text:
I never noticed this before, so I guess I'm a little late. But Wow I didn't know the koreans ripped the Japanese off too. I wonder what the lyrics are.

written by astrobunny \\


text:
# ProfileEye - browser-based visualization frontend¶

ProfileEye is a broswer-based visualization frontend for gprof2dot and graphviz.

The former is a great tool for parsing profiler outputs and building relevant statstics, and the latter contains excellent graph-placement algorithms. Unfortunately, their outputs are static images. Modern Javascript libraries, e.g., d3.js, do a superior job at dynamic information-rich visualizations. ProfileEye combines these three, with the former two serving as a backend, and the latter serving as a frontend.

Contents:

## Requirements¶

This tool requires the following:

## Installation¶

Install the package via pip, as so:

$pip install ProfileEye  or, depending on your system’s policy on installation, something like so: $ sudo pip install ProfileEye

## Invocation¶

The invocation is the same as that of gprof2dot, except, at the end, pipe the output through profile_eye and redirect the result to an html file which you can view via a browswer.

For example, to use pgprof, run

/path/to/your/executable arg1 arg2
gprof path/to/your/executable | gprof2dot | profile_eye > profile_output.html

to use the Python profiler, run

python -m profile -o output.pstats path/to/your/script arg1 arg2
gprof2dot -f pstats output.pstats | profile_eye --file-colon_line-colon-label-format > profile_output.html

Note

Note the use of the --file-colon_line-colon-label-format flag, which tells the script that gprof2dot’s output is in the format <file>:<line>:<label>. Omitting this flag is possible, as so:

python -m profile -o output.pstats path/to/your/script arg1 arg2
gprof2dot -f pstats output.pstats | profile_eye > profile_output.html

but the output of the former is more informative than that of the latter (place the mouse on a node to see the file and line).

Tip

Use gprof2dot’s options (e.g., --node-thres) to control the intermediate output passed to the visualization.

Tip

For gprof, use gprof2dot’s -s (strip) option to shorten function names.

## Output¶

The output is something like the following figure (which is shown in a small form; see here a more realistic output).

View Type Calls Internal time Total time Fisheye Spline edges Straight edges

The primary display element is a combination of a bar chart and heat map. Each function is represented by an external empty bar, within which is an internal filled bar, both sharing the same color.

• The external empty bar represents the total time spent in the function (e.g., in the above, main has the widest bar).
• The internal filled bar represents the net time spent in the function, i.e., excluding the time spent in its children (e.g., in the above, a lot of time was spent in main, but nearly all of it was spent in its children).
• The bars colors form a heat map of the number of calls, with a brighter color indicating a larger number of calls (e.g., in the above, main was called far fewer times than hash_find_slot).

Tip

Placing the mouse over an element opens a tooltip with its exact time fractions and calls.

### The Calls View¶

The Calls view emphasizes the call flow between the different functions. Select it using the Calls radio item.

Edges are colored by the same #calls heat map as the function bars (a brighter edge means a larger number of calls); edges’ opacity is determined by the total amount of time spent on the call (a more opaque edge means a larger total amount of time spent on the call). (E.g., in the above, main-> update_goal_chain was followed few times, but represents a large fraction of the total time; conversely, lookup_file -> hash_find_item was followed more times, but represents a lower fraction of the total time).

Function calls generally run top down. Self or back edges (each indicating some type of recursion) are dashed (e.g., in the above, see pattern_search for a self edge, and eval -> eval_makefile for a back edge).

This view has two options:

• Fisheye mode introduces a fisheye distortion; in this mode items under the mouse will be spaced wider apart, for improved clarity.
• Spline edges vs Straight edges toggle curved spline lines (as calculated by graphviz) vs. simpler straight line.

### The Internal and Total Times Views¶

These views emphasizes the ranking of functions by internal or total time. Select it using the Internal time or Total time radio item.

Each of these views has a bar chart to the left, and two pie charts more to the right. The bar chart is ordered by the selected criteria, and includes a selection brush marked by a greyed rectangle. The pie charts more to the right show the fraction of time covered by the selection rectangle. (E.g., when selecting the Internal time call, you can see that the five functions with the highest internal time take together about 55 percent of all internal time.)

Tip

Modify the brush selection to see the effect on the pie charts. Alternatively, click on a pie slice to change the corresponding brush selection.

# Issues¶

Bug reports can be made at https://bitbucket.org/atavory/profileeye/issues.

# Module Reference¶

Note

This is a reference of the library backend of profile_eye. While you can use this directly, unless you’re sure of what you’re doing, you should probably use the utility version.

class profile_eye.DotPlainParser(file_colon_line_colon_label_format=False)[source]

A parser for dot’s output.

add(l)[source]

Feeds a line into the parser.

Args:
l: Line fed.
prepare_all()[source]

Prepares a dictionary of all parsed and processed data.

Returns:
A dictionary mapping ‘nodes’ to the nodes, and ‘edges’ to the edges.
profile_eye.gprof2dot_to_dot_plain(gprof2dot_output)[source]

Pipes gprof2dot output through dot with plain output specified.

Args:
gprof2dot_output: Bytes consisting of gprof2dot output.
Returns:
Output of dot -Tplain operating on gprof2dot_output.
profile_eye.render_parsed`(d)[source]

Renders HTML for parsed gprof2dot + dot output.

Args:
d: the results of a DotPlainParser’s prepare_all()
Returns:
A bytes object of HTML.

format: plain


text:
## Update, 2018-12-11

With version 4 of nom a lot of things changed so the code snippets won’t work anymore and some descriptions might be outdated.

## Identifiers

We’re making good progress, booleans, numbers (in a simplified form), characters and strings already work, the only missing part is a parser for identifiers.

You might ask “But what about the keyword parser from Part 1”?

Turns out, we don’t even need it for the token parser, but it will come in handy once we start to parse expressions.

<token> →
<identifier> |
<boolean> |
<number> |
<character> |
<string> |
( | ) | #( | ’ |  | , | ,@ | .

## Peculiar Identifiers

named!(peculiar_identifier, alt!(tag!("+") | tag!("-") | tag!("...")));

named!(
identifier<String>,
map!(
peculiar_identifier,
|s| String::from_utf8_lossy(s).into_owned()
)
);

A combination of alt! and tag! matches each of the peculiar identifiers and we can use the same method as in the to_s function from earlier to convert &[u8] to String.

Next we need add a Identifier type to the Token enum and parser. Note that I removed the Keyword type, too.

#[derive(Debug, PartialEq)]
enum Token {
Number(i64),
Boolean(bool),
Character(char),
String(String),
Identifier(String),
}

named!(
token<Token>,
alt_complete!(
integer           => { |i| Token::Number(i) } |
boolean           => { |b| Token::Boolean(b) } |
character         => { |c| Token::Character(c) } |
string            => { |s| Token::String(s) } |
identifier        => { |s| Token::Identifier(s) }
)
);

Again it’s important to use alt_complete! instead of alt! to avoid conflicts between the number +1 and the identifier +.

## “Common” Identifiers

First we need some helper classes to match the different groups of characters. We can’t use nom::digit or nom::alpha here because they match multiple characters while we only want to match a single one.

named!(letter<char>, one_of!("abcdefghijklmnopqrstuvwxyz"));
named!(single_digit<char>, one_of!("0123456789"));
named!(special_initial<char>, one_of!("!$%&*/:<=>?^_~")); named!(special_subsequent<char>, one_of!("+-.@"));  I’m sure there is a more elegant way to do this but one_of! with a string of all characters is good enough for now. The result of these parsers is char, not &[u8] so we need to explicitely annotate their type. Like in Part 3 we can use a combination of recognize! and do_parse! to match “common” identifiers: named!( common_identifier, recognize!( do_parse!(initial >> many0!(subsequent) >> ()) ) );  Finally change identifier to support both types: named!( identifier<String>, map!( alt!(peculiar_identifier | common_identifier), |s| String::from_utf8_lossy(s).into_owned() ) );  And add the remaining tokens: #[derive(Debug, PartialEq)] enum Token { Number(i64), Boolean(bool), Character(char), String(String), Identifier(String), LBracket, RBracket, HashBracket, Quote, Quasiquote, Unquote, UnquoteSplicing, Dot } named!( token<Token>, alt_complete!( integer => { |i| Token::Number(i) } | boolean => { |b| Token::Boolean(b) } | character => { |c| Token::Character(c) } | string => { |s| Token::String(s) } | identifier => { |s| Token::Identifier(s) } | tag!("(") => { |_| Token::LBracket } | tag!(")") => { |_| Token::RBracket } | tag!("#(") => { |_| Token::HashBracket } | tag!("'") => { |_| Token::Quote } | tag!("") => { |_| Token::Quasiquote } | tag!(",@") => { |_| Token::UnquoteSplicing } | tag!(",") => { |_| Token::Unquote } | tag!(".") => { |_| Token::Dot } ) );  A quick test shows that everything works as expected and there don’t seem to be any strange conflicts between identifiers and numbers: >> test Parsed Done([], Identifier("test")) >> +1 Parsed Done([], Number(1)) >> + Parsed Done([], Identifier("+")) >> ... Parsed Done([], Identifier("...")) >>$foo123
Parsed Done([], Identifier("\$foo123"))
>> .
Parsed Done([], Dot)
>> (
Parsed Done([], LBracket)
>> #(
Parsed Done([], HashBracket)
>>

This was easier than I expected, but I’m sure things will get more exiting once we start parsing expressions.

Full source code: l3kn/r5rs-parser.


text:
# How to prove substring is a partial order

u is defined to be a substring of a string v if v = xuy for some string x and y. Either or both possibly empty.

How to you prove that a substring relation on any set of strings is a partial order?

• What have you tried? Where did you get stuck? Try using the definition of a partial order. – Yuval Filmus Oct 1 '13 at 4:35
• Do you know the proof for the subsets of a set $A$ ordered by $\subseteq$? If you do, what is different when you consider substrings? – frafl Oct 1 '13 at 7:18

Hint: show that if $u$ is a substring of $v$ and $v$ is a substring of $w$ then $u$ is a substring of $w$.


text:
# Is the directional derivative $f'(x;d)$ ALWAYS equal to $\nabla f^T d$? or only under certain conditions?

My professor always writes the directional derivative with the condition that the functionis $C^1$ smooth.
i see him write many times the directional derivative written as $f'(x;d)=\nabla f^T d$, but right next to it he writes "because $\beta - C^1 smooth$", where $\beta$ is the Lipschitz constant, and $C^1$ means that it is continuously differentiable up to and including the first derivative.

