Home » Math » On Ramanujan’s Nested Radicals

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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
$x+n+a =\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)}}} \ \ldots (8)$

Generalising 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 upto $k$ roots.

Some interesting points
As $x,n$ and $a$ all are real numbers, thus they can be interchanged with each other.
i.e.,
$\tiny \dpi{150} x+n+a = \\ \sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots+(x+(k-2)n)\\ \sqrt{a(x+(k-1)n)+{(n+a)}^2+x(x+(k+1)n+a)}}}}}) \\=\sqrt{an+{(x+a)}^2+n\sqrt{a(n+x)+{(x+a)}^2+(n+x)\sqrt{a(n+2x)+{(x+a)}^2+(n+2x)\sqrt{\ldots+(n+(k-2)x) \\ \sqrt{a(n+(k-1)x)+{(x+a)}^2+n(n+(k+1)x+a)}}}}}) \\=\sqrt{xa+{(n+x)}^2+a\sqrt{x(a+n)+{(n+x)}^2+(a+n)\sqrt{x(a+2n)+{(n+x)}^2+(a+2n)\sqrt{\ldots+(a+(k-2)n) \\ \sqrt{ x(a+(k-1)n)+{(n+x)}^2+a(a+(k+1)n+x)}}}}} \ \ldots (10)$
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:
$\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn+{(n+1)}^2+(k+2)n^2+n}}}}} \ \ldots (18)$
or,
$\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$

1. [...] are the pics for raw work done for Post On Ramanujan’s Nested Radicals #gallery-1 { margin: auto; } #gallery-1 .gallery-item { float: left; margin-top: 10px; text-align: [...]

2. 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…

3. rexantony says:

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

4. 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))))

5. utkarsh says:

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

6. Dear Utkarsh! Thanks for reading the post. Before I comment, I would like to mention that Ramanujan Nested Radical formulas are proposed for infinte 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).

• Akshay kumar says:

What is ramanujan redical ?

7. utkarsh says:

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