AP Phys C Mech Index

\begin{equation} \dv t \qty(e^{tA}) = \dv t \qty(I + \sum_{k=1}^{\infty} \frac{t^{k}}{k!}A^{k}) = \qty(\sum_{k=1}^{\infty }\frac{1}{k!}kt^{k-1}A^{k-1})A \end{equation}

We do this separation because \(k=0\) would’t make sense to raise \(A\) (\(k-1=-1\)) to as we are unsure about the invertability of \(A\). Obviously \(\frac{1}{k!}k = \frac{1}{(k-1)!}\). Therefore, we can shift our index back yet again:

\begin{equation} \qty(\sum_{k=1}^{\infty }\frac{1}{k!}kt^{k-1}A^{k-1})A = \qty(\sum_{j=0}^{\infty }\frac{1}{j!}t^{j}A^{j})A \end{equation}

approximate inference

Last edited: August 8, 2025

Direct Sampling

Direct Sampling is an approximate inference method where we pull samples from the given joint probability distribution.

Example

Suppose we are interested in:

where we dare \(P(B^{1}|D^{1},C^{1})\).

Step 1: sort

We obtain a topological sort of this network:

\begin{equation} B, S, E, D, C \end{equation}

Step 2: sample from \(B,S\)

We sample \(B\). We sampled that \(B=1\) today.
We sample \(S\). We sampled that \(S=0\) today.

Step 3: sample from \(E\)

We sample \(E\) GIVEN what we already sampled, that \(B=1, S=0\), we sampled that that \(E = 1\)

Step 4: sample from \(D, C\)

We sample \(D\) given that \(E=1\) as we sampled.
We sample \(C\) given that \(E=1\) as we sampled.

Repeat

Repeat steps 2-4