text:
\section{Results}

In this section, we introduce and comment the different tests realized on the BRAC CMDO and PMDO problems.
We also recall the aim here is to find the best solution to the optimization problem, the fastest, and the different optimizers are tested in this scope.
In the following, the six previously introduced optimizers (see Section~\ref{sec:opt}) are tested on the BRAC CMDO and PMDO.
For the sake of comparison, we will produce convergence and parallel plots to assess the efficiency of the SEGOMOE tool-box solvers compared to the Isight optimizers.

\subsection{BRAC CMDO results}

\subsubsection{Tests details}
\label{ssec:cmdo:details}

The optimizers, introduced in Section~\ref{sec:opt}, are compared with the following test plan.
We perform $10$ independent runs for each solver using $10$ different initial DoEs build with the Latin hyper-cube sampling method.
For each run, all the solvers are initiated with the same DoE.
The size of the initial DoEs is set to $n_{start} = 13$ (i.e $d+1$ where $d$ is the dimension of BRAC CMDO problem).
Note that Evol and Pointer-2 need a single point to launch the optimization process, we thus provide the point with the best valid value of the DoE.
If there is no valid value in the DoE, the point with the minimal constraints violation is provided.

The maximum number of evaluations of the SEGO-like solvers is set to $\text{max\_it\_nb}=227$ (i.e. $20d - n_{start}$ meaning a total of $20d = 240$ evaluations is performed).
Concerning Evol and Pointer-2, we used the historical options settings of the CMDO framework.
The maximum number of evaluations of Evol is set to $\text{max\_it\_nb} = 960$.
For each run, Evol thus performed $973$ function evaluations, which takes approximately 8 hours to perform.
We respectively fix the maximum allowable job time and the topography type options of Pointer-2 to $8$ hours and \textit{nonlinear}.
All the remaining options of Evol and Pointer-2 are kept as default.
Finally, the best solution among the ten optimizations performed for each optimizer is kept to initiate the BRAC PMDO problem as explained in Section~\ref{ssec:PMDO}.

\subsubsection{Convergence plots}
\label{sssec:cmdo:cv_plot}

To assess the performance of the introduced optimizers, we build two kinds of convergence plots.
The first one, named \textit{evaluation convergence plot}, displays the average and the standard deviation of the best valid value over the $10$ runs for increasing number of evaluations.
The best valid value is defined as the best valid value if there is, at least, one valid point in the DoE, otherwise, a penalization replaces the obtained invalid value.
The penalization is the highest feasible value of the objective function ever found.
Because of confidentiality reasons, we scale the value of the convergence plots between $0$ and $1$.
The second convergence plot, named \textit{time convergence plot}, also shows the average and the standard deviation of the best valid value over the $10$ runs along the optimization time.
The penalization is kept the same as in the evaluation convergence plot.

\begin{figure}[htp]
    \centering
    \subfloat[Evaluation convergence plot. \label{fig:CV_CMDO:Ite}]{\includegraphics[width=0.45\textwidth]{Figures/CV/CV_BRAC_Delta_it.pdf}}
    \subfloat[Time convergence plot. \label{fig:CMDO:time}]{\includegraphics[width=0.45\textwidth]{Figures/CV/CV_BRAC_Delta_time.pdf}}
    \caption{Convergence plots for the CMDO of BRAC. In the time convergence plot, the vertical red-dashed line indicates the maximum CPU time allowed for an industrial optimization process (8 hours for this test case).}
    \label{fig:CV_CMDO}
\end{figure}

Figure~\ref{fig:CV_CMDO} shows that SEGO-like solvers clearly outperform Evol and Pointer-2 in term of the optimum value.
In term of number of evaluations, Figure~\ref{fig:CV_CMDO:Ite} displays that SEGO is converging the fastest to the optimum value even if SEGOMOE has a fastest convergence rate at the beginning of the optimization.
Furthermore, note that SEGO-UTB and SEGOMOE-UTB are not converging as fast as SEGO and SEGOMOE because of their extensive exploratory behaviour.
When looking to the optimization time, Figure~\ref{fig:CMDO:time} reveals that the SEGO-like solvers converge in approximately $2$ hours meanwhile Evol and Pointer-2 does not converge after $8$ hours.

To conclude, these convergence plots have shown that the SEGO-like solvers provide a better solution in less running CPU-time than Evol and Pointer-2.

We also note that the use of the UTB option within SEGO and SEGOMOE seems to not lead to any improvement on the obtained performance which suggests that the exploration of the feasible domain was not a difficult task to handle on this test case.

\subsubsection{Parallel plots}
\label{sssec:cmdo:para}

The parallel plots introduced in the following depict the behaviour of the tested solvers along the optimization process in which targeted values are displayed (e.g., the explored design variables).
Here, the plotted values are the number of iterations, the design variables values, the objective function value and the constraints violation.
Furthermore, several colors are used to distinguish the reference design in red (i.e. the best feasible design found so far by all the tested optimizers, here SEGO), the optimum design found by the regarded solver in black, the feasible explored designs in green, and the unfeasible designs in blue.
Due to the stochastic nature of our tests, we build our parallel plots using a median run in the following way.
For each run of the optimizer, we store the best valid objective value; if none of the runs converges to a feasible point, we collect the minimal violation explored by the optimizer.
The median run is then selected based on the stored values for all runs.

