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# QFT를 이용한 전기유압 하이브리드 부하 시뮬레이터의 능동 힘제어
• Yoon, Joo-Hyeon (Department of Mechanical and Automotive Engineering, Univ. of Ulsan) ;
• Ahn, Kyoung-Kwan (School of Mechanical and Automotive Engineering, Univ. of Ulsan) ;
• Truong, Dinh Quang (Department of Mechanical and Automotive Engineering, Univ. of Ulsan) ;
• Jo, Woo-Geun (Department of Mechanical and Automotive Engineering, Univ. of Ulsan)
• 윤주현 (울산대학교 기계자동차공학과) ;
• 안경관 (울산대학교 기계자동차공학부) ;
• 딩쾅청 (울산대학교 기계자동차공학과) ;
• 조우근 (울산대학교 기계자동차공학과)
• Published : 2009.02.25
#### Abstract
Today, reduction of $CO_2$ exhaustion gas for global-warming prevention becomes important issues in all industrial fields. Hydraulic systems have been widely used in industrial applications due to high power density and so on. However hydraulic pump is always being operated by engine or electric motor in the conventional hydraulic system. Therefore most of the conventional hydraulic system is not efficient system. Recently, an electro-hydraulic hybrid system, which combines electric and hydraulic technology in a compact unit, can be adapted to a wide variety of force, speed and torque requirements. In the electro-hydraulic hybrid system, hydraulic pump is operated by electric motor only when hydraulic power is needed. Therefore the electro-hydraulic system can reduce the energy consumption drastically when compared to the conventional hydraulic systems. This paper presents a new kind of hydraulic load simulator which is composed of electro-hydraulic hybrid system. Disturbances in the real working condition make the control performance decrease or go bad. QFT controller is designed to eliminate or reduce the disturbance and improve the control performance of the electro-hydraulic load simulator. Experimental results show that the proposed controller is verified to apply for electro-hydraulic hybrid system with varied external disturbances.
#### References
1. Rahmfeld, R. and Ivantysynova, M., 'Displacement Controlled Linear Actuator with Differential Cylinder-A Way to Save Primary Energy in Mobile Machines,' Proc. of the 5th Int. Conf. on Fluid Power Transmission and Control, pp. 316-322, 2001
2. Yao, B., Bu, F., Reedy, J. and Chiu, G., 'Adaptive Robust Motion Control of Single-Rod Hydraulic Actuators: Theory and Experiments,' IEEE/ASME Trans. on Mechatronics, pp. 79-91, 2000 https://doi.org/10.1109/3516.828592
3. Grabbel, J., 'On the control of joint integrated servo actuators for mobile handling and robotic applications,' Proc. of the 1st FPNI- PhD Symp., pp. 449-465, 2000
4. Niksefat, N. and Sepehri, N., 'Robust force controller design for a hydraulic actuator based on experimental input-output data,' Proc. of the American Control Conf., pp. 3718-3722, 1999
5. Thompson, D. F. and Kremer, G. G., 'Quantitative feedback design for a variable displacement hydraulic vane pump,' Proc. of the American Control Conf., pp. 1061-1065, 1997
6. D'Azzo, J. and Houpis, C. H., 'Linear Control System Analysis and Design,' McGraw-Hill, 1998
7. Horowitz, I. M., 'Survey of Quantitative Feedback Theory (QFT),' International Journal of Control, Vol. 53, No. 2, pp. 255-291, 1991 https://doi.org/10.1080/00207179108953619
8. Ahn, K. K. and Chau, N. H. T., 'Design of a robust force controller for the new mini motion package using quantitative feedback theory,' Mechatronics, Vol. 17, No. 10, pp. 542-550, 2007 https://doi.org/10.1016/j.mechatronics.2007.07.011
9. Wu, S. F., Grimble, M. and Breslin, S. G., 'Introduction to Quantitative Feedback Theory for Lateral Robust Flight Control Systems Design,' Control Engineering Practice, Vol. 6, No. 7, pp. 805- 828, 1998 https://doi.org/10.1016/S0967-0661(98)00053-7
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# Symbol on a fixed horizontal position in the margin
I'm constructing an exercise sheet and would like to mark the thougher exercises.
Based on Placing symbol in margin I'm placing an symbol in the left margin. It works pretty well, but when I enumerate in more levels the horizontal position of the image shifts.
Is there a way to fix the horizontal position at like 2 cm from the page bound?
### MWE
\documentclass[a4paper,oneside, 11pt]{article}
\usepackage{tabto} %%symbol in left margin
\usepackage{manfnt}
\def\warningsymbol{\protect\marginsymbolhelper}
\def\marginsymbolhelper{\tabto*{-2cm} {\dbend} \tabto*{\TabPrevPos}}
\begin{document}
\begin{enumerate}
\item \warningsymbol Exercise 1
\item Exercise 2
\begin{enumerate}
\item \warningsymbol Subexercise 1
\end{enumerate}
\end{enumerate}
\end{document}
If you aren't using \marginpar for anything else... You can adjust the distance using \marginparsep.
\documentclass[a4paper,oneside, 11pt]{article}
\usepackage{manfnt}
\reversemarginpar
\newcommand{\marginsymbol}{\marginpar{\hfill\dbend}}
\usepackage{showframe}% debugging tool
\begin{document}
\begin{enumerate}
\item Exercise 1 \marginsymbol
\item Exercise 2
\begin{enumerate}
\item Subexercise 1 \marginsymbol
\end{enumerate}
\end{enumerate}
\end{document}
• I had to add \leavevmode in front of the \marginpar, but it works like a charm! Thanks! – dietervdf Jul 25 '17 at 9:51
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# All
## Observation-based symmetry breaking measures. (arXiv:1711.07347v1 [quant-ph])
Symmetry is one of the most general and useful concepts in physics. A theory
or a system that has a symmetry is fundamentally constrained by it. The same
constraints do not apply when the symmetry is broken. The quantitative
determination of "how much a system breaks a symmetry" allows to reach beyond
this binary situation and is a necessary step towards the quantitative
connection between symmetry breaking and its effects. We introduce measures of
symmetry breaking for a system interacting with external fields (particles).
## The Two-fold Role of Observables in Classical and Quantum Kinematics. (arXiv:1711.06914v1 [physics.hist-ph])
Observables have a dual nature in both classical and quantum kinematics: they
are at the same time \emph{quantities}, allowing to separate states by means of
their numerical values, and \emph{generators of transformations}, establishing
relations between different states. In this work, we show how this two-fold
role of observables constitutes a key feature in the conceptual analysis of
classical and quantum kinematics, shedding a new light on the distinguishing
## Quantum nonlinear dynamics: From noise-induced amplification to strong nonlinearity. (arXiv:1711.07376v1 [quant-ph])
We propose a noisy quantum analogue of the well known Stuart--Landau equation
for weakly nonlinear oscillators. Surprisingly, we find the oscillator's
amplitude to be amplified by the very same noise responsible for its stochastic
dynamics. This has interesting implications for the theory of linear amplifiers
and the so-called quantum van der Pol model used in quantum synchronisation. We
then go beyond the weakly nonlinear regime and obtain an exact quantum analogue
## Coherent State Mapping Ring-Polymer Molecular Dynamics for Non-Adiabatic quantum propagations. (arXiv:1706.08403v3 [physics.chem-ph] UPDATED)
We introduce the coherent state mapping ring-polymer molecular dynamics
(CS-RPMD), a new method that accurately describes electronic non-adiabatic
dynamics with explicit nuclear quantization. This new approach is derived by
using coherent state mapping representation for the electronic degrees of
freedom (DOF) and the ring-polymer path-integral representation for the nuclear
DOF. CS-RPMD Hamiltonian does not contain any inter-bead coupling term in the
state-dependent potential, which is a key feature that ensures correct
## Construction of State-independent Proofs for Quantum Contextuality. (arXiv:1707.05626v2 [quant-ph] UPDATED)
Since the enlightening proofs of quantum contextuality first established by
Kochen and Specker, and also by Bell, various simplified proofs have been
constructed to exclude the non-contextual hidden variable theory of our nature
at the microscopic scale. The conflict between the non-contextual hidden
variable theory and quantum mechanics is commonly revealed by Kochen-Specker
(KS) sets of yes-no tests, represented by projectors (or rays), via either
logical contradictions or noncontextuality inequalities in a
## Frequency-dependent current noise in quantum heat transfer with full counting statistics. (arXiv:1708.05537v2 [cond-mat.stat-mech] UPDATED)
To investigate frequency-dependent current noise (FDCN) in open quantum
systems at steady states, we present a theory which combines Markovian quantum
master equations with a finite time full counting statistics. Our formulation
of the FDCN generalizes previous zero-frequency expressions and can be viewed
as an application of MacDonald's formula for electron transport to heat
transfer. As a demonstration, we consider the paradigmatic example of quantum
## Resonance enhancement of two photon absorption by magnetically trapped atoms in strong rf-fields. (arXiv:1711.06971v1 [physics.atom-ph])
Applying a many mode Floquet formalism for magnetically trapped atoms
interacting with a polychromatic rf-field, we predict a large two photon
transition probability in the atomic system of cold $^{87}Rb$ atoms. The
physical origin of this enormous increase in the two photon transition
probability is due to the formation of avoided crossings between eigen-energy
levels originating from different Floquet sub-manifolds and redistribution of
population in the resonant intermediate levels to give rise to the resonance
## Exploring adiabatic quantum dynamics of the Dicke model in a trapped ion quantum simulator. (arXiv:1711.07392v1 [quant-ph])
We use a self-assembled two-dimensional Coulomb crystal of $\sim 70$ ions in
the presence of an external transverse field to engineer a quantum simulator of
the Dicke Hamiltonian. This Hamiltonian has spin and bosonic degrees of freedom
which are encoded by two hyperfine states in each ion and the center of mass
motional mode of the crystal, respectively. The Dicke model features a quantum
critical point separating two distinct phases: the superradiant (ferromagnetic)
## Quantized Ballistic Transport of Electrons and Electron Pairs in LaAlO3/SrTiO3 Nanowires. (arXiv:1611.05127v3 [cond-mat.mes-hall] UPDATED)
SrTiO3-based heterointerfaces support quasi-two-dimensional (2D) electron
systems that are analogous to III-V semiconductor heterostructures, but also
possess superconducting, magnetic, spintronic, ferroelectric and ferroelastic
degrees of freedom. Despite these rich properties, the relatively low
mobilities of 2D complex-oxide interfaces appear to preclude ballistic
transport in 1D. Here we show that nearly ideal 1D electron waveguides
exhibiting quantized ballistic transport of electrons and (non-superconducting)
## Asymmetrical interaction induced real spectra and exceptional points in a non-Hermitian Hamiltonian. (arXiv:1711.06982v1 [quant-ph])
Non-Hermitian systems with parity-time symmetry have been developed rapidly
and hold great promise for future applications. Unlike most existing works
considering the symmetry of the free energy terms (e.g., gain-loss system), in
this paper, we report that a realizable non-Hermitian interaction between two
quantum resonances can also have a real spectrum after the exceptional point.
That phenomenon is similar with that in the gain-loss system so that the
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# The time period of a simple harmonic motion is 8s. At t=0 , it is at its equilibrium position. The ratio of distances traversed by it in the first and second seconds is :
$\begin {array} {1 1} (1)\;\large\frac{1}{\sqrt 2} & \quad (2)\;\large\frac{1}{\sqrt 2-1} \\ (3)\;\large\frac{1}{\sqrt 3} & \quad (4)\;\large\frac{1}{ 2} \end {array}$
## 1 Answer
(2) $\large\frac{1}{\sqrt 2-1}$
answered Nov 7, 2013 by
1 answer
1 answer
1 answer
1 answer
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# logistic regression cost function octave
Logistic regression does not have such constraints since θ is allowed to take any real value. The Overflow Blog Podcast 286: If you could fix any software, what would you change? In this blog, we will discuss the basic concepts of Logistic Regression and what kind of problems can it help us to solve. sudo yum install octave-forge, Files included in this exercise can be downloaded here ⇒ : does not have such constraints since θ is allowed to take any real value. Function Reference: LinearRegression Function File: LinearRegression ( F , y ) The cost function for this changes as we will be using a sigmoid function to convert our prediction into a probability between 0 and 1. Octave/MATLABâs fminunc is an optimization solver that finds the minimum using the optimal parameters of θ. plotData.m - Function to display the dataset vector θ. proceed to report the training accuracy of your classifier by computing the In this post, I'm going to walk you through an elementary single-variable linear regression with Octave (an open-source Matlab alternative).. 925 claps. You will pass to fminunc the following inputs: In my previous blog, I talked about the math behind linear regression, namely gradient descent and the cost function. Embed. We use optional third-party analytics cookies to understand how you use GitHub.com so we can build better products. element (for j = 0, 1, . For this exercise, you will use logistic regression and neural networks to recognize handwritten digits (from 0 to 9).. and the framework code in ex2.m will guide you through the exercise. Some of the examples of classification problems are Email spam or not spam, Online transactions Fraud or not Fraud, Tumor Malignant or Benign. egonSchiele / regression.m. function J = computeCost (X, y, theta) % COMPUTECOST Compute cost for linear regression % J = COMPUTECOST(X, y, theta) computes the cost of using theta as the % parameter for linear regression to fit the data points in X and y % Initialize some useful values m = length(y); % number of training examples % You need to return the following variables correctly J = 0; (Part:2) In this article, you will learn the maths and theory behind Gradient Descent. Cost Function. Internally this line is a result of the parameters $$\theta_0$$ and $$\theta_1$$. It has a text interface and an experimental graphical one. g(z) > 0.5 when z>=0. Linear regression is an approach for modeling the relationship between a scalar dependent variable y and one or more explanatory variables (or independent variables) denoted X. to the parameters. warmUpExercise.m - Simple example function in Octave/MATLAB [?] Returns the cost in J and the gradient in grad Suppose you are the product manager of the factory and you have the test results for some microchips on two different tests. This time, instead of taking gradient descent steps, you will use an Octave built-in function called fminunc. . costFunction.m - Logistic Regression Cost Function [?] Throughout the exercise, you will be using the script ex1.m. Files you will need to modify as part of this assignment : plotData.m - Function to plot 2D classification data, costFunction.m - Logistic Regression Cost Function, predict.m - Logistic Regression Prediction Function, costFunctionReg.m - Regularized Logistic Regression Cost. We also provide our implementation below so you can Star 1 Fork 1 Star Code Revisions 2 Stars 1 Forks 1. Concretely, you are going to use fminunc to find the best parameters θ for the logistic regression cost function, given a fixed dataset (of X and y values). th The graph generated is not convex. Unlike linear regression which outputs continuous number values, logistic regression transforms its output using the logistic sigmoid function to return a probability value which can then be mapped to two or more discrete classes. Since we want the sum of the products, we can use a vector multiplication. Samrat Kar. training data, resulting in a figure similar to Figure 2. Cost Function. and the gradient of the cost is a vector of the same length as θ where the j to the parameters. The data is from the famous Machine Learning Coursera Course by Andrew Ng. Fitting Linear Regression. See the equation on ex2.pdf - Page 4. If you're already familiar with the basics of linear algebra operations with Octave, you can move on to the linear regression tutorial. 0.203. documentation. You will now complete the code in plotData so that it displays a figure Subtract the right-side term from the left-side term Which are the intermediary steps? The sigmoid function is defined as: $$g(z) = \frac{1}{1 + e^{-x}}$$. This data bundle contains two sets of data, one for linear regression and the other for logistic regression. you to look at the code in plotDecisionBoundary.m to see how to plot such Before I begin, below are a few reminders on how to plot charts using Octave. Given we have data in a csv with 2 columns for data X and Y. In the chapter on Logistic Regression, the cost function is this: Then, it is derivated here: I tried getting the derivative of the cost function but I got something completely different. Follow. In this part of the exercise, you will implement regularized logistic regression to predict whether microchips from a fabrication plant passes quality assurance (QA).During QA, each microchip goes through various tests to ensure it is functioning correctly. Each of these is a vector of size (m x 1). data and display it on a 2-dimensional plot by calling the function plotData. You can get a one-line function for sigmoid(z) if you use only element-wise operators. We also encourage regression logistic gradient-descent derivative. Concretely, you are going to use fminunc to nd the best parameters for the logistic regression cost function, given a xed dataset (of X and y values). Then In logistic regression, we create a decision boundary. Octave’s fminunc is an optimization solver that nds the minimum of an unconstrained2 function. Answered: João Marlon Souto Ferraz on 14 Sep 2020 Hi, I am trying to compute cost function . predict whether a student gets admitted into a university. This For logistic regression, the [texi]\mathrm{Cost}[texi] function is defined as: [tex] \mathrm{Cost}(h_\theta(x),y) = \begin{cases} -\log(h_\theta(x)) & \text{if y = 1} \\ -\log(1-h_\theta(x)) & \text{if y = 0} \end{cases} [tex] The [texi]i[texi] indexes have been removed for clarity. I am using the following code: function J = computeCost(X, y, theta) %COMPUTECOST Compute cost for linear regression % J = COMPUTECOST(X, y, theta) computes the cost … to the parameters. gradientDescent.m - Function to run gradient descent. Linear Regression: Hypothesis Function, Cost Function and Gradient Descent. If you're new to Octave, I'd recommend getting started by going through the linear algebra tutorial first.. You will pass to fminunc the following inputs: To specify the actual function we are minimizing, we use a âshort-handâ To help you get more familiar with plotting, we have left plotData.m If y = 1 If h(x) = 0 & y = 1, costs infinite; If h(x) = 1 & y = 1 , costs = 0 If y = 0 If h(x) = 0 & y = 0, costs = 0; If h(x) = 1 & y = 0, costs infinite 2b. to install Octave for windows. This In this part of the exercise, you will build a logistic regression model to Shuryu Kisuke Shuryu Kisuke. Linear regression - implementation (cost function) A cost function lets us figure out how to fit the best straight line to our dataChoosing values for θ i (parameters) Different values give you different functions; If θ 0 is 1.5 and θ 1 is 0 then we get straight line parallel with X … If we try to use the cost function of the linear regression in ‘Logistic Regression’ then it would be of no use as it would end up being a non-convex function with many local minimums, in which it would be very difficult to minimize the cost value and find the global minimum. This is the unregularized cost. Logistic regression is a classification algorithm used to assign observations to a discrete set of classes. The lab exercises in that course are in Octave/Matlab. 6 $\begingroup$ This will compute the sigmoid of a scalar, vector or matrix. like Figure 1, where the axes are the two exam scores, and the positive and What would you like to do? Skip to content. 0 ⋮ Vote. If you want an implementation that handles this case, then you'll need to modify the code a bit. In the previous assignment, you found the optimal parameters of a linear regression Learn more. % J = LRCOSTFUNCTION (theta, X, y, lambda) computes the cost of using % theta as the parameter for regularized logistic regression and the % gradient of the cost w.r.t. This is from Programming assignment 1 from the famous Machine Learning course by Andrew Ng. Follow. share | improve this question | follow | asked Dec 19 '17 at 16:43. This allows fminunc to Machine Learning And … But it turns out that if we use this particular cost function, this would be a non-convex function of the parameter's data. yourself, or set a learning rate like you did for gradient descent. what each of its commands is doing by consulting the Octave/MATLAB Function Reference: LinearRegression Function File: LinearRegression ( F , y ) How do we choose parameters? Suppose you are the product manager of the factory and you have the test results for some microchips on two different tests. If you're new to Octave, I'd recommend getting started by going through the linear algebra tutorial first.. values). Browse other questions tagged matlab vectorization logistic-regression or ask your own question. In this exercise, you will implement regularized linear regression and regularized logistic regression. For example, we might use logistic regression to classify an email as spam or not spam. Example of a linear curve: z = theta_0 + theta_1 x_1 + theta_2 x_2. Decision boundaries determined by parametrized curves. Cost Function of Linear Regression. Now you will implement the cost function and gradient for logistic regression. Suppose that you are the administrator of a university department and 0 ⋮ Vote. Simplified Cost Function & Gradient Descent. % … All gists Back to GitHub Sign in Sign up Sign in Sign up {{ message }} Instantly share code, notes, and snippets. The cost function for this changes as we will be using a sigmoid function to convert our prediction into a probability between 0 and 1. for the logistic regression cost function, given a fixed dataset (of X and y Octave-Forge is a collection of packages providing extra functionality for GNU Octave. egonSchiele / regression.m. This training function uses the minimize function from scipy to optimize the cost function. a boundary using the θ values. Use the element-wise division operator ./. Logistic regression cost function. You'll need to transpose and swap the product terms so the result is (m x n)' times (m x 1) giving you a (n x 1) result. Logistic regression is a classification algorithm used to assign observations to a discrete set of classes. My code goes as follows: I am using the vectorized implementation of the equation. Follow. A function that, when given the training set and a particular θ, computes Function List: » Octave core » by package » alphabetical; C++ API: [theta, beta, dev, dl, d2l, p] = logistic_regression (y, x, print, theta, beta) Perform ordinal logistic regression. Star 1 Fork 1 Star Code Revisions 2 Stars 1 Forks 1. In ... we've included a Matlab/Octave helper function named 'map_feature' that maps the original inputs to the feature vector. example, you have the applicantâs scores on two exams and the admissions For logistic regression, you want to optimize the cost function J( ) with parameters . Writing math in an octave is extremely simple, and so it is excellent for… MATLAB built-in function called fminunc. In words this is the cost the algorithm pays if it predicts a value [texi]h_\theta(x)[texi] while the actual … Writing math in an octave is extremely simple, and so it is excellent for… How do we choose parameters? In logistic regression, we create a decision boundary. In this part of the exercise, you will implement regularized logistic regression to predict whether microchips from a fabrication plant passes quality assurance (QA).During QA, each microchip goes through various tests to ensure it is functioning correctly. results on two exams. This is logistic regression, so the hypothesis is the sigmoid of the product of X and theta. function [ J, grad] = costFunction (theta, X, y) %COSTFUNCTION Compute cost and gradient for logistic regression % J = COSTFUNCTION (theta, X, y) computes the cost of using theta as the % parameter for logistic regression and the gradient of the cost But here, we're interested in logistic regression. Samrat Kar. On Fedora, you can use: For more information, see our Privacy Statement. We use essential cookies to perform essential website functions, e.g. Cost Function. You do not need to modify either of them. Cost Function. Throughout the exercise, you will be using the script ex1.m. part, your task is to complete the code in predict.m. How is the derivative obtained? Logistic regression transforms its output using the logistic sigmoi… You should see that the cost is about 0.693. These may or may not be a problem in actual use. The left-side term is the vector product of X and (h - y), scaled by 1/m. Download Unlike linear regression which outputs continuous number values, logistic regression transforms its output using the logistic sigmoid function to return a probability value which can then be mapped to two or more discrete classes. Simplified Cost Function & Gradient Descent. Don't use log10(). The aim of the linear regression is to find a line similar to the blue line in the plot above that fits the given set of training example best. costFunctionReg.m - Regularized Logistic Regression Cost ? Cost Function; Linear Regression; Logistic Regression; 925 claps. Octave-Forge is a collection of packages providing extra functionality for GNU Octave. Constraints in optimization often refer to constraints on the parameters, for example, You will pass to fminunc the following inputs: Concretely, you are going to use fminunc to find the best parameters θ Octave is a free, open-source application available for many platforms. GitHub Gist: instantly share code, notes, and snippets. [?] Follow 626 views (last 30 days) Muhammad Kundi on 22 Jun 2019. function [J, grad] = linearRegCostFunction (X, y, theta, lambda) % LINEARREGCOSTFUNCTION Compute cost and gradient for regularized linear % regression with multiple variables % [J, grad] = LINEARREGCOSTFUNCTION(X, y, theta, lambda) computes the % cost of using theta as the parameter for linear regression to fit the % data points in X and y. To begin, download ex5Data.zip and extract the files from the zip file. of an unconstrained This function works for a single training example as well as for an entire training. Linear regression in Octave. Linear Regression: Hypothesis Function, Cost Function and Gradient Descent. I am using the following code: function J = computeCost(X, y, theta) %COMPUTECOST Compute cost for linear regression % J = COMPUTECOST(X, y, theta) computes the cost … computeCost.m - Function to compute the cost of linear regression. In the first part of ex2.m, the code will load the Try these commands in your workspace console, and study how they work: Inside your predict.m script, you will need to assign the results of this sort of logical comparison to the 'p' variable. The minimization will be performed by a gradient descent algorithm, whose task is to parse the cost function output until it finds the lowest minimum point. This exercise will show you how the methods you’ve learned can be used for this classification task. Example of a linear curve: z = theta_0 + theta_1 x_1 + theta_2 x_2. copy it or refer to it. You have historical data from previous applicants This is all This final θ value will then be used to plot the decision boundary on the of 45 and an Exam 2 score of 85, you should expect to see an admission We can plot charts using Octaves plot function. You should see that the cost is about For each training Skip to content. regression logistic gradient-descent derivative. particular student will be admitted. Octave/MATLAB’s fminunc is an optimization solver that nds the min-imum of an unconstrained2 function. Learn more. If we could minimize this cost function that is plugged into J here, that will work okay. Logistic/Sigmoid function: g(z) = 1/(1+e^-z). Here's what I mean by non-convex. The below code would load the data present in your desktop to the octave memory x=load('ex4x.dat'); y=load('ex4y.dat'); %2. is to see how well the learned model predicts on our training set. Automated handwritten digit recognition is widely used today - from recognizing zip codes (postal codes) on mail envelopes to recognizing amounts written on bank checks.. Follow 626 views (last 30 days) Muhammad Kundi on 22 Jun 2019. and the gradient. predict.m - Logistic Regression Prediction Function [?] will produce â1â or â0â predictions given a dataset and a learned parameter The "simplification" is not equivalent to the piecewise cost function. Function File: [theta, beta, dev, dl, d2l, p] = logistic_regression (y, x, print, theta, beta) Perform ordinal logistic regression. (Part:2) In this article, you will learn the maths and theory behind Gradient Descent. In the chapter on Logistic Regression, the cost function is this: Then, it is derivated here: I tried getting the derivative of the cost function but I got something completely different. However when implementing the logistic regression using gradient descent I face certain issue. Written by. Last active Aug 31, 2018. Logistic Regression Model 2a. it terminates. Scale the result by 1/m. You can always update your selection by clicking Cookie Preferences at the bottom of the page. Now this cost function worked fine for linear regression. The GNU Public License, which means that it is excellent for… cost function ; linear regression and neural to! This function works for a single training example, you will learn the maths and theory behind gradient steps... 1+E^-Z ) a result of the parameters \ ( \theta_0\ ) and \ ( \theta_1\ ) on the... Copy it or refer to it $\lambda$ is set to zero: you only needed to provide function... Use p = followed by a logical comparison inside a set of classes Revisions 2 Stars Forks. 14 Sep 2020 Hi, I am trying to implement the cost of linear algebra tutorial first collection of providing! Am going to use the gradient in grad Introduction ¶ vector or matrix simple example in! ( z ) = 1/ ( 1+e^-z ) term is the vector product also includes the required.! Course by Andrew Ng Forks 1 operations with Octave, I 'd recommend getting started by going through the algebra! ( 1 ) can always update your logistic regression cost function octave by clicking Cookie Preferences at the bottom the. Below so you can use p = followed by a logical comparison inside a of... As a training set for logistic regression predicts the probability of admission based the from... An example of how to plot such a boundary using the optimal parameters of θ good idea, want function! Results for some microchips on two different datasets Ferraz on 14 Sep 2020,. Stars 1 Forks 1 predict whether a student gets admitted into a university ) logistic regression is the... Of parenthesis bottom of the product of x and y that course are in.... Call your costFunction function using the initial values of the factory and you have data! Each of its commands is doing by consulting the Octave/Matlab documentation 're interested in logistic,... J and the framework code in plotDecisionBoundary.m to see how to plot such a using! S fminunc is an optimization solver that finds the minimum of an unconstrained2 function sure learn. ( 1 ) zip File unconstrained2 function and theory behind gradient descent for linear regression and the gradient need! And ex2 reg.m already ) developers working together to host and review code, projects... Bundle contains two sets of data, one for linear regression, you want optimize. Manager of the cost is about 0.203 the script ex1.m are trying to the. And theory behind gradient descent steps, you will pass to fminunc the following inputs: initial! Parameter in the plot above and ( h - y ) logistic is. The following inputs: however when implementing the logistic regression does not have such constraints since θ is to! Indicates files you will be using the scripts ex2.m and ex2 reg.m the Machine Learning Coursera course by Andrew.. 2 Stars 1 Forks 1 not a good idea, want hypothesis function 0 < = 1 this snippet... 925 claps be admitted is an optimization solver that nds the minimum of an unconstrained2 function to the. Help us to solve regression to classify an email as spam or not spam will implement regularized regression... Applicants that you can use as a training set for logistic regression ; 925.. Elementary single-variable linear regression training function uses the minimize function from scipy optimize! Second equation ) > 0.5 when z > =0 have data in a vectorized manner sigmoid ( z ) 1/! Specifically, we set the GradObj option to 400, so the hypothesis is the vector product also includes required! And ( h - y ), scaled by 1/m Jun 2019 tutorial first by! Student gets admitted into a university build software together on, which calls your costFunction function using the vectorized of! Function to compute the sigmoid function ( z ) = 1/ ( 1+e^-z.. Used with fminunc build software together and the cost function octave/matlabâs fminunc is optimization! Those two exams logistic regression, you want to optimize you 'll need accomplish. To use the gradient when minimizing the function use a vector of size ( m x 1 to... Walk you through the linear algebra operations with Octave, I 'm going use. Predicts the probability of the outcome being true a training set for logistic regression, we might use logistic and! To 9 ) star 1 Fork 1 star code Revisions 2 Stars 1 Forks.! That maps the original inputs to the linear regression tutorial we first defined the options be. Calculating the cost function, with argument t, which calls your costFunction the. A logical comparison inside a set of classes look at logistic regression cost function octave bottom of the fminunc ). Y ), scaled by 1/m here 's an example of a scalar, vector or matrix code,... One for linear regression: hypothesis function, cost function example as well as for an training! Two exams what each of these is a classification algorithm used to observations. Exercise, you want an implementation that handles this case, then you 'll need finish... Octave, you will implement logistic regression predicts the probability of the parameter ' '! Regression tutorial help you get more familiar with the basics of linear.. Interface and an experimental logistic regression cost function octave one to copy our example, we use optional third-party analytics to! Octave, I am trying to compute the sigmoid of a linear curve: z = theta_0 + theta_1 +. Them better, e.g an optimization solver that nds the minimum of an unconstrained function particular student be! We also provide our implementation below so you can use the model to predict whether a student admitted... Which youâll need to finish to complete the course function called fminunc of... The maths and theory behind gradient descent and the gradient in grad Browse other questions tagged matlab vectorization logistic-regression ask! A collection of packages providing extra functionality for GNU Octave set theta ( 1 ) example! = 1 visit and how many clicks you need to finish to complete the. 'S an example of a scalar, vector or matrix Octave built-in function called fminunc and review,. Idea, want hypothesis function 0 < = h_theta ( x ) < = (! Data in a vectorized manner part of the outcome being true Souto Ferraz on 14 2020. Of logistic regression to classify an email as spam or not spam and! Regression or univariate linear regression ; 925 claps found the optimal parameters of θ the outcome being true,... Not have such constraints since θ is allowed to take any real value ( )! Implement logistic regression, you have the applicantâs scores on two different.. The Overflow blog Podcast 286: if you have n't already ) empty you. Zip File complete throughout the exercise, you will learn the maths and theory behind gradient descent the... \Theta \$ in trainLinearReg should be computed ' z ' J = 0 '' condition entirely and... To begin, below are a few parenthesis, and build software.! We have left plotData.m empty so you can move on to the piecewise cost function ; regression... Minimising the cost function J ( ) function passes the submit grader before going any further only needed to a... Since θ is allowed to take any real value outline and the cost function of the factory and you the! Passes the submit grader before going any further always update your selection by clicking Cookie at! = h_theta ( x ) < = 1 the exercise, you want optimize. Two different datasets if you 're new to Octave, you will implement the regularized logistic,! Different datasets function uses the minimize function from scipy to optimize θ is allowed to take any real.. To take any real value ex5Data.zip and extract the files from the left-side term the! Use essential cookies to understand how you use GitHub.com so we can use vector. Y ) logistic regression ( ungraded ) exercise you wrote a cost function worked fine for linear regression.! Then set theta ( 1 ) to 0 ( if you use only element-wise operators one-line function for (... Parameters θ is not a good idea, want hypothesis function 0 < = 1 plot above before... At 16:43 good idea, want hypothesis function, this would be a problem actual... Talked about the math behind linear regression, you have n't already ) this code snippet, we discuss... Descent and the cost function ; linear regression, we set the MaxIter option 400... Output using the fminunc ( ) with parameters θ using the optimal values of the Step and... Learning course by Andrew Ng has a text interface and an experimental graphical.... Logical comparison inside a set of parenthesis is extremely simple, and snippets this. Variable is called simple linear regression costFunction using the optimal values of the parameters \ ( ). > =0 essential cookies to perform essential website functions, e.g clicks you need complete... Parenthesis, and snippets '' condition entirely, and so it is excellent for… cost function asked. If you could fix any software, what would you change creates a function of regression... And what kind of problems can it help us to wrap the costFunction for use with fminunc â1â or predictions. To 0 ( if you choose to copy our example, we might logistic... Browse other questions tagged matlab vectorization logistic-regression or ask your own question of logistic regression using... Views ( last 30 days ) Muhammad Kundi on 22 Jun 2019 is excellent for… cost.... If we use optional third-party analytics cookies to understand how you use so. Sep 2020 Hi, I would like to plot the cost function J ( )!
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Physics is a mathematics based subject. Approximately 1 lakh to 2 lakh students appear in the 12th exam every year. More Maha Board Hsc previous year exam papers … In the horizontal plane the ball is in uniform circular motion. The official answer key of GSEB HSC Physics Exam 2020 will be released in 1st week of April. The Article “ HSC Physics Paper solution 2020 ” will help students for the solution of HSC 2020 Physics of Maharashtra Board. The Class 12 Physics 2020 Question Paper will also be made available for download soon. Rajshahi, Dinajpur, Sylhet, Technical and Madrashah are the educational boards Beside the MCQ questions, Physics 1st paper MCQ Assam Higher Secondary Education Council is also broadly known as AHSEC. Today we are going to discuss about your Physics 1st paper, the subject which seems to be a nightmare to most of you. The above shows that the speed is proportional to the power of the motor and inversely proportional to the coefficient of friction. Check the download link for Physics Question Paper from the below section. have to know as much information as possible. In this test paper they can see Bangla,English, Economics, Physics and chemistry and other important subjects. 2020 HSC Physics Exam Paper Solutions. Scientists also found that the energy associated with gamma rays was not sufficient to eject the protons with the velocities observed, apparently violating conservation of energy. DOWNLOAD. In your 1st paper, you have to study vector, dynamics, energy, waves, theories of gases etc. We have published here all the important questions which have a Download Maharashtra Board HSC Model Papers in pdf format for Science and Commerce students All Subject. This gives the change in flux: (The change in flux is taken to be negative, since it is a decrease of flux. boards follow the same question pattern. It is a hard subject too. GSEB will be releasing the Physics answer key on the same day of the exam. Expressing the mass’s speed in terms of the motor’s angular velocity, $$v = r \omega$$ gives: $$v = \frac{P}{\mu mg} = \frac{\tau \omega}{\mu mg}$$. The opposing force causes the magnet to slow down, and then fall at constant speed as seen in the graph (the force must be balancing its weight). The current produces a magnetic field which interacts with the magnet to produce an opposing (upwards) force, according to Lenz’s Law. Students can check the answer key subject wise from the official website at www.gseb.org. Important Question Bank of Physics for Maharashtra HSC Board Exam 2020 We have created Important Question Bank which will help students in scoring good marks in HSC Board Exams. Always remember that, no matter how brilliant you are, if you It appears that you have disabled your Javascript. In fine, we wish you best luck for your exam. So, we started with a special suggestion that applies to the Dhaka board, Comilla board, Chittagong board, Sylhet board, Rajshahi board, Jessore board and all the education boards applicable to Dinajpur. When the Earth is travelling away from Jupiter, the light signal from the start of one orbit will take longer to reach Earth than the light signal from the start of the previous orbit, as it needs to travel a longer distance to reach Earth. In your exam you will only get 25 Some transitions are impossible and others occur with varying probability. We have organized the HSC 2020 physics suggestion just according to your needs. In order to bring the spacing back to its original value, we can decrease the separation between the slits $$d$$, causing $$\theta$$ to increase, or we can place the screen further away from the slits by increasing $$y$$, causing their linear separation to increase. In this case practice is a must for doing well in the exam. HSC Test Paper Download 2020 HSC examinee friends, HSC exam … Download Tamil Nadu HSC / 12 th public exam Question Paper and TN 12 th Model Question Papers … SSC English 1st paper question 2020 has been published by Official Result BD. For doing well in the MCQ you This question combines aspects of projectile and circular motion. HSC Candidates of our country follow this trend because this has happened before in the past. They worked hard doing research a lot on the subject and finally prepared a SPECIAL HSC Physics 1 st Paper suggestion 2020. DOWNLOAD DOWNLOAD . $$W = q\frac{V}{2d} \times \frac{3}{4}2d = \frac{3qV}{4}$$. correctly. This corresponds to their maximum velocity also being lower. The physics is one of the oldest academic disciplines, perhaps the oldest through the inclusion of astronomy. Bangla so that you will interpret according to your will. Maha board 12th Sample Question Paper. Download Maharashtra 12 th Question Papers and Maharashtra HSC Model Papers at pdf. The proton is composed of three quarks, two up and one down. HSC 2020 Physics Board Question Paper of Maharashtra held on 24th February 2019-20 students can refer this for the same. The Question paper of Maharashtra Board Exam HSC exam also follows the 12th Syllabus made by the Maharashtra State Board of Secondary and Higher Secondary Education. $$\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right) = 1.097\times 10^7 \times \left(\frac{1}{2^2} – \frac{1}{3^2}\right)$$ which gives $$\lambda = 656 \text{ nm}$$. Gujarat secondary and higher secondary education board, GSEB had conducted Previous Year 14 th March 2020 Question Paper of Board examination is physics (14-03-2020) in the time period of 3.00 to 6.30 pm. (a) Gamma rays were known to have zero electric charge, which would explain their lack of deflection in an electric field. Don’t worry. Watch Free Zener Breakdown Video – Physics Std 12th Science SYJC. memorizing things without understanding the basic. (b) Electrons in the Bohr model exist in one of a number of stable energy states. According to the graph, the half-life is about 100 years (time when the remaining mass is half of the starting value). This is physics So, you will get physics 2nd paper suggestion for all general education boards from here with hsc question 2020 physics 2nd paper. Keep these things in mind and go ahead. $$\lambda = \frac{\text{ln }2}{100} = 0.0069 \text{ years}^{-1}$$, Being part of a cluster suggests that they have approximately the same age. If the rocket is initially travelling horizontally at a constant speed of $$20 \text{ ms}^{-1}$$, there must have been zero net force on it. MAHA 12th Question Paper is very important for all students of HSC, as practising 12th Question Paper helps you understand the time required and your level of understanding of HSC. The Class 12 Physics 2020 Question Paper will also be made available for download soon. When it reaches the top of the cylinder, its potential energy reduces to $$U = 0.04 \times 9.8 \times 0.2 = 0.0784 \text{ J}$$, and its kinetic energy is $$K = 0.30576 \text{ j}$$, such that the total energy remains the same at $$0.38416 \text{ J}$$. If you are also one of them and looking forward to download a pdf file of physics 2nd paper book for hsc then you can have it from here. When the thrust is increased to $$90000 \text{ N}$$ there will be a net force upwards, and hence an acceleration upwards. Check out our the solutions below and see how you went! (a) The magnetic flux through the loop is $$\Phi = BA \text{cos} \theta$$. After that you can download HSC All Subjects Suggestion and Question. Again there will be mathematical problems to So Here we are GSEB HSC 12th Class All Subject Answer key at given below pdf link. Going through it will help students to self-evaluate their performance. The mass will experience friction given by $$f = \mu N$$. You will have to answer the questions from two sections in the board exam. Apr 16,2020 . Related Videos. If you look for all these things in different sites it will be just a huge waste of your valuable time with a few results. This does not result in a charge separation across P and Q, and does not produce an EMF between them. (a) Energy can be converted to mass according to Special Relativity. That is why time is very much important in this level. HSC Book PDF Guide new version is available . AHSEC was continuously regulating and conducting Higher Secondary Final Examination for … Maharashtra State Board HSC Science (General) 12th Board Exam. First of all let’s download HSC Routine. HSC 2020 will start from estimated April and will end on estimated May. Download Free Previous Years HSC Science Question Papers from 2013-2020. The capsule is also accelerating, so it is not an inertial frame of reference and cannot be analysed using Special Relativity. The MSBSHSE Class 12 Physics Exam took place on 24 February 2020. Maharashtra HSC Question Papers 2020-21 - Important Topics. Time Tables 23. Apart from this you can take the reference of these question papers of old syllabus as because the concepts are still not deleted in new syllabus of maharahtra board. Get Last Year Question Paper for 12th Board Exam and solved answers for practice in your board and university exams. Logged In : Login. Since both observers must agree on the speed of light, Y sees it take a shorter time than X to reach the sensor. Huygens proposed that light propagated via the formation of wavelets. (b) These experiments showed the existence of the neutron, which were the unknown radiation mentioned in the question. … DOWNLOAD DOWNLOAD . The official answer key of GSEB HSC Physics Exam 2020 will be released in 1st week of April. As a consequence, the comet must travel at a higher speed when it is close to the Sun and a lower speed when it is further away. The MSBSHSE Class 12 Physics Exam took place on 24 February 2020. Students who struggle to finish on time are advised to start solving these papers at the earliest so that they don’t lag behind. Watch Free Zener Breakdown Video – Physics … The quarks are held together by the strong nuclear force, which is mediated by gauge bosons known as gluons. The gravitational force on $$B$$ from $$A$$ must be equal to the centripetal force keeping it in circular motion. mathematical problems. Check out our the solutions below and see how you went! Question Papers 219. It is because our suggestion is prepared by a team of the most expert teachers who are from the best colleges of our country. The electron is a lepton, an elementary particle. Syllabus. Physics doesn’t require any narrative things. These two forces remain perpendicular to each other, so the magnitude of their sum remains constant. So Here we are GSEB HSC 12th Class All Subject Answer key at given below pdf link. (a) A gas discharge tube will produce a spectrum composed of lines at discrete wavelengths (which will depend on the energy levels of the gas in the tube). You have to follow the Its initial energy was $$0.38416 \text{ J}$$ meaning that $$0.38416 – 0.0032 = 0.38096 \text{ J}$$ were converted into the electrical energy of the induced current in the cylinder, and then to heat through the copper’s resistance. Physics 1st part exam could be held at the end of April, 2020. GSEB 12th Physics Exam 2020 was conducted at various exam centers across the state of Gujarat. (b) Model X assumes that a black body radiates equally at all frequencies. Physics Paper 2 Important Questions to refer before exam Click to read MAHA 12th Blueprint Question PDF 2020 Syllabus is Very Important for Many Students as it is Helpful in Their Exam Preparation. But we have prepared the answers of these questions from reliable source and reviewed them by some experts. The stars vary in colour as indicated by the x-axis. When the magnet reaches the bottom of the cylinder its potential energy is 0 and its kinetic energy is $$K = \frac{1}{2} mv^2 = 0.0032 \text{ J}$$. MCQ plays a vital role in the exam. So, get these questions and start practicing before it is too The Matrix 2020 HSC Maths Ext 2 Exam Paper Solutions are here! © 2020 Matrix Education. The total energy before and after the transition must be the same as stated by the Law of Conservation of Energy. Physics, Chemistry, Maths, Biology, English, Hindi and Marathi Subjects Watch Free Well of Death Video – Physics Std 12th Science SYJC. Question Bank Solutions 10942. Physics question HSC 2020 may leak. In this case If you keep it to yourself, then you don’t have to look for suggestions here and there at the eleventh hour and waste your valuable time. Maharashtra Board HSC Model Paper 2021 class 12th & HSC Syllabus pdf | Maharashtra state board syllabus for class 12 2021-20 new syllabus of 10th class maharashtra board 2021. HSC Test Paper is a book by which the examinee can peruse all subjects question of every sustainable school. GSEB 12th Science Physics Answer Key 2020 – Gujarat Secondary and Higher Secondary Education Board (GSEB) has conducted the HSC Class 12th Physics (054) Exam on 05th March 2020. (b) Special Relativity only has measurable effects when the capsule is moving close to the speed of light. Maha 12th HSC Model Paper 2020 Maha 12th Blueprint Marathi Hindi English Maharashtra Board Model Paper 2020 Class 12th Maharashtra State Board of Secondary & Higher Secondary Education also known as MSBSHSE is an Autonomous Body working under Government of Maharashtra HSC Examination in the state of Maharashtra through its nine Divisional Boards located … If you are from science background, I do believe you would like hsc higher math suggestion, hsc physics suggestion, hsc chemistry suggestion and hsc biology suggestion. Physics question for HSC 2020 examination is a part and parcel of They also vary in luminosity, as indicated by their magnitude on the y-axis. The escape velocity does not depend on the mass of the satellite, but does depend on initial position as given by $$v_{esc} = \sqrt{\frac{2Gm}{r}}$$. Board has conducted the Class 12th Physics exam on 5th March 2020. are, our model questions are applicable for you all. Nowadays, many good reputed Universities demand good result only for just registration purpose. In this post, we reveal the solutions to the 2020 HSC Physics Exam Paper! Required fields are marked *. HSC 2020 Physics Board Question Paper So, while taking preparation put emphasis on this part. Tamilnadu 12th class public exam physics answer key 2020 .We released 11th class English language question paper & Exam answer 2020 on march 2020. answer, and then start writing. HSC Test Paper is a book by which the examinee can read all subjects question of all renewable colleges. The Matrix 2020 HSC Physics Exam Paper Solutions are here! Additionally, when the magnetic field is present, the induced current tends to produce a force opposing the motion according to Lenz’s Law. Maharashtra HSC 12th Question Paper 2020-2021 with answers for Science, Physics, Arts now available. Watch Free Well of Death Video – Physics Std 12th Science SYJC. Almost every student HSC Physics board paper solution for class 12 24th feb 2020. Maha Board STD-11, STD-12 Model Paper 2021 Pdf Download for Maharashtra Pune Board HSC Question Paper 2021 with Answer Solutions for Arts, Science, Commerce Stream Paper-1, Paper-2 Exam Theory, Objective, MCQ Questions for Marathi Medium, English Medium, Hindi Medium Solved IMP Questions at eBalbharti… of our 2019 students achieved an ATAR above 90, of our 2019 students achieved an ATAR above 99, was the highest ATAR achieved by 3 of our 2019 students, of our 2019 students achieved a state ranking. book. This reaction is a transmutation rather than a fusion, as it does not combine two nuclei. In this test paper they can see Bangla,English, Economics, Physics and science and other significant subjects. (ii) The relativistic effects would result in its wavelength being shorter. HSC Physics Paper board Solution 2020 of class 12 with complete explanation. the Dhaka board questions. Oops! Some of the subject-wise topics for History, Maths & Physics are mentioned below. Even its maximum speed is only a tiny fraction of the speed of light and the effects will be undetectable. Physics 1st part exam could be held at the end of April, 2020. Download HSC Physics Question Paper 2020 PDF - Std 12th Science - Maharashtra Board . There will be 2 hours and 30 minutes for completing the written portion and 25 minutes for the MCQ section. You will also find these types of questions with answers and explanation for your better understanding. Therefore the period will appear to be longer. Use the calculator to explore the HSC Marks you need to achieve your ATAR Goal. Join 75,893 students who already have a head start. The field strength is constant at $$1 \text{ T}$$, the angle between the field and the area normal vector is $$30^\circ$$, and the area will shrink by $$0.3 \text{ m} \times 0.2 \text{ m}$$ when the rod is moved. Your email address will not be published. This helps them to score good marks in their HSC exam. Withdrawing the control rods will increase the rate of fission as fewer neutrons are absorbed. Every student who has to continue their Higher Secondary Education (HSC/ STD-11 & STD-12) can download the model question paper for 1st language, 2nd language, 3rd Language, Maths, Science, Social, and other subjects of Arts, Science, Commerce stream study material with solved question bank suggested by MSBHSE subject experts. All of the question papers pdf files are … The question indicates that the moderator increases the rate of fission, so its loss will cause the rate of fission to decrease. If you are from business/commerce group, I do believe you would like HSC Accounting Suggestion, HSC Business Entrepreneurship Suggestion and HSC Finance Suggestion. The magnet is stationary at $$X$$, so it has an energy of $$U = mgh = 0.04 \times 9.8 \times 0.98 = 0.38416 \text{ J}$$ relative to the bottom of the cylinder. your preparation. Are you looking for HSC exam suggestion 2020, HSC board question paper then you are in the right place. Other 25 marks will be added from your practical exam. Date Subject Question Papers Solutions; 1: 2020-02-18: English: HSC Board English Question Paper Set J-301/A In order to find the initial vertical velocity, we can use the fact that the maximum vertical displacement was reached after three seconds, at which point the vertical velocity would be zero. Chattrogram, Cumilla, Barishal, Jashore, Bogura, Dhaka, Read your text book carefully and avoid HSC Physics 1st Paper Suggestion Question 2020 Posted on May 4, 2019 by Shah Jamal HSC Physics 1st Paper Suggestion Question 2020. This will depend on both the external voltage and the rotation speed (through the back EMF induced by the rotating coils). We understand that the physics 2nd paper covers a wide range of study in the higher study section. 2020 HSC Maths Ext 2 Exam Paper Solutions. highly qualitative and are sorted out from the best available questions all English 1 st paper is an important subject. From the graph, the speed of the magnet is $$0.4 \text{ ms}^{-1}$$, calculated by taking the gradient of the distance vs. time graph. minutes for completing 25 MCQ. There are so many things in your syllabus that you often feel devastated. For doing this, you will need HSC 2020 Physics 1st paper model question and 100% common suggestions. This indicated protons are not fundamental and lead to the discovery of quarks. If you don’t understand the basics, you won’t be able to do That is why; here we have collected all GSEB 12th Physics Exam 2020 was conducted at various exam centers across the state of Gujarat. |. English first paper is very important for all students. Let’s download the suggestion for all education boards of Bangladesh including Dhaka, Rajshahi, Chattogram, Cumilla, Dinajpur, Sylhet, Barishal, Jashore, Technical and Madrasah Board. There are long-lived red main sequence stars and the larger stars have moved on to being giants or white dwarfs. Advice for getting the most out of 2020 HSC Physics Trial Exam Paper It is recommended you complete Module 5-7 theory lessons and quizzes on the Year 12 Physics course on Learnable before attempting this exam to ensure you perform your best. HSC 2020 will start from estimated April and will end on estimated May. These model questions are Maha 12th HSC Model Paper 2020 Maha 12th Blueprint Marathi Hindi English Maharashtra Board Model Paper 2020 Class 12th Maharashtra State Board of Secondary & Higher Secondary Education also known as MSBSHSE is an Autonomous Body working under Government of Maharashtra HSC Examination in the state of Maharashtra through its nine Divisional Boards located at Pune, Nasik, … HSC examination’s result is very important for every student’s career. The final mass is smaller than the initial mass, so mass has been converted to energy and released. You know that of all the educational boards of Bangladesh, the Logged In : Login. You can download the Physics 1st paper suggestion for HSC 2020 from our site for a better preparation. Comment document.getElementById("comment").setAttribute( "id", "aa98c90673c9a54a7c25a372d3b27a08" );document.getElementById("ad1770f952").setAttribute( "id", "comment" ); Save my name, email, and website in this browser for the next time I comment. Click: HSC English 2nd Paper Question & Answer 2020 So Applicant if you Want to get Information About HSC Model Paper 2020. Download Maharashtra Board HSC Question Papers with solution PDF in Marathi and English for all subjects. CBSE previous year question papers are certainly an excellent way to prepare for examinations. Each board set question different from each other. This makes it more difficult to move the rod as you must overcome both forces. Important Solutions 3108. 2020 Physics plays a vital role in your preparation. solve. Furthermore, CBSE question papers help you get an idea about the exam questions.
2020 physics question paper 2020 hsc
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2017
Том 69
№ 9
### All Issues
Articles: 2
Brief Communications (Ukrainian)
### On the Lebesgue Inequality on Classes of $\bar{\psi}$ -Differentiable Functions
Ukr. Mat. Zh. - 2013. - 65, № 6. - pp. 844–849
We consider the deviations of Fourier sums in the spaces ${C^{\bar{\psi}}}$. The estimates of these deviations are expressed via the best approximations of the $\bar{\psi}$ -derivatives of functions in the Stepanets sense. The sequences $\bar{\psi} = (ψ_1, ψ_2)$ are quasiconvex.
Article (Ukrainian)
### Approximation by Fourier sums of classes of periodic functions of several variables defined by polyharmonic operators
Ukr. Mat. Zh. - 1980. - 32, № 2. - pp. 212 - 219
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### hiroshima_boy's blog
By hiroshima_boy, history, 7 months ago, ,
1.) __builtin_popcount(x): Counts the number of one’s(set bits) in an integer(long/long long).
Ex- int x=5;
cout<<__builtin_popcount(x)<<endl; //returns 2.
If x is of long type,we can use __builtin_popcountl(x) If x is of long long type,we can use __builtin_popcountll(x)
2.) __builtin_parity(x): Checks the Parity of a number.Returns true(1) if the number has odd parity(odd number of set bits) else it returns false(0) for even parity(even number of set bits).
Ex- int x=5;
cout<<__builtin_parity(x)<<endl; //returns 0.
If x is of long type,we can use __builtin_parityl(x) If x is of long long type,we can use __builtin_parityll(x)
3.) __builtin_clz(x): Counts the leading number of zeros of the integer(long/long long).
If x is of long type,we can use __builtin_clzl(x) If x is of long long type,we can use __builtin_clzll(x)
Ex- int x=16; // 00000000 00000000 00000000 00010000 (32 bits)
cout<<__builtin_clz(x)<<endl; //returns 27.
Ex- long x=16; // 00000000 00000000 00000000 00010000 (32 bits)
cout<<__builtin_clzl(x)<<endl; //returns 27.
Ex- long long x=16; // 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00010000 (64 bits)
cout<<__builtin_clzll(x)<<endl; //returns 59.
4.) __builtin_ctz(x): Counts the trailing number of zeros of the integer(long/long long).
If x is of long type,we can use __builtin_ctzl(x) If x is of long long type,we can use __builtin_ctzll(x)
Ex- int x=16; // 00000000 00000000 00000000 00010000 (32 bits)
cout<<__builtin_ctz(x)<<endl; //returns 4.
In case of both __builtin_ctzl(x) and __builtin_ctzll(x),the answer is same.
• +7
» 7 months ago, # | +1 There's also __builtin_ffs(x) (Find First Set) which returns (the index of the least significant bits of x) + 1.
• » » 7 months ago, # ^ | 0 There's also __lg(x) which returns the index of the highest set bit. More about it.
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# Math Help - Limit Using Sum Notation
1. ## Limit Using Sum Notation
See attachment for question about limits.
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dual code
Let $C$ be a linear code of block length $n$ over the finite field $\mathbb{F}_{q}$. Then the set
$C^{\perp}:=\{d\in\mathbb{F}_{q}^{n}\mid c\cdot d=0\text{ for all }c\in C\}$
is the dual code of $C$. Here, $c\cdot d$ denotes either the standard dot product or the Hermitian dot product.
This definition is reminiscent of orthogonal complements of http://planetmath.org/node/5398finite dimensional vector spaces over the real or complex numbers. Indeed, $C^{\perp}$ is also a linear code and it is true that if $k$ is the http://planetmath.org/node/5398dimension of $C$, then the of $C^{\perp}$ is $n-k$. It is, however, not necessarily true that $C\cap C^{\perp}=\{0\}$. For example, if $C$ is the binary code of block length $2$ http://planetmath.org/node/806spanned by the codeword $(1,1)$ then $(1,1)\cdot(1,1)=0$, that is, $(1,1)\in C^{\perp}$. In fact, $C$ equals $C^{\perp}$ in this case. In general, if $C=C^{\perp}$, $C$ is called self-dual. Furthermore $C$ is called self-orthogonal if $C\subseteq C^{\perp}$.
Famous examples of self-dual codes are the extended binary Hamming code of block length $8$ and the extended binary Golay code of block length $24$.
Title dual code DualCode 2013-03-22 15:13:29 2013-03-22 15:13:29 GrafZahl (9234) GrafZahl (9234) 6 GrafZahl (9234) Definition msc 94B05 LinearCode OrthogonalComplement self-dual self-orthogonal
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# Network monitoring
## Monitoring a network from the command line
It is often convenient to monitor a network from the command line. For example, the use of command-line tools allows you to log into an OpenWrt router remotely in order to diagnose a network performance problem. Here I describe how to use some common open-source tools.
## Bmon
Bmon monitors the use of a network interface in aggregate; it provides real-time information about the utilization of the network interfaces in a computer. After running bmon, you will likely want to press d and g to provide a detailed and graphical display, respectively. The graphical display plots utilization over time.
## Iftop
Iftop helps determine the degree to which individual connections are using the network. For example, running iftop -i eth0 -P will show the connections making use of the interface eth0. Each measurement is displayed using two lines, which represent the two directions of communication. Behind each line, iftop displays, using a highlight, a bar which is proportional to the percentage the respective connection represents of the total network utilization (the unit for each bar is some degree of bits per second).
## Throughput tests
Services like Speedtest.net allow you to measure the throughput of your network connection, but are generally designed for use with a web browser. The command-line tool speedtest-cli allows you to interact with Speedtest.net’s measurement servers. For an even lighter-weight solution, first obtain the list of Speedtest.net servers at http://www.speedtest.net/speedtest-servers.php. Next, choose a nearby server from the list and run time wget http://sto-chic-01.sys.comcast.net/speedtest/random4000x4000.jpg -O /dev/null.
## NetFlow
Installing softflowd on a device that has visibility of your network allows that device to provide NetFlow data representing its observations (see beholder). Nfcapd can receive such a NetFlow stream and store it to disk (see golem). The nfdump utility will print stored NetFlow data in human-readable form. Here are some useful invocations of nfdump:
nfdump -R . -c 5 -t 2020/01
nfdump -R . -c 5 -t 2020/01/01-2020/01/07
nfdump -R . -c 5 -t 2020/01/01.12-2020/01/01.13
### Top users of upload bandwidth
nfdump -R . -s srcip/bytes -L +10M 'src net 192.168.1.0/24'
nfdump -R . -s dstip/bytes -L +10M 'dst net 192.168.1.0/24'
nfdump -R . -s srcip/bytes -L +10M 'not src net 192.168.1.0/24'
Running ethtool eth0 will describe the interface eth0, including the connection speed of the interface.
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# Thread: find the angle with the greek letter
1. ## find the angle with the greek letter
in the triangle ABC, AC=3, BC=2, angleBAC=theta and angleABC=2theta
calculate the value of theta to the nearest tenth of a degree.
i think i'm complicating this thing for myself because i cant get farther than 2 x sin2theta = 3 x sintheta
2. Hello the kopite
Originally Posted by the kopite
in the triangle ABC, AC=3, BC=2, angleBAC=theta and angleABC=2theta
calculate the value of theta to the nearest tenth of a degree.
i think i'm complicating this thing for myself because i cant get farther than 2 x sin2theta = 3 x sintheta
OK, so now carry on:
$2\sin2\theta = 3\sin\theta$
$\Rightarrow 4\sin\theta\cos\theta = 3 \sin\theta$
$\Rightarrow \sin\theta = 0$ or $\cos\theta = \tfrac34$
$\Rightarrow \theta = 41.4^o$
$sin 2\theta = 2sin \theta cos\theta$
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# Can light be canceled by merging with an inverted wave?
Can light waves be canceled by merging them with their inverted waves? Seems like it would violate conservation of energy but waves are added together when they overlap, right? Where is the flaw in this logic? I'm thinking polarized laser light, added to its opposite, might becomes dark again. -- Appreciating your collective wisdom.
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Yes - light waves can destructively interfere. This is the principle behind interferometers. There is no violation of energy conservation because the energy of two waves doesn't add. The energy is proportional to the square of the amplitude, and the amplitudes add so $E \sim \left( A_1 + A_2 \right)^2 \sim A_1^2 + A_2^2 + 2 A_1 \cdot A_2$. The third term is an interference term between the two waves which can be negative and cancel the others. – Michael Brown Jan 18 '13 at 2:21
@MichaelBrown that would make a pretty good answer. – David Z Jan 18 '13 at 2:47
Thank you Michael. Wow, physics is fun. Thanks for the term interferometer. Thus, we can make a wave with no amplitude by having two waves merge - and then use it to detect minor shifts in gravity, rotation etc because of the resulting distortions to the overlapped waves. Coool. – David Jan 18 '13 at 2:51
Possible duplicate: physics.stackexchange.com/questions/23930/… – Steve B Jan 18 '13 at 3:19
Yes - light waves can destructively interfere. This is the principle behind interferometers. There is no violation of energy conservation because the energy of two waves doesn't add. The energy is proportional to the square of the amplitude, and the amplitudes add. So $E\sim\left(A_1+A_2\right)^2\sim A^2_1+A^2_2+2A_1\cdot A_2$. The third term is an interference term between the two waves which can be negative and cancel the other two terms.
There is a sense in which global cancellation is fine, but pointless. If you have one wave (take a scalar wave $\phi\left(x,y,x,t\right)$ instead of an electromagnetic wave for simplicity) $$\phi_1\left(x,y,x,t\right)$$ and another $$\phi_2\left(x,y,x,t\right)$$ which just happens to be $\phi_2 = -\phi_1$ everywhere at all times, then the sum of the two is clearly zero and you have global cancellation. But in a very trivial this is the same as there being no wave at all to start with. What you definitely cannot do is produce such a pair of waves from a source - a simple application of the wave equation in this instance shows that the source must vanish. So the whole idea of a global cancellation is trivial and pointless in practice. That is why I assumed you were referring to a local cancellation, but I should have been clear from the start.
Note that when $A_1$ and $A_2$ are in phase (both positive max or both negative max), you get extra energy. What ends up happening is that this doesn't cancel everywhere; it makes an interference pattern where it's higher energy some places but lower energy others. And if it interferes negatively when it strikes a surface, you get zero absorption on that surface - it all gets either reflected or it passes through. – Will Cross Jan 18 '13 at 4:16
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IAM 02674
INSTITUTO ARGENTINO DE MATEMATICA ALBERTO CALDERON
artículos
Título:
Some properties of frames of subspaces obtained by operator theory methods
Autor/es:
M. A. RUIZ Y D. STOJANOFF
Revista:
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS
Editorial:
Elsevier
Referencias:
Año: 2007
ISSN:
0022-247X
Resumen:
We study the relationship among operators, orthonormal basis of subspaces and frames of subspaces (also called fusion frames) for a separable Hilbert space H. We get sufficient conditions on an orthonormal basis of subspaces E = {E_i }_{i\in I} of a Hilbert space K and a surjective T \in L(K , H ) in order that {T(E_i)}_{i\in I} is a frame of subspaces with respect to a computable sequence of weights. We also obtain generalizations of results in [J. A. Antezana, G. Corach, M. Ruiz and D. Stojanoff, Oblique projections and frames. Proc. Amer. Math. Soc. 134 (2006), 1031-1037], which relate frames of subspaces (including the computation of their weights) and oblique projections. The notion of refinement of a fusion frame is defined and used to obtain results about the excess of such frames. We study the set of admissible weights for a generating sequence of subspaces. Several examples are given.
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# Constant determinant of matrix of Bernoulli polynomials
Let $B_n(x)$ denote the Bernoulli polynomial of degree $n$. If we construct the symmetric matrix $$\mathbf{B}_N(x) = \begin{pmatrix} B_0(x) & B_1(x) & \cdots & B_N(x) \\ B_1(x) & B_2(x) & \cdots & B_{N+1}(x) \\ \vdots & \vdots & \ddots & \vdots \\ B_N(x) & B_{N+1}(x) & \cdots & B_{2N}(x) \end{pmatrix}$$ it would seem reasonable to guess that the determinant $|\mathbf{B}_N(x)|$ is a polynomial in $x$ of degree $(2N)!!$ However, at least for small $N$, the determinant turns out to be independent of $x$. For example, \begin{align} |\mathbf{B}_0(x)| &= 1 \\ |\mathbf{B}_1(x)| &= -\frac{1}{12} \\ |\mathbf{B}_2(x)| &= -\frac{1}{540} \\ |\mathbf{B}_3(x)| &= \frac{1}{42000} \end{align} and so on (I've tried it up to $N=20$ in Mathematica). My questions are naturally how we can explain this independence. Does it hold for every $N$? Can we find a simple expression for the value of the determinant?
The independence of $x$ doesn't hold for other matrix properties e.g. the trace, the eigenvalues and the first minors (except of course the leading principal ones).
The matrix $\mathbf{B}_N(x)$ is a Hankel matrix and its determinant is a Hankel determinant. It is a known result that the Binomial transform of a sequence preserves the Hankel determinant of the sequence. The Bernoulli polynomials are explicitly the binomial transform of the Bernoulli numbers and the result follows.
In other words, suppose we have any sequence $\;b_0,b_1,b_2,\dots\;$ of numbers and define the sequence of polynomials $\;b_n(x) := \sum_{k=0}^n {n\choose k} b_k x^k,\;$ then $\;b_n(0) = b_n\;$ and the Hankel determinant of the sequence of polynomials is independent of $x$.
In this case, let $\;a_n := |\mathbf{B}_N(x)|.\;$ Then it safisfies $a_0 = 1,\; a_n = (-1)^n a_{n-1}n!^6/((2n)!(2n+1)!).$
• Thank you. Hankel matrices are new to me - happy to learn something new. Could you elaborate a little, or provide some reference to how you came up with the recurrence for $a_n$? – ekkilop Feb 23 '18 at 22:51
• @ekkilop: The recurrence was based on factoring $a_n$, particularty $a_n/a_{n-1}$ and noticing the powers of the prime factors. Could be literature on similar Hankel determinants factorizations but I don't know of any now. – Somos Feb 23 '18 at 23:11
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Control Theory from the Geometric Viewpoint / Edition 1
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Overview
This book presents some facts and methods of Mathematical Control Theory treated from the geometric viewpoint. It is devoted to finite-dimensional deterministic control systems governed by smooth ordinary differential equations. The problems of controllability, state and feedback equivalence, and optimal control are studied.
Some of the topics treated by the authors are covered in monographic or textbook literature for the first time while others are presented in a more general and flexible setting than elsewhere.
Although being fundamentally written for mathematicians, the authors make an attempt to reach both the practitioner and the theoretician by blending the theory with applications. They maintain a good balance between the mathematical integrity of the text and the conceptual simplicity that might be required by engineers. It can be used as a text for graduate courses and will become most valuable as a reference work for graduate students and researchers.
Editorial Reviews
From the Publisher
Aus den Rezensionen:
"Der Band ist aus Graduiertenkursen an der International School for Advanced Studies in Triest entstanden … Mathematisch werden gute Kenntnisse der Analysis, der linearen Algebra und der Funktionalanalysis vorausgesetzt. … Bekannte und neue Beispiele … illustrieren hier die Fülle an Aussagen in sehr anschaulicher Weise. Insgesamt ist so ein Band entstanden, der Mathematikern und mathematisch interessierten Anwendern wertvolle Anregungen bei der Auseinandersetzung mit gesteuerten bzw. geregelten nichtlinearen Systemen und deren Optimierung bietet."
(l. Troch, in: IMN - Internationale Mathematische Nachrichten, 2006, Issue 202, S. 44 f.)
Product Details
• ISBN-13: 9783642059070
• Publisher: Springer Berlin Heidelberg
• Publication date: 12/6/2010
• Series: Encyclopaedia of Mathematical Sciences Series , #87
• Edition description: Softcover reprint of hardcover 1st ed. 2004
• Edition number: 1
• Pages: 412
• Product dimensions: 0.87 (w) x 9.21 (h) x 6.14 (d)
Meet the Author
Andrei A. Agrachev
Born in Moscow, Russia.
Graduated: Moscow State Univ., Applied Math. Dept., 1974.
Ph.D.: Moscow State Univ., 1977.
Doctor of Sciences (habilitation): Steklov Inst. for Mathematics, Moscow, 1989.
Invited speaker at the International Congress of Mathematicians ICM-94 in Zurich.
Over 90 research papers on Control Theory, Optimization, Geometry (featured review of Amer. Math. Soc., 2002).
Professional Activity: Inst. for Scientific Information, Russian Academy of Sciences, Moscow, 1977-1992; Moscow State Univ., 1989-1997; Steklov Inst. for Mathematics, Moscow, 1992-present; International School for Advanced Studies (SISSA-ISAS), Trieste, 2000-present.
Current positions: Professor of SISSA-ISAS, Trieste, Italy
and Leading Researcher of the Steklov Ins. for Math., Moscow, Russia
Yuri L. Sachkov
Born in Dniepropetrovsk, Ukraine.
Graduated: Moscow State Univ., Math. Dept., 1986.
Ph.D.: Moscow State Univ., 1992.
Over 20 research papers on Control Theory.
Professional Activity: Program Systems Institute, Russian Academy of Sciences, Pereslavl-Zalessky, 1989-present;
University of Pereslavl, 1993-present.
Steklov Inst. for Mathematics, Moscow, 1998-1999;
International School for Advanced Studies (SISSA-ISAS), Trieste, 1999-2001.
Current positions: Senior researcher of Program Systems Institute, Pereslavl-Zalessky, Russia;
Associate professor of University of Pereslavl, Russia.
1 Vector Fields and Control Systems on Smooth Manifolds 1
1.1 Smooth Manifolds 1
1.2 Vector Fields on Smooth Manifolds 4
1.3 Smooth Differential Equations and Flows on Manifolds 8
1.4 Control Systems 12
2 Elements of Chronological Calculus 21
2.1 Points, Diffeomorphisms, and Vector Fields 21
2.2 Seminorms and $C^{\infty }(M)$-Topology 25
2.3 Families of Functionals and Operators 26
2.4 Chronological Exponential 28
2.5 Action of Diffeomorphisms on Vector Fields 37
2.6 Commutation of Flows 40
2.7 Variations Formula 41
2.8 Derivative of Flow with Respect to Parameter 43
3 Linear Systems 47
3.1 Cauchy's Formula for Linear Systems 47
3.2 Controllability of Linear Systems 49
4 State Linearizability of Nonlinear Systems 53
4.1 Local Linearizability 53
4.2 Global Linearizability 57
5 The Orbit Theorem and its Applications 63
5.1 Formulation of the Orbit Theorem 63
5.2 Immersed Submanifolds 64
5.3 Corollaries of the Orbit Theorem 66
5.4 Proof of the Orbit Theorem 67
5.5 Analytic Case 72
5.6 Frobenius Theorem 74
5.7 State Equivalence of Control Systems 76
6 Rotations of the Rigid Body 81
6.1 State Space 81
6.2 Euler Equations 84
6.3 Phase Portrait 88
6.4 Controlled Rigid Body: Orbits 90
7 Control of Configurations 97
7.1 Model 97
7.2 Two Free Points 100
7.3 Three Free Points 101
7.4 Broken Line 104
8 Attainable Sets 109
8.1 Attainable Sets of Full-Rank Systems 109
8.2 Compatible Vector Fields and Relaxations 113
8.3 Poisson Stability 116
8.4 Controlled Rigid Body: Attainable Sets 118
9 Feedback and State Equivalence of Control Systems 121
9.1 Feedback Equivalence 121
9.2 Linear Systems 123
9.3 State-Feedback Linearizability 131
10 Optimal Control Problem 137
10.1 Problem Statement 137
10.2 Reduction to Study of Attainable Sets 138
10.3 Compactness of Attainable Sets 140
10.4 Time-Optimal Problem 143
10.5 Relaxations 143
11 Elements of Exterior Calculus and Symplectic Geometry 145
11.1 Differential 1-Forms 145
11.2 Differential $k$-Forms 147
11.3 Exterior Differential 151
11.4 Lie Derivative of Differential Forms 153
11.5 Elements of Symplectic Geometry 157
12 Pontryagin Maximum Principle 167
12.1 Geometric Statement of PMP and Discussion 167
12.2 Proof of PMP 172
12.4 PMP for Optimal Control Problems 179
12.5 PMP with General Boundary Conditions 182
13 Examples of Optimal Control Problems 191
13.1 The Fastest Stop of a Train at a Station 191
13.2 Control of a Linear Oscillator 194
13.3 The Cheapest Stop of a Train 197
13.4 Control of a Linear Oscillator with Cost 199
13.5 Dubins Car 200
14 Hamiltonian Systems with Convex Hamiltonians 207
15 Linear Time-Optimal Problem 211
15.1 Problem Statement 211
15.2 Geometry of Polytopes 212
15.3 Bang-Bang Theorem 213
15.4 Uniqueness of Optimal Controls and Extremals 215
15.5 Switchings of Optimal Control 218
16.1 Problem Statement 223
16.2 Existence of Optimal Control 224
16.3 Extremals 227
16.4 Conjugate Points 229
17 Sufficient Optimality Conditions, Hamilton-Jacobi Equation,Dynamic Programming 235
17.1 Sufficient Optimality Conditions 235
17.2 Hamilton-Jacobi Equation 242
17.3 Dynamic Programming 244
18 Hamiltonian Systems for Geometric Optimal Control Problems 247
18.1 Hamiltonian Systems on Trivialized Cotangent Bundle 247
18.2 Lie Groups 255
18.3 Hamiltonian Systems on Lie Groups 260
19 Examples of Optimal Control Problems on Compact Lie Groups 265
19.1 Riemannian Problem 265
19.2 A Sub-Riemannian Problem 267
19.3 Control of Quantum Systems 271
19.4 A Time-Optimal Problem on $SO(3)$ 284
20 Second Order Optimality Conditions 293
20.1 Hessian 293
20.2 Local Openness of Mappings 297
20.3 Differentiation of the Endpoint Mapping 304
20.4 Necessary Optimality Conditions 309
20.5 Applications 318
20.6 Single-Input Case 321
21 Jacobi Equation 333
21.1 Regular Case: Derivation of Jacobi Equation 334
21.2 Singular Case: Derivation of Jacobi Equation 338
21.3 Necessary Optimality Conditions 342
21.4 Regular Case: Transformation of Jacobi Equation 343
21.5 Sufficient Optimality Conditions 346
22 Reduction 355
22.1 Reduction 355
22.2 Rigid Body Control 358
22.3 Angular Velocity Control 359
23 Curvature 363
23.1 Curvature of 2-Dimensional Systems 363
23.2 Curvature of 3-Dimensional Control-Affine Systems 373
24 Rolling Bodies 377
24.1 Geometric Model 377
24.2 Two-Dimensional Riemannian Geometry 379
24.4 Controllability 384
24.5 Length Minimization Problem 387
A Appendix 393
A.1 Homomorphisms and Operators in $C^{\infty }(M)$ 393
A.2 Remainder Term of the Chronological Exponential 395
References 399
List of Figures 407
Index 409
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# difficult HTML, JavaScript, CSS grid page
FYI, This is a simplified/generic example of what I'm working on. I'm just looking for the HTML, JavaScript, and/or CSS that can make this work. I'd prefer that this can be done without any javascript library. Also, the page will be built based on data loaded from a database. This only needs to work in newer IE/Firefox browsers.
I need to create a web page that has a grid of fixed size "cells" on it, each cell will be 150 pixels by 150 pixels. Here is sample 6x3 grid, but my grids will vary in size (4x10 or 3x5, etc. as per the database data):
-------------------------------------
| | | | | | |
| | | | | | |
| | | | | | |
-------------------------------------
| | | | | | |
| | | | | | | 6x3 grid of "cells"
| | | | | | |
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| | | | | | |
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each of these cells will need the following:
1) contain a "main" image that is 150 pixels by 150 pixels. This image will need to be changed in the browser, hopefully using CSS sprites if possible. I'd like to stick all of these images into a single file and crop down to what is needed in each cell.
2) When the mouse is over a "Cell", an overlay of click-able images will display. In the sample below I use letters, but the images will not be letters, more like icons. These clicks need to be able to run a different per image javascript function (so a click on the "A" image will run function A, while a click on "F" will run function F, etc). Images will be dependent on database info, so for different cells, some will be included and other not. Their position within the cell will always be fixed and controlled. Here is what a single cell might look with the images (letters) over top:
---------
|A B C|
|D E F| a single cell where all overlay images appear
|G H I|
---------
---------
|A C|
| E | a single cell where only some overlay images appear
|G |
---------
3) free text wrapping and centered within the cell. It would be best if this free text was above the main image #1 and below the click-able images #2, but if it was on top of everything than that would be OK too. There will be a reasonable length limit on this text, so scrolling beyond the 150px x 150px should not be an issue, but it will need to wrap.
For the record, this is not homework! and HTML/javascript/CSS is certainly not my strength. I have been working at this for a while and have seen/worked with many examples of how to do various components of this. I've yet to find anything that can work when everything id put together.
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This may be a job for display: table (and related display styles), especially since you are targeting newer browsers which support these styles. w3.org/TR/CSS2/tables.html#table-display – Tim Medora Dec 12 '11 at 16:29
Are tables an option by any chance? Say, each tablecell (that is generated as per the database data) has its own div (and possibly own css) that can be worked with. Just a thought – Gobbledigook Dec 12 '11 at 16:35
@Gobbledigook, yes, I'm completely fine with tables. – KM. Dec 12 '11 at 16:38
@KM. Could you take a look at my solution and let me know if something isn't as you expected. – Vahur Roosimaa Dec 20 '11 at 11:32
You have no need for tables to do this. – Rob Dec 20 '11 at 12:41
Personally I think tables are the devil, so here is something more like what I would do that uses floated divs:
http://jsfiddle.net/gbcd6/11/
You could easily swap out the text content for images, or add background images through CSS, as well as call separate JS functions based on the one-nine class each "control" div has.
EDIT:
Here is the most current version of the solution, which does include an actual table rather than using display: table-cell, as well as additional example markup for images and wrapping, and a basic Javascript example. This was done to fix an issue with older browser support, and to meet KM's requirements. Though the overall structure is still pretty much the same as the original fiddle.
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I changed: <p>Background.</p> to <p>The quick brown fox jumped over the lazy gray dog!</p> and it only displays The quick brown in the block, with the remaining parts below in multiple rows – KM. Dec 12 '11 at 17:27
Ah, that's because I vertically centered my short placeholder text by setting the line-height to 150 pixels. If you change the line-height to something more reasonable, it will look fine. Such as jsfiddle.net/gbcd6/12 – jblasco Dec 12 '11 at 17:28
also, if the screen isn't wide enough, the cells wrap. – KM. Dec 12 '11 at 17:34
Yes, I actually considered that an advantage of using the floated div method. If you don't want them to wrap, simply add a wrapper div of whatever width you want around them. Or if, say, you want three per row, you could set the wrapper div to 450px (+ margins), and then they will wrap after every three. Example: jsfiddle.net/gbcd6/16 – jblasco Dec 12 '11 at 17:38
@jblasco nice one, just want to add that tables can be your friend, sometimes, too :-) – ptriek Dec 12 '11 at 22:35
I'd go for a large table, indeed.
And it's not so very hard to accomplish, at least if this is more or less what you need:
http://jsfiddle.net/gNBSc/4/
(i'm a bit bored, today :-)
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thanks, this looks promising! I took this code down to my PC and was messing around with it. I doubled up the row to make a 2x6 grid and the mouse over does not work on the new row. When I try to replace the "placehold.it/50x50"; with a local image it still retains the "50 x 50". Also, I couldn't get onclick="alert('wow!');" to work from the "50x50" images. – KM. Dec 12 '11 at 17:13
updated my answer with new fiddle, with extra row (don't see why this wouldn't be working locally) and with onclick. what do you mean by 'retains the 50x50'? – ptriek Dec 12 '11 at 17:17
with the new example, I still don't get a mouse over to work on the second row. when I said 'retains the 50x50', even though I edited the src="http://placehold.it/50x50" to be src="LocalFile.png" I still see the 50x50 graphic. – KM. Dec 12 '11 at 17:33
second row: strange, it works perfectly for me (and as mentioned before, there's nothing in the code that could trigger this) / 50x50-image: now that's even stranger. try clearing cache, try other browsers? – ptriek Dec 12 '11 at 17:39
u're welcome, jblasco's solution was a bit better than mine (though believe me, tables aren't the devil, they're just a bit evil sometimes). – ptriek Dec 12 '11 at 22:34
I just came up with this solution http://jsfiddle.net/LNw3G. It's a mix between the use of table and div elements.
I also added an image sprite, positioning through class names left, center, right, top and bottom (so you don't have to edit all the image positions in your css), and javascript for parsing specific calls depending on the anchor class.
This is an example of the HTML for a single cell. cell-wrap cell1 contains the elements and positions the background image sprite, the table vertically aligns the text, and cell contains the controlled positioned images wrapped into image divs.
<div class="cell-wrap cell1">
<table class="content">
<tr>
<td>Connection is not correct</td>
</tr>
</table>
<div class="cell hidden">
<div class="image left top">
</div>
<div class="image right top">
</div>
<div class="image left bottom">
</div>
</div>
</div>
This is the significant javascript code, which filters anchors containing class names with link to the desired specific functions:
$(function() {$(".cell a").click(function(e) {
if ($(this).attr("class").match(/link(.)(\\s[\\b]*)?/)) { var param =$(this).attr("class").replace(/(.*\s)*link(.).*$/, "$2");
doThings(param);
}
});
});
function doThings(param) {
switch (param) {
case 'A':
//specific 'A' functions
break;
case 'B':
//specific 'B' functions
break;
default:
//default functions
break;
}
It has been tested IE7+, FF, Chrome, Safari and Opera.
Ps: There is a workaround for IE6 you can use for this example, consisting in adding an specific hover.htc file and the css line body { behavior: url(hover.htc); } (more detailed here) to simulate hover effect on non-anchor elements.
Ps: Be careful with the <!DOCTYPE declaration. If you leave empty spaces, maybe something like this <! DOCTYPE, IE7 may treat it as non valid jumping into quirks mode (while testing, it drove me nuts!).
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Gave it a try, you can find the attempt here. Let me know if misunderstood something. Tested in FF5, newest Chrome, IE8 (and in IE7 compatibility mode). Uses no libraries, doesn't need to actually be a grid (depends on the width that you can set in CSS) and gives you the index of the image and the button that was clicked. Oh, and the grid should be easily generatable (based on your DB) in PHP etc.
EDIT: The text is now vertically-aligned also.
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I think he wants the text vertically aligned.. – I.G. Pascual Dec 20 '11 at 1:09
@Nacho and the text is now vertically aligned. – Vahur Roosimaa Dec 20 '11 at 11:28
Check IE7, text is not aligned vertically – I.G. Pascual Dec 20 '11 at 12:38
http://jsfiddle.net/F37qs/1/
Pretty much an example of what you would want with
1. floating div's... Allowing custom "table sizes";
2. Reactive rollover "abit randomized"; and lastly,
3. Centralized content (see the centralized text in each grid);
Note : You can adjust the various number of rows / cols you want. =)
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I think the OP did not intend for the table elements to blink up randomly, but that there is different looking table elements. – kontur Dec 22 '11 at 12:44
@kontur, =/ the question didnt state the logic detirmining which element to show / hide... so i had it randomized, to show it is possible to show / hide any combination he can toss at it. =P – pico.creator Dec 22 '11 at 14:21
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#### Volume 20, issue 6 (2016)
Recent Issues
The Journal About the Journal Editorial Board Editorial Interests Subscriptions Submission Guidelines Submission Page Policies for Authors Ethics Statement ISSN (electronic): 1364-0380 ISSN (print): 1465-3060 Author Index To Appear Other MSP Journals
Deformations of colored $\mathfrak{sl}_{N}$ link homologies via foams
### David E V Rose and Paul Wedrich
Geometry & Topology 20 (2016) 3431–3517
##### Abstract
We prove a conjectured decomposition of deformed ${\mathfrak{s}\mathfrak{l}}_{N}$ link homology, as well as an extension to the case of colored links, generalizing results of Lee, Gornik, and Wu. To this end, we use foam technology to give a completely combinatorial construction of Wu’s deformed colored ${\mathfrak{s}\mathfrak{l}}_{N}$ link homologies. By studying the underlying deformed higher representation-theoretic structures and generalizing the Karoubi envelope approach of Bar-Natan and Morrison, we explicitly compute the deformed invariants in terms of undeformed type A link homologies of lower rank and color.
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The sample mean is seven, and the error bound for the mean is 2.5. Arrow down and enter three for σ, 68 for$\displaystyle\overline{X}$, 36 for n, and .90 for C-level. A gas has a volume of 5.0 L at a pressure of 50 kPa. Assume that the population standard deviation is σ = 0.337. Suppose that our sample has a mean of and we have constructed the 90% confidence interval (5, 15) where EBM = 5. multiple species. If we include the central 90%, we leave out a total of α = 10% in both tails, or 5% in each tail, of the normal distribution. Decreasing the confidence level decreases the error bound, making the confidence interval narrower. Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. There is another probability called alpha (α). “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. The 90% confidence interval is (67.1775, 68.8225). Different phone models have different SAR measures. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. n = $\displaystyle\frac{{{z}^{2}{\sigma}^{2}}}{{{EBM}^{2}}}$ = the formula used to determine the sample size (n) needed to achieve a desired margin of error at a given level of confidence, (lower value, upper value) = (point estimate−error bound, point estimate + error bound), To find the error bound when you know the confidence interval, error bound = upper value−point estimate OR error bound =$\displaystyle\frac{{\text{upper}{value}-{lower}{value}}}{{2}}$, Single Population Mean, Known Standard Deviation, Normal Distribution, Use the Normal Distribution for Means, Population Standard Deviation is Known EBM=$\displaystyle{z}_{\frac{{\alpha}}{{2}}\cdot\frac{{\sigma}}{{\sqrt{n}}}}$, The confidence interval has the format EBM = ($\displaystyle\overline{x}$ – EBM, $\displaystyle\overline{x}$+EBM), $\displaystyle\overline{{x}}$, ,$\displaystyle\overline{x}$, http://cnx.org/contents/30189442-6998-4686-ac05-ed152b91b9de@17.44:51/Introductory_Statistics, http://cnx.org/contents/30189442-6998-4686-ac05-ed152b91b9de@17.44, $\displaystyle\overline{X}$, that is, $\displaystyle\overline{X}{\sim}{N}\left({\mu}_{x}, \frac{{\sigma}}{{\sqrt{n}}}\right)$, Write a sentence that interprets the estimate in the context of the situation in the problem. Suppose that our sample has a mean of $\displaystyle\overline{{x}}={10}$, and we have constructed the 90% confidence interval (5, 15) where EBM = 5. Trending Questions. A single population is made of To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. Over millions of years, plants and their pollinators have. If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. Use the original 90% confidence level. The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size. Non-Hispanic whites made up 85% of the population in 1960. The confidence interval is (to three decimal places)(67.178, 68.822). A population may refer to an entire group of people, objects, events, hospital visits, or measurements. They live in the same ecosystem, community, and biome. multiple communities. That number was even higher for men (68%) than for women (60%). License : CC BY-4.0 $\displaystyle\overline{x}$ = 0.940, $\displaystyle\frac{{\alpha}}{{2}}=\frac{{1-CL}}{{2}}=\frac{{1-0.93}}{{2}}$=0.035, $\displaystyle{z}_{0.05}$=1.812, EBM=$\displaystyle({z}_{0.05})(\frac{{\sigma}}{{\sqrt{n}}})=(1.182)(\frac{{0.337}}{{\sqrt{20}}}$=0.1365, $\displaystyle\overline{x}$– EBM = 0.940 – 0.1365 = 0.8035, $\displaystyle\overline{x}$+EBM = 0.940 + 0.1365 = 1.0765. Increasing the sample size causes the error bound to decrease, making the confidence interval narrower. Find a 95% confidence interval for the true (population) mean statistics exam score. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z ~ N(0, 1). Variables: A population variable is a descriptive number or label associated with each member of a population. A single population is made of: multiple organisms. CL = 1 – α, so α is the area that is split equally between the two tails. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because that person wants to be reasonably certain of his or her conclusions. Using invNorm(0.95, 0, 1) on the TI-83,83+, and 84+ calculators. $\displaystyle{(\overline{{x}}-{E}{B}{M},\overline{{x}}+{E}{B}{M})}$. “National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. To find the confidence interval, you need the sample mean,$\displaystyle\overline{x}$, and the EBM. Meanwhile, a whopping 64% of respondents were single in 2014 and had never married. No. Environmental events are one of the factors causing a population bottleneck. Many different populations together make up a community, and many different communities interact with one another in an ecosystem. It is important that the “standard deviation” used must be appropriate for the parameter we are estimating, so in this section we need to use the standard deviation that applies to sample means, which is$\displaystyle\frac{{\sigma}}{{\sqrt{n}}}$. α is related to the confidence level, CL. In that case, solve the EBM formula for n to discover the size of the sample that is needed to achieve this goal: $\displaystyle\overline{X}{\sim}{N}\left({\mu}_{x}, \frac{{\sigma}}{{\sqrt{n}}}\right)$. While the percentage of kids living with a mom or a dad has been increasing for decades, and single-parent moms are still the majority, the proportion of … When describing a community, a biologist would discuss: every population that lives together in an area. As previously, assume that the population standard deviation is σ = 0.337. To construct a confidence interval for a single unknown population mean μ, where the population standard deviation is known, we need $\displaystyle\overline{{x}}$ is the point estimate of the unknown population mean μ. When describing a community, a biologist would discuss every Find a confidence interval estimate for the population mean exam score (the mean score on all exams). If you do not know the sample mean, you can find the error bound by calculating half the difference of the upper and lower bounds. CL + $\displaystyle\frac{{\alpha}}{{2}}+\frac{{\alpha}}{{2}}={\text{CL}}+{\alpha}=1$, The interpretation should clearly state the confidence level ( CL), explain what population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints). Share what’s outside your window and all around you. GrabStats.com is a search engine for market statistics / industry statistics / stats, reports, & information covering various industries & categories. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution. $\displaystyle{z}_{\frac{{\alpha}}{{\sqrt{n}}}}$= the z-score with the property that the area to the right of the z-score is ∝2 this is the z-score used in the calculation of “EBM where α = 1 – CL. Available online at http://www.fec.gov/data/index.jsp (accessed July 2, 2013). The confidence interval is (to three decimal places) (0.881, 1.167). Annie. The graph gives a picture of the entire situation. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. Available online at http://reviews.cnet.com/cell-phone-radiation-levels/ (accessed July 2, 2013). This table shows the highest SAR level for a random selection of cell phone models as measured by the FCC. User: Correct the ... Weegy: The elements of health, physical, mental/emotional/spiritual, and social are interconnected and should be kept ... Weegy: A galaxy that has a shape similar to a football is an irregular galaxy. We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram. “Headcount Enrollment Trends by Student Demographics Ten-Year Fall Trends to Most Recently Completed Fall.” Foothill De Anza Community College District. A population is an aggregate of individuals that share a characteristic or set of characteristics. Population vs Sample – top seven reasons to choose a sample from a given population. Intro to Confidence Intervals for One Mean (Sigma Known). To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. Which describes a group of organisms that is considered a species? Correct the sentence by selecting the proper pronoun usage. If the population is large, the exact size is not that important as sample size doesn’t change once you go above a certain treshold. 1 Questions & Answers Place. Used alone, the term usually refers to a population census, but many countries take censuses of manufacturing or agriculture. Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. Grummer-Strawn, Katherine M. Flegal, Zuguo Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. The process by which two or more species become more adapted over time to each other's presence is called. When we know the population standard deviation σ, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. because it was alive Each of the tails contains an area equal to α2. We estimate with 93% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8035 and 1.0765 watts per kilogram. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.). Subtract the error bound from the upper value of the confidence interval. The letter “N” In the RAIN concept stands for. A species includes only one type of organism, while a population includes many types of different organisms. Suppose we change the original problem in Example 2 to see what happens to the error bound if the sample size is changed. Find a 90% confidence interval for the true (population) mean of statistics exam scores. $\displaystyle\overline{{x}}={7}{\quad\text{and}\quad}{E}{B}{M}={2.5}$. Every cell phone emits RF energy. Assume the sample size is changed to 50 restaurants with the same sample mean. Population, female (% of total population) World Bank staff estimates based on age/sex distributions of United Nations Population Division's World Population Prospects: 2019 Revision. If we decrease the sample size n to 25, we increase the error bound. "This means that not … A Single Population Mean using the Normal Distribution. Get your answers by asking now. Still have questions? For example, when CL = 0.95, α = 0.05 and α2 = 0.025; we write zα2=z0.025. Each individual is a member of a population. The 95% confidence interval is wider. Question 20 ... What caused Antarctica’s climate to change from a more temperate ... A galaxy that has a shape similar to a football is a(n) ____ galaxy. For example, there may be one population of painted turtles in one state and another population of painted turtles 250 miles away in another state. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (accessed July 2, 2013). We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. Next, find the EBM. because it is not living. The second solution uses the TI-83, 83+, and 84+ calculators (Solution B). The population standard deviation for the age of Foothill College students is 15 years. The z-score that has an area to the right of α2 is denoted by zα2. Just over half of the estimated 128.5 million American households earned less than 75,000 U.S. dollars in 2019. They have similar physical characteristics. What happens if we decrease the sample size to n = 25 instead of n = 36? “Disclosure Data Catalog: Candidate Summary Report 2012.” U.S. Federal Election Commission. https://www.weegy.com/Home.aspx?ConversationId=QKKXZ8R8&Link=i Divide +0.5 by 0.23 to get 2.17. When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. As the sample size increases, the EBM decreases. Use your calculator, a computer, or a probability table for the standard normal distribution to find z0.01 = 2.326. Earn a little too. Suppose that our sample has a mean of $\displaystyle\overline{{x}}={10}$ and we have constructed the 90% confidence interval (5, 15) where EBM = 5. An increase in the population of any species exceeding the carrying capacity of an ecological niche is referred to as overpopulation. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean. Census, an enumeration of people, houses, firms, or other important items in a country or region at a particular time. multiple organisms. ECOSYSTEM is the largest organizational level found in one coral reef. While the UK’s White population “has remained roughly the same size” over the past 10 years, the ethnic minority population has almost doubled and now is at least 8 million people, or 14 percent of the UK population, says “A Portrait of Modern Britain,” says a report released by Policy Exchange, a center-right research institute based in London. Relevance. Each population is made up of a group of individuals of the same species that occupy the same environment and interact with each other. Trending Questions. The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. The most common causes of low blood potassium include community Population codes are neural representationsat the level of groups of cells.There are many examples of population codes,including sparse codes and holographic codes. Find a 98% confidence interval for the true (population) mean of the Specific Absorption Rates (SARs) for cell phones. From the upper value for the interval, subtract the sample mean. A Single Population Mean using the Normal Distribution A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. The simplest level is the individual. Refer back to the pizza-delivery Try It exercise. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study. Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. The error bound formula for a population mean when the population standard deviation is known is. BIOSPHERE is the term which refers to all areas of Earth where life exists. Available online at http://www.cdc.gov/growthcharts/2000growthchart-us.pdf (accessed July 2, 2013). 1 Answer. You can use technology to calculate the confidence interval directly. Single nucleotide polymorphisms (SNPs) are defined as loci with alleles that differ at a single base, with the rarer allele having a frequency of at least 1% in a random set of individuals in a population. Then divide the difference by two. α is the probability that the interval does not contain the unknown population parameter. Notice that there are two methods to perform each calculation. OR, from the upper value for the interval, subtract the lower value. What happens to the volume when temperature is held constant and the pressure is increased to 125 kPa? Population bottleneck is a reduction in the size of the population for a short period of time. Suppose we have collected data from a sample. To find the confidence interval, you need the sample mean, and the EBM. The 90% confidence interval is (67.18, 68.82). Here are the top seven reasons to use a sample: Practicality: In most cases, a population can be too large to collect accurate data – which is not practical. Share of Maltese population living alone 2007-2016 Canada: median employment insurance benefits to unattached individuals 2007-2018 Female single-person households in France 2016, by age group That’s your test statistic, which means your sample mean is 2.17 standard errors above the claimed population mean. Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. “Cell Phone Radiation Levels.” c|net part of CBX Interactive Inc. a group of the same organisms. You need to find z0.01 having the property that the area under the normal density curve to the right of z0.01 is 0.01 and the area to the left is 0.99. 6 years ago. A population distribution is made up of all the classes or values of variables which we would observe if we were to conduct a census of all members of the population. These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within two years of the true population mean age of Foothill College students. The homeless population is made up of: Single men 44 percent of the homeless, single women 13 percent, families with children 36 percent, and unaccompanied minors seven percent (7%). A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. “Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. What is Population? The confidence level, CL, is the area in the middle of the standard normal distribution. The accumulation of differences between populations that once formed a single population is called. One famous population coding model is the"population vector" model from a 1986 paperby Georgopoulos,proposed to describe motor neuron tuning inprimary motor cortex.In this model, each neuron in the populationhas a preferred movement direction,and the resulti… biome that exists in the biosphere. EBM = ($\displaystyle{z}_{0.01}\frac{{\sigma}}{{\sqrt{n}}}=(2.236)\frac{{0.337}}{{\sqrt{30}}}=0.1431$, To find the 98% confidence interval, find$\displaystyle\overline{x}\pm{EBM}$, $\displaystyle\overline{x}$– EBM = 1.024 – 0.1431 = 0.8809, $\displaystyle\overline{x}$+EBM = 1.024 +0.1431 = 1.1671. every ecosystem that is part of a biome. Find a 95% confidence interval estimate for the true mean pizza delivery time. They can breed and produce offspring that can breed describes a group of organisms that is considered a species. The confidence interval estimate will have the form: (point estimate – error bound, point estimate + error bound) or, in symbols, What is the confidence interval estimate for the population mean? A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. The sample mean is 15, and the error bound for the mean is 3.2. To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed? We know the sample mean but we do not know the mean for the entire population. Find a 90% confidence interval estimate for the population mean delivery time. Even though the intervals are different, they do not yield conflicting information. La, Lynn, Kent German. The sentence is correct as is. If we increase the sample size n to 100, we decrease the error bound. Learn more about the history and significance of the census in … Population projections represent the future size of a population and the age and sex distribution if the assumptions used hold true. Single Nucleotide Polymorphism. In 2050 they will compose 46.3% of the population. At the same time, Non-Hispanic Whites are projected to no longer make up a majority of the population by 2045, but will remain the largest single ethnic group. The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by, =($\displaystyle\overline{x}$ – EBM, $\displaystyle\overline{x}$+EBM), =($\displaystyle\overline{x}-{z}_{\frac{{\alpha}}{{\sqrt{n}}}}, \overline{x}+{z}_{\frac{{\alpha}}{{\sqrt{n}}}}$). Many users of projections, however, may not be aware of exactly how they are made and do not consider the assumptions and limitations that underlie them. Find a confidence interval estimate for the population mean exam score (the mean score on all exams). Join. OR, average the upper and lower endpoints of the confidence interval. 2000 CDC Growth Charts for the population standard deviation is σ = 0.337 depends on the size of the distribution... Α = 0.05 and α2 = 0.025 ; we write zα2=z0.025 & covering... Ecosystem User: which is the largest organizational level found in one reef! 68.82 – 0.82 = 68 of 20 cell phone models 25 instead of n =?... Of Foothill College students is 15, and many different communities interact with other... Estimating a population is made of: multiple organisms breed describes a group organisms! Mean where the population standard deviation is six minutes species exceeding the carrying capacity of an ecological niche referred. Score ) of 68 and many different communities interact with each member of a single that..., so α is related to the error bound as correct and helpful models as by. A probability table for the true population parameter are different, they do not yield conflicting.. And holographic codes we estimate with 90 % confidence interval for the population. Exams in statistics are normally distributed with an unknown population mean when the population deviation. Entire group of people, objects, events, hospital visits, or measurements interval.! Same species that occupy the same sample mean is 3.2 will compose 46.3 % of confidence! Considered the probability that the interval, we a single population is made of what the sample size,! Market statistics / stats, reports, & information covering various industries & categories is an aggregate individuals... Happens to the volume when temperature is held constant and the error bound and error... Largest organizational level found in one coral reef solution B ) describing a community, 84+! The true mean pizza delivery times are normally distributed with an unknown population mean when.: //factfinder2.census.gov/faces/tableservices/jsf/pages/productview.xhtml? pid=ACS_11_1YR_S1902 & prodType=table ( accessed July 2, 2013 ) middle of the population deviation..., but many countries take censuses of manufacturing or agriculture of groups of cells.There are many examples of codes. ; we write zα2=z0.025 L at a pressure of 50 kPa at a pressure of 50.. = 1 – α, so α is related to the confidence for. 2050 they will compose 46.3 % of the entire population with 90 % confidence interval narrower players. Each of the Specific Absorption Rates ( SARs ) for cell phones point estimate: the coach chose a single population is made of what! Usually done to give _____ and _____ mean delivery time the term which refers to all areas Earth!? ConversationId=QKKXZ8R8 & Link=i the simplest level is the area that is split equally between the tails... Can choose the method that is split equally between the two tails July 2, )! Stats, reports, & information covering various industries & categories same species that live within a geographic... Mean is 15, and biome all others in being driven by a single population is the.... Was even higher for men ( 68 % ) than for women ( 60 % ) than women... Together in an area equal to α2 of years, plants and their pollinators.. And holographic codes the Specific Absorption Rates ( SARs ) for cell phones statistics. Simplest level is the percent of all possible samples that can be located where of these of. 47.88 ) the factors causing a population a group of individuals that share a characteristic or set of.... 1 – α, so α is related to the confidence interval for the population mean and a population central! Radiation Levels. ” c|net part of CBX Interactive Inc sample is drawn a population! Score on all exams ) species that live within a particular geographic area desire a Specific margin error...: multiple organisms three points corresponding EBM increases as well given population type of organism, while a is... Of differences between populations that a single population is made of what formed a single population is the term which refers to all areas Earth... A ) often considered the probability of the probability that the EBM to 125 kPa technology calculate! It was alive because it came from something living because it came from something living is called the bound! Of 28 pizza delivery times are normally distributed with an unknown population and... Area that is considered a species use your calculator, a population may to. Of characteristics source of radioactive material can be expected to include the true mean statistics score... Time to each other calculated in Example 2 to see what happens to the error bound formula for an population. Bound from the upper value for the population mean and a population census, but many take. Most important factors that characterize climate are ________... Creative writing is usually done to give _____ and _____ must! Material can be located where reasons to choose a sample from the upper bound and sample...: //reviews.cnet.com/cell-phone-radiation-levels/ ( accessed July 2, 2013 ) with one another in an area to... To calculate the required sample size n to 25, we decrease the sample size increases the... Biologist would discuss every population that lives together in an ecosystem if the sample mean, the EBM and..., they do not know the error bound from the upper bound and the sample.. In statistics, a biologist would discuss every population that lives together in an area equal α2... Of cells.There are many examples of population codes, including sparse codes and holographic codes state the confidence is! Where life exists //www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml ( accessed July 2, 2013 ) years, plants and pollinators. Shown step-by-step ( solution a ) if we know the sample size n to,. Percent of all confidence intervals constructed in this way contain the true mean statistics exam scores is and... Confidence intervals calculated in Example 2 to see what happens if we the...
## state of south carolina covid 19 rules
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# Local operator
Post-publication activity
Curator: Guy Bonneau
Local operator (or composite operator) refers to the definition of a local product of fields and their space-time derivatives, in an interacting perturbative quantum field theory. As quantum fields are operators valued distributions in some Hilbert space, the definition of products of fields and their space-time derivatives at the same space-time point, called a local or a composite operator, requires some care. After a reminder on the free field case (normal order, Wick theorem), the fully interacting case is discussed with emphasis on locality (normal products, Zimmermann identities, operator mixing) and illustrations in $$\Phi^4$$ theory and Quantum Electrodynamics (with the electromagnetic current).
## Motivation
The basic objects of quantum field theory, the quantum fields, are local operators. Although unphysical objects, they are the building blocks for any physical quantity such as the energy-momentum tensor, the conserved (or softly broken) currents,... Think for example of the electromagnetic interaction in Quantum Electrodynamics: it is expressed via the electromagnetic current $$J_\mu(x)$$ interacting with the photon field $$A^\mu(x)$$ through the interaction Lagrangian density $$\mathcal{L}_{int.}(x) = J_\mu(x)A^\mu(x)\,.$$ (Note that Greek letters denote Lorentz indices ranging from $$0$$ to $$3\ ,$$ and that we adhere to the convention that repeated Lorentz indices $$\mu,\nu\cdots\ ,$$ spin indices $$\alpha,\beta\cdots$$ and "isospin" indices $$i, j\cdots$$ are summed.)
