text:
## pooja195 one year ago I need help with this word problem im completely lost....or just not reading it properly....but anywhoo please help! :) http://prntscr.com/8cd9mj And explain it so i understand..

1. chosenmatt

what part of this u ain't gettin?

2. pooja195

.-. the one thats in the LARGE red square .-.

3. chosenmatt

k

4. pooja195

Rectangle..

5. chosenmatt

i knew that im saying what part in the LARGE RED SQUARE do you not get?

6. chosenmatt

-.-

7. pooja195

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @pooja195 I need help with this word problem im completely lost....or just not reading it properly....but anywhoo please help! :) http://prntscr.com/8cd9mj And explain it so i understand.. $$\color{#0cbb34}{\text{End of Quote}}$$ All of it.

8. BloomLocke367

the entire path is 10.5 miles. it would help if you sketch it. at one end, there is bedford. The other end has Arlington. Do you understand that part?

9. BloomLocke367

|dw:1441334116658:dw| A is for Arlington, B is for bedford

10. BloomLocke367

Can you find the halfway point and where each biker is?

11. pooja195

lol i should've have drawn it out ._. now when i look at it,it makes so much more sense :#

12. pooja195

10.5/2=5.25 <----right?

13. BloomLocke367

Yes! now find where each biker is and subtract their position from the midpoint you just found

14. pooja195

wait what O_o?

15. BloomLocke367

okay. Do you know where Caitlin and Laurie are? It tells you where they start.

16. pooja195

3 miles

17. BloomLocke367

|dw:1441334437628:dw|

18. BloomLocke367

do you understand that? sorry it's kinda sloppy xp

19. pooja195

lol its gorgeous yes i get it now

20. pooja195

Ok i got it from hre thanks for the help! :D

21. BloomLocke367

Are you sure you understand? Because I don't mind explaining further

22. pooja195

Ye i got it from here thanks! :D

23. BloomLocke367

Of course! :D


text:
# Math Help - compact disc with diameter of 120mm

1. ## compact disc with diameter of 120mm

A compact disc is 120 mm across with a center hole of diameter 15mm. The center of the disc is at the origin.

1. What are the coordinates of the points at which the inner and outer edge intersect the positive x-axis?

2. What are the coordinates of the points at which the inner and outer edges cut a line making an angle theta with the positive x-axis?

Im having trouble with this and any help would be appreciated

2. 1. What is the inner and outer radius?
2. Have you studied learned the trig functions in terms of r, x, and y? If so, r is the radius, and (x, y) is your answer.

Let us know where you get stuck from here, and we'll be glad to keep you moving in the right direction!

format: plain


text:
# $\cot{(45^°)}$ value

The value of cot function when angle of right triangle equals to $45^°$ is called cot of angle $45$ degrees. As per sexagesimal system, it is written as $\cot{(45^°)}$ in mathematics.

$\cot{(45^°)} \,=\, 1$

The exact value of cot of angle $45$ degrees is equal to $1$ and it is also an integer.

## Alternative form

$\cot{(45^°)}$ can be written in alternative form as $\cot{\Big(\dfrac{\pi}{4}\Big)}$ in circular system and also written as $\cot{(50^g)}$ in centesimal system.

$(1) \,\,\,$ $\cot{\Big(\dfrac{\pi}{4}\Big)} \,=\, 1$

$(2) \,\,\,$ $\cot{(50^g)} \,=\, 1$

### Proof

Now, you learnt the exact value of cot of angle $45$ degrees and you can also learn how to derive $\cot{\Big(\dfrac{\pi}{4}\Big)}$ in trigonometry.


text:
# How do I detect revision of a particular PIC microcontroller?

I have a PIC18F4680. After reading the datasheet, it's still no clear to me how can I determine which revision the chip is.

So how do I determine the revision of the microcontroller other than asking the supplier to provide that information to me?

Answers relevant to other PIC series are welcome too.

If you look in the Flash Programming Specification for your device (document DS39622L in your case) you'll see this table which includes the device ID bits:

There are two Device ID bytes DEVID which you should be able to read along with the config bits in MPLAB. The device ID bits include a 5-bit revision code.

The errata sheets for the different revisions give a revision code, so you can match this bit string with the right revision code. e.g. the rev A3 code is 0 0100, whereas the rev B1 code is 0 0111.

If all else fails, or if you have a bunch of ICs in a tube + can't easily hook them up to a programmer, the IC should have a date code that's the last two digits of the year followed by the week number. (e.g. 1214 for 2012, April 1 which is week 14) You should be able to call up your Microchip rep and if you ask nicely they should be able to tell you what silicon revision that is. (I can't remember if they put the silicon rev # on the packaging. TI C2800 DSPs do.)

• Is there any way to read the data in MPLABX? As far as I can see, it only shows configuration bits up to 30000D using the window>pic memory views>configuration bits option. I'm using a PICkit 2 compatible programmer, if it matters. Also the revision number isn't on the packaging, only the week is. – AndrejaKo Apr 2 '12 at 20:43
• unfortunately it's been a while since I've run MPLAB so I'm not sure I can help with that. But if they don't provide you with a means to display the two DEVID bytes, this is a missing feature in MPLAB -- won't help you today but they should be able to provide that functionality. – Jason S Apr 2 '12 at 20:47
• Looks like the MPLABX doesn't have that functionality. Even the official article doesn't mention reading of configuration bits from the device itself. – AndrejaKo Apr 2 '12 at 20:51

Not sure about MPLABX, but in earlier versions, the programmers (certainly for ICD2,3 and PICkit3) display the rev in the output window when they connect.

• Unfortunately PICkit 2 doesn't seem to be able to read configuration bits that high. – AndrejaKo Apr 3 '12 at 6:14

It seems I managed to find the answer to my question.

In the PICkit 2 installation directory, there should be file named PICkit2.ini. This line: REVS: needs to be added to it in order to display the revision of the connected chip. The revision will appear to the right of the checksum value. This feature needs software version 2.10 or later.

Source


text:
• Measurement of the cross-section of $Z\gamma$ and limits on ADD models at the CMS with 7 TeV Large Hadron Collider data

• # Fulltext

https://www.ias.ac.in/article/fulltext/pram/079/04/0925-0928

• # Abstract

The measurement of the inclusive cross-section for $Z\gamma$ production at LHC with 7 TeV proton–proton collision is presented. The electron and muon decay modes are used to reconstruct the 𝑍 boson. The total cross-section is measured for photon transverse energy greater than 10 GeV and with photon and charged lepton separation in the pseudorapidity-azimuthal angle plane greater than 0.7. This study is extended by a measurement of $Z(\nu \bar{\nu})\gamma$ cross-section. A search is performed for extra dimensions in the Arkani-Hamed, Dimopoulos, Dvali framework using the final state of a graviton and photon. The limits are extended with MD $> 1.25–1.31$ TeV for $n = 2-6$. The measurement is found to be in agreement with the Standard Model prediction.

• # Author Affiliations

1. Saha Institute of Nuclear Physics, 1/AF Bidhan Nagar, Kolkata 700 064, India

• # Pramana – Journal of Physics

Volume 94, 2020
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
## Objectives

This post will guide you on how to take your data (in a CSV file) to a trained TensorFlow model of your choosing.

You’re not going to find any tricks or hacks here. The title to this blog post is so general because the TensorFlow developers have created a great API for importing data and training standard models. If you follow all the suggestions of the official TensorFlow docs, you should come to the same conclusions I do here.

It may tempting to quickly write a script that works for your current data and current task, but if you take a little extra time and write generalizable code, you will save yourself headaches in the future. The instructions here will help you easily scale to different datasets and different model architectures.

## Pre-requisites

1. A working, new version of TensorFlow installed.
2. Your data in CSV format. The reason I choose CSV data as the starting point is that almost any data can be formatted as a CSV file. Getting your raw data to a CSV file is on you, but once you get there, the rest is smooth sailing:) From CSV data, I show you how to get your data into tfrecords format, which is the prefered TF data format. So, if your data is already in tfrecords, you’re already ahead of the curve!

### Install TensorFlow

Just follow the official installation instructions!

### Get Data in CSV

To ground this post in a concrete example, below is my own labeled data in CSV. Each training data example is represented as a single row in the CSV file, where the first column represents the label (an integer), and all the following columns contain the features for that example (floating point numbers).

The labels in the first column represent categories of speech sounds, for example, label 45 might be the vowel [oh] and label 7 might be the consonant [k]. The features shown in column-two onward correspond to amplitudes at different frequency ranges. In a nutshell, this is speech data, where a snippet of audio (features) has been labeled with a language sound (labels).

