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# A hard definite integral with trigonometric functions

How could we get a closed form for this one? $$\displaystyle\int_{0}^{\frac{\pi }{2}}{{{x}^{2}}\sqrt{\tan x}\sin \left( 2x \right)\text{d}x}$$

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Why do you expect there to be one? A quick numeric computation and a look at the inverse symbolic calculator gives no match for the number (1.1057733282580962321535756282112900456642421892605) –  mrf Jan 23 '13 at 10:38
@mrf the answer is in terms of pai and logs. I dont know what substitution or other methods are involved. –  Ryan Jan 23 '13 at 10:47
Using @rlgordonma's method, the final result can be written in terms of elementary functions and is given by $\frac{\pi \left(5 \pi ^2-6 \pi (\log (4)-2)-3 (8+(\log (4)-4) \log (4))\right)}{96 \sqrt{2}}$. –  Fabian Jan 23 '13 at 11:21
@Fabian: I get precisely what you get. I take it we have Mathematica in common? And I can simplify what I have presented and present the actual result if it is wanted. –  Ron Gordon Jan 23 '13 at 11:52
@rigordonma I am really interested how to get these value :) –  Ryan Jan 23 '13 at 13:20

The general method of attack on an integral like this is to recognize that the $x^2$ could be ignored for the time being upon expressing the integral in terms of a suitable parameter which may then be twice differentiated. To wit, after a little manipulatin, I find:

$$\displaystyle \int_{0}^{\frac{\pi}{2}} dx \: {x}^2 \sqrt{\tan x} \sin \left( 2x \right) = 2 \Im{ \int_{0}^{\frac{\pi}{2}} dx \: x^2 e^{i x} \sqrt{\sin{x} \cos{x}}}$$

So consider the following integral:

$$J(\alpha) = \int_{0}^{\frac{\pi}{2}} dx \: e^{i \alpha x} \sqrt{\sin{x} \cos{x}}$$

Note that the integral we seek is

$$\displaystyle \int_{0}^{\frac{\pi}{2}} dx \: {x}^2 \sqrt{\tan x} \sin \left( 2x \right) = -2 \Im{\left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1}}$$

It turns out that there is, in fact, a closed for for $J(\alpha)$:

$$J(\alpha) = -\frac{\left(\frac{1}{16}+\frac{i}{16}\right) \sqrt{\frac{\pi }{2}} \left ( e^{\frac{i \pi a}{2}}-i\right) \Gamma \left(\frac{a-1}{4}\right)}{\Gamma \left ( \frac{a+5}{4} \right )}$$

Plugging this into the above expression, you will find terms including polylogs and harmonic numbers, which I will spare you unless explicitly asked for. But this is how you would get a closed-form expression for your integral.

EDIT

The integral is even nicer when we consider

$$J(\alpha) = \int_{0}^{\frac{\pi}{2}} dx \: \sin{\alpha x} \sqrt{\sin{x} \cos{x}}$$

Then

$$J(\alpha) = \frac{\pi ^{3/2} \sin \left(\frac{\pi a}{4}\right)}{8 \Gamma \left(\frac{5-a}{4}\right) \Gamma \left(\frac{5+a}{4}\right)}$$

and

$$\left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1} = \frac{\pi \left(-5 \pi ^2+6 \pi (\log (4)-2)+3 (8+(\log (4)-4) \log (4))\right)}{192 \sqrt{2}}$$

The integral we seek is then

$$\int_0^{\frac{\pi}{2}} dx \: x^2 \sqrt{\tan{x}} \sin{(2 x)} = -2 \left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1} = \frac{\pi \left(-24+12 \pi +5 \pi ^2-3 \log ^2(4)+12 \log (4)-6 \pi \log (4)\right)}{96 \sqrt{2}}$$

It turns out that the numerical value of the latter value is about $1.10577$, which agrees with the numerical approximation of the integral mentioned by @mrf and verified in Mathematica.

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Very nice! That was surprising, at least to me. –  mrf Jan 23 '13 at 13:03
@mrf: Thanks. That means a lot given what I've seen you do here. –  Ron Gordon Jan 23 '13 at 20:18

Let $\displaystyle f(z) = \log^{2}(z) \Big( \frac{z-1}{z+1} \Big)^{1/2}$ where $-\pi < \arg(z), \arg(z-1), \arg(z+1) \le \pi$.

Now integrate around a contour that consists of the segment just above $[-1,1]$ and the upper half of the unit circle, and is indented around the branch points at $z=0, z=1,$ and $z=-1$.

The contributions from the indentations around the branch points can be shown to be vanishing small.

Then going around the contour counterclockwise,

$$\int_{-1}^{0} \log^{2}(|x| + i \pi)\Big( \frac{(1-x)e^{\pi i}}{1+x} \Big)^{\frac{1}{2}}\ dx \ + \int_{0}^{1} \log^{2}(x) \Big(\frac{(1-x)e^{\pi i}}{1+x} \Big)^{1/2} \ dx$$

$$\ + i \int_{0}^{\pi} \log^{2}(e^{it}) \Big(\frac{e^{i t}-1}{1+e^{i t}} \Big)^{1/2} e^{it} \ dt = 0$$

Looking at each piece separately,

\begin{align} \int_{-1}^{0} \log^{2}(|x| + {i \pi})\Big( \frac{(1-x)e^{i \pi}}{1+x} \Big)^{\frac{1}{2}} \ dx &= i \int_{0}^{1} \Big(\log x + i \pi \Big)^{2} \Big(\frac{1+x}{1-x} \Big)^{1/2} \ dx \\ &= i \int_{0}^{1}\Big(\log^{2} (x) + 2 \pi i \log x - \pi^{2} \Big) \frac{1+x}{\sqrt{1-x^{2}}} \ dx \end{align}

$$\int_{0}^{1} \log^{2}(x) \Big(\frac{(1-x)e^{\pi i}}{1+x} \Big)^{1/2} \ dx = i \int_{0}^{1} \log^{2}(x) \frac{1-x}{\sqrt{1-x^{2}}} \ dx$$

\begin{align} i \int_{0}^{\pi} \log^{2}(e^{it}) \Big(\frac{e^{i t}-1}{1-e^{i t}} \Big)^{1/2} e^{it} \ dt &= -i \int_{0}^{\pi} t^{2} \sqrt{i \tan t/2} \ e^{it} \ dt \\ &= - i e^{i \pi/4} \int_{0}^{\pi} t^{2} \sqrt{\tan t/2} \ e^{it} \ dt \end{align}

 So we have

$$e^{i \pi /4} \int_{0}^{\pi} t^{2} \sqrt{\tan t/2} \ e^{it} \ dt = 2 \int_{0}^{1} \frac{\log^{2}(x)}{\sqrt{1-x^{2}}} \ dx + 2 \pi i \int_{0}^{1} \frac{\log x}{\sqrt{1-x^{2}}} \ dx$$

$$\ + 2 \pi i \int_{0}^{1} \frac{x \log x}{\sqrt{1-x^{2}}} - \pi^{2} \int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} \ dx - \pi^{2} \int_{0}^{1} \frac{x}{\sqrt{1-x^{2}}} \ dx$$

Now we can use the fact that

$$\int_{0}^{1} \frac{x^{a}}{\sqrt{1-x^{2}}} \ dx = \frac{1}{2} \int_{0}^{1} u^{a/2+1/2-1} (1-u)^{1/2-1} \ du = \frac{1}{2} B \left( \frac{1}{2}, \frac{a}{2}+ \frac{1}{2} \right)$$

to evaluate the integrals on the right.

\begin{align} \int_{0}^{1} \frac{\log^{2}(x)}{\sqrt{1-x^{2}}} &= \frac{d^{2}}{da^{2}} \frac{1}{2} B \left( \frac{1}{2}, \frac{a}{2}+ \frac{1}{2} \right) \Big|_{a=0} \\ &= \frac{1}{8} B \left( \frac{1}{2}, \frac{1}{2} \right) \Big[ \Big( \psi (\frac{1}{2} ) - \psi(1) \Big)^{2} + \psi_{1}(\frac{1}{2}) - \psi_{1}(1) \Big] \\ &= \frac{\pi^{3}}{24} + \frac{\log^{2} 4}{8} \end{align}

\begin{align} \int_{0}^{1} \frac{\log x}{\sqrt{1-x^{2}}} \ dx &= \frac{1}{2} \frac{d}{da} \frac{1}{2} B \left( \frac{1}{2}, \frac{a}{2}+ \frac{1}{2} \right) \Big|_{a=0} \\ &= \frac{1}{4} B \left( \frac{1}{2}, \frac{1}{2} \right) \left( \psi(\frac{1}{2}) - \psi(1) \right) \\ &= - \frac{\pi \log 2}{2} \end{align}

\begin{align} \int_{0}^{1} \frac{x \log x}{\sqrt{1-x^{2}}} \ dx &= \frac{1}{2} \frac{d}{da} \frac{1}{2} B \left( \frac{1}{2}, \frac{a}{2}+ \frac{1}{2} \right) \Big|_{a=1} \\ &= \frac{1}{4} B \left( \frac{1}{2}, 1 \right) \left( \psi(1) - \psi(\frac{3}{2}) \right) \\ &= \log 2 -1 \end{align}

$$\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} \ dx = \frac{\pi}{2}$$

$$\int_{0}^{1} \frac{x}{\sqrt{1-x^{2}}} \ dx = 1$$

Therefore,

$$\int_{0}^{\pi} t^{2} \sqrt{\tan t/2} \ e^{it} \ dt = e^{-i \pi /4} \Big( \frac{\pi^{3}}{12} + \frac{\pi \log^{2} 4}{4} - i \pi^{2} \log 2 + 2 \pi i \log 2 - 2 \pi i - \frac{\pi^{3}}{2} - \pi^{2} \Big)$$

or

$$\int_{0}^{\pi /2} t^{2} \sqrt{\tan t} \ e^{2it} \ dt = \frac{1}{8} e^{- i \pi /4} \Big( \frac{\pi^{3}}{12} + \frac{\pi \log^{2} 4}{4} - i \pi^{2} \log 2 + 2 \pi i \log 2 - 2 \pi i - \frac{\pi^{3}}{2} - \pi^{2} \Big)$$

