OFFSET
1,1
COMMENTS
Given z > 0, there exist positive real numbers x < y with x^y = y^x = z, if and only if z > e^e. In that case, (x,y) = ((1 + 1/t)^t,(1 + 1/t)^(t+1)) for some t > 0. For example, t = 1 gives 2^4 = 4^2 = 16 > e^e. When x^y = y^x = 17, at least one of x and y is transcendental. See Sondow and Marques 2010, pp. 155-157.
LINKS
J. Sondow and D. Marques, Algebraic and transcendental solutions of some exponential equations, Annales Mathematicae et Informaticae, 37 (2010), 151-164.
EXAMPLE
y=4.89536795554611347196719338722983584947273195280937244363084664929554121...
MATHEMATICA
x[t_] := (1 + 1/t)^t; y[t_] := (1 + 1/t)^(t + 1); t = t/. FindRoot[x[t]^y[t] == 17, {t, 1}, WorkingPrecision -> 120]; RealDigits[y[t], 10, 100] // First
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Jonathan Sondow, Aug 30 2011
STATUS
approved