10/10/2015, 03:08 AM
(This post was last modified: 10/10/2015, 03:21 AM by sheldonison.)
(10/09/2015, 11:56 AM)tommy1729 Wrote: 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.
- Sheldon
