1
$\begingroup$

This question concerns Kummers confluent Hypergeometric function $_1F_1$, sometimes denoted $M$. Recall that $$ M(a;b;z) = {_1F_1 (a;b;z)} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\frac{(a)_n}{(b)_n} $$

I am seeking to approximate $M(-\lfloor \alpha X\rfloor;-X;-1)$ when $X$ is a large positive real and $\alpha \in [0,1]$, perhaps by way of a series in $\alpha$.

Numerically, I find that:

$$ M(-\lfloor \alpha X\rfloor;-X;-1) \approx 1 - \left( \frac{e-1}{e}+\frac{\pi}{10}\right)\alpha + \left( \frac{\pi}{10}\right) \alpha^2 $$ is within 0.4% of the true value when $X\gg100$ and $\alpha \in [0,1]$. However, I see no combinatorial interpretation, or way to derive this from the hypergeometric series, or general approximation to higher orders.

$\endgroup$
1
$\begingroup$

Since $${_1\hspace{-1.5px}F_1}(-\lfloor\alpha X\rfloor;-X;-1)= \sum_{k=0}^{\infty}\frac{(-\lfloor\alpha X\rfloor)_k}{(-X)_k}\frac{(-1)^k}{k!}= \\\sum_{k=0}^{\infty} \frac{(\lfloor\alpha X\rfloor)!\,(X-k)!}{X!\,(\lfloor\alpha X\rfloor-k)!} \frac{(-1)^k}{k!}$$ and $$\lim_{X \to \infty} \frac{(\alpha X-\{\alpha X\})!\,(X-k)!}{X!\,(\alpha X-\{\alpha X\}-k)!}= \alpha^k,$$ taking the element-wise limit gives $$\lim_{X \to \infty} {_1\hspace{-1.5px}F_1}(-\lfloor\alpha X\rfloor;-X;-1)=e^{-\alpha}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.