Searching for an asymptotic to exp[0.5]
#85
(09/01/2014, 10:24 PM)tommy1729 Wrote: It seems my friend ( our ? ) mick has given us ( myself and sheldon in particular ) credit for the use of " fake function theory " Smile

here is the link :

http://math.stackexchange.com/questions/...for-real-x

As I mentioned before z - ln(2sinh(z)/z) is asymptotic to ln(z) near the real line, but it fails to be entire ( sinh is periodic ! ).

Guess we could consider the fake ln(x^2 + 1) here too.

Unlike the half-exp we are considering more standard functions which might give intresting closed form results !
...
Thanks for the link Tommy! I've been overseas vacationing... I was able to use the basic recipe from this post to interpolate \( \ln(x^2+1)\exp(x^2) \) as an even function, just as you suggested! I will post more later, either here or at Mathstack.

I have also been working on an example that should allow us to make the "fake function" theory more rigorous. A very interesting asymptotic function is below. It is interesting because the corresponding \( \ln(f(\exp(z))) \) function is simply \( \frac{x^2}{2} \), whose derivative is x, so \( h(n)=n \), so all of the Taylor series coefficients of the interpolating function have an exact closed form, as do all of the error equations.

\( f(x) = \exp \left\( \frac{(\ln(x))^2}{2} \right\) \)

\( a_n = \frac{1}{\exp(0.5n^2)\sqrt{2\pi}}\;\; \) for this function, the Gaussian approximation is also the best approximation.

\( f(x) \approx \sum_{n=-\infty}^{\infty} a_n x^n\;\; \) this is a converging Laurent series. We don't use the negative coefficients for the entire interpolant.

The Laurent series interpolant can also be expressed in a quickly converging closed form.
\( \sum_{n=-\infty}^{\infty} a_n x^n\;=\;\sum_{n=-\infty}^{\infty} \exp \left\( \frac{(\ln(x)+2n\pi i)^2}{2} \right\)\;\; \) f(x) corresponds to n=0
- Sheldon


Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/03/2014, 01:04 PM

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