Figures~\ref{fig:pp_cmdo} and \ref{fig:pp_cmdo_2} display the parallel plots of the median run of the tested optimizers.
First, note that all the SEGO-like optimizers are converging to the optimum.
On the contrary, Evol and Pointer-2 do not converge to the optimum, as depicted by Figures~\ref{fig:pp_cmdo:Evol} and \ref{fig:pp_cmdo:pointer}. Furthermore, Evol is not able to explore the entire domain as implied by the grouped green lines of Figure~\ref{fig:pp_cmdo:Evol}.
Secondly, note the different solutions of the SEGO-like solvers even if they all converge approximately to the same objective function value.
Indeed, the values for DV\_8 and DV\_11 (i.e. wing maximum thickness-to-chord ratios) are different from an optimizer to another.
This can be due to either a local optimum or an inactive design variable (i.e. the objective function does not change along the design variable).
From the expert point of view, the behavior observed on the BRAC CMDO problem is due to inactive design variables.

\begin{figure}[p]
    \centering
    \subfloat[Evol. \label{fig:pp_cmdo:Evol}]{\includegraphics[height=0.26\textheight]{Figures/PP/CMDO/EVOL.PNG}} \\
    \subfloat[Pointer-2. \label{fig:pp_cmdo:pointer}]{\includegraphics[height=0.26\textheight]{Figures/PP/CMDO/Pointer-2.PNG}} \\
    \subfloat[SEGO. \label{fig:pp_cmdo:SEGO}]{\includegraphics[height=0.26\textheight]{Figures/PP/CMDO/SEGO.PNG}}
    \caption{Parallel plots using the median run for the CMDO of BRAC for Evol, Pointer-2 and SEGO. In blue: the unfeasible designs; in green: the feasible designs; in black: the optimum; in red: the reference design.}
    \label{fig:pp_cmdo}
\end{figure}

\begin{figure}[p]
    \centering
    \subfloat[SEGO-UTB. \label{fig:pp_cmdo:SEGO-UTB}]{\includegraphics[height=0.26\textheight]{Figures/PP/CMDO/SEGO-UTB-DE.PNG}} \\
    \subfloat[SEGOMOE-UTB. \label{fig:pp_cmdo:SEGOMOE-UTB}]{\includegraphics[height=0.26\textheight]{Figures/PP/CMDO/SEGOMOE-UTB-DE.PNG}} \\
    \subfloat[SEGOMOE. \label{fig:pp_cmdo:SEGOMOE}]{\includegraphics[height=0.26\textheight]{Figures/PP/CMDO/SEGOMOE.PNG}}
    \caption{Parallel plots using the median run for the CMDO of BRAC for SEGOMOE, SEGOMOE-UTB and SEGO-UTB. In blue: the unfeasible designs; in green: the feasible designs; in black: the optimum; in red: the reference design.}
    \label{fig:pp_cmdo_2}
\end{figure}

\subsection{PMDO results}

\subsubsection{Tests details}

The optimizers are again compared on the BRAC PMDO problem.
Because of the computation time of the BRAC PMDO problem (ca. $25min$), we only performed one optimization for each solver.
Each of the optimizers is initiated with the same DoE of $n_{start}=20$ sample points using the Latin hyper-cube sampling method.
To mimic the introduced MDO process, we also add to the initial DoE the point corresponding to the best solution $\bm{x}^*$ of all the BRAC CMDO process.

As in Section~\ref{ssec:cmdo:details}, Evol and Pointer-2 only need a single point to launch the optimization, we thus provide the best solution $x^*$ of all the BRAC CMDO cases.

The maximum number of evaluations of the $4$ SEGO-like solvers is set to $\text{max\_it\_nb}=170$ (i.e. $10d - n_{start} - 1$ meaning a total of $10d = 190$ evaluations).
The maximum number of evaluations of Evol and Pointer-2 is set to $510$ to follow historical values used in the PMDO framework, and the parallel batch size is set to $4$ for both.
In addition, a \textit{smooth} topography type is selected for Pointer-2.
All the other settings are kept to default values.

\subsubsection{Convergence plots}

The convergence plots defined in this Section are slightly different from the ones introduced in Section~\ref{sssec:cmdo:cv_plot}.
Indeed, only one optimization is here performed for each of the $6$ optimizers.
Thus Figure \ref{fig:CV_PMDO} displays the best valid values of the regarded solvers for increasing time or number of evaluations.
\begin{figure}[htp]
    \centering
    \subfloat[Evaluation convergence plot. \label{fig:CV_PMDO:Ite}]{\includegraphics[width=0.45\textwidth]{Figures/CV/CV_BRAC_PMDO_Delta_it.pdf}}
    \subfloat[Time convergence plot. \label{fig:PMDO:time}]{\includegraphics[width=0.45\textwidth]{Figures/CV/CV_BRAC_PMDO_Delta_time.pdf}}
    \caption{Convergence plots for the PMDO of BRAC. The red dashed line is the value of the BRAC CMDO design in the BRAC PMDO process.}
    \label{fig:CV_PMDO}
\end{figure}
First, Figure \ref{fig:CV_PMDO:Ite} shows that none of the solvers converge to the same optimum value, with the best value obtained with SEGO.
Similar values are also achieved with Evol and SEGOMOE.
Furthermore, Pointer-2 provides the worst value of all the solvers with an optimum at $0.1$.
The other algorithms find better solution values than Pointer-2, but worse than SEGO, SEGOMOE and Evol.
Note that SEGO and SEGOMOE are able to find a similar solution as Evol with 30 fewer evaluations.

Secondly, in terms of convergence time, Evol and Pointer-2 benefit from the batch evaluation of the objective function which is not the case for the SEGO-like solvers.
Indeed, the 220 necessary evaluations of Evol to converge are done in less than 40 hours, whereas the SEGO-like solvers need 80 hours for 190 evaluations.