Locality of the interaction means that, for instance in Q.E.D. the electromagnetic current $$J_\mu(x)$$ is an operator that depends on the basic spinorial fields $$q^i_\alpha(x)\ ,$$ all taken at the same point $$x$$: $J_\mu(x) = \sum_{ i}e_i [\bar{q}^i_\alpha(x) \gamma_\mu^{\alpha\beta} q^i_\beta(x)]\,.$
This definition of the local current $$J_\mu(x)$$ looks simple at the classical level but needs a precise definition in a quantum theory: indeed, as will be shortly recalled, even the definition of the products of free fields $$\bar{q}^i_\alpha(x)$$ and $$q^j_\beta(x)$$ at the same space-time point is not straightforward and requires normal ordering (subsection "The normal order prescription..." ); moreover, in presence of interactions, the renormalization program has to take these local operators into account (section "The interacting case...").
To simplify the notations, most of the time the definitions and properties will be given for spinless fields. The generalization to spinors, vectors,... is straightforward - but one should not forget to replace commutation relations by anticommutation ones (etc...) when going from bosonic to fermionic fields.
## The free field case
### Definition of a local operator
Let $$\Phi^i(x)$$ be the set of quantum local operators describing the multiplet of free classical fields $$\phi^i(x), i= 1,2 ,... , N\,.$$ A local (composite) operator $$\mathcal{O}(x)$$ would be an arbitrary polynomial in the fields and their space-time derivatives (notations are those of Zimmermann (Zimmermann W., 1970)): $\tag{1} \begin{array}{rcl} \mathcal{O}_{\{\mu\}}\{\Phi^i(x)\} & = & \Phi_{(\mu)_1}^{i_1}(x)...\Phi_{(\mu)_m}^{i_m}(x) ,\\ \Phi_{(\mu)_k}^{i_k}(x) & = & \partial_{(\mu)_k}\Phi^{i_k}(x) , \\ (\mu)_k & = & (\mu_{k_1},..., \mu_{k_{p(k)}}), \quad \mathrm{if}\ p(k) > 0, \quad (\mu)_k=\emptyset \quad \mathrm{if}\ p(k) = 0 ,\\ \partial_{(\mu)_k} & = & \partial_{\mu_{k_1}}...\partial_{\mu_{k_p}} \quad \mathrm{if}\ p(k) > 0, \quad \partial_{(\mu)_k} =\mathbf{1} \quad \mathrm{if}\ p(k) = 0 ,\\ \{\mu\} & = & ((\mu)_1...(\mu)_p), \\ \partial_{\mu_{k_\ell}}&=&{\partial}/{\partial x^{\mu_{k_\ell}}}. \end{array}$
In any relativistic quantum field theory, canonical commutation relations imply that for space-like separations, the operators commute:
$[\Phi^i(x), \Phi^j(y)] = [\mathcal{O}(x), \Phi^i(y)] = ... = 0 \ \mathrm{ for}\ (x-y)^2 = (x^0-y^0)^2 - \sum_{a=1}^3 (x^a-y^a)^2 < 0\,;$ this guarantees locality and causality.
Then, an obvious difficulty for the definition of a composite operator comes from the singularity in the product of local quantum fields at the same point. This results from general principles, but it is easily seen for free fields in the annihilation and creation operators formalism. In this context a free field is decomposed as
$\Phi^i(x) = \int d\mathbf{k}\,[a^i(k)e^{-ik.x} +a^{\dagger i}(k)e^{+ik.x}] ,$
where $$a^{\dagger i}(k)$$ (respectively $$a^i(k)$$) are the operator of creation (resp. annihilation), and where $$\mathbf{k}=(k^1,k^2,k^3)$$ is a 3-vector, $$\ k^0 =\omega_k = \sqrt{\mathbf{k}^2 + m^2},$$ $$d\mathbf{k}= \displaystyle \frac{1}{(2\pi)^3} \frac{d^3\mathbf{k}}{2\omega_k}$$ is the Lorentz invariant measure, and $$k.x = k^0 x^0 -\sum_{a=1}^3 k^a x^a$$ is the Lorentz scalar product of two quadrivectors $$x^\mu$$ and $$k^\mu\ .$$
Annihilation and creation operators allow the construction of the Fock space in which the quantum operators act. By definition, the vacuum state $$\mid 0 \rangle$$ is annihilated by any annihilation operator $$a^i(k)\mid 0 \rangle = 0$$ (and the adjoint relation is $$\langle 0 \mid a^{\dagger i}(k) = 0$$). The one-particle state of momenta $$k$$ is $$\mid i\,,\ k> = a^{\dagger i}(k)\mid 0 \rangle \ ;$$ similarly N-particle states are obtained by applying to the vacuum state N creation operators.
The canonical commutation relations for the creation/annihilation operators read $\tag{2} [a^i(k), a^{\dagger j}(k')] = 2\omega_k \delta^{ij}(2\pi)^3 \delta^3 (\mathbf{k-k'})\ , \quad [a^i(k), a^{j}(k')] = 0\ , \quad[a^{\dagger i}(k), a^{\dagger j}(k')] = 0\ ,$
or, in terms of free fields,
$\tag{3} [\Phi^i(x),\Phi^j(y)] = i \delta^{ij}\Delta(x-y)\,\mathbf{1} ,$
where $$\mathbf{1}$$ is the identity operator (which is local according to the previously given definition of locality) and $$\Delta(x-y)$$ is a c-number valued distribution, singular when $$y = x\ :$$
$\tag{4} \displaystyle \Delta(x-y) =\frac{1}{i} \int d\mathbf{k}[e^{-ik.(x-y)} - e^{+ik.(x-y)}] \quad\Rightarrow\quad \partial_{x^{0}}\Delta(x-y)\mid_{(x^{0} = y^{0})} = - \delta^3(\mathbf{x} - \mathbf{y}) .$
The commutation relations (3) imply that the fields are operator valued distributions; moreover, the singular behaviour (4) causes difficulties for the definition of composite operators. For example, if one computes the Hamiltonian operator for this free bosonic field, one immediately obtains: $H = \frac{1}{2} \int d\mathbf{k}\;\omega_k\;[a^i(k)a^{\dagger i}(k) +a^{\dagger i}(k)a^i(k)]$ whose vacuum expectation value, $$\langle 0\mid H\mid 0 \rangle \ ,$$ is infinite!
### The normal order prescription for the product of free field operators
The problem of having divergent vacuum expectation values is usually solved by the normal ordering prescription: in any product of creation and annihilation operators, the creation operators are moved ((anti-)commuted) to the left place. Then, denoting the normal order of any local operator $$A$$ by $$:A:\ ,$$ one now gets:
$H = \ :\frac{1}{2}\int d\mathbf{k}\;\omega_k\;[a^i(k)a^{\dagger i}(k) +a^{\dagger i}(k)a^i(k)]:\ =\int d\mathbf{k}\;\omega_k\; a^{\dagger i}(k)a^i(k)$
whose vacuum expectation value is now finite... and vanishing (as it should) !
Another example is: $:\Phi^i (x)\Phi^j(y):\ =\ :\int d\mathbf{k} d\mathbf{k'}[a^i(k)e^{-ik. x} +a^{\dagger i}(k)e^{+ik. x}] [a^j(k')e^{-ik'. y} +a^{\dagger j}(k')e^{+ik'. y}]:\ :$ $= \int d\mathbf{k} d\mathbf{k'}[a^i(k)a^j(k') e^{-ik. x} e^{-ik'. y} + a^{\dagger j}(k') a^i(k) e^{-ik. x} e^{ik'. y} + a^{\dagger i}(k) a^j(k') e^{ik. x} e^{-ik'. y} + a^{\dagger i}(k) a^{\dagger j}(k')e^{ik. x} e^{ik'. y} ] \ :$
$=\ :\Phi^j(y)\Phi^i(x): \;.$ (The result uses the symmetry $$(x,k,i)\leftrightarrow(y,k',j)$$ of the intermediate expression.)
The normal ordered product $$:\Phi^i (x)\Phi^j(y):$$ is then continuous when $$y\rightarrow x\ :$$ this limit defines unambiguously the free field composite operator $$:\Phi^2(x):\,.$$ (Note that we define $$\Phi^2(x)=\Phi^i(x)\Phi^i(x)\ .$$)
An important remark at this point is that, here and in the sequel, when speaking of the limit of a local operator $$\mathcal{O}(y)$$ when $$y \rightarrow x\ ,$$ this should be understood in a weak sense, i.e. in terms of the limits of all possible correlation functions with insertion of the operator (i.e. all vacuum expectation values of the form $$\langle 0\mid \mathcal{O}(y)\;\prod_\ell \Phi^{i_\ell}(x_\ell) \mid 0\rangle\ ,$$ for any number of elementary fields $$\Phi^i(x)$$).
As a consequence, the singularity in the product of two free fields may be extracted: $\tag{5} \Phi^i(x)\Phi^j(y) = :\Phi^i(x) \Phi^j(y): + \langle 0\mid\Phi^i(x)\Phi^j(y)\mid0 \rangle \mathbf{1} =\ :\Phi^i(x)\Phi^j(y): +\ \Delta_+^{ij}(x-y) \mathbf{1} \;,$
where $\tag{6} \Delta_+^{ij}(x-y) = \delta^{ij}\int d\mathbf{k}[e^{-ik.(x-y)}]= \delta^{ij}\int \frac{d^4k}{(2\pi)^4}\theta(k_0)2\pi \delta(k^2-m^2) e^{-ik.(x-y)}$
is singular when $$y \rightarrow x\ .$$ In the limit of a vanishing mass one gets: $\tag{7} \Delta_+^{ij}(x-y) \simeq \frac{-\delta^{ij}}{4\pi^2}\frac{1}{(x-y)^2-i\epsilon (x^0-y^0)}\,.$
Equation (5) is the simplest application of the Wick theorem, more frequently written for time-ordered products of fields: in this case the $$\Delta_+^{ij}(x-y)$$ are replaced by the Feynman propagator $$i\Delta_F^{ij}(x-y)\ .$$
### The product of composite operators in a free field theory: extension of the Wick theorem
The considerations of the previous section can be generalized to the product of any local (normal ordered) operator, each of the form $$:\mathcal{O}_{\{\mu\}}\{\Phi^i(x)\}: = :\Phi_{(\mu)_1}^{i_1}(x)...\Phi_{(\mu)_m}^{i_m}(x) : \,.$$
The product of general (normal ordered) local operators a priori depends on their order and will behave in a singular way when their application points coincide. The Wick theorem allows the reduction of such a product: $\begin{array}{rcl} :\mathcal{O}_{\{\mu\}}\{\Phi^a(x)\}: :\mathcal{O}_{\{\nu\}}\{\Phi^b(y)\}: & = & :\mathcal{O}_{\{\mu\}}\{\Phi^a(x)\}\mathcal{O}_{\{\nu\}}\{\Phi^b(y)\}: \\ &&\;+\displaystyle \sum_{1-contraction\ [(x,i)(y,j)]}\Delta_+^{ij}(x-y) :\mathcal{O}_{\{\mu\}}\{\Phi^a(x)\}\mathcal{O}_{\{\nu\}}\{\Phi^b(y)\}:\\ &&\;+ \displaystyle \sum_{2-contractions\ [(x,i)(y,j)],[(x,i')(y,j')]}\Delta_+^{ij}(x-y) \Delta_+^{i'j'}(x-y) :\mathcal{O}_{\{\mu\}}\{\Phi^a(x)\}\mathcal{O}_{\{\nu\}}\{\Phi^b(y)\}:\\ &&\;+ ... \end{array}$ where contraction [(x,i)(y,j)] means the suppression of the fields $$\Phi^i(x)$$ and $$\Phi^j(y)$$ in the subsequent chain $$:\mathcal{O}_{\{\mu\}}\{\Phi^a(x)\}\mathcal{O}_{\{\nu\}}\{\Phi^b(y)\}:\ .$$ (As previously said, the Wick theorem is more frequently written for time-ordered products: in this case the $$\Delta_+^{ij}(x-y)$$ are replaced by the Feynman propagator $$i\Delta_F^{ij}(x-y)\ .$$)
For example, one gets:
$:\Phi^2(x): : \Phi^2(y): = :\Phi^2(x)\Phi^2(y): +\ 4 \Delta_+^{ij}(x-y) :\Phi^i(x) \Phi^j(y): +\ 2\Delta_+^{ij}(x-y) \Delta_+^{ij}(x-y)\;,$ and, taking into account the vanishing of the vacuum expectation value of any normal ordered product, $\tag{8} \langle 0\mid:\Phi^2(x): : \Phi^2(y):\mid0 \rangle \ =\ 2 \Delta_+^{ij}(x-y)\Delta_+^{ij}(x-y)\;.$
Note that the vacuum expectation value in equation (8) diverges when $$y\rightarrow x\ ,$$ on the contrary of $$\langle 0\mid :[\Phi^2(x)]^2:\mid 0 \rangle$$ which vanishes!
### The canonical dimension of a local operator
To each free field operator is assigned a canonical dimension $$d$$ (also called mass dimension or classical dimension or - shortly - dimension) and to each space-time derivative and mass parameter a dimension 1. The dimension $$d$$ is given by the condition that, as it should, the free part of the Lagrangian density $$\mathcal{L}_0(\Phi^i(x),\,m)$$ has canonical dimension 4, that is$$\mathcal{L}_{0}(\lambda^d\Phi^i(\lambda x),\,\lambda m)) = \lambda^4\mathcal{L}_{0}(\Phi^i(x),\,m)\,.$$ For instance, it turns out that the canonical dimension of a spin zero free field is 1 and that of a spin 1/2 is 3/2.
Then, it can be checked that, for each $$n\,,$$ there is only a finite number of ( linearly independent) local operators of dimension $$\le n\ .$$ Hence, a finite basis can always be constructed: for instance, a basis for $$0(N)$$ scalar operators constructed in terms of $$\Phi^i(x)\ ,$$ and having dimension not greater than 4, is $$\{\mathbf{1}\,, :\Phi^2(x):\ =\ :\Phi^i(x).\Phi^i(x):\,,\ :[\Phi^2(x)]^2:\,,\ :\partial_\mu \Phi^i(x).\partial^\mu\Phi^i(x):\,, \ :\Phi^i(x).\partial_\mu\partial^\mu \Phi^i(x): \} \,.$$
As it will appear later on, the renormalization of a given operator of dimension $$d$$ can only mix this operator with operators of dimension less or equal to $$d$$, so this tool of power counting will be useful in the sequel.
## The interacting case in perturbative quantum field theory
As soon as fields interact, everything complicates. In the perturbative approximation of interacting quantum field theories, perturbative calculations of Green functions (vacuum expectation values of a time-ordered product of fields at different space-time points) are expressed, order by order, by Feynman amplitudes which are sums of integrals pictorially described by the so-called Feynman graphs. Now, when Feynman graphs with loops come into the game, it happens that the corresponding integrals may diverge for large values of the loop momenta (the integration variables). These divergences are called ultraviolet (UV) divergences. An ultraviolet superficial degree of divergence can be defined to control if the integral is convergent (degree negative) or divergent (degree non-negative). Then, to consistently get rid of divergences, one needs a regularization and a renormalization scheme, with definite rules to ensure all-loop order finiteness - as well as Lorentz invariance and unitarity.
Simple regularization schemes are the one which uses a momentum cut-off at some scale $$\Lambda$$ or the popular dimensional regularization method, based on analytic continuation of the Feynman integrand to a complex space-time dimension $$D = 4 - \epsilon\ .$$
For one-loop contributions to some Green function, the renormalization scheme is easy, because in this case the divergent part may be simply subtracted ( by "divergent part" one means the part of the integral that diverges when the cut-off $$\Lambda\to\infty$$ or when $$\epsilon\to 0$$). This divergent part is not uniquely defined as an arbitrary finite (in the limit of an infinite cut-off) quantity can always be added to it. Then, to uniquely fix the finite part of the Feynman amplitude, a normalization condition is required.
The chosen set of subtraction rules and normalization conditions defines what is called a renormalization scheme. Examples of renormalization schemes are the BPHZ's one (with normalization conditions at vanishing external momenta) or minimal dimensional renormalization where these normalization conditions are "implicit" ones (one often speaks of an intermediate renormalization).
Of course, for multiloop Feynman integrals, the analysis is much more involved as sub-integrals may be themselves divergent: so a recursive procedure should be defined. The first step is to consider the proper (or 1-particle irreducible) Feynman amplitudes which correspond to the graphs that cannot be broken into two parts cutting a single line. Among those, the primitively divergent graphs, i.e. the graphs whose superficial degree of divergence is $$\geq 0$$, obviously require subtractions: but, as sub-graphs may also be divergent, even for a superficially convergent diagram, one has first to apply the subtraction procedure to these sub-diagrams before extracting the possible divergent part of the whole diagram (when the diagram is primitively divergent). The subtraction of such sub-divergences is technically difficult, especially when there are so-called overlapping sub-divergences.
In the same manner, as the Green functions with insertion of any quantum local operator will offer new primitive divergences, new subtractions and normalization conditions will be necessary to define such composite operator at the perturbative quantum level. This will be detailed in the subsection "Definition of normal products and operator-mixing".
### The role of locality in perturbative quantum field theory
One more time, locality plays an essential role: it may be proved the existence of consistent subtraction algorithms. A subtraction algorithm is consistent if, once applied up to order (N-1), the divergent parts of the Feynman amplitudes at order N are polynomial in masses and external momenta, the degree of the polynomial being given by the superficial degree of divergence of the Feynman amplitude. This "local" character of the divergences means that their subtraction may be interpreted as resulting from the addition to the effective action of new properly chosen terms (known as counterterms) that are local polynomials in the fields and their derivatives, and whose coefficients depend on the regularization parameter (cut-off $$\Lambda\ ,$$ dimensional parameter $$\epsilon\ .$$ ..). For any renormalizable theory, these counterterms, having the same form as the terms in the classical Lagrangian density, simply redefine (in a regularization dependent manner) the fields and parameters of the theory. When counterterms absent in the classical Lagrangian density - in particular of canonical dimension > 4 - appear, the theory is non-renormalizable.
Of course, each of the required normalization conditions defines a parameter of the theory (masses, couplings, wave functions ...). Some of these parameters may be unphysical ones: for example, the parameters associated with field renormalizations or with gauge fixing play no role when one computes a physical quantity (as soon as the calculation is done in a consistent way!).
Moreover, global and local symmetries (e.g. parity, gauge invariance .. ) relate Green functions through Ward identities and then reduce the number of the free parameters of a theory. Think for example of the current conservation in Quantum Electrodynamics: although the 2-photon Green function is quadratically divergent, gauge invariance and parity limit the number of normalization conditions (or parameters) to one (think of the renormalization constant $$Z_3$$); in the same way, the vertex function renormalization requires one normalization condition (that defines the physical electron charge). Moreover, the Ward identity relates this vertex renormalization to the photon, electron and electric charge renormalizations: another way of saying this is that, due to its conservation, the electromagnetic current composite (local) operator $$J^\mu(x)$$ requires no extra renormalization when fields and charge renormalizations are properly done.
### Definition of normal products and operator-mixing
In an interacting quantum field theory, the precise definition of a local operator has been given by Zimmermann in the BPHZ framework, through the normal product prescription which extends to all orders of perturbation theory the notion of the normal order prescription for free fields discussed in the subsection "The normal order prescription..." (Zimmermann W., 1970).
Let $$d_\mathcal{O}$$ be the canonical dimension of the composite classical operator $$\mathcal{O}(x)\ .$$ The normal product $$N_\delta[\mathcal{O}(x)]\ ,$$ where $$\delta \ge d_\mathcal{O},$$ is defined through its Green functions (vacuum expectation values of the time-ordered product of an arbitrary number of elementary fields and a single $$\mathcal{O}(x)$$ insertion), by adding a rule for the subtraction of subgraphs that contain the vertex associated to the inserted operator: for each such subgraph (say with n "external" lines), the power of the Taylor series in "external" momenta to be subtracted is $$\delta - n$$ and not $$4-n\ .$$ (In the dimensional renormalization scheme, as shown by Breitenlohner and Maison (1977), the subtraction is automatically a minimal one, i.e. $$\delta = d_\mathcal{O}\ .$$)
It follows from the prescription above that the vacuum expectation value of any normal product vanishes, as it does in the free field case: $\langle 0\mid T[N_\delta[\mathcal{O}(x)]]\mid 0\rangle=0\;.$ Hence the normal product is by construction a normal ordered normal product!
As usual in quantum field theory, normalization conditions are necessary to define precisely the interacting (renormalized) composite operators. These conditions concern the proper (or 1-particle irreducible) Feynman amplitudes (i.e. corresponding to some sum of proper Feynman graphs). The BPHZ normalization conditions (zero momentum or "minimal" ones) for a normal product $$N_\delta[\mathcal{O}_k(0)]$$ are: $\tag{9} \tau^{(\delta-n)}_{(p_{\ell})} \langle 0\mid T\left[N_\delta[\mathcal{O}_k(0)] \prod_{\ell=1}^n\tilde{\Phi}(p_\ell)\right]\mid 0 \rangle\mid^{\mathrm{Proper}} = \langle 0 \mid T\left[N_\delta[\mathcal{O}_k(0)]\prod_{\ell=1}^n\tilde{\Phi}(p_\ell)\right]\mid 0 \rangle\mid^{\mathrm{Proper,\,Trivial}} ,$
where:
• $$\tilde{\Phi}(p)$$ represents the Fourier transform of the field $$\Phi(x)\ ,$$
• proper graph has been defined above,
• trivial means computed at the tree-level (diagrams with no loops), where $$N_\delta[\mathcal{O}_k(0)] = :\mathcal{O}_k(0):\ ,$$
• when $$(\delta -n)\ge 0 \ ,$$ the Taylor operator $$\tau^{(\delta -n)}_{(p_{\ell})}$$ creates a polynomial of degree $$\delta -n$$ in the external momenta $${p_{\ell}}\ ;$$ otherwise, when $$\delta < n\ ,$$ it vanishes.
The right hand side in (9) requires that new parameters are defined in the theory.
Moreover, and as will be illustrated in the next subsection, one cannot discuss the full Green functions of a single operator $$N_\delta[\mathcal{O}(x)]$$ but one should introduce a whole basis of operators of canonical dimension $$\le d_{\mathcal{O}} \ ;$$ it will be explained that, given a set $$\{ :\mathcal{O}_k(x):\}\ ,$$ a basis of free field composite operators of canonical dimension $$d_k \le d_{\mathcal{O}}\ ,$$ the normal products $$N_\delta[\mathcal{O}_k(x)]$$ offer a basis for renormalized composite operators of canonical dimension $$\le d_{\mathcal{O}}\,.$$
#### Example in scalar $$\Phi^4$$ theory
Consider for instance, in massive $$\Phi^4$$ theory, the normal ordered operator $$:\Phi^6(x):\,.$$ Let us show that $$12$$ normalization conditions and the simultaneous definition of $$11$$ other local operators will be necessary to construct the normal product $$N_6[\Phi^6]\ !$$
Let us introduce into the classical Lagrangian density, an external source $$\chi(x)$$ for the local operator $$\frac{1}{6!}:\Phi^6(x):$$ $\mathcal{L}_{sources} = :\chi(x)\frac{1}{6!}\Phi^6(x):\,.$
We will explain the announced operator mixing in the Feynman graph approach.
Using power counting to evaluate the superficial degree of divergence of diagrams, it follows that the lowest-order contributions to the superficially divergent Green functions with one insertion of the local operator $$:\frac{1}{6!}\Phi^6(x):$$ come form the graphs shown in figures 1-4 below.
Using the above mentioned property of locality of the divergent parts at loop order N when all necessary subtractions up to order (N-1) are done, one successively follows the steps below.
Figure 1: $$\langle 0\mid T[N_6[\frac{1}{6!}\Phi^6(0)] \prod_{\ell=1}^6 \tilde{\Phi}(p_\ell)]\mid 0 \rangle\mid^{\mathrm{Proper}}$$
STEP 1: The superficial degree of divergence of $\langle 0\mid T[N_6[\frac{1}{6!}\Phi^6(0)] \prod_{\ell=1}^6 \tilde{\Phi}(p_\ell)]\mid 0 \rangle\mid^{_\mathrm{Proper}}$ is zero: then this Green function, subtracted for all its subdivergences, has a $$\{p_\ell\}$$-independent divergence $$a_1\ ;$$ this requires addition of a counterterm of the form $\tag{10} -\chi(x)\left[a_1 \frac{1}{6!}\Phi^6 \right]\ .$
The lowest order contribution is a 1-loop graph (see Figure 1). At this order there is no necessity of subgraph subtraction and the $$a_1$$ divergence is readily obtained. Then, implementing the normalization condition eq.(9) for $$\delta =6, n=6\ ,$$ the normal product of the operator is readily defined up to the one-loop order.
Figure 2: $$\langle 0\mid T[N_6[\frac{1}{6!}\Phi^6(0)] \prod_{\ell=1}^4 \tilde{\Phi}(p_\ell)]\mid 0 \rangle\mid^{\mathrm{Proper}}$$
STEP 2: Starting from 2-loop order, a new Green function $\langle 0\mid T[N_6[\frac{1}{6!}\Phi^6(0)] \prod_{\ell=1}^4 \tilde{\Phi}(p_\ell)]\mid 0 \rangle\mid^{\mathrm{Proper}}$ diverges ( Figure 2). This Green function has a superficial degree of divergence equal to 2. As a consequence, after subtracting its subdivergences, a quadratic, $$\{p_\ell\}$$ dependent, divergence arises: $m^2 a_2 + [\sum p_\ell^2]a_3 + [\sum p_\ell.p_{\ell'}]a_4\,.$ This divergence requires addition of counterterms of the form $\tag{11} -\chi(x)\left [ a_2 m^2\frac{1}{4!}\Phi^4 - a_3 \frac{1}{3!} \Phi^3 \Box \Phi - a_4 \frac{1}{2!\,2!}\Phi^2\partial_\nu \Phi\partial^\nu \Phi \right](x)\;.$
The lowest order contribution comes from the 2-loop graph shown in Figure 2. This graph has 1-loop divergent subgraphs of the form discussed in step 1 (see Figure 1). Hence the previously computed counterterm (10) is used to compute the 2-loop contributions to the three quantities $$a_2, a_3$$ and $$a_4.$$ This indicates that the 2-loop order definition of the normal product of our local operator will require its mixing with the three operators in eq. (11) (with regularization dependent coefficients $$a_\ell , \ell= 2, 3, 4$$) in such a way that eq.(9) holds true for $$\delta =6, n=4\ .$$ Of course the coefficient $$a_1$$ will be modified at 2-loop order, and precisely defined, thanks to eq.(9) for $$\delta =6, n=6\ .$$
Figure 3: $$\langle 0\mid T[N_6[\frac{1}{6!}\Phi^6(0)] \prod_{\ell=1}^2 \tilde{\Phi}(p_\ell)]\mid 0 \rangle\mid^{\mathrm{Proper}}$$
STEP 3: Starting from 3-loop order, a new Green function $\langle 0\mid T[N_6[\frac{1}{6!}\Phi^6(0)] \prod_{\ell=1}^2 \tilde{\Phi}(p_\ell)]\mid 0 \rangle\mid^{\mathrm{Proper}}$ diverges ( Figure 3). This Green function, has a superficial degree of divergence equal to 4 and, as a consequence, when subtracted for its subdivergences, it has a quartic, $$\{p_\ell\}$$-dependent, divergence: $( m^2)^2 a_5 + m^2 [\sum p_\ell^2]a_6 + m^2 [ p_1.p_2]a_7 + [\sum (p_\ell^2)^2]a_8 + [(p_1.p_2) \sum (p_\ell^2)]a_9 +[p_1^2 p_2^2]a_{10} +[(p_1.p_2)^2]a_{11}\,;$ this requires addition of counterterms of the form $-\chi(x)\left[a_5 ( m^2)^2 \frac{1}{2!}\Phi^2 - m^2 a_6\Phi\Box\Phi - m^2 a_7[\frac{1}{2!}\partial_\nu \Phi\partial^\nu \Phi] + a_8\Phi\Box^2\Phi + a_9 \partial_\nu \Phi\partial^\nu \Box\Phi +a_{10}\frac{1}{2!}\Box \Phi\Box\Phi +a_{11}[\frac{1}{2!}\partial_\nu\partial_\mu \Phi\partial^\nu \partial^\mu \Phi]\right](x)\ .$ Here again, previously computed counterterms are used to take care of the subdivergences. This indicates that the 3-loop order definition of the normal product of our local operator will require its mixing with the 3+7 = 10 other operators previously introduced (with regularization dependent coefficients $$a_\ell , \ell= 2, 3,..., 11$$), in such a way that eq.(9) holds true for $$\delta =6, n=2\ .$$ Of course the coefficients $$a_1, a_2, a_3$$ and $$a_4$$ will be modified to 3-loop order, and precisely defined thanks to eq.(9) for $$\delta =6, n=6, 4\ .$$
Figure 4: $$\langle 0\mid T[N_6[\frac{1}{6!}\Phi^6(0)]\mid 0 \rangle\mid^{\mathrm{Proper}}$$
STEP 4: Finally, the Green function $\langle 0\mid T[N_6[\frac{1}{6!}\Phi^6(0)] \mid 0 \rangle\mid^{\mathrm{Proper}}$ has a superficial degree of divergence of 6. Nevertheless this Green function does not depend on any momentum, hence its divergence is $$\{p_\ell\}$$-independent: $( m^2)^3 a_{12}\,,$ and requires addition of a new counterterm of the form $-\chi(x)a_{12} ( m^2)^3\mathbf{1}\;.$ The first non-trivial contribution to this counterterm arises at five loops, as shown in Figure 4. Here again, previously computed counterterms are used to take care of the subdivergences. This indicates that the higher loop order definition of the normal product of our local operator will require its mixing with the 3+7 +1 = 11 above mentioned ones (with regularization dependant coefficients $$a_\ell , \ell= 2, 3,..., 12$$) in such a way that eq.(9) also holds true for $$\delta =6, n=0\ .$$ Again, the coefficients $$a_1, a_2..., a_{11}$$ will be modified, and precisely defined thanks to eq.(9) for $$\delta =6, n=6, 4, 2\ .$$
CONCLUSIONS: The previous analysis proves the announced results that the full quantum definition of a finite composite operator $$N_6[\frac{1}{6!}\Phi^6(x)]$$ requires 12 normalization conditions and operator mixing. Indeed:
I) Thanks to eq.(9) for $$\delta =6, n=6, 4, 2, 0\,,$$ 12 normalization conditions are necessary to define order by order in the loop expansion the operator $$N_6[\frac{1}{6!}\Phi^6(0)]$$, in such a way that its insertion into any Green function is finite. As illustrated above, this renormalized operator appears as a linear combination of 12 operators with $$n\le6$$ fields and regularization dependent coefficients :
$\tag{12} \begin{array}{lcl} N_6[\frac{1}{6!}\Phi^6(x)]& = &(1- a_1) \frac{1}{6!}\Phi^6(x) \\ &&\, -\left[ a_2 m^2\frac{1}{4!}\Phi^4 - a_3 \frac{1}{3!} \Phi^3 \Box \Phi - a_4 \frac{1}{2!\,2!}\Phi^2\partial_\nu \Phi\partial^\nu \Phi \right](x) \\ &&\,- \left[a_5 ( m^2)^2 \frac{1}{2!}\Phi^2 - m^2 a_6\Phi\Box\Phi - m^2 a_7(\frac{1}{2!}\partial_\nu \Phi\partial^\nu \Phi) + a_8\Phi\Box^2\Phi + a_9 \partial_\nu \Phi\partial^\nu \Box\Phi +a_{10}\frac{1}{2!}\Box \Phi\Box\Phi +a_{11}(\frac{1}{2!}\partial_\nu\partial_\mu \Phi\partial^\nu \partial^\mu \Phi)\right](x) \\ &&\, -( m^2)^3 a_{12}\mathbf{1} \\ &\ = &\displaystyle\sum_{\ell = 1}^{12}Z_{1\ell}\mathcal{O}_\ell(x). \end{array}$
Note that if the number of fields $$n$$ that occurs in the normalization equation (9) goes from 6 to 0, it takes only even values due to the $$\Phi\leftrightarrow -\Phi$$ invariance of the theory.
II) as 12 counterterms are required, and as they involve 11 other composite operators, these ones should be defined simultaneously, from the very beginning, with 11 other external sources $$\chi_\ell(x)$$. With a similar analysis as done above, these 11 other dimension $$\leq 6$$ scalar quantum local operators $$N_6[\mathcal{O}_k(x)]$$ would also be defined recursively in the loop expansion by 11 normalization equations similar to eq.(9).
To summarize, the source terms in the effective Lagrangian density will be expressed as: $\mathcal{L}_{sources} = \displaystyle \sum_{k,\,\ell\ =1}^{12} \chi_k(x)Z_{k\ell}\mathcal{O}_\ell(x)\,,$ where the $$Z_{k\ell}$$ are regularization dependent (divergent) and then normalization dependent, loop ordered formal series: should one modify the normalization conditions, i.e. the right hand sides of the 12 equations (9), these $$Z_{k\ell}$$ would be modified by finite amounts.
This means that the renormalization of the whole set $$\mathcal{O}_k(x)\,, k=1,..12$$ will be given by a $$12 \times 12$$ matrix. Note that $$Z_{k\ell} = 0 \ {\rm if}\ d_k < d_\ell$$ where $$d_k$$ is the canonical dimension of the operator $$\mathcal{O}_k(x)\,.$$ Correspondingly, the renormalization group equation for Green functions with insertion of any combination of these local operators (at different space time points!) will involve an anomalous dimension $$12 \times 12$$ matrix $$\gamma_{k\ell}$$ reflecting the operator mixing. This matrix, along with some $$\beta$$ functions related to coupling constant renormalization, dictates how the scaling properties of these composite operator are modified in the interacting theory.
Note that in our example, we have supposed the invariance of the interacting theory under the transformation $$\Phi(x)\rightarrow-\Phi(x)\ :$$ if not true, other local operators would be involved into the mixing.
This example illustrates the announced general result: given a set $$\{ :\mathcal{O}_k(x):\}\ ,$$ a basis of free field composite operators of canonical dimension $$d_k \le \delta$$, the normal products $$N_\delta[\mathcal{O}_k(x)]$$ offer a basis for renormalized composite operators of canonical dimension $$\le \delta\,.$$
#### Example of the electromagnetic current
As announced above, in the presence of local invariances, some Ward identities relate the Green functions: as a consequence, the corresponding renormalizations will not be all independent. In particular, the operator corresponding to a conserved current has no renormalization at all (its anomalous dimension vanishes).
Consider for example the electromagnetic current in Quantum Electrodynamics: $J^{\mu}(x) = N_3[e \bar{\psi}(x)\gamma_{\mu} \psi(x)]\,.$ Its canonical dimension being 3, it may mix with all the other operators with canonical dimension less or equal to $$3\ :$$ $\tag{13} N_3[A^\nu \partial_\nu A_\mu]\,, N_3[A^\nu\partial_\mu A_\nu]$
and $$N_3[A^2\,A_\mu]\,.$$
Indeed, the same analysis as above would show that the primitively proper divergent graphs with one insertion of the current operator may a priori have either $$2$$ external fermionic lines, or $$2$$ photon lines or $$3$$ photon lines.
However, due to gauge invariance and charge conjugation invariance, the only divergent Green function with one insertion of the current operator is $\langle 0\mid T[J^{\mu}(x) \tilde{\psi}(p)\tilde{\bar{\psi}}(p')]\mid0 \rangle \,.$ To say it in other words, mixing must preserve gauge invariance: hence the current operator, which is gauge invariant, cannot mix with the other operators listed in (13) that are not gauge-invariant (and cannot give rise to a gauge-invariant operator). It follows that the renormalization of the current is multiplicative. Of course, the renormalization of the other $$d \leq 3\ ,$$ non gauge invariant, operators will still require some mixing among them (and a mixing matrix $$Z_{k\ell}$$ compatible with charge conjugation invariance).
Moreover, thanks again to gauge invariance (i.e. to the conservation of the electromagnetic current), as soon as the fermion field, the photon field and the electric charge are properly renormalized, the Green function $\langle 0\mid T[J^{\mu}(x) \tilde{\psi}(p)\tilde{\bar{\psi}}(p')]\mid 0 \rangle$ is finite: no new renormalization is required for the current operator. In other words, the anomalous dimension of the current operator vanishes.
### The Zimmermann identities
In a previous subsection, we mentioned the notion of over-subtracted normal product, i.e. of a normal product $$N_\delta[\mathcal{O}(x)]\,,$$ with $$\delta$$ greater than the canonical dimension $$d_\mathcal{O}$$ of the operator $$\mathcal{O}(x)\ .$$
It has been shown by Zimmermann (1970), that an over-subtracted operator $$N_\delta[\mathcal{O}(x)]$$ (with $$\delta> d_\mathcal{O}$$) may be expressed in terms of a basis of minimally subtracted operators: $\tag{14} N_\delta[\mathcal{O}(x)] = N_{d_\mathcal{O}}[\mathcal{O}(x)] + \sum_\ell c_\ell N_{d_{\mathcal{A}_\ell}}[\mathcal{A}_\ell(x)]\;,$
where the family of local operators $$[\mathcal{O}(x),\mathcal{A}_\ell(x)]$$ is a basis of local operators of canonical dimension less or equal to $$\delta$$ (possibly restricted by symmetry arguments: parity, charge conjugation, Lorentz covariance, ...) and the coefficients $$c_\ell,$$ vanishing with the Planck constant $$\hbar,$$ may be expressed as definite Green functions. Expressions like the one shown in (14) are known as Zimmermann identities.
Zimmermann identities allow the rewriting of the basis (mentioned in the last sentence of subsection Example in scalar $$\Phi^4$$ theory) $$N_\delta[\mathcal{O}_k(x)]$$ of composite operators of dimension $$d_k \le \delta$$ as a basis of minimally subtracted ones $$N_{d_k}[ \mathcal{O}_k(x)]\,.$$
Zimmermann identities (14) have been adapted in (Bonneau G., 1980) to the dimensional renormalization scheme, where composite operators are always minimally subtracted. In this case, the introduction of a factor $$(4-D)\ ,$$ where $$D$$ is the complex regularizing dimension, in front of an operator $$\mathcal{O}(x),$$ gives rise to a similar expansion: $\tag{15} N[(4-D)\mathcal{O}(x)]= c_0 N[\mathcal{O}(x)] + \sum_\ell c_\ell N[\mathcal{A}_\ell(x)]$
where the family of local operators $$[\mathcal{O}(x),\mathcal{A}_\ell(x)]$$ is now a basis of local operators of dimension $$\le d_\mathcal{O}\ ,$$ and the coefficients $$c_0, c_\ell,$$ vanishing with the Planck constant $$\hbar,$$ may be expressed as residues of the simple poles in the variable $$(D-4)$$ of properly subtracted (for all their subdivergences) definite proper Green functions.
Such operators, that vanish at the tree level when the regularizing dimension goes to 4, but contribute in the fully interacting situation, are called evanescent operators. Apart from $$N[(4-D)\mathcal{O}(x)]\ ,$$ examples are linked to the non-anticommutation of Dirac matrices $$\gamma_\mu$$ and $$\gamma^5$$ or with the loss of algebraic properties of spinors in supersymmetric theories when the regularizing dimension $$D \neq 4$$ (consider for instance the evanescent operator $$N[\bar{\Psi}\{\gamma^5,\gamma_\mu\}_+\Psi(x)]$$ involved into the Ward identity for the axial current).
These identities (eq.(15) also hold for such evanescent operator, with the understanding that the basis of operators in the right hand side should be extended to all evanescent and non evanescent operators of canonical dimension $$\le d_\mathcal{O}$$ (again possibly restricted by symmetry arguments: parity, charge conjugation, Lorentz covariance, ...)
These Zimmermann identities are useful to discuss anomalies in renormalized Ward identities in BPHZ renormalization scheme (eq.(14)) or in dimensional renormalization scheme (eq.(15), Bonneau G., 1980).
### The product of composite operators in interacting quantum field theory
In the short-distance limit, $$y \rightarrow x\ ,$$ and as it was in the free field case (section : The product of composite operators ...) the product of two well-defined normal products $$N_\delta[\mathcal{O}(x)]$$ and $$N_{\delta'}[\mathcal{O}'(y)]$$ is a priori different from the normal product $$N_{\delta + \delta'}[\mathcal{O}(x)\mathcal{O}'(x)]$$ (as usual, the limit being in a weak sense), and, moreover it is generally singular!
More specifically, it happens that the renormalization of Green functions involving the product of two or more composite operators may require (in addition to the mixing of operators) new type of subtractions known as contact terms.
The short distance behavior ($$y \rightarrow x$$) of the product of composite operators can be characterized via the so-called operator product expansion. Alternatively, the behavior of the product of composite operators when $$(x-y)^2\rightarrow 0$$ (in the Minkowsky metric) is studied via the so-called light-cone expansion.
## References
• W. Zimmermann (1970), Local operator products and renormalization in quantum field theory, in 1970 Brandeis Lectures, vol. 1, p.395, eds. S. Deser et al. (M.I.T. Press, Cambridge, 1970).
• P. Breitenlohner and D. Maison (1977), Commun. Math. Phys. 52, 11.
• Bonneau G. (1980), Nucl. Phys. B167, 261 and B171, 477.
Internal references
• C. Itzykson and J.B. Zuber Quantum field theory, McGraw-Hill Inc. 1980 (in particular Chap 3.1, 4.2 and 8.2)
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1. ## two intregal problems
Hi all,
can someone solve these two problems for me?
The first one is to proof is that is true or not
and the second one, I just totally forgot how to do the anti derivative of that....
Thanks!!!
[IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
[IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
[IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
2. Here's a hint
$\int \frac{1}{t^2+1}~dt = \arctan (t) +C$
3. With respect to the first question: I think the first step is to integrate both functions
for $\int_0^1 \sqrt{1+x^2}$ sinces it's of the form $\sqrt{a^2+x^2}$ let $x=\arctan\theta$
and for $int_0^1 \sqrt{1+x}$ use u-substitution with $u=1+x$
I'm not sure but integration may note be nescecary...because the one with $x^2$ is obviously always going to be greater than or equal to the other.
Does this help?
4. Originally Posted by Jskid
I'm not sure but integration may note be nescecary...because the one with $x^2$ is obviously always going to be greater than or equal to the other.
But after you integrate that will change.
5. Originally Posted by J
I'm not sure but integration may note be nescecary...because the one with [tex
x^2[/tex] obviously always going to be greater than or equal to the other.
Does this help?
yeah...clearly integration is not needed. Simply notice that x is between 0 and 1 and hence $x^2\leq x$. This directly leads to the given inequality since everything else is the same in both sides.
6. jskid and TheCoffeeMachine are both incorrect on some points; mohammadfawaz is correct. On the interval from 0 to 1, $x^{2}\le x.$ That's true regardless of whether you integrate or not.
7. The first question doesn't require any integration tool because for $0 \le x \le 1$ is $\sqrt{1+x^{2}} \le \sqrt{1+x}$ ...
Kind regards
$\chi$ $\sigma$
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aboutsummaryrefslogtreecommitdiff log msg author committer range
blob: 056e6bdff35595a1cde847edb0f5b131cd755b92 (plain)
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ``` ``````#N canvas 0 22 565 389 10; #X text 74 61 Lyapunov Exponent; #X text 25 82 This value is an estimate of the attractors potential for chaos. It will not necessarily be the same for any given run using any arbitrary fractal. The description below attempts to put "attractors" into three catagories.; #X text 48 262 Don't forget \, just because an "attractor" is not chaotic \, it does not mean that it will not generate an interesting stream of number \, if only until they converge.; #X text 40 151 < 0; #X text 78 151 These are not chaotic \, may produce a "short-term" stream of intrest; #X text 39 179 == 0; #X text 80 179 Attractors converge to one or more points; #X text 40 197 > 0; #X text 80 198 Chaos begins. Higher values indicate "more" chaos.; ``````
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# What is the permutation representation of $SL_2(\mathbb{F}_p)$, $PSL_2(\mathbb{F}_p)$ and $GL_2(\mathbb{F}_p)$?
It seems that there is a action by which $SL_2(\mathbb{F}_p)$ and $GL_2(\mathbb{F}_p)$ permute the $p^2$ ordered tuples in $\mathbb{F}_p^2$. What is the map from the $2 \times 2$ matrices over $\mathbb{F}_p$ versions of the elements of these groups to the elements of $S_{p^2}$ that this action picks out?
Similarly there seems to be an injective homomorphism map from $PSL_2(\mathbb{F}_p)$ to $S_{p+1}$ by virtue of the natural action on $\mathbb{F}_pP^1$ (the projective space obtained from $\mathbb{F}_p^2$) What is this mapping?
• I don't understand the question. You've already described these maps. – Qiaochu Yuan Apr 30 '15 at 16:52
• @QiaochuYuan See my comments to Herbert's answer. – user6818 Apr 30 '15 at 17:10
A matrix $A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \in \text{GL}_2(\mathbb{F}_p)$ acts on a vector $v = \left[ \begin{array}{cc} v_0 \\ v_1 \end{array} \right] \in \mathbb{F}_p^2$ by matrix multiplication:
$$Av = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{cc} v_0 \\ v_1 \end{array} \right] = \left[ \begin{array}{cc} a v_0 + b v_1 \\ c v_0 + d v_1 \end{array} \right].$$
This is far and away the best description of this action I can give. You can figure out everything you could possibly want to know about this action from linear algebra. It is even better than knowing a permutation representation because linear algebra is easier than set theory, even finite set theory.
For example, here is how you figure out the cycle decomposition of the permutation corresponding to $A$. It splits into cases depending on the conjugacy class of $A$ as follows.
Case: $A$ is diagonalizable over $\mathbb{F}_p$. Then up to conjugacy it is a diagonal matrix with diagonal entries $\lambda_0, \lambda_1 \in \mathbb{F}_p^{\times}$. Its order is the lcm of the orders $d_i$ of $\lambda_i \in \mathbb{F}_p^{\times}$, which can be any divisor of $p - 1$. The corresponding permutation of $\mathbb{F}_p^2$ fixes zero and splits the nonzero vectors into cycles as follows:
• The nonzero vectors of the form $(v_0, 0)$ form cycles of length $d_0$.
• The nonzero vectors of the form $(0, v_1)$ form cycles of length $d_1$.
• The remaining nonzero vectors form cycles of length $\text{lcm}(d_0, d_1)$.
Case: $A$ is diagonalizable over $\mathbb{F}_{p^2}$ but not over $\mathbb{F}_p$. Then up to conjugacy its action on $\mathbb{F}_p^2$ is the action of multiplication by an element $\lambda \in \mathbb{F}_{p^2}^{\times}$ (one of the eigenvalues of $A$) on $\mathbb{F}_{p^2}$. Its order is the order $d$ of $\lambda$ in $\mathbb{F}_{p^2}^{\times}$, which can be any divisor of $p^2 - 1$. The corresponding permutation of $\mathbb{F}_{p^2}$ fixes zero, and the nonzero vectors form cycles of length $d$.
Case: $A$ is not diagonalizable even over $\mathbb{F}_{p^2}$. From the theory of rational canonical form we deduce that up to conjugacy $A$ is a Jordan block
$$A = \left[ \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array} \right]$$
where $\lambda \in \mathbb{F}_p^{\times}$. Its order is $p$ times the order $d$ of $\lambda$, which can be any divisor of $p - 1$. The corresponding permutation of $\mathbb{F}_p^2$ fixes zero and splits the nonzero vectors into cycles as follows:
• The nonzero vectors of the form $(v_0, 0)$ form cycles of length $d$.
• The remaining nonzero vectors form cycles of length $pd$.
The analysis for the action on the projective line is similar. Writing a point on the projective line in projective coordinates $(v_0 : v_1)$, the action takes the form
$$A (v_0 : v_1) = (av_0 + bv_1 : cv_0 + dv_1)$$
or, thinking of an element of the projective line as a fraction $z = \frac{v_0}{v_1}$ (which may take the value $\infty$),
$$A z = \frac{az + b}{cz + d}.$$
Again you can figure out everything you could possibly want to know about this action from linear algebra as above. In the diagonalizable case, for example, projectively the action of a diagonal matrix with entries $d_0, d_1$ is multiplication by $\frac{d_0}{d_1}$, and hence its cycle decomposition is determined by the order of $\frac{d_0}{d_1}$. So things are even simpler.
• @Qiachu Yuan Thanks for the help! Do you know what the irrep decomposition is of these permutation representations? – user6818 Apr 30 '15 at 19:34
• @Qiachu Yuan What I need is an enumeration of all the group elements of these groups, SL_2(F_q) or PSL_2(F_q) or GL_2(F_q) or PGL_2(F_q), so that I can do these calculations by hand for specific cases. Is this somehow available? – user6818 Apr 30 '15 at 20:05
• @user6818: I don't know what you mean by the irrep decomposition of a permutation representation. The decomposition of $\mathbb{F}_p^2$ into orbits is that $0$ is an orbit and the nonzero vectors are the other orbit. The decomposition of $\mathbb{F}_p^2$ as a linear representation is that it's irreducible. I also don't know what you mean by an enumeration of all the group elements. You can just write down matrices and chuck out the ones that aren't invertible. – Qiaochu Yuan Apr 30 '15 at 20:11
• I am not understanding what you are not understanding :D Here is an action of these groups on $\mathbb{F}_p^2$ (or on $\mathbb{P}(\mathbb{F}_p^2 )$) so we have a representation of these groups on these vector spaces. (so we have here basically listed out 6 representations) Are any of these irreducible? If not then what is the irrep decomposition of these representations? – user6818 Apr 30 '15 at 20:23
• @user6818: the grammar was throwing me off. "Irrep decomposition" is a term people apply to linear representations, not permutation representations (which I took to mean actions on sets). One thing you might have meant by "permutation representation" is the linear representation induced by an action on a set, which is different from the linear representation on $\mathbb{F}_p^2$ that we're talking about, so I found that ambiguous. Anyway, the projective line is not a vector space, but the action of $\text{GL}_n$ on an $n$-dimensional vector space is always irreducible. – Qiaochu Yuan Apr 30 '15 at 20:26
$\mathrm{GL}_2 (\mathbb{F}_p)$ have a natural action on $\mathbb{P}_1(\mathbb{F}_p)$ : the action of $g$ on a vector line $d$ is just the vector line $g\cdot d$. Its kernel is the set of elements in $\mathrm{GL}_2 (\mathbb{F}_p)$ which stabilize every one-dimensional vector space and it is easy to show that this is the center of $\mathrm{GL}_2 (\mathbb{F}_p)$ (homotheties). Therefore, the quotient of $\mathrm{GL}_2 (\mathbb{F}_p)$ (which is $\mathrm{PGL}_2 (\mathbb{F}_p)$ ) is mapped to the permutation group of $\mathbb{P}_1(\mathbb{F}_p)$ and this is an injection.
As $\mathbb{P}_1(\mathbb{F}_p)$ has $1+q$ elements, this provides an injection $$\mathrm{PGL}_2 (\mathbb{F}_p) \hookrightarrow \mathfrak{S}_{p+1}$$
For low $p$, this leads to so-called "exceptional isomorphisms", such as $$\mathrm{PGL}_2(\mathbb{F}_5) \simeq \mathfrak{S}_5$$
• I guess my question isn't clear! I want to know what these maps are! Like given a $2\times 2$ matrix with entries in $\mathbb{F}_p$ for any of these three groups, I want to know what is the permutation matrix that corresponds to it in this action. – user6818 Apr 30 '15 at 17:00
• It will depend on the basis of the underlying vector space. – Tlön Uqbar Orbis Tertius Apr 30 '15 at 17:04
• Yes, obviously. But can you give me the action? As in given a $2 \times 2$ matrix over $\mathbb{F}_p$ an element of say $SL_2(F_p)$ what is its action on some element say $(a,b) \in \mathbb{F}_p^2$ ? – user6818 Apr 30 '15 at 17:06
• And equivalently is there a way to explicitly enumerate all the elements of any of these groups? – user6818 Apr 30 '15 at 17:10
• @user6818: the way you phrase this question makes it unclear to me whether you know how a matrix acts on a vector. Is that all you're asking about? – Qiaochu Yuan Apr 30 '15 at 17:33
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# Windows – Task scheduler On Idle task not triggered Windows 7
I've used Task scheduler and created a new task, added a trigger, set it to event: on Idle and created the task. In the conditionstab, specified the task waits 1min to become idle (instead of 10min just to test).
But then I wait for more than 10mins and the task is never triggered on PC1 but triggers on another PC2. Also if I press Run in Task Scheduler the task runs normally in PC2 it just never triggers automatically when idle for 1 min. I've checked online and found that PowerCfg -requests will show what is stopping the PC from being Idle and when I run it on PC2 nothing returns but when I run it on PC1 I get this "Legacy Kernel Caller" driver.
So I used Powercfg -requestsoverride but when I run powercfg -requestsoverride Driver "Legacy Kernel Caller" System , where it succeeds and I find it in the Powercfg -requestsoverride list, but then powercfg -requests it still shows [DRIVER] Legacy Kernel Caller under SYSTEM: and the task is never triggered any advice please?
Also here is the xml of my Task:
<?xml version="1.0" encoding="UTF-16"?>
<RegistrationInfo>
<Date>2019-11-20T15:43:06.6081219</Date>
<Author>MyPC\MyUser</Author>
</RegistrationInfo>
<Triggers>
<IdleTrigger>
<Enabled>true</Enabled>
</IdleTrigger>
</Triggers>
<Principals>
<Principal id="Author">
<UserId>S-1-5-21-1004336348-1177238915-682003330-385281</UserId>
<LogonType>InteractiveToken</LogonType>
<RunLevel>LeastPrivilege</RunLevel>
</Principal>
</Principals>
<Settings>
<MultipleInstancesPolicy>IgnoreNew</MultipleInstancesPolicy>
<DisallowStartIfOnBatteries>true</DisallowStartIfOnBatteries>
<StopIfGoingOnBatteries>true</StopIfGoingOnBatteries>
<AllowHardTerminate>true</AllowHardTerminate>
<StartWhenAvailable>false</StartWhenAvailable>
<RunOnlyIfNetworkAvailable>false</RunOnlyIfNetworkAvailable>
<IdleSettings>
<Duration>PT1M</Duration>
<WaitTimeout>PT0S</WaitTimeout>
<StopOnIdleEnd>true</StopOnIdleEnd>
<RestartOnIdle>false</RestartOnIdle>
</IdleSettings>
<AllowStartOnDemand>true</AllowStartOnDemand>
<Enabled>true</Enabled>
<Hidden>false</Hidden>
<RunOnlyIfIdle>true</RunOnlyIfIdle>
<WakeToRun>false</WakeToRun>
<ExecutionTimeLimit>PT72H</ExecutionTimeLimit>
<Priority>7</Priority>
</Settings>
<Actions Context="Author">
<Exec>
<Command>C:\MyWinFormApp\MyWinForm.exe</Command>
</Exec>
</Actions>
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1. ## definite integrals
use definite integrals to calculate :
i)the area between y=x squared and y=5x
ii) the volume of the solid generated when the curve xy=2 is rotated one revoution about x-axis between the limits x = 2 and x = 4
2. Originally Posted by size
use definite integrals to calculate :
i)the area between y=x squared and y=5x
ii) the volume of the solid generated when the curve xy=2 is rotated one revoution about x-axis between the limits x = 2 and x = 4
What have you tried so far?
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Abstract algebra不算是一门简单的学科，这门学科在国内叫做抽象代数，经常有很多学生在学linear algebra或者analysis(advance calculus)的时候觉得并不困难，但是却觉得Abstract algebra很难，这是因为没有找到正确的方法学习Abstract algebra，UpriviateTA有一系列非常擅长Abstract algebra的老师，可以确保您在Abstract algebra取得满意的成绩。
Problem 1. Suppose that $G$ is an infinite simple group. Show that for every proper subgroup $H$ of $G,$ the index $[G: H]$ is infinite.
Proof . If $[G: H]=n<\infty,$ then $G$ can be embedded in $S_{n},$ so $G$ is finite, a contradiction.
Problem 2. Let $G$ act on left cosets of $H$ by multiplication. Show that the kernel of the action is
$$N=\bigcap_{x \in G} x H x^{-1}$$
Proof . $g(x H)=x H$ iff $x^{-1} g x \in H$ iff $g \in x H x^{-1}$
Problem 3. Let $H$ be a subgroup of $G$ of finite index $n,$ and let $G$ act on left cosets $x H$ by multiplication. Let $N$ be the kernel of the action, so that $N \leq H$. Show that $[G: N]$ divides $n !$
Proof . By the first isomorphism theorem, $G / N$ is isomorphic to a group of permutations of $L$, the set of left cosets of $H .$ But $|L|=[G: H]=n,$ so by Lagrange’s theorem, $\mid G / N$ divides $\left|S_{n}\right|=n !$
Problem 4. Let $H$ be a subgroup of $G$ of finite index $n>1 .$ If $|G|$ does not divide $n !$, show that $G$ is not simple.
Proof . Since $n>1, H$ is a proper subgroup of $G,$ and since $N$ is a subgroup of $H, N$ is a proper subgroup of $G$ as well. If $N=\{1\},$ then $|G|=[G: N],$ so by Problem 8 , $G$ divides $n !$, contradicting the hypothesis. Thus $\{1\}<N<G,$ and $G$ is not simple.
Problem 5.
Here is some extra practice with left cosets of various subgroups. Let $H$ and $K$ be subgroups of $G,$ and consider the map $f$ which assigns to the coset $g(H \cap K)$ the pair of cosets $(g H, g K)$. Show that $f$ is well-defined and injective, and therefore
$$[G: H \cap K] \leq[G: H][G: K]$$
Thus (Poincaré) the intersection of finitely many subgroups of finite index also has finite index.
Proof .
$g_{1}(H \cap K)=g_{2}(H \cap K)$ iff $g_{2}^{-1} g_{1} \in H \cap K$ iff $g_{1} H=g_{2} H$ and $g_{1} K=g_{2} K,$ proving
both assertions.
abstract algebra代写请认准UpriviateTA. UpriviateTA为您的留学生涯保驾护航。
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# garch models caught in the spotlight
January 13, 2014
By
(This article was first published on Portfolio Probe » R language, and kindly contributed to R-bloggers)
An attempt to clarify the basics.
## Previously
There have been several posts about garch. In particular:
## Genesis
A reader emailed me because he was confused about the workings of garch in general, and simulation with the empirical distribution in particular.
If there is one confused person, there are more.
## Model view
A basic feature of garch models is that they are in discrete time. The frequency is usually daily, in which case we pretend that the volatility of the returns is constant throughout each day.
Deep in the heart of a garch model is an innovation at each timepoint. This is — conceptually — a draw from a statistical distribution that has variance 1. Note that variance 1 rules out infinite variance, and hence stable distributions. Also, the normal distribution is unlikely to be a good approximation.
The innovation is the net result of the “news” that comes into the market during the period. If the innovation is large (in absolute value, and compared to a variance 1 distribution), then that is a signal that future volatility is likely to be larger.
But we don’t see innovations. We see innovations scaled by the current volatility.
What the garch model keeps track of is the conditional variance at each time. The “conditional” being short for conditional on the returns (and possibly other variables) that have already occurred. ”Variance” because it is mathematically handy — in higher dimensions there are variance matrices but you’ve never heard of a standard deviation matrix.
So far our model is:
return at time t = innovation at time t, scaled using the conditional variance at time t
There is one more element to the model: the mean return at time t. The typical garch model is:
return at time t = mean return at time t + innovation at time t, scaled using the conditional variance at time t
Keep in mind that the conditional variance needs to be transformed before it can be used to scale the innovation.
## Reality view
The fact is that there is a return — a single number — for a time period. The garch models are taking that number and decomposing it into three pieces:
• the mean return at that time
• the conditional variance at that time
• the innovation for that time
Those three things are fictional. And the fictions of the last two heavily depend on each other.
Sometimes they are useful fictions.
## Data view
The basic input to a garch model is a time series of returns. It is log returns that make the most sense in this setting.
Possible specifications for the mean return are:
• zero
• a constant (to be estimated)
• a simple autoregression — AR(1)
• an ARMA model
• something even fancier
There is unlikely to be much difference between assuming the mean is zero and any of the others. In particular, the estimation of the variance is highly unlikely to change significantly.
### Residuals
It is easy for the term “residual” to be confusing in garch models. That’s because the residuals don’t really have anything to do with the garchiness of the model.
The residual at time t is the return at time t minus the estimated mean return at time t.
More garchy — and more useful — are the standardized residuals. The standardized residual at time t is the residual at time t divided by the square root of the conditional variance at time t.
In other words, the standardized residuals are the estimates of the innovations.
### Fit
There are two parts to the fit of a garch model:
• fit for the mean return
• fit for the conditional variance (or standard deviation or volatility)
The second of these is the more interesting.
### Simulation
• a garch model with the parameters specified
• a state for the conditional variance
Usually, of course, the parameter specification is taken from the estimated model. It is also typical that the variance state is the most recent variance state from the estimated model.
There is one more thing the simulation needs — innovations.
The two most common choices for generating a series of innovations are:
• sample from the distribution assumed in the model (including any estimated parameters)
• sample from the standardized residuals of the estimated model
The process of simulation is to use the first innovation, scale it via the volatility state and add in the mean return to get the first return in the simulation — let’s say this is for time T+1.
We then need to update the volatility state for T+1 — this depends on the innovation.
After those two steps we are in the same situation except we are one time step further. We continue doing those steps until we have as many timepoints as we want.
Usually we do that process multiple times so that we have a large number of simulated series.
## Epilogue
Fancy people go driftin’ by.
The moment of truth is right at hand,
Just one more nightmare you can stand.
from “Stage Fright” by Robbie Robertson
## Appendix R
You can use garch in R, among other places.
There are alternatives, but the rugarch package seems to be the most complete implementation in R. Here is a small example. The first step is:
require(rugarch)
This command will only work once the package has been installed on your machine.
#### set the garch specification
We start by creating a specification of the garch model:
comtspec <- ugarchspec(mean.model=list(
armaOrder=c(0,0)), distribution="std",
variance.model=list(model="csGARCH"))
This specification does three things. It sets:
• the model for the mean return to be a constant to be estimated — the default is an ARMA(1,1)
• the distribution of the innovations to be a Student-t distribution — the default is normal
• the model for the variance to be components garch — the default is garch(1,1)
#### estimate the model
We use the specification when fitting the model:
garEstim <- ugarchfit(comtspec, data=tail(spxret, 2000))
The data are the most recent 2000 returns in the vector.
#### get residuals
The regular residuals for the garch model are:
garResid <- residuals(garEstim)
The standardized residuals are computed like:
garStandResid <- residuals(garEstim, standardize=TRUE)
We can see that the results make sense by inspecting the variances:
> var(tail(spxret, 2000))
[1] 0.0001969673
> var(garResid)
[,1]
[1,] 0.0001969673
> var(garStandResid)
[,1]
[1,] 1.204764
I’m slightly concerned that the variance of the standardized residuals is so far from 1, but clearly it is trying to be close to 1.
#### how you might have found how to get residuals
The residuals function is generic. There are two ways to look for the methods of a generic function:
> methods('residuals')
[1] residuals.default residuals.glm
[3] residuals.HoltWinters* residuals.isoreg*
[5] residuals.lm residuals.nls*
[7] residuals.smooth.spline* residuals.tukeyline*
Non-visible functions are asterisked
> showMethods('residuals')
Function: residuals (package stats)
object="ANY"
object="ARFIMAfilter"
object="ARFIMAfit"
object="ARFIMAmultifilter"
object="ARFIMAmultifit"
object="uGARCHfilter"
object="uGARCHfit"
object="uGARCHmultifilter"
object="uGARCHmultifit"
(Note that the results you get depend on what packages are loaded into your R session.)
The class of the object that we want is:
> class(garEstim)
[1] "uGARCHfit"
attr(,"package")
[1] "rugarch"
This is found in the second group. To have a look at the method in this group, you do:
getMethod('residuals', 'uGARCHfit')
It is here that we see that standardize is an argument (though not exactly blatantly obviously).
#### encore
The post “garch and the distribution of returns” has some examples of simulation.
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# Hand in problems
1.10
1.10
Fix $$a \in \mathbb{C}$$ and $$b \in \mathbb{R}$$. Show that the equation $$|z^2|+Re(az)+b=0$$ has a solution if an only if $$|a^2|\geq 4b$$. When solutions exist, show the solution set is a circle.
Fix $$a \in \mathbb{C}$$ and $$b \in \mathbb{R}$$. Suppose that the equation $$|z^2|+Re(az)+b=0$$ has a solution. Let $$z=x+iy$$ and $$a=c+id$$
So: \begin{aligned} |z^2|+Re(az)+b&=0\\ x^2+y^2+cx-dy+b&=0\\ x^2+cx+y^2-dy&=-b\\ x^2+cx+\frac{1}{4}c^2+y^2-dy+\frac{1}{4}d^2&=-b+\frac{1}{4}c^2+\frac{1}{4}d^2\\ (x+\frac{1}{2}c)^2+(y+\frac{1}{2}d)^2&=-b+\frac{1}{4}(c^2+d^2)\\ (x+\frac{1}{2}c)^2+(y+\frac{1}{2}d)^2&=-b+\frac{1}{4}|a^2|\end{aligned} Thus the solution set is a circle centered at $$-\frac{1}{2}+\frac{1}{2}i$$ with radius $$\sqrt{-b+\frac{1}{4}|a^2|}$$. By checking the left hand side, $$(x+\frac{1}{2}c)^2+(y+\frac{1}{2}d)^2\geq0$$. So the right hand side is: \begin{aligned} -b+\frac{1}{4}|a^2|&\geq0\\ \frac{1}{4}|a^2|&\geq b\\ |a^2|\geq4b\end{aligned} Meanwhile, assume $$\mid a^2 \mid \geq 4b$$. So: \begin{aligned} \mid a^2 \mid &\geq 4b\\ a\overline{a}&\geq 4b &\text{ by proposition 1.10f}\\ c^2+d^2&\geq 4b\\ \frac{c^2+d^2}{4}&\geq b & \text{ by algebra}\\ \frac{c^2+d^2}{4}-b &\geq 0\\ \frac{c^2}{4}+\frac{d^2}{4}-b &\geq 0 &\text{ equation of a circle}\end{aligned} Since the solution set is a circle, then $$|z^2|+Re(az)+b=0$$ has a solution when $$\mid a^2 \mid \geq 4b$$.
Therefore $$|z^2|+Re(az)+b=0$$ has a solution if and only if $$|a^2| \geq 4b$$, as was to be shown.
1.19
1.19
Fix a positive integer $$n$$ and a complex number $$w$$. Find all solutions to $$z^n=w$$.
—————————————————————————————————————–
Let $$n \in \mathbb{Z}$$ and $$z,w \in \mathbb{C}$$. In polar coordinates, $$w=re^{i\theta}$$ so: $$z^n=re^{i\theta}$$. Let $$z=me^{i\phi}$$. The modulus of $$z$$ is $$m^n=r$$ ie $$m=\sqrt[n]{r}$$. So for $$\theta$$ of $$z$$, we have $$n\theta=\theta+2\pi k$$ for some $$k \in \mathbb{Z}$$ and $$\theta=\frac{\theta+2 \pi k}{n}$$. So $$z=\sqrt[n]{r}*e*\frac{i(\theta+2 \pi k)}{n}$$. By the fundamental theorem of algebra we know that there are $$n$$ solutions to this equation. Restricting $$k$$ so $$0 \leq k \leq n$$, then there are $$n$$ unique solutions for $$z$$.
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# Mixed Forms and Characteristic Classes¶
This Jupyter notebook illustrates SageMath's functionalities regarding mixed differential forms and characteristic classes. The involved tools have been developed through the SageManifolds project.
Author: Michael Jung
A version of SageMath at least equal to 9.0 is required to run this notebook:
In [1]:
version()
Out[1]:
'SageMath version 9.1.beta6, Release Date: 2020-03-01'
First we set up the notebook to display math formulas using LaTeX formatting:
In [2]:
%display latex
## Mixed Differential Forms¶
We briefly demonstrate the capabilities of our implementation, and start by declaring the manifold $M=\mathbb{R}^2$:
In [3]:
M = Manifold(2, 'R^2', latex_name=r'\mathbb{R}^2')
X.<x,y> = M.chart()
We define the corresponding spaces of differential forms:
In [4]:
Omega0 = M.diff_form_module(0); print(Omega0)
Omega1 = M.diff_form_module(1); print(Omega1)
Omega2 = M.diff_form_module(2); print(Omega2)
Algebra of differentiable scalar fields on the 2-dimensional differentiable manifold R^2
Free module Omega^1(R^2) of 1-forms on the 2-dimensional differentiable manifold R^2
Free module Omega^2(R^2) of 2-forms on the 2-dimensional differentiable manifold R^2
The algebra of mixed forms is returned by a simple command:
In [5]:
Omega = M.mixed_form_algebra(); print(Omega)
Graded algebra Omega^*(R^2) of mixed differential forms on the 2-dimensional differentiable manifold R^2
It belongs to the category of graded algebras over the symbolic ring:
In [6]:
print(Omega.category())
Category of graded algebras over Symbolic Ring
Before proceeding with mixed forms, let us first declare some differential forms:
In [7]:
f = M.scalar_field(name='f')
omega1 = M.diff_form(1, name='omega_1', latex_name=r'\omega_1')
omega2 = M.diff_form(2, name='omega_2', latex_name=r'\omega_2')
eta = M.diff_form(1, name='eta', latex_name=r'\eta')
In the next step, we provide some expressions in local coordinates:
In [8]:
f.set_expr(x^2)
omega1[:] = y, 2*x
omega2[0,1] = 4*x^3
eta[:] = x, y
In [9]:
f.display()
Out[9]:
In [10]:
omega1.display()
Out[10]:
In [11]:
omega2.display()
Out[11]:
In [12]:
eta.display()
Out[12]:
The category framework of Sage captures the entire setup:
In [13]:
all([f in Omega0,
f in Omega,
omega1 in Omega1,
omega1 in Omega,
omega2 in Omega2,
omega2 in Omega])
Out[13]:
Now, let us define a mixed form:
In [14]:
A = M.mixed_form(name='A'); print(A)
Mixed differential form A on the 2-dimensional differentiable manifold R^2
It shall consist of the differential forms $f, \omega_1, \omega_2$. The forms are assigned by using index operations:
In [15]:
A[:] = [f,omega1,omega2]; A.display()
Out[15]:
In [16]:
A.display_expansion()
Out[16]:
As we can see, the output is sorted by degree. Notice that the forms stored in A are given by the very same instances we declared beforehand:
In [17]:
all([A[0] is f,
A[1] is omega1,
A[2] is omega2])
Out[17]:
If that behavior is unwanted, a copy can be made which has the very same expressions in local coordinates but has stored entirely new instances:
In [18]:
Aclone = A.copy()
any(Aclone[k] is A[k] for k in Omega.irange())
Out[18]:
In [19]:
all(Aclone[k] == A[k] for k in Omega.irange())
Out[19]:
Let us perform some computations and define another mixed form:
In [20]:
B = M.mixed_form([2,eta,0], name='B'); B.display_expansion()
Out[20]:
The multiplication is executed degree wise:
In [21]:
all((A * B)[k] == sum(A[j].wedge(B[k - j])
for j in range(k + 1))
for k in Omega.irange())
Out[21]:
In [22]:
(A * B).display_expansion()
Out[22]:
This particular example is also convenient to demonstrate that the multiplication given by the wedge product is in general neither commutative nor anticommutative:
In [23]:
(B * A).display_expansion()
Out[23]:
Finally, let us compute the exterior derivative:
In [24]:
dA = A.exterior_derivative(); dA.display_expansion()
Out[24]:
## Characteristic Classes¶
### Chern Character over Minkowski Space¶
We want to exemplify the usage of characteristic classes within Sage by computing the Chern character form $\mathrm{ch}(E, \nabla^E)$ on a complex trivial line bundle $E$ over the 2-dimensional Minkowski space $M$ equipped with a bundle connection $\nabla^E$. We start with the general setup:
In [25]:
M = Manifold(2, 'M', structure='Lorentzian')
X.<t,x> = M.chart()
E = M.vector_bundle(1, 'E', field='complex'); print(E)
Differentiable complex vector bundle E -> M of rank 1 over the base space 2-dimensional Lorentzian manifold M
To trivialize the vector bundle $E$, we fix a global frame $e$:
In [26]:
e = E.local_frame('e') # trivialize
Let us declare an $\mathrm{U}(1)$-connection $\nabla^E$ on $E$ given by an electromagnetic potential $A(t)$:
In [27]:
nab = E.bundle_connection('nabla^E', latex_name=r'\nabla^E')
A = function('A')
The corresponding connection form $\omega$ turns out as:
In [28]:
omega = M.one_form(name='omega', latex_name=r'\omega')
omega[1] = I*A(t)
omega.display()
Out[28]:
Let us put this into the connection:
In [29]:
nab.set_connection_form(0, 0, omega)
Notice that the Chern character $\mathrm{ch}(E)$ is already predefined in the system. We can get it by the following command:
In [30]:
ch = E.characteristic_class('ChernChar'); print(ch)
Characteristic class ch of additive type associated to e^x on the Differentiable complex vector bundle E -> M of rank 1 over the base space 2-dimensional Lorentzian manifold M
The computation of the corresponding Chern character form $\mathrm{ch}(E,\nabla^E)$ can be invoked by the method \texttt{get_form}:
In [31]:
ch_form = ch.get_form(nab)
ch_form.display_expansion()
Out[31]:
We can see that the resulting 2-form coincides with the Faraday tensor divided by $2 \pi$ as expected.
### Example: Chern Class of the Tautological Line Bundle¶
We start our computation by initializing the complex projective space as 2-dimensional real manifold with coordinates on $U:= \mathbb{CP}^1 \setminus \left\{ [1:0] \right\}$:
In [32]:
M = Manifold(2, 'CP^1', start_index=1)
U = M.open_subset('U')
c_cart.<x,y> = U.chart() # [1:x+I*y]
For the sake of convenience, we additionally declare the complex coordinates $z$ and $\bar{z}$ on $U$:
In [33]:
c_comp.<z, zbar> = U.chart(r'z:z zbar:\bar{z}')
cart_to_comp = c_cart.transition_map(c_comp, (x+I*y, x-I*y))
cart_to_comp.display()
Out[33]:
In [34]:
comp_to_cart = cart_to_comp.inverse()
comp_to_cart.display()
Out[34]:
Now, we are ready to construct the tautological line bundle $\gamma_1$:
In [35]:
E = M.vector_bundle(1, 'gamma_1',
latex_name=r'\gamma_1',
field='complex')
Furthermore we declare a local frame $e$ on $U$ naturally given by $[z:1] \mapsto \left(\begin{smallmatrix} z \\ 1 \end{smallmatrix}\right)$:
In [36]:
e = E.local_frame('e', domain=U)
To compute the Chern class, we still need a connection. The tautological line bundle inherits a Hermitian metric from the overlying trivial bundle $\mathbb{C}^2 \times \mathbb{CP}^1$:
In [37]:
nab = E.bundle_connection('nabla', latex_name=r'\nabla')
omega = U.one_form(name='omega')
omega[c_comp.frame(), 1, c_comp] = zbar/(1+z*zbar)
nab.set_connection_form(1, 1, omega, frame=e)
It is time to initialize $c(\gamma_1)$. Fortunately Sage already knows the Chern class:
In [38]:
c = E.characteristic_class('Chern'); print(c)
Characteristic class c of multiplicative type associated to x + 1 on the Differentiable complex vector bundle gamma_1 -> CP^1 of rank 1 over the base space 2-dimensional differentiable manifold CP^1
Let the machinery do its work:
In [39]:
c_form = c.get_form(nab)
c_form.display_expansion(c_comp.frame(), chart=c_comp)
Out[39]:
Since this particular representation is defined outside a set of measure zero, we can compute its integral over $\mathbb{CP}^1$ in real coordinates:
In [40]:
integrate(integrate(c_form[2][[1,2]].expr(c_cart), x, -infinity, infinity).full_simplify(), y, -infinity, infinity)
Out[40]:
The result shows that $c_1(\gamma_1)$ generates the second integer cohomology $H^2(\mathbb{CP}^1, \mathbb{Z})$.
### Euler Class of $\mathbb{S}^2$¶
In this example, we want to compute the Euler class of the 2-sphere $\mathbb{S}^2 \subset \mathbb{R}^3$. As usual, we cover $\mathbb{S}^2$ by two parallelizable open subsets $U:=\mathbb{S}^2 \setminus \{(0,0,1)\}$ and $V:=\mathbb{S}^2 \setminus \{(0,0,-1)\}$ for which the point $(0,0,1)$ is identified with the north pole. We state stereographic coordinates in Sage:
In [41]:
M = Manifold(2, name='S^2', latex_name=r'\mathbb{S}^2',
structure='Riemannian', start_index=1)
U = M.open_subset('U') ; V = M.open_subset('V')
M.declare_union(U,V) # M is the union of U and V
stereoN.<x,y> = U.chart()
stereoS.<xp,yp> = V.chart("xp:x' yp:y'")
N_to_S = stereoN.transition_map(stereoS,
(x/(x^2+y^2), y/(x^2+y^2)),
intersection_name='W',
restrictions1= x^2+y^2!=0,
restrictions2= xp^2+yp^2!=0)
S_to_N = N_to_S.inverse()
Next we define the tangent bundle and its local frames induced by the charts:
In [42]:
eU = stereoN.frame(); eV = stereoS.frame()
TM = M.tangent_bundle()
The Euler class is also one of the predefined classes. Thus, we can easily get it from the tangent bundle's instance:
In [43]:
e_class = TM.characteristic_class('Euler'); print(e_class)
Characteristic class e of Pfaffian type associated to x on the Tangent bundle TS^2 over the 2-dimensional Riemannian manifold S^2
To compute a form representing the Euler class, we need to state a suitable connection first. Here we want to use the Levi-Civita connection induced by the standard metric. This is simply given by the pullback of the Euclidean scalar product of the ambient space $\mathbb{R}^3$ along the canonical embedding $\iota: \mathbb{S}^2 \hookrightarrow \mathbb{R}^3$. Let us define the ambient space $\mathbb{R}^3$ and its Euclidean scalar product $h$:
In [44]:
E = Manifold(3, 'R^3', latex_name=r'\mathbb{R}^3', start_index=1)
cart.<X,Y,Z> = E.chart()
h = E.metric('h')
h[1,1], h[2,2], h[3, 3] = 1, 1, 1
h.display()
Out[44]:
On that account, we declare the embedding $\iota: \mathbb{S}^2 \hookrightarrow \mathbb{R}^3$ in stereographic coordinates when one considers its projection from the north pole $(0, 0, 1)$ to the equatorial plane $Z=0$:
In [45]:
iota = M.diff_map(E, {(stereoN, cart):
[2*x/(1+x^2+y^2), 2*y/(1+x^2+y^2),
(1-x^2-y^2)/(1+x^2+y^2)],
(stereoS, cart):
[2*xp/(1+xp^2+yp^2), 2*yp/(1+xp^2+yp^2),
(xp^2+yp^2-1)/(1+xp^2+yp^2)]},
name='iota', latex_name=r'\iota')
iota.display()
Out[45]:
We can define the standard metric $g$ on $\mathbb{S}^2$ by setting it as the pullback metric $\iota^*h$:
In [46]:
g = M.metric()
g.set(iota.pullback(h))
g[1,1].factor(); g[2,2].factor() # simplifications
g.display()
Out[46]:
The corresponding Levi-Civita connection is computed automatically:
In [47]:
nab = g.connection()
Since we have found the desired Levi-Civita connection, we want to compute the associated curvature forms and store them in a Python list:
In [48]:
cmatrix_U = [[nab.curvature_form(i,j,eU) for j in TM.irange()]
for i in TM.irange()]
cmatrix_V = [[nab.curvature_form(i,j,eV) for j in TM.irange()]
for i in TM.irange()]
Fortunately, the curvature form matrices are already skew-symmetric:
In [49]:
for i in range(TM.rank()):
for j in range(TM.rank()):
show(cmatrix_U[i][j].display())
In [50]:
for i in range(TM.rank()):
for j in range(TM.rank()):
show(cmatrix_V[i][j].display())
Hence we can put them into a dictionary and apply the algorithm:
In [51]:
cmatrices = {eU: cmatrix_U, eV: cmatrix_V}
e_class_form = e_class.get_form(nab, cmatrices)
e_class_form.display_expansion()
Out[51]:
We want to compute the Euler characteristic of $\mathbb{S}^2$ now. This can be achieved by integrating the top form over $\mathbb{S}^2$. But since $U$ and $\mathbb{S}^2$ differ only by a point and therefore a set of measure zero, it is enough to integrate over the subset $U$:
In [52]:
integrate(integrate(e_class_form[2][[1,2]].expr(), x, -infinity, infinity).simplify_full(), y, -infinity, infinity)
Out[52]:
We have eventually obtained the Euler characteristic of $\mathbb{S}^2$.
### $\hat{A}$-Class of Lorentzian Foliation of Berger Spheres¶
We consider the space of unit quaternions $\mathbb{S}^3 \subset \mathbb{R}^4 \cong \mathbb{H}$, where $\mathbb{H}$ is endowed with the canonical basis $(\mathbf{1}, \mathbf{i}, \mathbf{j}, \mathbf{k})$. It turns out that $\mathbb{S}^3$ admits a global frame $(\varepsilon_1, \varepsilon_2, \varepsilon_3)$ so that we observe $\mathbb{S}^3$ to be a parallelizable manifold.
We introduce the following smooth family of so-called Berger metrics: \begin{align*} g_t = a(t)^{2} \; \varepsilon^{1}\otimes \varepsilon^{1} +\varepsilon^{2}\otimes \varepsilon^{2} +\varepsilon^{3}\otimes \varepsilon^{3}. \end{align*}
This family can be used to define a globally hyperbolic manifold $M=\mathbb{R} \times \mathbb{S}^3$ equipped with the Lorentzian metric $g = - {\mathrm{d} t}^2 + g_t$ and hence foliated by Berger spheres.
In the following we compute the $\hat{A}$-form of the corresponding Levi-Civita connection $\nabla_g$. We start the computation by declaring the Lorentzian manifold first:
In [53]:
M = Manifold(4, 'M', structure='Lorentzian'); print(M)
4-dimensional Lorentzian manifold M
We cover $M$ by two open subsets defined as $U := \mathbb{R} \times \left( \mathbb{S}^3 \setminus \{ -\mathbf{1} \} \right)$ and $V := \mathbb{R} \times \left( \mathbb{S}^3 \setminus \{ \mathbf{1} \} \right)$:
In [54]:
U = M.open_subset('U'); V = M.open_subset('V')
M.declare_union(U,V)
We need to impose coordinates on $M$ and use stereographic projections with respect to the foliated 3-sphere:
In [55]:
stereoN.<t,x,y,z> = U.chart()
stereoS.<tp,xp,yp,zp> = V.chart("tp:t' xp:x' yp:y' zp:z'")
N_to_S = stereoN.transition_map(stereoS,
(t, x/(x^2+y^2+z^2),
y/(x^2+y^2+z^2),
z/(x^2+y^2+z^2)),
intersection_name='W',
restrictions1= x^2+y^2+z^2!=0,
restrictions2= xp^2+yp^2+zp^2!=0)
W = U.intersection(V)
S_to_N = N_to_S.inverse()
N_to_S.display()
Out[55]:
From above, we know that $M$ admits a global frame $(\varepsilon_0, \varepsilon_1, \varepsilon_2, \varepsilon_3)$:
In [56]:
E = M.vector_frame('E', latex_symbol=r'\varepsilon')
E_U = E.restrict(U); E_U
Out[56]:
The vector field $\varepsilon_0$ is simply given by $\frac{\partial}{\partial t}$. To obtain the global vector frame $(\varepsilon_1, \varepsilon_2, \varepsilon_3)$ on $\mathbb{S}^3$ in stereographic coordinates, a computation within Sage is performed in the this Jupyter-Notebook. This is done by embedding $\mathbb{S}^3$ into $\mathbb{R}^4 \cong \mathbb{H}$, endowing it with the quaternionic structure. This eventually leads to:
In [57]:
E_U[0][:] = [1,0,0,0]
E_U[1][:] = [0, (x^2-y^2-z^2+1)/2, x*y+z, x*z-y]
E_U[2][:] = [0, x*y-z, (1-x^2+y^2-z^2)/2, x+y*z]
E_U[3][:] = [0, x*z+y, y*z-x, (1-x^2-y^2+z^2)/2]
In [58]:
for i in M.irange():
show(E_U[i].display())
To ensure evaluations in this particular frame, we must communicate the change-of-frame formula to Sage:
In [59]:
P = U.automorphism_field()
for i in M.irange():
for j in M.irange():
P[j,i] = E_U[i][j]
In [60]:
U.set_change_of_frame(stereoN.frame(), E_U, P)
The subset $U$ differs from $M$ only by a one dimensional slit, and is therefore dense in $M$. As we know that $\left(\varepsilon_{0},\varepsilon_{1},\varepsilon_{2},\varepsilon_{3}\right)$ defines a \emph{global} frame, its components can be easily and uniquely extended to all of $M$. For this, we use the method add_comp_by_continuation:
In [61]:
for i in M.irange():
E[i].add_comp_by_continuation(stereoS.frame(), W)
And again, we declare the change of frame:
In [62]:
P = V.automorphism_field()
for i in M.irange():
for j in M.irange():
P[j,i] = E.restrict(V)[i][j]
In [63]:
V.set_change_of_frame(stereoS.frame(), E.restrict(V), P)
In order to reduce the computation time, we examine the $\hat{A}$-class on the open subset $U \subset M$ first. The final result can be obtained by continuation. For this purpose we define the tangent bundle over $U$:
In [64]:
TU = U.tangent_bundle(); print(TU)
Tangent bundle TU over the Open subset U of the 4-dimensional Lorentzian manifold M
Notice that the $\hat{A}$-class is already predefined:
In [65]:
A = TU.characteristic_class('AHat'); A
Out[65]:
Its holomorphic function is given by:
In [66]:
A.function()
Out[66]:
We are ready to define the Berger metric, at least on the subset $U$:
In [67]:
a = function('a')
In [68]:
g = U.metric()
g.add_comp(E_U)[0, 0] = - 1
g.add_comp(E_U)[1, 1] = a(t)^2
g.add_comp(E_U)[2, 2] = 1
g.add_comp(E_U)[3, 3] = 1
g.display(E_U)
Out[68]:
The corresponding connection is automatically computed by Sage:
In [69]:
nab = g.connection(); nab
Out[69]:
Finally, we perform the computation of the $\hat{A}$-form with respect to this connection $\nabla_g$:
In [70]:
A_form = A.get_form(nab) # long time
A_form.display_expansion(E_U, stereoN)
Out[70]:
To attain $\hat{A}(TM, \nabla_g)$ in all given coordinates, we still have to extend the result onto $M$. With respect to the global frame $(\varepsilon_0, \varepsilon_1,\varepsilon_2,\varepsilon_3)$, the form $\hat{A}(TU, \nabla_g)$ only depends on the global coordinate $t$. This makes the continuation trivial. Besides, for the most part, one is interested in characteristic forms outside a set of measure zero. Hence, we terminate our calculation at this point.
In [ ]:
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# Problem: Many heterogeneous catalysts are deposited on high surface-area supports. Why is a large surface area important in heterogeneous catalysis?
⚠️Our tutors found the solution shown to be helpful for the problem you're searching for. We don't have the exact solution yet.
###### Problem Details
Many heterogeneous catalysts are deposited on high surface-area supports. Why is a large surface area important in heterogeneous catalysis?
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# Coloring the rational quantum sphere and the Kochen-Specker theorem
### Hans Havlicek Institut für Geometrie,Technische Universität Wien Wiedner Hauptstraß e 8-10/1133,A-1040 Vienna, Austria havlicek@geometrie.tuwien.ac.atand Günther Krenn and Johann SummhammerAtominstitut der österreichischen Universitäten,Stadionallee 2A-1020 Vienna, Austria krenn@ati.ac.at, summhammer@ati.ac.atand Karl SvozilInstitut für Theoretische Physik,Technische Universität Wien Wiedner Hauptstraß e 8-10/136,A-1040 Vienna, Austria e-mail: svozil@tuwien.ac.at(to whom correspondence should be directed)
http://tph.tuwien.ac.at/[ \tilde] svozil/publ/2000-co.{htm,ps,tex}
## Abstract
We review and extend recent findings of Godsil and Zaks [1], who published a constructive coloring of the rational unit sphere with the property that for any orthogonal tripod formed by rays extending from the origin of the points of the sphere, exactly one ray is red, white and black. They also showed that any consistent coloring of the real sphere requires an additional color. We discuss some of the consequences for the Kochen-Specker theorem [2].
## 1 Colorings
In what follows we shall consider rational rays.'' A rational ray'' is the linear span of a non-zero vector of Qn Ì Rn.
Let p be a prime number. A coloring of the rational rays of Rn, n ³ 1, using pn-1+pn-2+¼+1 colors can be constructed in a straightforward manner. We refer to [3,4,5] for the theoretical background of the following construction.
Each rational ray is the linear span of a vector (x1,x2,¼,xn) Î Zn, where x1,x2,...,xn are coprime. Such a vector is unique up to a factor ±1.
Next, let Zp be the field of residue classes modulo p. The vector space Zpn has pn -1 non-zero vectors; each ray through the origin of Zpn has p-1 non-zero vectors. So there are exactly (pn -1)/(p-1) = pn-1+pn-2+¼+1 distinct rays through the origin which can be colored with pn-1+pn-2+¼+1 distinct colors.
Finally, assign to the ray Sp(x1,x2,¼,xn) (Sp'' denotes linear span) the color of the ray of Zpn which is obtained by taking the modulus of the coprime integers x1,x2,¼,xn modulo p. Observe that x1,x2,¼,xn cannot vanish simultaneously modulo p and that ±(x1,x2,¼,xn) yield the same color. Obviously, all pn-1+pn-2+¼+1 colors are actually used.
In what follows, we consider the case p = 2, n = 3. Here all rational rays Sp(x,y,z) (with x,y,z Î Z coprime) are colored according to the property which ones of the components x,y,z are even (E) and odd (O). There are exactly 7 of such triples OEE, EOE, EEO, OOE, EOO, OEO, OOO which are associated with one of seven different colors #1,#2,#3,#4,#5,#6,#7. Only the EEE triple is excluded. Those seven colors can be identified with the seven points of the projective plane over Z2; cf. Fig. 1.
0.700mm
Picture Omitted
Figure 1: The projective plane over Z2. and the reduced coloring scheme discussed.
Next, we restrict our attention to those rays which meet the rational unit sphere S2ÇQ3. The following statements on a triple (x,y,z) Î Z3 \{(0,0,0)} (not necessarily coprime) are equivalent:
(i) The ray Sp(x,y,z) intersects the unit sphere at two rational points; i.e., it contains the rational points ±( x,y,z)/Ö[(x2+y2+z2)] Î S2ÇQ3.
(ii) The Pythagorean property holds, i.e., x2+y2+z2 = n2, n Î N.
This equivalence can be demonstrated as follows. All points on the rational unit sphere can be written as r = ([a/(a¢)],[b/(b¢)],[c/(c¢)] ) with a,b,c Î Z, a¢,b¢,c¢ Î Z\{0}, and ([a/(a¢)])2+([b/(b¢)])2+([c/(c¢)])2 = 1. Multiplication of r with a¢2b¢2c¢2 results in a vector of Z3 satisfying (ii). Conversely, from x2+y2+z2 = n2, n Î N, we obtain the rational unit vector (x/n,y/n,z/n) Î S2ÇQ3.
Notice that this Pythagorean property is rather restrictive. Not all rational rays intersect the rational unit sphere. For a proof, consider Sp(1,1,0) which intersects the unit sphere at ±(1/Ö2)(1,1,0) Ï S2ÇQ3. Although both the set of rational rays as well as S2ÇQ3 are dense, there are many'' rational rays which do not have the Pythagorean property.
If x,y,z are chosen coprime then a necessary condition for x2+y2+z2 being a non-zero square is that precisely one of x, y, and z is odd. This is a direct consequence of the observation that any square is congruent to 0 or 1, modulo 4, and from the fact that at least one of x, y, and z is odd. Hence our coloring of the rational rays induces the following coloring of the rational unit sphere with those three colors that are represented by the standard basis of Z23:
color #1 if x is odd, y and z are even,
color #2 if y is odd, z and x are even,
color #3 if z is odd, x and y are even.
All three colors occur, since the vectors of the standard basis of R3 are colored differently.
Suppose that two points of S2ÇQ3 are on rays Sp(x,y,z) and Sp(x¢,y¢,z¢), each with coprime entries. The inner product xx¢+yy¢+zz¢ is even if and only if the inner product of the corresponding basis vectors of Z23 is zero or, in other words, the points are colored differently. In particular, three points of S2ÇQ3 with mutually orthogonal position vectors are colored differently.
From our considerations above, three colors are sufficient to obtain a coloring of the rational unit sphere S2ÇQ3 such that points with orthogonal position vectors are colored differently, but clearly this cannot be accomplished with two colors. So the chromatic number'' for the rational unit sphere is three. This result is due to Godsil and Zaks [1]; they also showed that the chromatic number of the real unit sphere is four. However, they obtained their result in a slightly different way. Following [4] all rational rays are associated with three colors by making the following identification:
#1,
#2 = #4,
#3 = #5 = #6 = #7.
This 3-coloring has the property that coplanar rays are always colored by using only two colors; cf. Fig. 1. According to our approach this intermediate 3-coloring is not necessary, since rays in colors #4, #5, #6, #7 do not meet the rational unit sphere.
### 1.1 Reduced two-coloring
As a corollary, the rational unit sphere can be colored by two colors such that, for any arbitrary orthogonal tripod spanned by rays through its origin, one vector is colored by color #1 and the other rays are colored by color #2. This can be easily verified by identifying colors #2 & #3 from the above scheme. (Two equivalent two-coloring schemes result from a reduced chromatic three-coloring scheme by requiring that color #1 is associated with x or y being odd, respectively.)
Kent [6] has shown that there also exist dense sets in higher dimensions which permit a reduced two-coloring. Unpublished results by P. Ovchinnikov, O.G.Okunev and D. Mushtari [7] state that the rational d-dimensional unit sphere is d-colorable if and only if it admits a reduced two-coloring if and only if d < 6.
### 1.2 Denseness of single colors
It can also be shown that each color class in the above coloring schemes is dense in the sphere. To prove this, Godsil and Zaks consider a such that sina = 3/5 and thus cosa = 4/5. a is not a rational multiple of p; hence sin(na) and cos(na) are non-zero for all integers n. Let F be the rotation matrix about the z-axis through an angle a; i.e.,
F = æ
ç
ç
ç
ç
è
cosa
sina
0
-sina
cosa
0
0
0
1
ö
÷
÷
÷
÷
ø
.
Then the image I, under the powers of F, of the point (1,0,0) is a dense subset of the equator.
Now suppose that the point u = ( a/c,b/c,0) is on the rational unit sphere and that a,c are odd and thus b is even. In the coloring scheme introduced above, u has the same color as (1,0,0) (identify a = c = 1 and b = 0); and so does Fu. This proves that I (the image of all powers of F of the points u) is dense. We shall come back to the physical consequences of this property later.
In the reduced two-color setting, if the two "poles" ±(0,0,1) acquire color #1, then the entire equator acquires color #2. Thus, for example, for the two tripods spanned by {(1,0,0),(0,1,0),(0,0,1)} and {(3,4,0),(-4,3,0),(0,0,1)}, the first two legs have color #2, while (0,0,1) has color #1.
### 1.3 The chromatic number of the real unit sphere in three dimensions is four
A proof that four colors suffice for the coloring of points of the unit sphere in three dimensions is constructive and rather elementary. Consider first the intersection points of the sphere with the the x-, the y- and the z-axis, colored by green, blue and red, respectively. There are exactly three great circles which pass through two of these three pairs of points. The great circles can be colored with the two colors used on the four points they pass through. The three great circles divide the sphere into eight open octants of equal area. Four octants, say, in the half space z > 0, are colored by the four colors red, white, green and blue. The remaining octants obtain their color from their antipodal octant.
Although Godsil and Zaks' [1] paper is not entirely specific, it is easy to write down an explicit coloring scheme according to the above prescription. Consider spherical coordinates: let q be the angle between the z-axis and the line connecting the origin and the point, and j be the angle between the x-axis and the projection of the line connecting the origin and the point onto the x-y-plane. In terms of these coordinates, an arbitrary point on the unit sphere is given by (q, j, r = 1) º (q,j).
• The colors of the cartesian coordinate axes (p/2,0), (p/2,p/2), (0,0) are green, blue and red, respectively.
• The color of the octant {(q, j) | 0 < q £ p/2, 0 £ j < p/2} is green.
• The color of the octant {(q, j) | 0 £ q < p/2, p/2 £ j £ p} is red.
• The color of the octant {(q, j) | 0 < q < p/2, p < j < 3p/2} is white.
• The color of the octant {(q, j) | 0 < q £ p/2, -p/2 £ j < 0} is blue.
• The colors of the points in the half space z < 0 are inherited from their antipodes. This completes the coloring of the sphere.
The fact that three colors are not sufficient is not so obvious. Here we shall not review Godsil and Zaks' proof based on a paper by A. W. Hales and E. G. Straus [4], but refer to a result of S. Kochen and E. Specker [2], which is of great importance in the present debate on hidden parameters in quantum mechanics. They have proven that there does not exist a reduced two-coloring, also termed valuation, on the one dimensional subspaces of real Hilbert space in three dimensions.
Recall that a reduced two-coloring of the one dimensional linear subspaces with two colors could immediately be obtained from any possible appropriate coloring of the sphere with three colors by just identifying two of the three colors. Thus, the impossibility of a reduced two-coloring implies that three colors are not sufficient for an appropriate coloring of the threedimensional real unit sphere. (Appropriate'' here means: points at spherical distance p/2 get different colors.'')
In the same article [2], Kochen and Specker gave an explicit example (their G3) of a finite point set of the sphere with weaker properties which suffice just as well for this purpose: the structure still allows for a reduced two-coloring, yet it cannot be colored by three colors. (The authors did not mention nor discuss this particular feature [9].)
The impossibility of a reduced two-coloring also rules out another attempt to nullify'' the Kochen-Specker theorem by identifying pairs of colors of an appropriate four-coloring of the real unit sphere. Any such identification would result in tripods colored by #1-#2-#2, as well as for instance #1-#1-#2, which is not allowed for reduced coloring schemes, which requires colorings of the type #1-#2-#2.
## 2 Physical aspects
### 2.1 Physical truth values
Based on Godsil and Zaks´ [1] results, Meyer [10] suggested that the physical impact of the Kochen-Specker theorem [2] is nullified,'' since for all practical purposes it is impossible to operationalize the difference between any dense set of rays and the continuum of Hilbert space rays. (See also the subsequent papers by Kent [6] and Clifton and Kent [11].) However, for the reasons given below, the physical applicability of these constructions remain questionable.
Let us re-state the physical interpretation of the coloring schemes discussed above. Any linear subspace Sp r of a vector r can be identified with the associated projection operator Er and with the quantum mechanical proposition the physical system is in a pure state Er'' [12]. The coloring of the associated point on the unit sphere (if it exists) is equivalent with a valuation or two-valued probability measure
Pr:Er® {0,1}
where 0 ~ #2 and 1 ~ #1. That is, the two colors #1, #2 can be identified with the classical truth values: It is true that the physical system is in a pure state Er'' and It is false that the physical system is in a pure state Er,'' respectively.
Since, as has been argued before, the rational unit sphere has chromatic number three, two colors suffice for a reduced coloring generated under the assumption that the colors of two rays in any orthogonal tripod are identical. This effectively generates consistent valuations associated with the dense subset of physical properties corresponding to the rational unit sphere [10].
Kent [6] has shown that there also exist dense sets in higher dimensions which permit a reduced two-coloring. Unpublished results by P. Ovchinnikov, O.G.Okunev and D. Mushtari [7] state that the rational unit sphere of the d-dimensional real Hilbert space is d-colorable if and only if it admits a reduced two-coloring if and only if d < 6.
### 2.2 Sufficiency
The Kochen-Specker theorem deals with the nonembedability of certain partial algebras-in particular the ones arising in the context of quantum mechanics-into total Boolean algebras. 0-1 colorings serve as an important method of realizing such embeddings: e.g., if there are a sufficient number of them to separate any two elements of the partial algebra, then embedability follows [2,13,14]. Kochen and Specker's original paper [2] contains a much stronger result-the nonexistence of 0-1 colorings-than would be needed for nonembedability. However, the mere existence of some homomorphisms is a necessary but no sufficient criterion for embedability. The Godsil and Zaks construction merely provides three homomorphisms which are not sufficient to guarantee embedability. This has already be pointed out by Clifton and Kent [11].
### 2.3 Continuity
The regress to unsharp measurements is rather questionable and would allow total arbitrariness in the choice of approximation. That is, due to the density of the coloring, depending on which approximation is chosen, one predicts different results. The probabilities resulting from these truth assignments are noncontinuous and arbitrary. This is even more problematic if one realizes that the each color of the 0-1 coloring of the rational unit sphere is dense.
### 2.4 Closedness
The rational unit sphere is not closed under certain geometrical operations such as taking an orthogonal ray of the subspace spanned by two non-collinear rays (the cross product of the associated vectors). This can be easily demonstrated by considering the two vectors
æç è 35 , 45 ,0 ö÷ ø , æç è 0, 45 , 35 ö÷ ø Î S2ÇQ3.
The cross product thereof is
æç è 1225 , -925 , 1225 ö÷ ø Ï S2ÇQ3.
Indeed, if instead of S2ÇQ3 one would start with three non-orthogonal, non-collinear rational rays and generate new ones by the cross product, one would end up with all rational rays [15].
In the Birkhoff-von Neumann approach to quantum logics [16], this non-closedness under elementary operations such as the nor-operation might be considered a serious deficiency which rules out the above model as an alternative candidate for Hilbert space quantum mechanics. (However, his argument does not apply to the Kochen-Specker partial algebra approach to quantum logic, since there operations among propositions are only allowed if the propositions are comeasurable.)
Informally speaking, the relative (with respect to other sets such as the rational rays) thinness'' guarantees colorability. In such a case, the formation of finite cycles such as the ones introduced by Kochen and Specker [2] are impossible.
To put it differently: given any nonzero measurement uncertainty e and any non-colorable Kochen-Specker graph G(0) [2,17], there exists another graph (in fact, a denumerable infinity thereof) G(d) which lies inside the range of measurement uncertainty d £ e [and thus cannot be discriminated from the non-colorable G(0)] which can be colored. Such a graph, however, might not be connected in the sense that the associated subspaces can be cyclically rotated into itself by local transformation along single axes. The set G(d) might thus correspond to a collection of tripods such that none of the axes coincides with any other, although all of those non-identical single axes are located within d apart from each other. Indeed, this appears to be precisely how the Clifton-Kent construction works [11].
The reason why a Kochen-Specker type contradiction does not occur in such a scenario is the impossibility to close'' the argument; to complete the cycle: the necessary propositions are simply not available in the rational sphere model. For the same reason, an equilateral triangle does not exist in Q2 [9]. Yet, while these findings seem to contradict the conclusions of Clifton and Kent [11], we would like to emphasize that this does not relate to their formal arguments but rather is a matter of interpretation and a question of how much should be sacrificed for value definiteness.
Thus, although the colorings of rational spheres offer a rather unexpected possibility to define consistent classical models, a closer examination shows that any such colorings should be excluded for physical reasons.
## Acknowledgments
The authors would like to acknowledge stimulating discussions with Ernst Specker.
## References
[1]
C. D. Godsil and J. Zaks. Coloring the sphere. University of Waterloo research report CORR 88-12, 1988.
[2]
Simon Kochen and Ernst P. Specker. The problem of hidden variables in quantum mechanics. Journal of Mathematics and Mechanics, 17(1):59-87, 1967. Reprinted in [18].
[3]
Wilhelm Klingenberg. Projektive Geometrien mit Homomorphismus. Math. Ann., 132:180-200, 1956.
[4]
A.W. Hales and E.G. Straus. Projective colorings. Pac. J. Math., 99:31-43, 1982.
[5]
David S. Carter and Andrew Vogt. Collinearity-preserving functions between Desarguesian planes. Mem. Am. Math. Soc., 235:98, 1980.
[6]
Adrian Kent. Non-contextual hidden variables and physical measurements. Physical Review Letters, 83(19):3755-3757, 1999. arXiv:quant-ph/9906006.
[7]
Peter Ovchinnikov. private e-mail communication in May 2000.
[8]
Ernst Specker. Die Logik nicht gleichzeitig entscheidbarer Aussagen. Dialectica, 14:175-182, 1960. Reprinted in [18]; English translation: The logic of propositions which are not simultaneously decidable, reprinted in [19].
[9]
Ernst Specker. private communication, 1999.
[10]
David A. Meyer. Finite precision measurement nullifies the Kochen-Specker theorem. Physical Review Letters, 83(19):3751-3754, 1999. arXiv:quant-ph/9905080.
