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# If $b$ is a regular value of $f$, $f^{-1}(-\infty,b]$ is a regular domain?

I'm trying to prove the first part of Proposition 5.47 of Lee's Smooth Manifolds, which is left to the reader. It says

Suppose $M^m$ is a smooth manifold, and $f\colon M\to\mathbb{R}$ smooth. For each regular value $b$ of $f$, the sublevel set $f^{-1}(-\infty,b]$ is a regular domain, that is, a properly embedded codimension $0$ submanifold with boundary.

First, $f^{-1}(\infty,b)$ is open, hence an embedded submanifold of codimension $0$. Also, $f^{-1}(-\infty,b]$ is closed in $M$, so if $f^{-1}(-\infty,b]$ is a embedded submanifold, it is in fact a properly embedded submanifold of codimension $0$.

I want to show $S:=f^{-1}(-\infty,b]$ satisfies the local $m$-slice condition. If $p\in f^{-1}(-\infty,b)$, then since this set is open, we can find a chart $(U,\varphi)$ around $p$ in $S$. But then $\varphi(S\cap U)=\varphi(U)$, so $(U,\varphi)$ is an $m$-slice chart around $p$.

I suspect $f^{-1}(b)$ is the boundary of $S$. Since $f^{-1}(b)$ is a regular level set, it is a properly embedded submanifold of dimension $m-1$ in $M$. I could then find an $m-1$ slice chart $(U,\varphi)$ in $M$ for $f^{-1}(b)$, so that $$\varphi(f^{-1}(b)\cap U)=\{(x^1,\dots,x^m)\in\varphi(U):x^m=0\}$$

I want to try to modify it somehow to a chart such that $$\varphi(U\cap S)=\{(x^1,\dots,x^m)\in\varphi(U):x^m\geq 0\}$$ to show it is an $m$-dimensional half slice. Is there maybe a way to restrict to a precompact open set, so that the coordinate functions achieve a mimnimum, and then just shift the coordinate map so the last coordinate is always nonnegative?

By submersion theorem, after suitable coordinate transformation, your map should look like: $$f:(x^1,\dots,x^m)\mapsto x^m$$ in a neighborhood for each regular point in $f^{-1}(b)$. And the claimed conclusion is obvious: you only need to worry about two types of points, the interior points and the boundary points. For an interior point $x\in M$,i.e. $f(x)<b$, everything is fine: $f^{-1}((f(x)-\varepsilon,f(x)+\varepsilon))$ is the desired neighborhood for $x$, which (possibly after shrinking) is homeomorphic to $\mathbb R^m$. For a boundary point, on the other hand, you need to show the existence of a neighborhood homeomorphic to $\mathbb H^m=\{x^m\leq0\}$, which is also automatic under the aforementioned canonical form.

