text:
# Plastic Analysis with RFEM

## Technical Article on the Topic Structural Analysis Using Dlubal Software

### Technical Article

The elastic deformations of a structural component due to a load are based on Hooke's law, which describes a linear stress-strain relation. They are reversible: After the relief, the component returns to its original shape. However, plastic deformations lead to irreversible deformations. The plastic strains are usually considerably larger than the elastic deformations. For plastic stresses of ductile materials such as steel, yielding effects occur where the increase in deformation comes along with hardening. They lead to permanent deformations - and in extreme cases, to the destruction of the structural component.

Plastic deformations represent material properties that require a nonlinear analysis. Utilize RFEM together with the RF-MAT NL add-on module to consider elastic-plastic material behavior in the analysis. In addition to the yield condition according to von Mises, the approaches according to Tresca, Drucker-Prager, and Mohr-Coulomb are available as strain hypotheses. For ductile materials such as steel, we recommend the plasticity theory according to von Mises. For the general spatial stress state, it is as follows:

General stress state according to von Mises

$$σv = σx2 + σy2 + σz2 - σx· σy - σx · σz - σy · σz + 3 · (τxy2 + τxz2 + τyz2)$$

Read more about the stress hypotheses here:
https://en.wikipedia.org/wiki/Comparative stress

Based on a simple model with uniaxial loading, we show the options of plastic analyses with RFEM and RF-MAT NL.

#### Tensile Test Specimen Example

A tensile test specimen made of S 235 steel with the dimensions shown in Figure 01 is restrained at one end. The model is created as a surface with 2D elements.

Due to the very small dimensions of the test specimen, an FE mesh size of 1 mm is selected. Only the stresses due to tensile stress according to the linear static analysis are to be analyzed; a stability analysis is not performed.

#### Tensile Stress up to the Yield Strength

The force required to reach the yield strength is determined for the model as follows:
N = fy · A = 235 N/mm² · (10 mm · 3 mm) = 7,050 N

It is applied as a line load at the end of the test specimen:
7,050 N/14 mm = 503.57 N/mm = 503.57 kN/m

Load case 1 is first calculated with the standard isotropic linear-elastic material model according to the linear static analysis. The self-weight is not taken into account.

To display the equivalent stresses according to von Mises, the "Constant on Elements" result diagram is selected. Thus, there is no smoothing across the element boundaries. Image 02 shows that the yield stress of 235 N/mm² is exceeded in areas of the reduced cross-section. Therefore, the results do not reflect the stress distribution of a tensile test with yield effects.

For a realistic calculation, we recommend using the "Isotropic Plastic 2D/3D" material model, which is available with an RF-MAT NL license. This material model shows an isotropic material behavior in the elastic zone. The plastic area is based on the yield conditions of the von Mises strain hypothesis with a yield strength of the equivalent stress of 235 N/mm². The calculation is performed iteratively in several load steps. For this, the "Number of load increments for Newton-Raphson method to determine automatically" option is used, which can be activated in the RFEM calculation parameters for materials with a nonlinear model. This ensures the required accuracy of the results with a relatively short calculation time.

Image 03 shows the stresses and the degree of nonlinearity (proportion of elements with yields) of the nonlinear material model. The yield strength of 235 N/mm² is not exceeded in any element. The calculation diagram shows the deformation diagram during the iterations.

#### Load Increment and Yielding

In LC 2, the "yield load" is increased slightly from 503.57 kN/m to 505.00 kN/m. The calculation provides a strong total strain of the test specimen of 8.29 mm (for comparison: 0.10 mm in LC 1). However, there is still a convergence, which is due not least to the strain hardening modulus Ep = 2.1 N/mm² (see the technical article titled Hardening Parameters in Nonlinear Material Models). The weakened area of the cross-section is completely in the yield state.

In the elastic-plastic calculation, the total strain ε is divided into an elastic component εel and a plastic component εpl. ε = εel + εpl. However, this breakdown is only valid when assuming that the plastic strains are small (εpl < 0.1). If the plastic strains exceed this limit value, the plastic results should be evaluated with caution.

The change in length in the tapered area (assumption: l = 38 mm) due to elastic strain is:
εel = σ / E = 235 N/mm² / 210,000 N/mm² = 0.00119
∆lel = 0.00119 · 38 mm = 0.04 mm

This deformation can also be checked in Table "4.2 Nodes - Deformations" from the difference of the displacements uX of nodes 4 and 5 in LC 1.

In LC 2, the displacements of these nodes result in the following plastic strain:
εpl = (∆ltot - ∆lel ) / l = (8.24 mm - 0.05 mm - 0.04 mm) / 38 mm = 0.21 > 0.1

Thus, the limits of the nonlinear material model are exceeded. For comparison, further analyses are carried out on a 3D model of the tensile test specimen, which records the spatial component in a solid model.

#### Solid Model

The 3D model is also meshed with a mesh size of 1 mm. This results in three finite solid elements over the cross-section thickness.

The force of 7,050 N required to reach the yield strength (see above) is applied as a surface load at the end of the test specimen. It is determined as follows:
7,050 N / (14 mm · 3 mm) = 167.857 N/mm² = 167,857 kN/m²

As Figure 05 shows, the equivalent solid stresses and the nonlinearity ratio of LC 1 confirm the results of the surface model.

The calculation of LC 2 - with an equivalent surface load of 168,333 kN/m² - results in a smaller total deformation of 0.22 mm in the solid model than in the surface model with 8.29 mm. In the calculation diagram, however, the nonlinear effects of the material flow are also clearly visible.

As shown in Figure 06, the deformation figure (in an exaggerated representation) reflects the start of the reduction, including transverse deformation.

Thus, the solid model represents an interesting alternative to the representation of the yielding process. Only a load of about 177,900 kN/m² would lead to a deformation of 8.29 mm of LC 2 in the surface model. This corresponds to an increase of the yield strength load by 6% (surface model: 0.3%).

The comparison of the model with the real material behavior can ultimately be carried out by a corresponding experimental arrangement with a tensile test.

#### Summary

Using a simple example, we showed how nonlinear material laws can be represented and analyzed in RFEM. Both surface and solid elements can be modeled. The solid model is generally recommended if the influence of the element thickness is relevant. For relatively slender objects, however, the surface model is the first choice; it also requires less modeling and calculation effort.

Plastic material properties, such as flow behavior, can be displayed using the RF-MAT NL add-on module. In this case, user-defined stress-strain diagrams based on empirically determined data are also possible. In the presented example, the standard diagram with a preset strain hardening factor was used.

With the help of nonlinear material models, it is also possible to determine redistribution effects in the model that result, for example, from the formation of plastic hinges. For plastic effects, results with large deformations - especially according to the large deformation analysis - have to be evaluated with caution.

Find more explanations and examples in our Knowledge Base articles and FAQs, which are linked below this article.

#### Dipl.-Ing. (FH) Robert Vogl

Technical Editor, Product Engineering & Customer Support

Mr. Vogl creates and maintains the technical documentation. In addition, he is involved in the development of the SHAPE-THIN program and provides customer support.

#### Reference

 [1] Eurocode 3: Design of steel structures - Part 1‑1: General rules and rules for buildings; EN 1993‑1‑1:2010‑12 [2] Dlubal Software. (2018). Manual RFEM. Tiefenbach.

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# Geometric interpretation of the scalar triple product

The Wikipedia article for scalar triple product says the following:

Geometrically, the scalar triple product $$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})$$ is the (signed) volume of the parallelepiped defined by the three vectors given. Here, the parentheses may be omitted without causing ambiguity, since the dot product cannot be evaluated first. If it were, it would leave the cross product of a scalar and a vector, which is not defined.

$$(\mathbf{b}\times \mathbf{c})$$ is the vector normal to vectors $$\mathbf{b}$$ and $$\mathbf{c}$$. What is the geometric reasoning that leads us to understand that the dot product of $$\mathbf{a}$$ with this normal vector is equal to the volume of the parallelepiped defined by the three vectors? I would greatly appreciate it if people would please take the time to explain this.

• Does this answer your question? Scalar triple product - why equivalent to determinant? – Raskolnikov Jun 19 '20 at 5:38
• @Raskolnikov But this doesn't give any geometric explanation. – The Pointer Jun 19 '20 at 5:41
• The length of $b\times c$ is the area of the parallelogram spanned by $b$ and $c$. The scalar product with $a$ projects $a$ onto the line along $b\times c$ which is orthogonal to the parallelogram and, hence, gives the height of the parallelepiped. So, the triple product is nothing but $base \times height$ (up to a sign). – trancelocation Jun 19 '20 at 5:59
• @ThePointer : The dot-product does incorporate the magnitude of both vectors involved: $a \cdot b = |a||b|\cos \angle (a,b)$. – trancelocation Jun 19 '20 at 6:10

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2. For a further discussion of this distinction, see Edwin M. Truman (2010), "The International Monetary System and Global Imbalances " (Washington: Peterson Institute for International Economics, January). Return to text

We must also strengthen the international monetary system. We need a system characterized by more open capital accounts, flexible exchange rates, and independent monetary policies. Open capital accounts, supported by appropriate financial supervision and regulation, channel savings to their most productive uses, thereby enhancing welfare. Exchange rate flexibility improves domestic macroeconomic management, allowing countries to pursue independent monetary policies tailored to their individual needs, and limits unwelcome spillovers to other economies. Such a system can also flexibly adapt to changing economic and financial realities as countries develop, technology progresses, and shocks buffet the global economy.

