jlh@hou2e.UUCP (J.HEATWOLE) (04/30/85)
> To restate the problem: > > x^x^x^... = 2 What is x? > > x is the square root of 2 -- The test answer key says to use substitution: > > x^(x^x^x^...) = 2 Substitute 2 for the expression in parentheses, > > x^2 = 2 > Exponentiation is not associative. This solution depends on interpreting the problem definition as x^(x^(x^(x^... = 2, in which case x = sqrt(2) is the correct solution. This solution has been shown to be incorrect by several posters who have interpreted the problem definition as (((...((x^x)^x)^x)...)^x) = 2. In this case x = sqrt(2) => L.H.S. = oo. If interpreted this way, the problem has no solution, since (((...((x^x)^x)^x)...)^x) = oo for all x > 1. Now I have a question. x^(x^(x^(x^... = 2 implies x = sqrt(2) as shown earlier. Then in the equation x^(x^(x^(x^... = 4, we can perform a similar substitution to obtain x^4 = 4 => x = sqrt(2). How can the L.H.S. using x = sqrt(2) equal both 2 and 4? Jeff Heatwole ..!hou2e!jlh
wws@whuxlm.UUCP (Stoll W William) (05/01/85)
> Exponentiation is not associative. > This solution depends on interpreting the problem definition as > > x^(x^(x^(x^... = 2, > > in which case x = sqrt(2) is the correct solution. > > This solution has been shown to be incorrect by several posters > who have interpreted the problem definition as > > (((...((x^x)^x)^x)...)^x) = 2. > > In this case x = sqrt(2) => L.H.S. = oo. > If interpreted this way, the problem has no solution, since > > (((...((x^x)^x)^x)...)^x) = oo for all x > 1. x^(x^(x^(x^... = 2 is how the C compiler would interpret the equation (if it didn't choke on all the x's %-) ) > > Now I have a question. > > x^(x^(x^(x^... = 2 implies x = sqrt(2) as shown earlier. > > Then in the equation x^(x^(x^(x^... = 4, we can perform a similar > substitution to obtain > > x^4 = 4 => x = sqrt(2). > > How can the L.H.S. using x = sqrt(2) equal both 2 and 4? > > Jeff Heatwole > ..!hou2e!jlh I'm stumped. We should write to the Manufacturer. Bill Stoll, ..!whuxlm!wws
bs@faron.UUCP (Robert D. Silverman) (05/02/85)
> > Exponentiation is not associative. > > This solution depends on interpreting the problem definition as > > > > x^(x^(x^(x^... = 2, > > > > in which case x = sqrt(2) is the correct solution. > > > > This solution has been shown to be incorrect by several posters > > who have interpreted the problem definition as > > > > (((...((x^x)^x)^x)...)^x) = 2. > > > > In this case x = sqrt(2) => L.H.S. = oo. > > If interpreted this way, the problem has no solution, since > > > > (((...((x^x)^x)^x)...)^x) = oo for all x > 1. > > x^(x^(x^(x^... = 2 is how the C compiler would interpret the > equation (if it didn't choke on all the x's %-) ) > > > > > Now I have a question. > > > > x^(x^(x^(x^... = 2 implies x = sqrt(2) as shown earlier. > > > > Then in the equation x^(x^(x^(x^... = 4, we can perform a similar > > substitution to obtain > > > > x^4 = 4 => x = sqrt(2). > > > > How can the L.H.S. using x = sqrt(2) equal both 2 and 4? > > > > Jeff Heatwole > > ..!hou2e!jlh > > I'm stumped. We should write to the Manufacturer. > > Bill Stoll, ..!whuxlm!wws Apparantly you didn't get my note about CONVERGENCE!!! Anyone who's gotten beyond 1st year calculus should know that not all infinite sequences converge and of those that do some converge only for specific values. Since x^x^x^x... converges only in the OPEN interval: (e^(-1/e) , e^(1/e)) The exponentiation reaches a maximum value when x = e^(1/e) - epsilon (pick a suitable epsilon) and this value is less than 4. Thus, your question has no solution. Such problems frequently arise when your try to do formal manipulation of infinite sequences without regard to convergence.