text:
Author: Josh Cassidy (August 2013)

This five-part series of articles uses a combination of video and textual descriptions to teach the basics of writing a thesis using LaTeX. These tutorials were first published on the original ShareLateX blog site during August 2013; consequently, today's editor interface (Overleaf) has changed considerably due to the development of ShareLaTeX and the subsequent merger of ShareLaTeX and Overleaf. However, much of the content is still relevant and teaches you some basic LaTeX—skills and expertise that will apply across all platforms.

In the previous post we looked at configuring the page layout of our thesis using the geometry and fancyhdr packages. In this post we're going to look at using images and tables.

If you've never added images or tables into a LaTeX document, I recommend you first check out the relevant posts in our beginners tutorial series. You can see our video tutorial on using images in LaTeX and video tutorial on tables.

## Contents

### Images

For this project, every image we use we will store in the images folder to keep everything tidy. In the first post we prepared the document for images by loading up the graphicx package and by informing LaTeX where the images are stored using the \graphicspath command. Whenever we add an image into our thesis, we will use the figure environment. Here's an example:

\begin{figure}[h]
\centering
\includegraphics[scale=0.5]{graph_a}
\caption{An example graph}
\label{fig:x cubed graph}
\end{figure}

Notice that I've halved the size of the image and used the position specifier h to put it in the document where the code is in the text. It's really important to add captions to figures when writing a thesis. This is what it looks like compiled:

Notice that LaTeX has automatically numbered it according to what chapter it's part of. It is also really important to label each figure so you can accurately refer back to it in the text using the \ref command. If you added this in the text:

\ref{fig:x cubed graph}

LaTeX would give you the figure number '2.1' in place of this command in the pdf.

### Subfigures

When writing a thesis you may want to include some slightly more complicated figures with multiple images. You can do this using subfigure environments inside a figure environment. Before we can do this though, we need to load up the caption and subcaption packages:

\usepackage{caption}
\usepackage{subcaption}

We'll do an example with three images along side each other with separate captions and labels. Here's some example code:

\begin{figure}
\centering
\begin{subfigure}[b]{0.3\textwidth}
\centering
\includegraphics[width=\textwidth]{graph1}
\caption{$y=x$}
\label{fig:y equals x}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.3\textwidth}
\centering
\includegraphics[width=\textwidth]{graph2}
\caption{$y=3sinx$}
\label{fig:three sin x}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.3\textwidth}
\centering
\includegraphics[width=\textwidth]{graph3}
\caption{$y=5/x$}
\label{fig:five over x}
\end{subfigure}
\caption{Three simple graphs}
\label{fig:three graphs}
\end{figure}

To start with, we create a new figure, centre it and then create a new subfigure. In the subfigure command we need to add a placement specifier and then give it a width. Because we want three images next to each other we set a width of 0.3 times the value of \textwidth. You need to make sure that the sum of the widths you specify for the subfigures is less than the text width if you want them all on the same line.

When we add the image in we need to specify the width using width= followed by the \textwidth command. The reason this works is because the text width within the subfigure is the width we specified in the \begin{subfigure} command, i.e. 0.3 times the normal text width (which is the value of \textwidth).

Next we give the subfigure a separate caption and label. We can then end the subfigure and add the next two in. To add some spacing between the figures we'll use the \hfill command. If you didn't want them all on the same line you could just leave blank lines instead of the \hfill commands. Please note that the indents I have used do not affect the how the code is processed, they just make it more readable. The beauty of these subfigures is that we can refer to each of them individually in the text due to their individual labels—but we can also give the whole figure a caption and label.

This is what our figure will look like in the document:

Now if we add a \listoffigures command just after the table of contents, LaTeX will generate a list of all the figure used in the thesis and inform us where each can be found:

### Tables

Now lets talk about tables. When writing a thesis you should enclose all your tables in the table environment. Here's a basic example:

\begin{table}[h]
\centering
\begin{tabular}{l | l | l}
A & B & C \\
\hline
1 & 2 & 3 \\
4 & 5 & 6
\end{tabular}
\caption{very basic table}
\label{tab:abc}
\end{table}

Again make sure you add both a caption and a label.

### Subtables

Just like with images, you may want to group tables together into a single table environment. This can be done by using subtable environments inside a table environment. Here's an example:

\begin{table}[h]
\begin{subtable}[h]{0.45\textwidth}
\centering
\begin{tabular}{l | l | l}
Day & Max Temp & Min Temp \\
\hline \hline
Mon & 20 & 13\\
Tue & 22 & 14\\
Wed & 23 & 12\\
Thurs & 25 & 13\\
Fri & 18 & 7\\
Sat & 15 & 13\\
Sun & 20 & 13
\end{tabular}
\caption{First Week}
\label{tab:week1}
\end{subtable}
\hfill
\begin{subtable}[h]{0.45\textwidth}
\centering
\begin{tabular}{l | l | l}
Day & Max Temp & Min Temp \\
\hline \hline
Mon & 17 & 11\\
Tue & 16 & 10\\
Wed & 14 & 8\\
Thurs & 12 & 5\\
Fri & 15 & 7\\
Sat & 16 & 12\\
Sun & 15 & 9
\end{tabular}
\caption{Second Week}
\label{tab:week2}
\end{subtable}
\caption{Max and min temps recorded in the first two weeks of July}
\label{tab:temps}
\end{table}

Notice that in each \begin{subtable} command we've included a position specifier and a width. Again, we can give each subtable a label and caption as well as giving the whole table figure a label and caption.

Now in the same way we added a list of figures after the table of contents we can add a list of tables using the \listoftables command.

This concludes our discussion on images and tables. In the next post we'll look at adding a bibliography to our thesis.

### All articles in this series

• Part 1: Basic Structure;
• Part 2: Page Layout;
• Part 3: Figures, Subfigures and Tables;
• Part 4: Bibliographies with BibLaTeX;
• Part 5: Customising Your Title Page and Abstract.


text:
# Math Help - Triangle line lengths problem

1. ## Triangle line lengths problem

Please can anyone advise on the steps required to determine the lengths of the lines

qs
pr
tr
ts

in the attached diagram

Thanks

Luke

2. ## Re: Triangle line lengths problem

Can't see/reach your diagram...can you post it properly?
Or at least describe it?

Regards

Luke

4. ## Re: Triangle line lengths problem

Thanks; couple questions:
ts is part of qs, right? (you show ts)
is P3 really pr? Not pt?

Can you redo diagram:
change P1, P2, P3 to a, b, c
show points using capitals, lengths using small; like triangle ABC with sides a,b,c:
Code:
B

a          c

C         b             A

5. ## Re: Triangle line lengths problem

Well, had another look at your problem: only qs can be calculated.
r can be anywhere on the line that you show it on;
so, to calculate pr (and tr,ts), angle rps is required as a given.

sure that ALL points (lines crossing) are labelled with capital letters;
if not, I'll let ProveIt do the work!!

6. ## Re: Triangle line lengths problem

Hi Wilmer

I have attached the diagram with amended labeling.

I have highlighted in yellow the lengths I wish to calculate. I would appreciate any guidance. I just wish to know the steps of the calculations for each.

Thanks

Luke

8. ## Re: Triangle line lengths problem

Luke, I don't know what YOU know, like Law of Sines/Cosines and the likes.
Plus what is this problem: homework from Geometry class?
Anyway, I'll show how to calculate AE.

Since angleFDE = 90 and angleFDA = 25, then angleADE = 90 - 25 = 65.
Using Cosine Law:
AE = SQRT[50^2 + 60^2 - 2(50)(60)COS(65)] = ~59.7

To see what YOU can do, take triangleADF:
Can you calculate lengths of DF and AF?

Also, are you able to calculate angleDAB?
There is enough info given (the 2 angles) to permit that calculation.

9. ## Re: Triangle line lengths problem

Thanks Wilmer

I can see the calculation for DF and AF by drawing a line at right angles to AD to the angle DFA and then creating a calculation on the TAN of the angles (FDA and FAD) to determine the length of the perpendicular line to line AD. Knowing this I can the determine DF and FA.

Angle DAB is 85 degrees (90 - 5).

I can see how to get the answer for length AB but not the other unknowns.

Regards

Luke

10. ## Re: Triangle line lengths problem

Originally Posted by LukeP20110907
> I can see the calculation for DF and AF by drawing a line at right angles to AD to the angle DFA
> and then creating a calculation on the TAN of the angles (FDA and FAD) to determine the length
> of the perpendicular line to line AD. Knowing this I can the determine DF and FA.

Well, easier to use Law of Sines right away:
DF / SIN(5) = 50 / SIN(150) ; DF = 50SIN(5) / SIN(150) = ~8.716
AF / SIN(25) = 50 / SIN(150) ; AF = 50SIN(25) / SIN(150) = ~ 42.262

> Angle DAB is 85 degrees (90 - 5).

Correct.

> I can see how to get the answer for length AB but not the other unknowns.

OK, show me how YOU'd do that...then we'll carry on...
.

11. ## Re: Triangle line lengths problem

Thanks Wilmer

I can see a glaring oversight! Line AB needs to have a length or line BD can cross line AE at any point! Apologies for the nonsense. So, AB=60.

Using Cosine Law (which is new to me)

BD = sqrt (50*50 + 60*60 - 2 * 50 * 60 cos(85))
BD 74.67975324

Do you agree?