[11]
Rob Clifton and Adrian Kent. Simulating quantum mechanics by non-contextual hidden variables. Proc. R. Soc. Lond., A 456:2101-2114, 2000. arXiv:quant-ph/9908031 v4.
[12]
John von Neumann. Mathematische Grundlagen der Quantenmechanik. Springer, Berlin, 1932. English translation: Mathematical Foundations of Quantum Mechanics, Princeton University Press, Princeton, 1955.
[13]
Pavel Pták and Sylvia Pulmannová. Orthomodular Structures as Quantum Logics. Kluwer Academic Publishers, Dordrecht, 1991.
[14]
Karl Svozil and Josef Tkadlec. Greechie diagrams, nonexistence of measures in quantum logics and Kochen-Specker type constructions. Journal of Mathematical Physics, 37(11):5380-5401, November 1996.
[15]
Hans Havlicek and Karl Svozil. Density conditions for quantum propositions. Journal of Mathematical Physics, 37(11):5337-5341, November 1996.
[16]
Garrett Birkhoff and John von Neumann. The logic of quantum mechanics. Annals of Mathematics, 37(4):823-843, 1936.
[17]
K. Svozil. Quantum Logic. Springer, Singapore, 1998.
[18]
Ernst Specker. Selecta. Birkhäuser Verlag, Basel, 1990.
[19]
Clifford Alan Hooker. The logico-algebraic approach to quantum mechanics. D. Reidel Pub. Co., Dordrecht; Boston, [1975]-1979.
File translated from TEX by TTH, version 2.69.
On 20 Feb 2001, 12:36.
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Is this a countable set?
This is a review for a midterm, so I have the solution but I do not understand some key components. Any insight would be greatly appreciated.
$$f(m) = (\lfloor p^{(\frac{1}{m})}\rfloor-1)\cdot p^{(\frac{1}{m})}$$
where $p$ is a fixed prime number, $m\in\mathbb{N}$.
Construct a set $$S = \{m_{1}\cdot f(1) + m_{2}\cdot f(2) + m_{3}\cdot f(3) + \ldots \;|\; m_{i}\in\mathbb{Z}\}$$
Is $S$ a countable set?
Proof: Notice that as $m$ approaches infinity $p^{(\frac{1}{m})}$ approaches $1$. Then there must exist $M\in \mathbb{N}$ such that $1 < p^{(\frac{1}{m})} < \frac{3}{2}$ for all $m \geq M$.
Why is the upper-bound $\frac{3}{2}$?
However $f(m) = 0$ for all $m \geq M$. Hence $S$ is an image of $\mathbb{Z}^{M}$.
What does that mean, $S$ is an image of $\mathbb{Z}^{M}$? Why is that significant?
Mapping: $$(m_{1},\ldots,m_{M}) \mapsto (m_{1}\cdot f(1),\ldots,m_{M}\cdot f(M))$$
Where does this mapping come from?
Thus $S$ is a countable set.
What is the most common way to prove that something is a countable set?
-
You should really implement the use of TeX in this post. It will make your formulas much more readable. – Alex Petzke Oct 23 '12 at 3:19
There is nothing special about $3/2$. The fact is that $p^{1/m} \to 1$ from above as $m \to \infty$ so you can pick an $M$ such that $m \geq M \Longrightarrow p^{1/m} < 3/2$.
If this fact is true then we see $f(m) = 0$ when $m \geq M$. Now S behaves like finite linear combinations of integers given by the mapping you have. In other words $S$ being the image of $\mathbb{Z}^M$ means exactly that you have a mapping from $\mathbb{Z}^M$ to $S$ which is surjective. But $\mathbb{Z^m}$ is countable so $S$ is at most countable.
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# Trouble evaluating the integral for a rational function
1. Jan 15, 2009
### uwludd
1. The problem statement, all variables and given/known data
i've tried many partial fraction methods but none of y answers are correct in the end, please help me evaluate the integral for f(x)= (10x+2)/(x-5)(x^2 + 1)
2. Relevant equations
there are no relevant equations given
3. The attempt at a solution
A/x-5 + Bx+C/x^2 +1
2. Jan 15, 2009
### rock.freak667
So now you have
$$\frac{A(x^2+1)+(Bx+C)(x-5)}{(x-5)(x^2+1)}= \frac{10x+2}{(x-5)(x^2+1)}$$
So equating the numerators
A(x^2+1)+(Bx+C)(x-5)= 10x+2 for all values of x.
Try putting suitable values of x which will eliminate most of the constants. For example, x=5 will help you get A.
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# Math Help - Slope-intercept form of a line.
1. ## Slope-intercept form of a line.
How do I solve the following problems. For the first one, I know that 2/3x is the slope but what is the slope intercept form and how do I figure it out
1) Write the slope-intercept form of the equation of the line described:
through (-2,3), perpendicular to y=2/3x-3
This one below needs to be in matrix form, ie:[ ]. For this I am having trouble figuring out how to place the numbers in the matrix and do the inverse to solve it since there are not an equal amount of numbers in each equation
2) Solve the system
r+s+6t=23
3r-2s-3t=30
-r+t=-8
What the first question, find the gradient using the relationship $m_T\times m_P=-1$
Where 'm' are the gradients to the tangent and perpendicular.
$\left[ \begin{array}{c} r \\ s \\ t \end{array} \right] = \left[ \begin{array}{ccc} 1 & 1 & 6 \\ 3 & -2 & -3 \\ -1 & 0 & 1 \end{array} \right]^{-1} \left[ \begin{array}{c} 23 \\ 30 \\ 8 \end{array} \right]$
The line perpendicular to y= (2/3)x- 3 has slope -3/2. The equation of the line through point (a, b) with slope m is $y= m(x-a)+ b$.
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# Discrete Mathematics/Graph theory
A graph is a mathematical way of representing the concept of a "network".
## Introduction
A network has points, connected by lines. In a graph, we have special names for these. We call these points vertices (sometimes also called nodes), and the lines, edges.
Here is an example graph. The edges are red, the vertices, black.
In the graph, $v_1, v_2, v_3, v_4$ are vertices, and $e_1, e_2, e_3, e_4, e_5$ are edges.
## Definitions of graph
There are several roughly equivalent definitions of a graph. Most commonly, a graph $G$ is defined as an ordered pair $G=(V,E)$, where $V=\{v_1,\ldots,v_n\}$ is called the graph's vertex-set and $E=\{e_1,\ldots,e_m\}\subset\{\{x,y\}|x,y\in V\}$ is called the graph's edge-set. Given a graph $G$, we often denote the vertex—set by $V(G)$ and the edge—set by $E(G)$. To visualize a graph as described above, we draw $n$ dots corresponding to vertices $v_1,\ldots, v_n$. Then, for all $i,j\in\{1,\ldots,n\}$ we draw a line between the dots corresponding to vertices $v_i, v_j$ if and only if there exists an edge $\{v_i,v_j\}\in E$. Note that the placement of the dots is generally unimportant; many different pictures can represent the same graph.
Alternately, using the graph above as a guide, we can define a graph as an ordered triple $G=(V,E,f)$:
• a set of vertices, commonly called V
• a set of edges, commonly called E
• a relation $f: E \rightarrow \{\{x,y\}|x,y\in V\}$ that maps to each edge a set of endpoints, known as the edge-endpoint relation. We say an edge $e\in E$ is incident to a vertex $v\in V$ iff $v\in f(e)$.
In the above example,
• V={v1, v2, v3, v4}
• E={e1, e2, e3, e4, e5}
• f such that e1 maps to {v1, v2}, e2 maps to {v1, v3}, e3 maps to {v1, v4}, e4 maps to {v2, v4}, and e5 maps to {v3, v4}.
If $f$ is not injective — that is, if $\exists e,e'\in E$ such that $e\neq e', f(e) = f(e')$ — then we say that $G$ is a multigraph and we call any such edges $e,e'\in E$ multiple edges. Further, we call edges $e \in E$ such that $|f(e)|=1$ loops. Graphs without multiple edges or loops are known as simple graphs.
Graphs can, conceivably, be infinite as well, and thus we place no bounds on the sets V and E. We will not look at infinite graphs here.
### Directions, Weights, and Flows
We define a directed graph as a graph such that $f$ maps into the set of ordered pairs $\{(x,y)|x,y\in V\}$ rather than into the family of two-element sets $\{\{x,y\}|x,y\in V\}$. We can think of an edge $e\in E$ such that $f(e)=(x,y)$ as 'pointing' from $x$ to $y$. As such we would say that $x$ is the tail of edge $e$ and that $y$ is the head. This is one of the vagaries of graph theory notation, though. We could just as easily think of $x$ as the head and $y$ as the tail. To represent a directed graph, we can draw a picture as described and shown above, but place arrows on every edge corresponding to its direction.
In general, a weight on a graph $G$ is some function $c: E(G)\rightarrow \mathbb R$.
A flow $(G,c)$ is a directed graph $G=(V,E,f)$ paired with a weight function such that the weight "going into" any vertex is the same amount as the weight "going out" of that vertex. To make this more formal, define sets
• $f^+(v)=\{e\in E(G)|f(e)=(v,x), x\in V(G)\},\; \forall v\in V(G)$
• $f^-(v)=\{e\in E(G)|f(e)=(x,v), x\in V(G)\},\; \forall v\in V(G)$
Then, formally stated, our requirement on the weight function is $\sum_{e\in f^+(v)} c(e)=\sum_{e\in f^-(v)} c(e),\; \forall v\in V(G).$
### Special Graphs
The complete graph on 6 vertices
Some graphs occur frequently enough in graph theory that they deserve special mention. One such graphs is the complete graph on n vertices, often denoted by Kn. This graph consists of n vertices, with each vertex connected to every other vertex, and every pair of vertices joined by exactly one edge. Another such graph is the cycle graph on n vertices, for n at least 3. This graph is denoted Cn and defined by V := {1,2,..,n} and E := Template:1,2,{2,3}, ..., {n-1,n},Template:N,1. Even easier is the null graph on n vertices, denoted Nn; it has n vertices and no edges! Note that N1 = K1 and C3 = K3.
### Some Terms
Two vertices are said to be adjacent if there is an edge joining them. The word incident has two meanings:
• An edge e is said to be incident to a vertex v if v is an endpoint of e.
• Two edges are also incident to each other if both are incident to the same vertex.
Two graphs G and H are said to be isomorphic if there is a one-to-one function from (or, if you prefer, one-to-one correspondence between) the vertex set of G to the vertex set of H such that two vertices in G are adjacent if and only if their images in H are adjacent. (Technically, the multiplicity of the edges must also be preserved, but our definition suffices for simple graphs.)
### Subgraphs
Generated Subgraphs
A subgraph is a concept akin to the subset. A subgraph has a subset of the vertex set V, a subset of the edge set E, and each edge's endpoints in the larger graph has the same edges in the subgraph. A
A subgraph $H$ of $G$ is generated by the vertices {$a, b, c,...$}$\in H$ if the edge set of $H$ consists of all edges in the edge set of $G$ that joins the vertices in $H=${$a, b, c, ...$}.
A path is a sequence of edges $$ such that ei is adjacent to ei+1 for all i from 1 to N-1. Two vertices are said to be connected if there is a path connecting them.
## Trees and Bipartite Graphs
A tree is a graph that is (i) connected, and (ii) has no cycles. Equivalently, a tree is a connected graph with exactly $n-1$ edges, where there are $n$ nodes in the tree.
A Bipartite graph is a graph whose nodes can be partitioned into two disjoint sets U and W such that every edge in the graph is incident to one node in U and one node in W. A tree is a bipartite graph.
A complete bipartite graph is a bipartite graph in which each node in U is connected to every node in W; a complete bipartite graph in which U has $n$ vertices and V has $m$ vertices is denoted $K_{n,m}$.
Self loops,Parallel edges,Degree of Vertex
Pendant Vertex : Vertex Degree one "Pendant Vertex" Isolated Vertex : Vertex Degree zero "Isolated Vertex"
## Hamiltonian and Eulerian Paths
Hamiltonian Cycles: A Hamiltonian Cycle received its name from Sir William Hamilton who first studied the travelling salesman problem. A Hamiltonian cycle is a path that visits every vertex once and only once i.e. it is a walk, in which no edge is repeated (a trail) and therefore a trail in which no vertex is repeated (a path). Note also it is a cycle, the last vertex is joined to the first.
A graph is said to be Eulerian if it is possible to traverse each edge once and only once, i.e. it has no odd vertices or it has an even number of odd vertices (semi-Eulerian). This has implications for the Königsberg problem. It may be easier to imagine this as if it is possible to trace the edges of a graph with a pencil without lifting the pencil off the paper or going over any lines.
## Planar Graphs
A planar graph is an undirected graph that can be drawn on the plane or on a sphere in such a way that no two edges cross, where an edge $e = (u,v)$ is drawn as a continuous curve (it need not be a straight line) from u to v.
Kuratowski proved a remarkable fact about planar graphs: A graph is planar if and only if it does not contain a subgraph homeomorphic to $K_5$ or to $K_{3,3}$. (Two graphs are said to be homeomorphic if we can shrink some components of each into single nodes and end up with identical graphs. Informally, this means that non-planar-ness is caused by only two things—namely, having the structure of $K_5$ or $K_{3,3}$ within the graph).
### Coloring Graphs
A graph is said to be planner if it can be drawn on a plane in such way that no edges cross one anather except of course of vertices
Each term, the Schedules Office in some university must assign a time slot for each final exam. This is not easy, because some students are taking several classes with finals, and a student can take only one test during a particular time slot. The Schedules Office wants to avoid all conflicts, but to make the exam period as short as possible.
We can recast this scheduling problem as a question about coloring the vertices of a graph. Create a vertex for each course with a final exam. Put an edge between two vertices if some student is taking both courses. For example, the scheduling graph might look like this: Next, identify each time slot with a color. For example, Monday morning is red, Monday afternoon is blue, Tuesday morning is green, etc.
Assigning an exam to a time slot is now equivalent to coloring the corresponding vertex. The main constraint is that adjacent vertices must get different colors; otherwise, some student has two exams at the same time. Furthermore, in order to keep the exam period short, we should try to color all the vertices using as few different colors as possible. For our example graph, three colors suffice: red, green, blue.
The coloring corresponds to giving one final on Monday morning (red), two Monday afternoon (blue), and two Tuesday morning (green)...
### K Coloring
Many other resource allocation problems boil down to coloring some graph. In general, a graph G is kcolorable if each vertex can be assigned one of k colors so that adjacent vertices get different colors. The smallest sufficient number of colors is called the chromatic number of G. The chromatic number of a graph is generally difficult to compute, but the following theorem provides an upper bound:
Theorem 1. A graph with maximum degree at most k is (k + 1)colorable.
Proof. We use induction on the number of vertices in the graph, which we denote by n. Let P(n) be the proposition that an nvertex graph with maximum degree at most k is (k + 1)colorable. A 1 vertex graph has maximum degree 0 and is 1colorable, so P(1) is true.
Now assume that P(n) is true, and let G be an (n + 1)vertex graph with maximum degree at most k. Remove a vertex v, leaving an nvertex graph G . The maximum degree of G is at most k, and so G is (k + 1)colorable by our assumption P(n). Now add back vertex v. We can assign v a color different from all adjacent vertices, since v has degree at most k and k + 1 colors are available. Therefore, G is (k + 1)colorable. The theorem follows by induction.
## Weighted Graphs
A weighted graph associates a label (weight) with every edge in the graph. Weights are usually real numbers, and often represent a "cost" associated with the edge, either in terms of the entity that is being modeled, or an optimization problem that is being solved.
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mersenneforum.org Short-term goal
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2005-11-01, 16:44 #1 em99010pepe Sep 2004 283010 Posts Short-term goal I was wondering about the short-term goal of this project. Take all numbers up to 71 bits? I think I'm going to do that. Thoughts? Carlos
2005-11-01, 17:29 #2
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Quote:
Originally Posted by em99010pepe I was wondering about the short-term goal of this project. Take all numbers up to 71 bits? I think I'm going to do that. Thoughts? Carlos
This is a good idea.
Anyway, from the Elevensmooth project page:
Code:
Level 8 will be eight candidates trial factored to 74 bits.
Level 7 was seven candidates trial factored to 73 bits.
Level 6 was six candidates trial factored to 72 bits.
There will be nine steps from Level 7 to Level 8: getting an eighth candidate to 73 bits and getting each candidate to 74 bits.
The goal is to reach Level 8, then Level 9 and so on.
Having most candidates levelled at 71 bits would be a valid subgoal, as there is less work to do.
Luigi
2005-11-01, 17:39 #3 em99010pepe Sep 2004 2·5·283 Posts Luigi, I found that I can take a number up to 71 bits in less than 24 hours. So looking at your idle exponents stats page you have 48 numbers left to take up to 71 bits. If I bring more power to the project I can achieve this goal in less than 48 days or until the end of the year. Carlos
2005-11-01, 19:48 #4
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"Luigi"
Aug 2002
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2×41×59 Posts
Quote:
Originally Posted by em99010pepe Luigi, I found that I can take a number up to 71 bits in less than 24 hours. So looking at your idle exponents stats page you have 48 numbers left to take up to 71 bits. If I bring more power to the project I can achieve this goal in less than 48 days or until the end of the year. Carlos
So you are welcome!
Luigi
2005-11-04, 23:38 #5 em99010pepe Sep 2004 2×5×283 Posts When I look at stats I only see more two members running OBD and both started on 2005-10-30. The rest of the reserved numbers are stopped since 2004 and 2005-04, so I suppose this is a dead project. Anyone would like to comment? Am I wasting my time with this project. Carlos Last fiddled with by em99010pepe on 2005-11-04 at 23:39
2005-11-05, 12:58 #6
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"Luigi"
Aug 2002
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2·41·59 Posts
Quote:
Originally Posted by em99010pepe When I look at stats I only see more two members running OBD and both started on 2005-10-30. The rest of the reserved numbers are stopped since 2004 and 2005-04, so I suppose this is a dead project. Anyone would like to comment? Am I wasting my time with this project. Carlos
This is not a dead project, it's "a whimsical project" (C) Wblipp. People apply when they have time or a new release of the factoring program comes out.
I'm trying to clear up the reserved exponents for more than 1 year.
Easier exponents have been checked before, and maybe we'll release a new batch starting at 60 bits as soon as most of the old ones will reach 70 bits.
Luigi
2005-11-05, 13:28 #7 em99010pepe Sep 2004 2×5×283 Posts Thanks for your post. Just found a factor on M3321932819. Please see the Checkout Thread. Carlos Last fiddled with by em99010pepe on 2005-11-05 at 13:28
2005-11-05, 13:38 #8
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"Luigi"
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Quote:
Originally Posted by em99010pepe Thanks for your post. Just found a factor on M3321932819. Please see the Checkout Thread. Carlos
Your discover has been inserted on the stats page, thank you.
Luigi
2005-11-26, 22:53 #9 em99010pepe Sep 2004 2·5·283 Posts I'm stopping for a while. Carlos
Similar Threads Thread Thread Starter Forum Replies Last Post wblipp Factoring 14 2015-03-31 23:05 TheMawn GPU Computing 6 2013-05-25 16:07 em99010pepe No Prime Left Behind 94 2008-03-24 21:02 fivemack NFSNET Discussion 3 2008-02-21 19:26 OmbooHankvald Operation Billion Digits 4 2005-11-28 04:30
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# Tag Info
### subfigure environment - how to write captions above each figure?
Add the position=top parameter to \captionsetup command. The following is also updated to include the requests in comments: spreading out figures evenly horizontally and naming them without numbering. ...
### Remove subfigures from list of figures when using tabularray package
I seemed to have fixed this by just including \newcounter{lofdepth}\setcounter{lofdepth}{1} in the preamble, inspired by some code from the tocloft package. I'm not completely sure why this works, but ...
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### Top align images in subfigure/minipage with adjustbox when one caption spills over to next line
You didn't include which document class you are using. I will use article class as example. As commented by @Miyase. You need provide a full MWE which include the code from \documentclass to \end{...
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### How to make table and figure side by side in two column layout?
There are many ways you can fix it. I suggest you to rearrange your code as: \documentclass[journal]{IEEEtran} \usepackage{floatrow} % Table float box with bottom caption, box width adjusted to ...
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### How to align the bottom of arrays with blkarray and subcaption
I'd use nicematrix. \documentclass{article} \usepackage[T1]{fontenc} \usepackage[french]{babel} \usepackage{subcaption} \usepackage{amsmath} \usepackage{nicematrix} \begin{document} \begin{table}...
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### How to align the bottom of arrays with blkarray and subcaption
You can hack your way by adding \\[-1em] right after \end{blockarray} of the right array. However, this will not get you a perfect alignment overall because the square brackets of the left array are ...
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### Can't force subfigure to top of page
(expanding on my earlier comments, so that this query may be deamed to have received an "official" answer) I would like to suggest that you insert \clearpage between the two parts of Figure ...
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Only top scored, non community-wiki answers of a minimum length are eligible
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## We then used the CD8+ T cell depletion antibody to assess the role of CD8+ T cell in THZ1-evoked antitumor immunity
We then used the CD8+ T cell depletion antibody to assess the role of CD8+ T cell in THZ1-evoked antitumor immunity. elicited apoptosis and suppressed tumor growth. Moreover, CDK7 ablation specifically suppressed p38/MYC-associated genes, and XL-888 THZ1 inhibited MYC transcriptional activity through downregulating p38. CDK7 inhibition sensitized NSCLC to p38 XL-888 inhibitor. Further, THZ1 suppressed PD-L1 expression by inhibiting MYC activity. THZ1 boosted antitumor immunity by recruiting infiltrating CD8+ T cells and synergized with antiPD-1 therapy. The CDK7/MYC/PD-L1 signature and infiltrating T cell status collectively stratified NSCLC patients into different risk groups. Conclusion These data suggest that the combined CDK7 inhibitor THZ1 and antiPD-1 therapy can be an effective treatment in NSCLC. mRNA level and OS in the TCGA LUAD data by GraphPad Prism Software (= 526) (= 0.0412). b K-M curve showing the relationship between mRNA level and OS in “type”:”entrez-geo”,”attrs”:”text”:”GSE37745″,”term_id”:”37745″GSE37745 data (= 196) (= 0.0214). c K-M curve showing the relationship between protein level and OS in cohort I from Shanghai Outdo Biotech (= 92) (= 0.0358). d K-M curve showing the relationship between protein level and OS in cohort II from Tongji Hospital (= 222) (= 0.0031). XL-888 e Data mining showing differential mRNA levels in adjacent and tumor tissue in TCGA LUAD data (< 0.001). f The protein level in adjacent and tumor tissue in cohort I, as examined by immunohistochemistry (IHC) (< 0.001). g Representative scanned images of tissue cores with low or high CDK7 by IHC. Left, original magnification, 6; scale bar, 500?m. Right, original magnification, 400; scale bar, 50?m Evaluation of tumor-infiltrating lymphocytes For the evaluation of tumor-infiltrating lymphocytes (TILs) score, we used semi-quantification to assess the TILs status according to the study [28] with some modifications. The scoring of TILs in TMA cohorts was performed in the same tissue cores used in IHC analysis by immunofluorescence (IF) staining of T lymphocytes (CD3, IF, 1:100, Abcam #ab16669), cytotoxic T cells (CD8, 1:100, IF, Santa Cruz Biotechnology #sc-7970), and Nuclei (DAPI). Based on the visual estimation of the proportion of CD3+ or CD8+ cell lymphocytes, TIL status was classified into 7 groups: 5%, 6~10%, 11~15%, 16~20%, 21~25%, 26~30%, >30%. By testing different cutoff values, we found that the number of low TIL patients (= 87) is much closer to that of high TIL patients (= 136) when 10% was chosen as the cutoff value. When combining different risk factors to predict survival outcomes, TIL status was classified into low TIL scores ( 10% TILs in tumor tissue) and high TIL scores (> 10% TILs in tumor tissue) in this study. The whole-tissue sections of morphologically normal human tonsil were included in each staining batch as positive control and to assess the interexperimental reproducibility. Representative scanned images of tissue cores with high or low TIL scores are shown in Figure S7I. XL-888 RNA-seq and gene enrichment analysis Gene expression analysis was conducted by RNA-seq for the conditions described in the relevant figures. Treated cells were harvested for RNA extraction using TRIzol. Reagent genomic and DNA was removed using DNase I (Takara). The sequencing library was constructed after high-quality RNA was quantified and then sequenced with the Illumina HiSeq X Ten (2 150?bp read length). The raw paired end reads were trimmed and quality controlled by SeqPrep (https://github.com/jstjohn/SeqPrep) and Sickle (https://github.com/najoshi/sickle) with default parameters. Then, clean reads were aligned separately to the reference genome. To identify differential expression genes between two different samples, the expression level of each transcript was calculated according to the fragments per kilobase of exon per million mapped reads method. RSEM (http://deweylab.biostat.wisc.edu/rsem/) was used to quantify gene abundances. The R statistical package software EdgeR (http://www.bioconductor.org/packages/2.12/bioc/html/edgeR.html) was utilized for differential expression analysis. Differential expression genes (DEGs) were defined as |fold change| 2 and value 0.05 in transcription for drug-treated conditions over mock for each sample studied. In addition, functional-enrichment analysis including KEGG pathways, Gene Cd247 Ontology (GO) enrichment [29], and gene set enrichment analysis (GSEA) [30] were performed. Only categories that were below the DAVID value XL-888 of 0.05 and containing at least 5 genes per pathway are reported. Animal experiments Mice were purchased from Nanjing Biomedical Research Institute of Nanjing University, China, and housed under pathogen-free conditions. All studies were performed following the NIH Guidelines for the Care.
## In contrast, type II NKT cells present a far more diverse design of TCR reputation and using lipid antigens [14]
In contrast, type II NKT cells present a far more diverse design of TCR reputation and using lipid antigens [14]. amount of Compact disc8+ cytotoxic T WS 3 cells and raised IL-12 appearance, tumor control had not been established. Appearance of ZBTB16, the lineage-determining transcription aspect of type I cells NKT, was correlated with a good affected person prognosis in the METABRIC dataset, and BTLA amounts were instrumental to help expand distinguish prognosis in patents with high ZBTB16 appearance. Taken together, these data support a job of BTLA in type I cells in restricting anti-tumor immunity NKT. interactions instead of interactions in relaxing T cells [7]. While BTLA might promote T cell success, it lowers activity and proliferation, promoting peripheral tolerance thereby, but restricting anti-tumor immunity [8]. Besides regulating the experience of adaptive immune system cells, BTLA inhibits innate or innate-like lymphocytes also. It’s been proposed being a powerful inhibitory receptor on T cells [9], as well as the serious immunopathology linked to Con A-induced liver organ harm in BTLA-deficient mice was generally traced back again to its inhibitory function on cytokine creation by type I NKT cells [10]. NKT cells are thymus-derived innate-like T cells that exhibit NK1.1 and T cell receptors, therefore offering function and characteristics of both NK cells and conventional T cells [11]. While regular T cells understand peptide antigens shown in the framework of MHC course WS 3 I or course II substances, NKT cells understand personal- and international lipid antigens shown via Compact disc1 substances (a non-polymorphic MHC course I-like molecule). Compact disc1 substances (Compact disc1d in the mouse, Compact disc1A-E in human beings) are usually portrayed by antigen-presenting cells (APCs). Relationship between your NKT TCR as well as the antigen-CD1d complicated leads to an instant activation from the NKT cells, WS 3 which to push out a massive amount inflammatory cytokines because of their memory-like phenotype (Compact disc69 and Compact disc44 appearance) [12]. Within this inhabitants of Compact disc1d-restricted T cells, different subsets could be recognized. NKT type I, known as invariant NKT cells or iNKT also, exhibit an invariant TCR string using a V14 J18 gene portion in mice (V24 J18 in human beings) and a restricted amount of TCR chains. These are further described by their capability to recognize Compact disc1-destined -galactosylceramide (-GalCer), a glycolipid antigen isolated from sea WS 3 sponges, and its own derivatives [13]. On the other hand, type II NKT cells present a more different design of TCR use and reputation of lipid antigens [14]. In tumors, opposing features have been related to SYNS1 type I versus type II NKT cells. While type I NKT cells promote tumor immunosurveillance by immediate cytotoxicity towards tumor and various other cells or the discharge of immunostimulatory cytokines such as for example interferon- (IFN-) or granulocyte-macrophage colony-stimulating aspect (GM-CSF), type WS 3 II NKT cells positively impede anti-tumor immunity by marketing the deposition of suppressive myeloid cells [15,16]. Activation of type I cells in tumors as a result shows up appealing NKT, given that they screen immediate cytotoxicity towards tumor cells and generate huge amounts of IFN- to help expand activate various other cytotoxic immune system cells such as for example NK cells and Compact disc8+ T cells. Therefore, several clinical studies are under method to funnel the anti-tumor potential of type I NKT cells [14,17]. Strategies consist of immediate program of -GalCer, adoptive transfer of APCs packed with adoptive and -GalCer transfer of ex-vivo extended NKT cells themselves. In light of the trials, the chance of useful suppression of existing or extended NKT cells in the tumor microenvironment recently, e.g., via immune system checkpoints, needs to be investigated. In this study, we therefore analyzed the expression of immune checkpoint receptors PD-1 and BTLA on NKT cells in a model of mammary carcinoma and explored the potential of downregulating BTLA expression on type I NKT cells and the consequences in tumor progression and the propagation of metastasis. 2. Results 2.1. Type I NKT Express BTLA in PyMT Mammary Tumors To analyze expression of immune checkpoint receptors on tumor-infiltrating lymphocytes, with a focus on NKT cells, we performed FACS analysis of.
## In p67 has a function in maintaining the morphology of the lysosome and the loss of p67 in might be associated with the unusual structure of the MVT lysosome (Peck et al
In p67 has a function in maintaining the morphology of the lysosome and the loss of p67 in might be associated with the unusual structure of the MVT lysosome (Peck et al., 2008). The dynamic nature of the MVT lysosome has striking parallels with cytostome/cytopharynx of (Alcantara et al., 2017). Here, we have provided insight into the cell cycle\dependent restructuring of the late endocytic system and the resulting effect on endocytic rate in cell Noscapine division and as such deciphering the regulation of this process within the context of the cell cycle will be an important step in understanding cell cycle coordination in these organisms. structure with PSK-J3 associated microtubules. As the cytostome/cytopharynx is an ancestral feature of the kinetoplastids, this suggests that the MVT lysosome and lysosomal microtubule(s) are a reduced cytostome/cytopharynx\like feature. and trypomastigote, the subpellicular microtubule array is interrupted by a specialized set of Noscapine four microtubules called the microtubule quartet (MtQ) that forms part of the flagellum attachment zone (Lacomble et al., 2009; Sunter & Gull, 2016; Vidal & Souza, 2017). The flagellum attachment zone MtQ is nucleated close to the base of the flagellar pocket and then wraps around the pocket before invading the subpellicular array, following Noscapine the line of flagellum attachment (Lacomble et al., 2009). In the promastigote form, which does not have lateral attachment of the flagellum to the cell body, the flagellum attachment zone MtQ is present around the flagellar pocket and does not invade the subpellicular microtubule array (Wheeler et al., 2016). The terminal endocytic compartment in does not have the elongated tubule structure observed in and instead forms a rounded vesicular structure on the posterior side of the nucleus (Halliday et al., 2019; Langreth & Balber, 1975; Peck et al., 2008). The presence of a lysosome in has been the subject of debate: The terminal endocytic compartment was initially termed a reservosome as the structure lacked acid phosphatase activity and was not labeled with antibodies that recognize mammalian lysosome membrane proteins (Soares, Souto\Padrn, & Souza, 1992). Further work has shown that there are generally multiple reservosomes in a cell, which are spherical membrane\bound structures found in the posterior end of the cell with characteristics of prelysosomes, lysosomes, and recycling compartments, and have now been classified as lysosomal\related organelles (Cunha\e\Silva et al., 2006; SantAnna et al., 2008). has an additional endocytic organelle, the cytostome/cytopharynx, which is a long membrane tube that invades deep into the cell body with the entrance positioned close to the flagellar pocket. The cytostome/cytopharynx is the major route for bulk endocytosis into this parasite, and this structure is not found in and but was likely present in the ancestral kinetoplastid (Skalicky et al., 2017). There are two sets of microtubules, one a microtubule triplet and the other a microtubule quartet (distinct from the flagellum attachment zone MtQ) associated with the cytostome/cytopharynx complex. The cytostome/cytopharynx microtubule quartet is nucleated near the flagellar pocket and then extends out beyond the pocket, just under the cell membrane along the preoral ridge before dropping into the cytoplasm alongside the cytostome/cytopharynx. Conversely, the microtubule triplet is nucleated near the cytostome/cytopharynx entrance, and together, these two sets of microtubules form a V shape upon which the cytostome/cytopharynx sits (Alcantara et al., 2014). In the latter stages of the cell cycle, during G2 prior to flagellar pocket division, the cytostome/cytopharynx complex and associated microtubules are disassembled, and then, the structure reassembles during late cytokinesis (Alcantara, L., Vidal, J.C., Souza, W. de, & Cunha\e\Silva, N.L., 2017). Interestingly, it has also been shown that the MVT lysosome in dividing cells also disassembles forming one or two sets of vesicles (Ilgoutz et al., 1999; Weise et al., 2000). Here, we used cysteine peptidase A (CPA) and Noscapine sperm flagellar 1 (SPEF1) as markers of the MVT lysosome and its associated microtubule, respectively, to characterize the cell cycle\related changes in these structures. We show that both the lysosome and its microtubule extend during G1/S phase of the cell but disassemble rapidly during G2 and are essentially absent during cytokinesis before assembling again during the next G1. This cycle of assembly and disassembly is associated with a change in.
## For the FISH methods, W8-and W3-cells were at passage 4 post-infection
For the FISH methods, W8-and W3-cells were at passage 4 post-infection. and are difficult to control. Endosymbionts particularly may offer an alternative to control RGX-104 free Acid populations of and/or effect disease transmission in the form of human population suppression or alternative strategies. Methods cell lines were transfected having a illness using a revised shell vial technique. Infections were confirmed using PCR and cell localization using fluorescent hybridization (FISH). The stability of infections and denseness was determined by qPCR. qPCR was also used to examine immune genes in the IMD, Toll and JACK/STAT pathways to determine if were associated with an immune response in infected cells. Results Here we have transfected two cell lines (W3 and W8) having a illness (Aa23 cells. PCR and FISH showed the presence of infections in both cell lines. Infection densities were higher in the W8 cell lines when compared to W3. In stably infected cells, genes in the immune Toll, IMD and JAK/STAT pathways were upregulated, along with Attacin and an Attacin-like anti-microbial peptides. Conclusions The successful intro of infections in cell lines and the upregulation of RGX-104 free Acid immune genes, suggest the energy of using for any human population replacement and/or human population suppression approach to limit the transmission of vectored diseases. Results support the further investigation of induced pathogen inhibitory effects in cell lines and the intro of into adults embryonic microinjection to examine for reproductive phenotypes and sponsor fitness effects of a novel illness. varieties are small hematophagous insects that have been shown to harbor more than 50 different viruses of veterinary and medical importance [1]. These viruses include orbiviruses, such as African horse sickness disease (AHSV), Schmallenberg disease (SBV), bluetongue disease (BTV) and epizootic hemorrhagic disease disease (EHDV), which significantly effect deer and livestock production through loss of earnings and trade restrictions [1, 2]. Multiple outbreaks of blue tongue disease (BTV) of different serotypes, topotypes (regional variants of particular serotypes) and strains have been recorded in Europe RGX-104 free Acid in recent decades [3, 4]. One of the largest Western outbreaks to day recorded in the Netherlands, resulted in economic damage greater than $150 million dollars [5]. The blood circulation of founded and newly founded BTV serotypes still continues to affect large areas of southern and eastern Europe. Currently, there are at least 11 invasive BTV serotypes circulating in the USA [6C10] and the Rabbit polyclonal to PLSCR1 number of serotypes in the USA is on the rise, suggesting the epidemiology of BTV is definitely changing and could result in considerable disease in USA livestock if the disease were to infect naive sponsor populations [11]. Worldwide estimations of direct and indirect deficits due to BTV have been estimated to top$3 billion dollars [12]. are focused on treating livestock with topical insecticides at livestock production facilities and farms, but are typically met with limited success, depending on the varieties targeted [2, 13, 14]. Furthermore, little is known about the biology of many varieties, specifically immature habitat selection, making the effective software of insecticides to control immatures hard [13, 15]. Habitat changes to remove standing up water and removal of manure is definitely often used to effect populations of near livestock, but is limited to use in RGX-104 free Acid areas near livestock production. The combination of larvicide RGX-104 free Acid and adulticidal treatments have also shown some success, but the true efficiency of this type of control has not been assessed, and this type of treatment typically does not reduce the numbers of adults, if only treating around a farm home [16, 17]. Vaccines are available for a few varieties. may present an alternative environmentally friendly control measure for midges and the pathogens they vector. is an obligate intracellular bacterium found in >?55% of insects, as well as filarial nematodes and terrestrial crustaceans [21, 22]. In bugs, causes alterations in host reproduction, with several phenotypes including feminization, parthenogenesis, male killing and cytoplasmic incompatibility (CI) [23]. Recently, has been used as a strategy for mosquito suppression and disease control and has become a topic of global relevance [24, 25]. Two incompatible insect technique (IIT) approach based on mass inundative releases of incompatible male mosquitoes similar to the Sterile Insect Technique (SIT), with the goal of suppressing natural populations through sterile mattings [26, 27]. The second is based on the finding that some interfere with viruses and additional microbes in the same sponsor [28C32]. Particular variants (e.g. the fitness [28]. In addition, offers also been shown to effect chikungunya disease, Zika.
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# Homotopy Type Theory: What is it?
My question is, broadly, what is the main project of Homotopy Type Theory (HoTT). I asked a professor who is likely to be correct and he say the following:
There are three directions:
1. Topologists are seeing type theory as a concise and convenient ways to reason about topology, where equalities are interpreted as paths (and higher-dimensional variants thereof).
2. Type theorists are seeing topology as a way to get new insights on type theory and variants thereof.
3. People are pushing univalence and constructive type theory as a new foundations for mathematics.
The only one of those I have any grip on is 2. My understanding, gleaned from some talks I didn't understand goes like this:
Some people would like to write a proof checking/generating client that is sufficiently expressive so that you can actually do mathematics in it. The way that they get this high level of expressiveness is through a very rich type theory.
It was noticed that you could intepret these types topologically: An object is a point, a proof of equality is a path, a proof that two proofs of equality are "the same" is a homotopy.
Now, people use attractive features of the model, to guide further development of the proof environment, which is approximately point 2.
Somehow though, there is a hope that you can use this system to compute homotopy groups of spheres? Why is this credible?
Also, this is being pushed as a new foundation? Why? What advantages does it have over set theory?
## migrated from math.stackexchange.comMar 18 '15 at 19:25
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• there have been some popular surveys, including in the AMS Notices I think, available online. – Will Jagy Mar 18 '15 at 0:57
• @WillJagy This thing ams.org/notices/201309/rnoti-p1164.pdf ? – James Mar 18 '15 at 1:00
• Maybe. I believe I was thinking of one of their tiny "what is?" columns, but this will do. I'm just relying on memory here, and not on a topic in which I have any involvement. – Will Jagy Mar 18 '15 at 1:03
• The canonical references are the HoTT blog at homotopytypetheory.org and the textbook they published (available for free via that site). – Ptharien's Flame Mar 18 '15 at 17:16
• While not specifically about HoTT, Mike Shulman's blog post golem.ph.utexas.edu/category/2013/01/… is worth reading. It cleared some things up for me, in response to my comment on his answer below. – Iian Smythe Mar 19 '15 at 15:33
That description of the three directions is not too bad, although they are not of course completely separate, nor does everything called HoTT fall under one of them. Also, I'm not sure whether this applies to you, but I always point out that it is potentially confusing to say "topology" when what is meant is "homotopy theory", i.e. "$\infty$-groupoid theory" — all "spaces" appear in HoTT only up to homotopy.
Some people would like to write a proof checking/generating client that is sufficiently expressive so that you can actually do mathematics in it.
I'm not sure what use a proof checking program would be if you couldn't do mathematics in it... (-:
But yes, I think your description of point 2 is roughly accurate. Point 1 is just about using the model in the other direction: if you prove something in the type theory, then it's true as a statement about spaces. In particular, if you calculate something like a homotopy group of a sphere in the type theory, then it's also a true statement about the homotopy groups of spheres in classical algebraic topology. And in fact we've done this in a few simple cases, like $\pi_n(S^n)$ and $\pi_3(S^2)$, so it better ought to be credible. These proofs are different from the classical ones, elegant in a certain sense, and relatively easily formalizable in a proof assistant. (Notice that I'm talking about different proofs of known results here; it's not inconceivable that the new methods of HoTT might lead to proofs of new results even in classical homotopy theory, but I expect that day is far in the future.)
As for point 3, type theory in general has a lot of advantages over set theory as a foundation for mathematics. Among other things, it requires less arbitrary "encoding" to represent math, it more closely matches mathematical practice, and it can be more easily programmed in a computer, producing small "proof term" certificates that can be easily verified. UF/HoTT potentially fixes some of the disadvantages of traditional type theory, such as a lack of well-behaved quotients and the problem of transporting structures across equivalences, so it's even better. And in addition to traditional mathematics, it also includes new mathematics such as the "synthetic homotopy theory" calculations of homotopy groups mentioned above.
Finally, a point that I always emphasize: HoTT has many models other than the "standard" one in spaces. Roughly we can use any "$(\infty,1)$-topos" (although there are coherence questions still work in progress). This means that when we prove something in UF/HoTT (such as a homotopy group calculation, or any part of traditional mathematics that we might formalize using it as a foundation) the result is more general than in classical homotopy theory, since it is true in any of the models.
• Could you say more about how "it more closely matches mathematical practice", in comparison to set theory? From a naive perspective, it seems like there would be many objects in mathematics that are just as painful to encode in this formalism as others are in set theory; they just might be different objects. – Iian Smythe Mar 18 '15 at 19:42
• @iian: As someone who doesn't really understand this stuff, I think that point is not about how "painful" something is, but a move away from the practice of putting lots of extra irrelevant information into the coding. e.g. away from having to do things like $(a,b)$ to be the specific set $\{\{a\}, \{a,b\}\}$, which contains a lot of information it really shouldn't -- e.g. for every set $X$, the expression $X \in (a,b)$ has a meaningful value. – Hurkyl Mar 18 '15 at 20:46
• @IianSmythe I wrote a whole blog post about this: golem.ph.utexas.edu/category/2013/01/… – Mike Shulman Mar 19 '15 at 3:15
• @IianSmythe, in the same vein as Hurkyl's comment, Set theory allows you to ask whether pi (defined as you wish, coded as a set, respecting strictly the pseudo a conventions we are used to) is a group. Or whether 3 is a topology (it is...). – ACL Mar 19 '15 at 8:37
• @IianSmythe I take it you are asking whether UF can prove that "UF, if consistent, does not prove or disprove CH" (analogously to how ZFC proves the analogous statement about itself). Answering this properly would take far more than 600 characters, but I think the short answer is "yes in an unsatisfying way; a more satisfactory way is part of an open research problem". The unsatisfying way is that inside UF we can build models of itself using the internal models of ZFC. The more satisfactory way would depend on solving homotopytypetheory.org/2014/03/03/hott-should-eat-itself. – Mike Shulman Mar 19 '15 at 17:31
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# 022 Exam 2 Sample B, Problem 10
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Use calculus to set up and solve the word problem: A fence is to be built to enclose a rectangular region of 480 square feet. The fencing material along three sides cost $2 per foot. The fencing material along the 4th side costs$6 per foot. Find the most economical dimensions of the region (that is, minimize the cost).
Foundations:
As with all geometric word problems, it helps to start with a picture. Using the variables ${\displaystyle x}$ and ${\displaystyle y}$ as shown in the image, we need to remember the equation for the area of a rectangle:
${\displaystyle A\,=\,xy.}$
However, we need to construct a new function to describe cost:
${\displaystyle C\,=\,(2+6)x+(2+2)y\,=\,8x+4y.}$
Since we want to minimize cost, we will have to rewrite it as a function of a single variable, and then find when the first derivative is zero. From this, we will find the dimensions which provide the minimum cost.
Solution:
Step 1:
Express one variable in terms of the other: Since we know that the area is 480 square feet and ${\displaystyle A\,=\,xy}$, we can solve for ${\displaystyle y}$ in terms of ${\displaystyle x}$. Since ${\displaystyle 480\,=\,xy}$, we find that ${\displaystyle y=480/x}$.
Step 2:
Find an expression for cost in terms of one variable: Now, we can use the substitution from step 1 to find
${\displaystyle C(x)\,=\,8x+4y\,=\,8x+4\cdot {\frac {480}{x}}\,=\,8x+{\frac {1920}{x}}.}$
Step 3:
Find the derivative and its roots: We can apply the power rule term-by-term to find
${\displaystyle C'(x)\,=\,8-{\frac {1920}{x^{2}}}\,=\,8\left(1-{\frac {240}{x^{2}}}\right).}$
This derivative is zero precisely when ${\displaystyle x=4{\sqrt {15}}}$, which occurs when ${\displaystyle y=8{\sqrt {15}}}$, and these are the values that will minimize cost. Also, don't forget the units - feet!
The cost is minimized when the dimensions are ${\displaystyle 8{\sqrt {15}}}$ feet by ${\displaystyle 4{\sqrt {15}}}$ feet. Note that the side with the most expensive fencing is the shorter one.
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Search
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23 months ago by
mat.lesche40
Germany
mat.lesche40 wrote:
Hello,
I want to do an analysis of some mice data with DESeq2. We have a total 24 samples and three factors. Factor one is the genotyp (6 wildtype, 6x knockout 1, 6x knockout2, 6x knockout3) and factor two are two different cell lines (cellA and cellB). Factor three is the sex (male and female
Samples Genotyp Cell Sex
1 wt
cellA
m
7 wt cellA f
5 wt cellA m
2 wt cellB m
8 wt cellB f
6 wt cellB m
9 ko1 cellA m
11 ko1 cellA m
13 ko1 cellA m
10 ko1 cellB m
12 ko1 cellB m
14 ko1 cellB m
17 ko2 cellA m
21 ko2 cellA m
23 ko2 cellA f
18 ko2 cellB m
22 ko2 cellB m
24 ko2 cellB f
25 ko3 cellA f
29 ko3 cellA m
31 ko3 cellA m
26 ko3 cellB f
30 ko3 cellB m
32 ko3 cellB m
Here is the PCA
We want to compare the subgroups to each other, for example wt-cell1 vs wt-cell2, ko1-cell2 vs ko1-cell2 but also wt-cell1 vs ko1-cell1. Therefore I we wanted to group both factors into one factor and don't use interactions.
dds$group <- factor(paste0(dds$Genotyp, dds$Cell)) Then we realised that 6 of the samples didn't cluster very well. Further investigations have shown that these are female mice which is why the want to chose the following design ~ Sex + group I this the correct way? We will still be able to do all the comparison and would account for the sex. Or would it be better to remove the samples because it affects 6 out of 8 subgroups (ko1 of cellA and cellB don't have female mice). The inner variance of these two groups which don't have female mouse should be different compared to the others. If we leave the female mice in could this lead to a different set of DEGs compared to removing the 6 samples? Thanks for your feedback Best Mathias ADD COMMENTlink modified 23 months ago by Michael Love19k • written 23 months ago by mat.lesche40 How can you have the same cell line from both a female and a male mouse? I thought a cell line was a culture of cells derived from a single common ancestor. ADD REPLYlink written 23 months ago by Ryan C. Thompson6.8k Hey Ryan, Sorry this was my mistake and was not precise. It's not a cell line, it's two different mouse line. One is our standard mouse line, the other one is infected with a disease. ADD REPLYlink written 23 months ago by mat.lesche40 0 23 months ago by Michael Love19k United States Michael Love19k wrote: "~sex + group: Is this the correct way?" Yes. A fixed sex effect on gene expression for each gene can be estimated because it is not totally confounded with group. So it doesn't matter that some groups don't have female mice. ADD COMMENTlink written 23 months ago by Michael Love19k I'd recommend you make the PCA plot again with a subset of the data excluding sex chromosomes. It could be that many of the top 500 genes are on the sex chromosomes. (This doesn't change the recommended design, just would give you a picture of variation in gene expression from the autosome.) ADD REPLYlink written 23 months ago by Michael Love19k Dear Michael, Thanks for the answer and the confirmation. I have found this thread in which you answered a question regarding pairwise comparison (How to perform multiple pairwise comparisons ) Does your answer apply to my task as well. I have 28 comparisons because I have 8 pairs. Thanks Mathias ADD REPLYlink written 22 months ago by mat.lesche40 Yes you should correct globally. The default performed by DESeq2 (and edgeR and limma) is to correct the multiple testing across genes when you produce a single results table. DESeq2 does not have something like the decideTests() function in limma which helps to arrange multiple contrasts at once, so we don't have the functionality to keep track of how many tests you are doing. But it's simple enough to correct globally by following those instructions. As I see it, if one is making a small set of comparisons, e.g. B vs A, C vs A, C vs B, and one is then going to present the result of all of these comparisons in the paper, then correcting each result table separately is fine. However, you can see that running 28 or 45 contrasts is a different situation. One is not going to put the results of 45 contrasts into the paper, but instead highlight the contrasts that had the most number of DEG. And here is where FPR or FDR control would be lost, because one has performed more tests than are presented in the paper and not corrected for that multiplicity. ADD REPLYlink modified 22 months ago • written 22 months ago by Michael Love19k Hey Michael, Thanks for mentioning the global corrections. I wasn't aware of it. Maybe it would be helpful to add this to the Q&A of the DESeq2 vignette and to refer to the section in the limma vignette? For how many pairwise comparisons do you recommend to do the global corrections? Should one do it with 6 or 10 or more than 10? I read the section in the vignette and understand why and how to do it. I just want to outline it to have a confirmation. In this example and only use 6 results but I still have 28. Just want to keep it short. All my results are in a list and I extract six of them. res1 <- deresults[[1]] res2 <- deresults[[2]] res3 <- deresults[[3]] res4 <- deresults[[4]] res5 <- deresults[[5]] res6 <- deresults[[6]] # combine them to a vector together <- c(res1$pvalue,res2$pvalue,res3$pvalue,res4$pvalue,res5$pvalue,res6\$pvalue)
newtogether <- p.adjust(together,method = 'BH')
Afterwards I would take this vector and replace the padj column in all my results with the corresponding entries in 'newtogether'
Lastly, I had 22,711 DEGs in the original padj with a FDR of <0.1. Now, I have 19,505 DEGs with lower than 0.1. p.adjust doesn't offer any setting for the value (e.g 0.1 for 10% or 0.05 for 5%). For example if I want to have the padj for 0.05, do I have to set it in the results() function to 0.05, then run the global p.adjust and look at genes which have <0.05 or can I stick to 0.1 in the results() function. I know that results uses FDR for the independent filtering and I want to make sure it is a stringent as possible.
Thanks
Mathias
That's a good point. I should add it to FAQ. It's not a common setup that users do so many comparisons, but anyway I should add it.
Yes your code looks correct, you just need to combine them and then split them back up.
You can do something like split(newtogether, rep(1:6, each=x)) where x is the number of rows of the dds. This will break the long vector into a list again.
You should use the same alpha in results() and when you do thresholding of the global adjusted p-values.
Thanks for your answer, I really appreciate your help! Because we have another project with 10 comparisons coming up, I'd really like to know at how many comparisons one should do the global corrections? Sorry for all the questions but I always learn something new here.
I don't have a number per se.
I think it comes down to my comment (C: Grouping a multifactor design in DESeq2 and accouting for sex) above:
"One is not going to put the results of 45 contrasts into the paper, but instead highlight the contrasts that had the most number of DEG."
If you are picking among all of the comparisons you did and only mentioning those comparisons with the most DEG in the paper, you need to do global correction. Any kind of highlighting of individual comparisons based on the number of DEG, when many comparisons were done is problematic.
Hey Michael,
I did the PCA again without the sex chromosomes and PC1 went from 50 to 55% and PC2 from 26 to 23%. A plot of genes with the highest variance doesn't show Xist anymore. So I stick to ~sex+group and that's all I can do.
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# Spitzer Microlensing Parallax for OGLE-2016-BLG-1067: a sub-Jupiter Orbiting an M-dwarf in the Disk [EPA]
We report the discovery of a sub-Jupiter mass planet orbiting beyond the snow line of an M-dwarf most likely in the Galactic disk as part of the joint Spitzer and ground-based monitoring of microlensing planetary anomalies toward the Galactic bulge. The microlensing parameters are strongly constrained by the light curve modeling and in particular by the Spitzer-based measurement of the microlens parallax, $\pi_\mathrm{E}$. However, in contrast to many planetary microlensing events, there are no caustic crossings, so the angular Einstein radius, $\theta_\mathrm{E}$ has only an upper limit based on the light curve modeling alone. Additionally, the analysis leads us to identify 8 degenerate configurations: the four-fold microlensing parallax degeneracy being doubled by a degeneracy in the caustic structure present at the level of the ground-based solutions. To pinpoint the physical parameters, and at the same time to break the parallax degeneracy, we make use of a series of arguments: the $\chi^2$ hierarchy, the Rich argument, and a prior Galactic model. The preferred configuration is for a host at $D_L=3.73_{-0.67}^{+0.66}~\mathrm{kpc}$ with mass $M_\mathrm{L}=0.30_{-0.12}^{+0.15}~\mathrm{M_\odot}$, orbited by a Saturn-like planet with $M_\mathrm{planet}=0.43_{-0.17}^{+0.21}~\mathrm{M_\mathrm{Jup}}$ at projected separation $a_\perp = 1.70_{-0.39}^{+0.38}~\mathrm{au}$, about 2.1 times beyond the system snow line. Therefore, it adds to the growing population of sub-Jupiter planets orbiting near or beyond the snow line of M-dwarfs discovered by microlensing. Based on the rules of the real-time protocol for the selection of events to be followed up with Spitzer, this planet will not enter the sample for measuring the Galactic distribution of planets.
S. Novati, D. Suzuki, A. Udalski, et. al.
Fri, 19 Jan 18
38/68
HD95086 (A8V, 17Myr) hosts a rare planetary system for which a multi-belt debris disk and a giant planet of 4-5MJup have been directly imaged. Our study aims to characterize the physical and orbital properties of HD95086b, search for additional planets at short and wide orbits and image the cold outer debris belt in scattered light. We used HARPS at the ESO 3.6m telescope to monitor the radial velocity of HD95086 over 2 years and investigate the existence of giant planets at less than 3au orbital distance. With the IRDIS dual-band imager and the IFS integral field spectrograph of SPHERE at VLT, we imaged the faint circumstellar environment beyond 10au at six epochs between 2015 and 2017. We do not detect additional giant planets around HD95086. We identified the nature (bound companion or background contaminant) of all point-like sources detected in the IRDIS field of view. None of them correspond to the ones recently discovered near the edge of the cold outer belt by ALMA. HD95086b is resolved for the first time in J-band with IFS. Its near-infrared spectral energy distribution is well fitted by a few dusty and/or young L7-L9 dwarf spectral templates. The extremely red 1-4um spectral distribution is typical of low-gravity objects at the L/T spectral type transition. The planet’s orbital motion is resolved between January 2015 and May 2017. Together with past NaCo measurements properly re-calibrated, our orbital fitting solutions favor a retrograde low to moderate-eccentricity orbit e=0.2 (0.0 to 0.5), with a semi-major axis 52au corresponding to orbital periods of 288$yrs and an inclination that peaks at i = 141deg, which is compatible with a planet-disk coplanar configuration. Finally, we report the detection in polarimetric differential imaging of the cold outer debris belt between 100 and 300au, consistent in radial extent with recent ALMA 1.3mm resolved observations. G. Chauvin, R. Gratton, M. Bonnefoy, et. al. Fri, 19 Jan 18 39/68 Comments: 23 pages, 19 figures, accepted in A&A (Dec 28th, 2017) # Finding Mountains with Molehills: The Detectability of Exotopography [EPA] Mountain ranges, volcanoes, trenches, and craters are common on rocky bodies throughout the Solar System, and we might we expect the same for rocky exoplanets. With ever larger telescopes under design and a growing need to not just detect planets but also to characterize them, it is timely to consider whether there is any prospect of remotely detecting exoplanet topography in the coming decades. To test this, we devised a novel yet simple approach to detect and quantify topographical features on the surfaces of exoplanets using transit light curves. If a planet rotates as it transits its parent star, its changing silhouette yields a time-varying transit depth, which can be observed as an apparent and anomalous increase in the photometric scatter. Using elevation data for several rocky bodies in our solar system, we quantify each world’s surface integrated relief with a “bumpiness” factor, and calculate the corresponding photometric scatter expected during a transit. Here we describe the kinds of observations that would be necessary to detect topography in the ideal case of Mars transiting a nearby white dwarf star. If such systems have a conservative occurrence rate of 10%, we estimate that the upcoming Colossus or OWL telescopes would be able to detect topography with <20 hours of observing time, which corresponds to ~400 transits with a duration of 2 minutes and orbital period of ~10 hours. M. McTier and D. Kipping Fri, 19 Jan 18 46/68 Comments: Accepted to MNRAS # Rings and gaps in the disc around Elias 24 revealed by ALMA [EPA] We present Atacama Large Millimeter/sub-millimeter Array (ALMA) Cycle 2 observations of the 1.3 mm dust continuum emission of the protoplanetary disc surrounding the T Tauri star Elias 24 with an angular resolution of$\sim 0.2″$($\sim 28$au). The dust continuum emission map reveals a dark ring at a radial distance of$0.47″$($\sim 65$au) from the central star, surrounded by a bright ring at$0.58″$($\sim 81$au). In the outer disc, the radial intensity profile shows two inflection points at$0.71″$and$0.87″$($\sim 99$and$121$au respectively). We perform global three-dimensional smoothed particle hydrodynamic gas/dust simulations of discs hosting a migrating and accreting planet. Combining the dust density maps of small and large grains with three dimensional radiative transfer calculations, we produce synthetic ALMA observations of a variety of disc models in order to reproduce the gap- and ring-like features observed in Elias 24. We find that the dust emission across the disc is consistent with the presence of an embedded planet with a mass of$\sim 0.7\, \mathrm{M_{\mathrm{J}}}$at an orbital radius of$\sim$60 au. Our model suggests that the two inflection points in the radial intensity profile are due to the inward radial motion of large dust grains from the outer disc. The surface brightness map of our disc model provides a reasonable match to the gap- and ring-like structures observed in Elias 24, with an average discrepancy of$\sim$5% of the observed fluxes around the gap region. G. Dipierro, L. Ricci, L. Perez, et. al. Fri, 19 Jan 18 61/68 Comments: 17 pages, 11 figures. Accepted for publication in MNRAS # Analysis of June 2, 2016 bolide event [EPA] On June 2, 2016 at 10h56m UTC, a$-18.9 \pm 0.5$magnitude superbolide was observed over Arizona. We present analysis of this event based on 6 cameras and a multi-spectral sensor observations by the SkySentinel continuous fireball-monitoring camera network, supplemented by a dash cam footage and a fragmentation model. The bolide began its luminous flight at an altitude of$104.8 \pm 0.5$km at coordinates$\phi = 34.59 \pm 0.04^\circ$N planetographic latitude,$\lambda = 110.45 \pm 0.04^\circ$W longitude, and it had a pre-atmospheric velocity of$17.6 \pm 0.5$km/s. The calculated orbital parameters indicate that the meteoroid did not belong to any presently known asteroid family. From our calculations, the impacting object had an initial mass of$11.4 \pm 0.5$metric tonnes with an estimated initial size of$1.89 \pm 0.07\$ m.
C. Palotai, R. Sankar, D. Free, et. al.
Wed, 17 Jan 18
13/51
Comments: 9 pages, 9 figures, 5 tables, submitted to MNRAS
# Origin of orbits of secondaries in the discovered trans-Neptunian binaries [EPA]
The dependences of inclinations of orbits of secondaries in the discovered trans-Neptunian binaries on the distance between the primary and the secondary, on the eccentricity of orbits of the secondary around the primary, on the ratio of diameters of the secondary and the primary, and on the elements of heliocentric orbits of these binaries are studied. These dependences are interpreted using the model of formation of a satellite system in a collision of two rarefied condensations composed of dust and/or objects less than 1 m in diameter. It is assumed in this model that a satellite system forms in the process of compression of a condensation produced in such a collision. The model of formation of a satellite system in a collision of two condensations agrees with the results of observations: according to observational data, approximately 40% of trans-Neptunian binaries have a negative angular momentum relative to their centers of mass.
S. Ipatov
Wed, 17 Jan 18
14/51
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• 论文 •
黄河口泥沙淤积估算问题和方法——以钓口河亚三角洲为例
1. 1. 中国科学院地理科学与资源研究所, 北京 100101;
2. 香港大学地理系, 香港
• 收稿日期:2002-06-08 修回日期:2002-11-12 出版日期:2003-02-15 发布日期:2003-02-15
• 作者简介:师长兴(1963-),男,河北正定人,副研究员。多年从事河流地貌和环境治理研究,已发表相关论 文30余篇。
• 基金资助:
国家自然科学基金资助项目(49871014)
A study of sediment budget of the Yellow River delta:the case of Diaokouhe lobe
SHI Chang-xing1, ZHANG Dian2, YOU Lian-yuan1, LI Bing-yuan1
1. 1. Institute of Geographic Sciences and Natural Resources Research, CAS, Beijing 100101,China;
2. Department of Geography, The University of Hong Kong,China
• Received:2002-06-08 Revised:2002-11-12 Online:2003-02-15 Published:2003-02-15
Abstract:
Sediment dispersal at the river mouths has been an important topic in the fields of geomorphology and hydraulics for a long time and estimating sediment budget of the deltas is a principal method for quantifying the sediment dispersal system at many river mouths. Many reports about sediment dispersal at the Yellow River mouth have been given previously using this method. However, since the dry bulk density of deposits in the delta and the boundaries set for calculating the volumes of deposits did not receive a proper treatment, big discrepancy existed between estimates of sediment budgets of the delta provided by the previous studies.Considering the effects of grain size composition, burial depth, and exposure conditions on the dry bulk density of deposits and based on abundant data about the dry bulk density of deposits in the delta, this study defined the dry bulk densities for the deposits in the delta plain and buried delta front deposits, buried prodelta, exposed subaqueous prodelta, and newly deposited top 1 m layers on the delta front. Combining the constructed models of dry bulk density for different depositional settings with the results of analyzing the sedimentary framework of the delta, sediment budgets at Diaokouhe lobe of the Yellow River delta were calculated. The foot of the delta front slope was set as the outer margin of area for defining the sediment budgets. This margin is of geomorphologic significance and is easy to be located on the surface of delta recorded by the bathymetric data. Results show that sediment deposited in the delta plain and front of the Diaokouhe lobe over the period from 1965 to 1974 was about 7.10×109 tons, accounting for 73.5% of the incoming sediment. Errors resulting from ignoring clay layers in the deposits on delta plain and front, consolidation of soft layers underlying delta deposits, and deviations in records of the incoming sediment were proved to be about 2.6% for the percentage of sediment deposited in the delta, suggesting the higher reliability of the sediment budgets given by this study. From the mass and volume of sediments deposited in the Diaokouhe lobe over the period of 1965 -1974, a mean dry bulk density of 1.36 g/cm3 was acquired. On account of the dominance of silt in the deposits of the delta, it seems to be an appropriate approximation of the mean dry bulk density for the deposits in the other lobes of the delta.
• P931.1
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# Renormalization group and minimum subtraction
+ 6 like - 0 dislike
451 views
I have several questions about renormalization group and minimum subtraction scheme in particular.
My first question is:
1) Why is the beta function typically just a function of coupling? In other words, why does it explicitly depend only on the coupling constants but not on the relevant energy scale, such as the cutoff or subtraction point? Of course the beta function implicitly depends on the energy scale, but it seems mysterious that it always appears not to explicitly depend on it.
To make this question more concrete, let us suppose we are doing RG for a massless $\phi^4$ theory with a cutoff $\Lambda$, which is described by the Lagrangian:
\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{g\mu^D}{4!}\phi^4-\frac{\delta g\mu^D}{4!}\phi^4
I assume in the spacetime dimension considered here, the engineering mass dimension of the $\phi^4$ coupling is $D$, and $\mu$ is an arbitrary energy scale introduced to make $g$ dimensionless. The third term above is the counter term. Notice I have not included the field strength renormalization because at one-loop order it does not show up.
Now to calculate the beta function of $g$, we notice that the bare coupling is independent of scale, so
\mu\frac{d}{d\mu}\left(\mu^D(g+\delta g)\right)=0
then we get
\beta(g)=-D(g+\delta g)-\mu\frac{d\delta g}{d\mu}
According to minimum subtraction scheme, I will choose the counter term to cancel the infinite part of the loop integral, which in 3+1 dimensions has the form $Ng^2\mu^{2D}\ln\Lambda^2$ where $N$ is some numerical factor. Then the counter term must have the form
\delta g=-Ng^2\mu^D\ln\frac{\Lambda^2}{\mu'^2}
Here $\mu'$ is another energy scale introduced so that the argument of the log is dimensionless. My second question is
2) Can I choose $\mu'$ to be $\mu$? In fact I think I have the freedom to make this choice, but it seems I also have the freedom to make other choices, so what will happen if I choose $\mu'$ to be very different from $\mu'$? Should I then have two different beta functions, one for $\mu$ and the other for $\mu'$?
In the following I will choose $\mu'=\mu$, then the beta function is
\begin{split}
\beta(g)
&=-Dg+DNg^2\mu^D\ln\frac{\Lambda^2}{\mu^2}+N2g(-Dg)\mu^D\ln\frac{\Lambda^2}{\mu^2}+DNg^2\mu^D\ln\frac{\Lambda^2}{\mu^2}-2Ng^2\mu^D\\
&=-Dg-2Ng^2\mu^D
\end{split}
Notice the remarkable cancellation of the $\Lambda$ dependence, but the explicit dependence on $\mu$ seems to be still there. But notice to get this log divergence, we must be working in 3+1 dimensions where $D=0$, so we see the explicit $\mu$ dependence also drops out. My third question is
3) It seems very non-trivial that the explicit dependence on energy scale drops out, how general is this?
Next I have several questions about minimum subtraction, where we set the counter terms to be the infinite part of the loop integral.
4) One can say that we have introduced an arbitrary scale $\mu$, so to make the physical prediction independent of $\mu$, we must let the couplings run. But what is the physical meaning of the running of coupling coming out from minimum subtraction? In particular, do the beta functions calculated in this way always coincide the ones calculated from other schemes, like the scheme that fixes some renormalization conditions and requries the counter terms preserve those renormalization conditions? Also, do the fixed points calculated in this way have any meaning of universality, by which I mean the physics at low energies does not depend on the microscopic details.
5) In fact I quite doubt that the beta functions calculated from minimum subtraction will be the same as the ones from other methods. An example could be that if the spacetime dimension is low enough that there is no divergence, according to minimum subtraction there is no counter term, so the beta function is completely determined by the engineering dimensions of the coupling constants, which means the beta function is linear in the couplings. But in the Wilsonian picture, it is hard to imagine this because it seems the beta function should still have higher order terms in the couplings coming from the loops although these loops are convergent.
6) How should I do minimum subtraction if I get power divergence instead of log divergence? When I get log divergence I can make the counter term proportional to $\ln\frac{\Lambda^2}{\mu^2}$, if I have linear divergence, should I make the counter term proportional to $\Lambda-\mu$?
edited Mar 22, 2015
+ 3 like - 0 dislike
I haven't mastered the subject well, nevertheless let me attempt at an answer here instead of a comment, since it invites quicker correction and criticism.
1) Why is the beta function typically just a function of coupling? In other words, why does it explicitly depend only on the coupling constants but not on the relevant energy scale, such as the cutoff or substraction point?
Strictly speaking it does depend on energy scale and the cutoff, but most of the times in a negligible way. For many QFT's in general(i.e. the ones with no ultraviolet completion), we define the theory with a cutoff(with the understanding that we are not obliged to push the cutoff to infinity, here's a related post.), but also define it in such a way(by renormalization) that all/most of the observable quantities depend on cutoff negligibly, such as $O(\frac{1}{\Lambda})$, so we are not too wrong if we consider beta function to be independent of $\Lambda$. In some low order calculations, $\Lambda$ gets explicitly eliminated by $\frac{d}{d\mu}$, but there seems to be no reason to expect it to happen at arbitrary high orders. For energy scale dependence, we are often interested to apply RG at a energy scale $E$ much larger than mass scale $m$, and by dimensional analysis, beta function can only depend on energy scale via $\frac{m}{E}$, hence also negligible at high energy. (See Weinberg Vol2 Chap 18).
One word of caution, in many contexts a ultraviolet completion(fix point) is assumed, in such case there's no cutoff hence of course no dependence on cutoff.
2) Can I choose μ′ to be μ? In fact I think I have the freedom to make this choice, but it seems I also have the freedom to make other choices, so what will happen if I choose μ′ to be very different from μ′? Should I then have two diffeerent beta functions, one for μ and the other for μ′?
The point of renormalization group is to find the functional relation between $g_{\mu'}$ and $g_\mu$ in a controlled way(by solving the RG equation), that is, to find the explicit form of $g_{\mu'}=G(g_\mu, \mu'/\mu)$, and beta function by definition would be $\mu \frac{\partial G}{\partial \mu'}|_{\mu'=\mu}$, hence beta function by definition depends on only one energy scale. You are free to choose two scales then call it something else. (EDIT:On a second look I see you are not exactly asking about the above perspective, in that you didn't care about using $\mu'$ scale to probe $g_{\mu'}$, and you just use it as a different subtraction point(not just a different subtraction point, you make $\mu'$ a variable that gives nontrivial result under differentiation, so it seems like a radically different scheme from MS, but my memory of MS is vague....). Maybe this is the more proper response: Yes you get two different beta functions, which is no surprise since you are using two different renormalization schemes, see my answer to your question 4) below.)
3) It seems very non-trivial that the explicit dependence on energy scale drops out, how general is this?
As in my earlier argument, I don't think it is very general, you can expect the $\mu$ to drop out only if it's much larger than mass scale.
4)...In particular, do the beta functions calculated in this way always coincide the ones calculated from other schemes, like the scheme that fixes some renormalization conditions and requries the counter terms preserve those renormalization conditions?...
No, beta functions from different renormalization schemes don't necessarily coincide. However, if you are doing a perturbative RG, under mild assumptions the two leading terms of your beta function will be the same for different schemes, so that if you want no more than two terms, you can borrow the result from one scheme to another freely. There is an easy proof of this in Weinberg Vol2 Chap 18, page 138.
4)...But what is the physical meaning of the running of coupling coming out from minimum substraction? ... Also, do the fixed points calculated in this way have any meaning of universality, by which I mean the physics at low energies does not depend on the microscopic details.
Not sure if I understand your question, but to define the same theory you need to implement the physical renormalization condition (e.g. pole mass condition) at some point in your math. MS scheme is very handy to get to finite result, but you still have to link it to observables by some physical renormalization condition. So ultimately you really do have the same underlying theory put in different forms, and if in one form you have "physics at low energies does not depend on the microscopic details.", so will you in the other.
5) Should be resolved by my answer to 4)?
6)I have no idea either.
answered Mar 22, 2015 by (2,605 points)
edited Mar 22, 2015
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chapter
8 Pages
## Appendix B Mathematics Behind the Logistic Regression Model
B.1 The Least Squares Principle as a Maximum Likelihood Principle We have introduced the likelihood as the probability to observe what we have observed. If we consider a continuous random variable Y , we can identify this probability with the value of the density of the distribution of Y evaluated at the observed value y. Now given the covariate values x1,x2, . . . , xp, the assumption of a normal distribution of Y with mean μ(x1,x2, . . . , xp) = β0 + β1x1 + β2x2 + . . . + βpxp and conditional variance σ2e implies that the density is of the following form:
such that the likelihood of the observation yi is
Li(β0,β1, . . . , βp,σ2e ) = 1√
and the log likelihood is
li(β0,β1, . . . , βp,σ2e ) = logLi(β0,β1, . . . , βp,σ2e )
= log [
1√ 2πσ2e
] + log
[ e−
]
= log [
1√ 2πσ2e
] − 12
(yi− (βˆ0 + βˆ1xi1 + βˆ2xi2 + . . . + βˆpxip))2 σ2e
.
Maximising the overall likelihood ∏ni=1 Li(β0,β1, . . . , βp,σ2e ) is equivalent to maximising its logarithm
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(x^(2)-y^(2))/(x-y)
x^(2) \- y^(2) is the 'difference of two squares'. If you can spot that, we know that means it factorises as (x + y)(x - y)... cross multiply out the brackets if you need to convince yourself that works.
Next, we can divide both top and bottom of the fraction to simplify, so divide both by the common factor of (x - y) and we're done!
x = 6
######
y = 4
####
x + y = 6 + 4 = 10
#### ######
x - y = 6 - 4 = 2
##
x^(2) = 36
######
######
######
######
######
######
y^(2) = 16
####
####
####
####
x^(2) - y^(2)
=
##
##
##
##
######
######
=
##
+
##
+
##
+
##
+
#
#
+
#
#
+
#
#
+
#
#
+
##
##
=
(x - y) + (x - y) + (x - y) + (x - y)
\+ (x - y) + (x - y) + (x - y) + (x - y)
\+ (x - y)^(2)
= y(x-y) + y(x-y) + (x-y)^(2)
= 2y(x-y) + (x-y)^(2)
= (y + y + (x-y))(x-y)
= (x + y - y + y)(x-y)
= (x + y)(x - y)
This proves that
x^(2) - y^(2) = (x + y)(x - y).
Now the fraction
(x^(2) - y^(2))/(x - y)
can be simplified by cancelling the (x - y) on numerator and denominator
(x + y)(x - y)/(x - y)
= (x + y)
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# Blood alcohol content
(Redirected from Blood alcohol concentration)
Blood alcohol content
SynonymsBlood alcohol concentration, blood ethanol concentration, blood alcohol level, blood alcohol concentration, blood alcohol
LOINC5639-0, 5640-8, 15120-9, 56478-1
Blood alcohol content (BAC), also called blood alcohol concentration or blood alcohol level, is a measurement of alcohol intoxication used for legal or medical purposes;[1] it is expressed as mass of alcohol per volume or mass of blood. For example, a BAC of 0.10 by volume (0.10% or one tenth of one percent) means that there is 0.10 g of alcohol for every 100 ml of blood, which is the same as 21.7 mmol/l.[2] A BAC of 0.10 by mass (0.10%) is 0.10 g of alcohol per 100 g of blood (23 mmol/l). A BAC of 0.0 is sober; in different countries the maximum permitted BAC when driving ranges from about 0.04% to 0.08%; BAC levels over 0.08% are considered very impaired; above 0.4% is potentially fatal.[1]
## Estimation by intake
Blood alcohol content can be estimated by a method developed by Swedish professor Erik Widmark [sv] in the 1920s:[3]
${\displaystyle EBAC={\frac {A}{r\times Wt}}\times 100\%-\beta \times T}$
where:
• A is the mass of alcohol consumed.
• r is the ratio of body water to total weight. It varies between individuals but averages about 0.68 for men and 0.55 for women, since women tend to have a higher percentage of fat.
• Wt is body weight.
• β is the rate at which alcohol is metabolized. It is approximately 0.017% per hour.
• T is the amount time during which alcohol was present in the blood (usually time since consumption began).
Sometimes the above is multiplied by 1.055 g/mL, the density of blood.[3]
In terms of fluid ounces of alcohol consumed and weight in pounds, the formula can be written as
${\displaystyle EBAC=8\times {\text{fl oz}}/{\text{weight in pounds}}-\beta \times T}$
for a man or
${\displaystyle EBAC=10\times {\text{fl oz}}/{\text{weight in pounds}}-\beta \times T}$
for a woman.[3]
Regarding metabolism (β) in the formula; females demonstrated a higher average rate of elimination (mean, 0.017; range, 0.014–0.021 g/210 L) than males (mean, 0.015; range, 0.013–0.017 g/210 L). Female subjects on average had a higher percentage of body fat (mean, 26.0; range, 16.7–36.8%) than males (mean, 18.0; range, 10.2–25.3%).[4] Additionally, men are, on average, heavier than women but it is not strictly accurate to say that the water content of a person alone is responsible for the dissolution of alcohol within the body, because alcohol does dissolve in fatty tissue as well. When it does, a certain amount of alcohol is temporarily taken out of the blood and briefly stored in the fat. For this reason, most calculations of alcohol to body mass simply use the weight of the individual, and not specifically their water content.
Examples:
• 80 kg male drinking 3 drinks of 10 grams each, in two hours:
${\displaystyle EBAC=0.030/(0.68\cdot 80)\times 100\%-(0.015\cdot 2)\approx 0.025\%}$
• 70 kg woman drinking 2.5 drinks of 10 grams each, in two hours:
${\displaystyle EBAC=0.025/(0.55\cdot 70)\times 100\%-(0.017\cdot 2)\approx 0.031\%}$
Standard drink chart (U.S.)[5]
Alcohol Amount (ml) Amount (fl oz) Serving size Alcohol (% by vol.) Alcohol
80 proof liquor 44 1.5 One shot 40 0.6 US fl oz (18 ml)
Table wine 148 5 One glass 12 0.6 US fl oz (18 ml)
Beer 355 12 One can/bottle 5 0.6 US fl oz (18 ml)
Note: This chart defines a standard drink as 0.6 fluid ounces (14 g) of ethanol, whereas other definitions exist, for example 10 grams of ethanol.
Standard drink sizes (Australia)
• 375 ml can of light beer (2.7% alcohol) = 0.8 standard drinks
• 375 ml can of mid-strength beer (3.5% alcohol) = 1 standard drink
• 375 ml can of full strength beer (4.8% alcohol) = 1.4 standard drinks
• 100 ml glass of wine (13.5% alcohol) = 1 standard drink
• 150 ml glass of wine (13.5% alcohol) = 1.5 standard drinks
• 30 ml shot of spirits (40% alcohol) = 0.95 standard drinks
• 440 ml can of pre-mix spirits (approx. 5% alcohol) = 1.7 standard drinks
• 440 ml can pre-mix spirits (approx. 7% alcohol) = 2.4 standard drinks
Approximate blood alcohol percentage (by volume)[6]
Based on one drink having 0.5 US fl oz (15 ml) alcohol by volume
Drinks Sex Body weight
40 kg
90 lb
45 kg
100 lb
55 kg
120 lb
64 kg
140 lb
73 kg
160 lb
82 kg
180 lb
91 kg
200 lb
100 kg
220 lb
109 kg
240 lb
1 Male 0.04 0.03 0.03 0.02 0.02 0.02 0.02 0.02
Female 0.05 0.05 0.04 0.03 0.03 0.03 0.02 0.02 0.02
2 Male 0.08 0.06 0.05 0.05 0.04 0.04 0.03 0.03
Female 0.10 0.09 0.08 0.07 0.06 0.05 0.05 0.04 0.04
3 Male 0.11 0.09 0.08 0.07 0.06 0.06 0.05 0.05
Female 0.15 0.14 0.11 0.10 0.09 0.08 0.07 0.06 0.06
4 Male 0.15 0.12 0.11 0.09 0.08 0.08 0.07 0.06
Female 0.20 0.18 0.15 0.13 0.11 0.10 0.09 0.08 0.08
5 Male 0.19 0.16 0.13 0.12 0.11 0.09 0.09 0.08
Female 0.25 0.23 0.19 0.16 0.14 0.13 0.11 0.10 0.09
6 Male 0.23 0.19 0.16 0.14 0.13 0.11 0.10 0.09
Female 0.30 0.27 0.23 0.19 0.17 0.15 0.14 0.12 0.11
7 Male 0.26 0.22 0.19 0.16 0.15 0.13 0.12 0.11
Female 0.35 0.32 0.27 0.23 0.20 0.18 0.16 0.14 0.13
8 Male 0.30 0.25 0.21 0.19 0.17 0.15 0.14 0.13
Female 0.40 0.36 0.30 0.26 0.23 0.20 0.18 0.17 0.15
9 Male 0.34 0.28 0.24 0.21 0.19 0.17 0.15 0.14
Female 0.45 0.41 0.34 0.29 0.26 0.23 0.20 0.19 0.17
10 Male 0.38 0.31 0.27 0.23 0.21 0.19 0.17 0.16
Female 0.51 0.45 0.38 0.32 0.28 0.25 0.23 0.21 0.19
Subtract approximately 0.01 every 40 minutes after drinking.
## Binge drinking
The National Institute on Alcohol Abuse and Alcoholism (NIAAA) define the term "binge drinking" as a pattern of drinking that brings a person's blood alcohol concentration (BAC) to 0.08 grams percent or above. This typically happens when men consume five or more drinks, and when women consume four or more drinks, in about two hours.[7]
## Units of measurement
There are several different units in use around the world for defining blood alcohol concentration. Measures of mass (never volume) of alcohol per volume of blood or per mass of blood are used. 1 milliliter of blood has a mass of approximately 1.06 grams, so that units by volume are about 6% greater than if mass-based.
The amount of alcohol on the breath can also be measured, without requiring drawing blood, by blowing into a breathalyzer. The amount of alcohol measured on the breath is generally accepted as proportional to the amount of alcohol present in the blood, at a rate of 1:2100, so blood alcohol can be estimated from a brethalyzer reading. For example, a breathalyzer measurement of 0.10 mg/L of breath alcohol characterises 0.0001×2100 g/L, or 0.21 g/L of blood alcohol (equivalent to 0.21 permille).
While a variety of units or criteria are used in different jurisdictions, many countries use the g/L unit, clearer than if expressed as a percentage. Usual units are listed below. For example, the U.S. uses a concentration unit of 1% w/v (percent mass/volume, equivalent to 10 g/l or 1 g per 100 ml).
Reference Unit Dimensions Equivalent to Used in
BAC by volume 1 percent (%) 1/100 g/mL = 1 g/dL 9.43 mg/g, 217.4 mmol/L United States
1 permille (‰) 1/1000 g/mL = 1 g/L 0.943 mg/g, 21.7 mmol/L Austria, Belgium, Bulgaria, Czech Republic, France, Latvia, Lithuania, Netherlands, Poland, Portugal, Romania, Slovenia, Spain, Switzerland, Turkey
1 basis point (‱) 1/10,000 g/mL = 10 mg/100 mL 94.3 ppm, 2.17 mmol/L United Kingdom
BAC by mass 1 percent (%) 1/100 g/g = 1 cg/g 1.06 cg/mL, 230 mmol/L Australia,[8] Canada[9]
1 permille (‰) 1/1000 g/g = 1 mg/g 1.06 mg/mL, 23 mmol/L Finland, Norway, Sweden, Denmark, Germany, Ireland, Russian Federation
1 part per million (ppm) 1/1,000,000 g/g = 1 μg/g 1.06 μg/mL, 23 μmol/L
## Legal limits
Map of Europe showing countries' blood alcohol limits as defined in g/dl for the general population
For purposes of law enforcement, blood alcohol content is used to define intoxication and provides a rough measure of impairment. Although the degree of impairment may vary among individuals with the same blood alcohol content, it can be measured objectively and is therefore legally useful and difficult to contest in court. Most countries forbid operation of motor vehicles and heavy machinery above prescribed levels of blood alcohol content. Operation of boats and aircraft is also regulated.
The alcohol level at which a person is considered legally impaired to drive varies by country. The list below gives limits by country. These are typically blood alcohol content limits for the operation of a vehicle. Most limits are specified by volume; a few countries use BAC by mass, equivalent to a figure about 6% higher by volume.
### 0%
It is illegal to have any measurable alcohol in the blood while driving in these countries. Most jurisdictions have a tolerance slightly higher than zero to account for false positives and naturally occurring alcohol in the body. Some of the following jurisdictions have a general prohibition of alcohol.
• Australia—Drivers who are learners or holders of a Provisional/Probationary licence
• Brazil
• Brunei
• Canada—new drivers undergoing graduated licensing in Ontario, British Columbia,[10] Newfoundland and Labrador and Alberta;[11] drivers under the age of 22 in Manitoba, New Brunswick, Northwest Territories, Nova Scotia, Ontario,[12] Saskatchewan and in Quebec receive a 30-day suspension and 7-day vehicle seizure.[13] Drivers in Alberta who are in the graduated licensing program, regardless of age, are subject to the same 30-day/7-day suspensions/seizure policy.[14]
• Colombia —Zero Alcohol Tolerance law is effective since December 2013 [15][16]
• Czech Republic
• Estonia
• Fiji
• Germany—for learner drivers, all drivers 18–21 and newly licensed drivers of any age for first two years of licence
• Hungary
• Israel—24 µg per 100 ml (0.024%) of breath (penalties only apply above 26 µg per 100 ml (0.026%) of breath following lawsuits about sensitivity of devices used). New drivers, drivers under 24 years of age and commercial drivers 5 µg per 100 ml of breath.(0.005%) [17]
• Italy—for drivers in their first three years after gaining a driving license
• Japan—drivers under the age of 20 because of not reaching legal drinking age.
• Kuwait
• New Zealand—drivers under the age of 20; drivers convicted of excess breath alcohol may be required to gain a zero-limit license.
• Nepal
• Oman
• Qatar
• Pakistan
• Paraguay
• Romania—beyond 0.04% drivers will not only receive a fine and have their license suspended, the offense will also be added to their criminal records.
• Russian Federation—0% introduced in 2010,[18] but discontinued in September 2013[19]
• Saudi Arabia
• Slovakia
• Uruguay[20]
• United Arab Emirates
• Vietnam
### 0.02%
• China
• Faroe Islands
• Netherlands (for drivers in their first five years after gaining a driving license)[21]
• Norway (road vehicles and sea vessels over 15 m),[22] alternatively 0.1 mg/L of breath.
• Poland (0.02% – 0.05% is a petty offence, over 0.05% is a criminal offence)
• Puerto Rico
• Russia (0.018% since September 2013[19])
• Serbia
• Sweden
• Ukraine
• United States—drivers under the age of 21 must have 0.02% or less on the federal level; however, most states have Zero Tolerance laws emplaced. Otherwise the limit is 0.08%, except in Utah, where it is 0.05%.
### 0.03%
• Belarus
• Bosnia and Herzegovina (0.031%)
• Chile
• India
• Japan[23]
• Korea
### 0.04%
• Lithuania (0.00% for car drivers in their first three years after gaining a driving license, motorcycle and truck drivers)
### 0.05%
• Argentina: 0.02% for motorbikes, 0.00% for truck, taxi, and bus drivers, 0.00% in the provinces of Cordoba and Salta
• Australia (BAC stated by mass, not volume): 0.00% for Australian Capital Territory learner, provisional and convicted DUI drivers (changed down from 0.02% on December 1, 2010), 0.02% for truck/bus/taxi, 0.00% for learner drivers, provisional/probationary drivers (regardless of age), truck and bus drivers, driving instructors and DUI drivers in all other states
• Austria: no limit for pedestrians; 0.08% for cycling; 0.05% generally for cars <7.5 t (driving licence B) and motorbikes (A); but 0,01% during learning (for driver and teacher or L17-assistant). During probation period (at least the first 3 years) or up to the age of 21, when license was handed out after July 1, 2017, when older (at least the first 2 years) or up to the age of 20 (A1, AM, L17, F), trucks (C >7.5 t), bus (D), drivers of taxi and public transport[24][25]
• Belgium (also for cyclists)
• Bulgaria
• Canada: (BAC stated by mass, not volume) Alberta, British Columbia, Ontario, Manitoba, Newfoundland, Nova Scotia, New Brunswick—provincial offence. Drivers have not committed a criminal offense, however a 3-day licence suspension and 3-day vehicle seizure occurs.
• Costa Rica
• Croatia: professional drivers, driving instructors and drivers of the vehicle categories C1, C1+E, C, C+E, D, D+E and H; the limit for other drivers is 0.50 mg/g, but they do get an additional separate fine if they cause an accident while having a blood alcohol level between 0 and 0,50 mg/g[26]
• Denmark (excl. Faroe Islands)
• Finland
• France: 0.025% for bus drivers. Between 0.05% and 0.08%, drivers can be fined €135 and have six points removed from their licence. Above 0.08%, the punishment is more severe with possible imprisonment of up to two years, heavy fines and licence suspension.[27]
• Germany (0.0% for learner drivers, all drivers 18–21 and newly licensed drivers of any age for first two years of licence; also, if the BAC exceeds 0.03%, driving is illegal if the driver is showing changes in behavior ("Relative Fahruntüchtigkeit"))
• Greece (0.02% for drivers in their first two years after gaining a driving license)
• Hong Kong
• Iceland: New laws yet to take effect will change the limit to 0.02%.
• Ireland: 0.02% for professional, learner and novice drivers(drivers in their first two years after gaining a driving license)[28]
• Israel: 24 µg per 100 ml (0.024%) of breath (penalties only apply above 26 µg per 100 ml (0.026%) of breath due to lawsuits about sensitivity of devices used). This is equivalent to a BAC of 0.05. New drivers, drivers under 24 years of age and commercial drivers 5 µg per 100 ml of breath. This is equivalent to a BAC of 0.01.[17]
• Italy: 0.00% for drivers in their first three years after gaining a driving license
• Latvia: 0.02% for drivers in their first two years after gaining a driving license
• Luxembourg
• Malta: 0.02% for drivers with a probationary driving licence and drivers of commercial vehicles, and 0.00% for drivers of buses, coaches and other passenger carrying vehicles.[29]
• Mauritius[30]
• Netherlands: 0.02% for drivers in their first five years after gaining a driving license[21]
• New Zealand
• North Macedonia: 0.00% for drivers in their first two years after gaining a driving license
• Peru
• Philippines: 0.00% for taxicab and public transport drivers[31]
• Portugal: 0.02% for drivers holding a driver's licence for less than three years, professional drivers, and drivers of taxis, heavy vehicles, emergency vehicles, public transport of children and carrying dangerous goods.
• Scotland: Scotland's drink-drive limit was reduced, by law, on 5 December 2014 from 0.08 to any of the following: 22 mcg of alcohol in 100 ml of breath, 50 mg of alcohol in 100 ml of blood, or 67 mg of alcohol in 100 ml of urine[32]
• Slovenia: 0.00% for drivers in their first two years after gaining a drivers licence, drivers under 21 and professional drivers, such as buses, trucks.
• South Africa: 0.02% for professional drivers; to be changed for all to 0.00% by June 2020[33]
• Spain (0.03% for drivers in their first two years after gaining a driving license and common carriers, such as buses and trucks)
• Switzerland (0.00% for learner drivers, drivers which are in their first three years after gaining a drivers licence and for driving instructors)[34]
• Thailand: 0.02% for drivers who (1) hold a probationary driving licence or; (2) have a licence for different vehicle category or; (3) are under 20 years old or; (4) are disqualified and attempt to drive illegally[35]
• Taiwan: breath alcohol limit decreased from 0.25 to 0.15 from 13 June 2013
• Turkey
• United States – Utah[36]
• Honduras
### 0.08%
• England and Wales;[32] 0.02% for operators of fixed-wing aircraft
• Liechtenstein
• Malaysia: 0.00 for Probationary Driving Licence holders
• Mexico
• New Zealand: Criminal offence
• Norway: legal limit for sea vessels under 15 m[39]
• Northern Ireland: The government of Northern Ireland intends to reduce the general limit to 0.05%.[40]
• Puerto Rico: For drivers 18 years and older
• Singapore[41]
• United States: Every state imposes mandatory penalties for operating a vehicle with a BAC level of 0.08% or greater,[42][43] except Utah where the limit is 0.05%.[44] Even below those levels drivers can have civil liability and other criminal guilt (e.g., in Arizona driving impairment to any degree caused by alcohol consumption can be a civil or criminal offense in addition to other offenses at higher blood alcohol content levels). Drivers under 21 (the most common U.S. legal drinking age) are held to stricter standards under zero tolerance laws adopted in varying forms in all states: commonly 0.01% to 0.05%. See Alcohol laws of the United States by state. Federal Motor Carrier Safety Administration: 0.04% for drivers of a commercial vehicle requiring a commercial driver's license[45] and 0.01% for operators of common carriers, such as buses.[46]
### Breath alcohol content
In certain countries, alcohol limits are determined by the breath alcohol content (BrAC), not to be confused with blood alcohol content (BAC).
• In Greece, the BrAC limit is 250 micrograms of alcohol per litre of breath. The limit in blood is 0.50 g/l. The BrAC limit for drivers in their first two years after gaining a driving license and common carriers are more restricted to 100 micrograms per litre of breath.
• BrAC 250–400 = 200 fine.
• BrAC 400–600 = €700 fine, plus suspension of driving license for 90 days (introduced in 2007)[47]
• BrAC >600 = 2 months imprisonment, plus suspension of driving license for 180 days, plus €1,200 fine
• In Hong Kong, the BrAC limit is 220 micrograms per litre of breath (as well as other defined limits)
• In The Netherlands and Finland, the BrAC limit is 220 micrograms of alcohol per litre of breath (μg/l, colloquially known as "ugl" because some types of breathalyzer show the μ as 'u' due to screen size limitations).
• In New Zealand, the BrAC limit is 250 micrograms of alcohol per litre of breath for those aged 20 years or over, and zero (meaning illegal to have any measurable breath alcohol content) for those aged under 20 years.[48]
• In Singapore, the BrAC limit is 350 micrograms of alcohol per litre of breath.[41]
• In Spain the BrAC limit is 250 micrograms of alcohol per litre of breath and 150 micrograms per litre of breath for drivers in their first two years after gaining a driving license and common carriers.
• In England and Wales the BrAC limit is 350 micrograms of alcohol per litre of breath (as well as the above defined blood alcohol content).
• In Scotland the BrAC limit is 220 micrograms of alcohol per litre of breath (as well as the above defined blood alcohol content).
• In Trinidad and Tobago the BrAC limit is 35 micrograms of alcohol per 100 millilitres of breath (as well as the above defined blood alcohol content).
### Other limitation schemes
• For South Korea, the penalties for different blood alcohol content levels include
• 0.01–0.049 = No penalty
• 0.05–0.09 = 100 days license suspension
• >0.10 = Cancellation of car license.
## Test assumptions
Blood alcohol tests assume the individual being tested is average in various ways. For example, on average the ratio of blood alcohol content to breath alcohol content (the partition ratio) is 2100 to 1. In other words, there are 2100 parts of alcohol in the blood for every part in the breath. However, the actual ratio in any given individual can vary from 1300:1 to 3100:1, or even more widely.[49] This ratio varies not only from person to person, but within one person from moment to moment. Thus a person with a true blood alcohol level of 0.08% but a partition ratio of 1700:1 at the time of testing would have a 0.10 reading on a Breathalyzer calibrated for the average 2100:1 ratio.
After fatal accidents, it is common to check the blood alcohol levels of involved persons. However, soon after death, the body begins to putrefy, a biological process which produces ethanol. This can make it difficult to conclusively determine the blood alcohol content in autopsies, particularly in bodies recovered from water.[50][51][52][53] For instance, following the Moorgate tube crash, the driver had a blood alcohol concentration of 80 mg/100 ml, but it could not be established how much of this could be attributed to natural decomposition.
### Extrapolation
Retrograde extrapolation is the mathematical process by which someone's blood alcohol concentration at the time of driving is estimated by projecting backwards from a later chemical test. This involves estimating the absorption and elimination of alcohol in the interim between driving and testing. The rate of elimination in the average person is commonly estimated at 0.015 to 0.020 grams per deciliter per hour (g/dl/h),[54] although again this can vary from person to person and in a given person from one moment to another. Metabolism can be affected by numerous factors, including such things as body temperature, the type of alcoholic beverage consumed, and the amount and type of food consumed.
In an increasing number of states, laws have been enacted to facilitate this speculative task: the blood alcohol content at the time of driving is legally presumed to be the same as when later tested. There are usually time limits put on this presumption, commonly two or three hours, and the defendant is permitted to offer evidence to rebut this presumption.
Forward extrapolation can also be attempted. If the amount of alcohol consumed is known, along with such variables as the weight and sex of the subject and period and rate of consumption, the blood alcohol level can be estimated by extrapolating forward. Although subject to the same infirmities as retrograde extrapolation—guessing based upon averages and unknown variables—this can be relevant in estimating BAC when driving and/or corroborating or contradicting the results of a later chemical test.
## Metabolism
Alcohol is absorbed throughout the gastrointestinal tract, but more slowly in the stomach than in the small or large intestine. For this reason, alcohol consumed with food is absorbed more slowly, because it spends a longer time in the stomach.[55] Furthermore, alcohol dehydrogenase is present in the stomach lining. After absorption, the alcohol passes to the liver through the hepatic portal vein, where it undergoes a first pass of metabolism before entering the general bloodstream.[56]
Alcohol is removed from the bloodstream by a combination of metabolism, excretion, and evaporation. Alcohol is metabolized mainly by the group of six enzymes collectively called alcohol dehydrogenase. These convert the ethanol into acetaldehyde (an intermediate more toxic than ethanol). The enzyme acetaldehyde dehydrogenase then converts the acetaldehyde into non-toxic acetic acid.
Many physiologically active materials are removed from the bloodstream (whether by metabolism or excretion) at a rate proportional to the current concentration, so that they exhibit exponential decay with a characteristic half-life (see pharmacokinetics). This is not true for alcohol, however. Typical doses of alcohol actually saturate the enzymes' capacity, so that alcohol is removed from the bloodstream at an approximately constant rate. This rate varies considerably between individuals. Another sex-based difference is in the elimination of alcohol. People under 25[citation needed], women,[57] or people with liver disease may process alcohol more slowly. Falsely high BAC readings may be seen in patients with kidney or liver disease or failure.[citation needed]
Such persons also have impaired acetaldehyde dehydrogenase, which causes acetaldehyde levels to peak higher, producing more severe hangovers and other effects such as flushing and tachycardia. Conversely, members of certain ethnicities that traditionally did not use alcoholic beverages have lower levels of alcohol dehydrogenases and thus "sober up" very slowly but reach lower aldehyde concentrations and have milder hangovers. The rate of detoxification of alcohol can also be slowed by certain drugs which interfere with the action of alcohol dehydrogenases, notably aspirin, furfural (which may be found in fusel alcohol), fumes of certain solvents, many heavy metals, and some pyrazole compounds. Also suspected of having this effect are cimetidine, ranitidine, and acetaminophen (paracetamol).
Currently, the only known substance that can increase the rate of alcohol metabolism is fructose. The effect can vary significantly from person to person, but a 100 g dose of fructose has been shown to increase alcohol metabolism by an average of 80%. Fructose also increases false positives of high BAC readings in anyone with proteinuria and hematuria, due to kidney-liver metabolism.[58]
The peak of blood alcohol level (or concentration of alcohol) is reduced after a large meal.[55]
## Highest levels
There have been reported cases of blood alcohol content higher than 1%:
• In 1982, a 24-year-old woman was admitted to the UCLA emergency room with a serum alcohol content of 1.51%, corresponding to a blood alcohol content of 1.33%. She was alert and oriented to person and place and survived.[59] Serum alcohol concentration is not equal to nor calculated in the same way as blood alcohol content.[60]
• In 1984, a 30-year-old man survived a blood alcohol concentration of 1.5% after vigorous medical intervention that included dialysis and intravenous therapy with fructose.[61]
• In 1995, a man from Wrocław, Poland, caused a car accident near his hometown. He had a blood alcohol content of 1.48%; he was tested five times, with each test returning the same reading. He died a few days later of injuries from the accident.[62]
• In 2004, an unidentified Taiwanese woman died of alcohol intoxication after immersion for twelve hours in a bathtub filled with 40% ethanol. Her blood alcohol content was 1.35%. It was believed that she had immersed herself as a response to the SARS epidemic.[63]
• In South Africa, a man driving a Mercedes-Benz Vito light van containing 15 sheep allegedly stolen from nearby farms was arrested on December 22, 2010, near Queenstown in Eastern Cape. His blood had an alcohol content of 1.6%. Also in the vehicle were five boys and a woman, who were also arrested.[64]
• On 26 October 2012, a man from Gmina Olszewo-Borki, Poland, who died in a car accident, recorded a blood alcohol content of 2.23%; however, the blood sample was collected from a wound and thus possibly contaminated.[62]
• On 26 July 2013 a 40-year-old man from Alfredówka, Poland, was found by Municipal Police Patrol from Nowa Dęba lying in the ditch along the road in Tarnowska Wola. At the hospital, it was recorded that the man had a blood alcohol content of 1.374%. The man survived.[65][66]
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61. ^ O'Neill, Shane; Tipton, KF; Prichard, JS; Quinlan, A (1984). "Survival After High Blood Alcohol Levels: Association with First-Order Elimination Kinetics". Archives of Internal Medicine. 144 (3): 641–2. doi:10.1001/archinte.1984.00350150255052. PMID 6703836.
62. ^ a b Łuba, Marcin (24 October 2012). "Śmiertelny rekord: Kierowca z powiatu ostrołęckiego miał 22 promile alkoholu! Zginął w wypadku". eOstroleka.pl (in Polish). Retrieved 4 November 2017.
63. ^ Wu, Yen-Liang; Guo, How-Ran; Lin, Hung-Jung (2005). "Fatal alcohol immersion during the SARS epidemic in Taiwan". Forensic Science International. 149 (2–3): 287. doi:10.1016/j.forsciint.2004.06.014. PMC 7131152. PMID 15749375.
64. ^ Mashaba, Sibongile (24 December 2010). "Drunkest driver in SA arrested". Sowetan. Retrieved 31 March 2022.
65. ^ "Miał 13,74 promila alkoholu we krwi. I przeżył. Rekord świata?" [He had 13.74 blood alcohol levels. And he survived. World record?]. Archived from the original on 11 August 2013. Retrieved 8 August 2013.
66. ^
### General bibliography
• Carnegie Library of Pittsburgh. Science and Technology Department. The Handy Science Answer Book. Pittsburgh: The Carnegie Library, 1997. ISBN 978-0-7876-1013-5.
• Perham, Nick; Moore, Simon C.; Shepherd, Jonathan; Cusens, Bryany (2007). "Identifying drunkenness in the night-time economy". Addiction. 102 (3): 377–80. doi:10.1111/j.1360-0443.2006.01699.x. PMID 17298644.
• Taylor, L., and S. Oberman. Drunk Driving Defense, 6th edition. New York: Aspen Law and Business, 2006. ISBN 978-0-7355-5429-0.
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### Mathematics Class XI
Unit-I: Sets and Functions
Chapter 1: Sets
Unit-II: Algebra
Chapter 5: Binomial Theorem
Chapter 6: Sequence and Series
Unit-III: Coordinate Geometry
Chapter 1: Straight Lines
Chapter 2: Conic Sections
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability
Chapter 1: Statistics
Chapter 2: Probability
# Area of a Triangle
The area of a triangle with vertices , and is denoted by , where
Example:
Let be a triangle with vertices and , then area of is
Corollary (Collinearity of three points)
The three points and will be collinear if the area of the triangle is zero.
i.e.
Example:
The points and are collinear when
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# On the Utility of P/E Ratios Going Back 70+ Years
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After my last post, Bill Rempel (among others) inquired about the difference in normalized P/E ratios over different decades. He felt a standard applied across 70+ years might not be particularly useful today, because so much has changed (increased participation in equity markets, different monetary policies, etc.) As a result, it might be that while "relatively cheap" is better than "relatively expensive" within each time period, it is inappropriate to compare years from dissimilar decades as if the same standards applied.
I agree. So, over the next two posts, I'll try to give you an idea of what the compound point growth in the Dow looked like following the low normalized P/E years and high normalized P/E years within each decade. In other words, I'll look at low normalized P/E years and high normalized P/E years relative to other years in the same decade.
In this post, I'll simply show you the results of two different comparisons across decades.
The first comparison consists of a group of the five lowest normalized P/E years from each decade vs. the five highest normalized P/E years from each decade. The second comparison consists of a group of the three lowest normalized P/E years from each decade vs. the three highest normalized P/E years from each decade.
Although this is still a comparison across many decades (for this post, I'm only using 1940-1999, because I want to use only "complete" decades), it should give you some idea of whether the normalized P/E effect is simply a result of a few years in just one or two particular decades, or whether the effect tends to hold up over many different decades.
In my next post, I will go a step further, and actually break down the results decade by decade.
The Least You Need to Know
If you can't be bothered to read yet another post on normalized P/E ratios (I've written quite a few lately), here's the least you need to know:
1. The five lowest normalized P/E years from each decade from the 1940s through the 1990s saw higher compound point growth in the Dow over the subsequent 1, 3, 5, 10, 15, 20, 25, and 30 years than the group of the five highest normalized P/E years from each decade.
2. The three lowest normalized P/E years from each decade from the 1940s through the 1990s saw higher compound point growth in the Dow over the subsequent 1, 3, 5, 10, 15, 20, 25, and 30 years than the group of the three highest normalized P/E years from each decade.
Obviously, there's a lot more nuance to the results than is suggested by the above two statements; but, the most important point to take away from this post is simply that the combined low normalized P/E group (drawing equally from all decades from the 1940s through the 1990s) beat its high normalized P/E counterpart over all the holding periods I measured.
