# Limit of confluent Hypergeometric function [closed]

I have the following confluent Hypergeometric function : $$M(A,n+1,nx^2)$$ Here $$0\leq x \leq 1$$. $$n$$ is an integer and I wish to calculate the asymptotic behavior as $$n\rightarrow \infty$$. Also $$A \sim O(1)$$ does not scale with $$n$$. I wish to see it's dependence in the asymptotics so we don't set it to $$1$$. I wish to do this

1. At least close to the domain boundary $$x\sim0$$ and $$x\sim1$$.

2. For the entire domain $$[0,1]$$ to see if there is a simplification that happens and the resulting function in the limit $$n\rightarrow\infty$$ takes a simpler form than $$M(a,b,z)$$.

I have tried to look at the NIST digital library for appropriate asymptotic behavior, but I could not find an expansion for large $$b,z$$.

Any help in understanding the correct asymptotic behavior would be greatly appreciated !

A uniform asymptotic expansion can be obtained using the result linked in the comments. However, I will now derive simpler asymptotic approximations. I will set $$A = a$$, $$x^2 = x$$, and assume $$a > 0$$ and $$0 < x \leq 1$$. Substituting $$u = 1 - \mathrm{e}^{-t}$$ into the integral formula provided in Svyatoslav's answer gives $$M(a, n + 1, nx) = \frac{\Gamma(n+1)}{\Gamma(n+1-a)} \frac{1}{\Gamma(a)} \int_0^{+\infty} (\mathrm{e}^t - 1)^{a-1} \mathrm{e}^{-n(t - x(1 - \mathrm{e}^{-t}))} \, \mathrm{d}t.$$

Case 1. For $$0 \le x < 1$$ and $$(1 - x)^2 n \to +\infty$$, Laplace's method yields the asymptotic expansion: $$\frac{1}{\Gamma(a)} \int_0^{+\infty} (\mathrm{e}^t - 1)^{a-1} \mathrm{e}^{-n(t - x(1 - \mathrm{e}^{-t}))} \, \mathrm{d}t \sim \frac{1}{(1 - x)^a} \frac{1}{n^a} \sum_{k=0}^\infty \frac{c_k(x;a)}{n^k}$$ as $$n \to +\infty$$, where the coefficients $$c_k(x;a)$$ are given by $$c_k(x;a) = \frac{1}{(1 - x)^k} \frac{(a)_k}{k!} \left[ \frac{\mathrm{d}^k}{\mathrm{d}t^k} \left( \left( \frac{\mathrm{e}^t - 1}{t} \right)^{a-1} \left( \frac{(1 - x)t}{t - x(1 - \mathrm{e}^{-t})} \right)^{a+k} \right) \right]_{t=0}.$$ In particular, $$c_0(x;a) = 1, \quad c_1(x;a) = - \frac{a(2ax - a + 1)}{2(1 - x)^2}.$$ Using the known asymptotic expansion for the ratio of two gamma functions, we find $$\frac{\Gamma(n+1)}{\Gamma(n+1-a)} \sim n^a \sum_{k=0}^\infty \binom{a-1}{k} B_k^{(a)} \frac{1}{n^k}$$ as $$n \to +\infty$$, where $$B_k^{(a)}$$ are the generalised Bernoulli numbers. Multiplying these asymptotic expansions gives $$\tag{1} M(a, n+1, nx) \sim \frac{1}{(1-x)^a} \sum_{k=0}^\infty \frac{d_k(x;a)}{n^k}$$ as $$n \to +\infty$$, where $$d_k(x;a) = \sum_{j=0}^k \binom{a-1}{j} B_j^{(a)} c_{k-j}(x;a).$$ In particular, $$d_0(x;a) = 1, \quad d_1(x;a) = -\frac{ax((a-1)x + 2)}{2(1 - x)^2}.$$ The coefficients $$d_k(x;a)$$ have poles of order $$2k$$ at $$x = 1$$, which implies that we require $$(1-x)^2 n \to +\infty$$.

