wessels (07/18/82)
The problem was to solve for x, where x**x**x**x... = 2
[x**x**x == x**(x**x)].
By observing that x**x**x**x... = x**(x**x**x...)
= x**2
one might deduce that x=sqrt(2) is the(an) answer. In fact, my calculator
seems to agree that the series x,x**x,x**x**x,... where x=sqrt(2)
converges to a limit of 2.
By a simple extension to the above argument, one might show that the
solution to general problems of the form x**x**x... = n will be
x=n**(1/n). However, in the case of n=4, we find that x=4**(1/4)=sqrt(2)!!
I think I'm just going to go home and cry into my pillow ...
Ron Wessels
U. of Toronto
...!decvax!utzoo!utcsrgv!wesselsthomas (07/19/82)
You didn't go wrong. The function x^(1/x) (x>=0) has the values given below at specified points: x x^(1/x) 0 0 .1 1e-10 .5 .25 1 1 2 1.4142135623 2.7082818284 1.4446642425 2.7182818284 1.4446678610 [x = e, this is the maximum point] 2.7282818284 1.4446642866 3 1.4422495703 4 1.4142135623 10 1.2589254117 100 1.0471285480 (Thank goodness for bc!) Or, the calculus approach: x^1/x = exp(log(x)/x) d/dx(x^1/x) = exp(log(x)/x) * (1 - log(x)) / x^2 = (1 - log(x)) * x^(1/x - 2) The important thing about this result is that x^(1/x - 2) is always positive, thus the sign of the derivative is totally dependent on the sign of (1 - log(x)). This is 0 for x = e, positive for x < e and negative for x > e. The final question to be answered, then is "For what x,y (x != y) is x^(1/x) = y^(1/y)?" or "when is x^y = y^x?". It is "well known" that the only integer solutions to this equation are the pairs (2,4) and (4,2) (for x and y positive). =Spencer Thomas