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I have the following confluent hypergeometric function: $_1F_1\left(2(m+1),\frac{1}{2},-x^2\right)$.

By using Mathematica, I know that for values of $m=0,1,2,...$ this function expands into a power series involving $\operatorname{erfi}(x)$, that can be further simplified by introducing the Dawson function.

For example, for $m=0$ $$_1F_1\biggl(2,\frac{1}{2},-x^2\biggr)=1-\frac{3}{2}xe^{-x^2} \sqrt[]{\pi} \operatorname{erfi}(x)-x^2+x^3e^{-x^2} \sqrt[]{\pi} \operatorname{erfi}(x)$$ $$= 1-3xD_+(x)-x^2+2x^3D_+(x)$$

Here are two more examples, for $m=1$: enter image description here

and $m=2$: enter image description here

What is the general term of this expansion, and how can I obtain it by hand?

EDIT

As Aaron Hendrickson suggested in the comments, Mathematica expands the function by this formula, so I've found the general term of the sum. But how can I get this expansion starting from the definition of the particular confluent hypergeometric function $_1F_1\left(n,\frac{1}{2},z\right)$?

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    $\begingroup$ This should do it: functions.wolfram.com/07.20.03.0123.01 $\endgroup$ Commented Mar 13, 2021 at 22:33
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    $\begingroup$ In fact your have multiple options: functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/… $\endgroup$ Commented Mar 13, 2021 at 22:34
  • $\begingroup$ Since the top parameter in an integer you could probably derive these relations using integration by parts on an appropriate integral representation of your $_1F_1$ function. $\endgroup$ Commented Mar 13, 2021 at 22:42
  • $\begingroup$ @Aaron Hendrickson thank you for the link, it helped a lot. I've updated the question to include it. I am still not clear on how to get to this general term starting from the definition of the $_1F_1$ function. As far as I understand from what I've read online, the integral representation can be used only when $\Re(b)>\Re(a)>0$, which isn't the case for this version of the function, as $\Re(2(m+1))>\frac{1}{2}$. Could you please elaborate your comment intro an answer? $\endgroup$
    – Anthill
    Commented Mar 14, 2021 at 8:54

1 Answer 1

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So as mentioned in the comments this identity provides a "closed form" for your hypergeometric function in terms of Laguerre polynomials. Here is another option that may be considered a more elementary form:

Using DLMF 13.3.17 for $n=2m+1$ and $a=1$ you may write $$ \begin{aligned} {_1F}_1\left({2m+2\atop 1/2};z\right)% &=\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1} {_1F}_1\left({1\atop 1/2};z\right)\\ &=\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1} (1+\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z))\\ &=\frac{1}{(2m+1)!}\left((2m+1)!+\partial_z^{2m+1} z^{2m+1}\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z)\right)\\ &=1+\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1}\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z)\\ \end{aligned} $$ You may then apply the General Leibniz rule to evaluate the derivative. Once the derivatives are evaluated you may then make the substitution $z\mapsto-x^2$ to obtain the final result.

There are many other different forms you can put your hypergeometric function into using similar methods and variants of this method. What its going to come down to is what form is useful for your application.

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