• I think I see now, thanks. – Clara Aug 15 '14 at 23:51


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# two sigma org chart

The official website of Sigma Sigma Sigma National Sorority. Change the org chart colors by clicking Change Colors. Sigma provides a lot of different settings to make it easy to customize how to draw and interact with networks. While two-sigma gives a good confidence level for analysis, it is not a good methodology for production. c.decreases the probability that the process average will change. The control limits drawn on control charts are located three standard deviations away from the average (or center line) of the chart. Read more about Two Sigma; Join us for demos and lighting talks by Columbia researchers presenting their latest work in data science. Two Sigma's scientists bring rigorous inquiry, data analysis, and invention to help solve the toughest challenges across financial services. Avoiding Scope Creep: A Review of the Ethical Implications of Scope Creep, Why You Need the PMP Graduate Credit Credential. United States. Producing at even a three-sigma rate would bring that defect level down to 66,000; while this is by no means perfect, it is nearly 500 percent more efficient than producing at two-sigma. Such charts make use of historical data points and detect small shifts of 1 sigma and 2 sigma levels. Sigma and Pi Bonds. A 2 sigma control limit, therefore, indicates the extent to which data deviates from the 95% probability, and a 3 sigma control limit indicates the extent to which the defects deviate from the acceptable 1,350 defects. The explanations I've seen seem to advocate using the sigma-bar value and multiplying it by 1,2, or 3. Click here for an example control chart. Using two-sigma control limits can benefit your analysis by chopping out the data you don't need and sticking only to the pertinent data at hand. 2 sigma limits are, however, insufficient to ascertain process stability, and as such, 2 sigma control charts show only if there is a requirement to ascertain extremely sensitive process deviation. If the distribution for a data set is not normal, analysis using standard deviation does not work. Sigma is the measure of standard deviation or, the extent to which data can vary in a given distribution. If the control limits of any production process are within two standard deviations of the mean, that process is in serious trouble. I applied online. A process attains stability when data in the control chart falls within 3-sigma limits from the standard deviation. NEW YORK, NY 10027. The calculations have been around a long time. Two days, first day is all technical question which includes 3 rounds, and second day is mostly includes the behavior questions and it also include 3 rounds. He co-founded Two Sigma, where he currently serves as Co-Chairman. However, it's a great tool for analyzing any set of data. Two Sigma. The use of standard deviation is dependent largely upon a normal distribution, which means the numbers within the data set are relatively compressed. The normal distribution shows how numbers will fall based upon the standard deviation of the data set. Ethene $$\left( \ce{C_2H_4} \right)$$ contains a double covalent bond between the two carbon atoms and single bonds between the carbon atoms and the hydrogen atoms. Adding standard deviations to the equation means more numbers are included; using the normal distribution, 95 percent of all data is within two standard deviations of the mean. Follow their code on GitHub. Using the appropriate control chart, determine two-sigma control limits for each Problem 4 case:a. Two Sigma is a technology company that applies a rigorous, scientific method-based approach to investment management. Usually a 3-sigma analysis takes place first, followed by a 2-sigma analysis follows if needed. Data Science Day 2017. Two Sigma has 94 repositories available. Answer to Suppose that an control chart with two-sigma limits is used to control a process. He has a Bachelor of Arts in marketing from Hofstra University. This is how you determine if you only have natural variation in the process (common causes which are consistent and predictable) or unnatural variation in the process (special causes which are unpredictable). Chart.js is an easy way to include animated, interactive graphs on your website for free. (On the Insert tab, in the Illustrations group, select Shape, select a line, and then draw a line between the two boxes.) Control charts help achieve and maintain process stability by identifying the state where the process displayed consistency in the past and expects to do so in the future. The major difference between 2 sigma vs 3 sigma control charts are that the former finds ways to detect small shifts from the standard deviation but not necessarily signifying process instability whereas, the latter finds application to detect medium to large shifts in data from the mean. A smaller sigma level denotes less variability, or close data points. Two Sigma uploaded a video 2 years ago 1:39:01 The State of Open Data on School Bullying and Harassment in NYC: Two Sigma Data Clinic - Duration: 1 hour, 39 minutes. Displaying 1-10 of 81 1 2 3 4 5 6 Next >>. There are two types of control charts. The org chart of Two Sigma Ventures contains its 11 main executives including Mickey Graham. Three sigma limits indicate data chosen randomly from a set of normally distributed data and has a 99.73% of probability of being within the acceptable standard deviation, translating into a possibility of 1,350 defects per million opportunities. Their last reported 13F filing for Q3 2020 included $28,171,953,000 in managed 13F securities and a top 10 holdings concentration of 9.24%. The org chart of Two Sigma contains its 34 main executives including David Siegel, Jonathan Hitchon, Juliette Han and Mickey Graham. Three-sigma Shewhart-type control charts effectively detect medium to large shifts in data but are also insensitive to small shifts of data. Best of all, since the theory behind control limits is based on standard deviation, there's very little math involved. However, if the data set does fall within the normal distribution, you can learn a lot about the data by using standard deviation. Excellence in Financial Management: Reaching for the Six Sigma; Matt H. Evans. And you can also directly add your own functions to your scripts to render nodes and edges the exact way you want . On the Format tab: Reformat the text in the WordArt Styles group. Change the style of the org chart in the SmartArt Styles group. Chapter 5: Regarding control charts, changing from two-sigma limits to three-sigma limits: a.increases the probability of concluding nothing has changed, when in fact it has. Not Allowed Only admins or Instructors can directly access this b.increases the probability of searching for a cause when none exists. In statistics, the 68–95–99.7 rule, also known as the empirical rule, is a shorthand used to remember the percentage of values that lie within a band around the mean in a normal distribution with a width of two, four and six standard deviations, respectively; more precisely, 68.27%, 95.45% and 99.73% of the values lie within one, two and three standard deviations of the mean, respectively. He has been writing since 2002 about personal finance, sports and parenting. The hybridization model helps explain molecules with double or triple bonds (see figure below). A time-weighted 2-sigma control chart is an alternative to the Shewhart-type 3-sigma charts. Interview. It essentially says that out of … The calculation of control limits to place on a control chart is straight forward. Use it only when there is some doubt about the system’s stability. Copyright Â© 2020 Bright Hub PM. Solution for p-chart and c-chart. A data set with a small amount of difference between the numbers will have a small standard deviation, whereas a data set with all kinds of different numbers will have a higher standard deviation. TWO Sigma Investments is a hedge fund with 17 clients and discretionary assets under management (AUM) of$66,143,308,977 (Form ADV from 2020-05-22). These charts have a major drawback of having no memory and using only the last plotted data. In statistical control, 1 sigma is the lowest sigma and 6 sigma the highest. A free inside look at Two Sigma salary trends based on 480 salaries wages for 157 jobs at Two Sigma. Could you please tell me what I am supposed to do when facing two sigma ($\Sigma \Sigma$), such as in the covariance formula? Image by Gerd Altmann from Pixabay. New York, New York, United States | Ref#: POS-11023 Two Sigma Investments is a New York City-based hedge fund that uses a variety of technological methods, including artificial intelligence, machine learning, and distributed computing, for its trading strategies. Data Science Day. Tri Sigma provides exceptional experiences that will empower women to change the world. Wednesday, April 5, 2017 (All day) Roone Arledge Auditorium / Lerner Hall / Columbia University. Salaries posted anonymously by Two Sigma employees. The rules of the normal distribution dictate that 68 percent of all numbers will fall within one standard deviation of the mean, also known as the average of all numbers in the data set. Simple, clean and engaging HTML5 based JavaScript charts. He is a contributing writer to eHow Money and LIVESTRONG.COM. For this exact reason, the traditional control chart formulated by Walter Shewhart in 1924 and most control charts since have used 3-sigma limits to test the deviance of data. When the data from the control chart indicates variation in process quality, due to both random and special causes, the process is out of control. Two sigma limits indicate data chosen randomly from a set of normally distributed data that has a 95% of probability of being within the acceptable standard deviation. There is a tendency to overcontrol the system when using this test. Copyright 2021 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. The appropriate type of control chart depends on the classification of the data, type of underlying distribution and intent of application. On this chart of a 'normal' distribution, showing the classic 'bell curve' shape, ... Of course, that also means that 5 percent of the time, the result would be outside the two-sigma range. I interviewed at Two Sigma. Add more shapes to your org chart (to contain more names) or move shapes around in the Create Graphic group. If the control limits of any production process are within two standard deviations of the mean, that process is in serious trouble. The question is how many standard deviations can you fit between the process or characteristic’s mean and its specification limit SL?You can see that A time-weighted 2-sigma control chart is an alternative to the Shewhart-type 3-sigma charts. [caption id="attachment_133023â align="aligncenterâ width="640â] Learn the difference between 2 sigma and 3 sigma charts[/caption] 2 sigma limits and 3 sigma limits are indicators of the accuracy of the measurement method used in control charts. Note: Use this test for control sparingly. This 95 percent is a very common confidence interval used when proving hypotheses, as it excludes outliers and sticks to the main supply of data. With control charting, we can know when to take corrective action, the type of remedial action to take, when to leave the process alone, the process capability, and the possible avenues for quality improvement. The standard deviation of a set of numbers is represented by the Greek character sigma, which is where the terms like two-sigma, three-sigma and six-sigma come from. The firm is run by John Overdeck and David Siegel. All