Here’s what four lines of my data CSV file look like (where the delimiter is a single space):

95 21.50192 -2.16182 -1.591426 0.06965831 0.6690025  ...  -0.7368361 -1.385849 0.7551874 -0.8878949 -0.4799456
7 22.23698 -1.177924 -1.368747 -0.6289141 0.009075502  ... -0.9235415 -1.74792 0.2629939 -2.119366 -0.539937
45 22.83421 -0.9043457 -1.591426 -0.816999 -0.3035215  ...  -0.5301266 -1.456303 -0.1479924 -1.641482 -0.04098308
27 -0.9376022 -0.05841255 0.3308391 -0.7141842 -0.3867566  ...  -1.263647 23.4316 -0.0009118451 -1.035212 -1.635385

Pretty simple, right? One training example is one line in the CSV file.

If your data isn’t in this kind of CSV format, you’re going to have to spend a little time to get it here. The most important point is that you need one training example per line, and you should know exactly where each part of the example is located. For my example, I know that the label is the first column, and all the following columns are my features. You also must know how each label/feature is represented. For my case, all the labels are integers, and all the features are floating point numbers. You might have text-based labels or features (e.g. words from text), or you could have categorical features (e.g. you have a feature for "color" that you’ve coded as integers 1 through 5).

Whatever the case, you need to know exactly:

1. Where your data is (i.e. which column)
2. How your data is coded (e.g. float vs. integer vs. text)
3. What your data means (e.g. the integer entry 43 in column 5 corresponds to the color blue)

The last point is very important, because you might have integers whose numerical distance doesn’t correspond to anything meaningful (e.g. the distance between 3 and 7 means nothing if 3 is "orange" and 7 is "magenta"). On the other hand, you might have integers where the distance between them is very important (e.g. the score on a test 99 is much better than a grade of 59). The distance between test grades is meaningful, but the distance between colors is not.

In what follows, you have to decide how to represent your values, and whether or not their distances matter.

## Convert CSV to TFRecords

TFRecords is the preferred file format for TensorFlow. These tfrecords files take up a lot of space on disk, but they can be easily sharded and processed across machines, and the entire TensorFlow pipeline is optimized with tfrecords in mind.

To work with tfrecords data, you have to first format your CSV data using TensorFlow itself. We have to read in the CSV file one example at a time, and format it as a tf.train.Example example, and then print that example to a file on disk. Each tf.train.Example stores information about that particular example via so-called features, where these features can be anything (including the target label!). You will store each Example’s feature as an item in a dictionary, where the key should be descriptive. You can see in the following example I have chosen the keys "label" and "feats" to be make sure I won’t mix them up.

Below is an example Python script to read in a .csv data file and save to a .tfrecords file. You can find the original version of the following csv-to-records.py here. There are faster ways to do this (i.e. via parallelization), but I want to give you working code which is as readable as possible.

There’s a good amount of resources on tfrecords out there, check out the official docs on reading data, Python-IO, and importing data.

## Pat on the Back

If you’ve gotten to this point, you have successfully converted your data and saved it as TFRecords format. Take a pause and pat yourself on the back, because you’ve accomplished the most time-consuming and boring part of machine learning: data formatting.

Now that you have your data in a format TensorFlow likes, we can import that data and train some models. Before we jump straight into training code, you’ll want a little background on TensorFlow’s awesome APIs for working with data and models: tf.data and tf.estimator.

## Datasets and Estimators

The official TensorFlow docs push hard for you to use their Dataset and Estimator APIs. In general, if the docs explicitly tell you there is a preferred way to do something, you should do that because all the newest features will surely work for this format but maybe not others.

### tf.data.Dataset

The Dataset Class allows you to easily import, shuffle, transform, and batch your data. The Dataset API makes any pre-processing operation on your data just another part of the pipeline, and it’s optimized for large, distributed datasets. Your entire pre-processing pipeline can be as simple as this:

In the above definition of dataset, you can see there’s a line where you point TensorFlow to your data on disc, and read the data via tf.data.TFRecordDataset. The .shuffle() and .batch() functions are optional, but you will need the .map() function.

The .map() function provides the methods for parsing your data into meaningful pieces like “labels” and “features”. However, .map() is a super general function, and it doesn’t know anything about your data, so we have to pass a special parsing function which .map() then applies to the data. This parser function is probably the main thing you have to create for your own dataset, and it should exactly mirror the way you saved your data to TFRecords above with the tf.Example object (in the data formatting section above). Read more about parser functions in the official docs.

The above is one of the simplest ways to load, shuffle, and batch your data, but it is not the fastest way. For tips on speeding this stage up, take a look here and here.

Here’s an example of such a parser function:

To get into the details of this function and how you can define one for your data, take a look at the official parse function docs. Remember that if you have labeled training data, the features definition above includes the data features (feats) as well as the labels (label). If you’re doing something like k-means clustering (where labels aren’t used), you won’t return a label.

### tf.estimator.Estimator

The Estimator class gives you an API for interaction with your model. Here’s a good overview from the official docs. It’s like a wrapper for a model which allows you to train, evaluate, and export the model as well as make inferences on new data. Usually you won’t be interacting directly with the base class tf.estimator.Estimator, but rather with the Estimator Classes which directly inherit from it, such as the DNNClassifier Class. There are a whole set of pre-defined, easy to use Estimators which you can start working with out of the box, such as LinearRegressor or BoostedTreesClassifier

You can instantiate an Estimator object with minimal, readable code. If you decide to use the pre-existing Estimators from TensorFlow (i.e. “pre-canned” models), you can get started without digging any deeper than the __init__() function! I’ve defined a 4-layer Deep Neural Network which accepts as input my input data (377-dimensional feature vectors) and predicts one of my 96 classes as such:

We’ve just defined a new DNN Classifier with an input layer (feature_columns), four hidden layers (hidden_units), and an output layer (n_classes). Pretty easy, yeah?

You will probably agree that each of these three arguments is very clear expect for maybe the feature_columns argument. You can think of “feature_columns” as being identical to “input_layer”. However, feature_columns allows you to do a whole lot of pre-processing that a traditional input layer would never allow. The official documentation on feature_columns is really good, and you should take a look. In a nutshell, think of these feature_columns as a set of instructions for how to squeeze your raw data into the right shape for a neural net (or whatever model you’re training). Neural nets cannot take as input words, intergers, or anything else that isn’t a floating point number.

The feature_columns API helps you not only get your data into floats, but it helps you find floats that actually make sense for your task at hand. You can easily encode words or categories as one-hot vectors, but one-hot vectors are not practical if you have a billion different words in your data. Instead of using one-hot vector feature_columns, you can use the feature_column type embedding_column to find a lower-dimensional representation of your data. In the example above, I use the feature_column.numeric_column because my input data is already encoded as floating point numbers.

## Putting it All Together

Below is an example of minimal code you need for importing a tfrecords file, training a model, and making predictions on new data.

### parser_fn

The parser function will be the most data-specific part of your code. Learn about how to make a good function here.

### input_fn

This is an Estimator input function. It defines things like datasets and batches, and can perform operations such as shuffling. Both the dataset and dataset iterator are defined here. Read more about how to make a good input_fn on the official docs.

### Estimator

To get started fast, just choose an Estimator from the available pre-made Estimators. For more detail on how to use pre-made Estimators in general, check out the official docs.

If you want a custom architecture which is not pre-made, you can build your own Estimator.

### Train & Eval Specs

Defining the training and evaluation routine for your model is easy with TrainSpec and EvalSpec. These two classes allow you to combine your model with your data along with instructions on how to combine them.

After you’ve defined the specs, you feed them to the specialized function tf.estimator.train_and_evaluate which nicely handles all the heavy lifting. The Google Cloud folks wrote a very nice blog post on how to get best use of this function as well as your specs.

### Inference

Finally, to make predictions on new data, just use the .predict() method which is available to all Estimators.

## Conclusions

I hope you’ve found this post helpful!

Feel free to leave questions and comments below!


