Searching for an asymptotic to exp[0.5]

[jaydfox]

\( \exp(x)\sqrt{x} \approx \sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+0.5)}\;\; \) Have you seen this excellent asymptotic entire series? I can't explain it, other than it matches the "fakefunc" integral

Hey, this reminds me of the generalization I made for the (unscaled) asymptotic binary partition function:

\(
f(x) = \sum_{k=0}^{\infty}\frac{1}{2^{k(k-1)/2} k!} x^k \\
\Rightarrow f(x) = \sum_{k=0}^{\infty}\frac{1}{2^{k(k-1)/2} \Gamma(k+1)} x^k
\)

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 (well, any complex number, but the complex versions are less interesting).

\(
f_{\beta}(x) = \sum_{k=-\infty}^{\infty}\frac{2^{-(k+\beta)(k+\beta-1)/2}}{\Gamma(k+\beta+1)} x^{k+\beta}
\)

We can apply the same generalization to exp(x):

\(
\exp(x) \approx \sum_{k=-\infty}^{\infty}\frac{x^{k+\beta}}{\Gamma(k+\beta+1)}
\)

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).

Now, set beta = 1/2 and truncate the negative powers of x:
\(
\exp(x) \approx \sum_{k=0}^{\infty}\frac{x^{k+1/2}}{\Gamma(k+3/2)}
\)

...and comparing your power series, it easily follows that your power series is asymptotic to exp(x) x^(1/2). Hmm, oh dear, I think you missed a +1 in the Gamma function in your series? (LOL, it's okay, I usually forget the +1 as well.) If you set beta=-1/2, then all is well!

\(
\frac{\exp(x)}{\sqrt{x}} \approx \sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+1/2)}
\)

\(
\exp(x)\sqrt{x} \approx \sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+3/2)}
\)