Is the directional derivative $f'(x;d)$ ALWAYS equal to $\nabla f^T d$? or only under certain conditions?

EDIT And does "differentiable" mean with regards to a certain direction? Or if a function is differentiable, then does that mean it must be differentiable in all directions? If the former, then does $C^1$ mean it is differentiable in ALL directions?

That equality holds when $f$ is differentiable at $x$ (in the sense of having an appropriately good linear approximation at $x$). The standard theorem that one always proves is that $C^1$ implies differentiability. [And, of course, $(\nabla f(x))^\top$ will be the linear map that gives the good approximation.]
• So then which is a stronger form of the directional derivative? the one with the gradient dotted with $d$? or the one with the directional derivative $f'$? I often see the former used instead of the latter Apr 7 '18 at 18:03
• The directional derivative may well exist (and be computed from the definition) even when $f$ fails to be differentiable. (You might find some of the discussion in my lectures 22 and 23 helpful.) Apr 7 '18 at 18:08
• An example is $f(x,y) = x^2 + |y|$. It is not differentiable for $y=0$, but the directional derivative along the $x$ axis exists. Apr 7 '18 at 18:38
• Right, so then a few clarifications: 1) you're telling me that it is possible that the derivative can exist in ALL directions i.e. ($\forall d$), and yet there might not exist a linear operator to express it (i.e. $\nabla f^T*d$ is invalid/does not exist)? 2) C^1 smoothness implies differentiability in all d irections? If so, then can we say if a function is C^1 smooth, then that means that the linear operator $\nabla f$ DOES EXIST, and we can say $f'(x;d)=\nabla f^T d$ ? Apr 8 '18 at 21:41


text:
stationarity and fractional differencing

This is a methodology question. I would like to make the data stationary but not transform it "too much" (information loss), before it is fit for statistical/ML purposes such as regression or PCA.

Typical procedure:

1. We inspect the raw time series $$x_t$$. Assume we see trends or if the ADF test fails, so I seek to transform the data.

2. We check that differencing $$x_{t+1} - x_t$$ results in stationary series. (Assume it does.)

3. If possible, we seek alternatives to $$x_{t+1} - x_t$$: the transformation may excessively remove memory from the series. That may help stationarity but at the price of weaker correlations to other series. We therefore use fractional differencing, say with d=0.5, and check that the series is more stationary than the initial series.

Let's apply this to the following series:

Conclude that it is stationary with $$95\%$$ confidence when using ADF implemented by adfuller() from Python's statsmodels package. The default parameter is $$12(n/100)^{1/4}$$ where $$n$$ is the series length. However, we see some (noisy) trend between 2011 and 2016. For the first half of the data the values are almost all negative. I conclude that the data doesn't give enough "labels" for model training.

I am therefore not sure whether the ADF test can be trusted. Would you feel comfortable using it for regression/PCA? Now compare with the fractional differencing with $$d=0.5$$:

As expected this makes the series look more stationary. However, it also decreases the correlation to other similar dataset series transformed in the same way. Average correlation decreases as I increase $$d$$, as can be expected.

I therefore need to choose the smallest $$d$$ such that I feel comfortable that the transform is stationary. How do you know when to stop? I am reluctant to use the ADF pvalue as threshold to automatically set this considering the issue above.


text:
## Essential University Physics: Volume 1 (4th Edition)

We simplify the equation for spring energy: $U = \frac{1}{2}kx^2 \\ x = \sqrt{\frac{2U}{k}} \\ x = \sqrt{\frac{2(44,000)}{120,000}}= .86m$


text:
# Lemma in the Homotopy Axiom theorem

I have the following Lemma (from Rotman)

Let $X$ be a space and, for $i=0,1$, let $\lambda_i^X:X \to X \times I$ be defined by $x \mapsto (x,i)$. If $H_n \left(\lambda_0^X\right)=H_n\left(\lambda_1^X\right):H_n(x)\to H_n \left(X \times I\right)$, then $H_n(f)=H_n(g)$ whenever $f$ and $g$ are homotopic.

With the proof of this fact, this step is used:

$H_n\left(F\lambda_0^X\right)=H_n(F)H_n(\lambda_0^X)$

Maybe (probably) I am missing something obvious, but it is not immediately clear to me why that is true

-
This is exactly the functoriality of homology –  Exterior Jan 5 at 17:04

## 1 Answer

(Let's say you are working with singular homology.)

If you have maps of topological spaces $X \stackrel{f}{\to} Y \stackrel{g}{\to} Z$, then you have $H_n(g \circ f) = H_n(g) \circ H_n(f)$.

To prove this, check where a cycle $\gamma: I^n \to X$ is being sent to under $H_n(g \circ f)$ and $H_n(g) \circ H_n(f)$. Here $I^n$ is the standard $n$-simplex.

-
of course, thank you! –  Juan S Feb 28 '11 at 3:41


text:
# Networks and ranking in sports

Which is the best team in the league? How does my team fare compared to the rest of the league? To answer similar questions fans turn to the power rankings that are regularly provided by various sports networks or the rankings provided by the league standings (i.e., win-loss records). The rankings provided by sports networks are based on a variety of factors.  Some of them are objective, but some might also be very subjective, based on the opinion of the expert analysts.  Hence, me and Vagelis – also known as the “tensor whisperer” – set out to examine objective ways based on network science tools, alternative to the league standings, for ranking the teams!

To have network ranking we first need a network structure.  We define the ${\tt League}$ ${\tt Network}$ $\mathcal{G}=(\mathcal{V},\mathcal{E})$, as a weighted directed network where an edge $e_{i,j}\in \mathcal{E}$ points from node $i\in \mathcal{V}$ to node $j\in\mathcal{V}$ if team $j$ has won team $i$. The edge weight $w_{e_{i,j}}$ is equal to the point differential of the corresponding game. The following figure depicts an example of a ${\tt League}$ ${\tt Network}$ for the NFL 2015-2016 season.

The ${\tt League}$ ${\tt Network}$ captures the relations of who-wins-whom but also, and most importantly, it incorporates information about the schedule strength (who has played with whom).  Then important teams can be found using centrality metrics borrowed from the network theory.  Maybe the most well known centrality metric is Page Rank, which essentially emulates a random walker over the network.  In the above visualization the size of the nodes is proportional to their Page Rank score, while the width of the edges is proportional to their weight (only edges with weight greater than 7 are visualized for clarity).

Based on this network the worst five teams of last year were: Cleveland, San Diego, Dallas, Tenessee and Baltimore  –  which sounds about right 🙂 – while the best five performing teams were: Seattle, Atlanta, Arizona, Carolina and Pittsburgh – this sounds again about right except from Atlanta! Why is Atlanta so high when their actual record was only 8-8? Because they were the only team to beat the Carolina Panthers and essentially the centrality of the Panthers in the network was only diffused through the “random walker” to Atlanta.  One possible way to overcome similar glitches of Page Rank is by adjusting its parameters appropriately.  Nevertheless, in this blog post we just want to present the idea without getting into the nuts-and-bolts of the approach.

How good is this ranking? Is it better than the win-loss percentages? To answer this question we turned to a prediction task.  In particular, using the network up to week $\tau-1$ we used the ranking obtained from Page Rank (and the one from the standings) to predict the outcome of the match-ups of week $\tau$.  The prediction rule was simple; if team $i$ was ranked higher than team $j$ then $i$ was predicted to win! The following table provides the prediction accuracy for the Page Rank (termed with the fancy name ${\tt SportsNetRank}$) and the winning percentage.

As we can see the network-based ranking appears to capture the power of the teams much better as compared to the winning percentage – which is also actually used to decide which teams advance to the playoffs and which not! Just to make a point, the way more sophisticated Cortana system of Microsoft, had a performance of 62.9% with a standard error of 3% for the 2015 season and 66.4% for the 2014 season with a standard error of 3.2% (the system did not exist earlier)! Take that Cortana!

What’s next in power ranking? While Page Rank performs well – and certainly captures the power of each team better than the winning percentage – there is still room for improvement.  An idea we have put forward here with Vagelis is the use of tensors for ranking! A tensor is a multi-dimensional matrix and can capture more complex (network) relations.  In brief, we define the ${\tt League}$ ${\tt Tensor}$ $\underline{\mathbf{T}}$ a structure that captures the temporal evolution of the win-loss relationships, and in particular the element $\underline{\mathbf{T}}(i,j,k) = \delta$, if team $j$ won team $i$ by $\delta$ points.  One way to identify a ranking is through tensor factorization, where the tensor is decomposed to a sum of components as is visualized in the following figure.

The good thing about tensor factorization is that we can process the tensor together with other, side, information (e.g., performance information for every team such as yards per game, # of turnovers, # of interceptions etc.).  This information can be captured by a simple 2D matrix, where each row corresponds to a team and the columns correspond to the performance features.  The challenging problem is how one goes about combining the factors obtained in order to get a ranking of the teams! We will let you know as soon as we do it 🙂

In addition to team ranking, tensor decomposition is a very powerful technique that can help us identify latent groups of similarly performing teams across different dimensions; combining that with the 2D side information, we can correlate those team groups with performance statistics that mostly explain their behavior and use that to further enhance predictions. More on that in the near future! 🙂

It should be clear that the simple standings used by the various leagues for deciding which teams will play into the playoffs can be significantly improved.  Of course, it will be hard to convince team X with higher winning percentage but lower Page Rank (or other network rankings) than team Y, that it should give its playoff spot to Y.  Nevertheless, with the data analytics craze in the business today everything is possible!

## 2 thoughts on “Networks and ranking in sports”

1. […] a powerful set of tools for analyzing a variety of relationships between players – or even teams.  Just as an illustrative example, the following networks represent the substitution patterns for […]

Like

2. […] can then read the actual research paper.  Similar  to this post or this post or this post or this post. I will also be posting blogs that explain an analytical technique (using sports data), like this […]

Like


text:
My Math Forum Equation problem

 Algebra Pre-Algebra and Basic Algebra Math Forum

 June 6th, 2014, 07:31 AM #11 Newbie   Joined: Jun 2014 From: rhsfhb Posts: 24 Thanks: 0 Well, i understood what he said, but i thought that there are more solutions, beacause when i finished whole equation his solution was just one of many, anyway i think i got it now. Probably...
June 6th, 2014, 07:54 AM   #12
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,634
Thanks: 2620

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by skipjack If you still can't understand soroban's approach, read the first two paragraphs of this article.
An alternative (although longer) approach is to split the domain of $u$ into intervals in which each of the terms on the left is of constant sign.
\begin{align*}
(2-u) + (3-u) &= 1 & u &\le 2 \\
(u-2) + (3-u) &= 1 & 2\le u &\le 3 \\
(u-2) + (u-3) &= 1 & 3\le u \\
\end{align*}
The middle equation is valid for all values of $u$ (for which the equation applies) since $1 \equiv 1$.

