A reformist, counterculture movement during the ’80s lead by Ronald Reagan. Its a new response to the neoliberalism which aligned the blocks of Evangelical Christians (25% of voters) and Business leaders (powerful leaders.)

American liberalism expands under the new right as well.

President as a party leader: Reagan is often shown as shining beaken of the Republican Party Leadership—won every single state except Georgia .

\begin{equation} y’’=0 \end{equation}

that is, if \(F=0\), then the solution will travel along a straight line.

Putting something with a different temperature in a space with a constant temperature. The assumption underlying here is that the overall room temperature stays constant (i.e. the thing that’s cooling is so small that it doesn’t hurt room temperature).

\begin{equation} y’(t) = -k(y-T_0) \end{equation}

where, \(T_0\) is the initial temperature.

The intuition of this modeling is that there is some \(T_0\), which as the temperature \(y\) of your object gets closer to t. The result we obtain

\begin{equation} f(x) \approx f(x_{t-1}) + (x-x_{t-1}) f’(x_{t-1}) + \frac{(x-x_{t-1})^{2}}{2} f’’(x_{t-1}) \end{equation}

Taking a derivative with respect to this, we obtain:

\begin{equation} f’(x_{t-1}) + (x-x_{t-1}) f’’(x_{t-1}) \end{equation}

Solving the update equation for zero, we obtain that:

\begin{equation} x = x_{t-1} - \frac{f’(x_{t-1})}{f’’(x_{t-1})} \end{equation}

This converges quadratically!!

For gradients:

\begin{equation} x_{t} = x_{t-1} - \qty(\bold{H}_{g})^{-1}\bold{g}_{k} \end{equation}

Failure Case

If the function is near an inflection point (i.e. with bad quadratic approximation), you may converge at a point which doesn’t satisfy SONC (i.e. you will get an inflection but not a minima).

\begin{equation} \text{NL} = \text{NSPACE} \qty( \log n) \end{equation}

See also Certificates-Based Intepretation of NL

problems in \(NL\)

We can see \(L \subseteq NL\), because a TM is a NTM.

STCONN is in NL

On input \(\qty(G, s,t)\), if \(s = t\), accept; otherwise,

• currNode = 5
• numSteps = 0
• while steps <= n
  • non-deterministically choose a next node
  • update currNode = w
  • if w = t, accept
  • set numSteps ++
• reject

so we just have to remember the current node. So this whole thing is \(O\qty(\log n)\).