Reading notes

:claim: Mccarthyism was a process that de-politicized America

Since political activities could get you in trouble, prudent folk avoided them

Social conformaty became standard

middle-class Americans became social conformists

Communism serves as a form of balance checking, which Mccathyism lost

With their demise, the nation lost the institutional network that had created a public space where serious alternatives to the status quo could be presented.

Moderate-left was also diminished

Moreover, with the disappearance of a vigorous movement on their left, moderate reform groups were more exposed to right-wing attacks and thus rendered less effective.

lemmatization is the act of normalizing words into standard meaning irrespective of word variations in order to do a broader analysis

Any two basis of finite-dimensional vector space have the same length.

constituents

• A finite-dimensional vector space \(V\)
• Basis \(B_1\), \(B_2\) be bases in \(V\)

requirements

Given \(B_1\), \(B_2\) are basis in \(V\), we know that they are both linearly independent and spans \(V\). We have that the length of linearly-independent list \(\leq\) length of spanning list.

Let’s take first \(B_1\) as linearly independent and \(B_2\) as spanning:

We have then \(len(B_1) \leq len(B_2)\)

Swapping roles:

We have then \(len(B_2) \leq len(B_1)\)

Autonomously driving is really hard. How do we integrate planning + learning in a close-loop style. We’ll start from the current belief, and construct a tree of all reachable belief state.

Recall DESPOT.

Approach

1. learning (b): a neural network which maps the driving history into a policy and value
2. planning (a): we will use the neural network’s derived policy and value to run MCTS
3. execution (e): execute the actions in a simulator

The data which is obtained using the simulator is used to train the neural network.

occasionally, you can’t really get a specific solution.

\begin{equation} \dv{y}{t} = e^{t}\cos y \end{equation}

after doing the , you get:

\begin{equation} \ln (\sec y + \tan y) - e^{t} = C \end{equation}

you get sets of this function \(F(t,y)\) which shifts it up and down, by any constant C.

But at any given \((t,y)\), you get a slope \(e^{t}\cos y\).