Sometimes in systems like this (but not in this case), one or more of the equations produces a value for $u$ which is outside the interval for which the equation is valid. In this case the value of $u$ produced is not a valid solution. You also have to be careful about whether you use $\le$ or $\lt$.

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text:
# How to extract data from existing JPEG/TiFF graph?

• MATLAB
Gold Member
Is there any method to extract data from a graph in JPEG/TIFF format, using simple software such as MATLAB? I have a graph as shown below:

I only have this picture but no raw data and I want to plot the black curve in this graph. How to get approximate numerical data of the x and y axis? Thank you.

#### Attachments

• Untitled.jpg
17.6 KB · Views: 536

## Answers and Replies

DrClaude
Mentor
I don't know how to do it simply in Matlab. For such tasks, I use DataThief.

kelvin490
You can use getframe()

Code:
plot(1:10);
structFrame = getframe(); % Get structure
rgbImage = structFrame.cdata;
figure;
imshow(rgbImage);


text:
# Nonlinear gap modes and compactons in a lattice model for spin-orbit coupled exciton-polaritons in zigzag chains

@article{Johansson2019NonlinearGM,
title={Nonlinear gap modes and compactons in a lattice model for spin-orbit coupled exciton-polaritons in zigzag chains},
author={Magnus Johansson and Petra P. Beli{\vc}ev and Goran Gligori'c and Dmitry R Gulevich and Dmitry V. Skryabin},
journal={Journal of Physics Communications},
year={2019}
}
We consider a system of generalized coupled Discrete Nonlinear Schrodinger (DNLS) equations, derived as a tight-binding model from the Gross-Pitaevskii-type equations describing a zigzag chain of weakly coupled condensates of exciton-polaritons with spin-orbit (TE-TM) coupling. We focus on the simplest case when the angles for the links in the zigzag chain are $\pm \pi/4$ with respect to the chain axis, and the basis (Wannier) functions are cylindrically symmetric (zero orbital angular momenta…

## References

SHOWING 1-10 OF 60 REFERENCES
Nonlinear localized flat-band modes with spin-orbit coupling
• Physics
• 2016
We report the coexistence and properties of stable compact localized states (CLSs) and discrete solitons (DSs) for nonlinear spinor waves on a flat-band network with spin-orbit coupling (SOC). The
Symmetry breaking of localized discrete matter waves induced by spin–orbit coupling
• Physics
• 2015
Abstract We study localized nonlinear excitations of a dilute Bose–Einstein condensate (BEC) with spin–orbit coupling in a deep optical lattice (OL). For this we introduce a tight-binding model that
Kagome lattice from an exciton-polariton perspective
• Physics
• 2016
We study a system of microcavity pillars arranged into a kagome lattice. We show that polarization-dependent tunnel coupling of microcavity pillars leads to the emergence of the effective spin-orbit
Discrete and continuum composite solitons in Bose-Einstein condensates with the Rashba spin-orbit coupling in one and two dimensions.
• Physics, Medicine
Physical review. E, Statistical, nonlinear, and soft matter physics
• 2014
One- and two-dimensional continuum and discrete models for the two-component BEC are introduced, with the spin-orbit (SO) coupling of the Rashba type between the components, and attractive cubic interactions, and it is demonstrated that 1D and 2D discrete solitons are mobile, in a limited interval of velocities.
Flat bands and dynamical localization of spin-orbit-coupled Bose-Einstein condensates
• Physics
Physical Review A
• 2018
Flat bands and dynamical localization of binary mixtures of Bose-Einstein condensates, with spin-orbit coupling subjected to a deep optical lattice which is shaking in time and to a periodic time
Strongly correlated flat-band systems: The route from Heisenberg spins to Hubbard electrons
• Physics
• 2015
In this review we recapitulate the basic features of the flat-band spin systems and briefly summarize earlier studies in the field. Main emphasis is made on recent developments which include results
Coherent driving and freezing of bosonic matter wave in an optical Lieb lattice
• Physics, Medicine


text:
# UPPER LIMITS TO THE DETECTION OF AMMONIA FROM PROTOPLANETARY DISKS AROUND HL TAURI AND L1551-IRS 5

DOI:
10.1086/173079
Publication date:
01/09/1993
Main author:
GOMEZ, JF
IAA authors:
Authors:
GOMEZ, JF; TORRELLES, JM; HO, PTP; RODRIGUEZ, LF; CANTO, J
Journal:
ASTROPHYSICAL JOURNAL
Publication type:
Article
Volume:
414
Pages:
333-336
Number:
Abstract:
We present NH3(1, 1) and (2, 2) observations of the young stellar sources HL Tau and L1551-IRS 5 using the VLA in its B-configuration, which provides an angular resolution of approximately 0.4'' (approximately 50 AU at 140 pc) at 1.3 cm wavelength. Our goal was to detect and resolve circumstellar molecular disks with radius of the order of 100 AU around these two sources. No ammonia emission was detected toward either of them. The 3 sigma levels were 2.7 mJy beam-1 and 3.9 mJy beam-1 for HL Tau and L1551-IRS 5, respectively, with a velocity resolution of approximately 5 km s-1. With this nondetection, we estimate upper limits to the mass of the proposed protoplanetary molecular disks (within a radius of 10 AU from the central stars) on the order of 0.02 (X(NH3)/10(-8))-1 M. for HL Tau and 0.1 (X(NH3)/10(-8))-1 M. for L1551-IRS 5.
Database:
WOK
Keywords:
CIRCUMSTELLAR MATTER; STARS, INDIVIDUAL (HL TAURI, L1551 IRS 5); STARS, PRE-MAIN-SEQUENCE


text:
It is currently 17 Dec 2017, 00:25

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# Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a

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Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a [#permalink]

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Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a circle with center C such that each vertex of DEF lies on the circumference of C. If the shortest side of DEF has length x, which of the following represents the area of circle C?

A. 2πx

B. πx^2

C. sq. root 3πx^2

D. 2πx^2

E. 4 πx^2
[Reveal] Spoiler: OA

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Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a [#permalink]

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20 Sep 2014, 11:53
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neeti1813 wrote:
Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a circle with center C such that each vertex of DEF lies on the circumference of C. If the shortest side of DEF has length x, which of the following represents the area of circle C?

A. 2πx

B. πx^2

C. sq. root 3πx^2

D. 2πx^2

E. 4 πx^2

The triangle has angles in the ratio 1:2:3 --> a + 2a + 3a = 180° --> a = 30°. So, we have 30-60-90 right triangle.

This is one of the 'standard' triangles you should be able to recognize on sight. A fact you should commit to memory is: the sides are always in the ratio $$1 : \sqrt{3}: 2$$.

Since the shortest side is x, then the sides are $$x : \sqrt{3}x: 2x$$.

Next, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle (the reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle).

So, the largest side, the hypotenuse, 2x, must be the diameter of the circle --> radius = x --> $$area = \pi{r^2}=\pi{x^2}$$.

Hope it's clear.
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Re: Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a [#permalink]

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12 Oct 2016, 07:58
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Re: Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a [#permalink]

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06 Sep 2017, 21:57
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neeti1813 wrote:
Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a circle with center C such that each vertex of DEF lies on the circumference of C. If the shortest side of DEF has length x, which of the following represents the area of circle C?

A. 2πx

B. πx^2

C. sq. root 3πx^2

D. 2πx^2

E. 4 πx^2

Bunuel this is a PS in the DS directory please move. Kudos

Kudos [?]: 39 [1], given: 166

Math Expert
Joined: 02 Sep 2009
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Kudos [?]: 135910 [0], given: 12715

Re: Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a [#permalink]

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06 Sep 2017, 22:00
Nunuboy1994 wrote:
neeti1813 wrote:
Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a circle with center C such that each vertex of DEF lies on the circumference of C. If the shortest side of DEF has length x, which of the following represents the area of circle C?

A. 2πx

B. πx^2

C. sq. root 3πx^2

D. 2πx^2

E. 4 πx^2

Bunuel this is a PS in the DS directory please move. Kudos

Done. Thank you for noticing.
_________________

Kudos [?]: 135910 [0], given: 12715

Re: Right triangle DEF has angles in the ratio 1:2:3 and is inscribed in a   [#permalink] 06 Sep 2017, 22:00
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text:
# Nonlinear Structural Equation Modelling- Estimation via Pseudo Maximum Likelihood in R

I am trying to implement the pseudo maximum likelihood approach described in Wall and Amemiya (2007) (see link to article at the end) in estimating the parameters of a nonlinear structural model of the form:

$\eta=\gamma_1\xi_1+ \gamma_2\xi_2+ \gamma_3\xi_3+\omega_{12}\xi_1\xi_2+\omega_{13}\xi_1\xi_3+\omega_{23}\xi_{2}\xi_{3}+\zeta$

where $\zeta\sim N(0,\Delta)$ and the measurement model is given by:

$\boldsymbol{X}=\boldsymbol{\Lambda}_{x}\boldsymbol{\xi} + \boldsymbol{\delta}_{x}$

$\boldsymbol{Y}=\boldsymbol{\Lambda}_{y}\eta + \boldsymbol{\delta}_{y}$

which is combined into one model as:

$\boldsymbol{Z}=\boldsymbol{\Lambda}\boldsymbol{f} + \boldsymbol{\epsilon}$

where $\boldsymbol{Z}=(\boldsymbol{Y}',\boldsymbol{X}')'$ and $\boldsymbol{f}=(\eta,\boldsymbol{\xi}')'$ and $\boldsymbol{\epsilon}=( \boldsymbol{\delta}_{y}', \boldsymbol{\delta}_{x}')'$.

The method starts by first estimating the parameters in the measurement model through maximum likelihood. Using these parameter estimates, Bartlett scores for $\eta$ and $\boldsymbol{\xi}$ are computed. Using these factor scores, a Gaussian Mixture model is fitted to $\boldsymbol{\xi}$.