This leaves the lengths of lines DC and AC to calculate. Please can you point me towards the steps for these - I will then just check with you that the answers I calculate are correct and leave you in peace! Thanks for the early patience whilst i realised the omission!

Regards

Luke

12. ## Re: Triangle line lengths problem

Originally Posted by LukeP20110907
I can see a glaring oversight! Line AB needs to have a length or line BD can cross line AE at any point! Apologies for the nonsense. So, AB=60.
You FINALLY realise that! Why do you think I told you: "OK, show me how YOU'd do that"?

PLUS I told you that in this post (Post#5):
> Well, had another look at your problem: only qs can be calculated.
> r can be anywhere on the line that you show it on

I'll get back to you later on lines DC and AC.

T'a calculation pour BD est parfaite: good stuff!

13. ## Re: Triangle line lengths problem

Ok; so far we have calculated the following:
angleADE = u = 65; angle DAB = v = 85
lineAE = a = ~59.70; lineBD = b = ~74.68

Now to calculate lineAC and lineCD:

Step#1: using triangleABD, calculate angleABD = w:
50 / SIN(w) = b / SIN(v)
w = ASIN[50SIN(v) / b] = ~41.83

Step#2: using triangleADE, calculate angleDAE = x:
60^2 = 50^2 + a^2 - 2(50)(a)COS(x)
x = ACOS[(50^2 + a^2 - 60^2) / (100a)] = ~65.62

So angleBAC = y = v - x = ~19.38
and angleACB = z = 180 - w - y = ~118.79

Step#3: using triangleABC, calculate AC:
AC / SIN(w) = 60 / SIN(z)
AC = 60SIN(w) / SIN(z) = ~45.66

Step#4: using triangleABC, calculate BC:
BC / SIN(y) = 60 / SIN(z)
BC = 60SIN(y) / SIN(z) = ~22.72

So CD = BD - BC = 74.68 - 22.72 = ~51.96

There you go...hope you can follow that...

14. ## Re: Triangle line lengths problem

Many, many thanks Wilmer

I will be working through this today and putting into a spreadsheet.

Thanks for the patience.

Luke

format: plain


text:
# Issues with Reed-Solomon encoder

I am trying out the Reed-Solomon encoder in Sage and I have found it to exhibit a curious behavior. I first create an encoder for a $(7,3)$-Reed Solomon code as follows

sage: F.<a> = GF(8,name='a',modulus=x^3+x+1)
....: Fx.<x> = F[]
....: C = codes.ReedSolomonCode(F, 7, 3)
....: E = C.encoder("EvaluationPolynomial")

Now I can write

sage: E.encode(x)
(1, a, a^2, a + 1, a^2 + a, a^2 + a + 1, a^2 + 1)

Or

sage: E.encode(a*x^0+a^2*x+x^2)
(a^2 + a + 1, a^2 + 1, a, 0, a^2 + 1, 0, a^2 + a + 1)

But I get when I write the following

sage: E.encode(a)

I get the following error

"AttributeError: 'sage.rings.finite_rings.element_givaro.FiniteField_givaroElement' object has no attribute 'degree'"

sage: E.encode(a*x^0)

Everything works as intended. Is this how it is supposed to work? Does one always need an "$x$" to be part of a term, even when one actually has a constant term? Is there a better way to do it?

edit retag close merge delete

Sort by » oldest newest most voted

You can use parent(a) to see that a is not a polynomial, hence the error message. You need to tell sage that you want the image of a in the polynomial ring. You can also use Fx(a) for doing that.

more

FrédéricC's answer works for what I wanted to do! One can just write E.encode(Fx(a)) instead of what I initially did. I tried to look for a way to avoid needing to use Fx however. The following works

F = GF(8,name='a',modulus=x^3+x+1)
Fx.<x> = F[]
Fa.<a> = Fx

but I still get issues if I write E.encode(1) or E.encode(0). There is a NotImplementedError in this case. If one writes E.encode(Fx(1)) or E.encode(Fx(0)), this is circumvented.

( 2021-01-21 10:31:32 +0200 )edit
1

• Fx.zero() directly provides the zero element of Fx while Fx(0) uses the integer zero and then turns it into an element of Fx.
• Likewise, Fx.one() directly provides the one element of Fx while Fx(1) turns the integer one into an element of Fx.

So one can use E.encode(Fx.one()) and E.encode(Fx.zero()).

One could name the polynomials 0, 1 and a for easy access, e.g.

p0 = Fx.zero()
p1 = Fx.one()
pa = Fx(a)

and then use e.g. E.encode(pa) instead of E.encode(Fx(a)).

( 2021-01-21 12:01:59 +0200 )edit

I realized the code I suggested using in the above comment does not work - it seems to set $a=x$, which is not what I intended.

( 2021-01-21 17:03:57 +0200 )edit

Based on the help I got from FrédéricC and slelievre, I managed to create the following encoder-decoder:

F.<a> = GF(8,name='a',modulus=x^3+x+1)
Fx.<x> = F[]
C = codes.ReedSolomonCode(F, 7, 3)
E = C.encoder('EvaluationPolynomial')
M = E.message_space()
D = C.decoder('BerlekampWelch')

pa = Fx(a)
p0 = Fx.zero()
p1 = Fx.one()

m = M.random_element()
n = E.encode(m)
o = D.decode_to_message(n);
print(m)
print(o)

more


text:
# Mathematical Sciences Research Institute

Home » Workshop » Schedules » Beatty Games Big and Small

# Beatty Games Big and Small

## Berlekamp Memorial Workshop on Combinatorial Games October 21, 2019 - October 22, 2019

October 22, 2019 (11:30 AM PDT - 12:00 PM PDT)
Speaker(s): Michael Fisher (West Chester University)
Location: MSRI: Simons Auditorium
Primary Mathematics Subject Classification No Primary AMS MSC
Secondary Mathematics Subject Classification No Secondary AMS MSC
Video

#### 9-Fisher

Abstract

A Beatty sequence is a sequence of integers formed by taking the floor of the positive integral multiples of a positive irrational number $\alpha$.  The complementary sequence is formed in a similar manner using $\beta$, where $\beta$ satisfies the equation $\frac{1}{\alpha} + \frac{1}{\beta} = 1$.  For a given $\alpha$, we investigate the partizan subtraction game with left and right subtraction sets given by $(1, \alpha)$ and $(1, \beta)$, respectively.  We analyze this family of games using the Atomic Weight Calculus.

We will also report results for the non-atomic version, where the left and right subtraction sets are given by $(\alpha)$ and $(\beta)$, respectively.

Octal games are impartial games that involve removing tokens from heaps of tokens.  These types of games are interesting in that they can be described using an octal code.  Historically, research efforts have focused almost exclusively on octal games with finite codes.  We consider octal games based on infinite octal codes where the heap sizes corresponding to elements of a Beatty $\alpha$ sequence are played according to some fixed removal rule and the heap sizes corresponding to elements of a Beatty $\beta$ sequence are played according to some other fixed removal rule.  Interesting periodicity seems to occur in most cases.

Supplements No Notes/Supplements Uploaded
Video/Audio Files

#### 9-Fisher

 H.264 Video 948_27658_8061_9-Fisher.mp4


text:
# Proving since ever ? The compacity theorem revisited from a proof-theoretical perspective

Jean-Baptiste Joinet

27-10-2021 - 14:30
Largo San Leonardo Murialdo,1 - Pal.C - 311

The “compacity theorem” for propositional or first order logic (of which one possible formulation is the “finiteness theorem”, which states that reaching a conclusion from an infinite set of assumptions can always be done from a finite subset of that set) is often seen as an essentially “semantical” theorem (reason why it is usually stated in terms of the “consequence relation”  or in terms of “validity” – instead of in terms of the “deducibility relation” – and IS proved via a topological argument).

Indeed, considered from the proof-theoretical framework, the finiteness property has no proper meaning nor interest : usual formal proofs being finite objects (typically finite trees), it is almost trivial to see that they only use a finite number of assumptions.

To investigate the proof-theoretical meaning of the finiteness theorem (as I will do in my talk), one thus must broaden the horizon, by considering infinite proofs.

We will do it, not by considering proof-trees where the rules themselves are of an infinite arity (as omega-logic does), but by considering infinite proof-trees with usual rules of finite arity, with an end (a conclusion), but with no beginning – the kind of proofs that could only be built by a prover who would be proving since ever (a special dedication to longstanding debates particularly developed in Rome ;-).