Then finally equating the imaginary parts on both sides of the equation,

\begin{align} \int_{0}^{\pi /2} t^{2}\sqrt{\tan t} \sin(2t) \ dt &= \frac{1}{8 \sqrt{2}} \left(- \pi^{2} \log 2 + 2 \pi \log 2 - 2 \pi - \frac{\pi^{3}}{12} - \frac{\pi \log^{2} 4}{4} + \frac{\pi^{3}}{2} + \pi^{2}\right) \\ &=\frac{\pi }{96 \sqrt{2}} \left(5 \pi^{2} + 12 \pi -12 \pi \log 2 + 24 \log 2 - 3 \log^{2} 4 - 24 \right) \\ &\approx 1.1057733283 \end{align}

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# Soft-Dielectron Excess in Proton-Proton Collisions at s=13  TeV

@article{Collaboration2021SoftDielectronEI,
title={Soft-Dielectron Excess in Proton-Proton Collisions at
s=13  TeV},
author={Alice Collaboration},
journal={Physical Review Letters},
year={2021}
}
A measurement of dielectron production in proton–proton (pp) collisions at √ s = 13 TeV, recorded with the ALICE detector at the CERN LHC, is presented in this Letter. The data set was recorded with a reduced magnetic solenoid field. This enables the investigation of a kinematic domain at low dielectron invariant mass mee and pair transverse momentum pT,ee that was previously inaccessible at the LHC. The cross section for dielectron production is studied as a function of mee, pT,ee, and event…
2 Citations

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Soft-Dielectron Excess in Proton-Proton Collisions at sqrt[s]=13  TeV.
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A measurement of dielectron production in proton-proton (pp) collisions at sqrt[s]=13  TeV, recorded with the ALICE detector at the CERN LHC, enables the investigation of a kinematic domain at low dielectrons (ee) invariant mass m_{ee} and pair transverse momentum p_{T,ee} that was previously inaccessible at the LHC.
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We have measured the cross section for ${\ensuremath{\mu}}^{+}{\ensuremath{\mu}}^{\ensuremath{-}}$ production in the reactions $({\ensuremath{\pi}}^{+}, Low transverse momentum photon production in proton-nucleus collisions at 18 GeV/c. • Tincknell • Physics Physical review. C, Nuclear physics • 1996 The measured photon spectra are compared with the expected distribution of photons from {pi}{sup 0} decays and hadronic bremsstrahlung in order to search for new sources ofsoft photons at low transverse momentum and indicate no anomalous sources of soft photons in either {ital p}-Be or {italp}-W interactions at these energies. Hadronic Production of Massive Muon Pairs: Dependence on Incident Particle Type and on Target Nucleus. • Physics • 1977 Data from a study of muon pairs produced in hadron-nucleus collisions are compared with the Drell-Yan model. Comparison of dimuon production by${\ensuremath{\pi}}^{+}\$ and
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## Version 1.0.1

Recent Changes

This function generates uniformly distributed pseudo-random numbers. Random numbers are generated in the ranges $0<\xi <1,-1<\eta <1$ and random integers in $1\le k\le N$ where $N$ is speciﬁed by the user.
A multiplicative congruent method is used where a 31 bit generator word $g$ is maintained. On each call to the subroutine ${g}_{n}+1$ is updated to ${7}^{5}{g}_{n}mod\left({2}^{31}-1\right)$; the initial value of $g$ is 1. Depending upon the type of random number required the following are computed $\xi ={g}_{n+1}∕\left({2}^{31}-1\right)$; $\eta =2\xi -1$ or $k=int.part\left\{\xi N\right\}+1$.


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# Introduction

Bayesian optimisation (BO) is a sequential sampling technique for global optimisation. It has two appealing features: ability to handle complex objective functions and sample efficiency. The basic machinery of BO consists of:

• Gaussian process surrogate models for estimating objective functions, and
• Acquisition function for prescribing the next sampling location.

These two components dynamically interact with each other to achieve sample efficient global optimisation. Consequently, BO is particularly useful when function evaluations are expensive and the gradient information is not available.

This sequential “sampling technique” is different from Monte Carlo sampling methods, which are about how to sample from specific distributions. Here, some stochastic process is given, and we are assumed to have access to a sampler for that stochastic process no matter what input we use. The problem is what input from the index set to use. Imagine a noisy objective function, which is modelled as a stochastic process. In other words, the problem is where to evaluate the objective function.

## Global optimisation

Recall from the optimisation notes, global optimisation can be formulated as solving the following problem $\min_{x\in\mathcal{D}} f(x)$ where $$f$$ is the objective function and $$\mathcal{D}$$ is the choice set in which we try to find a minimiser $$x^*$$ such that $$f(x^*)\leq f(x)$$ for all $$x\in\mathcal{D}$$.

In these notes, we use minimisation to formulate optimisation problems.

The fundamental challenge arises from our limited knowledge of the shape of $$f$$. Unless $$f$$ is fairly simple such as low-degree polynomials and sinusoids, it is usually hard to get a global picture from the expression of $$f$$ even in 1-D cases.

Most of the times, therefore, we evaluate $$f$$ at a finite number of points and interpolate in-betweens. Using the same test function as in the optimisation notes, for example, the following graph is produced by first evaluating at 101 evenly spaced points from 0.26 to 3.6 and then connecting dots. The global minimum is approximately $$-1.025$$ at $$x=2.5$$.

With this graph, we can easily pick the best one (68th point) among 101 points as an approximate minimiser $$\hat{x}^*=2.498$$ yielding $$f(\hat{x}^*)=-1.025$$, which is essentially the global minimum.

Comparing to this simple yet effective approach, we notice something naïve in gradient descent; it does not utilise information that can be extracted from visited/evaluated points. Instead, it ignores the history but uses only local information (i.e, gradient) and blindly follows the direction at each step. As a result, if using random initial points to find the global minimum, we must repeat the process at least five times (very lucky) in the above case

## Surrogate models

Seeing those evaluated points as data, don’t you think the above interpolation is akin to statistical regression? In regression, we collect data $$(X,\mathbf{y})$$ and estimate the underlying data-generating process by fitting some models to the data. Above, using the evenly-spaced points $$X$$ and the corresponding objective values $$\mathbf{y}=f(X)$$, we essentially estimate the underlying objective function by connecting dots.

When we do regression, our models are more sophisticated than connecting data points (e.g., linear, polynomial, neural network, etc.). So, why not using statistical models for optimisation as well? It is reasonable to expect that suitable models could extract more information from evaluated points about the objective function and, in particular, promising regions of minimisers, leading to more efficient search. These models are commonly called surrogates.

Connecting dots can be seen as a special case of spline or local regression.

# Gaussian process

## Definition

Gaussian process (GP) is a collection of random variables (i.e., stochastic process) such that any finite sub-collection follows a multivariate normal distribution. GP is the most popular surrogate models in Bayesian optimisation.

Remember the Poisson process is a stochastic process specified by rate $$\lambda$$. For a counting process, we use a set of nonnegative real numbers $$\mathbb{R}_+$$ as an index set, with which for some epoch $$t\in\mathbb{R}_+$$ of interest, we have a counting random variable $$N(t)\sim\text{pois}(\lambda t)$$.

You may also find GP classified as Bayesian nonparametric in statistics and machine learning.

To satisfy the above condition, a GP is specified by a mean function $$m$$ and covariance function $$k$$. We write $$Y\sim\mathcal{GP}(m,k)$$ to denote that a stochastic process $$Y$$ follows a GP specified by $$m$$ and $$k$$. As the names suggest, the roles of $$m$$ and $$k$$ are as follows. For inputs $$x$$ and $$x'$$ of the index set, \begin{align} m(x) &= \mathbb{E}[Y(x)]\\ k(x,x') &= \mathbb{E}[(Y(x)-m(x))(Y(x')-m(x'))]. \end{align}

Intuitively, $$k(x,x')$$ measures “distance” between $$x$$ and $$x'$$ reflecting similarity between $$Y(x)$$ and $$Y(x')$$. If two inputs $$x$$ and $$x'$$ are close, the corresponding random variables $$Y(x)$$ and $$Y(x')$$ should be strongly correlated.

For concreteness, suppose the index set is the real line $$\mathbb{R}$$. Then, if we are interested in a collection of two random variables at $$\mathbf{x}=(x_1,x_2)$$, then a random vector $Y = \begin{bmatrix} Y(x_1)\\ Y(x_2) \end{bmatrix}$ is jointly distributed as $$\mathrm{N}(\boldsymbol{\mu},\Sigma)$$ where $\boldsymbol{\mu} = \begin{bmatrix} m(x_1)\\ m(x_2) \end{bmatrix} \quad\text{and}\quad \Sigma = \begin{bmatrix} k(x_1,x_1) & k(x_1,x_2)\\ k(x_2,x_1) & k(x_2,x_2) \end{bmatrix}.$

Remember the property of Normal distribution. To have the joint PDF, in practice, we define $$k$$ such that $$\Sigma$$ is positive definite.

### Demo

Let’s see what GP realisation looks like, using one of the most basic configurations: $$m(x) = 0$$ and $$k(x,x') = \exp(-\|x-x'\|^2)$$.

k <- function(x1,x2) exp(-(x1-x2)^2)

xx <- seq(.26, 3.6, length.out=101)
K <- outer(xx, xx, k) # 101×101 covariance matrix
Y <- mvrnorm(3, rep(0,m), K)
matplot(xx, t(Y), type="o", lwd=1.2, pch=20, cex=0.5, xlab="x", ylab="y")
title("Random functions from GP")

As clearly indicated in the above code, we drew three length-101 random vectors from the multivariate normal and plotted each set of 101 realisations separately. Since the points in each series are connected, they look like (random) functions.

