Thread Rating:
  • 2 Vote(s) - 4 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Searching for an asymptotic to exp[0.5]
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 !
Reply


Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/14/2014, 09:34 PM

Possibly Related Threads...
Thread Author Replies Views Last Post
  Merged fixpoints of 2 iterates ? Asymptotic ? [2019] tommy1729 1 718 09/10/2019, 11:28 AM
Last Post: sheldonison
  Another asymptotic development, similar to 2sinh method JmsNxn 0 2,619 07/05/2011, 06:34 PM
Last Post: JmsNxn



Users browsing this thread: 1 Guest(s)