Case 2. For $$x \approx 1$$, we examine $$M(a, n + 1, n + \tau n^{1/2})$$: \begin{align*} M(a,n + 1,n + \tau n^{1/2} ) =\; &\frac{{\Gamma (n + 1)}}{{\Gamma (n + 1 - a)}}\frac{1}{{\Gamma (a)}} \\ &\times \int_0^{ + \infty } {({\rm e}^t - 1)^{a - 1} \exp ( - n({\rm e}^{ - t}+t - 1) + n^{1/2} \tau (1 - {\rm e}^{ - t} ))\,{\rm d}t}. \end{align*} Here, $$\tau$$ is a fixed real number. Proceeding similarly to $$\S4.1$$ of this paper, an extension of Laplace's method yields \begin{align*} \tag{2} M(a,n + 1,n + \tau n^{1/2} ) \sim \; & n^{a/2}\exp \left( {\frac{{\tau ^2 }}{4}} \right) U\!\left( {a - \frac{1}{2}, - \tau } \right)\sum\limits_{k = 0}^\infty {\frac{{P_k (\tau ;a)}}{{n^{k/2} }}} \\ & + n^{(a - 1)/2} \exp \left( {\frac{{\tau ^2 }}{4}} \right)U\!\left( {a + \frac{1}{2}, - \tau } \right)\sum\limits_{k = 0}^\infty {\frac{{Q_k (\tau ;a)}}{{n^{k/2} }}} \end{align*} as $$n \to +\infty$$. Here, the $$U$$ denotes the parabolic cylinder function. The coefficients $$P_k(\tau; a)$$ and $$Q_k(\tau; a)$$ are polynomials in $$\tau$$ and $$a$$, with $$P_0(\tau; a)=1$$. By inspection, they are both $$\mathcal{O}(\tau^{3k+2})$$ for large $$\tau$$, which allows us to assume $$\tau = o(n^{1/6})$$. With this assumption, the two cases together cover $$0 \le x \leq 1$$.

Update. The leading order of the asymptotics in $$(2)$$ can be re-expressed as $$\tag{3} M(a, n + 1, nx) \sim n^{a/2} \exp\left( \frac{{(1 - x)^2}}{4}n \right) U\!\left( a - \frac{1}{2}, n^{1/2} (1 - x) \right)$$ as $$n \to +\infty$$, provided $$x = 1 + o(n^{-1/3})$$. However, when $$x = 1 + \mathcal{O}(n^{-1/2 + \varepsilon})$$, the right-hand side asymptotically behaves like $$(1 - x)^{-a}$$, aligning with $$(1)$$. Therefore, $$(3)$$ holds for $$0 \leq x \leq 1$$.

A numerical example with $$a = 2$$, $$n = 10^4$$, and $$0 \leq x \leq 1$$ is shown in the figure below. The red curve represents the exact values, while the dashed blue curve corresponds to the approximation in $$(3)$$.

• Thanks for the answer ! I'm still reading it but I have four questions. 1) In your answer where precisely did you use the fact that $a>0$ ? 2) If $0 \leq x<1$ and $n\rightarrow\infty$, isn't $(1-x)^2 n \rightarrow \infty$ true by definition ? 3) A notation question : Does $\square (n^{-1/2+\epsilon})$ mean Big O ? and 4) How did you figure out the various scalings for $x$ and $\tau$ ; for instance in the update $x\sim 1 + o(n^{-1/3})$ and close to equation (2), $\tau \sim o(n^{1/6})$ ? Apologies if these are basic questions. Appreciate the help ! Commented Sep 5 at 19:00
• 1) The integral representation we started with is valid only if $\operatorname{Re}(a) > 0$. However, this condition can be removed by following an argument similar to that given at the end of section 4.1 of the cited paper. 2) This holds only if $x$ is fixed. If $x$ scales with $n$, for example, $x = 1 - 1/n <1$, then it does not hold. 3) I use a calligraphic $O$ to denote big-$O$. It appears your device is mistakenly rendering it as a square.
– Gary
Commented Sep 6 at 1:35
• 4) For large $\tau$, the $k$th terms behave like $O(\tau^{3k + 2} / n^{k/2}) = \tau^2 O\left( (\tau / n^{1/6})^{3k} \right)$. To ensure the asymptotic behaviour, we must impose $\tau = o(n^{1/6})$ as $n \to +\infty$. Now, note that $\tau = o(n^{1/6})$ is equivalent to $x = 1 + o(n^{-1/3})$.
– Gary
Commented Sep 6 at 1:35

