Difference between revisions of "Derivatives of Hypergeometric pFq"

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(Created page with "<div class="toccolours mw-collapsible mw-collapsed" style="width:800px"> <strong>Proposition:</strong> Suppose that ${}_pF_q$ converges. Then $$\dfrac{d^n}{dt^n} {}_pF_q(\vec...")
 
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Latest revision as of 21:29, 26 June 2016

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 $n=N$: $$\begin{array}{ll} \dfrac{d^{N}}{dt^{N}} \left[ t^{\gamma} {}_pF_q(\vec{a};\vec{b};t)\right] &= \dfrac{d}{dt} \left[ \dfrac{d^{N-1}}{dt^{N-1}} \left[ t^{\gamma} {}_pF_q(\vec{a};\vec{b};t) \right] \right] \\ &=\dfrac{d}{dt} \left[ (\gamma-(N-1)+1)^{\overline{N-1}} t^{\gamma-(N-1)} {}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma+1-(N-1),\vec{b};t) \right] \\ &=(\gamma-N+2)^{\overline{N-1}}(\gamma-N+1)t^{\gamma-N}{}_{p+1}F_{q+1}(\gamma+1,\vec{a};\gamma-N+2,\vec{b};t) \\ &\hspace{4pt}+(\gamma-N+2)^{\overline{N-1}}t^{\gamma-N+1}\dfrac{(\gamma+1) \vec{a}}{(\gamma-N+2)\vec{b}} {}_{p+1}F_{q+1} (\gamma+2,\vec{a}+1;\gamma-N+3,\vec{b};t) \\ &= (\gamma-N+2)^{\overline{N-1}} \left\{ (\gamma-N+1)t^{\gamma-N}{}_{p+1}F_{q+1}(\gamma+1;\vec{a};\gamma-N+2,\vec{b};t) \right. \\ &\hspace{4pt}+ \left. t^{\gamma-N+1} \dfrac{(\gamma+1)\vec{a}}{(\gamma-N+2)\vec{b}} {}_{p+1}F_{q+1}(\gamma+2,\vec{a}+1;\gamma-N+3,\vec{b};t) \right\} NEEDSWORK \end{array}$$ █