As described in the paper attached, the parameters in the structural model are then estimated via a Monte-Carlo EM algorithm.

I have written a script in R to perform a simulation to assess parameter recovery properties of the method. However, the algorithm is not converging. I suspect that I've some problems in the code as I am still a novice in R programming. My gut feeling is that I have a mistake in the calculation of the weights but I cannot seem to find the problem. I've been going at it for weeks on end to no avail.

I am attaching a copy of the script I have written in the hope of getting some feedback or a new perspective as to where the problem with the code might be.

Many thanks!

Paper refered to in text: Wall and Amemiya (2007)- Nonlinear Structural Equation Modeling as a Statistical Method: Can be accessed on: https://pdfs.semanticscholar.org/590f/da1c7f216a110d4c11fff2d61eee2f451462.pdf

#Load Required Packages
library("lavaan", lib.loc="~/R/win-library/3.4")
library("mclust", lib.loc="~/R/win-library/3.4")
library("mixtools", lib.loc="~/R/win-library/3.4")
library("MASS", lib.loc="~/R/win-library/3.4")
library("MASS", lib.loc="C:/Program Files/R/R-3.4.1/library")
library("MixSim", lib.loc="~/R/win-library/3.4")
library("mvtnorm", lib.loc="~/R/win-library/3.4")
library("mixAK", lib.loc="~/R/win-library/3.4")
library("gdata", lib.loc="~/R/win-library/3.4")

#Data Simulation

n=1000 #sample size
s=100 #number of Monte-Carlo samples

#Generate population latent variables
mu_z<-c(rep(0,3))
sigma_z<-matrix(rep(0,9),3,3)
sigma_z[,1]<-c(1,0.5,0.6)
sigma_z[,2]<-c(0.5,1,0.4)
sigma_z[,3]<-c(0.6,0.4,1)

#Compute xi from zs (as is done in paper)
set.seed(28)
z<-matrix(0,ncol=3,nrow=n)
z<-mvrnorm(n, mu=mu_z, Sigma=sigma_z)
xi<-sqrt(exp(z))

#Error term of structual model
var_zeta<-1
zeta<-rnorm(n,0,sqrt(var_zeta))

#Setting Population parameters
gamma1<-1.5
gamma2<-1
gamma3<-2
omega12<--3
omega13<-1.25
omega23<-1.75

eta<-(gamma1*(xi[,1]))+(gamma2*(xi[,2]))+(gamma3*(xi[,3]))+(omega12*(xi[,1]*xi[,2]))+(omega13*(xi[,1]*xi[,3]))+(omega23*(xi[,2]*xi[,3]))+(zeta)

f<-cbind(eta,xi)

#Computing observable/indicator variables

#Variance-covariance matrix of error term in measurement model
cov_delta_epsilon<-diag(0.25,9)

#Measurement error
set.seed(281)
error_measurement_model<-mvrnorm(n,mu=c(rep(0,9)),Sigma=cov_delta_epsilon,empirical=FALSE)

#Matrix of observable variables
observable<-matrix(0,ncol=9,nrow=n)
observable[,1]<-f[,1]+error_measurement_model[,1]
observable[,2]<-f[,1]+error_measurement_model[,2]
observable[,3]<-f[,1]+error_measurement_model[,3]
observable[,4]<-f[,2]+error_measurement_model[,4]
observable[,5]<-f[,2]+error_measurement_model[,5]
observable[,6]<-f[,3]+error_measurement_model[,6]
observable[,7]<-f[,3]+error_measurement_model[,7]
observable[,8]<-f[,4]+error_measurement_model[,8]
observable[,9]<-f[,4]+error_measurement_model[,9]

as.data.frame(observable)
colnames(observable) <- c("y1","y2","y3","X1","X2","X3","X4","X5","X6")

#Estimation of Nuisance Parameters of measurment model using MLE
model<-'eta1=~y1+y2+y3
xi_1=~X1+X2
xi_2=~X3+X4
xi_3=~X5+X6'

cfa_model<-cfa(model,data=observable,estimator="ML")

#Extraction of nuisance parameters (measurement model parameters)

#matrix of estimated variances of error terms
psi_hat<-lavInspect(cfa_model,"est")$theta #matrix of estimated loadings lambda_hat<-lavInspect(cfa_model,"est")$lambda

#Computation of Bartlett factor scores
f_hat<-matrix(0,nrow=n,ncol=ncol(f))

#The following gives the matrix of estimated scores
#where the ith row is the score of the ith individual using equation 19.
for (i in 1:n){
f_hat[i,]<-solve(t(lambda_hat)%*%solve(psi_hat)%*%lambda_hat)%*%t(lambda_hat)%*%solve(psi_hat)%*%observable[i,]
}
f_hat<-as.data.frame(f_hat)

#Extract xi scores
xi_scores<-as.matrix(f_hat[,2:4])

#Computation of sigma_r_xi_hat (equation 21)
sigma_r_hat<-solve(t(lambda_hat)%*%solve(psi_hat)%*%lambda_hat)

#Extracting elements from sigma_r_hat which correspond to xi
sigma_r_xi_hat<-cbind(sigma_r_hat[2:4,2],sigma_r_hat[2:4,3],sigma_r_hat[2:4,4])

#Estimation of mixture parameters of xi_hat

#First check what type of mixture fits the xi scores best
mixture<-Mclust(xi_scores)

sigma_initial<-list(diag(1,3),diag(1,3),diag(1,3),diag(1,3))
mu_initial<-list(c(rep(0,3)),c(rep(0,3)),c(rep(0,3)),c(rep(0,3)))

#Fit mixture distribution to obtain pi_hat, mu_hat and sigma_hat_xi_hat
mixture_model<-mvnormalmixEM(xi_scores,k=4,mu=mu_initial,sigma=sigma_initial,arbmean=TRUE,arbvar=FALSE)
pi_hat<-mixture_model$lambda mu_1_hat<-mixture_model$mu[[1]]
mu_2_hat<-mixture_model$mu[[2]] mu_3_hat<-mixture_model$mu[[3]]
mu_4_hat<-mixture_model$mu[[4]] mu_hat_matrix<-rbind(mu_1_hat,mu_2_hat,mu_3_hat,mu_4_hat) sigma_hat_xi_hat<-mixture_model$sigma

#Obtain sigma_hat_xi through subtraction
sigma_hat_xi<-sigma_hat_xi_hat-sigma_r_xi_hat

#Setting initial parameter estimates for MCEM

#Initial parameter estimates to compute Monte-Carlo etas
set.seed(11)
gamma1_initial<-runif(1)
set.seed(12)
gamma2_initial<-runif(1)
set.seed(13)
gamma3_initial<-runif(1)
set.seed(14)
omega12_initial<-runif(1)
set.seed(15)
omega13_initial<-runif(1)
set.seed(16)
omega23_initial<-runif(1)

#Initial estimate of variance of zeta
set.seed(211)
est_var_zeta<-runif(1)

#Setting initial parameter vectors at iteration k and k+1
theta_k<-c(gamma1_initial,gamma2_initial,gamma3_initial,omega12_initial,omega13_initial,omega23_initial,sqrt(est_var_zeta))
theta_k_plus_1<-rep(0,length(theta_k))

diff_iter<-10

#xi MC samples

#Repeating variance-covariance matrix of xi_hat
s_mc<-array(0,dim=c(3,3,4))
for (i in 1:4){
s_mc[,,i]<-sigma_hat_xi
}

mc_samples_xi<-array(0,c(n,3,s))
for (j in 1:s){
mc_samples_xi[,,j]<-simdataset(n, Pi=pi_hat, Mu=mu_hat_matrix, S=s_mc)$X } rm(j) #Combine all xis into one matrix xi_reg<-mc_samples_xi[,,1] for (i in 2:s){ xi_reg<-rbind(xi_reg,mc_samples_xi[,,i]) } rm(i) #Define model for weighted least squares xi_one<-xi_reg[,1] xi_two<-xi_reg[,2] xi_three<-xi_reg[,3] denominator_3<-array(0,dim=c(n,s)) #Repeating variance-covariance matrix of mixture distribution as function requires a matrix for each mixture distribution sigma_hat_xi_list<-list(sigma_hat_xi,sigma_hat_xi,sigma_hat_xi,sigma_hat_xi_hat) #f(xi) from multivariate mixture of normal distributions for (i in 1:n){ for (m in 1:s){ denominator_3[i,m]<-dMVNmixture(x=mc_samples_xi[i,,m], weight=pi_hat, mean=mu_hat_matrix,Sigma=sigma_hat_xi_list, log=FALSE) } } rm(i) rm(m) ######################################################################################################################################################################################################################################################################################################################################## #MCEM Algorithm while (diff_iter>0.001){ mc_samples_zeta<-array(0,dim=c(n,1,s)) mc_samples_eta<-array(0,dim=c(n,1,s)) ##Generate Monte-Carlo samples of xis #xi monte-carlo samples were here #Generate Monte-Carlo samples of xi #MC samples for zeta for (m in 1:s){ mc_samples_zeta[,,m]<-rnorm(n,0,theta_k[7]) } rm(m) #MC samples for eta for (m in 1:s){ mc_samples_eta[,,m]<-(theta_k[1]*(mc_samples_xi[,1,m]))+(theta_k[2]*(mc_samples_xi[,2,m]))+(theta_k[3]*(mc_samples_xi[,3,m]))+(theta_k[4]*(mc_samples_xi[,1,m]*mc_samples_xi[,2,m]))+(theta_k[5]*(mc_samples_xi[,1,m]*mc_samples_xi[,3,m]))+(theta_k[6]*(mc_samples_xi[,2,m]*mc_samples_xi[,3,m]))+(mc_samples_zeta[,1,m]) } rm(m) f_i_hat<-array(0,dim=c(n,4,s)) for (m in 1:s){ f_i_hat[,,m]<-cbind(mc_samples_eta[,,m],mc_samples_xi[,,m]) } #E-Step: Computation of weights #Computation of numerator of weights from multivariate normal distribution numerator<-array(0,c(n,s)) for (i in 1:n){ for (m in 1:s){ numerator[i,m]<-dmvnorm(observable[i,], mean = lambda_hat%*%f_i_hat[i,,m], sigma = psi_hat, log = FALSE) } } rm(i) rm(m) #Computation of denominator of weights mc_samples_eta_hat<-array(0,dim=c(n,1,s)) for (m in 1:s){ mc_samples_eta_hat[,,m]<-mc_samples_eta[,,m]-mc_samples_zeta[,,m] } rm(m) denominator_1<-numerator denominator_2<-array(0,dim=c(n,s)) #f(eta|xi) from univariate normal distribution for (i in 1:n){ for (m in 1:s){ denominator_2[i,m]<-dnorm(mc_samples_eta[i,,m],mean=mc_samples_eta_hat[i,,m],sd=theta_k[7]) } } rm(i) rm(m) #Denominator 3 here #Final denominator as the product of the three terms making it up denominator<-denominator_1*denominator_2*denominator_3 denominator<-apply(denominator,1,sum)/s #Final weights W_i_m<-numerator/denominator #Perform weighted Least Squares #Combine all etas into one vector/matrix eta_reg<-mc_samples_eta[,,1] for (i in 2:s){ eta_reg<-c(eta_reg,mc_samples_eta[,,i]) } rm(i) #Combine all weights into one vector reg_weight<-W_i_m[,1] for (i in 2:s){ reg_weight<-c(reg_weight,W_i_m[,i]) } rm(i) reg_weight<-reg_weight/s reg_weight[which(is.nan(reg_weight))] = NA reg_weight[which(is.infinite(reg_weight))] = NA #Perform weighted least squares wls<-lm(eta_reg~0+xi_one+xi_two+xi_three+xi_one:xi_two+xi_one:xi_three+xi_two:xi_three,na.action=na.exclude) wls$coefficients