## GP regression

Suppose that a normal random vector is partitioned into $$Y_1$$ and $$Y_2$$ as follows: $\begin{bmatrix} Y_1\\ Y_2 \end{bmatrix} \sim \mathrm{N}(0,\Sigma),$ where $\Sigma = \begin{bmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} .$ Then, the conditional distribution of $$Y_1$$ given $$Y_2=\mathbf{y_2}$$ is also normal $Y_1|\mathbf{y_2} \sim \mathrm{N}(\bar{\boldsymbol{\mu}},\bar{\Sigma}),$ where $\bar{\boldsymbol{\mu}} = \Sigma_{12}\Sigma_{22}^{-1}\mathbf{y_2} \quad\text{and}\quad \bar{\Sigma} = \Sigma_{11} - \Sigma_{12} \Sigma_{22}^{-1} \Sigma_{21}.$

Based on the above property of multivariate normal distribution, in GP regression, we predict the distribution of $$Y(x)$$ at location $$x$$ using $$n$$ data points $$D_n = (X_n,\mathbf{y}_n) = \{(x_i,y_i)\}_{i=1}^n$$. That is, first, the joint distribution of $$Y(x)$$ and $$Y_n$$ is $\begin{bmatrix} Y(x)\\ Y_n \end{bmatrix} \sim \mathrm{N}(0,K),$ where $K = \begin{bmatrix} k(x,x) & \mathbf{k}(x,X_n)\\ \mathbf{k}(X_n,x) & \mathbf{k}(X_n,X_n) \end{bmatrix},$ and $$\mathbf{k}$$ denotes a matrix of appropriate size whose elements are obtained by applying $$k$$ separately for each element.

$$\mathbf{k}(x,X_n)$$ is in $$1\times n$$, $$\mathbf{k}(X_n,x)$$ is in $$n\times 1$$, and $$\mathbf{k}(X_n,X_n)$$ is in $$n\times n$$.

Then, the conditional distribution of $$Y(x)$$ given $$Y_n=\mathbf{y}_n$$ sampled at $$X_n$$ is $Y(x)|D_n \sim \mathrm{N}(\hat{\mu},\hat{K}),$ where \begin{align} \hat{\mu}(x) &= \mathbf{k}(x,X_n)\mathbf{k}(X_n,X_n)^{-1}\mathbf{y}_n\\ \hat{K}(x) &= k(x,x) - \mathbf{k}(x,X_n) \mathbf{k}(X_n,X_n)^{-1} \mathbf{k}(X_n,x). \end{align}

Observe that both $$\hat{\mu}$$ and $$\hat{K}$$ are functions of $$x$$. So, we may treat $$\hat{\mu}$$ as prediction of $$Y$$ at $$x$$ (i.e., a regression function) and $$\hat{K}$$ as prediction variance.

### Demo

Suppose we have evaluated the above objective function at evenly spaced 15 points from 0.26 to 3.5. The following graph traces $$\hat{\mu}$$ computed conditional on these data points.

Under this specific setting, with only 15 samples, we can estimate the function remarkably well.

# Acquisition function

## Efficient sequential sampling

BO is often used when function evaluations are expensive; otherwise, we could simply search over a fine grid just like the above case at the beginning. So, in these cases, we want to locate a good minimiser within an evaluation budget or simply as quickly as possible.

A story may be oil drilling. Before full production takes place, a company has to discover a good spot through test drilling. Suppose drilling a single hole and examining potential oil content under the ground cost some million dollars. So, the company wants to identify a good location conducting as few tests as possible. We may model the problem as follows:

• Index set. A testing site as a subset of $$\mathbb{R}^2$$.
• Objective function. A probability that a location $$(x,y)$$ contains a profitable amount of oil.
• Evaluation of the objective. Drilling.

To understand the task of efficient sampling, let’s sequentially plot both estimated mean function and 2-standard deviation band for increasing sample sizes.


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# Sensor

## 1. Introduction

Presentation of the basic for force measurement with strain gauges.

A strain gauge is a device used to measure strain on an object. The most common type of strain gauge consists of an insulating flexible backing which supports a metallic foil pattern. The gauge is attached to the object by a suitable adhesive. As the object is deformed, the foil is deformed, causing its electrical resistance to change. This resistance change, usually measured using a Wheatstone bridge, is related to the strain by the quantity known as the gauge factor.

CAD representation of a strain gauge

Representation of a strain gauge

Strain gauge

## 2. Running the case

### 2.1. Command line

The command line to run this case is

``!mpirun -np 4 feelpp_toolbox_solid --case "github:{repo:toolbox,path:examples/modules/csm/examples/sensor}"``
 The report of the execution of the command above is available here.

### 2.2. Python interface

``````from feelpp import *
from feelpp.toolboxes.core import *
from feelpp.toolboxes.solid import *

# create the application
# create a feelppdb subdirectory where the results are stored
app = Environment(['feelpp_toolbox_solid'], opts= toolboxes_options("solid"),config=localRepository(""))``````

``````sensorcfg=feelpp.download("github:{repo:toolbox,path:examples/modules/csm/examples/sensor/}", worldComm=app.worldCommPtr())[0] (1)
sensorcfg+='/sensor.cfg' (2)
if os.path.exists(sensorcfg): (3)
app.setConfigFile(sensorcfg) (4)
s = solid(dim=3) (5)
# get displacement and von-mises measures from the model
ok,meas=simulate(s) (6)
if ok:
# export in paraview format
s.exportResults() (7)``````
 1 download the configuration file 2 add the configuration file to the study 3 check if the configuration file exists 4 set the configuration file 5 create the solid mechanics model 6 simulate the model and get the resulting measures from the PostProcessing section 7 export the results
Results
```[ Starting Feel++ ] application feelpp_toolbox_solid version 0.1 date 2022-Nov-07
. feelpp_toolbox_solid files are stored in /scratch/jupyter/feelppdb/np_1
.. logfiles :/scratch/jupyter/feelppdb/np_1/logs
solid(3,1)
============================================================

time simulation: 0.05s/0.61s with step: 0.05
============================================================

0 solid SNES Function norm 4.324537e+02
1 solid SNES Function norm 3.894843e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
2 solid SNES Function norm 3.511776e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
3 solid SNES Function norm 3.103868e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
4 solid SNES Function norm 2.657841e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
5 solid SNES Function norm 2.165992e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
6 solid SNES Function norm 1.624799e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
7 solid SNES Function norm 1.621619e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
8 solid SNES Function norm 6.411767e-03
Linear solve converged due to CONVERGED_RTOL iterations 11
9 solid SNES Function norm 6.553165e-06
Linear solve converged due to CONVERGED_RTOL iterations 10
10 solid SNES Function norm 9.171302e-10
Linear solve converged due to CONVERGED_RTOL iterations 12
============================================================

time simulation: 0.1s/0.61s with step: 0.05
============================================================

0 solid SNES Function norm 4.056531e+02
1 solid SNES Function norm 3.653160e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
2 solid SNES Function norm 3.265140e+02
Linear solve converged due to CONVERGED_RTOL iterations 9
3 solid SNES Function norm 2.843344e+02
...

time simulation: 0.44999999999999996s/0.61s with step: 0.05```

## 3. Data files

The case data files are available in Github here

## 4. Model/Geometry

The first step is to create the model of the object, which we can simply do in the Creo Parametric program. With this program was the fastest and easiest way to create the model.

The finished geometry (Creo) and the meshed model (Gmsh):

CAD representation of a strain gauge

Mesh representation of a strain gauge

## 5. Materials and boundary conditions

### 5.1. Materials

The beam is made of steel, so we need to use the following data:

• $E = 205 GPa -> 205e3 N/mm^2$

• $nu = 0.3$

• $rho = 7850 kg/m^3 -> 7850e-9 kg/mm^3$

### 5.2. Boundary conditions

The object is fixed in the middle. Holes have opposite sinusodal forces.

CAD representation of a strain gauge

Mesh representation of a strain gauge

## 6. Results

 The result were run in time (half whole period, but on the pictures can be seen only a quarter period).
``````import pandas as pd
df=pd.DataFrame(meas)

# prepare for plotting
import plotly.graph_objects as go``````
Results
```Paraview files are in /scratch/jupyter/feelppdb/np_1/np_1/solid.exports
Statistics_disp_max  Statistics_disp_mean_0  Statistics_disp_mean_1  \
0             0.000000                0.000000                0.000000
1             0.423963                0.000030                0.000024
2             0.823267                0.000058                0.000047
3             1.147965                0.000077                0.000067
4             1.425199                0.000096                0.000084

Statistics_disp_mean_2  Statistics_disp_min  Statistics_von-mises_max  \
0                0.000000             0.000000                  0.000000
1               -0.000017            -0.424049                986.291965
2               -0.000033            -0.823410               1926.227644
3               -0.000046            -1.148120               2715.268042
4               -0.000058            -1.425366               3380.035626

Statistics_von-mises_mean  Statistics_von-mises_min  time
0                   0.000000                  0.000000  0.00
1                  76.661865                  0.000664  0.05
2                 149.031439                  0.003552  0.10
3                 208.581334                  0.002816  0.15
4                 259.019382                  0.004327  0.20```

### 6.1. Displacement

displacement at $t=0.1s$

displacement at $t=0.2s$

displacement at $t=0.3s$

displacement at $t=0.4s$

displacement at $t=0.5s$

Plot the max of the displacement over time
``````fig = go.Figure()
fig.update_layout(legend_title_text='Max displacement over time')
fig.show()``````
Results

### 6.2. Von-Mises Criterions

Von-Mises criterion at $t=0.1s$

Von-Mises criterion at $t=0.2s$

Von-Mises criterion at $t=0.3s$

Von-Mises criterion at $t=0.4s$

Von-Mises criterion at $t=0.5s$

On the displacement and stress diagrams, it can be clearly seen that this object is perfectly suited as a sensor.

