09/15/2014, 03:53 AM
(This post was last modified: 09/15/2014, 04:33 AM by sheldonison.)
(09/13/2014, 11:49 PM)jaydfox Wrote: ...
Treating Gamma(k) at negative non-positive integers as infinity, and the reciprocal of such as zero, we can take the limit from negative to positive infinity. And we can replace k with (k+b), where b is zero in the original solution, but can now be treated as any real...
\(
\exp(x) \approx \sum_{k=-\infty}^{\infty}\frac{x^{k+\beta}}{\Gamma(k+\beta+1)}
\)
Notice that I put "approximately equal". I haven't checked, but I assume it's exactly equal, but only in the sense that it should satisfy the functional equation exp'(x) = exp(x).
So we also have a fakeexp(z) function, and I especially like the simplicity of f'(x)=f(x) in the non-converging limit to explain why f(x)~exp(x). Here, going back to k=0.5, for simplicity.... We have to put some bounds on k, since the infinite Laurent series does not converge anywhere.
\(
\exp(x) \sim f(x) = \sum_{k=0}^{\infty}\frac{x^{k+0.5}}{\Gamma(k+0.5+1)} \)
\( \frac{d}{dx}f(x)=\sum_{k=-1}^{\infty} \frac{x^{k+0.5}}{\Gamma(k+1.5)} =f(x) + \frac{x^{-0.5}}{\Gamma(0.5)} \)
As \( \Re(x) \) gets arbitrarily large, the error term becomes more and more insignificant, relative to the exp(z) term, and the number of terms you can include also increases, until the \( \frac{x^{-n+0.5}}{\Gamma(-n+1.5)} \) starts growing in magnitude as n gets bigger negative ....
- Sheldon