To conclude, Figure \ref{fig:CV_PMDO} shows that the UTB feasibility criterion is not useful for the BRAC CMDO problem, whereas SEGO and SEGOMOE show great performances with a small budget.
Moreover, the batch evaluation capability of Evol speeds up the convergence time.

\subsubsection{Parallel plots}

Here, no run selections are needed for the parallel plots. Indeed, only a single run is performed for the BRAC PMDO problem for each solver.
They are produced in the same way as in Section \ref{sssec:cmdo:para}.
Figures \ref{fig:pp_pmdo} and \ref{fig:pp_pmdo_2} show the parallel plots for Evol, Pointer-2, SEGO, SEGOMOE, SEGO-UTB and SEGOMOE-UTB.

\begin{figure}[p]
    \centering
    \subfloat[Evol. \label{fig:pp_pmdo:Evol}]{\includegraphics[height=0.26\textheight]{Figures/PP/PMDO/EVOL.PNG}} \\
    \subfloat[Pointer-2. \label{fig:pp_pmdo:pointer}]{\includegraphics[height=0.26\textheight]{Figures/PP/PMDO/Pointer-2.PNG}} \\
    \subfloat[SEGO. \label{fig:pp_pmdo:SEGO}]{\includegraphics[height=0.26\textheight]{Figures/PP/PMDO/SEGO.PNG}}
    \caption{Parallel plots for the PMDO of BRAC for Evol, Pointer-2 and SEGO. In blue: the unfeasible designs; in green: the feasible designs; in black: the optimum; in red: the reference design.}
    \label{fig:pp_pmdo}
\end{figure}

\begin{figure}[p]
    \centering
    \subfloat[SEGO-UTB. \label{fig:pp_pmdo:SEGO-UTB}]{\includegraphics[height=0.26\textheight]{Figures/PP/PMDO/SEGO-UTB.PNG}} \\
    \subfloat[SEGOMOE-UTB. \label{fig:pp_pmdo:SEGOMOE-UTB}]{\includegraphics[height=0.26\textheight]{Figures/PP/PMDO/SEGOMOE-UTB.PNG}} \\
    \subfloat[SEGOMOE. \label{fig:pp_pmdo:SEGOMOE}]{\includegraphics[height=0.26\textheight]{Figures/PP/PMDO/SEGOMOE.PNG}}
    \caption{Parallel plots for the PMDO of BRAC for SEGO-UTB, SEGO-UTB and SEGOMOE. In blue: the unfeasible designs; in green: the feasible designs; in black: the optimum; in red: the reference design.}
    \label{fig:pp_pmdo_2}
\end{figure}

As mentioned, SEGO finds the best design, followed closely by Evol and SEGOMOE. These three solutions share similar main plan-form features, namely the wing span (DV\_0) and sweep (DV\_1), while there is some variation in the wing twist distribution (DV\_4-7) and maximum thickness-to-chord ratios (DV\_8-13), as shown in Figures~\ref{fig:pp_pmdo:Evol} and \ref{fig:pp_pmdo:SEGOMOE}. Only SEGO converges to the lower thickness bound for the two most outboard profiles (DV\_12 and DV\_13).
The winglet has a secondary impact on the MTOW, and as a result even greater variation is seen in the optimized winglet parameters (DV\_15-18).
The best solutions of the other three optimizers deviate even further from the SEGO optimum. It is possible that all the optimizers converge to different local optima, but the variability in the solutions is more likely due to incomplete convergence of the algorithms, since the number of iterations was limited to meet the time restrictions of an industrial application.


text:
\section{Straight Line has Zero Curvature}
Tags: Straight Lines, Curvature

\begin{theorem}
A [[Definition:Straight Line|straight lines]] has zero [[Definition:Curvature|curvature]].
\end{theorem}

\begin{proof}
From [[Equation of Straight Line in Plane/Slope-Intercept Form|Equation of Straight Line in Plane: Slope-Intercept Form]], a [[Definition:Straight Line|straight line]] has the [[Definition:Equation|equation]]:
:$y = m x + c$
[[Definition:Differentiation|Differentiating]] twice {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | l = \dfrac {\d y} {\d x}
      | r = m
      | c = [[Power Rule for Derivatives]]
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d^2 y} {\d x^2}
      | r = 0
      | c =
}}
{{end-eqn}}
By definition, the [[Definition:Curvature/Cartesian Form|curvature]] of a [[Definition:Curve|curve]] is defined as:
:$\kappa = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$
But we have that:
:$y'' := \dfrac {\d^2 y} {\d x^2} = 0$
and so, as in general $y' := \dfrac {\d y} {\d x} = m \ne 0$:
:$\kappa = \dfrac 0 {\paren {1 + m^2}^{3/2} }$
the [[Definition:Curvature|curvature]] is zero.
{{qed}}
\end{proof}

config: wiki


TITLE: How much load does an aqueduct support?
QUESTION [7 upvotes]: Recently, I found out about /r/InfrastructurePorn, and I found a particularly interesting photo of the Gouwe Aqueduct in Gouda, NE:

It seems like the bridge that is supporting the boat wouldn't be able to do it. Is the weight of the actual boat being supported by the aqueduct?

REPLY [1 votes]: Another way to look at this is to ask yourself "how is the water at the bottom to know about the ship above it?" It can't; a given water volume "communicates" only with its immediate neighbors. Now it's possible that the water volume at the bottom experiences different pressures from different sides. This will just push it away along the pressure gradient, until all over the bottom all water volumes experience the same pressure again. That may happen while the ship is passing over a spot, because of the waves created by the ship, but is only minor and temporary. Any water displaced by the ship just flows away until the water pressure is the same again at equal depths all over the length of the canal.
Since the water pressure of the bottommost layer is what the bridge must carry, and doesn't change, what the bridge must carry doesn't change either. Q.e.d. ;-).