Plot the max of the von-mises criterion over time
``````fig = go.Figure()


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45 views

If $a, b$ and $c$ are positive integers such that $ab=432, bc=96$ and $c<9,$ then the smallest possible value of $a+b+c$ is

1. $56$
2. $59$
3. $49$
4. $46$

Given that, $a,b,$ and $c$ are positive integers. That means $a,b,c > 0 \;(\text{or})\; a,b,c \in \mathbb{Z}^{+}$

And, $ab = 432, bc = 96, c<9.$
$$\require{cancel}\begin{array}{|c|c|c|} \hline b \cdot c = 96 & a \cdot b = 432 & a + b + c \\\hline 96 \cdot 1 & - \cdot \cancel{96} & \\ 48 \cdot 2 & 9 \cdot 48 & 9 + 48 + 2 = 56 \\ 32 \cdot 3 & – \cdot \cancel{32} & \\ 24 \cdot 4 & 18 \cdot 24 & 18 + 24 + 4 = 46 \\ 16 \cdot 6 & 27 \cdot 16 & 27 + 16 + 6 = 49 \\ 12 \cdot 8 & 36 \cdot 12 & 36 + 12 + 8 = 56 \\\hline \end{array}$$
$\therefore$ The smallest possible value of $a+b+c = 46.$

Correct Answer$: \text{D}$
by
7.7k points 3 8 30

1 vote
1
62 views
1 vote
2
87 views
1 vote


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# Tag Info

If you realize that the second term on the right-hand side of Eq. $(1)$ is only non-zero for negative frequencies, you note that you can remove that term using a phase splitter, i.e., a complex-valued filter with zero magnitude at negative frequencies and a constant magnitude (of value $2$) at positive frequencies. Such a system can be realized using a ...


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# pulsesep

Separation between bilevel waveform pulses

## Syntax

S = pulsesep(x)
S = pulsesep(x,Fs)
S = pulsesep(x,t)
[s,initcross] = pulsesep(___)
[s,initcross,finalcross] = pulsesep(___)
[s,initcross,finalcross,nextcross] = pulsesep(___)
[s,initcross,finalcross,nextcross,midlev] = pulsesep(___)
[s,initcross,finalcross,nextcross,midlev] = pulsesep(___,Name,Value)
pulsesep(___)

## Description

S = pulsesep(x) returns the differences between the mid-reference level instants of the final negative-going transitions of every positive-polarity pulse and the next positive-going transition. To determine the transitions that compose each pulse, the pulsesep function estimates the state levels of the input waveform by a histogram method. The function identifies all regions that cross the upper-state boundary of the low state and the lower-state boundary of the high state.
S = pulsesep(x,Fs) specifies the sample rate Fs.

example

S = pulsesep(x,t) specifies the sampling instants t.
[s,initcross] = pulsesep(___) returns the mid-reference level instants initcross of the first positive-polarity transitions. You can specify an input combination from any of the previous syntaxes.
[s,initcross,finalcross] = pulsesep(___) returns the mid-reference level instants finalcross of the final transition of each pulse.

example

[s,initcross,finalcross,nextcross] = pulsesep(___) returns the mid-reference level instants nextcross of the next detected transition after each pulse.
[s,initcross,finalcross,nextcross,midlev] = pulsesep(___) returns the mid-reference level midlev.
[s,initcross,finalcross,nextcross,midlev] = pulsesep(___,Name,Value) returns the pulse separations with additional options specified by one or more name-value arguments.

example

pulsesep(___) plots the signal and darkens the regions between each pulse where pulse separation is computed. The function marks the location of the mid crossings and their associated reference level. The function also plots the state levels and their associated lower and upper boundaries. You can adjust the boundaries using the 'Tolerance' name-value argument.

## Examples

collapse all

Compute the pulse separation in a bilevel waveform with two positive-polarity transitions. The sample rate is 4 MHz.

load('pulseex.mat','x','t') s = pulsesep(x,t)
s = 3.5014e-06

Plot the waveform and annotate the pulse separation.

pulsesep(x,t);

Determine the mid-reference level instants that define the pulse separation for a bilevel waveform.

load('pulseex.mat','x','t') [~,~,finalcross,nextcross] = pulsesep(x,t)
finalcross = 4.6256e-06
nextcross = 8.1270e-06

Return the pulse separation. Annotate the mid-reference level instants on a plot of the data.

pulsesep(x,t)

ans = 3.5014e-06

## Input Arguments

collapse all

Bilevel waveform, specified as a real-valued vector. If the waveform does not contain at least two transitions, the function outputs an empty matrix. The first time instant in x corresponds to t=0.

Sample rate, specified as a positive real scalar in hertz.

Sample instants, specified as a vector. The length of t must equal the length of the input bilevel waveform x.

### Name-Value Arguments

Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.

Example: p = pulsesep(x,t,Polarity="negative")

Before R2021a, use commas to separate each name and value, and enclose Name in quotes.

Example: p = pulsesep(x,t,'Polarity',"negative")

Mid-reference level as a percentage of the waveform amplitude, specified as scalar. For more information, see Mid-Reference Level.

Pulse polarity, specified as "positive" or "negative". If you specify "positive", the function looks for pulses with positive-going (positive polarity) initial transitions. If you specify "negative", the function looks for pulses with negative-going (negative polarity) initial transitions. For more information, see Pulse Polarity.

Low- and high-state levels, specified as a 1-by-2 real-valued vector. The first element is the low-state level and the second element is the high-state level. If you do not specify 'StateLevels', the function estimates the state levels from the input waveform using the histogram method.

Tolerance levels (lower- and upper-state boundaries), specified as a scalar and expressed as a percentage. The low-state and high-state boundaries are expressed as the state level plus or minus a multiple of the difference between the state levels. For more information, see State-Level Tolerances.

## Output Arguments

collapse all

Pulse separations in seconds, returned as a vector. The pulse separation is defined as the time between the mid-reference level instants of the final transition of one pulse and the initial transition of the next pulse. For more information, see Pulse Separation.

Note

Because pulsesep uses interpolation to determine the mid-reference level instants, s may contain values that do not correspond to sampling instants of the bilevel waveform x.

Mid-reference level instants of the initial transition, returned as a vector.

Mid-reference level instants of the final transition, returned as a vector.

Mid-reference level instants of the initial transition after the final transition of the preceding pulse, returned as a vector.

Waveform value that corresponds to the mid-reference level, returned as a scalar.

collapse all

### Mid-Reference Level

The mid-reference level in a bilevel waveform with low-state level S1 and high-state level S2 is

${S}_{1}+\frac{1}{2}\left({S}_{2}-{S}_{1}\right)$

### Mid-Reference Level Instant

• Let y50% denote the mid-reference level.

• Let t50%- and t50%+ denote the two consecutive sampling instants corresponding to the waveform values nearest in value to y50%.

• Let y50%- and y50%+ denote the waveform values at t50%- and t50%+.

The mid-reference level instant is

${t}_{50%}={t}_{50%}+\left(\frac{{t}_{50{%}_{+}}-{t}_{50{%}_{-}}}{{y}_{50{%}_{+}}-{y}_{50{%}_{-}}}\right)\left({y}_{50{%}_{+}}-{y}_{50{%}_{-}}\right)$

### Pulse Polarity

If the pulse has an initial positive-going transition, the pulse has positive polarity. This figure shows a positive-polarity pulse.

Equivalently, a positive-polarity (positive-going) pulse has a terminating state more positive than the originating state.

If the pulse has an initial negative-going transition, the pulse has negative polarity. This figure shows a negative-polarity pulse.

Equivalently, a negative-polarity (negative-going) pulse has a originating state more positive than the terminating state.

### State-Level Tolerances

You can specify lower- and upper-state boundaries for each state level. Define the boundaries as the state level plus or minus a scalar multiple of the difference between the high state and the low state. To provide a useful tolerance region, specify the scalar as a small number such as 2/100 or 3/100. In general, the region for the low state is defined as

where is the low-state level and is the high-state level. Replace the first term in the equation with to obtain the tolerance region for the high state.

This figure shows lower and upper 5% state boundaries (tolerance regions) for a positive-polarity bilevel waveform. The thick dashed lines indicate the estimated state levels.

### Pulse Separation

Pulse separation is the time difference between the mid-reference level instant of the final transition of one pulse and the mid-reference level instant of the initial transition of the next pulse. This figure illustrates pulse separation.

## References

[1] IEEE® Standard on Transitions, Pulses, and Related Waveforms, IEEE Standard 181, 2003.

## Version History

Introduced in R2012a


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# Definition of Abscissas. Meaning of Abscissas. Synonyms of Abscissas

Here you will find one or more explanations in English for the word Abscissas. Also in the bottom left of the page several parts of wikipedia pages related to the word Abscissas and, of course, Abscissas synonyms and on the right images related to the word Abscissas.

## Definition of Abscissas

Abscissas
Abscissa Ab*scis"sa, n.; E. pl. Abscissas, L. pl. Absciss[ae]. [L., fem. of abscissus, p. p. of absindere to cut of. See Abscind.] (Geom.) One of the elements of reference by which a point, as of a curve, is referred to a system of fixed rectilineal co["o]rdinate axes. Note: When referred to two intersecting axes, one of them called the axis of abscissas, or of X, and the other the axis of ordinates, or of Y, the abscissa of the point is the distance cut off from the axis of X by a line drawn through it and parallel to the axis of Y. When a point in space is referred to three axes having a common intersection, the abscissa may be the distance measured parallel to either of them, from the point to the plane of the other two axes. Abscissas and ordinates taken together are called co["o]rdinates. -- OX or PY is the abscissa of the point P of the curve, OY or PX its ordinate, the intersecting lines OX and OY being the axes of abscissas and ordinates respectively, and the point O their origin.
Abscissa
Abscissa Ab*scis"sa, n.; E. pl. Abscissas, L. pl. Absciss[ae]. [L., fem. of abscissus, p. p. of absindere to cut of. See Abscind.] (Geom.) One of the elements of reference by which a point, as of a curve, is referred to a system of fixed rectilineal co["o]rdinate axes. Note: When referred to two intersecting axes, one of them called the axis of abscissas, or of X, and the other the axis of ordinates, or of Y, the abscissa of the point is the distance cut off from the axis of X by a line drawn through it and parallel to the axis of Y. When a point in space is referred to three axes having a common intersection, the abscissa may be the distance measured parallel to either of them, from the point to the plane of the other two axes. Abscissas and ordinates taken together are called co["o]rdinates. -- OX or PY is the abscissa of the point P of the curve, OY or PX its ordinate, the intersecting lines OX and OY being the axes of abscissas and ordinates respectively, and the point O their origin.

## Meaning of Abscissas from wikipedia

- two-dimensional graph.[citation needed] In mathematics, the abscissa (/æbˈsɪs.ə/; plural abscissae or abscissas) and the ordinate are respectively the first and...
