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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.

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1 Answer 1

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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 $$\lim_{X \to \infty} {_1\hspace{-1.5px}F_1}(-\lfloor\alpha X\rfloor;-X;-1)=e^{-\alpha}.$$

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