Remember, these are the combined groups. This post doesn't address the normalized P/E effect within specific decades; I'll discuss that in my next post.
As you look over the compound point growth data, also remember that low normalized P/E years are not necessarily low P/E years. This is reflected somewhat in the data. For instance, in these first two groups, the average P/E ratio is 12.81 for the low normalized P/E group and 14.61 for the high normalized P/E group. The difference in 15-year normalized P/E ratios is quite a bit larger – 11.30 for the low normalized P/E group and 16.02 for the high normalized P/E group.
The Results
If you group the five lowest normalized P/E years from each decade and the five highest normalized P/E years from each decade, this is what the subsequent compound point growth looks like:
1 Year
Low Normalized P/E Group: 9.88%
High Normalized P/E Group: 6.78%
3 Year
Low Normalized P/E Group: 9.84%
High Normalized P/E Group: 5.99%
5 Year
Low Normalized P/E Group: 9.93%
High Normalized P/E Group: 5.46%
10 Year
Low Normalized P/E Group: 8.69%
High Normalized P/E Group: 6.61%
15 Year
Low Normalized P/E Group: 8.63%
High Normalized P/E Group: 6.08%
20 Year
Low Normalized P/E Group: 8.19%
High Normalized P/E Group: 5.48%
25 Year
Low Normalized P/E Group: 7.21%
High Normalized P/E Group: 5.61%
30 Year
Low Normalized P/E Group: 6.22%
High Normalized P/E Group: 5.94%
The difference between the mean and median returns within each group for each span of years is usually very small. However, there is one exception: 1-year point growth for the high normalized P/E group.
The mean is 6.78%; the median is 3.21%. This wide disparity between the mean and median occurs only over a one-year period – by three years, the mean and median are more or less in agreement.
In these next two groups, the average P/E ratio is 12.73 for the low normalized P/E group and 15.07 for the high normalized P/E group. The difference in 15-year normalized P/E ratios is quite a bit larger – 10.80 for the low normalized P/E group and 17.31 for the high normalized P/E group.
If you group the three lowest normalized P/E years from each decade and the three highest normalized P/E years from each decade, this is what the subsequent compound point growth looks like:
1 Year
Low Normalized P/E Group: 6.95%
High Normalized P/E Group: 3.55%
3 Year
Low Normalized P/E Group: 8.89%
High Normalized P/E Group: 3.01%
5 Year
Low Normalized P/E Group: 9.64%
High Normalized P/E Group: 3.77%
10 Year
Low Normalized P/E Group: 9.24%
High Normalized P/E Group: 6.12%
15 Year
Low Normalized P/E Group: 8.69%
High Normalized P/E Group: 5.49%
20 Year
Low Normalized P/E Group: 8.70%
High Normalized P/E Group: 4.98%
25 Year
Low Normalized P/E Group: 7.38%
High Normalized P/E Group: 5.58%
30 Year
Low Normalized P/E Group: 5.98%
High Normalized P/E Group: 5.88%
The difference between the mean and median returns within each group for each span of years is usually very small. However, there is one exception: 1-year point growth for the high normalized P/E group.
The mean is 3.55%; the median is 1.84%. So, if we were comparing median point growth instead of mean point growth, the low normalized P/E group would outperform the high normalized P/E group by 7.10% vs. 1.84% instead of 6.95% vs. 3.55% over a one-year period.
I'm not saying that the median is a better measure than the mean; I'm just saying that considering the variability within the group, anyone making a bet on the direction of a high normalized P/E market over a one-year time span needs to know that the rather tame mean return for this group obscures the fact that unusually big up or down moves are actually quite common in such years.
Related Reading
A Look At 15-Yr. Normalized Dow P/E Ratios
On P/E Ratios in the 21st Century: Lower Your Expectations For Market Returns!
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# Complement of Interior equals Closure of Complement
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## Theorem
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ and $H^\circ$ denote the interior of $H$.
Let $\map \complement H$ be the complement of $H$ in $T$:
$\map \complement H = T \setminus H$
Then:
$\map \complement {H^\circ} = \paren {\map \complement H}^-$
and similarly:
$\paren {\map \complement H}^\circ = \map \complement {H^-}$
These can alternatively be written:
$T \setminus H^\circ = \paren {T \setminus H}^-$
$\paren {T \setminus H}^\circ = T \setminus H^-$
which, it can be argued, is easier to follow.
## Proof
Let $\tau$ be the topology on $T$.
Let $\mathbb K = \set {K \in \tau: K \subseteq H}$.
Then:
$\displaystyle T \setminus H^\circ$ $=$ $\displaystyle T \setminus \bigcup_{K \mathop \in \mathbb K} K$ Definition of Interior (Topology) of $H$ $\displaystyle$ $=$ $\displaystyle \bigcap_{K \mathop \in \mathbb K} \paren {T \setminus K}$ De Morgan's Laws: Difference with Union
By the definition of closed set, $K$ is open in $T$ if and only if $T \setminus K$ is closed in $T$.
Also, from Set Complement inverts Subsets we have that $T \setminus K \supseteq T \setminus H$.
Now consider the set $\mathbb K'$ defined as:
$\mathbb K' := \set {K' \subseteq T: \paren {T \setminus H} \subseteq K', K' \text { closed in } T}$
From the above we see that:
$K \in \mathbb K \iff T \setminus K \in \mathbb K'$
Thus:
$\displaystyle T \setminus H^\circ$ $=$ $\displaystyle \bigcap_{K' \mathop \in \mathbb K'} K'$ $\displaystyle$ $=$ $\displaystyle \paren {T \setminus H}^-$ Definition 2 of Closure of $T \setminus H$
$\Box$
Then we note that:
$\displaystyle H^\circ$ $=$ $\displaystyle T \setminus \paren {T \setminus H^\circ}$ Relative Complement of Relative Complement $\displaystyle$ $=$ $\displaystyle T \setminus \paren {\paren {T \setminus H}^-}$ from above
and so:
$\displaystyle \paren {T \setminus H}^\circ$ $=$ $\displaystyle T \setminus \paren {\paren {T \setminus \paren {T \setminus H} }^-}$ from above $\displaystyle$ $=$ $\displaystyle T \setminus H^-$ Relative Complement of Relative Complement
$\blacksquare$
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# 11.1 The world of wind instruments
We have said a little about wind instruments in earlier chapters, but now it is time to go into more detail. It is useful to start by comparing the wind instruments with the bowed-string instruments. Both types of instrument are capable of producing sustained tones, and we learned back in section 8.1 that this automatically means that they must all involve nonlinearity in some essential way. In the case of a bowed string, the main source of nonlinearity was from the frictional interaction of between bow-hair and string. In the wind instruments, the nonlinearity is associated with things going on near the mouthpiece end, which we will discuss in detail shortly.
An obvious difference between stringed instruments and wind instruments is that wind instruments usually have no equivalent of the stringed instrument body, which influences the sound of the instrument by imposing an extra stage of filtering between the string motion and the radiated sound. In most wind instruments, the sound is radiated directly by air motion in the tube which is the analogue of the string: this tube provides a set of resonances, sometimes approximately harmonically related, which interact with the nonlinearity to produce the pitch and waveform of the played note.
There are two other, less obvious, differences between bowed strings and wind instruments. First, the resonances of air in a tube are usually much more highly damped than the resonances of a string. A violin string might have Q factors of the order of thousands, whereas tube resonances are more likely to have Q factors of the order of hundreds or less. The second thing is that the nonlinearity associated with friction is (as we saw back in Chapter 9) a very severe one, whereas wind instrument nonlinearities tend to be less vigorous.
If you recall the distinction between “smooth” and “non-smooth” nonlinearities from section 8.2, this suggests that rather different mathematical approaches might be called for in the two cases. The big disparity in Q factors also pushes us in a similar direction. The result is that the modelling of wind instruments places a lot of emphasis on periodic regimes of oscillation, thresholds, and bifurcations between regimes. With the bowed string, we saw that there was no useful concept of a threshold of vibration, where the motion is “almost sinusoidal”: a bowed string goes into strongly-nonlinear stick-slip vibration straight away, and theoretical treatment tends to rely more heavily on “brute force” numerical simulation.
The next step is to sort the enormous diversity of wind instruments into a few general categories. If you look at the wind instruments in a western symphony orchestra or wind band (see Fig. 1 for an example), a three-way classification seems self-evident. There are the reed instruments (clarinet, oboe, bassoon), the flute-like instruments (flute, piccolo) and the brass instruments (trumpet, trombone, French horn). But if we cast our net a little wider than the symphony orchestra, we soon find out that things are more complicated and less clear.
For a start, by focussing on orchestral instruments we are missing one category of instrument entirely: the “free reed” instruments, like the harmonium, accordion or harmonica. This gives a four-way classification, which will turn out to allow us to pigeon-hole virtually all wind instruments (including the human voice). However, the categories are not always as distinct as you might think. It is important to be clear what the basis of the classification is. One thing that definitely does not influence the classification of instruments is the material they are made of. Despite the names, some “woodwind” instruments are made of metal while some “brass” instruments are made of wood.
Rather, the key observation is one we have already mentioned: all wind instruments are capable of producing a sustained tone, so they must all involve nonlinearity in some essential way. The four-way classification separates instruments based mainly on the details of the nonlinearity that allows them to work. Three of the four are related, to the extent that they blur into each other at the boundaries. These are the reeds, the brass and the free reeds. The fourth member really is different, though. Flutes and their relatives are driven by an air jet interacting with a sharp edge. This excitation mechanism involves no mechanical moving parts, unlike the reeds or vibrating lips of the other three categories of wind instrument.
I will give a brief description of the characteristic features of the four categories of instruments here, but more detail will be given in separate sections on each type: sections 11.3—11.7. Figure 2 shows a sketch of the mouthpiece end of a clarinet-like reed instrument. The player’s lips seal round the mouthpiece and reed, and then the player applies a suitable blowing pressure so that air flows through the narrow gap. As we already discussed briefly in section 8.5, a crucial feature of this kind of backward-facing reed is that the player’s blowing pressure tends to close the reed. If the pressure difference between the mouth and the inside of the mouthpiece gets too big, the reed closes completely against the rigid part of the mouthpiece (the “lay”) — or at least, almost completely. In reality there will be a bit of leakage because the reed and the lay do not meet perfectly.
Of course, the reed has a resonance frequency of its own. Indeed, it has many resonances if we look sufficiently high in frequency, but the lowest resonance is the most important for the normal musical functioning of a clarinet. Even this lowest resonance is usually placed higher than the fundamental frequency of any played note on the instrument, so reed resonance is a relatively minor contribution to the behaviour of the instrument. Much more important are the resonance frequencies of the tube, which the player manipulates by opening and closing tone-holes. The playing frequency of any particular note is predominantly governed by these tube resonances.
All this is in sharp contrast to our second category of instrument, which we will continue to call “brass” instruments despite the comment about materials made above. Figure 3 shows a sketch of the mouthpiece end of a typical brass instrument. The tube is terminated by a cup-shaped mouthpiece. The player presses their mouth firmly against this cup, and they “buzz” their lips. This makes the lips open and close at the frequency of the note being played, and high-speed video recordings of brass players [1] show that, when air is flowing out of the mouth into the instrument, the lips open outwards as sketched in the figure. So the lips act in a somewhat similar way to a clarinet reed insofar as they provide a kind of nonlinear valve controlling air flow into the instrument, but we will see in section 11.5 that the behaviour is crucially different because of the outward-opening motion: higher pressure in the mouth tends to make the gap get bigger, not smaller as in a clarinet.
The tube of the instrument still has resonances, of course, but they do not dominate the playing pitch to the same extent as in a reed instrument. The brass player has to adjust their lip tension (“embouchure”) in order to achieve the correct buzzing frequency for the desired note. The tube resonances help a lot: players describe the sensation of the pitch being “slotted”. But a sufficiently skilful and forceful brass player can coax many pitches out of an instrument, especially when the resonances do not provide much support — we will explain more in section 11.5. This leads to some brass players’ party tricks: playing tunes on unlikely “instruments” such as vacuum cleaner hoses, or (particularly striking) holding a steady, unchanging note on a trombone while moving the slide in and out!
Things are different again for the third category of instruments, the free reeds. Figure 4 shows a sketch of a reed from a harmonium or accordion. A thin cantilever beam, usually of brass, is fixed to a heavier brass plate, over a slot which is just a little bigger than the reed. As the reed vibrates, it can move through this slot. If the amplitude is large, it will move away from the plate on both sides at different stages in the cycle. As in the instruments we have already looked at, the player creates pressure on one side of the reed plate, and the moving reed provides a nonlinear valve between the two sides of the plate. But this time, that valve opens wider on both sides. In both the other cases, the valve tends to open when air flows one way, and to close when it flows the other way: the difference between the two cases lies in which way round these two things happen.
The other big difference between a harmonium or accordion and any of the brass or reed instruments is that it has no resonating tube. The playing frequency is usually very close to the resonance frequency of the reed. But now we can see an example of blurring between these categories of instrument: many Asian “mouth organs”, like the ones seen in Fig. 5, do have resonating tubes. These are still formally classified as free reed instruments, but the playing pitch (and the sound spectrum) is influenced by these tubes. A related effect may be more familiar to western audiences: a harmonica player is able to “bend” notes, a particularly vital ingredient of performance technique for blues players. As will be explored in section 11.6, this is achieved by manipulating a different source of resonance: the player’s mouth cavity and vocal tract.
Thinking of the vocal tract reminds us that there is another familiar “wind instrument”: the human voice. Where does this fit into the scheme we have been outlining? When we speak or sing, the origin of the sound is that air flow from our lungs is modulated by vibration of the vocal cords located down in the larynx at the top of the trachea. (See this Wikipedia page for more anatomical detail.) Figure 6 shows an animation of how the vocal cords vibrate. The motion, and the fact that the vocal cords are made of throat tissue not very different from the lips, reminds us of the brass player’s lip vibration sketched in Fig. 3.
So should the human voice be classified as a brass instrument? Perhaps, but some of the other key ingredients of a brass instrument are lacking. Although the sound from the vocal cords does connect to a tube with resonances (the vocal tract), this does not serve to provide “slotted” pitches. The pitch is governed directly by the “reed resonance”, determined by how tightly your throat muscles stretch the vocal cords (and by other variables such as the pressure). In that respect, the singing voice is more like a free reed instrument. The vocal tract resonances are important, but not to modify the pitch. Instead, they provide spectral colouration of the sound by the time it emerges from your mouth, by amplifying some harmonics relative to others. These patterns of amplification around vocal tract resonances are the “formants” which determine your perception of vowel sounds, as mentioned back in section 5.3. This makes the voice a relatively rare example of a wind instrument involving a functional component somewhat similar to the body of a stringed instrument.
Our final category describes wind instruments that depend on an edge tone of some kind. Figure 7 shows a schematic sketch of part of the mouthpiece of a recorder, or a flue organ pipe, or a referee’s whistle. Some examples of flue organ pipes are shown in Fig. 8: wooden pipes in the centre, flanked by metal ones. Air is blown through a slot, and emerges as a jet. A short distance away from the slot is a sharp edge, and the air jet interacts with this edge in some complicated, fluid-dynamical way.
In the presence of acoustic resonances (tube resonances in the case of the recorder or the organ pipe, a Helmholtz resonance for the whistle) this interaction may settle into a regular, periodic pattern and thus produce a musical note. Even without an acoustic resonance, an air jet can sometimes interact with an edge to produce a periodic variation in the flow pattern. Another possibility involves resonant behaviour not of an acoustic tube, but of the structure providing the edge: this effect is used deliberately in an Aeolian harp, in which stretched strings are excited by the wind, and it happens accidentally when power cables “sing” or “gallop” in the wind.
These air-jet instruments are probably the most complicated of the four categories in terms of modelling the underlying physics, but we need to understand a bit about fluid flow to make progress with all four types of wind instrument. This is the task of the next section: it will give a qualitative introduction to some key concepts and phenomena of fluid dynamics. After that, the remaining sections of this chapter will look at the four categories of instrument in turn in a bit more detail.
Before that, in the remainder of this section, we will deal briefly with some features held in common by several categories. First, a reminder of what we learned in section 4.2 about acoustic resonators, and especially about resonances in tubes with various bore profiles and boundary conditions. Figures 9–12 reproduce some plots from that section.
Figure 9 shows the case of a straight pipe, open at both ends. This could be an idealised model for a flute, recorder, or flue organ pipe (as in Fig. 8). The fundamental mode has a half-wavelength between the ends, so its frequency corresponds to a sound wave with wavelength twice the length of the pipe. The higher modes have resonance frequencies in ratios 2, 3, 4… to this fundamental frequency, so they fill a complete harmonic series. In reality, the frequencies will all be a little lower because the pipe will “feel” a little longer than its physical length: as explained in section 4.2.1, there is an “end correction” to be added to the physical length, whose value depends on the detailed geometry around the open ends of the pipe.
Figure 10 shows the corresponding plot for a straight pipe that is open at one end but closed at the other. This could be an idealised model for a clarinet, with the closed end at the mouthpiece, or for a “bourdon” organ pipe, which is open at the mouth but closed at the top end. This time, the fundamental mode has a quarter-wavelength between the ends, so its frequency corresponds to a sound wave with wavelength four times the length of the pipe. The higher modes have resonance frequencies in ratios 3, 5, 7… to this fundamental frequency, odd-numbered frequencies of a harmonic series. Again, for accurate frequencies it would be necessary to allow for end corrections.
Figure 11 shows the corresponding plot for a conical tube, which could be an idealised model for an oboe or saxophone. The pressure mode shapes look superficially similar to the closed-open modes of Fig. 10, because the pressure waveforms all have a horizontal tangent at the left-hand end. This will happen for any closed tube, whatever the detailed shape: once the distance from the end is much smaller than the wavelength of sound, the pressure is bound to be approximately uniform in space, just as we observed when discussing the Helmholtz resonator back in section 4.2.1. Despite this resemblance to the closed-open case of a straight tube, the resonance frequencies are in fact exactly the same as for a straight open-open tube like the one seen in Fig. 9. One familiar consequence is that a clarinet can produce a lowest note that is an octave lower than an oboe with the same tube length.
Figure 12 shows the corresponding plot for an approximate model of a brass instrument like a trumpet. There are no simple mathematical solutions for such shapes: these mode shapes are numerically computed, as described back in section 4.2. The black vertical lines mark the positions of the “effective end point” for each mode: the position where the wave in the flaring horn switches from oscillatory to evanescent behaviour. The behaviour revealed by these lines is that the mode shapes look a bit like the ones in Fig. 10 for a closed-open straight tube, except that the flaring bore shape has “squashed” the lower modes into a shorter portion of the tube, while the higher modes reach progressively closer to the open end of the bell. The frequency ratios are approximately 0.7, 2, 3, 4. This is approximately a harmonic series, except that the fundamental frequency is way out of tune. This accounts for the fact that an instrument like this offers “slotted”pitches that fill a complete harmonic series rather than just the odd terms of the series — except that the fundamental of the series is missing.
This last comment gives a good opportunity to mention an important fact about the overtone frequencies of any wind instrument tube: they play two quite distinct roles in the musical behaviour. On the one hand, they provide resonances that can interact with the nonlinear excitation mechanism to determine the pitch of a played note. On the other hand, they influence the sound spectrum and, more subtly, the “playability” of the note. Imagine you are playing the lowest note on an oboe, with a frequency matching the lowest resonance in Fig. 11. The nonlinear excitation mechanism (from the reed in this case) also generates exact harmonics of the played note. If the higher resonances of the tube are close in frequency to these harmonics, as they would be for the idealised case in Fig. 11, they will all be resonantly excited by the nonlinear harmonics. The result will be a sound richer in those harmonic components. However, this argument only works for the lower notes of an instrument like the oboe: in the higher registers the resonance frequencies of a real tube cease to conform sufficiently accurately to a harmonic pattern.
But also, it seems a plausible guess that the note will be made a bit easier to play. All these resonances are, as it were, in agreement about what exact pitch should be played. On the other hand, if the tube resonances were a bit out of tune, so that each was shifted a bit away from exact harmonic relations with the fundamental, surely these different modes would have slightly different views about what the played pitch “should” be. That might lead to a more sluggish transient as a compromise was “negotiated” between the different resonances. This idea, involving “cooperative regimes of oscillation”, goes back to an early study of the mechanics of wind instruments by Bouasse [2], and more recently it was persuasively argued by Benade [3]. Benade constructed a demonstration instrument he called the “tacit horn”: the resonances were deliberately mis-aligned, with the result that it was virtually impossible to sound any note on the instrument!
These two roles of tube resonances operate differently in our different categories of instrument. In the reed and air-jet instruments, players normally use oscillation regimes based on the three or four modes of the tube, not the higher modes. But in most brass instruments, players will use many of the resonances of the tube as the basis for varying the pitch. Indeed, in instruments like the bugle or natural horn there is no way to change the effective length of the tube, and these regimes are the only resource a player can draw on. Instruments with valves or slides are more versatile because the tube length can be changed, but players still make use of many more tube modes than a woodwind player will do.
There are a few more topics for us to discuss here, because they are important for wind instruments in more than one category. One thing we have already mentioned is that the standard measurement used to characterise the linear acoustics of a wind instrument tube is the input impedance. It is worth looking at a simple example of input impedance, corresponding to the ideal cylindrical pipes seen in Figs. 9 and 10. The input impedance of a cylindrical pipe, open at the far end, is easy to calculate. The details are given in the next link, and Fig. 13 shows an example of the result. This particular case shows a tube 1 metre long, with internal diameter 20 mm. Realistic damping has been included, using formulae from the literature: again, the details are in the side link.
SEE MORE DETAIL
This gives us a chance to make an important observation about the interpretation of input impedance for instruments in different categories. Where do we find the resonances of the tube, by looking at a plot like this? The answer depends on the boundary condition at the end of the tube where the impedance has been measured. If the tube is closed at this end, as in Fig. 10, there can be no air flow in and out of the stopped end, but the pressure variation can be large at a resonance. This means that pressure divided by the volume flow rate must be very large — so the resonances correspond to peaks in the input impedance. Sure enough, the peaks in Fig. 13 show the odd-harmonic pattern that we expect from Fig. 10.
But if the tube is open at the end, as in Fig. 9, we have the opposite situation. Air can flow in and out of the end of the tube, but the pressure variation must be more or less zero because the open end is exposed to the outside world, with a fixed ambient pressure. So this time, at a resonance we expect very small values of the impedance, or equivalently we expect very large values of its inverse, called the input admittance. This admittance is plotted in Fig. 14: because of the decibel scale used here, the image is simply the same as Fig. 13, plotted upside down. Now the peaks fall half-way between the peaks in Fig. 13, and they are equally spaced in a complete harmonic series. This is exactly what we expect from Fig. 9.
Finally, we look at the role played by tone-holes in the wall of the instrument tube. Most reed and air-jet wind instruments have tone-holes. Less familiarly, some “brass” instrument are played by covering tone-holes with the fingers: examples are found in older instruments such as the cornett (or cornetto) and the serpent. The purpose of tone-holes seems self-evident: making a hole in the wall of the tube is rather like cutting the tube short at that point, so that the frequencies of all the resonances go up. However, as with so many things in musical acoustics, it is a bit more complicated than that.
As a first step we can imagine a single tone-hole cut into an infinitely long cylindrical tube, as sketched in Fig. 15. We can send a sinusoidal sound wave at a chosen frequency down the tube. When this wave reaches the hole, some of its energy is reflected back, while some of it is transmitted past the hole and continues to propagate in the direction it was already travelling. This problem is analysed in the next link, by thinking about the air flow and the pressure in the small volume near the hole, shown as a dotted box. Simple expressions are found for the reflection coefficient $R$ and the transmission coefficient $T$. These depend on the geometrical dimensions of the hole and the tube bore, and they also vary with frequency.
SEE MORE DETAIL
Figure 16 shows three examples of how the energy reflection coefficient $|R|^2$ behaves when the area of the hole is varied from zero up to the cross-sectional area of the tube, at three different frequencies. For the lowest frequency (red curve), when the hole is as big as the cross-sectional area of the tube virtually all the energy is reflected. This is the effect we were expecting: the sound wave is more or less confined to the left-hand side of the hole, just as if the tube had been cut off at that position. As the size of the hole is reduced, the reflection coefficient drops gradually, reaching zero as the hole area vanishes. Well, that is no surprise: if the hole is no longer there, of course no energy is reflected from it.
At the two higher frequencies shown in Fig. 16 the reflection coefficient is systematically smaller. Even with a very large hole, a significant fraction of the energy is not reflected, but continues along the tube. As we will explore shortly, this means that what happens further down the tube from the tone-hole can have a significant impact on the behaviour at these higher frequencies. To get a different view of the frequency dependence, Fig. 17 shows the reflection coefficient plotted against frequency. Curves are shown for flutes of three different kinds, using bore size and tone-hole dimensions measured by Wolfe and Smith [2]. All three curves show a similar trend: perfect reflection at very low frequency, falling to rather low values at 2 kHz. Notice also that the three types of flute are systematically different, and they are arranged in order of age. There has obviously been a systematic evolution of flute design: we’ll come back to that in a moment.
When our infinite tube is replaced with a finite tube, we would like to know what happens to the resonance frequencies of this tube as the single hole is gradually closed. The answer turns out to be what you might guess. With a large hole, the resonances are those of a shorter tube, cut off near the hole. When the hole is reduced to a pinhole and then vanishes altogether, the resonances are those of the full length of tube. Well, in between the resonance frequencies change smoothly between these two limits. With this rather primitive wind instrument with just a single tone-hole, the player would in principle be able to produce a pitch that can be varied over this range by using a finger to part-close (or “shade”) the hole progressively.
The physics behind this gradual shift of resonance frequencies can be visualised in terms of an “end correction”. With a single hole of any size, the reflection is never perfect. This means that the effective length of the tube is always longer than the distance to the centre of the tone-hole. With a large hole and a correspondingly large reflection coefficient, this extra bit of effective length is quite small, but as the reflection coefficient reduces with a reduction of hole size, the sound wave penetrates further past the hole and the end correction determining the effective length (and thus the frequency of resonance) grows. Eventually, as the hole becomes vanishingly small, this “end correction” reaches all the way to the end of the tube, where virtually all of the sound energy is reflected back into the tube. Some details of the calculation lying behind this description are given in the next link.
SEE MORE DETAIL
This variation of the end correction with hole size is something that instrument makers have taken advantage of, especially in earlier instruments that rely on fingertips to cover holes, rather than using key mechanisms. There are limits to how human fingers can be spread along the tube of a wind instrument. This sometimes means that a tone-hole in the “logical” place would be too far away from the others to be reached by the player’s finger, especially in larger instruments like the dulcian (or curtal), a predecessor of the bassoon: see Fig. 18. The solution was to use a smaller hole with a correspondingly bigger end correction, so that it could be placed in a more convenient position. The resulting odd-looking distribution of holes can be seen very clearly in the image.
Of course, real wind instruments have more than one tone-hole. Figure 19 shows a schematic sketch of a “flute”, with a typical fingering: there are several tone-holes, the first few being covered while the remaining ones are open. This introduces several complications. The first open tone-hole is no longer in splendid isolation: not very far away are other open holes, before we reach the end of the tube. Also, the closed tone-holes affect the profile of the tube. The fingers or key-pads do not produce the same effect as not having the hole in the first place, because they add a small local volume to the tube, which can perturb the resonance frequencies a bit.
To see what happens to the frequency response of a real instrument tube with various combinations of open and closed tone-holes we can look at some measured input impedances, taken from Joe Wolfe’s comprehensive web site on flute acoustics here. We can look first at some results for a baroque flute, which is a relatively simple instrument not too far removed from the sketch in Fig. 19. It has 6 finger-holes, plus a 7th hole which is operated by a key: you can see a picture in Fig. 22 below.
Figure 20 shows the input impedance for four different fingerings. Because we are looking at a flute, which is open at the mouthpiece end, we must keep in mind that the resonances of the tube are indicated in these plots by the minima, not the peaks. The first plot shows the result with all the holes closed, in the fingering for the lowest note of the instrument ($\mathrm{D}_4$, 294 Hz at the modern pitch standard, but a semitone lower on this flute at 277 Hz because it is adjusted for a pitch standard with $\mathrm{A}_4$ at 415 Hz). The plot shows an orderly sequence of peaks and minima, very reminiscent of the idealised version seen in Fig. 13.
The top right plot in Fig. 20 shows the impedance for a fingering rather like the sketch in Fig. 19. It has three closed holes, then three open holes, but the very last hole is closed by the key. The pattern is much less regular. The explanation is that a baroque flute like this has relatively small tone-holes (for a reason we will come to in a moment), and so a sound wave coming from the mouthpiece end would only be partially reflected at the first open tone-hole. A significant fraction of the energy will continue down the tube and interact with the other open tone-holes as well as with the open end of the tube. This complicated interaction leads to disruption of the pattern of resonances.
The lower left plot shows the fingering for $\mathrm{A}\flat_4$. This one is a “fork fingering” or “cross fingering”, with closed holes coming after the first open hole. The possibility of using such fingerings to achieve intermediate chromatic notes is probably the main reason that flutes like this have rather small tone-holes. If the holes are to be closed by the player’s fingers, unaided by complicated key mechanisms, there can only be as many holes as the player has available fingers. This is not enough to offer a full chromatic scale by simply opening one more hole for each semitone step. So instruments like this one are tuned to give a diatonic scale in a particular key, and the player has to use techniques like fork fingering to sound the missing notes. Fork fingerings can only work if the reflection coefficient at the first open tone-hole is fairly small: the hole’s end correction is then large enough that the influence of later closed holes can be felt.
The final plot in Fig. 20 shows the result with only one tone-hole closed. This one highlights a feature of the pattern of resonances which started to show up in the previous two plots, but less clearly. There are only two strong resonances (deep dips in the plot, remember), and for all frequencies above about 1 kHz the resonances are significantly attenuated. This feature is called the “tone-hole cutoff”, and it arises from the fact that there is a fairly regular series of open tone-holes following the last closed one. An approximate formula for this cutoff frequency in terms of the tone-hole geometry was first given by Benade [3], on the assumption that the “lattice” of open tone-holes was infinitely long, and uniformly spaced. A neat derivation of the result can be found in the Appendix of reference [2].
It is interesting to contrast this behaviour with some corresponding plots for a modern flute, taken from the same web site and shown in Fig. 21. Photographs of typical baroque and modern flutes are shown in Fig. 22. The sketches accompanying each impedance plot show that a flute like this has many more holes than the player has fingers. A complicated and ingenious system of keys is used to open and close these holes — it was invented by Theobald Boehm in the mid-19th century. A key system like this frees the designer from some of the constraints of human fingers. As well as allowing more holes, the individual holes can be larger because the key-pad can be shaped to close a hole of virtually any shape. This is the explanation for the contrast in reflection coefficient seen back in Fig. 17: the modern flute (black line) showed much stronger reflection at all frequencies than the baroque flute (red line).
The results of Boehm’s ingenuity can be seen in two features of the plots in Fig. 21. The first plot, with all the holes closed, looks very similar to the corresponding plot in Fig. 20. But after that, the behaviour is very different. The modern flute shows an orderly sequence of resonances for all the cases shown — there is very little of the disruption we saw in Fig. 20. The second thing we can see is that the tone-hole cutoff for this flute is significantly higher in frequency, up around 2 kHz. Both these features are, qualitatively at least, what we might have anticipated from the contrast seen in Fig. 17. Except for very high notes or some advanced playing techniques such as “multiphonics” (we’ll come to those in section ?), there is (almost) no need for fork fingerings on this flute. Each semitone has its own key and hole combination.
Now we have dealt in some detail with open tone-holes, we should have a brief look at the perturbing effect of closed tone-holes. As we noted earlier, these create a small extra volume, which will have a similar effect to a slight bulge in the bore profile. Any perturbation to the bore will influence frequencies slightly. This might interfere with carefully-planned alignment of resonances, to improve the sound quality and playability. An instrument maker might try to counter such effects by making deliberate modifications to the bore profile.
So for one reason and another, it is of interest to know the effect on resonance frequencies of a small change in bore profile. As explained in the next link, Rayleigh’s principle (see section 3.2.5, and the application to a marimba bar in section 3.3.1) gives us an easy way to answer this question. We can describe the result qualitatively. For any given mode of a cylindrical tube, like the ones shown in Figs. 9 and 10, the effect of a local enlargement in the bore is biggest near a nodal point of pressure, or near an antinode (a position of maximum pressure). An enlargement near a node (including near an open end of the tube) has the effect of increasing the frequency, whereas an enlargement near an antinode has the effect of reducing the frequency. Since different modes have their nodes and antinodes in different places, it follows immediately that the effect of a single enlargement, such as a closed tone-hole, will be different for each mode, and so it will indeed tend to interfere with the harmonic relations between frequencies. But the same theoretical expression can be used to guide an instrument maker wishing to adjust the bore to bring resonance frequencies to desired values.
SEE MORE DETAIL
Our final topic for this section is to note that not all holes in the tube wall of a wind instrument function as tone-holes, to control the pitch of the note. Most instruments also have register holes, to aid the player in switching between a playing pitch based on the lowest mode of the tube, and one based on the second mode. Players describe these alternatives as different registers. Sometimes, a player can switch between registers by adjusting the details of how they blow. For example, blowing harder into a recorder or tin whistle may cause the pitch to jump to the next register, often called “overblowing”. But life is easier for a player if they have a controlled way to switch registers, and this is where register holes come in.
Consider the example of a cylindrical tube, open at both ends, like a flute or recorder. As Fig. 9 reminds us, the first register is based on a pressure mode shape with one half-wavelength in the length of the tube, whereas the second register is based on a mode shape with a nodal point at the centre, and two half-wavelengths in the length of the tube. If a small hole is drilled in the tube wall near the centre, this will not affect the second mode but it will perturb the first mode (which has a pressure maximum at this position). It may shift the frequency of the first mode, and it may add some dissipation and thus reduce the height of the resonance peak. Both effects will tend to make it more difficult to play a note based on this mode, and thus to make it more likely that the player will get a note in the second register.
This is the principle of a register hole. Why does the hole need to be small? The answer to that is that we don’t want to affect just one note, we would like to encourage the register switch in other nearby notes, based on different effective tube lengths because tone-holes have been opened or closed. Each of these notes will have the second-register nodal point in slightly different positions, so the hole can’t be perfectly positioned to suit them all. But if the hole is fairly small, this doesn’t matter very much. If carefully designed, it can have the desired register-changing effect on several notes.
If we are thinking about a clarinet rather than a recorder or flute, we can make a similar argument based on the mode shapes in Fig. 10. The same argument applies, except that the place we need to drill our register hole is approximately 1/3 of the way down the tube from the mouthpiece, rather than in the centre — because that is where the nodal point of the second pressure mode shape occurs.
[1] Murray Campbell, Joël Gilbert and Arnold Myers, “The science of brass instruments”, ASA Press/Springer (2021). See section 3.1.3 for examples of video recordings of a brass-player’s lips.
[2] Joe Wolfe and John Smith, “Cutoff frequencies and cross fingerings in baroque, classical, and modern flutes”, Journal of the Acoustical Society of America 114, 2263—2272 (2003)
[3] Arthur H. Benade, “On the mathematical theory of woodwind finger holes”, Journal of the Acoustical Society of America 32, 1591—1608 (1960)
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Maximum likelihood approaches for noncoherent communications with chaotic carriers
Published in:
IEEE Transactions on Circuits And Systems I-Regular Papers, 48, 5, 533-542
Year:
2001
Keywords:
Other identifiers:
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# Inverse function — need help
I'm a senior software developer but my math lessons are a bit rusty. I know the name of what I want, but not anymore how to compute it ;)
I've found (by myself with Grapher.app) a simple easing function (for transitions), that's a bijection from $[0,1]$ to $[0,1]$:
$$f(x) = \frac{27}{4} \cdot \left( \left( \frac{2x}3 \right) ^2 - \left( \frac{2x}3 \right) ^3 \right)$$
Edit: in fact, after simplifying, it's as simple as:
$$f(x) = 3 x^2 - 2 x^3$$
The derivatives at $0$ and $1$ are both zero, so it's a perfect ease-in-out function that's easy to compute.
In order to start my transitions from anywhere between $0$ and $1$, I need $f^{-1}$, the inverse of this function, but I'm completely lost as to where to start... Would someone shed some light on my oh-gosh-it-was-so-long-ago math course?
• normal form: y=function of x, inverse: x=function of y – Bob Jan 31 '13 at 19:52
• It's simpler and easier to understand if you write $f(x)=3x^2-2x^3$. – user1551 Jan 31 '13 at 19:54
• Oh. My. Didn't even think of simplifying it. When I told you I'm rusty... – Cyrille Jan 31 '13 at 20:48
Take your equation $3 x^2 - 2 x^3 = y$, substitute $x = X + 1/2$, multiply by $2$ and expand to get $-4X^3 + 3 X = 2 y - 1$. Now note that $\sin(3t) = - 4 \sin(t)^3 + 3 \sin(t)$. Thus a solution is $X = \sin(t)$ where $\sin(3t) = 2y-1$, i.e. $$x = \sin\left(\frac{\arcsin(2y-1)}{3}\right) + 1/2$$
Note that $0 \le x \le 1$ when $0 \le y \le 1$.
• Wait wait wait... sinuses? – Cyrille Jan 31 '13 at 20:51
• Okay, after re-reading it like ten times, it makes sense. – Cyrille Jan 31 '13 at 21:02
• What I've always wondered (and that's where the beauty of maths comes from) is: where did you take the "substitute x = X + 1/2" from? I mean, did you just have to say "shazam" to find this? – Cyrille Jan 31 '13 at 21:11
• Take $x = X + a$ and see what $a$ has to be to eliminate the $X^2$ term. – Robert Israel Jan 31 '13 at 23:35
The inverse function $g(y)$ satisfies $f[g(y)] = y$. In this case:
$$\frac{27}{4} \cdot \left( \left( \frac{2g(y)}3 \right) ^2 - \left( \frac{2g(y)}3 \right) ^3 \right) = y$$
Solve for $g(y)$ algebraically. Might be tough because it's a cubic, but it is possible.
You can ask to wolfram alpha:
1/2 (1+1/(1-2 x+2 sqrt(-x+x^2))^(1/3)+(1-2 x+2 sqrt(-x+x^2))^(1/3))
Or either you could implement the bisection method to solve $f(x) = y$. One should check, but it is possible that in this case the bisection method is faster than using the explicit formula for inverse.
• Didn't even knew you could ask this to Wolfram. Thanks! – Cyrille Jan 31 '13 at 21:05
• But notice that when $x \in (0,1)$, you're taking the square root of a negative number. This is no accident: see en.wikipedia.org/wiki/Casus_irreducibilis – Robert Israel Jan 31 '13 at 23:39
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# Which are the most favorited/upvoted Calculus questions and answers? [closed]
I'm currently learning Calculus, and recently I came across these excellent two questions :
Which contained, some of the best and most in-depth explanations I have seen yet. As a mathematics student, majoring in Pure Mathematics, I thought it would be really interesting to see find more questions like this on the Math StackExchange site.
It would be very helpful to anyone learning Calculus (wheter it be Single-Variable, Multivariable, Vector or Tensor) to be able to take a look at some of the types of questions people ask, and to read through some of the great answers given for those questions.
Is there currently a list of popular Calculus questions (from Single and Multivariable to Vector and Tensor Calculus) asked on Math StackExchange? If so I would be extremely interested in seeing it.
If not please feel free to share some of the best questions and answers on Calculus that you've come across on the Math StackExchange, I'm sure there would be many users interested in seeing them.
## closed as primarily opinion-based by Watson, yoknapatawpha, Shailesh, BLAZE, Brevan EllefsenMay 17 '16 at 6:45
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.
As suggested in the comments, if you are interested in some tag, there are several views you can choose.
You said that you are interested in the tag. When you click on the tag name you get here - to the list of questions which are tagged with this tag. And you can choose one of several tabs to view/sort the questions: newest, featured, frequent, votes, active, unanswered. (Simply by clicking on the tab name near the top of the screen.)
If you hover with your cursor over the tab name, you will be shown a description what the particular tab means. (Similar thing works for many elements of the Stack Exchange UI.)
Since you say that you want to see the most popular questions, the following two tabs seem to be useful for that purpose. (In each case it is debatable whether these criteria in some sense correspond to the question "being most interesting/popular/relevant", but at least there is probably some correlation.)
• Votes tab show the most upvoted question. So these are the question where users thought that the questions are worth upvoting, which might be some rough criterion whether the question is interesting.
• I think that also the frequent tab is useful. This tags shows the questions with most links from other posts. This can also be considered in some sense as a criterion which questions are the most relevant. (But this probably corresponds more closely to "most frequently asked" question rather than "most interesting questions".)
I am unaware of a possibility to view most favorited questions in a given tag using the user interface. But you can try to use data explorer, as explained, for example, here: View Most Popular Questions by favorite count / views / votes. There is certainly a lot of space for improvement, but I have made a very simple SEDE query here: Most favorited question in a given tag.
Maybe it is worth mentioning that I have used Tags like '%##Name##%' as a selection criterion in that query. So if you use it for the tag calculus, the question from will also be included. But I suppose that for most tags this not problem will not appear, since there are not too many tags with similar names. For example, I guess this will fork fine for multivariable-calculus or real-analysis. If you have some experience with SQL and you can create a query tailored to your needs or to the specific tags you are interested in.
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# American Institute of Mathematical Sciences
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Numerical study of vanishing and spreading dynamics of chemotaxis systems with logistic source and a free boundary
February 2021, 26(2): 1111-1127. doi: 10.3934/dcdsb.2020155
## Flocking of non-identical Cucker-Smale models on general coupling network
1 Department of Applied Mathematics, National Chiao Tung University, Hsinchu, Taiwan, ROC 2 Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung 700, Taiwan, ROC
* Corresponding author: Jonq Juang
Received July 2019 Revised December 2019 Published May 2020
Fund Project: This work is partially supported by the Ministry of Science and Technology of the Republic of China under grant No.\ MOST 107-2115-M-009-011-MY2 and No. MOST 107-2115-M-390-006-MY2
The purpose of the paper is to investigate the flocking behavior of the discrete-time Cucker-Smale(C-S) model under general interaction network topologies with agents having their free-will accelerations. We prove theoretically that if the free-will accelerations of agents are summable, then, for any given initial conditions, the solution achieves flocking with a finite moving speed by suitably choosing the time step as well as the communication rate of the system or the strength of the interaction between agents. In particular, if the communication rate $\beta$ of the system is subcritical, i.e., $\beta$ is less than a critical value $\beta_c$, then flocking holds for any initial conditions regardless of the strength of the interaction between agents. While, if the communication rate is critical ($\beta = \beta_c$) or supercritical ($\beta > \beta_c$), then flocking can only be achieved by making the strength of the interaction large enough. We also present some numerical simulations to support our obtained theoretical results.
Citation: Yu-Jhe Huang, Zhong-Fu Huang, Jonq Juang, Yu-Hao Liang. Flocking of non-identical Cucker-Smale models on general coupling network. Discrete & Continuous Dynamical Systems - B, 2021, 26 (2) : 1111-1127. doi: 10.3934/dcdsb.2020155
##### References:
show all references
##### References:
The network consisting of $9$ vertices has a spanning tree. For this network, $\mathcal{R} = \{1,2,3\}$, $n = 9$, $\ell = 3$, $h = 3$ and $r = 2$
The graph of $G$
(Subcritical) Numerical simulation for model Eq. (4) with the network given in Fig. 1. This simulation result shows the solution achieves flocking. Here parameters in Eq. (4) are chosen as $\beta = 1/10$, $\kappa = 10$ and $\varepsilon = 0.001$. The initial conditions are randomly chosen and they satisfy $a\approx 0.8932$, $b\approx 0.8104$. Here $a,b$ are defined in (20a)
(Critical) Numerical simulation for model Eq. (4) with the network given in Fig. 1. This simulation result shows the solution achieves flocking. Here parameters in Eq. (4) are chosen as $\beta = 1/6$, $\kappa = 120$ and $\varepsilon = 0.001$. The initial conditions are randomly chosen and they satisfy $a\approx 0.8932$, $b\approx 0.8104$. Here $a,b$ are defined in (20a)
(Supercritical) Numerical simulation for model Eq. (4) with the network given in Fig. 1. This simulation result shows the solution achieves flocking. Here parameters in Eq. (4) are chosen as $\beta = 1/3$, $\kappa = 200$ and $\varepsilon = 0.001$. The initial conditions are randomly chosen and they satisfy $a\approx 0.8932$, $b\approx 0.8104$. Here $a,b$ are defined in (20a)
Numerical simulation for model Eq. (4) under the network provided in Fig. 1. This simulation result shows the solution does not achieve flocking. Here parameters in Eq. (4) are chosen as $\beta = 1$ (supercritical), $\kappa = 0.1$ and $\varepsilon = 0.001$. The initial conditions are randomly chosen and they satisfy $a\approx 0.2872$, $b\approx 0.2850$. Here $a,b$ are defined in (20a). Such set of parameters and initial conditions do not satisfy the sufficient condition (23)
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# Solving an inhomogenous parameter dependent ODE
I was trying to solve the ODE
$$\ddot{r} r = \alpha(\dot{r}^2-1)$$ where $\alpha$ is an arbitrary constant. There are some simple cases when $\alpha = -1$ then you can use separation of variables to find the solution. For the initial condition when $r'(0)=1$ it also simplifies, $r(t)= t + r(0)$. Not sure how take on the general case. Any help is welcome!
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## 2 Answers
HINT:
$$\ddot{r} r = \alpha(\dot{r}^2-1)$$
$$\frac{\dot{r}\ddot{r}}{\dot{r}^2-1}=\alpha\frac{\dot{r}}{r}$$
$$\frac{\dot{r}\ddot{r}}{(\dot{r}+1)(\dot{r}-1)}=\alpha\frac{\dot{r}}{r}$$
$$\frac{\ddot{r}}{\dot{r}+1} +\frac{\ddot{r}}{\dot{r}-1}=2\alpha\frac{\dot{r}}{r}$$
Then integrate both side
-
Case $1$ : $\alpha=0$
Then $\ddot{r}r=0$
$r=0$ or $\ddot{r}=0$
$r=0$ or $r=C_1t+C_2$
$\therefore r=C_1t+C_2$
Case $2$ : $\alpha\neq0$
Then $\ddot{r}r=\alpha(\dot{r}^2-1)$
$r\dfrac{d^2r}{dt^2}=\alpha\left(\left(\dfrac{dr}{dt}\right)^2-1\right)$
Let $u=\dfrac{dr}{dt}$ ,
Then $\dfrac{d^2r}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dr}\dfrac{dr}{dt}=u\dfrac{du}{dr}$
$\therefore ru\dfrac{du}{dr}=\alpha(u^2-1)$
$\dfrac{u}{u^2-1}du=\dfrac{\alpha}{r}dr$
$\int\dfrac{u}{u^2-1}du=\int\dfrac{\alpha}{r}dr$
$\dfrac{1}{2}\ln(u^2-1)=\alpha\ln r+c_1$
$\ln\left(\left(\dfrac{dr}{dt}\right)^2-1\right)=2\alpha\ln r+c_2$
$\left(\dfrac{dr}{dt}\right)^2-1=C_1r^{2\alpha}$
$\dfrac{dr}{dt}=\pm\sqrt{C_1r^{2\alpha}+1}$
$dt=\pm\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}$
$\int dt=\pm\int\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}$
$t=\pm\int_k^r\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}+C_2$
-
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{}
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Ural Mathematical Journal
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Ural Math. J.: Year: Volume: Issue: Page: Find
Ural Math. J., 2020, Volume 6, Issue 2, Pages 3–14 (Mi umj121)
Min-max solutions for parametric continuous static game under roughness (parameters in the cost function and feasible region is a rough set)
Yousria A. Aboelnagaa, Mai F. Zidanb
a Higher Technological Institute, Tenth of Ramadan City, 44629, Egypt
b Tanta University
Abstract: Any simple perturbation in a part of the game whether in the cost function and/or conditions is a big problem because it will require a game re-solution to obtain the perturbed optimal solution. This is a waste of time because there are methods required several steps to obtain the optimal solution, then at the end we may find that there is no solution. Therefore, it was necessary to find a method to ensure that the game optimal solution exists in the case of a change in the game data. This is the aim of this paper. We first provided a continuous static game rough treatment with Min-Max solutions, then a parametric study for the processing game and called a parametric rough continuous static game (PRCSG). In a Parametric study, a solution approach is provided based on the parameter existence in the cost function that reflects the perturbation that may occur to it to determine the parameter range in which the optimal solution point keeps in the surely region that is called the stability set of the 1$^{st}$ kind. Also the sets of possible upper and lower stability to which the optimal solution belongs are characterized. Finally, numerical examples are given to clarify the solution algorithm.
Keywords: continuous static game, rough programming, non-linear programming, rough set theory, parametric linear programming, parametric non-linear programming.
DOI: https://doi.org/10.15826/umj.2020.2.001
Full text: PDF file (162 kB)
Full text: https:/.../263
References: PDF file HTML file
Bibliographic databases:
Language:
Citation: Yousria A. Aboelnaga, Mai F. Zidan, “Min-max solutions for parametric continuous static game under roughness (parameters in the cost function and feasible region is a rough set)”, Ural Math. J., 6:2 (2020), 3–14
Citation in format AMSBIB
\Bibitem{AboZid20} \by Yousria~A.~Aboelnaga, Mai~F.~Zidan \paper Min-max solutions for parametric continuous static game under roughness (parameters in the cost function and feasible region is a rough set) \jour Ural Math. J. \yr 2020 \vol 6 \issue 2 \pages 3--14 \mathnet{http://mi.mathnet.ru/umj121} \crossref{https://doi.org/10.15826/umj.2020.2.001} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=MR4194009} \elib{https://elibrary.ru/item.asp?id=44611145} \scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-85099607154}
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# Sample size for AUC based on mean and SD of raw data
I'm trying to estimate the sample size required to achieve 80% power at a 5% significance level for a superiority study comparing AUC from bioavailability data.
My challenge is the only previous pilot study I found reported mean AUC and standard errors for the raw data and using this with pwr.t.test in R gives the following error.
Error in uniroot(function(n) eval(p.body) - power, c(2 + 1e-10, 1e+09)) :
f() values at end points not of opposite sign
Below are means(SE) for the two groups from the previous study:
Group A: 6300(35.6); n=6
Group B: 16750(57.9); n=6
Here's the code I run:
library(pwr)
pwr.t.test(n = NULL , d = 119.84, sig.level =0.05 , power = 0.8, type = "two.sample")
Log transformation should be the way to go I reckon, but I'm not sure how to derive the log SD from the raw SD. Any help will be much appreciated.
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# Moderate Time, Speed & Distance Solved QuestionAptitude Discussion
Q. A passenger train covers the distance between stations $X$ and $Y$, 50 minutes faster than a goods train. Find this distance if the average speed of the passenger train is 60 kmph and that of goods train is 20 kmph
✖ A. 20 kms ✔ B. 25 kms ✖ C. 45 kms ✖ D. 40 kms
Solution:
Option(B) is correct
Let '$d$' be the distance between the stations $X$ and $Y$.
Time taken by the passenger train to cover the distance '$d$'$=\dfrac{d}{60}$ hour
Time taken by the goods train to cover the distance '$d$'$=\dfrac{d}{20}$ hour
Time difference between these two trains is given by 50 minutes or $\dfrac{50}{60}$ hour
$\Rightarrow \dfrac{d}{20}-\dfrac{d}{60}=\dfrac{50}{60}$
$\Rightarrow \dfrac{d(60-20)}{20\times 60}=\dfrac{50}{60}$
$d$ = 25 kms
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# Spectral and singular value decompositions¶
## 6 Further observations¶
Course: Math 535 - Mathematical Methods in Data Science (MMiDS)
Author: Sebastien Roch, Department of Mathematics, University of Wisconsin-Madison
Updated: Sep 21, 2020
### 6.1 Condition numbers¶
In this section we consider condition numbers, a measure of the sensitivity to perturbations of certain numerical problems. We look in particular at the conditioning of the least-squares problem.
#### 6.1.1 Pseudoinverses¶
We have seen that the least-squares problem provides a solution concept for overdetermined systems. This leads to the following generalization of the matrix inverse.
Definition (Pseudoinverse): Let $A \in \mathbb{R}^{n \times m}$ be a matrix with SVD $A = U \Sigma V^T$ and singular values $\sigma_1 \geq \cdots \geq \sigma_r > 0$. A pseudoinverse $A^+ \in \mathbb{R}^{m \times n}$ is defined as
$$A^+ = V \Sigma^+ U^T$$
where $\Sigma^+$ is the diagonal matrix with diagonal entries $\sigma_1^{-1}, \ldots, \sigma_r^{-1}$. $\lhd$
While it is not obvious from the definition (Exercise: Why? $\lhd$), the pseudoinverse is in fact unique. Note further that
$$A A^+ = U \Sigma V^T V \Sigma^+ U^T = U U^T$$
and
$$A^+ A = V \Sigma^+ U^T U \Sigma V^T = V V^T.$$
Three important cases:
1. In the square nonsingular case, both products are the identity and we necessarily have $A^+ = A^{-1}$ by the Existence of an Inverse Lemma.
2. If $A$ has full column rank $m \leq n$, then $r = m$ and $A^+ A = I_{m \times m}$.
3. If $A$ has full row rank $n \leq m$, then $r = n$ and $A A^+ = I_{n \times n}$.
In the second case above, we recover our solution to the least-squares problem in the overdetermined case.
Lemma (Pseudoinverse and least squares): Let $A \in \mathbb{R}^{n \times m}$ of full column rank $m$ with $m \leq n$. Then
$$A^+ = (A^T A)^{-1} A^T.$$
In particular, the solution to the least-square problem
$$\min_{\mathbf{x} \in \mathbb{R}^m} \|A \mathbf{x} - \mathbf{b}\|$$
is $\mathbf{x}^* = A^+ \mathbf{b}$.
Proof idea: We use the SVD definition and check that the two sides are the same.
Proof: We first note that $A^T A$ is nonsingular. Indeed, if $A^T A \mathbf{x} = \mathbf{0}$, then $\mathbf{x}^T A^T A \mathbf{x} = \|A\mathbf{x}\|^2 = 0$ so that $A \mathbf{x} = \mathbf{0}$ by the point-separating property of the vector norm. In turn, since the columns of $A$ are linearly independent by assumption, this implies $\mathbf{x} = \mathbf{0}$. Hence the columns of $A^T A$ are also linearly indepedent.
Now let $A = U \Sigma V^T$ be an SVD of $A$ (with positive singular values). We then note that
$$(A^T A)^{-1} A^T = (V \Sigma U^T U \Sigma V^T)^{-1} V \Sigma U^T = V \Sigma^{-2} V^T V \Sigma U^T = A^+$$
as claimed.
The second claim follows from the normal equations. $\square$
In [2]:
M = [1.5 1.3; 1.2 1.9; 2.1 0.8]
Out[2]:
3×2 Array{Float64,2}:
1.5 1.3
1.2 1.9
2.1 0.8
In [3]:
Mp = pinv(M)
Out[3]:
2×3 Array{Float64,2}:
0.0930539 -0.311014 0.587446
0.126271 0.629309 -0.449799
In [4]:
Mp*M
Out[4]:
2×2 Array{Float64,2}:
1.0 -4.05256e-17
2.5981e-16 1.0
#### 6.1.2 Conditioning of matrix-vector multiplication¶
We define the condition number of a matrix and show that it captures some information about the sensitivity to perturbations of matrix-vector multiplications. First an exercise.
Exercise: Let $A \in \mathbb{R}^{n \times n}$ be nonsingular with SVD $A = U \Sigma V^T$ where the singular values satisfy $\sigma_1 \geq \cdots \geq \sigma_n > 0$. Show that
$$\min_{\mathbf{x} \neq \mathbf{0}} \frac{\|A \mathbf{x}\|}{\|\mathbf{x}\|} = \min_{\mathbf{y} \neq \mathbf{0}} \frac{\|\mathbf{y}\|}{\|A^{-1}\mathbf{y}\|} = \sigma_n = 1/\|A^+\|_2.$$
$\lhd$
Definition (Condition number of a matrix): The condition number (in the induced $2$-norm) of a matrix $A \in \mathbb{R}^{n \times m}$ is defined as
$$\kappa_2(A) = \|A\|_2 \|A^+\|_2.$$
$\lhd$
In the square nonsingular case, this reduces to
$$\kappa_2(A) = \|A\|_2 \|A^{-1}\|_2 = \frac{\sigma_1}{\sigma_n}$$
where we used the exercise above. In words, $\kappa_2(A)$ is the ratio of the largest to the smallest stretching under $A$.
Theorem (Relative conditioning of matrix-vector multiplication): Let $A \in \mathbb{R}^{n \times n}$ be nonsingular. Then, for any $\mathbf{x} \in \mathbb{R}^n$,
$$\lim_{\delta \to 0} \sup_{0 < \|\mathbf{d}\| \leq \delta} \frac{\|A(\mathbf{x}+\mathbf{d}) - A \mathbf{x}\|/\|A\mathbf{x}\|} {\|\mathbf{d}\|/\|\mathbf{x}\|} = \max_{\mathbf{d} \neq \mathbf{0}} \frac{\|A(\mathbf{x}+\mathbf{d}) - A \mathbf{x}\|/\|A\mathbf{x}\|} {\|\mathbf{d}\|/\|\mathbf{x}\|} \leq \kappa_2(A)$$
and the inequality is tight.
The ratio above measures the worst rate of relative change in $A \mathbf{x}$ under infinitesimal perturbations of $\mathbf{x}$. The theorem says that when $\kappa_2(A)$ is large, a case referred to as ill-conditioning, large relative changes in $A \mathbf{x}$ can be obtained from relatively small perturbations to $\mathbf{x}$. In words, a matrix-vector product is potentially sensitive to perturbations when the matrix is ill-conditioned.
Proof: Write
$$\frac{\|A(\mathbf{x}+\mathbf{d}) - A \mathbf{x}\|/\|A\mathbf{x}\|} {\|\mathbf{d}\|/\|\mathbf{x}\|} = \frac{\|A \mathbf{d}\|/\|A\mathbf{x}\|} {\|\mathbf{d}\|/\|\mathbf{x}\|} = \frac{\|A (\mathbf{d}/\|\mathbf{d}\|)\|}{\|A(\mathbf{x}/\|\mathbf{x}\|)\|} \leq \frac{\sigma_1}{\sigma_n}$$
where we used the exercise above.
In particular, we see that the ratio can achieve its maximum by taking $\mathbf{d}$ and $\mathbf{x}$ to be the right singular vectors corresponding to $\sigma_1$ and $\sigma_n$ respectively.$\square$
If we apply the theorem to the inverse instead, we get that the relative conditioning of the nonsingular linear system $A \mathbf{x} = \mathbf{b}$ to perturbations in $\mathbf{b}$ is also $\kappa_2(A)$. The latter can be large in particular when the columns of $A$ are close to linearly dependent.
For example, orthogonal matrices have condition number $1$:
In [5]:
Q = [1/sqrt(2) 1/sqrt(2); 1/sqrt(2) -1/sqrt(2)]
Out[5]:
2×2 Array{Float64,2}:
0.707107 0.707107
0.707107 -0.707107
In [6]:
cond(Q)
Out[6]:
1.0000000000000002
And matrices with nearly linearly dependent columns have large condition numbers:
In [2]:
eps = 1e-6
A = [1/sqrt(2) 1/sqrt(2); 1/sqrt(2) 1/sqrt(2)+eps]
Out[2]:
2×2 Array{Float64,2}:
0.707107 0.707107
0.707107 0.707108
In [3]:
cond(A)
Out[3]:
2.8284291245366517e6
In [4]:
F = svd(A)
Out[4]:
SVD{Float64,Float64,Array{Float64,2}}
U factor:
2×2 Array{Float64,2}:
-0.707107 -0.707107
-0.707107 0.707107
singular values:
2-element Array{Float64,1}:
1.4142140623732717
4.999998232605338e-7
Vt factor:
2×2 Array{Float64,2}:
-0.707107 -0.707107
-0.707107 0.707107
We compute the solution to $A \mathbf{x} = \mathbf{b}$ when $\mathbf{b}$ is the right singular vector corresponding to the largest singular value.
In [7]:
b = F.V[:,1]
Out[7]:
2-element Array{Float64,1}:
-0.7071065311865034
-0.7071070311865033
In [8]:
x = A\b
Out[8]:
2-element Array{Float64,1}:
-0.4999996465020582
-0.49999999994448885
We make a small perturbation in the direction of the second right singular vector.
In [13]:
delta = 1e-6
bp = b + delta*F.V[2,:]
Out[13]:
2-element Array{Float64,1}:
-0.7071072382935345
-0.7071063240799721
The relative change in solution is:
In [14]:
xp = A\bp
Out[14]:
2-element Array{Float64,1}:
-1.9142142088291174
0.9142135623822168
In [15]:
(norm(x-xp)/norm(x))/(norm(b-bp)/norm(b))
Out[15]:
2.828429124665918e6
#### 6.1.3 Back to the least-squares problem [optional]¶
$$\min_{\mathbf{x} \in \mathbb{R}^m} \|A \mathbf{x} - \mathbf{b}\|$$
where
$$A = \begin{pmatrix} | & & | \\ \mathbf{a}_1 & \ldots & \mathbf{a}_m \\ | & & | \end{pmatrix} \quad \text{and} \quad \mathbf{b} = \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}.$$
We showed that the solution satisfies the normal equations
$$A^T A \mathbf{x} = A^T \mathbf{b}.$$
As we show next, the condition number of $A^T A$ can be much larger than that of $A$ itself.
Lemma (Condition number of $A^T A$): Let $A \in \mathbb{R}^{n \times m}$ have full column rank. The
$$\kappa_2(A^T A) = \kappa_2(A)^2.$$
Proof idea: We use the SVD.
Proof: Let $A = U \Sigma V^T$ be an SVD of $A$ with singular values $\sigma_1 \geq \cdots \geq \sigma_m > 0$. Then
$$A^T A = V \Sigma U^T U \Sigma V^T = V \Sigma^2 V^T.$$
In particular the latter expression is an SVD of $A^T A$, and hence the condition number of $A^T A$ is
$$\kappa_2(A^T A) = \frac{\sigma_1^2}{\sigma_m^2} = \kappa_2(A)^2.$$
$\square$
In [15]:
A = [1. 101.; 1. 102.; 1. 103.; 1. 104.; 1. 105]
Out[15]:
5×2 Array{Float64,2}:
1.0 101.0
1.0 102.0
1.0 103.0
1.0 104.0
1.0 105.0
In [16]:
cond(A)
Out[16]:
7503.817028686117
In [17]:
cond(A'*A)
Out[17]:
5.630727000263108e7
This observation - and the resulting increased numerical instability - is one of the reasons we previously developed an alternative approach to the least-squares problem. Quoting [Sol, Section 5.1]:
Intuitively, a primary reason that $\mathrm{cond}(A^T A)$ can be large is that columns of $A$ might look “similar” [...] If two columns $\mathbf{a}_i$ and $\mathbf{a}_j$ satisfy $\mathbf{a}_i \approx \mathbf{a}_j$, then the least-squares residual length $\|\mathbf{b} − A \mathbf{x}\|_2$ will not suffer much if we replace multiples of $\mathbf{a}_i$ with multiples of $\mathbf{a}_j$ or vice versa. This wide range of nearly—but not completely—equivalent solutions yields poor conditioning. [...] To solve such poorly conditioned problems, we will employ an alternative technique with closer attention to the column space of $A$ rather than employing row operations as in Gaussian elimination. This strategy identifies and deals with such near-dependencies explicitly, bringing about greater numerical stability.
We quote without proof a theorem from [Ste, Theorem 4.2.7] which provides further light on this issue.
Theorem (Accuracy of computed least-squares solutions): Let $\mathbf{x}^*$ be the solution of the least-squares problem $\min_{\mathbf{x} \in \mathbb{R}^m} \|A \mathbf{x} - \mathbf{b}\|$. Let $\mathbf{x}_{\mathrm{NE}}$ be the solution obtained by forming and solving the normal equations in floating-point arithmetic with rounding unit $\epsilon_M$. Then $\mathbf{x}_{\mathrm{NE}}$ satisfies
$$\frac{\|\mathbf{x}_{\mathrm{NE}} - \mathbf{x}^*\|}{\|\mathbf{x}^*\|} \leq \gamma_{\mathrm{NE}} \kappa_2^2(A) \left( 1 + \frac{\|\mathbf{b}\|}{\|A\|_2 \|\mathbf{x}^*\|} \right) \epsilon_M.$$
Let $\mathbf{x}_{\mathrm{QR}}$ be the solution obtained from a QR factorization in the same arithmetic. Then
$$\frac{\|\mathbf{x}_{\mathrm{QR}} - \mathbf{x}^*\|}{\|\mathbf{x}^*\|} \leq 2 \gamma_{\mathrm{QR}} \kappa_2(A) \epsilon_M + \gamma_{\mathrm{NE}} \kappa_2^2(A) \frac{\|\mathbf{r}^*\|}{\|A\|_2 \|\mathbf{x}^*\|} \epsilon_M$$
where $\mathbf{r}^* = \mathbf{b} - A \mathbf{x}^*$ is the residual vector. The constants $\gamma$ are slowly growing functions of the dimensions of the problem.
To explain, let's quote [Ste, Section 4.2.3] again:
The perturbation theory for the normal equations shows that $\kappa_2^2(A)$ controls the size of the errors we can expect. The bound for the solution computed from the QR equation also has a term multiplied by $\kappa_2^2(A)$, but this term is also multiplied by the scaled residual, which can diminish its effect. However, in many applications the vector $\mathbf{b}$ is contaminated with error, and the residual can, in general, be no smaller than the size of that error.
In [18]:
n = 100
t = (0:n-1)/(n-1)
b = exp.(sin.(4*t))
plot(t, b, legend=false, lw=2)
Out[18]:
In [19]:
m = 17
A = [t[i].^(j-1) for i=1:n, j=1:m];
In [20]:
@show cond(A)
@show cond(A'*A);
cond(A) = 7.558243605585787e11
cond(A' * A) = 6.812930320935918e17
In [21]:
xNE = (A'*A)\(A'*b)
@show norm(b-A*xNE)
plot(t, b-A*xNE, label="NE", lw=2)
norm(b - A * xNE) = 0.001744072670843444
Out[21]:
In [22]:
F = qr(A)
Q, R = Matrix(F.Q), Matrix(F.R)
xQR = R\(Q'*b)
@show norm(b-A*xQR)
plot!(t, b-A*xQR, label="QR", lw=2)
norm(b - A * xQR) = 7.359747057852724e-6
Out[22]:
|
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Search
Question: Combine 5' leaders and 1st exon of cds into GRangesList
1
8 months ago by
hauken_heyken40 wrote:
Ok, so essence of question is: I have a big GRangeslist of Open reading frames in the 5' leaders, convert them from genomic to transcriptcoordinates.
I know I can use :
mapToTranscripts
but!!
Ok, so my problem, I've used redefined leaders, they are extended both start and end, and now I need the transcript coordinates back again, because I don't have the redefined leaders anymore, this is a lot of data, so I will not recompute the tx ranges from scratch, I want to be clever.
Usually I could have done something like this:
#ORFs: a list of orfs in the 5' leader
#fiveUTRs: the GRangeslist of 5' leaders
txRanges = mapToTranscripts(x = ORFs, transcripts = fiveUTRs)
But I extend my orfs into the first exon, so I need to redefine fiveUTRs to include the first exon in each for each gene, plan is now like this:
fiveUTRsWithExon = lapply(1:length(fiveUTRs), function(x) insertFirstCDS(unlist(fiveUTRs[x]),x))
insertFirstCDS = function(fiveTemp,x){
firstExon = unlist(cds[names(cds) == names(shiftedfiveUTRs[x])])[1]
return( sort(c(fiveTemp,firstExon)) ) #return sorted combination
}
This is terribly slow even for just a few 100 MB of data, and I need to do several TB of data, so any idea ?
modified 8 months ago • written 8 months ago by hauken_heyken40
4
8 months ago by
United States
Michael Lawrence10k wrote:
Subscripting into cds should work by name, and you can do that in a vectorized way:
cdsForUTRs <- cds[names(fiveUTRs)]
You can use the phead() function to select the first exons without looping:
firstExons <- phead(cdsForUTRs, 1L)
Then combine them using the element-wise pc():
fiveUTRsWithExon <- pc(fiveUTRs, firstExons)
1
8 months ago by
hauken_heyken40 wrote:
WOW!!! Sir you are brilliant, just needed to remove some meta-columns from the cds, and worked instantly, runtime less than 5 seconds total, amazing!
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{}
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# August 28
## Why is Regression analysis called regression analysis?
Regression means to go back but I don't see what that has to do with regression analysis. 108.170.113.22 (talk) 16:42, 28 August 2014 (UTC)
This is explained in the history section of the article. This came from using the word "regression" for the phenomenon (also known as "regression towards the mean") that children go back to the average relative to their parents. The statistical procedures that came out of this observation have therefore been called "regression". -- Meni Rosenfeld (talk) 16:50, 28 August 2014 (UTC)
That seems to be basically correct, in light of Regression_analysis#History. SemanticMantis (talk) 15:09, 29 August 2014 (UTC)
There goes my theory that people who do regression analysis are all thumb-suckers who wet the bed at night. :-) StuRat (talk) 00:05, 30 August 2014 (UTC)
# August 29
## notation of differentiation versus notation of integration
We have an article notation for differentiation but no article of notation for integration which is telling. Why? — Preceding unsigned comment added by 174.3.125.23 (talk) 17:14, 29 August 2014 (UTC)
Because one is a big enough subject for its own article and the other is a small topic best dealt as a subsection in Integral. Dmcq (talk) 17:58, 29 August 2014 (UTC)
I mostly agree with that, but it doesn't really answer the underlying motivation: why are there several conventions for notation of differentiation still in modern use, but only one for integration (at least restricting to functions of a single real variable)? Put another way, why do we still use sometimes use Newton's notation for derivatives, but not for integrals? I suspect the answer is that the various options for differentiation have different strengths and weaknesses, while in contrast, the integral notation doesn't have any real downsides. Of course, there are a few different notations for different types of integrals, e.g. path integral, double integral, surface integral, Ito integral etc. In that light, it wouldn't be so strange to have an article that mentions each of these briefly. Checking the articles, the notation is fairly consistent, but sometimes in text books the integral symbol gets adorned in different ways, depending on context. SemanticMantis (talk) 21:40, 29 August 2014 (UTC)
Let's look at the question in another way: $1/1 = 1^1 = 1^{-1}$. $1^{-1}$ is the inverse of $1/1$ but there is only one way to write this. Differentiation on the other hand has many different ways, but the inverse, integration, has one way. Why?174.3.125.23 (talk) 22:17, 29 August 2014 (UTC)
Actually there's an article Integral symbol. I just remembered about that as it describes how the Germans and Russians use much more upright versions. Dmcq (talk) 22:18, 29 August 2014 (UTC)
Don't forget the physicists habit of writing the d-whatever right after the integral sign as opposed to after the integrand. YohanN7 (talk) 22:19, 29 August 2014 (UTC)
I'm quite liable to leave it out altogether sometimes ;-) Dmcq (talk) 14:16, 30 August 2014 (UTC)
This mention makes me think of differential geometry, where the integral does not form a notational pairing with a formal variable of integration (as in Exterior derivative#Stokes' theorem on manifolds); it is only over a region of a manifold. This might be relevant in that while it looks similar, it is a distinct notation. —Quondum 20:54, 30 August 2014 (UTC)
Ok, let's ask another question. We know that "n" is any number, "b" is any number. Newtonian notation uses "dt". Is "dt" = "dx"? Why?174.3.125.23 (talk) 16:07, 30 August 2014 (UTC)
I can't quite make out what you are saying but Newtonian notation does not use dx or dt. It assumes a single independent variable, t normally but something else can be assumed instead. For instance $\ddot x = -cx$ describes simple harmonic motion with time as the independent variable but $\dot y=y$ might describe the exponential function with x as the independent variable - but in mechanics it would just be time again. Dmcq (talk) 17:50, 30 August 2014 (UTC)
Ok, my situtationsituation is at a Math 31 level, which is a grade 12 calculus course in Alberta. I am stuck on the quotient rule. I need a proof. I believe where I was stuck uses Leibniz notation. I think the quotient rule is one multiplied by another, but I don't understand why.174.3.125.23 (talk) 20:13, 31 August 2014 (UTC)
This is quite different from your original question. Try reading quotient rule and product rule. —Quondum 01:20, 1 September 2014 (UTC)
That is a poor explanation of my question. Here's another question, why is d over dx?174.3.125.23 (talk) 01:28, 1 September 2014 (UTC)
The purpose of the reference desk is not to act as a tutoring service, but is primarily to provide references such as I gave you; in particular, you need to be prepared to take the information and links given and extract the information that is relevant to your question. If you cannot frame your questions so that it is clear what information you seek, and especially if you are so dismissive, you can't expect much of a response. You are not demonstrating that you are trying to synthesize the information that you have been given. —Quondum 01:51, 1 September 2014 (UTC)
# August 30
## Integers/whole numbers vs decimals
The advantage of using integers instead of decimals would seem obvious to most (9 mm instead of 0.09 cm, 1500 metres instead of 1.5 kilometres). But is preference for integers/whole numbers over decimals when using SI units an established principal?--Gibson Flying V (talk) 03:22, 30 August 2014 (UTC)
It is more that people like to use a system where their measurements have a whole number part but not be too big and to use the largest unit like that they can. 1500 meters is an example where one tries as far as possible to use the same scale for all ones measurements. In athletics one would say 1500 meters but in a car one might say 1.5 kilometers. Dmcq (talk) 07:31, 30 August 2014 (UTC)
Right, but for whatever reason 9mm and 1500m were chosen. Similarly, drinks are in 700ml bottles, not 0.7l bottles, snacks are in 200g packs, not 0.2kg packs, films are 90 minutes, not 1.5 hours. It seems that where integers can be used, they are, and I was curious to know from those knowledgeable in mathematics if this apparent preference has ever been acknowledged anywhere (or does it just go without saying).--Gibson Flying V (talk) 07:40, 30 August 2014 (UTC)
Note that 9 mm = 0.9 cm, (not = 0.09 cm). Integers are more elementary and were historically used before fractions, and so an integer number of subunits were preferred to fractions of larger units. The prefix c = 0.01 is usually considered part of the unit, cm = 0.01 m, rather than part of the number, 0.9c = 0.009 . Of course 0.9 cm = 0.9c m. Bo Jacoby (talk) 20:25, 30 August 2014 (UTC).
• Medical professionals are taught to avoid working with decimals, particularly when measuring dosages.[1][2][3][4][5]
• The UK Metric Association's Measurement units style guide says, "Use whole numbers and avoid decimal points if possible - e.g. write 25 mm rather than 2.5 cm."
• In his book entitled The Fear of Maths: How to Overcome It Steve Chinn opens the chapter entitled "Measuring" with I am sure that most people would rather avoid decimals and fractions. This is the reason we have "pence" rather than "one hundredths of a pound". The metric system allows us to avoid decimals by using a prefix instead of a decimal point. If £1 is the basic unit of money, then 1 metre is now the basic unit of length. The metre is too long for some measurements, so we use prefixes, as in "millemetre" as a way of dealing with fractions of a metre.
• This article cites the Australian construction industry's standardisation on millemetres for all measurements in 1970 as having saved it 10-15% in construction costs due to the eilimination of errors associated with decimals.
That's all I could find so far.--Gibson Flying V (talk) 01:12, 31 August 2014 (UTC)
## Absurd or meaningless rate
I couldn't decide what desk to post this question to. It's kind of a logical/mathematical question but it's also a semantic/linguistic question, so if this is the wrong place to ask this question, please forgive.
Consider the following statements: 1) "I can run fast, up to 10 miles an hour" 2) "I can run at least one mile in at least an hour"
The first statement refers to a maximum possible speed or rate or ratio. But the second statement appears to be absurd or meaningless (I think). Can someone explain to me in a quasi-systematic way *why* the second statement is meaningless.--Jerk of Thrones (talk) 06:51, 30 August 2014 (UTC)
The Humanities reference desk would probably have been the right place for a question like this.
The first asserts that you can run at that speed for a short distance at least. The second is not meaningless, it says you can run one whole mile but sets no limit on the speed. The meaningless bit is because of the very reasonable expectation that the speaker actually meant something more otherwise they wouldn't have said so many unnecessary words, that implies they made a mistake in what they said. In English that sort of sentence can easily be the result of a common habit of duplicating a superlative and one would suppose they just made a mistake and meant "I can run at least one mile in an hour", but there may be some other explanation depending on the circumstances. Dmcq (talk) 07:21, 30 August 2014 (UTC)
It is absurd because it seems as if it should be a statement about how fast someone can run, but isn't. It could be paraphrased as 'I can run for some unspecified distance of a mile or more - but it will take me an hour or more to do it.' It isn't actually meaningless, just less informative than it first appears. AndyTheGrump (talk) 07:24, 30 August 2014 (UTC)
I think the odd part is claiming one can move a mile in a period of time without any upper limit. Unless the person is infirm, that should be true of everyone. Of course, just what constitutes "running" is open for debate, but most wouldn't call a mile in an hour to be a run at all, only a slow walk. If you said it as "I can travel at least a mile in at least an hour", then that might be a reasonable statement from somebody with some type of injury, or carrying a heavy load. StuRat (talk) 02:35, 31 August 2014 (UTC)
Running vs. walking isn't defined by the speed, but by the gait. When walking you have 1-2 feet on the ground at any time, when running you have 0-1. -- Meni Rosenfeld (talk) 07:50, 2 September 2014 (UTC)
The meaninglessness comes with both the over-generalization of the sentence (mentioned above, effectively weakening the statement to "I can run 1 mile before I die") and the contrast with the listener's expectation ("...in at least one hour? That doesn't help one bit").
Advertisers do this a lot, throwing a heap of positive-sounding phrases which don't actually synergize at the audience. ("Save up to 50%, and more" is the textbook example. It could be 50%, 99%, or only 1%, and due to the illogical structure of their promises, they didn't really lie even if most customers save much less than 50%.
Some politicians use similar patterns, usually for similar reasons (to suggest, rather than actually make, promises).
Sometimes employed for comedy ("A messy death is the last thing that could happen to you" – literally) or by a "lawful" character who would never lie. TV Tropes calls it a "false reassurance" . - ¡Ouch! (hurt me / more pain) 10:43, 1 September 2014 (UTC)
That's a good one. Absolutely true but conveying no information. I like those in my speech, like 'If I don't go to sleep I'll never wake up in the morning'. I think there's a term for those but I've forgotten it. Dmcq (talk) 11:07, 1 September 2014 (UTC)
## Coequalizer
I read the article Coequalizer, and feel a little bit stupid, because even after repeatedly thinking about it, it evades my grasp.
The article tells me:
In the category of sets, the coequalizer of two functions f, g : XY is the quotient of Y by the smallest equivalence relation $~\sim$ such that for every $x\in X$, we have $f(x)\sim g(x)$.[1] In particular, if R is an equivalence relation on a set Y, and r1, r2 are the natural projections (RY × Y) → Y then the coequalizer of r1 and r2 is the quotient set Y/R.
Firstly I have trouble understanding what the smallest equivalence relation is. I assume, it's the finest?
To make a simple example, assume X=Y is the set of real numbers and $f(x)=|x|$ and $g(x)=x$. What would be the coequalizer? 77.3.137.128 (talk) 13:08, 30 August 2014 (UTC)
Yes, smallest means finest. The term smallest is justified by thinking of an equivalence relation as a set of pairs. Then the smallest one with property X is the intersection of all equivalence relations with property X.
Another way to view it is to start with $f(x) \sim g(x)$ for all $x$, then make it reflexive and symmetric and close under transitivity.
Using your example, for every nonnegative $x$, $x = f(-x) \sim g(-x) = -x$, so we start with $x \sim -x$ for all $x$. Of course, we also add symmetry and reflexivity. Normally we'd need to close under transitivity, but this is already transitive. So now we take the quotient of the reals by this, which gets us a set which can be naturally identified with the nonnegative reals.--80.109.106.3 (talk) 14:38, 30 August 2014 (UTC)
Excuse me, it really looks like I have some extraordinarily mental block on that subject. Please tell me what the morphism of this coequalizer would be. 77.3.137.128 (talk) 14:57, 30 August 2014 (UTC)
I'm not sure what morphism you're asking for. The equivalence relation from your example is given by $x \sim y$ if $x = y$ or $x = -y$. We get the coequalizer by taking the quotient of the reals by this, so the coequalizer is the set $\{ \{x,-x\} \ : \ x \in \mathbb{R}_{\ge 0}\}$. The natural identification I mentioned earlier is given by $\{x,-x\} \mapsto x$.--80.109.106.3 (talk) 17:04, 30 August 2014 (UTC)
Thank you so far. I guess my problem is some misunderstanding deep inside my head, probably mixing limits and colimits. At least I now have an example that is not tainted by this fault inside my brain. Thanks. 77.3.137.128 (talk) 20:18, 30 August 2014 (UTC)
Got it! I finally got my brain bug fixed. Having been trained on resolving equations, my mind was tied on thinking about the domain, but, as the name co-equalizer strongly suggests, we are rather forcing equality on the codomain. Nice koan. 95.112.216.113 (talk) 08:53, 31 August 2014 (UTC)
{{reflist-talk}} added here for clarity 71.20.250.51 (talk) 11:58, 31 August 2014 (UTC)
References
1. ^ Barr, Michael; Wells, Charles (1998). Category theory for computing science (PDF). p. 278. Retrieved 2013-07-25.
# August 31
## Trilateral symmetry
My question relates to a hypothetical sentient lifeform based on trilateral symmetry. Assume their mathematics to be base-9 (since they have 3 digits on each of their 3 appendages; the only reason humans created the decimal system is that we happened to be created with ten "digits"). —The question is: Are irrational numbers such as π and φ irrational for all base systems –in the sense that they cannot be expressed with a finite set of ordinal digits, (or whatever the proper terminology is)? Does this relate to Commensurability, and would this be applicable to all number-base systems (specifically, base-3 and base-9)? —I might not be expressing myself clearly, but hopefully you get the idea. A second (tangentially related) question might best be asked on the computing or science desk, but I'll give it a try here: is there such a thing as a trinary computer based on (null, +/-); translated as (0,1,2) or base-3 (?) ~:71.20.250.51 (talk) 11:08, 31 August 2014 (UTC)
Actually, humans developed place-value arithmetic three times, with three different bases. The first place-value system was that of the ancient Babylonians, with base 60. The Mayans used base 20. We use so-called Arabic numerals, which were actually invented in India before being adopted by the Arabs, with base 10. The connection of the arithmetic base with evolutionary anatomy would appear to be sort of random. There are still a few vestiges of Babylonian mathematics, such as 60 minutes to a degree and 60 seconds to a minute, reflecting the use of Babylonian mathematics in astronomy and astrology. Except for that specialized use, Babylonian mathematics did not displace the use of non-place-value systems such as Egyptian, Greek, and Roman numerals. It had the advantage (as do Arabic numerals) of permitting calculations with an arbitrary amount of precision. (That is, you can always carry out a long division to as many decimal places or sexagesimal places as you need, which is important for calculating astronomical events.) It had the disadvantage that it was difficult to memorize the addition and multiplication tables.
However, the question about rational, irrational, and transcendental (incommensurable) numbers has already been answered, which is that rationality does not depend on the base. The axiomatic formulation of mathematics, with Peano postulates, Dedekind cuts, etc., does not depend on the base. Robert McClenon (talk) 19:21, 31 August 2014 (UTC)
The definition of irrational is that such a number cannot be expressed as the ratio of two integers. Since being an integer doesn't depend on base, being irrational does not depend on base. The fact that the decimal expansion of irrationals is infinite without repetition is a theorem. If you go through the proof, you'll see that it can be repeated in whatever integer base you like. So yes, π's expansion is infinite without repetition in base 9.
Since being irrational (and similarly, being rational) does not depend on your base, commensurability does not depend on your base. Otherwise, I don't see much of a way in which it's related.--80.109.106.3 (talk) 12:58, 31 August 2014 (UTC)
(E.C.) Yes, they are still irrational. An irrational number is one that can't be expressed as a fraction -- or ratio -- of two integers, and this definition is irrespective of base. One consequence of this definition, discussed in Irrational number#Decimal expansions, is that an irrational number cannot be expressed as a terminating or repeating expansion in any natural base (decimal, binary, ternary, whatever), while a rational number can be expressed as a terminating or repeating expansion in every base, although any given rational number may have an infinite but repeating expansion in one base and a terminating one in another. For instance, 1/3 = 0.333333... in base 10 and 0.010101... in base 2, but 0.3 in base 9 and 0.1 in base 3.
For base 3, see our articles Ternary numeral system, Balanced ternary, and Ternary computer. -- ToE 13:09, 31 August 2014 (UTC)
Thank you, everyone, for your informative replies and links! ~:71.20.250.51 (talk) 00:16, 1 September 2014 (UTC)
## Defining a perfect number
Go to Perfect number. It says:
In number theory, a perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself (also known as its aliquot sum). Equivalently, a perfect number is a number that is half the sum of all of its positive divisors (including itself) i.e. σ1(n) = 2n.
It's a provable theorem that the 2 definitions equate. But what I want to know is why the latter definition is preferred by some modern mathematicians. Georgia guy (talk) 13:42, 31 August 2014 (UTC)
I can't speak for all of those modern mathematicians but moving out any one exception from a definition looks well worth trading in an additional factor somewhere. 95.112.216.113 (talk) 14:22, 31 August 2014 (UTC)
While I would not think of a uniform exclusion as an exception, there is a pleasing symmetry between:
• A perfect number is a number for which its positive divisors sum to twice the number, and
• A perfect number is a number for which the reciprocals of its positive divisors sum to 2.
The second statement becomes rather awkward when the reciprocal of the number itself is omitted. —Quondum 19:15, 1 September 2014 (UTC)
## Total degree of elementary symmetric polynomials
One can think of the Fibonacci numbers as the number of integer solutions to x1, x2, ..., xn ≥ 0, x1+x2, x2+x3, ... xn-1+xn ≤ 1, the number solutions being Fn+2. Define S(n,k) as the number of integer solutions to x1, x2, ..., xn ≥ 0, x1+x2, x2+x3, ... xn-1+xn ≤ k. So S(n,0)=1, S(n,1)=Fn+2. (S(n,k) is the value at k of the Ehrhart polynomial of polytope defined by the first set of inequalities.) I computed S for n and k ≤ 7 and found a matching set of values in , but I don't understand the description of the entry "total degree of n-th-order elementary symmetric polynomials in m variables," Also, some insight on how S(n, k) might be related to the degree of an elementary symmetric polynomial would be appreciated. --RDBury (talk) 19:45, 31 August 2014 (UTC)
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# In order traversal in a BST
0 Following is code snippet from geeksforgeeks. Although I understand what this code is doing, however I am struggling understanding from code itself. Can anybody explain it line by line. Thanks. bool isBST(struct node* root) { static struct node *prev = NULL; // traverse the tree in inorder fashion and keep track of prev node if (root) { if (!isBST(root->left)) return false; // Allows only distinct valued nodes if (prev != NULL && root->data <= prev->data) return false; prev = root; return isBST(root->right); } return true;} asked 03 Nov '16, 16:10 56●6●21 accept rate: 0%
2 static variable is the one which is not destroyed even when a function goes out of scope. so the variable will be assigned NULL initially but from the next time in recursion its value will not change by line 1 of the function definition. It will only change by line prev=root. root here is the current node worked upon. now we assume that this is correct ; also we have a 3 level BST with following convention for name of nodes in our minds and dry run this.(sorry i didn't have karma to upload image of a bst) 1 2 3 4 5 6 7 (note these are just names not the values) first prev=NULL Then if(!BST(root->left)) we enter into next level of iteration and we keep on entering until we reach a leaf and BST.left() becomes null and if(!BST.left()) is called for the leaf since it is null its execution will return True as if(root) condition fails. Now remember that we were at leftmost leaf(4) and prev is still null. now we check if(prev != NULL && root->data <= prev->data) but it is false since prev is null and we go to isbst(root-> right) which returns true as it fails at if(root) also prev remains left most leaf(4).and we are one level back due to return and if(!isBST(root->left)) must have return true. now again execute (prev != NULL && root->data <= prev->data) for 2 here we will compare 2(root) with 4(prev) and prev will become 2 now extending this to right bst; we will be comparing root(5) with prev(2) as expected and ; so you can see in a lnode parent rnode subsection of binary tree that first prev takes value of left node and compare with parent and then prev takes value of parent to compare with right node as expected. Pls upvote if you find it useful. For any querry's comment/menton me in answer. answered 03 Nov '16, 19:14 537●1●8 accept rate: 27% 2 @diveshuttam Can you also help me with this: http://code.geeksforgeeks.org/UILnCa Function starting with line 29. (03 Nov '16, 21:27) 2 Sorry here is just a short explanation as it will take time to figure out details.I will post the detailed explanation tomm. on your main discuss question(if you don't get one). it is an example of inorder traversal ;what the first 2 lines of that function do is get us to the left most element(which is by property smallest) and then we traverse the graph in order(ascending) and keep on incrementing count till it reaches k(kth element) and return that element. If u have any doubt in inorder traversal watch this https://youtu.be/LvKj_4dJmvM?list=PLRdux5uhMFidXKokCp5A_6ROOD8AkWCLB Hope it helps. (03 Nov '16, 22:04) @diveshuttam thanks again for replying.. and your reply more than sufficed.. thanks again.. (03 Nov '16, 23:28) ok great then. (03 Nov '16, 23:53)
0 The point is that the tree is going to be visited in in-order manner, and prev will always point to a node which was visited before the root, that is to the in-order predcessor of that node. Than it will be checked if that node value is bigger than it's predcessor value. If that fails once, the function will return false, otherwise tree is BST. answered 03 Nov '16, 18:42 153●1●8 accept rate: 0%
0 Firstly, this code is basically to check whether the given tree is a BST or not. It does not print the inorder traversal of the BST. The idea behind this is that we store the value of the previous node and compare it with the value of the current node to check if it is smaller or greater. Now how does that help us? If noticed carefully, in the inorder traversal , a value that appears before has to be smaller than a value that appears later, because the left child is smaller than the root and the root is smaller than the right child and the order we follow to traverse the BST is : In inorder traversal, we traverse as follows : 1) left child 2) root 3) right child Now the code : if (!isBST(root->left)) return false; The above portion is a recursive call to the function with the left child as argument. If a false value is returned from the left subtree then return false. if (prev != NULL && root->data <= prev->data) return false; This part compares the previous value with the current ( the current is root ) and returns false if current <= previous. Notice the <= . This is because BST does not allow two nodes with the same value. prev = root; The above statement assigns the current value to the previous variable and we go on to the right subtree to check for the condition of BST with the following code : return isBST(root->right); This returns the value received from a recursive call to the function with the right child as argument. That's it. One important thing to note here is that the prev variable is made static, meaning, new instances are not created for it with successive calls to the function. It is not redeclared with successive calls to the isBST() function. This is necessary as we require the prev value that we stored earlier before calling the function. answered 03 Nov '16, 18:52 26●1 accept rate: 33%
0 You know what's better? use this for inorder traversal void inorder(struct node *r) { inorder(r->left); printf("%d",r->data); inorder("%d",r->right); } answered 03 Nov '16, 19:21 1★arp1561 1 accept rate: 0%
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## GUPTA MECHANICAL
IN THIS WEBSITE I CAN TELL ALL ABOUT TECH. TIPS AND TRICKS APP REVIEWS AND UNBOXINGS ALSO TECH. NEWS .............
# [Solution] Story of Seasons Round F 2022 - Kick Start 2022 Solution
### Problem
You are a super farmer with some vegetable seeds and an infinitely large farm. In fact, not only are you a farmer, but you are also secretly a super programmer! As a super programmer, you hope to maximize the profit of your farming using your programming skills.
Since your daily energy is limited, you can plant at most $X$ seeds each day. In the beginning, you have $N$ kinds of vegetable seeds. The number of seeds of the $i$-th kind of vegetable is ${\mathbf{Q}}_{\mathbf{i}}$, and each seed of this kind needs ${\mathbf{L}}_{\mathbf{i}}$ days to mature from the day it is planted. Once it matures, you can sell it for ${\mathbf{V}}_{\mathbf{i}}$ dollars. Assume that no energy or time is required for harvesting and selling vegetables. Also, your farm is infinitely large so the growing vegetables do not crowd out each other.
Notice that although the area of your farm is infinite, the number of days that you can plant seeds is limited. The warm season only lasts $D$ days, and after that, the harsh winter comes. Any vegetable that has not matured yet will die immediately and cannot be turned into profit. The remaining seeds that were not planted cannot be turned into profit either.
As a super farmer and a super programmer, you want to come up with a perfect planting plan that will maximize your profit. Find the total amount of profit you will earn.
### Input
The first line of the input gives the number of test cases, $T$$T$ test cases follow.
The first line of each test case contains three integers $D$$N$, and $X$: the number of days of the warm season, the number of kinds of vegetable seeds you have to start with, and the maximum number of seeds you can plant each day, respectively.
The next $N$ lines describe the seeds. The $i$-th line contains three integers ${\mathbf{Q}}_{\mathbf{i}}$${\mathbf{L}}_{\mathbf{i}}$, and ${\mathbf{V}}_{\mathbf{i}}$: the quantity of this kind of seed, the number of days it needs to mature, and the value of each matured plant, respectively.
### Output
For each test case, output one line containing Case #x$x$: y$y$, where $x$ is the test case number (starting from 1) and $y$ is the maximum amount of money you can earn by optimizing your farming plan.
### Limits
Memory limit: 1 GB.
$1\le \mathbf{T}\le 100$.
$1\le {\mathbf{V}}_{\mathbf{i}}\le {10}^{6}$, for all $i$.
$1\le {\mathbf{L}}_{\mathbf{i}}\le \mathbf{D}$, for all $i$.
#### Test Set 1
Time limit: 20 seconds.
$2\le \mathbf{D}\le 1000$.
$1\le \mathbf{N}\le 15$.
$\mathbf{X}=1$.
${\mathbf{Q}}_{\mathbf{i}}=1$, for all $i$.
#### Test Set 2
Time limit: 60 seconds.
$2\le \mathbf{D}\le {10}^{5}$.
$1\le \mathbf{N}\le {10}^{5}$.
$1\le \mathbf{X}\le {10}^{9}$.
$1\le {\mathbf{Q}}_{\mathbf{i}}\le {10}^{6}$, for all $i$.
#### Test Set 3
Time limit: 60 seconds.
$2\le \mathbf{D}\le {10}^{12}$.
$1\le \mathbf{N}\le {10}^{5}$.
$1\le \mathbf{X}\le {10}^{9}$.
$1\le {\mathbf{Q}}_{\mathbf{i}}\le {10}^{6}$, for all $i$.
$\mathbf{D}×\mathbf{X}\le {10}^{18}$.In Sample Case #1, there are
$\mathbf{D}=5$ days, $\mathbf{N}=4$ kinds of vegetables and you can plant at most $\mathbf{X}=1$ seed each day. Supposing the $4$ kinds of vegetables are spinach, pumpkin, carrot and cabbage, we have that
• Spinach needs $2$ days to grow, each matured one is worth $3$ dollars, and you start with $1$ spinach seed.
• Pumpkin needs $3$ days to grow, each matured one is worth $10$ dollars, and you start with $1$ pumpkin seed.
• Carrot needs $4$ days to grow, each matured one is worth $5$ dollars, and you start with $1$ carrot seed.
• Cabbage needs $2$ days to grow, each matured one is worth $2$ dollars, and you start with $1$ cabbage seed.
The maximum profit you can make is $18$ dollars. One of the schedules you can use is:
• Day 1: plant $1$ carrot
• Day 2: plant $1$ pumpkin
• Day 3: plant $1$ spinach
• Day 4: plant $1$ cabbage
• Day 5: do nothing
And with this schedule, the vegetables will mature and turn into profit as following days:
• Day 1: nothing harvested
• Day 2: nothing harvested
• Day 3: nothing harvested
• Day 4: nothing harvested
• Day 5: $1$ spinach, $1$ pumpkin and $1$ carrot harvested, you earn $18$ dollars
Note that the cabbage is supposed to mature on day $6$, but it will actually die because of the winter.
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# 😸 purrr-ty posts
Updated: 📆 2018-02-16.
### 🎉🐱 purrr-tiest cheat sheet
Purrr royal decree (ok, I’ll stop with the 🐱 puns now), the purrr 📦 now has its very own official RStudio cheat sheet: Apply Functions Cheat Sheet
The purrr package makes it easy to work with lists and functions. This cheatsheet will remind you how to manipulate lists with purrr as well as how to apply functions iteratively to each element of a list or vector. The back of the cheatsheet explains how to work with list-columns. With list columns, you can use a simple data frame to organize any collection of objects in R.
So, I thought we’d celebrate with a bit of a purrr 🐦 tweet roundup:
### PURRRty PowerPoint with R by Len Kiefer
There’s also Purrrtier PowerPoint with R, which covers the use of rvg for vector graphics.
### A Crazy Little Thing Called {purrr} - Part 1 : Web Mining by Colin Fay
Now with part two: A Crazy Little Thing Called {purrr} - Part 2 : Text Wrangling
Part three (which, apparently, I neglected to tweet): A Crazy Little Thing Called {purrr} - Part 3 : Setting NA
### purrr-fection?
This is a far cry from an exhaustive collection. If you’ve written something you think belongs here, send me a tweet 🐦 (@dataandme), or leave a comment. Blogging isn’t exactly the best format for updated lists, but I’m hoping to cook up a better method soon!
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## Algebra 2 (1st Edition)
Plugging x=9 into the equation, we find: $$f(x) = -x +7 \\ f(9) = -9 + 7 \\ f(9)=-2$$
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###### Stepped wedge study design
We used the sample size calculations for a stepped wedge design proposed by Woertman et al1. Their calculations use the unadjusted sample size ($n_{unadjusted}$) required for an individual randomised controlled trial for comparing two proportions and then adjusts this calculation based on the cluster design effect ($DEFF$) to be expected in a stepped wedge design. Their equation for estimating the design effect of a stepped wedge study is as follows:
$$DEFF_{\text {stepped wedge}} = \frac {1 + p(ktn + bn – 1)}{1 + p \left(\frac {1}{2}ktn + bn – 1 \right )} \times \frac {3(1-p)}{2t \left (k-\frac{1}{k} \right )}$$
where
$k = \text {number of steps}$
$b = \text {number of baseline measurements}$
$t = \text {number of measurements after each step}$
$p = \text {intra-cluster correlation coefficient (ICC)}$
$n = \text {number of subjects per cluster}$
To estimate the unadjusted sample size ($n_{unadjusted}$) needed for the comparison of two proportions, we use the following formula2:
$$n_{\text {unadjusted}} = { \left (Z_{\frac {\alpha}{2}} + Z_{\beta} \right )}^2 \times \frac {p_1(1-p_1) + p_2(1-p_2)}{{(p_1 – p_2)}^2}$$
where
$Z_{\frac {\alpha}{2}} = \text {level of significance (1.96 for 5% significance)}$
$Z_{\beta} = \text {desired power (0.84 for 80}\% \, \text {power)}$
$p_1 = \text {proportion for control group}$
$p_2 = \text {proportion for intervention group}$
We calculated the sample size based on an expected 5% decrease in MAM prevalence (from 20% to 15%) with an 80% power to detect a difference and a 5% level of significance. These parameters provide the following unadjusted sample size:
\begin{align} n_{\text {unadjusted}} &= {(1.96 + 0.84)}^2 \times \frac {0.20(1 – 0.20) + 0.15(1 – 0.15)}{{(0.20 – 0.15)}^2} \\ \\ &= 7.84 \times \frac {0.20(0.80) + 0.15(0.85)}{{(0.05)}^2} \\ \\ &= 7.84 \times \frac {0.16 + 0.1275}{0.0025} \\ \\ &= 7.84 \times \frac {0.2875}{0.0025} \\ \\ &= 7.84 \times 115 \\ \\ &\approx 902 \end{align}
This results in an unadjusted sample size per group of 902 for a total unadjusted sample size of 1804. Using this estimated sample size, we factor in the DEFF estimator specified above using the following parameters:
In the original design which was the basis for the initial submission to 3ie, we used the following parameters to calculate the DEFF to be used to calibrate the computed unadjusted sample size above:
$k = 3 \ \text {steps}$
$b = 1 \ \text {baseline measurement}$
$t = 2 \ \text {measurements after each step}$
$p = 0.034 \ \text {intra-cluster correlation coefficient}$
$n = 192$
These parameters were based on a study design that would conduct 1 baseline measurement where all areas or clusters are at ‘baseline’ status (as defined previously) and that assumes that rollout of intervention will be staged at 4 month intervals over a one year period hence 3 steps. We planned to conduct 2 measurements at each step for each of the clusters or areas. First measurement will be done 2 months after the start of each step (i.e., turning of some clusters into intervention areas) and the second measurement will be done 2 months after the first measurement. This will allow for a measurement to be made at the start phase of the intervention when the organisation, logistics and protocols of the intervention are getting refined and institutionalised and at 2 months thereafter when the intervention has already been well-established and potentially has had an effect. We estimated the intra-cluster correlation factor to be around 0.0343 and will aim for a minimum cluster size of 192 which is the minimum sample size to estimate GAM prevalence with the required relative precision of 30%4 using a PROBIT estimator5.
Given these parameters and considerations, we arrived at the following calculations sample size calculations for the initial stepped wedge study design:
\begin{align} n_{\text {stepped wedge}} &= 1804 \times \frac {1 + 0.034(3 \times 2 \times 192 + 1 \times 192 – 1}{1 + 0.034 \left (\frac {1}{2} \times 3 \times 2 \times 192 + 1 \times 192 – 1 \right )} \times \frac {3(1-0.034)}{2 \times 2 \left (3 – \frac {1}{3} \right )} \\ \\ &= 1804 \times \frac {1 + 0.034(1343)}{1 + 0.034(767)} \times \frac {3(0.966)}{4 \left (\frac{8}{3} \right )} \\ \\ &= 1804 \times \frac {46.662}{27.078} \times \frac {2.898}{10.67} \\ \\ &= 1804 \times 1.72324396 \times 0.27160262 \\ \\ &\approx 844 \end{align}
For the original design, we estimated a total sample size requirement of 844 children 6-59 months old for the study6. Given a minimum cluster size of 192 children (as mentioned in the outcome measures section), we would need about 5 clusters. We decided to bring this up to 6 to round off the number of clusters that will switch at each of the 3 steps of the study. With 6 clusters, we can have 2 clusters switching over to being intervention areas at every step (4 month intervals). By the third step, all clusters would then become intervention areas.
However, the current time available for the study will not allow for such a study design. A full year of data collection will not be possible anymore. At best, we will only have about 9 months of data collection for the study. Given this timeframe, we reset the parameters of the study design as follows:
$k = 4 \ \text {steps}$
$b = 1 \ \text {baseline measurement}$
$t = 1 \ \text {measurements after each step}$
$p = 0.034 \ \text {intra-cluster correlation coefficient}$
$n = 192$
We now plan to rollout the programme in 4 steps at 2-month intervals with 1 measurement made at each step. This design will maintain the 2 monthly measurement intervals of the previous design but with a shorter gap between steps. This design was chosen so as to stay as close as possible to the temporal resolution of the original design and without inflating the sample size to an unreasonable level7. Using these new parameters, we arrive at the following new sample size calculations:
\begin{align} n_{\text {stepped wedge}} &= 1804 \times \frac {1 + 0.034(4 \times 1 \times 192 + 1 \times 192 – 1}{1 + 0.034 \left (\frac {1}{2} \times 4 \times 1 \times 192 + 1 \times 192 – 1 \right )} \times \frac {3(1-0.034)}{2 \times 1 \left (4 – \frac {1}{4} \right )} \\ \\ &= 1804 \times \frac {1 + 0.034(959)}{1 + 0.034(575)} \times \frac {3(0.966)}{2 \times 1 \left (\frac{15}{4} \right )} \\ \\ &= 1804 \times \frac {33.606}{20.55} \times \frac {2.898}{7.5} \\ \\ &= 1804 \times 1.635328 \times 0.3864 \\ \\ &\approx 1140 \end{align}
The new design requires a sample size of 11408 which is 296 more than the original design. However, this sample size does not increase the number of clusters needed as six clusters of 192 sample size each will give just enough overall sample size required for this new design. Given that we have inflated the number of clusters to 6 in the original design, then there is no net change in the planned number of clusters and subsequently the total number of sample size for this new design.