- In mathematics, the spectral abscissa of a matrix or a bounded linear operator is the supremum among the real part of the elements in its spectrum, sometimes...
- _{n}=\log n} . It is possible to consider other abscissas of convergence for a Dirichlet series. The abscissa of bounded convergence σ b {\displaystyle \sigma...
- Typically, a Tauc plot shows the quantity hν (the photon energy) on the abscissa and the quantity (αhν)1/2 on the ordinate, where α is the absorption coefficient...
- distances measured along the diameter from the point of tangency are the abscissas, and the segments parallel to the tangent and intercepted between the...
- _{\mathrm {n} }} axis. The magnitude of the prin****l stresses are the abscissas of the points C {\displaystyle C} and E {\displaystyle E} (Figure 6) where...
- given its coordinates. The first and second coordinates are called the abscissa and the ordinate of P, respectively; and the point where the axes meet...
- he proves that if the abscissas of points are in geometric proportion, then the areas between a hyperbola and the abscissas are in arithmetic proportion...
- weights and abscissas, for n=2 through n=64, with Mathematica source code. Mathematica source code distributed under the GNU LGPL for abscissas and weights...
- were plotted as ordinates (i.e. vertically) and the atomic volumes as abscissas (i.e. horizontally)—the curve obtained a series of maximums and minimums—the...


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## Files in this item

FilesDescriptionFormat

application/pdf

9717290.pdf (6MB)
(no description provided)PDF

## Description

 Title: Low Energy Electronic Raman Scattering in Cuprate High Temperature Superconductors Author(s): Kang, Moonsoo Doctoral Committee Chair(s): Klein, M.V. Department / Program: Physics Discipline: Physics Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Physics, Condensed Matter Abstract: The temperature dependence of the 2$\Delta$-peak in overdoped Tl-2201 implies that the superconducting gap opens very rapidly below the critical temperature. The measured peak intensity as a function of temperature, however, shows an anomalous behavior. A magnetic field is observed to affect Raman spectra in way similar to the way temperature does. Nevertheless, the peak intensity change as a function of magnetic field is different from that of temperature. The upper critical fields were extracted by analyzing the field dependence of the 2$\Delta$-peak intensity. The upper critical fields obtained from Raman scattering in magnetic field show that superconducting order is present above the irreversibility field obtained from magnetization measurements. Issue Date: 1997 Type: Text Language: English Description: 131 p.Thesis (Ph.D.)--University of Illinois at Urbana-Champaign, 1997. URI: http://hdl.handle.net/2142/80641 Other Identifier(s): (MiAaPQ)AAI9717290 Date Available in IDEALS: 2015-09-25 Date Deposited: 1997


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# The inequation $\lfloor x\rfloor^2-\lfloor x\rfloor-6>0$

The question is to find the solutions to this inequation $$\lfloor x\rfloor^2 - \lfloor x\rfloor - 6 > 0$$

On factorising I got $$(\lfloor x\rfloor-3)(\lfloor x\rfloor+2)>0$$ so $\lfloor x\rfloor>3$ or $\lfloor x\rfloor>-2$ and so on

But in the text book the solution goes like this

$\lfloor x\rfloor>3$ so $x\ge 4$ and $\lfloor x\rfloor<-2$ so $x <-2$ therefore domain is...

Notice that the textbook has reversed the second inequality and it turns out to be correct as I checked it in wolfram alpha.

Can anyone please explain why the book reversed $\lfloor x\rfloor>-2$ into $\lfloor x\rfloor<-2$?

"On factorising I got $(\lfloor x\rfloor-3)(\lfloor x\rfloor +2)>0$, so $\lfloor x\rfloor>3$ or $\lfloor x\rfloor >-2$ and so on"
The bold "so" is wrong: for $a\le b\in\Bbb R$,
$$(t-a)(t-b)>0\iff t< a\vee t>b$$
You applied it incorrectly to the case $a=-2$, $b=3$
• @user453135 No, it is not the same thing. For instance, when $a<b$, all $t\in\Bbb R$ verify that condition. – user228113 Jul 26 '17 at 6:48


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## One for all and all for one:

### Linear regression from the mass of individual bones to assess human skeletal mass completeness

Objectives: Complete and accurate human skeletal inventory is seldom possible in archaeological and forensic cases involving severe fragmentation. In such cases, skeletal mass comparisons with published references may be used as an alternative to assess skeletal completeness but they are too general for a case-by-case routine analysis. The objective is to solve this issue by creating linear regression equations to estimate the total mass of a skeleton based on the mass of isolated bones.

Methodology: Total skeleton mass and individual mass of the clavicle, humerus, femur, patella, carpal, metacarpal, tarsal and metatarsal bones were recorded in a sample of 60 skeletons from the 21st Century Identified Skeletal Collection (University of Coimbra). The sample included 32 females and 28 males with ages ranging from 31 to 96 years old (mean = 76.4; sd = 14.8).

Results: The mass of isolated bones could be used to predict the approximate total mass of the skeleton. The femur, humerus and second metacarpal were the best predictors of total skeletal mass with root mean squared errors ranging from 292.9 to 346.1 gm.

Dicussion: Although relatively successful, the non-normal distribution of the sample in terms of mass may have reduced the predictive power of the equations. Mass regression can be more adequately applied to forensic than to archaeological remains as the latter are generally more affected by post-depositional mass loss. The impact for bioanthropology, especially forensic anthropology, is important since this method may improve methodology to calculate the completeness of the skeleton or the minimum number of individuals.

Keywords: bioarchaeology; forensic anthropology; bone mass; scattered remains; funerary practice

American Journal of Physical Anthropology. July 2016, Volume 160, Issue 3, pp 427–432

#### Data exploration through heatmap

All variables are in columns representing type of bone, ex. CLV.L = clavicle (left side). All the individuals of the sample are represented as rows and organized by clusters of mass values. Blank (white) spaces are non available values that for some reason were not measured.

As a pre-analysis procedure, significant bilateral asymmetries in each bone were investigated. This was carried out to determine if both sides of the skeleton had to be treated separately during statistical analysis. If differences from both sides were not substantial, then it would be possible to overcome the absence of a bone by replacing it by its antimere, if available. Bilateral asymmetry was investigated through a Wilcoxon signed ranks test. Also, relative directional asymmetry (%DA), which has been often used previously (Auerbach and Ruff, 2006), was calculated for each individual so that a better notion of asymmetry in a case by case basis could be attained.

The formula used was the following:

$$\%DA=\frac{M_R-M_L}{\frac{M_R+M_L}{2}}*100$$

where every M is an individual bone mass, while L and R are left and right side, respectively

References

Auerbach BM, and Ruff CB. 2006. Limb bone bilateral asymmetry: variability and commonality among modern humans. Journal of Human Evolution 50:203-218.

Current Value:

Total skeleton mass, based on your input value is:


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Solo MP

This MP, as well as all other MPs in CS 225, are to be completed without a partner.

You are welcome to get help on the MP from course staff, via open lab hours, or Piazza!

## Goals and Overview

In this MP you will:

• Implement the disjoint sets data structure.
• Create a program to generate random mazes.
• Understand graphs, graph distances, and graph traversal.
• Represent a maze and its solution on a PNG.

## Checking Out the Code

Get the code in the usual way.

The mp7 directory will contain sample output including all 4 possible 2x2 mazes and one 50x50 maze. We encourage you to test your code by writing your own main.cpp that uses your classes in different ways.

## Assignment Requirements

These are strict requirements that apply to both parts of the MP. Failure to follow these requirements may result in a failing grade on the MP.

• You are required to comment the MP as per the commenting standard described by the Coding Style Policy.
• You must name all files, public functions, public member variables (if any exist), and executables exactly as we specify in this document.
• Your code must produce the exact output that we specify: nothing more, nothing less. Output includes standard and error output and files such as Images.
• Your code must compile on the EWS machines using clang++. Being able to compile on a different machine is not sufficient.
• Your code must be submitted correctly by the due date and time. Late work is not accepted.
• Your code must not have any memory errors or leaks for full credit. ASAN tests will be performed separately from the functionality tests.
• Your public function signatures must match ours exactly for full credit. If using different signatures prevents compilation, you will receive a zero. Tests for const-correctness may be performed separately from the other tests (if applicable).

## Assignment Description

You will be implementing a Disjoint set data structure and then implementing a random maze generator and solver. The assignment is broken up into the two following parts:

• MP 7.1 — The DisjointSets data structure.
• MP 7.2 — The SquareMaze random maze generator and solver.

As usual, we recommend implementing, compiling, and testing the functions in MP 7.1 before starting MP 7.2. Submission information is provided for each part in the respective sections below.

## MP 7.1: The DisjointSets data structure

The DisjointSets class should be declared and defined in dsets.h and dsets.cpp, respectively. Each DisjointSets object will represent a family of disjoint sets, where each element has an integer index. It should be implemented with the optimizations discussed in lecture, as up-trees stored in a single vector of ints. Specifically, use path compression and union-by-size. Each element of the vector represents a node. (Note that this means that the elements in our universe are indexed starting at 0.) A nonnegative number is the index of the parent of the current node; a negative number in a root node is the negative of the set size.

Note that the default compiler-supplied Big Three will work flawlessly because the only member data is a vector<int> and this vector should initially be empty.

### The addelements function

See the Doxygen for this function.

### The find function

See the Doxygen for this function.

### The setunion function

See the Doxygen for this function.

### Testing MP 7.1

The following command can be used to compile the DisjointSets test executable:

make testdsets

The following command can be used to run the test executable:

./testdsets

Provided Catch test cases are available as well by running:

make test
./test

### Grading Information — MP 7.1

The following files are used to grade MP7:

• dsets.cpp
• dsets.h

All other files including your testing files will not be used for grading.

## MP 7.2: The SquareMaze random maze generator and solver

The SquareMaze class should be declared and defined in maze.h and maze.cpp, respectively. Each SquareMaze object will represent a randomly-generated square maze and its solution. Note that by “square maze” we mean a maze in which each cell is a square; the maze itself need not be a square. As always, we recommend reading the whole specification before starting.