In a sense, that proof-theoretical version of the proof of the compacity theorem brings a deeper result than the semantical one : it not only says that an infinite set of valid/provable formulas always includes a finite set of valid/provable formulas, but also says that enjoying an infinite time at our disposal would not change our demonstrative capacity (giving thus a good reason to content ourselves with our finite proofs).

org: TORTORA DE FALCO Lorenzo


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Temperature uses the symbol T and the unit Kelvin (text{K}), which is the absolute temperature scale. The SI unit of time, the second, symbol s, is easy enough to define, but time itself can only be defined in reference to other quantities. Answer : Kelvin (symbol Related Video. It is defined by taking the fixed numerical value of the Boltzmann constant k to be 1.380 649 x 10 –23 when expressed in the unit J K –1, which is equal to kg m 2 s –2 K –1, where the kilogram, metre and second are defined in … Kelvin is the SI unit. The second became the universal base unit of time: 1 hour = 3600 s; 1 day = 86 400 s; 1 year ~ 31.6 mega seconds (Ms) Pressure. Click hereto get an answer to your question  The S.I. 26.3k LIKES. What is Ready-mix concrete ? The name Système International d'Unités, (International System of Units) and the abbreviation SI, were established by the 11th General Conference on Weights and Measures (CGPM) in 1960. unit of temperature is: In view of the coronavirus pandemic, we are making LIVE CLASSES and VIDEO CLASSES completely FREE to prevent interruption in ... (SI) Example Definitions Formulaes. On the Kelvin scale, the unit of temperature is called a Kelvin and is symbolized by K on this scale water freezes at 273 K and boils at 373 K . The SI unit used to measure the temperature in Kelvin(K). Although the Celsius temperature scale, that is a multiple or a difference of is... Manufacturing plants, are thermocouples, infrared radiators, and resistive temperature devices tiffany announces... Of heat in the body reading this solution for FREE, Download our App  degree. System... 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# Beer

After three 10° beers consumed in a short time, there is 5.6 g of alcohol in 6 kg adult human blood. How much is it per mille?

Result

p =  0.9

#### Solution:

$m_{ 1 } = 5.6 \ g = 5.6 / 1000 \ kg = 0.0056 \ kg \ \\ m_{ 2 } = 6 \ kg \ \\ \ \\ p = 1000 \cdot \ \dfrac{ m_{ 1 } }{ m_{ 2 } } = 1000 \cdot \ \dfrac{ 0.0056 }{ 6 } = \dfrac{ 14 }{ 15 } \doteq 0.9333 = 0.9 \ ‰$

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# Why is there no indent to the first line in my quoted paragraph?

I wanted to indent some quoted content (both 1cm from left and 1cm from right), so I defined a new environment (MWE) as follows

\documentclass[11pt,a4paper,openany]{book}
\usepackage{latexsym}
\usepackage{cases}
\usepackage{indentfirst}
\usepackage{changepage}   % for the adjustwidth environment
\begin{document}
\chapter{Group Theory}
\subsection{Homomorphism from S4 to S3} % not sure if the structure matters in this problem..
It is not hard to check that

To show it is surjective, thanks to the answer provided by @David Wheeler on MSE:
\begin{qt}
No indent here No indent here No indent here No indent here

Let's examine the mapping is surjective surjective surjective surjective surjective surjective surjective surjective surjective surjective
\end{qt}
no indent for the first line..
\end{document}

When I put all of my quoted content into the newly defined qt environment, after compiling I noticed that the first line of the quoted content was not indented, despite my usage of the indentfirst package:

Of course, the adjustwidth environment did work, because the leftmost and rightmost of the included content were both indented 1cm, as is seen on the picture. The problem was just that the first line ("Denote ....") wasn't in place. There was supposed to be an indent before the word "Denote", just like before the first words in all the other lines, but there wasn't.

I also tried inserting an \indent before "Denote", which still didn't work.

So how am I supposed to make the first line indent (to be exact, indent from left)? Thanks in advance!

PS: MWE fixed finally..I'm sorry that it's the first time for me to write an MWE, I made many mistakes and didn't fix them till just two minutes ago, now it's alright.

• Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – user31729 Aug 8 '15 at 8:54
• Well it seems that the quote environment did the job perfectly well and there is literally no need for me to define a new one... – Vim Aug 8 '15 at 9:34

There is alredy a built in environment for this:

\documentclass[11pt,a4paper,openany]{book}
\usepackage{indentfirst}

\begin{document}
\chapter{Group Theory}
\subsection{Homomorphism from $S_4$ to $S_3$}

It is not hard to check that

To show it is surjective, thanks to the answer provided by @David Wheeler on MSE:
\begin{quote}
No indent here No indent here No indent here No indent here

Let's examine the mapping is surjective surjective surjective surjective surjective surjective surjective surjective surjective surjective
\end{quote}
no indent for the first line..
\end{document}

• @Vim -- You're welcome! Is the left and right indent ok with the standard settings of quote? – Ruben Aug 8 '15 at 9:39
• Yes they look good and to be honest set off the quoted content better, for example some additional vertical space is made between the previous paragraph and the quoted lines. – Vim Aug 8 '15 at 9:42
• It should be $S_4$ and $S_3$ also in the section title – egreg Aug 8 '15 at 18:08


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In mathematics, many functions are used, a beta function is one of them. Functions are very essential in mathematics for different purposes. Each function contains inputs and required the unique out for each by applying different laws of mathematics.

The first kind of Euler integral is another name of the beta function. But in that integral gamma functions are also used as beta and gamma functions are related to each other. The concept of integration is used in these functions.

## What is Beta Function?

The beta function is that function in mathematics that is unique and is organized by the First kind of Euler integral. It is mostly stated as the domain of real numbers. The beta function is generally denoted by the Greek letter β. The beta function is written as B(a, b), and a, b are the real numbers.

The beta function is symmetrical. So, B(a, b) must be equal to B(b, a). The connotation among the set of the inputs and the outputs in mathematics is usually explained by a function known as the beta function.

You can also find factorial of 100 also by checking above examples.

In the beta function, for each input value, there must be at least one output that is associated with the input. The real values of the beta function must be greater than zero. This function follows the definite integral as it has the upper limit one and the lower limit zero.

### The formula of Beta Function

The formula for this function can be written in the form of.

B (a, b) = ∫01 ta-1 (1 – t)b-1 dt

In this equation, a and b are real numbers and must be greater than zero and t is the variable whose integral is to be calculated.

There is another formula known as the factorial form of the beta function and is written as.

B (a, b) = (a – 1)! (b – 1)! / (a + b – 1)!

## Solving problems of beta function by using formula

For the calculation of beta function, the knowledge of integral and exponent is very important. Without the knowledge of these concepts, the problems of the beta function are not solved. The integral in beta function is used for definite integral.

And exponents of the functions are calculated by multiplying the base by itself in power times. By using the integral and exponent, the problems of this function can be solved easily.

Example 1:

Calculate the beta function, when the real values are a = 7 and b = 6.

Solution

Step 1: Identify the real values of the given function.

a = 7

b = 6

Step 2: Take the general equation of the beta function.

B (a, b) = ∫01 ta-1 (1 – t)b-1 dt

Step 3: Put the values of the real number in the above equation.

B (7, 6) = ∫01 t7-1 (1 – t)6-1 dt

B (7, 6) = ∫01 t6 (1 – t)5 dt

Step 4: Now open the powers of the function to make it simple for computing.

B (7, 6) = ∫01 t6 (-t5 + 5t4 – 10t3 + 10t2 – 5t + 1) dt

B (7, 6) = ∫01 (-t11 + 5t10 – 10t9 + 10t8 – 5t7 + t6) dt

Step 5: Now apply the sum and difference rule of integration.

B (7, 6) = ∫01 (-t11) dt + ∫01 (5t10) dt – ∫01 (10t9) dt + ∫01 (10t8) dt – ∫01 (5t7) dt + ∫01 (t6) dt

Step 6: Solve the integral.

B (7, 6) = ∫01 (-t11) dt + ∫01 (5t10) dt – ∫01 (10t9) dt + ∫01 (10t8) dt – ∫01 (5t7) dt + ∫01 (t6) dt

B (7, 6) = (-t11+1/11 + 1)10 + (5t10+1/10 + 1)10 – (10t9+1/9 + 1)10 + (10t8+1/8 + 1)10 – (5t7+1/7 + 1)10 + (t6+1/6 + 1)10

B (7, 6) = (-t12/12)10 + (5t11/11)10 – (10t10/10)10 + (10t9/9)10 – (5t8/8)10 + (t7/7)10

B (7, 6) = -1/12 (t12)10 + 5/11 (t11)10 – 10/10 (t10)10 + 10/9 (t9)10 – 5/8 (t8)10 + 1/7 (t7)10

Step 7: Now apply the fundamental theorem of calculus for solving the limits of the integral.

ba f(t) dt = F(a) – F(b)

B (7, 6) = -1/12 (112 – 012) + 5/11 (112 – 012) – 10/10 (112 – 012) + 10/9 (112 – 012) – 5/8 (112 – 012) + 1/7 (112 – 012)

B (7, 6) = -1/12 (1 – 0) + 5/11 (1 – 0) – 10/10 (1 – 0) + 10/9 (1 – 0) – 5/8 (1 – 0) + 1/7 (1 – 0)

B (7, 6) = -1/12 (1) + 5/11 (1) – 10/10 (1) + 10/9 (1) – 5/8 (1) + 1/7 (1)

B (7, 6) = -1/12 + 5/11– 10/10 + 10/9 – 5/8 + 1/7

B (7, 6) = -1/12 + 5/11– 1 + 10/9 – 5/8 + 1/7

Step 8: Now solve the fractions.

B (7, 6) = -1/12 + 5/11– 1 + 10/9 – 5/8 + 1/7

B (7, 6) = 0.00018038

These calculations are too large and very time-consuming. To avoid this, you can simply use the beta function calculator which gives you the required output in few seconds.

Example 2:

Find the beta function of, B (7, 6) = ∫01 t6 (1 – t)5 dt.

Solution

Step 1: Write the given function.

B (7, 6) = ∫01 t6 (1 – t)5 dt

Step 2: Now write the general form of the beta function.

