72 lines
2.1 KiB
TeX
72 lines
2.1 KiB
TeX
%--------------------------------------------------------------
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\begin{frame}
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\begin{center}
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\Huge Identities for Roots of Macdonald function
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\end{center}
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\end{frame}
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\begin{frame}
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\scriptsize
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\frametitle{Sketch of Main Results}
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\begin{itemize}
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\item
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Substitute $\color{eqncolor}\Psi_\ell(t,r) =
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\sum_{k=0}^\ell \frac{1}{r^k} c_{\ell k} f^{(\ell-k)}(t-r)$ into
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$$\color{eqncolor}
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\partial_t \Psi_\ell(t,r) + \partial_r \Psi_\ell(t,r)=
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\sum_{n=1}^\ell \frac{b_{\ell n}}{r^2}
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\int_0^t e^{\frac{b_{\ell n}}{r}(t-t')} \Psi_\ell(t',r)dt'
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$$
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\item Result:
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$$\color{eqncolor}
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-\sum_{k=1}^\ell \frac{k}{r^{k+1}} c_{\ell k} f^{(\ell-k)}(t-r)
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= \sum_{n=1}^\ell \frac{b_{\ell n}}{r^2}
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\sum_{k=0}^\ell
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\frac{1}{r^k} c_{\ell k}
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I^{(\ell-k)}[b_{\ell n},r,f]
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$$
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\item
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Here $\color{eqncolor}I^{(p)}[b,r,f] \equiv \int_0^t
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e^{\frac{b}{r}(t-t')} f^{(p)}(t'-r) dt'$
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\item
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Work into form (argument sketched later...integration by parts)
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$$\color{eqncolor}
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0 = \sum_{k=1}^\ell r^{-(k+1)} E_k f^{(\ell-k)}(u),\quad
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E_k = k c_{\ell k} + \sum_{n=1}^\ell b_{\ell n} \sum_{q=1}^k c_{\ell,q-1}
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(b_{\ell n})^{k-q}
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$$
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\end{itemize}
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\end{frame}
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\begin{frame}
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\scriptsize
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\frametitle{Sketch of Main Results}
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\begin{itemize}
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\item
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Starting with (last line of last slide)
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$$\color{eqncolor}
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0 = \sum_{k=1}^\ell r^{-(k+1)} E_k f^{(\ell-k)}(u),\quad
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E_k = k c_{\ell k} + \sum_{n=1}^\ell b_{\ell n} \sum_{q=1}^k c_{\ell,q-1}
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(b_{\ell n})^{k-q},
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$$
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isolate terms $\color{eqncolor}E_k f^{(\ell-k)}(u)$ with operator $\color{eqncolor}Q
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= (\partial_t + \partial_r)r^2$.
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\item Example ($\color{eqncolor}\ell = 3$):
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\begin{align*}\color{eqncolor}
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Q\big[
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\frac{1}{r^2} E_1 f''(u) +
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\frac{1}{r^3} E_2 f'(u) +
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\frac{1}{r^4} E_3 f(u) \big] \color{eqncolor} & =
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\color{eqncolor}-\frac{1}{r^2}E_2 f'(u)-\frac{2}{r^3}E_3 f(u)\\
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\color{eqncolor} Q^2\big[
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\frac{1}{r^2} E_1 f''(u) +
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\frac{1}{r^3} E_2 f'(u) +
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\frac{1}{r^4} E_3 f(u) \big] \color{eqncolor}& \color{eqncolor}= \frac{6}{r^2}E_3 f(u)
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\end{align*}
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\item
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Profile $\color{eqncolor}f$ arbitrary
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$\color{eqncolor}\implies E_3 = 0 \implies E_2 = 0 \implies E_1 = 0$.
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\end{itemize}
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\end{frame}
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