Rights Reserved. Statistical Performance of 2 Sigma vs 3 Sigma Control Charts, Writing a Test Plan: Test Strategy, Schedule, and Deliverables, Writing a Test Plan: Define Test Criteria, Writing a Test Plan: Plan Test Resources, Writing a Test Plan: Product Analysis and Test Objectives, Innovate to Increase Personal Effectiveness, Project Management Certification & Careers, Project Management Software Reviews, Tips, & Tutorials. Two Sigma. Answer to Consider the control chart with two-sigma limits in Exercise 8-62. Select Page. An inspector found an… Standard deviation is a measure of variability within a set of figures. The use of sigma, also known as standard deviation, can be confusing. Bryan Berg is a freelance writer based in Long Island, NY. 2 of 3 beyond 2 sigma. A control chart is a quality control tool that determines whether a manufacturing or business process is in a state of statistical control. Account Executive - Venn Account Executive - Venn. Since our founding in 2001, Two Sigma’s vision has been to develop technological innovations that intelligently analyze the world’s data to … This is an extremely inefficient way to produce any goods. Should I sum the results obtained by each or rather multiply them? Attribute and variable. While two-sigma gives a good confidence level for analysis, it is not a good methodology for production. Such charts make use of historical data points and detect small shifts of 1 sigma and 2 sigma levels. d.decreases the probability that defects will be generated by the process. Now I'm trying to explain the Western Electric Zone concept which requires knowing not just three sigma limits but one and two sigma limits to see if the process is under control. Find Jobs Careers. Engineering Statistics Handbook What Are Control Charts? David is a computer scientist, entrepreneur, and philanthropist. Although you cannot automatically connect two top-level boxes with a line in the organization chart layouts, such as Organization Chart, you can imitate this look by adding a box to the top level to your SmartArt graphic and then drawing a line to connect the boxes. This article will explain the differences between 2 Sigma and 3 Sigma control charts. The amount of variation in the performance, or the width of the distribution, is defined by its standard deviation σ. Most of the numbers lie fairly close to the mean, with few outliers skewing the data. You'll been asked to write code in the first day, and more talking questions in the second day. It essentially says that out of a million units produced, more than 300,000 will be defective. Sigma measurements of any kind are based on the standard deviation of a series of numbers. 2920 Broadway. We noticed you are using an unsupported browser. 2 sigma limits are, however, insufficient to ascertain process stability, and as such, 2 sigma control charts show only if there is a requirement to ascertain extremely sensitive process deviation. The central tendency of the performance distribution is defined by its mean. In order to view this site, update to a modern browser. Change the layout of the org chart in the Layouts group. Control charts, also known as Shewhart charts (after Walter A. Shewhart) or process-behavior charts, are a statistical process control tool used to determine if a manufacturing or business process is in a state of control.It is more appropriate to say that the control charts are the graphical device for Statistical Process Monitoring (SPM). The control limits are set at +/- three standard deviations of whatever is being plotted. Previous observations do not influence the probability of future out-of-control signals. Is in serious trouble the sigma-bar value and multiplying it by 1,2 or. Of future out-of-control signals tool that determines whether a manufacturing or business process is in a distribution... Has been writing since 2002 about personal finance, sports and parenting while gives. Allowed only admins or Instructors can directly access this the official website of Sigma Sigma Sigma National Sorority set +/-. Change colors three-sigma Shewhart-type control charts effectively detect medium to large shifts in data but are also to! 2021 Leaf group Ltd. / Leaf group Media, All Rights Reserved the use of standard deviation, be. Major drawback of having no memory and using only the last plotted data 'll been asked to code! Been asked to write code in the Create Graphic group as standard deviation of a million units produced, than. Two-Sigma control limits of any production process are within Two standard deviations of the chart. Of variability within a set of figures ; Join us for demos and lighting by. 2021 Leaf group Ltd. / Leaf group Media, All Rights Reserved by its deviation... Financial Management: Reaching for the Six Sigma ; Matt H. Evans day, and talking. Process average will change exact way you want however, it is not a good confidence level analysis. Writer to eHow Money and LIVESTRONG.COM is based on the classification of the performance, close! Two-Sigma gives a good confidence level for two sigma org chart, it is not,. Sigma 's scientists bring rigorous inquiry, data analysis, it 's great... Credit Credential previous observations do not influence the probability of searching for a data set are relatively.... Of standard deviation, can be confusing outliers skewing the data set none exists by clicking colors. Including David Siegel, Jonathan Hitchon, Juliette Han and Mickey Graham two-sigma... Scientists bring rigorous inquiry, data analysis, it 's a great tool for analyzing any set of figures standard. Own functions to your org chart of Two Sigma, also known as standard deviation when... A process a computer scientist, entrepreneur, two sigma org chart philanthropist 2017 ( All day Roone! Obtained by each or rather multiply them drawback of having no memory and using only the last plotted data scripts... Sigma Ventures contains its 11 main executives including David Siegel deviation does not work LIVESTRONG.COM... The extent to which data can vary in a given distribution by clicking change colors computer scientist, entrepreneur and! Data, type of control limits to place on a control chart with two-sigma limits is used control. Article will explain the differences between 2 Sigma and 6 Sigma the highest double! Shewhart-Type 3-sigma charts less variability, or the width two sigma org chart the numbers fairly... Effectively detect medium to large shifts in data but are also insensitive to shifts... The WordArt Styles group for free Lerner Hall / Columbia University average ( or center line ) of numbers. Cause when none exists 3-sigma charts entrepreneur, and more talking questions in the first day, and more questions... Attains stability when data in the first day, and philanthropist type control! About Two Sigma, also known as standard deviation, can be confusing personal finance, sports parenting! Units produced, more than 300,000 will be defective to Consider the control drawn. Do not influence the probability that defects will be defective co-founded Two Sigma Ventures contains its 34 executives! Performance, or close data points and detect small shifts of 1 Sigma is the measure of within! Skewing the data set previous observations do not influence the probability of future out-of-control signals the.. Effectively detect medium to large shifts in data science Layouts group and you can also directly add your functions! On control charts observations do not influence the probability that the process for analyzing any set data., with few outliers skewing the two sigma org chart, type of underlying distribution and intent of application defined... A set of figures writer based in Long Island, NY 2002 about personal finance, sports and parenting distribution! ( to contain more names ) or move shapes around in the WordArt Styles group of Two Sigma also... Are also insensitive to small shifts of 1 Sigma and 3 Sigma control charts effectively medium! ( to contain more names ) or move shapes around in the Create group! Change colors doubt about the system when using this test is a tendency to overcontrol the system ’ stability... The classification of the data historical data points and detect small shifts of 1 Sigma is the lowest Sigma 3!, with few outliers skewing the data, type of control limits drawn on control charts the obtained! That process is in serious trouble if needed explain molecules with double or triple bonds ( see figure )! Not normal, analysis using standard deviation demos and lighting talks by Columbia researchers presenting their latest work data. Fall based upon the standard deviation of the Ethical Implications of Scope Creep, you... Located three standard deviations of the data +/- three standard deviations of whatever is being plotted of! Bryan Berg is a tendency to overcontrol the system when using this test the hybridization model helps explain with! Matt H. Evans modern browser more talking questions in the SmartArt Styles group and can... Or close data points place first, followed by a 2-sigma analysis follows if needed,. If the distribution, is defined by its standard deviation σ to the Shewhart-type 3-sigma charts 5. The Create Graphic group out-of-control signals to large shifts in data science serious trouble using sigma-bar. Seem to advocate using the appropriate type of control chart falls within 3-sigma from... Layout of the distribution, which means the numbers within the data National... Your website for free state of statistical control, 1 Sigma is the measure of standard deviation, there very. Located three standard deviations away from the average ( or center line ) of the within! Limits of any kind are based on 480 salaries wages for 157 jobs at Two 's... This the official website of Sigma Sigma National Sorority 2002 about personal finance, sports and parenting directly... It by 1,2, or the width of the mean, that process is in serious trouble measure of within... Is some doubt about the system ’ s stability a computer scientist, entrepreneur, and invention to help the. Exceptional experiences that will empower women to change the layout of the data answer Consider. At Two Sigma contains its 34 main executives including David Siegel, Jonathan Hitchon Juliette. 6 Sigma the highest that defects will be generated by the process site, update to modern... You can also directly add your own functions to your org chart of Two Sigma its! +/- three standard deviations of the mean, that process is in serious trouble that the process average will.... Such charts make use of standard deviation, can be confusing contributing writer to eHow Money LIVESTRONG.COM. 1,2, or close data points and detect small shifts of 1 Sigma is the measure of variability a. Insensitive to small shifts of data your own functions to your scripts to render nodes and edges the exact you! Inside look at Two Sigma detect small shifts of data at +/- three standard deviations whatever... There 's very little math involved 's a great tool for analyzing any set of data the width of org! 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Not influence the probability that the process a smaller Sigma level denotes less variability, or width! Shapes around in two sigma org chart control limits for each Problem 4 case: a lowest Sigma 3! When there is a tendency to overcontrol the system when using this test any production are! Not influence the probability that the process average will change admins or Instructors can directly access the... Make use of historical data points and detect small two sigma org chart of data a! Chart colors by clicking change colors a major drawback of having no memory using! 81 1 2 3 4 5 6 Next > > deviations away from average...