theta_k_plus_1<-c(wls$coefficients,summary(wls)$sigma)
print(theta_k_plus_1)

diff_iter<-as.numeric(sqrt(sum((theta_k-theta_k_plus_1)^2)))
print(diff_iter)
theta_k<-theta_k_plus_1

rm(mc_samples_zeta)
rm(mc_samples_eta)
rm(numerator)
rm(denominator)
rm(denominator_1)
rm(denominator_2)
#rm(denominator_3)
rm(W_i_m)
rm(reg_weight)
rm(eta_reg)
rm(f_i_hat)
rm(mc_samples_eta_hat)
}

• Can you provide a link or a better reference to the papers you cite? – Jeremy Miles Aug 19 '17 at 18:51
• I have provided a link to the full article. Thanks – Linniker Aug 21 '17 at 15:14
• Any feedback would be highly appreciated – Linniker Aug 21 '17 at 16:13


text:
## LaTeX forum ⇒ Page Layout ⇒ Multiple pagestyles in one document Topic is solved

Information and discussion about page layout specific issues (e.g. header and footer lines, page formats, page numbers).
ispacewalker
Posts: 2
Joined: Sat Sep 04, 2021 9:43 am

### Multiple pagestyles in one document

When multiple \pagestyle in one document, the page styles become quite strange. How can I make the first work until the second, the 2nd until the 3rd, and so on? Please run the following code to view the effect.

\documentclass{report}

\begin{document}

\newcommand*{\name}{plain}
\pagestyle{\name}
\chapter{\name}
Some stuff.
\newpage
{\Huge \name}

\pagestyle{\name}
\chapter{\name}
Some stuff.
\newpage
{\Huge \name}

\renewcommand*{\name}{empty}
\pagestyle{\name}
\chapter{\name}
Some stuff.
\newpage
{\Huge \name}
\end{document}

Bartman
Posts: 240
Joined: Fri Jan 03, 2020 2:39 pm
In this one-sided document, insert the \clearpage command before changing the page style, so that the changed style will only take effect from the new chapter.

Since the page style of the first page of each chapter is set to plain, the style selected by pagestyle is not set until the second chapter page.

A suggestion:

\documentclass{report}
\usepackage{etoolbox}

\newcommand*{\name}{plain}
% If the selected style should also apply to the first chapter page.
%\patchcmd{\chapter}{plain}{\name}{}{}
% or
%\patchcmd{\chapter}{\thispagestyle{plain}}{}{}{}

\begin{document}
\pagestyle{\name}
\chapter{\name}
Some stuff.
\newpage
{\Huge \name}

\pagestyle{\name}
\chapter{\name}
Some stuff.
\newpage
{\Huge \name}

\renewcommand*{\name}{empty}
\pagestyle{\name}
\chapter{\name}
Some stuff.
\newpage
{\Huge \name}
\end{document}

If I am right with my assumption concerning the first chapter page, it should work even much shorter:

\documentclass{report}
\usepackage{etoolbox}

\newcommand*{\name}{plain}
% If the selected style should also apply to the first chapter page.
\patchcmd{\chapter}{\thispagestyle{plain}}{\pagestyle{\name}}{}{}

\begin{document}
\pagestyle{\name}
\chapter{\name}
Some stuff.
\newpage
{\Huge \name}

\chapter{\name}
Some stuff.
\newpage
{\Huge \name}

\renewcommand*{\name}{empty}
\chapter{\name}
Some stuff.
\newpage
{\Huge \name}
\end{document}

The source code I replaced can be found in the report.cls file.

ispacewalker
Posts: 2
Joined: Sat Sep 04, 2021 9:43 am
Thank you very much! This problem had stressed me out for days.


text:
# An example of finite, connected topological group

A finite Hausdorff topological group, has discrete topology and every discrete group is totally disconnected. I look for an example of a non-Abelian, finite, connected non-Hausdorff group . I think every non-Abelian, finite, connected groups must be discrete.

• A group with indiscrete topology. – Crostul Dec 6 '14 at 11:31
• Every finite Hausdorff space is discrete. A connected finite topological group is indiscrete. – Daniel Fischer Dec 6 '14 at 11:31
• Each $T_0$ topological group is Tychonoff. – Alex Ravsky Dec 8 '14 at 5:01

$G=\{0,1,2,3\}$ , $\tau=\{\emptyset, G, \{0,2\},\{1,3\}\}$. Then $(G,+_4,\tau)$ is disconnect toplogical group.
Every topological group is completely regular, as the topology is uniformizable. Now every countable connected completely regular space $X$ must be indiscrete: If $x$ is a point outside of the closed set $A$, then some map $f:X\to I$ sends $A$ to $0$ and $x$ to $1$, so by connectedness $f(X)=I$, implying that $X$ is uncountable. Hence any finite connected topological group is indiscrete.
Now just take any non-Abelian group, like $D_3$, and equip it with the indiscrete topology.


text:
# Probability of Obtaining A Particular Sum from Successive Dice Rolls

Suppose you have a regular die with 6 faces numbered 1 through 6, respectively, and roll the die 4 times. What is the probability that the sum of the 4 rolls is 14?

This problem is equivalent to asking the number of ways to partition the number 14 into 4 subsets using 6 different numbers. Hence we can consider 'multichoosing' 4 numbers with repetition out of 6 different numbers such that $14 = x_1 + x_2 + x_3 + x_4$. I suspect that this can be done using either using multinomial coefficients in some way or multisets but I am not entirely sure. One issue is that this does not give a way for removing those cases where the four number chosen $\ne$ 14.

The total number of outcomes from rolling a die 4 times is $6^4$, so our probability will be some number divided by this.

Is this likely a problem where I should just enumerate all possible options to find my numerator?

-
Have a look at the [this question][1] [1]: math.stackexchange.com/questions/68045/… – Abe Doe Apr 19 '13 at 17:09
Thank you, the generating function approach makes sense to me. However, I am curious as to how the 'stars-and-bars' approach works. I am familiar with the method but I do not understand how they applied it here. Moreover, are there any obvious modifications of this problem for which the generating function method would fail? – user73041 Apr 19 '13 at 17:22
In the linked question, Quixotic shows how to incorporate the constraint than none of the four can be greater than $6$. His approach would need some updating if you could have more than one part greater than $6$. But your case is the same-you can't have more than one part greater than 6 – Ross Millikan Apr 19 '13 at 17:29
If I am understanding his answer correctly, would the total in this case be $13\choose3$? – user73041 Apr 19 '13 at 17:34
The sum 14 can be represented by 14 stars: **************. To break this into four summands, you put three bars between stars. There are 13 in-between spots, so there are $\binom{13}{3}$ ways to place the bars. The problem is that you might have more than six stars in a row without a bar between them. That's where the trick in Quixotic's answer comes in. To remove solutions where the first summand, $x_1,$ is greater than $6$, let $x'_1=x_1+6,$ so that $x'_1+x_2+x_3+x_4=8$ and use stars-and-bars again to count solutions. Solutions where $x'_1\ge1$ are solutions where $x_1\ge7$. So ... – Will Orrick Apr 21 '13 at 10:04


text:
Volume 282 - 38th International Conference on High Energy Physics (ICHEP2016) - Quark and Lepton Flavor Physics
Measurements of radiative $B$ meson decays at Belle
H. Kim* On behalf of the BELLE Collaboration
*corresponding author
Full text: pdf
Pre-published on: February 06, 2017
Published on: April 19, 2017
Abstract
The radiative $B$ meson decays $\overline{B} \to X_s \gamma$, and $X_d \gamma$ provide good probes to new physics as new unknown particles can enter in the loop and change the branching fractions and the associated $CP$ asymmetries. We present the analyses of radiative $B$ decays for branching fractions and the direct $CP$ asymmetries. In particular, we report the inclusive $CP$ asymmetry of the $b\to s\gamma/d\gamma$ decays, recently obtained with the semi-inclusive and the lepton tagging method by the \textit{Belle} collaboration. HQE parameters in shape-function scheme is obtained using the measured spectrum as well. We also present the results of the exclusive $\mathcal{B}(B^{0}\to \phi\gamma)$, $\mathcal{B}(B^{0}_{s}\to \phi\gamma)$, and $\mathcal{B}(B^{0}_{s}\to \gamma\gamma)$.
DOI: https://doi.org/10.22323/1.282.0586
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
Submit Essay

Once you submit your essay, you can no longer edit it.

# US Min Wage If Biden President

### Question

Joe Biden claims he will increase the federal minimum wage to $15/hr, a figure notably promoted by the Fight for$15 movement, up from its current value of $7.25/hr. The$15 minimum wage movement has seen some successes on the local level, with six states having laws that will phase in a $15 minimum wage. Such a move is backed by the rest of the Democratic Party: the Raise the Wage Act, which includes a$15 minimum wage, has 205 cosponsors in the House, all Democratic (and cleared the House in the 116th Congress).

If Biden is elected president in the 2020 election and takes office, will the federal minimum wage be at the end of 2024 in nominal US dollars per hour?

