Compute Confluent Hypergeometric Function 1F1

I am attempting to compute the (Kummer's) confluent hypergeometric function (see also here)

\begin{align} M\left(\frac{n}{2}, n +\frac{3}{2}, -z\right) = {}_1F_1\left(\frac{n}{2};n +\frac{3}{2};-z\right) \end{align}

for real $z\gt0$ and many consecutive values of integer $n\geq0$. Since the first two arguments scale with $n$ at different rates, the usual recurrence relations don't apply. Additionally, it does not seem to fit into any of the usual special cases that relate $M(a,b,z)$ to other functions. Using Mathematica, we find that the first 7 terms are:

\begin{align} n&=0 &&1,\\ n&=1 &&\frac{3 \sqrt{\pi } (2 z-1) \text{erf}\left(\sqrt{z}\right)}{8 z^{3/2}}+\frac{3 e^{-z}}{4 z},\\ n&=2 &&\frac{15 \sqrt{\pi } e^{-z} \text{erfi}\left(\sqrt{z}\right)}{8 z^{5/2}}+\frac{5 (2 z-3)}{4 z^2},\\ n&=3 &&\frac{105 \sqrt{\pi } (4 (z-3) z+15) \text{erf}\left(\sqrt{z}\right)}{128 z^{7/2}}+\frac{105 e^{-z} (2 z-15)}{64 z^3},\\ n&=4 &&-\frac{945 \sqrt{\pi } e^{-z} (2 z+7) \text{erfi}\left(\sqrt{z}\right)}{64 z^{9/2}} + \frac{63 (8 (z-5) z+105)}{32 z^4},\\ n&=5 &&\frac{3465 \sqrt{\pi } \left(8 z^3-60 z^2+210 z-315\right) \text{erf}\left(\sqrt{z}\right)}{1024 z^{11/2}}+\frac{3465 e^{-z} \left(4 z^2+315\right)}{512 z^5},\\ n&=6 &&\frac{135135 \sqrt{\pi } e^{-z} (4 z (z+9)+99) \text{erfi}\left(\sqrt{z}\right)}{1024 z^{13/2}}+\frac{1287 (2 z (16 z (2 z-21)+1575)-10395)}{512 z^6} \end{align}

I would appreciate any tips/intuition for how to simplify this function.

• Just treat it as two sequences; odd and even n. Then you can apply normal recurrences. If you examine the items you printed you can see the interlacing happen. Dec 29, 2017 at 14:25

Here is a sketch: Please check and tell me if you want more or corrections.
Using the usual notation $\tilde{M}(1,2)\cong M(a+1,b+2)$
$\tilde{M}(1,2)=\frac{(2\cdot b^{2}+2\cdot b)\cdot z^{2}+(-b^{4}+(2\cdot a-2)\cdot b^{3}+3\cdot b^{2}+(4-2\cdot a)\cdot b)\cdot z+b^{5}-2\cdot b^{4}-b^{3}+2\cdot b^{2}}{(a\cdot b^{2}+(-a^{2}-2\cdot a)\cdot b+2\cdot a^{2})\cdot z^{2}}\tilde{M}(0,0)+\frac{(2\cdot b^{3}-2\cdot b)\cdot z-b^{5}+(a+1)\cdot b^{4}+b^{3}+(-a-1)\cdot b^{2}}{(a\cdot b^{2}+(a-2\cdot a^{2})\cdot b+a^{3}-a^{2})\cdot z^{2}}\tilde{M}(-1,-2)$