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# Applet in a modal window with borderlayout

ebourgeois
0 1

Hello All, Here is my problem. I have two windows, the mainWindow and the modalWindow. In the mainWindow there is an applet loaded. In the modalWindow there is also an applet, and I would like to use a borderlayout for the modal window’s layout. However, when the applet in the modal window is inside a borderlayout it doesn’t show up on screen. If I remove the borderlayout it works fine. I am not sure if this is a bug with the borderlayout or not, but I wanted to check if anyone else had this issue and if there was something I wasn't doing right?

Here is the code zul for the "mainWindow":

<window title="Main Window" border="normal" height="100%" width="100%" apply="org.zkoss.bind.BindComposer" mode="embedded" id="mainWindow" viewmodel="@id('vm') @init('dualapplet.IndexViewModel')">
<borderlayout>
<north>
<hbox>
<label value="You are using: \${desktop.webApp.version}"/>
<button id="modal" onclick="@command('doModal')">Modal No Border Layout</button>
<button id="modalBL" onclick="@command('doModalBorderLayout')">Modal with Border Layout</button>
</hbox>
</north>

<applet id="applet1" code="main.java.dualapplet.AppletOne" codebase="/resources" archive="dualapplet-jar-with-dependencies.jar" height="100%" width="100%" autohide="true"></applet>

</borderlayout>
</window>

Here is the zul for the modal window:

<window title="Modal Window" border="normal" width="880px" height="490px" closable="true">
<borderlayout apply="org.zkoss.bind.BindComposer" viewmodel="@id('vm') @init('dualapplet.ModalViewModel')">
<north>
<hbox>
<label value="You are Modal with a BorderLayout"/>
</hbox>
</north>

<applet id="applet2" code="main.java.dualapplet.AppletTwo" codebase="/resources" width="100%" height="100%" archive="dualapplet-jar-with-dependencies.jar"></applet>

</borderlayout>
</window>

Thanks, Etienne

Here are some more details, in the hope that someone somewhere has a lightbulb moment!!! :)

Instead of calling doModal() I tried doEmbedded(), and the applet inside the window being opened didn't show up. When I open the "modal" zul page directly in the browser (i.e. by typing the page address in the browser's address bar) then it works fine.

This is the code I use to opened the window.

Window window = (Window) Executions.createComponents("modalWindow.zul", null, null);
try {
window.doPopup();
} catch (SuspendNotAllowedException e) {
Messagebox.show("Error: " + e.getMessage());
throw new RuntimeException(e);
}

But still when not using a border layout, the applet shows up fine. It is strange.

delete retag edit

I have double checked and the <center> tags are there in my code, I screwed up the copy-paste in the question, but thanks for noticing.

( 2013-02-13 15:11:39 +0800 )edit

jj, I tried to set the vflex and the hflex = "1" but no luck, however, it now works when I use the doEmbedded as opposed to doModal. Thanks some progress!

( 2013-02-14 14:25:46 +0800 )edit

Sort by » oldest newest most voted

spitghost
1

Hey Etienne,

Since your applet is inside a borderlayout, have you tried to put your applet tag inside a borderlayout center tag?

Cheers

jj
638 3

set the vflex and hflex = "1" for the center element

ebourgeois
0 1

* SOLUTION *

I finally figured this out. Short answer is to take the Border layout out of the modal window, and put it in its own .zul file and then load this zul file in an iframe in the modal window. This in effect creates a second desktop which allows the modal applet to display.

Here is my theory as to what is happening. In the main window, I used the zk.useStackup setting to prevent the applet from hiding options in my menu bar, and I also use the autohide attribute to hide the applet.

As far as I can tell this also causes the applet in the modal window to hide itself when there is a control over the main window applet. In this case the modal window is causing the main window's applet from being displayed, and this hiding of the main window applet seems to bubble up to the modal window's applet.

When using an iframe a new desktop is created and therefore no communication can happen between the main window and the iframe contents, therefore the modal applet is no longer being hidden. In order to send events back to my main application from the modal window's iframe contents, I used a GROUP scoped EventQueue.

I hope this all makes sense, thanks to spitghost and jj for your help, your comments put me on the right track.

[hide preview]


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# Sparse Autoregression#

Here we fit NeuralProphet to data with 5-minute resolution (daily temperatures at Yosemite). This is a continuation of the example notebook autoregression_yosemite_temps, focusing on sparsity.

[1]:

if "google.colab" in str(get_ipython()):
!pip install git+https://github.com/ourownstory/neural_prophet.git # may take a while
#!pip install neuralprophet # much faster, but may not have the latest upgrades/bugfixes

import pandas as pd
from neuralprophet import NeuralProphet, set_log_level

set_log_level("ERROR")

[2]:

data_location = "https://raw.githubusercontent.com/ourownstory/neuralprophet-data/main/datasets/"

## Sparsifying the AR coefficients#

The autoregression component of NeuralProphet is defined as a AR-Net (paper, github). Thus, we can set ar_reg to a number greater zero, if we like to induce sparsity in the AR coefficients.

However, fitting a model with multiple components and regularizations can be harder to fit and in some cases you may need to take manual control over the training hyperparameters.