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R Dataset / Package Stat2Data / TwinsLungs

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Visual Summaries

Embed
<iframe src="https://embed.picostat.com/r-dataset-package-stat2data-twinslungs.html" frameBorder="0" width="100%" height="307px" />
Attachment Size
257 bytes
Dataset License
GNU General Public License v2.0
Documentation License
GNU General Public License v2.0
Documentation

On this Picostat.com statistics page, you will find information about the TwinsLungs data set which pertains to TwinsLungs. The TwinsLungs data set is found in the Stat2Data R package. You can load the TwinsLungs data set in R by issuing the following command at the console data("TwinsLungs"). This will load the data into a variable called TwinsLungs. If R says the TwinsLungs data set is not found, you can try installing the package by issuing this command install.packages("Stat2Data") and then attempt to reload the data. If you need to download R, you can go to the R project website. You can download a CSV (comma separated values) version of the TwinsLungs R data set. The size of this file is about 257 bytes.

TwinsLungs

Description

Comparing lung function for twins between rural and urban environments

Format

A dataset with 14 observations on the following 3 variables.

 Pair Code for the twin pair: A - G Environ Living environment: Rural or Urban Percent Percentage of radioactivity remaining in lungs

Details

This dataset is from a study to compare the effect of living environment (rural or urban) on human lung function, where the researchers were able to locate seven pairs of twins with one twin in each pair living in the country, the other in a city. To measure lung function, twins inhaled an aerosol of radioactive Teflon particles. By measuring the level of radioactivity immediately and then again after an hour, the scientists could measure the rate of “tracheobronchial clearance." The percentage of radioactivity remaining in the lungs after an hour told how quickly subjects' lungs cleared the inhaled particles.

Source

“Urban factor and tracheobronchial clearance" by Per Camner and Klas Philipson in Archives of Environmental Health, V. 27 (1973), page 82. Data can be found in Introduction to Mathematical Statistics and its Applications, 2nd Edition by Richard J. Larson and Morris L. Marx. Englewood Cliffs, NJ: Prentice Hall, p. 548.

--

Dataset imported from https://www.r-project.org.

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text:
# Control Problems with Vanishing Lie Bracket Arising from Complete Odd Circulant Evolutionary Games

We study an optimal control problem arising from a generalization of rock-paper-scissors in which the number of strategies may be selected from any positive odd number greater than 1 and in which the payoff to the winner is controlled by a control variable γ. Using the replicator dynamics as the equations of motion, we show that a quasi-linearization of the problem admits a special optimal control form in which explicit dynamics for the controller can be identified. We show that all optimal controls must satisfy a specific second order differential equation parameterized by the number of strategies in the game. We show that as the number of strategies increases, a limiting case admits a closed form for the open-loop optimal control. In performing our analysis we show necessary conditions on an optimal control problem that allow this analytic approach to function.

## Authors

• 7 publications
• 4 publications
• ### Error estimates for a pointwise tracking optimal control problem of a semilinear elliptic equation

We consider a pointwise tracking optimal control problem for a semilinea...
09/01/2020 ∙ by Alejandro Allendes, et al. ∙ 0

• ### Deep learning as optimal control problems: models and numerical methods

We consider recent work of Haber and Ruthotto 2017 and Chang et al. 2018...
04/11/2019 ∙ by Martin Benning, et al. ∙ 0

• ### Fast Mesh Refinement in Pseudospectral Optimal Control

Mesh refinement in pseudospectral (PS) optimal control is embarrassingly...
04/29/2019 ∙ by N. Koeppen, et al. ∙ 0

• ### On the Sample Complexity of the Linear Quadratic Regulator

This paper addresses the optimal control problem known as the Linear Qua...
10/04/2017 ∙ by Sarah Dean, et al. ∙ 0

• ### Learning Minimum-Energy Controls from Heterogeneous Data

In this paper we study the problem of learning minimum-energy controls f...
06/18/2020 ∙ by Giacomo Baggio, et al. ∙ 0

• ### Analysis, Design, and Generalization of Electrochemical Impedance Spectroscopy (EIS) Inversion Algorithms

We introduce a framework for analyzing and designing EIS inversion algor...
06/12/2020 ∙ by Surya Effendy, et al. ∙ 0

• ### Exploring optimal institutional incentives for public cooperation

Prosocial incentive can promote cooperation, but providing incentive is ...
07/25/2019 ∙ by Shengxian Wang, et al. ∙ 0

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

Cyclic competition occurs frequently in nature JB75 ; SL96 ; KRFB02 ; GHS03 ; KR04 ; KNS05 ; NHK11 . Cyclic competition in coral reef populations are studied in JB75 . Sinervo and Lively first characterized rock-paper-scissors like competition in lizards SL96 , while Gilg, Hanski and Sittler GHS03 study this behavior in rodents. Cyclic behavior in microbial populations is studied in KRFB02 ; KR04 ; KNS05 ; NHK11 . In classical and evolutionary game theory, cyclic dominance (e.g., matching pennies, rock-paper-scissors) games are commonly studied Mor94 ; Wei95 . Biologically speaking, in an idealized cyclic game, the absolute fitness measure (payoff) resulting from species interaction can be represented by a circulant matrix, in which row is a rotation of row for each . Games with circulant payoff matrices have been studied extensively in evolutionary game theory Ze80 ; SSW80 ; HS98 ; HS03 ; DG09 ; HS00 ; SMJS14 ; GK16 and provide some of the most interesting behaviors SMJS14 .

In early work, cyclic interaction is considered without explicit reference to games. Cyclic (chemo-biological) interactions are studied extensively in SSW78 ; HSSW80 ; SSW79 in which both competitive and cooperative behaviors are identified. Analysis of the replicator in which a circulant matrix emerges is studied in SSW80 as a result of cyclic mass interaction kinetics. Zeeman Ze80 made an early study of the dynamics of cyclic games showing that in rock-paper-scissors a degenerate Hopf bifurcation leads to the emergence of a non-linear center with no limit cycle possible in any 3 strategy game under the replicator dynamics. Since this early work, several authors have investigated various cyclic games and games characterized by circulant matrices. Among many other works: Hofbauer and Schlag HS00 consider imitation in cyclic games; Diekmann and Gils specifically study the cyclic replicator dynamics and focus on the properties of low-dimensional cyclic games DG09 ; Ermentrout et al. consider a transition matrix evolutionary dynamic in which a limit cycle emerges in the rock-paper-scissors game EGB16 ; and Griffin and Belmonte GB17 study a triple public goods game and show that is is diffeomorphic to generalized rock-paper-scissors. Each of these works focuses explicitly on classes of circulant games, while recent work by Granić and Kerns GK16 characterizes the Nash equilibria of arbitrary circulant games, but does not focus on the evolutionary game context.

There has also been extensive work on spatial games with circulant payoff matrices. Peltomäki and Alvara PA08 consider both 3 and 4 state rock-paper-scissors. Other papers consider rock-paper-scissors with variations on reaction rate HMT10 or study the basins of attraction SWYL10 . DeForest and Belmonte dB13 study a fitness gradient variation on the spatial replicator and show rock-paper-scissors can exhibit spatial chaos under these dynamics. Finally more recent work by Szczesny et. al SMR14 considers spiral formations in rock-paper-scissors.

In this paper, we extend work in GB17 by studying an optimal control problem defined on a -strategy () generalization of rock-paper-scissors. Odd cardinality interactions are interesting because they model specific biological cases SL96 ; GHS03 ; SSW78 ; HSSW80 ; SSW79 . Additionally, when is very large, these have the potential to model systems in which individuals a variety of individuals with strengths and weaknesses interact.

In particular, the payoff matrix is (i) defined by the sum of two circulant matrices, and (ii) admits a single control parameter. Thus we consider the general class of control problems first studied in a specific case in GB17 . Our payoff matrix is inspired by the generalized rock-paper-scissors matrix defined in Wei95 . Since every pair of heterogeneous strategic interactions (e.g., rock vs. scissors) results in a non-zero payoff, we refer to this class of games as complete odd circulant games.

The major results of this paper are:

1. We show that the complete odd circulant games admit only a unique interior fixed point under the replicator dynamics. We also characterize the fixed points of the -strategy complete odd circulant game in terms of the fixed points of the complete odd circulant games.

2. We show that the replicator dynamics can be written as the sum of an uncontrolled component and a controlled component both of which have circulant Jacobian matrices.

3. As a consequence, we completely characterize the stability of the interior fixed point and use this to define an optimal control problem with objective to drive the system trajectories toward this interior fixed point.

4. We describe the properties of a general class of control problems with a vanishing Lie Bracket that will be used to analyze the control problem we define. This suggests interesting geometric and algebraic connections between evolutionary games and optimal control theory. As a by-product, we generalize recent control theoretic results in FG17 .

