Difference between revisions of "Convergence of Hypergeometric pFq"

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=Convergence=
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__NOTOC__
If any of the $a_j$'s is a a nonpositive integer, then the series terminates and is a polynomial.
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==Theorem==
 
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Consider the series ${}_pF_q(a_1,a_2,\ldots,a_p;b_1,b_2,\ldots,b_q;z)$, where ${}pF_q$ denotes [[hypergeometric pFq]]. Then,
If any of the $b_{\ell}$'s is a nonpositive integer, the series diverges because of divison by zero.
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*If any of the $a_j$'s is a a nonpositive integer, then the series terminates and is a polynomial.  
 
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*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:
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*The remaining convergence of the series can be split into three cases:
 
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**$p<q+1$
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**$p=q+1$
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**$p>q+1$
 
==Case I: $p<q+1$==
 
==Case I: $p<q+1$==
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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===Proposition:===
<strong>Proposition:</strong> The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.<br />
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The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.<br />
<div class="mw-collapsible-content">
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===Proof:===
<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
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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}
 
$$\begin{array}{ll}
 
L &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\alpha_{k+1}}{\alpha_k} \right| \\
 
L &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\alpha_{k+1}}{\alpha_k} \right| \\

Latest revision as of 14:32, 7 October 2016

Theorem

Consider the series ${}_pF_q(a_1,a_2,\ldots,a_p;b_1,b_2,\ldots,b_q;z)$, where ${}pF_q$ denotes hypergeometric pFq. Then,

  • 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:
    • $p<q+1$
    • $p=q+1$
    • $p>q+1$

Case I: $p<q+1$

Proposition:

The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.

Proof:

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 $t \in \mathbb{C}$. █ </div></div>

Case II: $p=q+1$

Proposition: The hypergeometric pFq ${}_pF_q$ 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: █