The minimum wage will be decided by the US Department of Labor website. This question is about the active minimum wage at the end of 2024, so if a bill is passed in 2024 that mandates a certain minimum wage in 2025, that will not be considered. If that website does not give information on the minimum wage (or lack thereof), this question will resolve based on credible media reports.

If Joe Biden is not elected or does not take office, this question resolves as ambiguous.

If there is no minimum wage, the minimum wage will be considered to be \$0.

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text:
National Curriculum
• Mathematical Methods Unit 1

### Description

This unit begins with a review of the basic algebraic concepts and techniques required for a successful introduction to the study of calculus. The basic trigonometric functions are then introduced. Simple relationships between variable quantities are reviewed, and these are used to introduce the key concepts of a function and its graph. The study of inferential statistics begins in this unit with a review of the fundamentals of probability and the introduction of the concepts of conditional probability and independence. Access to technology to support the computational aspects of these topics is assumed.

#### Learning Outcomes - By the end of this unit, students:

$\large\cdot$ understand the concepts and techniques in algebra, functions, graphs, trigonometric functions and probability
$\large\cdot$ solve problems using algebra, functions, graphs, trigonometric functions and probability
$\large\cdot$ apply reasoning skills in the context of algebra, functions, graphs, trigonometric functions and probability
$\large\cdot$ interpret and evaluate mathematical information and ascertain the reasonableness of solutions to problems
$\large\cdot$ communicate their arguments and strategies when solving problems.
• Mathematical Methods Unit 2

### Description

The algebra section of this unit focuses on exponentials and logarithms. Their graphs are examined and their applications in a wide range of settings are explored. Arithmetic and geometric sequences are introduced and their applications are studied. Rates and average rates of change are introduced, and this is followed by the key concept of the derivative as an ‘instantaneous rate of change’. These concepts are reinforced numerically, by calculating difference quotients both geometrically, as slopes of chords and tangents, and algebraically. Calculus is developed to study the derivatives of polynomial functions, with simple applications of the derivative to curve sketching, calculating slopes and equations of tangents, determining instantaneous velocities and solving optimisation problems.

Access to technology to support the computational aspects of these topics is assumed.

### Learning Outcomes

By the end of this unit, students:

$\large\cdot$ understand the concepts and techniques used in algebra, sequences and series, functions, graphs and calculus
$\large\cdot$ solve problems in algebra, sequences and series, functions, graphs and calculus
$\large\cdot$ apply reasoning skills in algebra, sequences and series, functions, graphs and calculus
$\large\cdot$ interpret and evaluate mathematical and statistical information and ascertain the reasonableness of solutions to problems
$\large\cdot$ communicate arguments and strategies when solving problems.
• Mathematical Methods Unit 3

### Description

In this unit the study of calculus continues with the derivatives of exponential and trigonometric functions and their applications, together with some differentiation techniques and applications to optimisation problems and graph sketching. It concludes with integration, both as a process that reverses differentiation and as a way of calculating areas. The fundamental theorem of calculus as a link between differentiation and integration is emphasised. In statistics, discrete random variables are introduced, together with their uses in modelling random processes involving chance and variation.

This supports the development of a framework for statistical inference.

Access to technology to support the computational aspects of these topics is assumed.

#### Learning Outcomes

By the end of this unit, students:

$\large\cdot$ understand the concepts and techniques in calculus, probability and statistics
$\large\cdot$ solve problems in calculus, probability and statistics
$\large\cdot$ apply reasoning skills in calculus, probability and statistics
$\large\cdot$ interpret and evaluate mathematical and statistical information and ascertain the reasonableness of solutions to problems.
$\large\cdot$ communicate their arguments and strategies when solving problems.
• Mathematical Methods Unit 4

### Description

The calculus in this unit deals with derivatives of logarithmic functions. In probability and statistics, continuous random variables and their applications are introduced and the normal distribution is used in a variety of contexts.

The study of statistical inference in this unit is the culmination of earlier work on probability and random variables. Statistical inference is one of the most important parts of statistics, in which the goal is to estimate an unknown parameter associated with a population using a sample of data drawn from that population. In Mathematical Methods statistical inference is restricted to estimating proportions in two-outcome populations.

Access to technology to support the computational aspects of these topics is assumed.

#### Learning Outcomes

By the end of this unit, students:

$\large\cdot$ understand the concepts and techniques in calculus, probabilty and statistics
$\large\cdot$ solve problems in calculus, probability and statistics
$\large\cdot$ apply reasoning skills in calculus, probability and statistics
$\large\cdot$ interpret and evaluate mathematical and statistical information and ascertain the reasonableness of solutions to problems.
$\large\cdot$ communicate their arguments and strategies when solving problems.
• At university i would like to study...
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Precalculus (10th Edition)

$f$ and $g$ are inverses of each other. The vaules $x\lt2$ are to be excluded.
$f(g(x))=f(\sqrt{ x}+2)=((\sqrt{ x}+2)-2)^2=(\sqrt{ x})^2=x$ $g(f(x))=g((x-2)^2)=\sqrt{(x-2)^2}+2=x-2+2=x$ Since for $f(x)$, $x\geq2$, then $x\lt2$ are values to be excluded.


text:
# rmgpy.cantherm.PressureDependenceJob¶

class rmgpy.cantherm.PressureDependenceJob(network, Tmin=None, Tmax=None, Tcount=0, Tlist=None, Pmin=None, Pmax=None, Pcount=0, Plist=None, maximumGrainSize=None, minimumGrainCount=0, method=None, interpolationModel=None, maximumAtoms=None, activeKRotor=True, activeJRotor=True, rmgmode=False)

A representation of a pressure dependence job. The attributes are:

Attribute Description
Tmin The minimum temperature at which to compute $$k(T,P)$$ values
Tmax The maximum temperature at which to compute $$k(T,P)$$ values
Tcount The number of temperatures at which to compute $$k(T,P)$$ values
Pmin The minimum pressure at which to compute $$k(T,P)$$ values
Pmax The maximum pressure at which to compute $$k(T,P)$$ values
Pcount The number of pressures at which to compute $$k(T,P)$$ values
Emin The minimum energy to use to compute $$k(T,P)$$ values
Emax The maximum energy to use to compute $$k(T,P)$$ values
maximumGrainSize The maximum energy grain size to use to compute $$k(T,P)$$ values
minimumGrainCount The minimum number of energy grains to use to compute $$k(T,P)$$ values
method The method to use to reduce the master equation to $$k(T,P)$$ values
interpolationModel The interpolation model to fit to the computed $$k(T,P)$$ values
maximumAtoms The maximum number of atoms to apply pressure dependence to (in RMG jobs)
activeKRotor A flag indicating whether to treat the K-rotor as active or adiabatic
activeJRotor A flag indicating whether to treat the J-rotor as active or adiabatic
rmgmode A flag that toggles “RMG mode”, described below
network The unimolecular reaction network
Tlist An array of temperatures at which to compute $$k(T,P)$$ values
Plist An array of pressures at which to compute $$k(T,P)$$ values
Elist An array of energies to use to compute $$k(T,P)$$ values

In RMG mode, several alterations to the k(T,P) algorithm are made both for speed and due to the nature of the approximations used:

• Densities of states are not computed for product channels
• Arbitrary rigid rotor moments of inertia are included in the active modes; these cancel in the ILT and equilibrium expressions
• k(E) for each path reaction is computed in the direction A -> products, where A is always an explored isomer; the high-P kinetics are reversed if necessary for this purpose
• Thermodynamic parameters are always used to compute the reverse k(E) from the forward k(E) for each path reaction

RMG mode should be turned off by default except in RMG jobs.

Plist

The pressures at which the k(T,P) values are computed.

Pmax

The maximum pressure at which the computed k(T,P) values are valid, or None if not defined.

Pmin

The minimum pressure at which the computed k(T,P) values are valid, or None if not defined.

Tlist

The temperatures at which the k(T,P) values are computed.

Tmax

The maximum temperature at which the computed k(T,P) values are valid, or None if not defined.

Tmin

The minimum temperature at which the computed k(T,P) values are valid, or None if not defined.

copy()

Return a copy of the pressure dependence job.

draw(outputDirectory, format='pdf')

Generate a PDF drawing of the pressure-dependent reaction network. This requires that Cairo and its Python wrapper be available; if not, the drawing is not generated.

You may also generate different formats of drawings, by changing format to one of the following: pdf, svg, png.

generatePressureList()

Returns an array of pressures based on the interpolation model, minimum and maximum pressures Pmin and Pmax in Pa, and the number of pressures Pcount. For Chebyshev polynomials a Gauss-Chebyshev distribution is used; for all others a linear distribution on an logarithmic pressure domain is used. Note that the Gauss-Chebyshev grid does not place Pmin and Pmax at the endpoints, yet the interpolation is still valid up to these values.

generateTemperatureList()

Returns an array of temperatures based on the interpolation model, minimum and maximum temperatures Tmin and Tmax in K, and the number of temperatures Tcount. For Chebyshev polynomials a Gauss-Chebyshev distribution is used; for all others a linear distribution on an inverse temperature domain is used. Note that the Gauss-Chebyshev grid does not place Tmin and Tmax at the endpoints, yet the interpolation is still valid up to these values.

maximumGrainSize

The maximum allowed energy grain size, or None if not defined.

saveInputFile(path)

Save a CanTherm input file for the pressure dependence job to path on disk.


text:
# Lets get insane

Algebra Level 4

$\begin{cases}x^4+2x^3-y=-\frac{1}{4}+\sqrt{3}\\ y^4+2y^3-x=-\frac{1}{4}-\sqrt{3} \end{cases}$

All the ordered pairs of real numbers that satisfy the system of equations above are $$(x_1,y_1),(x_2,y_2),...(x_n,y_n)$$. Find the value of $$x_1+x_2+...+x_n+y_1+...y_n$$ correct upto two decimal places.

• If you think there are infinite solutions then answer 777 and if you think no real solutions answer 666.
• Ordered pair means (11,12),(12,11) are considered different.
×


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# Cost Minimization

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Have you ever wondered how successful businesses get the right ratio of workers and machines? These businesses must be very precise when calculating how many people to hire and how many machines to rent so that their resources produce optimal outputs. This concept is called cost minimization. Let's keep reading to learn more about it and find out what it means to be cost-effective.

