08/01/2014, 11:36 PM
Some further thought :
\( a_n = \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1}) + ln(n-1)) \)
There is good news and bad news.
Basicly both are that this is in a way very similar to the Original estimate for a_n combined with our estimate for a_(n+1).
\( a_n = (n-1) \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1}) \)
\( a_n / (n-1) = \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1}) \)
Just as our estimate for a_(n+1) from a_n and the Original estimate of a_n.
But maybe a refinement of the previous post helps.
Am I wrong in assuming we prefer to solve for one variable ?
regards
tommy1729
\( a_n = \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1}) + ln(n-1)) \)
There is good news and bad news.
Basicly both are that this is in a way very similar to the Original estimate for a_n combined with our estimate for a_(n+1).
\( a_n = (n-1) \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1}) \)
\( a_n / (n-1) = \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1}) \)
Just as our estimate for a_(n+1) from a_n and the Original estimate of a_n.
But maybe a refinement of the previous post helps.
Am I wrong in assuming we prefer to solve for one variable ?
regards
tommy1729