setWall and canTravel

You should triple check that setWall and canTravel function exactly according to spec, as an error in these functions will not be caught by making your own mazes but can cost you a majority of the points during grading.

See the Doxygen.

See the Doxygen.

See the Doxygen.

See the Doxygen.

See the Doxygen.

See the Doxygen.

### Testing MP 7.2

#### Square Maze Testing

Since your mazes will be randomly generated, we cannot provide you with any sample images to diff against. However, we have provided you with all four possible 2x2 mazes. If you have your program create and solve a 2x2 maze, the resulting image (with solution) should match one (and only one) of the provided images m0.png, m1.png, m2.png, and m3.png. We strongly suggest that you diff against these to make sure that you have formatted the output image correctly.

We provide some basic code to test the functionality of SquareMaze.

The following command can be used to compile the SquareMaze test executable:

make testsquaremaze

The following command can be used to run the test executable:

./testsquaremaze

You can compare the console output of your program with the expected by comparing it with the file soln_testsquaremaze.out.

#### Catch Test Suite

Additional unit tests similar to how we will grade your MP are provided for you in tests/test_part2.cpp. Once you have completed Part 2, you may uncomment the Part 2 tests and re-compile to include Part 2 tests:

// uncomment test_part2.cpp
make test
./test

### Runtime Concerns

You should strive for the best possible implementation. This MP can be implemented so that the given testsquaremaze.cpp runs in less than a quarter of a second on the EWS linux machines. To have a high probability of finishing within the time constraints of the grading script, make sure you can run the given testsquaremaze.cpp in under 3 seconds on an unencumbered machine. You can time MP7 by running the command time ./testsquaremaze.

### Grading Information — MP 7.2

• We will use canTravel to reconstruct your randomly generated maze in our own maze class, to check that it’s a tree and to compare your implementation to a correct implementation of this MP. This will require height*width*2 calls to canTravel. Therefore, it is very important that canTravel works, and works quickly (constant time). If it doesn’t work, you will lose a lot of points.
• We will use setWall to replace your maze with our own, for the purpose of testing all of your other functions independently of your createMaze.
setWall and canTravel

You should triple check that setWall and canTravel function exactly according to spec, as an error in these functions will not be caught by making your own mazes but can cost you a majority of the points during grading.

The following files are used to grade MP7:

• dsets.cpp
• dsets.h
• maze.cpp
• maze.h

All other files (including your main.cpp) will not be used for grading.


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# Definition of the number of microstates accessible to a system

by Sam_Goldberg
Tags: accessible, definition, microstates, number
 P: 46 Hi guys, I'm going through Reif's book on statistical mechanics and have a question on the parameter omega, the number of microstates accessible to a system. Let's imagine a big box, and divide it into two equal halves. Now place a gas with a macroscopic amount of molecules in the left half, and say that it has an energy between E and E + dE. There is an initial value here of omega characterized by the energy and the volume. Now remove the division in the box. Reif would say that since the volume is now increased, the number of accessible states of the system instantly increases; however, the probability of the system of being in most of them is zero, at least instantly after the division is removed (after an amount of time this is not the case). He says: since the probabilities of being in most of the accessible states are zero, the system will tend to evolve so that the probabilities become equal (due to the fundamental postulate of statistical mechanics). But this means that when the division is removed, the inverse of the absolute temperature instantly increases, for it is the partial derivative of log(omega) with respect to energy. Is it correct that the absolute temperature instantly increases? Here is an alternative: maybe the value omega stays the same instantly after the division is removed; it then increases more or less continuously, approaching a maximum, final, value. Then, since the definition of temperature uses a partial derivative, the value would not change instantly after the division is removed. Rather, it would just be the same. Which view of the amount of microstates is correct? I want to be sure I understand this book and the author's message. Thanks.
 PF Gold P: 864 What is important to keep in mind is that when one talks about a number of accessible states, one is always speaking about a specific ensemble of states. In the most straightfoward way of thinking about the situation you describe the number of accessible states instantaneously increases because you instantly change the ensemble you are talking about. What happens immediately after the partition is physically removed is not part of the subject of equilibrium statistical mechanics because the system moves into a non-equilibrium state and does not sample from the equilibrium distribution of states. You could analyze the situation in different ways and come to different conclusions. You could consider the ensemble of states to include the entire container from the beginning, for instance. In that case, when the gas is confined to half the container by the partition, it is in a non-equilibrium state of this ensemble because it is kinetically trapped by the barrier from reaching certain states. This would not be a particularly useful way of analyzing the system, but my point is that in statistical mechanics, one has to keep in mind the ensemble one is talking about. As far as the inverse of the temperature goes, I think you're making a mistake. For each particle the number of positions it can be in doubles. If there are N particles, then the number of accessible states increases by a factor of $$2^N$$. Thus $$\Omega_{new} = 2^N\Omega_{old}$$ $$\rightarrow S_{new} = k\ln(\Omega_{new}) = k\ln(\Omega_{old})+k\ln{2^N} = S_{old} + k\ln(2^N)$$ $$\rightarrow \frac{1}{T_{new}} = \frac{dS_{new}}{dE}=\frac{dS_{old}}{dE}=\frac{1}{T_{old}}$$ If there was a similar change that did bring about a temperature change by passing through non-equilibrium states, then whether there is a gradual change in temperature or not depends on what definition of temperature you use. If you use a definition from the kinetic theory of gases that relies only on the velocities of the particles, then it will probably change gradually. If you use a statistical mechanical definition that depends on a derivative of the entropy, then it doesn't make sense to talk about temperature outside of equilibrium in the first place.
 P: 46 Euler, that makes perfect sense. Thanks very much for your help.

 Related Discussions Introductory Physics Homework 1 General Math 30 Linear & Abstract Algebra 15


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# Properties

 Label 735.4.a.i Level $735$ Weight $4$ Character orbit 735.a Self dual yes Analytic conductor $43.366$ Analytic rank $1$ Dimension $1$ CM no Inner twists $1$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$735 = 3 \cdot 5 \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 735.a (trivial)

## Newform invariants

 Self dual: yes Analytic conductor: $$43.3664038542$$ Analytic rank: $$1$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 15) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$

## $q$-expansion

 $$f(q)$$ $$=$$ $$q + 3 q^{2} + 3 q^{3} + q^{4} + 5 q^{5} + 9 q^{6} - 21 q^{8} + 9 q^{9} + O(q^{10})$$ $$q + 3 q^{2} + 3 q^{3} + q^{4} + 5 q^{5} + 9 q^{6} - 21 q^{8} + 9 q^{9} + 15 q^{10} - 24 q^{11} + 3 q^{12} - 74 q^{13} + 15 q^{15} - 71 q^{16} - 54 q^{17} + 27 q^{18} + 124 q^{19} + 5 q^{20} - 72 q^{22} - 120 q^{23} - 63 q^{24} + 25 q^{25} - 222 q^{26} + 27 q^{27} - 78 q^{29} + 45 q^{30} - 200 q^{31} - 45 q^{32} - 72 q^{33} - 162 q^{34} + 9 q^{36} - 70 q^{37} + 372 q^{38} - 222 q^{39} - 105 q^{40} - 330 q^{41} + 92 q^{43} - 24 q^{44} + 45 q^{45} - 360 q^{46} + 24 q^{47} - 213 q^{48} + 75 q^{50} - 162 q^{51} - 74 q^{52} + 450 q^{53} + 81 q^{54} - 120 q^{55} + 372 q^{57} - 234 q^{58} - 24 q^{59} + 15 q^{60} + 322 q^{61} - 600 q^{62} + 433 q^{64} - 370 q^{65} - 216 q^{66} - 196 q^{67} - 54 q^{68} - 360 q^{69} - 288 q^{71} - 189 q^{72} + 430 q^{73} - 210 q^{74} + 75 q^{75} + 124 q^{76} - 666 q^{78} - 520 q^{79} - 355 q^{80} + 81 q^{81} - 990 q^{82} - 156 q^{83} - 270 q^{85} + 276 q^{86} - 234 q^{87} + 504 q^{88} - 1026 q^{89} + 135 q^{90} - 120 q^{92} - 600 q^{93} + 72 q^{94} + 620 q^{95} - 135 q^{96} + 286 q^{97} - 216 q^{99} + O(q^{100})$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
 0
3.00000 3.00000 1.00000 5.00000 9.00000 0 −21.0000 9.00000 15.0000
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Atkin-Lehner signs

$$p$$ Sign
$$3$$ $$-1$$
$$5$$ $$-1$$
$$7$$ $$-1$$

## Inner twists

This newform does not admit any (nontrivial) inner twists.

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 735.4.a.i 1
3.b odd 2 1 2205.4.a.c 1
7.b odd 2 1 15.4.a.b 1
21.c even 2 1 45.4.a.b 1
28.d even 2 1 240.4.a.f 1
35.c odd 2 1 75.4.a.a 1
35.f even 4 2 75.4.b.a 2
56.e even 2 1 960.4.a.l 1
56.h odd 2 1 960.4.a.bi 1
63.l odd 6 2 405.4.e.d 2
63.o even 6 2 405.4.e.k 2
77.b even 2 1 1815.4.a.a 1
84.h odd 2 1 720.4.a.r 1
105.g even 2 1 225.4.a.g 1
105.k odd 4 2 225.4.b.d 2
140.c even 2 1 1200.4.a.o 1
140.j odd 4 2 1200.4.f.m 2

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
15.4.a.b 1 7.b odd 2 1
45.4.a.b 1 21.c even 2 1
75.4.a.a 1 35.c odd 2 1
75.4.b.a 2 35.f even 4 2
225.4.a.g 1 105.g even 2 1
225.4.b.d 2 105.k odd 4 2
240.4.a.f 1 28.d even 2 1
405.4.e.d 2 63.l odd 6 2
405.4.e.k 2 63.o even 6 2
720.4.a.r 1 84.h odd 2 1
735.4.a.i 1 1.a even 1 1 trivial
960.4.a.l 1 56.e even 2 1
960.4.a.bi 1 56.h odd 2 1
1200.4.a.o 1 140.c even 2 1
1200.4.f.m 2 140.j odd 4 2
1815.4.a.a 1 77.b even 2 1
2205.4.a.c 1 3.b odd 2 1

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{4}^{\mathrm{new}}(\Gamma_0(735))$$:

 $$T_{2} - 3$$ $$T_{11} + 24$$ $$T_{13} + 74$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$-3 + T$$
$3$ $$-3 + T$$
$5$ $$-5 + T$$
$7$ $$T$$
$11$ $$24 + T$$
$13$ $$74 + T$$
$17$ $$54 + T$$
$19$ $$-124 + T$$
$23$ $$120 + T$$
$29$ $$78 + T$$
$31$ $$200 + T$$
$37$ $$70 + T$$
$41$ $$330 + T$$
$43$ $$-92 + T$$
$47$ $$-24 + T$$
$53$ $$-450 + T$$
$59$ $$24 + T$$
$61$ $$-322 + T$$
$67$ $$196 + T$$
$71$ $$288 + T$$
$73$ $$-430 + T$$
$79$ $$520 + T$$
$83$ $$156 + T$$
$89$ $$1026 + T$$
$97$ $$-286 + T$$


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Coxeter triangulations have good quality - Archive ouverte HAL Access content directly
Preprints, Working Papers, ... Year :

## Coxeter triangulations have good quality

Aruni Choudhary
• Function : Author
Siargey Kachanovich
• Function : Author
• PersonId : 1025210
Mathijs Wintraecken
• Function : Author

#### Abstract

Coxeter triangulations are triangulations of Euclidean space based on a simple simplex. By this we mean that given an individual simplex we can recover the entire triangulation of Euclidean space by inductively reflecting in the faces of the simplex. In this paper we establish that the quality of the simplices in all Coxeter triangulations is $O(1/ √ d)$ of the quality of regular simplex. We further investigate the Delaunay property (and an extension thereof) for these triangulations. In particular, one family of Coxeter triangulations achieves the protection $O(1/d 2)$. We conjecture that both bounds are optimal for triangulations in Euclidean space.

