Difference between revisions of "Hypergeometric pFq"
(→Case I: $p<q+1$) |
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<strong>Proposition:</strong> The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.<br /> | <strong>Proposition:</strong> The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.<br /> | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
| − | <strong>Proof: We will apply the [[ratio_test | ratio test]]. Let $ | + | <strong>Proof:</strong> Notice if $t=0$ then the series converges trivially, so suppose $t \neq 0$. We will apply the [[ratio_test | ratio test]]. Let $\alpha_k=\dfrac{\vec{a}^{\overline{k}}t^k}{\vec{b}^{\overline{k}}k!}$. Then |
| + | $$\begin{array}{ll} | ||
| + | L &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\alpha_{k+1}}{\alpha_k} \right| \\ | ||
| + | &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\dfrac{\vec{a}^{\overline{k}}t^k}{\vec{b}^{\overline{k}}k!}}{\dfrac{\vec{a}^{\overline{k+1}}t^{k+1}}{\vec{b}^{\overline{k+1}}(k+1)!}} \right| \\ | ||
| + | &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\vec{a}^{\overline{k}} \vec{b}^{\overline{k+1}}(k+1)!t^k }{\vec{b}^{\overline{k}} \vec{a}^{\overline{k+1}}k!t^{k+1}} \right| \\ | ||
| + | &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{k(\vec{b}+k)}{(\vec{a}+k)t} \right| \\ | ||
| + | &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{O(k^{q+1})}{O(k^{p})}\right| \\ | ||
| + | &= 0 < 1, | ||
| + | \end{array}$$ | ||
| + | therefore the series converges for all $t \in \mathbb{C}$. █ | ||
</div></div> | </div></div> | ||
Revision as of 06:46, 4 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 $\vec{a}=\displaystyle\prod_{j=1}^p a_j$ and $\vec{b}=\displaystyle\prod_{\ell=1}^q b_{\ell}$ 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 $${}_pF_q(a_1,a_2,\ldots,a_p;b_1,\ldots,b_q;t)={}_pF_q(\vec{a};\vec{b};t)=\displaystyle\sum_{k=0}^{\infty}\dfrac{\displaystyle\prod_{j=1}^p a_j^{\overline{k}}}{\displaystyle\prod_{\ell=1}^q b_{\ell}^{\overline{k}}} \dfrac{t^k}{k!}.$$
Contents
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: Notice if $t=0$ then the series converges trivially, so suppose $t \neq 0$. We will apply the ratio test. Let $\alpha_k=\dfrac{\vec{a}^{\overline{k}}t^k}{\vec{b}^{\overline{k}}k!}$. Then $$\begin{array}{ll} L &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\alpha_{k+1}}{\alpha_k} \right| \\ &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\dfrac{\vec{a}^{\overline{k}}t^k}{\vec{b}^{\overline{k}}k!}}{\dfrac{\vec{a}^{\overline{k+1}}t^{k+1}}{\vec{b}^{\overline{k+1}}(k+1)!}} \right| \\ &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\vec{a}^{\overline{k}} \vec{b}^{\overline{k+1}}(k+1)!t^k }{\vec{b}^{\overline{k}} \vec{a}^{\overline{k+1}}k!t^{k+1}} \right| \\ &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{k(\vec{b}+k)}{(\vec{a}+k)t} \right| \\ &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{O(k^{q+1})}{O(k^{p})}\right| \\ &= 0 < 1, \end{array}$$ therefore the series converges for all $t \in \mathbb{C}$. █
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: █
Derivatives
Proposition: Suppose that ${}_pF_q$ converges. Then $$\dfrac{d^n}{dt^n} {}_pF_q(\vec{a};\vec{b};t)=\dfrac{\vec{a}^{\overline{n}}}{\vec{b}^{\overline{n}}} {}_pF_q(\vec{a}+n;\vec{b}+n;t).$$
Proof: The computation $$\begin{array}{ll} \dfrac{d^n}{dt^n} {}_pF_q(\vec{a};\vec{b};t) &= \dfrac{d^n}{dt^n}\displaystyle\sum_{k=0}^{\infty} \dfrac{ \vec{a}^{\overline{k}} }{ \vec{b}^{\overline{k}} } \dfrac{t^{\underline{k}}}{k!} \\ &= \displaystyle\sum_{k=n}^{\infty} \dfrac{ \vec{a}^{\overline{k}} }{ \vec{b}^{\overline{k}} } \dfrac{t^{\underline{k-n}}}{(k-n)!} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{ \vec{a}^{\overline{k+n}} }{ \vec{b}^{\overline{k+n}} } \dfrac{t^{\underline{k}}}{k!} \\ &=\dfrac{ \vec{a}^{\overline{n}} }{ \vec{b}^{\overline{n}} } \displaystyle\sum_{k=0}^{\infty} \dfrac{ (\vec{a}+n)^{\overline{k}} }{ (\vec{b}+n)^{\overline{k}} } \dfrac{t^{\underline{k}}}{k!} \\ &=\dfrac{ \vec{a}^{\overline{n}} }{ \vec{b}^{\overline{n}} } {}_pF_q(\vec{a}+n;\vec{b}+n;t) \end{array}$$ proves the claim. █
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 $$(\dagger) \hspace{35pt} \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta + b_j-1) - t \displaystyle\prod_{i=1}^p (\vartheta+a_i) \right]y=0.$$
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 $$\begin{array}{ll} \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta + b_j -1) \right]y(t) &= \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta+b_j-1) \right]\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!} \\ &=\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}k!} \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta + b_j -1) \right] t^k \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{k!} \left[ \dfrac{\displaystyle\prod_{j=1}^q (k + b_j -1)}{b^{\overline{k}}} \right] \vartheta t_k \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{k!} \left[ k\displaystyle\prod_{j=1}^q \dfrac{k+b_j-1}{b_j(b_j+1)\ldots(b_j+k-1)} \right] t^k \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{k!} \left[ \displaystyle\prod_{j=1}^q \dfrac{1}{b_j(b_j+1)\ldots(b_j+k-2)} \right] t^k \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k-1}}(k-1)!} t^k \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k+1}}}{\vec{b}^{\overline{k}}k!}t^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^{k+1}}{k!} \\ &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!} \\ &= \left[ t \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] y(t) \end{array}$$ proves the claim. █
References
Rainville's Special Functions
