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