\section{Floquet Fermi Goldern Rule}

In this section we are going to derive the Floquet Fermi goldern rule for above derived quantum Floquet states using $t-t'$ formalism.

\vspace{5mm}
\noindent
The Floquet states \eqref{3.17} fullfills the $t-t'$ Schrödinger equation [*Ref:myReport] as follows
\begin{equation} \label{4.1}
  i \hbar \pdv{t}\ket{\psi_{\alpha}(t,t')} =
  H_F(t') \ket{\psi_{\alpha}(t,t')}
\end{equation}
where Floquet Hamiltonian given by
\begin{equation} \label{4.2}
  H_F(t') \equiv
  H_e(t) - i\hbar \dv{t}
\end{equation}
and
\begin{equation} \label{4.3}
  \ket{\psi_{\alpha}(t,t')} =
  \exp(-\frac{i}{\hbar}\varepsilon_{\alpha} t)\ket{\phi_{\alpha}(t')}
\end{equation}
Now for the Eq. \eqref{4.1} corresponding time evolution operator satisfy the Schrödinger equation
\begin{equation} \label{4.4}
  U_0(t,t_0;t') = \exp(-\frac{i}{\hbar}H_F(t')\qty[t-t_0])
\end{equation}
Consider a time-independent total perturbation $V(\mb{r})$ switched on at the reference time $t=t_0$, then Schrödinger equation becomes
\begin{equation} \label{4.5}
  i \hbar \pdv{t}\ket{\Psi_{\alpha}(t,t')} =
  \qty[H_F(t') + V(\mb{r})]\ket{\Psi_{\alpha}(t,t')}
\end{equation}
and when $t\leq t_0$ both solutions of the Schrödinger equation coincide
\begin{equation} \label{4.6}
  \ket{\psi_{\alpha}(t,t')} =\ket{\Psi_{\alpha}(t,t')} \quad
  \text{when} \quad
  t \leq t_0
\end{equation}
Now, we can introduce the interaction picture representation of the $t-t'$ Floquet state as
\begin{equation} \label{4.7}
  \ket{\Psi_{\alpha}(t,t')}_I = U_0^{\dagger}(t,t_0;t')
  \ket{\Psi_{\alpha}(t,t')}
\end{equation}
and the perturbation in the interaction picture will be
\begin{equation} \label{4.8}
  V_I(\mb{r}) = U_0^{\dagger}(t,t_0;t')V(\mb{r})U_0(t,t_0;t') =
  V(\mb{r}).
\end{equation}
This leads to the Schrödinger  eqution in the interction picture
\begin{equation} \label{4.9}
  i \hbar \pdv{t}\ket{\Psi_{\alpha}(t,t')}_I =
  V_I(\mb{r})\ket{\Psi_{\alpha}(t,t')}_I
\end{equation}
with the recursive solution
\begin{equation} \label{4.10}
  \ket{\Psi_{\alpha}(t,t')}_I = \ket{\Psi_{\alpha}(t_0,t')}_I +
  \frac{1}{i\hbar}
  \int_{t_0}^t dt_1 \;
  V_I(\mb{r}) \ket{\Psi_{\alpha}(t_1,t')}_I
\end{equation}
Iterating the solution only upto first order (Born approximation) this leads to
\begin{equation} \label{4.11}
  \ket{\Psi_{\alpha}(t,t')}_I \approx \ket{\psi_{\alpha}(t_0,t')} +
  \frac{1}{i\hbar}
  \int_{t_0}^t dt_1 \;
  V_I(\mb{r}) \ket{\psi_{\alpha}(t_0,t')}
\end{equation}
and multiply it by $\bra{\psi_{\beta}(t_0,t')}$ and we will get
\begin{equation} \label{4.12}
  \braket{\psi_{\beta}(t_0,t')}{\Psi_{\alpha}(t,t')}_I = \braket{\psi_{\beta}(t_0,t')}{\psi_{\alpha}(t_0,t')} +
  \frac{1}{i\hbar}
  \int_{t_0}^t dt_1 \;
  \bra{\psi_{\beta}(t_0,t')}
  V_I(\mb{r}) \ket{\psi_{\alpha}(t_0,t')}.
\end{equation}
Then introdusing unitory operator $U_0$ we can re-write this as
\begin{equation} \label{4.13}
  \begin{aligned}
    \mel{\psi_{\beta}(t_0,t')}{U_0^{\dagger}(t,t_0;t')}{\Psi_{\alpha}(t,t')} & = \mel{\psi_{\beta}(t_0,t')}{U_0^{\dagger}(t,t_0;t')U_0(t,t_0;t')}{\psi_{\alpha}(t_0,t')} \\
    & +
    \frac{1}{i\hbar}
    \int_{t_0}^t dt_1 \;
    \bra{\psi_{\beta}(t_0,t')}
    U_0^{\dagger}(t_1,t_0;t')
    V(\mb{r})
    U_0(t_1,t_0;t')
    \ket{\psi_{\alpha}(t_0,t')}
  \end{aligned}
\end{equation}
and this can be simplied as
\begin{equation} \label{4.14}
  \begin{aligned}
    \braket{\psi_{\beta}(t,t')}{\Psi_{\alpha}(t,t')} = \braket{\psi_{\beta}(t,t')}{\psi_{\alpha}(t,t')} +
    \frac{1}{i\hbar}
    \int_{t_0}^t dt_1 \;
    \bra{\psi_{\beta}(t_1,t')}
    V(\mb{r}) \ket{\psi_{\alpha}(t_1,t')}.
  \end{aligned}
\end{equation}
Since our $t-t'$ Floquet states are orthonormal [*Ref:myReport- t-t' formalism] we can derive that
\begin{equation} \label{4.15}
  \begin{aligned}
    \braket{\psi_{\beta}(t,t')}{\Psi_{\alpha}(t,t')} =
    \delta_{\alpha\beta}\exp(i\omega\qty[t'-t]) +
    \frac{1}{i\hbar}
    \int_{t_0}^t dt_1 \;
    \bra{\psi_{\beta}(t_1,t')}
    V(\mb{r}) \ket{\psi_{\alpha}(t_1,t')}.
  \end{aligned}
\end{equation}
Now, set $t_0 = 0$ and for a case $\alpha \neq \beta$ where we can represent $\alpha = (n_{\alpha},m_{\alpha})$ and $\beta = (n_{\beta},m_{\beta})$ and this will simplied to
\begin{equation} \label{4.16}
  \begin{aligned}
    \braket{\psi_{\beta}(t,t')}{\Psi_{\alpha}(t,t')} =
    -
    \frac{i}{\hbar}
    \int_{0}^t dt_1 \;
    \bra{\psi_{\beta}(t_1,t')}
    V(\mb{r}) \ket{\psi_{\alpha}(t_1,t')}.
  \end{aligned}
\end{equation}
In addition, since our Floquet states create a basis for composite space we can represent any solution using our Floquet states
\begin{equation} \label{4.17}
  \ket{\Psi_{\alpha}(t,t')} = \sum_{\beta} a_{\alpha\beta}(t,t')
  \ket{\psi_{\beta}(t,t')}.
\end{equation}
Therefore we can derive a equation for this \textit{scattering amplitude} as
\begin{equation} \label{4.18}
  a_{\alpha\beta}(t,t') =
  \braket{\psi_{\beta}(t,t')}{\Psi_{\alpha}(t,t')} =
  -
  \frac{i}{\hbar}
  \int_{0}^t dt_1 \;
  \bra{\psi_{\beta}(t_1,t')}
  V(\mb{r}) \ket{\psi_{\alpha}(t_1,t')}.
\end{equation}

\vspace{5mm}
\noindent
Now lets assume a scattering event from a $t-t'$ Floquet state $\ket{\psi_{\beta}(t,t')}$ into another $t-t'$ Floquet state $\ket{\Psi_{\alpha}(t,t')}$ with constant quansienergy $\varepsilon$ given as follows
\begin{equation} \label{4.19}
  \ket{\Psi_{\alpha}(t,t')} =
  \exp(-\frac{i}{\hbar}\varepsilon t)
  \ket{\Phi_{\alpha}(t')}
\end{equation}
Now consider a scattering event
\begin{equation} \label{4.20}
  \psi_{\beta}(\mb{k'},t,t') = \exp(-\frac{i}{\hbar}\varepsilon_{\beta} t)
  \phi_{\beta}(\mb{k'},t')
  \longrightarrow
  \Psi_{\alpha}(\mb{k},t,t') = \exp(-\frac{i}{\hbar}\varepsilon t)
  \Phi_{\alpha}(\mb{k},t')
\end{equation}
\begin{figure}[ht!]
  \centering
  \includegraphics[scale=1.0]{figures/fig2.pdf}
  \caption{Scattering from $\ket{\psi_{\beta}(t,t')}$ to constant energy state $\ket{\Psi_{\alpha}(t,t')}$ due to scattering potential created by impurities.}
  \label{fig:2}
\end{figure}

\noindent
Here we need to undestand a state of this considering system only be represented by two independent quantum numbers which are $n$ energy eigen
states and $m$ quantum number which represents the qunatized momentum in $x$ direction values. Lets calculate the scattering amplitutde of the above mentioned scattering scenario using the equation derived in \eqref{4.18}.
\begin{equation} \label{4.21}
  \begin{aligned}
    a_{\alpha\beta}(t,t') & =
    -
    \frac{i}{\hbar}
    \int_{0}^t dt_1 \;
    \bra{\psi_{\beta,\mb{k'}}(t_1,t')}
    V(\mb{r}) \ket{\psi_{\alpha,\mb{k}}(t_1,t')} \\
    & =
    -
    \frac{i}{\hbar}
    \int_{0}^t dt_1 \;
    e^{\frac{i}{\hbar}\qty(\varepsilon_{\beta} - \varepsilon)t_1}
    \bra{\phi_{\beta,\mb{k'}}(t')}
    V(\mb{r}) \ket{\phi_{\alpha,\mb{k}}(t')}
  \end{aligned}
\end{equation}
Next assuimg this scenario for long time $t \rightarrow \infty$ we can turn this integral into a delta distrubution as follows
\begin{equation} \label{4.22}
  \begin{aligned}
    a_{\alpha\beta}(\mb{k'},\mb{k},t,t') & =
    -
    \frac{i}{\hbar}
    \lim_{t \rightarrow \infty}\qty[
      \int_{-t/2}^{t/2} dt_1 \;
      e^{\frac{i}{\hbar}\qty(\varepsilon_{\beta} - \varepsilon)t_1}
      \bra{\phi_{\beta,\mb{k'}}(t')}
      V(\mb{r}) \ket{\phi_{\alpha,\mb{k}}(t')}
    ] \\
    & =
    -2\pi i \delta(\varepsilon_{\beta} - \varepsilon)
    \bra{\phi_{\beta,\mb{k'}}(t')}
    V(\mb{r}) \ket{\phi_{\alpha,\mb{k}}(t')}
  \end{aligned}
\end{equation}

\noindent
Now lets consider about the inner product of the above derivation. Using completeness properties we can write that as follows
\begin{equation} \label{4.23}
  \begin{aligned}
    Q & \equiv
    \bra{\phi_{\beta,\mb{k'}}(t')}
    V(\mb{r}) \ket{\phi_{\alpha,\mb{k}}(t')} \\
    & =
    \sum_{\mb{k}}\sum_{\mb{k'}}
    \braket{\phi_{\beta,\mb{k'}}(t')}{\mb{k'}}
    \mel{\mb{k'}}{V(\mb{r})}{\mb{k}}
    \braket{\mb{k}}{\phi_{\alpha,\mb{k}}(t')}
  \end{aligned}
\end{equation}
and seperating $x$ and $y$ directional momentums we can modify this as follows (Assuming $L_y \rightarrow \infty$) and since we assumed that $L_y \rightarrow \infty$
\begin{equation} \label{4.24}
  \begin{aligned}
    Q & \equiv
    \bra{\phi_{\beta,\mb{k'}}(t')}
    V(\mb{r}) \ket{\phi_{\alpha,\mb{k}}(t')} \\
    & =
    \sum_{k_x}\sum_{{k'}_x}
    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} dk_y d{k'}_y \;
    \phi_{\beta}^{\dagger}(\mb{k'},t')
    \mel{\mb{k'}}{V(\mb{r})}{\mb{k}}
    \phi_{\alpha}(\mb{k},t').
  \end{aligned}
\end{equation}
For a random white scattering potential we can represent the inner product of scattering potential with momentum as a constant value as
\begin{equation} \label{4.25}
  V_{\mb{k'},\mb{k}} \equiv \mel{\mb{k'}}{V(\mb{r})}{\mb{k}}.
\end{equation}

\noindent
In this study, the perturbation potential is assumed to be formed by an ensemble of randomly distributed impurities, since randomimpurities in a disorded metal is a better approximation for experimental results.

