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:
$$ 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn+{(n+1)}^2+(k+2)n^2+n}}}}} \ \ldots (18) $$
or,
$$ 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$$
Comments?

Total
0
Shares


Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word. How to write better comments?
7 comments
  1. 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?

  2. 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).

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

You May Also Like

TeXStudio is the most complete LaTeX Editor

A detailed and practical review of one of the most amazing LaTeX editors available for all desktop platforms like Windows, Mac and Linux. TeXStudio is a team project of Benito van der Zander, Jan Sundermeyer, Daniel Braun and Tim Hoffmann, forked from the TeXMaker application, a non-open source application, which open-source development stalled in 2009.

Free Online Calculus Text Books

Once I listed books on Algebra and Related Mathematics in this article, Since then I was receiving emails for few more related articles. I have tried to list almost all freely available Calculus texts. Here we go: Elementary Calculus : An approach using infinitesimals by H. J. Keisler Multivariable Calculus by Jim Herod and George Cain Calculus by Gilbert Strang…

Test your counting skills with Branifyd game for Android

Without basic operations of counting, like Addition, Subtraction, Multiplication and Division, it is not possible to imagine math problems. Counting is the base of human life. A student, whether he’s a math major or not, must be good at counting numbers.  The counting ability builds from experience and is definitely a time taking process.   Larger you have given time…

Just another way to Multiply

Multiplication is probably the most important elementary operation in mathematics; even more important than usual addition. Every math-guy has its own style of multiplying numbers. But have you ever tried multiplicating by this way? Exercise: $ 88 \times 45$ =? Ans: as usual :- 3960 but I got this using a particular way: 88            45…

Irrational Numbers and The Proofs of their Irrationality

“Irrational numbers are those real numbers which are not rational numbers!” Def.1: Rational Number A rational number is a real number which can be expressed in the form of where $ a$ and $ b$ are both integers relatively prime to each other and $ b$ being non-zero. Following two statements are equivalent to the definition 1. 1. $ x=\frac{a}{b}$…

#ThankYou2014

2014 was an amazing year for me. Things didn’t went that wrong, like they did in 2013 and year before that. Both my online and offline lives went smooth. This post is a summary of things I did/learned in 2014. It has primarily a personal approach and doesn’t intend to force any show-off to readers. 2013 : Learning with troubles 2013 had…