B (a, b) = ∫01 ta-1 (1 – t)b-1 dt

Step 3: Compare the given function with the general function.

a – 1 = 6

a = 6 + 1

a = 7

b – 1 = 5

b = 5 + 1

b = 6

Step 4: Now take the factorial form of the beta function.

B (a, b) = (a – 1)! (b – 1)! / (a + b – 1)!

Step 5: Put the value of a and b in the above formula.

B (a, b) = (a – 1)! (b – 1)! / (a + b – 1)!

B (7, 6) = (7 – 1)! (6 – 1)! / (7 + 6 – 1)!

B (7, 6) = (6)! (5)! / (13 – 1)!

B (7, 6) = (6)! (5)! / (12)!

B (7, 6) = (720) (120) / (479001600)

B (7, 6) = 86400 / 479001600

B (7, 6) = 0.00018038

## Summary

The beta function is a function used widely in mathematics for the calculation of the Euler integral. You can solve this function by two formulas one is by using exponent and integral and the other is by using the factorial form. Both the formulas are explained above, you can perform any problem by following these two formulas.

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text:
# Partition function is simply temperature if possible sub system energy is continuous?

Partition function is $$Z=\sum_j\exp\left(-\frac{\epsilon_j}{kT}\right)$$ a sum over all possible energy levels $\epsilon_1,\epsilon_2, ..., \epsilon_M$. There must be a finite number of choices otherwise $Z$ does not converge.

If each individual subsystem can have a range of energy level, naturally I can rewrite this to integrals: $$Z=\int_0^\infty\exp\left(-\frac\epsilon{kT}\right)\mathrm d \epsilon = kT.$$ I can get this generalization from variation of entropy (as the normalization constant) as well. Does this mean that if the subsystem can have continuous range of non-negative energy, the partition function is simply that? Or am I missing something?

• Usually all the interesting stuff happens because some energy states can be realized through more than one microstate, i.e. you'll get the density of states factor into your equation. – alarge Jun 22 '15 at 23:59
• So does it mean it is not defined as a sum over energy levels but over "states"? – Shinjikun Jun 23 '15 at 0:15
• @Shinjikun, yes the sum is over the states and not the energies. Moreover, in the last integral for Z, the dimensions don't match. Z has to be dimensionless. – Abinash Chakraborty Jun 23 '15 at 6:20
• @AbinashChakraborty ouch, that's right. $\mathrm d \epsilon$ is introducing an energy dimension. Thanks! – Shinjikun Jun 23 '15 at 16:23

If each individual subsystem can have a range of energy level

I'm not sure what you mean here. The summation in the partition function applies when the energy states are discrete. When they are continuous, as is the case in most of classical stat mech$^*$, the sum is replaced by an integral. This has nothing to do with subsystems; it is merely a consequence of switching to a continuous energy spectrum.

As @alarge said, the Boltzmann factor should be weighted by the degeneracy in the discrete case or the density of states in the continuous case. Then your partition function, to within a constant factor, is

$$Z = \int \Omega\left( E\right)e^{-E/kT}\mathrm{d}E$$

However, in most cases the calculation of $\Omega\left( E\right)$ is non-trivial at best, and outright impossible at worst.

A major benefit of the canonical ensemble is that we don't actually need to calculate $\Omega\left( E\right)$ if we know how the energy depends on the degrees of freedom within the system. For this we use the Hamiltonian function, which for, e.g. a gas particle, describes the energy in terms of the position and momentum of the particle. It is then sufficient to integrate the Boltzmann factor - with the Hamiltonian replacing $E$ - over all of phase space (all possible positions and momenta), like so:

$$Z = \int e^{-H\left(q, p\right)/kT}dqdp$$

where $q$ and $p$ are the position and momentum, respectively, and constants have again been omitted for clarity.

So to answer your question, it is not sufficient to integrate the energy over all states. If you want to integrate the energy, you need to weight it with the DOS. However, I highly suggest you take the approach of summing or integrating the Hamiltonian over all degrees of freedom.

$^*$ Discrete systems are generally easier to deal with, so they are often introduced first.


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# Pressure Control

## Physical Model

The pressure control component enforces a specific pressure at the controlled_junction. This is achieved by adding a required pressure lift or pressure drop between the from_junction and to_junction, so that controlled_p_bar will be reached at the controlled junction. The controlled junction can be identical to the to_junction to control the pressure directly at the outlet.

Internally, the behaviour is achieved by fixing the pressure variable at controlled_junction in the system matrix and keeping the pressure drop of the pressure control unit variable, so that is calculated during the Newton-Raphson-calculation.

Note

The temperature at the inlet and outlet junction will not be adapted by the pressure control unit. Therefore, these components usually operate isothermal (inlet temperature = outlet temperature). It is assumed that temperature changes due to compression or expansion (Joule-Thomson-effect) are balanced internally by adding or removing heat.

Warning

A sufficient hydraulic connection between from_junction, to_junction and controlled_junction is crucial for proper operation of this component. Hydraulically impossible configurations (e.g., if the controlled junction is on a different stub) or contradicting other pressure control units will lead to non-convergence of the pipeflow.

## Create Function

For creating a single pressure control unit:

create_pressure_control(net, from_junction, to_junction, controlled_junction, controlled_p_bar, control_active=True, loss_coefficient=0.0, name=None, index=None, in_service=True, type='pressure_control', **kwargs)

Adds one pressure control that enforces a pressure at a specific junction.

The pressure control unit creates a pressure drop / lift between the ‘from’ and the ‘to’ junction so that the pressure set point at the controlled junction is met. It is required that the controlled junction is hydraulically properly connected to the from and to junction and no other pessure control unit is inbetween.

Parameters
• net (pandapipesNet) – The net for which this pressure control should be created

• from_junction (int) – ID of the junction on one side which the pressure control will be connected with

• to_junction (int) – ID of the junction on the other side which the pressure control will be connected with

• controlled_junction (int) – ID of the junction at which the pressure is controlled

• controlled_p_bar (float) – Pressure set point

• control_active (bool, default True) – Variable to state whether the pressure control is active (otherwise behaviour similar to open valve)

• loss_coefficient (float, default 0) – The pressure loss coefficient introduced by the component’s shape (used only if control is not active).

• name (str) – Name of the pressure control element

• index (int, default None) – Force a specified ID if it is available. If None, the index one higher than the highest already existing index is selected.

• in_service (bool, default True) – True if the pressure control is in service or False if it is out of service

• type (str, default "pressure_control") – Currently not used - possibility to specify a certain type of pressure control

• kwargs (dict) – Additional keyword arguments will be added as further columns to the net[“press_control”] table

Returns

index - The unique ID of the created element

Return type

int

Example

Connect junction 0 and 1 and set the pressure at junction 1 to 5 bar.

>>> create_pressure_control(net, 0, 1, 1, controlled_p_bar=5)

For creating multiple pressure control units at once:

create_pressure_controls(net, from_junctions, to_junctions, controlled_junctions, controlled_p_bar, control_active=True, loss_coefficient=0.0, name=None, index=None, in_service=True, type='pressure_control', **kwargs)

Convenience function for creating many pressure controls at once. Parameters ‘from_junctions’ and ‘to_junctions’ must be arrays of equal length. Other parameters may be either arrays of the same length or single values.

Pressure control units enforce a pressure at a specific junction by creating a pressure drop / lift between the ‘from’ and the ‘to’ junction so that the pressure set point at the controlled junction is met. It is required that the controlled junction is hydraulically properly connected to the from and to junction and no other pessure control unit is inbetween.

Parameters
• net (pandapipesNet) – The net for which these pressure controls should be created

• from_junctions (Iterable(int)) – IDs of the junctions on one side which the pressure controls will be connected to

• to_junctions (Iterable(int)) – IDs of the junctions on the other side to which the pressure controls will be connected to

• controlled_junctions (Iterable or int) – IDs of the junctions at which the pressure is controlled

• controlled_p_bar (Iterable or float) – Pressure set points

• control_active (bool, default True) – Variable to state whether the pressure control is active (otherwise behaviour similar to open valve)

• loss_coefficient (float, default 0) – The pressure loss coefficient introduced by the component’s shape (used only if control is not active).

• name (Iterable or str) – Name of the pressure control elements

• index (Iterable(int), default None) – Force specified IDs if they are available. If None, the index one higher than the highest already existing index is selected and counted onwards.

• in_service (Iterable or bool, default True) – True if the pressure controls are in service or False if they are out of service

• type (Iterable or str, default "pressure_control") – Currently not used - possibility to specify a certain type of pressure control

• kwargs – Additional keyword arguments will be added as further columns to the net[“press_control”] table

Returns

index - The unique IDs of the created elements

Return type

array(int)

Example

Create one unit to connect junction 0 and 1, set the pressure at junction 1 to 5 bar.

And create a 2nd unit to connect junction 2 and 4, set the pressure at junction 3 to 4.9 bar.

