Conference rusure::math

    Letting y = x**y = 2, and solving gives x = sqrt(2)

    - Jim

    Surely there are other solutions to this equation.

    Isn't it ambiguous ?  I don't think the following two equations
    have the same solutions:
    
    x**(x**(x** . . . ))) = 2
    
    and
    
    ((((((. . . (x**x)**x)**x). . .**x = 2
    
    /Eric

    Since (x^x)^x reduces to x^(x*x), exponentiation is assumed to bind
    to the right, i.e. x^x^x means x^(x^x).
    
    However, x=-sqrt(2) appears also to be a solution.

Regarding the equation

    	    ad infinitum
    	   X
    	  X
         X
        X = 2,

reply .4 is correct that the proper grouping is 

    x**(x**(x** . . . ))) = 2,

and reply .1 is correct that we can set y = 2 = x**y = x**2, 
yielding x = sqrt(2).

There are some subtleties to this problem.  Consider the equation

    x**(x**(x** . . . ))) = 4,

which by the above analysis reduces to x**4 = 4, or x = sqrt(2).

So for x = sqrt(2), what is the value of x**(x**(x** ... ))) ?  2 or 4?

Answer in next reply (to be labeled "spoiler").

David Ellis -- Secure Systems Group -- LTN2-2/C08 -- DTN 226-6784 

*** SPOILER ***

The expression x**(x**(x** ... )) does not yield a well-defined value.

If we let y = x**(x**(x** ... )), then y = x**y WHEREVER THE SUBSTITUTION
IS MEANINGFUL.

If we solve for x, we obtain x = y**(1/y), which is a well-defined function
yielding a unique positive real value of x for each positive real number y.  

This function is not one-to-one; in particular, its values for y=2 and y=4 are 
x = sqrt(2).

Differential calculus tells us that the function has an absolute
maximum point at y = e.  The range of the function is {x: 0 < x <= e**(1/e)}.

The only valid interpretation of the expression x**(x**(x** ... )) is
as an inverse to the above function.  The expression is single-valued for
0 < x <= 1 or x = e**(1/e), double-valued for 1 < x < e**(1/e), and
undefined elsewhere.

David Ellis -- Secure Systems Group -- LTN2-2/C08 -- DTN 226-6784 

    This is a cute problem and I think Gilbert's reply has to win the
    "subtlety of the year" award.
    
    Regarding the previous spoiler, isn't it true that
    
    if you were to sit down and code up a program that calculated
    x^(x^(x^(...))) for x = sqrt(2), then you would get 2?  (I did).
    OK, so I only did the first 1000 exponentiations, but convergence
    was rapid and there was no oscillation or other indication of
    multiple values.
    
    I haven't done enough thinking about this, but it seems that if
    we can calculate a deterministic value for an expression, then one
    should be able to produce a theory consistent with observed results.

      John

    John, you claim you did the "first 1000 exponentiations", but
    I'm not sure.  Perhaps you actually did the LAST exponentiations,
    or the WRONG exponentiations.  You're faced with:
    
    	x^(x^(x^(...)))
    
    You say you used x=sqrt(2).  I bet you also used "..."=sqrt(2), right ?
    Perhaps this is an invalid starting condition.
    
    /Eric

    Eric, my VT241 is blushing.  Yes, I only did the LAST 1000 iterations.
    I computed x^(x^(x^(...^y))) for y = sqrt(2).  Interestingly enough,
    the exponentiation seems to converge to 2 for all real y <= 4.   (!)

    The exponentiation diverges for y > 4.  That would seem to imply that
    the starting point for y was well-chosen, seeing as how any good value
    (one that results in convergence) produces the same result as any other.

    Hence, again, it seems reasonable to assume the exponentiation is
    well-defined, and that one should be able to deal with it theoretically.

      John

    Echoing .7:
                    .
                   .
                  x
                 x
    	 A(x) =	x
    
    seems to make most sense as the limit of {Aj}, where A1 = x and
                        Aj
    		Aj+1 = x
    
    Then A( sqrt(2) ) = 2.  Note also A(-1) = -1.
    
    An inverse function, B(y) = y ^ (1/y), doesn't seem to help as much
    as it looks.  For instance, B(-1/2) = 4.  And B(4) = B(2).

    John

If x may be complex, are there any other solutions?

Certainly, 2 = x^2 yields only x = +/- sqrt(2), but 2 = x^(x^2) might
have other solutions, right?  Note that if there are other solutions
then x^x^... 'oscillates'.