\noindent
Consider $N_{imp}$ identical impurities positioned at the randomly distributed but fixed potions $\mb{r}_i$. The elastic scattering potential $V(\mb{r})$ is then given by the sum over uncorrelated single impurity potentials $\upsilon(\mb{r})$
\begin{equation} \label{4.26}
  V(\mb{r}) \equiv
  \sum_{i=1}^{N_{imp}}
  \upsilon (\mb{r}-\mb{r}_i).
\end{equation}
Now assume that the perturbation $V(\mb{r})$ is a Gaussian random potential where one can choose the zero of enerrgy such that the potential is zero on average. This model characterized by [*Ref: e.Akkermans G. Montambaux]
\begin{equation} \label{4.27}
  \expval{\upsilon(\mb{r})}_{imp} =0
\end{equation}
\begin{equation} \label{4.28}
  \expval{\upsilon(\mb{r})\upsilon(\mb{r'})}_{imp} = \Upsilon(\mb{r}-\mb{r'})
\end{equation}
where $\expval{\cdot}_{imp}$ denoted the average over realizations of the impurity disorder. In addition, this model assume that $\upsilon (\mb{r}-\mb{r'})$ only dependes on the position difference $|\mb{r}-\mb{r'}|$ and it decays with a characteristic leangth $r_c$. Since the study considers the case where the waveleagth of radiation or scattering electrons is much faster than $r_c$, it is good approximation to make two-point correlation function to be
\begin{equation} \label{4.29}
  \expval{\upsilon(\mb{r})\upsilon(\mb{r'})}_{imp} = \Upsilon_{imp}^2\delta(\mb{r}-\mb{r'})
\end{equation}
and a random potential $V(\mb{r})$ with this property is called white noise [*Ref: e.Akkermans G. Montambaux]. Then we can choose approximately total scattering potential as
\begin{equation} \label{4.30}
  V(\mb{r}) =
  \sum_{i=1}^{N_{imp}}
  \Upsilon_{imp} \delta(\mb{r}-\mb{r}_i).
\end{equation}

\noindent
Since $\braket{\mb{r}}{\mb{k}} = \frac{1}{\sqrt{L_xL_y}} \exp(-i[k_xx + k_yy]),$ we can calculate the Eq. \eqref{4.25} using this assumption as follows
\begin{equation} \label{4.31}
  \begin{aligned}
    V_{\mb{k'},\mb{k}}
    & = \mel{\mb{k'}}{V(\mb{r})}{\mb{k}} \\
    & = \mel**{\mb{k'}}{\sum_{i=1}^{N_{imp}}
    \Upsilon_{imp} \delta(\mb{r}-\mb{r}_i)}{\mb{k}} \\
    & = \mel**{\mb{k'}}{\sum_{i=1}^{N_{imp}}
    \Upsilon_{imp} \delta(x-x_i)\delta(y-y_i)}{\mb{k}} \\
    & =
    \sum_{i=1}^{N_{imp}}
    \int_{-\infty}^{\infty} dy\;
    \frac{1}{\sqrt{L_xL_y}} e^{ik'_y y} \delta(y-y_i) \frac{1}{\sqrt{L_xL_y}} e^{-i{k}_y y}
    \mel**{k'_x}{\Upsilon_{imp} \delta(x-x_i)}{k_x} \\
    &=
    \sum_{i=1}^{N_{imp}} \frac{1}{{L_xL_y}}
    e^{i(k'_y - k_y )y}
    \mel**{k'_x}{\Upsilon_{imp} \delta(x-x_i)}{k_x}
  \end{aligned}
\end{equation}
Assuming the total  umber of scatterers $N_{imp}$ is macroscopically large and for each impurity will produce same impurity potential($V_{k'_x,k_x}$) (\textcolor{red}{this is not sure and we need to prove this somehonw or we need to bring the $V_{\mb{k},\mb{k'}}$ without calculations})for every $x$-directional momentum pairs. Because $\upsilon (\mb{r})$ in momentum space is a constant value. Then we can achieve following expression
\begin{equation} \label{4.32}
  \begin{aligned}
    V_{\mb{k'},\mb{k}}
    & =
    V_{k'_x,k_x}
    \frac{N_{imp}}{L_y L_x} \int_{-\infty}^{\infty} dy_i\;
    e^{i\qty({k'}_y - k_y)y_i} \\
    & =
    \eta_{imp} V_{k'_x,k_x} \delta(k'_y - k_y)
  \end{aligned}
\end{equation}
where
\begin{equation} \label{4.33}
  V_{{k'}_x,k_x} \equiv
  \mel**{k'_x}{\Upsilon_{imp} \delta(x-x_i)}{k_x}
\end{equation}
is constant value for every $i$ impurity and $\eta_{imp}$ is number of impurities in a unit area. It is important to notice that $\ket{k_x} = e^{-ik_xx}$.\\

\noindent
Therefore, using the Eq. \eqref{3.36}, the Eq. \eqref{4.24} modified to (we can change varable $t' \rightarrow t$)
\begin{equation} \label{4.34}
  \begin{aligned}
    Q & =
    \sum_{k_x}\sum_{{k'}_x}
    {\eta_{imp} V_{{k'}_x,k_x}}
    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} dk_y d{k'}_y \;
    \delta({k'}_y - k_y)
    \\
    & \times
    \sqrt{L_x}
    \exp(
      -ib\sin(2\omega t)
    )
    \exp(
      i{k'}_y  \qty[d\sin(\omega t) + {y'}_0]
    )
    \tilde{\chi}_{n_{\beta}}\qty({k'}_y -g\cos(\omega t))
    \\
    & \times
    \sqrt{L_x}
    \exp(
      ib\sin(2\omega t)
    )
    \exp(
      -ik_y  \qty[d\sin(\omega t) + y_0]
    )
    \tilde{\chi}_{n_{\alpha}}\qty(k_y -g\cos(\omega t))
  \end{aligned}
\end{equation}
and we can simplify this as
\begin{equation} \label{4.35}
  \begin{aligned}
    Q =
    \sum_{k_x}\sum_{{k'}_x} &
    {\eta_{imp} L_x V_{{k'}_x,k_x}}
    \int_{-\infty}^{\infty} dk_y \;
    \\
    & \times
    \exp(
      i{k}_y {y'}_0
    )
    \tilde{\chi}_{n_{\beta}}\qty({k}_y -g\cos(\omega t))
    \exp(
      -ik_y y_0
    )
    \tilde{\chi}_{n_{\alpha}}\qty(k_y -g\cos(\omega t))
  \end{aligned}
\end{equation}
and this can re-write as
\begin{equation} \label{4.36}
  \begin{aligned}
    Q =
    \sum_{k_x}\sum_{{k'}_x}
    {\eta_{imp} L_x V_{{k'}_x,k_x}} I
  \end{aligned}
\end{equation}
where
\begin{equation} \label{4.37}
  \begin{aligned}
    I \equiv
    \int_{-\infty}^{\infty} dk_y \;
    \tilde{\chi}_{n_{\beta}}\qty(k_y -g\cos(\omega t))
    \tilde{\chi}_{n_{\alpha}}\qty(k_y -g\cos(\omega t))
    \exp(
      -ik_y  \qty[y_0 - {y'}_0  ]
    ).
  \end{aligned}
\end{equation}

\noindent
To avoid the energy tramision from external high-frequency field and electrons in the system, the applied radiation should be purely dressing field. Therefore, the only effect of the dressing field on 2DEG is the renormalization of the probability of elastic electron scattering within same Landau level $(n_{\alpha} = n_{\beta})$. Therefore Eq. \eqref{4.37} can be modified to
\begin{equation} \label{4.38}
  \begin{aligned}
    I \equiv
    \int_{-\infty}^{\infty} dk_y \;
    \tilde{\chi}_{n_{\beta}}^2 \qty(k_y -g\cos(\omega t))
    \exp(
      -ik_y  \qty[y_0 - {y'}_0  ]
    ).
  \end{aligned}
\end{equation}

\noindent
Lets consider about this integral and we can calculate it as using the following subtitution. Let
\begin{equation} \label{4.39}
  {k}_y -g\cos(\omega t) = \bar{k}_y \longrightarrow d{k}_y = d\bar{k}_y
\end{equation}
and this leads to
\begin{equation} \label{4.40}
    I \equiv
    2\pi \times \frac{1}{2\pi}
    \int_{-\infty}^{\infty} d\bar{k}_y \;
    \tilde{\chi}_{n_{\alpha}}^2 \qty(\bar{k}_y)
    \exp(
      -i\qty(\bar{k}_y + g\cos(\omega t)) \qty(y_0 - {y'}_0)
    ).
\end{equation}
Using Fourier transform of Gauss-Hermite functions and convolution theorem we can write this as
\begin{equation} \label{4.41}
    I \equiv
    {2\pi}
    \exp(g[{y'}_0 - y_0]\cos(\omega t))
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y).
\end{equation}
Therefore the scatterng amplitude \eqref{4.22} will modified to
\begin{equation} \label{4.42}
  \begin{aligned}
    a_{\alpha\beta}({k'}_x,k_x,t)  =
    -2\pi i
    \delta(\varepsilon_{\beta} - \varepsilon)
    \sum_{k_x}\sum_{{k'}_x}
    {\eta_{imp} L_x V_{{k'}_x,k_x}} I
  \end{aligned}
\end{equation}
Consideirng qunatized momentum given in $x$ direction derived in Eq. \eqref{1.53}, we can identify the non-zero values for scattering amplitude using following conditions
\begin{equation} \label{4.43}
    {k'}_x = \frac{p_{x_{\beta}}}{\hbar} = m' \frac{2\pi}{L_x}
    \quad \text{and} \quad
    k_x = \frac{p_{x_{\alpha}}}{\hbar} = m \frac{2\pi}{L_x}.
\end{equation}
Then we can simplfied scattering amplitude for given ${k'}_x$ and $k_x$ as (details of $k_x$ and $k'_x$ are included $\alpha$ and $\beta$ quantum numbers.)
\begin{equation} \label{4.44}
  \begin{aligned}
    a_{\alpha\beta}(t)  =
    -2\pi i
    \delta(\varepsilon_{\beta} - \varepsilon)
    \eta_{imp} L_x V_{{k'}_x,k_x}&
    \exp(g[{y'}_0 - y_0]\cos(\omega t)) \\
    & \times
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y)
  \end{aligned}
\end{equation}
Since this scattering amplitude is time-periodic we can write this as a Fourier series expansion
\begin{equation} \label{4.45}
    a_{\alpha\beta}(t) =
    \sum_{l=-\infty}^{\infty} a^l_{\alpha\beta} e^{-il\omega t}.
\end{equation}
In addition, using Jacobi-Anger expansion
\begin{equation} \label{4.46}
    e^{iz\cos(\theta)} = \sum_{l=-\infty}^{\infty} i^l J_l\qty(z) e^{-il\theta}
\end{equation}
we can re-write the Eq.\eqref{4.44} as folllows
\begin{equation} \label{4.47}
  \begin{aligned}
    a_{\alpha\beta}(t)  =
    -2\pi i
    \delta(\varepsilon_{\beta} - \varepsilon)
    \eta_{imp} L_x V_{{k'}_x,k_x}&
    \sum_{l=-\infty}^{\infty} i^l J_l\qty(g[{y'}_0 - y_0]) e^{-il\omega t}\\
    & \times
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y)
  \end{aligned}
\end{equation}
\begin{equation} \label{4.48}
  \begin{aligned}
    a_{\alpha\beta}(t)  =
    \sum_{l=-\infty}^{\infty}
    -2\pi i^{l+1} &
    \delta(\varepsilon_{\beta} - \varepsilon)
    \eta_{imp} L_x V_{{k'}_x,k_x}
    J_l\qty(g[{y'}_0 - y_0]) \\
    & \times
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y) e^{-il\omega t}
  \end{aligned}
\end{equation}
Then we can identified the Fourier series component as
\begin{equation} \label{4.49}
    a^l_{\alpha\beta} =
    -2\pi i^{l+1}
    \delta(\varepsilon_{\beta} - \varepsilon)
    \eta_{imp} L_x V_{{k'}_x,k_x}
    J_l\qty(g[{y'}_0 - y_0])
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y)
\end{equation}
Now one can introduce the definition of the \textit{transition probability matrix} as
\begin{equation} \label{4.50}
    \qty(A_{\alpha\beta})_{l,l'} \equiv
    a^l_{\alpha\beta}\qty[a^{l'}_{\alpha\beta}]^{*}
\end{equation}
and this becomes
\begin{equation} \label{4.51}
  \begin{aligned}
      \qty(A_{\alpha\beta})_{l,l'} & =
      \qty[{ 2 \pi \eta_{imp} L_x |V_{{k'}_x,k_x}|}]^2
      J_l\qty(g[{y'}_0 - y_0]) J_{l'}\qty(g[{y'}_0 - y_0])
      \delta^2(\varepsilon_{\beta} - \varepsilon) \\
      & \times
      \int_{-\infty}^{\infty} dy \;
      {\chi}_{n_{\beta}}\qty(y)
      {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y)
      \int_{-\infty}^{\infty} d\bar{y} \;
      {\chi}_{n_{\beta}}\qty(\bar{y})
      {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - \bar{y}).
  \end{aligned}
\end{equation}

