On Ramanujan’s Nested Radicals

Ramanujan (1887-1920) discovered some formulas on algebraic nested radicals. This article is based on one of those formulas. The main aim of this article is to discuss and derive them intuitively. Nested radicals have many applications in Number Theory as well as in Numerical Methods .
The simple binomial theorem of degree 2 can be written as:
$ {(x+a)}^2=x^2+2xa+a^2 \ \ldots (1)$
Replacing $ a$ by $ (n+a)$ where $ x, n, a \in \mathbb{R}$ , we can have
$ {(x+(n+a))}^2= x^2+2x(n+a)+{(n+a)}^2$
or, $ {(x+n+a)}^2 =x^2+2xn+2ax+{(n+a)}^2$
Arranging terms in a way that
$ {(x+n+a)}^2 =ax+{(n+a)}^2+x^2+2xn+ax=ax+{(n+a)}^2+x(x+2n+a)$
Taking Square-root of both sides
or,
$ x+n+a=\sqrt{ax+{(n+a)}^2+x(x+2n+a)} \ \ldots (2)$

Take a break. And now think about $ (x+2n+a)$ in the same way, as:
$ x+2n+a =(x+n)+n+a$ .
Therefore, in equation (2), if we replace $ x$ by $ x+n$ , we get
$ x+2n+a=(x+n)+n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)((x+n)+2n+a)}$
or, $ x+2n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)} \ \ldots (3)$
Similarly, $ x+3n+a=\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)} \ \ldots (4)$
and also, $ x+4n+a=\sqrt{a(x+3n)+{(n+a)}^2+(x+3n)(x+5n+a)} \ \ldots (5)$
Similarly,
$ x+kn+a=\sqrt{a(x+(k-1)n)+{(n+a)}^2+(x+(k-1)n)(x+(k+1)n+a)} \ \ldots (6)$ where, $ k \in \mathbb{N}$ .

Putting the value of $ x+2n+a$ from equation (3) in equation (2), we get:
$ x+n+a=\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)}} \ \ldots (7)$
Again, putting the value of $ x+3n+a$ from equation (4) in equation (7), we get
$ x+n+a =\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)}}} \ \ldots (8)$

Generalizing$ the result for $ k$ -nested radicals:
$ x+n+a =\\ \sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots+(x+(k-2)n)\sqrt{a(x+(k-1)n)+{(n+a)}^2+x(x+(k+1)n+a)}}}}} \ \ldots (9)$
This is the general formula of Ramanujan Nested Radicals up-to $ k$ roots.

Some interesting points
As $ x,n$ and $ a$ all are real numbers, thus they can be interchanged with each other.
i.e.,
x+n+a = \\ \sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots+(x+(k-2)n)\\ \sqrt{a(x+(k-1)n)+{(n+a)}^2+x(x+(k+1)n+a)}}}}}) \\=\sqrt{an+{(x+a)}^2+n\sqrt{a(n+x)+{(x+a)}^2+(n+x)\sqrt{a(n+2x)+{(x+a)}^2+(n+2x)\sqrt{\ldots+(n+(k-2)x) \\ \sqrt{a(n+(k-1)x)+{(x+a)}^2+n(n+(k+1)x+a)}}}}}) \\=\sqrt{xa+{(n+x)}^2+a\sqrt{x(a+n)+{(n+x)}^2+(a+n)\sqrt{x(a+2n)+{(n+x)}^2+(a+2n)\sqrt{\ldots+(a+(k-2)n) \\ \sqrt{ x(a+(k-1)n)+{(n+x)}^2+a(a+(k+1)n+x)}}}}} \ \ldots (10)
etc.

Putting $ n=0$ in equation (9)
we have
$ x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots+x\sqrt{ax+{a}^2+x(x+a)}}}}} \ \ldots (11)$
or just, $ x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots}}}} \ \ldots (12)$

Again putting $ x=1 \ a=0$ in (9)

$ 1+n =\sqrt{{n}^2+\sqrt{n^2+(1+n)\sqrt{{n}^2+(1+2n)\sqrt{\ldots+(1+(k-2)n)\sqrt{{n}^2+1+(k+1)n}}}}} \ldots (13)$

Putting $ x=1 \ a=0$ in equation (8)
$ 1+n =\sqrt{{n}^2+\sqrt{{n}^2+(1+n)\sqrt{{n}^2+(1+2n)(1+4n)}}} \ \ldots (14)$

Again putting $ x=a=n$ =n(say) then
$ 3n=\sqrt{n^2+4{n}^2+n\sqrt{2n^2+4{n}^2+2n\sqrt{3n^2+4{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn^2+4{n}^2+(k+3)n^2}}}}}$
or, $ 3n=\sqrt{5{n}^2+n\sqrt{6{n}^2+2n\sqrt{7{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+4)n^2+(k+3)n^2}}}}} \ \ldots (15)$

Putting $ n=1$ in (15)
$ 3=\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{\ldots+(k-1)\sqrt{(2k+7)}}}}} \ \ldots (16)$

Putting $ x=n \in \mathbb{N}$ and $ a=0$ in (9) we get even numbers
$ 2n =\sqrt{{n}^2+n\sqrt{{n}^2+2n\sqrt{{n}^2+3n)\sqrt{\ldots+(k-1)n\sqrt{(k-1)n)+{n}^2+(k+2)n^2}}}}} \ \ldots (17)$

Similary putting $ x=n \in \mathbb{N}$ and $ a=1$ in (9) we get a formula for odd numbers:

or,

Comments?