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?

Gaurav Tiwari

A designer by profession, a mathematician by education but a Blogger by hobby. With an experience of over seven years with WordPress, PHP and CSS3, Gaurav is capable of doing almost anything related to these. Beyond that, He is a mathematics graduate & a civil service aspirant.

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7 Responses

  1. hugmamma says:

    wow! i’m impressed!…and you’re reading…my blog?

    uh…i’m flattered…hope 2012 brings you more math puzzles to solve… 🙂

  2. rexantony says:

    this is very helpful to me and you are doing a great job . thank you

  3. utkarsh says:

    hi..i am utkarsh.i have been working on a formula and i am stuck in nested radicals.
    basically, i want to find out value of sqrt(2+sqrt(2+sqrt(2…………….sqrt(2)
    for x of times,for example, for x=3, i want value of sqrt(2+sqrt(2+sqrt(2+sqrt(2))))
    would you please help me?

  4. utkarsh says:

    by the way,are you left-handed,your hand writing is similiar to mine!

  5. Dear Utkarsh! Thanks for reading the post. Before I comment, I would like to mention that Ramanujan Nested Radical formulas are proposed for infinite number of radicals in a number. When, there are finite number of nested radicals, the exact numerical value is calculated by an advanced calculator.
    Let me be clear. $ \sqrt {2}$ always means $ \sqrt {2}$ or approximately 1.4142… Similarly $ \sqrt {2+\sqrt{2}}$ has its own numerical value. And so on. As we increases the number of squareroots, the value tends to 2 (not exactly 2).
    But when infinite terms are considered, the numerical values cam be easily calculated using algebraic equations.
    Let $ N= \sqrt {2+\sqrt{2+\sqrt{2+ \ldots +\sqrt{2}}}}$ upto infinte terms
    $ N= \sqrt {2+N}$
    or, $ N^2-N-2=0$.
    The non-negative solution of above quadratic equation is the numerical value of the nested radical (i.e., N=2).

  6. utkarsh says:

    thanks for the answer!i guess i will really have to use calculators!

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