\noindent
We can reduce these intragal into one variable and derive
\begin{equation} \label{4.52}
  \begin{aligned}
      \qty(A_{\alpha\beta})_{l,l'} =
      \qty[{ 2 \pi \eta_{imp} L_x |V_{{k'}_x,k_x}|}]^2 &
      J_l\qty(g[{y'}_0 - y_0]) J_{l'}\qty(g[{y'}_0 - y_0])
      \delta^2(\varepsilon_{\beta} - \varepsilon) \\
      & \times
      \qty|
      \int_{-\infty}^{\infty} dy \;
      {\chi}_{n_{\beta}}\qty(y)
      {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y)|^2.
  \end{aligned}
\end{equation}
Then desribing the square of the delta distribution using following procedure
\begin{equation} \label{4.53}
    \delta^2(\varepsilon) =
    \delta(\varepsilon)\delta(0) =
    \frac{\delta(\varepsilon)}{2\pi \hbar}
    \int_{-t/2}^{t/2} e^{i0\times t'/\hbar} dt'\; =
    \frac{\delta(\varepsilon)t}{2\pi \hbar}
\end{equation}
one can modify our derivation in Eq. \eqref{4.51} as
\begin{equation} \label{4.54}
  \begin{aligned}
      \qty(A_{\alpha\beta})_{l,l'} =
      \qty[{ 2 \pi \eta_{imp} L_x |V_{{k'}_x,k_x}|}]^2 &
      J_l\qty(g[{y'}_0 - y_0]) J_{l'}\qty(g[{y'}_0 - y_0])
      \delta(\varepsilon_{\beta} - \varepsilon)
      \frac{t}{2\pi \hbar}\\
      & \times
      \qty|
      \int_{-\infty}^{\infty} dy \;
      {\chi}_{n_{\beta}}\qty(y)
      {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y)|^2.
  \end{aligned}
\end{equation}
Then performing thetime derivation of each matrix element yeild the \textit{transition amplitude matrix} as folllows
\begin{equation} \label{4.55}
  \begin{aligned}
    \Gamma_{\alpha\beta}^{ll'} &  \equiv
    \frac{d \qty(A_{\alpha\beta})_{l,l'}}{dt} \\
    & =
    \Lambda |V_{{k'}_x,k_x}|^2
    \delta(\varepsilon_{\beta} - \varepsilon)
    J_l\qty(g[{y'}_0 - y_0]) J_{l'}\qty(g[{y'}_0 - y_0])
    \qty|
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(y_0 - {y'}_0 - y)|^2
  \end{aligned}
\end{equation}
where
\begin{equation} \label{4.56}
    \Lambda \equiv
    \frac { 2\pi \eta_{imp}^2 L_x^2}{ \hbar}
\end{equation}

\noindent
Now using defintion of $y_0$ given in Eq. \eqref{1.11} we can write that
\begin{equation} \label{4.57}
    y_0 - {y'}_0 =
    - \frac{p_{x_{\alpha}}}{eB} + \frac{p_{x_{\beta}}}{eB} =
    \frac{\hbar {k'}_x}{eB} - \frac{\hbar {k}_x}{eB} =
    \frac{\hbar}{eB} \qty[{k'}_x - {k}_x]
\end{equation}
and this leads Eq. \eqref{4.56} to
\begin{equation} \label{4.58}
  \begin{aligned}
    \Gamma_{\alpha\beta}^{ll'}({k'}_x,k_x) =
    \Lambda |V_{{k'}_x,k_x}|^2
    \delta(\varepsilon_{\beta} - \varepsilon) &
    J_l\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x])
    J_{l'}\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x]) \\
    & \times
    \qty|
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(\frac{\hbar}{eB} \qty[{k'}_x - {k}_x] - y)|^2
  \end{aligned}
\end{equation}

\noindent
An impurity average of white noise potential allows to identify $\expval{|V_{{k'}_x,k_x}|^2} = V_{imp}$ and the inverse scattering time matrix is the sum over all momentum over the transition probability matrix
\begin{equation} \label{4.59}
    \qty(\frac{1}{\tau(\varepsilon,k_x)})^{ll'}_{\alpha\beta} \equiv
    \frac{1}{L_x} \sum_{{k'}_x}
    \expval**{\Gamma_{\alpha\beta}^{ll'}({k'}_x,k_x)}_{imp}
\end{equation}
and this implies
\begin{equation} \label{4.60}
  \begin{aligned}
    \Gamma_{\alpha\beta}^{ll'}({k'}_x,k_x) =
    \frac{\Lambda V_{imp}}{L_x} \sum_{{k'}_x}
    \delta(\varepsilon_{\beta} - \varepsilon) &
    J_l\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x])
    J_{l'}\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x]) \\
    & \times
    \qty|
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(\frac{\hbar}{eB} \qty[{k'}_x - {k}_x] - y)|^2
  \end{aligned}
\end{equation}
For the 1-dimentional case introduce the momentum continuum limit as folllows
\begin{equation} \label{4.61}
    \frac{1}{L_x} \sum_{{k'}_x} \longrightarrow
    \frac{1}{2\pi}\int d {k'}_x
\end{equation}
and this leads to
\begin{equation} \label{4.62}
  \begin{aligned}
    \qty(\frac{1}{\tau(\varepsilon,k_x)})^{ll'}_{\alpha\beta} =
    \frac{\Lambda V_{imp}}{2\pi}
    \delta(\varepsilon_{\beta} - \varepsilon)
    \int_{-\infty}^{\infty} d {k'}_x
    &
    J_l\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x])
    J_{l'}\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x]) \\
    & \times
    \qty|
    \int_{-\infty}^{\infty} dy \;
    {\chi}_{n_{\beta}}\qty(y)
    {\chi}_{n_{\beta}}\qty(\frac{\hbar}{eB} \qty[{k'}_x - {k}_x] - y)|^2
  \end{aligned}
\end{equation}
Using following subtitution
\begin{equation} \label{4.63}
    y = \frac{\hbar \bar{k}}{eB} \longrightarrow
    dy = \frac{\hbar }{eB} d\bar{k}
\end{equation}
we can modify above derivation as
\begin{equation} \label{4.64}
  \begin{aligned}
    \qty(\frac{1}{\tau(\varepsilon,k_x)})^{ll'}_{\alpha\beta} =
    \frac{\Lambda V_{imp}}{2\pi}
    \delta(\varepsilon_{\beta} - \varepsilon)
    \int_{-\infty}^{\infty} d {k'}_x
    &
    J_l\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x])
    J_{l'}\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x]) \\
    & \times
    \qty(\frac{\hbar }{eB})^2
    \qty|
    \int_{-\infty}^{\infty} d\bar{k} \;
    {\chi}_{n_{\beta}}\qty(\frac{\hbar}{eB}\bar{k})
    {\chi}_{n_{\beta}}\qty(\frac{\hbar}{eB} \qty[{k'}_x - {k}_x - \bar{k}])|^2,
  \end{aligned}
\end{equation}
% Since squared Guess-Hermite functions are even function around zero we can re-write above derived expression as
% \begin{equation} \label{4.65}
%   \begin{aligned}
%     \qty(\frac{1}{\tau(\varepsilon,k_x)})^{ll'}_{\alpha\beta} =
%     \frac{\Lambda V_{imp}}{2\pi}
%     \delta(\varepsilon_{\beta} - \varepsilon)
%     \int_{-\infty}^{\infty} d {k'}_x
%     &
%     J_l\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x])
%     J_{l'}\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x]) \\
%     & \times
%     \frac{\hbar }{eB}
%     \int_{-\infty}^{\infty} d\bar{k} \;
%     \tilde{\chi}^2_{n_{\beta}}\qty(\frac{\hbar}{eB}\bar{k})
%     \tilde{\chi}^2_{n_{\beta}}\qty(\frac{\hbar}{eB}
%     \qty[\bar{k} - ({k}_x  - {k'}_x)])
%   \end{aligned}
% \end{equation}
and finally we can derive our expression for the \textit{inverse
scattering time matrix} for $N$th Landau level (let $n_{\alpha} = n_{\beta} = N$)
\begin{equation} \label{4.65}
  \begin{aligned}
    \qty(\frac{1}{\tau(\varepsilon,k_x)})^{ll'}_{N} =
    \frac { \eta_{imp}^2 L_x^2 \hbar V_{imp}}{\qty(eB)^2}
    \delta(\varepsilon - \varepsilon_{N})
    \int_{-\infty}^{\infty} d {k'}_x
    &
    J_l\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x])
    J_{l'}\qty(\frac{g\hbar}{eB}[{k}_x - {k'}_x]) \\
    & \times
    \qty|
    \int_{-\infty}^{\infty} d\bar{k} \;
    {\chi}_{N}\qty(\frac{\hbar}{eB}\bar{k})
    {\chi}_{N}\qty(\frac{\hbar}{eB} \qty[{k'}_x - {k}_x - \bar{k}])|^2.
  \end{aligned}
\end{equation}
\hfill$\blacksquare$


%% LyX 2.1.1 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[english]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{luainputenc}
\usepackage{amsmath}
\usepackage{esint}
\usepackage{babel}
\begin{document}

\title{Calculus 2}

\author{Professor Glaser}

\maketitle

\part{Review}

\section{Basic Antiderivatives}

Below are some antiderivatives you should be able to list at the drop
of a hat. These are fundamental to any serious problem solving.

\subsection{Exponents}

\[
\int x^{n}dx=\frac{x^{n+1}}{x+1}+C\, n\neq-1\qquad\int\frac{1}{x}dx=\ln|x|+C
\]

\[
\int e^{x}dx=e^{x}+C\qquad\int a^{x}dx=\frac{a^{x}}{\ln a}+C\,\left(a>0,a\neq1\right)
\]

\subsection{Trig Functions}

\[
\int\cos xdx=\sin x+C\qquad\int\sin xdx=-\cos x+C
\]

\[
\int\sec^{2}xdx=\tan x+C\qquad\int\csc^{2}xdx=-\cot x+C
\]

\[
\int\sec x\tan xdx=\sec x+C\qquad\int\csc x\cot xdx=-\csc x+C
\]

\[
\int\tan xdx=\ln|\sec x|+C\quad\left(=\ln|\cos x|+C\right)\qquad\int\cot xdx=\ln|\sin x|+C
\]

\[
\int\sec xdx=\ln|\sec x+\tan x|+C\qquad\int\csc xdx=-\ln|\csc x+\cot x|+C
\]

\subsection{Fractions}

\[
\int\frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}x+C
\]

\[
\int\frac{dx}{1+x^{2}}=\tan^{-1}x+C
\]

\[
\int\frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C
\]

\[
\int\frac{1}{2\sqrt{x}}dx=\sqrt{x}+C\qquad\int\frac{dx}{x^{2}}=-\frac{1}{x}+C
\]

\section{U-Substitution}

Again, you should know this stuff. Like, you should be able to do
all of this without thinking. These problems are designed to be simple.

