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| − | 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
| + | __NOTOC__ |
| − | $$\vec{a}^{\overline{k}} = \displaystyle\prod_{j=1}^p a_j^{\overline{k}},$$ | + | The generalized hypergeometric function ${}_pF_q$ is defined by |
| − | and
| + | $${}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(a_1)_k(a_2)_k\ldots(a_p)_k}{(b_1)_k(b_2)_k\ldots(b_q)_k} \dfrac{z^k}{k!},$$ |
| − | $$\vec{a}+k = \displaystyle\prod_{j=1}^p (a_j+k),$$
| + | where $(a_1)_k$ denotes the [[Pochhammer]] symbol. |
| − | (and similar for $\vec{b}^{\overline{k}}$).
| |
| − | Define the generalized hypergeometric function
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| − | $${}_pF_q(a_1,a_2,\ldots,a_p;b_1,\ldots,b_q;t)={}_pF_q(\vec{a};\vec{b};t)=\displaystyle\sum_{k=0}^{\infty}\dfrac{\displaystyle\prod_{j=1}^p a_j^{\overline{k}}}{\displaystyle\prod_{\ell=1}^q b_{\ell}^{\overline{k}}} \dfrac{t^k}{k!}.$$ | |
| | | | |
| − | =Convergence= | + | =Properties= |
| − | If any of the $a_j$'s is a a nonpositive integer, then the series terminates and is a polynomial.
| + | [[Convergence of Hypergeometric pFq]]<br /> |
| | + | [[Hypergeometric pFq terminates to a polynomial if an a_j is a nonpositive integer]]<br /> |
| | + | [[Hypergeometric pFq diverges if a b_j is a nonpositive integer]]<br /> |
| | + | [[Hypergeometric pFq converges for all z if p less than q+1]]<br /> |
| | + | [[Hypergeometric pFq converges in the unit disk if p=q+1]]<br /> |
| | + | [[Hypergeometric pFq diverges if p greater than q+1]]<br /> |
| | | | |
| − | If any of the $b_{\ell}$'s is a nonpositive integer, the series diverges because of divison by zero.
| + | [[Derivatives of Hypergeometric pFq]]<br /> |
| | + | [[Differential equation for Hypergeometric pFq]]<br /> |
| | | | |
| − | The remaining convergence of the series can be split into three cases:
| + | =Videos= |
| | + | [https://www.youtube.com/watch?v=l8udH-Zb5Vs Special functions - Hypergeometric series (9 March 2011)]<br /> |
| | | | |
| − | ==Case I: $p<q+1$== | + | =External links= |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| + | [http://www.johndcook.com/HypergeometricFunctions.pdf Notes on hypergeometric functions]<br /> |
| − | <strong>Proposition:</strong> The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.<br />
| + | [http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=PPN600494829_0016%7CLOG_0038 Note on a hypergeometric series - Cayley]<br /> |
| − | <div class="mw-collapsible-content">
| |
| − | <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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| − | $$\begin{array}{ll}
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| − | L &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\alpha_{k+1}}{\alpha_k} \right| \\
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| − | &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\dfrac{\vec{a}^{\overline{k}}t^k}{\vec{b}^{\overline{k}}k!}}{\dfrac{\vec{a}^{\overline{k+1}}t^{k+1}}{\vec{b}^{\overline{k+1}}(k+1)!}} \right| \\
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| − | &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\vec{a}^{\overline{k}} \vec{b}^{\overline{k+1}}(k+1)!t^k }{\vec{b}^{\overline{k}} \vec{a}^{\overline{k+1}}k!t^{k+1}} \right| \\
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| − | &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{k(\vec{b}+k)}{(\vec{a}+k)t} \right| \\
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| − | &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{O(k^{q+1})}{O(k^{p})}\right| \\
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| − | &= 0 < 1,
| |
| − | \end{array}$$
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| − | therefore the series converges for all $t \in \mathbb{C}$. █
| |
| − | </div></div>
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| − | | |
| − | ==Case II: $p=q+1$==
| |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| |
| − | <strong>Proposition:</strong> The series ${}_pF_q$ converges for all $t\in \mathbb{C}$ with $|t|<1$.<br />
| |
| − | <div class="mw-collapsible-content">
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| − | <strong>Proof: █</strong>
| |
| − | </div></div>
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| − | | |
| − | ==Case III: $p>q+1$==
| |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| |
| − | <strong>Proposition:</strong> The series ${}_pF_q$ diverges for all $t \in \mathbb{C}$.<br />
| |
| − | <div class="mw-collapsible-content">
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| − | <strong>Proof: █</strong>
| |
| − | </div></div>
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| − | | |
| − | =Derivatives=
| |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| |
| − | <strong>Proposition:</strong> Suppose that ${}_pF_q$ converges. Then
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| − | $$\dfrac{d^n}{dt^n} {}_pF_q(\vec{a};\vec{b};t)=\dfrac{\vec{a}^{\overline{n}}}{\vec{b}^{\overline{n}}} {}_pF_q(\vec{a}+n;\vec{b}+n;t).$$