5. We show that a quasi-linearization of the control problem (as done in GB17 ) has special form admitting a complete characterization of the dynamics of the optimal control. We also derive a sufficient condition on control optimality and thus completely generalize the results in GB17 to arbitrary complete odd circulant games.

6. As a part of the generalization, we find a limiting second order ordinary differential equation (as

grows large) that the optimal control must obey and show that it has a natural closed form solution.

The remainder of this paper is organized as follows: In Section 2 we present preliminary results and notation. In Section 3 we introduce the control problem of interest and study a general class of optimal control problems that will assist in the derivation of our main results. Our main results on control of complete odd circulant games are found in Section 4. Conclusions and future directions are presented in Section 5.

## 2 Notation and Preliminary Results

A circulant matrix is a square matrix with form:

 A=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣a0an−1an−2⋯a1a1a0an−1⋯a2⋮⋮⋮⋱⋮an−1an−2an−3⋯a0⎤⎥ ⎥ ⎥ ⎥ ⎥⎦.

A circulant matrix is entirely characterized by its first row and all other rows are cyclic permutations of this first row. The set of

circulant matrices forms a commutative algebra, a fact that will be used frequently in this paper. Moreover, the eigenvalues of these matrices have special form. If

is an circulant matrix and are the roots of unity, then eigenvalue () is given by the expression:

 λj=a0+an−1ωj+an−2ω2j+⋯+a1ωN−1j.

Further details on this class of matrices is available in D12 .

Let:

 ΔN={u∈RN:1Tu=1,u≥0}

be the unit -simplex embedded in -dimensional Euclidean space. Here

is an appropriately sized vector of

’s and is a zero vector.

We consider a family of control problems defined on parametrized cyclic games with strategies, where . If and , define:

 μ(k)={kmodNif kmodN≠0,Notherwise. (1)

For the remainder of this paper, define so that:

 LNij={(−1)μ(N(i−1)+j−i)if i≠j,0otherwise. (2)

and

 MNij={1if i≠j and (−1)μ(N(i−1)+j−i)=1,0otherwise. (3)

By way of example, we illustrate the matrices and for the 5-strategy cyclic game.

 L5=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0−11−1110−11−1−110−111−110−1−11−110⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦,M5=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0010110010010011010001010⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦.

From a game-theoretic perspective, we can think of as being the traditional payoff matrix of the complete cyclic game with strategies; e.g., the payoff matrix of rock-paper-scissors. On the other hand, can be thought of an actuating matrix that will determine whether the interior fixed point of the complete cyclic game is stable or unstable, as we show in the sequel.

In the remainder of this paper, we will consider the generalized cyclic game with strategies where , and we note that both and are circulant matrices. The payoff matrix for the generalized cyclic game with strategies and parameter is:

 AN(γ)=LN+γMN.

If and , then is just the rock-paper-scissors matrix. Without loss of generality, we assume . Otherwise, the natural winning precedence in the cyclic game is reversed.

The matrix is the adjacency matrix of an vertex circulant directed graph , whose edge direction determines the winning precedence between two strategies in the underlying cyclic game. This is illustrated in Figure 1.

In particular, if , then Strategy defeats Strategy and yields payoff in .

In the control problem defined in the sequel, the replicator dynamics are the nonlinear equations of motion with control parameter :

 SN={˙ui=ui((ei−u)TAN(γ)u),u(0)=u0. (4)

Here is the vector denoting the proportion of the population playing each of the strategies. It is well known HS98 ; Wei95 that if , then is confined to for all time. For the remainder of this paper, we assume .

Assume and let and . The point-to-set mapping maps a point to the set of points in so that:

1. elements of the vector consist of the elements of . The remaining elements are .

2. If the rows and columns of (resp. ) corresponding to the zero-entries in are deleted to form the matrix (resp. ), then (resp. .

The second condition is equivalent to stating that the vertices corresponding to the non-zero strategies in induce a sub-graph of that is isomorphic to .

###### Lemma 1.

Let and , then:

1. has among its fixed points () and .

2. Furthermore, if , , and and is a fixed point of , then are fixed points of .

###### Proof.

To prove Statement 1, note first that pure strategies are always fixed points of the replicator dynamics Wei95 . To see that is a fixed point, note that:

 LNu∗=0, γMNu∗=γnN1.

Consequently:

 γ(u∗)TMNu∗=γnN=γeTiMNu∗.

The fact that is a fixed point follows immediately.

To prove Statement 2, suppose that for some fixed point of . Let and be the square sub-matrices of and obtained by removing the rows and columns corresponding to the zero entries in . By assumption, it follows at once that:

 (eTi−u)TAN(γ)u=(eTi−u+)TAM(γ)u+=0.

This completes the proof. ∎

###### Corollary 2.

The fixed point is the unique interior fixed point for .

###### Proof.

If is any interior fixed point, then necessarily it must satisfy the equation:

 AN(γ)u=uTAN(γ)u⋅1.

Note that for any , . Therefore, must satisfy:

 AN(γ)u=γ⋅uTMNu⋅1.

It is straight forward to compute:

 uTMNu=N∑i=1∑j>iuiuj.

We now proceed in cases. If , then we solve:

 LNu =0, 1Tu =1.

This system has rank with unknowns and therefore admits a unique solution. Elementary row reduction shows that is the unique fixed point in this case.

On the other hand, if , then let:

 α=N∑i=1∑j>iuiuj.

We can determine a relationship between the variables by solving:

 AN(γ)u=α1.

Row-reduction on the system shows that when we obtain the relationship:

 u=αnγ1. (5)

Consequently, and necessarily is again the unique interior fixed point. This completes the proof. ∎

For the remainder of this paper, assume and let be defined component-wise as:

 Fi(u) =ui((ei−u)LNu), (6) Gi(u) =ui((ei−u)MNu). (7)

The replicator dynamics are then:

 ˙ui=F(u)+γG(u),

which are the dynamics that will be used in the control problem of interest. We note that and are the functional imprints of the standard payoff matrix and the actuation matrix within the replicator framework.

###### Lemma 3.

The Jacobian matrix of evaluated at is:

 J≜DuF=1NLN.
###### Proof.

We prove the result for Row 1 of . The remainder of the argument follows from the circulant structure of . We have:

 F1(u)=u1eT1LNu,

because . Note:

 u1eT1LNu=u1(N∑j=2(−1)j−1uj). (8)

Differentiating with respect to and evaluating at yields:

 [DuF]1,1=1NN∑j=2(−1)j−1=0,

since is odd. Differentiating Expression 8 with respect to and evaluating at yields:

 [DuF]1,j=(−1)j−1N=1NLN1,j.

The result now follows from the fact that is a circulant matrix. ∎

###### Lemma 4.

The Jacobian matrix of evaluated at is:

 H≜DuG=1N2MN−2nN2(1N−MN)=1N2(N(MN−1N)+1N),

where is an matrix of ’s.

###### Proof.

As in the proof of Lemma 3, we show the result for Row 1 of . The remainder of the argument follows from the circulant structure of . We have already noted in Corollary 2 that:

 uTMNu=N∑i=1∑j>iuiuj.

We compute:

 eT1MNu=(N−1)/2∑j=1u2j+1.

Differentiate with respect to for , corresponding to a non-zero index in the first row of . We have:

 ∂G∂uk=u1⎛⎝1−⎛⎝∑j≠kuj⎞⎠⎞⎠.

Evaluating at we obtain:

 [DuG]1,k=1N(1−N−1N)=1N2,

for with . Differentiate now with respect to to obtain:

 ∂G∂u1=(N−1)/2∑j=1u2j+1−2N∑j=2u1uj−N∑i=2∑j>iuiuj. (9)

Evaluating at we obtain:

 [DuG]1,1=nN−22nN2−1N2(12(N−2)(N−1))=−N2−2nN+N−2N2=−2nN,

when is substituted for in the numerator. Finally, consider and for . Differentiating with respect to we have:

 ∂G∂uk=−u1⎛⎝∑j≠kuj⎞⎠.

Evaluating at we obtain:

 [DuG]1,k=−1N2(N−1)=−2nN.

The result now follows from the fact that is a circulant matrix. ∎

###### Theorem 5.

If , then the fixed point is asymptotically stable. If , then the fixed point is asymptotically unstable.

###### Proof.

From Lemmas 3 and 4, we note that the Jacobian matrix of at has the following form:

 J=J+γH=1NLN+γ(1N2MN−2nN2(1N−MN)).