\subsection{Basic U-Sub}

\[
\int2x\left(x^{2}+10\right)^{98}dx
\]
We assign $u$ to the pain in the ass, so $u=x^{2}+10$ and $du=2xdx$.
Rewrite the equation so it's blantantly obvious and solve:

\[
\int2xdx\left(x^{2}+10\right)^{98}=\int u^{98}du=\frac{u^{99}}{99}+C
\]
Then you substitute in the original like so:

\[
\boxed{\frac{\left(x^{2}+10\right)^{99}}{99}+C}
\]
Not so bad, right?

\subsection{A bit harder}

All right, what if the answer isn't glaringly obvious?

\[
\int2x\left(x^{2}+10\right)^{98}dx
\]

This is also really easy. $u=x^{2}+10$ and $du=2xdx$, so just divide
until $dx$ fits in nicely. Since $\frac{1}{2}du=xdx$, we can do
the exct same thing, but with a new constant.

\[
\int xdx\left(x^{2}+10\right)^{98}=\int\frac{1}{2}u^{98}du=\frac{1}{2}*\frac{u^{99}}{99}+C
\]

Substitute again:

\[
\boxed{\frac{\left(x^{2}+10\right)^{99}}{198}+C}
\]

Again, not hard at all.

\subsection{Bounded Integrals}

What is there are bounds on the integral? This isn't hard, just more
tedious.

\[
\int_{0}^{1}2x\left(x^{2}+2\right)^{5}dx
\]

Do the same thing with $u$:

\[
u=x^{2}+2\quad du=2xdx
\]

\[
\int_{?}^{?}u^{5}du
\]

Great. Now that the equation is in terms of $u$, we have to recalculate
the bounds. Since we already know the equation for $u$, shoudln't
be that bad. Plug in 0 and 1 in this case and solve it out:

\[
\int_{2}^{3}u^{5}du=\boxed{\frac{3^{6}}{6}-\frac{2^{6}}{6}}
\]

Boom. Done.

\subsection{Fractions}

\[
\int\frac{2xdx}{x^{2}+34}
\]

If you set $u=x^{2}+34$ and $du=2xdx$, this problem is actually
stupidly simple. But you have to remember some log rules.

\[
\int\frac{du}{u}=\ln|u|+C=\boxed{\ln|x^{2}+34|+C}
\]

...Let's keep going.

\subsection{$e$ and Friends}

\[
\int xe^{x^{2}}dx
\]

Remember, $u$ handles the pain in the ass part. $u=x^{2}$and $\frac{1}{2}du=xdx$.

\[
\frac{1}{2}\int e^{u}du=\frac{1}{2}e^{u}+C=\boxed{\frac{1}{2}e^{x^{2}}+C}
\]

\subsection{Trig}

I'm too lazy to write commentary for this:

\[
\int\frac{\cos x}{\sqrt{\sin x}}
\]

\[
u=\sin x\quad u=\cos xdx
\]

\[
\int\frac{du}{\sqrt{u}}=\int u^{-\frac{1}{2}}du=2u^{\frac{1}{2}}=\boxed{2\sqrt{\sin x}+C}
\]

\section{Final Words}

I swear, Alex, if you don't remember any part of this after learning
it 4 times in your life, I want you to call Miguel and Carrie and
Alexandra so they can yell at you. A lot.
\end{document}


%
% STAT 100: Chance and Data Analysis - A Course Overview
% Section: Analysis of Multi-Variable Data
%
% Author: Jeffrey Leung
%

\section{Analysis of Multi-Variable Data}
	\label{sec:analysis-of-multi-variable-data}
\subsection{Relationships between Two Quantitative Variables}
	\label{subsec:analysis-of-multi-variable-data:relationships-between-two-quantitative-variables}
\begin{easylist}

	& Relationships between two quantitative variables:
	& \emph{Dependent/respondent variable:} Variable to be predicted
		&& Placed on the vertical (y) axis
	& \emph{Independent variable/predictor:} Variable used to predict the dependent variable
		&& Placed on the horizontal (x) axis
	& \emph{Scatterplot:} Visual comparisons of the values of two quantitative variables (see figure %TODO in bullet point?

	& \emph{positive/negative:}

	& Forms and strength of a relationship:
		&& \emph{Linear:} Approximate change in values can be summarized in a single 2-d direction
			&&& I.e. Points which follow a straight line
		&& \emph{Non-linear:} Values change in multiple 2-d directions
			&&& E.g. Exponential functions, trigonometric functions

		&& \emph{Strength:} Denseness of distribution which shows a clear form/relation
			&&

	& \emph{Correlation coefficient:} Value which denotes the strength of the data
			&&& Denoted by $r$
			&&& Formula: $r = $
			&&& Value:
				&&&& $-1 \leq r \leq 1$
				&&&& Closer to -1, the data is more negatively related
				&&&& Closer to 0, the data is more weakly related
				&&&& Closer to 1, the data is more positively related
			&&& Interpretation: The closer to 1 $r$ is, the stronger the correlation
			&&& Only applies to linear data
				&&&& 0 for non-linear data
			&&& Outliers may need to be removed before calculation
			&&& E.g. The correlation coefficient of a midterm grade and a final grade was found to be 0.8747. The relationship between the two grades is positive, strong, and linear.

		&& For an interpretation of the magnitude, see table~\ref{tab:strength-of-a-correlation-coefficient}

		\Deactivate
		\begin{table}[!htb]
			\centering
			\caption{Strength of a Correlation Coefficient}
			\label{tab:strength-of-a-correlation-coefficient}
			\begin{tabular}{ c l }
				Magnitude of the Correlation Coefficient & Strength \\
				\hline
				$0 < |r| < 0.4$ & Weak \\
				$0.4 \leq |r| < 0.6$ & Moderately weak \\
				$0.6 \leq |r| < 0.8$ & Moderately strong \\
				$0.8 \leq |r| < 1.0$ & Strong \\
				$|r| = 1$ & Perfect
			\end{tabular}
		\end{table}
		\Activate

		%TODO incomplete

	& Correlation:
		&& Declares an association between two variables
		&& Does not imply causation - i.e. an existing linear relationship does not mean a change in the independent variable causes a change in the dependent variable (see figure~\ref{fig:xkcd-552})

	\begin{figure}[!htb]
		\centering
		\caption{XKCD Webcomic - 552: Correlation}
		\label{fig:xkcd-552}
		\href{https://xkcd.com/552/}{\includegraphics[scale=.75]{xkcd-552-correlation}}
	\end{figure}

	& \emph{Lurking variable:} Variable outside of the tested variables which explains an association between the two tested variables

	& \emph{Regression analysis:}
		&& \emph{Regression line:} Best-fit line
			&&& Before using the regression line to predict data, check data ranges
		&& %TODO
		&& \emph{Regression equation:} Mathematical equation of a regression line
			&&& May have multiple $x$ variables
			&&& Formula: $y = slope_1 \cdot x_1 + slope_2 \cdot x_2 + \ldots + slope_n \cdot x_n + y_{intercept}$
				&&&& Where $y =$ dependent variable, and $x_1, x_2, \ldots, x_n =$ independent variables
				&&&& Use a scientific calculator to determine the intercept and slope %TODO how
			&&& Cannot be used to predict data less than the minimum or greater than the maximum because there is no data to support the analysis
			&&& Slope of a regression equation: $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
				&&&& Per 1 unit change in $x$, $y$ should change by the value of the slope
				&&&& Example of interp: For every 1 additional $x$, we predict $y$ will change by $slope$.
		&& \emph{Coefficient of determination:}
			&& Denoted by $R^{2}$
			&& Value:
				&&& $R^{2} = r^2$ where $r =$ the correltion coefficient
				&& $0 \leq R^{2} \leq 1$
			&& Interpretation: $R^{2}$ represents the variability in percent which can be explained by the regression line.
			%TODO notes

	& Multiple regression:
		&& See video
		&& Use excel

\end{easylist}
\clearpage


% !TEX root = ./Basilisk-MRPROTATION-20180522.tex

\begin{figure}[htb]
	\centerline{
	\includegraphics[]{Figures/moduleIO}
	}
	\caption{Illustration of the mrpRotation() input and output messages.  Required messages are shown with a solid line, while optional message have a dashed line.}
	\label{fig:moduleIO}
\end{figure}

\section{Model Description}
The purpose of this {\tt mrpRotation} module is to add a constant rotation relative to the input frame $\mathcal{R}_{0}$.  The output reference frame is called $\mathcal{R}$.  The initial orientation is specified through an MRP\cite{schaub} set $\bm\sigma_{R/R0}$, while the  $\mathcal{R}$-frame angular velocity vector $\leftexp{R}{\bm\omega}_{R/R0}$ is held constant in this module.

Assume that the input reference frame $\mathcal{R}_{0}$ is given through an attitude state input message containing $\bm\sigma_{R_{0}/N}$, $\leftexp{N}{\bm\omega}_{R_{0}/N}$ and $\leftexp{N}{\dot{\bm\omega}}_{R_{0}/N}$ as illustrated in Figure~\ref{fig:moduleIO}.  The MRP set is mapped into the corresponding Direction Cosine Matrix or DCM\cite{schaub} using
\begin{equation}
	[R_{0}N] = [R_{0}N ( \bm\sigma_{R_{0}/N})]
\end{equation}

The goal of the motion is to compute the attitude of $\mathcal{R}$ relative to input frame $\mathcal{R}_{0}$ such that
\begin{align}
	\label{eq:mRot1}
	\dot{\bm\sigma}_{R/R_{0}} &= \frac{1}{4} [B(\bm\sigma_{R/R_{0}})] \leftexp{R}{\bm\omega}_{R/R_{0}}
	\\
	\label{eq:mRot2}
	\frac{\leftexp{R}{\D} {\bm\omega}_{R/R_{0}}}{\D t} &= \bm 0
\end{align}
Assume the initial $\bm\sigma_{R/R_{0}}(t_{0})$ set and the $\mathcal{R}$-frame relative invariant $\leftexp{R}{\bm\omega}_{R/R_{0}}$ vector are provided to the module.  The current $\bm\sigma_{R/R_{0}}(t_{0})$ value is then obtained by Eq.~\eqref{eq:mRot1}.  The current DCM of the $\mathcal{R}$-frame is thus
\begin{equation}
	\label{eq:mRot3}
	[RN] = [RR_{0}(\bm\sigma_{R/R_{0}}(t) ] [R_{0}N]
\end{equation}

Next, the angular velocity vector is transformed to inertial frame $\mathcal{N}$-frame components using
\begin{equation}
	\label{eq:mRot4}
	\leftexp{N}{\bm\omega}_{R/R_{0}} = [RN]^{T}\ \leftexp{R}{\bm\omega}_{R/R_{0}}
\end{equation}
to find the inertial angular velocity of the output reference frame:
\begin{equation}
	\label{eq:mRot5}
	\leftexp{N}{\bm\omega}_{R/N} = \leftexp{N}{\bm\omega}_{R/R_{0}} + \leftexp{N}{\bm\omega}_{R_{0}/N}
\end{equation}

Finally, the inertial angular acceleration of the output reference frame is found using the transport theorem:
\begin{equation}
	\label{eq:mRot6}
	\dot{\bm\omega}_{R/N} =
	\frac{\leftexp{R}{\D} {\bm\omega}_{R/R_{0}}}{\D t}
	 + \bm\omega_{R/N} \times {\bm\omega}_{R/R_{0}}
	+ \dot{\bm\omega}_{R_{0}/N}
	= \bm\omega_{R_{0}/N} \times {\bm\omega}_{R/R_{0}}
	+ \dot{\bm\omega}_{R_{0}/N}
\end{equation}
where $\bm\omega_{R/N} \times {\bm\omega}_{R/R_{0}} = (\bm\omega_{R/R_{0}} +\bm\omega_{R_{0}/N}) \times {\bm\omega}_{R/R_{0}} = \bm\omega_{R_{0}/N} \times {\bm\omega}_{R/R_{0}}$ is used.  Expressed in $\mathcal{N}$ frame components, this vector equation is numerically evaluated using:
\begin{equation}
	\label{eq:mRot7}
	\leftexp{N}{\dot{\bm\omega}}_{R/N} = \leftexp{N}{\bm\omega}_{R_{0}/N} \times \leftexp{N}{\bm\omega}_{R/R_{0}}
	+ \leftexp{N}{\dot{\bm\omega}}_{R_{0}/N}
\end{equation}


\section{Bandits}
The \emph{multi-armed bandits problem} is a simplified setting of RL where actions do not affect the world state. In other words, the current state does not depend on previous actions, and the reward is immediate. Like any RL problem with unknown MDP, a successful agent should solve the \emph{exploration-exploitation dilemma}, i.e. find a balance between exploiting what it has already learned to improve the reward and exploring in order to find the best actions.