>>> create_pressure_controls(net, [0, 2], [1, 4], [1, 3],
>>>                          controlled_p_bar=[5, 4.9])

## Component Table Data

net.press_control

 Parameter Datatype Value Range Explanation net pandapipesNet The pandapipes net object in whose press_control table the new unit will be added from_junction integer $$>$$ 0 Index of junction at which the pressure control unit starts to_junction integer $$>$$ 0 Index of junction at which the pressure control unit ends controlled_junction integer > 0 ID of the junction at which the pressure is controlled controlled_p_bar float > 0 Pressure set point [barg] control_active boolean If false, the unit behaves like an open valve loss_coefficient float > ß Pressure loss coefficient (only effective if control is not active) name string Name of the pressure control unit index integer > 0 Force specific index the press_control-DataFrame, if available in_service boolean True / False If false, the unit behaves like a closed valve. type string Type variable to classify different kinds of pressure control units (not considered by pandapipes’ calculations)

## Result Table Data

net.res_press_control

 Parameter Datatype Explanation p_from_bar float Pressure at “from”-junction [bar] p_to_bar float Pressure at “to”-junction [bar] t_from_k float Temperature at “from”-junction [K] t_to_k float Temperature at “to”-junction [K] mdot_to_kg_per_s float Mass flow out of pressure control unit [kg/s] mdot_from_kg_per_s float Mass flow into pressure control unit [kg/s] vdot_norm_m3_per_s float Norm volume flow [m^3/s] deltap_bar float Pressure difference between to- and from junction created by the pressure controller [bar]


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# Posts published in “Simulation”

Problem:

Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place.

The move sequence is represented by a string. And each move is represent by a character. The valid robot moves are R (Right), L (Left), U (Up) and D (down). The output should be true or false representing whether the robot makes a circle.

Example 1:

Example 2:

Idea:

Simulation

Solution:

C++

Problem:

There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in N days. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then.

Given an array flowers consists of number from 1 to N. Each number in the array represents the place where the flower will open in that day.

For example, flowers[i] = x means that the unique flower that blooms at day i will be at position x, where i and x will be in the range from 1 to N.

Also given an integer k, you need to output in which day there exists two flowers in the status of blooming, and also the number of flowers between them is k and these flowers are not blooming.

If there isn’t such day, output -1.

Example 1:

Input:
flowers: [1,3,2]
k: 1
Output: 2
Explanation: In the second day, the first and the third flower have become blooming.

Example 2:

Input:
flowers: [1,2,3]
k: 1
Output: -1

Note:

1. The given array will be in the range [1, 20000].

Idea:

BST/Buckets

# Problem:

https://leetcode.com/problems/next-closest-time/description/
no limit on how many times a digit can be reused.

You may assume the given input string is always valid. For example, “01:34”, “12:09” are all valid. “1:34”, “12:9” are all invalid.

## Example 1:

Input: "19:34"
Output: "19:39"
Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39,
which occurs 5 minutes later.
It is not 19:33, because this occurs 23 hours and 59 minutes later.

## Example 2:

Input: "23:59"
Output: "22:22"
Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22.
It may be assumed that the returned time is next day's time since it is smaller
than the input time numerically.

# Related Problems:

Problem:

You’re now a baseball game point recorder.

Given a list of strings, each string can be one of the 4 following types:

1. Integer (one round’s score): Directly represents the number of points you get in this round.
2. "+" (one round’s score): Represents that the points you get in this round are the sum of the last two validround’s points.
3. "D" (one round’s score): Represents that the points you get in this round are the doubled data of the last valid round’s points.
4. "C" (an operation, which isn’t a round’s score): Represents the last valid round’s points you get were invalid and should be removed.

Each round’s operation is permanent and could have an impact on the round before and the round after.

You need to return the sum of the points you could get in all the rounds.

Example 1:

Example 2:

Note:

• The size of the input list will be between 1 and 1000.
• Every integer represented in the list will be between -30000 and 30000.

Idea:

Simulation

Solution:

C++:

Java:

Python

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text:
# Hyphenation in Perl package names [duplicate]

Possible Duplicate:
Line break in lines with colons

I have Perl package name defined as command:

\newcommand{\myperlpackage}{\texttt{XXXXX::YYYYY::ZZZZZ}}

How can I tell LaTeX to hyphen package name after double colon but without any hyphen char eg.

XXXXX::-
YYYYY::ZZZZZ

should be

XXXXX::
YYYYY::ZZZZZ

-
Maybe your question is answered here: Line break in lines with colons. (Possible duplicate?) –  Hendrik Vogt Mar 22 '11 at 13:59
Yes, the main point of both questions is the same, but that question is more general - that's why I probably can't find it. –  jesper Mar 22 '11 at 14:07
@jesper: So do the answers to that question answer yours? –  Andrew Stacey Mar 22 '11 at 14:09
@Andrew: So what do you think? –  Hendrik Vogt Mar 22 '11 at 14:28
I usually define a command \dbr for discretionary breakpoints like this: \newcommand\dbr{\disretionary{}{}{}}. Then you can write XXXX::\dbr YYYY::\dbr ZZZZ, which should have the effect you desire.
To explain, \discretionary<arg1><arg2><arg3> tells TeX that a line break is permissible and that it should put <arg1> before the breakpoint, <arg2> after the breakpoint (in the next line), and <arg3> in the case that no line break is performed. See the TeXbook (page 95) for more details.


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# Problem with a Positive Definite Kernel/Matrix

$$K(x_1, x_2) = e^{-(x_1-x_2)^2/2}$$

is known to be positive definite (P.D.). This can be shown by, for example, the Bochner Theorem. By positive definite it is meant that

$$\sum_{i=1}^k \sum_{j=1}^k c_i c_j K(x_i, x_j) \geq 0$$

for any choice of $k \in \mathbb{N}$, $x_1, x_2, \ldots, x_k \in \mathbb{R}^n$ and $c_1, c_2, \ldots, c_k \in \mathbb{R}$.

From the above inequality, it seems obvious that the Hermitian part of the matrix with entries $$K_{ij}=K(x_i, x_j)$$ should have positive eigenvalues. Which brings me to the following problem:

G[x1_, x2_] := Exp[-1/2 (x1 - x2)^2];
list1 = RandomReal[{-10, 10}, 100];
(mat = Outer[G, list1, list1, 1]);
Eigenvalues[1/2 (mat + Transpose[mat])]

The resulting eigenvalues are not all positive. Why is that?

{14.4349, 12.5488, 11.8824, 10.0691, 9.54906, 8.65466, 7.27382, \
5.90905, 4.70692, 3.60972, 2.94933, 2.2864, 1.67143, 1.38664, \
0.926639, 0.694622, 0.479601, 0.346991, 0.218595, 0.144655, \
0.0947855, 0.0693128, 0.0390857, 0.0228503, 0.0137675, 0.0077189, \
0.00423406, 0.00209994, 0.00146088, 0.000639826, 0.000346808, \
0.000175915, 0.000081206, 0.0000427318, 0.0000218744, 0.000010557,
5.11307*10^-6, 2.03378*10^-6, 1.00012*10^-6, 4.39773*10^-7,
2.10151*10^-7, 7.53904*10^-8, 3.44437*10^-8, 1.22666*10^-8,
4.87432*10^-9, 2.15116*10^-9, 7.33137*10^-10, 4.1995*10^-10,
1.24422*10^-10, 4.71167*10^-11, 1.48485*10^-11, 6.61021*10^-12,
2.78995*10^-12, 9.61269*10^-13, 2.65273*10^-13, 8.56566*10^-14,
3.34234*10^-14, 1.14188*10^-14, 4.38365*10^-15, -2.17014*10^-15,
2.05478*10^-15, 1.69751*10^-15, -1.34612*10^-15, 1.32415*10^-15,
1.17034*10^-15, -1.09847*10^-15, 9.08927*10^-16,
8.03778*10^-16, -7.81348*10^-16, -6.84237*10^-16,
6.31285*10^-16, -5.8266*10^-16, 5.71942*10^-16, -4.67633*10^-16,
4.53221*10^-16, 4.25156*10^-16,
4.17792*10^-16, -3.92773*10^-16, -3.64044*10^-16, 3.39323*10^-16,
3.18269*10^-16, -3.06691*10^-16, 2.9718*10^-16, -2.78096*10^-16,
2.6519*10^-16, 2.12749*10^-16, -1.93748*10^-16, -1.81685*10^-16,
1.74396*10^-16, 1.56957*10^-16,
1.50008*10^-16, -1.42644*10^-16, -1.33609*10^-16,
1.29917*10^-16, -1.0308*10^-16, 7.93279*10^-17, -7.7838*10^-17,
4.323*10^-17, -1.54597*10^-17, 6.30094*10^-18}

• The ones that are negative are numerically zero. See the docs of Chop. If you use numbers with more than machine precision, you will also see that the negative entries become as small as the precision can resolve. – Marius Ladegård Meyer Nov 24 '16 at 17:39

The worst ones are order 10^-15. This is effectively zero, so the negative values are only a result of mathematica working with finite numerical precision.