TITLE: Prove that $\mathbb{Z}[i]/(3)$ and $\mathbb{Z}[i]/(7)$ are finite fields and find their cardinality
QUESTION [2 upvotes]: In my rings subject's test I had to prove that $\mathbb{Z}[i]/(21)$ was decomposed as a product of two finite fields, and that was easy to prove for me because $21 = 3\cdot 7$, and $\mathbb{Z}[i]$ is a PID and $3$ and $7$ are irreducible, so by the Chinese Remainder Theorem, and because $(3)$ and $(7)$ are maximal, we have the decomposition into two fields.
What I am unable to do is to find why they are finite fields and in such case, I had to find their cardinality. Can you help me/give some hints?

REPLY [1 votes]: As $\mathbf Z[i]\simeq\mathbf Z[X]/(X^2+1)$, we have that, for instance,
$$\mathbf Z[i]/(3)\simeq\mathbf Z[X]/(X^2+1)\big/3\mathbf Z[X]/(X^2+1)\simeq\mathbf Z/3\mathbf Z[X]\big/(X^2+1).$$
Now $X^2+1$ is irreducible over $\mathbf Z/3\mathbf Z=\mathbf F_3$ ($3$ is inert), so  we have a quadratic extension of $\mathbf F_3$, i.e., up to an isomorphism, $\mathbf F_9$.
Note: $\mathbf Z[i]$ is not only a U.F.D., but a Euclidean domain, for the euclidean function $N(a+ib)=a^2+b^2$.


TITLE: Single predictor soft thresholding
QUESTION [1 upvotes]: I have found in this book the following problem
$$\underset{\beta}{\mathrm{minimize}}\left\{\frac{1}{2N}\sum\limits_{i=1}^N(y_i-z_i\beta)^2+\lambda|\beta|\right\}$$
for parameter $\beta>0$ known real sequences $y_i$s, $z_i$s and known $N\in\mathbb{N}$. The non-differentiability at zero and the fact that this cannot be decomposed into $N$ minimizations since $\beta$ is common for all measurements are my problems here. So (by inspection) they propose the following optimal solution for $\beta$:
$$
\hat{\beta}=
\left\{
\begin{array}{ll}
\frac{1}{N}\langle\mathbf{z},\mathbf{y}\rangle-\lambda,&\quad\frac{1}{N}\langle{\mathbf{z}},\mathbf{y}\rangle>\lambda\\
0,&\quad\left|\frac{1}{N}\langle{\mathbf{z}},\mathbf{y}\rangle\right|\leq\lambda\\
\frac{1}{N}\langle\mathbf{z},\mathbf{y}\rangle+\lambda,&\quad\frac{1}{N}\langle{\mathbf{z}},\mathbf{y}\rangle<-\lambda
\end{array}
\right.
$$
But the proof of this is not clear to me. Can you give any hints?

REPLY [0 votes]: The exercise in question (2.2 of SLS) actually instructs you not to make use of subgradients, so here is such a proof. Let $\hat{\beta}_o=\mathbf{z}^\intercal\mathbf{y}/N$ be the usual OLS estimator, and define $\ell(x)$ to be the loss function in the minimization problem. We have
$$
\ell(b)-\ell(a)=\frac{b^{2}-a^{2}}{2}+\hat{\beta}_{o}(a-b)+\lambda(|b|-|a|).
$$
If $|\hat{\beta}_{o}|<\lambda$ then
$$
\ell(b)-\ell(0)\ge\frac{b^{2}}{2}-\hat{\beta}_{o}b+|\hat{\beta}_{o}b|\ge0,
$$
so the minimum is achieved when $b=0$.
Otherwise, suppose $\hat{\beta}_{o}>\lambda$, so that
$$\ell(b)-\ell(\hat{\beta}_{o}-\lambda)\ge\frac{1}{2}\left(2\lambda|b|+b^{2}-2\hat{\beta}_{o}b+(\hat{\beta}_{o}-\lambda)^{2}\right).
$$
If $b<0$ then this is obviously non-negative, while if
$b>0$ then
$$
\ell(b)-\ell(\hat{\beta}_{o}-\lambda)\ge \frac{1}{2}(b-\hat{\beta}_{o}+\lambda)^{2}.
$$
Thus the loss function is minimized when $b=\hat\beta_o-\lambda$. The case $\hat\beta_o<-\lambda$ follows from applying the above result to the predictor $-\mathbf{z}.$


TITLE: Counting integers related to Bézout's identity
QUESTION [1 upvotes]: I am reading a proof on a double sums and I don't understand. Here is the context:
Let $p,q$ two fixed integers and $k\in[1,\lfloor\frac{N}{p}\rfloor-1],l\in[1,\lfloor\frac{N}{q}\rfloor-1].$
Now he said that there is for all $n\in[-\lfloor\frac{N}{pq}\rfloor,\lfloor\frac{N}{pq}\rfloor]$ (interval of integers) $$\gcd(p,q)\times \lfloor\frac{N}{pq}\rfloor$$ couples $(k,l)$ such that $$n\gcd(p,q)=pk-ql$$
It's seems related to Bezout's identity but I don't see how to prove this.

