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I have a series of the form

\begin{equation} \sum_{n=0}^{\infty}\frac{(1)_{n}\,(\alpha_{3}+2)_{n}}{(\alpha_{1}+2)_{n}\, n!}\left(\frac{\beta_{1}}{\beta_{3}}\right)^{n}\, {_{1}}F_{1}(-\alpha_{1}-1-n;\,-\alpha_{3}-1-n;\,\beta_{3}x) \end{equation}

where,

$\alpha_{1}<\alpha_{3}, \quad \{\alpha_{1},\alpha_{3}\}\in \mathbb{Z}^{+} \ \text{or} \ \in \frac{\mathbb{Z}^{+}}{2}$

$\beta_{1}<\beta_{3}, \quad \{\beta_{1},\beta_{3}\}\in \mathbb{R}^{+}$

$0\leq x$

which looks very similar to some of the addition theorems for the confluent hypergeometric function ${_{1}}F_{1}(a,b,x)$. Also, take away the ${_{1}}F_{1}$ function from the series and it can be seen that the sum would the the Gauss hypergeometric function ${_{2}}F_{1}$. Been trying to manipulate this for a while with no clear path forward. Any thoughts are appreciated.

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  • $\begingroup$ Well, there is a simplification by doing a "Metamorphosis" on the front end. For the sake of simplicity: are the a,d real or integer. Integers are substantially easier. $\endgroup$
    – rrogers
    Aug 3, 2016 at 14:17
  • $\begingroup$ @rrogers I updated the question with the exact series and clarified the assumptions. All parameters are real. $\alpha_{1}$ and $\alpha_{3}$ are either both positive integers, both positive half integers, or some combination of the two. I know the integer case reduces the 1F1 to a polynomial which should as you said make it a simpler sum. I am certainly interested in the special case of $\alpha_{1}$ and $\alpha_{3}$ being both integers. That said, if $\alpha_{1}$ and $\alpha_{3}\in \frac{\mathbb{Z}^{+}}{2}$, the kummer transformation can make 1F1 into poly with exp term as a constant. $\endgroup$ Aug 3, 2016 at 15:00
  • $\begingroup$ @rrogers I also happen to have a closed-form solution for the derivative of the above function w.r.t. $x$. If that helps. $\endgroup$ Aug 3, 2016 at 15:12
  • $\begingroup$ I would like to see the derivative. I finally got around to running tests in reduce-algebra ( a symbolic math program) and my result seems correct; although I didn't carry it all the way through. reduce-algebra also coughed up a closed form of hypergeometric function in terms of pochammer(x,n/2) (although I can usually change the n/2 results to n). Be aware you can't completely trust these answers and the answer, for general expressions, comes with caveats. $\endgroup$
    – rrogers
    Sep 15, 2016 at 11:07

1 Answer 1

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Please check and make (constructive) comments and corrections. I usually get some strange dyslexia while thinking.

We start with a few simple formula:

$\left(\gamma+\psi\right)_{n}=\frac{\Gamma\left(\gamma+\psi+n\right)}{\Gamma\left(\gamma+\psi\right)}$

$\Gamma(z)\cdot\Gamma\left(1-z\right)=\frac{\pi}{sin\left(\pi\cdot z\right)}$

And for future compression/representation:

Lemma 1. $e^{\beta\cdot t}\cdot_{p}F_{q}((a);(b);\lambda\cdot t)={\displaystyle \sum_{k=0}^{\infty}\frac{t^{k}}{k!}\cdot\beta^{k}\cdot _{p+1}F_{q}\left(-k,\left(a\right);\left(b\right);-\frac{\lambda}{\beta}\right)}$

Proof. Consider a general convolution term and evaluate the coefficient of $\frac{t^{k}}{k!}$ $e^{\beta\cdot t}\cdot_{p}F_{q}((a);(b);\lambda\cdot t)={\displaystyle \sum_{k=0}^{\infty}}{\displaystyle \sum_{l=0}^{k}\frac{\left(a\right)_{l}}{\left(b\right)_{l}}\frac{\left(\beta\cdot t\right)^{k-l}}{l!}\cdot\frac{\left(\lambda\cdot t\right)^{l}}{\left(k-l\right)!}}$

${\displaystyle \sum_{k=0}^{\infty}\frac{t^{k}}{k!}}{\displaystyle \sum_{l=0}^{k}}\frac{k!}{l!\cdot\left(k-l\right)!}\cdot\beta^{k}\cdot\frac{\left(a\right)_{l}}{\left(b\right)_{l}}\cdot\left(\frac{\lambda}{\beta}\right)^{l}$

$ {\displaystyle \sum_{k=0}^{\infty}\frac{t^{k}}{k!}}{\displaystyle \sum_{l=0}^{k}}\beta^{k}\cdot\frac{\left(-k,\left(a\right)\right)_{l}}{\left(b\right)_{l}}\cdot\frac{\left(-\frac{\lambda}{\beta}\right)^{l}}{l!}$

QED

Note that the spliting of $\beta$ is soley for the purpose of standard generalized Hypergeometric representation; i.e. don't let $\beta^{k}$ wander off unless you realize the risks.

