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[AMR] More fixes
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@@ -3,5 +3,5 @@ $\vec{f}(\vec{x}) \approx \vec{f}(\vec{\overline{x}}) + \mat{J}_{\vec{f}} \big|_
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\shortdefinition[Jac.] $\mat{J}_{\vec{f}}$ rows for eq of $\vec{f}$ cols for vars of each eq.
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Approximation using finite differences $\frac{f(\overline{x} + h) - f(\overline{x})}{h}$,\\
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Part. diff; Approx. using finite differences $\frac{f(\overline{x} + h) - f(\overline{x})}{h}$,\\
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or central differences (vector of $\frac{\vec{f}(\vec{\overline{x}}) + h_i \vec{e_i}}{h_i}$, with $\vec{e_i}$ unit vec)
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@@ -3,8 +3,8 @@
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\bi{Man. sol.}: comp. $M = A^\top A$, $b' = A^\top b$, then $Mx = b'$.
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\bi{Prob. sol.}: $\text{argmax} p(\vec{x} \divider \vec{z})$ with
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\bi{Prob. sol.}: $\text{argmax} \; \P(\vec{x} \divider \vec{z})$ with Maximum \dots
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\begin{itemize}
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\item \bi{Max. Like} $p(\vec{x} | \vec{z}) \propto p(\vec{z} | \vec{x}) = \prod_{i = 1}^N p(\vec{z}_i | \vec{x})$
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\item \bi{M a Post} $p(\vec{x} | \vec{z}) \propto p(\vec{z} | \vec{x}) p(\vec{x}) = p(\vec{x}) \prod_{i = 1}^N p(\vec{z}_i | \vec{x})$
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\item \bi{Likelihood} $\P(\vec{x} | \vec{z}) \propto \P(\vec{z} | \vec{x}) = \prod_{i = 1}^N \P(\vec{z}_i | \vec{x})$
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\item \bi{a Post} $\P(\vec{x} | \vec{z}) \propto \P(\vec{z} | \vec{x}) \P(\vec{x}) = \P(\vec{x}) \prod_{i = 1}^N \P(\vec{z}_i | \vec{x})$
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\end{itemize}
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@@ -1,5 +1,5 @@
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\subsection{Non-Linear Least Squares}
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Find $\vec{x}^* = \text{argmax} p(\vec{x} | \vec{z}) = \argmin{}(-\log(p(\vec{x}|\vec{z})))$
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Find $\vec{x}^* = \text{argmax}\; \P(\vec{x} | \vec{z}) = \argmin{}(-\log(\P(\vec{x}|\vec{z})))$
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\bi{Gauss-Newton} % TODO: Do we really need these? If so, use from NumCS (much simpler notation)
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@@ -1,4 +1,4 @@
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\subsection{Bayes Filter}
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\subsection{Bayes Filter (in DAG)}
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$\vec{x}_k^R$ state at time k, $\vec{z}_k^p$ dist. meas., $\vec{u}^p_k$ wheel odometry (= meas.).
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Typ. care ab. curr. state: altern. pred. \& update.
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% TODO: Do we really need the below?
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@@ -10,3 +10,5 @@ Non-l. state trans. model $\vec{x}_k = \vec{f}(\vec{x}_{k - 1}, \vec{u}_k, \vec{
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\item \bi{Meas. residual}: $\vec{y}_k = \tilde{\vec{z}}_k - \vec{h}(\hat{\vec{x}}_{k | k - 1})$
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\end{itemize}
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Difference to above: $\mat{H}$ becomes $\mat{H}_k$, and $\mat{H}^\top$ is $\mat{H}_k^\top$
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% TODO: Consider adding examples
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