# Asymptotics of hyperrgeometric 2F1 for large integer parameters

I have stumbled upon $$\frac{\Gamma(n+\frac{1}{2})\Gamma(n-\frac{1}{2})}{\Gamma(2n+1)} \ {}_2F_1\left(n+\frac{1}{2}, \ n-\frac{1}{2}; \ 2n+1, \ z\right)$$

Where the integer $$n \to \infty$$. I need to determine the asymptotic expression of this with this condition. I can do that for the gamma function part using Stirling's approximation, but I haven't found any way to determine

$${}_2F_1\left(n+\frac{1}{2}, \ n-\frac{1}{2}; \ 2n+1, \ z\right) \ \ \text{for } \ \ n \to \infty.$$

How do I obtain this?

Any derivation or reference to the asymptotics of this specific parameters would be very helpful.

Some edits:

– Gary
Commented Jul 25 at 2:35
– Gary
Commented Aug 6 at 22:40
• I am very sorry for a late response because I was into obtaining a few results along this line and I had an easier solution which I am adding as the edit. But this has numerical instability near h=1 as I implement it in mathematica. For that I had to revisit your answer and although it hasn't answered it completely but it is heading towards that. Commented 2 days ago
• Your argument is completely formal and non-rigorous. Note also that $\frac{{\Gamma (N)}}{{\Gamma \left( {N - \frac{1}{2}} \right)\sqrt N }} \sim 1$ as $N\to+\infty$.
– Gary
Commented 2 days ago
• You are also missing a factor of $(1-z)^{1/4}$ (which should be due to the non-rigorous argument). Take a look at the final result in my answer. I also did some numerical experiments and the formula I provided works.
– Gary
Commented 2 days ago

With $$z = \cosh \zeta$$, $$\lambda = n$$, $$\alpha = \frac{1}{2}$$, $$\beta = \frac{1}{2}$$, and $$\gamma = 2$$, the first result in $$\S9$$ of G. N. Watson's paper Asymptotic expansions of hypergeometric functions gives \begin{align*} \left( {\operatorname{csch} \!\big( \tfrac{\zeta }{2} \big)} \right)^{2n + 1} & {}_2F_1\!\left( {n + \tfrac{1}{2},n - \tfrac{1}{2};2n + 1; - \operatorname{csch}^2\! \big( \tfrac{\zeta }{2} \big)} \right) \\ &\sim \frac{{2\Gamma (2n + 1)}}{{\Gamma \!\left( {n - \frac{1}{2}} \right)\Gamma \!\left( {n + \frac{3}{2}} \right)}}{\rm e}^{ - \left( {n + \frac{1}{2}} \right)\zeta } (1 - {\rm e}^{ - \zeta } )^{ - \frac{3}{2}} (1 + {\rm e}^{ - \zeta } )^{\frac{1}{2}} \sqrt {\frac{\pi }{n}} \\ &\sim 2^{2n+1}{\rm e}^{ - \left( {n + \frac{1}{2}} \right)\zeta } (1 - {\rm e}^{ - \zeta } )^{ - \frac{3}{2}} (1 + {\rm e}^{ - \zeta } )^{\frac{1}{2}} , \end{align*} as $$n\to+\infty$$. Here $$\zeta \in \left\{\xi: \operatorname{Re}(\xi)\ge 0,\, |\operatorname{Im}(\xi)|\le \pi \right\}$$. Consequently, \begin{align*} {}_2F_1\!\left( n + \tfrac{1}{2},n - \tfrac{1}{2};2n + 1; - \operatorname{csch}^2 \!\big( \tfrac{\zeta }{2}\big) \right) & \sim \Big( {2\,{\rm e}^{ - \frac{\zeta }{2}} \sinh\! \big( \tfrac{\zeta }{2} \big)} \Big)^{2n + 1} (1 - {\rm e}^{ - \zeta } )^{ - \frac{3}{2}} (1 + {\rm e}^{ - \zeta } )^{\frac{1}{2}} \\ & = \coth ^{\frac{1}{2}} \!\big( \tfrac{\zeta }{2} \big)(1 - {\rm e}^{ - \zeta } )^{2n} \end{align*} as $$n\to+\infty$$. If we set $$w=- \operatorname{csch}^2 \!\big( \tfrac{\zeta }{2}\big)$$, then the domain of validity corresponds to $$\left| \arg ( - w) \right| \le \pi$$: $${}_2F_1\!\left( n + \tfrac{1}{2},n - \tfrac{1}{2};2n + 1; w \right) \sim (1 - w)^{1/4} \left( \tfrac{1}{2} + \tfrac{1}{2}\sqrt {1 - w} \right)^{ - 2n}.$$
• Thank you username Gary for your insightful input but I think something easier is possible. Commented 2 days ago