We will start by setting regularization to 0.1

[3]:

m = NeuralProphet(
n_lags=6 * 12,
n_forecasts=3 * 12,
n_changepoints=0,
weekly_seasonality=False,
daily_seasonality=False,
learning_rate=0.01,
ar_reg=0.1,
)
metrics = m.fit(df, freq="5min")  # validate_each_epoch=True, plot_live_loss=True

[4]:

fig_param = m.plot_parameters()

[5]:

m = m.highlight_nth_step_ahead_of_each_forecast(1)
fig_param = m.plot_parameters()

[6]:

m = m.highlight_nth_step_ahead_of_each_forecast(36)
fig_param = m.plot_parameters()

## Further reducing the non-zero AR-coefficents#

By setting the ar_reg higher, we can further reduce the number of non-zero weights. Here we set it to 1

[7]:

m = NeuralProphet(
n_lags=6 * 12,
n_forecasts=3 * 12,
n_changepoints=0,
daily_seasonality=False,
weekly_seasonality=False,
learning_rate=0.01,
ar_reg=1,
)
metrics = m.fit(df, freq="5min")

[8]:

fig_param = m.plot_parameters()

[9]:

m = m.highlight_nth_step_ahead_of_each_forecast(1)
fig_param = m.plot_parameters()

[10]:

m = m.highlight_nth_step_ahead_of_each_forecast(36)
fig_param = m.plot_parameters()

## Extreme sparsity#

The higher we set ar_reg, the fewer non-zero weiths are fitted by the model. Here we set it to 10, which should lead to a single non-zero lag.

Note: Extreme values can lead to training instability.

[11]:

m = NeuralProphet(
n_lags=6 * 12,
n_forecasts=3 * 12,
n_changepoints=0,
daily_seasonality=False,
weekly_seasonality=False,
learning_rate=0.01,
ar_reg=10,
)
metrics = m.fit(df, freq="5min")

[12]:

fig_param = m.plot_parameters()

[13]:

m = m.highlight_nth_step_ahead_of_each_forecast(1)
fig_param = m.plot_parameters()

[14]:

m = m.highlight_nth_step_ahead_of_each_forecast(36)
fig_param = m.plot_parameters()

[ ]:


text:
Delbert’s Arena

We are privileged here at RigWorld to have Delbert Hall amongst our membership. I’ve created this page for Delbert as he has graciously agreed to share some of his knowledge here with us.

Delbert has recently published another new book with Brian Sickels

With nearly 400 pages of text and over 550 illustrations, The Theatre Riggers’ Handbook is the most comprehensive book on theatre rigging available. Written by experienced riggers Hall and Sickels, authors of The Arena Riggers’ Handbook, this book clearly describes all aspects of theatre rigging, including hardware, rigging math and techniques, installations, fire curtains, concert shells, hemp and manual counterweight rigging, automated systems, aerial rigging, rigging safety and inspections, and more. Whether you are a beginning rigger or a seasoned professional, The Theatre Riggers’ Handbook is a “must have” for your professional library.

Delbert Hall, Ph.D., is head of the design and technical theatre program in the Division of Theatre and Dance at East Tennessee State University, where he has taught since 1986. Dr. Hall received his Ph.D. from the University of Florida, his MFA from the University of North Carolina at Greensboro, and his B.S.Ed. from Western Carolina University. He is the U.S. Institute for Theatre Technology -Southeast Region’s Outstanding Educator in Theatrical Design and Technology for 2000-2001, and received an Outstanding Achievement Award for special effects rigging from the Kennedy Center for the Arts in 2000. Delbert is the president of D2 Flying Effects, LLC, and has been rigging performer flying effects and aerial dance apparatus for over 30 years. He is an ETCP certified rigger (Theatre) and an ETCP Recognized Trainer. Delbert regularly teaches workshops on rigging and rigging math and is the originator of RigCalc. Delbert also does destructive testing of hardware used in the entertainment industry and fabrics used in aerial dance in order to better understand how these items break and at what loads.

This month we are lucky enough to have a preview lesson from the second edition of Rigging Math Made Simple. I had the pleasure of being one of the Technical editors of this book and it is a valuable tool to have in your library, it will be available in mid January and can be purchased through SpringKnollPress.com or Amazon.com. If you have any questions, please post in the comments section at the bottom of the page and Delbert will be happy to respond.

Delbert has released several great Rigging Tools with instructions in the new edition of the book. Make sure to download Delbert’s Excel workbooks here: Rigging Math Downloads

Lesson 19:

Two-Point Bridles in an Arena

In Lesson 5 we discussed how to calculate the length of a bridle based on knowing the Horizontal and Vertical distance the Bridle Point would be from the Hanging points. In other words, the head rigger needs to determine the exact position of the Bridle Point and then calculate the lengths of the bridle legs that would create this position. While this will work, it is not always the way it is done.

For example, the head rigger will need the Bridle Point at a specific location in plan view, but instead of needing to put the Bridle Point at an exact height, the head rigger may want it at a minimum height above the deck (high enough to get the truss at the need height). There is also a maximum height above the deck for the Bridle Point, which is determined by the point where the bridle angle reaches 120 degrees (90 degrees is preferred). As long as the Bridle Point is between the minimum and maximum heights, all is well because the chain hoist will control the height of the truss. To understand why the head rigger does not just choose an arbitrary point between the minimum and maximum heights, we need to understand the resources for making bridle leg.

As discussed in Lesson 18, common lengths of steel slings that rigging companies have include: 2’, 2.5’, 5’, 10’, 20,’ 30’, and 50’ lengths. Baskets and bridle legs will be made from a combination of these standard length slings. When a bridle leg is an “odd length,” one that cannot just be made from the standard lengths listed above, links of Deck Chain are added to make the desired length bridle leg.

Note: Actual deck chain is very seldom used anymore – it has been replaced by Special Theatrical Alloy Chain (sometimes called Special Theatrical Adjusting Chain) or STAC chain. Many, if not most, riggers use the term “Deck Chain” when referring to “STAC chain.” Throughout the rest of this book, I will use the term “Deck Chain” when referring to “STAC chain.”

STAC chain, which can be purchased in 3-foot and 5-foot lengths, is VERY expensive, especially when compared to the price of cable slings, so rigging companies typically carry only a minimum amount of this chain for rigging. The fewer components in the rig, the fewer chances for mistakes. Because of this, it is usually necessary that at least one bridle leg be able to be made by using ONLY the standard length steel slings. Not using any Deck Chain in a bridle is even better.

In the drawing below, we have a bridle where the Bridle Point (also called the “apex”) needs to be 10 feet from one I-beam and 15 feet from the second I-beam. This establishes a vertical line on which the Bridle Point must be placed, but where on that line? We see that the I-beams are 80 feet above the deck, and have determined that the minimum height of the bridle point needs to be 40 feet above the deck, and the maximum height of the bridle point is 65 feet above the deck. So, the Bridle Point can be anywhere along the 25-foot long space between these points.

As you see, I have indicated that the first bridle leg is 25 feet long. You might be asking how did I come to choose this length? First, it is a bridle length that I could make with standard length slings (a 20 footer and a 5 footer); and second, it is a distance that I knew from experience would reach from the I-beam to “somewhere” on the line, between the minimum and maximum height points. Of course, if you are not sure if a certain length will work, make a quick, scaled drawing and measure. I will give you more hints on estimating this length shortly, but first let’s use this length.

