Difference between revisions of "Hypergeometric pFq"

Revision as of 22:49, 3 July 2014

Let $p,q \in \{0,1,2,\ldots\}$ and $a_j,b_{\ell} \in \mathbb{R}$ for $j=1,\ldots,p$ and $\ell=1,\ldots,q$. We will use the notation $\vec{a}=(a_1,\ldots,a_p)$ and $\vec{b}=(b_1,\ldots,b_q)$ and we define the notations $$\vec{a}^{\overline{k}} = \displaystyle\prod_{j=1}^p a_j^{\overline{k}},$$ and $$\vec{a}+k = \displaystyle\prod_{j=1}^p (a_j+k),$$ (and similar for $\vec{b}^{\overline{k}}$). Define the generalized hypergeometric function $${}_pF_q(a_1,a_2,\ldots,a_p;b_1,\ldots,b_q;t)={}_pF_q(\vec{a};\vec{b};t)=\displaystyle\sum_{k=0}^{\infty}\dfrac{\displaystyle\prod_{j=1}^p a_j^{\overline{k}}}{\displaystyle\prod_{\ell=1}^q b_{\ell}^{\overline{k}}} \dfrac{t^k}{k!}.$$

Convergence

If any of the $a_j$'s is a a nonpositive integer, then the series terminates and is a polynomial.

If any of the $b_{\ell}$'s is a nonpositive integer, the series diverges because of divison by zero.

The remaining convergence of the series can be split into three cases:

Case I: $p<q+1$

Proposition: The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.
Proof: █

Case II: $p=q+1$

Proposition: The series ${}_pF_q$ converges for all $t\in \mathbb{C}$ with $|t|<1$.
Proof: █

Case III: $p>q+1$

Proposition: The series ${}_pF_q$ diverges for all $t \in \mathbb{C}$.
Proof: █

Differential equation

Define the derivative operator $\vartheta=t \dfrac{d}{dt}$.Then $$\vartheta t^k = t \dfrac{d}{dt} t^k = t(kt^{k-1})=kt^k.$$

Proposition: The operator $\vartheta$ is a linear operator.
Proof: █

Theorem: Define $y(t)={}_pF_q(\vec{a};\vec{b};t)$. Then $y$ satisfies $$\left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta + b_j-1) - t \displaystyle\prod_{i=1}^p (\vartheta+a_i) \right]y=0.$$

Proof: First compute $$\begin{array}{ll} \left[ t \displaystyle\prod_{i=1}^p (\vartheta+a_i) \right] y(t) &= \left[ t \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] \displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!} \\ &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \left[ \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] \dfrac{t^k}{k!} \\ &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}k!} \left[ \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] t^k \\ &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \left[ \displaystyle\prod_{i=1}^p (k+a_i) \right] \dfrac{t^k}{k!} \\ &=t\displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!}. \\ \end{array}$$ Now the computation $$\begin{array}{ll} \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta + b_j -1) \right]y(t) &= \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta+b_j-1) \right]\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!} \\ &=\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}k!} \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta + b_j -1) \right] t^k \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{k!} \left[ \dfrac{\displaystyle\prod_{j=1}^q (k + b_j -1)}{b^{\overline{k}}} \right] \vartheta t_k \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{k!} \left[ k\displaystyle\prod_{j=1}^q \dfrac{k+b_j-1}{b_j(b_j+1)\ldots(b_j+k-1)} \right] \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{k!} \left[ \displaystyle\prod_{j=1}^q \dfrac{1}{b_j(b_j+1)\ldots(b_j+k-2)} \right] t^k \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k-1}}(k-1)!} t^k \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k+1}}}{\vec{b}^{\overline{k}}k!}t^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^{k+1}}{k!} \\ &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!} \\ &= \left[ t \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] y(t) \end{array}$$ proves the claim. █

References

Rainville's Special Functions