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| =Properties= | | =Properties= |
− | [[Convergence of Hypergeometric pFq]] | + | [[Convergence of Hypergeometric pFq]]<br /> |
− | [[Derivatives of Hypergeometric pFq]] | + | [[Derivatives of Hypergeometric pFq]]<br /> |
− | | + | [[Differential equation for Hypergeometric pFq]]<br /> |
− | =Differential equation=
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− | Define the derivative operator $\vartheta=t \dfrac{d}{dt}$.Then
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− | $$\vartheta t^k = t \dfrac{d}{dt} t^k = t(kt^{k-1})=kt^k.$$
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− | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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− | <strong>Proposition:</strong> The operator $\vartheta$ is a [[Linear_operator | linear operator]]. <br />
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− | <div class="mw-collapsible-content">
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− | <strong>Proof: █</strong>
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− | </div></div>
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− | <br /> | |
− | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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− | <strong>Theorem:</strong> Define $y(t)={}_pF_q(\vec{a};\vec{b};t)$. Then $y$ satisfies
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− | $$(\dagger) \hspace{35pt} \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta + b_j-1) - t \displaystyle\prod_{i=1}^p (\vartheta+a_i) \right]y=0.$$
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− | <div class="mw-collapsible-content">
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− | <strong>Proof:</strong>
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− | First compute
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− | $$\begin{array}{ll}
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− | \left[ t \displaystyle\prod_{i=1}^p (\vartheta+a_i) \right] y(t) &= \left[ t \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] \displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!} \\
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− | &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \left[ \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] \dfrac{t^k}{k!} \\
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− | &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}k!} \left[ \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] t^k \\
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− | &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \left[ \displaystyle\prod_{i=1}^p (k+a_i) \right] \dfrac{t^k}{k!} \\
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− | &=t\displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!}. \\
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− | \end{array}$$
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− | Now the computation
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− | $$\begin{array}{ll}
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− | \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!} \\
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− | &=\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 \\
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− | &= \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 \\
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− | &= \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 \\
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− | &= \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 \\
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− | &= \displaystyle\sum_{k=1}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k-1}}(k-1)!} t^k \\
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− | &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k+1}}}{\vec{b}^{\overline{k}}k!}t^{k+1} \\
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− | &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^{k+1}}{k!} \\
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− | &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!} \\
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− | &= \left[ t \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] y(t)
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− | \end{array}$$
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− | proves the claim. █
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− | </div></div>
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| =Examples= | | =Examples= |
Revision as of 21:30, 26 June 2016
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
$${}_pF_q(a_1,a_2,\ldots,a_p;b_1,\ldots,b_q;t)={}_pF_q(\vec{a};\vec{b};t)=\displaystyle\sum_{k=0}^{\infty}\dfrac{\displaystyle\prod_{j=1}^p a_j^{\overline{k}}}{\displaystyle\prod_{\ell=1}^q b_{\ell}^{\overline{k}}} \dfrac{t^k}{k!}.$$
Properties
Convergence of Hypergeometric pFq
Derivatives of Hypergeometric pFq
Differential equation for Hypergeometric pFq
Examples
${}_0F_0$
Exponential in terms of hypergeometric 0F0
${}_0F_1$
Relationship between cosine and hypergeometric 0F1
Relationship between sine and hypergeometric 0F1
Relationship between cosh and hypergeometric 0F1
Relationship between sinh and hypergeometric 0F1
Relationship between Bessel J sub nu and hypergeometric 0F1
${}_1F_0$
${}_1F_1$
${}_1F_2$
Relationship between Struve function and hypergeometric pFq
${}_2F_0$
Bessel polynomial generalized hypergeometric
${}_2F_1$
z2F1(1,1;2,-z) equals log(1+z)
Relationship between arcsin and hypergeometric 2F1
Relationship between arctan and hypergeometric 2F1
Relationship between Chebyshev T and hypergeometric 2F1
Relationship between Chebyshev U and hypergeometric 2F1
Relationship between Legendre polynomial and hypergeometric 2F1
Relationship between incomplete beta and hypergeometric 2F1
Videos
Special functions - Hypergeometric series
References
Notes on hypergeometric functions
Rainville's Special Functions
Abramowitz and Stegun
Note on a hypergeometric series - Cayley