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Inflexible exchange rates in these countries have tended to inhibit adjustment of unsustainable global imbalances in trade and capital flows. Indeed, as I noted, such imbalances appear to have fostered the buildup of vulnerabilities in the run-up to the recent financial crisis. Countries with current account surpluses and restricted capital flows have been able to resist currency appreciation for prolonged periods, even when justified by underlying fundamentals. In principle, adjustment of imbalances could occur if countries permitting relatively limited movements in nominal exchange rates allow their national price levels to adjust over time. But sterilization operations and other policy tools can, and often have, restrained such adjustment. Meanwhile, countries with current account deficits should take steps to increase national saving, including by putting in place credible plans to reduce their fiscal deficits in the longer run.

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text:
# How is the general linear group a topological group?

How to see if the general linear group GL($n$), of non-singular $n$-square matrices over the real (or complex) numbers under matrix multiplication, is a topological group? How to show that matrix multiplication and inversion are continuous mappings?

-

Hint: Matrix multiplication, entry wise, is just a polynomial! Inversion can be seen to be continuous using the same logic and Cramer's theorem: $\displaystyle A^{-1}=\frac{1}{\det(A)}\text{ adj}(A)$.
Along with Alex's answer I think this might help you too. Try to think of these,$\phi: M_n(\mathbb{R})\rightarrow \mathbb{R}$ we define $\phi(A)=\det(A)$ is continuous map because $1.$ Polynomials are continuous $2.$ Sum of Two Continuous function is again a continuous. $3.$ Product of two continuous function is again a continuous function.


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# Head style and Contenst problem

I have two problems in one. I am writing my thesis using the package fancyhdr, and all my chapters look great. However I have to add some parts, which are not chapter but I would like to have the title as one.

The problem is my results do not look like as I expected.

All chapters look like:

So my idea is that no-chapter parts should look like:

But in real they look like:

and it gets worse, because my Contents looks awful.

I am using this preamble for the style of my heading.

\fancyhead{} \fancyhead[]{}

\fancyfoot[]{}

And in the none-Chapter part I wrote

\renewcommand{\chaptername}{}

%\renewcommand\chaptername{\fontsize{50}{50}\selectfont\Large{Curriculum Vitae}}

\renewcommand\thechapter{\fontsize{30}{30}\selectfont\Huge{Curriculum Vitae}}

I was trying too many solutions posted here, but none of those work for me. If I use \chaptername instead of \thecapter (which I know is the chapter number), the nice rectangle frame becomes into a line. if I use just \chapter no eve a line is generate and it is equivalent if I use \chapter*{}. I know the problem in the contents is because the \fontsize and the \selectfont, but those are the size which give me a properly size in the title.

Does anybody knows what to do? I would really appreciate any help. If you need to know my preamble it is here

Regards,

Flavia

• You seem to use the fncychap package that you don't mention. If that's the case, you could take the style used for numbered chapters as a basis and make an own style for unnumbered chapters. Alternatively, modify your re-definition of \thechapter as follows : \renewcommand\thechapter{{\fontsize{30}{30}\selectfont\Huge{Curriculum Vitae}}\vphantom{0}}, but I'm not sure there won't be unpredictable effects. – Bernard Mar 12 '14 at 21:56
• Please post a complete Minimal Working Example and not just fragments of code which leave people guessing what is used or what is relevant. This should be code which people can compile to reproduce the issue - not just the preamble - and it should contain only what is necessary to reproduce that issue - not the entire preamble. – cfr Mar 13 '14 at 0:04

You are using the Lenny chapter style from fncychap (this style is, in my opinion, a little ugly and perhaps you should reconsider using it). Redefining \thechapter to produce a title and its formatting is not the thing to do; \thechapter is only for the representation of the chapter counter; misusing it as you did in your example code will produce the undesired results you already experienced.

With the help of titlesec and TikZ I defined a command \specialchapter to produce a style for chapter headings like the one you want; the special chapter is produced using \chapter* and the ToC entry is added used \addcontentsline:

\documentclass{book}
\usepackage[Lenny]{fncychap}
\usepackage[explicit]{titlesec}
\usepackage{tikzpagenodes}

\newcommand\specialchapter{%
\titleformat{\chapter}[display]
{\normalfont\usefont{OT1}{phv}{m}{n}\selectfont}
{}
{0pt}
{\begin{tikzpicture}[remember picture]
\node[inner ysep=0pt] (title)
{\hspace*{15pt}\Huge\rmfamily##1\hspace*{15pt}};
\draw[overlay]
([xshift=10pt]title.south west) --
(current page text area.west|-title.south west) --
++(0pt,40pt) --
++(\textwidth,0pt) --
(current page text area.east|-title.south east) --
([xshift=-10pt]title.south east);
\end{tikzpicture}%
}
}
\begin{document}

\tableofcontents
\chapter{Discussion and Perspectives}
{
\specialchapter
\chapter*{Curriculum Vitae}
}
\chapter{Another regular chapter}

\end{document}

An image of the ToC:

A regular chapter:

The special chapter:

The next regular chapter:

• @cfr Ah, yes. I'll fox it. Thanks for noticing! – Gonzalo Medina Mar 13 '14 at 0:02
• Now I'm really intrigued... I'm certainly glad I'm not a chicken! – cfr Mar 13 '14 at 0:05
• @cfr LOL! Sorry for the typo :) In the meantime, I "foxed" the code. – Gonzalo Medina Mar 13 '14 at 0:06
• Yes you are right, I use fancycahp, sorry about the package. Thanks for the code, I were never think something like that, however it works partially. I mean it solve the first problem I posted, but now other parts of my thesis are damaged. I am using \appendix an before the code they were looking with the [Lenny] style, but now they are treated as \specialchapter and look terrible. – Flavia Mar 13 '14 at 12:28
• @Flavia You need to keep the effect of \specialchapter local (the extra pair of braces in my example code). In your file CV.tex you then need: { \specialchapter \chapter*{Curriculum Vitae} \addcontentsline{toc}{chapter}{Curriculum Vitae} } And the content of the chapter goes here... (again, notice the braces before \specialchapter and after the \addcontententsline command). – Gonzalo Medina Mar 13 '14 at 12:29


text:
# What is wrong with the tag excerpts?

Looking at the list of tags https://mathoverflow.net/tags it seems that all the tags are surrounded by html
...
. This seems to be happening because the html
...
 was put directly into the tag excerpts themselves (see https://mathoverflow.net/edit-tag-wiki/443 for example). Why is that? Is it supposed to be this way?

I added some of those tag descriptions earlier today, and they definitely did not have the
's then. I baselessly speculate this might be related to this fix.


text:
# Math mode in document title with R Markdown

I was trying to put $\LaTeX$ in the document title but can't find a way to do so. I suppose that this is because I can't use latex packages with the document title. Does anyone know how to solve this?

Here's what I have:

title: "Welcome to $\LaTeX$ Workshop"
output:
pdf_document:
citation_package: natbib
number_sections: yes
bibliography: sample.bib
classoption: twocolumn
- \usepackage{lscape}
- \usepackage[english]{babel}
- \usepackage{multicol}
- \setlength{\columnsep}{1cm}
- \setlength{\parindent}{1cm} # paragraph indentation
- \setlength{\parskip}{8pt} # paragraph spacing
- \usepackage{setspace}
- \doublespacing

The error I'm getting looks like this:

! Package inputenc Error: Unicode char
     (U+2028)
(inputenc)                not set up for use with LaTeX.

See the inputenc package documentation for explanation.
Type  H <return>  for immediate help.
...

l.96 \maketitle

Try running pandoc with --latex-engine=xelatex.
pandoc: Error producing PDF
Error: pandoc document conversion failed with error 43
Execution halted

• U+2028 is unicode LINE SEPARATOR an unusual end-of-line marker, you shoudl probably delete the line and retype it making sure it just has a normal U+000A end of line marker. – David Carlisle Feb 3 at 20:31
• \LaTeX isn't intended to be math mode so normally it would be \LaTeX not $\LaTeX$ unless the \$ here are part of some non-tex markdown convention? – David Carlisle Feb 3 at 20:33

You need to escape the backslash using

title: "Welcome to \\LaTeX{} Workshop"
...

Or you can use a work-around:

---
title: "Welcome to LaTeX Workshop"
output:
pdf_document

The above approach removes all formatting from the title and resets it via the traditional way using \title after pandoc's \title insertion. This effectively overwrites it without the processing used by pandoc.


text:
# 0% Нас он CH3 CH3 H20 Br however, does not result from the addition of HO-Br, for example. Instea...