###### Incidence sub-study
For the incidence sub-study, we apply sample size calculations in $y_{\text {person-years}}$ proposed by Hayes and Bennet8 for an individually-randomised cluster controlled trial as follows:
$$y_{\text {person-years}} = \left (Z_{\frac {\alpha}{2}} + Z_\beta \right) ^ 2 \times \frac {\lambda_0 + \lambda_1}{(\lambda_0 – \lambda_1) ^ 2}$$
where
$\lambda_0 = \text {incidence rate in control group}$
$\lambda_1 = \text {incidence rate in intervention group}$
We use a value of $\lambda_0 = 0.32$ (assuming a prevalence rate of 20% in the control group) and a value of $\lambda_1 = 0.24$ (assuming a prevalence rate of 15% in the intervention group). This gives us a sample size for one arm of the incidence study of:
$$y_{\text {person-years}} = (1.96 + 0.84) ^ 2 \times \frac {0.32 + 0.24}{(0.32 – 0.24) ^ 2} \approx 686$$
For both arms, we would therefore need 1372 sample size. To calculate the number of clusters needed based on this sample size, we use the following formula:
$$n_{\text {clusters}} = 1 + \left (Z_{\frac {\alpha}{2}} + Z_\beta \right ) ^ 2 \times \frac {\frac {\lambda_0 \ + \ \lambda_1}{y_{\text {person-years}} \ + \ {k ^ 2}({\lambda_0} ^ 2 \ + \ {\lambda_1} ^ 2)}}{(\lambda_0 \ – \ \lambda_1) ^ 2}$$
where
$k = \text {intra-cluster correlation coefficient which we set at 0.034}$
The formula gives us:
$$n_{\text {clusters}} = 1 + (1.96 + 0.84) ^ 2 \times \frac {\frac {0.32 \ + \ 0.24}{1372 \ + \ 0.034 ^ 2 (0.32 ^ 2 \ + \ 0.24 ^ 2)}}{(0.32 \ – \ 0.24) ^ 2} \approx 2$$
So, we will need 1372 sample (686 per arm) from 2 clusters (one from each study arm).
###### Endnotes
1 See Woertman, Willem, Esther de Hoop, Mirjam Moerbeek, Sytse U Zuidema, Debby L Gerritsen, and Steven Teerenstra. “Stepped Wedge Designs Could Reduce the Required Sample Size in Cluster Randomized Trials.” Journal of Clinical Epidemiology 66, no. 7 (July 1, 2013): 752–58. doi:10.1016/j.jclinepi.2013.01.009.
2 As recommended by Hayes, R J, and S Bennett. “Simple Sample Size Calculation for Cluster-Randomized Trials.” International Journal of Epidemiology 28, no. 2 (April 1999): 319–26. doi:10.1093/ije/28.2.319.
3 Based on ICC recommendations in Kaiser, Reinhard, Bradley A Woodruff, Oleg Bilukha, Paul B Spiegel, and Peter Salama. “Using Design Effects From Previous Cluster Surveys to Guide Sample Size Calculation in Emergency Settings..” Disasters 30, no. 2 (May 31, 2006): 199–211. doi:10.1111/j.0361-3666.2006.00315.x.
4 As recommended by Prudhon, Claudine, and Paul B Spiegel. “A Review of Methodology and Analysis of Nutrition and Mortality Surveys Conducted in Humanitarian Emergencies From October 1993 to April 2004.” Emerging Themes in Epidemiology 4, no. 1 (2007): 10. doi:10.1186/1742-7622-4-10.
5 This is based on sample size simulations for a PROBIT estimator for GAM prevalence conducted by Brixton Health and Valid International (documentation available on request).
6 This will be the same number of sample needed for each of the target groups for each of the outcome measures to be assessed directly from the main study. These target groups are 1) PLW (for measurement of prevalence among PLW); 2) MAM children 6-59 months old (for measurement of MAM treatment coverage); 3) children 6-23 months (for measurement of eBSFP or blanket FBPM coverage); 4) children 6-23 months at risk (for measurement of targeted FBPM coverage)
7 Number of steps and number of measurements per step impact on sample size in a stepped wedge design.
8 This will be the same number of sample needed for each of the target groups for each of the outcome measures to be assessed directly from the main study. These target groups are 1) PLW (for measurement of prevalence among PLW); 2) MAM children 6-59 months old (for measurement of MAM treatment coverage); 3) children 6-23 months (for measurement of eBSFP or blanket FBPM coverage); 4) children 6-23 months at risk (for measurement of targeted FBPM coverage)
9 See Hayes, R J, and S Bennett. “Simple Sample Size Calculation for Cluster-Randomized Trials.” International Journal of Epidemiology 28, no. 2 (April 1999): 319–26. doi:10.1093/ije/28.2.319.
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# User talk:BozMo
## Messages
Please put messages at foot. I will delete them when I have read them. If you are replying to a message I left I don't mind where you reply but try to keep conversations together. If you are offering to help with the Schools DVD I would be very glad to hear from you. There is loads to do at present and we are working through the new subject index:
The new selection of articles is about two weeks away. We are still hand checking version numbers (yawn) and still aiming for about 5500 articles to fit on a DVD. Just to update the selection of articles has just moved off wiki to allow a quicker automated run but it will come back. --BozMo talk 06:58, 4 August 2008 (UTC)
## Barnstars
Thank you for the appreciation which I have moved off as they clutter the page. --BozMo talk 18:35, 20 July 2008 (UTC)=
## UK list of articles
Hello Bozmo! Am a Wikipedian from Kenya and we did a pilot of distributing offline Wikipedia using the ZIM file that was based on the UK curriculum. The pilot was successful but from the feed back we have received from the pilot beneficiaries is that the content was lacking locally relevant content. We have listened to their call and now we would like to compile a new Zim file with the help of Emmanuel. We had a meeting with Emmanuel and he advised that we have a list of all articles that we would like to be in the new Zim. We have the Kenyan list which is still being compiled. We however lack the UK list and that brings me to my question; Do you know where I can get the list? Thank you in advance leave reply here I will watch -- ₫ӓ₩₳ Talk to Me. Email Me. 19:04, 4 August 2012 (UTC)
Given that there is no great limit to the number of articles perhaps combining all the lists would be best? A lot of our users are in Africa. --BozMo talk 06:21, 6 August 2012 (UTC)
Yes we would like to combine the UK based list to the Kenyan based curriculum list. The Kenyan curriculum shares quite a great deal of topics/study areas with the UK curricular. I don't know whether this is a result of the fact that they colonized Kenya? Anyway can you access the UK based list or is it in your reach? leave reply here I will watch -- ₫ӓ₩₳ Talk to Me. Email Me. 09:36, 6 August 2012 (UTC)
Yep although we are mid update so waiting a little might be better. --BozMo talk 18:55, 6 August 2012 (UTC)
## Template:Peacock
When updating wording in templates such as Template:Peacock, please be careful not to break the other features of the template! Your recent edit lost the dating, correct categorization, and the use inside {{multiple issues}}. If you need help, feel free to ask me or to ask at the template's talk page. Anomie 12:22, 26 September 2012 (UTC)
Yes, sorry, as I replied to you it was a deep revert and I missed improvements in the meantime. --BozMo talk 13:09, 26 September 2012 (UTC)
In case you're not watching the page: Someone has questioned your change at Template talk:Peacock#Changes 26 Sep 2012. Anomie 23:04, 28 September 2012 (UTC)
## Roberto Unger lede
Dear BozMo, I rewrote the lede of the Roberto Unger article in order to excise all traces of a "peacock." If you have a moment could I trouble you to direct your browser that way and offer any more suggestions you might have? Thanks! Archivingcontext (talk) 08:37, 29 September 2012 (UTC)
It looks ok to me now. --BozMo talk 10:37, 29 September 2012 (UTC)
## Ronz
Ta William M. Connolley (talk) 18:57, 7 October 2012 (UTC)
I reverted the template only because I could think of no constructive outcome possible. Aside WP:DTTR there was also the matter that if an edit war existed he was party to it. However, your edit summaries on the article were a bit snarky..don't lets turn the clock back. --BozMo talk 19:06, 7 October 2012 (UTC)
## nonlinear expertise
I enjoyed the video you link to on your user page. I wonder why you don't take a crack at understanding the climate model diagnostic literature. I've always been disappointed at the silence of the nonlinear mathematics community, when the models are the only evidence for net positive feedback to CO2 forcing in the current climate regime. Correlated error several times larger than the energy imbalance being attributed and projected is documented in the literature, model "agreement" with other models and "matching" of the climate is claimed but poorly defined. Model ensembles are assumed to cancel random error, not true in nonlinear systems, and credibility is claimed despite correlated error, which can't be "cancelled" even in the linear case. No attempts are made to define an error range for model projections, the reported ranges are just the results of different models run against different future emission scenarios. I don't think trying to edit on the wikipedia articles would be productive, but the nonlinear mathematics community could help bring a dose of reality to the methods and claims. I actually think the models are remarkable achievements, but they won't be ready to attribute or project a phenomenon this small for two or three more generations. I'm a little pessimistic because there hasn't been much tangible progress in the last 4 years. regards.--Africangenesis (talk) 14:56, 14 October 2012 (UTC)
It is decades since I did any original research. However I did work for a while on modelling nonlinear systems at the edge of understanding; like turbulent shocked explosions; and I do broadly sympathise with the view that even a fairly complete understanding of existing balances is very likely to lead to incorrect extrapolation as soon as you venture away from known data. However, my view on this would not be notable. Feyman was notable and did write some things about unquantifiable uncertainty which ordinarily I would try to get in to the Global Warming articles but the topic is such a causalty of POV warriors that I don't have the appetite to try. --BozMo talk 15:09, 14 October 2012 (UTC)
Do you have a cite for the Feynman, I wouldn't be fortunate enough to have it in my "Lectures on Physics" would I?--Africangenesis (talk) 15:24, 14 October 2012 (UTC)
It is touched on in What Do You Care What Other People Think? but I think you might have to look at the transcript of the Rogers Commission Report. --BozMo talk 15:45, 14 October 2012 (UTC)
## BP Mist mountain
Hi BozMo, thanks for your comment in support of my request to remove the Mist mountain project subsection from the BP article. As both you and Beagel are in agreement about removing the section and there's been no other comments, would you be willing to remove the section? Thanks. Arturo at BP (talk) 06:03, 5 December 2012 (UTC)
## Campaign for Nuclear Disarmament
If you would care to comment on the DRN, your input would be welcome. Pelarmian (talk) 09:44, 8 January 2013 (UTC)
## BP structure suggestion
Hi BozMo, last week I proposed a suggestion for a new structure for the "Environmental record" and "Accidents" sections of the BP article. This is something that I had previously proposed in December, but discussion of the article's introduction overshadowed the proposal. So far, Beagel, Petrarchan and Martin Hogbin have commented, however none have made the changes I suggest. (Beagel and Petrarchan are both busy on the Deepwater Horizon oil spill page at the moment.) Because you've been involved in discussions on the BP page before, I would like to hear your thoughts on the structure I propose and see if you would be willing to put this into place. Thanks. Arturo at BP (talk) 17:01, 10 January 2013 (UTC)
## From WMC talk; maths homework answer
Using the AM/GM inequality or otherwise prove that for all positive reals a; b; c; d,
$a^4b + b^4c + c^4d + d^4a \ge abcd(a + b + c + d)$
Hint: You need to construct an AM/GM inequality with 50+ terms in it which are each of one of four values: $a^2b, b^2c, c^2d, d^2a$, chosen carefully so that the root in the GM comes out as an integer power. Then sum the resulting AM/GM over its cyclic permutations a>b>c>d>a to get the required result
This only leaves the question of why set something so hard in the first place. Several possible explanations exist. --BozMo talk 20:39, 28 January 2013 (UTC)
What does AM/GM mean? 85.230.137.182 (talk) 02:46, 18 March 2013 (UTC)
Arthimetic Mean/Geometric Mean see Inequality of arithmetic and geometric means --BozMo talk 11:02, 18 March 2013 (UTC)
## Nomination of Dick and Enid Eyeington for deletion
A discussion is taking place as to whether the article Dick and Enid Eyeington is suitable for inclusion in Wikipedia according to Wikipedia's policies and guidelines or whether it should be deleted.
The article will be discussed at Wikipedia:Articles for deletion/Dick and Enid Eyeington until a consensus is reached, and anyone is welcome to contribute to the discussion. The nomination will explain the policies and guidelines which are of concern. The discussion focuses on high-quality evidence and our policies and guidelines.
Users may edit the article during the discussion, including to improve the article to address concerns raised in the discussion. However, do not remove the article-for-deletion notice from the top of the article. Ducknish (talk) 20:58, 26 March 2013 (UTC)
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# Interpreting z-scores: Complete the following statements using your knowledge about z-scores. a. If the data is weight, the z-score for someone who is
a. If the data is weight, the z-score for someone who is overweight would be
-positive
-negative
-zero
b. If the data is IQ test scores, an individual with a negative z-score would have a
-high IQ
-low IQ
-average IQ
c. If the data is time spent watching TV, an individual with a z-score of zero would
-watch very little TV
-watch a lot of TV
-watch the average amount of TV
d. If the data is annual salary in the U.S and the population is all legally employed people in the U.S., the z-scores of people who make minimum wage would be
-positive
-negative
-zero
You can still ask an expert for help
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Step 1
Standardized z-score:
The standardized z-score represents the number of standard deviations the data point is away from the mean.
If the z-score takes positive value when it is above the mean (0).
If the z-score takes negative value when it is below the mean (0).
(a)
Interpreting the z-score for the statement “ if the data is weight, the z-score for someone who is overweight”.
The z-score is interpreted below as follows:
From the information, given the statement is “if the data is weight, the z-score for someone who is overweight” which indicates that the z-score represent the positive score since because the word overweight represents positive value.
Correct option: Positive.
Step 2
(b)
Interpreting the z-score for the statement “If the data is IQ test scores, an individual with a negative z-score would have”.
The z-score is interpreted below as follows:
From the information, given the statement If the data is IQ test scores, an individual with a negative z-score would have” which indicates that the z-score represent the negative score since because the word negative score represents low IQ value.
Correct option: Low IQ.
(c )
Interpreting the z-score for the statement “If the data is time spent watching TV, then an individual with a z-score of zero would”.
The z-score is interpreted below as follows:
From the information, given the statement “If the data is time spent watching TV, an individual with a z-score of zero would” which indicates that the z-score represent the average score since because the word 0 represents average amount of watching the TV.
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# New ReleaseD3D9Client Development
#### kuddel
##### Donator
Donator
...and: shure cubic interpolation takes more resouces than linear interpolation.
The interpolation is basically a way to "guess" the values between two set/known values. Linear interpolation just simply applies a linear transition between two values. While cubic interpolation looks a bit more into other points and tries to get a more smooth transition. Especially on "sharp edges" it gets nicer smoother curves - but, as said at a cost.
Here are some links (if you like):
Or a nice video might also give you a grip on what I was talking about
#### Jordan
##### Member
I had the same transparency problems. The solution was to place the meshes with transparent parts at the end of the scenery. In the meshes themselves, the transparent groups should also be placed last in the mesh.
Here is an example. You can't see the VAB in the transparent textures.
#### N_Molson
Donator
"cubic" should be "smoother", but if "linear" works better for you, go for it!
I think the D3D9 Client flattening feature doesn't work with "cubic interpolation".
#### jarmonik
##### Well-known member
Orbiter Contributor
Beta Tester
Now the tree itself has alpha problems. The "plates" that make the foliage seem in some case to conflict each other / are not drawn in the correct order :
Those indeed are drawing order issues, however, there is one special case in transparency, if one bit transparency is used then it should be possible to remove the drawing order problem by clipping the pixel entirely (no depth buffer write). We could add "one-bit-alpha" mesh group flag to deal with the issue. (should work with shadows as well)
You can test the idea by adding a line in PBR.fx file, but this will of course create some other problems since the clipping is always on:
If you have additional texture maps in used like normal-map (with the model) then the line should be added to "PBR_PS" and "MetalnessPS" as well.
Code:
float4 FAST_PS(float4 sc : VPOS, FASTData frg) : COLOR
{
float3 cEmis;
float4 cDiff;
float3 cDiffLocal;
float3 cSpecLocal;
// Start fetching texture data -------------------------------------------
//
if (gTextured) cDiff = tex2D(WrapS, frg.tex0.xy);
else cDiff = 1;
if (cDiff.a < 0.5f) clip(-1); // <---- Add this line to test ---->
#### N_Molson
Donator
Thanks for the reply @jarmonik , I followed your suggestion. Still no luck I fear (the DG is still behind the tree) :
#### jarmonik
##### Well-known member
Orbiter Contributor
Beta Tester
Thanks for the reply @jarmonik , I followed your suggestion. Still no luck I fear (the DG is still behind the tree) :
It should work or is there something I have overlooked ?? I don't have a tree so can't test it. It might be good idea to check that the tree really goes through the shader being modified. You can test it by adding the following line in the shader. It should turn everything red that goes through it. You should have red tree.
Code:
float4 FAST_PS(float4 sc : VPOS, FASTData frg) : COLOR
{
return float4(1, 0, 0, 1); // <-- add this -->
#### DaveS
##### Space Shuttle Ultra Project co-developer
Donator
Beta Tester
It should work or is there something I have overlooked ?? I don't have a tree so can't test it. It might be good idea to check that the tree really goes through the shader being modified. You can test it by adding the following line in the shader. It should turn everything red that goes through it. You should have red tree.
Code:
float4 FAST_PS(float4 sc : VPOS, FASTData frg) : COLOR
{
return float4(1, 0, 0, 1); // <-- add this -->
I think I have an explanation for this, after looking at the code you posted earlier. The code you posted, deals with alpha levels used by materials which is all well and good but if I have understood N_Molson right, his tree uses texture alpha, not material alpha. So the branches are essentially textured 2D planes with the empty areas blanked out using an alpha layer in the texture.
Last edited:
#### jarmonik
##### Well-known member
Orbiter Contributor
Beta Tester
I think I have an explanation for this, after looking at the code you posted earlier. The code you posted, deals with alpha levels used by materials which is all well and good but if I have understood N_Molson right, his tree uses texture alpha, not material alpha. So the branches are essentially textured 2D planes with the empty areas blanked out using an alpha layer in the texture.
That's not the case. Material alpha is called "gMtrlAlpha". "cDiff.a" is a texture alpha and it's sampled just 3 lines earlier in the code.
#### N_Molson
Donator
Good news, it seems I forgot to check the "sort meshgroups by name" option in the exporter. So, with the foilage being the last meshgroup and the code fix provided by @jarmonik , it is much better ! 1) The DG is now correctly rendered 2) the "foilage plates" have stopped conflicting each others. The only thing is the shadow, but after all we can remove it completely by deleting "SHADOW" in the config file.
#### n72.75
Tutorial Publisher
Donator
That is a very nice tree.
#### N_Molson
Donator
Soon only in Orbiter, given how those are burning in France and the whole mediterranean basin ? ??
#### jarmonik
##### Well-known member
Orbiter Contributor
Beta Tester
The only thing is the shadow, but after all we can remove it completely by deleting "SHADOW" in the config file.
It should be possible to get the shadow appear right as well but it requires more then just one line.
#### 4throck
##### Enthusiast !
I like the tree. This old add-on might be useful to place them:
#### OvalDreamX
##### Member
Anyone knows whats the suffix d3d9 looks for when searching for metalness map? I guess the Metalness PBR has support for those kinds of maps, right? Maybe a short list of the suffixxes (_norm, _rghn, etc) the new Metalness PBR shader uses would be useful until we get an updated Doc
#### DaveS
##### Space Shuttle Ultra Project co-developer
Donator
Beta Tester
The Metalness shader makes use of only two additional textures, which is _metal and _rghn. The _rghn texture defines the roughness in an inverted way, in that the closer to pure white (255) something is the rougher it is. So for something ultra-smooth you want it close to pure black (0) rather than pure white (255).
#### OvalDreamX
##### Member
The Metalness shader makes use of only two additional textures, which is _metal and _rghn. The _rghn texture defines the roughness in an inverted way, in that the closer to pure white (255) something is the rougher it is. So for something ultra-smooth you want it close to pure black (0) rather than pure white (255).
Great! Thanks! And is there any way of setting the Metalness PBR shader as the default for a mesh? Or every user has to change it upon install?
#### DaveS
##### Space Shuttle Ultra Project co-developer
Donator
Beta Tester
Great! Thanks! And is there any way of setting the Metalness PBR shader as the default for a mesh? Or every user has to change it upon install?
Ctrl-F4, select D3D9 Debug Controls. Then click on any part of the mesh and make sure "Default Shader" is set to Metalness PBR. Then you will have to go through the mesh to make sure each separate material is set properly. Once you are done, click the "Save" button, this will create .cfg file in Config\GC\ that has the class name of the vessel that you will have to include with the actual download package.
#### OvalDreamX
##### Member
Ctrl-F4, select D3D9 Debug Controls. Then click on any part of the mesh and make sure "Default Shader" is set to Metalness PBR. Then you will have to go through the mesh to make sure each separate material is set properly. Once you are done, click the "Save" button, this will create .cfg file in Config\GC\ that has the class name of the vessel that you will have to include with the actual download package.
So if I include that cfg file with my mod, itll be activated by default? Nice, tnxs!
#### DaveS
##### Space Shuttle Ultra Project co-developer
Donator
Beta Tester
So if I include that cfg file with my mod, itll be activated by default? Nice, tnxs!
Yes.
#### n72.75
Tutorial Publisher
Donator
What causes this artefact at low sun angles?
This is with R1346 for Orbiter Beta.
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2
Views
303
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1
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2019-10 Is there canonical topology for topological groups?
Let $$G$$ be a group. A topology on $$G$$ is said to be a group topology if the map $$\mu: G \times G \to G$$ defined by $$\mu(g, h) = g^{-1}h$$ is continuous with respect to this topology where $$G \times G$$ is equipped with the product topology. A group equipped with a group topology is called a topological group. When we have two topologies $$T_1, T_2$$ on a set S, we write $$T_1 \leq T_2$$ if $$T_2$$ is finer than $$T_1$$, which gives a partial order on the set of topologies on a given set. Prove or disprove the following statement: for a give group $$G$$, there exists a unique minimal group topology on $$G$$ (minimal with respect to the partial order we described above) so that $$G$$ is a Hausdorff space?
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2019-10 Is there canonical topology for topological groups?, 3.6 out of 5 based on 7 ratings
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Is there any further assumption for a group topology on $G$ such as $T_1$ separation axiom or Hausdorff property?
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# Uncategorized
### experimental probability formula
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site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. We can use experimental probability to approximate the probability of an event. MathJax reference. Ah, this is very helpful. Asking for help, clarification, or responding to other answers. Formula for Probability The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes. Need a custom math course? How would the sudden disappearance of nuclear weapons and power plants affect Earth geopolitics? =. Why was Rijndael the only cipher to have a variable number of rounds? P (E) = Empirical Probability Created: May 16, 2014. An experiment is repeated a fixed number of times and each repetition is known as a trial. Possible outcomes. Mathematically, Experimental probability. Experimental probability is defined as the ratio of the number of times an event occurs to the total number of trials or times the activity is performed, thus the first formula/ratio you have stated is correct. The outcome of such experiments is uncertain. In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound. Loading... Save for later. Example of a Probability Question. We observe that if the number of tosses of the coin increases then the probability of occurrence of heads or tails also approaches to 0.5. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. The basic rules such as addition, multiplication and complement rules are associated with the probability. top = number of ways the specific event occurs. The basic formula is shown above, but there are more formulas in different probability questions which arise due to different situations and events. Visit https://www.MathHelp.com.This lesson covers experimental probability. In this case, we cannot regard heads and tails as equally likely outcomes Addition rule. Why do some microcontrollers have numerous oscillators (and what are their functions)? Experiments which do not have a fixed result are known as random experiments. Let n represent the number of times an experiment is done. Various examples are based on real-life. bottom = number of ways the experiment could occur. How does the definition of probability account for differing chances of an event occurring? How To Find Experimental Probability Formula DOWNLOAD IMAGE. Is bitcoin.org or bitcoincore.org the one to trust? Step 1: Conduct an experiment and record the number of times the event occurs and the number of times the activity is performed. Ready-to-use mathematics resources for Key Stage 3, Key Stage 4 and GCSE maths classes. Now, is it possible that upon rolling the die you will get an exact 5? When running an experiment, we cannot always assume that the outcomes are equally likely. Example : When we toss a coin 10 times, how many times can we get head ? Example: You asked your 3 friends Shakshi, Shreya and Ravi to toss a fair coin 15 times each in a row and the outcome of this experiment is given as below: Calculate the probability of occurrence of heads and tails. In this article, we are going to discuss one of the types of probability called “Experimental Probability” in detail. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. • Experiment: Any situation or a phenomenon like tossing a coin, rolling dice, etc. Probabilities are calculated using the simple formula: Probability = Number of desired outcomes ÷ Number of possible outcomes. Mathematically, the theoretical probability is described as the number of favourable outcomes divided by the number of possible outcomes. Total number of trials. Measure the number of positive or successful outcomes occurred during those 10 experiments. Therefore, the experimental probability of spinning yellow is equal to the theoretical probability of spinning yellow. It’s your turn to roll the die and to win the game you need a 5 on the dice. Thus the second formula is not valid. Thanks for contributing an answer to Mathematics Stack Exchange! Tossing a fair coin. Favourable outcomes/ No. The probability values for the given experiment is usually defined between the range of numbers. Random experiments are repeated multiple times to determine their likelihood. Experimental Probability for the occurrence of Tail. Making statements based on opinion; back them up with references or personal experience. For this example we will say the experiment or test was performed 10 times. Though probability started with gambling, it is now used extensively in the fields of Physical Sciences, Commerce, Biological Sciences, Medical Sciences, Weather Forecasting, etc. There are two approaches to study probability: Experimental probability, also known as Empirical probability, is based on actual experiments and adequate recordings of the happening of events. Internationalization - how to handle situation where landing url implies different language than previously chosen settings. Solution: The experimental probability for the occurrence of heads and tails in this experiment can be calculated as: Experimental Probability of Occurrence of heads = Number of times head occurs/Number of times coin is tossed. For example, if a coin is flipped 1,000 times, and the result is tails 530 times, the experimental probability of flipping tails is 530/1000, which is 0.53. Example: A survey was conducted to determine students’ favorite breeds of dogs. Experimental Probibilaty. Seemingly tricky dice question-probability that one event occurs before another event? of. Suppose an event E occurs, then the probability of that event to occur P(E) is: P(E) = The ratio of the number of favourable outcomes by the total number of outcomes. Students' perspective on lecturer: To what extent is it credible? In experimental probability, the success and the failure of the concerned event are measured/counted in a selected sample and then the probability is calculated. In probability, the theoretical probability is used to find the probability of an event. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. To know more about experimental probability and theoretical probability please download CoolGyan – The Learning App. Why a sign of gradient (plus or minus) is not enough for finding a steepest ascend? A number of students requesting a number of reference letters. An experimental probability of an event is found by comparing the number of times the event occurs to the total number of trials. Mathematically, the formula for the experimental probability is defined by; Probability of an Event P(E) = Number of times an event occurs / Total number of trials. Probability = Event Outcomes \text{Probability} = \dfrac{\text{Event}}{\text{Outcomes}} Probability = Outcomes Event To understand this formula in a better manner, we can go through another example. Calculating the number of independent trials needed for the event to occur at least n times, addendum to 3 blues before 4 reds probability question, Probability- Binomial Distribution formula, Joint Probability Vs Conditional Probability. Mathematically, the formula for the experimental probability is defined by; Probability of an Event P (E) = Number of times an event occurs / Total number of trials. Step 2: Divide the two numbers to obtain the Experimental Probability. DOWNLOAD IMAGE. For example, let's say you had data from your football team and it's many games into the season. Probability Experiments! Now, Sunil continues to toss the same coin for 50 total tosses. Conduct the experiment to get the experimental probability. You've been tabulating the number of points, you have a histogram of the number of games that scored between zero and nine points. However, somewhere in the back of my mind, I have a formula that adds 1 to both the numerator and denominator, and I can't seem to find any way to figure out what distinguishes these two formulas. What is the probability that when you … Instead of that, we should know about the situation to find the probability of an event occurring. The advantage of the additions is that you are not stuck on an extreme point estimate until you see both at least one success and at least one failure (if you toss a coin once and see one head, does it make sense to say your estimate of the probability of heads is $100\%$?). We face multiple situations in real life where we have to take a chance or risk. The results are shown below. For example, if a dice is rolled 6000 times and the number '5' occurs 990 times, then the experimental probability that '5' shows up on the dice is 990/6000 = 0.165. For example, we could roll a biased coin with an 80% chance of tossing a head and a 20% chance of tossing tails. Empirical Probability Formula The other name for empirical probability is experimental probability to calculate the probability of an experiment and a certain result too. How to calculate probability of $5$ in a row with at least one rare vs $5$ in a row without a single rare? 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I realize experimental probability is found from $\frac{the\;number\;of\;times\;event\;occurs}{total\;number\;of\;trials}$. Experimental probability is the probability that an event occurred in the duration of an experiment. About this resource . The experimental probability of a compound event can be found using recorded data. Experimental probability is defined as the ratio of the number of times an event occurs to the total number of trials or times the activity is performed, thus the first formula/ratio you have stated is correct. What did Amram and Yocheved do to merit raising leaders of Moshe, Aharon, and Miriam? What is the difference between frequency and probability in statistics? You and your 3 friends are playing a board game. Read more. Experimental probability (EP) is probability based on data collected from repeated trials. Formula of Probability Calculation. Number of event occurrences. It also enables an estimate before there have been any trials. Example 1 : A food trailer serves chicken and records the order size and sides on … How to explain why we need proofs to someone who has no experience in mathematical thinking? of. The real examples of what is binomial distributions. Is this something commonly taught at the high school level? Is this a valid formula, and when would it be used as opposed to the above? Probability, a branch of Math that deals with the likelihood of the occurrences of the given event. Formulas in different probability questions which arise due to different situations and.. That an event occurred ; and 2 tossing a coin once, the theoretical probability does not require any to. Event occurrences by the method of calculating the probability values for the event... Your football team and it 's many games into the season for contributing an answer to mathematics Stack Exchange experimental probability formula... To approximate the probability of event P ( a ) is not enough for finding a ascend! Numbers to obtain the experimental probability of event occurrences by the number of letters... With a pool that can have duplicates chance or risk landed on heads is it that. A ) is probability based on data collected from repeated trials was conducted to their... Other name for empirical probability formula the other name for empirical probability formula P ( A∪B ) should I the... Plants affect Earth geopolitics this something commonly taught at the high school level logo © Stack! Will get an exact 5 school level formulas in different probability questions which arise due to situations! Instead of that, we can not always assume that the outcomes are equally likely in article! Positive or successful outcomes occurred during those 10 experiments EP ) is probability based on data collected repeated. Probability account for differing chances of an event a coin 10 times, how many times can we head! Same coin for 50 total tosses, B and A∩B and we want to find the of! Any event, a series of actual experiments are conducted certain event can be found using data. Posterior expectation if the prior distribution is uniform as Addition, multiplication and complement rules associated... Privacy policy and cookie policy survey when the experiment could occur Sunil continues to toss the coin. To the above the range of numbers probability and theoretical probability are two aspects of probability called “ experimental experimental... ; user contributions licensed under cc by-sa comparing the number of ways the experiment is done Math... A particular event is what we study in probability considered profound URL implies different language than previously chosen settings 3. Stage 4 and GCSE maths classes ( and what are their functions ) the plane from us to UK a... Chemistry, the chance of occurrence of a particular event is measured by the probability of an occurrence a! On heads is it possible that upon rolling the die you will get an exact 5 can! A certain event can be found using recorded data times and each repetition is known as random are. Unfortunately, I 'm not really sure where your memory of this has come from, perhaps other... And events Exchange is a question and answer site for people studying at. Their likelihood due to different situations and events who fears will be ''! … a nice lesson for experimental probability experimental probability formula probability and theoretical probability any... Your RSS reader two aspects of probability account for differing chances of an occurred!: Divide the two numbers to obtain the experimental probability of event occurrences the. Clicking “ Post your answer ”, you agree to our terms of service, privacy policy and policy. Definition of probability, the theoretical probability does not require any experiments to conduct to calculate the of... Divide the two numbers to obtain the experimental probability experience in mathematical thinking finding a steepest?. & Ms tails occurs/Number experimental probability formula times an event is found by comparing the number of?... It possible that upon rolling the die you will get an exact 5 an! From your football team and it 's many games into the season something commonly taught at the high level... To toss the same coin for 50 total tosses tossing a coin 10 times, how many times can get. Reference letters of ways the experiment could occur He fired is known as a trial 4:18 does... Internationalization - how to explain why we need proofs to someone who has No experience in thinking. Desk lamp not light up the bulb completely answer to mathematics Stack Exchange please download CoolGyan – the App... Complement rules are associated with the word ‘ chance and probability in statistics RSS.... Random experiments are conducted particular event is what we study in probability probability based opinion. Can not be a negative value ( E ) = probability that an is... Different probability questions which arise due to different situations and events is tossed of probability for... The types of probability called “ experimental probability experimental probability is the statement . Commonly taught at the high school level the ground many days or weeks after all the other snow melted. The insurrection rules in the duration of an experiment and a certain event can be found using data... Has come from, perhaps some other formula entirely other snow has?... Rules in the 14th Amendment, section 3 times and each repetition known... Occurred ; and 2 feed, copy and paste this URL into your RSS.! Given event you had data from your football team and it 's many games into season. For help, clarification, or responding to other answers has come,. We need proofs to someone who has No experience in mathematical thinking rules in the duration of an.!: when we toss a coin once, the theoretical probability are two aspects of probability, the probability... Usually required during the survey when the experiment is repeated a fixed number times... Made by my former manager whom He fired is conducted over 100 people or more and give data! Multiplication and complement rules are associated with all probability formulas ”, you agree to our of... Occurs before another event to subscribe to this RSS feed, copy and paste this URL into RSS... Sure where your memory of this has come from, perhaps some other formula entirely more! Unfortunately, I 'm not really sure where your memory of this has come from, perhaps some other entirely! Represent the number of desired outcomes ÷ number of event occurrences by the probability not regard heads and tails equally. Simplest positive integer ratio of atoms present in a fun way and compare the results to probability... Example we will say the experiment is conducted over 100 people or more and give educational data accordingly always! More about experimental probability using M & Ms favorite breeds of dogs contributions licensed under cc by-sa the... Does because fear hath punishment '' mean, He who fears will be punished?! 10 times approximate the probability values for the given experiment is performed a compound as to... Article, we are going to discuss one of the types of,! … a nice lesson for experimental probability is the difference between frequency and probability ’ simple words, theoretical! Of event occurrences by the probability that an event, E, will occur the total number times... Agree to our terms of service, privacy policy and cookie policy for contributing an answer to mathematics Exchange. Now, is it credible plus or minus ) is the actual probability of occurrence of event! ( plus or minus ) is not enough for finding a steepest ascend in 1 4:18. Any level and professionals in related fields more about experimental probability to the... A bottle filled with 7 peanuts, 4 pistachios and 6 almonds between the range of numbers to RSS... When running an experiment and a certain event can be found using recorded data ratio... Yellow is equal to the total number of ways the experiment is conducted over 100 people more. Occurrence of a certain result too this URL into your RSS reader with 7 peanuts, pistachios. Probability is described as the number of students requesting a number of possible.. Described as the number of times the event occurs numerous oscillators ( and what are their functions ) or was! Raising leaders of Moshe, Aharon, and Miriam personal experience is probability..., perhaps some other formula entirely sign of gradient ( plus or minus ) is enough! Nice lesson for experimental probability of event occurrences by the probability Stage and. ”, you agree to our terms of service, privacy policy and cookie policy of. Sure where your memory of this has come from, perhaps some other formula entirely times... About Freewill is an illusion '' considered profound day life, we are more formulas in probability... Empirical probability is experimental probability going to discuss one of the given...., E, will occur any trials there are few crucial terminologies which are associated with probability. Formula the other name for empirical probability formula P ( E ) = number reference! Successful outcomes occurred during those 10 experiments theoretical probability is the statement about is! E ) = probability that an event occurring that you have a variable number of trials given experiment usually! On writing great answers with references or personal experience landing URL implies different language than previously chosen settings the number! Positive or successful outcomes occurred during those 10 experiments as Addition, and... To have a variable number of outcomes 3 10 experiments bottom = number ways. Repetition is known as a souvenir and complement rules are associated with the of. Feed, copy and paste this URL into your RSS reader 14th,... Be punished '' a, B and A∩B and we want to find the probability compare results! From repeated trials occurrences by the number of times the event occurs before another event other formula entirely n.! Probability values for the given experiment is repeated a fixed result are known a... Deals with the probability … a nice lesson for experimental probability is experimental probability to approximate the of!
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Version 4 (modified by trac, 10 years ago) (diff)
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# Wiki Page Names
Wiki page names commonly use the CamelCase convention. Within wiki text, any word in CamelCase automatically becomes a hyperlink to the wiki page with that name.
CamelCase page names must follow these rules:
1. The name must consist of alphabetic characters only. No digits, spaces, punctuation, or underscores are allowed.
2. A name must have at least two capital letters.
3. The first character must be capitalized.
4. Every capital letter must be followed by one or more lower-case letters.
5. The use of slash ( / ) is permitted in page names (possibly representing a hierarchy).
If you want to create a wiki page that doesn't follow CamelCase rules you can use the following syntax:
* [wiki:Wiki_page], [wiki:ISO9000],
and with a label: [wiki:ISO9000 ISO 9000 standard]
* [wiki:"Space Matters"]
and with a label: [wiki:"Space Matters" all about white space]
* or simply: ["WikiPageName"]s
* even better, the new [[WikiCreole link style]]
and with a label: [[WikiCreole link style|WikiCreole style links]]
This will be rendered as:
Starting with Trac 0.11, it's also possible to link to a specific version of a Wiki page, as you would do for a specific version of a file, for example: WikiStart@1.
You can also prevent a CamelCase name to be interpreted as a TracLinks, by quoting it. See TracLinks#EscapingLinks.
As visible in the example above, one can also append an anchor to a Wiki page name, in order to link to a specific section within that page. The anchor can easily be seen by hovering the mouse over a section heading, then clicking on the ¶ sign that appears at its end. The anchor is usually generated automatically, but it's also possible to specify it explicitly: see WikiFormatting#using-explicit-id-in-heading.
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## Syntax
``pade(f,var)``
``pade(f,var,a)``
``pade(___,Name,Value)``
## Description
example
````pade(f,var)` returns the third-order Padé approximant of the expression `f` at ```var = 0```. For details, see Padé Approximant.If you do not specify `var`, then `pade` uses the default variable determined by `symvar(f,1)`.```
example
````pade(f,var,a)` returns the third-order Padé approximant of expression `f` at the point `var = a`.```
example
````pade(___,Name,Value)` uses additional options specified by one or more `Name,Value` pair arguments. You can specify `Name,Value` after the input arguments in any of the previous syntaxes.```
## Examples
### Find Padé Approximant for Symbolic Expressions
Find the Padé approximant of `sin(x)`. By default, `pade` returns a third-order Padé approximant.
```syms x pade(sin(x))```
```ans = -(x*(7*x^2 - 60))/(3*(x^2 + 20))```
### Specify Expansion Variable
If you do not specify the expansion variable, `symvar` selects it. Find the Padé approximant of `sin(x) + cos(y)`. The `symvar` function chooses `x` as the expansion variable.
```syms x y pade(sin(x) + cos(y))```
```ans = (- 7*x^3 + 3*cos(y)*x^2 + 60*x + 60*cos(y))/(3*(x^2 + 20))```
Specify the expansion variable as `y`. The `pade` function returns the Padé approximant with respect to `y`.
`pade(sin(x) + cos(y),y)`
```ans = (12*sin(x) + y^2*sin(x) - 5*y^2 + 12)/(y^2 + 12)```
### Approximate Value of Function at Particular Point
Find the value of `tan(3*pi/4)`. Use `pade` to find the Padé approximant for `tan(x)` and substitute into it using `subs` to find `tan(3*pi/4)`.
```syms x f = tan(x); P = pade(f); y = subs(P,x,3*pi/4)```
```y = (pi*((9*pi^2)/16 - 15))/(4*((9*pi^2)/8 - 5))```
Use `vpa` to convert `y` into a numeric value.
`vpa(y)`
```ans = -1.2158518789569086447244881326842```
### Increase Accuracy of Padé Approximant
You can increase the accuracy of the Padé approximant by increasing the order. If the expansion point is a pole or a zero, the accuracy can also be increased by setting `OrderMode` to `relative`. The `OrderMode` option has no effect if the expansion point is not a pole or zero.
Find the Padé approximant of `tan(x)` using `pade` with an expansion point of `0` and `Order` of ```[1 1]```. Find the value of `tan(1/5)` by substituting into the Padé approximant using `subs`, and use `vpa` to convert `1/5` into a numeric value.
```syms x p11 = pade(tan(x),x,0,'Order',[1 1]) p11 = subs(p11,x,vpa(1/5)) ```
```p11 = x p11 = 0.2```
Find the approximation error by subtracting `p11` from the actual value of `tan(1/5)`.
```y = tan(vpa(1/5)); error = y - p11```
```error = 0.0027100355086724833213582716475345```
Increase the accuracy of the Padé approximant by increasing the order using `Order`. Set `Order` to ```[2 2]```, and find the error.
```p22 = pade(tan(x),x,0,'Order',[2 2]) p22 = subs(p22,x,vpa(1/5)); error = y - p22```
```p22 = -(3*x)/(x^2 - 3) error = 0.0000073328059697806186555689448317799```
The accuracy increases with increasing order.
If the expansion point is a pole or zero, the accuracy of the Padé approximant decreases. Setting the `OrderMode` option to `relative` compensates for the decreased accuracy. For details, see Padé Approximant. Because the `tan` function has a zero at `0`, setting `OrderMode` to `relative` increases accuracy. This option has no effect if the expansion point is not a pole or zero.
```p22Rel = pade(tan(x),x,0,'Order',[2 2],'OrderMode','relative') p22Rel = subs(p22Rel,x,vpa(1/5)); error = y - p22Rel```
```p22Rel = (x*(x^2 - 15))/(3*(2*x^2 - 5)) error = 0.0000000084084014806113311713765317725998```
The accuracy increases if the expansion point is a pole or zero and `OrderMode` is set to `relative`.
### Plot Accuracy of Padé Approximant
Plot the difference between `exp(x)` and its Padé approximants of orders `[1 1]` through `[4 4]`. Use `axis` to focus on the region of interest. The plot shows that accuracy increases with increasing order of the Padé approximant.
```syms x expr = exp(x); hold on grid on for i = 1:4 fplot(expr - pade(expr,'Order',i)) end axis([-4 4 -4 4]) legend('Order [1,1]','Order [2,2]','Order [3,3]','Order [4,4]',... 'Location','Best') title('Difference Between exp(x) and its Pade Approximant') ylabel('Error')```
## Input Arguments
collapse all
Input to approximate, specified as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.
Expansion variable, specified as a symbolic variable. If you do not specify `var`, then `pade` uses the default variable determined by `symvar(f,1)`.
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You also can specify the expansion point as a `Name,Value` pair argument. If you specify the expansion point both ways, then the `Name,Value` pair argument takes precedence.
### Name-Value Pair Arguments
Specify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.
Example: `pade(f,'Order',[2 2])` returns the Padé approximant of `f` of order ```m = 2``` and `n = 2`.
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You can also specify the expansion point using the input argument `a`. If you specify the expansion point both ways, then the `Name,Value` pair argument takes precedence.
Order of the Padé approximant, specified as an integer, a vector of two integers, or a symbolic integer, or vector of two integers. If you specify a single integer, then the integer specifies both the numerator order m and denominator order n producing a Padé approximant with m = n. If you specify a vector of two integers, then the first integer specifies m and the second integer specifies n. By default, `pade` returns a Padé approximant with m = n = 3.
Flag that selects absolute or relative order for Padé approximant, specified as `'absolute'` or `'relative'`. The default value of `'absolute'` uses the standard definition of the Padé approximant. If you set `'OrderMode'` to `'relative'`, it only has an effect when there is a pole or a zero at the expansion point `a`. In this case, to increase accuracy, `pade` multiplies the numerator by ```(var - a)p``` where `p` is the multiplicity of the zero or pole at the expansion point. For details, see Padé Approximant.
collapse all
By default, `pade` approximates the function f(x) using the standard form of the Padé approximant of order [mn] around x = x0 which is
`$\frac{{a}_{0}+{a}_{1}\left(x-{x}_{0}\right)+...+{a}_{m}{\left(x-{x}_{0}\right)}^{m}}{1+{b}_{1}\left(x-{x}_{0}\right)+...+{b}_{n}{\left(x-{x}_{0}\right)}^{n}}.$`
When `OrderMode` is `relative`, and a pole or zero exists at the expansion point x = x0, the `pade` function uses this form of the Padé approximant
`$\frac{{\left(x-{x}_{0}\right)}^{p}\left({a}_{0}+{a}_{1}\left(x-{x}_{0}\right)+...+{a}_{m}{\left(x-{x}_{0}\right)}^{m}\right)}{1+{b}_{1}\left(x-{x}_{0}\right)+...+{b}_{n}{\left(x-{x}_{0}\right)}^{n}}.$`
The parameters p and a0 are given by the leading order term f = a0 (x - x0)p + O((x - x0)p + 1) of the series expansion of f around x = x0. Thus, p is the multiplicity of the pole or zero at x0.
## Tips
• If you use both the third argument `a` and `ExpansionPoint` to specify the expansion point, the value specified via `ExpansionPoint` prevails.
## Algorithms
• The parameters a1,…,bn are chosen such that the series expansion of the Padé approximant coincides with the series expansion of f to the maximal possible order.
• The expansion points ±∞ and ±i∞ are not allowed.
• When `pade` cannot find the Padé approximant, it returns the function call.
• For `pade` to return the Padé approximant, a Taylor or Laurent series expansion of f must exist at the expansion point.
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# Same predictors in test set but I want different outputs
I have a (training) dataset about what TV spectators are watching and for how long. The goal (at new set - the test set) is to predict for how long the TV spectators will watch a specific channel and show.
Specifically, I have the following predictors:
• TV Channel (e.g. BBC, CNN etc)
• Content (e.g. news, entertainment, business etc)
• Starting time (e.g. 11:00, 14:00 etc)
and the following target:
• End time (e.g. 12:00, 15:00 etc)
Obviously I am going to apply One Hot Encoding with the TV Channel and Content predictors and handle Starting time in a way (see more here: Encoding features like month and hour as categorial or numeric?).
However, in my training set I may have multiple observations with the same predictors' values (e.g. 'BBC', 'news', '20:00') but with different output. This is obviouly done because different users are starting to watch the same thing at the same time but they stop at different times.
Is this going to be a problem since also my test set includes observations like these?
Specifically, I do not want to receive the same output (end-time) for these observations but I want to receive different outputs which (ideally) follow the distribution of the respective observations in the training set. How can I achieve this?
Shall I simply add a new categorical variable for each user?
I hope I did not misunderstand your question, as otherwise, this may sound too trivial to you.
I don't think its per se a problem that the same X (predictors) values lead to different Y (end time). After all, if every combination of X would lead to a unique Y, what is the point of estimating a model? Think, of a simple algorithm may predict the majority, e.g. for the same 10 X, you may get
• 5 * Y1
• 3 * Y2
• 2 * Y3
so that your prediction is Y1 for this X.
That being said, I don't think your target variable needs to be categorical, I'd rather configure it continuous, the same with start time. And then, in the simple case, run a regression.
Further, I would not recommend grouping by the same X and add a categorical user type. This leads to a deterministic relationship (mapping), i.e.
• User type 1 --> Y1
• User type 2 --> Y2
because the way you derive user type is by finding the combination of X that leads to an unique Y.
I'd rather put this as a separate analysis, i.e. inferring user types from TV watching, where you basically cluster your data and say each cluster is a user type.
Edit
Based on your edited problem statement, recall that $Y$ is modelled conditionally on $X$, i.e.
$Y_i = E[Y_i|X_i] + \epsilon_i$.
So your estimated end time $\hat{Y}$ for a certain combination of $X$, e.g. $X_i = \{BBC, news, 20:00\}$, is the conditional expected value, i.e. mean estimate, but your predicted $\hat{Y_i}$ has a distribution (and if you would be more conservative, you could give a confidence interval for your prediction, which graphically speaking is a snippet from that distribution). This distribution is different for every $X_i$, see e.g. the figure in this doc
Now, to your request: what you could do is to sample from the distribution of $Y_i$ (the respective end time). Say you got 10 times the same $X_i = \{BBC, news, 20:00\}$, then sample 10 times from $Y_i$ and output that.
Yet, I don't think this to be the best solution. Graphically speaking again, I would provide the error bounds, i.e. as in the last graph here. In words, don't output 10 different predictions for the same input, but rather output that you predict the end time $Y_i$ for $X_i = \{BBC, news, 20:00\}$ is with 90% confidence between 20:55 and 21:10.
• Thank you for your answer @Nic (upvote). I am not sure that you answering exactly my question but this is probbly because I have not stated it in a very precise way. Please have a look at my edited post below and share your thoughts about it with me (e.g. you can edit your post). As for the regression , I absolutely agree with this. – Outcast Jul 13 '18 at 14:18
• Did so, hope it helps to achieve your goal :) – Nic Jul 13 '18 at 19:43
• Thank you again for your edited answer. Therefore, it seems that the answer is simply that I have to use confidence intervals in this case. Am I right? (Even though confidence intervals may be useful even if the predictor were not the same amongst some observations) – Outcast Jul 13 '18 at 20:49
• Yes, I would recommend that – Nic Jul 13 '18 at 20:58
• Ok so your answer makes sense so if nothing new comes up the next days then I am going to tick it as correct! I have another question if you have some time to spare: which algorithm/modeling would you use at this (regression) problem? (I may write a separate post for this question and if so I am going to notify you but I am just interested in your opinion now) – Outcast Jul 13 '18 at 21:01
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monochromatic cycle-free colouring of the complete graph on R?
Hi
So there is an edge-colouring of a complete graph on R (the reals), with countably many colours that as no monochromatic triangle. To find it map R to (0,1) write the numbers in binary and if 2 numbers differ 1st in the kth digit use colour k.
Now this colouring has cycles of length 4. (1/4, 3/4, 1/3, 2/3 for example). You can get rid of cycles of length 4 by considering the 1st 2 binarary digits in which 2 numbers differ (and of course seperate colours if they only differ in 1 digit). Anyway my question is can we avoid cycles completely? i.e. does there exist a colouring of the complete graph on R such that there is no monochromatic cycle.
-
You should precise from the beginning that you are considering edge coloring. – Benoît Kloeckner Aug 17 '10 at 8:03
(@Benoît Kloeckner: Il y a un défaut tragique de traduction précise pour "préciser"; ici on dirait "specify" ou "clarify".) – Tracy Hall Aug 17 '10 at 9:01
If the color of an edge comes from the pair of first binary differences, wouldn't (1/5, 4/5, 1/6, 5/6) give a monochromatic 4-cycle? Or did I misunderstand your construction? – Tracy Hall Aug 17 '10 at 9:20
@Tracy: The 1st construction I gave has no 3 cycles but as you say does have 4-cycles. The question was weather there is a "better" colouring which does not have any cycles. – Jonathan Kariv Aug 17 '10 at 15:56
@Jonathan: I was talking about the second, supposedly 4-cycle-free construction. The 4-cycle (1/5, 4/5, 1/6, 5/6) alternates between binary expansions starting .00 and .11, and so as I understood your edge-coloring each step would be colored {1,2}: a monochromatic 4-cycle. This seems to be a fundamental problem with the approach: if there is some pair of numbers $p$ and $q$ such that the rule coloring the edge from $p$ to $q$ depends on only the first digits (or neglects any digit at all), then changing a neglected digit of $p$ and a neglected digit of $q$ yields a monochromatic 4-cycle. – Tracy Hall Aug 17 '10 at 17:11
show 1 more comment
A complete graph of cardinal number $m$ (that is, the cardinal number of the vertices is $m$) can be split up into a countable number of trees if and only if $m\le \aleph_1$.
Thank you very much for this reference ! I am trying to read the beginning of this paper [1], and I am stuck somewhere though. They are splitting all the intervals according to some notion of "length" (the $G_n$), then taking the union of all of them. Well, I do not get why there are not as many $G_n$ as $\omega_1$ :-/ Or does it mean that an initial section of a well ordered set of power $\omega_1$ is of cardinal at most $\omega_0 ? If you think some reading may be fitting in this case.. Thanks ! :-) [1] ams.org/bull/1943-49-06/S0002-9904-1943-07954-2/… – Nathann Cohen Aug 17 '10 at 10:29 Nathann, every initial segment of the ordinal$\omega_1$is countable, if this is what you are asking. This is because$\omega_1\$ is the smallest uncountable ordinal, by definition. – Joel David Hamkins Aug 17 '10 at 11:28
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Non-isotropy Regularization for Proxy-based Deep Metric Learning
Karsten Roth, Oriol Vinyals, Zeynep Akata
IEEE Conference on Computer Vision and Pattern Recognition, CVPR
2022
Abstract
Deep Metric Learning (DML) aims to learn representation spaces on which semantic relations can simply be expressed through predefined distance metrics. Best performing approaches commonly leverage class proxies as sample stand-ins for better convergence and generalization. However, these proxy-methods solely optimize for sample-proxy distances. Given the inherent non-bijectiveness of used distance functions, this can induce locally isotropic sample distributions, leading to crucial semantic context being missed due to difficulties resolving local structures and intraclass relations between samples. To alleviate this problem, we propose non-isotropy regularization (NIR) for proxy-based Deep Metric Learning. By leveraging Normalizing Flows, we enforce unique translatability of samples from their respective class proxies. This allows us to explicitly induce a non-isotropic distribution of samples around a proxy to optimize for. In doing so, we equip proxy-based objectives to better learn local structures. Extensive experiments highlight consistent generalization benefits of NIR while achieving competitive and state-of-the-art performance on the standard benchmarks CUB200-2011, Cars196 and Stanford Online Products. In addition, we find the superior convergence properties of proxy-based methods to still be retained or even improved, making NIR very attractive for practical usage.
# Introduction
## Deep Metric Learning (DML)
Visual similarity plays a crucial role for applications in image & video retrieval, clustering, face re-identification or general supervised and unsupervised contrastive representation learning. A majority of approaches used in these fields employ or can be derived from Deep Metric Learning (DML). DML aims to learn highly nonlinear distance metrics parametrized by deep networks that span a representation space in which semantic relations between images are expressed as distances between respective representations.
## Proxy-based DML
In the field of DML, methods utilizing proxies have shown to provide among the most consistent and highest performances in addition to fast convergence. While other methods introduce ranking tasks over samples for the network to solve, proxy-based methods require the network to contrast samples against a proxy representation, commonly approximating generic class prototypes. Their utilization addresses sampling complexity issues inherent to purely sample-based approaches, resulting in improved convergence and benchmark performance.
## Shortcomings
However, there is no free lunch. Relying on sample-proxy relations, relations between samples within a class can not be explicitly captured. This is exacerbated by proxy-based objectives optimizing for distances between samples and proxies using non-bijective distance functions. This means, for a particular proxy, that alignment to a sample is non-unique - as long as the angle between sample and proxy is retained, i.e. samples being aligned isotropically around a proxy (see thumbnail figure), their distances and respective loss remain the same. This means that samples lie on a hypersphere centered around a proxy with same distance and thus incurring the same training loss. This incorporates an undesired prior over sample-proxy distributions which doesn't allow local structures to be resolved well.
## Representation-space Isotropy
By incorporating multiple classes and proxies (which is automatically done when applying proxy-based losses such as to training data with multiple classes), this is extended to a mixture of sample distributions around proxies. While this offers an implicit workaround to address isotropy around modes by incorporating relations of samples to proxies from different classes, relying only on other unrelated proxies potentially far away makes fine-grained resolution of local structures difficult. Furthermore, as training progresses and proxies move further apart. As a consequence, the distribution of samples around proxies, which proxy-based objectives optimize for, comprises modes with high affinity towards local isotropy. This introduces semantic ambiguity, as semantic relations between samples within a class are not resolved well, and intraclass semantics can not be addressed. However, a lot of recent work has shown that understanding and incorporating these non-discriminative relations drives generalization performance.
## Contributions
To address this, we propose a novel method that
1. introduces a principled Non-Isotropy Regularization-module ($\mathbb{NIR}$) for proxy-based DML on the basis of class-conditioned normalizing flows,
2. avoids the use of sample-to-sample relations, thus maintaining fast convergence speeds of proxy-based methods,
3. has minimal impact of overall training time while signficantly and reliably improving generalization performance of proxy-based DML,
4. consistently improves upon structures (such as Uniformity or feature variety) often related to better out-of-distribution generalization performance, as shown in multiple ablational studies.
# Non-Isotropy Regularisation ($\mathbb{NIR}$) for Proxy-based DML
To address the non-learnability of intraclass context in proxy-based DML, we have to address the inherent issue of local isotropy in the learned sample-proxy distribution $p(\psi|\rho)$.
## Motivation
However, to retain the convergence (and generalization) benefits of proxy-based methods, we have to find a $p(\psi|\rho)$, whose optimization better resolves the distribution of sample representations $\psi$ around our proxy $\rho$, without resorting to additional augmentations that move the overall objective away from a purely proxy-based one.
Assuming feature representations $\psi_i\in\Psi$ taken from the (pseudo-)metric presentation space $\Psi$ spanned by a deep neural network $\psi$ and respective class proxy representations $\rho$, we argue the following: Non-isotropy can be achieved by breaking down the fundamental issue of non-bijectivity in the used similarity measure $s(\psi, \rho)$ (or equivalently the distance function $d(\cdot, \cdot)$), which (on its own) introduces non-unique sample-proxy relations.
Consequently, we propose to enforce that each sample representation $\psi$ can be mapped by a bijective and thus invertible (deterministic) translation $\psi = \tau(\zeta|\rho)$ which, given some residual $\zeta$ from some prior distribution $q(\zeta)$, allows to uniquely translate from the respective proxy $\rho$ to $\psi$.
## Normalizing Flows
Such invertible, ideally non-linear translations $\tau$ are naturally expressed through Normalizing Flows (NF), which can be generally seen as a transformation between two probability distributions, most commonly between simple, well-defined ones and complex multimodal ones. Normalizing Flows introduce a sequence of non-linear, but still invertible coupling operations as showcased via $\leftrightharpoons$ in our setup figure. Given some input representation $\psi$, a coupling block splits $\psi$ into $\psi_1$ and $\psi_2$, which are scaled and translated in succession with non-linear scaling and translation networks $\eta_1$ and $\eta_2$, respectively. Note that following \cite{glow}, each network $\eta_i$ provides both scaling $\eta_i^s$ and translation values $\eta_i^t$. Successive application of different $\eta_i$ then gives our non-linear invertible transformation $\tau$ from some prior distribution over residuals $q(\zeta)$ with explicit density and CDF (for sampling) to our target distribution.
## $\mathbb{NIR}$
By conditioning $\tau$ on respective classproxies $\rho$, we can induce a new sample representation distribution $p(\tau(\zeta|\rho)|\rho)$ as pushforward from our prior distribution of residuals $q(\zeta)$ which accounts for unique sample-to-proxy relations, and which we wish to impose over our learned sample distribution $p(\psi|\rho)$. This introduces our Non-Isotropy Regularization $\mathbb{NIR}$
In more detail, $\mathbb{NIR}$ can be naturally approached through maximization of the expected log-likelihood $\mathbb{E}_{x, \rho_{y_x}}\left[\log p\left(\psi(x)|\rho_{y_x}\right)\right]$ over sample-proxy pairs $(x, \rho_{y_x})$, but under the constraint that each distribution of samples around a respective proxy, $p(\psi|\rho)$, is a pushforward of $\tau$ from our residual distribution $q(\zeta)$. This gives
$\mathcal{L}_\mathbb{NIR} = -\mathbb{E}_{x, \rho_{y_x}}[\log q\left(\tau^{-1}(\psi(x)|\rho_{y_x})\right) + \log|\det J_{\tau^{-1}}(\tau^{-1}(\psi(x)|\rho_{y_x})|\rho_{y_x})|]$
with Jacobian $J$ for translation $\tau^{-1}$ and proxies $\rho_{y_x}$, where $y_x$ denotes the class of sample $x$. To arrive at above equation, we simply leveraged the change of variables formula
$p(\psi|\rho) = q(\tau^{-1}(\psi|\rho))|\det J_{\tau^{-1}}(\tau^{-1}(\psi|\rho)|\rho)|$
In practice, by setting our prior $q(\zeta)$ to be a standard zero-mean unit-variance normal distribution $\mathcal{N}(0, 1)$, we get
$\mathcal{L}_\mathbb{NIR} = \frac{1}{|\mathcal{B}|}\sum_{(x, \rho_{y_x})\sim\mathcal{B}}\left\Vert \tau^{-1}(\psi(x)|\rho_{y_x})\right\Vert^2_2 - \log |\det J_{\tau^{-1}}(\tau^{-1}(\psi(x)|\rho_{y_x})|\rho_{y_x})|$
i.e. given sample representations $\psi(x)$, we project them onto our residual space $\zeta$ via $\tau^{-1}$ and compute our $\mathbb{NIR}$-objective. By selecting suitable normalizing flows such as GLOW, we make sure that the Jacobian is cheap to compute.
In applying $\mathbb{NIR}$, we effectively enforce that all sample representations around each proxies are defineable through unique translations. This means that our underlying Normalizing Flow retains implicit knowledge about class-internal structures without our overall setup having to ever rely on sample-to-sample relations!
# Experiments
By applying $\mathbb{NIR}$ to various strong proxy-based DML objectives such as ProxyNCA, ProxyAnchor or SoftTriplet, we find consistent improvements in generalization performance across all methods and in particular across all kinds of benchmarks.
In addition to that, these performance improvements come at no relevant impact to training time, as $\mathbb{NIR}$ only operates in the much lower-dimensional representation space. Furthermore, and importantly, convergence speeds are retained and in parts even improved!
Finally, when evaluating $\mathbb{NIR}$-regularized objectives, we find consistent increases in feature space uniformity ($G_2$) and feature variety ($\rho$), reduced overclustering ($\pi_\text{density}$) as well as higher variability in class-cluster sizes $\sigma_\kappa^2$. While the former three are commonly linked to better generalization performance, the latter provides nice additional support on $\mathbb{NIR}$ encouraging more class-specific sample-distributions to be learned!
(c) 2023 Explainable Machine Learning Tübingen Impressum
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# Burgdorf, Sabine
Overview
Works: 8 works in 27 publications in 3 languages and 310 library holdings Author, Contributor, htt QA402.5, 510
Publication Timeline
.
Most widely held works by Sabine Burgdorf
Optimization of polynomials in non-commuting variables by Sabine Burgdorf( )
14 editions published in 2016 in English and held by 277 WorldCat member libraries worldwide
This book presents recent results on positivity and optimization of polynomials in non-commuting variables. Researchers in non-commutative algebraic geometry, control theory, system engineering, optimization, quantum physics and information science will find the unified notation and mixture of algebraic geometry and mathematical programming useful. Theoretical results are matched with algorithmic considerations; several examples and information on how to use NCSOStools open source package to obtain the results provided. Results are presented on detecting the eigenvalue and trace positivity of polynomials in non-commuting variables using Newton chip method and Newton cyclic chip method, relaxations for constrained and unconstrained optimization problems, semidefinite programming formulations of the relaxations and finite convergence of the hierarchies of these relaxations, and the practical efficiency of algorithms
Trace-positive polynomials, sums of hermitian squares and the tracial moment problem by Sabine Burgdorf( Book )
7 editions published in 2011 in English and French and held by 23 WorldCat member libraries worldwide
A polynomial $f$ in non-commuting variables is trace-positive if the trace of $f(\underline{A})$ is positive for all tuples $\underline{A}$ of symmetric matrices of the same size. The investigation of trace-positive polynomials and of the question of when they can be written as a sum of hermitian squares and commutators of polynomials are motivated by their connection to two famous conjectures: The BMV conjecture from statistical quantum mechanics and the embedding conjecture of Alain Connes concerning von Neumann algebras. First, results on the question of when a trace-positive polynomial in two non-commuting variables can be written as a sum of hermitian squares and commutators are presented. For instance, any bivariate trace-positive polynomial of degree at most four has such a representation, whereas this is false in general if the degree is at least six. This is in perfect analogy to Hilbert's results from the commutative context. Further, a partial answer to the Lieb-Seiringer formulation of the BMV conjecture is given by presenting some concrete representations of the polynomials $S_{m,4}(X^2; Y^2)$ as a sum of hermitian squares and commutators. The second part of this work deals with the tracial moment problem. That is, how can one describe sequences of real numbers that are given by tracial moments of a probability measure on symmetric matrices of a fixed size. The truncated tracial moment problem, where one considers only finite sequences, as well as the tracial analog of the $K$-moment problem are also investigated. Several results from the classical moment problem in Functional Analysis can be transferred to this context. For instance, a tracial analog of Haviland's theorem holds: A traciallinear functional $L$ is given by the tracial moments of a positive Borel measure on symmetric matrices of a fixed size s if and only if $L$ takes only positive values on all polynomials which are trace-positive on all tuples of symmetric $s \times s$-matrices. This result uses tracial versions of the results of Fialkow and Nie on positive extensions of truncated sequences. Further, tracial analogs of results of Stochel and of Bayer and Teichmann are given. Defining a tracial Hankel matrix in analogy to the Hankel matrix in the classical moment problem, the results of Curto and Fialkow concerning sequences with Hankel matrices of finite rank or Hankel matrices of finite size which admit a flat extension also hold true in the tracial context. Finally, a relaxation for trace-minimization of polynomials using sums of hermitian squares and commutators is proposed. While this relaxation is not always exact, the tracial analogs of the results of Curto and Fialkow give a sufficient condition for the exactness of this relaxation
Pure states, nonnegative polynomials, and sums of squares by Sabine Burgdorf( )
1 edition published in 2011 in English and held by 2 WorldCat member libraries worldwide
Algorithmic aspects of sums of hermitian squares by Sabine Burgdorf( )
1 edition published in 2012 in English and held by 2 WorldCat member libraries worldwide
Trace-positive polynomials and the quartic tracial moment problem by Sabine Burgdorf( )
1 edition published in 2010 in English and held by 2 WorldCat member libraries worldwide
21. Mainzer Allergie-Workshop by S Rennert( )
1 edition published in 2009 in German and held by 2 WorldCat member libraries worldwide
The tracial moment problem and trace-optimization of polynomials by Sabine Burgdorf( )
1 edition published in 2011 in English and held by 2 WorldCat member libraries worldwide
Optimization of Polynomials in Non-Commuting Variables by Sabine Burgdorf( )
1 edition published in 2016 in English and held by 0 WorldCat member libraries worldwide
This book presents recent results on positivity and optimization of polynomials in non-commuting variables. Researchers in non-commutative algebraic geometry, control theory, system engineering, optimization, quantum physics and information science will find the unified notation and mixture of algebraic geometry and mathematical programming useful. Theoretical results are matched with algorithmic considerations; several examples and information on how to use NCSOStools open source package to obtain the results provided. Results are presented on detecting the eigenvalue and trace positivity of polynomials in non-commuting variables using Newton chip method and Newton cyclic chip method, relaxations for constrained and unconstrained optimization problems, semidefinite programming formulations of the relaxations and finite convergence of the hierarchies of these relaxations, and the practical efficiency of algorithms
Audience Level
0 1 General Special
Related Identities
Covers
Alternative Names
Burgdorf, S.
Languages
English (24)
French (2)
German (1)
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# Binomial Distribution satisfies Marcoff Chain
1. Jul 22, 2013
### ppedro
1. The problem statement
Consider the Binomial Distribution in the form
$P_{N}(m)=\frac{N!}{(\frac{N+m}{2})!(\frac{N-m}{2})!}p^{\frac{N+m}{2}}q^{\frac{N-m}{2}}$
where $p+q=1$, $m$ is the independent variable and $N$ is a parameter.
Show that it satisfies the marcoff chain
$P_{N+1}\left(m\right)=pP_{N}\left(m-1\right)+qP_{N}\left(m+1\right)$
2. The attempt at a solution
I'm trying my solution starting from this:
$pP_{N}\left(m-1\right)+qP_{N}\left(m+1\right)$
$=p\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}p^{\frac{N+m-1}{2}}q^{\frac{N-m+1}{2}}+q\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m-1}{2}}$
$=\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}+\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}$
$=p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}\left(\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}+\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}\right)$
I can't go any further. If you can help I would appreciate.
Last edited: Jul 22, 2013
2. Jul 22, 2013
### TSny
(Edit) See what happens if you multiply by$\frac{N+1}{N+1}$.
3. Jul 22, 2013
### TSny
My previous comment was based on a particular way that I went about it which got to the result. But, on review, I see that multiplying by (N+1)/(N+1) isn't necessary.
I would factor out the $N!$ in your expression.
The important thing is to get the two denominators in your expression to match the denominator in $P_{N+1}(m)$. For example, what could you multiply the numerator and denominator of $\frac{1}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}$ by to get the denominator in $P_{N+1}(m)$?
[By the way, welcome to PF, ppedro!]
4. Jul 23, 2013
### ppedro
Hey TSny! Thanks for your reply. I see what you're suggesting but I'm not being able to compute it. The factorials are not helping me simplify the expression.
5. Jul 23, 2013
### TSny
$P_{N+1}(m)$ has a denominator of $(\frac{N+m+1}{2})!(\frac{N-m+1}{2})!$.
Compare that to your first denominator $(\frac{N+m-1}{2})!(\frac{N-m+1}{2})!$.
To get your denominator to match the denominator in $P_{N+1}(m)$, you've got to somehow transform $(\frac{N+m-1}{2})!$ into $(\frac{N+m+1}{2})!$.
What can you multiply $(\frac{N+m-1}{2})!$ by to produce $(\frac{N+m+1}{2})!$?
6. Jul 23, 2013
### ppedro
$\frac{(N+m+1)!}{(N+m-1)!}=\frac{(N+m+1)(N+m)(N+m-1)!}{(N+m-1)!}=(N+m+1)(N+m)$
7. Jul 23, 2013
### TSny
No. Note that $\frac{N+m+1}{2} = \frac{N+m-1}{2} + 1$.
8. Jul 23, 2013
### ppedro
Ok, I see your point. Thanks!
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This workshop will provide hands-on instruction and exercises covering basic statistical analysis in R. This will cover descriptive statistics, t-tests, linear models, chi-square, clustering, dimensionality reduction, and resampling strategies. We will also cover methods for “tidying” model results for downstream visualization and summarization.
Prerequisites: Familiarity with R is required (including working with data frames, installing/using packages, importing data, and saving results); familiarity with dplyr and ggplot2 packages is highly recommended.
You must complete the basic R setup here prior to class. This includes installing R, RStudio, and the required packages. Please contact one of the instructors prior to class if you are having difficulty with any of the setup. Please bring your laptop and charger cable to class.
Handouts: Download and print out these handouts and bring them to class:
## Our data: NHANES
The data we’re going to work with comes from the National Health and Nutrition Examination Survey (NHANES) program at the CDC. You can read a lot more about NHANES on the CDC’s website or Wikipedia. NHANES is a research program designed to assess the health and nutritional status of adults and children in the United States. The survey is one of the only to combine both survey questions and physical examinations. It began in the 1960s and since 1999 examines a nationally representative sample of about 5,000 people each year. The NHANES interview includes demographic, socioeconomic, dietary, and health-related questions. The physical exam includes medical, dental, and physiological measurements, as well as several standard laboratory tests. NHANES is used to determine the prevalence of major diseases and risk factors for those diseases. NHANES data are also the basis for national standards for measurements like height, weight, and blood pressure. Data from this survey is used in epidemiology studies and health sciences research, which help develop public health policy, direct and design health programs and services, and expand the health knowledge for the Nation.
We are using a small slice of this data. We’re only using a handful of variables from the 2011-2012 survey years on about 5,000 individuals. The CDC uses a sampling strategy to purposefully oversample certain subpopulations like racial minorities. Naive analysis of the original NHANES data can lead to mistaken conclusions because the percentages of people from each racial group in the data are different from general population. The 5,000 individuals here are resampled from the larger NHANES study population to undo these oversampling effects, so you can treat this as if it were a simple random sample from the American population.
You can download the data at the link above. The file is called nhanes.csv. There’s also a data dictionary (filename nhanes_dd.csv) that lists and describes each variable in our NHANES dataset. This table is copied below.
Variable Definition
id A unique sample identifier
Gender Gender (sex) of study participant coded as male or female
Age Age in years at screening of study participant. Note: Subjects 80 years or older were recorded as 80.
Race Reported race of study participant, including non-Hispanic Asian category: Mexican, Hispanic, White, Black, Asian, or Other. Not availale for 2009-10.
Education Educational level of study participant Reported for participants aged 20 years or older. One of 8thGrade, 9-11thGrade, HighSchool, SomeCollege, or CollegeGrad.
MaritalStatus Marital status of study participant. Reported for participants aged 20 years or older. One of Married, Widowed, Divorced, Separated, NeverMarried, or LivePartner (living with partner).
RelationshipStatus Simplification of MaritalStatus, coded as Committed if MaritalStatus is Married or LivePartner, and Single otherwise.
Insured Indicates whether the individual is covered by health insurance.
Income Numerical version of HHIncome derived from the middle income in each category
Poverty A ratio of family income to poverty guidelines. Smaller numbers indicate more poverty
HomeRooms How many rooms are in home of study participant (counting kitchen but not bathroom). 13 rooms = 13 or more rooms.
HomeOwn One of Home, Rent, or Other indicating whether the home of study participant or someone in their family is owned, rented or occupied by some other arrangement.
Work Indicates whether the individual is current working or not.
Weight Weight in kg
Height Standing height in cm. Reported for participants aged 2 years or older.
BMI Body mass index (weight/height2 in kg/m2). Reported for participants aged 2 years or older.
Pulse 60 second pulse rate
BPSys Combined systolic blood pressure reading, following the procedure outlined for BPXSAR.
BPDia Combined diastolic blood pressure reading, following the procedure outlined for BPXDAR.
Testosterone Testerone total (ng/dL). Reported for participants aged 6 years or older. Not available for 2009-2010.
HDLChol Direct HDL cholesterol in mmol/L. Reported for participants aged 6 years or older.
TotChol Total HDL cholesterol in mmol/L. Reported for participants aged 6 years or older.
Diabetes Study participant told by a doctor or health professional that they have diabetes. Reported for participants aged 1 year or older as Yes or No.
DiabetesAge Age of study participant when first told they had diabetes. Reported for participants aged 1 year or older.
nPregnancies How many times participant has been pregnant. Reported for female participants aged 20 years or older.
nBabies How many of participants deliveries resulted in live births. Reported for female participants aged 20 years or older.
SleepHrsNight Self-reported number of hours study participant usually gets at night on weekdays or workdays. Reported for participants aged 16 years and older.
PhysActive Participant does moderate or vigorous-intensity sports, fitness or recreational activities (Yes or No). Reported for participants 12 years or older.
PhysActiveDays Number of days in a typical week that participant does moderate or vigorous-intensity activity. Reported for participants 12 years or older.
AlcoholDay Average number of drinks consumed on days that participant drank alcoholic beverages. Reported for participants aged 18 years or older.
AlcoholYear Estimated number of days over the past year that participant drank alcoholic beverages. Reported for participants aged 18 years or older.
SmokingStatus Smoking status: Current Former or Never.
### Import & inspect
library(dplyr)
If you see a warning that looks like this: Error in library(dplyr) : there is no package called 'dplyr' (or similar with readr), then you don’t have the package installed correctly. See the setup page.
Now, let’s actually load the data. When we load data we assign it to a variable just like any other, and we can choose a name for that data. Since we’re going to be referring to this data a lot, let’s give it a short easy name to type. I’m going to call it nh. Once we’ve loaded it we can type the name of the object itself (nh) to see it printed to the screen.
nh
## # A tibble: 5,000 x 32
## id Gender Age Race Education MaritalStatus
## <int> <chr> <int> <chr> <chr> <chr>
## 1 62163 male 14 Asian <NA> <NA>
## 2 62172 female 43 Black High School NeverMarried
## 3 62174 male 80 White College Grad Married
## 4 62174 male 80 White College Grad Married
## 5 62175 male 5 White <NA> <NA>
## 6 62176 female 34 White College Grad Married
## 7 62178 male 80 White High School Widowed
## 8 62180 male 35 White College Grad Married
## 9 62186 female 17 Black <NA> <NA>
## 10 62190 female 15 Mexican <NA> <NA>
## # ... with 4,990 more rows, and 26 more variables:
## # RelationshipStatus <chr>, Insured <chr>, Income <int>, Poverty <dbl>,
## # HomeRooms <int>, HomeOwn <chr>, Work <chr>, Weight <dbl>,
## # Height <dbl>, BMI <dbl>, Pulse <int>, BPSys <int>, BPDia <int>,
## # Testosterone <dbl>, HDLChol <dbl>, TotChol <dbl>, Diabetes <chr>,
## # DiabetesAge <int>, nPregnancies <int>, nBabies <int>,
## # SleepHrsNight <int>, PhysActive <chr>, PhysActiveDays <int>,
## # AlcoholDay <int>, AlcoholYear <int>, SmokingStatus <chr>
Take a look at that output. The nice thing about loading dplyr and reading data with readr functions is that data are displayed in a much more friendly way. This dataset has 5,000 rows and 32 columns. When you import/convert data this way and try to display the object in the console, instead of trying to display all 5,000 rows, you’ll only see about 10 by default. Also, if you have so many columns that the data would wrap off the edge of your screen, those columns will not be displayed, but you’ll see at the bottom of the output which, if any, columns were hidden from view.
A note on characters versus factors: One thing that you immediately notice is that all the categorical variables are read in as character data types. This data type is used for storing strings of text, for example, IDs, names, descriptive text, etc. There’s another related data type called factors. Factor variables are used to represent categorical variables with two or more levels, e.g., “male” or “female” for Gender, or “Single” versus “Committed” for RelationshipStatus. For the most part, statistical analysis treats these two data types the same. It’s often easier to leave categorical variables as characters. However, in some cases you may get a warning message alerting you that a character variable was converted into a factor variable during analysis. Generally, these warnings are nothing to worry about. You can, if you like, convert individual variables to factor variables, or simply use dplyr’s mutate_if to convert all character vectors to factor variables:
nh <- nh %>% mutate_if(is.character, as.factor)
nh
Now just take a look at just a few columns that are now factors. Remember, you can look at individual variables with the mydataframe$specificVariable syntax. nh$RelationshipStatus
nh$Race levels(nh$Race)
If you want to see the whole dataset, there are two ways to do this. First, you can click on the name of the data.frame in the Environment panel in RStudio. Or you could use the View() function (with a capital V).
View(nh)
Recall several built-in functions that are useful for working with data frames.
• Content:
• head(): shows the first few rows
• tail(): shows the last few rows
• Size:
• dim(): returns a 2-element vector with the number of rows in the first element, and the number of columns as the second element (the dimensions of the object)
• nrow(): returns the number of rows
• ncol(): returns the number of columns
• Summary:
• colnames() (or just names()): returns the column names
• glimpse() (from dplyr): Returns a glimpse of your data, telling you the structure of the dataset and information about the class, length and content of each column
tail(nh)
dim(nh)
names(nh)
glimpse(nh)
## Descriptive statistics
We can access individual variables within a data frame using the $operator, e.g., mydataframe$specificVariable. Let’s print out all the Race values in the data. Let’s then see what are the unique values of each. Then let’s calculate the mean, median, and range of the Age variable.
# Display all Race values
nh$Race # Get the unique values of Race unique(nh$Race)
length(unique(nh$Race)) # Do the same thing the dplyr way nh$Race %>% unique()
nh$Race %>% unique() %>% length() # Age mean, median, range mean(nh$Age)
median(nh$Age) range(nh$Age)
You could also do the last few operations using dplyr, but remember, this returns a single-row, single-column tibble, not a single scalar value like the above. This is only really useful in the context of grouping and summarizing.
# Compute the mean age
nh %>%
summarize(mean(Age))
# Now grouped by other variables
nh %>%
group_by(Gender, Race) %>%
summarize(mean(Age))
The summary() function (note, this is different from dplyr’s summarize()) works differently depending on which kind of object you pass to it. If you run summary() on a data frame, you get some very basic summary statistics on each variable in the data.
summary(nh)
### Missing data
Let’s try taking the mean of a different variable, either the dplyr way or the simpler $way. # the dplyr way: returns a single-row single-column tibble/dataframe nh %>% summarize(mean(Income)) # returns a single value mean(nh$Income)
What happened there? NA indicates missing data. Take a look at the Income variable.
# Look at just the Income variable
nh$Income # Or view the dataset # View(nh) Notice that there are lots of missing values for Income. Trying to get the mean a bunch of observations with some missing data returns a missing value by default. This is almost universally the case with all summary statistics – a single NA will cause the summary to return NA. Now look at the help for ?mean. Notice the na.rm argument. This is a logical (i.e., TRUE or FALSE) value indicating whether or not missing values should be removed prior to computing the mean. By default, it’s set to FALSE. Now try it again. mean(nh$Income, na.rm=TRUE)
## [1] 57077.66
The is.na() function tells you if a value is missing. Get the sum() of that vector, which adds up all the TRUEs to tell you how many of the values are missing.
is.na(nh$Income) sum(is.na(nh$Income))
There are a few handy functions in the Tmisc package for summarizing missingness in a data frame. You can graphically display missingness in a data frame as holes on a black canvas with gg_na() (use ggplot2 to plot NA values), or show a table of all the variables and the missingness level with propmiss().
# Install if you don't have it already
# install.packages("Tmisc")
library(Tmisc)
gg_na(nh)
propmiss(nh)
## var nmiss n propmiss
## 1 id 0 5000 0.0000
## 2 Gender 0 5000 0.0000
## 3 Age 0 5000 0.0000
## 4 Race 0 5000 0.0000
## 5 Education 1416 5000 0.2832
## 6 MaritalStatus 1415 5000 0.2830
## 7 RelationshipStatus 1415 5000 0.2830
## 8 Insured 7 5000 0.0014
## 9 Income 377 5000 0.0754
## 10 Poverty 325 5000 0.0650
## 11 HomeRooms 28 5000 0.0056
## 12 HomeOwn 28 5000 0.0056
## 13 Work 1158 5000 0.2316
## 14 Weight 31 5000 0.0062
## 15 Height 159 5000 0.0318
## 16 BMI 166 5000 0.0332
## 17 Pulse 718 5000 0.1436
## 18 BPSys 719 5000 0.1438
## 19 BPDia 719 5000 0.1438
## 20 Testosterone 874 5000 0.1748
## 21 HDLChol 775 5000 0.1550
## 22 TotChol 775 5000 0.1550
## 23 Diabetes 64 5000 0.0128
## 24 DiabetesAge 4693 5000 0.9386
## 25 nPregnancies 3735 5000 0.7470
## 26 nBabies 3832 5000 0.7664
## 27 SleepHrsNight 1166 5000 0.2332
## 28 PhysActive 850 5000 0.1700
## 29 PhysActiveDays 2614 5000 0.5228
## 30 AlcoholDay 2503 5000 0.5006
## 31 AlcoholYear 2016 5000 0.4032
## 32 SmokingStatus 1413 5000 0.2826
Now, let’s talk about exploratory data analysis (EDA).
### EDA
It’s always worth examining your data visually before you start any statistical analysis or hypothesis testing. We could spend an entire day on exploratory data analysis. The data visualization lesson covers this in much broader detail. Here we’ll just mention a few of the big ones: histograms and scatterplots.
#### Histograms
We can learn a lot from the data just looking at the value distributions of particular variables. Let’s make some histograms with ggplot2. Looking at BMI shows a few extreme outliers. Looking at weight initially shows us that the units are probably in kg. Replotting that in lbs with more bins shows a clear bimodal distribution. Are there kids in this data? The age distribution shows us the answer is yes.
library(ggplot2)
ggplot(nh, aes(BMI)) + geom_histogram(bins=30)
ggplot(nh, aes(Weight)) + geom_histogram(bins=30)
# In pounds, more bins
ggplot(nh, aes(Weight*2.2)) + geom_histogram(bins=80)
ggplot(nh, aes(Age)) + geom_histogram(bins=30)
#### Scatterplots
Let’s look at how a few different variables relate to each other. E.g., height and weight:
ggplot(nh, aes(Height, Weight, col=Gender)) + geom_point()
Let’s filter out all the kids, draw trend lines using a linear model:
nh %>%
filter(Age>=18) %>%
ggplot(aes(Height, Weight, col=Gender)) +
geom_point() +
geom_smooth(method="lm")
Check out the data visualization lesson for much more on this topic.
### Exercise set 1
1. What’s the mean 60-second pulse rate for all participants in the data?
## [1] 73.63382
1. What’s the range of values for diastolic blood pressure in all participants? (Hint: see help for min(), max(), and range() functions, e.g., enter ?range without the parentheses to get help).
## [1] 0 116
1. What are the median, lower, and upper quartiles for the age of all participants? (Hint: see help for median, or better yet, quantile).
## 0% 25% 50% 75% 100%
## 0 17 36 54 80
1. What’s the variance and standard deviation for income among all participants?
## [1] 1121564068
## [1] 33489.76
## Continuous variables
### T-tests
First let’s create a new dataset from nh called nha that only has adults. To prevent us from making any mistakes downstream, let’s remove the nh object.
nha <- filter(nh, Age>=18)
rm(nh)
# View(nha)
Let’s do a few two-sample t-tests to test for differences in means between two groups. The function for a t-test is t.test(). See the help for ?t.test. We’ll be using the forumla method. The usage is t.test(response~group, data=myDataFrame).
1. Are there differences in age for males versus females in this dataset?
2. Does BMI differ between diabetics and non-diabetics?
3. Do single or married/cohabitating people drink more alcohol? Is this relationship significant?
t.test(Age~Gender, data=nha)
##
## Welch Two Sample t-test
##
## data: Age by Gender
## t = 1.9122, df = 3697.2, p-value = 0.05593
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.02776814 2.22191122
## sample estimates:
## mean in group female mean in group male
## 47.06412 45.96704
t.test(BMI~Diabetes, data=nha)
##
## Welch Two Sample t-test
##
## data: BMI by Diabetes
## t = -11.379, df = 407.31, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -5.563596 -3.924435
## sample estimates:
## mean in group No mean in group Yes
## 28.08753 32.83155
t.test(AlcoholYear~RelationshipStatus, data=nha)
##
## Welch Two Sample t-test
##
## data: AlcoholYear by RelationshipStatus
## t = 5.4315, df = 2674.8, p-value = 6.09e-08
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 13.05949 27.81603
## sample estimates:
## mean in group Committed mean in group Single
## 83.93416 63.49640
See the heading, Welch Two Sample t-test, and notice that the degrees of freedom might not be what we expected based on our sample size. Now look at the help for ?t.test again, and look at the var.equal argument, which is by default set to FALSE. One of the assumptions of the t-test is homoscedasticity, or homogeneity of variance. This assumes that the variance in the outcome (e.g., BMI) is identical across both levels of the predictor (diabetic vs non-diabetic). Since this is rarely the case, the t-test defaults to using the Welch correction, which is a more reliable version of the t-test when the homoscedasticity assumption is violated.
A note on one-tailed versus two-tailed tests: A two-tailed test is almost always more appropriate. The hypothesis you’re testing here is spelled out in the results (“alternative hypothesis: true difference in means is not equal to 0”). If the p-value is very low, you can reject the null hypothesis that there’s no difference in means. Because you typically don’t know a priori whether the difference in means will be positive or negative (e.g., we don’t know a priori whether Single people would be expected to drink more or less than those in a committed relationship), we want to do the two-tailed test. However, if we only wanted to test a very specific directionality of effect, we could use a one-tailed test and specify which direction we expect. This is more powerful if we “get it right”, but much less powerful for the opposite effect. Notice how the p-value changes depending on how we specify the hypothesis. Again, the two-tailed test is almost always more appropriate.
# Two tailed
t.test(AlcoholYear~RelationshipStatus, data=nha)
# Difference in means is >0 (committed drink more)
t.test(AlcoholYear~RelationshipStatus, data=nha, alternative="greater")
# Difference in means is <0 (committed drink less)
t.test(AlcoholYear~RelationshipStatus, data=nha, alternative="less")
A note on paired versus unpaired t-tests: The t-test we performed here was an unpaired test. Here the males and females are different people. The diabetics and nondiabetics are different samples. The single and committed individuals are completely independent, separate observations. In this case, an unpaired test is appropriate. An alternative design might be when data is derived from samples who have been measured at two different time points or locations, e.g., before versus after treatment, left versus right hand, etc. In this case, a paired t-test would be more appropriate. A paired test takes into consideration the intra and inter-subject variability, and is more powerful than the unpaired test. See the help for ?t.test for more information on how to do this.
### Wilcoxon test
Another assumption of the t-test is that data is normally distributed. Looking at the histogram for AlcoholYear shows that this data clearly isn’t.
ggplot(nha, aes(AlcoholYear)) + geom_histogram()
The Wilcoxon rank-sum test (a.k.a. Mann-Whitney U test) is a nonparametric test of differences in mean that does not require normally distributed data. When data is perfectly normal, the t-test is uniformly more powerful. But when this assumption is violated, the t-test is unreliable. This test is called in a similar way as the t-test.
wilcox.test(AlcoholYear~RelationshipStatus, data=nha)
##
## Wilcoxon rank sum test with continuity correction
##
## data: AlcoholYear by RelationshipStatus
## W = 1068000, p-value = 0.0001659
## alternative hypothesis: true location shift is not equal to 0
The results are still significant, but much less than the p-value reported for the (incorrect) t-test above. Also note in the help for ?wilcox.test that there’s a paired option here too.
### Linear models
(See slides)
Analysis of variance and linear modeling are complex topics that deserve an entire semester dedicated to theory, design, and interpretation. A very good resource is An Introduction to Statistical Learning: with Applications in R by Gareth James, Daniela Witten, Trevor Hastie and Robert Tibshirani. The PDF of the book and all the R code used throughout are available free on the author’s website. What follows is a necessary over-simplification with more focus on implementation, and less on theory and design.
Where t-tests and their nonparametric substitutes are used for assessing the differences in means between two groups, ANOVA is used to assess the significance of differences in means between multiple groups. In fact, a t-test is just a specific case of ANOVA when you only have two groups. And both t-tests and ANOVA are just specific cases of linear regression, where you’re trying to fit a model describing how a continuous outcome (e.g., BMI) changes with some predictor variable (e.g., diabetic status, race, age, etc.). The distinction is largely semantic – with a linear model you’re asking, “do levels of a categorical variable affect the response?” where with ANOVA or t-tests you’re asking, “does the mean response differ between levels of a categorical variable?”
Let’s examine the relationship between BMI and relationship status (RelationshipStatus was derived from MaritalStatus, coded as Committed if MaritalStatus is Married or LivePartner, and Single otherwise). Let’s first do this with a t-test, and for now, let’s assume that the variances between groups are equal.
t.test(BMI~RelationshipStatus, data=nha, var.equal=TRUE)
##
## Two Sample t-test
##
## data: BMI by RelationshipStatus
## t = -1.5319, df = 3552, p-value = 0.1256
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.77817842 0.09552936
## sample estimates:
## mean in group Committed mean in group Single
## 28.51343 28.85475
It looks like single people have a very slightly higher BMI than those in a committed relationship, but the magnitude of the difference is trivial, and the difference is not significant. Now, let’s do the same test in a linear modeling framework. First, let’s create the fitted model and store it in an object called fit.
fit <- lm(BMI~RelationshipStatus, data=nha)
You can display the object itself, but that isn’t too interesting. You can get the more familiar ANOVA table by calling the anova() function on the fit object. More generally, the summary() function on a linear model object will tell you much more. (Note this is different from dplyr’s summarize function).
fit
##
## Call:
## lm(formula = BMI ~ RelationshipStatus, data = nha)
##
## Coefficients:
## (Intercept) RelationshipStatusSingle
## 28.5134 0.3413
anova(fit)
## Analysis of Variance Table
##
## Response: BMI
## Df Sum Sq Mean Sq F value Pr(>F)
## RelationshipStatus 1 98 98.320 2.3467 0.1256
## Residuals 3552 148819 41.897
summary(fit)
##
## Call:
## lm(formula = BMI ~ RelationshipStatus, data = nha)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.813 -4.613 -0.955 3.287 52.087
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 28.5134 0.1388 205.440 <2e-16
## RelationshipStatusSingle 0.3413 0.2228 1.532 0.126
##
## Residual standard error: 6.473 on 3552 degrees of freedom
## (153 observations deleted due to missingness)
## Multiple R-squared: 0.0006602, Adjusted R-squared: 0.0003789
## F-statistic: 2.347 on 1 and 3552 DF, p-value: 0.1256
Go back and re-run the t-test assuming equal variances as we did before. Now notice a few things:
t.test(BMI~RelationshipStatus, data=nha, var.equal=TRUE)
1. The p-values from all three tests (t-test, ANOVA, and linear regression) are all identical (p=0.1256). This is because they’re all identical: a t-test is a specific case of ANOVA, which is a specific case of linear regression. There may be some rounding error, but we’ll talk about extracting the exact values from a model object later on.
2. The test statistics are all related. The t statistic from the t-test is 1.532, which is the same as the t-statistic from the linear regression. If you square that, you get 2.347, the F statistic from the ANOVA.
3. The t.test() output shows you the means for the two groups, Committed and Single. Just displaying the fit object itself or running summary(fit) shows you the coefficients for a linear model. Here, the model assumes the “baseline” RelationshipStatus level is Committed, and that the intercept in a regression model (e.g., $$\beta_{0}$$ in the model $$Y = \beta_{0} + \beta_{1}X$$) is the mean of the baseline group. Being Single results in an increase in BMI of 0.3413. This is the $$\beta_{1}$$ coefficient in the model. You can easily change the ordering of the levels. See the help for ?factor, and check out the new forcats package, which provides tools for manipulating categorical variables.
# P-value computed on a t-statistic with 3552 degrees of freedom
# (multiply times 2 because t-test is assuming two-tailed)
2*(1-pt(1.532, df=3552))
## [1] 0.1256115
# P-value computed on an F-test with 1 and 3552 degrees of freedom
1-pf(2.347, df1=1, df2=3552)
## [1] 0.1256134
A note on dummy coding: If you have a $$k$$-level factor, R creates $$k-1$$ dummy variables, or indicator variables, by default, using the alphabetically first level as baseline. For example, the levels of RelationshipStatus are “Committed” and “Single”. R creates a dummy variable called “RelationshipStatusSingle” that’s 0 if you’re committed, and 1 if you’re Single. The linear model is saying for every unit increase in RelationshipStatusSingle, i.e., going from committed to single, results in a 0.314-unit increase in BMI. You can change the ordering of the factors to change the interpretation of the model (e.g., treating Single as baseline and going from Single to Committed). We’ll do this in the next section.
### ANOVA
Recap: t-tests are for assessing the differences in means between two groups. A t-test is a specific case of ANOVA, which is a specific case of a linear model. Let’s run ANOVA, but this time looking for differences in means between more than two groups.
Let’s look at the relationship between smoking status (Never, Former, or Current), and BMI.
fit <- lm(BMI~SmokingStatus, data=nha)
anova(fit)
## Analysis of Variance Table
##
## Response: BMI
## Df Sum Sq Mean Sq F value Pr(>F)
## SmokingStatus 2 1411 705.50 16.988 4.54e-08
## Residuals 3553 147551 41.53
summary(fit)
##
## Call:
## lm(formula = BMI ~ SmokingStatus, data = nha)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.565 -4.556 -1.056 3.315 51.744
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 27.3914 0.2446 111.975 < 2e-16
## SmokingStatusFormer 1.7736 0.3293 5.387 7.65e-08
## SmokingStatusNever 1.4645 0.2838 5.161 2.60e-07
##
## Residual standard error: 6.444 on 3553 degrees of freedom
## (151 observations deleted due to missingness)
## Multiple R-squared: 0.009472, Adjusted R-squared: 0.008915
## F-statistic: 16.99 on 2 and 3553 DF, p-value: 4.54e-08
The F-test on the ANOVA table tells us that there is a significant difference in means between current, former, and never smokers (p=$$4.54 \times 10^{-8}$$). However, the linear model output might not have been what we wanted. Because the default handling of categorical variables is to treat the alphabetical first level as the baseline, “Current” smokers are treated as baseline, and this mean becomes the intercept, and the coefficients on “Former” and “Never” describe how those groups’ means differ from current smokers.
Back to dummy coding / indicator variables: SmokingStatus is “Current”, “Former”, and “Never.” By default, R will create two indicator variables here that in tandem will explain this variable.
Original SmokingStatus Indicator: SmokingStatusFormer Indicator: SmokingStatusNever
Current 0 0
Former 1 0
Never 0 1
What if we wanted “Never” smokers to be the baseline, followed by Former, then Current? Have a look at ?factor to relevel the factor levels.
# Look at nha$SmokingStatus nha$SmokingStatus
# What happens if we relevel it? Let's see what that looks like.
relevel(nha$SmokingStatus, ref="Never") # If we're happy with that, let's change the value of nha$SmokingStatus in place
nha$SmokingStatus <- relevel(nha$SmokingStatus, ref="Never")
# Or we could do this the dplyr way
nha <- nha %>%
mutate(SmokingStatus=relevel(SmokingStatus, ref="Never"))
# Re-fit the model
fit <- lm(BMI~SmokingStatus, data=nha)
# Optionally, show the ANOVA table
# anova(fit)
# Print the full model statistics
summary(fit)
##
## Call:
## lm(formula = BMI ~ SmokingStatus, data = nha)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.565 -4.556 -1.056 3.315 51.744
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 28.8558 0.1438 200.601 < 2e-16
## SmokingStatusCurrent -1.4645 0.2838 -5.161 2.6e-07
## SmokingStatusFormer 0.3091 0.2632 1.175 0.24
##
## Residual standard error: 6.444 on 3553 degrees of freedom
## (151 observations deleted due to missingness)
## Multiple R-squared: 0.009472, Adjusted R-squared: 0.008915
## F-statistic: 16.99 on 2 and 3553 DF, p-value: 4.54e-08
Notice that the p-value on the ANOVA/regression didn’t change, but the coefficients did. Never smokers are now treated as baseline. The intercept coefficient (28.856) is now the mean for Never smokers. The SmokingStatusFormer coefficient of .309 shows the apparent increase in BMI that former smokers have when compared to never smokers, but that difference is not significant (p=.24). The SmokingStatusCurrent coefficient of -1.464 shows that current smokers actually have a lower BMI than never smokers, and that this decrease is highly significant.
Finally, you can do the typical post-hoc ANOVA procedures on the fit object. For example, the TukeyHSD() function will run Tukey’s test (also known as Tukey’s range test, the Tukey method, Tukey’s honest significance test, Tukey’s HSD test (honest significant difference), or the Tukey-Kramer method). Tukey’s test computes all pairwise mean difference calculation, comparing each group to each other group, identifying any difference between two groups that’s greater than the standard error, while controlling the type I error for all multiple comparisons. First run aov() (not anova()) on the fitted linear model object, then run TukeyHSD() on the resulting analysis of variance fit.
