\sqrt{\frac{(y-2)(y+z)z}{2y}}\tag{4}$$


$$z =\pm\sqrt{y^2+4}\tag{5}$$

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 $n=5$, the $32$-deg factors into a quadratic and a $30$-deg. The latter can be decomposed similar to what Ramanujan did, but I found it now involves a sextic which, for general $a$, was not solvable.) The exception is $a=2$ where the solution involves roots of unity. See this related post.

share|cite|improve this answer

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

share|cite|improve this answer
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 5 '13 at 7:50
@wendy.krieger that will be wrong too.:) – SHOBHIT GAUTAM Oct 5 '13 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. – SHOBHIT GAUTAM Oct 5 '13 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 – wendy.krieger Oct 5 '13 at 10:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.