REPLY [0 votes]: I started trying to prove this when I realized it's not always true as currently stated. For a fairly simple example, let $p = q = 2$ and $N = 4$. Then $\gcd(p,q) = 2$, $\left\lfloor\frac{N}{p}\right\rfloor = 2$, $\left\lfloor\frac{N}{q}\right\rfloor = 2$ and $\left\lfloor\frac{N}{pq}\right\rfloor = 1$. Also,
$$k\in \left[1,\left\lfloor\frac{N}{p}\right\rfloor-1\right] = [1,2-1] = [1,1] \tag{1}\label{eq1}$$
$$l\in \left[1,\left\lfloor\frac{N}{q}\right\rfloor-1\right] = [1,2-1] = [1,1] \tag{2}\label{eq2}$$
Next, note that
$$n\gcd(p,q)=pk-ql \; \Rightarrow \; 2n = 2k - 2l \tag{3}\label{eq3}$$
The range for $n$ becomes
$$n\in\left[-\left\lfloor \frac{N}{pq} \right\rfloor,\left\lfloor \frac{N}{pq} \right\rfloor\right] = [-1,1] \tag{4}\label{eq4}$$
The stated total # of solutions becomes
$$\gcd(p,q) \times \left\lfloor \frac{N}{pq} \right\rfloor = 2 \times 1 = 2 \tag{5}\label{eq5}$$
However, as $k = l = 1$ are the only allowed values to choose from, the RHS of \eqref{eq3} has a sole value of $0$, giving just $n = 0$ as the only value allowed in the range given in \eqref{eq4}. However, this means there is just $1$ couple of $(k,l)$, which doesn't match the value of $2$ given in \eqref{eq5} as the expected number of solutions.
It seems either something is incorrect in the provided details or additional conditions are required for what you're trying to prove to always be true. As you only say that $p$ and $q$ are fixed, perhaps you're meant to show some $N$ exists which satisfies all of the conditions? I believe this may be the case, but I won't try to prove it unless you confirm this is actually what is required.


TITLE: Convex continuous piecewise linear function form
QUESTION [0 upvotes]: Let $n\in \mathbb{N}$ and $\mathcal{E}_n$ the set of functions $f$ of the form
$$f(x)=b_0+b_1x+\sum_{i=1}^n a_i\left\lvert x-r_i\right\rvert$$
for $b_0,b_1\in\mathbb{R},a_1,\dots,a_n>0$ and $r_1,\dots,r_n\in\mathbb{R}$.
Show that if $f$ is a convex continuous piecewise linear function, then there exists $n\in\mathbb{N}$ such that $f\in\mathcal{E}_n$.
This result does not seem clear to me, and I do not know how to prove it.

REPLY [0 votes]: Here is my proof for this.
Let $k_1<\dots k_n\in\mathbb{R}$ such that $f$ is linear on $\left]k_i,k_{i+1}\right[$ for all $1\leqslant i\leqslant n-1$ and $f$ is linear on $\left]-\infty,k_1\right[$ and on $\left]k_n,+\infty\right[$.
Let $c_i$ be the slope of $f$ on $\left[ k_i,k_{i+1}\right[$ for all $0\leqslant i\leqslant n$ with the convention $k_0=-\infty$ and $k_{n+1}=+\infty$.
As $f$ is convex, we have $c_0\leqslant c_1\leqslant\dots\leqslant c_{n+1}$. Let
$$a_i=\frac{1}{2}\left(c_i-c_{i-1}\right)$$
for $1\leqslant i\leqslant n$. Indeed, in the formula $x\mapsto b_0+b_1+\sum_i\left\lvert x-k_i\right\rvert$, the jump of the derivative in $x=k_i$ is $2a_i$.
Let $b_1$ such that
$$b_i+\sum_{i=1}^na_i=c_n$$
that is
$b_1=\frac{1}{2}\left(c_n+c_o\right)$.
As such, the two applications $f$ and $x\mapsto b_1x+\sum_{i=1}^na_i\left\lvert x-k_i\right\rvert$ are two linear functions on every intervals $\left]k_{i-1},k_i\right[$ for $1\leqslant i\leqslant n+1$, continuous and with the same slope on every interval by construction.
That is the two applications differ by a constant, named $b_0$. Thus, $f\in \mathcal{E}_n$.


TITLE: Find the sum with induction
QUESTION [0 upvotes]: Find the sum
$\sum_{j=1}^n 2^{-j}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}$
for all $n=3,4,...$. I need to use induction.
I don't know how to start, can you show me?

REPLY [0 votes]: Let
$$s_n=\sum_{j=1}^n 2^{-j}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}$$
$$\begin{align}
s_n&=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^n}\\
\frac{s_n}{2}&=\quad\quad\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^n}+\frac{1}{2^{n+1}}
\end{align}$$
$$\frac{s_n}{2}-s_n=\frac{1}{2^{n+1}}-\frac{1}{2}$$
$$s_n\left(\frac12-1\right)=\frac{1-2^n}{2^{n+1}}$$
$$s_n=\frac{2^n-1}{2^{n}}$$
Now for $n=1$
$$s_1=\frac{2^1-1}{2^{1}}=\frac12$$
for $n=k$
$$s_k=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^k}=\frac{2^k-1}{2^{k}}$$
for $n=k+1$
$$s_{k+1}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^k}+\frac{1}{2^{k+1}}=\frac{2^k-1}{2^{k}}+\frac{1}{2^{k+1}}=\frac{2^{k+1}-2}{2^{k+1}}+\frac{1}{2^{k+1}}\\\implies s_{k+1}=\frac{2^{k+1}-1}{2^{k+1}}$$
Now if above formula is true for $k$ then it's also true for $k+1$ and then using Principle of Mathematical Induction..


