Is there a asymtopic bound to a confluent hypergeometric function of ${_1F_1}(n,1.5n,nz)$?

Question

As the title mentioned, is there a good way to approximate

$${_1F_1}(n,1.5n,nz)$$

where $n \in \mathbb{N}$, and $z$ is a real positive number. If the direct bound is difficult, an asymptopic bound (as $n$ tends to infinity) is also welcomed.

Answer 1

We can use the integral representation of $_1F_1$ function to find the desiredt asymptotics: $$_1F_1(a,b,z)=\frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\int_0^1e^{zt}t^{a-1}(1-t)^{b-a-1}dt$$ We will use the Laplace method to find the asymptotics at $n\to\infty$

First, we consider the case of $_1F_1(n,kn,z)$, where $z$ is fixed and $k>1$ $$_1F_1(n,kn,z)=\frac{\Gamma(kn)}{\Gamma(kn-n)\Gamma(n)}\int_0^1e^{zt}t^{n-1}(1-t)^{kn-n-1}dt$$ $$=\frac{\Gamma(kn)}{\Gamma(kn-n)\Gamma(n)}\int_0^1\frac{e^{zt}}{t(1-t)}e^{n\big(\ln t+(k-1)\ln(1-t)\big)}dt$$ We find the extremum (maximum) of $f(t)=\ln t+(k-1)\ln(1-t)$ $$f'(t)=\frac1t-\frac{k-1}{1-t}=0\,\,\Rightarrow\,\,t_0=\frac1k$$ $$f''\big(\frac1k\big)=-k^2-\frac{k^2}{k-1}=-\frac{k^3}{k-1}$$ Therefore, $$_1F_1(n,kn,z)\sim\frac{\Gamma(kn)}{\Gamma(kn-n)\Gamma(n)}\frac{e^{zt_0}}{t_0(1-t_0)}e^{n\big(\ln t_0+(k-1)\ln(1-t_0)\big)}\bigg|_{t_0=\frac1k}\int_{-\infty}^\infty e^{-n\frac{k^3}{k-1}\frac{x^2}2}dx$$ $$=\frac{\Gamma(kn)\,e^{\frac zk}}{\Gamma(kn-n)\Gamma(n)}\frac{k^2}{k-1}\left(\frac{k-1}k\right)^{n(k-1)}\frac1{k^n}\sqrt{\frac{2\pi}n}\sqrt{\frac{k-1}{k^3}}$$ $$=\frac{\Gamma(kn)\,e^{\frac zk}}{\Gamma\big(n(k-1)\big)\Gamma(n)}\frac{(k-1)^{n(k-1)}}{k^{nk}}\sqrt{\frac{2\pi}n}\sqrt{\frac k{k-1}}$$

Using the Stirling formula I got a rather simple result:

$$\frac{\Gamma(kn)}{\Gamma\big(n(k-1)\big)\Gamma(n)}=\frac{\Gamma(kn+1)}{\Gamma\big(n(k-1)+1\big)\Gamma(n+1)}\frac{n^2(k-1)}{nk}$$ $$\sim\sqrt{\frac n{2\pi}}\frac{k^{kn}}{(k-1)^{n(k-1)}}\sqrt{\frac{k-1}k}$$ $$\boxed{\,\,\Rightarrow\,\,_1F_1(n,kn,z)\sim e^{\frac zk}\,\,}$$ what can be also checked numerically.

Now we can consider a more exotic case of $_1F_1(n,kn,nz)$: $$_1F_1(n,kn,nz)=\frac{\Gamma(kn)}{\Gamma\big(n(k-1)\big)\Gamma(n)}\int_0^1\frac1{t(1-t)}e^{n\big(zt+\ln t+(k-1)\ln (1-t)\big)}dt$$ $$f(t)=zt+\ln t+(k-1)\ln(1-t)$$ At $k\in(1;2)$ and $z\in(0;\infty) \,\, f(t)$ has the maximum at $$t_0=\frac{\sqrt{(\frac kz-1)^2+\frac4z}+1-\frac kz}2;\,\,t_0\in(0;1)$$ $$f''(t_0)=-\frac{kt^2_0-2t_0+1}{t^2_0(1-t_0)^2}$$ $$_1F_1(n,kn,nz)\sim\frac{\Gamma(kn)}{\Gamma\big(n(k-1)\big)\Gamma(n)}\frac{e^{nzt_0}t_0^n(1-t_0)^{n(k-1)}}{t_0(1-t_0)}\int_{-\infty}^\infty e^{-n\frac{kt^2_0-2t_0+1}{t^2_0(1-t_0)^2}\frac{x^2}2}dx$$ $$\sim\frac{k^{kn}}{(k-1)^{n(k-1)}}\sqrt{\frac{k-1}k}\frac{e^{nzt_0}\big(t_0(1-t_0)^{k-1}\big)^n}{\sqrt{kt^2_0-2t_0+1}}$$ This seems a bulky expression, but probably some simplifications are possible.