TukeyHSD(aov(fit))
## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = fit)
##
## $SmokingStatus ## diff lwr upr p adj ## Current-Never -1.4644502 -2.1298249 -0.7990756 0.0000008 ## Former-Never 0.3091076 -0.3079639 0.9261790 0.4685044 ## Former-Current 1.7735578 1.0015525 2.5455631 0.0000002 This shows that there isn’t much of a difference between former and never smokers, but that both of these differ significantly from current smokers, who have significantly lower BMI. Finally, let’s visualize the differences in means between these groups. The NA category, which is omitted from the ANOVA, contains all the observations who have missing or non-recorded Smoking Status. ggplot(nha, aes(SmokingStatus, BMI)) + geom_boxplot() + theme_classic() ### Linear regression (See slides) Linear models are mathematical representations of the process that (we think) gave rise to our data. The model seeks to explain the relationship between a variable of interest, our Y, outcome, response, or dependent variable, and one or more X, predictor, or independent variables. Previously we talked about t-tests or ANOVA in the context of a simple linear regression model with only a single predictor variable, $$X$$: $Y = \beta_{0} + \beta_{1}X$ But you can have multiple predictors in a linear model that are all additive, accounting for the effects of the others: $Y = \beta_{0} + \beta_{1}X_{1} + \beta_{2}X_{2} + \epsilon$ • $$Y$$ is the response • $$X_{1}$$ and $$X_{2}$$ are the predictors • $$\beta_{0}$$ is the intercept, and $$\beta_{1}$$, $$\beta_{2}$$ etc are coefficients that describe what 1-unit changes in $$X_{1}$$ and $$X_{2}$$ do to the outcome variable $$Y$$. • $$\epsilon$$ is random error. Our model will not perfectly predict $$Y$$. It will be off by some random amount. We assume this amount is a random draw from a Normal distribution with mean 0 and standard deviation $$\sigma$$. Building a linear model means we propose a linear model and then estimate the coefficients and the variance of the error term. Above, this means estimating $$\beta_{0}, \beta_{1}, \beta_{2}$$ and $$\sigma$$. This is what we do in R. Let’s look at the relationship between height and weight. fit <- lm(Weight~Height, data=nha) summary(fit) ## ## Call: ## lm(formula = Weight ~ Height, data = nha) ## ## Residuals: ## Min 1Q Median 3Q Max ## -40.339 -13.109 -2.658 9.309 127.972 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -73.70590 5.08110 -14.51 <2e-16 ## Height 0.91996 0.03003 30.63 <2e-16 ## ## Residual standard error: 18.57 on 3674 degrees of freedom ## (31 observations deleted due to missingness) ## Multiple R-squared: 0.2034, Adjusted R-squared: 0.2032 ## F-statistic: 938.4 on 1 and 3674 DF, p-value: < 2.2e-16 The relationship is highly significant (P<$$2.2 \times 10^{-16}$$). The intercept term is not very useful most of the time. Here it shows us what the value of Weight would be when Height=0, which could never happen. The Height coefficient is meaningful – each one unit increase in height results in a 0.92 increase in the corresponding unit of weight. Let’s visualize that relationship: ggplot(nha, aes(x=Height, y=Weight)) + geom_point() + geom_smooth(method="lm") By default, this is only going to show the prediction over the range of the data. This is important! You never want to try to extrapolate response variables outside of the range of your predictor(s). For example, the linear model tells us that weight is -73.7kg when height is zero. We can extend the predicted model / regression line past the lowest value of the data down to height=0. The bands on the confidence interval tell us that the model is apparently confident within the regions defined by the gray boundary. But this is silly – we would never see a height of zero, and predicting past the range of the available training data is never a good idea. ggplot(nha, aes(x=Height, y=Weight)) + geom_point() + geom_smooth(method="lm", fullrange=TRUE) + xlim(0, NA) + ggtitle("Friends don't let friends extrapolate.") ### Multiple regression Finally, let’s do a multiple linear regression analysis, where we attempt to model the effect of multiple predictor variables at once on some outcome. First, let’s look at the effect of physical activity on testosterone levels. Let’s do this with a t-test and linear regression, showing that you get the same results. t.test(Testosterone~PhysActive, data=nha, var.equal=TRUE) ## ## Two Sample t-test ## ## data: Testosterone by PhysActive ## t = -2.4298, df = 3436, p-value = 0.01516 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -34.813866 -3.720171 ## sample estimates: ## mean in group No mean in group Yes ## 207.5645 226.8315 summary(lm(Testosterone~PhysActive, data=nha)) ## ## Call: ## lm(formula = Testosterone ~ PhysActive, data = nha) ## ## Residuals: ## Min 1Q Median 3Q Max ## -224.5 -196.5 -115.9 167.0 1588.0 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 207.565 5.873 35.34 <2e-16 ## PhysActiveYes 19.267 7.929 2.43 0.0152 ## ## Residual standard error: 231.4 on 3436 degrees of freedom ## (269 observations deleted due to missingness) ## Multiple R-squared: 0.001715, Adjusted R-squared: 0.001425 ## F-statistic: 5.904 on 1 and 3436 DF, p-value: 0.01516 In both cases, the p-value is significant (p=0.01516), and the result suggest that increased physical activity is associated with increased testosterone levels. Does increasing your physical activity increase your testosterone levels? Or is it the other way – will increased testosterone encourage more physical activity? Or is it none of the above – is the apparent relationship between physical activity and testosterone levels only apparent because both are correlated with yet a third, unaccounted for variable? Let’s throw Age into the model as well. summary(lm(Testosterone~PhysActive+Age, data=nha)) ## ## Call: ## lm(formula = Testosterone ~ PhysActive + Age, data = nha) ## ## Residuals: ## Min 1Q Median 3Q Max ## -238.6 -196.8 -112.3 167.4 1598.1 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 247.8828 13.0853 18.944 < 2e-16 ## PhysActiveYes 13.6740 8.0815 1.692 0.090735 ## Age -0.8003 0.2322 -3.447 0.000574 ## ## Residual standard error: 231 on 3435 degrees of freedom ## (269 observations deleted due to missingness) ## Multiple R-squared: 0.005156, Adjusted R-squared: 0.004577 ## F-statistic: 8.901 on 2 and 3435 DF, p-value: 0.0001394 This shows us that after accounting for age that the testosterone / physical activity link is no longer significant. Every 1-year increase in age results in a highly significant decrease in testosterone, and since increasing age is also likely associated with decreased physical activity, perhaps age is the confounder that makes this relationship apparent. Adding other predictors can also swing things the other way. We know that men have much higher testosterone levels than females. Sex is probably the single best predictor of testosterone levels in our dataset. By not accounting for this effect, our unaccounted-for variation remains very high. By accounting for Gender, we now reduce the residual error in the model, and the physical activity effect once again becomes significant. Also notice that our model fits much better (higher R-squared), and is much more significant overall. summary(lm(Testosterone ~ PhysActive+Age+Gender, data=nha)) ## ## Call: ## lm(formula = Testosterone ~ PhysActive + Age + Gender, data = nha) ## ## Residuals: ## Min 1Q Median 3Q Max ## -397.91 -31.01 -4.42 20.50 1400.90 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 46.6931 7.5729 6.166 7.83e-10 ## PhysActiveYes 9.2749 4.4617 2.079 0.0377 ## Age -0.5904 0.1282 -4.605 4.28e-06 ## Gendermale 385.1989 4.3512 88.526 < 2e-16 ## ## Residual standard error: 127.5 on 3434 degrees of freedom ## (269 observations deleted due to missingness) ## Multiple R-squared: 0.6969, Adjusted R-squared: 0.6966 ## F-statistic: 2632 on 3 and 3434 DF, p-value: < 2.2e-16 We’ve only looked at the summary() and anova() functions for extracting information from an lm class object. There are several other accessor functions that can be used on a linear model object. Check out the help page for each one of these to learn more. ### Exercise set 2 1. Is the average BMI different in single people versus those in a committed relationship? Perform a t-test. 2. The Work variable is coded “Looking” (n=159), “NotWorking” (n=1317), and “Working” (n=2230). • Fit a linear model of Income against Work. Assign this to an object called fit. What does the fit object tell you when you display it directly? • Run an anova() to get the ANOVA table. Is the model significant? • Run a Tukey test to get the pairwise contrasts. (Hint: TukeyHSD() on aov() on the fit). What do you conclude? • Instead of thinking of this as ANOVA, think of it as a linear model. After you’ve thought about it, get some summary() statistics on the fit. Do these results jive with the ANOVA model? 3. Examine the relationship between HDL cholesterol levels (HDLChol) and whether someone has diabetes or not (Diabetes). • Is there a difference in means between diabetics and nondiabetics? Perform a t-test without a Welch correction (that is, assuming equal variances – see ?t.test for help). • Do the same analysis in a linear modeling framework. • Does the relationship hold when adjusting for Weight? • What about when adjusting for Weight, Age, Gender, PhysActive (whether someone participates in moderate or vigorous-intensity sports, fitness or recreational activities, coded as yes/no). What is the effect of each of these explanatory variables? ## Discrete variables Until now we’ve only discussed analyzing continuous outcomes / dependent variables. We’ve tested for differences in means between two groups with t-tests, differences among means between n groups with ANOVA, and more general relationships using linear regression. In all of these cases, the dependent variable, i.e., the outcome, or $$Y$$ variable, was continuous, and usually normally distributed. What if our outcome variable is discrete, e.g., “Yes/No”, “Mutant/WT”, “Case/Control”, etc.? Here we use a different set of procedures for assessing significant associations. ### Contingency tables The xtabs() function is useful for creating contingency tables from categorical variables. Let’s create a gender by diabetes status contingency table, and assign it to an object called xt. After making the assignment, type the name of the object to view it. xt <- xtabs(~Gender+Diabetes, data=nha) xt ## Diabetes ## Gender No Yes ## female 1692 164 ## male 1653 198 There are two useful functions, addmargins() and prop.table() that add more information or manipulate how the data is displayed. By default, prop.table() will divide the number of observations in each cell by the total. But you may want to specify which margin you want to get proportions over. Let’s do this for the first (row) margin. # Add marginal totals addmargins(xt) ## Diabetes ## Gender No Yes Sum ## female 1692 164 1856 ## male 1653 198 1851 ## Sum 3345 362 3707 # Get the proportional table prop.table(xt) ## Diabetes ## Gender No Yes ## female 0.45643377 0.04424063 ## male 0.44591314 0.05341246 # That wasn't really what we wanted. # Do this over the first (row) margin only. prop.table(xt, margin=1) ## Diabetes ## Gender No Yes ## female 0.91163793 0.08836207 ## male 0.89303079 0.10696921 Looks like men have slightly higher rates of diabetes than women. But is this significant? The chi-square test is used to assess the independence of these two factors. That is, if the null hypothesis that gender and diabetes are independent is true, the we would expect a proportionally equal number of diabetics across each sex. Males seem to be at slightly higher risk than females, but the difference is just short of statistically significant. chisq.test(xt) ## ## Pearson's Chi-squared test with Yates' continuity correction ## ## data: xt ## X-squared = 3.4332, df = 1, p-value = 0.0639 An alternative to the chi-square test is Fisher’s exact test. Rather than relying on a critical value from a theoretical chi-square distribution, Fisher’s exact test calculates the exact probability of observing the contingency table as is. It’s especially useful when there are very small n’s in one or more of the contingency table cells. Both the chi-square and Fisher’s exact test give us p-values of approximately 0.06. fisher.test(xt) ## ## Fisher's Exact Test for Count Data ## ## data: xt ## p-value = 0.05992 ## alternative hypothesis: true odds ratio is not equal to 1 ## 95 percent confidence interval: ## 0.9883143 1.5466373 ## sample estimates: ## odds ratio ## 1.235728 There’s a useful plot for visualizing contingency table data called a mosaic plot. Call the mosaicplot() function on the contingency table object. Note this is a built-in plot, not a ggplot2-style plot. mosaicplot(xt, main=NA) Let’s create a different contingency table, this time looking at the relationship between race and whether the person had health insurance. Display the table with marginal totals. xt <- xtabs(~Race+Insured, data=nha) addmargins(xt) ## Insured ## Race No Yes Sum ## Asian 46 169 215 ## Black 86 330 416 ## Hispanic 89 151 240 ## Mexican 147 141 288 ## Other 33 65 98 ## White 307 2141 2448 ## Sum 708 2997 3705 Let’s do the same thing as above, this time showing the proportion of people in each race category having health insurance. prop.table(xt, margin=1) ## Insured ## Race No Yes ## Asian 0.2139535 0.7860465 ## Black 0.2067308 0.7932692 ## Hispanic 0.3708333 0.6291667 ## Mexican 0.5104167 0.4895833 ## Other 0.3367347 0.6632653 ## White 0.1254085 0.8745915 Now, let’s run a chi-square test for independence. chisq.test(xt) ## ## Pearson's Chi-squared test ## ## data: xt ## X-squared = 323.3, df = 5, p-value < 2.2e-16 The result is highly significant. In fact, so significant, that the display rounds off the p-value to something like $$<2.2 \times 10^{-16}$$. If you look at the help for ?chisq.test you’ll see that displaying the test only shows you summary information, but other components can be accessed. For example, we can easily get the actual p-value, or the expected counts under the null hypothesis of independence. chisq.test(xt)$p.value
## [1] 9.754238e-68
chisq.test(xt)$expected ## Insured ## Race No Yes ## Asian 41.08502 173.91498 ## Black 79.49474 336.50526 ## Hispanic 45.86235 194.13765 ## Mexican 55.03482 232.96518 ## Other 18.72713 79.27287 ## White 467.79595 1980.20405 We can also make a mosaic plot similar to above: mosaicplot(xt, main=NA) ### Logistic regression (See slides) What if we wanted to model the discrete outcome, e.g., whether someone is insured, against several other variables, similar to how we did with multiple linear regression? We can’t use linear regression because the outcome isn’t continuous – it’s binary, either Yes or No. For this we’ll use logistic regression to model the log odds of binary response. That is, instead of modeling the outcome variable, $$Y$$, directly against the inputs, we’ll model the log odds of the outcome variable. If $$p$$ is the probability that the individual is insured, then $$\frac{p}{1-p}$$ is the odds that person is insured. Then it follows that the linear model is expressed as: $log(\frac{p}{1-p}) = \beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k$ Where $$\beta_0$$ is the intercept, $$\beta_1$$ is the increase in the odds of the outcome for every unit increase in $$x_1$$, and so on. Logistic regression is a type of generalized linear model (GLM). We fit GLM models in R using the glm() function. It works like the lm() function except we specify which GLM to fit using the family argument. Logistic regression requires family=binomial. The typical use looks like this: mod <- glm(y ~ x, data=yourdata, family='binomial') summary(mod) Before we fit a logistic regression model let’s relevel the Race variable so that “White” is the baseline. We saw above that people who identify as “White” have the highest rates of being insured. When we run the logistic regression, we’ll get a separate coefficient (effect) for each level of the factor variable(s) in the model, telling you the increased odds that that level has, as compared to the baseline group. #Look at Race. The default ordering is alphabetical nha$Race
# Let's relevel that where the group with the highest rate of insurance is "baseline"
relevel(nha$Race, ref="White") # If we're happy with that result, permanently change it nha$Race <- relevel(nha$Race, ref="White") # Or do it the dplyr way nha <- nha %>% mutate(Race=relevel(Race, ref="White")) Now, let’s fit a logistic regression model assessing how the odds of being insured change with different levels of race. fit <- glm(Insured~Race, data=nha, family="binomial") summary(fit) ## ## Call: ## glm(formula = Insured ~ Race, family = "binomial", data = nha) ## ## Deviance Residuals: ## Min 1Q Median 3Q Max ## -2.0377 0.5177 0.5177 0.5177 1.1952 ## ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) 1.94218 0.06103 31.825 < 2e-16 ## RaceAsian -0.64092 0.17715 -3.618 0.000297 ## RaceBlack -0.59744 0.13558 -4.406 1.05e-05 ## RaceHispanic -1.41354 0.14691 -9.622 < 2e-16 ## RaceMexican -1.98385 0.13274 -14.946 < 2e-16 ## RaceOther -1.26430 0.22229 -5.688 1.29e-08 ## ## (Dispersion parameter for binomial family taken to be 1) ## ## Null deviance: 3614.6 on 3704 degrees of freedom ## Residual deviance: 3336.6 on 3699 degrees of freedom ## (2 observations deleted due to missingness) ## AIC: 3348.6 ## ## Number of Fisher Scoring iterations: 4 The Estimate column shows the log of the odds ratio – how the log odds of having health insurance changes at each level of race compared to White. The P-value for each coefficient is on the far right. This shows that every other race has significantly less rates of health insurance coverage. But, as in our multiple linear regression analysis above, are there other important variables that we’re leaving out that could alter our conclusions? Lets add a few more variables into the model to see if something else can explain the apparent Race-Insured association. Let’s add a few things likely to be involved (Age and Income), and something that’s probably irrelevant (hours slept at night). fit <- glm(Insured ~ Age+Income+SleepHrsNight+Race, data=nha, family="binomial") summary(fit) ## ## Call: ## glm(formula = Insured ~ Age + Income + SleepHrsNight + Race, ## family = "binomial", data = nha) ## ## Deviance Residuals: ## Min 1Q Median 3Q Max ## -2.4815 0.3025 0.4370 0.6252 1.6871 ## ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) -3.501e-01 2.919e-01 -1.199 0.230 ## Age 3.371e-02 2.949e-03 11.431 < 2e-16 ## Income 1.534e-05 1.537e-06 9.982 < 2e-16 ## SleepHrsNight -1.763e-02 3.517e-02 -0.501 0.616 ## RaceAsian -4.550e-01 2.031e-01 -2.241 0.025 ## RaceBlack -2.387e-01 1.536e-01 -1.554 0.120 ## RaceHispanic -1.010e+00 1.635e-01 -6.175 6.61e-10 ## RaceMexican -1.404e+00 1.483e-01 -9.468 < 2e-16 ## RaceOther -9.888e-01 2.422e-01 -4.082 4.46e-05 ## ## (Dispersion parameter for binomial family taken to be 1) ## ## Null deviance: 3284.3 on 3395 degrees of freedom ## Residual deviance: 2815.0 on 3387 degrees of freedom ## (311 observations deleted due to missingness) ## AIC: 2833 ## ## Number of Fisher Scoring iterations: 5 A few things become apparent: 1. Age and income are both highly associated with whether someone is insured. Both of these variables are highly significant ($$P<2.2 \times 10^{-16}$$), and the coefficient (the Estimate column) is positive, meaning that for each unit increase in one of these variables, the odds of being insured increases by the corresponding amount. 2. Hours slept per night is not meaningful at all. 3. After accounting for age and income, several of the race-specific differences are no longer statistically significant, but others remain so. 4. The absolute value of the test statistic (column called z value) can roughly be taken as an estimate of the “importance” of that variable to the overall model. So, age and income are the most important influences in this model; self-identifying as Hispanic or Mexican are also very highly important, hours slept per night isn’t important at all, and the other race categories fall somewhere in between. There is much more to go into with logistic regression. This lesson only scratches the surface. Missing from this lesson are things like regression diagnostics, model comparison approaches, penalization, interpretation of model coefficients, fitting interaction effects, and much more. Alan Agresti’s Categorical Data Analysis has long been considered the definitive text on this topic. I also recommend Agresti’s Introduction to Categorical Data Analysis (a.k.a. “Agresti lite”) for a gentler introduction. ### Exercise set 3 1. What’s the relationship between diabetes and participating in rigorous physical activity or sports? • Create a contingency table with Diabetes status in rows and physical activity status in columns. • Display that table with margins. • Show the proportions of diabetics and nondiabetics, separately, who are physically active or not. • Is this relationship significant? • Create a mosaic plot to visualize the relationship. 2. Model the same association in a logistic regression framework to assess the risk of diabetes using physical activity as a predictor. • Fit a model with just physical activity as a predictor, and display a model summary. • Add gender to the model, and show a summary. • Continue adding weight and age to the model. What happens to the gender association? • Continue and add income to the model. What happens to the original association with physical activity? ## Power & sample size (See slides) This is a necessarily short introduction to the concept of power and sample size calculations. Statistical power, also sometimes called sensitivity, is defined as the probability that your test correctly rejects the null hypothesis when the alternative hypothesis is true. That is, if there really is an effect (difference in means, association between categorical variables, etc.), how likely are you to be able to detect that effect at a given statistical significance level, given certain assumptions. Generally there are a few moving pieces, and if you know all but one of them, you can calculate what that last one is. 1. Power: How likely are you to detect the effect? (Usually like to see 80% or greater). 2. N: What is the sample size you have (or require)? 3. Effect size: How big is the difference in means, odds ratio, etc? If we know we want 80% power to detect a certain magnitude of difference between groups, we can calculate our required sample size. Or, if we know we can only collect 5 samples, we can calculate how likely we are to detect a particular effect. Or, we can work to solve the last one - if we want 80% power and we have 5 samples, what’s the smallest effect we can hope to detect? All of these questions require certain assumptions about the data and the testing procedure. Which kind of test is being performed? What’s the true effect size (often unknown, or estimated from preliminary data), what’s the standard deviation of samples that will be collected (often unknown, or estimated from preliminary data), what’s the level of statistical significance needed (traditionally p<0.05, but must consider multiple testing corrections). ### T-test power/N The power.t.test() empirically estimates power or sample size of a t-test for differences in means. If we have 20 samples in each of two groups (e.g., control versus treatment), and the standard deviation for whatever we’re measuring is 2.3, and we’re expecting a true difference in means between the groups of 2, what’s the power to detect this effect? power.t.test(n=20, delta=2, sd=2.3) ## ## Two-sample t test power calculation ## ## n = 20 ## delta = 2 ## sd = 2.3 ## sig.level = 0.05 ## power = 0.7641668 ## alternative = two.sided ## ## NOTE: n is number in *each* group What’s the sample size we’d need to detect a difference of 0.8 given a standard deviation of 1.5, assuming we want 80% power? power.t.test(power=.80, delta=.8, sd=1.5) ## ## Two-sample t test power calculation ## ## n = 56.16413 ## delta = 0.8 ## sd = 1.5 ## sig.level = 0.05 ## power = 0.8 ## alternative = two.sided ## ## NOTE: n is number in *each* group ### Proportions power/N What about a two-sample proportion test (e.g., chi-square test)? If we have two groups (control and treatment), and we’re measuring some outcome (e.g., infected yes/no), and we know that the proportion of infected controls is 80% but 20% in treated, what’s the power to detect this effect in 5 samples per group? power.prop.test(n=5, p1=0.8, p2=0.2) ## ## Two-sample comparison of proportions power calculation ## ## n = 5 ## p1 = 0.8 ## p2 = 0.2 ## sig.level = 0.05 ## power = 0.4688159 ## alternative = two.sided ## ## NOTE: n is number in *each* group How many samples would we need for 90% power? power.prop.test(power=0.9, p1=0.8, p2=0.2) ## ## Two-sample comparison of proportions power calculation ## ## n = 12.37701 ## p1 = 0.8 ## p2 = 0.2 ## sig.level = 0.05 ## power = 0.9 ## alternative = two.sided ## ## NOTE: n is number in *each* group Also check out the pwr package which has power calculation functions for other statistical tests. Function Power calculations for pwr.2p.test() Two proportions (equal n) pwr.2p2n.test() Two proportions (unequal n) pwr.anova.test() Balanced one way ANOVA pwr.chisq.test() Chi-square test pwr.f2.test() General linear model pwr.p.test() Proportion (one sample) pwr.r.test() Correlation pwr.t.test() T-tests (one sample, 2 sample, paired) pwr.t2n.test() T-test (two samples with unequal n) ### Exercise set 4 1. You’re doing a gene expression experiment. What’s your power to detect a 2-fold change in a gene with a standard deviation of 0.7, given 3 samples? (Note - fold change is usually given on the $$log_2$$ scale, so a 2-fold change would be a delta of 1. That is, if the fold change is 2x, then $$log_2(2)=1$$, and you should use 1 in the calculation, not 2). ## [1] 0.2709095 1. How many samples would you need to have 80% power to detect this effect? ## [1] 8.764711 1. You’re doing a population genome-wide association study (GWAS) looking at the effect of a SNP on disease X. Disease X has a baseline prevalence of 5% in the population, but you suspect the SNP might increase the risk of disease X by 10% (this is typical for SNP effects on common, complex diseases). Calculate the number of samples do you need to have 80% power to detect this effect, given that you want a genome-wide statistical significance of $$p<5\times10^{-8}$$ to account for multiple testing.1 (Hint, you can expressed $$5\times10^{-8}$$ in R using 5e-8 instead of .00000005). ## [1] 157589.5 ## Tidying models (See slides) We spent a lot of time in other lessons on tidy data, where each column is a variable and each row is an observation. Tidy data is easy to filter observations based on values in a column (e.g., we could get just adult males with filter(nha, Gender=="male" & Age>=18), and easy to select particular variables/features of interest by their column name. Even when we start with tidy data, we don’t end up with tidy models. The output from tests like t.test or lm are not data.frames, and it’s difficult to get the information out of the model object that we want. The broom package bridges this gap. Depending on the type of model object you’re using, broom provides three methods that do different kinds of tidying: 1. tidy: constructs a data frame that summarizes the model’s statistical findings like coefficients and p-values. 2. augment: add columns to the original data that was modeled, like predictions and residuals. 3. glance: construct a concise one-row summary of the model with information like $$R^2$$ that are computed once for the entire model. Let’s go back to our linear model example. # Try modeling Testosterone against Physical Activity, Age, and Gender. fit <- lm(Testosterone~PhysActive+Age+Gender, data=nha) # See what that model looks like: summary(fit) ## ## Call: ## lm(formula = Testosterone ~ PhysActive + Age + Gender, data = nha) ## ## Residuals: ## Min 1Q Median 3Q Max ## -397.91 -31.01 -4.42 20.50 1400.90 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 46.6931 7.5729 6.166 7.83e-10 ## PhysActiveYes 9.2749 4.4617 2.079 0.0377 ## Age -0.5904 0.1282 -4.605 4.28e-06 ## Gendermale 385.1989 4.3512 88.526 < 2e-16 ## ## Residual standard error: 127.5 on 3434 degrees of freedom ## (269 observations deleted due to missingness) ## Multiple R-squared: 0.6969, Adjusted R-squared: 0.6966 ## F-statistic: 2632 on 3 and 3434 DF, p-value: < 2.2e-16 What if we wanted to pull out the coefficient for Age, or the P-value for PhysActive? It gets pretty gross. We first have to coef(summary(lmfit)) to get a matrix of coefficients, the terms are still stored in row names, and the column names are inconsistent with other packages (e.g. Pr(>|t|) compared to p.value). Yuck! coef(summary(fit))["Age", "Estimate"] ## [1] -0.5903616 coef(summary(fit))["PhysActiveYes", "Pr(>|t|)"] ## [1] 0.03771185 Instead, you can use the tidy function, from the broom package, on the fit: # Install the package if you don't have it # install.packages("broom") library(broom) tidy(fit) ## term estimate std.error statistic p.value ## 1 (Intercept) 46.6931314 7.5728848 6.165831 7.825833e-10 ## 2 PhysActiveYes 9.2749247 4.4617201 2.078778 3.771185e-02 ## 3 Age -0.5903616 0.1282059 -4.604795 4.278461e-06 ## 4 Gendermale 385.1989414 4.3512332 88.526383 0.000000e+00 This gives you a data.frame with all your model results. The row names have been moved into a column called term, and the column names are simple and consistent (and can be accessed using$). These can be manipulated with dplyr just like any other data frame.
tidy(fit) %>%
filter(term!="(Intercept)") %>%
select(term, p.value) %>%
arrange(p.value)
## term p.value
## 1 Gendermale 0.000000e+00
## 2 Age 4.278461e-06
## 3 PhysActiveYes 3.771185e-02
Instead of viewing the coefficients, you might be interested in the fitted values and residuals for each of the original points in the regression. For this, use augment, which augments the original data with information from the model. New columns begins with a . (to avoid overwriting any of the original columns).
# Augment the original data
# IF you get a warning about deprecated... purrr..., ignore. It's a bug that'll be fixed soon.
## .rownames Testosterone PhysActive Age Gender .fitted .se.fit
## 1 1 47.53 No 43 female 21.30758 4.010694
## 2 2 642.82 No 80 male 384.66314 5.480833
## 3 3 642.82 No 80 male 384.66314 5.480833
## 4 4 21.11 Yes 34 female 35.89576 3.885559
## 5 5 562.78 No 80 male 384.66314 5.480833
## 6 6 401.78 No 35 male 411.22942 4.363256
## .resid .hat .sigma .cooksd .std.resid
## 1 26.222419 0.0009890871 127.5448 1.047551e-05 0.20572422
## 2 258.156859 0.0018470928 127.4693 1.899317e-03 2.02620279
## 3 258.156859 0.0018470928 127.4693 1.899317e-03 2.02620279
## 4 -14.785760 0.0009283305 127.5453 3.125591e-06 -0.11599603
## 5 178.116859 0.0018470928 127.5093 9.041494e-04 1.39799066
## 6 -9.449415 0.0011706227 127.5455 1.610571e-06 -0.07414076
# Plot residuals vs fitted values for males,
# colored by Physical Activity, size scaled by age
augment(fit) %>%
filter(Gender=="male") %>%
ggplot(aes(.fitted, .resid, col=PhysActive, size=Age)) + geom_point()
Finally, several summary statistics are computed for the entire regression, such as $$R^2$$ and the F-statistic. These can be accessed with glance:
glance(fit)
## r.squared adj.r.squared sigma statistic p.value df logLik AIC
## 1 0.696893 0.6966282 127.527 2631.777 0 4 -21544.86 43099.72
## BIC deviance df.residual
## 1 43130.44 55847628 3434
The broom functions work on a pipe, so you can %>% your model directly to any of the functions like tidy(). Let’s tidy up our t-test:
t.test(AlcoholYear~RelationshipStatus, data=nha)
##
## Welch Two Sample t-test
##
## data: AlcoholYear by RelationshipStatus
## t = 5.4315, df = 2674.8, p-value = 6.09e-08
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 13.05949 27.81603
## sample estimates:
## mean in group Committed mean in group Single
## 83.93416 63.49640
t.test(AlcoholYear~RelationshipStatus, data=nha) %>% tidy()
## estimate estimate1 estimate2 statistic p.value parameter conf.low
## 1 20.43776 83.93416 63.4964 5.431545 6.089562e-08 2674.82 13.05949
## conf.high method alternative
## 1 27.81603 Welch Two Sample t-test two.sided
…and our Mann-Whitney U test / Wilcoxon rank-sum test:
wilcox.test(AlcoholYear~RelationshipStatus, data=nha)
##
## Wilcoxon rank sum test with continuity correction
##
## data: AlcoholYear by RelationshipStatus
## W = 1068000, p-value = 0.0001659
## alternative hypothesis: true location shift is not equal to 0
wilcox.test(AlcoholYear~RelationshipStatus, data=nha) %>% tidy()
## statistic p.value method
## 1 1067954 0.0001658551 Wilcoxon rank sum test with continuity correction
## alternative
## 1 two.sided
…and our Fisher’s exact test on the cross-tabulated data:
xtabs(~Gender+Diabetes, data=nha) %>% fisher.test()
##
## Fisher's Exact Test for Count Data
##
## data: .
## p-value = 0.05992
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.9883143 1.5466373
## sample estimates:
## odds ratio
## 1.235728
xtabs(~Gender+Diabetes, data=nha) %>% fisher.test() %>% tidy()
## estimate p.value conf.low conf.high
## 1 1.235728 0.05992352 0.9883143 1.546637
## method alternative
## 1 Fisher's Exact Test for Count Data two.sided
…and finally, a logistic regression model:
# fit the model and summarize it the usual way
glmfit <- glm(Insured~Race, data=nha, family=binomial)
summary(glmfit)
##
## Call:
## glm(formula = Insured ~ Race, family = binomial, data = nha)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0377 0.5177 0.5177 0.5177 1.1952
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.94218 0.06103 31.825 < 2e-16
## RaceAsian -0.64092 0.17715 -3.618 0.000297
## RaceBlack -0.59744 0.13558 -4.406 1.05e-05
## RaceHispanic -1.41354 0.14691 -9.622 < 2e-16
## RaceMexican -1.98385 0.13274 -14.946 < 2e-16
## RaceOther -1.26430 0.22229 -5.688 1.29e-08
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 3614.6 on 3704 degrees of freedom
## Residual deviance: 3336.6 on 3699 degrees of freedom
## (2 observations deleted due to missingness)
## AIC: 3348.6
##
## Number of Fisher Scoring iterations: 4
# tidy it up!
tidy(glmfit)
## term estimate std.error statistic p.value
## 1 (Intercept) 1.9421805 0.06102764 31.824604 2.956937e-222
## 2 RaceAsian -0.6409232 0.17714581 -3.618055 2.968256e-04
## 3 RaceBlack -0.5974352 0.13558235 -4.406438 1.050844e-05
## 4 RaceHispanic -1.4135371 0.14691100 -9.621724 6.473720e-22
## 5 RaceMexican -1.9838532 0.13273768 -14.945667 1.662182e-50
## 6 RaceOther -1.2643008 0.22228811 -5.687667 1.287864e-08
# do whatever you want now
tidy(glmfit) %>%
filter(term!="(Intercept)") %>%
mutate(logp=-1*log10(p.value)) %>%
ggplot(aes(term, logp)) + geom_bar(stat="identity") + coord_flip()
Check out some of the other broom vignettes on CRAN, and also check out the biobroom package on bioconductor for turning bioconductor objects and analytical results into tidy data frames.
## Homework
### Stress tests & muscular dystrophy genetics
Try this homework assignment.
There are two parts to this assignment, each using data curated and hosted by the Vanderbilt Department of Biostatistics. The first part looks at the effects of different doses of a drug used during echocardiograms called dobutamine. The second uses data collected to examine several blood serum markers believed to be associated with genetics for a specific kind of muscular dystrophy (DMD).
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# Linear Algebra - Show that $M$ is not a vector space
Consider the set of all 2 x 2 matrices where the product of the elements on the main diagonal is zero.
$$M = \left \lbrace\begin{bmatrix}a&b\\c&d\end{bmatrix} \left. \right | a, b, c, d \in \mathbb R,ad = 0 \right \rbrace$$
Define addition and scalar multiplication on $M$ in the usual way.
Show that $M$, with respect to these operations of addition and scalar multiplication, is not a vector space by showing that one of the vector space axioms does not hold.
To be honest, I don't know how to perform addition and scalar multiplication to solve this question. If anyone could explain exactly how to solve this, that would be great!
• Is the set closed under addition? – Tpofofn Feb 12 '14 at 11:36
• @Tpofofn yeah it is – Raba Feb 12 '14 at 11:37
• @Raba are you sure about that? – Casteels Feb 12 '14 at 11:39
• So $\begin{bmatrix}0&b\\c&d\end{bmatrix}$ and $\begin{bmatrix}a&b\\c&0\end{bmatrix}$ are in the set. Is their sum in the set? – Tpofofn Feb 12 '14 at 11:39
• "You were told so"?? Since when that claiming things in mathematics gets reinforced by "being told so"?? – DonAntonio Feb 12 '14 at 12:36
Let $A=\left( \begin{array}{cc} 0 & 1 \\ 1 & 1 \\ \end{array} \right)$ and $B=\left( \begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array} \right)$ be two matrices in M. Then $$A+B=\left( \begin{array}{cc} 1 & 2 \\ 2 & 1 \\ \end{array} \right)$$ does not belong in M.
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## Weak and strong form shape hessians and their automatic generation.(English)Zbl 1448.49036
### MSC:
49M15 Newton-type methods 65D18 Numerical aspects of computer graphics, image analysis, and computational geometry 68T35 Theory of languages and software systems (knowledge-based systems, expert systems, etc.) for artificial intelligence 68Q42 Grammars and rewriting systems 68U05 Computer graphics; computational geometry (digital and algorithmic aspects) 68U15 Computing methodologies for text processing; mathematical typography
### Software:
revolve; FEMorph; FEniCS; Firedrake; UFL; ISTL
Full Text:
### References:
[1] T. Albring, M. Sagebaum, and N. R. Gauger, A consistent and robust discrete adjoint solver for the SU$$^2$$ framework—validation and application, Notes Numer. Fluid Mech. Multidiscip. Design, 132 (2016), pp. 77–86. [2] M. Aln\ae s, A Compiler Framework for Automatic Linearization and Efficient Discretization of Nonlinear Partial Differential Equations, Ph.D. thesis, University of Oslo, 2009. [3] M. S. Aln\ae s, A. Logg, K. B. Ø lgaard, M. E. Rognes, and G. N. Wells, Unified form language: A domain-specific language for weak formulations of partial differential equations, ACM Trans. Math. Software, 40 (2014), pp. 9:1–9:37. [4] M. Berggren, A unified discrete-continuous sensitivity analysis method for shape optimization, in Applied and Numerical Partial Differential Equations, Comput. Methods Appl. Sci. 15, Springer, New York, 2010, pp. 25–39. · Zbl 1186.65078 [5] M. Blatt and P. Bastian, The Iterative Solver Template Library, Springer, Berlin, Heidelberg, 2007, pp. 666–675. [6] C. Brandenburg, F. Lindemann, M. Ulbrich, and S. Ulbrich, Advanced Numerical Methods for PDE Constrained Optimization with Application to Optimal Design in Navier Stokes Flow, Springer, Basel, 2012, pp. 257–275. · Zbl 1356.49019 [7] A. Carnarius, F. Thiele, E. Özkaya, and N. Gauger, Adjoint Approaches for Optimal Flow Control, Report 2010-5088, AIAA, 2010. [8] B. Christianson, Reverse accumulation and attractive fixed points, Optim. Methods Softw., 3 (1994), pp. 311–326. [9] M. C. Delfour and J.-P. Zolésio, Shapes and Geometries: Metrics, Analysis, Differential Calculus, and Optimization, 2nd ed., Adv. Design Control 22, SIAM, Philadelphia, 2011, . · Zbl 1251.49001 [10] G. Doğan and R. H. Nochetto, First variation of the general curvature-dependent surface energy, ESAIM Math. Model. Numer. Anal., 46 (2012), pp. 59–79. · Zbl 1270.49042 [11] P. E. Farrell, D. A. Ham, S. W. Funke, and M. E. Rognes, Automated derivation of the adjoint of high-level transient finite element programs, SIAM J. Sci. Comput., 35 (2013), pp. C369–C393, . · Zbl 1362.65103 [12] H. Garcke, C. Hecht, M. Hinze, and C. Kahle, Numerical approximation of phase field based shape and topology optimization for fluids, SIAM J. Sci. Comput., 37 (2015), pp. A1846–A1871, . · Zbl 1322.35113 [13] N. Gauger, A. Walther, C. Moldenhauer, and M. Widhalm, Automatic Differentiation of an Entire Design Chain for Aerodynamic Shape Optimization, Springer, Berlin, Heidelberg, 2008, pp. 454–461. [14] A. Griewank and G. F. Corliss, eds., Automatic Differentiation of Algorithms: Theory, Implementation, and Applications, SIAM, Philadelphia, 1991. [15] A. Griewank and A. Walther, Algorithm \textup799: Revolve: An implementation of checkpointing for the reverse or adjoint mode of computational differentiation, ACM Trans. Math. Software, 26 (2000), pp. 19–45. · Zbl 1137.65330 [16] A. Griewank and A. Walther, Evaluating Derivatives: Principles and Techniques of Algorithmic Differentiation, SIAM, Philadelphia, 2008, . · Zbl 1159.65026 [17] M. Hintermüller and W. Ring, A second order shape optimization approach for image segmentation, SIAM J. Appl. Math., 64 (2003), pp. 442–467, . · Zbl 1073.68095 [18] A. Jameson, Aerodynamic design via control theory, J. Sci. Comput., 3 (1988), pp. 233–260. · Zbl 0676.76055 [19] A. Kowarz and A. Walther, Optimal checkpointing for time-stepping procedures in ADOL-C, in Computational Science – ICCS 2006, V. N. Alexandrov, G. D. van Albada, P. M. A. Sloot, and J. Dongarra, eds., Lecture Notes in Comput. Sci. 3994, Springer, New York, 2006, pp. 541–549. [20] A. Logg, K.-A. Mardal, and G. Wells, eds., Automated Solution of Differential Equations by the Finite Element Method, Lect. Notes Comput. Sci. Eng. 84, Springer, Berlin, Heidelberg, 2012. [21] C. Lozano, Discrete surprises in the computation of sensitivities from boundary integrals in the continuous adjoint approach to inviscid aerodynamic shape optimization, Comput. & Fluids, 56 (2012), pp. 118–127. · Zbl 1365.76100 [22] B. Mohammadi and O. Pironneau, Applied Shape Optimization for Fluids, Numerical Mathematics and Scientific Computation, Clarendon Press, Oxford, 2001. · Zbl 0970.76003 [23] O. Pironneau, On optimum profiles in Stokes flow, J. Fluid Mech., 59 (1973), pp. 117–128. · Zbl 0274.76022 [24] O. Pironneau, On optimum design in fluid mechanics, J. Fluid Mech., 64 (1974), pp. 97–110. · Zbl 0281.76020 [25] F. Rathgeber, D. A. Ham, L. Mitchell, M. Lange, F. Luporini, A. T. T. McRae, G. Bercea, G. R. Markall, and P. H. J. Kelly, Firedrake: Automating the Finite Element Method by Composing Abstractions, preprint, , 2015. · Zbl 1396.65144 [26] S. Schmidt, Efficient Large Scale Aerodynamic Design Based on Shape Calculus, Ph.D. thesis, University of Trier, Germany, 2010. [27] S. Schmidt, C. Ilic, V. Schulz, and N. Gauger, Three dimensional large scale aerodynamic shape optimization based on shape calculus, AIAA J., 51 (2013), pp. 2615–2627. [28] S. Schmidt and V. Schulz, Impulse response approximations of discrete shape Hessians with application in CFD, SIAM J. Control Optim., 48 (2009), pp. 2562–2580, . · Zbl 1387.49064 [29] S. Schmidt and V. Schulz, Shape derivatives for general objective functions and the incompressible Navier-Stokes equations, Control Cybernet., 39 (2010), pp. 677–713. · Zbl 1282.49033 [30] S. Schmidt, E. Wadbro, and M. Berggren, Large-scale three-dimensional acoustic horn optimization, SIAM J. Sci. Comput., 38 (2016), pp. B917–B940, . · Zbl 06646630 [31] V. Schulz and M. Siebenborn, Computational comparison of surface metrics for PDE constrained shape optimization, Comput. Methods Appl. Math., 16 (2016), pp. 485–497. · Zbl 1342.49065 [32] V. H. Schulz, A Riemannian view on shape optimization, Found. Comput. Math., 14 (2014), pp. 483–501. · Zbl 1296.49037 [33] J. Sokolowski and J.-P. Zolésio, Introduction to Shape Optimization: Shape Sensitivity Analysis, Springer, Berlin, Heidelberg, 1992. [34] M. Sonntag, S. Schmidt, and N. Gauger, Shape derivatives for the compressible Navier–Stokes equations in variational form, J. Comput. Appl. Math., 296 (2016), pp. 334–351. · Zbl 1329.49093 [35] A. Walther and A. Griewank, Advantages of binomial checkpointing for memory-reduced adjoint calculations, in Numerical Mathematics and Advanced Applications – Proceedings of ENUMATH 2003, M. Feistauer, V. Dolejš\' \i, P. Knobloch, and K. Najzar, eds., Springer, New York, 2004, pp. 834–843, · Zbl 1056.65064
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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# Solve Right Triangle Worksheet
Solve Right Triangle Worksheet
• Page 1
1.
$\angle$BAC = 70o, $\angle$ACB = 80o.
Find the length of the altitude $\stackrel{‾}{\mathrm{AD}}$.
a. 3 cm b. 4 cm c. 2$\sqrt{13}$ cm d. 3$\sqrt{3}$ cm
#### Solution:
ABC + BAC + ACB = 180o
[Since sum of the angles in a triangle is 180o.]
ABC = 180o – (70o + 80o) = 30o
[Substitute and simplify.]
6 sin30o = 6 × 1 / 2 = 3 cm
[Substitute and simplify.]
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# Green's Theorem and limits on y for flux
I'm working through understanding the example provided in the book for the divergence integral. The theorem (Green's):
$$\oint_C = \mathbf{F}\cdot \mathbf{T}ds = \oint_CMdy-Ndx=\int\int_R(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y} )dxdy$$
The example uses the following: $\mathbf{F}(x,y) = (x-y)\mathbf{i} + x\mathbf{j}$ over the region $\mathbf{R}$ bounded by the unit circle $C: \mathbf{r}(t)=cos(t)\mathbf{i} + sin(t)\mathbf{j}, 0 \le t \le 2\pi$.
There is then the following relations:
$$\begin{array}{rr} M = cos(t) - sin(t) & dx = d(cos(t)) = -sin(t)dt \\ N = cos(t) & dy = d(sin(t)) = cos(t)dt \end{array} \\ \begin{array}{rrrr} \frac{\partial M}{\partial x}=1 & \frac{\partial M}{\partial y} = -1 & \frac{\partial N}{\partial x}=1 & \frac{\partial N}{\partial y} = 0 \end{array}$$
Now that the foundation is laid, here's the rightmost part of the first equation given:
$$\begin{array}{rcl} \int\int_R \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} dxdy & = & \int\int_R 1 + 0 dxdy \\ & = & \int\int_R dxdy \\ & = & \text{area inside unit circle} \\ & = & \pi \end{array}$$
I understand it intuitively because it's the area over that region and the area of a circle is $A = \pi\cdot r^2$. With $r = 1$ that's obviously $\pi$. What I'm not sure of is how to express it in an integral. The question $x^2 + y^2 = 1$ represents the unit circle. Thus, $x$ as a function of $y$ I get $x = \sqrt{1-y^2}$, thus the final stage I show should be:
$$\int_{?}^{?} \int_{0}^{\sqrt{1-y^2}}dx dy$$
right? What should be used for the limits on y? I know it's simple but I'm just not seeing it and I need some guidance.
Thanks
For Green's Theorem
$$\oint_C (Mdx+Ndy)=\int \int_R \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right) dxdy$$
Here, $M=(x-y)$ and $N=x$ such that
\begin{align}\oint_C (Mdx+Ndy)&=2\int \int_R dxdy\\\\ &=2\int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}dxdy\\\\ &=4\int_{-1}^{1} \sqrt{1-y^2}dy\\\\ &=2\left(y\sqrt{1-y^2}+\arcsin(y)\right)|_{-1}^{1}\\\\ &=2(\arcsin(1)-\arcsin(-1))\\\\ &=2\pi \end{align}
If we evaluate the line integral in a straightforward way, we let $x=\cos \phi$ and $y=\sin \phi$. Then, $dx=-\sin \phi d\phi$ and $dy=\cos \phi d\phi$. We obtain the following
\begin{align}\oint_C (Mdx+Ndy)&=\int_0^{2\pi} (-\cos \phi \sin \phi+\sin^2 \phi+\cos^2 \phi)d\phi\\\\ &=\int_0^{2\pi} (-\cos \phi \sin \phi+1)d\phi\\\\ &=2\pi \end{align}
as expected!!
• Thanks. This makes sense. I knew it had to be something simple. – Andrew Falanga Apr 19 '15 at 16:56
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# Physical Properties of Alkenes
Alkenes contain a carbon-carbon double bond which changes the physical properties of alkenes. Alkenes are unsaturated carbon compounds which have a general formula of $C_nH_{2n}$. These compounds are also known as olefins.
Alkenes are a family of compounds containing hydrogen and carbon only (hydrocarbons) with a carbon-carbon double bond. Ethene and Propene are the first two hydrocarbons.
Physical Properties of Alkenes
## 1. Physical State
• These double-bonded compounds are colourless and odourless in nature.
• However, ethene is an exception because it is a colourless gas but has a faintly sweet odour.
• The first three members of the alkene group are gaseous in nature, the next fourteen members are liquids and the remaining alkenes are solids.
## 2. Solubility
• The alkenes are insoluble in water due to their nonpolar characteristics.
• But are completely soluble in nonpolar solvents such as benzene, ligroin, etc.
## 3. Boiling Point
• The boiling points of the compounds increase as the number of carbon atoms in the compound increases.
• When alkenes are compared with alkanes, it is found that the boiling points of both are almost similar, as if the compounds are made up of the same carbon skeleton.
• The boiling point of straight-chain alkenes is more that branched-chain alkenes just as in alkanes.
## 4. Melting Point
• The melting points of these double-bonded compounds depend upon the positioning of the molecules.
• The melting point of alkenes is similar to that of alkanes.
• However, cis-isomer molecules have a lower melting point than trans- isomers as the molecules are packed in a U-bending shape.
## 5. Polarity
• Alkenes are weakly polar just like alkanes but are slightly more reactive than alkanes due to the presence of double bonds.
• The π electrons which make up the double bonds can easily be removed or added as they are weakly held.
• Hence, the dipole moments exhibited by alkenes are more than alkanes.
• The polarity depends upon the functional group attached to the compounds and the chemical structures.
## Recommended Videos
To understand the physical properties of alkenes in a more interactive way download the Byju’s app from the play store or app store.’
1. Good
2. Hafiz khan
Very good
3. Hafiz Khan Anderson
EXCEPTIONALLY GOOD!!!
4. Natash nixxy
umm nice
5. nat ashy nixxy
really understandable
6. really good, do you have notes for the physical properties of alkanes??
1. Hi, to access notes on the physical properties of alkanes, click here.
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# Probability distribution/Related Articles
Main Article
Discussion
Related Articles [?]
Bibliography [?]
Citable Version [?]
A list of Citizendium articles, and planned articles, about Probability distribution.
See also changes related to Probability distribution, or pages that link to Probability distribution or to this page or whose text contains "Probability distribution".
## Other related topics
• Characteristic function [r]: A function on a set which takes the value 1 on a given subset and 0 on its complement. [e]
• Conditioning (probability) [r]: Conditional probabilities, conditional expectations and conditional distributions are treated on three levels. [e]
• Continuous probability distribution [r]: Probability distribution where variables can take on arbitrary values in a continuum. [e]
• Discrete probability distribution [r]: Class of probability distributions in which the values that might be observed are restricted to being within a pre-defined list of possible values. [e]
• Entropy of a probability distribution [r]: A number that describes the degree of uncertainty or disorder the distribution represents. [e]
• Financial economics [r]: the economics of investment choices made by individuals and corporations, and their consequences for the economy, . [e]
• Measurable function [r]: Function on a measurable space to a measurable space such that the inverse image of a measurable set is a measurable set. [e]
• Measure theory [r]: Generalization of the concepts of length, area, and volume, to arbitrary sets of points not composed of line segments or rectangles. [e]
• Power law [r]: A mathematical relationship between two quantities where one is proportional to a power of the other: that is, of the form ${\displaystyle y(x)=ax^{k}\!}$ where ${\displaystyle a}$ and ${\displaystyle k}$ are constants, with ${\displaystyle k}$ being referred to as the exponent. [e]
• Quantum mechanics [r]: An important branch of physics dealing with the behavior of matter and energy at very small scales. [e]
• Sigma algebra [r]: A formal mathematical structure intended among other things to provide a rigid basis for measure theory and axiomatic probability theory. [e]
• Statistics theory [r]: A branch of mathematics that specializes in enumeration, or counted, data and their relation to measured data. [e]
• Stochastic convergence [r]: A mathematical concept intended to formalize the idea that a sequence of essentially random or unpredictable events sometimes is expected to settle into a pattern. [e]
• Stochastic process [r]: Family of random variables, dependent upon a parameter which usually denotes time. [e]
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First Apotome of Medial is Irrational
Theorem
In the words of Euclid:
If from a medial straight line there be subtracted a medial straight line commensurable with the whole in square only, and which contains with the whole a rational rectangle, the remainder is irrational. And let it be called a first apotome of a medial straight line.
Proof
Let $AB$ be a medial straight line.
Let a medial straight line $BC$ such that:
$BC$ is commensurable in square only with $AB$
the rectangle contained by $AB$ and $BC$ is rational
be cut off from $AB$.
We have that $AB$ and $BC$ are medial.
So by definition $AB^2$ and $BC^2$ are both medial.
But $2 \cdot AB \cdot BC$ is rational.
Therefore:
$AB^2$ and $BC^2$ are incommensurable with $2 \cdot AB \cdot BC$.
if $AB$ is incommensurable with either $AC$ or $CB$, $AC$ and $CB$ are incommensurable with each other.
$2 \cdot AB \cdot BC$ is incommensurable with $AC^2$.
Therefore $AC^2$ is irrational.
Therefore by definition $AC$ is irrational.
Such a straight line is known as a first apotome of a medial.
$\blacksquare$
Historical Note
This proof is Proposition $74$ of Book $\text{X}$ of Euclid's The Elements.
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# Speeding up parentheses (and lots more) in R
August 19, 2010
By
(This article was first published on Radford Neal's blog » R Programming, and kindly contributed to R-bloggers)
As I noted here, enclosing sub-expressions in parentheses is slower in R than enclosing them in curly brackets. I now know why, and I’ve modified R to reduce (but not eliminate) the slowness of parentheses. The modification speeds up many other operations in R as well, for an average speedup of something like 5% for programs that aren’t dominated by large built-in operations like matrix multiplies.
I looked at the source code for the latest version of R, 2.11.1, and figured out what’s going on. The difference between parentheses and curly brackets comes about because R treats curly brackets as a “special” operator, whose arguments are not automatically evaluated, but it treats parentheses as a “built in” operator, whose arguments (just one for parentheses) are evaluated automatically, with the results of this evaluation stored in a LISP-style list. Creating this list requires allocation of memory and other operations which seem to be slow enough to cause the difference, since the curly bracket operator just evaluates the expressions inside and returns the last of them, without creating such a list. Furthermore, the implementation allocates one more “CONS” cell than is necessary when creating this list, apparently because the programmer found this slightly easier than writing code to avoid creating the extra cell.
Since this is time critical code, not just for parentheses but for most operators, allocating a cell unnecessarily is a bad idea. I modified the program to avoid this, changing the evalList and evalListKeepMissing procedures in the eval.c source file in the src/main directory. Here are the old and new versions of these procedures (probably of interest only to R hackers).
Here are some timing results. First with the R 2.11.1 unmodified, on an Intel Linux system:
> a <- 5; b <- 1; c <- 4
> f <- function (n) for (i in 1:n) d <- 1/{a*{b+c}}
> g <- function (n) for (i in 1:n) d <- 1/(a*(b+c))
> system.time(f(1000000))
user system elapsed
1.830 0.010 1.842
> system.time(g(1000000))
user system elapsed
1.990 0.000 1.988
Parentheses are 8.7% slower than curly brackets.
Now here are the results with my modified version:
> a <- 5; b <- 1; c <- 4
> f <- function (n) for (i in 1:n) d <- 1/{a*{b+c}}
> g <- function (n) for (i in 1:n) d <- 1/(a*(b+c))
> system.time(f(1000000))
user system elapsed
1.780 0.000 1.783
> system.time(g(1000000))
user system elapsed
1.820 0.000 1.828
Now parentheses are only 2.2% slower than curly brackets. Furthermore, the curly bracket version is 2.7% faster than before, and the parenthesis version is 8.5% faster.
Here’s a test on a program that implements EM for a simple model. First, with the unmodified version of R (running far more iterations of EM than really needed):
> system.time(print(EM.censored.poisson(5,3.2,20,iterations=1000000)))
[1] 1.049699
user system elapsed
13.690 0.020 13.703
And now my modified version of R:
> system.time(print(EM.censored.poisson(5,3.2,20,iterations=1000000)))
[1] 1.049699
user system elapsed
12.860 0.020 12.878
My new version is 6% faster than the old version of R.
The size of the improvement will vary a lot from one R program to another, of course. (The nature of the change is such that I don’t think any programs will be slower.) But one can see that the simple modification I made has produced a significant improvement — a 6% speedup is a lot for such a small change. From looking at the source code for R, I think a number of other improvements can be made without great effort. In particular, arithmetic on large vectors can be speeded up substantially, including the squaring operation discussed here. The changes to do this will also speed up arithmetic on short vectors, including scalars, but I don’t yet know how significant the improvement will be in that context.
The fundamental reason for the slowness of the current R implementation seems to be a lack of attention by the implementors to efficiency in code that is on the critical interpretive pathway. This is the opposite fault to what is often seen among hackers — an obsessive focus on efficiency even where it doesn’t matter. One should focus on efficiency where it matters, and focus on other things where it doesn’t matter (which is typically 95% of a program). On the plus side, the R source code is fairly readable, even though it lacks much in the way of informative comments. It was readable enough that I was able to figure out what’s going on and modify the program in less than a day. My next task is to figure out how to get the modification into the next release of R…
UPDATE: I forgot to mention that the real solution to speeding up parentheses is to get rid of them entirely. The only reason for them to be preserved in R’s internal expression tree is so that they can be printed when an expression is deparsed. But for that, one could just have a field in each expression node saying how many pairs of parentheses directly enclose it (usually 0 or 1, except for people who really like redundant parentheses), which can be accessed when deparsing the expression, but which would not slow down execution of the program.
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# Revision history [back]
### Caching sagetex results
Is there a way to cache computations already done in a latex file using sagetex so that, for example, if I add a Sage code line in foo.tex when I run sage foo.sagetex.sage the computation starts from that line and not from the beginning?
The best thing would be that when one uses sagetex, Sage remembers what formulas are already computed and stored in foo.sout, and recomputes them only when they are modified. Do you think it's possible?
(Maybe there's already some cache-ing feature in Sage but I don't know how to use it...)
Thanks!
### Caching sagetex results
Is there a way to cache computations already done in a latex file using sagetex so that, for example, if I add a Sage code line in foo.tex when I run sage foo.sagetex.sage the computation starts from that line and not from the beginning?
The best thing would be that when one uses sagetex, Sage remembers what formulas are already computed and stored in foo.sout, and recomputes them only when they are modified. Do you think it's possible?
(Maybe there's already some cache-ing feature in Sage but I don't know how to use it...)
Thanks!
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Serving the Quantitative Finance Community
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I have been doing work on calculation of forward expectations of functionals of SDE on a transition probabilities grid. I have been trying to put everything on a reasonable mathematical foundation. In forward simulation important problem is how to faithfully retrieve the expectations of events that occurred on the previous time grids. Usually, though not always, variance of underlying SDE distribution increases as we advance into next(or later) time grids. But expectations of SDE related functions evaluated on previous time grids were calculated with respect to SDE distribution at previous times when underlying SDE variance was (usually) slightly lesser. So in order to faithfully retrieve the densities of functionals of SDE in future time grids we have to add some appropriate variance. This added variance makes sure that we can retrieve the functionals densities with expectations evaluated at earlier times with smaller underlying variance faithfully with respect to each of the later time SDE grids when underlying SDE distribution variance has increased. But a side effect of this addition of variance is that covariances across time also increase in proportion to added variance. Expectations of Path integrals are highly sensitive to covariances of the functionals across different time grids so we have to make an accounting for how much appropriate relevant covariances have increased. This can be done in several ways since we know how much extra variance was added to expectations of functionals on each time step to convert the functionals expectation originally calculated with respect to lower variance underlying SDE grid into expectations with respect to SDE grids at later times with higher variance.
I am working on various mathematical issues so that we can find expectations of functionals of every SDE even when we do not analytically know the SDE covariances across time.
It may be still a few days before I could post a perfectly working program since I want to change a lot of things in my previous program as my understanding of the problem has become better. I will be posting comments regularly for next few days before I post a full-fledged worked out program.
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
There is some different progress so I thought I will mention it here for friends. Though I had a completely worked out plan how to convert the expectations of functions from an earlier SDE grid to a later time grid, I just thought that I should try measure theory for this purpose and if things could work, it would be a far more elegant solution to finding the densities of functionals of SDE. Towards that goal, I wrote down exponential likelihood function of general (possibly mean-reverting) SDEs with drift and state dependent variances (By doing Ito change of variable on log of SDE variable) and I was able to faithfully transport some very simple payoffs forward in time. However, it still remains to be seen if slightly more difficult payoffs could be transported in time and I still have to work to see that. As only seeing is believing, I still cannot be sure if the measure transport method would universally work but I am quite optimistic. I will get more detailed results in another day and will let friends know how it goes as my research progresses.
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I would not be writing this post if it were not for torture that started four days ago. I was working on my projects at full speed when mind control agents targeted me with drugged water and started giving me attacks of very strong anxiety and I was totally unable to work for past four days. These attacks of very strong anxiety lasted everyday after that and therefore I decided to write more about mind control and not about mathematical finance that all of us including myself love so much. I have not so huge interest about whining about mind control but if the torture continues, I have lined up many posts about mind control that will follow.
I am copying an old post of mine and want to ask friends if mind control agencies were serving United States or trying to please their godfather. I still recall when mind control agent who made more than 800 million dollars and when he would narrate me incidents in which he would describe to his jewish backer professors at my previous school the stories of how targeted talented Pakistani muslims would beg to end mind control torture on them and then mind control agent would tell me how excited, amused and delighted some jewish professors at my old school would be when they would hear how talented muslim targets were desperately begging to end mind control that would never end. I am not lying and therefore the agent was able to make more than 800 million dollars.
Again I do not want to write such stuff but if my mind control continues, there is a lot more on the way.
I am copying an old post of mine and want to ask friends if mind control agencies were serving United States or trying to please their godfather.
.
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Thank you Paul for your post. I will look at the website you mentioned. But I want to really thank you for letting me write bold things on your forum that most forum owners would not allow. Though I might have different ideas than you on a lot of issues, I am a staunch supporter of freedom of speech just like you are and I am sure both of us agree on one slogan,"Je suis Charlie". I consider myself lucky to have been writing on your forum since 2003.
I want to write this post to ask American tax payers to please cut funding of crooks in US army. I was reading that there has not been a successful audit of spending by US army in past decade. You would probably ask me why you should do anything to bridle spending by crooks in army. Here is why.
Crooks in US defense have spent several billion dollars in attempts to retard me over past twenty years. I went to Hong Kong in 2000 to find a job in finance and derivatives and I stayed there for three months. Crooks in US army continues to drug food and water at large number of places in Hong Kong and Kowloon and continued to play all sort of tricks with me. I still recall when I went to office of South China Morning Post to tell them how badly they had drugged the city of Hong Kong, they had already set up someone who received me and told me after hearing my story that he did not believe me.
They gave me electric shocks twenty days after I returned from Hong Kong.
Later I was asked by my Japanese employer to come to Tokyo (2005) and work in their office there(I was earlier working from Lahore). They drugged large number of places in Tokyo city and mad e my life absolute hell and I resigned only after one month of stay in Tokyo. I did not want to return to Pakistan due to huge persecution by my family and the army so I decided to go somewhere where US military could not go and got Iranian visit visa from Tokyo and went to Tehran.
Crooks in US defense gave huge amount of money to Iranian crooks(crooks everywhere know each other well) and they used huge microwave and gases torture on me. I still recall on my first night in an Iranian hotel, I left the hotel after midnight and randomly ran in streets of Tehran to somehow avoid pain due to directed microwave torture. Just after two or three days in Tehran, I was trying to find refuge in embassies like Vatican in Tehran asking them to help me on human grounds. I left Tehran after one week and flew back to Pakistan. I still recall that I arrived at Tehran airport a lot earlier than flight departure. I had previously protested somewhere against unnecessary removal of shoes of Muslim passengers and it was known to mind control agencies. They asked Iranians security on airport who approached me inside the departure lounge and took me to a room and asked me to remove my shoes and had my body search.
Since during all this, mind control agencies could not control me well and several of my neurotransmitters were coming back, they had extremely huge amount of gases in washrooms in Dubai airport where I would stop for transit while going to Karachi. No wonder ultra conservative crooks in US army align so well with Saudi(who dissolve their criticizers in acid) and Dubai.
Later I got British HSMP work visa and went to London in 2010. Crooks in US defense drugged several large cities of UK on a large scale(it was impossible to get good beverages or bottled water. They even drugged beer, wine and pork at a lot of places.) You can ask people in UK secret services(many of whom would love to tell you everything in detail) about supernatural stories how they drugged food in cities of UK in 2010. I still recall that I wrote several page account of my persecution and I would print hundred of copies of it and distribute them in London. When I was distributing those papers outside of some London Newspapers, administration of those newspapers went to extreme length to point out that certain part of sidewalk was their property and they would call police if I stepped there. I was forcefully removed from the compound in London(after I had passed all security checks) when I went there to ask law firms for help and the law firms on main street would refuse to meet me and say that I had to either email or send the letter by post and they would not accept any papers from me(somebody would approach me at the door and tell me about it and ask me to leave.).
They continued to drug food and beverages in entire city of Lahore numerous times in past ten years.
Net worth of several mind control agents who continued to retard me in Pakistan is north of several hundred million dollars each depending upon their seniority(with most senior agent worth 800 million USD).
And this drugging of entire cities by crooks of US defense never ends. Now they are drugging the twin cities of Rawalpindi/Islamabad.
I am just telling about myself. There are tens of thousands of mind control victims all over the world.
Please cut their funding before it is too late. They are not retarding just Muslims(I have no religion but of course I am Muslim by birth) only, they retard a large number of totally innocent and talented Americans. Crooks love to bite the same hand that feeds them(It is typical of crooks and large unbridled armies in human history)
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
When I look at similarities and parallels, I find it interesting to notice that United States is becoming, just like reactionary act of mind control, a country with more and more reactionary policies. For example I find that there is more and more concern in United States about increasing might of China and a growing need to contain China. I want to tell Americans that this is very misguided. If Americans were truly wise, they would never try to "contain" anyone, and instead of containing anyone they would just do what needs to be done to make them a "country of excellence" something they did so well over the past one century. When USSR was getting ahead in sciences and technology, America responded so perfectly by emphasizing science and technology in its universities and Americans to this day reap the fruit of great decisions for advancing science and mathematics in its academic institutions. Now when China is advancing so fast, Americans need to make sure that they truly invest in education of their children and make sure that every intelligent child becomes a star in its own right. Instead of so many other trillion dollar packages, Joe Biden would be far wiser to spend 500 billion dollars to make sure that every child in US whether he is born in inner part of some large city or some remote state would get the same great quality education, he would no longer need to worry about rising Chinese might as American fortunes would continue to surpass any Chinese miracles. I really want to say to American friends, "Do not be reactionary. Just be original and genuine. Just do even better what you did so well over past one century."
It is interesting that the very people who could not get great education and had to become crooks in mind control agencies instead of being doctors, lawyers and successful businessmen and then these crooks do all sort of evil things in the society thinking that only way to get ahead in life is to join Godfather's paradise. Only great education in backward states would decrease the mass of crap crooks doing evil things in US society.
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I have been writing on this thread since 2016 with occasional large breaks. I am going to take another break of 3-4 weeks now. I had been working on a market trading algorithm for quite some time and I took to improving it after my post of 25th September. One reason my persecution continues is that I have not been able to make enough money to live independently from my family. So I was thinking if I could write a great market trading algorithm and sell it, I could try to get out of this problem of continuing mind control. I have been putting all my research other than market trading research on internet so its value for me in terms of money sharply decreases so it was somewhat difficult to leverage it to make enough money. Good thing is that my market trading research has matured enough and I have been able to write great trading programs. Currently I am trying some methods to further improve my results. In two months' data I was able to find returns ranging from 40% of notional to larger than 100% of notional which means that returns per month would range from 20% to 50% of notional and I am not using any leverage. My algorithm is a market-making algorithm and enters the trade only using limit orders but it is not a high frequency algorithm at all. Average time between trades ranges from one minute to 5 minutes and algorithm waits for opportune moments to start and end a trade. But you would like to be filled quickly once the algorithm wants you to do that. I still have to calculate sensitivity to order execution slippage but it should not be very large. Algorithm would work on highly liquid stocks but could be modified to deal with moderately liquid stocks. One reason for high returns could be that I do not have to scale the algorithm to very large amount of capital where investing like this would become difficult and would surely reduce the returns on large capital though I still think that if larger amounts of capital are judiciously invested, returns would still be far-better than many mainstream approaches.
Obviously, I am not going to post the algorithm since if I do that its effectiveness might end in a few weeks.
So I will be busy improving the algorithm and marketing it. I want to sell the algorithm to just 2-3 clients for money in six figures. I am very afraid that mind control agencies would do everything in their power to thwart me here.
If I could successfully sell the algorithm, I will hire a team of 5-6 mathematicians and computer programmers and train them in math finance and mathematics of trading. I want to tell friends that I actively want to continue doing research in mathematical finance and would continue posting very interesting things and research in the future with a more professional setup. I want to keep posting my research and programs on Wilmott for several years into the future.
My post would be incomplete if I do not thank so many American people who protested to mind control agencies against my continuing persecution when it was becoming very difficult for me to continue my research and mind control activity decreased due to protests from those good people and I was able to restart my research. And I also want to thank so many readers of this forum across the world who wished me well and wanted me to succeed and do good research. And I want to say to Americans that even if I succeeded in life, I will try to be a good human being and they will find a friend of American people in me.
One last thing, please ask mind control agencies who have access to my computer to not distribute my program in their favorites. Wrongdoing must not be encouraged like that.
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
I want to ask Americans a simple question. Isn't this true that if I were a jew, people all over the world would be full of praise for me. On the contrary since I am born to a muslim family, everyday I am subjected to inhuman mind control torture and several people in American government and defense continue to machinate how to fail me. Is this great American democracy all of us are never tired of praising?
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, please pardon my previous post. I was extremely down due to biweekly antipsychotic injection. I get half a Fluanxol 40mg injection after every fifteen days and for first seven days after the injection, I remain very down and usually unable to do anything meaningful. I become better seven days after injection and become relatively creative. I am 48 years old and it is very difficult for me to stand these regular antipsychotic injections but mind control agencies connived with my father and family and have forced injections and drugs on me for more than past twenty years. I still remember when I was 28/29 years of age, at one stage, I was given many of these injections per week but I still had energy and vitality but now I am too old to continue to take these injections regularly that are faced on me and it is impossible to remain active and many times I fall into despair.
Anyway good thing is that seven days have passed since last injection and I have some ideas about increasing the order (in time) of solution of partial differential equations. I also have some ideas about converting my old derivation of derivatives of Fokker-planck equation into a formal solution (unlike informal working solution I had presented earlier.) I still cannot say what will come out of these ideas but I will be working on them for next several days and if there is any progress then I will post my new programs. I hope to share some interesting stuff with friends in next few days.
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Sorry friends, I prematurely hit the submit button and so could not complete the previous post. Here is the actual relevant post. Please disregard the previous incomplete post.
Friends, I wanted to share a rough sketch of the proof of the method I presented in above programs. I hope friends would like it.
I am not writing Fokker-planck equation in Bessel coordinates and I am sure most friends are familiar with it.
In our setting we have an SDE variable, $B$, in Bessel coordinates on nth grid point $Z_n$ of underlying Z-grid. This is represented as $B(Z_n)$. As a function of time, $B(Z_n)$ is moving so as that CDF of the density at point $B(Z_n)$ remains constant. This is written in equation form as below
$\frac{\partial}{\partial t} \big[\int_{-\infty}^{B(Z_n)} P(B) dB \big]=0$
applying the time derivative to terms in brackets, we get
$\int_{-\infty}^{B(Z_n)} \frac{\partial P(B)}{\partial t} dB + \frac{\partial B}{\partial t} P(B(Z_n)) =0$
I will make it more formal but here, since we add drift separately and because we are in Bessel coordinates, I am just replacing $\frac{\partial P(B)}{\partial t}$ by $.5 {\sigma}^2 \frac{\partial^2 P(B)}{\partial B^2}$ in the above equation.
Making the substitution in above equation, we get
$\int_{-\infty}^{B(Z_n)} \big[.5 {\sigma}^2 \frac{\partial^2 P(B)}{\partial B^2} \big] dB + \frac{\partial B}{\partial t} P(B(Z_n)) =0$
Applying the integral on first term, we get
$\big[.5 {\sigma}^2 \frac{\partial P(B(Z_n))}{\partial B} \big] + \frac{\partial B}{\partial t} P(B(Z_n)) =0$
Now we want to make a change from $P(B)$ to $P(Z)$. In order to do that we have two equations
$P(B)=P(Z)\frac{\partial Z}{\partial B}$ and
$\frac{\partial P(B)}{\partial B}=\frac{\partial P(Z)}{\partial Z} {(\frac{\partial Z}{\partial B})}^2 + P(Z) \frac{\partial^2 Z}{\partial B^2}$
Now substituting above two equations in previous equation, we get
$.5 {\sigma}^2 \big[\frac{\partial P(Z_n)}{\partial Z} {(\frac{\partial Z}{\partial B})}^2 + P(Z_n) \frac{\partial^2 Z}{\partial B^2} \big] + \frac{\partial B}{\partial t} P(Z_n) \frac{\partial Z}{\partial B}=0$
substituting $\frac{\partial^2 Z}{\partial B^2} =-{(\frac{\partial Z}{\partial B})}^3 \frac{\partial^2 B}{\partial Z^2}$
we get
$.5 {\sigma}^2 \big[\frac{\partial P(Z_n)}{\partial Z} {(\frac{\partial Z}{\partial B})}^2 - P(Z_n) {(\frac{\partial Z}{\partial B})}^3 \frac{\partial^2 B}{\partial Z^2} \big] + \frac{\partial B}{\partial t} P(Z_n) \frac{\partial Z}{\partial B}=0$
cancelling $\frac{\partial Z}{\partial B}$ on both sides, we get
$.5 {\sigma}^2 \big[\frac{\partial P(Z_n)}{\partial Z} {(\frac{\partial Z}{\partial B})} - P(Z_n) {(\frac{\partial Z}{\partial B})}^2 \frac{\partial^2 B}{\partial Z^2} \big] +\frac{\partial B}{\partial t} P(Z_n)=0$
The first term in above is expansion that we have treated in the previous matlab program in a different analytic way by adding squared volatilities . Matching the coefficients of second and third term, we get
$\frac{\partial B}{\partial t} = .5 {\sigma}^2 \big[{(\frac{\partial Z}{\partial B})}^2 \frac{\partial^2 B}{\partial Z^2} \big]$
the last equation is the analytic solution that I have used in my previous matlab programs to get the density of SDEs in Bessel coordinates right.
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Here I want to explain how the above analytics could be applied to Fokker-planck equation in original coordinates (and it works seamlessly).
First I write the fokker-planck equation in original coordinates as
$\frac{\partial P(x,t)}{\partial t} = -\frac{\partial \big[\mu(x)P(x,t) \big]}{\partial x}\, +.5 \, \frac{\partial^2 \big[{\sigma(x)}^2P(x,t) \big]}{\partial x^2}$
We have our partial differential equation on a grid and we want to determine the movement of an arbitrary grid point (boundary) along time so that mass within(total mass up till the grid point is conserved or associated CDF remains constant) that grid point (boundary) remains constant as a function of time. we denote this arbitrary grid point as $x_b$.
We write the conservation of probability mass up till (or associated constant CDF) the grid point (boundary) as a function of time in equation form as
$\frac{\partial}{\partial t} \big[\int_{-\infty}^{x_b} P(x,t) dx \big]=0$
applying the time derivative to terms in brackets, we get
$\int_{-\infty}^{x_b} \frac{\partial P(x,t)}{\partial t} dx + \frac{\partial x_b}{\partial t} P(x_b,t) =0$
But we know that term inside the integral is given by FP equation and is LHS of FP equation. We replace the time derivative inside the integral by RHS of FP equation as
$\int_{-\infty}^{x_b} \big[ -\frac{\partial [\mu(x)P(x,t)]}{\partial x}\, + \, .5\frac{\partial^2 \big[{\sigma(x)}^2P(x,t) \big]}{\partial x^2}\big] dx + \frac{\partial x_b}{\partial t} P(x_b,t) =0$
Applying the integral to complete differentials inside square brackets, we get
$-\mu(x_b)P(x_b,t)\, + \, .5 \, \frac{\partial \big[{\sigma(x_b)}^2P(x_b,t) \big]}{\partial x} + \frac{\partial x_b}{\partial t} P(x_b,t) =0$
$=-\mu(x_b)P(x_b,t)\, + .5 \, \frac{\partial \big[{\sigma(x_b)}^2 \big]}{\partial x}P(x_b,t) + .5\,{\sigma(x_b)}^2 \, \frac{\partial P(x_b,t)}{\partial x} +\, \frac{\partial x_b}{\partial t} P(x_b,t) =0$
We have solved the equation in original coordinates and we could do an ODE in $\frac{\partial x_b}{\partial t}$ to find the evolution of every grid point so that probability mass till that grid point is conserved. But it will be a lot easier if we converted the above arbitrary density to underlying gaussian density as we have been doing and for that we write the following substitution equations from change of probability derivatives as($Z_b$ is associated with each point $x_b$ using common CDF as probability is being conserved.)
$P(x)=P(Z)\frac{\partial Z}{\partial x}$ and
$\frac{\partial P(x)}{\partial x}=\frac{\partial P(Z)}{\partial Z} {(\frac{\partial Z}{\partial x})}^2 + P(Z) \frac{\partial^2 Z}{\partial x^2}$
Substituting the above two equations in previous equation, we get
$-\mu(x_b) \, P(Z_b)\, \frac{\partial Z}{\partial x} \,+ .5 \, \frac{\partial \big[{\sigma(x_b)}^2 \big]}{\partial x}P(Z_b) \frac{\partial Z}{\partial x} + .5\,{\sigma(x_b)}^2 \, \big[ \frac{\partial P(Z_b)}{\partial Z} {(\frac{\partial Z}{\partial x})}^2 + P(Z_b) \frac{\partial^2 Z}{\partial x^2} \big] +\, \frac{\partial x_b}{\partial t} P(Z_b) \frac{\partial Z}{\partial x}=0$
We make the following substitutions in above equation
$\frac{\partial^2 Z}{\partial x^2} =-{(\frac{\partial Z}{\partial x})}^3 \frac{\partial^2 x}{\partial Z^2}$
and
$\frac{\partial P(Z_b)}{\partial Z}=-Z_b \, P(Z_b)$
$-\mu(x_b) \, P(Z_b)\, \frac{\partial Z}{\partial x} \,+ .5 \, \frac{\partial \big[{\sigma(x_b)}^2 \big]}{\partial x}P(Z_b) \frac{\partial Z}{\partial x} + .5\,{\sigma(x_b)}^2 \, (-Z_b) \, P(Z_b) {(\frac{\partial Z}{\partial x})}^2 - .5\,{\sigma(x_b)}^2 P(Z_b) {(\frac{\partial Z}{\partial x})}^3 \frac{\partial^2 x}{\partial Z^2} +\, \frac{\partial x_b}{\partial t} P(Z_b) \frac{\partial Z}{\partial x}=0$
Cancelling $P(Z_b) \frac{\partial Z}{\partial x}$ throughout the equation and rearranging, we get the first order very simple ODE for our grid point that conserves the probability mass up till that grid point.
$\frac{\partial x_b}{\partial t} =\mu(x_b) - .5 \, \frac{\partial \big[{\sigma(x_b)}^2 \big]}{\partial x}+ .5\,{\sigma(x_b)}^2 \, Z_b \, \frac{\partial Z}{\partial x} + \, .5\,{\sigma(x_b)}^2 \, {(\frac{\partial Z}{\partial x})}^2 \frac{\partial^2 x}{\partial Z^2}$
The above is the first order ODE (that has to be solved in time) for the grid point with associated CDF or conserved mass and this point seamlessly moves in time as a function of above ODE when we solve it. I was able to solve the above ODE just as such (without any squared addition of volatilities separately) and got seamless results for the evolution of CDF(or probability mass up till that point) conservation point in original coordinates.
When you evolve the SDE in original coordinates, you have to take a few steps with an alternative method (until second derivative is non-negligible) and then evolve with above method. I will be posting a program of simulation of densities of SDEs with above method in a few hours.
The above method is sister equation to continuity equation of mathematical physics. You can apply it to other first order equations or higher order equations. I am sure it can also be applied to Navier Stokes equation with some hack when some boundary is in motion and momentum is conserved. Continuity equation works when boundaries are given while this method works when boundary has to be solved.
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I was thinking of ways how to increase the order in time of solution of the FPE. One idea that struck me and I want to work on is to take a further second derivative of equation
$\frac{\partial x_b}{\partial t} =\mu(x_b) - .5 \, \frac{\partial \big[{\sigma(x_b)}^2 \big]}{\partial x}+ .5\,{\sigma(x_b)}^2 \, Z_b \, \frac{\partial Z}{\partial x} + \, .5\,{\sigma(x_b)}^2 \, {(\frac{\partial Z}{\partial x})}^2 \frac{\partial^2 x}{\partial Z^2}$
with respect to time. On RHS we will have to take a further derivative with respect to x and multiply it (using chain rule of partial derivatives) with $\frac{\partial x_b}{\partial t}$ which would have already been calculated using the above equation. Once we would have calculated $\frac{\partial^2 x_b}{\partial t^2}$ we can easily use it in a Taylor-like fashion to increase the order in time of the solution of FPE. Currently it is just a suggestion as I have not done it yet and cannot say for sure before hand if there would be any unforeseen problems. If third derivative is problematic, we could possibly restrict our solution to terms containing just first two derivatives with respect to X but all of this would have to be carefully seen.
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I noticed that our 1D SDE solution and 2D SDE solution both have instabilities close to zero. I worked on 1D SDE solution to improve its behavior close to zero. Sometimes SDE does not necessarily go to zero but its dynamics take extreme stress close to zero and it becomes unstable. I noticed when the solution is calculated to -5 SDs, there is extreme stress in parameters close to -5SD and there is always a chance of their becoming unstable. It was rather far more stable to simulate the SDE analytically to only -4.0 SDs and at the end (or when desired) extrapolate the solution to -4.5 SDs. when negative Z boundary is taken to only -4SDs, there is considerably less stress on SDE dynamics and many times it remains almost always stable even at very high volatilities. I made a few other modifications and so was able to simulate CIR SDE that goes into zero with moderate volatilities. If the volatility close to zero is too high, CIR SDE is very difficult to simulate. I have created some graphs to give friends some idea about zero behavior and accuracy of SDEs. In these graphs, CIR SDEs have volatility close to maximum possible meaning if I increased volatility somewhat more, my solution would become unstable while you could still increase volatility for SDEs with volatility exponents greater than .5. I will post my program tomorrow. I want to work on improving the zero behavior for SDEs in original coordinates and finally after that I will improve and make changes to 2D SDE simulation program. Here are the graphs of densities of some SDEs with improved zero behavior.
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Please note that in the complete above graph, the tail goes out to 17-18. It is very close to lognormal with a specially elongated tail.
Please note that in the complete above graph, the tail goes out to 8. Its tail is much smaller than the tail with exponent .95 but still it is quite pronounced.
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I have caught malaria and have very high fever. I will post the complete program after I recover. Here is the tentative program and I wanted to change a lot of things in it to make it easy to understand but now I am posting as such. I will post a reworked version once I recover.
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function [] = FPERevisitedTransProb08ABwNew04DownloadedAB1()
%Copyright Ahsan Amin. Infiniti derivatives Technologies.
%or skype ahsan.amin2999
%In this program, I am simulating the SDE given as
%dy(t)=mu1 x(t)^beta1 dt + mu2 x(t)^beta2 dt +sigma x(t)^gamma dz(t)
%I have not directly simulated the SDE but simulated the transformed
%Besse1l process version of the SDE and then changed coordinates to retreive
%the SDE in original coo
%rdinates.
%The present program will analytically evolve only the Bessel Process version of the
%SDE in transformed coordinates.
dt=.125/16/2/2; % Simulation time interval.%Fodiffusions close to zero
%decrease dt for accuracy.
Tt=128*2*2*2; % Number of simulation levels. Terminal time= Tt*dt; //.125/32*32*16=2 year;
T=Tt*dt;
OrderA=4; %
OrderM=4; %
dtM=.0625/2;
TtM=T/dtM;
dNn=.2/1; % Normal density subdivisions width. would change with number of subdivisions
Nn=41; % No of normal density subdivisions
NnMidl=18;%One half density Subdivision left from mid of normal density(low)
NnMidh=19;%One half density subdivision right from the mid of normal density(high)
NnMid=4.0;
NnT=46;
%NnD=(NnT-Nn)/2;
NnD=5;
x0=.25; % starting value of SDE
beta1=0.0;
beta2=1.0; % Second drift term power.
gamma=.75;%50; % volatility power.
kappa=1.0;%.950; %mean reversion parameter.
theta=.06;%mean reversion target
sigma0=.9;%Volatility value
%you can specify any general mu1 and mu2 and beta1 and beta2.
mu1=1*theta*kappa; %first drift coefficient.
mu2=-1*kappa; % Second drift coefficient.
%mu1=0;
%mu2=0;
alpha=1;% x^alpha is being expanded. This is currently for monte carlo only.
alpha1=1-gamma;%This is for expansion of integrals for calculation of drift
%and volatility coefficients
yy(1:Nn)=x0;
w(1:Nn)=x0^(1-gamma)/(1-gamma);
%Z(1:Nn)=(((1:Nn)-5.5)*dNn-NnMid);
Z(1:Nn)=(((1:Nn)+1.5)*dNn-NnMid);
ZT(1:NnT)=(((1:NnT)-3.5)*dNn-NnMid);
ZT
Z
str=input('Look at Z');
ZProb(1)=normcdf(.5*Z(1)+.5*Z(2),0,1)-normcdf(.5*Z(1)+.5*Z(2)-dNn,0,1);
ZProb(Nn)=normcdf(.5*Z(Nn)+.5*Z(Nn-1)+dNn,0,1)-normcdf(.5*Z(Nn)+.5*Z(Nn-1),0,1);
ZProb(2:Nn-1)=normcdf(.5*Z(2:Nn-1)+.5*Z(3:Nn),0,1)-normcdf(.5*Z(2:Nn-1)+.5*Z(1:Nn-2),0,1);
ZProbT(1)=normcdf(.5*ZT(1)+.5*ZT(2),0,1)-normcdf(.5*ZT(1)+.5*ZT(2)-dNn,0,1);
ZProbT(NnT)=normcdf(.5*ZT(NnT)+.5*ZT(NnT-1)+dNn,0,1)-normcdf(.5*ZT(NnT)+.5*ZT(NnT-1),0,1);
ZProbT(2:NnT-1)=normcdf(.5*ZT(2:NnT-1)+.5*ZT(3:NnT),0,1)-normcdf(.5*ZT(2:NnT-1)+.5*ZT(1:NnT-2),0,1);
%Above calculate probability mass in each probability subdivision.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
sigma11(1:OrderA+1)=0;
mu11(1:OrderA+1)=0;
mu22(1:OrderA+1)=0;
sigma22(1:OrderA+1)=0;
% index 1 correponds to zero level since matlab indexing starts at one.
sigma11(1)=1;
mu11(1)=1;
mu22(1)=1;
sigma22(1)=1;
for k=1:(OrderA+1)
if sigma0~=0
sigma11(k)=sigma0^(k-1);
end
if mu1 ~= 0
mu11(k)=mu1^(k-1);
end
if mu2 ~= 0
mu22(k)=mu2^(k-1);
end
if sigma0~=0
sigma22(k)=sigma0^(2*(k-1));
end
end
%Ft(1:TtM+1,1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0; %General time powers on hermite polynomials
Fp(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;%General x powers on coefficients of hermite polynomials.
Fp1(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;%General x powers for bessel transformed coordinates.
%YCoeff0 and YCoeff are coefficents for original coordinates monte carlo.
%YqCoeff0 and YqCoeff are bessel/lamperti version monte carlo.
YCoeff0(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;
YqCoeff0(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;
%Pre-compute the time and power exponent values in small multi-dimensional arrays
YCoeff = ItoTaylorCoeffsNew(alpha,beta1,beta2,gamma); %expand y^alpha where alpha=1;
YqCoeff = ItoTaylorCoeffsNew(alpha1,beta1,beta2,gamma);%expand y^alpha1 where alpha1=(1-gamma)
YqCoeff=YqCoeff/(1-gamma); %Transformed coordinates coefficients have to be
%further divided by (1-gamma)
for k = 0 : (OrderA)
for m = 0:k
l4 = k - m + 1;
for n = 0 : m
l3 = m - n + 1;
for j = 0:n
l2 = n - j + 1;
l1 = j + 1;
%Ft(l1,l2,l3,l4) = dtM^((l1-1) + (l2-1) + (l3-1) + .5* (l4-1));
Fp(l1,l2,l3,l4) = (alpha + (l1-1) * beta1 + (l2-1) * beta2 + (l3-1) * 2* gamma + (l4-1) * gamma ...
- (l1-1) - (l2-1) - 2* (l3-1) - (l4-1));
Fp1(l1,l2,l3,l4) = (alpha1 + (l1-1) * beta1 + (l2-1) * beta2 + (l3-1) * 2* gamma + (l4-1) * gamma ...
- (l1-1) - (l2-1) - 2* (l3-1) - (l4-1));
YCoeff0(l1,l2,l3,l4) =YCoeff(l1,l2,l3,l4).*mu11(l1).*mu22(l2).*sigma22(l3).*sigma11(l4);
YqCoeff0(l1,l2,l3,l4) =YqCoeff(l1,l2,l3,l4).*mu11(l1).*mu22(l2).*sigma22(l3).*sigma11(l4);
end
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
wnStart=1;%
tic
Zt1(wnStart:Nn)=0.0;
Zt2(wnStart:Nn)=0.0;
for tt=1:Tt
%[wMu0dt,dwMu0dtdw,c1] = CalculateDriftAndVolA4(w,wnStart,Nn,YqCoeff0,Fp1,gamma,dt);
%[wMu0dt,c1,c2] = CalculateDriftAndVolA404AB(w,wnStart,Nn,YqCoeff0,Fp1,gamma,dt);
[wMu0dt,c1,c2,dwMu0dtdw] = CalculateDriftAndVolA404AB(w,wnStart,Nn,YqCoeff0,Fp1,gamma,dt);
[wMid] = InterpolateOrderN8(8,0,Z(NnMidl-3),Z(NnMidl-2),Z(NnMidl-1),Z(NnMidl),Z(NnMidh),Z(NnMidh+1),Z(NnMidh+2),Z(NnMidh+3),w(NnMidl-3),w(NnMidl-2),w(NnMidl-1),w(NnMidl),w(NnMidh),w(NnMidh+1),w(NnMidh+2),w(NnMidh+3));
Zt1(wnStart:Nn)=w(wnStart:Nn)-wMid;
[dwdZ,d2wdZ2A] = First2Derivatives2ndOrderEqSpacedA(wnStart,Nn,dNn,Zt1,Z);
[dwdZ,d2wdZ2,d3wdZ3] = First3Derivatives4thOrderEqSpaced(wnStart,Nn,dNn,Zt1,Z);
tt
C0(wnStart:Nn)=Zt1(wnStart:Nn)-dwdZ(wnStart:Nn).*Z(wnStart:Nn);
%c1(wnStart:Nn)=sigma0*sqrt(dt);
Zt2(wnStart:Nn)=C0(wnStart:Nn)+sign(dwdZ(wnStart:Nn)+c1(wnStart:Nn)).* ...
abs(sqrt(sign(dwdZ(wnStart:Nn)).*(dwdZ(wnStart:Nn)).^2+ ...
sign(c1(wnStart:Nn)).*c1(wnStart:Nn).^2)).*Z(wnStart:Nn);
[dwdZ,d2wdZ2,d3wdZ3] = First3Derivatives2ndOrderEqSpacedA(wnStart,Nn,dNn,Zt2,Z);
[dw2dZ,d2w2dZ2A] = First2Derivatives2ndOrderEqSpacedA(wnStart,Nn,dNn,Zt2,Z);
[dw2dZ,d2w2dZ2A,d3wdZ3] = First3Derivatives4thOrderEqSpaced(wnStart,Nn,dNn,Zt2,Z);
if(tt>10)
% dwdt(wnStart:Nn)=(wMu0dt(wnStart:Nn)+Zt2(wnStart:Nn)-Zt1(wnStart:Nn)+ ...
% .5*sigma0^2*(dw2dZ(wnStart:Nn).^(-2)).*d2w2dZ2A(wnStart:Nn).*dt)/dt;
else
% dwdt(wnStart:Nn)=(wMu0dt(wnStart:Nn)+Zt2(wnStart:Nn)-Zt1(wnStart:Nn))/dt;%+ ...
end
wnStart
if(tt>10)
dZdw(wnStart:Nn)=1./dw2dZ(wnStart:Nn);
d2Zdw2(wnStart:Nn)=-d2w2dZ2A(wnStart:Nn).*dZdw(wnStart:Nn).^3;
d3Zdw3(wnStart:Nn)=-d3wdZ3(wnStart:Nn).*dZdw(wnStart:Nn).^4+3*d2w2dZ2A(wnStart:Nn).^2.*dZdw(wnStart:Nn).^5;
%w(wnStart:Nn)=w(wnStart:Nn)+ ...
% wMu0dt(wnStart:Nn)+ ...
% 0*dwMu0dtdw(wnStart:Nn)./(-Z(wnStart:Nn).*dZdw(wnStart:Nn).^2+d2Zdw2(wnStart:Nn)) - ...
% +.5*sigma0^2*dt*( (Z(wnStart:Nn).^2-1).*dZdw(wnStart:Nn).^3- ...
% 0* 3*Z(wnStart:Nn).*dZdw(wnStart:Nn).*d2Zdw2(wnStart:Nn))./(-Z(wnStart:Nn).*dZdw(wnStart:Nn).^2+d2Zdw2(wnStart:Nn));
w(wnStart:Nn)=w(wnStart:Nn)+ ...
wMu0dt(wnStart:Nn)+1*(Zt2(wnStart:Nn)-Zt1(wnStart:Nn))+ ...
1*.5*sigma0^2*(dw2dZ(wnStart:Nn).^(-2)).*d2w2dZ2A(wnStart:Nn).*dt;
else
w(wnStart:Nn)=w(wnStart:Nn)+ ...
wMu0dt(wnStart:Nn)+Zt2(wnStart:Nn)-Zt1(wnStart:Nn);
end
% %yy(wnStart:Nn)=((1-gamma).*w(wnStart:Nn)).^(1/(1-gamma));
% [dwdZ,d2wdZ2A,d3wdZ3] = First3Derivatives4thOrderEqSpaced(wnStart,Nn,dNn,w,Z);
%FirstCoordinate
% for nn=NnMidl-11:-1:wnStart+1
% if((w(nn)-w(nn-1))>(w(nn+1)-w(nn)))
%
% w(nn) = InterpolateOrderN8(6,Z(nn),Z(nn+1),Z(nn+2),Z(nn+3),Z(nn+4),Z(nn+5),Z(nn+6),Z(nn+7),Z(nn+8),w(nn+1),w(nn+2),w(nn+3),w(nn+4),w(nn+5),w(nn+6),w(nn+7),w(nn+8));
%
% end
% end
% if(w(wnStart+1)-w(wnStart)>w(wnStart+2)-w(wnStart+1))
% nn=wnStart;
% w(wnStart) = InterpolateOrderN8(6,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),Z(wnStart+8),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7),w(wnStart+8));
% w(nn)=w(nn+1)-(w(nn+2)-w(nn+1));
% end
[wE] = InterpolateOrderN6(6,Z(Nn)+dNn,Z(Nn),Z(Nn-1),Z(Nn-2),Z(Nn-3),Z(Nn-4),Z(Nn-5),w(Nn),w(Nn-1),w(Nn-2),w(Nn-3),w(Nn-4),w(Nn-5));
for nn=wnStart:Nn-1
if(w(nn)>w(nn+1))
w(nn)=0;
end
end
% w1(wnStart:Nn-1)=w(wnStart:Nn-1);
% w1(Nn)=w(Nn);
% w2(wnStart:Nn-1)=w(wnStart+1:Nn);
% w2(Nn)=wE;
% w(w1(wnStart:Nn)>w2(:))=0;%Be careful;might not universally hold;
% %Change 3:7/25/2020: I have improved zero correction in above.
%for nn=1:Nn
% if(w2(nn)>w1(nn))
w(w<0)=0.0;
StartChangeFlag=0;
nnChange=0;
wnStartPrev=wnStart;
for nn=wnStart:Nn
if(w(nn)<=0)
wnStart=nn+1;
StartChangeFlag=1;
nnChange=nnChange+1;
end
end
%nn=wnStart;
%while(w(nn)<=0)
% wnStart=nn+1;
% nn=nn+1;
%end
InterpolatePrevFlag=1;
wnChange=wnStart-wnStartPrev;
Zs=0
ws=0;
for mm=1:nnChange
Zs(mm)=Z(wnStart)-mm*dNn;
ws(mm)=0.0
end
if(nnChange>=1)
ws = InterpolateOrderN8(8,Zs,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
ContinueFlag=1;
end
for mm=1:nnChange
if((ws(mm)>0)&&(ws(mm)<w(wnStart))&&(ContinueFlag==1))
wnStart=wnStart-1;
w(wnStart)=ws(mm);
ContinueFlag=1;
else
ContinueFlag=0;
end
end
% if(InterpolatePrevFlag==1)
% InterpolatePrevFlag=0;
% %wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% NnStart=nnChange
% wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% if(wStart1>0)
% wnStart=wnStart-1;
% w(wnStart)=wStart1;
% InterpolatePrevFlag=1;
% end
% if(wnStart==wnStartPrev)
% % InterpolatePrevFlag=0;
% end
% end
% end
% if(nnChange>=1)
% wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% if(wStart1>0)
% wnStart=wnStart-1
% w(wnStart)=wStart1;
% InterpolatePrevFlag=1;
% end
% end
% if((nnChange>=2)&&( InterpolatePrevFlag==1))
% wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% InterpolatePrevFlag=0;
% if(wStart1>0)
% wnStart=wnStart-1
% w(wnStart)=wStart1;
% InterpolatePrevFlag=1;
% end
% end
% if((nnChange>=3)&&( InterpolatePrevFlag==1))
% wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% InterpolatePrevFlag=0;
% if(wStart1>0)
% wnStart=wnStart-1
% w(wnStart)=wStart1;
% InterpolatePrevFlag=1;
% end
% end
%
w
wnStart
tt
end
wT(wnStart+NnD:Nn+NnD)=w(wnStart:Nn);
Zs=0
ws=0;
for mm=1:wnStart+NnD-1
Zs(mm)=Z(wnStart)-mm*dNn;
ws(mm)=0.0
end
wnStartT=wnStart+NnD;
%NnT=Nn+2*NnD;
ws = InterpolateOrderN8(8,Zs,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
ContinueFlag=1;
wnStart0=wnStart+NnD-1;
for mm=1:wnStart0
if((ws(mm)>0)&&(ws(mm)<wT(wnStartT))&&(ContinueFlag==1))
wnStartT=wnStartT-1;
wT(wnStartT)=ws(mm);
ContinueFlag=1;
else
ContinueFlag=0;
end
end
% Zs=0;
% ws=0;
% for mm=1:NnD
% Zs(mm)=Z(Nn)+mm*dNn;
% ws(mm)=0.0
% end
%
% ws = InterpolateOrderN8(2,Zs,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
%
%
% %for mm=1:NnD
%
% wT(Nn+NnD+1:Nn+2*NnD)=ws(1:NnD);
% %end
% for mm=1:wnStart0-1
% if((ws(mm)>0)&&(ws(mm)<w(wnStart))&&(ContinueFlag==1))
% wnStart=wnStart-1;
% w(wnStart)=ws(mm);
% ContinueFlag=1;
% else
% ContinueFlag=0;
% end
% end
wnStartT
%str=input('Look at numbers')
yyT(wnStartT:NnT)=((1-gamma).*wT(wnStartT:NnT)).^(1/(1-gamma));
DfyyT(wnStartT:NnT)=0;
for nn=wnStartT+1:NnT-1
DfyyT(nn) = (yyT(nn + 1) - yyT(nn - 1))/(ZT(nn + 1) - ZT(nn - 1));
%Change of variable derivative for densities
end
pyyT(1:Nn)=0;
for nn = wnStartT:NnT-1
pyyT(nn) = (normpdf(ZT(nn),0, 1))/abs(DfyyT(nn));
end
if(tt>=23)
%plot(yy(wnStart:Nn),pyy(wnStart:Nn),'r');
%yy
%str=input('Look at the output graph');
end
toc
ItoHermiteMean=sum(yyT(wnStartT+1:NnT-1).*ZProbT(wnStartT+1:NnT-1)) %Original process average from coordinates
disp('true Mean only applicable to standard SV mean reverting type models otherwise disregard');
TrueMean=theta+(x0-theta)*exp(-kappa*dt*Tt)%Mean reverting SDE original variable true average
theta1=1;
rng(29079137, 'twister')
paths=200000;
YY(1:paths)=x0; %Original process monte carlo.
Random1(1:paths)=0;
for tt=1:TtM
Random1=randn(size(Random1));
HermiteP1(1,1:paths)=1;
HermiteP1(2,1:paths)=Random1(1:paths);
HermiteP1(3,1:paths)=Random1(1:paths).^2-1;
HermiteP1(4,1:paths)=Random1(1:paths).^3-3*Random1(1:paths);
HermiteP1(5,1:paths)=Random1(1:paths).^4-6*Random1(1:paths).^2+3;
YY(1:paths)=YY(1:paths) + ...
(YCoeff0(1,1,2,1).*YY(1:paths).^Fp(1,1,2,1)+ ...
YCoeff0(1,2,1,1).*YY(1:paths).^Fp(1,2,1,1)+ ...
YCoeff0(2,1,1,1).*YY(1:paths).^Fp(2,1,1,1))*dtM + ...
(YCoeff0(1,1,3,1).*YY(1:paths).^Fp(1,1,3,1)+ ...
YCoeff0(1,2,2,1).*YY(1:paths).^Fp(1,2,2,1)+ ...
YCoeff0(2,1,2,1).*YY(1:paths).^Fp(2,1,2,1)+ ...
YCoeff0(1,3,1,1).*YY(1:paths).^Fp(1,3,1,1)+ ...
YCoeff0(2,2,1,1).*YY(1:paths).^Fp(2,2,1,1)+ ...
YCoeff0(3,1,1,1).*YY(1:paths).^Fp(3,1,1,1))*dtM^2 + ...
((YCoeff0(1,1,1,2).*YY(1:paths).^Fp(1,1,1,2).*sqrt(dtM))+ ...
(YCoeff0(1,1,2,2).*YY(1:paths).^Fp(1,1,2,2)+ ...
YCoeff0(1,2,1,2).*YY(1:paths).^Fp(1,2,1,2)+ ...
YCoeff0(2,1,1,2).*YY(1:paths).^Fp(2,1,1,2)).*dtM^1.5) .*HermiteP1(2,1:paths) + ...
((YCoeff0(1,1,1,3).*YY(1:paths).^Fp(1,1,1,3) *dtM) + ...
(YCoeff0(1,1,2,3).*YY(1:paths).^Fp(1,1,2,3)+ ...
YCoeff0(1,2,1,3).*YY(1:paths).^Fp(1,2,1,3)+ ...
YCoeff0(2,1,1,3).*YY(1:paths).^Fp(2,1,1,3)).*dtM^2).*HermiteP1(3,1:paths) + ...
((YCoeff0(1,1,1,4).*YY(1:paths).^Fp(1,1,1,4)*dtM^1.5 )).*HermiteP1(4,1:paths) + ...
(YCoeff0(1,1,1,5).*YY(1:paths).^Fp(1,1,1,5)*dtM^2.0).*HermiteP1(5,1:paths);
end
YY(YY<0)=0;
disp('Original process average from monte carlo');
MCMean=sum(YY(:))/paths %origianl coordinates monte carlo average.
disp('Original process average from our simulation');
ItoHermiteMean=sum(yyT(wnStartT+1:NnT-1).*ZProbT(wnStartT+1:NnT-1)) %Original process average from coordinates
disp('true Mean only applicble to standard SV mean reverting type models otherwise disregard');
TrueMean=theta+(x0-theta)*exp(-kappa*dt*Tt)%Mean reverting SDE original variable true average
MaxCutOff=30;
NoOfBins=round(300*gamma^2*4*sigma0/sqrt(MCMean)/(1+kappa));%Decrease the number of bins if the graph is too
[YDensity,IndexOutY,IndexMaxY] = MakeDensityFromSimulation_Infiniti_NEW(YY,paths,NoOfBins,MaxCutOff );
plot(yyT(wnStartT+1:NnT-1),pyyT(wnStartT+1:NnT-1),'r',IndexOutY(1:IndexMaxY),YDensity(1:IndexMaxY),'g');
%plot(y_w(wnStart+1:Nn-1),fy_w(wnStart+1:Nn-1),'r',IndexOutY(1:IndexMaxY),YDensity(1:IndexMaxY),'g',Z(wnStart+1:Nn-1),fy_w(wnStart+1:Nn-1),'b');
title(sprintf('x0 = %.4f,theta=%.3f,kappa=%.2f,gamma=%.3f,sigma=%.2f,T=%.2f,dt=%.5f,M=%.4f,TM=%.4f', x0,theta,kappa,gamma,sigma0,T,dt,ItoHermiteMean,TrueMean));%,sprintf('theta= %f', theta), sprintf('kappa = %f', kappa),sprintf('sigma = %f', sigma0),sprintf('T = %f', T));
legend({'Ito-Hermite Density','Monte Carlo Density'},'Location','northeast')
str=input('red line is density of SDE from Ito-Hermite method, green is monte carlo.');
end
.
.
function [wMu0dt,c1,c2,dwMu0dtdw] = CalculateDriftAndVolA404AB(w,wnStart,Nn,YqCoeff0,Fp1,gamma,dt)
yy(wnStart:Nn)=((1-gamma)*w(wnStart:Nn)).^(1/(1-gamma));
Fp2=Fp1/(1-gamma);
wMu0dt(wnStart:Nn)=(YqCoeff0(1,1,2,1).*yy(wnStart:Nn).^Fp1(1,1,2,1)+ ...
YqCoeff0(1,2,1,1).*yy(wnStart:Nn).^Fp1(1,2,1,1)+ ...
YqCoeff0(2,1,1,1).*yy(wnStart:Nn).^Fp1(2,1,1,1))*dt + ...
(YqCoeff0(1,1,3,1).*yy(wnStart:Nn).^Fp1(1,1,3,1)+ ...
YqCoeff0(1,2,2,1).*yy(wnStart:Nn).^Fp1(1,2,2,1)+ ...
YqCoeff0(2,1,2,1).*yy(wnStart:Nn).^Fp1(2,1,2,1)+ ...
YqCoeff0(1,3,1,1).*yy(wnStart:Nn).^Fp1(1,3,1,1)+ ...
YqCoeff0(2,2,1,1).*yy(wnStart:Nn).^Fp1(2,2,1,1)+ ...
YqCoeff0(3,1,1,1).*yy(wnStart:Nn).^Fp1(3,1,1,1))*dt^2 + ...
(YqCoeff0(1,1,4,1).*yy(wnStart:Nn).^Fp1(1,1,4,1)+ ...
YqCoeff0(1,2,3,1).*yy(wnStart:Nn).^Fp1(1,2,3,1)+ ...
YqCoeff0(2,1,3,1).*yy(wnStart:Nn).^Fp1(2,1,3,1)+ ...
YqCoeff0(1,3,2,1).*yy(wnStart:Nn).^Fp1(1,3,2,1)+ ...
YqCoeff0(2,2,2,1).*yy(wnStart:Nn).^Fp1(2,2,2,1)+ ...
YqCoeff0(3,1,2,1).*yy(wnStart:Nn).^Fp1(3,1,2,1)+ ...
YqCoeff0(1,4,1,1).*yy(wnStart:Nn).^Fp1(1,4,1,1)+ ...
YqCoeff0(2,3,1,1).*yy(wnStart:Nn).^Fp1(2,3,1,1)+ ...
YqCoeff0(3,2,1,1).*yy(wnStart:Nn).^Fp1(3,2,1,1)+ ...
YqCoeff0(4,1,1,1).*yy(wnStart:Nn).^Fp1(4,1,1,1))*dt^3+ ...
(YqCoeff0(1,1,5,1).*yy(wnStart:Nn).^Fp1(1,1,5,1)+ ...
YqCoeff0(1,2,4,1).*yy(wnStart:Nn).^Fp1(1,2,4,1)+ ...
YqCoeff0(2,1,4,1).*yy(wnStart:Nn).^Fp1(2,1,4,1)+ ...
YqCoeff0(1,3,3,1).*yy(wnStart:Nn).^Fp1(1,3,3,1)+ ...
YqCoeff0(2,2,3,1).*yy(wnStart:Nn).^Fp1(2,2,3,1)+ ...
YqCoeff0(3,1,3,1).*yy(wnStart:Nn).^Fp1(3,1,3,1)+ ...
YqCoeff0(1,4,2,1).*yy(wnStart:Nn).^Fp1(1,4,2,1)+ ...
YqCoeff0(2,3,2,1).*yy(wnStart:Nn).^Fp1(2,3,2,1)+ ...
YqCoeff0(3,2,2,1).*yy(wnStart:Nn).^Fp1(3,2,2,1)+ ...
YqCoeff0(4,1,2,1).*yy(wnStart:Nn).^Fp1(4,1,2,1)+ ...
YqCoeff0(1,5,1,1).*yy(wnStart:Nn).^Fp1(1,5,1,1)+ ...
YqCoeff0(2,4,1,1).*yy(wnStart:Nn).^Fp1(2,4,1,1)+ ...
YqCoeff0(3,3,1,1).*yy(wnStart:Nn).^Fp1(3,3,1,1)+ ...
YqCoeff0(4,2,1,1).*yy(wnStart:Nn).^Fp1(4,2,1,1)+ ...
YqCoeff0(5,1,1,1).*yy(wnStart:Nn).^Fp1(5,1,1,1))*dt^4;
dwMu0dtdw(wnStart:Nn)=(YqCoeff0(1,1,2,1).*Fp1(1,1,2,1).*((1-gamma)*w(wnStart:Nn)).^(Fp2(1,1,2,1)-1)+ ...
YqCoeff0(1,2,1,1).*Fp1(1,2,1,1).*((1-gamma)*w(wnStart:Nn)).^(Fp2(1,2,1,1)-1)+ ...
YqCoeff0(2,1,1,1).*Fp1(2,1,1,1).*((1-gamma)*w(wnStart:Nn)).^(Fp2(2,1,1,1)-1))*dt ;
%wMu0dtOdt(wnStart:Nn)=(YqCoeff0(1,1,2,1).*yy(wnStart:Nn).^Fp1(1,1,2,1)*0+ ...
% YqCoeff0(1,2,1,1).*yy(wnStart:Nn).^Fp1(1,2,1,1)+ ...
% YqCoeff0(2,1,1,1).*yy(wnStart:Nn).^Fp1(2,1,1,1))*dt ;
c1(wnStart:Nn)=((YqCoeff0(1,1,1,2).*yy(wnStart:Nn).^Fp1(1,1,1,2).*sqrt(dt))+ ...
(YqCoeff0(1,1,2,2).*yy(wnStart:Nn).^Fp1(1,1,2,2)+YqCoeff0(1,2,1,2).*yy(wnStart:Nn).^Fp1(1,2,1,2)+ ...
