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I am trying to find the asymptotic expansion of $_1F_1\left(-m;\frac{1}{2};-\frac{1}{2}\right)$ for large $m$ where $_1F_1\left(a;b;z\right)$ is the Kummer confluent hypergeometric function, also denoted as $M(a,b,z)$ in Chapter 13 Confluent Hypergeometric Functions of the Digital Library of Mathematical Functions (http://dlmf.nist.gov/13). In the standard notation I am interested in the limit $a \to -\infty$ with the following values fixed $b=1/2$ and $z=-1/2$. In section 13.8 of DLFM they give approximations for:

  1. $a \to \infty$ and $b \leq 1$ (Eq. 13.8.8)

  2. $a \to -\infty$ and $b \geq 1$ (Eq. 13.8.9)

  3. $a \to -\infty$ and $(b-1)/|a|$ positive

  4. Finally $a \pm \infty$ and $\text{ph}(a) \leq \pi -\delta$ i.e. $a$ cannot be purely real negative. (See Eq. 13.8.13)

As you can see none of these cases apply to my problem. Any suggestions on how to get the scaling?

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Don't know if the question is still relevant, but the asymptotics can be obtained as follows. Use the identity $${_1F_1}\left(-m;\frac{1}{2};-\frac{1}{2}\right)= e^{-1/2}\,{_1F_1}\left(m+\frac{1}{2};\frac{1}{2};\frac{1}{2}\right)$$ Then use the integral representation $${_1F_1}\left(m+\frac{1}{2};\frac{1}{2};\frac{1}{2}\right)= \frac{1}{\Gamma\left(m+\frac{1}{2}\right)} \int_{0}^{\infty}e^{-t}t^{m-1/2}\,{_0F_1}\left(;\frac{1}{2};\frac{t}{2}\right)dt= \\\frac{1}{\Gamma\left(m+\frac{1}{2}\right)} \int_{0}^{\infty}e^{-t}t^{m-1/2}\cosh(\sqrt{2\,t})dt$$ The exponent is $-t+(m-1/2)\ln(t)+\sqrt{2\,t}$; the other term coming from $\cosh$ contributes subdominant terms and can be neglected. Apply (a variation of) Laplace's method to get $${_1F_1}\left(-m;\frac{1}{2};-\frac{1}{2}\right)\sim e^{\sqrt{2\,m}-1/4} \left(\frac{1}{2}+\frac{7\sqrt{2}}{96\sqrt{m}}-\frac{23}{4608m}+\ldots\right)$$

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  • $\begingroup$ Don't worry, questions are forever relevant here. Until your answer answers the question and significantly differs from the already posted answers, it will be okay. Welcome on the Math SE! :-) $\endgroup$ – peterh Jan 16 '18 at 17:30
  • $\begingroup$ Dear @Maxim -- Thank you so much. This precisely what I was looking for! $\endgroup$ – Nicolás Quesada Jan 17 '18 at 18:25

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