Ramanujan, The greatest ever Indian mathematician.

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 Ax = [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

Ax = [ax +(n+a)2 +x Ax+ n]

 

i.e.       Ax2 = [ax +(n+a)2 +x Ax+ n]

 

 

i.e.       Ax2 – (n+a)2  = [ax + x Ax+ n] = [ a + Ax+ n ] x

 

 

i.e.       [Ax – (n+a)] . [Ax + (n+a)] = [ a + Ax+ n ] x

 

 

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

 

Hence,             Ax = kx + n + a;   And so, Ax+ 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        xk2 + (2n + 2a – x – n) k – (n + 2a) = 0

i.e.       xk2 + (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)

            Ax = x + n + a , or Ax = – a, (not admissible)

Hence proved

 

 

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

 

Derivation

(x +n+ a)    =  [ (x + n + a)2 ]

= [ x2 + ( n + a)2 + 2x(n + a) ]

= [ ax + x2 + ( 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 = n2 + 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

 

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3 Responses to “Ramanujan, The greatest ever Indian mathematician.”

  1. Alex Says:

    Your blog is interesting!

    Keep up the good work!

  2. Nagarajan Says:

    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

  3. S.Nagaraj Says:

    Thanks for the book titles.

    I will follow them up.

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