08/16/2010, 11:28 PM
with \( b=t^{\frac1t} \), t in the range 1<t<exp(1)
it seems that
\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{h(b)}{b\^\^ ^j })^{\frac1j} -> \log(t) \)
i think i can show this to be correct as well.
that would be 2 proofs in 1 day
i believe the above is a koenigs like form in disguise.
if we take blog instead of log we get :
\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{h(b)}{b\^\^ ^j })^{\frac1j} -> \log(t) \)
yes , exactly the same , because the blog = log / log(b) and the log(b) gets canceled by the power ^(1/j).
then we can reduce the fraction to a difference !
that difference grows like the (de)nominator of the koenigs function does.
hence log(t) is simply the absolute value of the derivate of the fixpoint in the analogue koenigs function.
and thus the identity is explained and proved by koenigs function.
Q.E.D.
regards
tommy1729
it seems that
\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{h(b)}{b\^\^ ^j })^{\frac1j} -> \log(t) \)
i think i can show this to be correct as well.
that would be 2 proofs in 1 day
i believe the above is a koenigs like form in disguise.
if we take blog instead of log we get :
\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{h(b)}{b\^\^ ^j })^{\frac1j} -> \log(t) \)
yes , exactly the same , because the blog = log / log(b) and the log(b) gets canceled by the power ^(1/j).
then we can reduce the fraction to a difference !
that difference grows like the (de)nominator of the koenigs function does.
hence log(t) is simply the absolute value of the derivate of the fixpoint in the analogue koenigs function.
and thus the identity is explained and proved by koenigs function.
Q.E.D.
regards
tommy1729