### Dates and versions

hal-01667404 , version 1 (19-12-2017)

### Identifiers

• HAL Id : hal-01667404 , version 1

### Cite

Aruni Choudhary, Siargey Kachanovich, Mathijs Wintraecken. Coxeter triangulations have good quality. 2017. ⟨hal-01667404⟩

### Export

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Posted by: atri | September 11, 2009

## Lect 4: Gale-Shapley Algorithm

(Guest Post by Devanshu Pandey)

At the beginning of Wednesday’s class we went over some of the definitions that we discussed last class on Friday. These included:

The set $M$ of men and the set $W$ of Women.

Preferences: Rankings given to every individual in the other set by a man or a woman.

Matching: A subset of the $M\times W$ such that there is no polygamy.

Perfect Matching: A matching which ensures that everyone gets married.

Instability: A situation in which $(m ,w)$ and $(m^{\prime} ,w^{\prime})$ are pairs but $m$ prefers $w^{\prime}$ to $w$ and $w^{\prime}$ prefers $m$ to $m^{\prime}$,  which gives them an incentive to break up with their current partners and join up with each other.

Stable Matching: A perfect matching in which everyone is perfectly happy with their current matches.

After this was done we moved on to covering two examples which would help  clear up the situation a little. The examples we covered were very trivial  compared to what the real world problem looks like. However, we did so with a small number of men and women because it is easier to work with. We picked $n = 2$ i.e. the number of men (which is also the number of women) is equal to $2$. We noted that the two perfect matching possible were $\{ (m, w), (m^{\prime}, w^{\prime})\}$ and $\{ (m, w^{\prime}), (m^{\prime}, w)\}$.  I will call these $P_1$ and $P_2$  respectively:

1. Example 1: Both men prefer $w$ over $w^{\prime}$ and both women prefer $m$ over $m^{\prime}$. When asked which of the two matchings were stable, the class shouted out all the possible answers except the correct one which was $P_1$ is stable and $P_2$  is not. This is because in $P_1$  neither $m$ nor $w$ have any intention of leaving each other (since they rank highest on each other’s preference lists) and hence there is no instability. Where as in $P_2$$(m, w)$ would be an instability. This led us to a new and a more intuitive way of predicting when an instability occurs. That was: “An instability occurs only if there is someone in both the parties ($M$ and $W$) who wants to switch”. To make this clearer and more specific if w wants to switch to m then for there to be an instability, $m$ should also want to switch to $w$ (i.e. the will to switch should be on both sides).  This Dr. Rudra pointed out was similar to saying that “You cannot force someone to get married unless they really want to get married”. I’d like to make an addition, the same is true about forcing someone to break up.
2. In the second example, the preference lists were: $L_m =\{w, w^{\prime}\}, L_{m^{\prime}} =\{w^{\prime}, w\}, L_w =\{m^{\prime}, m\}$ and $L_{w^{\prime}} =\{m, m^{\prime}\}$. Here both the perfect matchings were stable because in either case there was a pair which had no incentive whatsoever to break up.

Then we tried picking a very informal and “naïve” algorithm to solve the generalization of this problem (i.e. with an  arbitrary number of men and women). Andrew suggested Guess and Check.  However we analyzed that this would be very inefficient because there were $n!$ perfect matchings possible when $|M| =|W| = n$. This meant that the algorithm would have a best case running time of $n!\cdot n^3$ ( i.e. $\Omega(n!\cdot n^3 )$). The $n^3$ factor is added because we also have to verify if these perfect matchings are actually stable. Also it isn’t necessary that a stable matching would be the output in every case. This also resulted in the question: “Does a stable matching always exist?”

Hence, we turned to the Gale-Shapely algorithm developed in 1962 by David Gale and Lloyd Shapely. Following is the algorithm in my own words:

1. In the beginning all men and women are free (single).
2. As long as there is a woman (say $w$) who is allowed to propose (i.e. she is single and has someone on her list that she hasn’t proposed to already):
• $w$ proposes to the man highest on her preference list that she hasn’t proposed to (say $m$).
• if $m$ is free then $w$ and $m$ enter an engagement which like in real life is an intermediate (and possibly temporary) stage. On the other hand if he isn’t free then he checks his preference list. If $w$ is higher up on his preference list that his partner, $(m, w)$ is formed and $m$ breaks up with his current partner. If $w$ isn’t higher on the preference list then she stays free and $m$ remains engaged.
3. This algorithm runs till there are no women left who can propose.

In the next class we plan to check that the algorithm runs in $O(n^2)$ and that it always outputs a stable matching.

## Responses

1. Hello everyone. This was written by me and I’d appreciate any comments that anyone has to make here. Thank You.

2. That was a good summary. The thing that makes the Gale-Shapley so effective is the fact that there are checks and balances. The woman is allowed to assert her preference while the man is allowed the option to change the woman he eventually marries based on his preference. I suspect this algorithm always results in a stable matching because of this balance.

• John,

Yes the check and balance you mention works in the algorithm’s favor. Also the fact that every woman proposes in one direction (and never “goes back”) is crucial for the GS algorithm to work.

3. […] of exactly what we’re trying to do with tokens! So, the solution to this problem is called the Gale-Shapley algorithm, and get this, it’s only worst case O(M*W) where M is the number of men tokens and W is the […]


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Parameter Discussion¶

The RSS papers uses a few constants required for the safety calculations. The values for these constants are not yet defined and open for discussion/regulation. Nevertheless, the implementation of the ad-rss-lib needs to define initial values for these functions.

Note

To be as flexible as possible, the parameters are implemented as configuration values so these can be easily adjusted during evaluation or after the release. In other words, the values are not hard coded in the library, but can be changed via the provided inputs.

In the following, the key parameters and the decision for possible initial values are discussed. The used parameters based on the RSS paper are:

• Response time $\rho$

It is assumed that an AV vehicle has a shorter response time than a human driver. Therefore, there is a need to have two different parameters. As it might not be possible to determine whether another object is an AV vehicle or has a human driver, the ad-rss-lib will safely assume that all other objects are driven by humans. Hence, two parameters for the response time are used:

• $\rho_{ego}$: for the ego vehicle
• $\rho_{other}$: for all other objects
• Acceleration $\alpha$

RSS proposes several different acceleration/deceleration values. One could argue that acceleration/deceleration differs with the type of vehicle. Also at least the acceleration is dependent on the current vehicle speed. As it cannot be assured that the individual acceleration of each and every car can be known and the specific car can be reliably detected, the ad-rss-lib will assume fixed constants for those values. These could be either the maximum physically possible values or restrictions that are imposed by regulation. Also there will not be different values for the ego vehicle and the other vehicles. It could be argued that for the ego vehicle e.g. desired acceleration might be known. Therefore, a shorter safety distance would be sufficient. But as all other vehicles do not know about the intention of the ego vehicle this would lead to a violation of their safe space. So the ad-rss-lib will need to calculate its checks with the globally defined accelerations values even if the vehicle does not intend to utilize them to its limits. The parameters used for acceleration are:

• $\alpha_{accel,max}$ maximum possible acceleration
• $\alpha_{brake,min}$ minimum allowed braking deceleration in longitudinal direction for most scenarios
• $\alpha_{brake,max}$ maximum allowed deceleration in longitudinal direction
• $\alpha_{brake,min,correct}$ minimum allowed deceleration in longitudinal direction for a car on its lane with another car approaching on the same lane in wrong driving direction
• $\alpha^{lat}_{brake,min}$ minimum allowed braking deceleration in lateral direction
• $\alpha^{lat}_{accel,max}$ maximum allowed acceleration in lateral direction
• $\delta^{lat}_{min}$ fluctuation margin for that needs to be respected when calculating lateral safe distance

Decision on Selected Parameter Values¶

Note

The following parameter values are only suggestions and open for discussion. These can be changed at anytime, if it is required.

Response time

For the response times a common sense value for human drivers is about 2 seconds. For an AV vehicle the response time could be way lower. In order to be not too restrictive the initial value for the ego vehicle response time will be assumed as 1 second. Hence, $\rho_{other} = 2s$ and $\rho_{ego} = 1s$. If we assume a case of AD vehicles only, the response time may be reduced.

Longitudinal Acceleration

Finding meaningful acceleration values is more complicated. At the one hand the values should be as close as possible or even exceed the maximum physically possible values. The minimum deceleration values must also not exceed normal human driving behavior. So assuming a too high deceleration for other cars may lead to a false interpretation of the situation.

On the other hand a too big difference between the minimum and maximum acceleration values will lead to a very defensive driving style. As a result, participating in dense traffic, will not be possible (see figure below). A rule of thumb for deceleration in German driving schools is: $\alpha_{brake,min} = 4m/s^2$ and $\alpha_{brake,max} = 8m/s^2$.

But on the other hand, modern cars are able to decelerate with up to $12m/s^2$. Especially for deceleration, it is questionable whether it is possible and tolerable to restrict maximum braking below physically possible braking force.

For the maximum acceleration at low speeds a standard car will be in the range of $3.4m/s^2$ to $7m/s^2$. But there are also sport cars that can go faster than that. But for acceleration a regulation to a maximum value seems to be more likely than for deceleration.

Restricting velocity to the current speed limit
Required safety distance for cars driving at 50 km/h (city speed) in same direction with $\alpha_{brake,min} = 4m/s^2$ and $\alpha_{brake,max} = 8m/s^2$ and $\rho_{ego} = \rho_{other} = 2s$

The assumption that a car can always accelerate at $\alpha_{accel,max}$ during the reponse time, leads to a significant increase of the required safety distance. The figure above shows the required safety distance for different acceleration values. So acceleration about $4m/s^2$ doubles the required safety distance from $40m$ to about $80m$ at city speeds.

Therefore, it might be advisable to add a restriction that a car is only allowed to accelerate up to the allowed speed limit. In addition, the common behavior (in Germany) is to respect a safety distance of speed/2, e.g. in a city with a speed limit of $50km/h$ the safety distance shall be $25m$. Hence, it is obvious that the parameters may require some adjustments to allow reasonable driving.

