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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!

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  • $\begingroup$ Likely a bug in the documentation. $\endgroup$
    – user64494
    Sep 26 at 10:35
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    $\begingroup$ 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. $\endgroup$
    – Roman
    Sep 26 at 11:49
  • $\begingroup$ 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] $\endgroup$
    – user293787
    Sep 26 at 11:50
  • $\begingroup$ @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? $\endgroup$
    – user64494
    Sep 26 at 12:08
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    $\begingroup$ @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}]. $\endgroup$
    – user293787
    Sep 26 at 12:46

1 Answer 1

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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}]]

enter image description here

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

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  • $\begingroup$ 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. $\endgroup$
    – user64494
    Sep 26 at 12:13
  • $\begingroup$ 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 $\endgroup$
    – user64494
    Sep 26 at 12:24
  • $\begingroup$ Unfortunately, your "Summing analytically" is built on the sand. $\endgroup$
    – user64494
    Sep 26 at 12:26
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    $\begingroup$ @user64494 you are only adding noise. Do you have anything correct and/or constructive to say? $\endgroup$
    – Roman
    Sep 26 at 12:28
  • $\begingroup$ 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 ". $\endgroup$
    – user64494
    Sep 26 at 12:36

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