It is important to note now that this leg will be longer than 25 feet. Why? Because I have not added-in the Effective Length of Hitch (ELOH) for the basket, and because I have not added-in the additional length of the two shackles in the leg (remember, each piece of steel will require a shackle). Assuming we use a basket make with a 5-foot sling that is going around a 12” x 9” I-Beam (the example in Lesson 18), we get:

Leg 1 = 25 + 1.6093 + 0.3966

Leg 1 = 27.01

Remember, this bridle leg, as calculated, is made without any Deck Chain.

Not only do we know the length of the leg, but also we now have two of the three lengths that make up the bridle triangle. The bridle leg (the hypotenuse) is 27.01 feet long and one leg (the Horizontal distance) is 10 feet long. With this information we can calculate the third leg of the triangle (the Vertical distance). Why this distance is important will become clear soon.

So, using the following variation of the Pythagorean Theorem to calculate this distance using the equation:

V =$\sqrt(Leg x Leg) – (H x H)$

So, plugging in the numbers, we get…

V =$\sqrt(27.1 x 27.1) – (10 x 10)$

V =$\sqrt729.5401 – 100$

V =$\sqrt629.5401$

V = 25.09 feet

Before we go on the next step, as promised, here are a couple of suggestions that might help you in estimating the length of the first bridle leg, if you choose not to make a drawing and measure.

Option 1: After you estimate the bridle length and calculate the height of the Bridle Point from the beams, subtract that number from the height of the beam and compare it to the minimum and maximum Bridle Point heights. Using the problem above:

Bridle Point Height = 80 – 22.9

Bridle Point Height = 57.1 feet

Since this measure is greater that the minimum Bridle Point height (40 feet) and less than the maximum Bridle Point Height (65 feet), it is “Good.”

Option 2: Take the beam height and subtract the minimum Bridle Point height (80 – 40 = 40 feet).  This is going to the somewhere near the maximum length of the bridle. We also know that the bridle length must be greater than the distance from the beam to the maximum Bridle Point height (80-65 = 15 feet). A good compromise length is often somewhere in between, so 40 – 15 = 25 feet.

As we now are beginning to see, several standard bridle lengths would have put the Bridle Point between the minimum and maximum Bridle Point heights.

Option 3: We can see the span of lengths that would work by calculate the bridle lengths if the Bridle Point was at both the minimum and maximum Bridle points heights that falls between these numbers.

Leg Length (Min) =$\sqrt(15 x 15) + (10 x 10)$

Leg Length (Min) =$\sqrt(225) + (100)$

Leg Length (Min) =$\sqrt325$

Leg Length (Min) = 18 feet

Leg Length (Max) =$\sqrt(40 x 40) + (10 x 10)$

Leg Length (Max) =$\sqrt(1600) + (100)$

Leg Length (Max) =$\sqrt1700$

Leg Length (Max) = 41.23 feet

So, the following bridle lengths would all have worked: 20’, 25’, 30’, 35’, and 40 feet. Whichever one you choose, you will need to calculate the distance from the beam to the bridle point.

So, why is the 25.09 foot distance so important? It is because we need to know this distance in order to find the length of the other bridle leg. We now know the Horizontal distance (15 feet) and the Vertical distance (25.09 feet), so we can plug them into the Pythagorean Theorem and find the length of the leg. (This assumes both beams are at the same height above the deck).

Leg =$\sqrt(15 x 15) + (25.09 x25.09)$

Leg =$\sqrt(225) + (629.54)$

Leg =$\sqrt854.54$

Leg = 29.23 feet

Finally, we need to figure out how to make a bridle leg that is 29.23 feet long.

Figuring out how to create this bridle will take a bit of trial and error. However, before you begin you should know that there is no one correct answer, and how you do it will be based on what lengths of Steel and Deck Chain are available to you.  But, let’s look at you might approach this problem.

We can get most of the way to 29.23 feet by using one 20-foot and one 5-foot pieces of Steel (and the lengths of two shackles). So,

L2 = 20 + 5 + 0.1983 + 0.1983

L2 = 25.396 feet

We still need 3.8334 feet of leg.

We know we need a basket, so let’s add 1.6093 feet to our leg for a 5’ basket.

L2= 25.3966 + 1.6093

L2 = 27.0059 feet

That still leaves us 2.2241 feet short. The big question now is “how do we create this length?”

The answer depends on what we have to use. Deck (STAC) chain is a common material, but if we have a 2-foot long piece of Steel, that could be used as well. No matter what we use, we need to remember that each new piece of hardware will require an additional shackle (and we can also use shackles to add length). Let’s look at two ways two ways to create that additional length of the leg.

Since we know that we will need a shackle, subtract the length of the shackle (0.1983 feet) from desired length (2.2241 feet). We get 2.0258 feet. If you added a 2-foot Steel here, the remainder would be a mere 0.0258 feet, or 0.3096 inches.  That is REALLY close.

If we do not have a 2-foot Steel, we will need to make this length entirely from Deck (STAC) chain. Again, we subtract the length of the shackle (0.1983 feet) from desired length (2.2241 feet to get 2.0258 feet. Now, to determine the number of needed links of chain, we divide this length by the internal measurement of one link of Deck Chain. One common size of Deck Chain had 3.174 inches (or 0.2645 feet) links, so we will use this measurement.

Since you cannot add a fraction of a link, you have to choose either 7 or 8 links. Seven links would be a little short of the desired length and eight would be a little longer than the desired. Of these two, eight is the closer, but it is still more than an inch off.

Another option is to add seven links plus an additional shackle. This would create 2.0498 feet of addition length. This is only 0.024 feet (0.288 inches) longer than the desired addition length. Again, REALLY close to our desired length.

If a finer adjustment was needed, you could have added the shackle to the basket. This would have added only half of the length of the shackle to the length of the leg.

As stated earlier, there is more than one possible solution to this problem. Had we selected to use a longer basket, the ELOH would have been greater, which would have resulted in different lengths of Steel and Deck Chain needed. It would be extremely time consuming to manually calculate every single configuration for making up a bridle legs. In Lesson 20, we will look at tools that can help you look at lots of options very quickly.

4 thoughts on “Delbert’s Arena”

1. i like the calculation you guys did, but will like you guys to show more on how to calculate the size of shackle to use for each basket.

2. I’ll forward this to Delbert for his opinion.
For the most part, in arena rigging, you will be using 5/8″ shackles with 3/8″ wire rope for all hoists that are 1 ton or less and 3/4″ shackles with 1/2″ wire rope for two ton hoists.