First we define two new sequences:
Starting at n=0
$(n,n+1,n+2,n+3)\rightarrow\left(m,l,m+1,l+1\right)$
n even
$m=\frac{n}{2},M\left(m,2\cdot m+\frac{3}{2};-z\right)\rightarrow M\left(m+1,2\cdot m+\frac{3}{2}+2;-z\right)$
n odd
$l=\frac{n+1}{2} ,M\left(l,2\cdot l+\frac{3}{2};-z\right)\rightarrow M\left(l+1,2\cdot l+\frac{3}{2}+2;-z\right)$
Rearranging DLMF 13.3.2 we can bump $b\rightarrow b+2$
$M(a,b+2;-z)=\frac{\left(b-1\right)\cdot\left(b+z\right)}{z\cdot\left(b+1-a\right)}M\left(a,b+1;-z\right)+\frac{b\cdot\left(b+1\right)}{z\cdot\left(b+1-a\right)}M\left(a,b;-z\right)$
And subsequently 13.3.3 $a\rightarrow a+1$
$M\left(a+1,b+2;-z\right)=\frac{\left(a-\left(b+2\right)+1\right)}{a}M\left(a,b+2;-z\right)+\frac{\left(b+1\right)}{z}M\left(a,b+1;-z\right)$
Where I haven't reduced expressions in order to make tracking DLMF easier
Using obvious substitutions to get j,k ;we have the single recursion
$M\left(a+1,b+2;-z\right)=k\cdot M\left(a,b+1;-z\right)+j\cdot M\left(a,b;-z\right)$
I will fill in j,k if needed.
This can be applied to both n even (m) or odd (l) as above.

The calculation of the real requested recursion follows.
Alternate ways to find a recursion
I recomend skipping to "Obvious solution" ; I left the "Simple Matrix algebra" for the record. It does use some higher power tools (and less programming) but is less obvious.

$r\cdot M\left(a,b,z\right)+s\cdot M(a+1,b+2,z)=M(a+2,b+4)$

Simple Matrix algebra
With a modicum of identity selection we can select from the contiguous relations, like so:

Make a matrix expression.
$$\left[\begin{array}{ccccc} r & s & t & u & v\end{array}\right]\begin{pmatrix}\mathit{a1} & \mathit{a2} & \mathit{a3} & 0 & 0 & 0 & 0\\ 0 & \mathit{b2} & 0 & 0 & \mathit{b5} & \mathit{b6} & 0\\ 0 & 0 & \mathit{c3} & \mathit{c4} & 0 & 0 & \mathit{c7}\\ \mathit{d1} & 0 & 0 & 0 & 0 & \mathit{d6} & \mathit{d7}\\ 0 & 0 & 0 & \mathit{e4} & 0 & \mathit{e6} & \mathit{e7} \end{pmatrix}\left[\begin{array}{c} \tilde{M}(0,0)\\ \tilde{M}\left(1,1\right)\\ \tilde{M}\left(-1,-1\right)\\ \tilde{M}\left(-1,-2\right)\\ \tilde{M}\left(1,2\right)\\ \tilde{M}\left(1,0\right)\\ \tilde{M}\left(0,-1\right) \end{array}\right]$$
$$r,s,t,u,v$$ are the weights to combine rows/identities for required output.

The rows correspond to DLMF
13.3.13 (a,b offset by -1,-1)
$M(a,b,z)\cdot b\cdot(-z+b-1)+a\cdot z\cdot M(a+1,b+1,z)-M(a-1,b-1,z)\cdot(b-1)\cdot b$

13.3.2 (a,b offset by 1,1) $M(a+1,b+1,z)\cdot(b+1)\cdot(-z-b)-M(a+1,b+2,z)\cdot(b-a)\cdot z+M(a+1,b,z)\cdot b\cdot(b+1)$

13.3.3 (a,b offset by -1,-1)
$M(a-1,b-1,z)\cdot(-b+a+1)+M(a-1,b-2,z)\cdot(b-2)-(a-1)\cdot M(a,b-1,z)$

13.3.3
$M(a,b,z)\cdot(-b+a+1)-a\cdot M(a+1,b,z)+M(a,b-1,z)\cdot(b-1)$

13.3.13 ( a,b offset by -1,-2)
$-M(a-1,b-2,z)\cdot\left(b-2\right)\cdot(b-1)+a\cdot z\cdot M(a+1,b,z)+M(a,b-1,z)\cdot\left(b-1\right)\cdot(-z+b-2)$

