Only recently I read a biography “The Man Who Knew Infinity”, on Ramanujan, the greatest ever Indian mathematician, . It was amazing and poignant at the same time.

Some of the problems mentioned in the book, revved up my aging mind. I tried to prove one of his equations. I am sure this being one of the simpler equations of Ramanujan, many would have solved it during the last century itself. Still I feel elated that I was able to share a very small part of his genious. Ofcourse, unlike Ramanujan, I had a formal education upto master’s level in engineering in one of the IITs. If only we had IITs in those days ……..?

I am attaching two proofs for Ramanujan’s equation, one of which I call as Derivation. I feel very happy to share the same with my readers.

Happy to be associated with great minds of India,

regards to all,

L V Nagarajan

**Solution for the Problem by Sri Ramanujan**

**1. ****To Prove**

**(x + n + a) = ****√**[ax +(n+a)^{2} +x√[a(x+n) +(n+a)^{2} + (x+n)√[a(x+2n) +(n+a)^{2} + (x+2n) √etc ….

**Proof**

Let A_{x} = **√**[ax +(n+a)^{2} +x√[a(x+n) +(n+a)^{2} + (x+n)√[a(x+2n) +(n+a)^{2} + (x+2n) √etc ….

Then, We may write

A_{x} = **√**[ax +(n+a)^{2} +x A_{x+ n}]

i.e. A_{x}^{2} = [ax +(n+a)^{2} +x A_{x+ n}]

i.e. A_{x}^{2} – (n+a)^{2} = [ax + x A_{x+ n}] = [ a + A_{x+ n }] x

i.e. [A_{x} – (n+a)] . [A_{x} + (n+a)] = [ a + A_{x+ n }] x

i.e. [A_{x} – (n+a)] / x = [ a + A_{x+ n }] / [A_{x} + (n+a)] = k (say) —– (1)

Hence, A_{x} = kx + n + a; And so, A_{x+ n }= k(x + n) + n + a ———(2)

Substituting in the second part of eqn(1) above,

[a + k(x + n) + n + a] / [kx +n +a + n+ a] = k

i.e xk^{2} + (2n + 2a – x – n) k – (n + 2a) = 0

i.e. xk^{2} + (n + 2a – x) k – (n + 2a) = 0

i.e. (xk + n + 2a) (k – 1) = 0, giving the values, k =1 or kx = – n – 2a

Substituting in (2)

**A _{x} = x + n + a** , or

**A**, (not admissible)

_{x}= – aHence proved

** **

**2. ****The above is a proof. Let us call the following as a derivation.**

** **

**Derivation**

(x +n+ a) = **√ [ **(x + n + a)^{2} ]

= **√ **[ x^{2} + ( n + a)^{2} + 2x(n + a) ]

= **√ **[ ax + x^{2} + ( n + a)^{2} + x(2n + a) ]

= **√ **[ ax + ( n + a)^{2} + x(x + n +n + a) ] ——–(i)** **

Following as per Eqn (i) above, we may write:

(x+n +n + a) = **√ **[ a(x+n) + ( n + a)^{2} + (x+n)(x + 2n + n + a) ] ——(ii)

From Eqns (i) and (ii), we can recursively write:

**(x +n +a) = ****√[ax + ( n + a) ^{2} + x**

**√[a(x+n) + ( n + a)**

^{2}+ (x+n)**√[a(x + 2n) + (n+a)**

^{2}+ (x+2n)**√…**

** **

**Hence Derived**

**3. ****To Find the value of √[1+2√[1+3√[1+4√[1+5√[1+ …….**

(n+1)^{2} = n^{2} + 2n + 1

= 1 + n(n+2)

i.e.

n + 1 = √ [1 + n(n+2]]

Hence we may write,

3 = √ [1 + (2 x 4)] and

4 = √ [1 + (3 x 5)] and

5= √ [1 + (4 x 6)], etc

Hence,

**3= √[1+2√[1+3√[1+4√[1+5√[1+ …….**

** **

L V Nagarajan

01 July 2008

August 17, 2008 at 4:11 am |

Your blog is interesting!

Keep up the good work!

April 14, 2009 at 12:39 pm |

fromProf. K. Ramachandra

to

Nagarajan LV

dateWed, Aug 13, 2008 at 1:01 PM

subjectRe: Ramanujan’s Equation, Proof and Derivation

mailed-bymath.tifrbng.res.in

Dear Dr.Nagarajan,

The numerical result at the end suggests the generalisation in the beginning.

All the results of Ramanujan are very nice and these results are also nice.

Please convey my regards to Professor Narasimha.

With regards,

K.Ramachandra.

–

TIFR Centre

P.O.Box 1234,

IISc Campus,

Bangalore – 560 012

India

February 10, 2017 at 9:46 pm |

Thanks for the book titles.

I will follow them up.