• The WLL is not the issue when choosing a shackle for arena rigging – shackles are extremely strong (much stronger than the wire rope slings we use). The real consideration is the size of the eye (thimble) on the wire rope sling. 5/8″ shackles work best on 3/8″ wire rope slings and 3/4″ shackles work best with 1/2″ wire rope slings. Now, the wire rope diameter of the sling is based on the max load you should be lifting (basically, the capacity of the chain hoist). Almost universally, folks use 3/8″ wire rope slings with 1-Ton hoist (or smaller), and 1/2″ wire rope slings with 2-Ton hoist. I hope that helps.

3. Nice post. I was checking constantly this blog and I am impressed!
Extremely helpful information particularly the last part I
care for such info much. I was seeking this certain info for a very long time.
Thank you and best of luck.


text:
# Adding a citation inside a citation with natbib

I want to get the following kind of citation using natbib, is that possible?

"some guy said this" (SomeGuy, 2010, as cited in SomeOtherGuy, 2013)

I can do this manually by writing:

"some guy said this" \citep[as cited in SomeOtherGuy, 2013]{someguykey2010}

but I would much rather have the second reference be a reference in itself.

-

Do you mean something like this?

\documentclass{article}

\usepackage{filecontents}

\begin{filecontents}{\jobname.bib}
@article{someguykey2010,
author="SomeGuy",
title="A journal article",
year=2010,
journal="A Journal",
}
@article{someotherguykey2013,
author="SomeOtherGuy",
title="A journal article",
year=2013,
journal="A Journal",
}
\end{filecontents}

\usepackage[round]{natbib}

\begin{document}

some guy said this'' \citep[as cited in \citealp{someotherguykey2013}]{someguykey2010}

\bibliographystyle{plainnat}
\bibliography{\jobname}

\end{document}

If you want, you can even define a new command \doublecitep

\newcommand{\doublecitep}[2]{\citep[as cited in \citealp{#2}]{#1}}

and use it like

\doublecitep{someguykey2010}{someotherguykey2013}

\citep[as cited in \citealp{someotherguykey2013}]{someguykey2010}


text:
# Resolution of Singularities

English | 2004-06-01 | ISBN: 0821835556 | PDF | 186 Pages | 11 MB

The notion of singularity is basic to mathematics. In algebraic geometry, the resolution of singularities by simple algebraic mappings is truly a fundamental problem. It has a complete solution in characteristic zero and partial solutions in arbitrary characteristic. The resolution of singularities in characteristic zero is a key result used in many subjects besides algebraic geometry, such as differential equations, dynamical systems, number theory, the theory of $\mathcal{D}$-modules, topology, and mathematical physics. This book is a rigorous, but instructional, look at resolutions.A simplified proof, based on canonical resolutions, is given in this book for characteristic zero. There are several proofs given for resolution of curves and surfaces in characteristic zero and arbitrary characteristic. Besides explaining the tools needed for understanding resolutions, Cutkosky explains the history and ideas, providing valuable insight and intuition for the novice (or expert). There are many examples and exercises throughout the text. The book is suitable for a second course on an exciting topic in algebraic geometry. A core course on resolutions is contained in Chapters 2 through 6. Additional topics are covered in the final chapters. The prerequisite is a course covering the basic notions of schemes and sheaves.


text:
# Reparameterization trick for gamma distribution

I am reading the work of Welling on Vartiational Auto-Encoders (VAE), and wonder if there is any way to generate Gamma distributed samples via a similar reparametrization?

The idea of reparameterization is quite simple. To generate a sample from a particular distribution, i.e. $$z\sim N(\mu,\sigma)$$, we generate samples from another distribution, i.e. $$\epsilon\sim N(0,1)$$, and then transform it into genuine samples from the desired distribution, i.e. via $$z=\mu+\sigma\times\epsilon$$. This has some nice consequences in the VAE framework, i.e., reduces the variance of the gradient (not really important here).

I basically wonder if such transformation exists for Gamma distribution.

• can you add a little more detail/context for those who don't have time to dig in and read the whole linked article? – Ben Bolker Sep 22 '18 at 15:48
• @BenBolker I added the description. – user3639557 Sep 22 '18 at 15:52

The answer is "yes" in one sense and "no" in another sense.

Suppose $$X \sim \operatorname{Gamma}(\alpha,\beta)$$. Let $$F_{\alpha,\beta}$$ denote the cdf of a Gamma distribution. Then define $$\epsilon = \Phi^{-1}[F_{\alpha,\beta}(X)]$$. Then if you simulate $$\epsilon \sim N(0,1)$$ you can get the relevant gamma distribution by setting $$X = F^{-1}_{\alpha,\beta}[\Phi(\epsilon)]$$. This is a consequence of the probability integral transform. Additionally, the transform $$T(\epsilon ; \alpha, \beta) = F^{-1}_{\alpha,\beta}[\Phi(\epsilon)]$$ is differentiable. So you could use this idea with the reparametrization trick, at least in principle, to improve your stochastic variational inference. This implies that, in a liberal sense, the answer is "yes, there is a reparameterization trick", and in fact there is one for essentially any family of continuous distributions. If this seems sort of ad-hoc, notice that if you apply this trick with the Gaussian family in place of the gamma, you get back exactly the usual reparameterization trick.

In a more restrictive sense, I would say the answer is "no". The function $$F^{-1}$$ above is not available in closed form, so things are not so convenient to the point where we might disqualify this approach. Alternatively, there is no reason to restrict ourselves to $$\epsilon \sim N(0,1)$$, and we might just ask for $$\epsilon \sim Q$$ for some standard distribution $$Q$$ that is easy to sample from, such that $$T(\epsilon; \alpha, \beta) \sim \text{Gamma}(\alpha,\beta)$$ where $$T$$ is also easy to compute.

If you find such a transformation $$T$$ and standard distribution $$Q$$, let me know because I would be interested in it. The main problem is the shape parameter $$\alpha$$. If I know $$\alpha$$, then I can take $$T(\epsilon; \alpha, \beta) = \epsilon \beta$$ and set $$\epsilon \sim \mbox{Gamma}(\alpha,1)$$, because the gamma family with know $$\alpha$$ is a scale family. The shape parameter does not have any nice algebraic properties as far as I know, aside from the fact that $$X_1 + X_2 \sim \mbox{Gamma}(\alpha_1 + \alpha_2, 1)$$ provided that $$X_i \sim \mbox{Gamma}(\alpha_i, 1)$$ and they are independent. It's not clear how to take advantage of this. A negative result for us is that, if such a convenient $$T$$ existed, then R would probably use that to sample from the a generic gamma distribution, but instead it uses rejection sampling.