Too long for a comment

The general answer can be found in the link posted by @Gary; in some cases the asymptotic can be simplified. Using the integral representation of confluent hypergeometric function $$M(a, n+1, nx^2)=\frac{\Gamma(n+1)}{\Gamma(a)\Gamma(n+1-a)}\int_0^1u^{a-1}(1-u)^{-a}e^{n\big(x^2u+\ln(1-u)\big)}du$$ The Laplace methodis the convenient tool to find the asymptotic at $$n\to\infty$$. Let's consider some particular cases.

$$\displaystyle 1.\,\,a>0,\, x^2\in[0;1), \,n(1-x^2)\gg1$$ $$M(a, n+1, nx^2)\sim\frac{n^a}{\Gamma(a)}\int_0^1u^{a-1}e^{-nu(1-x^2)}du\sim\frac1{(1-x^2)^a}$$

$$\displaystyle 2.\,\,a>0,\, x^2=1$$ $$M(a, n+1, n)\sim\frac{n^a}{\Gamma(a)}\int_0^1u^{a-1}e^{-\frac{nu^2}2}du\sim\frac{\Gamma\big(\frac a2\big)}{2\Gamma(a)}(2n)^\frac a2$$

$$\displaystyle 3.\,\,a>0,\, x^2>1$$ $$M(a, n+1, n)\sim\frac{n^a}{\Gamma(a)}\int_0^1u^{a-1}(1-u)^{-a}e^{n\big(x^2u+\ln(1-u)\big)}du$$ $$=\frac{n^a}{\Gamma(a)}\int_0^1u^{a-1}(1-u)^{-a}e^{nf(u)}du$$ Applying again the Laplace' method, $$f(u)=x^2u+\ln(1-u);\,f'(u_0)=x^2-\frac1{1-u_0}=0\,\,\text{at}\,\,u_0=\frac{x^2-1}{x^2}\in[0;1]$$ $$f''(u_0)=-x^4$$ $$M(a, n+1, n)\sim\frac{n^a}{\Gamma(a)}\left(\frac{x^2-1}{x^2}\right)^{a-1}(x^2)^{a-n}e^{n(x^2-1)}\int_0^1e^{-nx^4\frac{(u-u_0)^2}2}du$$ $$\sim\sqrt{\frac{2\pi}n}\frac{n^a}{\Gamma(a)}e^{n(x^2-1)}\frac{(x^2-1)^{a-1}}{x^{2n}}$$

• this is precisely what I wished to do over the entire domain. My case of interest is $0<x<1$. In this regime of $x$, I don't think I can use Laplace's method, since the critical point is not inside the requisite domain . Maybe I'm doing this wrong. But consider the exponent $n(x^2 u + \log{(1-u)})$. The critical point is $u_0 = 1-1/x^2$ and $u_0 \notin (0,1)$ for $x\in(0,1)$. So I cannot successfully use the Laplace method. Am I wrong ? Commented Sep 3 at 18:43
• I'll try and use the general expression in the link @Gary posted. I wish to convert the confluent Hypergeometric or the parabolic cylinder functions (which appear in the general expression from the link) into something more tractable in the $n\rightarrow \infty$ limit. Either at the edges of the domain or if possible over the entire domain. Commented Sep 3 at 18:49
• You are write. In the case $x^2\in[0;1)$ you will can use the Watson' lemma - en.wikipedia.org/wiki/Watson%27s_lemma - denoting $t=x^2u+\ln(1-u)$ as a new variable and making the transformations of the integrand accordingly Commented Sep 4 at 2:10
• @Fragglerock In general, the critical point does not need to lie within the interval. See dlmf.nist.gov/2.3.iii
– Gary
Commented Sep 5 at 0:37
• Thanks @Gary ! I'll take a look. Commented Sep 5 at 19:00