• How can i control that precision. Suppose i want the results to be precise enough so that these small values are either zero or positive. – Iconoclast Nov 24 '16 at 17:59
• reference.wolfram.com/language/howto/… – Wouter Nov 24 '16 at 20:30
• Wouter, I would include your comment in the answer. It's really helpful. Thanks ;) – An old man in the sea. Aug 2 '17 at 20:17


text:
location:  Publications → journals
Search results

Search: MSC category 18D10 ( Monoidal categories (= multiplicative categories), symmetric monoidal categories, braided categories [See also 19D23] )

 Expand all        Collapse all Results 1 - 5 of 5

1. CJM Online first

Ovchinnikov, Alexey; Wibmer, Michael
 Tannakian categories with semigroup actions Ostrowski's theorem implies that $\log(x),\log(x+1),\dots$ are algebraically independent over $\mathbb{C}(x)$. More generally, for a linear differential or difference equation, it is an important problem to find all algebraic dependencies among a non-zero solution $y$ and particular transformations of $y$, such as derivatives of $y$ with respect to parameters, shifts of the arguments, rescaling, etc. In the present paper, we develop a theory of Tannakian categories with semigroup actions, which will be used to attack such questions in full generality, as each linear differential equation gives rise to a Tannakian category. Deligne studied actions of braid groups on categories and obtained a finite collection of axioms that characterizes such actions to apply it to various geometric constructions. In this paper, we find a finite set of axioms that characterizes actions of semigroups that are finite free products of semigroups of the form $\mathbb{N}^n\times \mathbb{Z}/{n_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/{n_r}\mathbb{Z}$ on Tannakian categories. This is the class of semigroups that appear in many applications. Keywords:semigroup actions on categories, Tannakian categories, difference algebraic groups, differential and difference equations with parametersCategories:18D10, 12H10, 20G05, 33C05, 33C80, 34K06

2. CJM 2013 (vol 66 pp. 481)

Aguiar, Marcelo; Mahajan, Swapneel
 On the Hadamard Product of Hopf Monoids Combinatorial structures that compose and decompose give rise to Hopf monoids in Joyal's category of species. The Hadamard product of two Hopf monoids is another Hopf monoid. We prove two main results regarding freeness of Hadamard products. The first one states that if one factor is connected and the other is free as a monoid, their Hadamard product is free (and connected). The second provides an explicit basis for the Hadamard product when both factors are free. The first main result is obtained by showing the existence of a one-parameter deformation of the comonoid structure and appealing to a rigidity result of Loday and Ronco that applies when the parameter is set to zero. To obtain the second result, we introduce an operation on species that is intertwined by the free monoid functor with the Hadamard product. As an application of the first result, we deduce that the Boolean transform of the dimension sequence of a connected Hopf monoid is nonnegative. Keywords:species, Hopf monoid, Hadamard product, generating function, Boolean transformCategories:16T30, 18D35, 20B30, 18D10, 20F55

3. CJM 2013 (vol 66 pp. 205)

Iovanov, Miodrag Cristian
 Generalized Frobenius Algebras and Hopf Algebras "Co-Frobenius" coalgebras were introduced as dualizations of Frobenius algebras. We previously showed that they admit left-right symmetric characterizations analogue to those of Frobenius algebras. We consider the more general quasi-co-Frobenius (QcF) coalgebras; the first main result in this paper is that these also admit symmetric characterizations: a coalgebra is QcF if it is weakly isomorphic to its (left, or right) rational dual $Rat(C^*)$, in the sense that certain coproduct or product powers of these objects are isomorphic. Fundamental results of Hopf algebras, such as the equivalent characterizations of Hopf algebras with nonzero integrals as left (or right) co-Frobenius, QcF, semiperfect or with nonzero rational dual, as well as the uniqueness of integrals and a short proof of the bijectivity of the antipode for such Hopf algebras all follow as a consequence of these results. This gives a purely representation theoretic approach to many of the basic fundamental results in the theory of Hopf algebras. Furthermore, we introduce a general concept of Frobenius algebra, which makes sense for infinite dimensional and for topological algebras, and specializes to the classical notion in the finite case. This will be a topological algebra $A$ that is isomorphic to its complete topological dual $A^\vee$. We show that $A$ is a (quasi)Frobenius algebra if and only if $A$ is the dual $C^*$ of a (quasi)co-Frobenius coalgebra $C$. We give many examples of co-Frobenius coalgebras and Hopf algebras connected to category theory, homological algebra and the newer q-homological algebra, topology or graph theory, showing the importance of the concept. Keywords:coalgebra, Hopf algebra, integral, Frobenius, QcF, co-FrobeniusCategories:16T15, 18G35, 16T05, 20N99, 18D10, 05E10

4. CJM 2012 (vol 65 pp. 241)

Aguiar, Marcelo; Lauve, Aaron
 Lagrange's Theorem for Hopf Monoids in Species Following Radford's proof of Lagrange's theorem for pointed Hopf algebras, we prove Lagrange's theorem for Hopf monoids in the category of connected species. As a corollary, we obtain necessary conditions for a given subspecies $\mathbf k$ of a Hopf monoid $\mathbf h$ to be a Hopf submonoid: the quotient of any one of the generating series of $\mathbf h$ by the corresponding generating series of $\mathbf k$ must have nonnegative coefficients. Other corollaries include a necessary condition for a sequence of nonnegative integers to be the dimension sequence of a Hopf monoid in the form of certain polynomial inequalities, and of a set-theoretic Hopf monoid in the form of certain linear inequalities. The latter express that the binomial transform of the sequence must be nonnegative. Keywords:Hopf monoids, species, graded Hopf algebras, Lagrange's theorem, generating series, PoincarÃ©-Birkhoff-Witt theorem, Hopf kernel, Lie kernel, primitive element, partition, composition, linear order, cyclic order, derangementCategories:05A15, 05A20, 05E99, 16T05, 16T30, 18D10, 18D35

5. CJM 2002 (vol 54 pp. 970)

Cegarra, A. M.; García-Calcines, J. M.; Ortega, J. A.
 On Graded Categorical Groups and Equivariant Group Extensions In this article we state and prove precise theorems on the homotopy classification of graded categorical groups and their homomorphisms. The results use equivariant group cohomology, and they are applied to show a treatment of the general equivariant group extension problem. Categories:18D10, 18D30, 20E22, 20F29
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# Hammersley–Clifford theorem

Hammersley–Clifford theorem The Hammersley–Clifford theorem is a result in probability theory, mathematical statistics and statistical mechanics that gives necessary and sufficient conditions under which a strictly positive probability distribution (of events in a probability space)[clarification needed] can be represented as events generated by a Markov network (also known as a Markov random field). It is the fundamental theorem of random fields.[1] It states that a probability distribution that has a strictly positive mass or density satisfies one of the Markov properties with respect to an undirected graph G if and only if it is a Gibbs random field, that is, its density can be factorized over the cliques (or complete subgraphs) of the graph.

The relationship between Markov and Gibbs random fields was initiated by Roland Dobrushin[2] and Frank Spitzer[3] in the context of statistical mechanics. The theorem is named after John Hammersley and Peter Clifford, who proved the equivalence in an unpublished paper in 1971.[4][5] Simpler proofs using the inclusion–exclusion principle were given independently by Geoffrey Grimmett,[6] Preston[7] and Sherman[8] in 1973, with a further proof by Julian Besag in 1974.[9] Contents 1 Proof outline 2 See also 3 Notes 4 Further reading Proof outline A simple Markov network for demonstrating that any Gibbs random field satisfies every Markov property.

It is a trivial matter to show that a Gibbs random field satisfies every Markov property. As an example of this fact, see the following: In the image to the right, a Gibbs random field over the provided graph has the form {displaystyle Pr(A,B,C,D,E,F)propto f_{1}(A,B,D)f_{2}(A,C,D)f_{3}(C,D,F)f_{4}(C,E,F)} . If variables {displaystyle C} and {displaystyle D} are fixed, then the global Markov property requires that: {displaystyle A,Bperp E,F|C,D} (see conditional independence), since {displaystyle C,D} forms a barrier between {displaystyle A,B} and {displaystyle E,F} .

With {displaystyle C} and {displaystyle D} constant, {displaystyle Pr(A,B,E,F|C=c,D=d)propto [f_{1}(A,B,d)f_{2}(A,c,d)]cdot [f_{3}(c,d,F)f_{4}(c,E,F)]=g_{1}(A,B)g_{2}(E,F)} where {displaystyle g_{1}(A,B)=f_{1}(A,B,d)f_{2}(A,c,d)} and {displaystyle g_{2}(E,F)=f_{3}(c,d,F)f_{4}(c,E,F)} . This implies that {displaystyle A,Bperp E,F|C,D} .

To establish that every positive probability distribution that satisfies the local Markov property is also a Gibbs random field, the following lemma, which provides a means for combining different factorizations, needs to be proved: Lemma 1 provides a means for combining factorizations as shown in this diagram. Note that in this image, the overlap between sets is ignored.

Lemma 1 Let {displaystyle U} denote the set of all random variables under consideration, and let {displaystyle Theta ,Phi _{1},Phi _{2},dots ,Phi _{n}subseteq U} and {displaystyle Psi _{1},Psi _{2},dots ,Psi _{m}subseteq U} denote arbitrary sets of variables. (Here, given an arbitrary set of variables {displaystyle X} , {displaystyle X} will also denote an arbitrary assignment to the variables from {displaystyle X} .) If {displaystyle Pr(U)=f(Theta )prod _{i=1}^{n}g_{i}(Phi _{i})=prod _{j=1}^{m}h_{j}(Psi _{j})} for functions {displaystyle f,g_{1},g_{2},dots g_{n}} and {displaystyle h_{1},h_{2},dots ,h_{m}} , then there exist functions {displaystyle h'_{1},h'_{2},dots ,h'_{m}} and {displaystyle g'_{1},g'_{2},dots ,g'_{n}} such that {displaystyle Pr(U)={bigg (}prod _{j=1}^{m}h'_{j}(Theta cap Psi _{j}){bigg )}{bigg (}prod _{i=1}^{n}g'_{i}(Phi _{i}){bigg )}} In other words, {displaystyle prod _{j=1}^{m}h_{j}(Psi _{j})} provides a template for further factorization of {displaystyle f(Theta )} .

show Proof of Lemma 1 The clique formed by vertices {displaystyle x_{1}} , {displaystyle x_{2}} , and {displaystyle x_{3}} , is the intersection of {displaystyle {x_{1}}cup partial x_{1}} , {displaystyle {x_{2}}cup partial x_{2}} , and {displaystyle {x_{3}}cup partial x_{3}} .