\paragraph{Setting}
Bandits problems can be stationary or non-stationary, and the setting can be stochastic or adversarial. Agents typically learn in an online setting and, in the case of a non-associative task (no need to associate different actions with different situations), they try to find a single best action out of a finite number of actions (also called “arms”). For associative tasks, \emph{contextual bandits} make use of additional information which can be global or individual context (i.e. per arm). Context can be fixed (e.g. $x\in \mathbb{R}^d$) or variable (e.g. $x_t\in \mathbb{R}^d, \forall t$).

The objective is to find a policy $\pi$ that maximizes the cumulated reward:
\[
    \pi^* = \argmax_\pi \sum_{t=0}^T r_{\pi_t, t}
\]
Where $\pi_t \in \{1, \dots, K\}$ is the arm selected by $\pi$ at time $t$, and $r_{i,t}$ is the reward obtained at time $t$ after selecting the arm $i \in \{1, \dots, K\}$. At any point $t$, the expected reward of an arm $i \in \{1, \dots, K\}$ is:
\[
    \hat{\mu}_{i,t} = \E[r_{i,t}] = \frac{1}{T_i} \sum_{\substack{s=0 \\ \pi_s=i}}^t r_{i,s}
\]
Where $T_i = \sum_{s=0}^t \mathbbm{1}_{\pi_s=i}$ is the number of times arm $i$ was played up to time $t$.

\paragraph{$\epsilon$-greedy}
The greedy strategy simply consists in selecting the arm that has given the best rewards and therefore has the best expected reward:
\[
    \forall t, \pi_t^\text{greedy} = \argmax_i \hat{mu}_{i,t}
\]
The $\epsilon$-greedy sometimes acts greedily, and sometimes selects a random action to improve exploration:
\[
    \forall t, \pi_t^\text{$\epsilon$-greedy} = \begin{cases}
        \argmax_i \hat{mu}_{i,t} & \text{with probability $1-\epsilon$} \\
        i \sim \mathcal{U}\{1,K\} & \text{with probability $\epsilon$}
    \end{cases}
\]

\paragraph{Upper Confidence Bounds (UCB)}
UCB \cite{auer2002using} follows an optimistic strategy and selects the best arm in the best case scenario, i.e. according to the upper bounds $B_t(i)$ on the arms value estimates:
$$
\pi_t = \argmax_i B_t(i)
\quad \text{with} \quad
B_t(i) = \hat{\mu}_{i,t} + \sqrt{\frac{2 \log t}{T_i}}
$$
LinUCB \cite{li2010contextual} follows the UCB strategy but considers a linear and individual context $x_{i,t}$. We have $\mathbb{E}[r_{i,t} \mid x_{i,t}] = \theta_i^T x_{i,t}$ and parameters $\theta_i$ are estimated with Ridge Regression on previously observed contexts and rewards.

\paragraph{Thompson Sampling}
Thompson Sampling \cite{kaufmann2012thompson} follows a Bayesian approach and considers a parametric model $P(\mathcal{D} \mid \theta)$ with a prior $P(\theta)$. For instance, in the linear case \cite{agrawal2013thompson}: $P(r_{i,t} \mid \theta) = \mathcal{N}(\theta^T x_{i,t}, v^2)$ and $P(\theta) = \mathcal{N}(0, \sigma^2)$. Then, at each iteration $t$, we sample $\theta$ from $P(\theta \mid \mathcal{D}) \propto P(\mathcal{D}\mid\theta) P(\theta)$ and select the arm $\pi_t = \argmax_i \mathbb{E}[r_{i,t} \mid x_{i,t}, \theta]$.


\documentclass[10pt,a4paper]{article}
\title{Bubble Sort}
\date{}

\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
% For algorithms
\usepackage[linesnumbered,ruled,lined,boxed,commentsnumbered]{algorithm2e}

\usepackage{titling}

\setlength{\droptitle}{-10em}

\begin{document}
\maketitle

\begin{algorithm}[H]
 \KwData{An array A[0 ... n-1] of size n}
 \KwResult{Sorted array A in non decreasing order i.e $A[0]<=A[1]<=...<=A[n-1]$ }
  \BlankLine
  \Begin {
  \For{$i \leftarrow 0$ to $n-2$} {
	\tcc{Shift the maximum element of sub-array A[0..n-i-1] to A[n-i-1] by adjacent pairwise swapping}
  	\For{$j \leftarrow 0$ to $n-i-2$} {
  		\If{$A[j] > A[j+1]$}{
			\BlankLine
			\tcc{Swap A[j] and A[j+1]}
			$tmp \leftarrow A[j]$\;
			$A[j] \leftarrow A[j+1]$\;
			$A[j+1] \leftarrow tmp$\;
  		}
  		$j \leftarrow j+1$\;
  	}
  	\BlankLine
  	$i \leftarrow i+1$\;
 }
 }
 \caption{Bubble Sort}
\end{algorithm}

\section{Proof of Correctness}
\subsection{Invariant}
At the beginning of each outer for loop, the sub array $A[n-i \text{ .. } n-1]$ consists of the first $(n-1)-(n-i)+1 = i$ largest elements of the entire array A, in sorted order i.e $A[n-i] <= A[n-i+1] <= \text{ .. } <= A[n-1]$.

\subsection{Initialization}
Initially, $i = 0$ and hence the sub array $A[n \text{ .. } n-1]$ is an empty list and consists of 0
elements. As the list is empty, the invariant trivially holds.

\subsection{Maintenance}
The inner for loop runs from $j \leftarrow 0$ through $n-i-2$. By performing adjacent comparisons and swaps between $A[j]$ and $A[j+1]$, we ensure that the the maximum element of sub array $A[0 \text{ .. } n-i-1]$ is shifted and placed at $A[n-i-1]$ giving us the first $(n-1) - (n-i-1) + 1 = i+1$ largest elements in $A[n-i-1 \text{ .. } n-1]$, that too in sorted order. Incrementing $i$ to $i+1$ then makes the invariant hold at the start of the next iteration.

\subsection{Termination}
The procedure terminates when $ i > n-2$. As $i$ is always incremented by 1, when the
outer while loop terminates $i$ will always be equal to $n-1$. According to the invariant,
when $i = n-1$, the sub array $A[1 \text{ .. } n-1]$ must contain the first $(n-1)$ largest elements
of A in sorted order. By elimination, $A[0]$ must now contain the $n^{th}$ largest element (the smallest element).
In other words, the array $A$ is now sorted. $\blacksquare$

\end{document}


\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb,amsfonts}
\usepackage{enumitem}
\usepackage{tabu}
\usepackage{xcolor}
\usepackage{mathtools}
\usepackage{tcolorbox}
\usepackage{changepage}
\usepackage{kpfonts}
\usepackage{picture}
\usepackage{venndiagram}
\usepackage{graphicx}

\newcommand{\prob}[1]{\mathbb{P}(#1)}
\newcommand{\condprob}[2]{\mathbb{P}(#1 \text{ } \lvert \text{ } #2)}
\newcommand{\expected}[1]{\text{E}(#1)}
\newcommand{\variance}[1]{\text{V}(#1)}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\field}{\mathcal{F}}

\begin{document}
% ===================================================================================
% ===================================================================================

\section*{CHAPTER 8: Law of Large Numbers}
% ================ SECTION 8.1 ================
\subsection*{Law of Large Numbers for Discrete Random Variables}
\noindent
We have also defined probability mathematically as a value of a distribution function for the random variable representing the experiment. The Law of Large Numbers, which is a theorem proved about the mathematical model of probability, shows that this model is consistent with the frequency interpretation of  probability.

\subsubsection*{ChebyShev Inequality}
\noindent
\textbf{Theorem 8.1 (ChebyShev Inequality)} Let $X$ be a discrete random variable with expected value $\mu \expected{X}$, and let $\epsilon > 0$ be any positive real number. Then,

\begin{equation*}
\prob{\lvert X - \mu \rvert \geq \epsilon} \leq \frac{\variance{X}}{\epsilon^2}.
\end{equation*}

\begin{tcolorbox}
\noindent
Let X by any random variable with $\expected{X} = \mu$ and $\variance{X} = \sigma^2$. Then, if $\epsilon = k \sigma$, Chebyshev's Inequality states that

\begin{equation*}
\prob{\lvert X - \mu \rvert \geq k \sigma} \leq \frac{\sigma^2}{k^2 \sigma^2} = \frac{1}{k^2}.
\end{equation*}

\noindent
Thus, for any random variable, the probability of a deviation from the mean of
more than k standard deviations is $\leq \frac{1}{k^2}$. If, for example, $k = 5$, $\frac{1}{k^2} = .04$.

\vspace*{.5cm}
\noindent
Chebyshev’s Inequality is the best possible inequality in the sense that, for any
$\epsilon > 0$, it is possible to give an example of a random variable for which Chebyshev’s Inequality is in fact an equality. To see this, given $\epsilon > 0$, choose X with distribution

\begin{equation*}
p_X = {- \epsilon \qquad  + \epsilon \choose 1/2 \qquad 1/2}
\end{equation*}

Then $\expected{X} = 0$, $\variance{X} = \epsilon^2$, and

\begin{equation*}
\prob{\lvert X - \mu \rvert \geq \epsilon} \leq \frac{\variance{X}}{\epsilon^2} = 1.
\end{equation*}

We are now prepared to state and prove the Law of Large Numbers.
\end{tcolorbox}

% https://www.youtube.com/watch?v=uMgK000XFhA
\noindent
Alternatively, Chebyshev's theorem says that the proportion of any distribution that lies within $k$ standard deviations of the mean is at least: $1-\frac{1}{k^2}$, where $k$ is any positive number greater than $1$. This theorem applies to \textbf{all distributions}.

\begin{tcolorbox}
For example, Chebyshev's theorem says that within $2$ standard deviations of the mean, you will find at least

\begin{equation*}
1 - \frac{1}{k^2} = 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4},
\end{equation*}

\noindent
or at least $75\%$ of data will lie within $2$ standard deviations from the mean.
\end{tcolorbox}

\vspace*{.5cm}
\begin{tcolorbox}
Let $X_1, X_2, \ldots, X_n$ be a Bernoulli trials process with probability $.3$ for success and $.7$ for failure. Then, $\expected{X_j} = .3$ and $\variance{X_j} = .3 \cdot .7 = .21$. If

\begin{equation*}
A_n = \frac{S_n}{n} = \frac{X_1, X_2, \ldots, X_n}{n}
\end{equation*}

\noindent
is the average of the $X_j$, then $\expected{A_n} = .3$ and $\variance{A_n} = \frac{\variance{S_n}}{n^2} = \frac{.21}{n}$.