By its construction, it is a circulant matrix with first row given by:

 J1j=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩−γ2nN2if j=1,−1N−γ2nN2if j>1 and j−1 is odd,1N+γ1N2otherwise. (10)

Letting for be the roots of unity,111For details see D12 , we know that the eigenvalue of is:

 N∑k=1J1,kωk−1j.

It now remains to show that the sign of the real-part of is entirely dependent on . The real part of the eigenvalue is given by:

 Re(λj)=N∑k=1J1,kcos(2πj(k−1)N). (11)

The first eigenvalue () is real and readily computed:

 λ0=n(γ1N2−γ2nN2)−γ2nN2=−γn(1+2n)N2.

It is clear at once that the sign of this eigenvalue is entirely controlled by the sign of .

For , note that the periodicity of the cosine function (and the fact that the roots of unity are the vertices of the regular unit -gon) implies that the coefficient of is identical to the coefficient of if . From this fact and the Expression 10, the sum in Equation 11 becomes:

 Re(λj)=γ(−2nN2+n+1∑k=2cos(2πj(k−1)N)(1N2−2nN2)).

Factoring further we see:

The roots of unity are evenly distributed on the vertices of the unit -gon in and therefore the sum of the real parts must be zero. It follows that:

 n+1∑k=2cos(2πj(k−1)N)=−12.

We now obtain an exact value for the real parts of the eigenvalues:

 Re(λj)=γN2(−2n−12(1−2n))=−γN2(n+12). (12)

Thus we have proved that when , then for all and if , then . The asymptotic stability (resp. instability) of the fixed point follows immediately. ∎

We illustrate the attractive interior point of the five and seven strategy cyclic games by projecting them into regular -gons (), shown in Figure 2. A pure strategy corresponds to a vertex, while a mixed strategy is located in the interior. Note, these are not true trajectories as would be seen in a classic ternary plot on the three rock-paper-scissors, but they are similarly representative.

## 3 The Control Problem and Some General Results

We now state our control problem of interest:

 CN=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩min∫tf012∥u−u∗∥2+r2γ2dts.t.˙u=F(u)+γG(u),u(0)=u0. (13)

where . Such a problem arises naturally if we consider species interacting in a cyclic manner and is a costly control mechanism by which an external manager may control species populations. In GB17 , arises naturally as a tax in a public-goods game, which is shown to be diffeomorphic to a three stragegy cyclic game. As in GB17 , we will show that a quasi-linearization of this control problem has special structure. In this case, however, we show this special structure holds for all cyclic games (i.e., for all ). Furthermore, we discuss the limiting behavior of the control as grows large. To do this, we first consider a very general optimal control problem and obtain necessary conditions for simplifying the Euler-Lagrange necessary conditions. We then use these simplifications to generalize the results in GB17 .

### 3.1 Control Problems with One Control and Vanishing Lie Bracket

In the remainder of this section, the functions are arbitrary smooth functions, rather than the functions specific to the replicator dynamics for cyclic games given in Equations 6 and 7, is a state vector, and is the control function to be determined.

Consider the general optimal control problem with form:

 ⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩minΨ(x(tf))+∫tf0F0(x)+γG0(x)+r2γ2dts.t.˙x=F(x)+γG(x),x(0)=x0. (14)

The functions are smooth. Let , be the terminal time, and be convex. Expression 13 has this structure, so we are simply considering a more general case of our problem of interest.

The Euler-Lagrange necessary conditions for control are simple to derive for this problem and have an almost linear behavior. Note the Hamiltonian is:

 H(x,γ,λ)=F0(x)+γG0(x)+r2γ2+λTF(x)+γλTG(x). (15)

The Hamiltonian is (strictly) convex in the control , and thus we propose the following:

###### Lemma 6.

Any solution to satisfies the necessary conditions:

1. , and

2. , the strong Legendre-Clebsch condition;

therefore, it minimizes the Hamiltonian at all times. ∎

Deriving the optimal control by solving for to obtain:

 γ∗=−1r(λTG(x)+G0(x)). (16)

The two conditions in Lemma 6, along with the fact that and solve the resulting Euler-Lagrange two-point boundary value problem (see Expression 18), form the complete set of necessary conditions for the optimal control problem. Adding in the additional requirement that the corresponding matrix Riccati equation is bounded on , these form sufficient conditions for a weak local minimal optimal controller BY65 ; J70 . We discuss this sufficient condition in the sequel.

For simplicity, we refer to the optimal control as (rather than ) in the remainder of this paper and assume it is given by Equation 16. The adjoint dynamics are:

 ˙λT=−(∇xF0(x))T−γ(∇xG0(x))T−λTDxF−γλTDxG, (17)

where is the Jacobian (with respect to ). Thus we have the Euler-Lagrange two-point boundary value problem:

 (18)
###### Proposition 7.

If is an optimal control, then:

 γ(tf)=−1r(∇xΨ(x(tf))TG(x(tf))+G0(x(tf))). (19)
###### Proof.

This follows from the transversality condition. ∎

From Equation 16, note that:

 r˙γ=−˙λTG(x)−λT(DxG)˙x−(∇xG0)˙x. (20)

Then:

 r˙γ=((∇xF0(x))T+γ(∇xG0(x))T+λT(DxF)+γλT(DxG))G(x)−λT(DxG)(F(x)+γG(x))−(∇xG0)(F(x)+γG(x)). (21)

Simplifying we have:

 r˙γ=(∇xF0(x))TG(x)−(∇xG0(x))TF(x)+λT((DxF)G(x)−(DxG)F(x)).

If the Lie Bracket vanishes, i.e.,:

 [F,G]=(DxF)G−(DxG)F=0, (22)

then this simplifies to:

 ˙γ=1r((∇xF0(x))TG(x)−(∇xG0(x))TF(x)), (23)

and all co-state variables are eliminated. We have shown the following:

###### Theorem 8.

Consider the general optimal control problem given in Expression 14. If and is an optimal control, then the pair is a solution of the two point boundary value problem:

 ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩˙x=F(x)+γG(x),˙γ=1r((∇xF0(x))TG(x)−(∇xG0(x))TF(x)),x(0)=x0,γ(tf)=−1r(∇xΨ(x(tf))TG(x(tf))+G0(x(tf))). (24)

Geometrically, Equation 22 implies that the flows derived by the vector fields and commute locally. From a game-theoretic view, this means that locally evolutionary motion caused by competition in uncontrolled game commutes with evolutionary motion caused by the actuation payoffs on local space/time scales. As we see in the sequel, this is not true for actuated cyclic games, but is true for quasi-linear approximations of the evolutionary dynamics as in GB17 .

It is worth noting that a differential equation for the control function is derived in B78 , without the assumption of the vanishing Lie Bracket. However, without this assumption the system does not simplify in as useful a way and, in fact, in B78 the relevant Lie Bracket is not considered. Note, in formulating Theorem 8, we are assuming that solving the Euler-Lagrange equations will yield an optimal control. We can use the well known fact that a sufficient condition for optimality is the boundedness of the solution to the matrix Ricatti equation BY65 ; J70 to derive a complete necessary and sufficient condition for optimality of the control. Let:

 ˙x=F(x)+γG(x)=f(x,γ).

Then the Matrix Ricatti equation is:

 (25)

Here is the second differential operator with respect to the state and is an ordinary partial derivative, since there is only one control variable. When taken together with Lemma 6, the system of differential equations in Theorem 8 and the co-state dynamics, Equation 17, we have a complete characterization of the necessary and sufficient conditions for the optimal control. This yields the corollary:

###### Corollary 9 (Corollary to Theorem 8).

Let in the optimal control problem in Expression 14, with Hamiltonian . Any solution to the system of differential equations:

 ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩˙x=f(x,γ),˙γ=1r((∇xF0(x))TG(x)−(∇xG0(x))TF(x)),˙λ=−∇xF0(x)−γ∇xG0(x)−(DxF)Tλ−u(DxG)Tλ,−˙S=DxxH+(Dxf)TS+S(Dxf)−1r(DγxH+(∂γf)TS)T(DγxH+(∂γf)TS),x(0)=x0,γ(tf)=−1r(∇xΨ(x(tf))TG(x(tf))+G0(x(tf))),λ(tf)=∇xΨ(x[tf]),S(tf)=∇2xΨ(x[tf]), (26)

in which and is bounded for all constitutes a weak local optimal solution for Expression 14.

We note that this is the general analog of Proposition 2 in FG17 , which is specialized to a one-dimensional control problem. In general, checking the boundedness of the solution to the Matrix Ricatti equation must be done numerically. In the sequel we develop a simpler test for optimality using Mangasarian’s sufficiency condition; i.e., by checking that the Hamiltonian is jointly convex.