Now to the derivation

$\Gamma\left(\alpha+2+n\right)=\frac{\pi}{\Gamma\left(1-\left(\alpha+2+n\right)\right)\cdot sin\left(\pi\cdot\left(\alpha+2+n\right)\right)}=\frac{\pi}{\Gamma\left(-\left(\alpha+1+n\right)\right)\cdot sin\left(\pi\cdot\left(\alpha+2+n\right)\right)}$

Thus the front end is:

$\frac{sin\left(\pi\cdot\left(\alpha_{1}+2+n\right)\right)}{sin\left(\pi\cdot\left(\alpha_{3}+2+n\right)\right)}\cdot\frac{\Gamma\left(-\left(\alpha_{1}+1+n\right)\right)}{\Gamma\left(-\left(\alpha_{3}+1+n\right)\right)}$

And the Confluent Hypergeometric Function is $$_{1}F_{1}\left(-\left(\alpha_{1}+1+n\right);-\left(\alpha_{3}+1+n\right);\beta_{3}\cdot x\right)={\displaystyle \sum_{k=0}^{\infty}\frac{\frac{\Gamma\left(-\left(\alpha_{1}+1+n\right)+k\right)}{\Gamma\left(-\left(\alpha_{1}+1+n\right)\right)}}{\frac{\Gamma\left(-\left(\alpha_{3}+1+n\right)+k\right)}{\Gamma\left(-\left(\alpha_{3}+1+n\right)\right)}}\cdot\frac{\left(\beta_{3}\cdot x\right)^{k}}{k!}}$$

Thus we have: $${\displaystyle \sum_{n=0}^{\infty}}{\displaystyle \sum_{k=0}^{\infty}}\frac{sin\left(\pi\cdot\left(\alpha_{1}+2+n\right)\right)}{sin\left(\pi\cdot\left(\alpha_{3}+2+n\right)\right)}\cdot\frac{\Gamma\left(k-\alpha_{1}-1-n\right)}{\Gamma\left(k-\alpha_{3}-1-n\right)}\cdot\left(\frac{\beta_{1}}{\beta_{3}}\right)^{n}\cdot\frac{\left(\beta_{3}\cdot x\right)^{k}}{k!}$$

For integer $\alpha$ the sin()'s basically cancel and the ratio can be replaced with something like $(-1)^{\alpha_{1}-\alpha_{3}}$

A standard representation:

This part is incomplete/incorrect. Please finish or delete it: or I will later. (Hint: you have to add $\frac{(1)}{(p-k)!}$ and some other shaping (I think). Now we start to rewrite it and use the Lemma to get a standard (more or less) convolution representation: Let: $p=k-n;n=k-p$

${\displaystyle \sum_{k=0}^{\infty}}{\displaystyle \sum_{p=0}^{\infty}}\frac{sin\left(\pi\cdot\left(\alpha_{1}+2+n\right)\right)}{sin\left(\pi\cdot\left(\alpha_{3}+2+n\right)\right)}\cdot\frac{\Gamma\left(p-\alpha_{1}-1\right)}{\Gamma\left(p-\alpha_{3}-1\right)}\cdot\left(\frac{\beta_{1}}{\beta_{3}}\right)^{p-k}\cdot\frac{\left(\beta_{3}\cdot x\right)^{k}}{k!}$

${\displaystyle \frac{\Gamma\left(-\alpha_{1}-1\right)}{\Gamma\left(-\alpha_{3}-1\right)}\cdot\left(-1\right)^{\left(\alpha_{1}-\alpha_{3}\right)}\cdot\sum_{k=0}^{\infty}}{\displaystyle \frac{t^{k}}{k!}\sum_{n=0}^{\infty}\beta_{3}^{k}\cdot}\frac{\left(-\alpha_{1}-1\right)_{\left(p\right)}}{\left(-\alpha_{3}-1\right)_{\left(p\right)}}\cdot\left(\frac{\beta_{1}}{\beta_{3}}\right)^{p-k}$

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  • $\begingroup$ If you could please show how you got to this point that would be great. I have a few series similar to this one that probably can be simplified the same way you did here. Thank you for your time. $\endgroup$ Aug 3, 2016 at 22:03
  • $\begingroup$ Thank you for your answer. The explanation of your derivation really helps. As for the last bit of your post, I would leave it as is (or maybe bold the bit about it being incomplete or incorrect). It's still a potentially useful approach for moving forward with a final solution. Thank you again for your time. $\endgroup$ Aug 4, 2016 at 15:21

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