## The Cost Minimization Problem

Firms perform cost minimization in order to determine the most efficient way of producing goods and services from at least two inputs. For example, if a firm invests too much on its workers and not enough on its capital, or vice versa, the cost for both will not be efficient.

Cost minimization is the rule in which producers seek to calculate the right balance between two inputs in order to have the most cost-effective productivity.

Before we discuss cost minimization in depth, it is important to go over the concepts of substitutes and complements in a factor market. These are types of factors of production whose price affects the demand of similar or related factors of production, which will in turn, affect how to calculate cost minimization.

In general, substitutes are goods that can be replaced with similar goods. For example, a consumer might prefer buying e-books to physical books. However, if the price of a desired e-book rises, or is more expensive than the hard copy, then that consumer may buy the physical book instead. The same applies to the factors of production.

Substitutes in a factor market are factors of production that can be replaced with similar factors of production.

Two goods are complements if consumers tend to buy those two goods together. Think of the relationship between video games and a game console, such as a Playstation or Xbox. If there is an increase in the price of a game console, then the demand to buy video games will decrease. The same applies to the factors of production.

Complements in a factor market are factors of production that are jointly utilized with other factors of production.

### Cost Minimization and Profit Maximization

Reflecting on substitute and complementary factors of production can help the way we think about cost minimization and profit maximization. As you can imagine, there are different combinations of inputs a firm can calculate to minimize the cost of its capital and labor, in turn maximizing the firm's profits. Let's look at a specific example and see how it might be solved in the real world.

Think of a modern office warehouse with machine operators and automated forklifts that pick up products and move them from one place to another without a driver. In order to run most efficiently, the business needs to figure out how many machine operators (labor) and forklifts (capital) they require. If there are more forklifts than operators, there might be unused machines in the warehouse, not maximizing the total productivity. On the other hand, if there are more operators than forklifts, there may be slower movement of products across the warehouse floor. In both situations, the cost is not minimized.

## Cost Minimization Examples

Figuring out the best cost-effective approach for a firm can be challenging. However, when a firm needs to decide which combination of labor and capital inputs to use, calculating cost minimization may be relatively straightforward.

Let's refer to the example of a modern office warehouse with machine operators and automated forklifts.

Assume the following: if the firm with the office warehouse rents 40 forklifts, they will need to hire 1 machine operator to oversee 5 forklifts, requiring a total of 8 operators. In another scenario, more experienced operators may be more efficient by using other tools to help them move large quantities of product across the warehouse floor. The warehouse can move around the same amount of products with 20 automated forklifts and 20 machine operators.

Let's suppose the cost to rent a forklift is $2,000 per month and hiring a machine operator costs$1,500 per month. The cost of these inputs would look like the following:

 Scenario 1 Scenario 2 Cost of capital: 40 x $2,000 =$80,000 Cost of capital: 20 x $2,000 =$40,000 Cost of labor: 8 x $1,500 =$12,000 Cost of labor: 20 x $1,500 =$30,000 Total cost: $92,000 Total cost:$70,000

Table 1. Scenarios of cost minimization - StudySmarter

We see that the second scenario, where the warehouse hires 20 machine operators and rents 20 automated forklifts, is the more cost-effective combination. But these are only two scenarios. How do we know there isn't another better scenario that finds the most cost-effective combination of capital and labor? We use the cost minimization formula to check for the most cost-effective combination in such cases.

## Cost Minimization Formula

It can be difficult for firms with thousands of employees and even more units of capital to use the cost minimization approach among seemingly infinite scenarios. That is why there is a formula to find the optimal inputs of labor and capital. This formula illustrates the marginal product of labor over the wage rate to equal the marginal product of capital over the rental price of capital.

or

The marginal product of an input refers to the additional output that results from additional input.

Let's apply this formula to an example. Similar to the relationship between machine operators and automated forklifts, the relationship between bank tellers and ATMs (Automated Teller Machines) may be observed to find the most cost-effective combination between the two.

Suppose the marginal product of labor from bank tellers is 100 units, and the marginal product of capital from ATMs is 200 units. What happens when the wage is $10, and the rental rate of capital is$5?

We use the formula:

The output per dollar of labor would be 100/$10 = 10 units, while the output per dollar of capital would be 200/$5 = 40 units. In other words, the business receives 10 units of output that is spent on labor, and 40 units of output spent on capital.

According to these inputs, the bank should invest more in ATMs than in bank tellers. However, the law of diminishing returns reminds us that as the bank rents more ATMs, the marginal product of capital declines. Additionally, as the bank hires less tellers, the marginal product of labor rises.

The bank therefore will change the amount of labor and capital until the marginal product per dollar spent on each input is the same.

## Cost Minimization Approach: The Two Conditions

When representing the cost-minimizing value on a graph, the value must uphold two conditions:

1. The value is on the y-isoquant
2. No other value on the y-isoquant is on a lower isocost line

These two conditions are essential in illustrating a value whose cost is minimized. This puts into perspective what it means to have the most advantageous combination of labor and capital.

Let's take what we learned above and put it into a graph to see how it translates.

In Figure 1, we see the graph drawn with respect to labor (L) and capital (K). The two dark blue lines are isocost lines, which represent all combinations of labor and capital that cost the same total amount. The curved line, labeled as the y-isoquant shows the optimal combination of labor and capital that produces the maximum output at minimum cost.

Fig 1. - The cost minimization problem

Now focus on point A, which touches the y-isoquant and the first isocost line. This point shows the value of cost minimization. If you were to go to point B, there would be a higher input of capital than labor. Point C similarly illustrates a scenario where there is too much of one input. In this case, there is a higher input of labor and not enough capital for it to be considered a cost-effective alternative.

## Cost Minimization Analysis - Key takeaways

• Cost minimization is the rule in which producers seek to calculate the right balance between two inputs in order to have the most cost-effective productivity.
• Substitutes in a factor market are factors of production that can be replaced with similar factors of production. Complements in a factor market are factors of production that are jointly utilized with other factors of production.
• In the cost minimization formula, the marginal product of labor divided by the wage rate equals the marginal product of capital divided by the rental price of capital.
• Depending on the output per dollar of the marginal product of capital or labor, a firm would need to increase capital and decrease labor, or vice versa, until the marginal product per dollar on each input is equal.

Cost minimization is the rule in which producers seek to calculate the right balance between two inputs in order to have the most cost-effective productivity.

A supermarket wants to implement the right balance of self-checkout machines and cashiers. Using the cost minimization formula, one can find the optimal level of the marginal product of labor with respect to wage until it equals the marginal product of capital with respect to rental rate.

To determine cost minimization between two scenarios, add the products of the cost of capital and the cost of labor. For most cases, use the cost minimization formula to calculate the most cost-effective productivity.

(1) The value is on the y-isoquant, and (2) no other value on the y-isoquant is on a lower isocost line.

## Final Cost Minimization Quiz

Question

What is the definition of cost minimization?

The rule in which producers seek the most cost-effective balance between inputs of capital and labor.

Show question

Question

What is the law of diminishing returns?

It is an economic law stating that if one input increases, then at a certain point, the resulting output will start to increase at a lesser rate.

Show question

Question

What are the two conditions for cost minimization?

(1) The value is on the y-isoquant, and (2) no other value on the y-isoquant is on a lower isocost line.

Show question

Question

What does the term "returns to scale" mean?

It is the rate at which output changes if all inputs are changed by a proportionate factor.

Show question

Question

Is there only one approach to calculate cost minimization?

The cost minimization formula may be used to find the most cost-effective combination of inputs. However, if there are given scenarios, one also can find the cost-minimized scenario by comparing total cost of capital and labor.

Show question

Question

What is an isocost line?

It represents all combinations of labor and capital that cost the same total amount.

Show question

Question

What is the y-isoquant?

It shows the optimal combination of labor and capital that produces the maximum output at minimum cost.

Show question

Question

If output per dollar of capital is greater than output per dollar of labor, what would a firm have to do in order to reach cost minimization?

The firm would need to rent more capital and hire less labor until the decreasing marginal product of capital per dollar reaches the increasing marginal product of labor per dollar.

Show question

Question

If output per dollar of labor is greater than output per dollar of capital, what would a firm have to do in order to reach cost minimization?

The firm would need to hire more workers and rent less capital until the decreasing marginal product of labor per dollar reaches the increasing marginal product of capital per dollar.

Show question

Question

Which comparison of labor and capital might use the cost minimization approach?

All of these comparisons would use the cost minimization approach.

Show question

Question

What is a substitute?

It is a good that can be replaced with similar goods. For example, if someone doesn't have access to a car, then that person may use a bus or a bicycle instead.

Show question

Question

What is a complement?

It is a good that is typically bought with another good. For instance, hot dogs and hot dog buns are considered complements.

Show question

Question

What is an example of substitute goods?

tea and coffee

Show question

Question

What is an example of complementary goods?

a suit and a tie

Show question

Question

Which economic term is used to describe

the rule in which producers seek to calculate the right balance between two inputs in order to have the most cost-effective productivity.

cost minimization

Show question

Question

Highlight the incorrect portion of the text:

Complements in a factor market are factors of production that can be replaced with similar factors of production.

Complements

Show question

Question

Fill the blanks in the following statement

... are goods that can be replaced with similar goods. For example, a consumer might prefer buying e-books to physical books. However, if the price of the desired e-book ..., then that consumer may buy the physical book instead.

substitutes/rises

Show question

Question

Highlight the incorrect portion of the text:

Highlight the incorrect portion of the text:

Substitutes in a factor market are factors of production that are jointly utilized with other factors of production.

Substitutes

Show question

Question

Suppose the marginal product of labor from bank tellers is 100 units, and the marginal product of capital from ATMs is 200 units. Which of the following is not true if the wage is $10, and the rental rate of capital is$5?

The output per dollar of labor would be 100/$10 = 10 units Show question Question Think of the relationship between video games and a game console, such as a Playstation or Xbox. If there is an increase in the price of a game console, then the demand to buy video games will decrease. This is an example of ... Show answer Answer complements Show question Question Explain the cost minimization formula. Show answer Answer In the cost minimization formula, the marginal product of labor divided by the wage rate equals the marginal product of capital divided by the rental price of capital. Show question Question Depending on the output per dollar of the marginal product of capital or labor, a firm would need to increase capital and decrease labor, or vice versa, until ... Show answer Answer ... the marginal product per dollar on each input is equal. Show question Question ... in a factor market are factors of production that can be replaced with similar factors of production. ... in a factor market are factors of production that are jointly utilized with other factors of production. Show answer Answer Substitutes/Complements Show question Question When representing the cost-minimizing value on a graph, the value must uphold two conditions. What are they? Show answer Answer The value is on the y-isoquant 1. No other value on the y-isoquant is on a lower isocost line Show question Question Suppose the marginal product of labor from bank tellers is 100 units, and the marginal product of capital from ATMs is 200 units. What happens when the wage is$10, and the rental rate of capital is $5? Show answer Answer The output per dollar of labor would be 100/$10 = 10 units, while the output per dollar of capital would be 200/\$5 = 40 units. In other words, the business receives 10 units of output that is spent on labor, and 40 units of output spent on capital.

