- #1
converting1
- 65
- 0
which is greater as n gets large, [itex] f(n) = 2^{2^{2^n}} [/itex] or [itex] g(n) = 100^{100^n} [/itex]
instinctively I'd go with f(n) but I have no idea of actually showing that f(n) would indeed get larger, obviously sticking in values of n doesn't particularly work. A method I was thinking was to show many digits would f(n) be when n=100, and same with g(n), I attempted this but to no avail,
any hints would be appreciated
thanks.
instinctively I'd go with f(n) but I have no idea of actually showing that f(n) would indeed get larger, obviously sticking in values of n doesn't particularly work. A method I was thinking was to show many digits would f(n) be when n=100, and same with g(n), I attempted this but to no avail,
any hints would be appreciated
thanks.