Further possible restrictions

Another possibility to decrease the required safety distance to the leading vehicle would be to take the intention of the ego vehicle into account. E.g. if the ego vehicle is following another vehicle and is not intending to accelerate, then there is no need to assume that the ego vehicle is accelerating during its response time. Nevertheless, there are several issues with that approach:

1. It needs to be assured that all intended and unintended accelerations (e.g. driving down a slope) are known to RSS.
2. If RSS formulas are regarded as regulations, the safety distance must be kept regardless to the intent of the vehicle.

Therefore, in the current implementation this approach will not be applied.

Lateral Acceleration

When defining the parameters for lateral acceleration and deceleration, it is import to keep in mind that the definition must allow bypassing of vehicles. Physically high lateral accelerations are possible. In order to be able to bypass a vehicle that is driving on a parallel lane, the safe lateral distance needs to be safe during the complete response time of the other vehicle.

Let us consider two identical vehicles driving on the centerline of two adjacent lanes with zero lateral velocity. There is no lateral conflict, if the distance between the border of the car and the adjacent lane is bigger than the distance that the vehicle will cover when accelerating laterally at maximum during its response time and then decelerating to zero lateral velocity.

Distance a vehicle will cover when applying the "Stated Braking Pattern" with $\rho_{vehicle} = 2s$

The figure above shows the required safety distance, without considering the fluctuation margin, each car needs to keep to the lane border so the vehicles can pass without lateral conflict. With an assumed minimal lane width of 3 meters and an assumed vehicle width of 2 meters, the distance from vehicle edge to lane border is 0.5 meter, if the car is driving exactly in the middle of the lane.

Hence, the required safety distance must be at most 0.5 meter. When using the same values for acceleration and deceleration this will lead to $\alpha^{lat}_{accel,max} < 0.1m/s^2$. But when restricting the acceleration to that value a lane change will take almost 8 seconds.

As a result it is advisable, to use a higher deceleration than acceleration to keep the required safety margin and allow for faster lane changes. E.g. $\alpha^{lat}_{brake,min} = 0.8m/s^2$ and $\alpha^{lat}_{accel,max} = 0.2m/s^2$ will fulfill the given safety distance requirement. An increase to higher acceleration values is for the given constraints not possible, as the distance covered during response time is already 0.4 meters.

It is obvious that given the lateral safety definition a lane change will at least have a duration of two times the response time.

The lateral distance requirement is very strict, therefore it is required to also come up with a desirably small value for the required lateral safety margin $\delta^{lat}_{min}$. As this should only cover for fluctuations, there is also no need for a huge margin. Thus initially this value will be set to $\delta^{lat}_{min} = 10 cm$. This value should be able to cover small fluctuations, but will not have a big impact on the safety distance.

Note

As a conservative starting point the values can be set to:

Parameter Value
$\rho_{ego}$ $1s$
$\rho_{other}$ $2s$
$\alpha_{accel,max}$ $3.5m/s^2$
$\alpha_{brake,min}$ $4m/s^2$
$\alpha_{brake,max}$ $8m/s^2$
$\alpha_{brake,min,correct}$ $3m/s^2$
$\alpha^{lat}_{brake,min}$ $0.8m/s^2$
$\alpha^{lat}_{accel,max}$ $0.2m/s^2$
$\delta^{lat}_{min}$ $10cm$


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# Find the 4th term in the expansion of (x-2y)^12.

Updated On: 27-06-2022


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Does forcing with a proper class of conditions require Global Choice?

When Jech, in his Set Theory, deals with forcing with a class of forcing conditions (with the aim of proving Easton's theorem), he starts with the assumption that there is a well-ordering of the ground model, i.e. the ground model satisfies the Axiom of Global Choice.

Having examined and, to a degree, understood the development in this section, I can't figure out where he actually uses this assumption. The only place I can see where this might be relevant is in the discussion on the existence of a generic set and even here it seems Global Choice isn't needed in every possible solution. For example, if you justify forcing via a reflection theorem argument or a countable ground model, I don't believe you need Global Choice.

On the other hand, you can take the Boolean-valued semantics approach and define the canonical name for the generic set as $\dot{G}(\check{p})=p$ for a forcing condition $p$ (assume here that the forcing notion is a proper class Boolean algebra). So far we're fine, $\dot{G}$ is a class in the Boolean-valued model, everything is rosy. Conceivably, if we were to define a class $\check{M}$, representing the ground model in the Boolean-valued model, via $$\|x\in\check{M}\|=\bigvee_{y\in M}\|x=\check{y}\|$$ the Boolean-valued model would see itself as the generic extension of $\check{M}[\dot{G}]$, since this holds when forcing with a set of conditions. Of course, there is a problem in defining $\check{M}$ this way, since we can't generally take sups of a proper class of (different) Boolean values.

I expect this approach should be salvageable, using Global Choice. In particular, I think we should be able to take the offending sup along the given well-ordering of $M$ and somehow "stagger" it, so it becomes well defined.

I'm not at all sure if this is legitimate or if it even leads anywhere, so I would appreciate comments and an explanation of what is really going on. Additionally, can anyone suggest another reference for class forcing? I generally enjoy Jech's book, but I found this section to be somewhat opaque and hard to understand.

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The forcing is done internally even if you use the CTM argument, it will not be countable internally, and you want to make the choice within the model. –  Asaf Karagila Feb 4 '12 at 10:17
@AsafKaragila Surely, the CTM argument is done externally. You can't argue inside the model that there are only countably many dense sets. On the other hand, I seem to unconsciously believe that the proof that there is a generic set for countably many dense sets should work for any cardinal. Silly inner voice. –  Miha Habič Feb 4 '12 at 11:01
True, it is done externally. However depending on the forcing you could argue it only have so-many dense sets. As for the generalizations, I asked my advisor once and he thought about it and said it's not immediate and you'd have to require more on the forcing itself and on the cardinal. –  Asaf Karagila Feb 4 '12 at 11:04
@AsafKaragila I had a thought. Since the CTM argument doesn't seem to work as is, would it be possible to cheat a bit and temporarily move the discussion to a countable model of NBG? There we can reproduce the argument verbatim and get our generic class. Since ZFC and NBG are equiconsistent, this should be adequate for relative consistency proofs. I'm beginning to suspect Easton had the right idea, originally working in NBG. Doing these things in ZFC looks really tedious. –  Miha Habič Feb 4 '12 at 18:55
It is indeed cheating, the fact NBG is a conservative extension of ZFC means only that proofs which only contain statements about sets (and maybe definable classes) can be done in ZFC; but class forcing is a whole other thing. Equiconsistency is of no help either, ZF is equiconsistent with ZFC but this doesn't mean you can just go about using the axiom of choice if you did not assume it at first, nor it will help you prove things in pure ZF. –  Asaf Karagila Feb 4 '12 at 19:04

For class forcing I'd suggest to start with Sy Friedman's work which can be found here.

In particular, his chapter from the Handbook of Set Theory (available on the above site) can be used as a good start.

The problem with class forcing is that classes are not "real" objects in the universe of set theory. They are formulas interpreted in the model as definable collections. This means that arguments which you can get "for free" from sets are now very expensive in the sense that you need to verify things. Global choice makes things easier because it keeps all classes in the same size and allows us to choose from everything at once.

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Could you elaborate how Global Choice actually helps here? In particular, as I gather from Friedman's papers and other places, the problem seems to be identifying the ground model in an extension. –  Miha Habič Feb 5 '12 at 18:36
@Miha: No, I cannot really tell you because I am unfamiliar with the intimate details of class forcing. Whenever I bring this up I mostly get the reaction "Oh this is very complicated... you should check out Sy Friedman's work. He's done some cool things but they are very hard to understand to the most part." You asked for a reference and I gave you one. The little intuition I have is that classes are not actual objects in ZFC therefore require a much more careful handling. –  Asaf Karagila Feb 5 '12 at 22:10

There exists $L$-definable class forcings $P_0$ and $P_1$ such that whenever $G_0$ and $G_1$ are $P_0$-generic, $P_1$-generic over $L$, respectively.

(a) $ZFC$ holds in $\langle L[G_0],G_0\rangle$ and in $\langle L[G_1],G_1\rangle$.

(b) ZFC fails in $\langle L[G_0,G_1],G_0,G_1\rangle$.

Thus, we cannot preserve ZFC and have generics for all ZFC preserving $L$-definable class forcings. So the existence of the generic is not a trivial matter.

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You could at least say that you copied this from Friedman's Fine structure and class forcing (Preface, Fact 1)... –  Asaf Karagila Feb 5 '12 at 22:06

Global choice can be skipped since we can argue within some CTM, $M$, and noticing that the rest of the construction can be carried only assuming that $M$ is a model of $\mathsf{ZFC+GCH}$ and that there exists a generic filter; notice that $M^B$ is not a Boolean-Valued model, and thus we don't need choice to prove the truth lemma; we don't have to show $M^B$ is a full Boolean-valued model, and we define $p\Vdash\exists x\psi(x,\ldots)$ iff $\forall q\leq p\exists r\leq q\exists a\in M^B(p\Vdash \varphi(a,\ldots)),$ and it is easy to prove the truth lemma using only the genericity of the given $G$.

However, if you use the canonical name argument approach, you do need global choice to hold in the ground model, as you need to turn $M^B$ into a 2-valued model inside $M$, to accomplish this you construct a non-principal ultrafilter of the Boolean algebra class $B$; as done here.

To prove the existence of $U$ inside $M$ you use global choice as follows: as $B=\bigcup_\lambda B_\lambda$, where $\lambda$ goes through all regular cardinals. By transifinite induction, choose a non-principal ultrafilter $U_\omega$ of $B_\omega$. For $\lambda>\omega$ regular, choose a non-principal ultrafilter $U_\lambda$ of $B_\lambda$ extending all of the $U_\kappa$ for regular $\kappa<\lambda$, then put $U=\bigcup_\lambda U_\lambda$.

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You know, in just one year you seem to have studied a lot of set theory. Nice work! –  Asaf Karagila Dec 10 '13 at 0:36
@AsafKaragila, How not to do it, it's so much fun! :-) –  Camilo Arosemena Dec 10 '13 at 0:38