Problem 13 () does not satisfy the necessary condition that . However, a quasi-linearization of the problem does satisfy this condition (as in GB17 ). We now discuss a special case of Theorem 8 as well as extensions that apply to this quasi-linearized form.

### 3.2 The Quasi-Linear Case

For the remainder of this section, let and be arbitrary matrices of appropriate size, rather than the Jacobian matrices derived in Lemmas 3 and 4.

In Problem 14, let:

 (27)

where is a (symmetric) positive definite matrix of appropriate size. We will add additional criteria to and as we proceed. We refer to this as a quasi-linear case because the only non-linearity arises from the interaction of the state and control variables. The following Corollary is immediate from Theorem 8:

###### Corollary 10.

If and is an optimal control, then the pair is a solution of the two point boundary value problem:

 ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩˙x=Jx+γHx,˙γ=1rxTQHx,x(0)=x0,γ(tf)=−1r∇xΨ(x(tf))THx(tf). (28)

The condition that and commute is exactly the statement that the Lie Bracket of the vector fields in the dynamics vanishes. Therefore, Theorem 8 can be applied to any linear quadratic control problem where the state equation satisfies this condition.

We now derive some special results on and the optimal control in this quasi-linear case. Let and assume . Computing the second derivative of yields:

 r¨γ=˙xTKx+xTK˙x=(xTJT+γxTHT)Kx+xTK(Jx+γHx)=xT(JTK+KJ+γ(HTK+KH))x.

To simplify this, we will add an additional assumption to ; suppose that (i.e.,

is skew-symmetric) and

. Then:

 r¨γγ=xT(HTK+KH)x. (29)

Before proceeding note that:

 ddt(xTQx)= (xTJT+γxTHT)Qx+xTQ(Jx+γHx)= xT(−JQ+QJ)x+γxT(HTQ+QH)x= xT(−JQ+QJ)x+γxT(HTQT+QH)x= xT(−JQ+QJ)x+2γxTKx= xT(−JQ+QJ)x+2rγ˙γ.

Thus, we have the following proposition and its corollary:

###### Proposition 11.

If , and and , then:

1.  JK=KJ,
2.  ddt(xTQx)=2rγ˙γ, (30)
3.  r¨γγ=xT(HTK+KH)x. (31)
###### Corollary 12.

For some constant ,

 xTQx=rγ2+C (32)

is the implicit closed-loop control, where must satisfy:

 C=xT(tf)Qx(tf)−r(1r∇xΨ(x(tf))THx(tf))2. (33)

Furthermore the optimal control exists at time just in case:

 xT(t)Qx(t)−C≥0. (34)

## 4 Application to Complete Odd Circulant Games

We now return to the study of cyclic games with strategies and specifically to the control problem in Expression 13. As noted already, we cannot apply Theorem 8 directly to Problem 13 because the appropriate Lie Bracket does not vanish. However, we can construct the quasi-linearized form of the problem. Let . The quasi-linearized problem is:

 ~CN=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩min∫tf012∥x∥2+r2γ2dts.t.˙x=Jx+γHx,x(0)=x0. (35)

In Expression 35, and are the Jacobian matrices of and as defined in Lemmas 3 and 4. Problem 35 is an instance of the general control problem studied in Section 3.

The following useful fact follows at once from Lemma 3.

###### Corollary 13.

The Jacobian matrix is skew-symmetric. ∎

###### Lemma 14.

Let be the optimal control for Problem 13. Then:

1. The (open-loop) optimal control obeys the following differential equations:

 ˙γ =1rxTHx,γ(tf)=0, (36) r¨γγ =xT(HTH+HH)x. (37)
2. The following identity holds:

 xTx=∥x∥2=rγ2+C, (38)

where:

 C=∥∥x(tf)∥∥2.
###### Proof.

Problem 35 is an instance of Problem 14, but with quasi-linear system dynamics and quadratic objective as given in the quasi-linear conditions in Expression 27. In particular, Problem 35 sets . As a consequence the matrix . From Corollary 13, we know is skew-symmetric. Further, since the circulant matrices form a commutative algebra, we have . The lemma follows at once from Corollary 10, Proposition 11 and Corollary 12. ∎

Expression 38 is the closed-loop control law for the controlled cyclic game. Furthermore, Equation 36 allows us to determine some structural properties of .

###### Proposition 15.

The matrix is negative definite and therefore for all .

###### Proof.

Consider any vector . Then:

Let be the upper-triangular matrix defined as:

 Sij=⎧⎨⎩−N−1N2if i=j,−N−2N2</


text:
## anonymous one year ago Joey is going to paint his room. His ceiling is 8ft high. The room is 12ft wide and 15ft long. What is the total area of the walls? (Calculate as if there were no windows or doors.) A. 1440 ft3 B. 1440 ft2 C. 432 ft3 D. 432 ft2

1. anonymous

I think B, I just need someone to check it please

2. anonymous

Can someone please tell me if this is right, ill fan and medal

3. ybarrap

|dw:1433003944297:dw|

4. anonymous

I know that, I just needed to know if my answer was right

5. ybarrap

This is the situation. Area is the sum of all sides: There are 2 walls that are 8 ft high and 12 ft wide: 2 * 8*12 Two walls that are 8 ft high and 15 ft long: 2 * 8 * 15 One ceiling that is 15 ft long and 12 ft wide: 15 * 12 So to get total area we sum all these to get $$\text{Area = }2\times8\times 12+2\times 8\times 15+15\times 12$$ Is that what you did?


text:
Verifying continuity of the deformation retraction of the mapping cylinder

Given $f : X \rightarrow Y$ I have the mapping cylinder $M_f = (X \times I) \sqcup Y / \sim$ where $I = [0,1]$ and $(x,0)_1$ is glued to $f(x)_2$ (I'd like to be pedantic for this verification so I'm adding subscripts resulting from the disjoint union). It is usually said that the mapping cylinder deformation retracts to the subspace $Y$; again being pedantic what must be meant is that $M_f$ deformation retracts to the subspace consisting of the points $\{y_2\}$ for $y \in Y-f(X)$ and $\{(x,0)_1,f(x)_2\}$ for $x \in X$, I'll call this subspace $C = \{\{y_2\}, \{(x,0)_1,f(x)_2\} \: | \: y \in Y - f(X), x \in X\}$.

My first question is that do we not have to be pedantic here because I assume that we could only loosely say $Y$ instead of this subspace if these two spaces were homeomorphic and for the obvious map between them to be a homeomoprhism in general we would require that $f$ be an injection?

Secondly, I'd like to verify that the expected deformation retraction is indeed one. $M_f$ consists of three types of points, $\{y_2\}$'s, $\{(x,s)_1\}$'s and $\{(x,0)_1,f(x)_2\}$'s; and the relevant map $M_f \times I \rightarrow M_f$ is $$\left(\{y_2\}, t\right) \mapsto \{y_2\} \:\:\: \forall y \in Y - f(X), t \in I,$$ $$\left(\{(x,0)_1,f(x)_2\},t\right) \mapsto \{(x,0)_1,f(x)_2\} \:\:\: \forall x \in X, t \in I,$$ $$\left(\{(x,s)_1\},t\right) \mapsto \{(x,(1-t)s)_1\} \:\:\: \forall x \in X, s \in (0,1], t \in [0,1),$$ $$\left(\{(x,s)_1\},1\right) \mapsto \{(x,0)_1,f(x)_2\} \:\:\: \forall x \in X, s \in (0,1].$$

That the map $M_f \rightarrow M_f$ retrieved by setting $t = 0$ is the identity map is clear, and so is that the map retrieved by setting $t = 1$ maps all of $M_f$ onto $C$ and finally that each map for a fixed $t$ restricted to $C$ is the identity map. The only concern is continuity. I was able to show that each map for a fixed $t$ is continuous (in a manner similar to how I attempted this problem), but not this map from $M_f \times I \rightarrow M_f$ that we have here. My attempt is as follows:

Let $U$ be any open subset of $M_f$; $U$ consists of points $\{y_2\}$ for $y$ in some $B \subseteq Y$, of points $\{(x,0)_1,f(x)_2\}$ for $x$ in some $A \subseteq X$ and of points $\{(x,s)_1\}$ for $(x,s)$ in some $Q \subseteq X \times (0,1]$. Using the definitions of the quotient and disjoint union topologies this tells us that $B \cup f(A)$ is open in $Y$ and that $A \times \{0\} \cup Q$ is open in $X \times I$. The preimage of $U$ under our map above is the union of the following

$$\{\{y_2\} \: | \: y \in B\} \times I,$$

(where the $\{y_2\}$'s come from)

$$\{\{(x,0)_1,f(x)_2\} \: | \: x \in A\} \times I,$$

$$\{\{(x,s)_1\} \: | \: x \in A, s \in (0,1]\} \times \{1\},$$

(where the $\{(x,0)_1,f(x)_2\}$'s come from), and

$$\{(\{(x,s)_1\},t) \: | \: (x,s,t) \in \varphi^{-1}[Q]\}$$

(where the $\{(x,s)_1\}$'s come from)

where $\varphi : X \times (0,1] \times [0,1) \rightarrow X \times I : (x,s,t) \mapsto (x,(1-t)s)$.

