top | item 42505688 (no title) rak1507 | 1 year ago What do you mean by x^x^x^... = 2? Isn't the solution to that sqrt(2)? discuss order hn newest 77pt77|1 year ago f(x,n) = f(x,n-1) ^ xf(x,1) = xc_n is such that f(c_n,n) = 2c = lim c_n rak1507|1 year ago Do you mean f(x,n-1)^x or x^f(x,n-1)? With the first definition, c=1, the second, c=sqrt(2). I'm still not seeing the connection to omega.
77pt77|1 year ago f(x,n) = f(x,n-1) ^ xf(x,1) = xc_n is such that f(c_n,n) = 2c = lim c_n rak1507|1 year ago Do you mean f(x,n-1)^x or x^f(x,n-1)? With the first definition, c=1, the second, c=sqrt(2). I'm still not seeing the connection to omega.
rak1507|1 year ago Do you mean f(x,n-1)^x or x^f(x,n-1)? With the first definition, c=1, the second, c=sqrt(2). I'm still not seeing the connection to omega.
77pt77|1 year ago
f(x,1) = x
c_n is such that f(c_n,n) = 2
c = lim c_n
rak1507|1 year ago