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 Three threes, or two twos? (Posted on 2003-12-09)
You can represent ANY positive integer n using only three twos by writing

n= -log2(log2(√√...√2))

where log2 is the logarithm base 2, and there are n square roots.

Can you manage to do the same (that is, represent all positive integers) using three THREEs instead? And by using only TWO twos?

 See The Solution Submitted by Federico Kereki Rating: 4.1667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 f(x) = x+1 | Comment 1 of 3
Using only two 2's:

sec (arctan (sqrt x)) = sqrt (x + 1) is true for all positive values of x, so starting at x=2:

(sec (arctan (sqrt 2)))^2 = 3

(sec (arctan (sec (arctan (sqrt 2)))))^2 = 4

(sec (arctan (sec (arctan (sec (arctan (sqrt 2)))))))^2 = 5

(sec (arctan (sec (arctan (sec (arctan (sec (arctan (sqrt 2)))))))))^2 = 6

Using only two 3's:

The exponent 2 can be written as (sec (arctan (sqrt 3))) = 2, so starting at x=3:

(sec (arctan (sqrt 3)))^(sec (arctan (sqrt 3))) = 4

(sec (arctan (sec (arctan (sqrt 3)))))^(sec (arctan (sqrt 3))) = 5

(sec (arctan (sec (arctan (sec (arctan (sqrt 3)))))))^(sec (arctan (sqrt 3))) = 6

(sec (arctan (sec (arctan (sec (arctan (sec (arctan (sqrt 3)))))))))^(sec (arctan (sqrt 3))) = 7
Edited on December 10, 2003, 11:19 am
 Posted by Brian Smith on 2003-12-09 15:43:35

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