###### Question:

0% Нас он CH3 CH3 H20 Br however, does not result from the addition of HO-Br, for example. Instead the addition is done indirectly by reaction of the alkene with Br2 in the presence of water. The reaction also works with Cl2 to give chlorohydrins instead of bromohydrins. The reaction proceeds through a cyclic intermediate known as a bromonium ion. In the second step of the reaction, water is the nucleophile and reacts with the bromonium ion to yield the product The reaction can also proceed using an alcohol as the solvent instead of water, to give a 1,2-haloether Draw curved arrows to show the movement of electrons in this step of the mechanism An CH3 CH3 Br CH3 CH3 H3C Br Br. :Br.
Draw s al formulas for all alkenes that could be used to prepare the alcohol shown below by H3C OH 2. NaBH4 CH3 You do not have to consider stereochemistry Include only alkenes that will give the alcohol as the single major product. Separate structures with signs from the drop-down menu. . .
он 1. Hg(OAch, H20 CH3 H3C c Ht adds to the sp? carbon with the most hydrogens to yield the most stable carbocation to an alkene yields an rearrangements can occur prior to the addition of watet. intermediate, which then adds water to give the product alcohol. Because a carbocation intermediate is To avoid the possibility of rearangement and still give a Markovnıkov alcohol, alkenes can instead be treated with m acetate in aqueous THF and then subsequently reduced with sodium through a cyclic mercurinium ion irnt hich cannot rearrange. Water adds to the cyclic intermediate at the most substituted carbon to borohydride. This reaction alcohol. The reduction step with sodium borohydride is complex and involves radicals. Draw curved atrows to show the movement of electrons in this step of the m OAc OAc Hg Hg H2C ње H2C CH3 CH3 PreviousN
Draw a for the formed upon hydrob ation of the alkene below CH3 CH3CHCH= CH2 . Use wedge and hash bonds ONLY for rings. . Do not show stereochemistry in other cases. . If the reaction s a racemic mixture, just draw one stereoisomer
CH3 CH3 1. BH3-THF 2. NaOH, H20, H2o2 он CH2 H3C Alkenes can be hydrated via the addition of borane to yield alcohols with non-Markovnikov regsochemistry The boron atom is an electrophile and thus follows Markovnikov's rule in adding to the carbon alcohol Draw curved arrows to show the movement of electrons in this step of the mechanism An CH3 CH3 H H-BH2 BH CH2 H3C 10 iterm attempts remaining
select the s that could be used to carry out the on, giving the al as the single major In each case trans The procedures are Hydroborationoxidation alkene + ВНЗ; then H2O2, OH. Oxymercuration: alkene + Hg(OAc)2, H20; then NaBH4 CH3 CH3 CH3 CH3CHCH CCH3CHCHCHCH3 CH3 CH3 он он CH3CH2 CH2CH3 CH2CH2CHCH2CH2CH3 Iry Another Version 10 item attempts remaining
Either the reactant (X) or the major o product is missing from the below Draw the missing 1S XH2Pd/C CH2CH(CH3)2 . You do not have to consider stereochemistry . In the case of a missing reactant, there may be more than one answer. If so, draw all possible reactants, . Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate multiple reactants using the +sign from the drop-down menu.

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text:
# Exponential integral En(x) Calculator

## Calculates the exponential integral En(x).

 $E_n(x)={\large\int_{\small 1}^{\hspace{25}\small\infty}\frac{\large e^{-xt}}{\large t^n}}dt$

 n x 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit
 $\normal Exponential\ integral\ E_n(x)\\[10](1)\ E_n(x)={\large\int_{\small 1}^{\hspace{25}\small\infty}\frac{\large e^{-xt}}{\large t^n}}dt\\(2)\ E_n(0)={\large\frac{1}{n-1}}\\[10](3)\ E_1(x)=-E_i(-x)\\$

Exponential integral En(x)
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at home without handbook tables
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I am a rewtired Physical Chemistry Professor and am using Physical Chemistry to do calculation of greenhouse gas fluxes. I need values of the exponential integral function to evaluate the fluxes.
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Exponential Integral function has no short-term expansion for values around 1 or .75. This was useful for that.
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Special functions calc
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text:
# Characteristics of the injected ion beam in the ECR charge breeder $1^{+}\to n^{+}$

Abstract : Different ion species (rare gases, alkali, metallic) have been injected on the axis of the MINIMAFIOS - 10 GHz - Electron Cyclotron Resonance Ion Source which is the basics of the 1+ -> n+ method, special attention have been paid to the optics of the incoming beam for the validation of the 1+ -> n+ method for the SPIRAL project (Radioactive Ion Beam facility). The capture of the incoming ion beam by the ECR plasma depends, first, on the relative energy of the incoming ions with respect to the average ion energy in the plasma, and secondly, on the optics of the injection line. The efficiency of the process when varying the potential V n+ of the MINIMAFIOS source with respect to the potential V 1+ applied to the 1+ source (DV=V n+ -V 1+ ) is an image of the energy dispersion of the 1+ beam. 1+ -> n+ spectra efficiencies, DV efficiency dependence for the most efficient charge state obtained, and measured primary beam emittances are given for the Ar, Rb, Pb, Cr. Highest efficiencies obtained are respectively Ar1+ -> Ar8+ : 8.7 %, Rb1+ -> Rb15+ : 5.5 %, Pb 1+ -> Pb 22+ : 4.8 % , Cr 1+ -> Cr 12+ : 3.5 %. Last results obtained are given for Sulfur and Uranium.
Document type :
Conference papers

Cited literature [4 references]

http://hal.in2p3.fr/in2p3-00006012
Contributor : LPSC bibliotheque Connect in order to contact the contributor
Submitted on : Thursday, September 7, 2000 - 4:46:04 PM
Last modification on : Thursday, November 19, 2020 - 12:58:27 PM
Long-term archiving on: : Friday, May 29, 2015 - 5:10:59 PM

### Identifiers

• HAL Id : in2p3-00006012, version 1

### Citation

T. Lamy, J.F. Bruandet, N. Chauvin, J.C. Curdy, R. Geller, et al.. Characteristics of the injected ion beam in the ECR charge breeder $1^{+}\to n^{+}$. International Workshop on ECR Sources 14, 1999, Geneva, Switzerland. pp.107-110. ⟨in2p3-00006012⟩

Record views


text:
Wikipedia - Potenziale di Morse Morse potential

https://doi.org/10.1351/goldbook.MT07073
The empirical function relating the potential energy of a molecule to the inter-atomic distance r accounting for the anharmonicity of bond stretching: $E\left(r\right) = D_{\text{e}}\ (1- \text{e}^{-a\ (r- r_{\text{e}})})^{2}$ where $$D_{\text{e}}$$ is the @B00699@, $$r_{\text{e}}$$ is the equilibrium @BT07003@, and $$a$$ is a parameter characteristic of a given molecule.
See:
harmonic approximation
Source:
PAC, 1999, 71, 1919. (Glossary of terms used in theoretical organic chemistry) on page 1953 [Terms] [Paper]


text:
# new language accepted by FA when new transitions are added to a FA?

found this question online and I am trying to solve this question. I have solved this question but I think I might be missing some cases. Could someone verify if my answers are correct?

Let N = (Σ ∪ {λ}, Q, δ, q0, {qf }) be a λ − NFA and let L = L(N) be the language accepted by N. For N assume q0 has no incoming transitions and qf has no outgoing transitions. For each of the following FAs that are modifications of N describe the language accepted by each in terms of L.

(a) A λ-transition is added from qf to q0.

(b) Add a λ-transition from q0 to every state reachable from q0 along a path with labels that may be λ or symbols from Σ.

(c) Add a λ-transition to qf from every state that reach qf along some path.

(d) The FA where both (b) and (c) are done.

a) L+

b) For this one I have defined a function suff, this takes all the valid strings in the original Language and generates new stings from them. For example, if L = {112,12} where alphabet is {1,2}

Then, suff(L) means suff(112) + suff(12)

where suff(112) = {112,12,2}, keeps chopping off the first symbol until u get empty string

Similarly, suff(12) = {12,2}

So, the final answer would be suff(L) + the empty string. (Is this correct?)

C) for this one, I defined pre which is similar to suff but chops off the last symbol.

pre(123) = {123,12,1}
pre(112) = {112,11,1}

So, the final ans is pre(L) + empty string

d) I need help with this one..

• What does L+ mean? – kotu Feb 18 at 19:43
• Kleene plus operator applied to language L – ray an Feb 18 at 19:53
• You mean this thing, right? en.wikipedia.org/wiki/Kleene_star – kotu Feb 18 at 19:55
• – ray an Feb 18 at 19:58
• You should look at the first three categories of en.wikipedia.org/wiki/Substring – Nathaniel Feb 18 at 21:57

Hint for (d): Let's say $$s = s_1s_2...s_k \in L$$, and on input $$s$$ the original transition function follows the path $$P = q_0, q_1... q_f$$. Let $$q_{s_i}$$ denote the state $$q_i$$ closest to $$q_f$$ in $$P$$ after only reading up to $$s_i$$ in the original NFA. What are the strings can I generate by doing the following in the NFA defined by (d)? $$\lambda$$-transition from $$q_0$$ to $$q_{s_i}$$, follow $$P$$ up to $$q_{s_j}$$ (for any $$1 < i < j < k$$), and then $$\lambda$$-transition to $$q_f$$ and accept.


text:
HILBERT 2-CLASS FIELD TOWERS OF REAL QUADRATIC FUNCTION FIELDS

Title & Authors
HILBERT 2-CLASS FIELD TOWERS OF REAL QUADRATIC FUNCTION FIELDS
Jung, Hwanyup;

Abstract
In this paper we study the infiniteness of Hilbert 2-class field towers of real quadratic function fields over $\small{\mathbb{F}_q(T)}$, where q is a power of an odd prime number.
Keywords
Hilbert 2-class field tower;real quadratic function field;
Language
English
Cited by
References
1.
S. Bae, S. Hu, and H. Jung, The generalized R'edei matrix for function fields, Finite Fields Appl. 18 (2012), no 4, 760-780.

2.
F. Gerth, Quadratic fields with infinite Hilbert 2-class field towers, Acta Arith. 106 (2003), no. 2, 151-158.

3.
H. Jung, Imaginary bicyclic function fields with the real cyclic subfield of class number one, Bull. Korean Math. Soc. 45 (2008), no. 2, 375-384.

4.
H. Jung, Hilbert 2-class field towers of imaginary quadratic function fields, J. Chungcheong Math. Soc. 23 (2010), no. 3, 547-553.

5.
F. Lemmermeyer, The 4-class group of real quadratic number fields, Preprint, 1998.

6.
C. Maire, Un raffinement du theoreme de Golod-Safarevic, Nagoya Math. J. 150 (1998), 1-11.

7.
J. Martinet, Tours de corps de classes et estimations de discriminants, Invent. Math. 44 (1978), no. 1, 65-73.

8.
M. Rosen, The Hilbert class field in function fields, Exposition. Math. 5 (1987), no. 4, 365-378.

9.
R. Schoof, Algebraic curves over \${\mathbb{F}}_2\$ with many rational points, J. Number Theory 41 (1992), no. 1, 6-14.


text:
source:docs/PACT2011/05-corei3.tex@1048

Last change on this file since 1048 was 1048, checked in by ksherdy, 8 years ago

General edits.

