11/05/2014, 11:58 PM
POST 102 correction !
ln(a_n x^n) = ln(exp(x)sqrt(x))
ln(a_n) + n ln(x) = x + 0.5 ln(x)
ln(a_n) = min ( x - (n-1/2) ln(x) )
d/dx ( x - (n-1/2) ln(x) ) = 1 - (n-1/2) / x
1 = (n-1/2) / x
x = n-1/2
ln(a_n) = n - 1/2 - (n - 1/2) ln(n - 1/2)
NOTICE this is equivalent to the solution for exp , but n replaced by n - 1/2 ...
which is the analogue of the mittag-leffler function solution !!
BTW notice Im aware of sheldon's scaling from post 9 , however working with that is not so easy yet.
regards
tommy1729
ln(a_n x^n) = ln(exp(x)sqrt(x))
ln(a_n) + n ln(x) = x + 0.5 ln(x)
ln(a_n) = min ( x - (n-1/2) ln(x) )
d/dx ( x - (n-1/2) ln(x) ) = 1 - (n-1/2) / x
1 = (n-1/2) / x
x = n-1/2
ln(a_n) = n - 1/2 - (n - 1/2) ln(n - 1/2)
NOTICE this is equivalent to the solution for exp , but n replaced by n - 1/2 ...
which is the analogue of the mittag-leffler function solution !!
BTW notice Im aware of sheldon's scaling from post 9 , however working with that is not so easy yet.
regards
tommy1729
