WARNING : CORRECTED IN POST 117 !

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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 !