File size: 8.9 KB
RevLine
[1042]1\section{Baseline Evaluation on \CI{}}
[969]2
3%some of the numbers are roughly calculated, needs to be recalculated for final version
4\subsection{Cache behavior}
[1039]5\CI\ has a three level cache hierarchy.  The miss penalty for each
[1048]6level is approximately 4, 11, and 36 cycles respectively.  Figure
[1001]7\ref{corei3_L1DM}, Figure \ref{corei3_L2DM} and Figure
8\ref{corei3_L3TM} show the L1, L2 and L3 data cache misses of all the
9four parsers.  Although XML parsing is not a memory intensive
10application, the cost of cache miss for Expat and Xerces can be about
[1048]11half a cycle per byte while the performance of Parabix is essentially
12unaffected by cache misses.  Cache misses are not just a problem for
[1001]13performance but also energy consumption.  L1 cache miss cost about
148.3nJ; L2 cache miss cost about 19nJ; L3 cache miss cost about 40nJ.
[1048]15With a 1GB input file, Expat and Xerces would consume over 0.6J and 0.9J due to cache misses alone respectively.
16%With a 1GB input file, Expat would consume more than 0.6J and Xerces
17%would consume 0.9J on cache misses alone.
[969]18
19
20\begin{figure}
21\begin{center}
[1004]22\includegraphics[width=0.5\textwidth]{plots/corei3_L1DM.pdf}
[969]23\end{center}
[1039]24\caption{L1 Data Cache Misses on \CI\ (y-axis: Cache Misses per KByte)}
[969]25\label{corei3_L1DM}
26\end{figure}
27
28\begin{figure}
29\begin{center}
[1004]30\includegraphics[width=0.5\textwidth]{plots/corei3_L2DM.pdf}
[969]31\end{center}
[1039]32\caption{L2 Data Cache Misses on \CI\ (y-axis: Cache Misses per KByte)}
[969]33\label{corei3_L2DM}
34\end{figure}
35
36\begin{figure}
37\begin{center}
[1004]38\includegraphics[width=0.5\textwidth]{plots/corei3_L3CM.pdf}
[969]39\end{center}
[1039]40\caption{L3 Cache Misses on \CI\ (y-axis: Cache Misses per KByte)}
[969]41\label{corei3_L3TM}
42\end{figure}
43
44\subsection{Branch Mispredictions}
[1001]45Despite years of improvement, branch misprediction is still a
[1048]46significant bottleneck when it comes to performance.  The cost of a branch
47misprediction is generally over 10 CPU cycles.  As shown in
48Figure \ref{corei3_BM}, the cost of branch mispredictions per byte of XML for Expat
49can be over 7 cycles---which is approximately the number of cycles
50required by Parabix2 to process a byte of XML data using the same workload.
[969]51
[1048]52But reducing the branch misprediction rate is difficult for text-based
[1001]53applications due to the variable-length nature of syntactic elements.
[1048]54Therefore, the goal is to reduce the total number of branches.  However, traditional byte-at-a-time XML
55parsing requires a large number of inevitable branches.  As
[1001]56shown in Figure \ref{corei3_BR}, Xerces can have an average of 13
57branches for each byte it processed on the high markup density file.
[1048]58Parabix1 minimizes the branches by using parallel bit streams for each 128-bit block but still requires a few
59branches for sequential scanning. Utilizing the new parallel scanning technique, Parabix2 is relatively branch-free, as shown in Figure \ref{corei3_BR}. As a result, Parabix2 has minimal
[1001]60dependency on the markup density of the workloads.
[1048]61% Parabix1 minimize the branches by using parallel bit
62% streams.  Parabix1 still have a few branches for each block of 128
63% bytes (SSE) due to the sequential scanning.  But with the new parallel
64% scanning technique, Parabix2 is essentially branch-free as shown in
65% the Figure \ref{corei3_BR}.  As a result, Parabix2 has minimal
66% dependency on the markup density of the workloads.
[969]67
68\begin{figure}
69\begin{center}
[1004]70\includegraphics[width=0.5\textwidth]{plots/corei3_BR.pdf}
[969]71\end{center}
[1039]72\caption{Branches on \CI\ (y-axis: Branches per KByte)}
[969]73\label{corei3_BR}
74\end{figure}
75
76\begin{figure}
77\begin{center}
[1004]78\includegraphics[width=0.5\textwidth]{plots/corei3_BM.pdf}
[969]79\end{center}
[1039]80\caption{Branch Mispredictions on \CI\ (y-axis: Branch Mispredictions per KByte)}
[969]81\label{corei3_BM}
82\end{figure}
83
[1048]84\subsection{SIMD Instructions vs. Total Instructions}
[969]85
[1001]86Parabix gains its performance by using parallel bitstreams, which are
87mostly generated and calculated by SIMD instructions.  The ratio of
88executed SIMD instructions over total instructions indicates the
[1048]89amount of parallel processing we were able to achieve.
90Using Intel PIN, a dynamic binary instrumentation tool, we gathered the running instruction mix of each XML workload and classified the instructions as either vector (SIMD-based) instructions or non-vector (Non-SIMD-based) instructions.
91Figure \ref{corei3_INS_p1} and Figure \ref{corei3_INS_p2} shows the
92percentage of SIMD instructions of Parabix1 and Parabix2
93%(Expat and Xerce do not use any SIMD instructions)
94.  For Parabix1, 18\% to 40\%
[1001]95of the executed instructions consists of SIMD instructions.  By using
96bistream addition for parallel scanning, Parabix2 uses 60\% to 80\%
[1048]97SIMD instructions.  Although the resulting ratios are (negatively) proportional to the markup density
98for both Parabix1 and Parabix2, the degradation rate of
99Parabix2 is much lower and thus the performance penalty incurred by
100increasing the markup density is reduced.
101%Expat and Xerce do not use any SIMD instructions and were not included in this portion of the study.
[969]102
[1048]103% Parabix gains its performance by using parallel bitstreams, which are
104% mostly generated and calculated by SIMD instructions.  The ratio of
105% executed SIMD instructions over total instructions indicates the
106% amount of parallel processing we were able to achieve.  We use Intel
107% pin, a dynamic binary instrumentation tool, to gather instruction mix.
108% Then we adds up all the vector instructions that have been executed.
109% Figure \ref{corei3_INS_p1} and Figure \ref{corei3_INS_p2} show the
110% percentage of SIMD instructions of Parabix1 and Parabix2 (Expat and
111% Xerce do not use any SIMD instructions).  For Parabix1, 18\% to 40\%
112% of the executed instructions consists of SIMD instructions.  By using
113% bistream addition for parallel scanning, Parabix2 uses 60\% to 80\%
114% SIMD instructions.  Although the ratio decrease as the markup density
115% increase for both Parabix1 and Parabix2, the decreasing rate of
116% Parabix2 is much lower and thus the performance degradation caused by
117% increasing markup density is smaller.
118
119
[969]120\begin{figure}
121\begin{center}
[1004]122\includegraphics[width=0.5\textwidth]{plots/corei3_INS_p1.pdf}
[969]123\end{center}
[1004]124\caption{Parabix1 SIMD Instruction Ratio (y-axis: percent)}
[969]125\label{corei3_INS_p1}
126\end{figure}
127
128\begin{figure}
129\begin{center}
[1004]130\includegraphics[width=0.5\textwidth]{plots/corei3_INS_p2.pdf}
[969]131\end{center}
[1004]132\caption{Parabix2 SIMD Instruction Ratio (y-axis: percent)}
[969]133\label{corei3_INS_p2}
134\end{figure}
135
136\subsection{CPU Cycles}
137
[1001]138Figure \ref{corei3_TOT} shows the result of the overall performance
[1048]139evaluated as CPU cycles per thousand input bytes.  Parabix1 is 1.5 to
[1001]1402.5 times faster on document-oriented input and 2 to 3 times faster on
141data-oriented input compared with Expat and Xerces.  Parabix2 is 2.5
142to 4 times faster on document-oriented input and 4.5 to 7 times faster
143on data-oriented input.  Traditional parsers can be dramatically
144slowed down by higher markup density while Parabix with parallel
145processing is less affected.  The comparison is not entirely fair for
146Xerces that transcodes input into UTF-16, which typically takes
147several cycles per byte.  However, transcoding using parallel
148bitstreams can be much faster and it takes less than a cycle per byte
149to transcode ASCII files such as road.gml, po.xml and soap.xml
150\cite{Cameron2008}.
[969]151
152\begin{figure}
153\begin{center}
[1004]154\includegraphics[width=0.5\textwidth]{plots/corei3_TOT.pdf}
[969]155\end{center}
[1039]156\caption{Processing Time on \CI\ (y-axis: Total CPU Cycles per KByte)}
[969]157\label{corei3_TOT}
158\end{figure}
159
160\subsection{Power and Energy}
161There is a growing concern of power consumption and energy efficiency.
[1001]162Chip producers not only work on improving the performance but also
[1048]163have worked hard to develop power efficient chips. We studied the
[1001]164power and energy consumption of Parabix in comparison with Expat and
[1042]165Xerces on \CI{}
[969]166
[1001]167Figure \ref{corei3_power} shows the average power consumed by the four
[1042]168different parsers.  The average power of \CI{} 530 is about 21 watts.
[1001]169This model released by Intel last year has a good reputation for power
170efficiency.  Parabix2 dominated by SIMD instructions uses only about
1715\% higher power than the other parsers.
[969]172
173\begin{figure}
174\begin{center}
[1004]175\includegraphics[width=0.5\textwidth]{plots/corei3_power.pdf}
[969]176\end{center}
[1039]177\caption{Average Power on \CI\ (watts)}
[969]178\label{corei3_power}
179\end{figure}
180
[1001]181The more interesting trend is energy, Figure \ref{corei3_energy} shows
182the energy consumption of the four different parsers.  Although
[1048]183Parabix2 requires slightly more power (per instruction), its processing time is significantly lower
184and therefore consumes substantially less energy than the other parsers. Parabix2 consumes 50 to 75
[1001]185nJ per byte while Expat and Xerces consumes 80nJ to 320nJ and 140nJ to
186370nJ per byte seperately.
[969]187
188\begin{figure}
189\begin{center}
[1004]190\includegraphics[width=0.5\textwidth]{plots/corei3_energy.pdf}
[969]191\end{center}
[1039]192\caption{Energy Consumption on \CI\ ($\mu$J per KByte)}
[969]193\label{corei3_energy}
194\end{figure}
195
Note: See TracBrowser for help on using the repository browser.


