Limit of Ai$(x)$ from above

Question

I wonder how to show that $$ \text{Ai}(x)>0\text{ for }x=x_{1}+\epsilon,\epsilon>0, $$ where $x_{1}$ is the first real zero (nearest to $x=0$) of Ai$(x)$. I want to know which formula for Ai$(x)$ I should use here. A hint should be sufficient, thanks!

Answer 1

Using Mathematica:

N@AiryAiPrime[AiryAiZero[1]]

(* 0.701211 *)

Or avoiding any explicit numerics:

Element[AiryAiPrime[AiryAiZero[1]], PositiveReals]

(* TRUE *)

AiryAiZero1 is the first zero of AiryAi ($\approx -2.3381074104597670385$), and AiryAiPrime is the derivative, which we evaluate at AiryAiZero1.

The positive derivative shows what you need, and can be confirmed by a simple plot.