11/07/2009, 11:30 PM
If we set \( f(x+1)=b^{f(x)} \) which is to say
\( \lim_{k\to \infty} (log_{b}^{ok}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (log_{b}^{o(k-1)}(x({}^k b- {}^{(k-1)} b)+{}^k b) ) \)
then reduce it to
\( \lim_{k\to \infty} (log_{b}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (x({}^k b- {}^{(k-1)} b)+{}^k b) \)
Then just strait up plug in infinity for k
we get \( log_{b} {}^\infty b = {}^\infty b \) which is the same as \( {}^\infty b = {}^\infty b \)
This is really weird because if i do \( \lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \) for \( b= sqrt2 \)
i get about 1.558 which is substantially larger then \( sqrt2 \)
Can anyone else confirm that \( \lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \approx 1.558 \) for \( b= sqrt2 \) ?
\( \lim_{k\to \infty} (log_{b}^{ok}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (log_{b}^{o(k-1)}(x({}^k b- {}^{(k-1)} b)+{}^k b) ) \)
then reduce it to
\( \lim_{k\to \infty} (log_{b}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (x({}^k b- {}^{(k-1)} b)+{}^k b) \)
Then just strait up plug in infinity for k
we get \( log_{b} {}^\infty b = {}^\infty b \) which is the same as \( {}^\infty b = {}^\infty b \)
This is really weird because if i do \( \lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \) for \( b= sqrt2 \)
i get about 1.558 which is substantially larger then \( sqrt2 \)
Can anyone else confirm that \( \lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \approx 1.558 \) for \( b= sqrt2 \) ?