Substituting the identity coefficients we get:
$\begin{pmatrix}\mathit{b\cdot(-z+b-1)} & a\cdot z & \mathit{-b\cdot\left(b-1\right)} & 0 & 0 & 0 & 0\\ 0 & \mathit{\left(b+1\right)\cdot\left(-b-z\right)} & 0 & 0 & -z\cdot\left(b-a\right) & \mathit{b\cdot\left(b+1\right)} & 0\\ 0 & 0 & \mathit{\left(a-b-1\right)} & \mathit{\left(b-2\right)} & 0 & 0 & \mathit{-\left(a-1\right)}\\ \mathit{\left(a-b-1\right)} & 0 & 0 & 0 & 0 & \mathit{-a} & \mathit{\left(b-1\right)}\\ 0 & 0 & 0 & \mathit{-\left(b-2\right)\cdot\left(b-1\right)} & 0 & \mathit{z\cdot a} & \mathit{\left(b-1\right)\cdot\left(b-z-2\right)} \end{pmatrix}$
We can work against either representation. I cheated a little and used elim() in Maxima computer algebra to eliminate $[a2,a3,b6,b7]$ to compute $[r,s,t,u,v]$ . The correct columns went to zero as indicated by the multiplication $saam\,.\,aa$ below. I really should make a cleaner explanation for this choice.
Weights (r is left free)
$$saam= \begin{pmatrix}r & -\frac{\mathit{a2} r}{\mathit{b2}} & -\frac{\mathit{a3} r}{\mathit{c3}} & \frac{\left( \mathit{a2}\, \mathit{b6}\, \mathit{c3}\, \mathit{e7}-\mathit{a3}\, \mathit{b2}\, \mathit{c7}\, \mathit{e6}\right) r}{\mathit{b2}\, \mathit{c3}\, \mathit{d6}\, \mathit{e7}-\mathit{b2}\, \mathit{c3}\, \mathit{d7}\, \mathit{e6}} & -\frac{\left( \mathit{a2}\, \mathit{b6}\, \mathit{c3}\, \mathit{d7}-\mathit{a3}\, \mathit{b2}\, \mathit{c7}\, \mathit{d6}\right) r}{\mathit{b2}\, \mathit{c3}\, \mathit{d6}\, \mathit{e7}-\mathit{b2}\, \mathit{c3}\, \mathit{d7}\, \mathit{e6}}\end{pmatrix}$$
Substituting for the $a_{m},b_{n},c_{o},d_{p},e_{q}$ we have
$saam2=$
$\begin{array}{ccc} [r, & \frac{arz}{(b+1)z+b^{2}+b} & ,-\frac{\left(b^{2}-b\right)\cdot r}{b-a+1},-\frac{b^{2}\cdot r\cdot z^{2}+(-b^{3}+2a\cdot b^{2}+(2-2a)b)r\cdot z}{(b^{2}+(-a-1)\cdot b+2a-2)\cdot z+b^{3}+(-a-1)\cdot b^{2}+(2a-2)\cdot b},\,\frac{b^{2}\cdot r\cdot z+(a-1)\cdot b^{2}\cdot r}{(b^{2}+(-a-1)\cdot b+2\cdot a-2)\cdot z+b^{3}+(-a-1)\cdot b^{2}+(2\cdot a-2)\cdot b}]\end{array}$
Result of matrix multiplication
$saam\,.\,aa$
$[\frac{(2\cdot b\cdot r\cdot z^{2}+(-b^{3}+(2\cdot a-1)\cdot b^{2}+(4-2\cdot a)\cdot b)\cdot r\cdot z+(b^{4}-3\cdot b^{3}+2\cdot b^{2})\cdot r)}{((b-2)\cdot z+b^{2}-2\cdot b)},0,0,\frac{(2\cdot b^{2}-2\cdot b)\cdot r\cdot z+(-b^{4}+(a+2)\cdot b^{3}+(-a-1)\cdot b^{2})\cdot r}{(b-a+1)\cdot z+b^{2}+(1-a)\cdot b},\frac{(a\cdot b-a^{2})\cdot r\cdot z^{2}}{(b+1)\cdot z+b^{2}+b},0,0]$
Using the substitution we have
$\tilde{M}(1,2)=\frac{(2\cdot b^{2}+2\cdot b)\cdot z^{2}+(-b^{4}+(2\cdot a-2)\cdot b^{3}+3\cdot b^{2}+(4-2\cdot a)\cdot b)\cdot z+b^{5}-2\cdot b^{4}-b^{3}+2\cdot b^{2}}{(a\cdot b^{2}+(-a^{2}-2\cdot a)\cdot b+2\cdot a^{2})\cdot z^{2}}\tilde{M}(0,0)+\frac{(2\cdot b^{3}-2\cdot b)\cdot z-b^{5}+(a+1)\cdot b^{4}+b^{3}+(-a-1)\cdot b^{2}}{(a\cdot b^{2}+(a-2\cdot a^{2})\cdot b+a^{3}-a^{2})\cdot z^{2}}\tilde{M}(-1,-2)$
Where the third term is the coefficient of $M(1,2)$ and the first term is the coefficient of $M(0,0)$ and the fourth term is the coefficient of $M(-1,-2)$.