| |
| − | <div class="mw-collapsible-content">
| |
| − | <strong>Proof:</strong> The computation
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| − | $$\begin{array}{ll}
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| − | \dfrac{d^n}{dt^n} {}_pF_q(\vec{a};\vec{b};t) &= \dfrac{d^n}{dt^n}\displaystyle\sum_{k=0}^{\infty} \dfrac{ \vec{a}^{\overline{k}} }{ \vec{b}^{\overline{k}} } \dfrac{t^{\underline{k}}}{k!} \\
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| − | &= \displaystyle\sum_{k=n}^{\infty} \dfrac{ \vec{a}^{\overline{k}} }{ \vec{b}^{\overline{k}} } \dfrac{t^{\underline{k-n}}}{(k-n)!} \\
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| − | &= \displaystyle\sum_{k=0}^{\infty} \dfrac{ \vec{a}^{\overline{k+n}} }{ \vec{b}^{\overline{k+n}} } \dfrac{t^{\underline{k}}}{k!} \\
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| − | &=\dfrac{ \vec{a}^{\overline{n}} }{ \vec{b}^{\overline{n}} } \displaystyle\sum_{k=0}^{\infty} \dfrac{ (\vec{a}+n)^{\overline{k}} }{ (\vec{b}+n)^{\overline{k}} } \dfrac{t^{\underline{k}}}{k!} \\
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| − | &=\dfrac{ \vec{a}^{\overline{n}} }{ \vec{b}^{\overline{n}} } {}_pF_q(\vec{a}+n;\vec{b}+n;t)
| |
| − | \end{array}$$
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| − | proves the claim. █
| |
| − | </div></div>
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| − | <br /> | |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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| − | <strong>Proposition:</strong> Suppose that ${}_pF_q$ converges. Then
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| − | $$\dfrac{d^n}{dt^n} \left[ t^{\gamma} {}_pF_q(\vec{a};\vec{b};t) \right] = (\gamma-n+1)^{\overline{n}}t^{\gamma-n} {}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma+1-n,\vec{b};t).$$
| |
| − | <div class="mw-collapsible-content">
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| − | <strong>Proof:</strong> First we suppose $n=0$ yielding the formula
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| − | $$\begin{array}{ll}
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| − | t^{\gamma}{}_pF_q(\vec{a};\vec{b};t) &= t^{\gamma-n} \displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}}{\vec{b}} \dfrac{t^k}{k!} \\
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| − | &= t^{\gamma-n} \displaystyle\sum_{k=0}^{\infty} \dfrac{(\gamma+1)\vec{a}}{(\gamma+1)\vec{b}}\dfrac{t^k}{k!} \\
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| − | &= t^{\gamma-n} {}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma+1-0,\vec{b};t),
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| − | \end{array}$$
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| − | obeying the formula. Now suppose that the formula is satisfied for $n=1,2,\ldots,N-1$. We will show now that the formula holds for $n=N$:
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| − | $$\begin{array}{ll}
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| − | \dfrac{d^{N}}{dt^{N}} \left[ t^{\gamma} {}_pF_q(\vec{a};\vec{b};t)\right] &= \dfrac{d}{dt} \left[ \dfrac{d^{N-1}}{dt^{N-1}} \left[ t^{\gamma} {}_pF_q(\vec{a};\vec{b};t) \right] \right] \\
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| − | &=\dfrac{d}{dt} \left[ (\gamma-(N-1)+1)^{\overline{N-1}} t^{\gamma-(N-1)} {}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma+1-(N-1),\vec{b};t) \right] \\
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| − | &=(\gamma-N+2)^{\overline{N-1}}(\gamma-N+1)t^{\gamma-N}{}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma-N+2,\vec{b};t) \\
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| − | &\hspace{4pt}+(\gamma-N+2)^{\overline{N-1}}t^{\gamma-N+1}\dfrac{(\gamma+1) \vec{a}}{(\gamma-N+2)\vec{b}} {}_{p+1}F_{q+1} (\gamma+2,\vec{a}+1;\gamma-N+3,\vec{b};t) \\
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| − | &= (\gamma-N+2)^{\overline{N-1}} \left\{ (\gamma-N+1)t^{\gamma-N}{}_{p+1}F_{q+1}(\gamma+1;\vec{a};\gamma-N+2,\vec{b};t) \right. \\
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| − | &\hspace{4pt}+ \left. t^{\gamma-N+1} \dfrac{(\gamma+1)\vec{a}}{(\gamma-N+2)\vec{b}} {}_{p+1}F_{q+1}(\gamma+2,\vec{a}+1;\gamma-N+3,\vec{b};t) \right\} NEEDSWORK
| |
| − | \end{array}$$ █
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| − | </div></div>
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| − | | |
| − | =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 />
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| − | <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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| − | $$(\dagger) \hspace{35pt} \left[ \vartheta \displaystyle\prod_{j=1}^q (\vartheta + b_j-1) - t \displaystyle\prod_{i=1}^p (\vartheta+a_i) \right]y=0.$$
| |
| − | <div class="mw-collapsible-content">
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| − | <strong>Proof:</strong>
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| − | First compute
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| − | $$\begin{array}{ll}