According to these inputs, the bank should invest more in ATMs than in bank tellers. However, the law of diminishing returns reminds us that as the bank rents more ATMs, the marginal product of capital declines. Additionally, as the bank hires less tellers, the marginal product of labor rises.

The bank therefore will change the amount of labor and capital until the marginal product per dollar spent on each input is the same.

Show question

Question

Give an example for substitute goods.

For example, a consumer might prefer buying e-books to physical books. However, if the price of the desired e-book rises, or is more expensive than the hard copy, then that consumer may buy the physical book instead.

Show question

Question

Give an example of complements.

Think of the relationship between video games and a game console, such as a Playstation or Xbox. If there is an increase in the price of a game console, then the demand to buy video games will decrease. The same applies to the factors of production.

Show question

Question

Firms perform ... in order to determine the most efficient way of producing goods and services from at least two inputs.

cost minimization

Show question

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A few years ago I wrote a post arguing that it is time to end ordered authorship. However that time has not yet arrived, and it appears that it is unlikely to arrive anytime soon. In the meantime, if one is writing a paper with 10 authors, a choice for authorship ordering and equal contribution designation must be made from among the almost 2 billion possibilities (1857945600 to be exact). No wonder authorship arguments are commonplace! The purpose of this short post is to explain the number 1857945600.

At first glance the enumeration of authorship orderings seems to be straightforward, namely that in a paper with n authors there are n! ways to order the authors. However this solution fails to account for designation of authors as “equal contributors”. For example, in the four author paper Structural origin of slow diffusion in protein folding, the first two authors contributed equally, and separately from that, so did the last two (as articulated via a designation of “co-corresponding” authorship). Another such example is the paper PRDM/Blimp1 downregulates expression of germinal center genes LMO2 and HGAL. Equal contribution designations can be more complex. In the recent preprint Connect-seq to superimpose molecular on anatomical neural circuit maps the first and second authors contributed equally, as did the third and fourth (though the equal contributions of the first and second authors was distinct from that of the third and fourth). Sometimes there are also more than two authors who contributed equally. In SeqVis: Visualization of compositional heterogeneity in large alignments of nucleotides the first eight authors contributed equally. A study on “equal contribution” designation in biomedical papers found that this type of designation is becoming increasingly common and can be associated with nearly every position in the byline.

To account for “equal contribution” groupings, I make the assumption that a set of authors who contributed equally must be consecutive in the authorship ordering. This assumption is certainly reasonable in the biological sciences given that there are two gradients of “contribution” (one from the front and one from the end of the authorship list), and that contributions for those in the end gradient are fundamentally distinct from those in the front. An authorship designation for a paper with n authors therefore consists of two separate parts: the n! ways to order the authors, and then the $2^{n-1}$ ways of designating groups of equal contribution for consecutive authors. The latter enumeration is simple: designation of equal authorship is in one-to-one correspondence with placement of dividers in the n-1 gaps between the authors in the authorship list. In the extreme case of placement of no dividers the corresponding designation is that all authors contributed equally. Similarly, the placement of dividers between all consecutive pairs of authors corresponds to all contributions being distinct. Thus, the total number of authorship orderings/designations is given by $n! \cdot 2^{n-1}$. These numbers also enumerate the number of ways to lace a shoe. Other examples of objects whose enumeration results in these numbers are given in the Online Encyclopedia of Integer Sequences entry for this sequence (A002866). The first twenty numbers are:

1, 4, 24, 192, 1920, 23040, 322560, 5160960, 92897280, 1857945600, 40874803200, 980995276800, 25505877196800, 714164561510400, 21424936845312000, 685597979049984000, 23310331287699456000, 839171926357180416000, 31888533201572855808000, 1275541328062914232320000.

In the case of a paper with 60 authors, the number of ways to order authors and designate equal contribution is much larger than the number of atoms in the universe. Good luck with your next consortium project!


text:
# How to refer to labels in other documents? [duplicate]

I have two documents file1.tex and file2.tex in two different folds fold1 and fold2, which are in the fold father.

In file1, I defined \label{file1:hello}, how do I use it in file2?

• I think the xr package would help. Feb 12 '14 at 20:51
• If you wish to cross-reference with the aid of hyperref, then see Does hyperref work between two files? Feb 12 '14 at 20:53
• @Werner how to set the relative path correct? Feb 12 '14 at 20:58
• @MaMing: You should read the linked post. It requires you to define (say) \externaldocument{../fold2/file2} in file1.tex. Then you would use the label as-is: \label{hello}. Otherwise, if you may have duplicate labels (hello in file1.tex and hello in file2.tex), then you should use \externaldocument[<prefix>]{...}. Feb 12 '14 at 21:01

I replace here the same answer given for this question: in this case solutions are for files in the same folder.

# Solution 1

You can use the xr package to reference to other LaTeX document.

In the file supplementmain.tex you say:

\usepackage{xr}
\externaldocument{supplement}
...
\ref{ext:supplement}

Note: normally xr doesn't add hyperlinks. Use zref package for this. You only need to replace \externaldocument with \zexternaldocument.

# Solution 2

A minimal working example taken from the following site: GUIT

First.tex

\documentclass[a4paper]{book}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}

\usepackage{xr-hyper}
\usepackage{hyperref}
\externaldocument[S-]{Second}[Second.pdf]
\begin{document}
\chapter{One}\label{chp:one}
\chapter{Two}\label{chp:two}
Bla Bla Bla.
\end{document}

Second.tex

\documentclass[a4paper]{book}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}

\usepackage{xr-hyper}
\usepackage{hyperref}
\externaldocument[F-]{First}[First.pdf]
\begin{document}
\chapter{Beginning}\label{chp:One}
Hello!
\chapter{Continuation}
I read chapters~\ref{F-chp:one}~and~\ref{F-chp:two} of the first file.

\end{document}

GUIT also says that in this way you have the hyperlink to the first file.

If files are in different folders, you can add paths into \externaldocument


text:
Current Page
Ellipsoid

# Ellipsoid

Object type: Surface

### Definition

In $\mathbb{R}^3$, an ellipsoid is a set of points $(x,y,z)$ satisfying $$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 + \left(\frac{z}{c}\right)^2 = 1$$ for some constants $a, b, c > 0$.

• If $a = b = c =: r$, we have a sphere of radius $r$.
• If two of the three constants are equal, and the third one differs from the two equal constants, then the surface is a (non-spherical) ellipsoid of revolution, that is, the surface obtained as a (non-circular) ellipse is rotated about one of its axes. If the exceptional constant is smaller than the two equal constants, the ellipsoid is oblate; otherwise, it is prolate.
• If all three constants are distinct, we have a tri-axial ellipsoid.

Below, a prolate ellipsoid of revolution with $a = c = 3, b = 6$ is shown.

### Parameterisation

The ellipsoid is the image $\mathbf{r}(D)$ where $$\mathbf{r}(\theta,\varphi) = \underline{ \mathbf{e}}\begin{pmatrix}a\sin\theta\cos\varphi\\b\sin\theta\sin\varphi\\c\cos\theta\end {pmatrix}$$ and $D := \left[0,\pi\right]\times\left[0,2\pi\right[$.

### Properties

The ellipsoid is a subset of $\left[-a, a\right]\times\left[-b, b\right]\times\left[-c, c\right]$. The properties below are with respect to the parameterisation given above.

#### Parameter-curve tangent vectors

The parameter-curve tangent vectors are $$\mathbf{r}_\theta(\theta,\varphi) = \underline{ \mathbf{e}}\begin{pmatrix}a\cos\theta\cos\varphi\\b\cos\theta\sin\varphi\\-c\sin\theta\end {pmatrix}, \quad\quad \mathbf{r}_\varphi(\theta,\varphi) = \underline{\mathbf{e}}\begin{pmatrix} -a\sin\theta\sin\varphi\\b\sin\theta\cos\varphi\\0\end{pmatrix}.$$

#### Standard unit normal

The standard unit normal vector field is $$\mathbf{\hat{N}}(\theta,\varphi) = \frac{1}{\sqrt{ b^2 c^2 \sin^2 \theta\cos^2 \varphi+a^2 c^2 \sin^2\theta\sin^2 \varphi+a^2 b^2 \cos^2 \theta}} \underline{\mathbf{e}}\begin{pmatrix}bc\sin\theta\cos\varphi\\ac\sin\theta\sin\varphi\\ab\cos\theta \end{pmatrix}.$$

#### Area element

The area element is $$dA = \sin\theta~\sqrt{b^2c^2\sin^2\theta\cos^2\varphi+a^2c^2\sin^2\theta \sin^2\varphi+a^2b^2\cos^2\theta}~d\theta d\varphi.$$

#### First fundamental form

The first fundamental form of the ellipsoid is $$\mathcal{F}(\theta,\varphi) = \begin{pmatrix} a^2\cos^2\theta\cos^2\varphi+b^2\cos^2\theta\sin^2\varphi+c^2\sin^2\theta&&-a^2\sin\theta\cos\theta \sin\varphi\cos\varphi+b^2\sin\theta\cos\theta\sin\varphi\cos\varphi\\-a^2\sin\theta\cos\theta\sin \varphi\cos\varphi+b^2\sin\theta\cos\theta\sin\varphi\cos\varphi&&a^2\sin^2\theta\sin^2\varphi+b^2\sin^2 \theta\cos^2\varphi\end{pmatrix}.$$

#### Second fundamental form

The second fundamental form is $$\mathcal{M}(\theta,\varphi) = \frac{-abc}{\sqrt{b^2c^2\sin^2\theta \cos^2\varphi+a^2c^2\sin^2\theta\sin^2\varphi+a^2b^2\cos^2\theta}}\begin{pmatrix}1&&0\\0&&\sin^2\theta \end{pmatrix}.$$