Consecutive neighbours of Hypergeometric ${}_1 F_1(a,b,z)$

Question

I am working with the hypergeometric function ${}_1 F_1(a,b;z)$, where $a\in \mathbb{N^+}$, $b=2$, and $z\in \mathbb{C}$. The Wolfram function repository lists the following relation

$$\begin{equation} {}_1 F_1(a,b;z)=\frac{(1-b)(b+z-2)}{(a-b+1)z}F_1(a,b-1;z)+\frac{(1-b)(2-b)}{(a-b+1)z}F_1(a,b-2;z) \end{equation}$$

with no restrictions on $a,b$ and $z$. My question is how to implement this relation for $b=2$. I see the limit of the second term is well defined and it is 0, but the first term alone does not numerically agree with the left hand side.

Thanks in advance!

Answer 1

For $c\approx0$ we can use a series-expansion for the second term on the right-hand side: $$ _1F_1(a,c;z) = \frac{a z}{c}{_1}F_1(a+1,2;z)+O(1) $$ which turns the OP's expression into an approximation for $b\approx2$: $$ _1 F_1(a,b;z)=\frac{(1-b)(b+z-2)}{(a-b+1)z}{_1}F_1(a,b-1;z)+\frac{(1-b)(2-b)}{(a-b+1)z}\frac{a z}{b-2}{_1}F_1(a+1,2;z)+O(b-2)\\ =-\frac{1}{a-1}{_1}F_1(a,1;z)+\frac{a}{a-1}{_1}F_1(a+1,2;z)+O(b-2)\\ $$ and from this we get the $b=2$ case: $$ _1 F_1(a,2;z)=-\frac{1}{a-1}{_1}F_1(a,1;z)+\frac{a}{a-1}{_1}F_1(a+1,2;z) $$

derivation of the approximation used

For $c\approx0$, $$ _1F_1(a,c;z) = 1+\sum_{k=1}^{\infty}\frac{(a)_k}{(c)_k}\frac{z^k}{k!} = 1+\sum_{k=1}^{\infty}\frac{(a)_k}{(k-1)!c}\frac{z^k}{k!}+O(1) $$ where I have used the approximation $(c)_k=(k-1)!c+O(c^2)$ for $k\ge1$:

Series[Pochhammer[c, k], {c, 0, 1}]
(*    Gamma[k] c + O[c]^2    *)

Summing analytically,

1 + Sum[Pochhammer[a,k]/((k - 1)! c) z^k/k!, {k, 1, ∞}]
(*    1 + (a z Hypergeometric1F1[1 + a, 2, z])/c    *)

gives the approximation used, to order $O(1)$. Let's test it with random parameters:

With[{a = 1.3, z = 0.47},
  Plot[{c*Hypergeometric1F1[a, c, z], 
        a*z*Hypergeometric1F1[1 + a, 2, z]}, {c, -0.1, 0.1}]]

Seems to work for the limit $c\to0$.