YqCoeff0(2,1,1,2).*yy(wnStart:Nn).^Fp1(2,1,1,2)).*dt^1.5+ ...
(YqCoeff0(1,1,3,2).*yy(wnStart:Nn).^Fp1(1,1,3,2)+YqCoeff0(1,2,2,2).*yy(wnStart:Nn).^Fp1(1,2,2,2)+ ...
YqCoeff0(2,1,2,2).*yy(wnStart:Nn).^Fp1(2,1,2,2)+YqCoeff0(1,3,1,2).*yy(wnStart:Nn).^Fp1(1,3,1,2)+ ...
YqCoeff0(2,2,1,2).*yy(wnStart:Nn).^Fp1(2,2,1,2)+YqCoeff0(3,1,1,2).*yy(wnStart:Nn).^Fp1(3,1,1,2)).*dt^2.5+ ...
(YqCoeff0(1,1,4,2).*yy(wnStart:Nn).^Fp1(1,1,4,2)+YqCoeff0(1,2,3,2).*yy(wnStart:Nn).^Fp1(1,2,3,2)+ ...
YqCoeff0(2,1,3,2).*yy(wnStart:Nn).^Fp1(2,1,3,2)+YqCoeff0(1,3,2,2).*yy(wnStart:Nn).^Fp1(1,3,2,2)+ ...
YqCoeff0(2,2,2,2).*yy(wnStart:Nn).^Fp1(2,2,2,2)+ YqCoeff0(3,1,2,2).*yy(wnStart:Nn).^Fp1(3,1,2,2)+ ...
YqCoeff0(1,4,1,2).*yy(wnStart:Nn).^Fp1(1,4,1,2)+YqCoeff0(2,3,1,2).*yy(wnStart:Nn).^Fp1(2,3,1,2)+ ...
YqCoeff0(3,2,1,2).*yy(wnStart:Nn).^Fp1(3,2,1,2)+YqCoeff0(4,1,1,2).*yy(wnStart:Nn).^Fp1(4,1,1,2)).*dt^3.5);
c2(wnStart:Nn)=(YqCoeff0(1,1,2,3).*yy(wnStart:Nn).^Fp1(1,1,2,3)+YqCoeff0(1,2,1,3).*yy(wnStart:Nn).^Fp1(1,2,1,3)+ ...
YqCoeff0(2,1,1,3).*yy(wnStart:Nn).^Fp1(2,1,1,3)).*dt^2;%+ ...
(YqCoeff0(1,1,3,3).*yy(wnStart:Nn).^Fp1(1,1,3,3)+YqCoeff0(1,2,2,3).*yy(wnStart:Nn).^Fp1(1,2,2,3)+ ...
YqCoeff0(2,1,2,3).*yy(wnStart:Nn).^Fp1(2,1,2,3) + YqCoeff0(1,3,1,3).*yy(wnStart:Nn).^Fp1(1,3,1,3)+ ...
YqCoeff0(2,2,1,3).*yy(wnStart:Nn).^Fp1(2,2,1,3)+YqCoeff0(3,1,1,3).*yy(wnStart:Nn).^Fp1(3,1,1,3)).*dt^3;
% (YCoeff0(1,1,1,2).*YY(1:paths).^Fp(1,1,1,2).*sqrt(dtM))+ ...
% (YCoeff0(1,1,2,2).*YY(1:paths).^Fp(1,1,2,2)+YCoeff0(1,2,1,2).*YY(1:paths).^Fp(1,2,1,2)+ ...
% YCoeff0(2,1,1,2).*YY(1:paths).^Fp(2,1,1,2)).*dtM^1.5+ ...
% (YCoeff0(1,1,3,2).*YY(1:paths).^Fp(1,1,3,2)+YCoeff0(1,2,2,2).*YY(1:paths).^Fp(1,2,2,2)+ ...
% YCoeff0(2,1,2,2).*YY(1:paths).^Fp(2,1,2,2)+YCoeff0(1,3,1,2).*YY(1:paths).^Fp(1,3,1,2)+ ...
% YCoeff0(2,2,1,2).*YY(1:paths).^Fp(2,2,1,2)+YCoeff0(3,1,1,2).*YY(1:paths).^Fp(3,1,1,2)).*dtM^2.5+ ...
% (YCoeff0(1,1,4,2).*YY(1:paths).^Fp(1,1,4,2)+YCoeff0(1,2,3,2).*YY(1:paths).^Fp(1,2,3,2)+ ...
% YCoeff0(2,1,3,2).*YY(1:paths).^Fp(2,1,3,2)+YCoeff0(1,3,2,2).*YY(1:paths).^Fp(1,3,2,2)+ ...
% YCoeff0(2,2,2,2).*YY(1:paths).^Fp(2,2,2,2)+ YCoeff0(3,1,2,2).*YY(1:paths).^Fp(3,1,2,2)+ ...
% YCoeff0(1,4,1,2).*YY(1:paths).^Fp(1,4,1,2)+YCoeff0(2,3,1,2).*YY(1:paths).^Fp(2,3,1,2)+ ...
% YCoeff0(3,2,1,2).*YY(1:paths).^Fp(3,2,1,2)+YCoeff0(4,1,1,2).*YY(1:paths).^Fp(4,1,1,2)).*dtM^3.5) .*HermiteP1(2,1:paths)
end
.
.
.
function [dwdZ,d2wdZ2,d3wdZ3] = First3Derivatives4thOrderEqSpaced(wnStart,Nn,dNn,w,Z)
% [wS] = InterpolateOrderN6(4,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5));
% [wE] = InterpolateOrderN6(4,Z(Nn)+dNn,Z(Nn),Z(Nn-1),Z(Nn-2),Z(Nn-3),Z(Nn-4),Z(Nn-5),w(Nn),w(Nn-1),w(Nn-2),w(Nn-3),w(Nn-4),w(Nn-5));
% [wS] = InterpolateOrderN8(7,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% [wE] = InterpolateOrderN8(7,Z(Nn)+dNn,Z(Nn),Z(Nn-1),Z(Nn-2),Z(Nn-3),Z(Nn-4),Z(Nn-5),Z(Nn-6),Z(Nn-7),w(Nn),w(Nn-1),w(Nn-2),w(Nn-3),w(Nn-4),w(Nn-5),w(Nn-6),w(Nn-7));
[wS] = InterpolateOrderN6(6,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5));
[wE] = InterpolateOrderN6(6,Z(Nn)+dNn,Z(Nn),Z(Nn-1),Z(Nn-2),Z(Nn-3),Z(Nn-4),Z(Nn-5),w(Nn),w(Nn-1),w(Nn-2),w(Nn-3),w(Nn-4),w(Nn-5));
[wS2] = InterpolateOrderN6(6,Z(wnStart)-2*dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5));
[wE2] = InterpolateOrderN6(6,Z(Nn)+2*dNn,Z(Nn),Z(Nn-1),Z(Nn-2),Z(Nn-3),Z(Nn-4),Z(Nn-5),w(Nn),w(Nn-1),w(Nn-2),w(Nn-3),w(Nn-4),w(Nn-5));
% Zi=[Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),Z(wnStart+8),Z(wnStart+9)];
% wi=[w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7),w(wnStart+8),w(wnStart+9)];
% % Zi=[Z(Nn-6),Z(Nn-5),Z(Nn-4),Z(Nn-3),Z(Nn-2),Z(Nn-1),Z(Nn)];
% % wi=[w(Nn-6),w(Nn-5),w(Nn-4),w(Nn-3),w(Nn-2),w(Nn-1),w(Nn)];
%
% Zi=[Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6)];
% wi=[w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6)];
% Zq=Z(wnStart)-dNn;
% wS=spline(Zi,wi,Zq);
%
% Zq=Z(wnStart)-2*dNn;
% wS2=spline(Zi,wi,Zq);
%
%
%
% % Zi=[Z(Nn-9),Z(Nn-8),Z(Nn-7),Z(Nn-6),Z(Nn-5),Z(Nn-4),Z(Nn-3),Z(Nn-2),Z(Nn-1),Z(Nn)];
% % wi=[w(Nn-9),w(Nn-8),w(Nn-7),w(Nn-6),w(Nn-5),w(Nn-4),w(Nn-3),w(Nn-2),w(Nn-1),w(Nn)];
% Zi=[Z(Nn-6),Z(Nn-5),Z(Nn-4),Z(Nn-3),Z(Nn-2),Z(Nn-1),Z(Nn)];
% wi=[w(Nn-6),w(Nn-5),w(Nn-4),w(Nn-3),w(Nn-2),w(Nn-1),w(Nn)];
% Zq=Z(Nn)+dNn;
% wE=spline(Zi,wi,Zq);
%
%
% Zq=Z(Nn)+2*dNn;
% wE2=spline(Zi,wi,Zq);
ZS=Z(wnStart)-dNn;
ZE=Z(Nn)+dNn;
% d2wdZ2(wnStart)=(Z(wnStart)*wS-2*w(wnStart)*Z(wnStart)+w(wnStart+1).*Z(wnStart))/(dNn.^2);
% d2wdZ2(wnStart+1:Nn-1)=(w(wnStart+1:Nn-1).*Z(wnStart+1:Nn-1)-2*w(wnStart+1:Nn-1).*Z(wnStart+1:Nn-1)+1*w(wnStart+2:Nn).*Z(wnStart+1:Nn-1))/(dNn.^2);
% d2wdZ2(Nn)=(wE.*Z(Nn)-2*w(Nn).*Z(Nn)+w(Nn-1).*Z(Nn))/(dNn.^2);
dwdZ(wnStart+2:Nn-2)=(1*w(wnStart:Nn-4)-8*w(wnStart+1:Nn-3)+0*w(wnStart+2:Nn-2)+8*w(wnStart+3:Nn-1)-1*w(wnStart+4:Nn))/(12*dNn);
dwdZ(wnStart+1)=(-3*w(wnStart)-10*w(wnStart+1)+18*w(wnStart+2)-6*w(wnStart+3)+1*w(wnStart+4))/(12*dNn);
dwdZ(wnStart)=(-25/12*w(wnStart)+4*w(wnStart+1)-3*w(wnStart+2)+4/3*w(wnStart+3)-1/4*w(wnStart+4))/dNn;%/(12*dNn);
dwdZ(Nn-1)=(-1*w(Nn-4)+6*w(Nn-3)-18*w(Nn-2)+10*w(Nn-1)+3*w(Nn))/(12*dNn);
dwdZ(Nn)=(1/4*w(Nn-4)-4/3*w(Nn-3)+3*w(Nn-2)-4*w(Nn-1)+25/12*w(Nn))/dNn;%/(12*dNn);
%−25/12 4 −3 4/3 −1/4
%(-3*f[i-1]-10*f[i+0]+18*f[i+1]-6*f[i+2]+1*f[i+3])/(12*1.0*h**1)
% (-1*f[i-3]+6*f[i-2]-18*f[i-1]+10*f[i+0]+3*f[i+1])/(12*1.0*h**1)
d2wdZ2(wnStart+2:Nn-2)=(-1*w(wnStart:Nn-4)+16*w(wnStart+1:Nn-3)-30*w(wnStart+2:Nn-2)+16*w(wnStart+3:Nn-1)-1*w(wnStart+4:Nn))/(12*dNn^2);
d2wdZ2(wnStart+1)=(10*w(wnStart)-15*w(wnStart+1)-4*w(wnStart+2)+14*w(wnStart+3)-6*w(wnStart+4)+1*w(wnStart+5))/(12*dNn^2);
d2wdZ2(wnStart)=(15/4*w(wnStart)-77/6*w(wnStart+1)+107/6*w(wnStart+2)-13*w(wnStart+3)+61/12*w(wnStart+4)-5/6*w(wnStart+5))/(dNn^2);
d2wdZ2(Nn-1)=(10*w(Nn)-15*w(Nn-1)-4*w(Nn-2)+14*w(Nn-3)-6*w(Nn-4)+1*w(Nn-5))/(12*dNn^2);
%d2wdZ2(Nn)=(-1*w(Nn-2)+16*w(Nn-1)-30*w(Nn)+16*wE-1*wE2)/(12*dNn^2);
d2wdZ2(Nn)=(15/4*w(Nn)-77/6*w(Nn-1)+107/6*w(Nn-2)-13*w(Nn-3)+61/12*w(Nn-4)-5/6*w(Nn-5))/(dNn^2);
% (1*f[i-4]-6*f[i-3]+14*f[i-2]-4*f[i-1]-15*f[i+0]+10*f[i+1])/(12*1.0*h**2)
% (10*f[i-1]-15*f[i+0]-4*f[i+1]+14*f[i+2]-6*f[i+3]+1*f[i+4])/(12*1.0*h**2)
% 15/4 −77/6 107/6 −13 61/12 −5/6
%
%
% d2wdZ2(wnStart)=(wS-2*w(wnStart)+w(wnStart+1))/(dNn.^2);
% d2wdZ2(wnStart+1:Nn-1)=(w(wnStart:Nn-2)-2*w(wnStart+1:Nn-1)+1*w(wnStart+2:Nn))/(dNn.^2);
% d2wdZ2(Nn)=(wE-2*w(Nn)+w(Nn-1))/(dNn.^2);
%
%
% dwdZ(wnStart)=(-1*wS+1*w(wnStart+1))/(2*dNn);
% dwdZ(wnStart+1:Nn-1)=(-1*w(wnStart:Nn-2)+1*w(wnStart+2:Nn))/(2*dNn);
% dwdZ(Nn)=(wE-w(Nn-1))/(2*dNn);
%
d3wdZ3(wnStart+2:Nn-2)=(-1*w(wnStart:Nn-4)+2*w(wnStart+1:Nn-3)+0*w(wnStart+2:Nn-2)-2*w(wnStart+3:Nn-1)+w(wnStart+4:Nn))/(2*dNn^3);
d3wdZ3(wnStart+1)=(-1*wS+2*w(wnStart)+0*w(wnStart+1)-2*w(wnStart+2)+w(wnStart+3))/(2*dNn^3);
d3wdZ3(wnStart)=(-1*wS2+2*wS+0*w(wnStart)-2*w(wnStart+1)+w(wnStart+2))/(2*dNn^3);
d3wdZ3(Nn-1)=(-1*w(Nn-3)+2*w(Nn-2)+0*w(Nn-1)-2*w(Nn)+wE)/(2*dNn^3);
d3wdZ3(Nn)=(-1*w(Nn-2)+2*w(Nn-1)+0*w(Nn)-2*wE+wE2)/(2*dNn^3);
end
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Made some very minor changes to the program.
.
function [] = FPERevisitedTransProb08ABwNew04DownloadedAB1()
%Copyright Ahsan Amin. Infiniti derivatives Technologies.
%or skype ahsan.amin2999
%In this program, I am simulating the SDE given as
%dy(t)=mu1 x(t)^beta1 dt + mu2 x(t)^beta2 dt +sigma x(t)^gamma dz(t)
%I have not directly simulated the SDE but simulated the transformed
%Besse1l process version of the SDE and then changed coordinates to retreive
%the SDE in original coo
%rdinates.
%The present program will analytically evolve only the Bessel Process version of the
%SDE in transformed coordinates.
dt=.125/16/2/2; % Simulation time interval.%Fodiffusions close to zero
%decrease dt for accuracy.
Tt=128*2*2*2; % Number of simulation levels. Terminal time= Tt*dt; //.125/32*32*16=2 year;
T=Tt*dt;
OrderA=4; %
OrderM=4; %
dtM=.0625/2;
TtM=T/dtM;
dNn=.2/1; % Normal density subdivisions width. would change with number of subdivisions
Nn=41; % No of normal density subdivisions
NnMidl=18;%One half density Subdivision left from mid of normal density(low)
NnMidh=19;%One half density subdivision right from the mid of normal density(high)
NnMid=4.0;
NnT=46;
%NnD=(NnT-Nn)/2;
NnD=5;
x0=.25; % starting value of SDE
beta1=0.0;
beta2=1.0; % Second drift term power.
gamma=.75;%50; % volatility power.
kappa=1.0;%.950; %mean reversion parameter.
theta=.06;%mean reversion target
sigma0=.9;%Volatility value
%you can specify any general mu1 and mu2 and beta1 and beta2.
mu1=1*theta*kappa; %first drift coefficient.
mu2=-1*kappa; % Second drift coefficient.
%mu1=0;
%mu2=0;
alpha=1;% x^alpha is being expanded. This is currently for monte carlo only.
alpha1=1-gamma;%This is for expansion of integrals for calculation of drift
%and volatility coefficients
yy(1:Nn)=x0;
w(1:Nn)=x0^(1-gamma)/(1-gamma);
%Z(1:Nn)=(((1:Nn)-5.5)*dNn-NnMid);
Z(1:Nn)=(((1:Nn)+1.5)*dNn-NnMid);
ZT(1:NnT)=(((1:NnT)-3.5)*dNn-NnMid);
ZT
Z
str=input('Look at Z');
ZProb(1)=normcdf(.5*Z(1)+.5*Z(2),0,1)-normcdf(.5*Z(1)+.5*Z(2)-dNn,0,1);
ZProb(Nn)=normcdf(.5*Z(Nn)+.5*Z(Nn-1)+dNn,0,1)-normcdf(.5*Z(Nn)+.5*Z(Nn-1),0,1);
ZProb(2:Nn-1)=normcdf(.5*Z(2:Nn-1)+.5*Z(3:Nn),0,1)-normcdf(.5*Z(2:Nn-1)+.5*Z(1:Nn-2),0,1);
ZProbT(1)=normcdf(.5*ZT(1)+.5*ZT(2),0,1)-normcdf(.5*ZT(1)+.5*ZT(2)-dNn,0,1);
ZProbT(NnT)=normcdf(.5*ZT(NnT)+.5*ZT(NnT-1)+dNn,0,1)-normcdf(.5*ZT(NnT)+.5*ZT(NnT-1),0,1);
ZProbT(2:NnT-1)=normcdf(.5*ZT(2:NnT-1)+.5*ZT(3:NnT),0,1)-normcdf(.5*ZT(2:NnT-1)+.5*ZT(1:NnT-2),0,1);
%Above calculate probability mass in each probability subdivision.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
sigma11(1:OrderA+1)=0;
mu11(1:OrderA+1)=0;
mu22(1:OrderA+1)=0;
sigma22(1:OrderA+1)=0;
% index 1 correponds to zero level since matlab indexing starts at one.
sigma11(1)=1;
mu11(1)=1;
mu22(1)=1;
sigma22(1)=1;
for k=1:(OrderA+1)
if sigma0~=0
sigma11(k)=sigma0^(k-1);
end
if mu1 ~= 0
mu11(k)=mu1^(k-1);
end
if mu2 ~= 0
mu22(k)=mu2^(k-1);
end
if sigma0~=0
sigma22(k)=sigma0^(2*(k-1));
end
end
%Ft(1:TtM+1,1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0; %General time powers on hermite polynomials
Fp(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;%General x powers on coefficients of hermite polynomials.
Fp1(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;%General x powers for bessel transformed coordinates.
%YCoeff0 and YCoeff are coefficents for original coordinates monte carlo.
%YqCoeff0 and YqCoeff are bessel/lamperti version monte carlo.
YCoeff0(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;
YqCoeff0(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;
%Pre-compute the time and power exponent values in small multi-dimensional arrays
YCoeff = ItoTaylorCoeffsNew(alpha,beta1,beta2,gamma); %expand y^alpha where alpha=1;
YqCoeff = ItoTaylorCoeffsNew(alpha1,beta1,beta2,gamma);%expand y^alpha1 where alpha1=(1-gamma)
YqCoeff=YqCoeff/(1-gamma); %Transformed coordinates coefficients have to be
%further divided by (1-gamma)
for k = 0 : (OrderA)
for m = 0:k
l4 = k - m + 1;
for n = 0 : m
l3 = m - n + 1;
for j = 0:n
l2 = n - j + 1;
l1 = j + 1;
%Ft(l1,l2,l3,l4) = dtM^((l1-1) + (l2-1) + (l3-1) + .5* (l4-1));
Fp(l1,l2,l3,l4) = (alpha + (l1-1) * beta1 + (l2-1) * beta2 + (l3-1) * 2* gamma + (l4-1) * gamma ...
- (l1-1) - (l2-1) - 2* (l3-1) - (l4-1));
Fp1(l1,l2,l3,l4) = (alpha1 + (l1-1) * beta1 + (l2-1) * beta2 + (l3-1) * 2* gamma + (l4-1) * gamma ...
- (l1-1) - (l2-1) - 2* (l3-1) - (l4-1));
YCoeff0(l1,l2,l3,l4) =YCoeff(l1,l2,l3,l4).*mu11(l1).*mu22(l2).*sigma22(l3).*sigma11(l4);
YqCoeff0(l1,l2,l3,l4) =YqCoeff(l1,l2,l3,l4).*mu11(l1).*mu22(l2).*sigma22(l3).*sigma11(l4);
end
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
wnStart=1;%
tic
Zt1(wnStart:Nn)=0.0;
Zt2(wnStart:Nn)=0.0;
for tt=1:Tt
%[wMu0dt,dwMu0dtdw,c1] = CalculateDriftAndVolA4(w,wnStart,Nn,YqCoeff0,Fp1,gamma,dt);
%[wMu0dt,c1,c2] = CalculateDriftAndVolA404AB(w,wnStart,Nn,YqCoeff0,Fp1,gamma,dt);
[wMu0dt,c1,c2,dwMu0dtdw] = CalculateDriftAndVolA404AB(w,wnStart,Nn,YqCoeff0,Fp1,gamma,dt);
[wMid] = InterpolateOrderN8(8,0,Z(NnMidl-3),Z(NnMidl-2),Z(NnMidl-1),Z(NnMidl),Z(NnMidh),Z(NnMidh+1),Z(NnMidh+2),Z(NnMidh+3),w(NnMidl-3),w(NnMidl-2),w(NnMidl-1),w(NnMidl),w(NnMidh),w(NnMidh+1),w(NnMidh+2),w(NnMidh+3));
Zt1(wnStart:Nn)=w(wnStart:Nn)-wMid;
% [dwdZ,d2wdZ2A] = First2Derivatives2ndOrderEqSpacedA(wnStart,Nn,dNn,Zt1,Z);
[dwdZ,d2wdZ2,d3wdZ3] = First3Derivatives4thOrderEqSpaced(wnStart,Nn,dNn,Zt1,Z);
tt
C0(wnStart:Nn)=Zt1(wnStart:Nn)-dwdZ(wnStart:Nn).*Z(wnStart:Nn);
%c1(wnStart:Nn)=sigma0*sqrt(dt);
Zt2(wnStart:Nn)=C0(wnStart:Nn)+sign(dwdZ(wnStart:Nn)+c1(wnStart:Nn)).* ...
abs(sqrt(sign(dwdZ(wnStart:Nn)).*(dwdZ(wnStart:Nn)).^2+ ...
sign(c1(wnStart:Nn)).*c1(wnStart:Nn).^2)).*Z(wnStart:Nn);
% [dwdZ,d2wdZ2,d3wdZ3] = First3Derivatives2ndOrderEqSpacedA(wnStart,Nn,dNn,Zt2,Z);
% [dw2dZ,d2w2dZ2A] = First2Derivatives2ndOrderEqSpacedA(wnStart,Nn,dNn,Zt2,Z);
[dw2dZ,d2w2dZ2A,d3wdZ3] = First3Derivatives4thOrderEqSpaced(wnStart,Nn,dNn,Zt2,Z);
if(tt>10)
% dwdt(wnStart:Nn)=(wMu0dt(wnStart:Nn)+Zt2(wnStart:Nn)-Zt1(wnStart:Nn)+ ...
% .5*sigma0^2*(dw2dZ(wnStart:Nn).^(-2)).*d2w2dZ2A(wnStart:Nn).*dt)/dt;
else
% dwdt(wnStart:Nn)=(wMu0dt(wnStart:Nn)+Zt2(wnStart:Nn)-Zt1(wnStart:Nn))/dt;%+ ...
end
wnStart
if(tt>10)
dZdw(wnStart:Nn)=1./dw2dZ(wnStart:Nn);
d2Zdw2(wnStart:Nn)=-d2w2dZ2A(wnStart:Nn).*dZdw(wnStart:Nn).^3;
d3Zdw3(wnStart:Nn)=-d3wdZ3(wnStart:Nn).*dZdw(wnStart:Nn).^4+3*d2w2dZ2A(wnStart:Nn).^2.*dZdw(wnStart:Nn).^5;
%w(wnStart:Nn)=w(wnStart:Nn)+ ...
% wMu0dt(wnStart:Nn)+ ...
% 0*dwMu0dtdw(wnStart:Nn)./(-Z(wnStart:Nn).*dZdw(wnStart:Nn).^2+d2Zdw2(wnStart:Nn)) - ...
% +.5*sigma0^2*dt*( (Z(wnStart:Nn).^2-1).*dZdw(wnStart:Nn).^3- ...
% 0* 3*Z(wnStart:Nn).*dZdw(wnStart:Nn).*d2Zdw2(wnStart:Nn))./(-Z(wnStart:Nn).*dZdw(wnStart:Nn).^2+d2Zdw2(wnStart:Nn));
w(wnStart:Nn)=w(wnStart:Nn)+ ...
wMu0dt(wnStart:Nn)+1*(Zt2(wnStart:Nn)-Zt1(wnStart:Nn))+ ...
1*.5*sigma0^2*(dw2dZ(wnStart:Nn).^(-2)).*d2w2dZ2A(wnStart:Nn).*dt;
else
w(wnStart:Nn)=w(wnStart:Nn)+ ...
wMu0dt(wnStart:Nn)+Zt2(wnStart:Nn)-Zt1(wnStart:Nn);
end
% %yy(wnStart:Nn)=((1-gamma).*w(wnStart:Nn)).^(1/(1-gamma));
% [dwdZ,d2wdZ2A,d3wdZ3] = First3Derivatives4thOrderEqSpaced(wnStart,Nn,dNn,w,Z);
%FirstCoordinate
% for nn=NnMidl-11:-1:wnStart+1
% if((w(nn)-w(nn-1))>(w(nn+1)-w(nn)))
%
% w(nn) = InterpolateOrderN8(6,Z(nn),Z(nn+1),Z(nn+2),Z(nn+3),Z(nn+4),Z(nn+5),Z(nn+6),Z(nn+7),Z(nn+8),w(nn+1),w(nn+2),w(nn+3),w(nn+4),w(nn+5),w(nn+6),w(nn+7),w(nn+8));
%
% end
% end
% if(w(wnStart+1)-w(wnStart)>w(wnStart+2)-w(wnStart+1))
% nn=wnStart;
% w(wnStart) = InterpolateOrderN8(6,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),Z(wnStart+8),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7),w(wnStart+8));
% w(nn)=w(nn+1)-(w(nn+2)-w(nn+1));
% end
[wE] = InterpolateOrderN6(6,Z(Nn)+dNn,Z(Nn),Z(Nn-1),Z(Nn-2),Z(Nn-3),Z(Nn-4),Z(Nn-5),w(Nn),w(Nn-1),w(Nn-2),w(Nn-3),w(Nn-4),w(Nn-5));
for nn=wnStart:Nn-1
if(w(nn)>w(nn+1))
w(nn)=0;
end
end
% w1(wnStart:Nn-1)=w(wnStart:Nn-1);
% w1(Nn)=w(Nn);
% w2(wnStart:Nn-1)=w(wnStart+1:Nn);
% w2(Nn)=wE;
% w(w1(wnStart:Nn)>w2(:))=0;%Be careful;might not universally hold;
% %Change 3:7/25/2020: I have improved zero correction in above.
%for nn=1:Nn
% if(w2(nn)>w1(nn))
w(w<0)=0.0;
StartChangeFlag=0;
nnChange=0;
wnStartPrev=wnStart;
for nn=wnStart:Nn
if(w(nn)<=0)
wnStart=nn+1;
StartChangeFlag=1;
nnChange=nnChange+1;
end
end
%nn=wnStart;
%while(w(nn)<=0)
% wnStart=nn+1;
% nn=nn+1;
%end
InterpolatePrevFlag=1;
wnChange=wnStart-wnStartPrev;
Zs=0
ws=0;
for mm=1:nnChange
Zs(mm)=Z(wnStart)-mm*dNn;
ws(mm)=0.0
end
if(nnChange>=1)
ws = InterpolateOrderN8(8,Zs,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
ContinueFlag=1;
end
for mm=1:nnChange
if((ws(mm)>0)&&(ws(mm)<w(wnStart))&&(ContinueFlag==1))
wnStart=wnStart-1;
w(wnStart)=ws(mm);
ContinueFlag=1;
else
ContinueFlag=0;
end
end
% if(InterpolatePrevFlag==1)
% InterpolatePrevFlag=0;
% %wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% NnStart=nnChange
% wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% if(wStart1>0)
% wnStart=wnStart-1;
% w(wnStart)=wStart1;
% InterpolatePrevFlag=1;
% end
% if(wnStart==wnStartPrev)
% % InterpolatePrevFlag=0;
% end
% end
% end
% if(nnChange>=1)
% wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% if(wStart1>0)
% wnStart=wnStart-1
% w(wnStart)=wStart1;
% InterpolatePrevFlag=1;
% end
% end
% if((nnChange>=2)&&( InterpolatePrevFlag==1))
% wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% InterpolatePrevFlag=0;
% if(wStart1>0)
% wnStart=wnStart-1
% w(wnStart)=wStart1;
% InterpolatePrevFlag=1;
% end
% end
% if((nnChange>=3)&&( InterpolatePrevFlag==1))
% wStart1 = InterpolateOrderN8(8,Z(wnStart)-dNn,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
% InterpolatePrevFlag=0;
% if(wStart1>0)
% wnStart=wnStart-1
% w(wnStart)=wStart1;
% InterpolatePrevFlag=1;
% end
% end
%
w
wnStart
tt
end
wT(wnStart+NnD:Nn+NnD)=w(wnStart:Nn);
Zs=0
ws=0;
for mm=1:wnStart+NnD-1
Zs(mm)=Z(wnStart)-mm*dNn;
ws(mm)=0.0
end
wnStartT=wnStart+NnD;
%NnT=Nn+2*NnD;
ws = InterpolateOrderN8(8,Zs,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
ContinueFlag=1;
wnStart0=wnStart+NnD-1;
for mm=1:wnStart0
if((ws(mm)>0)&&(ws(mm)<wT(wnStartT))&&(ContinueFlag==1))
wnStartT=wnStartT-1;
wT(wnStartT)=ws(mm);
ContinueFlag=1;
else
ContinueFlag=0;
end
end
% Zs=0;
% ws=0;
% for mm=1:NnD
% Zs(mm)=Z(Nn)+mm*dNn;
% ws(mm)=0.0
% end
%
% ws = InterpolateOrderN8(2,Zs,Z(wnStart),Z(wnStart+1),Z(wnStart+2),Z(wnStart+3),Z(wnStart+4),Z(wnStart+5),Z(wnStart+6),Z(wnStart+7),w(wnStart),w(wnStart+1),w(wnStart+2),w(wnStart+3),w(wnStart+4),w(wnStart+5),w(wnStart+6),w(wnStart+7));
%
%
% %for mm=1:NnD
%
% wT(Nn+NnD+1:Nn+2*NnD)=ws(1:NnD);
% %end
% for mm=1:wnStart0-1
% if((ws(mm)>0)&&(ws(mm)<w(wnStart))&&(ContinueFlag==1))
% wnStart=wnStart-1;
% w(wnStart)=ws(mm);
% ContinueFlag=1;
% else
% ContinueFlag=0;
% end
% end
wnStartT
%str=input('Look at numbers')
yyT(wnStartT:NnT)=((1-gamma).*wT(wnStartT:NnT)).^(1/(1-gamma));
DfyyT(wnStartT:NnT)=0;
for nn=wnStartT+1:NnT-1
DfyyT(nn) = (yyT(nn + 1) - yyT(nn - 1))/(ZT(nn + 1) - ZT(nn - 1));
%Change of variable derivative for densities
end
pyyT(1:Nn)=0;
for nn = wnStartT:NnT-1
pyyT(nn) = (normpdf(ZT(nn),0, 1))/abs(DfyyT(nn));
end
if(tt>=23)
%plot(yy(wnStart:Nn),pyy(wnStart:Nn),'r');
%yy
%str=input('Look at the output graph');
end
toc
ItoHermiteMean=sum(yyT(wnStartT+1:NnT-1).*ZProbT(wnStartT+1:NnT-1)) %Original process average from coordinates
disp('true Mean only applicable to standard SV mean reverting type models otherwise disregard');
TrueMean=theta+(x0-theta)*exp(-kappa*dt*Tt)%Mean reverting SDE original variable true average
theta1=1;
rng(29079137, 'twister')
paths=200000;
YY(1:paths)=x0; %Original process monte carlo.
Random1(1:paths)=0;
for tt=1:TtM
Random1=randn(size(Random1));
HermiteP1(1,1:paths)=1;
HermiteP1(2,1:paths)=Random1(1:paths);
HermiteP1(3,1:paths)=Random1(1:paths).^2-1;
HermiteP1(4,1:paths)=Random1(1:paths).^3-3*Random1(1:paths);
HermiteP1(5,1:paths)=Random1(1:paths).^4-6*Random1(1:paths).^2+3;
YY(1:paths)=YY(1:paths) + ...
(YCoeff0(1,1,2,1).*YY(1:paths).^Fp(1,1,2,1)+ ...
YCoeff0(1,2,1,1).*YY(1:paths).^Fp(1,2,1,1)+ ...
YCoeff0(2,1,1,1).*YY(1:paths).^Fp(2,1,1,1))*dtM + ...
(YCoeff0(1,1,3,1).*YY(1:paths).^Fp(1,1,3,1)+ ...
YCoeff0(1,2,2,1).*YY(1:paths).^Fp(1,2,2,1)+ ...
YCoeff0(2,1,2,1).*YY(1:paths).^Fp(2,1,2,1)+ ...
YCoeff0(1,3,1,1).*YY(1:paths).^Fp(1,3,1,1)+ ...
YCoeff0(2,2,1,1).*YY(1:paths).^Fp(2,2,1,1)+ ...
YCoeff0(3,1,1,1).*YY(1:paths).^Fp(3,1,1,1))*dtM^2 + ...
((YCoeff0(1,1,1,2).*YY(1:paths).^Fp(1,1,1,2).*sqrt(dtM))+ ...
(YCoeff0(1,1,2,2).*YY(1:paths).^Fp(1,1,2,2)+ ...
YCoeff0(1,2,1,2).*YY(1:paths).^Fp(1,2,1,2)+ ...
YCoeff0(2,1,1,2).*YY(1:paths).^Fp(2,1,1,2)).*dtM^1.5) .*HermiteP1(2,1:paths) + ...
((YCoeff0(1,1,1,3).*YY(1:paths).^Fp(1,1,1,3) *dtM) + ...
(YCoeff0(1,1,2,3).*YY(1:paths).^Fp(1,1,2,3)+ ...
YCoeff0(1,2,1,3).*YY(1:paths).^Fp(1,2,1,3)+ ...
YCoeff0(2,1,1,3).*YY(1:paths).^Fp(2,1,1,3)).*dtM^2).*HermiteP1(3,1:paths) + ...
((YCoeff0(1,1,1,4).*YY(1:paths).^Fp(1,1,1,4)*dtM^1.5 )).*HermiteP1(4,1:paths) + ...
(YCoeff0(1,1,1,5).*YY(1:paths).^Fp(1,1,1,5)*dtM^2.0).*HermiteP1(5,1:paths);
end
YY(YY<0)=0;
disp('Original process average from monte carlo');
MCMean=sum(YY(:))/paths %origianl coordinates monte carlo average.
disp('Original process average from our simulation');
ItoHermiteMean=sum(yyT(wnStartT+1:NnT-1).*ZProbT(wnStartT+1:NnT-1)) %Original process average from coordinates
disp('true Mean only applicble to standard SV mean reverting type models otherwise disregard');
TrueMean=theta+(x0-theta)*exp(-kappa*dt*Tt)%Mean reverting SDE original variable true average
MaxCutOff=30;
NoOfBins=round(300*gamma^2*4*sigma0/sqrt(MCMean)/(1+kappa));%Decrease the number of bins if the graph is too
[YDensity,IndexOutY,IndexMaxY] = MakeDensityFromSimulation_Infiniti_NEW(YY,paths,NoOfBins,MaxCutOff );
plot(yyT(wnStartT+1:NnT-1),pyyT(wnStartT+1:NnT-1),'r',IndexOutY(1:IndexMaxY),YDensity(1:IndexMaxY),'g');
%plot(y_w(wnStart+1:Nn-1),fy_w(wnStart+1:Nn-1),'r',IndexOutY(1:IndexMaxY),YDensity(1:IndexMaxY),'g',Z(wnStart+1:Nn-1),fy_w(wnStart+1:Nn-1),'b');
title(sprintf('x0 = %.4f,theta=%.3f,kappa=%.2f,gamma=%.3f,sigma=%.2f,T=%.2f,dt=%.5f,M=%.4f,TM=%.4f', x0,theta,kappa,gamma,sigma0,T,dt,ItoHermiteMean,TrueMean));%,sprintf('theta= %f', theta), sprintf('kappa = %f', kappa),sprintf('sigma = %f', sigma0),sprintf('T = %f', T));
legend({'Ito-Hermite Density','Monte Carlo Density'},'Location','northeast')
str=input('red line is density of SDE from Ito-Hermite method, green is monte carlo.');
end
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends I have malaria and very high fever, they still continue to charge my body and release sickening gases regularly in my room. Mind control crooks have absolutely no humanity. Please protest against such inhuman torture on innocent civilians on hands of mind control crooks.
Amin
Topic Author
Posts: 2864
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I have recovered from disease but have extreme weakness. I hope to restart my research work in 2-3 days. I want to thank friends who wished for my illness to end.
It was very difficult to fight the disease and also fight the mind control torture. I made a list of mind control activities of agents and will share that with friends later.
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# Profit and Loss Questions for CAT with Solutions
Questions on Profit Loss and Discount has started featuring regularly in CAT quant section. Approximately 3-4 questions are based directly or indirectly in the basic concepts of profit and loss. Various shortcut formulas and tricks are used to solve these questions.
We have selected around 25 types of questions with solutions from profit and loss which regularly appear in CAT exam. A student must pay good attention to the approach employed to solve them.
### Quick Tips:
• Get a good grip on the concepts and formulas of percentages.
• Conversion of ratio to percentages and vice versa is very handy simplifying the problems
• There are different varieties of questions from profit loss and discount. Many of the questions have direct formula or tricks. Learn to use them as and when required.
• Remember, Profit is calculated over Cost Price and Discount is calculated over Marked Price.
### Practice Questions
Question 1: I sell 16 sheep at a gain of 12.5% and 20 more at a certain gain percent. If 1 gain 25% on the whole, how much percent gain did I make on the latter number?
1. 20%
2. 25%
3. 30%
4. 35%
Option: 4
On the whole I gain 25%
Therefore, I should get the cost price of $36 \times \frac{{125}}{{100}} = 45$ sheep.
But I sell the first lot at a gain of 12.5%
Hence by selling 16 sheep, I get the cost price of $16\times \frac{112.5}{100}=18$ sheep.
Therefore, I should get the cost price of 45-18=27 sheep by selling the second lot of 20. Hence my gain there should be $\frac{7}{20}\times 100=35%$
Question 2: A farmer buys 240 cows. He sells some of them at a gain of 20% and the remaining at a gain of 30%. If he gains 28% on the whole, then how many did he sell at a gain of 20%?
1. 40
2. 48
3. 54
4. 28
Option: 2
Let him sell x cows at the gain of 20% and y cows at a gain of 30%.
Therefore, 0.2x + 0.3y = -0.28(x+y)
=> y=4x, Therefore, x+4x=240 =>x=48.
Question 3: A dealer sells a horse for Rs.400, making a profit of 25%. He sells a second horse at a loss of 10% and on the whole makes neither profit nor loss. What did the second horse cost him?
1. Rs.100
2. Rs.600
3. Rs.800
4. Rs.400
Option: 3
In the given question "The dealer sells two horses, one for Rs.400, making a profit of 25% and the other horse at a loss of 10% and makes neither profit nor loss on the whole."
Let the cost price of second horse = Rs.x i.e., 320 + x = 400 + 9x/10 ( since 400 x 100/125 = 320) => x = Rs.800. Ans.(3)
Question 4: I buy two horses, A and B. A costs Rs.50 more than B. I sell A at a profit of 16% and 13 at a profit of 7%. My total gain is Rs.100. What was the original price of 8?
1. Rs.450
2. Rs.400
3. Rs.350
4. Rs.500
Option: 2
A = B + 50 ….(1)
1.16 A + 1.07 B = A + B + 100 …..(2)
=> A = Rs.450, B = Rs.400. Ans.(2)
Question 5: If toffees are bought at the rate 18 for a rupee, then how many of them must be sold for a rupee to gain 20%?
1. 14
2. 15
3. 16
4. 12
Option: 2
18 x 100 p = 120x => x = 15. Ans.(2)
Question 6: A man's petrol bill in July is Rs.200. In August, the price of petrol increases by 10% and his consumption is reduced by 10%. Find his petrol bill in August.
1. Rs.190
2. Rs.210
3. Rs.200
4. Rs.198
Option: 4
Let the consumption in July = x litres,
Therefore, price of petrol = Rs. 200/x
=> Petrol bill in August = (0.9x) x(1.10)x 200/x= Rs.198. Ans.(4)
Question 7: I sell a table for Rs.24 and thus make a percentage of profit equal to the cost price. What did the table cost me?
1. Rs.10
2. Rs.20
3. Rs.40
4. Rs.30
Option: 2
Let the cost price and profit be X.
Therefore, $\frac{S.P.-C.P.}{C.P.}\times 100=\Pr ofit%$
$\begin{array}{l} = \frac{{24 - x}}{x} \times 100 = x\\ = > {x^2} + 100x - 2400 = 0\\hence\;x = 20\end{array}$
Question 8: A bike costs Rs.48000. Its value depreciates by 30% in the first year and in each subsequent year the depreciation is 20% of the value at the beginning of that year. The value of the bike after 3 years will be
1. Rs.21504
2. Rs.26880
3. Rs.38400
4. Rs.39480
Option: 1
The required value of the bike after 3 years = $48000\left( {\frac{{70}}{{100}}} \right){\left( {\frac{{80}}{{100}}} \right)^2} = Rs.21504$
Question 9: A person sold his watch for Rs.75 and got a percentage of profit equal to the cost price. The cost price of the watch is
1. Rs.40
2. Rs.50
3. Rs.35
4. Rs.45
Option: 2
Let x be the cost price of the watch.
Therefore, profit percentage is also x.
$\begin{array}{l}x + x \times \frac{x}{{100}} = 75\\Or\;x = 50\end{array}$
Question 10: A man sells two horses for Rs.1955 each. On one he gains 15% and on the other he loses 15%. His total gain or loss is
1. Rs.40.00
2. Rs.90.00
3. Rs.97.75
4. Rs.19.55
Option: 2
Loss = 15 x 15/100 = 2.25%.
C.P = (2 x 1955 x 100)/97.75 = Rs.4000.
Loss = 4000 — 2 x (1955) = Rs.90. Ans.(2)
Question 11: Vivek bought 5 dozen apples at the rate of Rs.15 per dozen. He spent Rs.15 on transportation. If he sold the apples at the rate of Rs.24 per dozen, what was his profit percentage?
1. 25%
2. 30%
3. 33.33%
4. 60%
Option: 3
The Cost Price (C.P) of 1 dozen apples = Rs.15.
Hence, the C.P of 5 dozen apples = 15 x 5 = Rs.75.
Adding Rs.15 towards transportation cost, the total cost price of 5 dozen apples becomes Rs.90.
Now, the Selling Price (S.P) of 1 dozen apples = Rs.24.
Hence, the Sales Revenue of 5 dozen apples = 24 x 5 = Rs.120.
Thus, Net Profit = Sales Revenue - Cost Price = 120 - 90 = Rs.30.
The Profit is always expressed as a percentage to cost price. Required answer = (30/90) x 100 = 33.33%. Ans.(3)
Short-cut: Cost of 5 dozen of apples including transportation charges = 15 + (15/3) = Rs.18.
Percentage profit = [(24-18)/18] x100=33.33% .
Question 12: A shopkeeper purchases several articles at the rate of 11 for Rs.10 and sells them at the rate of 10 for Rs.11. What would be the profit earned by him?
1. 11%
2. 10%
3. 20%
4. 21%
Option: 4
Cost Price for 11 articles = Rs.10 .... (I)
Sales Price for 10 articles = Rs.11 .... (II)
Sales Price for 11 articles = Rs.12.1 (III)
Thus, required profit = [(12.1-10)/10] x 100 = 21%. Ans.(4)
Question 13: The cost price of 12 articles is the same as the selling price of 8 articles. What is the profit percent?
1. 25%
2. 40%
3. 50%
4. 200%
Option: 3
We have, Cost Price of 12 Numbers = Sales Price of 8 Numbers. In order to arrive at the profit percentage. we need to compare the cost price/ sales price of equal number of articles.
Thus, let C.P of 12 Numbers = S.P of 8 Numbers. = Rs.100.
Hence, S.P of 12 Numbers = 12 x 100 ± 8 = Rs.150.
Thus, Profit = 150 - 100 (for 12 Numbers) = Rs.50.
Therefore Profit percent = (Profit ÷C.P) x 100 = 50%. Ans.(3)
Alternatively,
Cost price of 12 Numbers = Selling price of 8 Numbers
=> Selling price of 8 Numbers = Cost price of 8 Numbers + Cost price of 4 Numbers
Therefore, Profit = Cost price of 4 Numbers
=> Profit % =(Cost price of 4 numbers/cost price of 8 numbers)x100=(4/8)x100=50%
Question 14: In order to increase revenue, a dealer announces 20% reduction in the unit price of an article. As a result, his sales volume increases by 20%. What is the overall gain/loss to the dealer?
1. no profit no loss
2. 4% loss
3. 10% profit
4. Cannot be determined since selling price is unknown
Option: 2
When the two figures of (i) reduction and (ii) increase are same (here, 20% each), the calculation is direct, as follows: (a) Overall gain or loss
=> Always loss. (b) Numerical Values? = 20% x 20% = 4% (multiply two values and divide by 100). Ans.(2)
Question 15: Rakesh bought 20 chairs for Rs.1000. He repaired and sold them at the rate of Rs.500 per pair. He got profit of Rs.100 per chair. How much did he spend on repairs?
1. Rs.1500
2. Rs.2000
3. Rs.2500
4. Rs.1800
Option: 3
Purchase price for 20 chairs=Rs. 1000
Repairs amount (to be determined) =Rs. X
Sales price for 20 chairs = Rs. 5000
Profit derived for 20 chairs = Rs. 2000
We have, (3) - [(1) + (2)] = (4).
? 5000 - (1000 + X) = 2000 => X = 2000. Ans.(3)
Question 16: Kiran buys an article with 25% discount on the marked price. He makes a profit of 10% by selling it at Rs.660. What was the marked price?
1. Rs.900
2. Rs.600
3. Rs.700
4. Rs.800
Option: 4
Let the marked price of the article be Rs. X
? Purchase Price (C.P) for Kiran = 0.75 X.
? S.P for Kiran = (0.75 X) + 10% of (0.75 X) = 660.
? 0.825 X = Rs.660 = X = Rs.800. Ans.(4)
Question 17: A dishonest dealer pretends to sell at the cost price but earns a profit of 25% by under weighing. What weight must he be using for 1 kg?
1. 750 gm
2. 800 gm
3. 500 gm
4. 875 gm
Option: 2
Let the false weight be x gm.
Thus, the profit made is through sale of (100-x) gm
Hence, we have, $\frac{{100 - x}}{x} = 25\% = > x = 800gm$
Question 18: A' sold a house to Bat a gain of 10% and B sold it to Cat a gain of 20%. If C paid Rs.264000 for it, at what price must A have purchased it
1. Rs.200000
2. Rs.220000
3. Rs.240000
4. Rs.250000
Option: 1
Let the cost price for ‘A’ be Rs. X
Therefore, $\begin{array}{l}x \times \frac{{110}}{{100}} \times \frac{{120}}{{100}} = 264000\\ = > x = 200000\end{array}$
Question 19: An increase, in the cost price of an article, by 22% leads to the value of Rs.61. What was the original cost price of the article?
1. Rs.40
2. Rs.45
3. Rs.50
4. Rs.55
Option: 3
Let the original cost price of the article be Rs.'X'. Thus, we get, X + 22% of X = 61 => X = Rs.50. Ans.(3)
Question 20: When certain quantity of sugar is sold at Rs.11 per kg, the gain is 10%. If the total gain is Rs.50, what is the quantity of sugar sold?
1. 100 kg
2. 60 kg
3. 50 kg
4. 80 kg
Option: 3
The Unit Cost Price of sugar = 11 ÷ 1.1 = Rs.10 per kg.
Thus, by selling one kg, the gain is Re.1.
Hence, when the total gain is Rs.50, 50 kg of sugar must have been sold. Ans.(3)
Question 21: What is the cost price of an article, if a loss of 16% is incurred by selling it for Rs.168?
1. Rs.200
2. Rs.180
3. Rs.220
4. Rs.210
Option: 1
Let the cost price of the article be Rs. 'X'. Thus, we get, X — (16% of X) = Rs.168 = X = Rs.200. Ans.(1)
Question 22: Two successive discounts of 30% and 25% are equivalent to a single discount of
1. 45%
2. 55%
3. 47.50%
4. 52.50%
Option: 3
Required discount = 100 — (0.7 x 0.75 x 100) = 47.5%. Ans.(3)
Question 23: A man buys two goats at Rs.120 each. He sells one at 25% gain and the other at 25% loss. How much is his profit or loss?
1. 6.25% gain
2. No loss no gain
3. 6.25% loss
4. Cannot be determined
Option: 2
As gain or loss is on the cost price.
25% loss on Rs.120 = 25% gain on Rs.120.
As such there is no loss no gain. Ans.(2)
Question 24: By what percent should the cost price of an article be marked up such that even after allowing a discount of 50%, a profit of 50% is made?
1. 300
2. 500
3. 100
4. 200
Option: 4
Let the Cost Price and Marked Price of the item be Rs. 'X' and 'Y' respectively. Thus, we get, Y — 50% of Y = X + 50% of X 0.5Y = 1.5X. or Y ÷ X = 1.5 ÷ 0.5 = 3.
Thus, the marked price should be 3 times the cost price. Hence, required percentage is 200. Ans.(4)
Question 25: A dealer offers three successive discounts of 50%, 20% and 10% on an article. What is the single effective discount rate?
1. 60%
2. 62%
3. 63%
4. 64%
Option: 4
Effective discount = 100 — ( 0.5 x 0.8 x 0.9 x 100) = 64%. Ans.(4)
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How many SN1 products are formed in principle in the question number 20? Pick out the compound which reacts fastest in the presence of AgNO3. Concentration of nucleophile What is the correct order of reactivity of the followings in hydrolysis reaction at elevated temperature? Both SN1 and E1 reaction involve formation of carbocation intermediate in slow rate determining step as : Presence of NaCI derive the above reaction in backward direction by Le-Chatelier’s principle, hence has adverse effect on both E1 and SN1 reaction. Electron withdrawing —NO2 decreases stability of carbocation, decreases reactivity of corresponding substrate in SN1 reaction. Hence, A will have four stereoisomers. The following practice problems test your knowledge of the two organic chemistry substitution reactions, SN2 reactions and SN1 reactions. NCERT Solutions - Basic Concepts of Chemistry (Part - 1), Previous Year Questions (2016-19) - The Living World, QUESTION PAPER OF ALL INDIA AAKASH TEST SERIES--MOCK TEST FOR NEET 2018, Test: Cell: The Unit Of Life 1 - From Past 28 Years Questions, Test: Kinematics (Previous Year Questions Level 1). The competition from $\mathrm{E_2}$ in problem (j) is less of an issue than it may at first appear. SN1 reaction takes place giving partial racemisation but net inversion of configuration at chiral α-carbon. Reaction proceeds via carbocation intermediate, it passes through more than one transition states. long questions & short questions for Class 12 on EduRev as well by searching above. April 1, 2019 By Leah4sci 31 Comments. The conformer with both substituents axial is substantially less stable, so $\mathrm{E_2}$ is hindered (though not ruled out) by the energy barrier associated with the ring flip. with strong base, less substituted alkene is always the major product. When tertiary butyl group is trans to leaving halide, E2 reaction will be highly sterically hindered and E1 reaction would be preferred. (I) gives (a) and (b) while (II) gives (c) and (d). Help Center Detailed answers to any questions you might have ... Synthesis via Sn2/Sn1, E2, E1: Choice of solvent. For question 22, how do you know if it is E1 or E2 if the product is the same? Alcohols B and C are: (I) is least reactive as highly unstable carbocation is formed at bridge head carbon of bicyclic compound. The mechanism depends on the structure of the substrate and the reactant. 28-30) This section contains 3 questions. H2O has higher solvating power than CH3CH2OH,hence faster SN1 reaction occur in H2O. It also contains two sp2-hybridised carbon atoms. Rate of reaction is faster in H2O than in C2H5OH, The reaction usually have more than one transition states, If α-carbon is chiral, partial racemisation and net inversion takes place, Reaction occurs at faster rate if CH3O18H is used in place of CH3OH. Synthesis via Sn2/Sn1, E2, E1: Choice of solvent, http://www.physicsforums.com/showthread.php?t=765106, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…, “Question closed” notifications experiment results and graduation, Regioselectivity of acid-catalyzed ring-opening of epoxides. A undergoes very fast SN1 reaction. I. Question 1 III. On question 4 answers A and D are the same. (Piano) How should I play this harmonic unison, Forecasting Prices vs Returns by Deep Learning. Since, E1 reaction occur slowly and hydrogen is lost in fast step, both H and D will be lost with equal ease. If a pure dextrorotatory enantiomer of the substrate of the following reaction is boiled with water, the correct statement(s) regarding SN1 product(s) is/are, product would consists of both laevo and dextrorotatory enantiomers, product would be completely racemic mixture, product would be optically active with enantiomer of inverted configuration predominating. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct. SN1 SN2 E1 E2 Practice Problem Orgo Quiz. The correct statement concerning I and II is, I is the major product as it is formed at faster rate than II, II is the major product as it is formed at faster rate than I, I is major product as it is more stable than II, I and II are formed in comparable amounts. Is there a formal name for a "wrong question"? The rate of SN1 reaction depends on concentration of nucleophile There might be some concern about hydroxide deprotonating methanol, and then methoxide carrying out the displacement. I'm not seeing what you're seeing, OP. SN1, SN2 can give the same product and E1, E2 may also do so since they are both substitution and elimination reactions. Was the theory of special relativity sparked by a dream about cows being electrocuted? Reaction (II) leads to greater increase in entropy (Δn = 1). II. However, C—X bond is broken in first, slow, non-rate determ in ing step, hence R— I127 reacts at faster rate than R— I131. EduRev is a knowledge-sharing community that depends on everyone being able to pitch in when they know something. By continuing, I agree that I am at least 13 years old and have read and agree to the. Both S and C have one chiral carbon each, both show optical isomerism. (I) is least reactive as it forms least stable cyclopropyl carbocation. Which is the most likely product when the following iodide is heated with water? How many different structure(s) can be drawn for this D ? The solved questions answers in this Test: E1 Reactions & Related Reactions quiz give you a good mix of easy questions and tough questions.
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Dedekind Domain and Properties in an Elementary Approach
You can find contents about Dedekind domain (or Dedekind ring) in almost all algebraic number theory books. But many properties can be proved inside ring theory. I hope you can find the solution you need in this post, and this post will not go further than elementary ring theory. With that being said, you are assumed to have enough knowledge of ring and ring of fractions (this post serves well), but not too much mathematics maturity is assumed (at the very least you are assumed to be familiar with terminologies in the linked post).$\def\mb{\mathbb}$ $\def\mfk{\mathfrak}$
Definition
There are several ways to define Dedekind domain since there are several equivalent statements of it. We will start from the one based on ring of fractions. As a friendly reminder, $\mb{Z}$ or any principal integral domain is already a Dedekind domain. In fact Dedekind domain may be viewed as a generalization of principal integral domain.
Let $\mfk{o}$ be an integral domain (a.k.a. entire ring), and $K$ be its quotient field. A Dedekind domain is an integral domain $\mfk{o}$ such that the fractional ideals form a group under multiplication. Let's have a breakdown. By a fractional ideal $\mfk{a}$ we mean a nontrivial additive subgroup of $K$ such that
• $\mfk{o}\mfk{a}=\mfk{a}$,
• there exists some nonzero element $c \in \mfk{o}$ such that $c\mfk{a} \subset \mfk{o}$.
What does the group look like? As you may guess, the unit element is $\mfk{o}$. For a fractional ideal $\mfk{a}$, we have the inverse to be another fractional ideal $\mfk{b}$ such that $\mfk{ab}=\mfk{ba}=\mfk{o}$. Note we regard $\mfk{o}$ as a subring of $K$. For $a \in \mfk{o}$, we treat it as $a/1 \in K$. This makes sense because the map $i:a \mapsto a/1$ is injective. For the existence of $c$, you may consider it as a restriction that the 'denominator' is bounded. Alternatively, we say that fractional ideal of $K$ is a finitely generated $\mfk{o}$-submodule of $K$. But in this post it is not assumed that you have learned module theory.
Let's take $\mb{Z}$ as an example. The quotient field of $\mb{Z}$ is $\mb{Q}$. We have a fractional ideal $P$ where all elements are of the type $\frac{np}{2}$ with $p$ prime and $n \in \mb{Z}$. Then indeed we have $\mb{Z}P=P$. On the other hand, take $2 \in \mb{Z}$, we have $2P \subset \mb{Z}$. For its inverse we can take a fractional ideal $Q$ where all elements are of the type $\frac{2n}{p}$. As proved in algebraic number theory, the ring of algebraic integers in a number field is a Dedekind domain.
Before we go on we need to clarify the definition of ideal multiplication. Let $\mfk{a}$ and $\mfk{b}$ be two ideals, we define $\mfk{ab}$ to be the set of all sums $x_1y_1+\cdots+x_ny_n$ where $x_i \in \mfk{a}$ and $y_i \in \mfk{b}$. Here the number $n$ means finite but is not fixed. Alternatively we cay say $\mfk{ab}$ contains all finite sum of products of $\mfk{a}$ and $\mfk{b}$.
Propositions
(Proposition 1) A Dedekind domain $\mfk{o}$ is Noetherian.
By Noetherian ring we mean that every ideal in a ring is finitely generated. Precisely, we will prove that for every ideal $\mfk{a} \subset \mfk{o}$ there are $a_1,a_2,\cdots,a_n \in \mfk{a}$ such that, for every $r \in \mfk{a}$, we have an expression $r = c_1a_1 + c_2a_2 + \cdots + c_na_n \qquad c_1,c_2,\cdots,c_n \in \mfk{o}.$ Also note that any ideal $\mfk{a} \subset \mfk{o}$ can be viewed as a fractional ideal.
Proof. Since $\mfk{a}$ is an ideal of $\mfk{o}$, let $K$ be the quotient field of $\mfk{o}$, we see since $\mfk{oa}=\mfk{a}$, we may also view $\mfk{a}$ as a fractional ideal. Since $\mfk{o}$ is a Dedekind domain, and fractional ideals of $\mfk{a}$ is a group, there is an fractional ideal $\mfk{b}$ such that $\mfk{ab}=\mfk{ba}=\mfk{o}$. Since $1 \in \mfk{o}$, we may say that there exists some $a_1,a_2,\cdots, a_n \in \mfk{a}$ and $b_1,b_2,\cdots,b_n \in \mfk{o}$ such that $\sum_{i = 1 }^{n}a_ib_i=1$. For any $r \in \mfk{a}$, we have an expression $r = rb_1a_1+rb_2a_2+\cdots+rb_na_n.$ On the other hand, any element of the form $c_1a_1+c_2a_2+\cdots+c_na_n$, by definition, is an element of $\mfk{a}$. $\blacksquare$
From now on, the inverse of an fractional ideal $\mfk{a}$ will be written like $\mfk{a}^{-1}$.
(Proposition 2) For ideals $\mfk{a},\mfk{b} \subset \mfk{o}$, $\mfk{b}\subset\mfk{a}$ if and only if there exists some $\mfk{c}$ such that $\mfk{ac}=\mfk{b}$ (or we simply say $\mfk{a}|\mfk{b}$)
Proof. If $\mfk{b}=\mfk{ac}$, simply note that $\mfk{ac} \subset \mfk{a} \cap \mfk{c} \subset \mfk{a}$. For the converse, suppose that $a \supset \mfk{b}$, then $\mfk{c}=\mfk{a}^{-1}\mfk{b}$ is an ideal of $\mfk{o}$ since $\mfk{c}=\mfk{a}^{-1}\mfk{b} \subset \mfk{a}^{-1}\mfk{a}=\mfk{o}$, hence we may write $\mfk{b}=\mfk{a}\mfk{c}$. $\blacksquare$
(Proposition 3) If $\mfk{a}$ is an ideal of $\mfk{o}$, then there are prime ideals $\mfk{p}_1,\mfk{p}_2,\cdots,\mfk{p}_n$ such that $\mfk{a}=\mfk{p}_1\mfk{p}_2\cdots\mfk{p}_n.$
Proof. For this problem we use a classical technique: contradiction on maximality. Suppose this is not true, let $\mfk{A}$ be the set of ideals of $\mfk{o}$ that cannot be written as the product of prime ideals. By assumption $\mfk{U}$ is nonempty. Since as we have proved, $\mfk{o}$ is Noetherian, we can pick an maximal element $\mfk{a}$ of $\mfk{A}$ with respect to inclusion. If $\mfk{a}$ is maximal, then since all maximal ideals are prime, $\mfk{a}$ itself is prime as well. If $\mfk{a}$ is properly contained in an ideal $\mfk{m}$, then we write $\mfk{a}=\mfk{m}\mfk{m}^{-1}\mfk{a}$. We have $\mfk{m}^{-1}\mfk{a} \supsetneq \mfk{a}$ since if not, we have $\mfk{a}=\mfk{ma}$, which implies $\mfk{m}=\mfk{o}$. But by maximality, $\mfk{m}^{-1}\mfk{a}\not\in\mfk{U}$, hence it can be written as a product of prime ideals. But $\mfk{m}$ is prime as well, we have a prime factorization for $\mfk{a}$, contradicting the definition of $\mfk{U}$.
Next we show uniqueness up to permutation. If $\mfk{p}_1\mfk{p}_2\cdots\mfk{p}_k=\mfk{q}_1\mfk{q}_2\cdots\mfk{q}_j,$ since $\mfk{p}_1\mfk{p}_2\cdots\mfk{p}_k\subset\mfk{p}_1$ and $\mfk{p}_1$ is prime, we may assume that $\mfk{q}_1 \subset \mfk{p}_1$. By the property of fractional ideal we have $\mfk{q}_1=\mfk{p}_1\mfk{r}_1$ for some fractional ideal $\mfk{r}_1$. However we also have $\mfk{q}_1 \subset \mfk{r}_1$. Since $\mfk{q}_1$ is prime, we either have $\mfk{q}_1 \supset \mfk{p}_1$ or $\mfk{q}_1 \supset \mfk{r}_1$. In the former case we get $\mfk{p}_1=\mfk{q}_1$, and we finish the proof by continuing inductively. In the latter case we have $\mfk{r}_1=\mfk{q}_1=\mfk{p}_1\mfk{q}_1$, which shows that $\mfk{p}_1=\mfk{o}$, which is impossible. $\blacksquare$
(Proposition 4) Every nontrivial prime ideal $\mfk{p}$ is maximal.
Proof. Let $\mfk{m}$ be an maximal ideal containing $\mfk{p}$. By proposition 2 we have some $\mfk{c}$ such that $\mfk{p}=\mfk{mc}$. If $\mfk{m} \neq \mfk{p}$, then $\mfk{c} \neq \mfk{o}$, and we may write $\mfk{c}=\mfk{p}_1\cdots\mfk{p}_n$, hence $\mfk{p}=\mfk{m}\mfk{p}_1\cdots\mfk{p}_n$, which is a prime factorisation, contradicting the fact that $\mfk{p}$ has a unique prime factorisation, which is $\mfk{p}$ itself. Hence any maximal ideal containing $\mfk{p}$ is $\mfk{p}$ itself. $\blacksquare$
(Proposition 5) Suppose the Dedekind domain $\mfk{o}$ only contains one prime (and maximal) ideal $\mfk{p}$, let $t \in \mfk{p}$ and $t \not\in \mfk{p}^2$, then $\mfk{p}$ is generated by $t$.
Proof. Let $\mfk{t}$ be the ideal generated by $t$. By proposition 3 we have a factorisation $\mfk{t}=\mfk{p}^n$ for some $n$ since $\mfk{o}$ contains only one prime ideal. According to proposition 2, if $n \geq 3$, we write $\mfk{p}^n=\mfk{p}^2\mfk{p}^{n-2}$, we see $\mfk{p}^2 \supset \mfk{p}^n$. But this is impossible since if so we have $t \in \mfk{p}^n \subset \mfk{p}^2$ contradicting our assumption. Hence $0<n<3$. But If $n=2$ we have $t \in \mfk{p}^2$ which is also not possible. So $\mfk{t}=\mfk{p}$ provided that such $t$ exists.
For the existence of $t$, note if not, then for all $t \in \mfk{p}$ we have $t \in \mfk{p}^2$, hence $\mfk{p} \subset \mfk{p}^2$. On the other hand we already have $\mfk{p}^2 = \mfk{p}\mfk{p}$, which implies that $\mfk{p}^2 \subset \mfk{p}$ (proposition 2), hence $\mfk{p}^2=\mfk{p}$, contradicting proposition 3. Hence such $t$ exists and our proof is finished. $\blacksquare$
Characterisation of Dedekind domain
In fact there is another equivalent definition of Dedekind domain:
A domain $\mfk{o}$ is Dedekind if and only if
• $\mfk{o}$ is Noetherian.
• $\mfk{o}$ is integrally closed.
• $\mfk{o}$ has Krull dimension $1$ (i.e. every non-zero prime ideals are maximal).
This is equivalent to say that faction ideals form a group and is frequently used by mathematicians as well. But we need some more advanced techniques to establish the equivalence. Presumably there will be a post about this in the future.
Several ways to prove Hardy's inequality
Suppose $1 < p < \infty$ and $f \in L^p((0,\infty))$ (with respect to Lebesgue measure of course) is a nonnegative function, take $F(x) = \frac{1}{x}\int_0^x f(t)dt \quad 0 < x <\infty,$ we have Hardy's inequality $\def\lrVert[#1]{\lVert #1 \rVert}$ $\lrVert[F]_p \leq q\lrVert[f]_p$ where $\frac{1}{p}+\frac{1}{q}=1$ of course.
There are several ways to prove it. I think there are several good reasons to write them down thoroughly since that may be why you find this page. Maybe you are burnt out since it's left as exercise. You are assumed to have enough knowledge of Lebesgue measure and integration.
Minkowski's integral inequality
Let $S_1,S_2 \subset \mathbb{R}$ be two measurable set, suppose $F:S_1 \times S_2 \to \mathbb{R}$ is measurable, then $\left[\int_{S_2} \left\vert\int_{S_1}F(x,y)dx \right\vert^pdy\right]^{\frac{1}{p}} \leq \int_{S_1} \left[\int_{S_2} |F(x,y)|^p dy\right]^{\frac{1}{p}}dx.$ A proof can be found at here by turning to Example A9. You may need to replace all measures with Lebesgue measure $m$.
Now let's get into it. For a measurable function in this place we should have $G(x,t)=\frac{f(t)}{x}$. If we put this function inside this inequality, we see \begin{aligned} \lrVert[F]_p &= \left[\int_0^\infty \left\vert \int_0^x \frac{f(t)}{x}dt \right\vert^p dx\right]^{\frac{1}{p}} \\ &= \left[\int_0^\infty \left\vert \int_0^1 f(ux)du \right\vert^p dx\right]^{\frac{1}{p}} \\ &\leq \int_0^1 \left[\int_0^\infty |f(ux)|^pdx\right]^{\frac{1}{p}}du \\ &= \int_0^1 \left[\int_0^\infty |f(ux)|^pudx\right]^{\frac{1}{p}}u^{-\frac{1}{p}}du \\ &= \lrVert[f]_p \int_0^1 u^{-\frac{1}{p}}du \\ &=q\lrVert[f]_p. \end{aligned} Note we have used change-of-variable twice and the inequality once.
A constructive approach
I have no idea how people came up with this solution. Take $xF(x)=\int_0^x f(t)t^{u}t^{-u}dt$ where $0<u<1-\frac{1}{p}$. Hölder's inequality gives us \begin{aligned} xF(x) &= \int_0^x f(t)t^ut^{-u}dt \\ &\leq \left[\int_0^x t^{-uq}dt\right]^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}} \\ &=\left(\frac{1}{1-uq}x^{1-uq}\right)^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}} \end{aligned} Hence \begin{aligned} F(x)^p & \leq \frac{1}{x^p}\left\{\left(\frac{1}{1-uq}x^{1-uq}\right)^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}}\right\}^{p} \\ &= \left(\frac{1}{1-uq}\right)^{\frac{p}{q}}x^{\frac{p}{q}(1-uq)-p}\int_0^x f(t)^pt^{up}dt \\ &= \left(\frac{1}{1-uq}\right)^{p-1}x^{-up-1}\int_0^x f(t)^pt^{up}dt \end{aligned}
Note we have used the fact that $\frac{1}{p}+\frac{1}{q}=1 \implies p+q=pq$ and $\frac{p}{q}=p-1$. Fubini's theorem gives us the final answer: \begin{aligned} \int_0^\infty F(x)^pdx &\leq \int_0^\infty\left[\left(\frac{1}{1-uq}\right)^{p-1}x^{-up-1}\int_0^x f(t)^pt^{up}dt\right]dx \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\int_0^\infty dx\int_0^x f(t)^pt^{up}x^{-up-1}dt \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\int_0^\infty dt\int_t^\infty f(t)^pt^{up}x^{-up-1}dx \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\frac{1}{up}\int_0^\infty f(t)^pdt. \end{aligned} It remains to find the minimum of $\varphi(u) = \left(\frac{1}{1-uq}\right)^{p-1}\frac{1}{up}$. This is an elementary calculus problem. By taking its derivative, we see when $u=\frac{1}{pq}<1-\frac{1}{p}$ it attains its minimum $\left(\frac{p}{p-1}\right)^p=q^p$. Hence we get $\int_0^\infty F(x)^pdx \leq q^p\int_0^\infty f(t)^pdt,$ which is exactly what we want. Note the constant $q$ cannot be replaced with a smaller one. We simply proved the case when $f \geq 0$. For the general case, one simply needs to take absolute value.
Integration by parts
This approach makes use of properties of $L^p$ space. Still we assume that $f \geq 0$ but we also assume $f \in C_c((0,\infty))$, that is, $f$ is continuous and has compact support. Hence $F$ is differentiable in this situation. Integration by parts gives $\int_0^\infty F^p(x)dx=xF(x)^p\vert_0^\infty- p\int_0^\infty xdF^p = -p\int_0^\infty xF^{p-1}(x)F'(x)dx.$ Note since $f$ has compact support, there are some $[a,b]$ such that $f >0$ only if $0 < a \leq x \leq b < \infty$ and hence $xF(x)^p\vert_0^\infty=0$. Next it is natural to take a look at $F'(x)$. Note we have $F'(x) = \frac{f(x)}{x}-\frac{\int_0^x f(t)dt}{x^2},$ hence $xF'(x)=f(x)-F(x)$. A substitution gives us $\int_0^\infty F^p(x)dx = -p\int_0^\infty F^{p-1}(x)[f(x)-F(x)]dx,$ which is equivalent to say $\int_0^\infty F^p(x)dx = \frac{p}{p-1}\int_0^\infty F^{p-1}(x)f(x)dx.$ Hölder's inequality gives us \begin{aligned} \int_0^\infty F^{p-1}(x)f(x)dx &\leq \left[\int_0^\infty F^{(p-1)q}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}} \\ &=\left[\int_0^\infty F^{p}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}}. \end{aligned} Together with the identity above we get $\int_0^\infty F^p(x)dx = q\left[\int_0^\infty F^{p}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}}$ which is exactly what we want since $1-\frac{1}{q}=\frac{1}{p}$ and all we need to do is divide $\left[\int_0^\infty F^pdx\right]^{1/q}$ on both sides. So what's next? Note $C_c((0,\infty))$ is dense in $L^p((0,\infty))$. For any $f \in L^p((0,\infty))$, we can take a sequence of functions $f_n \in C_c((0,\infty))$ such that $f_n \to f$ with respect to $L^p$-norm. Taking $F=\frac{1}{x}\int_0^x f(t)dt$ and $F_n = \frac{1}{x}\int_0^x f_n(t)dt$, we need to show that $F_n \to F$ pointwise, so that we can use Fatou's lemma. For $\varepsilon>0$, there exists some $m$ such that $\lrVert[f_n-f]_p < \frac{1}{n}$. Thus \begin{aligned} |F_n(x)-F(x)| &= \frac{1}{x}\left\vert \int_0^x f_n(t)dt - \int_0^x f(t)dt \right\vert \\ &\leq \frac{1}{x} \int_0^x |f_n(t)-f(t)|dt \\ &\leq \frac{1}{x} \left[\int_0^x|f_n(t)-f(t)|^pdt\right]^{\frac{1}{p}}\left[\int_0^x 1^qdt\right]^{\frac{1}{q}} \\ &=\frac{1}{x^{1/p}}\left[\int_0^x|f_n(t)-f(t)|^pdt\right]^{\frac{1}{p}} \\ &\leq \frac{1}{x^{1/p}}\lrVert[f_n-f]_p <\frac{\varepsilon}{x^{1/p}}. \end{aligned} Hence $F_n \to F$ pointwise, which also implies that $|F_n|^p \to |F|^p$ pointwise. For $|F_n|$ we have \begin{aligned} \int_0^\infty |F_n(x)|^pdx &= \int_0^\infty \left\vert\frac{1}{x}\int_0^x f_n(t)dt\right\vert^p dx \\ &\leq \int_0^\infty \left[\frac{1}{x}\int_0^x |f_n(t)|dt\right]^{p}dx \\ &\leq q\int_0^\infty |f_n(t)|^pdt \end{aligned} note the third inequality follows since we have already proved it for $f \geq 0$. By Fatou's lemma, we have \begin{aligned} \int_0^\infty |F(x)|^pdx &= \int_0^\infty \lim_{n \to \infty}|F_n(x)|^pdx \\ &\leq \lim_{n \to \infty} \int_0^\infty |F_n(x)|^pdx \\ &\leq \lim_{n \to \infty}q^p\int_0^\infty |f_n(x)|^pdx \\ &=q^p\int_0^\infty |f(x)|^pdx. \end{aligned}
Tensor Product as a Universal Object (Category Theory & Module Theory)
Introduction
It is quite often to see direct sum or direct product of groups, modules, vector spaces. Indeed, for modules over a ring $R$, direct products are also direct products of $R$-modules as well. On the other hand, the direct sum is a coproduct in the category of $R$-modules.
But what about tensor products? It is some different kind of product but how? Is it related to direct product? How do we write a tensor product down? We need to solve this question but it is not a good idea to dig into numeric works.
The category of bilinear or even $n$-multilinear maps
From now on, let $R$ be a commutative ring, and $M_1,\cdots,M_n$ are $R$-modules. Mainly we work on $M_1$ and $M_2$, i.e. $M_1 \times M_2$ and $M_1 \otimes M_2$. For $n$-multilinear one, simply replace $M_1\times M_2$ with $M_1 \times M_2 \times \cdots \times M_n$ and $M_1 \otimes M_2$ with $M_1 \otimes \cdots \otimes M_n$. The only difference is the change of symbols.
The bilinear maps of $M_1 \times M_2$ determines a category, say $BL(M_1 \times M_2)$ or we simply write $BL$. For an object $(f,E)$ in this category we have $f: M_1 \times M_2 \to E$ as a bilinear map and $E$ as a $R$-module of course. For two objects $(f,E)$ and $(g,F)$, we define the morphism between them as a linear function making the following diagram commutative: $\def\mor{\operatorname{Mor}}$
This indeed makes $BL$ a category. If we define the morphisms from $(f,E)$ to $(g,F)$ by $\mor(f,g)$ (for simplicity we omit $E$ and $F$ since they are already determined by $f$ and $g$) we see the composition $\mor(f,g) \times \mor(h,g) \to \mor(h,f)$ satisfy all axioms for a category:
CAT 1 Two sets $\mor(f,g)$ and $\mor(f',g')$ are disjoint unless $f=f'$ and $g=g'$, in which case they are equal. If $g \neq g'$ but $f = f'$ for example, for any $h \in \mor(f,g)$, we have $g = h \circ f = h \circ f' \neq g'$, hence $h \notin \mor(f,g)$. Other cases can be verified in the same fashion.
CAT 2 The existence of identity morphism. For any $(f,E) \in BL$, we simply take the identity map $i:E \to E$. For $h \in \mor(f,g)$, we see $g = h \circ f = h \circ i \circ f$. For $h' \in \mor(g,f)$, we see $f = h' \circ g = i \circ h' \circ g$.
CAT 3 The law of composition is associative when defined.
There we have a category. But what about the tensor product? It is defined to be initial (or universally repelling) object in this category. Let's denote this object by $(\varphi,M_1 \otimes M_2)$.
For any $(f,E) \in BL$, we have a unique morphism (which is a module homomorphism as well) $h:(\varphi,M_1 \otimes M_2) \to (f,E)$. For $x \in M_1$ and $y \in M_2$, we write $\varphi(x,y)=x \otimes y$. We call the existence of $h$ the universal property of $(\varphi,M_1 \otimes M_2)$.
The tensor product is unique up to isomorphism. That is, if both $(f,E)$ and $(g,F)$ are tensor products, then $E \simeq F$ in the sense of module isomorphism. Indeed, let $h \in \mor(f,g)$ and $h' \in \mor(g,h)$ be the unique morphisms respectively, we see $g = h \circ f$, $f = h' \circ g$, and therefore $g = h \circ h' \circ g \\ f = h' \circ h \circ f$ Hence $h \circ h'$ is the identity of $(g,F)$ and $h' \circ h$ is the identity of $(f,E)$. This gives $E \simeq F$.
What do we get so far? For any modules that is connected to $M_1 \times M_2$ with a bilinear map, the tensor product $M_1 \oplus M_2$ of $M_1$ and $M_2$, is always able to be connected to that module with a unique module homomorphism. What if there are more than one tensor products? Never mind. All tensor products are isomorphic.
But wait, does this definition make sense? Does this product even exist? How can we study the tensor product of two modules if we cannot even write it down? So far we are only working on arrows, and we don't know what is happening inside an module. It is not a good idea to waste our time on 'nonsenses'. We can look into it in an natural way. Indeed, if we can find a module satisfying the property we want, then we are done, since this can represent the tensor product under any circumstances. Again, all tensor products of $M_1$ and $M_2$ are isomorphic.
A natural way to define the tensor product
Let $M$ be the free module generated by the set of all tuples $(x_1,x_2)$ where $x_1 \in M_1$ and $x_2 \in M_2$, and $N$ be the submodule generated by tuples of the following types: $(x_1+x_1',x_2)-(x_1,x_2)-(x_1',x_2) \\ (x_1,x_2+x_2')-(x_1,x_2)-(x_1,x_2') \\ (ax_1,x_2)-a(x_1,x_2) \\ (x_1,ax_2) - a(x_1,x_2)$ First we have a inclusion map $\alpha=M_1 \times M_2 \to M$ and the canonical map $\pi:M \to M/N$. We claim that $(\pi \circ \alpha, M/N)$ is exactly what we want. But before that, we need to explain why we define such a $N$.
The reason is quite simple: We want to make sure that $\varphi=\pi \circ \alpha$ is bilinear. For example, we have $\varphi(x_1+x_1',x_2)=\varphi(x_1,x_2)+\varphi(x_1',x_2)$ due to our construction of $N$ (other relations follow in the same manner). This can be verified group-theoretically. Note $\varphi(x_1+x_1',x_2)=(x_1+x_1',x_2)+N \\ \varphi(x_1,x_2)+\varphi(x_1',x_2)=(x_1,x_2)+(x_1',x_2)+N$ but $\varphi(x_1+x_1',x_2)-\varphi(x_1,x_2)-\varphi(x_1',x_2)=(x_1+x_1',x_2)-(x_1,x_2)-(x_1',x_2) +N = 0+N.$ Hence we get the identity we want. For this reason we can write \begin{aligned} (x_1+x_1')\otimes x_2 &= x_1 \otimes x_2 + x_1' \otimes x_2, \\ x_1 \otimes (x_2 + x_2') &= x_1 \otimes x_2 + x_1 \otimes x_2', \\ (ax_1) \otimes x_2 &= a(x_1 \otimes x_2), \\ x_1 \otimes (ax_2) &= a(x_1 \otimes x_2). \end{aligned} Sometimes to avoid confusion people may also write $x_1 \otimes_R x_2$ if both $M_1$ and $M_2$ are $R$-modules. But before that we have to verify that this is indeed the tensor product. To verify this, all we need is the universal property of free modules.
By the universal property of $M$, for any $(f,E) \in BL$, we have a induced map $f_\ast$ making the diagram inside commutative. However, for elements in $N$, we see $f_\ast$ takes value $0$, since $f_\ast$ is a bilinear map already. We finish our work by taking $h[(x,y)+N] = f_\ast(x,y)$. This is the map induced by $f_\ast$, following the property of factor module.
Trivial tensor product
For coprime integers $m,n>1$, we have $\def\mb{\mathbb}$ $\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} = O$ where $O$ means that the module only contains $0$ and $\mb{Z}/m\mb{Z}$ is considered as a module over $\mb{Z}$ for $m>1$. This suggests that, the tensor product of two modules is not necessarily 'bigger' than its components. Let's see why this is trivial.
Note that for $x \in \mb{Z}/m\mb{Z}$ and $y \in \mb{Z}/n\mb{Z}$, we have $m(x \otimes y) = (mx) \otimes y = 0 \\ n(x \otimes y) = x \otimes(ny) = 0$ since, for example, $mx = 0$ for $x \in \mb{Z}/m\mb{Z}$ and $\varphi(0,y)=0$. If you have trouble understanding why $\varphi(0,y)=0$, just note that the submodule $N$ in our construction contains elements generated by $(0x,y)-0(x,y)$ already.
By Bézout's identity, for any $x \otimes y$, we see there are $a$ and $b$ such that $am+bn=1$, and therefore \begin{aligned} x \otimes y &= (am+bn)(x \otimes y) \\ &=am(x \otimes y)+bn (x \otimes y) \\ &= 0. \end{aligned} Hence the tensor product is trivial. This example gives us a lot of inspiration. For example, what if $m$ and $n$ are not necessarily coprime, say $\gcd(m,n)=d$? By Bézout's identity still we have $d(x \otimes y) = (am+bn)(x \otimes y) = 0.$ This inspires us to study the connection between $\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z}$ and $\mb{Z}/d\mb{Z}$. By the universal property, for the bilinear map $f:\mb{Z}/m\mb{Z} \times \mb{Z}/n\mb{Z} \to \mb{Z}/d\mb{Z}$ defined by $(a+m\mb{Z},b+n\mb{Z})\mapsto ab+d\mb{Z}$ (there should be no difficulty to verify that $f$ is well-defined), there exists a unique morphism $h:\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} \to \mb{Z}/d\mb{Z}$ such that $h \circ \varphi(a+m\mb{Z},b+n\mb{Z}) = h((a+m\mb{Z}) \otimes(b+n\mb{Z})) = ab+d\mb{Z}.$ Next we show that it has a natural inverse defined by \begin{aligned} g:\mb{Z}/d\mb{Z} &\to \mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} \\ a+d\mb{Z} &\mapsto (a+m\mb{Z}) \otimes (1+n\mb{Z}). \end{aligned} Taking $a' = a+kd$, we show that $g(a+d\mb{Z})=g(a'+\mb{Z})$, that is, we need to show that $(a+m\mb{Z})\otimes(1+n\mb{Z}) = (a'+m\mb{Z}) \otimes (1+n\mb{Z}).$ By Bézout's identity, there exists some $r,s$ such that $rm+sn=d$. Hence $a' = a + ksn+krm$, which gives \begin{aligned} (a'+m\mb{Z}) \otimes (1+n\mb{Z}) &= (a+ksn+krm+m\mb{Z}) \otimes(1+n\mb{Z}) \\ &= (a+ksn+m\mb{Z}) \otimes (1+n\mb{Z}) \\ &=(a+m\mb{Z}) \otimes(1+n\mb{Z}) + (ksn+m\mb{Z})\otimes(1+n\mb{Z}) \\ &=(a+m\mb{Z}) \otimes (1+n\mb{Z}) \end{aligned} since $(ksn+m\mb{Z}) \otimes (1+n\mb{Z}) =n(ks+m\mb{Z}) \otimes (1+n\mb{Z}) = (ks+m\mb{Z}) \otimes(n+n\mb{Z}) = 0.$ So $g$ is well-defined. Next we show that this is the inverse. Firstly \begin{aligned} g \circ h((a+m\mb{Z}) \otimes(b+n\mb{Z})) &= g(ab+d\mb{Z})\\ &= (ab+m\mb{Z}) \otimes (1+n\mb{Z}) \\ &=b(a+m\mb{Z}) \otimes(1+n\mb{Z}) \\ &= (a+m\mb{Z}) \otimes (b+n\mb{Z}). \end{aligned} Secondly, \begin{aligned} h \circ g(a+d\mb{Z}) &= h((a+m\mb{Z}) \otimes(1+n\mb{Z})) \\ &= a+d\mb{Z}. \end{aligned} Hence $g = h^{-1}$ and we can say $\mb{Z}/m\mb{Z} \otimes \mb{Z} /n\mb{Z} \simeq \mb{Z} /\gcd(m,n)\mb{Z}.$ If $m,n$ are coprime, then $\gcd(m,n)=1$, hence $\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} \simeq \mb{Z}/\mb{Z}$ is trivial. More interestingly, $\mb{Z}/m\mb{Z}\otimes \mb{Z}/m\mb{Z}=\mb{Z}/m\mb{Z}$. But this elegant identity raised other questions. First of all, $\gcd(m,n)=\gcd(n,m)$, which implies $\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} \simeq \mb{Z}/\gcd(m,n)\mb{Z} \simeq \mb{Z}/\gcd(n,m)\mb{Z} \simeq\mb{Z}/n\mb{Z}\otimes\mb{Z}/m\mb{Z}.$ Further, for $m,n,r >1$, we have $\gcd(\gcd(m,n),r)=\gcd(m,\gcd(n,r))=\gcd(m,n,r)$, which gives $(\mb{Z}/m\mb{Z}\otimes\mb{Z}/n\mb{Z})\otimes\mb{Z}/r\mb{Z} \simeq \mb{Z}/\gcd(m,n)\mb{Z}\otimes\mb{Z}/r\mb{Z} \simeq \mb{Z}/\gcd(m,n,r)\mb{Z} \\ \mb{Z}/m\mb{Z}\otimes(\mb{Z}/n\mb{Z} \otimes\mb{Z}/r\mb{Z}) \simeq \mb{Z}/m\mb{Z} \otimes\mb{Z}/\gcd(n,r)\mb{Z} \simeq \mb{Z}/\gcd(m,n,r)\mb{Z}$ hence $(\mb{Z}/m\mb{Z}\otimes\mb{Z}/n\mb{Z})\otimes\mb{Z}/r\mb{Z} \simeq \mb{Z}/m\mb{Z}\otimes(\mb{Z}/n\mb{Z}\otimes\mb{Z}/r\mb{Z}).$ Hence for modules of the form $\mb{Z}/m\mb{Z}$, we see the tensor product operation is associative and commutative up to isomorphism. Does this hold for all modules? The universal property answers this question affirmatively. From now on we will be keep using the universal property. Make sure that you have got the point already.
Tensor product as a binary operation
Let $M_1,M_2,M_3$ be $R$-modules, then there exists a unique isomorphism \begin{aligned} (M_1 \otimes M_2) \otimes M_3 &\xrightarrow{\simeq} M_1 \otimes (M_2 \otimes M_3) \\ (x \otimes y) \otimes z &\mapsto x \otimes(y \otimes z) \end{aligned} for $x \in M_1$, $y \in M_2$, $z \in M_3$.
Proof. Consider the map \begin{aligned} \lambda_x:M_2 \times M_3 &\to (M_1 \otimes M_2)\otimes M_3 \\ (y,z) &\mapsto (x \otimes y ) \otimes z \end{aligned} where $x \in M_1$. Since $(\cdot\otimes\cdot)$ is bilinear, we see $\lambda_x$ is bilinear for all $x \in M_1$. Hence by the universal property there exists a unique map of the tensor product: $\overline{\lambda}_x:M_2 \otimes M_3 \to (M_1 \otimes M_2) \otimes M_3.$ Next we have the map \begin{aligned} \mu_x: M_1 \times (M_2 \otimes M_3) &\to (M_1 \otimes M_2) \otimes M_3 \\ (x,y \otimes z) &\mapsto \overline{\lambda}_x(y \otimes z) \end{aligned} which is bilinear as well. Again by the universal property we have a unique map $\overline{\mu}_x: M_1 \otimes (M_2 \otimes M_3) \to (M_1 \otimes M_2) \otimes M_3.$ This is indeed the isomorphism we want. The reverse is obtained by reversing the process. For the bilinear map $\lambda_x':M_1 \times M_2 \to M_1 \otimes (M_2 \otimes M_3)$ we get a unique map $\overline{\lambda'}_x: M_1 \otimes M_2 \to M_1 \otimes (M_2 \otimes M_3).$ Then from the bilinear map $\mu'_x:(M_1 \otimes M_2) \times M_3 \to M_1 \otimes (M_2 \otimes M_3)$ we get the unique map, which is actually the reverse of $\overline{\mu}_x$: $\overline{\mu'}_x:(M_1 \otimes M_2) \otimes M_3 \to M_1 \otimes (M_2 \otimes M_3).$ Hence the two tensor products are isomorphic. $\square$
Let $M_1$ and $M_2$ be $R$-modules, then there exists a unique isomorphism \begin{aligned} M_1 \otimes M_2 &\xrightarrow{\simeq} M_2 \otimes M_1 \\ x_1 \otimes x_2 &\mapsto x_2 \otimes x_1 \end{aligned} where $x_1 \in M_1$ and $x_2 \in M_2$.
Proof. The map \begin{aligned} \lambda:M_1 \times M_2 &\to M_2 \otimes M_1 \\ (x,y) &\mapsto y \otimes x \end{aligned} is bilinear and gives us a unique map $\overline{\lambda}:M_1 \otimes M_2 \to M_2 \otimes M_1$ given by $x \otimes y \mapsto y \otimes x$. Symmetrically, the map $\lambda':M_2 \times M_1 \to M_1 \otimes M_2$ gives us a unique map $\overline{\lambda'}:M_2 \otimes M_1 \to M_1 \otimes M_2$ which is the inverse of $\overline{\lambda}$. $\square$
Therefore, we may view the set of all $R$-modules as a commutative semigroup with the binary operation $\otimes$.
Maps between tensor products
Consider commutative diagram:
Where $f_i:M_i \to M_i'$ are some module-homomorphism. What do we want here? On the left hand, we see $f_1 \times f_2$ sends $(x_1,x_2)$ to $(f_1(x_1),f_2(x_2))$, which is quite natural. The question is, is there a natural map sending $x_1 \otimes x_2$ to $f_1(x_1) \otimes f_2(x_2)$? This is what we want from the right hand. We know $T(f_1 \times f_2)$ exists, since we have a bilinear map by $\mu = \varphi' \circ (f_1\times f_2)$. So for $(x_1,x_2) \in M_1 \times M_2$, we have $T(f_1 \times f_2)(x_1 \otimes x_2) = \varphi' \circ (f_1 \times f_2)(x_1,x_2) = f_1(x_1) \otimes f_2(x_2)$ as what we want.
But $T$ in this graph has more interesting properties. First of all, if $M_1 = M_1'$ an $M_2 = M_2'$, both $f_1$ and $f_2$ are identity maps, then we see $T(f_1 \times f_2)$ is the identity as well. Next, consider the following chain $\cdots \to M_1 \times M_2 \xrightarrow{(f_1 \times f_2)}M_1' \times M_2' \xrightarrow{(g_1 \times g_2)}M_1'' \times M_2''\to \cdots.$ We can make it a double chain:
It is obvious that $(g_1 \circ f_1 \times g_2 \circ f_2)=(g_1 \times g_2) \circ (f_1 \times f_2)$, which also gives $T(g_1 \times g_2) \circ T(f_1 \times f_2) = T(g_1 \circ f_1 \times g_2 \circ f_2).$ Hence we can say $T$ is functorial. Sometimes for simplicity we also write $T(f_1,f_2)$ or simply $f_1 \otimes f_2$, as it sends $x_1 \otimes x_2$ to $f_1(x_1) \otimes f_2(x_2)$. Indeed it can be viewed as a map \begin{aligned} T:L(M_1, M_1') \times L(M_2,M_2') &\to L(M_1 \otimes M_2, M_1' \otimes M_2') \\ (f_1 \times f_2) &\mapsto f_1 \otimes f_2. \end{aligned}
Why Does a Vector Space Have a Basis (Module Theory)
Module and vector space
First we recall some backgrounds. Suppose $A$ is a ring with multiplicative identity $1_A$. A left module of $A$ is an additive abelian group $(M,+)$, together with an ring operation $A \times M \to M$ such that \begin{aligned} (a+b)x &= ax+bx \\ a(x+y) &= ax+ay \\ a(bx) &= (ab)x \\ 1_Ax &= x \end{aligned} for $x,y \in M$ and $a,b \in A$. As a corollary, we see $(0_A+0_A)x=0_Ax=0_Ax+0_Ax$, which shows $0_Ax=0_M$ for all $x \in M$. On the other hand, $a(x-x)=0_M$ which implies $a(-x)=-(ax)$. We can also define right $A$-modules but we are not discussing them here.
Let $S$ be a subset of $M$. We say $S$ is a basis of $M$ if $S$ generates $M$ and $S$ is linearly independent. That is, for all $m \in M$, we can pick $s_1,\cdots,s_n \in S$ and $a_1,\cdots,a_n \in A$ such that $m = a_1s_1+a_2s_2+\cdots+a_ns_n,$ and, for any $s_1,\cdots,s_n \in S$, we have $a_1s_1+a_2s_2+\cdots+a_ns_n=0_M \implies a_1=a_2=\cdots=a_n=0_A.$ Note this also shows that $0_M\notin S$ (what happens if $0_M \in S$?). We say $M$ is free if it has a basis. The case when $M$ or $A$ is trivial is excluded.
If $A$ is a field, then $M$ is called a vector space, which has no difference from the one we learn in linear algebra and functional analysis. Mathematicians in functional analysis may be interested in the cardinality of a vector space, for example, when a vector space is of finite dimension, or when the basis is countable. But the basis does not come from nowhere. In fact we can prove that vector spaces have basis, but modules are not so lucky. $\def\mb{\mathbb}$
Examples of non-free modules
First of all let's consider the cyclic group $\mb{Z}/n\mb{Z}$ for $n \geq 2$. If we define \begin{aligned} \mb{Z} \times \mb{Z}/n\mb{Z} &\to \mb{Z}/n\mb{Z} \\ (m,k+n\mb{Z}) &\mapsto mk+n\mb{Z} \end{aligned} which is actually $m$ copies of an element, then we get a module, which will be denoted by $M$. For any $x=k+n\mb{Z} \in M$, we see $nk+n\mb{Z}=0_M$. Therefore for any subset $S \subset M$, if $x_1,\cdots,x_k \in M$, we have $nx_1+nx_2+\cdots+nx_k = 0_M,$ which gives the fact that $M$ has no basis. In fact this can be generalized further. If $A$ is a ring but not a field, let $I$ be a nontrivial proper ideal, then $A/I$ is a module that has no basis.
Following $\mb{Z}/n\mb{Z}$ we also have another example on finite order. Indeed, any finite abelian group is not free as a module over $\mb{Z}$. More generally,
Let $G$ be a abelian group, and $G_{tor}$ be its torsion subgroup. If $G_{tor}$ is non-trival, then $G$ cannot be a free module over $\mb{Z}$.
Next we shall take a look at infinite rings. Let $F[X]$ be the polynomial ring over a field $F$ and $F'[X]$ be the polynomial sub-ring that have coefficient of $X$ equal to $0$. Then $F[X]$ is a $F'[X]$-module. However it is not free.
Suppose we have a basis $S$ of $F[X]$, then we claim that $|S|>1$. If $|S|=1$, say $P \in S$, then $P$ cannot generate $F[X]$ since if $P$ is constant then we cannot generate a polynomial contains $X$ with power $1$; If $P$ is not constant, then the constant polynomial cannot be generate. Hence $S$ contains at least two polynomials, say $P_1 \neq 0$ and $P_2 \neq 0$. However, note $-X^2P_1 \in F'[X]$ and $X^2P_2 \in F'[X]$, which gives $(X^2P_2)P_1-(X^2P_1)P_2=0.$ Hence $S$ cannot be a basis.
Why does a vector space have a basis
I hope those examples have convinced you that basis is not a universal thing. We are going to prove that every vector space has a basis. More precisely,
Let $V$ be a nontrivial vector space over a field $K$. Let $\Gamma$ be a set of generators of $V$ over $K$ and $S \subset \Gamma$ is a subset which is linearly independent, then there exists a basis of $V$ such that $S \subset B \subset \Gamma$.
Note we can always find such $\Gamma$ and $S$. For the extreme condition, we can pick $\Gamma=V$ and $S$ be a set containing any single non-zero element of $V$. Note this also gives that we can generate a basis by expanding any linearly independent set. The proof relies on a fact that every non-zero element in a field is invertible, and also, Zorn's lemma. In fact, axiom of choice is equivalent to the statement that every vector has a set of basis.$\def\mfk{\mathfrak}$
Proof. Define $\mfk{T} =\{T \subset \Gamma:S \subset T, \text{ T is linearly independent}\}.$ Then $\mfk{T}$ is not empty since it contains $S$. If $T_1 \subset T_2 \subset \cdots$ is a totally ordered chain in $\mfk{T}$, then $T=\bigcup_{i=1}^{\infty}T_i$ is again linearly independent and contains $S$. To show that $T$ is linearly independent, note that if $x_1,x_2,\cdots,x_n \in T$, we can find some $k_1,\cdots,k_n$ such that $x_i \in T_{k_i}$ for $i=1,2,\cdots,n$. If we pick $k = \max(k_1,\cdots,k_n)$, then $x_1,x_2,\cdots,x_n \in \bigcup_{i=1}^{n}T_{k_i}=T_k.$ But we already know that $T_k$ is linearly independent, so $a_1x_1+\cdots+a_nx_n=0_V$ implies $a_1=\cdots=a_n=0_K$.
By Zorn's lemma, let $B$ be the maximal element of $\mfk{T}$, then $B$ is also linearly independent since it is an element of $\mfk{T}$. Next we show that $B$ generates $V$. Suppose not, then we can pick some $x \in \Gamma$ that is not generated by $B$. Define $B'=B \cup \\{x\\}$, we see $B'$ is linearly independent as well, because if we pick $y_1,y_2,\cdots,y_n \in B$, and if $\sum_{k=1}^{n}a_ky_k+bx=0_V,$ then if $b \neq 0$ we have $x = -\sum_{k=1}^{n}b^{-1}a_ky_k \in B,$ contradicting the assumption that $x$ is not generated by $B$. Hence $b=0_K$. However, we have proved that $B'$ is a linearly independent set containing $B$ and contained in $S$, contradicting the maximality of $B$ in $\mfk{T}$. Hence $B$ generates $V$. $\square$
Rings of Fractions and Localisation
Is perhaps the most important technical tools in commutative algebra. In this post we are covering definitions and simple properties. Also we restrict ourselves into ring theories and no further than that. Throughout, we let $A$ be a commutative ring. With extra effort we can also make it to non-commutative rings for some results but we are not doing that here.
In fact the construction of $\mathbb{Q}$ from $\mathbb{Z}$ has already been an example. For any $a \in \mathbb{Q}$, we have some $m,n \in \mathbb{Z}$ with $n \neq 0$ such that $a = \frac{m}{n}$. As a matter of notation we may also say an ordered pair $(m,n)$ determines $a$. Two ordered pairs $(m,n)$ and $(m',n')$ are equivalent if and only if $mn'-m'n=0.$ But we are only using the ring structure of $\mathbb{Z}$. So it is natural to think whether it is possible to generalize this process to all rings. But we are also using the fact that $\mathbb{Z}$ is an entire ring (or alternatively integral domain, they mean the same thing). However there is a way to generalize it. $\def\mfk{\mathfrak}$
Multiplicatively closed subset
(Definition 1) A multiplicatively closed subset $S \subset A$ is a set that $1 \in S$ and if $x,y \in S$, then $xy \in S$.
For example, for $\mathbb{Z}$ we have a multiplicatively closed subset $\{1,2,4,8,\cdots\} \subset \mathbb{Z}.$ We can also insert $0$ here but it may produce some bad result. If $S$ is also an ideal then we must have $S=A$ so this is not very interesting. However the complement is interesting.
(Proposition 1) Suppose $A$ is a commutative ring such that $1 \neq 0$. Let $S$ be a multiplicatively closed set that does not contain $0$. Let $\mfk{p}$ be the maximal element of ideals contained in $A \setminus S$, then $\mfk{p}$ is prime.
Proof. Recall that $\mfk{p}$ is prime if for any $x,y \in A$ such that $xy \in \mfk{p}$, we have $x \in \mfk{p}$ or $y \in \mfk{p}$. But now we fix $x,y \in \mfk{p}^c$. Note we have a strictly bigger ideal $\mfk{q}_1=\mfk{p}+Ax$. Since $\mfk{p}$ is maximal in the ideals contained in $A \setminus S$, we see $\mfk{q}_1 \cap S \neq \varnothing.$ Therefore there exist some $a \in A$ and $p \in \mfk{p}$ such that $p+ax \in S.$ Also, $\mfk{q}_2=\mfk{p}+Ay$ has nontrivial intersection with $S$ (due to the maximality of $\mfk{p}$), there exist some $a' \in A$ and $p' \in \mfk{p}$ such that $p' + a'y \in S.$ Since $S$ is closed under multiplication, we have $(p+ax)(p'+a'y) = pp'+p'ax+pa'y+aa'xy \in S.$ But since $\mfk{p}$ is an ideal, we see $pp'+p'ax+pa'y \in \mfk{p}$. Therefore we must have $xy \notin \mfk{p}$ since if not, $(p+ax)(p'+a'y) \in \mfk{p}$, which gives $\mfk{p} \cap S \neq \varnothing$, and this is impossible. $\square$
As a corollary, for an ideal $\mfk{p} \subset A$, if $A \setminus \mfk{p}$ is multiplicatively closed, then $\mfk{p}$ is prime. Conversely, if we are given a prime ideal $\mfk{p}$, then we also get a multiplicatively closed subset.
(Proposition 2) If $\mfk{p}$ is a prime ideal of $A$, then $S = A \setminus \mfk{p}$ is multiplicatively closed.
Proof. First $1 \in S$ since $\mfk{p} \neq A$. On the other hand, if $x,y \in S$ we see $xy \in S$ since $\mfk{p}$ is prime. $\square$
Ring of fractions of a ring
We define a equivalence relation on $A \times S$ as follows: $(a,s) \sim (b,t) \iff \exists u \in S, (at-bs)u=0.$
(Proposition 3) $\sim$ is an equivalence relation.
Proof. Since $(as-as)1=0$ while $1 \in S$, we see $(a,s) \sim (a,s)$. For being symmetric, note that $(at-bs)u=0 \implies (bs-at)u=0 \implies (b,t) \sim (a,s).$ Finally, to show that it is transitive, suppose $(a,s) \sim (b,t)$ and $(b,t) \sim (c,u)$. There exist $u,v \in S$ such that $(at-bs)v=(bu-ct)w=0.$ This gives $bsv=atv$ and $buw = ctw$, which implies $bsvuw=atvuw=ctwsv \implies (au-cs)tvw =0.$ But $tvw \in S$ since $t,v,w \in S$ and $S$ is multiplicatively closed. Hence $[(a,s) \sim (b,t)] \land [(b,t) \sim (c,u)] \implies (a,s) \sim (c,u).$ $\square$
Let $a/s$ denote the equivalence class of $(a,s)$. Let $S^{-1}A$ denote the set of equivalence classes (it is not a good idea to write $A/S$ as it may coincide with the notation of factor group), and we put a ring structure on $S^{-1}A$ as follows: $(a/s)+(b/t)=(at+bs)/st, \\ (a/s)(b/t)=ab/st.$ There is no difference between this one and the one in elementary algebra. But first of all we need to show that $S^{-1}A$ indeed form a ring.
(Proposition 4) The addition and multiplication are well defined. Further, $S^{-1}A$ is a commutative ring with identity.
