Searching for an asymptotic to exp[0.5]

[tommy1729]

So gaussian can not be exp ... / sqrt( 2 pi g" (h_n)).

It has to be exp( g(h_n) - n h_n) * sqrt( 2 pi g " (h_n) ).

Start with \( \int_{-\infty}^{+\infty}\exp\(\frac{-g''x^2}{2}\) = \sqrt{\frac{2\pi}{g''}} \)

I think your just missing a little bit of algebra. We are interested an approximation for
\( a_n \; = \; \frac{1}{2\pi i} \int_{h_n-\pi i}^{h_n+\pi i} \exp \( g(x) - n\cdot x \) \;\;\; \) This is an exact equation if \( g(x)=\ln(f(\exp(x))) \)

\( a_n \approx \; \frac{1}{2\pi i} \cdot \exp \( g(h_n) - n\cdot h_n \) \cdot \int_{h_n- i\infty}^{h_n+ i\infty} \exp\(\frac{g''x^2}{2}\) \;\;\; \) see post#16 for more details on getting to this step

\( a_n \approx \; \exp \( g(h_n) - n\cdot h_n \) \cdot \frac{1}{2\pi} \cdot \sqrt{\frac{2\pi}{g''}} \)

\( a_n \approx \; \exp \( g(h_n) - n\cdot h_n \) \cdot \sqrt{\frac{1}{2\pi g''}} \)

\( a_n \approx \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }} \)

All of the numerical approximations I have posted, use this version of the Gaussian approximation, and get excellent results. In most cases, the Gaussian approximation is also an over-approximation of an entire function with all positive derivatives.

Hmm.

The reason I got confused is

1) i incorrectly thought 1/n! ~ sqrt(2 pi n) (e/n)^n.
See the fake exp.


2) I assumed the gaussian was bigger then S9 because

2 a) the Gaussian is better then S9.

2 b) You Said in post 9 that the method ( S9 ) always gives An underestimate.

So apparantly 2 b) is false.

This was not mentioned before !

Im sorry. Big misunderstanding.

Did I understand my misunderstanding Well ?

Since 2 b) is false, does this imply there exist functions equal to their fake , independent of the method for fake ?
I guess so.
So we could solve S9( f(x) ) = f(x) ?

Intresting.

I need to read the entire thread again.

Regards

Tommy1729