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Topic: Alternative solution to ramanujan's sqrt(1 + 2*sqrt(1 + 3*sqrt(1 +
4*sqrt(............

Replies: 2   Last Post: Jan 24, 2008 10:33 AM

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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
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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 + (n-1)*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 + (n-1)*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 + (n-1)*sqrt(1 + n*sqrt(1 + (n+1)*sqrt(..........

So n^2 = 1 + (n-1)*sqrt(1 + n*sqrt(1 + (n+1)*sqrt(..........

So n^2 - 1 = (n-1)*sqrt(1 + n*sqrt(1 + (n+1)*sqrt(..........

So n+1 = sqrt(1 + n*sqrt(1 + (n+1)*sqrt(.......... for n-1<>0 and
our case is proven

Is this solution not easier to understand?

Carel




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