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Revision as of 18:23, 20 May 2015
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 $${}_pF_q(a_1,a_2,\ldots,a_p;b_1,\ldots,b_q;t)={}_pF_q(\vec{a};\vec{b};t)=\displaystyle\sum_{k=0}^{\infty}\dfrac{\displaystyle\prod_{j=1}^p a_j^{\overline{k}}}{\displaystyle\prod_{\ell=1}^q b_{\ell}^{\overline{k}}} \dfrac{t^k}{k!}.$$
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. █
Proposition: Suppose that ${}_pF_q$ converges. Then $$\dfrac{d^n}{dt^n} \left[ t^{\gamma} {}_pF_q(\vec{a};\vec{b};t) \right] = (\gamma-n+1)^{\overline{n}}t^{\gamma-n} {}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma+1-n,\vec{b};t).$$
Proof: First we suppose $n=0$ yielding the formula $$\begin{array}{ll} t^{\gamma}{}_pF_q(\vec{a};\vec{b};t) &= t^{\gamma-n} \displaystyle\sum_{k=0}^{\infty} \dfrac{\vec{a}}{\vec{b}} \dfrac{t^k}{k!} \\ &= t^{\gamma-n} \displaystyle\sum_{k=0}^{\infty} \dfrac{(\gamma+1)\vec{a}}{(\gamma+1)\vec{b}}\dfrac{t^k}{k!} \\ &= t^{\gamma-n} {}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma+1-0,\vec{b};t), \end{array}$$ 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 $n=N$: $$\begin{array}{ll} \dfrac{d^{N}}{dt^{N}} \left[ t^{\gamma} {}_pF_q(\vec{a};\vec{b};t)\right] &= \dfrac{d}{dt} \left[ \dfrac{d^{N-1}}{dt^{N-1}} \left[ t^{\gamma} {}_pF_q(\vec{a};\vec{b};t) \right] \right] \\ &=\dfrac{d}{dt} \left[ (\gamma-(N-1)+1)^{\overline{N-1}} t^{\gamma-(N-1)} {}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma+1-(N-1),\vec{b};t) \right] \\ &=(\gamma-N+2)^{\overline{N-1}}(\gamma-N+1)t^{\gamma-N}{}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma-N+2,\vec{b};t) \\ &\hspace{4pt}+(\gamma-N+2)^{\overline{N-1}}t^{\gamma-N+1}\dfrac{(\gamma+1) \vec{a}}{(\gamma-N+2)\vec{b}} {}_{p+1}F_{q+1} (\gamma+2,\vec{a}+1;\gamma-N+3,\vec{b};t) \\ &= (\gamma-N+2)^{\overline{N-1}} \left\{ (\gamma-N+1)t^{\gamma-N}{}_{p+1}F_{q+1}(\gamma+1;\vec{a};\gamma-N+2,\vec{b};t) \right. \\ &\hspace{4pt}+ \left. t^{\gamma-N+1} \dfrac{(\gamma+1)\vec{a}}{(\gamma-N+2)\vec{b}} {}_{p+1}F_{q+1}(\gamma+2,\vec{a}+1;\gamma-N+3,\vec{b};t) \right\} NEEDSWORK \end{array}$$ █
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. █
Examples
${}_0F_0$
Theorem
The following formula holds: $$e^z={}_0F_0(;;z),$$ where ${}_0F_0$ denotes the hypergeometric 0F0 and $e^z$ denotes the exponential.
Proof
References
${}_0F_1$
Theorem
The following formula holds: $$\sin(az)=az{}_0F_1 \left(;\dfrac{3}{2};-\dfrac{(az)^2}{4} \right),$$ where $\sin$ denotes the sine function and ${}_0F_1$ denotes the hypergeometric pFq.
Proof
References
Theorem
The following formula holds: $$\cos(z)={}_0F_1 \left(;\dfrac{1}{2};-\dfrac{z^2}{4} \right),$$ where $\cos$ denotes the cosine and ${}_0F_1$ denotes the hypergeometric pFq.
Proof
References
${}_1F_0$
Theorem: The following formula holds: $$(1-z)^a = {}_1F_0(-a;;z).$$
Proof: █
${}_1F_1$
${}_2F_0$
Theorem: The following formula holds: $$y_n(x)={}_2F_0 \left( -n, 1+n;-; -\dfrac{1}{2}x \right),$$ where $y_n(x)$ denotes a Bessel polynomial and ${}_2F_0$ denotes the hypergeometric pFq.
Proof: █
${}_2F_1$
Theorem
The following formula holds: $$\log(1+z)=z{}_2F_1(1,1;2;-z),$$ where $\log$ denotes the logarithm and ${}_2F_1$ denotes the hypergeometric pFq.
Proof
Calculate $$\begin{array}{ll} z{}_2F_1(1,1;2;-z) &= z\displaystyle\sum_{k=0}^{\infty} \dfrac{1^{\overline{k}}1^{\overline{k}}}{2^{\overline{k}}k!} (-z)^k \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\left( \frac{\Gamma(k+1)}{\Gamma(1)} \right)^2}{\left( \frac{\Gamma(2+k)}{\Gamma(2)} \right)k!}(-1)^k z^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(k!)^2(-1)^k}{(k+1)!k!} z^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k+1} z^{k+1} \\ &= -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k z^k}{k} \\ &= \log(1+z), \end{array}$$ by the Taylor series of $\log(1+z)$. █
References
Theorem: The following formula holds: $$2{}_2F_1(a,a+\frac{1}{2};\frac{1}{2};z)=(1+\sqrt{z})^{-2a}+(1-\sqrt{z})^{-2a}.$$
Proof: █
Theorem: The following formula holds: $$e^{-az}=(2\cosh z)^{-a} \tanh(z) {}_2F_1(1+\frac{a}{2},\frac{1}{2}+\frac{a}{2};1+a;(\cosh z)^{-2}).$$
Proof: █
Relationship between arcsin and hypergeometric 2F1
Theorem: The following formula holds: $$\arctan z = z {}_2F_1(\frac{1}{2},1;\frac{3}{2};-z^2).$$
Proof: █
Theorem: The following formula holds: $$\log \left( \dfrac{1+z}{1-z} \right) = 2z {}_2F_1(\frac{1}{2},1;\frac{3}{2};z^2).$$
Proof: █
Theorem: The following formula holds: $$T_n(x) = {}_2F_1 \left( -n,n ; \dfrac{1}{2}; \dfrac{1-x}{2} \right),$$ where $T_n$ denotes a Chebyshev T polynomial and ${}_2F_1$ denotes the hypergeometric pFq.
Proof: █
Theorem: The following formula holds: $$U_n(x) = (n+1){}_2F_1 \left( -n,n+2 ; \dfrac{3}{2}; \dfrac{1-x}{2} \right),$$ where $U_n$ denotes a Chebyshev U polynomial and ${}_2F_1$ denotes the hypergeometric pFq.
Proof: █
Theorem: The following formula holds: $$P_n(x) = {}_2F_1 \left( -n, n+2; \dfrac{3}{2}; \dfrac{1-x}{2} \right),$$ where $P_n$ denotes a Legendre polynomial and ${}_2F_1$ denotes the hypergeometric pFq.
Proof: █
Videos
Special functions - Hypergeometric series
References
Notes on hypergeometric functions
Rainville's Special Functions
Abramowitz and Stegun