TITLE: Give counterexample to the following claim about invertible homomorphic posets
QUESTION [0 upvotes]: In general, it is known that all invertible (bijective) group homomorphisms are group isomorphisms. However the same reasoning need not hold true for invertible poset homomorphisms. That is, not all invertible poset homomorphisms are poset isomorphisms. I'm having trouble coming up with an example that validates this claim - since I'm relatively new to Abstract Algebra. Any help would be appreciated.

REPLY [1 votes]: "In general, it is known that all invertible (bijective) group homomorphisms are group isomorphisms"

This is true for sets and groups, but not for abritrary objects on a category. For example, not all bijective continuous functions have a continuous inverse.
As for the posets counterexample: let $a \to b$ be a poset of two elements with $a \leq b$, and $\{c,d\}$ be a poset where no elements are comparable. Let $f(a) = c, f(b) = d$ and $g(c) = a, g(d) = b$ be inverse functions. In particular, both are bijective. Since $g$ is a homomorphism (the poset has no comparable objects so the condition is satisfied trivially), by your claim its inverse should be a morphism, but $f$ is not: $a \leq b$ would imply $f(a) = c \leq d = f(b)$, but once again, $c$ and $d$ are not comparable.


text:
In this section we present some numerical studies of the proposed
class of algorithms.
We carry out simulations and discuss
how various choices of $T_d$,
of the singular values of $W$, of the sequence length (period) $m$,
or of the parameters $\alpha_1,\alpha_2$ influence the qualitative behavior of the algorithm.
Further, a numerical approach to  construct  the exploration
sequences using nonlinear programming is presented.

We start by summarizing the
proposed optimization algorithm and the involved parameters.

\subsection{Algorithm and Parameters}
\label{sec:algo_params}

\corrected{
The design parameters and functions involved in the proposed algorithm are
\begin{enumerate}[leftmargin=*]
	\item map parameters $\alpha_1,\alpha_2 \in \mathbb{R}$ with $\alpha_1 + \alpha_2 \not = 0$,
	\item gradient generating functions $f,g: \mathbb{R} \to \mathbb{R}$,
	\item matrix $T_d$ and exploration sequence matrix $W$, in particular
	singular values $\sigma_{i},\ i = 1,\ldots,\text{rk}(T_d)$ of $W$,
	\item step size $h>0$.
\end{enumerate}

Hence, algorithm \eqref{eq:algo} with \eqref{eq:TM} is defined in terms of $W$ as follows:

\begin{algorithm}[h]
	\corrected{
	\caption{Derivative-free optimization algorithm with non-commutative maps}
	\begin{algorithmic}[1]

		\State \textbf{Input:} $x_0$, $h$, $\alpha_1, \alpha_2$, $f(J(\cdot)), g(J(\cdot))$, $T_d$, $\sigma_i\ (i = 1,\ldots,\text{rk}(T_d))$, stop criterion
		\State Calculate $W$ and $m$ as described in \cref{sec:construction_svd}
		\State $k=0$
		\While {stop criterion is not fulfilled}
		\State $\ell = k \bmod (m) + 1$
		\State $e_\ell = [0_{\ell},1,0_{n-1-\ell}]^\top$
		\State $Y(J(x_k)) = [f(J(x_k))I \ g(J(x_k))I]$
		\State $\hat{x}_{k} = x_k + \sqrt{h}Y\big(J(x_k)\big)W e_{\ell}$
		\State $Y(J\big(\hat{x}_{k})\big) = [f(J(\hat{x}_k))I \ g(J(\hat{x}_k))I]$
		\State $x_{k+1} = x_k + \sqrt{h}\big(\alpha_1 Y\big(J(x_k)\big)+\alpha_2 Y(J\big(\hat{x}_{k})\big)\big) We_{\ell} $
		\State $k\leftarrow k+1$
		\EndWhile
		\State \textbf{return} \textit{$[x_{0},x_1,\ldots]$}
	\end{algorithmic}
	\label{alg:alg}}
\end{algorithm}

Notice that $Y(J(x_k))$ is a $n \times 2n$ matrix. Moreover, if $\alpha_2 = 0$, then line \correct{8} and \correct{9} in \cref{alg:alg} can be skipped.}

In the following we discuss the influence of the design parameters on the algorithm's behavior. Additionally, a set of parameters working well in generic situations, which can then be used as a starting point to obtain optimized parameters for a particular application, is provided.

\textit{1. Map parameters ${\alpha_1,\, \alpha_2}$.}

The parameters weigh $Y(J(x_k)$ and  $Y(J(\hat{x}_k)$ in \cref{alg:alg}, respectively.
In particular, they can be utilized to choose between a single-point ($\alpha_2 = 0$) or a two-point algorithm.
They are to be normalized according to $\alpha_1+\alpha_2 = 1$ and to tune
 according to the ratio of $\alpha_1,\alpha_2$, while utilizing the step size $h$ to tune the convergence speed.

Moreover, the choice of $\alpha_1,\alpha_2$ restricts the choice of $T_d$ to be skew-symmetric for $2\alpha_2 \corrected{-} (\alpha_1+\alpha_2)^2 = 0$ and otherwise normal (cf. \cref{thm:structure_W_H} and \cref{thm:structure_W_E}).

In practice, the parameter sets we found providing the best performance were $[\alpha_1\ \alpha_2] = [1\ 0]$ for the single-point and $[\alpha_1\ \alpha_2] = [\nicefrac{1}{2}\ \nicefrac{1}{2}]$ for the two-point gradient-approximation scheme.