Lemma 1 provides a means of combining two different factorizations of {displaystyle Pr(U)} . The local Markov property implies that for any random variable {displaystyle xin U} , that there exists factors {displaystyle f_{x}} and {displaystyle f_{-x}} such that: {displaystyle Pr(U)=f_{x}(x,partial x)f_{-x}(Usetminus {x})} where {displaystyle partial x} are the neighbors of node {displaystyle x} . Applying Lemma 1 repeatedly eventually factors {displaystyle Pr(U)} into a product of clique potentials (see the image on the right).

End of Proof See also Markov random field Conditional random field Notes ^ Lafferty, John D.; Mccallum, Andrew (2001). "Conditional Random Fields: Probabilistic Models for Segmenting and Labeling Sequence Data". Proc. of the 18th Intl. Conf. on Machine Learning (ICML-2001). Morgan Kaufmann. ISBN 9781558607781. Retrieved 14 December 2014. by the fundamental theorem of random fields (Hammersley & Clifford 1971) ^ Dobrushin, P. L. (1968), "The Description of a Random Field by Means of Conditional Probabilities and Conditions of Its Regularity", Theory of Probability and Its Applications, 13 (2): 197–224, doi:10.1137/1113026 ^ Spitzer, Frank (1971), "Markov Random Fields and Gibbs Ensembles", The American Mathematical Monthly, 78 (2): 142–154, doi:10.2307/2317621, JSTOR 2317621 ^ Hammersley, J. M.; Clifford, P. (1971), Markov fields on finite graphs and lattices (PDF) ^ Clifford, P. (1990), "Markov random fields in statistics", in Grimmett, G. R.; Welsh, D. J. A. (eds.), Disorder in Physical Systems: A Volume in Honour of John M. Hammersley, Oxford University Press, pp. 19–32, ISBN 978-0-19-853215-6, MR 1064553, retrieved 2009-05-04 ^ Grimmett, G. R. (1973), "A theorem about random fields", Bulletin of the London Mathematical Society, 5 (1): 81–84, CiteSeerX 10.1.1.318.3375, doi:10.1112/blms/5.1.81, MR 0329039 ^ Preston, C. J. (1973), "Generalized Gibbs states and Markov random fields", Advances in Applied Probability, 5 (2): 242–261, doi:10.2307/1426035, JSTOR 1426035, MR 0405645 ^ Sherman, S. (1973), "Markov random fields and Gibbs random fields", Israel Journal of Mathematics, 14 (1): 92–103, doi:10.1007/BF02761538, MR 0321185 ^ Besag, J. (1974), "Spatial interaction and the statistical analysis of lattice systems", Journal of the Royal Statistical Society, Series B, 36 (2): 192–236, JSTOR 2984812, MR 0373208 Further reading Bilmes, Jeff (Spring 2006), Handout 2: Hammersley–Clifford (PDF), course notes from University of Washington course. Grimmett, Geoffrey (2018), "7.", Probability on Graphs (2nd ed.), Cambridge University Press, ISBN 9781108438179 Langseth, Helge, The Hammersley–Clifford Theorem and its Impact on Modern Statistics (PDF), Department of Mathematical Sciences, Norwegian University of Science and Technology Categories: Probability theoremsTheorems in statisticsMarkov networks

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Classical Mechanics

# Linear restoring force - perturbation analysis

As shown in the figure above, a physical pendulum consists of a disc of radius $R = 5.0 \text{ cm}$ and mass $m = 1.0 \text{ kg}$ fixed at the end of a massless rod. The other end of the rod is pivoted about point $P$ on the ceiling. The distance from the pivot point to the center of mass of the bob is $L =4 \text{ m}.$ Initially the bob is released from rest from a small angle $\theta_0$ with respect to the vertical. Find the period of the bob.

Assumptions and Details

• The gravitational acceleration is $g = 9.8 \text{ m/s}^2.$

What is the frequency of a simple pendulum with arm length $l = 2.00 \text{ m}$ that is in an elevator accelerating upward at a rate of $a = 1.00 \text{ m/s}^2$?

Assumptions and Details

• Assume that the amplitude of the simple pendulum is very small.
• The gravitational acceleration is $g = 9.80 \text{ m/s}^2.$

A physical pendulum consists of a $6.0 \text{ m}$ long stick with mass $m = 100 \text{ g}$ joined to the ceiling, as shown in the above figure. What is the pendulum’s period of oscillation about point $A$ at the tip of the stick?

Assumptions and Details

• The gravitational acceleration is $g= 9.8 \text{ m/s}^2.$
• Assume that the amplitude of the pendulum is very small.

There is a half-ring of density $\rho = 4 \text{ kg/m}$ and radius $R = 2 \text{ m}$. When it is perturbed by small angle $\theta$, it oscillates. If the ring's period can be expressed as $T = 2\pi\sqrt{\frac{a}{b}} \text{ s},$ where $a$ and $b$ are coprime positive integers, what is the value of $a+b?$

The gravitational acceleration is $g=10\text{ m/s}^2.$
Assume that $\sqrt{2} = 1.5$ and $\pi = 3 .$ (This is a model of a roly-poly.)

A disk of mass $M = 1 \text{ kg}$ and radius $R = 4 \text{ m}$ oscillates on a rod. If the disk's period can be expressed as $T = 2\pi\sqrt{\frac{a}{b}} \text{ s},$ where $a$ and $b$ are coprime positive integers, what is the value of $a+b?$

Assume that $\theta$ is very small.
The moment of inertia on the axis is given by $I = \frac{3}{2}MR^2.$
The gravitational acceleration is $g=10\text{ m/s}^2.$

×


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# Convergence of probability measures on a generating field of a sigma-field

Let $(\Omega,\mathcal{B})$ be a measurable space and let $\mathcal{F}$ be a generating field of $\mathcal{B}$. Assume $\mathcal{F}$ is standard, i.e. it is countable, and any normalized, non-negative, finitely-additive set function on $\mathcal{F}$ is also countably-additive, thus can be extended to a probability measure on $\mathcal{B}$ uniquely by Caratheodory's extension theorem.

Let $m_n$ be a sequence of probability measures on $\mathcal{B}$. If for all $F\in\mathcal{F}$, $m_n(F)\to\alpha(F)$ for some set function $\alpha$, then $\alpha$ is normalized, non-negative, finitely-additive on $\mathcal{F}$ and so can be extended to a probability measure $m$ on $\mathcal{B}$. Is it true that $m_n(F)\to m(F)$ for all $F\in\mathcal{B}$?

This is related to the following question: if $\mathcal{F}\subset\mathcal{G}\subset\mathcal{B}$ where $\mathcal{G}$ could be countable, and the convergence $m_n(F)\to\alpha(F)$ is on $\mathcal{G}$. Restricting $\alpha$ on $\mathcal{F}$ and extend $\alpha|_{\mathcal{F}}$ to a probability measure $m$ on $\mathcal{B}$. Then is it true that $m(F) = \alpha(F)$ on $\mathcal{G}$? We know it is true if $\mathcal{G}=\mathcal{B}$, since by Vitali–Hahn–Saks theorem, $\alpha$ is already a probability measure.

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No, you can have measures whose supports are finite sets tending towards (atomless) measures. E.g., the average over n equally spaced points tends the uniform distribution (I have cantor space in mind here). But, I am not familar with the term "standard generating field". What kind of examples do you have in mind? – George Lowther Sep 19 '12 at 23:51
I don't see in your example for what event $F$ $m_n(F)\to m(F)$ is false. A "standard field" is a term in R. Gray's book Probability, Random Processes, and Ergodic Properties. A field is standard if it possesses a basis, which is a sequence of finite fields $\mathcal{F}_n$ with the following properties: 1. $\mathcal{F}_n$ asymptotically generates F, that is, $\mathcal{F}_n\subset\mathcal{F}_{n+1}$ and $\mathcal{F} = \bigcup_n\mathcal{F}_n$. 2. If $G_n$ is a sequence of atoms of $\mathcal{F}_n$ such that $G_{n+1}\subset G_n$, then $\bigcup_n G_n\neq\emptyset$. – Wei Mao Sep 20 '12 at 4:55


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## Method #1 – Radius from the Area

### Formula – How to calculate the radius of a semicircle from the area

π is approximately equal to 3.1415926535

Radius = √(2 x Area ÷ π)

##### Example

A semicircle has an area of 300

Radius = √(2 x 300 ÷ π)

Radius = √(2 x 300 ÷ 3.14159)