“Obvious solution”

Rather than belabor the details I will just generally show the principle; which is simply a form of Gaussian Elimination/Back Substitution.

Consider 5 independent constraints/identities applied to 7 functions M1, M2, M3, M5, M6, M7.
$$\begin{pmatrix}\mathit{M6}\,\mathit{a6}+\mathit{M4}\,\mathit{a4}+\mathit{M1}\,\mathit{a1}\\ \mathit{M5}\,\mathit{b5}+\mathit{M4}\,\mathit{b4}+\mathit{M2}\,\mathit{b2}\\ \mathit{M7}\,\mathit{c7}+\mathit{M5}\,\mathit{c5}+\mathit{M3}\,\mathit{c3}\\ \mathit{M7}\,\mathit{d7}+\mathit{M6}\,\mathit{d6}+\mathit{M2}\,\mathit{d2}\\ \mathit{M7}\,\mathit{e7}+\mathit{M6}\,\mathit{e6}+\mathit{M3}\,\mathit{e3} \end{pmatrix}=\left(\begin{array}{c} 0\\ 0\\ 0\\ 0\\ 0 \end{array}\right)$$

We represent this as $A\left(5,7\right)$ and our goal will be $A\left(1,3\right)$, that is the trilinear $f(M1,M2,M3)$.

But this is obvious: for instance we start from the bottom and solve for M7 and substitute back into the equations. We now have 4 equations in 6 unknowns: i.e.

$A(5,7)\rightarrow A(4,6)$

and so on for M6, M5, M4 and having eliminated everything but M1, M2, M3 we have our answer. This is three lines of Maxima code.

The only obstruction is if a back substitution leaves an additional zero constraint. In this case the constraints/identities were redundant and one of the constraints is a linear combination the others and you have to find a new constraint or settle for having another $MX$ in your result.

The Independence can also be calculated beforehand by exterior calculus or in matrix minors terms.

Exterior Calculus

I am leaving this out because I am not satisfied with my explanations. The factoring of linear dependence in the language of the “/\” operations seems intuitively obvious but a little complicated to put into words (and it shouldn't be). In addition I am not satisfied that it really adds to the previous process; i.e. it's just a rehash.

• I agree that two sequences are needed. The ideal recurrence would be $M(a+2, b+4, -z) = j \cdot M(a+1, b+2, -z) + k \cdot M(a, b, -z)$, as no terms are computed that are not themselves members of the series. The solution you proposed above requires computation of a "transition" term $M(a, b+1, -z)$. Is there a simplification that eliminates such terms? Dec 29, 2017 at 20:40
• I realized afterwards that I had left one recurrence out. I will try to fill it in. Dec 29, 2017 at 21:22
• Back to basics: Yes, your stated relationship exists! All contiguous relations exist. Thus a+2,a+1,a are linearly (with variable coefficients!) related. And b+2 is linearly related to b+1 and b+3 which in turn are related to b,b+2 and b+4,b+2. I will work on it; unless somebody is faster (please save me!). It always seems to me that this type of thing must be amenable to a simple computer program, but I have never crossed the gap. Dec 30, 2017 at 20:45
• Or at least I think; there is a diagonalization problem involved but it's amenable to a diagram/tree expansion. Dec 30, 2017 at 21:52
• I have a form of what you requested with derivation but the whole thing is sketchy and disorganized. It doesn't resemble the next answer but I haven't imposed the 2*n condition on b, and I don't know if that helps. Having said that I have to review the whole things carefully to check signs and identity transcription errors (and maybe simplify things). If you want I will put the material up on dropbox for review, suggestions, and corrections. There are maxima (.mac), pdf, and lyx files. Otherwise, it will be a couple of days. The technique is basically elementary algebra. Jan 2, 2018 at 18:17