\vspace*{.3cm}
\noindent
Let $\epsilon = .1$. Then, Chebyshev's Inequality states that

\begin{equation*}
\prob{\lvert A_n - .3 \rvert \geq .1 } \leq \frac{.21}{n(.1)^2} = \frac{21}{n}.
\end{equation*}

\noindent
If $n = 100$, this can be rewritten as

\begin{equation*}
\prob{.2 <  A_{100} < .4 } \geq \mathbb{P}_{A_{100}}
\end{equation*}

\noindent
where $\mathbb{P}_{A_{100}}$ is the probability that the mean of $A_n$ lies between $.2$ and $.4$, which is $.1$ away from $\expected{A_n} = .3$. By Chebyshev, we can calculate this by finding how many standard deviations $.1$ is. The standard deviation, $\sigma = \sqrt{\frac{.21}{100}} = .0458$. Thus, $k = \frac{.1}{.0458} = 2.182$. Thus,

\begin{equation*}
\mathbb{P}_{A_{100}} = 1 - \frac{1}{2.182^2} = .79 = 79\%.
\end{equation*}
\end{tcolorbox}

\subsubsection*{Law of Large Numbers}
\noindent
\textbf{Theorem 8.2: Law of Large Numbers} Let $X_1, X_2, \ldots, X_n$ be an independent trial process, with finite expected value $\mu = \expected{X_j}$, and finite variance $\sigma = \variance{X_j}$. Then for any $\epsilon > 0$,

\begin{equation*}
\mathbb{P} \Bigg ( \Bigg \lvert \frac{S_n}{n} - \mu \Bigg \rvert \geq \epsilon \Bigg ) \rightarrow 0
\end{equation*}

\noindent
as $n \rightarrow \infty$. Equivalently,

\begin{equation*}
\mathbb{P} \Bigg ( \Bigg \lvert \frac{S_n}{n} - \mu \Bigg \rvert < \epsilon \Bigg ) \rightarrow 1
\end{equation*}

\noindent
as $n \rightarrow \infty$.

\begin{center}
\includegraphics[width=13cm, height=10cm]{LLN_proof_pic}
\end{center}

% https://www.youtube.com/watch?v=MntX3zWNWec
\noindent
The \textbf{Law of Large Numbers} states that as the number of observations increases, the actual or observed probability approaches the theoretical or expected probabilities.

\vspace*{.3cm}
\noindent
\textbf{Weak Law of Large Numbers} vs. \textbf{Strong Law of Large Numbers} \\
\noindent
The \textbf{WLLN} says that for a large sample, there is a very high probability that the mean of the sample will be very close to the expected value, due to \textit{weak convergence}. However, the probability of convergence is still very high.

\vspace*{.3cm}
\noindent
The \textbf{SLLN} says that for a large sample, the mean of the sample will almost surely be the expected value, due to \textit{strong convergence}. The proof of the \textbf{SLLN} is much more complex and much harder to prove.

\begin{center}
\includegraphics[width=13cm, height=6cm]{DIE_ROLLING_EXAMPLE}
\end{center}

\noindent
There are exercises for \textbf{LLN} on page $312$.

% ================ SECTION 8.2 ================

\subsection*{Law of Large Numbers for Discrete Random Variables}
This law has a natural analogue for continuous probability distributions, which we consider somewhat more briefly here.

\subsubsection*{ChebyShev Inequality}
\noindent
\textbf{Theorem 8.3 (ChebyShev Inequality)} Let $X$ be a continuous random variable with density function $f(x)$. Suppose $X$ has a finite expected value $\mu \expected{X}$ and finite variance $\variance{X} = \sigma^2$. Then, for any positive number $\epsilon > 0$ we have

\begin{equation*}
\prob{\lvert X - \mu \rvert \geq \epsilon} \leq \frac{\variance{X}}{\epsilon^2}.
\end{equation*}

\noindent
Note that this theorem says nothing if $\variance{X} = \sigma^2$ is infinite.

\subsubsection*{Law of Large Numbers}
\noindent
\textbf{Theorem 8.4: Law of Large Numbers} Let $X_1, X_2, \ldots, X_n$ be an independent trial process with a continuous density function $f$, finite expected value $\mu$, and finite variance $\sigma^2$. Let $S_n = X_1, X_2, \ldots, X_n$ be the sum of the $X_i$. Then for any real number $\epsilon > 0$ we have

\begin{equation*}
\lim_{n \rightarrow \infty} \mathbb{P} \Bigg ( \Bigg \lvert \frac{S_n}{n} - \mu \Bigg \rvert \geq \epsilon \Bigg ) = 0,
\end{equation*}

\noindent
or equivalently,

\begin{equation*}
\lim_{n \rightarrow \infty} \mathbb{P} \Bigg ( \Bigg \lvert \frac{S_n}{n} - \mu \Bigg \rvert < \epsilon \Bigg ) = 1.
\end{equation*}

\noindent
There are exercises for \textbf{LLN} on page $320$.

% ========================================================
% ========================================================

\section*{CHAPTER 9: Central Limit Theorem}
\subsubsection*{Central Limit Theorem for Bernoulli Trials}
\noindent
The second fundamental theorem of probability is the \textit{Central Limit Theorem}. This theorem says that if Sn is the sum of n mutually independent random variables, then the distribution function of Sn is well-approximated by a certain type of continuous function known as a normal density function, which is given by the formula

\begin{equation*}
f_{\mu, \sigma} = \frac{1}{\sqrt{2 \pi} \sigma} e^{-(x - \mu)^2 / (2 \sigma^2)}.
\end{equation*}

In this section, we will deal only with the case that $\mu = 0$ and $\sigma = 1$. We will call this particular normal density function the standard normal density, and we will denote it by $\phi (x)$:

\begin{equation*}
\phi (x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2 / 2}.
\end{equation*}

\noindent
\textbf{Definition 9.1} The \textit{standardized sum} of $S_n$ is given by

\begin{equation*}
S_n^* = \frac{S_n - np}{\sqrt{npq}},
\end{equation*}

\noindent
where $S_n^*$ has expected value $0$ and variance $1$.

\vspace*{.3cm}
\noindent
\textbf{Approximating Binomial Distributions}
\noindent
We can approximate a binomial distribution function as follows

\begin{align*}
b(n,p,j) &\sim \frac{\phi(x)}{\sqrt{npq}} \\
&= \frac{1}{\sqrt{npq}} \phi \Bigg ( \frac{j - np}{\sqrt{npq}} \Bigg ).
\end{align*}

\noindent
An example of a situation where this can be used is when there are $100$ flips of a fair coin and we want to know the probability of seeing exactly $55$ heads.

%\begin{tcolorbox}
%\end{tcolorbox}
\end{document}


% declare document class and geometry
\documentclass[12pt]{article} % use larger type; default would be 10pt
\usepackage[margin=1in]{geometry} % handle page geometry

\input{../header2.tex}

\title{Phys 220A -- Classical Mechanics -- Lec08}
\author{UCLA, Fall 2014}
\date{\formatdate{28}{10}{2014}} % Activate to display a given date or no date (if empty),
         % otherwise the current date is printed

\begin{document}
\setlength{\unitlength}{1mm}
\maketitle

\section{Quick note on higher dimensional Legendre Transform}

Here we will have positions $\v{x} \in \R^n$ and a convex function $f : \R^n \rightarrow R$, convex here meaning the Hessian $H_{ij} = \partial_i \partial_j f$ is positive definite, i.e. has positive determinant. Then we will define the function
\begin{eqn}
F(\v p, \v x) = \v{p} \cdot \v{x} - f(\v x)
\end{eqn}
and define the Legendre transform as
\begin{eqn}
\mathcal{L}f (\v p) = \sup_{\v{x} \in \R^n} F(\v p, \v x).
\end{eqn}
Again we can write this as
\begin{eqn}
\mathcal{L}f(\v p) = F(\v p, \v{x}(\v p)),
\end{eqn}
where $p_i = \pd{f}{x_i}$ determines $\v{x}(\v p)$.

\begin{example}
Consider the function
\begin{eqn}
f(\v x) = \frac{1}{2} \v{x}^\top M \v{x}
\end{eqn}
where $M$ is a symmetric matrix. Then $\v{x}(\v p)$ is determined by
\begin{eqn}
\v p = M \v{x} \quad \implies \quad \v{x}(\v p) = M^{-1} \v{p}.
\end{eqn}
Then we find the Legendre transform
\begin{align}
\mathcal{L}f (\v p) &= F(\v p, \v{x}(\v p)) \\
	&= \v{p} \cdot \v{x} - \frac{1}{2} \v{p}^\top (M^\top)^{-1} M M^{-1} \v{p} \\
	&= \frac{1}{2} \v{p}^\top M^{-1} \v{p}.
\end{align}
\end{example}

\section{Hamiltonian formulation}

Euler-Lagrange formulation gives us $n$ 2nd order differential equations, which can be a bit of a mess. In the Hamiltonian formulation, they are replaced by $2n$ 1st order differential equations,
\begin{eqn}
\dot{p}_i = -\pd{H}{q_i}, \qquad \dot{q}_i = \pd{H}{p_i},
\end{eqn}
where
\begin{eqn}
H(\v q, \v p, t) = \sum_i p_i \dot{q}_i - L(\v q, \vd q, t)
\end{eqn}
is the Legendre transform of $L$ with respect to $\dot{q}_i$, i.e. $\dot{q}_i(p)$ is determined by $p_i = \pd{L}{\dot{q}_i}$.

\begin{proof}
Looking at the differential of $H$, we have
\begin{align}
\dif H &= \sum_i \pd{H}{p_i} \dif{p_i} + \sum_i \pd{H}{q_i} \dif{q_i} + \pd{H}{t} \dif{t} \\
	&= \dot{q}_i \dif{p_i} + \cancel{p_i \dif{\dot{q}_i}} - \pd{L}{q_i} \dif{q_i} - \cancel{\pd{L}{q_i} \dif{q_i}} - \pd{L}{t} \dif{t},
\end{align}
so equating the differentials, we see that
\begin{eqn}
\dot{p}_i = -\pd{H}{q_i}, \qquad \dot{q}_i = \pd{H}{p_i}, \qquad \pd{H}{t} = -\pd{L}{t}.
\end{eqn}
The last term of course is only important when there is explicit time dependence, but it is trivially true by definition.
\end{proof}

\begin{example}[1-dim. particle in potential]
Consider the Lagrangian
\begin{eqn}
L = \frac{1}{2} m \dot{q}^2 - V(q),
\end{eqn}
then we find $p = m \dot{q}$, so that
\begin{align}
H = p \dot{q} - L &= \frac{p^2}{m} - \frac{1}{2m} p^2 + V(q) \\
	&= \frac{p^2}{2m} + V(q).
\end{align}
Thus $H$ is just the energy.
\end{example}

\begin{example}[System of particles]
Consider the Lagrangian
\begin{eqn}
L = \frac{1}{2} \sum_{ij} M_{ij}(q) \dot{q}^i \dot{q}^j - V(q),
\end{eqn}
then we have momentum $p_i = \sum_j M_{ij} \dot{q}^j$ and
\begin{eqn}
H(p,q) = \frac{1}{2} \sum_{ij} M_{ij}^{-1} p^i p^j + V(q).
\end{eqn}
\end{example}

This is simple for diagonal $M$s, for example given Lagrangian
\begin{eqn}
L = \frac{1}{2} m (r^2 + r^2 \theta^2 + r^2 \sin^2 \theta \, \dot{\phi}^2 ) - V(r,\theta,\phi),
\end{eqn}
we have
\begin{eqn}
H = \frac{1}{2m} \left( p_r^2 + \frac{1}{r^2} p_\theta^2 + \frac{1}{r^2 \sin^2 \theta} p_\phi^2 \right) + V(r,\theta,\phi).
\end{eqn}
We must be more careful if there are terms like $N_i (q) \dot{q}^i$ in the Lagrangian, because then $p_i = \pd{L}{\dot{q}_i}$ will pick up a $N_i(q)$ term.

\begin{example}[particle in EM field]
Consider the Lagrangian
\begin{eqn}
L = \frac{1}{2} m \vd{x}^2 - e \phi(\v x) + e \vd{x} \cdot \v{A}(\v x).
\end{eqn}
Then
\begin{eqn}
\v p = m \vd{x} + e \v{A}(\v x) \qquad \implies \qquad \vd{x} = \frac{1}{m} (\v{p} - e \v{A}),
\end{eqn}
which results in a Hamiltonian (check this)
\begin{eqn}
H = \frac{1}{2m} (\v{p} - e \v{A})^2 + e \phi(\v x).
\end{eqn}
This is interesting because we have a first order term in the Lagrangian ending up squared in the Hamiltonian.
\end{example}

\subsection{Why Hamiltonian formalism?}

We will see that the classical Hamiltonian formalism gives us a close correspondence to quantum mechanics. It turns out that many interesting calculations are the same or almost the same, to some order, in both quantum and classical mechanics when we use the Hamiltonian formulation along with the appropriate language.

