lspirkov@udenva.UUCP (Goldilocks) (04/30/85)
x^x^x^x ... = x^(x^x^...) = x^2 = 2 , therefore, x=sqrt(2); someone posted a comment about how x couldn't equal sqrt(2) because that goes to infinity so x should probably equal sqrt (-2). no, it doesn't go to infinity because exponentiation associates right to left so you wouldn't get something like: sqrt(2) ^ sqrt(2) ^ sqrt (2) = 2 because that's really sqrt(2) ^ (sqrt(2) ^ sqrt(2)) ~ 1.76. -- Goldi email: udenva!lspirkov
cjh@petsd.UUCP (Chris Henrich) (05/01/85)
[] John Woods and Jeff Sonntag have posted articles about this problem, containing the statement that x ^ x ^ x ^ ... ^ x \________ _______/ \/ n times tends to infinity with n if x = sqrt(2). This seems to spring from associating the operators the "wring" way. Messr Woods and Sonntag are parsing x ^ x ^ x as ((x ^ x) ^ x), whereas the "right" (i.e., conventional, conformist, wimpy) way to do it is as (x ^ (x ^ x)). With the latter method, if x = sqrt(2), then the value of the iterated exponentiation must be less than 2. Proof is by induction: if its true for n-1 operands, then you can prove it for n. (Try it, it is not very hard.) And it is true for one operand, since x < 2. On the other hand, the sequence of values does increase as n increases. Again, it's easy to prove inductively. Therefore, there is a limit of the sequence; call it y. That is, x ^ y = 2. The only possilbe value of y is 2. Problem: how fast does the sequence approach its limit? That is, what is a rough estimate for 2 - x ^ x ^ ...^ x with n x's, in terms of n? Regards, Chris -- Full-Name: Christopher J. Henrich UUCP: ..!(cornell | ariel | ukc | houxz)!vax135!petsd!cjh US Mail: MS 313; Perkin-Elmer; 106 Apple St; Tinton Falls, NJ 07724 Phone: (201) 758-7288
gjk@talcott.UUCP (Greg Kuperberg) (05/02/85)
> x ^ x ^ x ^ ... ^ x > \________ _______/ > \/ > n times > > tends to infinity with n if x = sqrt(2). This seems to spring > from associating the operators the "wring" way. Messr Woods > and Sonntag are parsing x ^ x ^ x as ((x ^ x) ^ x), whereas > the "right" (i.e., conventional, conformist, wimpy) way to do > it is as (x ^ (x ^ x)). There's a reason to assume that the inner parentheses are on the right: If they were on the left, then one could write the above expression much more simply as x^(x^n). Not only would there be a simple way of stating the problem, the problem would be trivial: If x>1 then x^(x^n) clearly goes to infinity. > Therefore, there is a limit of the sequence; call it > y. That is, x ^ y = 2. The only possilbe value of y is 2. > Problem: how fast does the sequence approach its > limit? That is, what is a rough estimate for > > 2 - x ^ x ^ ...^ x > > with n x's, in terms of n? ... > Full-Name: Christopher J. Henrich Let y(n)=x^y(n-1) and y(0)=1. Eventually, y(n) will get fairly close to 2. At this point we can approximate the function x^y by its derivative, i.e. sqrt(2)^y is close to ln(2)*(y-2)+2, where ln() is the natural logarithm. From then on 2 - y(n) is a geometric sequence with ratio ln(2). Thus 2-y(n) is close to C*ln(2)^n, where "C" is some constant which I don't wish to find. There is no doubt a rigorous proof of this, but I don't wish to find it, on account of I'm lazy. I guess I might as well at least ask the following question: What is the exact (as opposed to numerical) value of this constant C which I mentioned above? -- Greg Kuperberg harvard!talcott!gjk "The eerily accurate drawing of Goetz showed the face of the 'before' figure in comic-book ads for body-building devices."-Time Magazine, April 8