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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.

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7  
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
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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 5 '13 at 5:31
1  
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
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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 10 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,

$$(x^2-x-a)(x^2+x-a+1)=0\tag{2}$$

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 $n=5$, 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.

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

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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 5 '13 at 7:50
    
@wendy.krieger that will be wrong too.:) –  Ramanujan Oct 5 '13 at 9:33
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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. –  Ramanujan 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

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