In the Lagrangian formalism, our canonical variables are the coordinates $\v{q}(t)$. The trajectory in $q$-space can intersect itself because the trajectory also depends on $\vd{q}(t)$. On the other hand, in the Hamiltonian formalism, our canonical variables are the coordinates $\v q, \v p$ and in $qp$-space trajectories cannot intersect themselves because the flow is uniquely determined by $\v q(t), \v p(t)$.

The space of all $\set{\v q, \v p}$ is called the \textit{phase space} of the Hamiltonian system. This gives us a special kind of geometry called \textit{symplectic geometry}. In quantum mechanics, we will see that, crudely speaking, the phase space has fundamental cells of volume $\hbar^3$. The phase space here is analogous to the phase space in kinetic theory, where a system of $N$ particles in $\R^3$ is $6N$-dimensional.

We can also make more general coordinate changes in phase space---these are called \textit{canonical transformations}. Furthermore, Hamiltonian systems have the nice property of \textit{integrability} (i.e. action angle variable). Later, the Hamiltonian formulation will lead us to the \textit{Hamilton-Jacobi} formulation, which is very nice.

\subsection{Variational principle for Hamilton equations}

Looking at the action in the Hamiltonian formalism, we have
\begin{eqn}
S[q(t), p(t)] = \int \dif{t} \left[ \dot{q} p - H(q,p) \right],
\end{eqn}
and taking the variation we find
\begin{align}
\delta{S} &= \int \dif{t} \left[ p \delta{\dot q} + \dot{q} \delta{p} - \pd{H}{q} \delta{q} - \pd{H}{p} \delta{p} \right] \\
	&= \int \dif{t} \left[ (-\dot{p} - \pd{H}{q}) \delta{q} + (\dot{q} - \pd{H}{p}) \delta{p} \right] + \eval[3]{(p \delta{q})}_{t_1}^{t_2}.
\end{align}
We see that the boundary term vanishes under the action principle assumption that $\delta{q}(t_1) = \delta{q}(t_2) = 0$, and furthermore demanding that the variation vanishes gives us Hamilton's equations.

\subsection{Conservation and preservation}

\begin{remark}[Simple conservation laws]
Taking the total time derivative we have
\begin{eqn}
\od{H}{t} = \pd{H}{q} \dot{q} + \pd{H}{p} \dot{p} + \pd{H}{t} = -\cancel{\dot{p} \dot{q}} + \cancel{\dot{q} \dot{p}} + \pd{H}{t}
\end{eqn}
[???] where $\pd{H}{t} = -\pd{L}{t}$ so if there is no time dependence then $H$ is conserved.

Furthermore, if $q_i$ is a cyclic variable then the canonical momentum $p_i$ is conserved, since
\begin{eqn}
\pd{L}{q_i} = 0 \qquad \implies \qquad \dot{p}_i = \pd{H}{q_i} = 0.
\end{eqn}
\end{remark}

Next we'll look at Liouville's theorem. We can motivate this by looking at a harmonic oscillator. If we plot trajectories of the SHO in phase space, i.e. a plot of $p$ vs $q$, the trajectories are ALL simply circles around the origin. In this case it is easy to see that if we take a volume in phase space and flow it along the Hamiltonian trajectories, the volume is simply rotated about the origin---thus the volume is preserved. It turns out that this is true for all Hamiltonian systems.

\begin{theorem}[Liouville's theorem]
The Hamiltonian flow is volume-preserving. In other words, if we denote the flow on phase space $\Phi_t(\v q, \v p) = (\v{q}(t), \v{p}(t))$, and we have a volume $V \subseteq \set{\text{phase space}}$ then
\begin{eqn}
\od{\Phi_t(V)}{t} = 0 \qquad \implies \qquad \Phi_t(V) = V \quad \text{for all $t$}.
\end{eqn}
\end{theorem}

\begin{proof}

[continued in next lecture]
\end{proof}

\end{document}


\documentclass[../main.tex]{subfiles}
\begin{document}
\chapter{Introduction to Combinatorics}

In discrete optimization, some or all of the variables in a model are required to belong to a discrete set;
this is in contrast to continuous optimization in which the variables are allowed to take on any value within a range of values.
There are two branches of discrete optimization: integer programming and  combinatorial optimization where the discrete set is a set of objects,
or combinatorial structures, such as assignments, combinations, routes, schedules, or sequences.
Combinatorial optimization is the process of searching for maxima (or minima) of an objective function $F$ whose domain is a discrete but
large configuration space (as opposed to an N-dimensional continuous space).  Typical combinatorial optimization problems are the travelling
salesman problem (``TSP''), the minimum spanning tree problem (``MST''), and the knapsack problem.
We start with basic combinatorics which is able to enumerate the all solutions exhaustively.
Later on, other chapters we will dive into different combinatorial/disrete optimization problems.

%https://www.cs.cmu.edu/afs/cs.cmu.edu/project/learn-43/lib/photoz/.g/web/glossary/comb.html
%https://en.wikipedia.org/wiki/Combinatorial_optimization

%%%%%%%%%%%%%%%%%%%%%%%%%Combinatorics%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Combinatorics, as a branch in mathematics that  mainly concerns with counting and enumerating, is a means in obtaining results, and certain properties of finite structures. Combinatorics is used frequently in computer science to obtain formulas and estimates in both the design and analysis of algorithms.  It is a broad and thus seemingly hard to define topic that can solve the following types of questions:
\begin{itemize}
    \item  The counting or enumerating  of specified structures, sometimes referred to as arrangements or
    configurations in a very general sense, associated with finite systems,
    \item the existence of such structures that satisfy certain given criteria,
    this is usually called Contraint Restricted Problems (CSPs).
    \item optimization, finding the ``best'' structure or solution among several possibilities,
    be it the ``largest'',``smallest'' or satisfying some other optimality criterion.
\end{itemize}
In this section, we introduce common combinatorics that can help us come up with the simplest which potentially be quite large state space. At least, this is the first step, and solving a small problem in this way might offer us more insights on continuing finding a better solution.

When the situation is easy, we can mostly figure out the counting with some logic and get a closed-form solution; when the situation is more complex such as in the \textit{partition} section, we detour by using recurrence relation and math induction.
\section{Permutation}
Given a list of integer $[1,2,3]$, how many way can we order these three numbers? Imagine that we have three positions for these three integers.  For the first position, it can choose 3 integers, leaving the second position with 2 options. Further, when it reaches to the last position, it can only choose whatever that is left, we have 1. The total count will be $3\times2\times1$.

Similarly, for $n$ distinct numbers,  we will get the number of permutation  easily as $n\times(n-1)\times...\times1$. A factorial, denoted as as $n!$, is used to abbreviate it. Worth to notice, the factorial sequence grows even quicker than the exponential sequence, such as $2^n$.

\subsection{$n$ Things in $m$ positions}  Permutation of $n$ things on $n$ positions is denoted as $p(n,n)$. Think about what if we have $m\in[1, n-1]$ positions instead? How to get a closed-form function for  $p(n,m)$.  The process is the same: we fix each position and consider the number of choice of things each one has.
\begin{align}
   p(n,m) &= n\times(n-1)\times...\times (n-(m-1)) \\
   &=\frac{n\times(n-1)\times...\times (n-m+1)\times (n-m)\times...\times 1}{(n-m)\times...\times 1}\\
   &=\frac{n!}{(n-m)!}
\end{align}
If we want $p(n,n)$ to follow the same form, it would require us to define $0!=1$.
\begin{bclogo}[couleur = blue!30, arrondi=0.1,logo=\bccrayon,ombre=true]{What if there are repeated things, that things are not distinct? }
\end{bclogo}
% \paragraph{What if things are not distinct?} For example, how to get the permutation of [1,2,2,3].

\subsection{Recurrence Relation and Math Induction} The number or the full set of permuations can be generated incrementally. We demonstrate how with recurrence relation and math induction. We start from $P(0,0)=1$. Easily, we get $P(i, 0)=1, i\in[1, n]$. With math induction, now assume we know $P(n-1, m-1)$, for the m-th position, what choice does it have? First, we need to pick this thing from the $n-(m-1)$ things. Then,  we have $m-1$ things lined up linearly, there are $m$ positions to insert the m-th item, resulting $P(n, m) = (n-m+1)*mP(n-1, m-1)$.

Now, we can use iterative method to obtain the closed-form solution:
\begin{align}
    P(n, m) &= (n-m+1)*m*P(n-1, m-1)\\
    &=(n-m+1)*m*(P(n-2, m-2) \\
    &...\\
    &=m!P(n-m+1, 1
\end{align}

\subsection{See Permutation in Problems} Suppose we want to sort an array of integers incrementally, say the array is $A=[4,8,2]$. The right order is $[2,4,8]$, which is trivial to obtain in this case. If we are about to form it as a search problem, we need to define a \textit{search space}. Using our knowledge in combinatorics, we know all possible ordering of these numbers are [4,8,2],[4,2,8],[2,4,8],[2,8,4],[8,2,4],[8,4,2]. Generating all possible ordering and save it in an array maybe. Then this sorting problem is converted into checking which array is incrementally sorted. However, it comes with large price on space usage, since for $n$ numbers there, the number of possible orderings are $n!$. A smarter way to do it is to check the ordering as we are generating the ordering set.

% Another problem is the one we mentioned, the subarray sum problem. The subarray if be represented as $a_{i,j}$, it is like we need to select two $n$ items and put them at two positions: $i$ and $j$. For position $i$ we have $n$ options, for $j$, there is only
\section{Combination}
%\paragraph{Problem Definition}
Same as before, we have to choose $m$ things out of $n$ but with one difference--the order does not matter, how many ways we have?  This problem is called \textbf{combination}, and it is denoted as $C(n,m)$. For example, for [1,2,3], $C(3, 2)=[1, 2], [2, 3], [1, 3]$. Comparatively, P(3, 2) = [1, 2], [2, 1], [2, 3], [3, 2], [1, 3], [3, 1].

To get combination, we can leverage and apply permutation first. However, this results over-counting. As shown in our example, when there are two things in the combination, a permutation would double count it. If there are $m$ things, we over count by $m!$ times. Therefore, if we  divide the permutation by all permutation of $m$ things, we get out formula for combination:
\begin{align}
    C(n,m)&=\frac{P(n,m)}{P(m,m)}\\
    &=\frac{n!}{(n-m)!m!}
\end{align}
\begin{bclogo}[couleur = blue!30, arrondi=0.1,logo=\bccrayon,ombre=true]{Back to the last question, when there are repeats in the permutation. We can use the same idea. Assume we have $n,m$ that $n$ things in total but only $m$ types and $a_i, i\in[1,m]$ to denote the number of each type, this means $a_1+ a_2+ ... + a_m= n$. The number of ways to linearly order these objects is $\frac{n!}{a_1!a_2!...a_m!}$. }
\end{bclogo}

The combination of k things out of n, will be the same as choosing (n-k) things.
    \begin{align}
        C(n,k)=C(n, n-k)
    \end{align}

\subsection{Recurrence Relation and Math Induction} We also show how the combination can be generated incrementally. We start from $C(0,0)=1$, and easily we get $C(n, 0)=1$. Assume we know $C(n-1, k-1)$, now we need to add the k-th item into the combination? :
\begin{itemize}
    \item Use k-th item, then we just need to put the the k-th item into any sets in $C(n-1,k-1)$, resulting $C(n-1,k-1)$.
    \item Not use k-th item, this means we need to pick $k$ items from the other $n-1$ items, resulting $C(n-1,k)$.
\end{itemize}
Thus, we have $C(n, k) = C(n-1, k-1) + C(n-1, k)$, this is called \textbf{Pascal's Identity}.

\paragraph{What if things are not distinct?}

%%%%%%%%%%%%%%%%%%Partition%%%%%%%%%%%%%%%
\section{Partition}
We discuss three types of partitions: (1) integer partition, (2) set partition, and (3) array partition. In this section, counting becomes less obvious compared with combination and permutation, this is where we rely more on \textbf{recurrence relation} and \textbf{math induction}.

% {14, 23},{1, 234},{1, 23, 4},{14,2,3},{1,24,3},{1,2,34}, and {1,2,3,4} for parts from 1 to 4. Vetex or edge coloring, and connected components are one of the set partition examples. Set partition is actually a combination problem. We will show how we can do set parititon with backtrack and dynamic programming sometimes.

