12/15/2009, 01:01 AM
(This post was last modified: 12/15/2009, 01:02 AM by dantheman163.)
Upon closer study i think i have found a formula that actually works.
\( {}^ x b = \lim_{k\to \infty} \log_b ^k({}^ k b (ln(b){}^ \infty b)^x-{}^ \infty b(ln(b){}^ \infty b)^x+{}^ \infty b) \)
Also i have noticed that this can be more generalized to say,
if \( f(x)=b^x \)
then
\( f^n(x)= \lim_{k\to \infty} \log_b ^k( Exp_b^k(x) (ln(b){}^ \infty b)^n-{}^ \infty b(ln(b){}^ \infty b)^n+{}^ \infty b) \)
numerical evidence shows this to be true.
some plots
\( {}^x sqrt2 \)
half iterite of \( sqrt2^x \)
The graph falls apart at x=4 because \( Exp_b^k(4.01) \) diverges as k goes to infinity.
thanks.
Edit: sorry for the huge pictures
\( {}^ x b = \lim_{k\to \infty} \log_b ^k({}^ k b (ln(b){}^ \infty b)^x-{}^ \infty b(ln(b){}^ \infty b)^x+{}^ \infty b) \)
Also i have noticed that this can be more generalized to say,
if \( f(x)=b^x \)
then
\( f^n(x)= \lim_{k\to \infty} \log_b ^k( Exp_b^k(x) (ln(b){}^ \infty b)^n-{}^ \infty b(ln(b){}^ \infty b)^n+{}^ \infty b) \)
numerical evidence shows this to be true.
some plots
\( {}^x sqrt2 \)
half iterite of \( sqrt2^x \)
The graph falls apart at x=4 because \( Exp_b^k(4.01) \) diverges as k goes to infinity.
thanks.
Edit: sorry for the huge pictures
