robertm@dartvax.UUCP (Robert P. Munafo) (12/03/83)
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If it is given that x^x = 2, is it possible to obtain an exact
expression for x which contains no variables? ( Or, general case:
if x^x = y, what is y in terms of x?
No one here at Dartmouth has been able to solve this. Is it
unsolvable?
Robert P. Munafo ...|decvax|dartvax|robertmdebray@sbcs.UUCP (Saumya Debray) (12/06/83)
> If it is given that x^x = 2, is it possible to obtain an > exact expression for x which contains no variables? ( Or, > general case: if x^x = y, what is y in terms of x? For values of x > 1, here's a solution: for x^x = 2, we have x log x = log 2, whence log 2 x = ----- log x which leads to a continued fraction representation for x: log 2 x = -------------------------------- log 2 log ( ------------------------ ) log 2 log ( ------------------ ) log 2 log ( ------------ ) . . . The general case (x^x = y) has a similar solution. Questions: (1) Does anyone have a general solution that would hold for all values of x ? (2) It seems to me that since "x^x = 2" is a polynomial of order x, it ought to have x roots, of which the above is just one. Any suggestions for a general solution? -- Saumya Debray SUNY at Stony Brook {philabs, ogcvax}!sbcs!debray
leimkuhl@uiuccsb.UUCP (12/10/83)
#R:dartvax:-46400:uiuccsb:9700014:000:290 uiuccsb!leimkuhl Dec 9 13:16:00 1983 In response to statement 2, x^x is not a "polynomial of order x." A polynomial has fixed (integral) order, and the fundamental theorem of algebra does not even have analogy for such functions as x^(3/2). Thus the statement really has no meaning. Ben Leimkuhler (uiucdcs!leimkuhl)