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| − | \left[ t \displaystyle\prod_{i=1}^p (\vartheta+a_i) \right] y(t) &= \left[ t \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] \displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!} \\
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| − | &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \left[ \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] \dfrac{t^k}{k!} \\
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| − | &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}k!} \left[ \displaystyle\prod_{i=1}^p (\vartheta + a_i) \right] t^k \\
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| − | &= t\displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \left[ \displaystyle\prod_{i=1}^p (k+a_i) \right] \dfrac{t^k}{k!} \\
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| − | &=t\displaystyle\sum_{k=0}^{\infty} \dfrac{(\vec{a}+k)\vec{a}^{\overline{k}}}{\vec{b}^{\overline{k}}} \dfrac{t^k}{k!}. \\
| |
| − | \end{array}$$
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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= | + | =References= |
| | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=findme|next=Pochhammer}}: $4.1 (1)$ (note: typo in the text, the sum there starts at $1$ but should start at $0$) |
| | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=findme|next=Pochhammer}}: $5.1 (2)$ |
| | + | * {{BookReference|Special Functions of Mathematical Physics and Chemistry|1956|Ian N. Sneddon|prev=findme|next=findme}}: $\S 12 (12.4)$ |
| | + | * {{BookReference|Special Functions for Scientists and Engineers|1968|W.W. Bell|prev=findme|next=findme}}: $(9.1)$ |
| | | | |
| − | ==${}_0F_0$==
| + | {{:Hypergeometric functions footer}} |
| − | #${}_0F_0(;;z)=e^z$
| |
| | | | |
| − | ==${}_0F_1$==
| + | [[Category:SpecialFunction]] |
| − | #${}_0F_1 \left(;\dfrac{1}{2};-\dfrac{z^2}{4} \right)=\cos(z)$
| |
| − | #$z{}_0F_1 \left(;\dfrac{3}{2};-\dfrac{z^2}{4} \right)=\sin(z)$
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| − | | |
| − | ==${}_1F_0$==
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| − | #${}_1F_0(-a;;z)=(1-z)^a$
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| − | | |
| − | ==${}_1F_1$==
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| − | ==${}_2F_1$==
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| − | #$z{}_2F_1(1,1;2;-z)=\log(1+z)$<br />
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| − | <strong>Proof:</strong> Calculate
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| − | $$\begin{array}{ll}
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| − | z{}_2F_1(1,1;2;-z) &= z\displaystyle\sum_{k=0}^{\infty} \dfrac{1^{\overline{k}}1^{\overline{k}}}{2^{\overline{k}}k!} (-z)^k \\
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| − | &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\left( \frac{\Gamma(k+1)}{\Gamma(1)} \right)^2}{\left( \frac{\Gamma(2+k)}{\Gamma(2)} \right)k!}(-1)^k z^{k+1} \\
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| − | &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(k!)^2(-1)^k}{(k+1)!k!} z^{k+1} \\
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| − | &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k+1} z^{k+1} \\
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| − | &= -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k z^k}{k} \\
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| − | &= \log(1+z),
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| − | \end{array}$$
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| − | using a well-known formula for the [[logarithm|Taylor series of $\log(1+z)$]].
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| − | #${}_2F_1(1,1;2;z)=-\dfrac{\log(1-z)}{z}$
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| − | #${}_2F_1 \left( \dfrac{1}{2}, 1; \dfrac{3}{2}; z^2 \right) = \dfrac{\log \left(\dfrac{1+z}{1-z} \right)}{2z}$
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| − | #${}_2F_1 \left( \dfrac{1}{2},1 ; \dfrac{3}{2};-z^2 \right) = \dfrac{\arctan z}{z}$
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| − | #${}_2F_1 \left( \dfrac{1}{2}, \dfrac{1}{2}; \dfrac{3}{2}; z^2 \right) = \sqrt{1-z^2} {}_2F_1 \left( 1,1;\dfrac{3}{2};z^2 \right)=\dfrac{\arcsin z}{z}$
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| − | | |
| − | ==References==
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| − | [http://www.johndcook.com/HypergeometricFunctions.pdf Notes on hypergeometric functions]<br />
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| − | Rainville's Special Functions<br />
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| − | [http://dualaud.net/specialfunctionswiki/abramowitz_and_stegun-1.03/page_555.htm Abramowitz and Stegun]
| |
The generalized hypergeometric function ${}_pF_q$ is defined by
$${}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(a_1)_k(a_2)_k\ldots(a_p)_k}{(b_1)_k(b_2)_k\ldots(b_q)_k} \dfrac{z^k}{k!},$$
where $(a_1)_k$ denotes the Pochhammer symbol.