text:
# Field Extension

• Dec 30th 2008, 11:26 AM
peteryellow
Field Extension
How can I show that there is no finite field extension of \$\mathbb C.\$
• Dec 30th 2008, 11:32 AM
peteryellow
I thought to prove it by contradiction. but then I am having problem to prove it.
• Dec 30th 2008, 01:52 PM
ThePerfectHacker
Quote:

Originally Posted by peteryellow
How can I show that there is no finite field extension of \$\mathbb C.\$

Since $\mathbb{C}$ is algebraically closed it has no proper algebraic extension let alone a finite extension.


text:
Example A-2-3 - Maple Help

Appendix

A-2: Arithmetic Calculations

Example A-2.3

 Determine if ${2}^{9}-1$ is a composite number or a prime. If it is composite, find its integer factors.

For more information on Maplesoft products and services, visit www.maplesoft.com


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Formulas for Problem Solving

## The d = rt formula

Estimated9 minsto complete
%
Progress
Practice Formulas for Problem Solving
Progress
Estimated9 minsto complete
%
MPH to KPH
Teacher Contributed

## Real World Applications – Algebra I

### Topic

Converting from MPH to KPH

### Student Exploration

There are major differences driving in the United States and driving in the United Kingdom (and possibly other countries, too!). For example, people drive on different sides of the road. In the US, people usually drive on the right side of the road. In the UK, people drive on the left side of the road.

Speed limit is also different. The units are different. In the US, speed limit signs are given in miles per hour. See below.

In the UK, these speed limit signs mean are talking about kilometers per hour. See below

Even though these two signs have the same number, they don’t represent the same thing. If you’re from the United States, would you want to go just as fast or slower than what the sign says? Which speed limit is higher?

We can find out by using a formula to convert the number of kilometers to miles. We know that \begin{align*}1\ mile = 1.609\ kilometers\end{align*}. With this information, we can multiply any number of miles by 1.609 to find its equivalence in kilometers.

For the example above, to find out how many kilometers is equal to 50 miles, we use the formula. Let \begin{align*}k\end{align*} represent the number of kilometers, and \begin{align*}m\end{align*} represent the number of miles we’re going to convert.

So, \begin{align*}k = 1.609 m\end{align*}

\begin{align*}& k = 1.609 ^* 50\\ & k = 80.45\ kilometers\end{align*}

So, if we saw the “50kph” sign in the UK, we probably don’t want to drive 50 miles per hour. We’d probably get a speeding ticket!

### Extension Investigation

See if you can figure out a formula to convert the number of kilometers to miles, given the information above. What is the conversion factor and why?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes


text:
# [texhax] How do I find out what my page layout defaults are?

Steve Schwartz s.schwartz at imperial.ac.uk
Fri Apr 23 18:10:38 CEST 2010

Mark,

On Fri, 2010-04-23 at 14:31 +0100, Mark Woodard wrote:
> "I would like to know the margin measurements as well as the font,
> including the size used for Chapter title, solutions, section labels,

The margins and such are easy:

typeset the following (your preamble fragment plus 4 lines)

\documentclass[twoside]{book}
\usepackage[margin=1in,twoside]{geometry}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}

\usepackage{layout}
\begin{document}
\layout
\end{document}

which will result in a document that has pictures that show the various
lengths and their values.

I don't know of something similar to show the font sizes, but if you
look at the log file when you typeset the above you will see what class
file(s) were used, e.g.,

(/usr/share/texmf/tex/latex/base/book.cls
Document Class: book 2005/09/16 v1.4f Standard LaTeX document class
(/usr/share/texmf/tex/latex/base/bk10.clo))

The book.cls file defines all the formatting in terms of latex font
sizes, e.g.:

\newcommand\section{\@startsection {section}{1}{\z@}%
{-3.5ex \@plus -1ex \@minus -.2ex}%
{2.3ex \@plus.2ex}%
{\normalfont\Large\bfseries}}

tells you that the section head will be in the normal font, using
boldface, and size Large. Then bk10.clo tells you what Large means for
the 10pt book class (10pt is the default):

\newcommand\Large{\@setfontsize\Large\@xivpt{18}}

So Large is 14pt (set on 18pt baselines). This is a bit more digging
than you might like, but it should give you everything you need.

If you're in any doubt, just add a few test lines to the empty document,
e.g.

\bf{\Large{Large font}}

and compare the two lines for size.

HTH

Steve

--
+-------------------------------------------------------------------+
Professor Steven J Schwartz        Phone: +44-(0)20-7594-7660
Head, Space & Atmospheric Physics  Fax:   +44-(0)20-7594-7772
The Blackett Laboratory            E-mail: s.schwartz at imperial.ac.uk
Imperial College London            Office: Huxley 6M67A
London SW7 2AZ, U.K.               Web: www.sp.ph.ic.ac.uk/~sjs
+-------------------------------------------------------------------+


text:
# ROC and AUC curve for CNN multi-class classification problem

I have produced a convolutional neural network to classify images (malware images) into different classes/families. I have managed to produce a confusion matrix and classification report. My supervisor for my project wants me to output a ROC and AUC graph, but ROC and AUC graphs are only really for binary classification problems when reading online articles; this is what it tells me.

I am new to this area but thought I'd spend my weekend giving this a go. So far, nothing online has been helpful.

My model predicts on a dataset with 9,339 images; however, the dataset is unbalanced, but I have addressed this in the model. In total, there are 25 different classes/Families an image could belong to. I have read up that you can produce a ROC graph for every possible class.

path_to_data = "dataset"

train_test_generate_batches = ImageDataGenerator()
batch = train_test_generate_batches.flow_from_directory(directory=path_to_data, target_size=(64,64), batch_size=10000)
imgs, labels = next(batch)
X_train, X_test, y_train, y_test = train_test_split(imgs/255.,labels, train_size=0.7, test_size=0.3)

num_classes = 25

model = Sequential()

#input layer
model.add(Conv2D(32, kernel_size=(3, 3), activation='relu', input_shape=(64,64,3)))
model.add(Conv2D(16, (3, 3), activation='relu'))


text:
Activity >

### GAIA Seminar on Complex Analytic Geometry - Dr. Thomas Pawlaschyk

posted Mar 13, 2016, 6:50 PM by Hyemin Shin
 - Title: Polynomial convexity for unbounded closed sets in C^n- Speaker: Dr. Thomas Pawlaschyk (Universität Wuppertal) - Abstract: I will introduce different types of polynomially convex hulls for arbitrary closed sets A in C^n. I will give some of their properties and examples. One of these hulls is the closure of the union of the polynomially convex hulls of all compact sets lying inside A. Another one is the collection of points which satisfy the following property: if there is a non-negative continuous entire plurisubharmonic function which vanishes on A, then it already vanishes at this point.  These definitions arise naturally from the characterization of polynomially convex hulls in the compact case. In fact, in contrary to the compact case, it seems that they are not the same. To see this, I will give an example in C^2. At the end, I will explain the situation in \$\cbb\$ and discuss our recent observations on unbounded hulls. The study of these hulls is a joint work with T. Harz and Y. Nagata.  - Place: GAIA Seminar Room (#106, Math Science bldg.) - Date: 2 p.m. - 4 p.m on Thursday, 17 March 2016Poster <---- click