In the cases of the maps for each fixed $t$, the preimages worked out to be similar but simpler and openness wasn't too difficult to establish. Here all I have been able to say is that because $A \times \{0\} \cup Q$ is open in $X \times I$, its preimage under $\varphi$, which is in fact just $\varphi^{-1}[Q]$ because the second coordinate of any image point cannot be zero, is open in $X \times (0,1] \times [0,1)$ and as a result in $X \times [0,1] \times [0,1]$, this then implies that our final set $\{(\{(x,s)_1\},t) \: | \: (x,s,t) \in \varphi^{-1}[Q]\}$ is open in $M_f \times I$. From here however I haven't been able to get further. I've tried picking points in one of the four sets and then trying to combine subsets of the four sets in various ways to find an open set containing the point within our four sets and I've noted some small things like the first two of our four sets may be considered together in the form $E \cup I$ with $E \subseteq M_f$ and the pullback across the quotient projection map and then the disjoint union inclusion map of $E$ gives $B \cup f(A)$ in $Y$ which is known to be open but we get $A \times \{0\}$ in $X \times I$.

I decided to do this verification, and in this pedantic a manner, for practice. I'd very much like to move on but it's bugging me and I'd really appreciate some help.

• Not a complete answer, but a few points that might help simplify your task. First, given any quotient space $E/{\sim}$, you have $E/{\sim}\times I \cong (E \times I)/{\sim}$, where the second $\sim$ is, by abuse of notation, the obvious equivalence relation on $E \times I$. (Local) compactness of $I$ is essential here. Second, wherever possible, use the fact that cont. maps from a quotient space $E/{\sim}$ to a space $F$ correspond precisely to cont. maps $E \to F$ that respect the equivalence relation $\sim$. Third, a cont. map $E \sqcup F \to G$ is the same as maps $E \to G$ and $F \to G$. Jan 13 '14 at 12:06

First note that we can indeed regard $Y$ as a subspace of $M_f$. This is because the map $\bar i:Y\to M_f$ which is the composition $Y\hookrightarrow X\times I\sqcup Y\xrightarrow q M_f$ is

• injective: Assume that $y_1,y_2$ are two points in $Y$. Then each $y_k$ is identified with all the points $(x,0)\in X\times I$ such that $f(x)=y_k$. The only way that $y_1$ and $y_2$ could be identified is by means of a finite sequence of identification "$(x,0)\sim y$ whenever $f(x)=y$". Induction over the number of such basic identifications shows that all $y$ which appear in this sequence must denote the same element in $Y$. So $\bar i(y_1)=\bar i(y_2)$ only if $y_1=y_2$.
• closed: Take a closed $F\subseteq Y$. Then its saturation is the set $\hat F=F\sqcup f^{-1}(F)\times\{0\}$, which is closed in $X\times I\sqcup Y$. Thus its image $q(\hat F)$ is closed in $M_f$ and $q(\hat F)\cap q(Y)=q(F)=\bar i(F)$ is thus a closed subset of $\bar i(Y)$

That means that $\bar i:Y\to\bar i(Y)$ is a homeomorphism. For convenience we talk about a point $y\in M_f$ when we actually mean the class $\{y,(x,0)\mid f(x)=y\}$.

Now you want a homotopy $H:M_f\times I\to M_f$. Put $h_t=H(-,t)$. Since $M_f$ is a quotient space, each $h_t:M_f\to M_f$ is determined by a map $g_t:Z:=Y\sqcup X\times I\to M_f$ which respects the relation. Now define $\sim'$ to be the relation on $Z\times I$ such that $(a,t)\sim'(a',t')\iff a\sim a'\wedge t=t'$. We see that if $G(a,t)=g_t(a)$ is continuous on $Z\times I$, then it respects $\sim'$ iff each $g_t$ respects $\sim$. So it induces a continuous map $H:Z\times I/\sim'\to M_f$. However, $(Z\times I)/\sim'$ is not the same as $(Z/\sim)\times I$. We merely have a continuous bijection $(Z\times I)/\sim'\to(Z/\sim)\times I$ induced by $q\times\text{id}:Z\times I\to Z/\sim\times I$, and there is no reason why it should be open. However, in this situation it is, and this is due to the local compactness of $I$.

• Sorry I just realised I completely forgot to accept your answer and thank you! This is much delayed but your answered helped me very much, so again thank you
– user34832
May 11 '14 at 11:27
• “So it induces a continuous map H”. What the hell are you talking about? You never define what H or G is! Jul 24 '20 at 2:26


text:
# Sanic JWT¶

Sanic JWT adds authentication protection and endpoints to Sanic.

It is both easy to get up and running, and extensible for the developer. It can act to protect endpoints and also provide authentication scoping, all wrapped into a nice JWT.

Pick your favorite user management system, run a single class to initialize, and you are all set.

Open source code on GitHub

## What is new in Version 1.2?¶

Version 1.2 saw a a few minor fixes and changes. Some new features were added to allow custom claims and additional payload verifications. Version 1.1.4 is still stable, so there is no need to upgrade to 1.2.

In addition, there was a change that will (unfortunately) have some potential impact on users. Expired and invalid tokens will report as HTTP 401.

## What is new in Version 1.1?¶

The biggest changes are under the hood relating to how configuration settings are implemented. They are now fully dynamic allowing you to not only dynamically set them at run time, but also have them evaluated at the last minute to give you flexibility when needed.

Flexibility is really the name of the game for v. 1.1. Most of the features are to enable the developer that wants to dig deeper and gain more control. For example, the Authentication now has a number of new renamed methods. Checkout the source code to see what they are (hint: they are the ones NOT with an _ at the beginning.)

Checkout the changelog for a more detailed description.

Note

It is recommended that you are at least using version 1.1.2. You can upgrade to version 1.1.4 for improved exception handling and some bug fixes.

## What is new in Version 1.0?¶

If you have been using Sanic JWT, there should really not be that much different, although under the hood a lot has changed. For starters, the initialize method still works. But, the new recommended way to start Sanic JWT is to use the new Initialize class as seen above.

Using this class allows you to subclass it and really dive deep into modifying and configuring your project just the way you need it. Want to change the authentication responses? No problem. Want to add some new authentication endpoints? Easy.

One of the bigger changes is that we have enabled a new way to add configuration settings. You can of course continue to set them as recommended by Sanic by making them in all capital letters, and giving it a SANIC_JWT_ prefix.

app.config.SANIC_JWT_ACCESS_TOKEN_NAME = 'mytoken'

Or, you can simply pass your configurations into the Initialize class as keyword arguments.

Initialize(
app,
access_token_name='mytoken'
)

Do you need some more complicated logic, or control? Then perhaps you want to subclass the Configuration class.

class MyConfig(Configuration):
access_token_name='mytoken'
def get_refresh_token_name(self):
return some_crazy_logic_to_get_token_name()

Initialize(
app,
configuration_class=MyConfig
)

The point is, with Version 1, we made the entire package extremely adaptable and extensible for you to get done what you need without making decisions for you.