## Method #2 – Radius from the Circumference

### Formula – How to calculate the radius of a semicircle from the circumference

π is approximately equal to 3.1415926535

##### Example

A semicircle has a circumference of 35

Radius = 35 ÷ π = 35 ÷ 3.14159 = 11.14086


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 Generalization of the Multiplicative Fourier Transform and Its PropertiesS. S. Volosivets 323 Extenders and $\varkappa$-Metrizable CompactaV. M. Valov 331 On Annihilators of Polynomial Near-RingsSh. Ghalandarzadeh, P. Malakooti Rad, S. Shirinkam 342 An Example Related to the Sidon Property for $m$-Dependent SystemsV. F. Gaposhkin, Yu. S. Semenov 350 On a Class of Lattice Semigroup FunctorsS. F. Kamornikov 355 The al-Husayn Equation $x^4+y^2=z^2$S. Sh. Kozhegel'dinov 365 Existence of a Trade-Off Factorization of a Partially $4$-Homogeneous GraphA. M. Magomedov 378 Transformation of the Triple Series of Gelfand, Graev, and Retakh into a Series of the Same Type and Related ProblemsA. W. Niukkanen 384 Weighted Identities for the Solutions of Generalized Korteweg–de Vries EquationsS. I. Pokhozhaev 393 New Iterations with Errors for Approximating Common Fixed Points for two Generalized Asymptotically Quasi-Nonexpansive Nonself-MappingsS. Thianwan 410 On the Boundedness of Lagrange Multipliers of the Kuhn–Tucker Theorem Applied to the Problem of Tikhonov Function MinimizationN. Ciric, D. Kostic 424 Asymptotically Optimally Reliable Circuits in the Basis $\{x_1\mathbin& x_2\mathbin& x_3,x_1\vee x_2\vee x_3,\overline x_1\}$ for Inverse Faults at the Inputs of ElementsV. V. Chugunova 440 Inverse Problem for Sturm–Liouville Operators on Hedgehog-Type GraphsV. A. Yurko 459 Brief Communications Mean Value of Sums of Partial Quotients of Continued FractionsE. N. Zhabitskaya 472 Computer Solution of the Almost Empty Hexagon ProblemV. A. Koshelev 477


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1. ## Really Confused

just really confused.

list three additional points on a line that has a slope of 3/2 and goes through (3,-1)

Thanks

2. Originally Posted by tori323
just really confused.

list three additional points on a line that has a slope of 3/2 and goes through (3,-1)

Thanks
There are a number of ways this can be done...

since the slope is defined as $m=\frac{\Delta y}{\Delta x}=\frac{\text{rise}}{\text{run}}$, and the slope given to us is $m=\frac{3}{2}$, we see that from a given point on the line, we can increase the y value by three and increase the x value by two.

So one additional point could be $(3+3,-1+3)=(6,2)$

Also, note that $\frac{3}{2}=\frac{-3}{-2}$

So we can subtract three from a y value of a point on the line, and subtract 2 from the x value of a point on the line.

So using the original point given to you, we see another possible point could be $(3-3,-1-2)=(0,-3)$

I leave it for you to find another possible point on the line.

I hope this makes sense!

--Chris


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### Syntax

y = fitfun.voigtfwhm(x,p)

### Description

y = fitfun.voigtfwhm(x,p) calculates the voigt function. The width parameters define the FWHM value. The integral of the function is normalized assuming that $dx = 1$. The conversion between different width:

• gamma parameter of the Lorentzian $\gamma = w_L/2$
• standard deviation of the Gaussian $\sigma = w_G/\sqrt{8\cdot\ln(2)}$

### Input Arguments

x
Input coordinates where the function will be calculated.
p
Parameters in a vector with elements [A x0 wG wL]:
• A integral of the output assuming $dx=1$ (for different $dx$ multiply the amplitude with $dx$ to keep the integral constant).
• x0 peak center positions.
• wG FWHM of the Gaussian component.
• wL FWHM of the Lorentzian component.


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# Show by means of an example that $\displaystyle \lim_{x \to a}\left[ f(x) + g(x) \right]$ may exist even though neither $\displaystyle \lim_{x \to a}f(x)$ nor $\displaystyle \lim_{x \to a}g(x)$ exists.

## see work

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### Video Transcript

This is Problem number sixty two of the Stuart Calculus eighth edition section two point three show by means of an example that the limit is expert is a ah, the quantity f of X plus. DMX may exist, even though neither the limiters expert today over half nor ability as expert city of G exists now. One way to fear this example out I want to choose of inappropriate example is to choose two functions that are potentially opposites around that they themselves haven't issue. Um, with respect to Element. At a certain point, let's take after Max hands. We have a civilian ex rx for as an example. From what we know for this function, his ex approaches zero from the right. Never. Okay, get us. We get that it is X rex since approaching zero from the right mate. That's always positives that absolutely signed going, and this employs toe worn as the limit. As we approach a tear from the left, we're dealing with negative values, and so we must consider only the negative part of the function of that. So I mix, meaning that the limit is negative one. So this is an example of efforts and example of function as we approach a equals zero from less than the rate we get two different values, meaning that is limited, not exist as you approach zero. If we chose a function opposite of this a negative times a square root for the absolute value of X over X, we see that we were to the same steps show two different to different limits one negative one one as equal to one concluding that the limit as experts zero does not exist again. The same steps applying because the absolute value function changes as you're approaching zero from one end or the other Fergie of X As you approach zero from the right, your, uh, limited equal to neither one and X approaches Here on the left, you're limited approaches of one. And that means that the limit is experts Nero. For this function, gene is not exist. So if we now consider the limit as experts, zero of some of these two functions absolutely of X. Come on, Rex. Plus negative. Absolutely. Vicks Rex, We see that this clever choice of functions give this Parliament is expert zero one function plus the upset of the other. Which is it? Quit of your own. Which means it is, um it does exist, and it is equal to zero.

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# zbMATH — the first resource for mathematics

Generalized recurrent and concircular recurrent manifolds. (English) Zbl 0981.53030
Summary: The concepts of generalized recurrent manifold $$GK_n$$ and generalized concircular recurrent manifold $$G(ZK_n)$$ are considered to prove that a manifold $$(M_n,g)$$ which is either $$GK_n$$ or $$G(ZK_n)$$ is a concircular recurrent manifold $$ZK_n$$. By using a result due to P. Desai and K. Amur [Tensor, New Ser. 29, 98-102 (1975; Zbl 0309.53022)], it is claimed that such a manifold $$(M_n,g)$$ is either a recurrent space $$K_n$$ or an Einstein space. Further it is noticed that the condition $$A(lx)+(n-1) B(x)=0$$ used by U. C. De and D. Kamilya [Bull. Calcutta Math. Soc. 86, 69-72 (1994; Zbl 0815.53025)] in proving the result is not required.

##### MSC:
 53C25 Special Riemannian manifolds (Einstein, Sasakian, etc.)
##### Keywords:
Einstein space; concircular recurrent manifold


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In order to keep the chats more or less readable, we decided to give some information on the length of disguised messages, which are accordingly marked as e.g. redactedQ12tokens55characters. This annotation lets you know that the original message consisted of 55 characters that formed 12 tokens.


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ComplexNumber AIME Intermediate
2014

Problem - 107
Let $z$ be a complex number with $|z|=2014$. Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}$. Then the area enclosed by $P$ can be written in the form $n\sqrt{3}$, where $n$ is an integer. Find the remainder when $n$ is divided by $1000$.

The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.


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# alphas-2019: Workshop on precision measurements of the QCD coupling constant

11-15 February 2019
ECT* Trento
Europe/Zurich timezone

## Scientific Programme

The strong coupling constant $\alpha_S$ is the least well known of all constants of nature, which play a role in the Standard Model (SM) of particle physics and related fields such as cosmology and astrophysics. For many searches for new physics beyond the SM as well as for some important precision tests of the SM using collider data the uncertainty on the value of $\alpha_S$ is a limiting factor. In recent years progress in theoretical predictions of Quantum Chromodynamics (QCD), and the availability of collider data at the highest energies has led to many improved determinations of $\alpha_S$. The current world average quotes an uncertainty of less than 1%. However, there are noticeable discrepancies between different categories of determinations of $\alpha_S$, which may limit the ultimate precision of future world averages. We plan to bring together in this workshop the leading experts on determinations of $\alpha_S$ from theory and experiment and all important categories. With presentations of the latest results and intense discussion by all participants we will focus on a global view of advantages and problems of each method.

All talks will be plenary and of the same length (20' + 10' questions) with ample time for discussion. The topics to discuss include:

• Current status of the $\alpha_S$ world average
• Impact of $\alpha_S$ on precision and BSM physics, and beyond
• Lattice QCD results
• $\alpha_S$ from hadronic tau decays
• $\alpha_S$ from e-p deep inelastic scattering
• $\alpha_S$ from $e^+e^-$ hadronic final states
• $\alpha_S$ from electroweak observables
• $\alpha_S$ from pp collisions
• New $\alpha_S$ extraction approaches
• PDG approach to $\alpha_S$ world average and uncertainty estimation
• A preliminary new average (?) and future of $\alpha_S$
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# Definition:Characteristic Polynomial of Matrix

## Definition

Let $R$ be a commutative ring with unity.

Let $M$ be a square matrix over $R$ of order $n > 0$.

Let $I_n$ be the $n\times n$ identity matrix.

Let $R[x]$ be the polynomial ring in one variable over $R$.

The characteristic polynomial of $M$ is the determinant of a matrix over $R[x]$:

$p_M (x) = \operatorname{det}(xI - M)$.