text:
# American Institute of Mathematical Sciences

January  2011, 10(1): 327-338. doi: 10.3934/cpaa.2011.10.327

## Critical points of solutions to elliptic problems in planar domains

 1 Departamento de Matemáticas, Universidad del Valle, Cali, Colombia, Colombia

Received  April 2009 Revised  October 2009 Published  November 2010

Given a planar domain $\Omega$, and an analytic function $f$, we describe the set of critical points for the solution $u$ of the semilinear elliptic problem $\Delta u = f(u)$ in $\Omega$, $u=0$ on $\partial\Omega$. For simply connected domains we establish that the set of critical points is finite while for non--simply connected domains we show that this set is made up of finitely many isolated points and finitely many analytic Jordan curves. Further results are given in the case that $\Omega$ is an annular domain whose border has nonzero curvature.
Citation: Jaime Arango, Adriana Gómez. Critical points of solutions to elliptic problems in planar domains. Communications on Pure & Applied Analysis, 2011, 10 (1) : 327-338. doi: 10.3934/cpaa.2011.10.327
##### References:
 [1] G. Alessandrini and R. Magnanini, The index of isolated critical points and solutions of elliptic equations in the plane,, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 19 (1992), 567. Google Scholar [2] J. Arango, Uniqueness of critical points for semi-linear elliptic problems in convex domains,, Electron. J. Differential Equations, 2005 (2005), 1. Google Scholar [3] J. Arango and O. Perdomo, Morse theory for analytic functions on surfaces,, J. Geom., 84 (2005), 13. doi: doi:10.1007/s00022-005-0027-8. Google Scholar [4] V. I. Arnold, "The Theory of Singularities and its Applications,'', Cambridge University Press, (1991). Google Scholar [5] X. Cabré, and S. Chanillo, Stable solutions of semilinear elliptic problems in convex domains,, Selecta Math. (N.S.), 4 (1998), 1. Google Scholar [6] S.Y. Cheng, Eigenfunctions and nodal sets,, Comment. Math. Helv., 51 (1976), 43. doi: doi:10.1007/BF02568142. Google Scholar [7] D.L. Finn, Convexity of level curves for solutions to semilinear elliptic equations,, Commun. Pure Appl. Anal., 7 (2008), 1335. doi: doi:10.3934/cpaa.2008.7.1335. Google Scholar [8] G. Gilbarg and N. Trudinger, "Elliptic Partial Differential Equations of Second Order,'', Springer-Verlag, (1983). Google Scholar [9] B. Kawohl, When are solutions to nonlinear elliptic boundary value problems convex?,, Comm.\ Partial Differential Equations, 10 (1985), 1213. doi: doi:10.1080/03605308508820404. Google Scholar [10] Xi-Nan Ma, Concavity estimates for a class of nonlinear elliptic equations in two dimensions,, Math. Z., 240 (2002), 1. doi: doi:10.1007/s002090100341. Google Scholar [11] L.G. Makar-Limanov, Solution of Dirichlet's problem for the equation $\Delta u=-1$ in a convex region,, Math. Notes Acad. Sci. U.S.S.R., 9 (1971), 52. doi: doi:10.1007/BF01405053. Google Scholar [12] F. Müller, On the continuation of solutions for elliptic equations in two variables,, Ann. Inst. H. Poincaré Anal. Non Linéaire, 19 (2002), 745. Google Scholar

show all references

##### References:
 [1] G. Alessandrini and R. Magnanini, The index of isolated critical points and solutions of elliptic equations in the plane,, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 19 (1992), 567. Google Scholar [2] J. Arango, Uniqueness of critical points for semi-linear elliptic problems in convex domains,, Electron. J. Differential Equations, 2005 (2005), 1. Google Scholar [3] J. Arango and O. Perdomo, Morse theory for analytic functions on surfaces,, J. Geom., 84 (2005), 13. doi: doi:10.1007/s00022-005-0027-8. Google Scholar [4] V. I. Arnold, "The Theory of Singularities and its Applications,'', Cambridge University Press, (1991). Google Scholar [5] X. Cabré, and S. Chanillo, Stable solutions of semilinear elliptic problems in convex domains,, Selecta Math. (N.S.), 4 (1998), 1. Google Scholar [6] S.Y. Cheng, Eigenfunctions and nodal sets,, Comment. Math. Helv., 51 (1976), 43. doi: doi:10.1007/BF02568142. Google Scholar [7] D.L. Finn, Convexity of level curves for solutions to semilinear elliptic equations,, Commun. Pure Appl. Anal., 7 (2008), 1335. doi: doi:10.3934/cpaa.2008.7.1335. Google Scholar [8] G. Gilbarg and N. Trudinger, "Elliptic Partial Differential Equations of Second Order,'', Springer-Verlag, (1983). Google Scholar [9] B. Kawohl, When are solutions to nonlinear elliptic boundary value problems convex?,, Comm.\ Partial Differential Equations, 10 (1985), 1213. doi: doi:10.1080/03605308508820404. Google Scholar [10] Xi-Nan Ma, Concavity estimates for a class of nonlinear elliptic equations in two dimensions,, Math. Z., 240 (2002), 1. doi: doi:10.1007/s002090100341. Google Scholar [11] L.G. Makar-Limanov, Solution of Dirichlet's problem for the equation $\Delta u=-1$ in a convex region,, Math. Notes Acad. Sci. U.S.S.R., 9 (1971), 52. doi: doi:10.1007/BF01405053. Google Scholar [12] F. Müller, On the continuation of solutions for elliptic equations in two variables,, Ann. Inst. H. Poincaré Anal. Non Linéaire, 19 (2002), 745. Google Scholar
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2018 Impact Factor: 0.925


text:
Question

# If emf induced in a coil is $$2V$$ by changing the current in it from $$8A$$ to $$6A$$ in $$2\times {10}^{-3}s$$, then the coefficient of self induction is

A
2×103H
B
103H
C
0.5×103H
D
4×103H

Solution

## The correct option is C $$2\times { 10 }^{ -3 }H\quad$$Given :      $$\mathcal{E} = 2$$ V               $$\Delta t = 2\times 10^{-3}$$ sChange in current       $$\Delta I = 6-8 = -2$$ AUsing      $$\mathcal{E} = -L\dfrac{\Delta I}{\Delta t}$$$$\therefore$$    $$2 = -L\times \dfrac{-2}{2\times 10^{-3}}$$                 $$\implies L = 2\times 10^{-3}H$$          Physics

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text:
# Question 53001

Feb 4, 2016

${P}_{\text{sol" = "23.46 torr}}$

#### Explanation:

Notice that the problem provides you with the volume of water and its density at ${25}^{\circ} \text{C}$.

This should automatically tell you that you can find its mass by using the definition of density

$\textcolor{b l u e}{\text{density" = "mass"/"unit of volume}}$

Water is said to have a density of $\text{0.9971 g/mL}$ at that temperature, which means that every milliliters of water will have a mass of $\text{0.9971 g}$.

If that's the case, then $\text{643.5 mL}$ will have a mass of

643.5 color(red)(cancel(color(black)("mL"))) * overbrace("0.9971 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("density")) = "641.634 g"

Now, the vapor pressure of the solution will depend on the mole fraction of the solvent in the solution, ${\chi}_{\text{solvent}}$, and on the vapor pressure of the pure solvent, ${P}_{\text{solvent}}^{\circ}$ - think Raoult's Law.

$\textcolor{b l u e}{{P}_{\text{solvent" = chi_"solvent" xx P_"solvent}}^{\circ}}$

In order to find the mole fraction of the solvent, you need to know the number of moles of solvent and the total number of moles present in solution.

Since you have the mass of water used to make this solution, use water's molar mass to help you find the number of moles it contains

641.634 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "35.617 moles H"_2"O"

To find the total number of moles present in solution, find the number of moles of solute first

158.0 color(red)(cancel(color(black)("g"))) * "1 mole sucrose"/(342.3 color(red)(cancel(color(black)("g")))) = "0.4616 moles sucrose"

The solution will thus contain a total number of

${n}_{\text{total" = "35.617 moles" + "0.4616 moles" = "36.079 moles}}$

The mole fraction of water will be

chi_"water" = (35.617color(red)(cancel(color(black)("moles"))))/(36.079color(red)(cancel(color(black)("moles")))) = 0.9872

Therefore, the vapor pressure of the solution will be

${P}_{\text{sol" = 0.9872 * "23.76 torr}}$

P_"sol" = color(green)("23.46 torr") -># rounded to four sig figs


text:
# Post AT&T tryst, T-Mobile’s decline continues

It’s now been two full quarters since its merger with AT&T was scuttled, but T-Mobile USA isn’t showing many signs of recovery. In the first quarter, T-Mobile looked like it might have reversed course, posting a modest 187,000 net customer gain, but in the second quarter it lost 205,000 subscribers, T-Mobile revealed Thursday in its earnings report.

As has been the case for the last two years, T-Mobile’s contract customers are departing. The problem is T-Mo couldn’t replace with them prepaid customers fast enough. T-Mobile lost 557,000 contract subscribers between April and June, offset by 227,000 net new prepaid subscribers.

The U.S. subsidiary of Deutsche Telekom also added 95,000 machine-to-machine connections, linking to its network to everything from point-of-sale payment devices to ice machines, and it gained 30,000 wholesale customers from its mobile virtual network operator partners such as TracFone’s (s amx) Straight Talk. But both numbers were far below the M2M and wholesale gains it reported in previous quarters.

In all, T-Mobile is a smaller operator today than it was a year ago, having shrunk by about 400,000 subscribers since June 2011.

The bright spot for T-Mobile is that its customers are taking to T-Mobile’s ultra-cheap data plans. It sold 2.1 million smartphones in the last quarter, a 31 percent increase over last year’s second quarter. Those new smartphones drove up the  average revenue per user (ARPU) spent on data by 14.6 percent for contract customers, though overall ARPU in postpaid increased only slightly to $57.35 a month. The continued shift of smartphones to prepaid plans, however, increased prepaid ARPU by 13.6 percent to$26.81 a month.

The lack of the iPhone(s aapl) is still weighing heavily on T-Mobile. Though it isn’t working outright miracles for Sprint(s s) since it arrived on its network last fall, the carrier has leaned on the iPhone to deliver quarterly subscribers boosts. The Apple device is responsible for many of the defections to Sprint from other carriers.

T-Mobile has made no official announcement on the iPhone, but it is in the process of overhauling its network, which will make its network completely iPhone-compatible by next year. The sooner it gets the device, though, the sooner it has a key weapon in fighting off its much larger competitors.

If you want to learn more about T-Mobile’s aggressive network plans, check out GigaOM’s Mobilize conference next month where T-Mobile CTO Neville Ray is speaking on Sept 21.

T-Mobile image courtesy of Flickr user swruler9284


text:
# Q-Puzzle on $\mathbb F_7$

My high school math student had to solve this puzzle for his homework. I made it harder by not telling you what is the definition of $$\mathbb{F}_7$$. Neither why letter Q is used and which of his chapter this is.

