05/04/2014, 08:30 AM
I was about to post that f^[-1](x) for x > e is Always smaller than x^x. This is easy to show.
Also f^[-1] = (approximate) = (ln(x)+ln^[2](x) + ln(1+1/x))^[-1].
But this now seems trivial after sheldon's post.
(ln(x)+ln^[2](x) + ln(1+1/x))^[-1] < exp(x)
and that can be improved.
The approximate value for x_n is thus
e^^(n) < x_n < e^^(n+1)
regards
tommy1729
Also f^[-1] = (approximate) = (ln(x)+ln^[2](x) + ln(1+1/x))^[-1].
But this now seems trivial after sheldon's post.
(ln(x)+ln^[2](x) + ln(1+1/x))^[-1] < exp(x)
and that can be improved.
The approximate value for x_n is thus
e^^(n) < x_n < e^^(n+1)
regards
tommy1729