\subsection{Integer Partition}
\paragraph{Integer Partition Definition} Integer partitions is to partition a given integer $n$ into distinct subsets that add up to $n$.
\begin{lstlisting}[numbers=none]
For example, given n=5, the resulting partitioned subsets are these 7 subsets:
{5}
{4, 1},
{3, 2}
{3, 1, 1},
{2, 2, 1},
{2, 1, 1, 1},
{1, 1, 1, 1, 1}
\end{lstlisting}

\paragraph{Analysis} Let us assume the resulting sequence is $(a_1, a_2, ..., a_k)$, and $ a_1\geq a_2\geq...\geq a_k\geq 1$, and  $a_1+ a_2+ ... + a_k= n$. The ordering is simply to help us to track the sequence.  We use

The easiest way to generate integer partition is to construct them incrementally. We first start from the partition of {n}. For n=5, we get {5} first. Then we subtract one from the largest item that is larger than 1, and add it to the smallest item if it exists and that the resulting s+1 < l, s < l-1 , and other option is to put it aside.  For {5}, there is no other item, so that it becomes {4, 1}. For {4,1}, following the same rule, we get {3, 2},  for {3, 2}, we get {3,1,1}.
\begin{lstlisting}
{5}, no other smaller item, put it aside
{4, 1}, satisfy s<l-1, become{3,2}
{3, 2}, not satisfy s<l-1, put it aside
{3, 1, 1}, satisfy s<l-1, add it to
    {2, 2, 1}, not satisfy, put it aside
    {2, 1, 1, 1}, not satisfy, put it aside
{1, 1, 1, 1, 1}
\end{lstlisting}
\begin{bclogo}[couleur = blue!30, arrondi=0.1,logo=\bccrayon,ombre=true]{Try to generate the partition when n=6.}
\end{bclogo}
If we draw out the transfer graph, we can see a lot of overlapping of some state. Therefore, we add one more limitation on the condition, s>1.

\subsection{Set Partition}
\paragraph{Set Partition Problem Definition} How many ways exist to partition a set of $n$ distinct items ${a_1, a_2, ..., a_n}$ into $k$ nonempty subsets, $k<=n$.
\begin{lstlisting}[numbers=none]
Here are 7 ways that we can partition the set {a1, a2, a3,a4} into 2 nonempty subsets. They are
{a1}, {a2,a3,a4};
{a2}, {a1,a3,a4};
{a3}, {a1, a2, a4};
{a4}, {a1, a2, a3}
{a1, a2}, {a3, a4};
{a1, a3}, {a2, a4};
{a1, a4}, {a2, a3};
\end{lstlisting}
Let us denote the total ways as $s(n,k)$. As seen in the example, given 2 groups and 4 items, there are two combination of each group's size: 1+3 and 2+2. For combination {1,3}, this is equivalent to choose one item from the set to put at the first subset $C(n,1)$, and then choose 3  items for the other subset $C(3, 3)$. For combination {2, 2}, we have $C(4,2)$ for one subset and $C(2,2)$ for the other subset. However, because the ordering of the subsets does not matter, we need to divide it by $2!$. The set partition problem thus consists of two steps:
\begin{itemize}
    \item Partition $n$ into $k$ integers: This subrountine can be solved with integer partition we just learned. We have $b_1+b_2+...+b_k = n$.
    \item For each combination of integer partition, we compute the number of ways choosing  $b_i$ items for that set, we get $C(n, b_1)\times C(n-b_1, b_2)\times C(n-b_1-b_2, b_3)\times...\times C(b_k, b_k)$. Now, we find the distinct $b_i$ and its number of appearance in the sequence. If we have $m$ distinct number denoted as $b_i$, and its  count $c_i$, then we divide the above ways by $c_1!c_2!...c_m!$.
\end{itemize}

From this solution, it is hard to get a closed form for $s(n,k)$.

\paragraph{Find Recurrence Relation} There is just one way to handle this problem, let us try the incremental method--find a recurrence relation. We first start with $s(0, 0)=0$, and we can also easily get $s(n,0)=0$. Now, with the mathematical induction, we assume we solved a subproblem, say $s(n-1, k-1)$, can we induce $s(n,k)$? What do we need?

Now we have n-1 items in k-1 groups, now there is one addition group and one additional item. There are two ways:
\begin{itemize}
    \item put the additional item into the additional group. In this way, $s(n,k)$ is simply the same as of $s(n-1,k-1)$.
    \item spread the n-1 items from the original k-1 groups into k groups, that is $s(n-1, k)$ and our additional item has k options now, making $k\times s(n-1,k)$ in total
\end{itemize}
Combing together the count of these two ways, we get a recurrence relation that
\begin{align}
    s(n,k)=s(n-1,k-1)+k s(n-1,k)
\end{align}

\section{Array Partition}
\paragraph{Problem Definition} How many ways exist to partition an array of $n$ items ${a_1, a_2, ..., a_n}$ into subarrays. There are different subtypes depending on the number of subarrays, say $m$:
\begin{enumerate}
    \item When the number $m$ is as flexible as $m\in[1, n-1]$.
    \item When the number $m$ is fixed as a number in range $[2, n-1]$.
\end{enumerate}

\paragraph{When the number of subarray is fixed} For example, it is common to partition an array into 2 or 3 subarrays. First, we find an item $a_i$ as a  partition point,  getting the last subarray $a[i:n]$ and left an array to further consider $a[0:i]$. If $m=2$, $a[0:i]$ results the first subarray and the partition process is done. This gives out $n$ ways of parition. When $m=3$, we need to further partition $a[0:i]$ into two parts. This can be represented with recurrence relation:
\begin{align}
    d(n, m) = (d(i, m-1), a[i:n]), i \in[0, n-1]
\end{align}
Further, for $d(i, m-1)$:
\begin{align}
    d(i, m-1) = (d(j, m-2), a[j:n]), j \in[0, i-1]
\end{align}
This can be done recursively: we will have a recursive function with depth $m$.

\paragraph{When the number of subarray is flexible} The process is the same other than $m$ can be as large as $n-1$. If we are about to use dynamic programming, for all these states, we need to come up with an ordering of the state $(i, j)$, where $i$ is the subproblem $a[0:i]$, and $j$ is the number of partitions. We imagine it as a matrix with $i, j$ as row and column respectively:
\begin{lstlisting}[numbers=none]
   0  1  2     n-1: partition
0  X  -  -    -
1  X  X  -   -
2  X  X  X

n-1
n X  X  X   X  X
\end{lstlisting}
Does the ordering of the \texttt{for} loop matter? Actually it does not.

\paragraph{Applications} There are many applications that involve splitting an array/string or cutting a rod.  This relates to spliting type of dynamic programming.

\section{Merge}

\section{More Combinatorics}
Combinatorics is about enuemrating specified structures, there are some structures are of our main interests through this book and often appears in the interviews, they are: \textit{subarray}, \textit{subsequence}, and \textit{subsets}.
\paragraph{Subarray} We have solved one example with subarray. Subarray is defined as a contigious sequence in the array, which can be represented as $a[i,...,j]$. The number of subarray exist in an array of size $n$ will be:
\begin{align}
    sa=\sum_{i=1}^{i=n} i = n*(n+1)/2
\end{align}
A substring is a contiguous sequence of characters within a string. For instance, "the best of" is a substring of "It was the best of times". This is not to be confused with subsequence, which is a generalization of substring. For example, "Itwastimes" is a subsequence of "It was the best of times", but not a substring.

Prefix and suffix are special cases of substring. A prefix of a string S S is a substring of S that occurs at the beginning of  S. A suffix of a string S is a substring that occurs at the end of S.
\paragraph{Subsequence} For a subsequence means any sequence we can find the array, which is not required to be contiguous, but the ordering still matters. For example, in the array of [ABCD], the subsequence will be
\begin{lstlisting}
               [],
[A],    [B], [C],[D],
[AB],[AC],[AD],  [BC],[BD],  [CD],
[ABC],[ABD], [ACD],  [BCD],
[ABCD]
\end{lstlisting}
You would actually see for $n=4$, there are 16 possible subsequence, which is $2^4$. This is not coincidence. Imagine for each item in the array, they have two options, either be chosen into the possible sequence or not chosen, which make it to $2^n$.
\begin{align}
    ss=2^n
\end{align}
\paragraph{Subset} The Subset B of a set A is defined as a set within all elements of this subset are from set A. In other words, the subset B is contained inside the set A, $B \in A$. There are two kinds of subsets: if the order of the subset does'nt matter, it is a combination problem, otherwise, it is a permutation problem.

If it is the case that ordering does not matter, for $n$ distinct things, the number of possible subsets, also called \textit{the power set} will be:
\begin{align}
    power_set = C(n,0)+C(n,1)+...+C(n,n)
\end{align}

\end{document}


\section{Cylindrical Coordinates}
\noindent
Cylindrical coordinates are the expansion of polar coordinates by including a third term that represents the height from the xy-plane, $z$. All cylindrical coordinates have the form $(r, \theta, z)$. This system is called cylindrical because it’s easy to describe shapes with cylindrical symmetry because integrations have constant, independent bounds.

[INSERT IMAGE]

\input{./curvilinearCoordinates/conversions_cylindrical}
\input{./curvilinearCoordinates/integration_cylindrical}


\chapter{Data Sets}
\label{appendix:data_sets}

All of the data sets used are defined such that the dependent variable is the output of a particular mathematical function for a given set of inputs.  They are taken from a paper entitled \textit{Genetic Programming Needs Better Benchmarks} by McDermott et al. \cite{benchmarks}  All of the inputs are taken to form a grid on some interval.  Let $E[a, b, c]$ denote $c$ samples equally spaced in the interval $[a,b]$. (Note that McDermott et al. defines $E[a, b, c]$ slightly differently.)  Below is a list of all of the data sets that are used:

\begin{enumerate}[noitemsep]
\item \textbf{Keijzer1}: $0.3x \sin(2 \pi x);$ $x \in E[-1,1,20]$
\item \textbf{Keijzer11}: $x y+\sin((x-1)(y-1));$ $x, y \in E[-3,3,5]$
\item \textbf{Keijzer12}: $x^{4}-x^{3}+\frac{y^{2}}{2}-y;$ $x, y \in E[-3,3,5]$
\item \textbf{Keijzer13}: $6 \sin(x) \cos(y);$ $x, y \in E[-3,3,5]$
\item \textbf{Keijzer14}: $\frac{8}{2 + x^{2} + y^{2}};$ $x,y \in E[-3,3,5]$
\item \textbf{Keijzer15}: $\frac{x^{3}}{5} - \frac{y^{3}}{2} - y - x;$ $x, y \in E[-3,3,5]$
\item \textbf{Keijzer4}: $x^{3} e^{-x} \cos(x) \sin(x) (\sin^{2}(x) \cos(x) - 1);$ $x \in E[0,10,20]$
\item \textbf{Keijzer5}: $\frac{3 x z}{(x - 10) y^{2}};$ $x,y \in E[-1,1,4]; z \in E[1,2,4]$
\item \textbf{Nguyen10}: $2 \sin(x) \cos(y);$ $x,y \in E[0,1,5]$
\item \textbf{Nguyen12}: $x^{4} - x^{3} + \frac{y^{2}}{2} - y;$ $x,y \in E[0,1,5]$
\item \textbf{Nguyen3}: $x^{5} + x^{4} + x^{3} + x^{2} + x;$ $x \in E[-1,1,20]$
\item \textbf{Nguyen4}: $x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x;$ $x \in E[-1,1,20]$
\item \textbf{Nguyen5}: $\sin(x^{2}) \cos(x) - 1;$ $x \in E[-1,1,20]$
\item \textbf{Nguyen6}: $\sin(x) + \sin(x + x^{2});$ $x \in E[-1,1,20]$
\item \textbf{Nguyen7}: $\ln(x + 1) + \ln(x^{2} + 1);$ $x \in E[0,2,20]$
\item \textbf{Nguyen9}: $\sin(x) + \sin(y^{2});$ $x,y \in E[0,1,5]$
\item \textbf{Sext}: $x^{6} - 2 x^{4} + x^{2};$ $x \in E[-1,1,20]$
\end{enumerate}