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carel
Posts:
161
Registered:
12/12/04


Alternative solution to ramanujan's sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(............
Posted:
Jan 24, 2008 6:42 AM


Dear group
While reading up on Ramanujan in the Wikipedia I saw he once published a problem in the Indian Mathematical journal. Let sqrt be square root of His question was what is the solution to z = sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(.........
He waited for more than 6 months, but no one came forward with a solution. He then published the following solution.
x + n + a = sqrt(ax + (n+a)^2 + x*sqrt( a(x+n) + (n+a)^2 + (x +n)sqrt(.........
I propose the following solution n = sqrt(1 + (n1)*sqrt(1 + n*sqrt(1 + (n+1)*sqrt(...........
So for n = 4 we have 4 = sqrt(1 + 3*sqrt(1 + 4*sqrt(......
so that z = sqrt(1 + 2*4) = sqrt(9) = 3
The solution is therefore 3
Proof follows.
It is easy to approximate the solution by calculating the first few sqrt terms out. The solution tends to 3
So we assume the solution is 3 and we have
3 = sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(...... equation 1
9 = 1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(......
8 = 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(....
4 = sqrt(1 + 3*sqrt(1 + 4*sqrt(... equation 2
16 = 1 + 3*sqrt(1 + 4*sqrt(.....
15 = 3*sqrt(1 + 4*sqrt(.....
5 = sqrt(1 + 4*sqrt(..... equation 3
So in general we have that n = sqrt(1 + (n1)*sqrt(1 + n*sqrt(1 + (n +1)*sqrt(........... and we must thus prove that
n+1 = sqrt(1 + (n)*sqrt(1 + (n+1)*sqrt(1 + (n+2)*sqrt(........... in order to prove by induction
Now n = sqrt(1 + (n1)*sqrt(1 + n*sqrt(1 + (n+1)*sqrt(..........
So n^2 = 1 + (n1)*sqrt(1 + n*sqrt(1 + (n+1)*sqrt(..........
So n^2  1 = (n1)*sqrt(1 + n*sqrt(1 + (n+1)*sqrt(..........
So n+1 = sqrt(1 + n*sqrt(1 + (n+1)*sqrt(.......... for n1<>0 and our case is proven
Is this solution not easier to understand?
Carel



