119 lines
2.0 KiB
Markdown
119 lines
2.0 KiB
Markdown
# Landau distribution
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## Landau PDF
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:::: {.columns align=center}
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::: {.column width=50%}
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$$
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L(x) = \frac{1}{\pi} \int \limits_{0}^{+ \infty}
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dt \, e^{-t \ln(t) -xt} \sin (\pi t)
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$$
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:::
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::: {.column width=50%}
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:::
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::::
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. . .
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\begin{center}
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No closed form for \alert{ANYTHING}
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\end{center}
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## Landau median
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::::: {.columns}
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:::: {.column width=50%}
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::: incremental
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- The median of $f$ is defined by
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$$
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F(m) = \int_{-\infty}^m fdx = \frac{1}{2}
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$$
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- Equivalently
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$$
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m = F^{-1}\left(\frac{1}{2}\right)
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$$
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- Numerical integration or QDF is needed
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:::
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::::
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::: {.column width=50%}
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:::
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:::::
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## Landau median
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- CDF computed by numerical integration
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- QDF computed by numerical root-finding
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\setbeamercovered{}
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\begin{center}
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\begin{tikzpicture}[remember picture]
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\node at (0,0) (here) {$m_L\ex = 1.3557804...$};
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\pause
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\node [opacity=0.5, xscale=0.35, yscale=0.25 ] at (here) {\includegraphics{images/high.png}};
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\end{tikzpicture}
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\end{center}
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\setbeamercovered{transparent}
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## Landau mode
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- Maximum $\hence \partial_x L(\mu) = 0$
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. . .
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- Computed by numerical minimization (Brent)
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\setbeamercovered{}
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\begin{center}
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\begin{tikzpicture}[remember picture]
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\node at (0,0) (here) {$\mu_L\ex = − 0.22278...$};
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\pause
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\node [opacity=0.5, xscale=0.32, yscale=0.25 ] at (here) {\includegraphics{images/high.png}};
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\end{tikzpicture}
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\end{center}
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\setbeamercovered{transparent}
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## Landau FWHM
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We need to compute the maximum:
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$$
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L_{\text{max}} = L(\mu_L)
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$$
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$$
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\text{FWHM} = w = x_+ - x_- \with L(x_{\pm}) = \frac{L_{\text{max}}}{2}
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$$
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. . .
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- Computed by numerical root finding (Brent)
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\setbeamercovered{}
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\begin{center}
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\begin{tikzpicture}[remember picture]
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\node at (0,0) (here) {$w_L\ex = 4.018645...$};
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\pause
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\node [opacity=0.5, xscale=0.32, yscale=0.25 ] at (here) {\includegraphics{images/high.png}};
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\end{tikzpicture}
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\end{center}
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\setbeamercovered{transparent}
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