\textit{2. Generating functions ${f,\, g}$.}

The generating functions comprise a scaling of the  objective evaluated at $x_k$ and $\hat{x}_k$ as stated in \cref{alg:alg}.

Various choices are presented in \cref{thm:structure_W_H} and \cref{thm:structure_W_E}; depending on $\alpha_1,\alpha_2$. Often we have chosen $f,g$ as sinusoidal functions, since the algorithm showed a very stable behavior
for that cases.
Note that high function values of $f,g$ or if $f,g$ scale arbitrarily large with $J(x_k)$, the algorithm performs large steps which may cause instabilities.
In the case of bounded functions, such as sinusoidal functions, (arbitrarily) large steps sizes are avoided.
Further, if $J,f,g$ vanish at a minimum $x^*$, asymptotic convergence to $x^*$ (instead of practical convergence) has been observed in our studies.

\textit{3. Exploration sequence matrix $W$ and $T_d$.}

The exploration sequence matrix $W$ depends on the choice of $T_d$, specifically on the eigenvalues of $T_d$.

A step-by-step construction of $W$ based on the algorithm parameters is presented in \cref{sec:construction_svd}.

As explained in this construction, the singular values of $W$ can be chosen (see \Cref{thm:Td_normal} below), hence this degree of freedom can be used in the algorithm tuning. As shown in numerical examples below, smaller singular values lead to smoother trajectories, but more steps (larger $m$) are needed to perform one gradient approximation step.

There exists a set of optimal singular values in the sense of minimal number of steps $m$, which is $m=\text{rk}(T_d)+1$. Therewith, the choice of $T_d$ influences the lower bound on $m$ (see \Cref{thm:min_steps} below).

Note that minimal sequence length $m$ does not always lead to the fastest convergence behaviour.

\textit{4. Step size $h$.}
The approximated gradient is scaled with the step size $h$, and hence, $h$ influences the speed of convergence, as well as the area of exploration around $x_k$. As stated in \Cref{thm:conv} there exists an upper bound on $h$ such that \correct{semi-global practical asymptotic convergence (if (A1) and (A2) holds)} is ensured. In our numerical studies, $h$ is often chosen as $0.001 \le h \le 0.5$.

\begin{corollary}

    \label{thm:Td_normal}

    In the case $2\alpha_2-(\alpha_1+\alpha_2)^2 = 0$, the singular values $\sigma_{2\ell-1},\sigma_{2\ell}$, $\ell = 1,\ldots,\lceil \text{rk}(T_d)/2 \rceil$, of $W$ can  be chosen arbitrarily. Otherwise, the singular values of $W$ have to satisfy $\sigma_{2\ell-1}=\sigma_{2\ell}$ and
    \begin{align}
    	T_d =
    	\begin{bmatrix}
    		\text{diag}([\gamma_1\ \cdots \gamma_n]) & - I \\
    		I & \text{diag}([\gamma_1\ \cdots \gamma_n])
    	\end{bmatrix},
    	\label{eq:Td_E}
    \end{align}

    with

    \begin{align}
    	\gamma_\ell = \Big(\alpha_2-\frac{1}{2}(\corrected{\alpha_1+\alpha_2})^2\Big)\sigma^2_{2\ell-1}.
    	\label{eq:eval_Td_normal}
    \end{align}
    \end{corollary}

\begin{proof}
    Following directly from the proof of \Cref{thm:existence_svd} in \Cref{sec:proof_existence_svd}. Specifically, $T_d$ in \eqref{eq:Td_E} satisfies \eqref{eq:cases_E2} and \eqref{eq:eval_Td_normal} corresponds to \eqref{pf:2_cond1}.
\end{proof}

\corrected{
\begin{corollary}
	\label{thm:min_steps}
	The minimal number of steps to approximate a gradient according to \eqref{eq:algo} - \eqref{eq:TM} and using the $T_d$'s in \cref{thm:structure_W_H} and \cref{thm:structure_W_E} is $m = n+1$.
	\begin{proof}
		Since the first $n$ rows of each $T_d$ in \cref{thm:structure_W_H} and \cref{thm:structure_W_E} are linearly independent, we know $\min\{\text{rk}(T_d)\} \ge n$. This implies with \Cref{thm:existence_svd} that $m \ge n + 1$.
	\end{proof}
\end{corollary}}

\corrected{
\begin{rem}
	A gradient step is a approximated in $m = n+1$ steps
	for $T_d$ as specified in \eqref{eq:cases_H2} and $Q$ with elements $\{q_{ij}\}_{i,j=1}^{2n}$ such that
	\begin{align}
	q_{ij} = \begin{cases}
	~~~1\quad \text{ if } i+j = 2n+1,\ i>j \\
	-1\quad \text{ if } i+j = 2n+1,\ i<j \\
	~~~0\quad \text{ else}
	\end{cases}
	\end{align}
	holds, while the singular values $\sigma_{2\ell-1},\sigma_{2\ell}$ of $W$ satisfy \eqref{pf:2_cond_skew} for $\ell = 1,\ldots,\lceil n/2 \rceil$.
\end{rem}}

config: arxiv


TITLE: Find width and length of rectangle given diagonal and area
QUESTION [0 upvotes]: The diagonal of rectangle is 25, its area is 168, find width and length. I tried solving this problem using trigonometry since diagonal and two sides forms a right triangle, from area i got that a=168/b, considering that area is a*b=168, I reached to a quadratic equation but i got a negative root, does anyone have any idea? Besides the solutions are a=7 and b=24.