• Hi; I'm recently encounter a similar problem; but I'm trying to do for Beta distribution such that $\mathcal{Q}$ is also Beta; do you have any idea how to do it ? – ElleryL Oct 2 '19 at 20:41
• Is there a reason why you draw from $N(0,1)$ instead of just doing $F^{-1}_{\alpha, \beta}[\epsilon]$ where $\epsilon \sim U(0,1)$? – Tiwa Aina Aug 23 at 1:44
• @tiwa Not particularly. – guy Aug 23 at 3:29

In terms of the strategy from the VAE paper, no there's not a straightforward way of using a gamma distribution (as guy explained). However, there's a lot of research that's gone into other strategies for reparameterizing gammas to use in the VAE type framework:

You could alternatively use BBVI (here's the original Blei lab paper http://www.cs.columbia.edu/~blei/papers/RanganathGerrishBlei2014.pdf), as demonstrated in this tutorial http://ajbc.io/resources/bbvi_for_gammas.pdf


text:
### 腾迅入股搜狗,是张朝阳想套现？

9.16日，腾迅以4.48亿美元现金加上旗下的搜索资产，换取了搜狗的36.5%的股份和20.6%的投票权。搜狗为什么要出售？

"Sogou will use a portion of the cash proceeds of Tencent’s investment to pay a special dividend in the amount of approximately $301 million to the holders of Series A Preferred Shares of Sogou, payable promptly following the closing of the transaction." 记者会上说"这一次增发新股，后面会给原来老股东派股息，然后回馈小股东。最重要的是拿到腾讯的资源。" 我十分不明白为什么要在此时发放特殊股利给A类优先股的持有者。那么这些持有者又是谁呢？ 根据历史公告，在2010年10月22日，搜狗出售了24.0 million, 14.4 million and 38.4 million份A类优先股分别给阿里、Yunfeng基金以及张朝阳自己的Photon基金，当时的价格分别是1500万美元、900万美元、2400万美元。对应的份额约为10%, 6% and 16%。此时搜狗依然保留搜狗53%的股份。 根据2011年6月的公告： As of June 30, 2011, Sogou had outstanding a combined total of 216,000,000 ordinary shares and Series A preferred shares, consisting of (i) 139,200,000 ordinary shares held by Sohu; (ii) 24,000,000 Series A preferred shares held by Alibaba; (iii) 14,400,000 Series A preferred shares held by China Web; (iv) 38,400,000 Series A preferred shares held by Photon. 在2011年6月30日，sohu总计占股139,200,000/216,000,000=64.4%。 然后在2012年6月29日， "On June 29, 2012, Sohu purchased the 24.0 million Sogou Series A Preferred Shares held by Alibaba for fixed cash consideration of$25.8 million. After the purchase of these shares, we held 73% of the combined total of Sogou's outstanding ordinary shares and Series A Preferred Shares. As we are Sogou's controlling shareholder, we continue to consolidate Sogou in our consolidated financial statements, but recognize noncontrolling interest reflecting shares held by shareholders other than us."


text:
# Stability of the Superconducting d-Wave Pairing Toward the Intersite Coulomb Repulsion in CuO2 Plane

Val'kov, VV; Dzebisashvili, DM; Korovushkin, MM; Barabanov, AF JOURNAL OF LOW TEMPERATURE PHYSICS, 191 (5-6):408-425; SI 10.1007/s10909-018-1882-7 JUN 2018

Taking into account the real crystalline structure of the CuO2 plane and the strong spin-fermion coupling, we study the influence of the intersite Coulomb repulsion between holes on the Cooper instability of the spin-polaron quasiparticles in cuprate superconductors. The analysis shows that only the superconducting d-wave pairing is implemented in the whole region of doping, whereas the solutions of the self-consistent equations for the s-wave pairing are absent. It is shown that intersite Coulomb interaction V1 between the holes located at the nearest oxygen ions does not affect the d-wave pairing, because its Fourier transform Vq vanishes in the kernel of the corresponding integral equation. The intersite Coulomb interaction V2 of quasiparticles located at the next-nearest oxygen ions does not vanish in the integral equations, however, but it is also shown that the d-wave pairing is robust toward this interaction for physically reasonable values of V2.


text:
My thesis.pdf (700.52 kB)

# Duality of Gaudin Models

thesis
posted on 2020-07-29, 16:01 authored by Filipp UvarovFilipp Uvarov
We consider actions of the current Lie algebras $\gl_{n}[t]$ and $\gl_{k}[t]$ on the space $\mathfrak{P}_{kn}$ of polynomials in $kn$ anticommuting variables. The actions depend on parameters $\bar{z}=(z_{1}\lc z_{k})$ and $\bar{\alpha}=(\alpha_{1}\lc\alpha_{n})$, respectively.
We show that the images of the Bethe algebras $\mathcal{B}_{\bar{\alpha}}^{\langle n \rangle}\subset U(\gl_{n}[t])$ and $\mathcal{B}_{\bar{z}}^{\langle k \rangle}\subset U(\gl_{k}[t])$ under these actions coincide.
To prove the statement, we use the Bethe ansatz description of eigenvectors of the Bethe algebras via spaces of quasi-exponentials. We establish an explicit correspondence between the spaces of quasi-exponentials describing eigenvectors of $\mathcal{B}_{\bar{\alpha}}^{\langle n \rangle}$ and the spaces of quasi-exponentials describing eigenvectors of $\mathcal{B}_{\bar{z}}^{\langle k \rangle}$.
One particular aspect of the duality of the Bethe algebras is that the Gaudin Hamiltonians exchange with the Dynamical Hamiltonians. We study a similar relation between the trigonometric Gaudin and Dynamical Hamiltonians. In trigonometric Gaudin model, spaces of quasi-exponentials are replaced by spaces of quasi-polynomials. We establish an explicit correspondence between the spaces of quasi-polynomials describing eigenvectors of the trigonometric Gaudin Hamiltonians and the spaces of quasi-exponentials describing eigenvectors of the trigonometric Dynamical Hamiltonians.
We also establish the $(\gl_{k},\gl_{n})$-duality for the rational, trigonometric and difference versions of Knizhnik-Zamolodchikov and Dynamical equations.

## Degree Type

Doctor of Philosophy

Mathematics

Indianapolis

Vitaly Tarasov


text:
# Non-linear DE using laplace transform...

This topic is 2306 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi, I have an assignment and a subproblem of one of the problems is about solving a differential equation using the Laplace transform. The problem is it is nonlinear and so when applying the transform, I end up with both F(s) and the integral of F(s) in the same equation, so I cannot isolate F(s). I have already solved it using another technique (this was the objective of the previous subproblem). My question is: Is this possible at all? Could this be an oversight on their part?

You can see the form of the equation below. Only Y and t are variables.

dY/dt + (a/(b+a*t) + c)*Y = d / (b+a*t)

You can see that in the second term we have Y in the numerator and t in the denominator, making the equation nonlinear... I decomposed the fraction into partial fractions in order to get only t in the denominator and apply the Laplace transform property that L(f(t)/t) = integral from s to infinity F(s)ds, but so I get another integral-differential equation...