Your aim is to reconstitute the original image and discover the pattern. Would you use your technique for any random picture and $$\mathbb F_{17}$$?

Note: the LaTeX mathbb font and color (blue) are different from the ones in puzzling.se, but that's not part of the puzzle (it was a URL link in the image).

Note bis: while it could be easier to solve it thanks to computers, I will favor no computers solutions and math computations. That's why no-computers tag for this puzzle.

• While many image stitching algorithms require partial overlap, I suspect numerical methods could get you most if not all the way to the reconstruction. – Galen May 12 '20 at 17:18
• @Galen You made me add the no-computers tag :) – JKHA May 12 '20 at 17:22

I get the picture

A toucan

I think the maths connection relates to the Farey sequence of order $$7$$.

Some sequence can be see in the skip-1 vertical sequence in the original arrangement which translates to horizontal placings in the solved picture.

Thanks to @MacGyver88 for pointing the way.

• Well! That's a correct no-computers answer. However, would you use your technique if it was $\mathbb F_{17}$? You still get my +1 ;) – JKHA May 12 '20 at 18:35
• I decided to change the puzzle's aim tanks to your answer ;) – JKHA May 12 '20 at 18:37
• Of course, F7 is rot13(Tnybvf svyrq bs beqre 7 (v.r. svryq bs vagrtref zbqhyb 7)). So, the table may be rot13(n xvaq bs nqqvgvba/zhygvcyvpngvba gnoyr bire S7). – trolley813 May 12 '20 at 18:53
• Ah thanks. I was trying to be stealthy. :) – MacGyver88 May 12 '20 at 18:56
• Alright, I understand the issue. I've decided to select your answer as first answer to original puzzle :) – JKHA May 12 '20 at 20:06

The subimages are permuted using an affine map in $$\mathbb F_7^2$$. I wrote a quick program that lets me adjust the affine map using keypresses until the image is correct, and this is the result:

You can read off the exact affine map by hand if you are so inclined, by comparing the positions of tiles that are supposed to be adjacent, but I figured this was more fun and also provided a more visual solution.

• Congrats! That is the answer I wanted to be found ;) if I could give two selected answers, I would for yours – JKHA May 13 '20 at 13:31
• You can change your selection if you really want to do that. – Magma May 13 '20 at 13:54
• That would no be fair according to the comments on @WeatherVane's answer, that's partially my fault, I admit – JKHA May 13 '20 at 16:23


text:
# Properties

 Label 648.4.a.i Level $648$ Weight $4$ Character orbit 648.a Self dual yes Analytic conductor $38.233$ Analytic rank $1$ Dimension $4$ CM no Inner twists $1$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$648 = 2^{3} \cdot 3^{4}$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 648.a (trivial)

## Newform invariants

 Self dual: yes Analytic conductor: $$38.2332376837$$ Analytic rank: $$1$$ Dimension: $$4$$ Coefficient field: 4.4.72153.1 Defining polynomial: $$x^{4} - x^{3} - 8 x^{2} + 3 x + 12$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$2^{3}\cdot 3^{2}$$ Twist minimal: no (minimal twist has level 72) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( 1 + \beta_{2} ) q^{5} + ( -1 - \beta_{3} ) q^{7} +O(q^{10})$$ $$q + ( 1 + \beta_{2} ) q^{5} + ( -1 - \beta_{3} ) q^{7} + ( -4 + \beta_{1} - 2 \beta_{2} - \beta_{3} ) q^{11} + ( -6 - \beta_{1} + 4 \beta_{3} ) q^{13} + ( 1 - 5 \beta_{1} - 10 \beta_{2} + 6 \beta_{3} ) q^{17} + ( -28 + 6 \beta_{1} - \beta_{2} + 2 \beta_{3} ) q^{19} + ( -10 - \beta_{1} - 3 \beta_{2} - 7 \beta_{3} ) q^{23} + ( -26 - 4 \beta_{1} + 9 \beta_{2} - 2 \beta_{3} ) q^{25} + ( -4 + 17 \beta_{1} + 2 \beta_{2} ) q^{29} + ( -70 - 17 \beta_{1} - 25 \beta_{2} + 9 \beta_{3} ) q^{31} + ( -41 - 2 \beta_{1} - 2 \beta_{2} + 3 \beta_{3} ) q^{35} + ( -103 + 13 \beta_{1} + 17 \beta_{2} - 16 \beta_{3} ) q^{37} + ( 86 - 7 \beta_{1} + 21 \beta_{2} - 20 \beta_{3} ) q^{41} + ( -141 + 10 \beta_{1} + 19 \beta_{2} - 29 \beta_{3} ) q^{43} + ( -43 - 6 \beta_{1} + 10 \beta_{2} + 3 \beta_{3} ) q^{47} + ( -211 + 11 \beta_{1} + 2 \beta_{2} - 2 \beta_{3} ) q^{49} + ( 119 - 25 \beta_{1} - 9 \beta_{2} + 32 \beta_{3} ) q^{53} + ( -280 + 5 \beta_{1} - 35 \beta_{2} + 13 \beta_{3} ) q^{55} + ( -149 + 8 \beta_{1} + 13 \beta_{2} + 41 \beta_{3} ) q^{59} + ( -308 - 27 \beta_{1} - 44 \beta_{2} + 28 \beta_{3} ) q^{61} + ( 194 + 9 \beta_{1} + 12 \beta_{2} - 18 \beta_{3} ) q^{65} + ( -300 - 9 \beta_{1} + 56 \beta_{2} + 9 \beta_{3} ) q^{67} + ( -121 + 37 \beta_{1} + 47 \beta_{2} - 56 \beta_{3} ) q^{71} + ( -321 - 39 \beta_{1} + 26 \beta_{2} + 10 \beta_{3} ) q^{73} + ( 175 + 12 \beta_{1} + 9 \beta_{2} - 20 \beta_{3} ) q^{77} + ( -470 + 29 \beta_{1} + 13 \beta_{2} + 15 \beta_{3} ) q^{79} + ( -333 - 60 \beta_{1} - 24 \beta_{2} + 5 \beta_{3} ) q^{83} + ( -539 + 57 \beta_{1} - 3 \beta_{2} - 28 \beta_{3} ) q^{85} + ( 181 + 41 \beta_{1} + 15 \beta_{2} - 36 \beta_{3} ) q^{89} + ( -478 - 41 \beta_{1} - 11 \beta_{2} + 31 \beta_{3} ) q^{91} + ( -286 + 2 \beta_{1} - 118 \beta_{2} + 32 \beta_{3} ) q^{95} + ( -606 + 21 \beta_{1} - 39 \beta_{2} + 64 \beta_{3} ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q + 5q^{5} - 3q^{7} + O(q^{10})$$ $$4q + 5q^{5} - 3q^{7} - 16q^{11} - 29q^{13} - 17q^{17} - 109q^{19} - 37q^{23} - 97q^{25} + 3q^{29} - 331q^{31} - 171q^{35} - 366q^{37} + 378q^{41} - 506q^{43} - 171q^{47} - 829q^{49} + 410q^{53} - 1163q^{55} - 616q^{59} - 1331q^{61} + 815q^{65} - 1162q^{67} - 344q^{71} - 1307q^{73} + 741q^{77} - 1853q^{79} - 1421q^{83} - 2074q^{85} + 816q^{89} - 1995q^{91} - 1292q^{95} - 2506q^{97} + O(q^{100})$$

Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{4} - x^{3} - 8 x^{2} + 3 x + 12$$:

 $$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$3 \nu^{3} - 3 \nu^{2} - 12 \nu + 4$$ $$\beta_{2}$$ $$=$$ $$-\nu^{3} + 5 \nu^{2} - 17$$ $$\beta_{3}$$ $$=$$ $$-\nu^{3} + 5 \nu^{2} + 6 \nu - 19$$
 $$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$($$$$\beta_{3} - \beta_{2} + 2$$$$)/6$$ $$\nu^{2}$$ $$=$$ $$($$$$2 \beta_{3} + \beta_{2} + \beta_{1} + 51$$$$)/12$$ $$\nu^{3}$$ $$=$$ $$($$$$10 \beta_{3} - 7 \beta_{2} + 5 \beta_{1} + 51$$$$)/12$$

## 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
 1.51317 −1.26386 2.90825 −2.15756
0 0 0 −8.01626 0 0.937231 0 0 0
1.2 0 0 0 −5.99447 0 15.5776 0 0 0
1.3 0 0 0 1.69184 0 −17.1413 0 0 0
1.4 0 0 0 17.3189 0 −2.37353 0 0 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Atkin-Lehner signs

$$p$$ Sign
$$2$$ $$-1$$
$$3$$ $$1$$

## 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 648.4.a.i 4
3.b odd 2 1 648.4.a.h 4
4.b odd 2 1 1296.4.a.ba 4
9.c even 3 2 72.4.i.a 8
9.d odd 6 2 216.4.i.a 8
12.b even 2 1 1296.4.a.y 4
36.f odd 6 2 144.4.i.e 8
36.h even 6 2 432.4.i.e 8

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
72.4.i.a 8 9.c even 3 2
144.4.i.e 8 36.f odd 6 2
216.4.i.a 8 9.d odd 6 2
432.4.i.e 8 36.h even 6 2
648.4.a.h 4 3.b odd 2 1
648.4.a.i 4 1.a even 1 1 trivial
1296.4.a.y 4 12.b even 2 1
1296.4.a.ba 4 4.b odd 2 1

## Hecke kernels

This newform subspace can be constructed as the kernel of the linear operator $$T_{5}^{4} - 5 T_{5}^{3} - 189 T_{5}^{2} - 503 T_{5} + 1408$$ acting on $$S_{4}^{\mathrm{new}}(\Gamma_0(648))$$.