REPLY [1 votes]: You have two equations $$ab = 168 \implies a = \frac{168}{b}$$ and $$a^2 + b^2 = 25^2$$
Substituting the first into the second and multiplying throughout by $b^2$ yields  $$\frac{168^2}{b^2} + b^2 = 25^2 \implies 168^2 + b^4 = 25^2b^2$$ This is a quadratic in $b^2$ that gives us solutions $$b^2 = 49 \quad \text{ or } \quad 576$$
Hence $b = \pm 7$ and $b = \pm 24$. We neglect the negative solutions to get $$a  = 24, b=7 \quad \text{or} \quad a=7, b= 24$$
Which is just symmetric. So you can simply say that one side is $24$ and the other is $7$.

REPLY [0 votes]: We have
$$ab=168\tag 1$$
$$a^2+b^2=25^2\tag2$$
So, from $(1)(2)$, we have
$$(a+b)^2=a^2+b^2+2ab=25^2+2\cdot 168=961\Rightarrow a+b=31.\tag3$$
Now from $(1)(3)$, we know that $a,b$ are the solutions of
$$x^2-31x+168=0,$$
i.e.
$$(x-7)(x-24)=0.$$

REPLY [0 votes]: It normal to find negative solution, they do mathematics. In real world of course lengh is positive !
Just look $$7^2+24^2=25^2=(-7)^2+24^2=7^2+(-24)^2$$
Hope it helps you ;)


TITLE: Confirmation of a concept under General Relativity and Frames of Reference
QUESTION [1 upvotes]: I would like to preface this by saying that this isn't necessarily meant to be a full question, though it may become that, but it rather a confirmation of my understanding of a concept.
Newton would say that an individual standing on the earth represents an Inertial frame of Reference. In this scenario, if an object, let's say an apple, is released above the ground, the observer would suggest that the apple accelerated towards the ground due to gravity.
Einstein on the other hand would suggest that an individual standing on the earth is NOT an Inertial frame of reference, and that the apple itself represents an inertial frame of reference. More broadly, an inertial frame of reference is one in which an object has no forces acting on it, such as an object in deep space, or in free fall, not being PULLED by gravity, but rather following geodesic a through curved spacetime, just as the apple is doing as it follows a geodesic a towards the Earth.
If we accept this to be the Inertial frame of reference, the logical conclusion is that an apple located above the Earth does not accelerate towards Earth, but rather, from the Inertial frame of reference, from which we measure true acceleration, the Earth is accelerating upwards (radially outwards).
My understanding of the radially outwards acceleration, which is the concept that I would like to be confirmed or denied, is as follows:
Due to the curvature of spacetime, the 'future' of the Earth in spacetime is bent due to it's own matter, inwardly, or to put it simply, it should be collapsing into itself. However, due to the rigidity (may not be the correct term, but I assume it is evident what I mean... Can clarify) of the planet, it does not collapse inward. This causes the Earth to be, relative to the inertial frame, accelerating away from it's future path through spacetime, at 9.8m/s/s, exactly the rate at which it would collapse inward if this wasn't the case.
**Note: I apologize for the rather long-winded question, but this was a very broad concept for me, and I'm afraid that my education isn't refined enough to be able to concisely explain this.
My thanks for clarification on this matter.

REPLY [4 votes]: Newton would not say that an individual standing on the Earth represents an inertial frame of reference. An inertial frame is one in which Newton first law applies i.e. an object moves in a straight line at constant speed. Since the observer dropping the apple observers the apple to accelerate that observer's frame is non-inertial.
However you are correct, or rather almost correct, to say that the apple's frame is inertial. If you were falling alongside that apple (ignoring air resistance) then you would observe the apple remaining stationary beside you. A good example of this is the astronauts in the International Space Station. They are falling freely so, as countless videos show, anything they release remains next to them, or moves away at constant velocity.
I say almost correct because in general relativity inertial frames are normally only locally inertial. The apple appears to be stationary relative to you, but actually it is slowly approaching you because you and the apple are both moving along different lines towards the centre of the Earth. The farther away from you the apple is the greater will be this relative motion.
But I'm guessing that your real interest is in describing the acceleration of the Earth's surface, and you are quite correct that it is accelerating. Technically its proper acceleration is non-zero. The proper acceleration is calculated using a somewhat complicated equation but basically it is the acceleration relative to a freely falling object, and as you say relative to the apple the surface of the Earth is accelerating outwards at $9.81$ m/s$^2$.
Another convenient definition is that the proper acceleration is the acceleration felt by an observer in their rest frame. For example I weigh 68 kg and right now in my rest frame I'm experiencing a force of $9.81 \times 68$ Newtons, which is just the upwards force exerted on my by my chair. My proper acceleration acceleration is just this force divided by my mass i.e. $9.81$ m/s$^2$.
Likewise the surface of the Earth is being pushed up by the rock/whatever immediately under the surface, which is being pushed up by the stuff immediately under it and so on down to the centre of the Earth (where the gravity is zero so in fact the centre of the Earth is approximately an inertial frame).
It seems weird to be talking about something that is stationary as accelerating, but this is normal in general relativity. In everyday life when we talk about acceleration we normally mean coordinate acceleration i.e. if we measure the position of an object using some coordinate system (e.g. latitude, longitude and altitude) then acceleration means the position is changing as measured using those coordinates. However GR tells us that the latitude/longitude/altitude are actually accelerating coordinates not inertial ones, and to be stationary in an accelerating coordinate system is to be accelerating in an inertial coordinate system.
If you're interested in What is the weight equation through general relativity? there is a detailed calculation of the proper acceleration in a spherical gravitational field like the Earth's. However you might find this a bit hard going if you're new to GR.