Conference rusure::math

1654.0. "Infinite continued square roots" by XFMV01::POWALE () Fri Aug 14 1992 08:30

I have a problem in evaluating continued square roots.
Consider the following ..
a+b+c = sqrt((a+b+c)^2)
      = sqrt(2ab+2bc+2ca + sqrt((a^2+b^2+c^2)^2)
      = sqrt(2ab+2bc+2ca + sqrt(2a^2b^2 + 2b^2c^2 + 2c^2a^2 
      + sqrt((a^4+b^4+c^4)^2))))
proceeding in this way one can get an infinite continued square
roots where the first `Term` contains the sum of ab, bc and ca
second `Term` contains the sum of (ab)^2, (bc)^2 and (ca)^2
In general the Nth term will have the following form
(ab)^(2^(N-1))+ (bc)^(2^(N-1)) + (ca)^(2^(N-1))

Now consider infinite continued sqrare roots form for a+b+c
    
Putting a=0, b=0 and c=1 one gets
0+0+1 = sqrt(0 + sqrt(0 + sqrt(0 + sqrt(0 + ......

Now putting a=0, b=0, and c=2 one gets
0+0+2 = sqrt(0 + sqrt(0 + sqrt(0 + sqrt(0 + ......

Thus the L.H.S. can assume different values whereas the R.H.S.
is the same infinite continued sqrare root series.

Can someone tell me what has gone wrong ???
I have taken care to see that a+b+c > 0 and a>=0, b>=0 c>=0
also the sqrt() function takes the positive (or non-negative)
values.



- Pravin Powale

Similarly,

	c = 0 + sqrt(c^2)
	  = 0 + sqrt(0 + sqrt(c^4))
	  = 0 + sqrt(0 + sqrt(0 + sqrt(c^8)))
	  = 0 + sqrt(0 + sqrt(0 + sqrt(0 + sqrt(c^16))))

Here's another, simpler, example:

	d = 0 + d
	  = 0 + 0 + d
	  = 0 + 0 + 0 + d

The non-zero term that's being pushed to the right is *significant*.