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

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

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

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.

• Likely a bug in the documentation. Sep 26, 2022 at 10:35
• For $b=2$ you need to take the limit $b\to2$ in the second term on the right-hand side. Simply setting $b=2$ won't work because $_1F_1(a,0;z)$ is infinite. Sep 26, 2022 at 11:49
• Mathematica expression: Hypergeometric1F1[a,b,z]-(1-b)*(b+z-2)/((a-b+1)*z)*Hypergeometric1F1[a,b-1,z]-(1-b)*(2-b)/((a-b+1)*z)*Hypergeometric1F1[a,b-2,z] Sep 26, 2022 at 11:50
• @Roman: Limit[(1 - b)*(2 - b)/((a - b + 1)*z)*Hypergeometric1F1[a, b - 2, z], b -> 2] results in 0. Did you read "see the limit of the second term is well defined and it is 0" in the question? Sep 26, 2022 at 12:08
• @user64494 That limit is not actually zero. Take say f[a_,b_,z_]:=(1-b)*(2-b)/((a-b+1)*z)*Hypergeometric1F1[a,b-2,z]; Plot[f[1/2,b,1/2],{b,1.5,2.5}]. Sep 26, 2022 at 12:46

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 $$c\to0$$.

• Can you explain $$_1F_1(a,c;z) = \frac{a z}{c}{_1}F_1(a+1,2;z)+O(1)$$ and "which turns the OP's expression into an approximation for b≈2" in details? TIA. Sep 26, 2022 at 12:13
• Limit[(1 - b)*(2 - b)/((a - b + 1)*z)*Hypergeometric1F1[a, b - 2, z], b -> 2 results in 0. Did you read "see the limit of the second term is well defined and it is 0" in the question Sep 26, 2022 at 12:24
• Unfortunately, your "Summing analytically" is built on the sand. Sep 26, 2022 at 12:26
• @user64494 you are only adding noise. Do you have anything correct and/or constructive to say? Sep 26, 2022 at 12:28
• Roman (@ does not work.): Please, be correct. Your " you are only adding noise" is not grounded and polite. I repeat "Limit[(1 - b)*(2 - b)/((a - b + 1)*z)*Hypergeometric1F1[a, b - 2, z], b -> 2] results in 0. Did you read "see the limit of the second term is well defined and it is 0" in the question ". Sep 26, 2022 at 12:36