WARNING : CORRECTED IN POST 117 !
---
Using post 9 for exp(x) sqrt(x) :
ln(a_n x^n) < ln(exp(x)sqrt(x))
...
ln(a_n) = min ( exp(x) - (n-1/2) x )
--
d/dx [exp(x) - (n-1/2) x] = exp(x) - (n - 1/2)
=>
exp(x) = n - 1/2
x = ln(n - 1/2)
--
ln(a_n) = exp(ln(n - 1/2)) - (n - 1/2) ln(n - 1/2)
=>
a_n = exp( n - 1/2 - (n - 1/2) ln(n - 1/2) )
a_n = exp( (n - 1/2) (1 - ln(n - 1/2)) )
...
Gamma(n + 1/2) vs exp( - (n - 1/2) (1 - ln(n - 1/2)) )
=>
Loggamma(n + 1/2) vs - (n - 1/2) (1 - ln(n - 1/2))
=>
Loggamma(z) vs - z ( 1 - ln(z) )
Loggamma(z) vs z ( ln(z) - 1 )
=>
Lim loggamma(z) / ( z ( ln(z) - 1 ) ) < Constant ?
From the Stirling series we know that the limit equals 1.
This implies that using post 9 also gives the correct solution up to a multiplicative constant !
So the post 9 method does a good job to estimate the fake exp(x)sqrt(x).
regards
tommy1729
---
WARNING : CORRECTED IN POST 117 !
---
Using post 9 for exp(x) sqrt(x) :
ln(a_n x^n) < ln(exp(x)sqrt(x))
...
ln(a_n) = min ( exp(x) - (n-1/2) x )
--
d/dx [exp(x) - (n-1/2) x] = exp(x) - (n - 1/2)
=>
exp(x) = n - 1/2
x = ln(n - 1/2)
--
ln(a_n) = exp(ln(n - 1/2)) - (n - 1/2) ln(n - 1/2)
=>
a_n = exp( n - 1/2 - (n - 1/2) ln(n - 1/2) )
a_n = exp( (n - 1/2) (1 - ln(n - 1/2)) )
...
Gamma(n + 1/2) vs exp( - (n - 1/2) (1 - ln(n - 1/2)) )
=>
Loggamma(n + 1/2) vs - (n - 1/2) (1 - ln(n - 1/2))
=>
Loggamma(z) vs - z ( 1 - ln(z) )
Loggamma(z) vs z ( ln(z) - 1 )
=>
Lim loggamma(z) / ( z ( ln(z) - 1 ) ) < Constant ?
From the Stirling series we know that the limit equals 1.
This implies that using post 9 also gives the correct solution up to a multiplicative constant !
So the post 9 method does a good job to estimate the fake exp(x)sqrt(x).
regards
tommy1729
---
WARNING : CORRECTED IN POST 117 !