Have fun, and happy coding.


text:
# Indefinite integral of (x+(1/x))^4

• Sep 13th 2012, 11:00 PM
fbarawr
Indefinite integral of (x+(1/x))^4
Hello, I'm Fahad and I'm new here
I'm here to get help with simple math problems for my IGCSES
Plese help me.
The following math is from the binomial expansions part of my syllabus:

⌠ (x+(1/x))^4

Also, please refer to the techniques of integration you are using.
• Sep 13th 2012, 11:06 PM
MaxJasper
Re: Indefinite integral of (x+(1/x))^4
$\int \left(x+\frac{1}{x}\right)^4 \, dx$ = $\int \left(6+\frac{1}{x^4}+\frac{4}{x^2}+4 x^2+x^4\right) \, dx$
• Sep 13th 2012, 11:08 PM
fbarawr
Re: Indefinite integral of (x+(1/x))^4
• Sep 13th 2012, 11:12 PM
MarkFL
Re: Indefinite integral of (x+(1/x))^4
We are given:

$\int \left(x+\frac{1}{x} \right)^4\,dx$

Using the binomial theorem on the integrand, we find:

$\left(x+\frac{1}{x} \right)^4=\left(x+x^{-1} \right)^4=\sum_{k=0}^4{4 \choose k}x^{4-k}x^{-k}$

After simplification:

$x^4+4x^2+6+4x^{-2}+x^{-4}$

So then we have:

$I=\int x^4+4x^2+6+4x^{-2}+x^{-4}\,dx$

We will integrate term by term, using the power rule for integration $\int k\cdot x^n\,dx=\frac{k}{n+1}x^{n+1}+C$

where k is a constant and $n\ne-1$

So we have:

$I=\frac{1}{5}x^5+\frac{4}{3}x^3+6x-4x^{-1}-\frac{1}{3}x^{-3}+C$


text:
# Quantum mechanics in ten ideas for people on the move

• 6
Shares

The last few posts on quantum mechanics have been quite extensive and at times rather deep for those who are on the move. So, here are ten important ideas about particles and wave functions for when you’re en route in slightly more normal English.

## 1: Subatomic particles

To describe objects in our everyday world, such as rocks, buildings, and cars, Newton’s laws suffice. To describe subatomic, elementary particles, such as electrons, protons, neutrons, and photons, however, there is a whole different type of physics: quantum mechanics.

## 2: Wave functions

The most complete description of an elementary particle is called the wave function. Actually, the word ‘particles’ seems to incorrectly refer to tiny points, balls or spheres or something, which they are absolutely not. They aren’t waves either. ‘Particles’ are wave functions with wave-like properties (emphasis on ‘like’). Upon interaction with other particles and/or measurement, they exhibit particle-like behaviour, however. The wave function contains all physically possible states a ‘particle’ can be in at the moment we measure its state. The wave function can be seen as a mathematical description of the probabilities of the states that the particle will snap into as soon as you measure it. It’s often visualised as a ‘cloud’ even though that’s not what it actually looks like. It’s just a visual metaphor for a mathematical object that actually lives in complex space as it is complex valued.

## 3: Quantum state

Once measured, particles show one specific quantum state out of a whole range of possible quantum states prior measurement. Position is the most intuitive to understand example of a quantum state. Momentum is another (momentum is a measure of the amount of motion of a particle). Then there are states such as polarity, spin, and a bunch of others. The wave function encapsulates all these possible states and yields a probability-value for actually measuring a particular state. In other words, even before you measure it, the wave function allows you to calculate the chances of encountering this particular quantum state.

## 4: Measurement problem

As long you don’t measure a particle, and as long as it doesn’t interact with other particles, the particle is not in a specific quantum state yet. Instead, its wave function just describes all these possible states as though they are mathematically added on top of each other. This ‘adding of quantum states’ is what is meant when physicists talk about superposition. The term is from the mathematics of waves and linear algebra in general, not quantum mechanics in particular. While the situation is often portrayed as particles being in multiple states all at once (such as being in two positions at the same time), it’s more accurate to say that the particle does not have a specific state at all. There’s just the wave function with all the probabilities of future quantum states. As soon as you perform a measurement, the particle snaps out of its wave function full of possibilities into a single possibility. In other words, what you see is not what it was. What you observe is just a sliver of its total prior existence. How this happens, nobody knows. It’s called the measurement problem. There’s a Nobel Prize waiting for you.

## 5: Schrödinger equation

Wave functions obey the Schrödinger equation. You could say that what Newton’s second law is for objects in our everyday world, is what the Schrödinger equation is for the subatomic world. It gives us the ability to predict how the wave function evolves in time. This is a completely classical equation; it is 100% deterministic. Where the wave function captures a range of probabilities, the Schrödinger equation tells us how this range of probabilities changes over time perfectly predictably so. In other words, it doesn’t predict the exact state of a particle once measured, but it does accurately predict the probability-value of an exact state once measured at any given time.

## 6: Uncertainty

There is a fundamental informational trade-off between certain possible states such as between position and momentum, energy and time, and time and frequency. The origin for this does not lie in quantum mechanics. It’s due to the way they are related to each other. Mathematically, these variables are called Fourier transform pairs or conjugate variables. To calculate one from the other, you have to execute a mathematical procedure called a Fourier transform. The trade-off is that Fourier transforming a variable whose range of possible values is smaller leads to the other variable having a larger range of possible values. And if a range of possible values becomes larger, then the exact outcome of measurement is less certain (the probability of a specific state after measurement becomes more uncertain). Heisenberg showed that this uncertainty principle also applies to the wave function in quantum mechanics, hence, there the principle is called Heisenberg’s uncertainty principle.

## 7: Certainty

That same principle predicts that, while very valid at the scale of subatomic particles, this uncertainty becomes utterly meaningless at our large-scale world of everyday objects. A bowling ball whose range of possible positions is very limited (locked in a very tight enclosure with little to no leeway), will never portray any uncertainty values pertaining to its motion (momentum), for instance. By Heisenberg’s uncertainty principle, upon measurement, it might show to have the speed of $3.283 \times 10^{-35} \text{ m/s}.$ This means that after 965.9 billion years it will have travelled the distance of the width of a proton. So, no, uncertainty effects play no role in our everyday world, unless you are doing experiments with a running time of seventy times the age of our current Universe. In that case, you will have to deal with the uncertainty of the width of a proton1. We do note that extraordinarily sensitive larger-scale equipment such as the the mirrors at the LIGO and Virgo experiments are capable of measuring quantum effects, however, this isn’t really unexpected nor is it the same as saying a human body is in a quantum superposition. Measuring quantum effects is one thing, brains supposedly being in two places on Earth (‘based on principles from quantum mechanics’) is a whole other thing.

## 8: Quantum entanglement

When two or more particles can only be described by one wave function – not as separate wave functions – those particles are said to be quantum entangled, either partly or completely. A measurement performed on one particle immediately determines the measurement outcome on the other entangled particle, irrespective of the spatial distance between them. This is why this phenomenon is said to be non-local. How this happens, is unknown. This effect dissipates to zero when entangled particles interact with yet other particles. At the large scale of our everyday world, the number of particles inside an object to be interacted with is so great, quantum entanglement completely fades away. In very special conditions, however, such as in our labs, entanglement can be sustained for quite some time.

## 9: Quantum Field Theory

Over the years, the mathematical and physical theory of (quantum) wave mechanics has been extended to describe quantum fields as the fundamental building blocks of our Universe. The Universe is made of quantum fields. The most complete description of fields are wave functions. This is called Quantum Field Theory (QFT). The most successful version of QFT is called the Standard Model of quantum physics. ‘Particles’ are here some kind of disturbance in their field: an electron is a disturbance in the electron field. The particle’s description is here part of the wave function of its entire field. The challenge is now to extend this quantum field theory into its next form, encapsulating something called quantum gravity. What we don’t know yet, for example, is how to have space and time in extreme regions such as black holes, naturally appear out of a quantum theory.

## 10: Applications

While the famous physicist and Nobel Prize winner Richard Feynman is known for having said, ‘I think I can safely say that nobody understands quantum mechanics’, this is sometimes incorrectly taken to be a reason to state that, therefore, physicists don’t know what they’re talking about. Feynman alluded to the fact that there is much we don’t know about the foundations of quantum mechanics. There is still much employment in solving hard problems such as quantum gravity, the measurement problem, the strong CP problem, the interpretation of quantum mechanics, non-locality, and so forth.

On the other hand, we now have WiFi, internet, touchscreens, lasers, MRI scanners, LEDs, flash memory, solid state disks, the old crunchy hard disks, transistors, and CPUs or integrated chips (ICs) in general.

I think I can safely say that the fact that you’ve plucked this article out of the air to have it displayed on your (touch)screen is at least an indication of the level at which ‘nobody understands quantum mechanics’.

Nevertheless, we’re far from done. There is still much to discover in this Universe with a bit of maths and physics.

1. When people state or think that everyday objects (our bodies, brains, tennis balls, animals) can exhibit quantum effects such as being at multiple places at the same time, I suspect this is because they have no well-defined idea how small subatomic particles really are and no inkling as to how large the everyday world is in those terms. Also, they didn’t do the calculations.[]