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{4}$$
$3$ $$T^{4}$$
$5$ $$1408 - 503 T - 189 T^{2} - 5 T^{3} + T^{4}$$
$7$ $$594 - 387 T - 267 T^{2} + 3 T^{3} + T^{4}$$
$11$ $$38929 + 2056 T - 1866 T^{2} + 16 T^{3} + T^{4}$$
$13$ $$141262 + 33335 T - 3771 T^{2} + 29 T^{3} + T^{4}$$
$17$ $$2567224 - 723268 T - 14562 T^{2} + 17 T^{3} + T^{4}$$
$19$ $$-5081408 - 1246544 T - 11772 T^{2} + 109 T^{3} + T^{4}$$
$23$ $$37951678 - 275669 T - 18855 T^{2} + 37 T^{3} + T^{4}$$
$29$ $$1572741522 - 2926449 T - 94407 T^{2} - 3 T^{3} + T^{4}$$
$31$ $$-3182697596 - 34618859 T - 67149 T^{2} + 331 T^{3} + T^{4}$$
$37$ $$-215981856 - 7645176 T - 18492 T^{2} + 366 T^{3} + T^{4}$$
$41$ $$-360819333 + 38061882 T - 135996 T^{2} - 378 T^{3} + T^{4}$$
$43$ $$-567289217 - 45189154 T - 68328 T^{2} + 506 T^{3} + T^{4}$$
$47$ $$-310694022 - 7460379 T - 37059 T^{2} + 171 T^{3} + T^{4}$$
$53$ $$-15963093536 + 137282536 T - 239580 T^{2} - 410 T^{3} + T^{4}$$
$59$ $$1415131633 - 240406592 T - 485994 T^{2} + 616 T^{3} + T^{4}$$
$61$ $$-48115815704 - 133690159 T + 349659 T^{2} + 1331 T^{3} + T^{4}$$
$67$ $$-137320199621 - 658601234 T - 329940 T^{2} + 1162 T^{3} + T^{4}$$
$71$ $$9762389248 - 196817728 T - 654384 T^{2} + 344 T^{3} + T^{4}$$
$73$ $$-88005243128 - 613828588 T - 217110 T^{2} + 1307 T^{3} + T^{4}$$
$79$ $$-98670843044 - 91124605 T + 875259 T^{2} + 1853 T^{3} + T^{4}$$
$83$ $$67888747828 - 495031429 T - 280773 T^{2} + 1421 T^{3} + T^{4}$$
$89$ $$22003976592 + 86780160 T - 307656 T^{2} - 816 T^{3} + T^{4}$$
$97$ $$-818649939857 - 1529506394 T + 865152 T^{2} + 2506 T^{3} + T^{4}$$


text:
# prove that $n(n+1)$ is even using induction

Prove that $n(n+1)$ is even using induction

The base case of $n=1$ gives us $2$ which is even.

Assuming $n=k$ is true,

$n=(k+1)$ gives us $k^2 +2k +k +2$

while $k(k+1) + (k+1)$ gives us $k^2+2k+1.$

whats is the next step to prove this by induction? I can't seem to show

$k^2 +2k +k +2$ = $k^2+2k+1$

• $n(n+1) = n(n-1)+2n$. By induction, $n(n-1)$ is even. Adding another even number $2n$ gives another even number. – angryavian Nov 27 '15 at 3:19
• $k(k+1) = k^2 + k$ and $(k+1)(k+2) = k^2 + 3k + 2= k^2 + k + (2k+2) = k(k+1) + 2(k+1)$. "while k(k+1)+(k+1) gives us..." where did k(k+1)+ (k+1) (which is $(k+1)^2$) come from? – fleablood Dec 11 '17 at 17:01

Hint: $$(n+1)((n+1)+1)=(n+1)(n+2)=n(n+1)+2(n+1)$$

What you wrote in the second line is incorrect.

To show that $n(n+1)$ is even for all nonnegative integers $n$ by mathematical induction, you want to show that following:

Step 1. Show that for $n=0$, $n(n+1)$ is even;

Step 2. Assuming that for $n=k$, $n(n+1)$ is even, show that $n(n+1)$ is even for $n=k+1$.

[Added:] In Step 2, what you really need to show is the following implication:

if $k(k+1)$ is even, then $(k+1)(k+2)$ is even.

• I see, so simply show that n=k+1 is even? – Noah Deng Nov 27 '15 at 3:22
• Exactly because that would show $n=1$ is also even, which would show $n=2$ is also even, which would show $n=3$ is also even,.... – Ahmed S. Attaalla Nov 27 '15 at 3:23
• Well, it is not that "$n=k+1$ is even", but that "when $n=k+1$, $n(n+1)$ is even". – Jack Nov 27 '15 at 3:23
• I've ended up with $k^2 +k +2(k+1)$, we know that 2(k+1) is even, however it is ok to simply assume $k^2 +k$ is also even? – Noah Deng Nov 27 '15 at 3:29
• (use your inductive hypothesis) – Morgan Rodgers Nov 27 '15 at 7:38

From $k^2$ + $3k$ + 2, you could do cases for k.

Case 1: If k is even, then let k = 2c for some integer c. Then you get $(2c)^2$+ $3(2c)$ +2 , which could be written as 2(2cc) + 2(3c) +2 or 2(2cc + 3c + 1), which is even.

Case 2: If k is odd, then let k = 2c+1 for some integer c. Then you get $(2c+1)^2$+ $3(2c+1)$ +2 , which could be written as $4c^2$ + $4c$ + 1 + $2c$+ + 1 + 2 , where you can simplify the expression down to 2 ($2c^2$ + $3c$ + 2). The factored out 2 indicates that it is even.

• To finish of the proof, you would say that since n=1 makes n(n+1) even and since n=k+1 makes the expression n(n+1) even, then by the principle of induction, for all n greater than equal to 1, n(n+1) is even. – sdggsgehse Nov 27 '15 at 3:53
• A proof by cases applied to $n(n+1)$ is essentially the best proof all by itself. Your answer tacks on induction only because the OP was required to use induction. (I disapprove of questions that force you to use an inappropriate technique just to practice the technique.) – Ethan Bolker Dec 11 '17 at 17:00

Claim:

P(n): n(n+1) is even $\forall n \in \mathbb{N}$

Base case:

P(0): 0(1) = 0. Since 0 can be written in the form 2t, t $\in \mathbb{Z}$, 0 is even. The base case holds.

Alternative base case if you want to start at 1:

P(1): 1(2) = 2. Since 2 can be written in the form 2t, t $\in \mathbb{Z}$, 0 is even. The base case holds.

inductive hypothesis:

Suppose P(n) is true for some k $\in \mathbb{N}$. That is k(k+1) is even. Or $k^{2} + k$ is even.

inductive step:

P(k+1): $$(k+1)((k+1)+1) = (k+1)(k+2)$$ $$= k^{2} + 2k + k + 2$$ $$= (k^{2} + k) + (2k+2)$$ $$= (k^{2} + k) + 2(k+1)$$

By IH $k^{2} + k$ is even and since 2(k+1) is even by definition then P(k+1) is even. By principle of induction P(n) holds $\forall n \in \mathbb{N}$.

Note that it may require proof that even + even = even. Very simple proof however.


text:
IPPP-14-58
DCPT-14-116
arXiv:1406.3030
Comput.Phys.Commun. 197 (2015) 239-262

by: Degrande, Celine (Durham U., IPPP)

Abstract:
The computation of renormalized one-loop amplitudes in quantum field theory requires not only the knowledge of the Lagrangian density and the corresponding Feynman rules, but also that of the ultraviolet counterterms. More in general, and depending also on the method used in the actual computation of the one-loop amplitudes, additional interactions might be needed. One example is that of the R2 rational terms in the OPP method. In this paper, we argue that the determination of all elements necessary for loop computations in arbitrary models can be automated starting only from information on the Lagrangian at the tree-level. In particular, we show how the R2 rational and ultraviolet counterterms for any renormalizable model can be computed with the help of a new package, which we name NLOCT and builds upon FeynRules  and FeynArts . To show the potential of our approach, we calculate all additional rules that are needed to promote a Two Higgs Doublet Model Lagrangian to one-loop computations in QCD and electroweak couplings.


text:
# Proof of the Hamiltonian slice theorem

Theorem(Guillemin Sternberg Marle) Let $$(M, \omega, \mu)$$ be a symplectic manifold together with a Hamiltonian group action. Let $$p$$ be a point in $$M$$ such that $$O_p$$ is contained in the zero level set of the moment map. Denote $$G_p$$ the stabilizer and $$O_p$$ the orbit of $$p$$. There is a $$G$$-equivariant symplectomorphism from a neighborhood of the zero section of the bundle $$T^*G ×_{G_p} V_p$$ equiped with a symplectic model to a neighborhood of the orbit $$O_p$$.

I'm looking for a reference of this theorem. ( The references that people recommend are the articles of Marle which is written in french and the article The normal form for the moment map by Guillemin and Sternberg which I can't download from internet). Could you please give some references where I can find the proof of this theorem.

• I think it is covered in section 41, Symplectic slices and moment reconstruction, starting on p. 324 of Guillemin and Sternberg, Symplectic Techniques in Physics. Dec 9, 2021 at 13:24


text: Python lists are still hash tables containing Python objects even if you declare them as cdef list foo. In particuar, it is not possible to have a memory-contiguous array of a particular C datatype that way. You shoud use C arrays or C++ std::vector if you need arrays of a fixed C datatype.