Searching for an asymptotic to exp[0.5]

[sheldonison]

\( \text{dexphalf}(x)=\frac{d}{dx} \exp^{0.5}(x) \;\; h_n = \text{dexphalf}^{-1}(n) \)

\( a_n = \exp(\exp^{0.5}(h_n) - n h_n)\;\;\; \ln(a_n) = \exp^{0.5}(h_n) - n h_n \)

I have an improvement in the equation for a_n
\( a_n = \frac{\exp(\exp^{0.5}(h_n) - n h_n)}{\sqrt{2 \pi \frac{d^2}{dx^2}\exp^{0.5}(h_n) }} \; \; a_0 = \exp^{0.5}(0) \)

This is from post#16. Start with a generic function f(x) with f(x) and f'(x) increasing, and positive for all x>0 and eventually increasing faster than any x^n. f(x) may or may not be entire. Then we can also use this approximation. Surprisingly, this approximation leads to Stirling's approximation for \( a_n \) when used for f(x)=exp(x). For a generic f(x), first we generate g(x) which changes the contour integral to a line integral between \( \pm \pi i \).

\( g(x) = \ln\( f( e^x) \) \;\;\; \text{dg}(x)=\frac{d}{dx} g(x) \;\;\; h_n = \text{dg}^{-1}(n) \)

\( a_n \; = \; \frac{1}{2\pi} \oint \frac{f(x)}{x^{n+1}}\; = \; \frac{1}{2\pi i} \int_{x-\pi i}^{x+\pi i} {\exp \( g(x) - n\cdot x \)} \)

The line integral is exact, for any value of x if f(x) is entire. We may have to cancel out some logarithmic singularities in g(x). The minimum of \( \frac{f(x)}{x^{-n}} \) occurs at \( x=e^{h_n} \). Then we have the Gaussian approximation for \( a_n \) for a generic function f(x) is as follows.

\( a_n \approx \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }} \;\;\; \) This is the generic approximation for \( a_n \) using \( h_n = \text{dg}^{-1}(n) \) from above

\( f(x) \approx f(0) + \sum_{n=1}^{\infty} a_n \cdot x^n\;\;\; \) You can refer back to post#16 to see the derivation of this approximation if interested. This is the best approximation I've found without using integration. If f(x) is entire, then integration allows for an exact value of \( a_n \). If f(x) is not entire, then variants on the integration technique can also be used, which leads to some of the other more accurate approximations I generated for \( exp^{0.5}(x) \) later in this thread after post#16.

So, what do you think the Gaussian approximation would be for \( a_n \) for \( f(x)=\exp(x) \)? If you go through the arithmetic, then you get Stirling's approximation!

\( g(x) = \ln \( \exp(f(x)) \) = \exp(x) \;\; g'(x)=\exp(x) \;\; g''(x)=\exp(x) \)
\( h_n = \ln(n) \)

\( \frac{1}{a_n} \; = \; n! \; \approx \; \sqrt{(2\pi n)(\frac{n}{e})^n} \;\;\; \) The Gaussian approximation \( a_n \) for exp(x) turns out to be exactly Stirling's approximation

This value of a_n does work rather well since Stirling's approximation has a known error term of \( \frac{1}{12n} \). It it does not converge quite as quickly as the Gaussian approximation for \( \exp^{0.5}(x) \), which was the original problem motivating this thread. This leads to the conjecture that in general, the Gaussian approximation for the Taylor series coefficients probably works better for functions that grow slower than exponential functions. One can conjecture that for many functions the ratio of f(x) over the Gaussian approximation for f(x) approaches arbitrarily close to 1 as x gets arbitrarily large, but its an open question as to for what class of entire functions the ratio Gaussian approximation does not converge to 1.