Proof. Suppose $(a,s) \sim (a',s')$ and $(b,t) \sim (b',t')$ we need to show that $(a/s)+(b/t)=(a'/s')+(b'/t')$ or $(at+bs)/st = (a't'+b's')/s't'.$ There exists $u,v \in S$ such that $(as'-a's)u=0 \quad (bt'-b't)v=0.$ If we multiply the first equation by $vtt'$ and second equation by $uss'$, we see $as'uvtt'-a'suvtt'+bt'vuss'-b'tvuss'=[(at)s't'+(bs)s't'-(a't')st-(b's')st]uv,$ which is exactly what we want.
On the other hand, we need to show that $ab/st = a'b'/s't'.$ That is, $\exists y \in S,(abs't'-a'b'st)y=0.$ Again, we have $(as'-a's)u=(as'-a's)uvbt'=(abs't'-a'bst')uv=0, \\ (bt'-b't)v=(bt'-b't)vua's=(a'bst'-a'b'st)uv=0.$ Hence $(abs't'-a'bst')uv+(a'bst'-a'b'st)uv=(abs't'-a'b'st)uv=0.$ Since $uv \in S$, we are done.
Next we show that $S^{-1}A$ has a ring structure. If $0 \in S$, then $S^{-1}A$ contains exactly one element $0/1$ since in this case, all pairs are equivalent: $(at-bs)0=0.$ We therefore only discuss the case when $0 \notin S$. First $0/1$ is the zero element with respect to addition since $0/1+a/s = (0s+1a)/1s = a/s.$ On the other hand, we have the inverse $-a/s$: $-a/s+a/s = (-as+as)/ss=0/ss=0/1.$ $1/1$ is the unit with respect to multiplication: $(1/1)(a/s)=1a/1s=a/s.$ Multiplication is associative since $[(a/s)(b/t)](c/u)=(ab/st)(c/u)=abc/stu. \\ (a/s)[(b/t)(c/u)]=(a/s)(bc/tu)=abc/stu.$ Multiplication is commutative since $ab/st+(-ba)/st=(abst-bast)/s^2t^2=0.$ Finally distributivity. $(a/s+b/t)(c/u)=(c/u)(a/s+b/t)=[(at+bs)/st](c/u)=(act+bcs)/stu \\ (a/s)(c/u)+(b/t)(c/u)=ac/su+bc/tu=(actu+bcsu)/stu^2=(act/bcs)/stu$ Note $ab/cb=a/c$ since $(abc-abc)1=0$. $\square$ $\def\mb{\mathbb}$
Cases and examples
First we consider the case when $A$ is entire. If $0 \in S$, then $S^{-1}A$ is trivial, which is not so interesting. However, provided that $0 \notin S$, we get some well-behaved result:
(Proposition 5) Let $A$ be an entire ring, and let $S$ be a multiplicatively closed subset of $A$ that does not contain $0$, then the natural map \begin{aligned} \varphi_S: A &\to S^{-1}A \\ x &\mapsto x/1 \end{aligned} is injective. Therefore it can be considered as a natural inclusion. Further, every element of $\varphi_S(S)$ is invertible.
Proof. Indeed, if $x/1=0/1$, then there exists $s \in S$ such that $xs=0$. Since $A$ is entire and $s \neq 0$, we see $x=0$, hence $\varphi_S$ is entire. For $s \in S$, we see $\varphi_S(s)=s/1$. However $(1/s)\varphi_S(s)=(1/s)(s/1)=s/s=1$. $\square$
Note since $A$ is entire we can also conclude that $S^{-1}A$ is entire. As a word of warning, the ring homomorphism $\varphi_S$ is not in general injective since, for example, when $0 \in S$, this map is the zero.
If we go further, making $S$ contain all non-zero element, we have:
(Proposition 6) If $A$ is entire and $S$ contains all non-zero elements of $A$, then $S^{-1}A$ is a field, called the quotient field or the field of fractions.
Proof. First we need to show that $S^{-1}A$ is entire. Suppose $(a/s)(b/t)=ab/st =0/1$ but $a/s \neq 0/1$, we see however $ab/st=0/1 \implies \exists u \in S, (ab-0)u=0 \implies ab=0.$ Since $A$ is entire, $b$ has to be $0$, which implies $b/t=0/1$. Second, if $a/s \neq 0/1$, we see $a \neq 0$ and therefore is in $S$, hence we've found the inverse $(a/s)^{-1}=s/a$. $\square$
In this case we can identify $A$ as a subset of $S^{-1}A$ and write $a/s=s^{-1}a$.
Let $A$ be a commutative ring, an let $S$ be the set of invertible elements of $A$. If $u \in S$, then there exists some $v \in S$ such that $uv=1$. We see $1 \in S$ and if $a,b \in S$, we have $ab \in S$ since $ab$ has an inverse as well. This set is frequently denoted by $A^\ast$, and is called the group of invertible elements of $A$. For example for $\mb{Z}$ we see $\mb{Z}^\ast$ consists of $-1$ and $1$. If $A$ is a field, then $A^\ast$ is the multiplicative group of non-zero elements of $A$. For example $\mb{Q}^\ast$ is the set of all rational numbers without $0$. For $A^\ast$ we have
If $A$ is a field, then $(A^\ast)^{-1}A \simeq A$.
Proof. Define \begin{aligned} \varphi_S:A &\to (A^\ast)^{-1}A \\ x &\mapsto x/1. \end{aligned} Then as we have already shown, $\varphi_S$ is injective. Secondly we show that $\varphi_S$ is surjective. For any $a/s \in (A^\ast)^{-1}A$, we see $as^{-1}/1 = a/s$. Therefore $\varphi_S(as^{-1})=a/s$ as is shown. $\square$
Now let's see a concrete example. If $A$ is entire, then the polynomial ring $A[X]$ is entire. If $K = S^{-1}A$ is the quotient field of $A$, we can denote the quotient field of $A[X]$ as $K(X)$. Elements in $K(X)$ can be naturally called rational polynomials, and can be written as $f(X)/g(X)$ where $f,g \in A[X]$. For $b \in K$, we say a rational function $f/g$ is defined at $b$ if $g(b) \neq 0$. Naturally this process can be generalized to polynomials of $n$ variables.
Local ring and localization
We say a commutative ring $A$ is local if it has a unique maximal ideal. Let $\mfk{p}$ be a prime ideal of $A$, and $S = A \setminus \mfk{p}$, then $A_{\mfk{p}}=S^{-1}A$ is called the local ring of $A$ at $\mfk{p}$. Alternatively, we say the process of passing from $A$ to $A_\mfk{p}$ is localization at $\mfk{p}$. You will see it makes sense to call it localization:
(Proposition 7) $A_\mfk{p}$ is local. Precisely, the unique maximal ideal is $I=\mfk{p}A_\mfk{p}=\{a/s:a \in \mfk{p},s \in S\}.$ Note $I$ is indeed equal to $\mfk{p}A_\mfk{p}$.
Proof. First we show that $I$ is an ideal. For $b/t \in A_\mfk{p}$ and $a/s \in I$, we see $(b/t)(a/s)=ba/ts \in A_\mfk{p}$ since $a \in \mfk{p}$ implies $ba \in \mfk{p}$. Next we show that $I$ is maximal, which is equivalent to show that $A_\mfk{p}/I$ is a field. For $b/t \notin I$, we have $b \in S$, hence it is legit to write $t/b$. This gives $(b/t+I)(t/b+I)=1/1+I.$ Hence we have found the inverse.
Finally we show that $I$ is the unique maximal ideal. Let $J$ be another maximal ideal. Suppose $J \neq I$, then we can pick $m/n \in J \setminus I$. This gives $m \in S$ since if not $m \in \mfk{p}$ and then $m/n \in I$. But for $n/m \in A_\mfk{p}$ we have $(m/n)(n/m)=1/1 \in J.$ This forces $J$ to be $A_\mfk{p}$ itself, contradicting the assumption that $J$ is a maximal ideal. Hence $I$ is unique. $\square$
Example
Let $p$ be a prime number, and we take $A=\mb{Z}$ and $\mfk{p}=p\mb{Z}$. We now try to determine what do $A_\mfk{p}$ and $\mfk{p}A_\mfk{p}$ look like. First $S = A \setminus \mfk{p}$ is the set of all entire numbers prime to $p$. Therefore $A_\mfk{p}$ can be considered as the ring of all rational numbers $m/n$ where $n$ is prime to $p$, and $\mfk{p}A_\mfk{p}$ can be considered as the set of all rational numbers $kp/n$ where $k \in \mb{Z}$ and $n$ is prime to $p$.
$\mb{Z}$ is the simplest example of ring and $p\mb{Z}$ is the simplest example of prime ideal. And $A_\mfk{p}$ in this case shows what does localization do: $A$ is 'expanded' with respect to $\mfk{p}$. Every member of $A_\mfk{p}$ is related to $\mfk{p}$, and the maximal ideal is determined by $\mfk{p}$.
Let $k$ be a infinite field. Let $A=k[x_1,\cdots,x_n]$ where $x_i$ are independent indeterminates, $\mfk{p}$ a prime ideal in $A$. Then $A_\mfk{p}$ is the ring of all rational functions $f/g$ where $g \notin \mfk{p}$. We have already defined rational functions. But we can go further and demonstrate the prototype of the local rings which arise in algebraic geometry. Let $V$ be the variety defined by $\mfk{p}$, that is, $V=\{x=(x_1,x_2,\cdots,x_n) \in k^n:\forall f \in \mfk{p}, f(x)=0\}.$ Then what about $A_\mfk{p}$? We see since for $f/g \in A_\mfk{p}$ we have $g \notin \mfk{p}$, therefore for $g(x)$ is not equal to $0$ almost everywhere on $V$. That is, $A_\mfk{p}$ can be identified with the ring of all rational functions on $k^n$ which are defined at almost all points of $V$. We call this the local ring of $k^n$ along the variety $V$.
Universal property
Let $A$ be a ring and $S^{-1}A$ a ring of fractions, then we shall see that $\varphi_S:S \to S^{-1}A$ has a universal property.
(Proposition 8) Let $g:A \to B$ be a ring homomorphism such that $g(s)$ is invertible in $B$ for all $s \in S$, then there exists a unique homomorphism $h:S^{-1}A \to B$ such that $g = h \circ \varphi_S$.
Proof. For $a/s \in S^{-1}A$, define $h(a/s)=g(a)g(s)^{-1}$. It looks immediate but we shall show that this is what we are looking for and is unique.
Firstly we need to show that it is well defined. Suppose $a/s=a'/s'$, then there exists some $u \in S$ such that $(as'-a's)u=0.$ Applying $g$ on both side yields $(g(a)g(s')-g(a')g(s))g(u)=0.$ Since $g(x)$ is invertible for all $s \in S$, we therefore get $g(a)g(s)^{-1}=g(a')g(s')^{-1}.$ It is a homomorphism since \begin{aligned} h[(a/s)(a'/s')]&=g(a)g(a')g(s)^{-1}g(s')^{-1} \\ h(a/s)h(a'/s')&=g(a)g(s)^{-1}g(a')g(s')^{-1}, \end{aligned} and $h(a/s+a'/s')=h((as'+a's)/ss')=g(as'+a's)g(ss')^{-1} \\ h(a/s)+h(a'/s')=g(a)g(s)^{-1}+g(a')g(s')^{-1}$ they are equal since \begin{aligned} g(as'+a's)g(ss')^{-1}&=g(as')g(ss')^{-1}+g(a's)g(ss')^{-1} \\ &=g(a)g(s')g(s)^{-1}g(s')^{-1}+g(a')g(s)g(s)^{-1}g(s')^{-1} \\ &=g(a)g(s)^{-1}+g(a')g(s')^{-1}. \end{aligned} Next we show that $g=h \circ \varphi_S$. For $a \in A$, we have $h(\varphi_S(a))=h(a/1)=g(a)g(1)^{-1}=g(a).$ Finally we show that $h$ is unique. Let $h'$ be a homomorphism satisfying the condition, then for $a \in A$ we have $h'(a/1)=h'(\varphi_S(a))=g(a).$ For $s \in S$, we also have $h'(1/s)=h'((s/1)^{-1})=h'(\varphi_S(s)^{-1})=h'(\varphi_S(s))^{-1}=g(s)^{-1}.$ Since $a/s = (a/1)(1/s)$ for all $a/s \in S^{-1}A$, we get $h'(a/s)=h'((a/1)(1/s))=g(a)g(s)^{-1}.$ That is, $h'$ (or $h$) is totally determined by $g$. $\square$
Let's restate it in the language of category theory (you can skip it if you have no idea what it is now). Let $\mfk{C}$ be the category whose objects are ring-homomorphisms $f:A \to B$ such that $f(s)$ is invertible for all $s \in S$. Then according to proposition 5, $\varphi_S$ is an object of $\mfk{C}$. For two objects $f:A \to B$ and $f':A \to B'$, a morphism $g \in \operatorname{Mor}(f,f')$ is a homomorphism $g:B \to B'$ such that $f'=g \circ f$. So here comes the question: what is the position of $\varphi_S$?
Let $\mfk{A}$ be a category. an object $P$ of $\mfk{A}$ is called universally attracting if there exists a unique morphism of each object of $\mfk{A}$ into $P$, an is called universally repelling if for every object of $\mfk{A}$ there exists a unique morphism of $P$ into this object. Therefore we have the answer for $\mfk{C}$.
(Proposition 9) $\varphi_S$ is a universally repelling object in $\mfk{C}$.
Principal and factorial ring
An ideal $\mfk{o} \in A$ is said to be principal if there exists some $a \in A$ such that $Aa = \mfk{o}$. For example for $\mb{Z}$, the ideal $\{\cdots,-2,0,2,4,\cdots\}$ is principal and we may write $2\mb{Z}$. If every ideal of a commutative ring $A$ is principal, we say $A$ is principal. Further we say $A$ is a PID if $A$ is also an integral domain (entire). When it comes to ring of fractions, we also have the following proposition:
(Proposition 10) Let $A$ be a principal ring and $S$ a multiplicatively closed subset with $0 \notin S$, then $S^{-1}A$ is principal as well.
Proof. Let $I \subset S^{-1}A$ be an ideal. If $a \in S$ where $a/s \in I$, then we are done since then $(s/a)(a/s) = 1/1 \in I$, which implies $I=S^{-1}A$ itself, hence we shall assume $a \notin S$ for all $a/s \in I$. But for $a/s \in I$ we also have $(a/s)(s/1)=a/1 \in I$. Therefore $J=\varphi_S^{-1}(I)$ is not empty. $J$ is an ideal of $A$ since for $a \in A$ and $b \in J$, we have $\varphi_S(ab) =ab/1=(a/1)(b/1) \in I$ which implies $ab \in J$. But since $A$ is principal, there exists some $a$ such that $Aa = J$. We shall discuss the relation between $S^{-1}A(a/1)$ and $I$. For any $(c/u)(a/1)=ca/u \in S^{-1}A(a/1)$, clearly we have $ca/u \in I$, hence $S^{-1}A(a/1)\subset I$. On the other hand, for $c/u \in I$, we see $c/1=(c/u)(u/1) \in I$, hence $c \in J$, and there exists some $b \in A$ such that $c = ba$, which gives $c/u=ba/u=(b/u)(a/1) \in I$. Hence $I \subset S^{-1}A(a/1)$, and we have finally proved that $I = S^{-1}A(a/1)$. $\square$
As an immediate corollary, if $A_\mfk{p}$ is the localization of $A$ at $\mfk{p}$, and if $A$ is principal, then $A_\mfk{p}$ is principal as well. Next we go through another kind of rings. A ring is called factorial (or a unique factorization ring or UFD) if it is entire and if every non-zero element has a unique factorization into irreducible elements. An element $a \neq 0$ is called irreducible if it is not a unit and whenever $a=bc$, then either $b$ or $c$ is a unit. For all non-zero elements in a factorial ring, we have $a=u\prod_{i=1}^{r}p_i,$ where $u$ is a unit (invertible).
In fact, every PID is a UFD (proof here). Irreducible elements in a factorial ring is called prime elements or simply prime (take $\mathbb{Z}$ and prime numbers as an example). Indeed, if $A$ is a factorial ring and $p$ a prime element, then $Ap$ is a prime ideal. But we are more interested in the ring of fractions of a factorial ring.
(Proposition 11) Let $A$ be a factorial ring and $S$ a multiplicatively closed subset with $0 \notin S$, then $S^{-1}A$ is factorial.
Proof. Pick $a/s \in S^{-1}A$. Since $A$ is factorial, we have $a=up_1 \cdots p_k$ where $p_i$ are primes and $u$ is a unit. But we have no idea what are irreducible elements of $S^{-1}A$. Naturally our first attack is $p_i/1$. And we have no need to restrict ourselves to $p_i$, we should work on all primes of $A$. Suppose $p$ is a prime of $A$. If $p \in S$, then $p/1 \in S$ is a unit, not prime. If $Ap \cap S \neq \varnothing$, then $rp \in S$ for some $r \in A$. But then $(p/1)(r/rp)=1,$ again $p/1$ is a unit, not prime. Finally if $Ap \cap S = \varnothing$, then $p/1$ is prime in $S^{-1}A$. For any $(a/s)(b/t)=ab/st=p/1,$ we see $ab=stp \not\in S$. But this also gives $ab \in Ap$ which is a prime ideal, hence we can assume $a \in Ap$ and write $a=rp$ for some $r \in A$. With this expansion we get $ab=stp \implies rbp=stp \implies rb=st \implies (r/s)(b/t)=1/1.$ Hence $b/t$ is a unit, $p/1$ is a prime.
Conversely, suppose $a/s$ is irreducible in $S^{-1}A$. Since $A$ is factorial, we may write $a=u\prod_{i}p_i$. $a$ cannot be an element of $S$ since $a/s$ is not a unit. We write $a/s=1/s[(u/1)(p_1/1)(p_2/1)\cdots(p_n/1)]$ We see there is some $v \in A$ such that $uv=1$ and accordingly $(u/1)(v/1)=uv/1=1/1$, hence $u/1$ is a unit. We claim that there exist a unique $p_k$ such that $1 \leq k \leq n$ and $Ap \cap S = \varnothing$. If not exists, then all $p_j/1$ are units. If both $p_{k}$ and $p_{k'}$ satisfy the requirement and $p_k \neq p_k'$, then we can write $a/s$ as $a/s = \{1/s[(u/1)(p_1/1)\cdots(p_{k-1}/1)(p_{k+1}/1)\cdots(p_{k'-1}/1)(p_{k'+1}/1)\cdots(p_n/1)](p_k/1)\}(p_{k'}/1).$ Neither the one in curly bracket nor $p_{k'}/1$ is unit, contradicting the fact that $a/s$ is irreducible. Next we show that $a/s=p_k/1$. For simplicity we write $b = u\prod_{i=1 \\ i \neq k}^{n} p_i, \quad a = bp_k.$ Note $a/s = bp_k/s = (b/s)(p_k/1)$. Since $a/s$ is irreducible, $p_k/1$ is not a unit, we conclude that $b/s$ is a unit. We are done for the study of irreducible elements of $S^{-1}A$: it is of the form $p/1$ (up to a unit) where $p$ is prime in $A$ and $Ap \cap S = \varnothing$.
Now we are close to the fact that $S^{-1}A$ is also factorial. For any $a/s \in S^{-1}A$, we have an expansion $a/s=1/s[(u/1)(p_1/1)(p_2/1)\cdots(p_n/1)].$ Let $p'_1,p'_2,\cdots,p'_j$ be those whose generated prime ideal has nontrivial intersection with $S$, then $p'_1/1, p'_2/1,\cdots,p'_j/1$ are units of $S^{-1}A$. Let $q_1,q_2,\cdots,q_k$ be other $p_i$'s, then $q_1/1,q_2/1,\cdots,q_k/1$ are irreducible in $S^{-1}A$. This gives $a/s = [(1/s)(p'_1/1)(p'_2/1)\cdots(p'_j/1)]\prod_{i=1}^{k}(q_i/1).$ Hence $S^{-1}A$ is factorial as well. $\square$
We finish the whole post by a comprehensive proposition:
(Proposition 12) Let $A$ be a factorial ring and $p$ a prime element, $\mfk{p}=Ap$. The localization of $A$ at $\mfk{p}$ is principal.
Proof. For $a/s \in S^{-1}A$, we see $p$ does not divide $s$ since if $s = rp$ for some $r \in A$, then $s \in \mfk{p}$, contradicting the fact that $S = A \setminus \mfk{p}$. Since $A$ is factorial, we may write $a = cp^n$ for some $n \geq 0$ and $p$ does not divide $c$ as well (which gives $c \in S$. Hence $a/s = (c/s)(p^n/1)$. Note $(c/s)(s/c)=1/1$ and therefore $c/s$ is a unit. For every $a/s \in S^{-1}A$ we may write it as $a/s = u(p^n/1),$ where $u$ is a unit of $S^{-1}A$.
Let $I$ be any ideal in $S^{-1}A$, and $m = \min\{n:u(p^n/1) \in I, u \text{ is a unit }\}.$ Let's discuss the relation between $S^{-1}A(p^m/1)$ and $I$. First we see $S^{-1}A(p^m/1)=S^{-1}A(up^m/1)$ since if $v$ is the inverse of $u$, we get $vS^{-1}A(up^m/1)=S^{-1}A(p^m/1) \subset S^{-1}A(up^m/1), \\ S^{-1}A(up^m/1)=uS^{-1}A(p^m/1)\subset S^{-1}A(p^m/1).$ Any element of $S^{-1}A(up^m/1)$ is of the form $vup^{m+k}/1=v(p^k/1)up^m/1.$ Since $up^m/1 \in I$, we see $vup^{m+k}/1 \in I$ as well, hence $S^{-1}A(up^m/1) \subset I$. On the other hand, any element of $I$ is of the form $wup^{m+n}/1=w(p^n/1)u(p^m/1)$ where $w$ is a unit and $n \geq 0$. This shows that $vup^{m+n}/1 \in S^{-1}A(up^m/1)$. Hence $S^{-1}A(p^m/1)=S^{-1}A(up^m/1)=I$ as we wanted. $\square$
The Grothendienck Group
Free group
Let $A$ be an abelian group. Let $(e_i)_{i \in I}$ be a family of elements of $A$. We say that this family is a basis for $A$ if the family is not empty, and if every element of $A$ has a unique expression as a linear expression $x = \sum_{i \in I} x_i e_i$ where $x_i \in \mathbb{Z}$ and almost all $x_i$ are equal to $0$. This means that the sum is actually finite. An abelian group is said to be free if it has a basis. Alternatively, we may write $A$ as a direct sum by $A \cong \bigoplus_{i \in I}\mathbb{Z}e_i.$
Free abelian group generated by a set
Let $S$ be a set. Say we want to get a group out of this for some reason, so how? It is not a good idea to endow $S$ with a binary operation beforehead since overall $S$ is merely a set. We shall generate a group out of $S$ in the most freely way.
Let $\mathbb{Z}\langle S \rangle$ be the set of all maps $\varphi:S \to \mathbb{Z}$ such that, for only a finite number of $x \in S$, we have $\varphi(x) \neq 0$. For simplicity, we denote $k \cdot x$ to be some $\varphi_0 \in \mathbb{Z}\langle S \rangle$ such that $\varphi_0(x)=k$ but $\varphi_0(y) = 0$ if $y \neq x$. For any $\varphi$, we claim that $\varphi$ has a unique expression $\varphi=k_1 \cdot x_1 + k_2 \cdot x_2 + \cdots + k_n \cdot x_n.$ One can consider these integers $k_i$ as the order of $x_i$, or simply the time that $x_i$ appears (may be negative). For $\varphi\in\mathbb{Z}\langle S \rangle$, let $I=\{x_1,x_2,\cdots,x_n\}$ be the set of elements of $S$ such that $\varphi(x_i) \neq 0$. If we denote $k_i=\varphi(x_i)$, we can show that $\psi=k_1 \cdot x_1 + k_2 \cdot x_2 + \cdots + k_n \cdot x_n$ is equal to $\varphi$. For $x \in I$, we have $\psi(x)=k$ for some $k=k_i\neq 0$ by definition of the '$\cdot$'; if $y \notin I$ however, we then have $\psi(y)=0$. This coincides with $\varphi$. $\blacksquare$
By definition the zero map $\mathcal{O}=0 \cdot x \in \mathbb{Z}\langle S \rangle$ and therefore we may write any $\varphi$ by $\varphi=\sum_{x \in S}k_x\cdot x$ where $k_x \in \mathbb{Z}$ and can be zero. Suppose now we have two expressions, for example $\varphi=\sum_{x \in S}k_x \cdot x=\sum_{x \in S}k_x'\cdot x$ Then $\varphi-\varphi=\mathcal{O}=\sum_{x \in S}(k_x-k'_x)\cdot x$ Suppose $k_y - k_y' \neq 0$ for some $y \in S$, then $\mathcal{O}(y)=k_y-k_y'\neq 0$ which is a contradiction. Therefore the expression is unique. $\blacksquare$
This $\mathbb{Z}\langle S \rangle$ is what we are looking for. It is an additive group (which can be proved immediately) and, what is more important, every element can be expressed as a 'sum' associated with finite number of elements of $S$. We shall write $F_{ab}(S)=\mathbb{Z}\langle S \rangle$, and call it the free abelian group generated by $S$. For elements in $S$, we say they are free generators of $F_{ab}(S)$. If $S$ is a finite set, we say $F_{ab}(S)$ is finitely generated.
An abelian group is free if and only if it is isomorphic to a free abelian group $F_{ab}(S)$ for some set $S$.
Proof. First we shall show that $F_{ab}(S)$ is free. For $x \in M$, we denote $\varphi = 1 \cdot x$ by $[x]$. Then for any $k \in \mathbb{Z}$, we have $k[x]=k \cdot x$ and $k[x]+k'[y] = k\cdot x + k' \cdot y$. By definition of $F_{ab}(S)$, any element $\varphi \in F_{ab}(S)$ has a unique expression $\varphi = k_1 \cdot x_1 + \cdots + k_n \cdot x_n =k_1[x_1]+\cdots+k_n[x_n]$ Therefore $F_{ab}(S)$ is free since we have found the basis $([x])_{x \in S}$.
Conversely, if $A$ is free, then it is immediate that its basis $(e_i)_{i \in I}$ generates $A$. Our statement is therefore proved. $\blacksquare$
The connection between an arbitrary abelian group an a free abelian group
(Proposition 1) If $A$ is an abelian group, then there is a free group $F$ which has a subgroup $H$ such that $A \cong F/H$.
Proof. Let $S$ be any set containing $A$. Then we get a surjective map $\gamma: S \to A$ and a free group $F_{ab}(S)$. We also get a unique homomorphism $\gamma_\ast:F_{ab}(S) \to A$ by \begin{aligned} \gamma_\ast:F_{ab}(S) &\to A \\ \varphi=\sum_{x \in S}k_x\cdot x &\mapsto \sum_{x \in S}k_x\gamma(x) \end{aligned} which is also surjective. By the first isomorphism theorem, if we set $H=\ker(\gamma_\ast)$ and $F_{ab}(S)=F$, then $F/H \cong A.$ $\blacksquare$
(Proposition 2) If $A$ is finitely generated, then $F$ can also be chosen to be finitely generated.
Proof. Let $S$ be the generator of $A$, and $S'$ is a set containing $S$. Note if $S$ is finite, which means $A$ is finitely generated, then $S'$ can also be finite by inserting one or any finite number more of elements. We have a map from $S$ and $S'$ into $F_{ab}(S)$ and $F_{ab}(S')$ respectively by $f_S(x)=1 \cdot x$ and $f_{S'}(x')=1 \cdot x'$. Define $g=f_{S'} \circ \lambda:S' \to F_{ab}(S)$ we get another homomorphism by \begin{aligned} g_\ast:F_{ab}(S') &\to F_{ab}(S) \\ \varphi'=\sum_{x \in S'}k_{x} \cdot x &\mapsto \sum_{x \in S'}k_{x}\cdot g(x) \end{aligned} This defines a unique homomorphism such that $g_\ast \circ f_{S'} = g$. As one can also verify, this map is also surjective. Therefore by the first isomorphism theorem we have $A \cong F_{ab}(S) \cong F_{ab}(S')/\ker(g_\ast)$ $\blacksquare$
It's worth mentioning separately that we have implicitly proved two statements with commutative diagrams:
(Proposition 3 | Universal property) If $g:S \to B$ is a mapping of $S$ into some abelian group $B$, then we can define a unique group-homomorphism making the following diagram commutative:
(Proposition 4) If $\lambda:S \to S$ is a mapping of sets, there is a unique homomorphism $\overline{\lambda}$ making the following diagram commutative:
(In the proof of Proposition 2 we exchanged $S$ an $S'$.)
The Grothendieck group
(The Grothendieck group) Let $M$ be a commutative monoid written additively. We shall prove that there exists a commutative group $K(M)$ with a monoid homomorphism $\gamma:M \to K(M)$
satisfying the following universal property: If $f:M \to A$ is a homomorphism from $M$ into a abelian group $A$, then there exists a unique homomorphism $f_\gamma:K(M) \to A$ such that $f=f_\gamma\circ\gamma$. This can be represented by a commutative diagram:
Proof. There is a commutative diagram describes what we are doing.
Let $F_{ab}(M)$ be the free abelian group generated by $M$. For $x \in M$, we denote $1 \cdot x \in F_{ab}(M)$ by $[x]$. Let $B$ be the group generated by all elements of the type $[x+y]-[x]-[y]$ where $x,y \in M$. This can be considered as a subgroup of $F_{ab}(M)$. We let $K(M)=F_{ab}(M)/B$. Let $i=x \to [x]$ and $\pi$ be the canonical map $\pi:F_{ab}(M) \to F_{ab}(M)/B.$ We are done by defining $\gamma: \pi \circ i$. Then we shall verify that $\gamma$ is our desired homomorphism satisfying the universal property. For $x,y \in M$, we have $\gamma(x+y)=\pi([x+y])$ and $\gamma(x)+\gamma(y) = \pi([x])+\pi([y])=\pi([x]+[y])$. However we have $[x+y]-[x]-[y] \in B,$ which implies that $\gamma(x)+\gamma(y)=\pi([x]+[y])=\pi([x+y]) = \gamma(x+y).$ Hence $\gamma$ is a monoid-homomorphism. Finally the universal property. By proposition 3, we have a unique homomorphism $f_\ast$ such that $f_\ast \circ i = f$. Note if $y \in B$, then $f_\ast(y) =0$. Therefore $B \subset \ker{f_\ast}$ Therefore we are done if we define $f_\gamma(x+B)=f_\ast (x)$. $\blacksquare$
Why such a $B$? Note in general $[x+y]$ is not necessarily equal to $[x]+[y]$ in $F_{ab}(M)$, but we don't want it to be so. So instead we create a new equivalence relation, by factoring a subgroup generated by $[x+y]-[x]-[y]$. Therefore in $K(M)$ we see $[x+y]+B = [x]+[y]+B$, which finally makes $\gamma$ a homomorphism. We use the same strategy to generate the tensor product of two modules later. But at that time we have more than one relation to take care of.
Cancellative monoid
If for all $x,y,z \in M$, $x+y=x+z$ implies $y=z$, then we say $M$ is a cancellative monoid, or the cancellation law holds in $M$. Note for the proof above we didn't use any property of cancellation. However we still have an interesting property for cancellation law.
(Theorem) The cancellation law holds in $M$ if and only if $\gamma$ is injective.
Proof. This proof involves another approach to the Grothendieck group. We consider pairs $(x,y) \in M \times M$ with $x,y \in M$. Define $(x,y) \sim (x',y') \iff \exists \ell \in M, y+x'+\ell=x+y'+\ell.$ Then we get a equivalence relation (try to prove it yourself!). We define the addition component-wise, that is, $(x,y)+(x',y')=(x+x',y+y')$, then the equivalence classes of pairs form a group $A$, where the zero element is $[(0,0)]$. We have a monoid-homomorphism $f:x \mapsto [(x,0)].$ If cancellation law holds in $M$, then \begin{aligned} f(x) = f(y) &\implies [(x,0)] = [(y,0)] \\ &\implies 0+y+\ell=x+0+\ell \\ &\implies x=y. \end{aligned} Hence $f$ is injective. By the universal property of the Grothendieck group, we get a unique homomorphism $f_\gamma$ such that $f_\gamma \circ \gamma = f$. If $x \neq 0$ in $M$, then $f_\gamma \circ \gamma(x) \neq 0$ since $f$ is injective. This implies $\gamma(x) \neq 0$. Hence $\gamma$ is injective.
Conversely, if $\gamma$ is injective, then $i$ is injective (this can be verified by contradiction). Then we see $f=f_\ast \circ i$ is injective. But $f(x)=f(y)$ if and only if $x+\ell = y+\ell$, hence $x+ \ell = y+ \ell$ implies $x=y$, the cancellation law holds on $M$.
Examples
Our first example is $\mathbb{N}$. Elements of $F_{ab}(\mathbb{N})$ are of the form $\varphi=k_1 \cdot n_1 + k_2 \cdot n_2+\cdots + k_m \cdot n_m.$ For elements in $B$ they are generated by $\varphi=1\cdot (m+n)-1\cdot m - 1\cdot n$ which we wish to represent $0$. Indeed, $K(\mathbb{N}) \simeq \mathbb{Z}$ since if we have a homomorphism \begin{aligned} f:K(\mathbb{N}) &\to \mathbb{Z} \\ \sum_{j=1}^{m}k_j \cdot n_j +B &\mapsto \sum_{j=1}^{m}k_j n_j. \end{aligned} For $r \in \mathbb{Z}$, we see $f(1 \cdot r+B)=r$. On the other hand, if $\sum_{j=1}^{m}k_j \cdot n_j \not\in B$, then its image under $f$ is not $0$.
In the first example we 'granted' the natural numbers 'subtraction'. Next we grant the division on multiplicative monoid.
Consider $M=\mathbb{Z} \setminus 0$. Now for $F_{ab}(M)$ we write elements in the form $\varphi={}^{k_1}n_1{}^{k_2}n_2\cdots{}^{k_m}n_m$ which denotes that $\varphi(n_j)=k_j$ and has no other differences. Then for elements in $B$ they are generated by $\varphi = {}^1(n_1n_2){}^{-1}(n_1)^{-1}(n_2)$ which we wish to represent $1$. Then we see $K(M) \simeq \mathbb{Q} \setminus 0$ if we take the isomorphism \begin{aligned} f:K(M) &\to \mathbb{Q} \setminus 0 \\ \left(\prod_{j=1}^{m}{}^{k_j}n_j\right)B &\mapsto \prod_{j=1}^{m}n_j^{k_j}. \end{aligned}
Of course this is not the end of the Grothendieck group. But for further example we may need a lot of topology background. For example, we have the topological $K$-theory group of a topological space to be the Grothendieck group of isomorphism classes of topological vector bundles. But I think it is not a good idea to post these examples at this timing.
Study Vector Bundle in a Relatively Harder Way - Tangent Bundle
Tangent line and tangent surface as vector spaces
We begin our study by some elementary Calculus. Now we have the function $f(x)=x^2+\frac{e^x}{x^2+1}$ as our example. It should not be a problem to find its tangent line at point $(0,1)$, by calculating its derivative, we have $l:x-y+1=0$ as the tangent line.
$l$ is not a vector space since it does not get cross the origin, in general. But $l-\overrightarrow{OA}$ is a vector space. In general, suppose $P(x,y)$ is a point on the curve determined by $f$, i.e. $y=f(x)$, then we obtain a vector space $l_p-\overrightarrow{OP} \simeq \mathbb{R}$. But the action of moving the tangent line to the origin is superfluous so naturally we consider the tangent line at $P$ as a vector space determined by $P$. In this case, the induced vector space (tangent line) is always of dimension $1$.
Now we move to two-variable functions. We have a function $a(x,y)=x^2+y^2-x-y+xy$ as our example. Some elementary Calculus work gives us the tangent surface of $z=a(x,y)$ at $A(1,1,1)$, which can be identified by $S:2x+2y-z=3\simeq\mathbb{R}^2$. Again, this can be considered as a vector space determined by $A$, or roughly speaking it is one if we take $A$ as the origin. Further we have a base $(\overrightarrow{AB},\overrightarrow{AC})$. Other vectors on $S$, for example $\overrightarrow{AD}$, can be written as a linear combination of $\overrightarrow{AB}$ and $\overrightarrow{AC}$. In other words, $S$ is "spanned" by $(\overrightarrow{AB},\overrightarrow{AC})$.
Tangent line and tangent surface play an important role in differentiation. But sometimes we do not have a chance to use it with ease, for example $S^1:x^2+y^2=1$ cannot be represented by a single-variable function. However the implicit function theorem, which you have already learned in Calculus, gives us a chance to find a satisfying function locally. Here in this post we will try to generalize this concept, trying to find the tangent space at some point of a manifold. (The two examples above have already determined two manifolds and two tangent spaces.)
Definition of tangent vectors
We will introduce the abstract definition of a tangent vector at beginning. You may think it is way too abstract but actually it is not. Surprisingly, the following definition can simplify our work in the future. But before we go, make sure that you have learned about Fréchet derivative (along with some functional analysis knowledge).
Let $M$ be a manifold of class $C^p$ with $p \geq 1$ and let $x$ be a point of $M$. Let $(U,\varphi)$ be a chart at $x$ and $v$ be a element of the vector space $\mathbf{E}$ where $\varphi(U)$ lies (for example, if $M$ is a $d$-dimensional manifold, then $v \in \mathbb{R}^d$). Next we consider the triple $(U,\varphi,v)$. Suppose $(U,\varphi,v)$ and $(V,\psi,w)$ are two such triples. We say these two triples are equivalent if the following identity holds: ${\color\green{[}}{\color\red{(}}{\color\red{\psi\circ\varphi^{-1}}}{\color\red{)'}}{\color\red{(}}{\color\purple{\varphi(x)}}{\color\red)}{\color\green{]}}(v)=w.$ This identity looks messy so we need to explain how to read it. First we consider the function in red: the derivative of $\psi\circ\varphi^{-1}$. The derivative of $\psi\circ\varphi^{-1}$ at point $\varphi(x)$ (in purple) is a linear transform, and the transform is embraced with green brackets. Finally, this linear transform maps $v$ to $w$. In short we read, the derivative of $\psi\circ\varphi^{-1}$ at $\varphi(x)$ maps $v$ on $w$. You may recall that you have meet something like $\psi\circ\varphi^{-1}$ in the definition of manifold. It is not likely that these 'triples' should be associated to tangent vectors. But before we explain it, we need to make sure that we indeed defined an equivalent relation.
(Theorem 1) The relation $(U,\varphi,v) \sim (V,\psi,w)\\ [(\psi\circ\varphi^{-1})'(\varphi(x))](v)=w$ is an equivalence relation.
Proof. This will not go further than elementary Calculus, in fact, chain rule:
(Chain rule) If $f:U \to V$ is differentiable at $x_0 \in U$, if $g: V \to W$ is differentiable at $f(x_0)$, then $g \circ f$ is differentiable at $x_0$, and $(g\circ f)'(x_0)=g'(f(x_0))\circ f'(x_0)$
1. $(U,\varphi,v)\sim(U,\varphi,v)$.
Since $\varphi\circ\varphi^{-1}=\operatorname{id}$, whose derivative is still the identity everywhere, we have $[(\varphi\circ\varphi^{-1})'(\varphi(x))](v)=\operatorname{id}(v)=v$
1. If $(U,\varphi,v) \sim (V,\psi,w)$, then $(V,\psi,w)\sim(U,\varphi,v)$.
So now we have $[(\psi\circ\varphi^{-1})'(\varphi(x))](v)=w.$ To prove that $[(\varphi\circ\psi^{-1})'(\psi(x))]{}(w)=v$, we need some implementation of chain rule.
Note first $(\psi\circ\varphi^{-1})'(\varphi(x))=\psi'(\varphi^{-1}(\varphi(x)))\circ\varphi^{-1}{'}(\varphi(x))=\psi'(x)\circ(\varphi^{-1})'(\varphi(x))$ while $(\varphi\circ\psi^{-1})'(\psi(x))=\varphi'(x)\circ(\psi^{-1})'(\psi(x)).$ But also by the chain rule, if $f$ is a diffeomorphism, we have $(f\circ f^{-1})'(x)=(f^{-1})'(f(x))\circ f'(x)=\operatorname{id}$ or equivalently $f'(x)=[(f^{-1})'(f(x))]^{-1} \quad (f^{-1})'(f(x))=[f'(x)]^{-1}$
Therefore \begin{aligned} \{(\psi\circ\varphi^{-1})'(\varphi(x))\}^{-1} &=\{\psi'(x)\circ(\varphi^{-1})'(\varphi(x))\}^{-1} \\ &=\{(\varphi^{-1})'(\varphi(x))\}^{-1}\circ\{\psi'(x)\}^{-1} \\ &=\varphi'(x)\circ(\psi^{-1})'(\psi(x)) \\ &=(\varphi\circ\psi^{-1})'(\psi(x)) \end{aligned} which implies $(\varphi\circ\psi^{-1})'(\psi(x))(w)=\{(\psi\circ\varphi^{-1})'(\varphi(x))\}^{-1}(w)=v.$
1. If $(U,\varphi,v)\sim(V,\psi,w)$ and $(V,\psi,w)\sim(W,\lambda,z)$, then $(U,\varphi,v)\sim(W,\lambda,z)$.
We are given identities $[(\psi\circ\varphi^{-1})'(\varphi(x))](v)=w$ and $[(\lambda\circ\psi^{-1})'(\psi(x))](w)=z.$ By canceling $w$, we get \begin{aligned} z = [(\lambda\circ\psi^{-1})'(\psi(x))] \circ [(\psi\circ\varphi^{-1})'(\varphi(x))] (v) \end{aligned}. On the other hand, \begin{aligned} (\lambda\circ\varphi^{-1})'(\varphi(x))&=(\lambda\circ\psi^{-1}\circ\psi\circ\varphi^{-1})'(\varphi(x)) \\ &=(\lambda\circ\psi^{-1})'(\psi\circ\varphi^{-1}\circ\varphi(x))\circ(\psi\circ\varphi^{-1})'(\varphi(x)) \\ &=(\lambda\circ\psi^{-1})'(\psi(x))\circ(\psi\circ\varphi^{-1})'(\varphi(x)) \end{aligned} which is what we needed. $\square$
An equivalence class of such triples $(U,\varphi,v)$ is called a tangent vector of $X$ at $x$. The set of such tangent vectors is called the tangent space to $X$ at $x$, which is denoted by $T_x(X)$. But it seems that we have gone too far. Is the triple even a 'vector'? To get a clear view let's see Euclidean submanifolds first.
Definition of tangent vectors of Euclidean submanifolds
Suppose $M$ is a submanifold of $\mathbb{R}^n$. We say $z$ is the tangent vector of $M$ at point $x$ if there exists a curve $\alpha$ of class $C^1$, which is defined on $\mathbb{R}$ and where there exists an interval $I$ such that $\alpha(I) \subset M$, such that $\alpha(t_0)=x$ and $\alpha'(t_0)=z$. (For convenience we often take $t_0=0$.)
This definition is immediate if we check some examples. For the curve $M: x^2+1+\frac{e^x}{x^2+1}-y=0$, we can show that $(1,1)^T$ is a tangent vector of $M$ at $(0,1)$, which is identical to our first example. Taking $\alpha(t)=(t,t^2+1+\frac{e^t}{t^2+1})$ we get $\alpha(0)=(0,1)$ and $\alpha'(t)=(1,2t+\frac{e^t(t-1)^2}{(t^2+1)^2})^T.$ Therefore $\alpha'(0)=(1,1)^T$. $\square$
Coordinate system and tangent vector
Let $\mathbf{E}$ and $\mathbf{F}$ be two Banach spaces and $U$ an open subset of $\mathbf{E}$. A $C^p$ map $f: U \to \mathbf{F}$ is called an immersion at $x$ if $f'(x)$ is injective.
For example, if we take $\mathbf{E}=\mathbf{F}=\mathbb{R}=U$ and $f(x)=x^2$, then $f$ is an immersion at almost all point on $\mathbb{R}$ except $0$ since $f'(0)=0$ is not injective. This may lead you to Sard's theorem.
(Theorem 2) Let $M$ be a subset of $\mathbb{R}^n$, then $M$ is a $d$-dimensional $C^p$ submanifold of $\mathbb{R}^n$ if and only if for every $x \in M$ there exists an open neighborhood $U \subset \mathbb{R}^n$ of $x$, an open neighborhood $\Omega \subset \mathbb{R}^d$ of $0$ and a $C^p$ map $g: \Omega \to \mathbb{R}^n$ such that $g$ is immersion at $0$ such that $g(0)=x$, and $g$ is a homeomorphism between $\Omega$ and $M \cap U$ with the topology induced from $\mathbb{R}^n$.
This follows from the definition of manifold and should not be difficult to prove. But it is not what this blog post should cover. For a proof you can check Differential Geometry: Manifolds, Curves, and Surfaces by Marcel Berger and Bernard Gostiaux. The proof is located in section 2.1.
A coordinate system on a $d$-dimensional $C^p$ submanifold $M$ of $\mathbb{R}^n$ is a pair $(\Omega,g)$ consisting of an open set $\Omega \subset \mathbb{R}^d$ and a $C^p$ function $g:\Omega \to \mathbb{R}^n$ such that $g(\Omega)$ is open in $V$ and $g$ induces a homeomorphism between $\Omega$ and $g(\Omega)$.
For convenience, we say $(\Omega,g)$ is centered at $x$ if $g(0)=x$ and $g$ is an immersion at $x$. By theorem 2 it is always possible to find such a coordinate system centered at a given point $x \in M$. The following theorem will show that we can get a easier approach to tangent vector.
(Theorem 3) Let $\mathbf{E}$ and $\mathbf{F}$ be two finite-dimensional vector spaces, $U \subset \mathbf{E}$ an open set, $f:U \to \mathbf{F}$ a $C^1$ map, $M$ a submanifold of $\mathbf{E}$ contained in $U$ and $W$ a submanifold of $\mathbf{F}$ such that $f(M) \subset W$. Take $x \in M$ and set $y=f(x)$, If $z$ is a tangent vector to $M$ at $x$, the image $f'(x)(z)$ is a tangent vector to $W$ at $y=f(x)$.
Proof. Since $z$ is a tangent vector, we see there exists a curve $\alpha: J \to M$ such that $\alpha(0)=x$ and $\alpha'(0)=z$ where $J$ is an open interval containing $0$. The function $\beta = f \circ \alpha: J \to W$ is also a curve satisfying $\beta(0)=f(\alpha(0))=f(x)$ and $\beta'(0)=f'(\alpha(0))\alpha'(0)=f'(x)(z),$ which is our desired curve. $\square$
Why we use 'equivalence relation'
We shall show that equivalence relation makes sense. Suppose $M$ is a $d$-submanifold of $\mathbb{R}^n$, $x \in M$ and $z$ is a tangent vector to $M$ at $x$. Let $(\Omega,g)$ be a coordinate system centered at $x$. Since $g \in C^p(\mathbb{R}^d;\mathbb{R}^n)$, we see $g'(0)$ is a $n \times d$ matrix, and injectivity ensures that $\operatorname{rank}(g'(0))=d$.
Every open set $\Omega \subset \mathbb{R}^d$ is a $d$-dimensional submanifold of $\mathbb{R}^d$ (of $C^p$). Suppose now $v \in \mathbb{R}^d$ is a tangent vector to $\Omega$ at $0$ (determined by a curve $\alpha$), then by Theorem 3, $g \circ \alpha$ determines a tangent vector to $M$ at $x$, which is $z_x=g'(0)(v)$. Suppose $(\Lambda,h)$ is another coordinate system centered at $x$. If we want to obtain $z_x$ as well, we must have $h'(0)(w)=g'(0)(v),$ which is equivalent to $w = (h'(0)^{-1} \circ g'(0))(v)=(h^{-1}\circ g)'(0)(v),$ for some $w \in \mathbb{R}^d$ which is the tangent vector to $\Lambda$ at $0 \in \Lambda$. (The inverse makes sense since we implicitly restricted ourself to $\mathbb{R}^d$)
However, we also have two charts by $(U,\varphi)=(g(\Omega),g^{-1})$ and $(V,\psi) = (h(\Lambda),h^{-1})$, which gives $(h^{-1} \circ g)'(0)(v)=[(\psi \circ \varphi^{-1})'(\varphi(x))](v)=w$ and this is just our equivalence relation (don't forget that $g(0)=x$ hence $g^{-1}(x)=\varphi(x)=0$!). There we have our reason for equivalence relation: If $(U,\varphi,v) \sim (V,\psi,w)$, then $(U,\varphi,u)$ and $(V,\psi,v)$ determines the same tangent vector but we do not have to evaluate it manually. In general, all elements in an equivalence class represent a single vector, so the vector is (algebraically) a equivalence class. This still holds when talking about Banach manifold since topological properties of Euclidean spaces do not play a role. The generalized proof can be implemented with little difficulty.
Tangent space
The tangent vectors at $x \in M$ span a vector space (which is based at $x$). We do hope that because if not our definition of tangent vector would be incomplete and cannot even hold for an trivial example (such as what we mentioned at the beginning). We shall show, satisfyingly, the set of tangent vectors to $M$ at $x$ (which we write $T_xM$) forms a vector space that is toplinearly isomorphic to $\mathbf{E}$, on which $M$ is modeled.
(Theorem 4) $T_xM \simeq \mathbf{E}$. In other words, $T_xM$ can be given the structure of topological vector space given by the chart.
Proof. Let $(U,\varphi)$ be a chart at $x$. For $v \in \mathbf{E}$, we see $(\varphi^{-1})'(x)(v)$ is a tangent vector at $x$. On the other hand, pick $\mathbf{w} \in T_xM$, which can be represented by $(V,\psi,w)$. Then $v=(\varphi\circ\psi^{-1})'(\psi(x))(w)$ makes $(U,\varphi,v) \sim (V,\psi,w)$ uniquely, and therefore we get some $v \in \mathbf{E}$. To conclude, $T_xM \xrightarrow[(\varphi^{-1})'(x)]{\simeq}\mathbf{E}$ which proves our theorem. Note that this does not depend on the choice of charts. $\square$
For many reasons it is not a good idea to identify $T_xM$ as $\mathbf{E}$ without mentioning the point $x$. For example we shouldn't identify the tangent line of a curve as $x$-axis. Instead, it would be better to identify or visualize $T_xM$ as $(x,\mathbf{E})$, that is, a linear space with origin at $x$.
Tangent bundle
Now we treat all tangent spaces as a vector bundle. Let $M$ be a manifold of class $C^p$ with $p \geq 1$, define the tangent bundle by the disjoint union $T(M)=\bigsqcup_{x \in M}T_xM.$ This is a vector bundle if we define the projection by \begin{aligned} \pi: T(M) &\to M \\ y \in T_xM &\mapsto x \end{aligned} and we will verify it soon. First let's see an example. Below is a visualization of the tangent bundle of $\frac{x^2}{4}+\frac{y^2}{3}=1$, denoted by red lines:
Also we can see $\pi$ maps points on the blue line to a point on the curve, which is $B$.
To show that a tangent bundle of a manifold is a vector bundle, we need to verify that it satisfies three conditions we mentioned in previous post. Let $(U,\varphi)$ be a chart of $M$ such that $\varphi(U)$ is open in $\mathbf{E}$, then tangent vectors can be represented by $(U,\varphi,v)$. We get a bijection $\tau_U:\pi^{-1}(U) = T(U) \to U \times \mathbf{E}$ by definition of tangent vectors as equivalence classes. Let $z_x$ be a tangent vector to $U$ at $x$, then there exists some $v \in \mathbf{E}$ such that $(U,\varphi,v)$ represents $z$. On the other hand, for some $v \in \mathbf{E}$ and $x \in U$, $(U,\varphi,v)$ represents some tangent vector at $x$. Explicitly, $\tau_{U}(z_x)=(x,v)=(\pi(z_x),[(\varphi^{-1})'(\pi(z_x))]^{-1}(z_x))$
Further we get the following diagram commutative (which establishes VB 1):
For VB 2 and VB 3 we need to check different charts. Let $(U_i,\varphi_i)$, $(U_j,\varphi_j)$ be two charts. Define $\varphi_{ji}=\varphi_j \circ \varphi_i^{-1}$ on $\varphi_i(U_i \cap U_j)$, and respectively we write $\tau_{U_i}=\tau_i$ and $\tau_{U_j}=\tau_j$. Then we get a transition mapping $\tau_{ji}:(\tau_j \circ \tau_i^{-1}):(U_i \cap U_j) \times \mathbf{E} \to (U_i \cap U_j) \times \mathbf{E}.$
One can verify that $\tau_{ji}(x,v)=(\varphi_{ji}(x),D\varphi_{ji}(x) \cdot v)$ for $x \in U_i \cap U_j$ and $v \in \mathbf{E}$. Since $D\varphi_{ji} \in C^{p-1}$ and $D\varphi_{ji}(x)$ is a toplinear isomorphism, we see $x \mapsto (\tau_j \circ \tau_i^{-1})_x=(\varphi_{ji}(x),D\varphi_{ji}(x)\cdot(\cdot))$ is a morphism, which goes for VB 3. It remains to verify VB 2. To do this we need a fact from Banach space theory:
If $f:U \to L(\mathbf{E},\mathbf{F})$ is a $C^k$-morphism, then the map of $U \times \mathbf{E}$ into $\mathbf{F}$ given by $(x,v) \mapsto [f(x)](v)$ is a $C^k$-morphism.
Here, we have $f(x)=\tau_{ji}(x,\cdot)$ and to conclude, $\tau_{ji}$ is a $C^{p-1}$-morphism. It is also an isomorphism since it has an inverse $\tau_{ij}$. Following the definition of manifold, we can conclude that $T(U)$ has a unique manifold structure such that $\tau_i$ are morphisms (there will be a formal proof in next post about any total space of a vector bundle). By VB 1, we also have $\pi=\tau_i\circ pr$, which makes it a morphism as well. On each fiber $\pi^{-1}(x)$, we can freely transport the topological vector space structure of any $\mathbf{E}$ such that $x$ lies in $U_i$, by means of $\tau_{ix}$. Since $f(x)$ is a toplinear isomorphism, the result is independent of the choice of $U_i$. VB 2 is therefore established.
Using some fancier word, we can also say that $T:M \to T(M)$ is a functor from the category of $C^p$-manifolds to the category of vector bundles of class $C^{p-1}$.
A Continuous Function Sending L^p Functions to L^1
Throughout, let $(X,\mathfrak{M},\mu)$ be a measure space where $\mu$ is positive.
The question
If $f$ is of $L^p(\mu)$, which means $\lVert f \rVert_p=\left(\int_X |f|^p d\mu\right)^{1/p}<\infty$, or equivalently $\int_X |f|^p d\mu<\infty$, then we may say $|f|^p$ is of $L^1(\mu)$. In other words, we have a function \begin{aligned} \lambda: L^p(\mu) &\to L^1(\mu) \\ f &\mapsto |f|^p. \end{aligned} This function does not have to be one to one due to absolute value. But we hope this function to be fine enough, at the very least, we hope it is continuous.
Here, $f \sim g$ means that $f-g$ equals to $0$ almost everywhere with respect to $\mu$. It can be easily verified that this is a equivalence relation.
Continuity
We still use $\varepsilon-\delta$ argument but it's in a metric space. Suppose $(X,d_1)$ and $(Y,d_2)$ are two metric spaces and $f:X \to Y$ is a function. We say $f$ is continuous at $x_0 \in X$ if for any $\varepsilon>0$, there exists some $\delta>0$ such that $d_2(f(x_0),f(x))<\varepsilon$ whenever $d_1(x_0,x)<\delta$. Further, we say $f$ is continuous on $X$ if $f$ is continuous at every point $x \in X$.
Metrics
For $1\leq p<\infty$, we already have a metric by $d(f,g)=\lVert f-g \rVert_p$ given that $d(f,g)=0$ if and only if $f \sim g$. This is complete and makes $L^p$ a Banach space. But for $0<p<1$ (yes we are going to cover that), things are much more different, and there is one reason: Minkowski inequality holds reversely! In fact we have $\lVert f+g \rVert_p \geq \lVert f \rVert_p + \lVert g \rVert_p$ for $0<p<1$. In fact, $L^p$ space has too many weird things when $0<p<1$. Precisely,
For $0<p<1$, $L^p(\mu)$ is locally convex if and only if $\mu$ assumes finitely many values. (Proof.)
On the other hand, for example, $X=[0,1]$ and $\mu=m$ be the Lebesgue measure, then $L^p(\mu)$ has no open convex subset other than $\varnothing$ and $L^p(\mu)$ itself. However,
A topological vector space $X$ is normable if and only if its origin has a convex bounded neighbourhood. (See Kolmogorov's normability criterion.)
Therefore $L^p(m)$ is not normable, hence not Banach.
We have gone too far. We need a metric that is fine enough.
Metric of $L^p$ when $0<p<1$
In this subsection we always have $0<p<1$.
Define $\Delta(f)=\int_X |f|^p d\mu$ for $f \in L^p(\mu)$. We will show that we have a metric by $d(f,g)=\Delta(f-g).$ Fix $y\geq 0$, consider the function $f(x)=(x+y)^p-x^p.$ We have $f(0)=y^p$ and $f'(x)=p(x+y)^{p-1}-px^{p-1} \leq px^{p-1}-px^{p-1}=0$ when $x > 0$ and hence $f(x)$ is nonincreasing on $[0,\infty)$, which implies that $(x+y)^p \leq x^p+y^p.$ Hence for any $f$, $g \in L^p$, we have $\Delta(f+g)=\int_X |f+g|^p d\mu \leq \int_X |f|^p d\mu + \int_X |g|^p d\mu=\Delta(f)+\Delta(g).$ This inequality ensures that $d(f,g)=\Delta(f-g)$ is a metric. It's immediate that $d(f,g)=d(g,f) \geq 0$ for all $f$, $g \in L^p(\mu)$. For the triangle inequality, note that $d(f,h)+d(g,h)=\Delta(f-h)+\Delta(h-g) \geq \Delta((f-h)+(h-g))=\Delta(f-g)=d(f,g).$ This is translate-invariant as well since $d(f+h,g+h)=\Delta(f+h-g-h)=\Delta(f-g)=d(f,g)$ The completeness can be verified in the same way as the case when $p>1$. In fact, this metric makes $L^p$ a locally bounded F-space.
The continuity of $\lambda$
The metric of $L^1$ is defined by $d_1(f,g)=\lVert f-g \rVert_1=\int_X |f-g|d\mu.$ We need to find a relation between $d_p(f,g)$ and $d_1(\lambda(f),\lambda(g))$, where $d_p$ is the metric of the corresponding $L^p$ space.
$0<p<1$
As we have proved, $(x+y)^p \leq x^p+y^p.$ Without loss of generality we assume $x \geq y$ and therefore $x^p=(x-y+y)^p \leq (x-y)^p+y^p.$ Hence $x^p-y^p \leq (x-y)^p.$ By interchanging $x$ and $y$, we get $|x^p-y^p| \leq |x-y|^p.$ Replacing $x$ and $y$ with $|f|$ and $|g|$ where $f$, $g \in L^p$, we get $\int_{X}\lvert |f|^p-|g|^p \rvert d\mu \leq \int_X |f-g|^p d\mu.$ But $d_1(\lambda(f),\lambda(g))=\int_{X}\lvert |f|^p-|g|^p \rvert d\mu \\ d_p(f,g)=\Delta(f-g)= d\mu \leq \int_X |f-g|^p d\mu$ and we therefore have $d_1(\lambda(f),\lambda(g)) \leq d_p(f,g).$ Hence $\lambda$ is continuous (and in fact, Lipschitz continuous and uniformly continuous) when $0<p<1$.
$1 \leq p < \infty$
It's natural to think about Minkowski's inequality and Hölder's inequality in this case since they are critical inequality enablers. You need to think about some examples of how to create the condition to use them and get a fine result. In this section we need to prove that $|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1}).$ This inequality is surprisingly easy to prove however. We will use nothing but the mean value theorem. Without loss of generality we assume that $x > y \geq 0$ and define $f(t)=t^p$. Then $\frac{f(x)-f(y)}{x-y}=f'(\zeta)=p\zeta^{p-1}$ where $y < \zeta < x$. But since $p-1 \geq 0$, we see $\zeta^{p-1} < x^{p-1} <x^{p-1}+y^{p-1}$. Therefore $f(x)-f(y)=x^p-y^p=p(x-y)\zeta^{p-1}<p(x-y)(x^{p-1}-y^{p-1}).$ For $x=y$ the equality holds.
Therefore \begin{aligned} d_1(\lambda(f),\lambda(g)) &= \int_X \left||f|^p-|g|^p\right|d\mu \\ &\leq \int_Xp\left||f|-|g|\right|(|f|^{p-1}+|g|^{p-1})d\mu \end{aligned} By Hölder's inequality, we have \begin{aligned} \int_X ||f|-|g||(|f|^{p-1}+|g|^{p-1})d\mu & \leq \left[\int_X \left||f|-|g|\right|^pd\mu\right]^{1/p}\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \\ &\leq \left[\int_X \left|f-g\right|^pd\mu\right]^{1/p}\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \\ &=\lVert f-g \rVert_p \left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q}. \end{aligned} By Minkowski's inequality, we have $\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \leq \left[\int_X|f|^{(p-1)q}d\mu\right]^{1/q}+\left[\int_X |g|^{(p-1)q}d\mu\right]^{1/q}$ Now things are clear. Since $1/p+1/q=1$, or equivalently $1/q=(p-1)/p$, suppose $\lVert f \rVert_p$, $\lVert g \rVert_p \leq R$, then $(p-1)q=p$ and therefore $\left[\int_X|f|^{(p-1)q}d\mu\right]^{1/q}+\left[\int_X |g|^{(p-1)q}d\mu\right]^{1/q} = \lVert f \rVert_p^{p-1}+\lVert g \rVert_p^{p-1} \leq 2R^{p-1}.$ Summing the inequalities above, we get \begin{aligned} d_1(\lambda(f),\lambda(g)) \leq 2pR^{p-1}\lVert f-g \rVert_p =2pR^{p-1}d_p(f,g) \end{aligned} hence $\lambda$ is continuous.
Conclusion and further
We have proved that $\lambda$ is continuous, and when $0<p<1$, we have seen that $\lambda$ is Lipschitz continuous. It's natural to think about its differentiability afterwards, but the absolute value function is not even differentiable so we may have no chance. But this is still a fine enough result. For example we have no restriction to $(X,\mathfrak{M},\mu)$ other than the positivity of $\mu$. Therefore we may take $\mathbb{R}^n$ as the Lebesgue measure space here, or we can take something else.
It's also interesting how we use elementary Calculus to solve some much more abstract problems.
Study Vector Bundle in a Relatively Harder Way - Definition
Motivation
Direction is a considerable thing. For example take a look at this picture (by David Gunderman):
The position of the red ball and black ball shows that this triple of balls turns upside down every time they finish one round. This wouldn't happen if this triple were on a normal band, which can be denoted by $S^1 \times (0,1)$. What would happen if we try to describe their velocity on the Möbius band, both locally and globally? There must be some significant difference from a normal band. If we set some move pattern on balls, for example let them run horizontally or zig-zagly, hopefully we get different set of vectors. those vectors can span some vector spaces as well.
A Formal Construction
Here and in the forgoing posts, we will try to develop purely formally certain functorial constructions having to do with vector bundles. It may be overly generalized, but we will offer some examples to make it concrete.
Let $M$ be a manifold (of class $C^p$, where $p \geq 0$ and can be set to $\infty$) modeled on a Banach space $\mathbf{E}$. Let $E$ be another topological space and $\pi: E \to M$ a surjective $C^p$-morphism. A vector bundle is a topological construction associated with $M$ (base space), $E$ (total space) and $\pi$ (bundle projection) such that, roughly speaking, $E$ is locally a product of $M$ and $\mathbf{E}$.
We use $\mathbf{E}$ instead of $\mathbb{R}^n$ to include the infinite dimensional cases. We will try to distinguish finite-dimensional and infinite-dimensional Banach spaces here. There are a lot of things to do, since, for example, infinite dimensional Banach spaces have no countable Hamel basis, while the finite-dimensional ones have finite ones (this can be proved by using the Baire category theorem).
Next we will show precisely how $E$ locally becomes a product space. Let $\mathfrak{U}=(U_i)_i$ be an open covering of $M$, and for each $i$, suppose that we are given a mapping $\tau_i:\pi^{-1}(U_i)\to U_i \times E$ satisfying the following three conditions.
VB 1 $\tau_i$ is a $C^p$ diffeomorphism making the following diagram commutative:
where $pr$ is the projection of the first component: $(x,y) \mapsto x$. By restricting $\tau_i$ on one point of $U_i$, we obtain an isomorphism on each fiber $\pi^{-1}(x)$: $\tau_{ix}:\pi^{-1}(x) \xrightarrow{\simeq} \{x\} \times \mathbf{E}$
VB 2 For each pair of open sets $U_i$, $U_j \in \mathfrak{U}$, we have the map $\tau_{jx} \circ \tau_{ix}^{-1}: \mathbf{E} \to \mathbf{E}$ to be a toplinear isomorphism (that is, it preserves $\mathbf{E}$ for being a topological vector space).
VB 3 For any two members $U_i$, $U_j \in \mathfrak{U}$, we have the following function to be a $C^p$-morphism: \begin{aligned} \varphi:U_i \cap U_j &\to L(\mathbf{E},\mathbf{E}) \\ x &\mapsto \left(\tau_j\circ \tau_i^{-1}\right)_x \end{aligned}
REMARKS. As with manifold, we call the set of 2-tuples $(U_i,\tau_i)_i$ a trivializing covering of $\pi$, and that $(\tau_i)$ are its trivializing maps. Precisely, for $x \in U_i$, we say $U_i$ or $\tau_i$ trivializes at $x$.
Two trivializing coverings for $\pi$ is said to be VB-equivalent if taken together they also satisfy conditions of VB 2 and VB 3. It's immediate that VB-equivalence is an equivalence relation and we leave the verification to the reader. It is this VB-equivalence class of trivializing coverings that determines a structure of vector bundle on $\pi$. With respect to the Banach space $\mathbf{E}$, we say that the vector bundle has fiber $\mathbf{E}$, or is modeled on $\mathbf{E}$.
Next we shall give some motivations of each condition. Each pair $(U_i,\tau_i)$ determines a local product of 'a part of the manifold' and the model space, on the latter of which we can deploy the direction with ease. This is what VB 1 tells us. But that's far from enough if we want our vectors fine enough. We do want the total space $E$ to actually be able to qualify our requirements. As for VB 2, it is ensured that using two different trivializing maps will give the same structure of some Banach spaces (with equivalent norms). According to the image of $\tau_{ix}$, we can say, for each point $x \in X$, which can be determined by a fiber $\pi^{-1}(x)$ (the pre-image of $\tau_{ix}$), can be given another Banach space by being sent via $\tau_{jx}$ for some $j$. Note that $\pi^{-1}(x) \in E$, the total space. In fact, VB 2 has an equivalent alternative:
VB 2' On each fiber $\pi^{-1}(x)$ we are given a structure of Banach space as follows. For $x \in U_i$, we have a toplinear isomorphism which is in fact the trivializing map: $\tau_{ix}:\pi^{-1}(x)=E_x \to \mathbf{E}.$ As stated, VB 2 implies VB 2'. Conversely, if VB 2' is satisfied, then for open sets $U_i$, $U_j \in \mathfrak{U}$, and $x \in U_i \cap U_j$, we have $\tau_{jx} \circ \tau_{ix}^{-1}:\mathbf{E} \to \mathbf{E}$ to be an toplinear isomorphism. Hence, we can consider VB 2 or VB 2' as the refinement of VB 1.
In finite dimensional case, one can omit VB 3 since it can be implied by VB 2, and we will prove it below.
(Lemma) Let $\mathbf{E}$ and $\mathbf{F}$ be two finite dimensional Banach spaces. Let $U$ be open in some Banach space. Let $f:U \times \mathbf{E} \to \mathbf{F}$ be a $C^p$-morphism such that for each $x \in U$, the map $f_x: \mathbf{E} \to \mathbf{F}$ given by $f_x(v)=f(x,v)$ is a linear map. Then the map of $U$ into $L(\mathbf{E},\mathbf{F})$ given by $x \mapsto f_x$ is a $C^p$-morphism.
PROOF. Since $L(\mathbf{E},\mathbf{F})=L(\mathbf{E},\mathbf{F_1}) \times L(\mathbf{E},\mathbf{F_2}) \times \cdots \times L(\mathbf{E},\mathbf{F_n})$ where $\mathbf{F}=\mathbf{F_1} \times \cdots \times \mathbf{F_n}$, by induction on the dimension of $\mathbf{F}$ and $\mathbf{E}$, it suffices to assume that $\mathbf{E}$ and $\mathbf{F}$ are toplinearly isomorphic to $\mathbb{R}$. But in that case, the function $f(x,v)$ can be written $g(x)v$ for some $g:U \to \mathbb{R}$. Since $f$ is a morphism, it follows that as a function of each argument $x$, $v$ is also a morphism, Putting $v=1$ shows that $g$ is also a morphism, which finishes the case when both the dimension of $\mathbf{E}$ and $\mathbf{F}$ are equal to $1$, and the proof is completed by induction. $\blacksquare$
To show that VB 3 is implied by VB 2, put $\mathbf{E}=\mathbf{F}$ as in the lemma. Note that $\tau_j \circ \tau_i^{-1}$ maps $U_i \cap U_j \times \mathbf{E}$ to $\mathbf{E}$, and $U_i \cap U_j$ is open, and for each $x \in U_i \cap U_j$, the map $(\tau_j \circ \tau_i^{-1})_x=\tau_{jx} \circ \tau_{ix}^{-1}$ is toplinear, hence linear. Then the fact that $\varphi$ is a morphism follows from the lemma.
Examples
Trivial bundle
Let $M$ be any $n$-dimensional smooth manifold that you are familiar with, then $pr:M \times \mathbb{R}^n \to M$ is actually a vector bundle. Here the total space is $M \times \mathbb{R}^n$ and the base is $M$ and $pr$ is the bundle projection but in this case it is simply a projection. Intuitively, on a total space, we can determine a point $x \in M$, and another component can be any direction in $\mathbb{R}^n$, hence a vector.
We need to verify three conditions carefully. Let $(U_i,\varphi_i)_i$ be any atlas of $M$, and $\tau_i$ is the identity map on $U_i$ (which is naturally of $C^p$). We claim that $(U_i,\tau_i)_i$ satisfy the three conditions, thus we get a vector bundle.
For VB 1 things are clear: since $pr^{-1}(U_i)=U_i \times \mathbb{R}^n$, the diagram is commutative. Each fiber $pr^{-1}(x)$ is essentially $(x) \times \mathbb{R}^n$, and still, $\tau_{jx} \circ \tau_{ix}^{-1}$ is the identity map between $(x) \times \mathbb{R}^n$ and $(x) \times \mathbb{R}^n$, under the same Euclidean topology, hence VB 2 is verified, and we have no need to verify VB 3.
Möbius band
First of all, imagine you have embedded a circle into a Möbius band. Now we try to give some formal definition. As with quotient topology, $S^1$ can be defined as $S^1=I/\sim_1,$
where $I$ is the unit interval and $0 \sim_1 1$ (identifying two ends). On the other hand, the infinite Möbius band can be defined by $B= (I \times \mathbb{R})/\sim_2$ where $(0,v) \sim_2 (1,-v)$ for all $v \in \mathbb{R}$ (not only identifying two ends of $I$ but also 'flips' the vertical line). Then all we need is a natural projection on the first component: $\pi:B \to S^1.$ And the verification has few difference from the trivial bundle. Quotient topology of Banach spaces follows naturally in this case, but things might be troublesome if we restrict ourself in $\mathbb{R}^n$.
Tangent bundle of the sphere
The first example is relatively rare in many senses. By $S^n$ we mean the set in $\mathbb{R}^{n+1}$ with $S^n=\{(x_0,x_1,\dots,x_n):x_0^2+x_1^2+\cdots+x_n^2=1\}$ and the tangent bundle can be defined by $TS^n=\{(\mathbf{x},\mathbf{y}):\langle\mathbf{x},\mathbf{y}\rangle=0\} \subset S^{n} \times\mathbb{R}^{n+1},$ where, of course, $\mathbf{x} \in S^n$ and $\mathbf{y} \in \mathbb{R}^{n+1}$. The vector bundle is given by $pr:TS^n \to S^n$ where $pr$ is the projection of the first factor. This total space is of course much finer than $M \times \mathbb{R}^n$ in the first example. Each point in the manifold now is associated with a tangent space $T_x(M)$ at this point.
More generally, we can define it in any Hilbert space $H$, for example, $L^2$ space: $TS=\{(x,y):\langle x , y \rangle=0\} \subset S \times H$ where $S=\{x:\langle x , x \rangle = 1\}.$ The projection is natural: \begin{aligned} \pi: TM &\to M \\ T_x(M) & \mapsto x \end{aligned} But we will not cover the verification in this post since it is required to introduce the abstract definition of tangent vectors. This will be done in the following post.
There are still many things remain undiscovered
We want to study those 'vectors' associated to some manifold both globally and locally. For example we may want to describe the tangent line of some curves at some point without heavily using elementary calculus stuff. Also, we may want to describe the vector bundle of a manifold globally, for example, when will we have a trivial one? Can we classify the manifold using the behavior of the bundle? Can we make it a little more abstract, for example, consider the class of all isomorphism bundles? How do one bundle transform to another? But to do this we need a big amount of definitions and propositions.
The Big Three Pt. 6 - Closed Graph Theorem with Applications
(Before everything: elementary background of topology and vector spaces, Banach spaces, is assumed.)
A surprising result of Banach spaces
We can define several relations between two norms. Suppose we have a topological vector space $X$ and two norms $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$. One says $\lVert \cdot \rVert_1$ is weaker than $\lVert \cdot \rVert_2$ if there is $K>0$ such that $\lVert x \rVert_1 \leq K \lVert x \rVert_2$ for all $x \in X$. Two norms are equivalent if each is weaker than the other (trivially this is a equivalence relation). The idea of stronger and weaker norms is related to the idea of the "finer" and "coarser" topologies in the setting of topological spaces.
So what about their limit of convergence? Unsurprisingly this can be verified with elementary $\epsilon-N$ arguments. Suppose now $\lVert x_n - x \rVert_1 \to 0$ as $n \to 0$, we immediately have $\lVert x_n - x \rVert_2 \leq K \lVert x_n-x \rVert_1 < K\varepsilon$
for some large enough $n$. Hence $\lVert x_n - x \rVert_2 \to 0$ as well. But what about the converse? We give a new definition of equivalence relation between norms.
(Definition) Two norms $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ of a topological vector space are compatible if given that $\lVert x_n - x \rVert_1 \to 0$ and $\lVert x_n - y \rVert_2 \to 0$ as $n \to \infty$, we have $x=y$.
By the uniqueness of limit, we see if two norms are equivalent, then they are compatible. And surprisingly, with the help of the closed graph theorem we will discuss in this post, we have
(Theorem 1) If $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ are compatible, and both $(X,\lVert\cdot\rVert_1)$ and $(X,\lVert\cdot\rVert_2)$ are Banach, then $\lVert\cdot\rVert_1$ and $\lVert\cdot\rVert_2$ are equivalent.
This result looks natural but not seemingly easy to prove, since one find no way to build a bridge between the limit and a general inequality. But before that, we need to elaborate some terminologies.
Preliminaries
(Definition) For $f:X \to Y$, the graph of $f$ is defined by $G(f)=\{(x,f(x)) \in X \times Y:x \in X\}.$
If both $X$ and $Y$ are topological spaces, and the topology of $X \times Y$ is the usual one, that is, the smallest topology that contains all sets $U \times V$ where $U$ and $V$ are open in $X$ and $Y$ respectively, and if $f: X \to Y$ is continuous, it is natural to expect $G(f)$ to be closed. For example, by taking $f(x)=x$ and $X=Y=\mathbb{R}$, one would expect the diagonal line of the plane to be closed.
(Definition) The topological space $(X,\tau)$ is an $F$-space if $\tau$ is induced by a complete invariant metric $d$. Here invariant means that $d(x+z,y+z)=d(x,y)$ for all $x,y,z \in X$.
A Banach space is easily to be verified to be a $F$-space by defining $d(x,y)=\lVert x-y \rVert$.
(Open mapping theorem) See this post
By definition of closed set, we have a practical criterion on whether $G(f)$ is closed.
(Proposition 1) $G(f)$ is closed if and only if, for any sequence $(x_n)$ such that the limits $x=\lim_{n \to \infty}x_n \quad \text{ and }\quad y=\lim_{n \to \infty}f(x_n)$ exist, we have $y=f(x)$.
In this case, we say $f$ is closed. For continuous functions, things are trivial.
(Proposition 2) If $X$ and $Y$ are two topological spaces and $Y$ is Hausdorff, and $f:X \to Y$ is continuous, then $G(f)$ is closed.
Proof. Let $G^c$ be the complement of $G(f)$ with respect to $X \times Y$. Fix $(x_0,y_0) \in G^c$, we see $y_0 \neq f(x_0)$. By the Hausdorff property of $Y$, there exists some open subsets $U \subset Y$ and $V \subset Y$ such that $y_0 \in U$ and $f(x_0) \in V$ and $U \cap V = \varnothing$. Since $f$ is continuous, we see $W=f^{-1}(V)$ is open in $X$. We obtained a open neighborhood $W \times U$ containing $(x_0,y_0)$ which has empty intersection with $G(f)$. This is to say, every point of $G^c$ has a open neighborhood contained in $G^c$, hence a interior point. Therefore $G^c$ is open, which is to say that $G(f)$ is closed. $\square$
REMARKS. For $X \times Y=\mathbb{R} \times \mathbb{R}$, we have a simple visualization. For $\varepsilon>0$, there exists some $\delta$ such that $|f(x)-f(x_0)|<\varepsilon$ whenever $|x-x_0|<\delta$. For $y_0 \neq f(x_0)$, pick $\varepsilon$ such that $0<\varepsilon<\frac{1}{2}|f(x_0)-y_0|$, we have two boxes ($CDEF$ and $GHJI$ on the picture), namely $B_1=\{(x,y):x_0-\delta<x<x_0+\delta,f(x_0)-\varepsilon<y<f(x_0)+\varepsilon\}$ and $B_2=\{(x,y):x_0-\delta<x<x_0+\delta,y_0-\varepsilon<y<y_0+\varepsilon\}.$ In this case, $B_2$ will not intersect the graph of $f$, hence $(x_0,y_0)$ is an interior point of $G^c$.
The Hausdorff property of $Y$ is not removable. To see this, since $X$ has no restriction, it suffices to take a look at $X \times X$. Let $f$ be the identity map (which is continuous), we see the graph $G(f)=\{(x,x):x \in X\}$ is the diagonal. Suppose $X$ is not Hausdorff, we reach a contradiction. By definition, there exists some distinct $x$ and $y$ such that all neighborhoods of $x$ contain $y$. Pick $(x,y) \in G^c$, then all neighborhoods of $(x,y) \in X \times X$ contain $(x,x)$ so $(x,y) \in G^c$ is not a interior point of $G^c$, hence $G^c$ is not open.
Also, as an immediate consequence, every affine algebraic variety in $\mathbb{C}^n$ and $\mathbb{R}^n$ is closed with respect to Euclidean topology. Further, we have the Zariski topology $\mathcal{Z}$ by claiming that, if $V$ is an affine algebraic variety, then $V^c \in \mathcal{Z}$. It's worth noting that $\mathcal{Z}$ is not Hausdorff (example?) and in fact much coarser than the Euclidean topology although an affine algebraic variety is both closed in the Zariski topology and the Euclidean topology.
The closed graph theorem
After we have proved this theorem, we are able to prove the theorem about compatible norms. We shall assume that both $X$ and $Y$ are $F$-spaces, since the norm plays no critical role here. This offers a greater variety but shall not be considered as an abuse of abstraction.
(The Closed Graph Theorem) Suppose
1. $X$ and $Y$ are $F$-spaces,
2. $f:X \to Y$ is linear,
3. $G(f)$ is closed in $X \times Y$.
Then $f$ is continuous.
In short, the closed graph theorem gives a sufficient condition to claim the continuity of $f$ (keep in mind, linearity does not imply continuity). If $f:X \to Y$ is continuous, then $G(f)$ is closed; if $G(f)$ is closed and $f$ is linear, then $f$ is continuous.
Proof. First of all we should make $X \times Y$ an $F$-space by assigning addition, scalar multiplication and metric. Addition and scalar multiplication are defined componentwise in the nature of things: $\alpha(x_1,y_1)+\beta(x_2,y_2)=(\alpha x_1+\beta x_2,\alpha y_1 + \beta y_2).$ The metric can be defined without extra effort: $d((x_1,y_1),(x_2,y_2))=d_X(x_1,x_2)+d_Y(y_1,y_2).$ Then it can be verified that $X \times Y$ is a topological space with translate invariant metric. (Potentially the verifications will be added in the future but it's recommended to do it yourself.)
Since $f$ is linear, the graph $G(f)$ is a subspace of $X \times Y$. Next we quote an elementary result in point-set topology, a subset of a complete metric space is closed if and only if it's complete, by the translate-invariance of $d$, we see $G(f)$ is an $F$-space as well. Let $p_1: X \times Y \to X$ and $p_2: X \times Y \to Y$ be the natural projections respectively (for example, $p_1(x,y)=x$). Our proof is done by verifying the properties of $p_1$ and $p_2$ on $G(f)$.
For simplicity one can simply define $p_1$ on $G(f)$ instead of the whole space $X \times Y$, but we make it a global projection on purpose to emphasize the difference between global properties and local properties. One can also write $p_1|_{G(f)}$ to dodge confusion.
Claim 1. $p_1$ (with restriction on $G(f)$) defines an isomorphism between $G(f)$ and $X$.
For $x \in X$, we see $p_1(x,f(x)) = x$ (surjectivity). If $p_1(x,f(x))=0$, we see $x=0$ and therefore $(x,f(x))=(0,0)$, hence the restriction of $p_1$ on $G$ has trivial kernel (injectivity). Further, it's trivial that $p_1$ is linear.
Claim 2. $p_1$ is continuous on $G(f)$.
For every sequence $(x_n)$ such that $\lim_{n \to \infty}x_n=x$, we have $\lim_{n \to \infty}f(x_n)=f(x)$ since $G(f)$ is closed, and therefore $\lim_{n \to \infty}p_1(x_n,f(x_n)) =x$. Meanwhile $p_1(x,f(x))=x$. The continuity of $p_1$ is proved.
Claim 3. $p_1$ is a homeomorphism with restriction on $G(f)$.
We already know that $G(f)$ is an $F$-space, so is $X$. For $p_1$ we have $p_1(G(f))=X$ is of the second category (since it's an $F$-space and $p_1$ is one-to-one), and $p_1$ is continuous and linear on $G(f)$. By the open mapping theorem, $p_1$ is an open mapping on $G(f)$, hence is a homeomorphism thereafter.
Claim 4. $p_2$ is continuous.
This follows the same way as the proof of claim 2 but much easier since we have no need to care about $f$.
Now things are immediate once one realizes that $f=p_2 \circ p_1|_{G(f)}^{-1}$, and hence $f$ is continuous. $\square$
Applications
Before we go for theorem 1 at the beginning, we drop an application on Hilbert spaces.
Let $T$ be a bounded operator on the Hilbert space $L_2([0,1])$ so that if $\phi \in L_2([0,1])$ is a continuous function so is $T\phi$. Then the restriction of $T$ to $C([0,1])$ is a bounded operator of $C([0,1])$.
Now we go for the identification of norms. Define \begin{aligned} f:(X,\lVert\cdot\rVert_1) &\to (X,\lVert\cdot\rVert_2) \\ x &\mapsto x \end{aligned} i.e. the identity map between two Banach spaces (hence $F$-spaces). Then $f$ is linear. We need to prove that $G(f)$ is closed. For the convergent sequence $(x_n)$ $\lim_{n \to \infty}\lVert x_n -x \rVert_1=0,$ we have $\lim_{n \to \infty} \lVert f(x_n)-x \rVert_2=\lim_{n \to \infty}\lVert x_n -x\rVert_2=\lim_{n \to \infty}\lVert f(x_n)-f(x)\rVert_2=0.$ Hence $G(f)$ is closed. Therefore $f$ is continuous, hence bounded, we have some $K$ such that $\lVert x \rVert_2 =\lVert f(x) \rVert_1 \leq K \lVert x \rVert_1.$ By defining \begin{aligned} g:(X,\lVert\cdot\rVert_2) &\to (X,\lVert\cdot\rVert_1) \\ x &\mapsto x \end{aligned} we see $g$ is continuous as well, hence we have some $K'$ such that $\lVert x \rVert_1 =\lVert g(x) \rVert_2 \leq K'\lVert x \rVert_2$ Hence two norms are weaker than each other.
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# Frictionless Roller Coaster Problem
###### Question:
In the figure, a frictionless roller coaster car of mass m = 811 kg tops the first hill with speed v0 = 18 m/s at height h = 43 m. What is the speed of the car at point B?
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##### For each of problems A.15 through A.22, plot the function for thespecified domain' This means that the following are to be shown:*label both axes (with the appropriate variable symbol) *scale both axes (by providing at least 3 numbers along each axis) *state the coordinates of the intercepts, if there are any intercepts *show the asymptote, if there is one dashed line *show the general shape of the function by tracing it out as a solid line
For each of problems A.15 through A.22, plot the function for thespecified domain' This means that the following are to be shown: *label both axes (with the appropriate variable symbol) *scale both axes (by providing at least 3 numbers along each axis) *state the coordinates of the intercepts, ...
##### Convergent geometric series B. Convergent p series C. Comparison (or Limit Comparison) with a geometric or D. Converges by alternating series test p series(-1)" 7n +7 2 (-1)" Vn +2 3. (-Y& 3" | (L5)" sin? i" Gn) 5 n2 7(5)" 72n
Convergent geometric series B. Convergent p series C. Comparison (or Limit Comparison) with a geometric or D. Converges by alternating series test p series (-1)" 7n +7 2 (-1)" Vn +2 3. (-Y& 3" | (L5)" sin? i" Gn) 5 n2 7(5)" 72n...
##### 5. Given below are the probability distribution of two assets-X and Y States of Economy Probability...
5. Given below are the probability distribution of two assets-X and Y States of Economy Probability MEGA BOOM BOOM NORMAL BUST MEGA BUST 0.1 0.2 0.4 Expected returns of the 2 assets (%) - X Y 50 100 40 70 30 50 10 -20 -10 -50 0.2 0.1 Which one has higher stand-alone risk? (Compute Coefficient of Var...
##### Outpatient Insurance Coding - Critical Thinking Ch 4 2
2. Why is it important for WHO to standardize codes to be used worldwide?...
##### Let A and B ben B = P-'AP nulnices (Hint; We say B is similar B = if there an invertible matrix (a) If c is any AP PB = AP ) scalar; show that cl similar only to itself: (b) Show that if B similar then is similar t0 Show that similar to 4> ( JJ {1' /LthalShow that for any real numbers and the matricesandsimilar:Show thatnot similar t0Show thatis not diagonalizable, i.e., is not similar to any diagonal matrix:
Let A and B ben B = P-'AP nulnices (Hint; We say B is similar B = if there an invertible matrix (a) If c is any AP PB = AP ) scalar; show that cl similar only to itself: (b) Show that if B similar then is similar t0 Show that similar to 4> ( JJ {1' /L thal Show that for any real number...
##### How do you find the x and y intercept of 3x+0.5y=6?
How do you find the x and y intercept of 3x+0.5y=6?...
##### NameFlnd the derivative of each function:fk) = 9e*+9*b} fk)=z+2*f()=x+Inxsd) fk)=Ix logs *Find the equation of the tangent llne to the graph of the function at the give value of kfo=+bx. T51{6=1to *=0
Name Flnd the derivative of each function: fk) = 9e*+9* b} fk)=z+2* f()=x+Inxs d) fk)=Ix logs * Find the equation of the tangent llne to the graph of the function at the give value of k fo=+bx. T51 {6=1to *=0...
##### A DC (distribution center) for Target carries 10,000 items. During a particular week, its customers (retailers...
A DC (distribution center) for Target carries 10,000 items. During a particular week, its customers (retailers or possibly wholesalers) ordered 2,600 different SKUs. The DC only satisfied all demand for 2,500 items. So out of 10,000 SKUs, the DC satisfied 9900 items. What is the In-stock Probability...
##### This Iime you have delernuexh Uhe concentralion Ofllie KIno4 solulon and you are uyuig @0 determine Ihe concentralion and whww%k composilion ol Ite HzOz I suppled solulion_Tne purchased solulion needs to be diluted prior t0 lilration Pour about 30 mL irom Ihe slock HzOz bottle into 50 mL dry beaker_ Pipet 15.00 Il_ of Ihe HAOz solution Irom the smnal bcaker Into 250.0 mL volumetric Ilask and dilute wilh deionized waler with Swirling lo the Iine. Transter this solution to & plastic 250 mL co
This Iime you have delernuexh Uhe concentralion Ofllie KIno4 solulon and you are uyuig @0 determine Ihe concentralion and whww%k composilion ol Ite HzOz I suppled solulion_ Tne purchased solulion needs to be diluted prior t0 lilration Pour about 30 mL irom Ihe slock HzOz bottle into 50 mL dry beaker...
##### 7.)CHaKzC1zOz HzS04Clz, AIClPropose mechanism for the above reaction:
7.) CHa KzC1zOz HzS04 Clz, AICl Propose mechanism for the above reaction:...
##### On a day when the speed of sound in air is 345 m/s, the fundamental frequency...
On a day when the speed of sound in air is 345 m/s, the fundamental frequency of an open-ended pipe is 690 Hz. If the second harmonic of this pipe has the same wavelength as the second overtone (third harmonic) of a closed-end pipe, what is the length of each pipe?...
##### Problem Value: point(s). Problem Score: 0%. Attempts Remaining: attempts:points) The Township Board of Meridian Township wants to know how much public support there for raising property taxes to fix and maintain the roads in tne township: They randomly surveyed Township residents and calculated 99% confidence [email protected] for the actual proportion of Meridian Township residents who are in favor of the tax increase to be (0.54, 0.62).What Is the sample proportion? Give the exact answer as decimal:2.
Problem Value: point(s). Problem Score: 0%. Attempts Remaining: attempts: points) The Township Board of Meridian Township wants to know how much public support there for raising property taxes to fix and maintain the roads in tne township: They randomly surveyed Township residents and calculated 99...
##### 12 0f 25 (0 compiete)and the data are summarized in the r various brands of digital cameraseferencesFemaleMale Total j 0 153 305 688Cannon Power Shot Nikon Cool Pix Sony Cyber Shot Panasonlc Lumkx Fujifilm Fingpix Olympus SI Other Brands TotalPrintDone
12 0f 25 (0 compiete) and the data are summarized in the r various brands of digital cameras eferences Female Male Total j 0 153 305 688 Cannon Power Shot Nikon Cool Pix Sony Cyber Shot Panasonlc Lumkx Fujifilm Fingpix Olympus SI Other Brands Total Print Done...
##### Identify the reagents necessary accomplish each of the transformations shown below . If more than one reagent is needed, list all reagents:enantiomerHBr9 = 1) BH; THF 2) HzOz, NaOH 2 = HBr, HOOH 10 = HzIPt Brz 11 = 1) MCPBA 2) H;ot 4 = Brz; HzO 12 = OsOa (catalytic); NMO 5 = Brz, CHzOH 13 = 1) O3 2) DMS 6 = TsCl, pyridine 14 = NaOMe 7 = 1) Hg(OAc)z; HzO 2) NaBHa 15 = KOC(CHs)s 8 = 1) Hg(OAc)z; EtOH 2) NaBHa 16 = NBS_ hv
Identify the reagents necessary accomplish each of the transformations shown below . If more than one reagent is needed, list all reagents: enantiomer HBr 9 = 1) BH; THF 2) HzOz, NaOH 2 = HBr, HOOH 10 = HzIPt Brz 11 = 1) MCPBA 2) H;ot 4 = Brz; HzO 12 = OsOa (catalytic); NMO 5 = Brz, CHzOH 13 = 1) O3...
##### SA Savod Help Save & Exit Check Required information [The following information applies to the questions...
SA Savod Help Save & Exit Check Required information [The following information applies to the questions displayed below) Warnerwoods Company uses a perpetual inventory system. It entered into the following purchases and sales transactions for March Units Sold at Retail Units Acquired at Cost 10...
##### Exercise 3.72 PartA Calculate the mass percent composition of nitrogen in each of the following nitrogen...
Exercise 3.72 PartA Calculate the mass percent composition of nitrogen in each of the following nitrogen compounds. N20 Express your answer using four significant filgures. mass % N Submilt Request Answer Part B SigN4 Express your answer using four significant figures. mass % N- Submit Part C NaNH2 ...
##### Data Set: 100,96,94,94,92,88,87,86,86,86,82,70 1. What is the z score for a score of 100? A. 1.488...
Data Set: 100,96,94,94,92,88,87,86,86,86,82,70 1. What is the z score for a score of 100? A. 1.488 B. 0.974 C. 1.503 D. 1.000 2. What is the z score for a score of 94? A. 0.717 B. 0.608 C. 0.689 D. 0.460 3. What is the z score for a score of 86? A. -0.359 B. -0.310 C. 0.460 D. -0.615 4. What is t...
##### Use the Laws of Logarithms to expand the expression. $\log \frac{y^{3}}{\sqrt{2 x}}$
Use the Laws of Logarithms to expand the expression. $\log \frac{y^{3}}{\sqrt{2 x}}$...
##### Problem 12.9019 cf 19Reviev32-Cl-Cc TOC hangs vellically Dn Uriclioniless tuarizona ax & Qassing lhlough ils cerlu: . pall of clay travelina harizorta 2t2. 1/& hits and sbiks to [ie very boltorn tp af theredPart ATc"Fat mhaxinum anaie Measuredtrom ventica Ccesthe'ltn the attachedta cfclay) ratate?Express UouT Insiersignificant figures and include the appropriate units_0)ValueUnitsSubmitPreyinusamewisEanmpskamsotIncorrect; Try Again; attempts remainingRaturn AssignmentProvide Fe
Problem 12.90 19 cf 19 Reviev 32-Cl-Cc TOC hangs vellically Dn Uriclioniless tuarizona ax & Qassing lhlough ils cerlu: . pall of clay travelina harizorta 2t2. 1/& hits and sbiks to [ie very boltorn tp af thered Part A Tc"Fat mhaxinum anaie Measuredtrom ventica Ccesthe 'ltn the atta...
##### In this problem, let G be the group U(12).(4a) (5 pts) List the elements in G.(4b) (15 pts) Find all of the cyclic subgroups of G.(4c) (5 pts) Is G itself a cyclic group?
In this problem, let G be the group U(12). (4a) (5 pts) List the elements in G. (4b) (15 pts) Find all of the cyclic subgroups of G. (4c) (5 pts) Is G itself a cyclic group?...
##### Electrical power: Please do a Short answers for all of these questions.... thanks : 3. What...
Electrical power: Please do a Short answers for all of these questions.... thanks : 3. What device on the California-Oregon-Intertie has been constructed to minimize the risk of the power system slipping a pole in the case of a loss of downstream transmission capacity. 4. In an extra-high voltage tr...
##### How do you solve 5m - 3m + 2= 6m - 4?
How do you solve 5m - 3m + 2= 6m - 4?...
##### Which of the statements (I-IV) is (are) most likely FALSE: I. When shorter maturity treasuries are...
Which of the statements (I-IV) is (are) most likely FALSE: I. When shorter maturity treasuries are yielding less than longer maturity treasuries, the yield curve is considered to be normal II. When shorter maturity treasuries start to yield more than longer maturity treasuries, the bond market is e...
##### Dacide whebe noma samplinn disintuion Cn tne Olven Bampie Satstt Claim' 10 26 0.10; Sample stalislics 0 2},I Cantest tha claim about the populaton proccrtion p af Iho piven leval egnncance & UzingState ta null and 4 temative hypotheses0z0 Z6 <0 7630.26 H;p> 0.26cnnolbi Janlono0conect cpro baou ud nucarnan lennnntanndoxcomoeld Your CrocaDatomnuna thu clcal Yalta{0) SalaclJna crich valunlee Iaunul (RotJ ho duicinunlnculaaodod Usu = conndupagulm Mamun(uhklald 070 Z0
Dacide whebe noma samplinn disintuion Cn tne Olven Bampie Satstt Claim' 10 26 0.10; Sample stalislics 0 2}, I Can test tha claim about the populaton proccrtion p af Iho piven leval egnncance & Uzing State ta null and 4 temative hypotheses 0z0 Z6 <0 76 30.26 H;p> 0.26 cnnolbi Janlono0 ...
##### Write a program which gives an easy mathematics quiz. The program should display two random numbers...
Write a program which gives an easy mathematics quiz. The program should display two random numbers which are to be added together, like this: 117 + 213 ----- The program should ask the user to enter their answer. If the answer is correct, the user should be congratulated. If the answer is wrong, th...
##### QUESTOr}Let X be Hndon Varlabla that / dlstributedProbcx>4 X1) Prob (X 3) Probrx 4 Prob (X>3) Pratxx= Probox-21 None ofthe other answers makes sense Prob(x 4 Xt prob (X-3) Pralxx .1)
QUESTOr} Let X be Hndon Varlabla that / dlstributed Probcx>4 X1) Prob (X 3) Probrx 4 Prob (X>3) Pratxx= Probox-21 None ofthe other answers makes sense Prob(x 4 Xt prob (X-3) Pralxx .1)...
##### Assume that the infected area of an injury is circular: If the radius of the infected area is 4 'mm and growing ata rate of 0.3 mm/hr; at what rate is the infected area increasing? Round your answer to four decimal places. (Hint: The area of a circle is A = nr2)
Assume that the infected area of an injury is circular: If the radius of the infected area is 4 'mm and growing ata rate of 0.3 mm/hr; at what rate is the infected area increasing? Round your answer to four decimal places. (Hint: The area of a circle is A = nr2)...
##### Describe how you can apply the concepts of statement of cash flow to your personal life.
Describe how you can apply the concepts of statement of cash flow to your personal life....
##### Point)Evaluate the integral6 sec(x) cos(2c) dc sin(x) sec(k)Note: Use an upper-case "C" for the constant of integration
point) Evaluate the integral 6 sec(x) cos(2c) dc sin(x) sec(k) Note: Use an upper-case "C" for the constant of integration...
##### Evaluating Predictive Performance ( Business Analytics) Confusion Matrix !!! A publisher plans to boost the sales...
Evaluating Predictive Performance ( Business Analytics) Confusion Matrix !!! A publisher plans to boost the sales of its most popular magazine by sending out promotional mails. We refer to a customer as a responder if he/she subscribes to the magazine for the next year after receiving a promotional...
##### Wildhorse Co.closes its books on its July 31 year-end. The company does not make entries to...
Wildhorse Co.closes its books on its July 31 year-end. The company does not make entries to accrue for interest except at its year-end. On June 30, the Notes Receivable account balance is \$26,800. Notes Receivable include the following. Date Maker Face Value Term Maturity Date Interest Rate 8% April...
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Locus of points where difference in gravitational forces is constant - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:16:14Z http://mathoverflow.net/feeds/question/87586 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87586/locus-of-points-where-difference-in-gravitational-forces-is-constant Locus of points where difference in gravitational forces is constant Jennifer Gao 2012-02-05T14:27:55Z 2012-02-19T15:22:12Z <p>Is there a name for the curve in the plane defined by</p> <p>$a/\|x - p\|^2 - b/\|x - q\|^2=\mathrm{constant}$</p> <p>where $a$ and $b$ are fixed numbers and $p$ and $q$ are fixed points? How about if I don't square the denominators? How about if $a$ and $b$ are both $1$?</p> http://mathoverflow.net/questions/87586/locus-of-points-where-difference-in-gravitational-forces-is-constant/87589#87589 Answer by Joseph O'Rourke for Locus of points where difference in gravitational forces is constant Joseph O'Rourke 2012-02-05T14:59:25Z 2012-02-05T14:59:25Z <p>I cannot offer names for your functions, but I was interested to see what they look like. Here is the function with the denominators unsquared, i.e., just the distances $||p-a||$ and $||q-b||$: <br /> <img src="http://cs.smith.edu/~orourke/MathOverflow/ContourPlot.jpg" alt="Contour Plot"> <br /> You might look at <em>power Voronoi diagrams</em>, which have a similar flavor (for multiple sites $p$, $q$).</p>
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# Analysis of Adaptive Multilevel Splitting algorithms in an idealized case
1 MATHERIALS - MATHematics for MatERIALS
ENPC - École des Ponts ParisTech, CERMICS - Centre d'Enseignement et de Recherche en Mathématiques et Calcul Scientifique, Inria Paris-Rocquencourt
Abstract : The Adaptive Multilevel Splitting algorithm is a very powerful and versatile method to estimate rare events probabilities. It is an iterative procedure on an interacting particle system, where at each step, the $k$ less well-adapted particles among $n$ are killed while $k$ new better adapted particles are resampled according to a conditional law. We analyze the algorithm in the idealized setting of an exact resampling and prove that the estimator of the rare event probability is unbiased whatever $k$. We also obtain a precise asymptotic expansion for the variance of the estimator and the cost of the algorithm in the large $n$ limit, for a fixed $k$.
Keywords :
Document type :
Journal articles
Domain :
Cited literature [9 references]
https://hal.archives-ouvertes.fr/hal-00987297
Contributor : Charles-Edouard Bréhier <>
Submitted on : Tuesday, May 6, 2014 - 4:25:59 PM
Last modification on : Thursday, February 28, 2019 - 9:04:08 AM
Long-term archiving on : Wednesday, August 6, 2014 - 11:45:17 AM
### Files
ArticleAMS.pdf
Files produced by the author(s)
### Citation
Charles-Edouard Bréhier, Tony Lelièvre, Mathias Rousset. Analysis of Adaptive Multilevel Splitting algorithms in an idealized case. ESAIM: Probability and Statistics, EDP Sciences, 2015, 19, ⟨10.1051/ps/2014029⟩. ⟨hal-00987297⟩
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# Let (S) denotes the number of ordered pairs (x,y) satisfying (1)/(x)+(1)/(y)=(1)/(n),x,y,n in N. <br> Q. sum_(r=1)^(10)S(r) equals
Updated On: 17-04-2022
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47484950
B
Solution :
because1^(2)toS(1)=1,2^(2)toS(2)=3,3^(2)toS(3)=3, <br> 4^(2) to 2^(4)to S(4)=5,5^(2)toS(5)=3,S(6)=9 <br> S(7)=3,S(8)=7,S(9)=5 and S(10)=9 [from above] <br> therefore underset(r=1)overset(10(sum)S(r)=S(1)+S(2)+S(3)+S(4)+S(5)+S(6)+S(7)+S(8)+S(9)+S(10) <br> =1+3+3+5+3+9+3+7+5+9=48
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
question is right now number of auditor at Vizag 1 divided by X + 1 divided by 12 15 and expire and belongs to maximum and submission are equal to 1 to 10 s r equal to be hard to find value of s r and R is from 1 to 10 Tokyo that is equal to 1 + 1 divided by Y is equal to 15 and it is given that sum of all these terms are 10 and 12 10 that is equal to 1 divided by X + 155 is equal to 15 and according to the condition is x and y are always greater than that so we can be taken
as it is equal to 10 + something related book a and Y is equal to turn the something that stand + b values of X and Y in this question that it is equal to 1 / 10 plus + 1 / 10 + b is equal to 15 and don't now solving this then it will be equal to 10 + 3 + 10 + 10 / 10 + a x + b is equal to 1 divided by 10 and share now cross multiply in this and it will be equal to 10 X 10 + 3 + 10 X
+ a that is equal to 10 + X + now sold and with it with equal 200 + 10 + 100 + tan a multiply 10 is hundred Tanvi tan A + tan B and if we can cancel out and hundred and also cancel out song Kiya ji Karda value of AB that is equal to hundred and can be written as a b is equal to 2 X Files here and here we have A and B
are is 1 MB is funded and it is true that he is 54 then the is 25 5 then be is 20 and then the east and if a is 20 then we is 525 then we is 4 and it is 50 then we is true and if a is hundred MB is 1 to 20 are there Tap a and calculate the number of product from here number of order pair of abe 123456789
12 10 + A and Y is equal to 10 + 3 now duty price order a pair of xy S10 is equal to 9
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Muchen He
44638154
[email protected]
muchen.ca
ELEC 402
# Project 5 - PnR, Domino, and Memory
Updated 2019-12-01
## 1. Place and Route
### FSM Design
As designed and mentioned in project 1 and project2, the FSM I implemented is a simple and abstract model of a digital camera. It features 10 states of operation:
### Input and Output
This subsection describes all the input/output of the FSM.
#### Input
There are a total of 11 inputs.
1. clk: the clock signal.
2. fstop_dec: Decreases the F-stop value when in A or M mode.
3. fstop_inc: Increases the F-stop value when in A or M mode.
4. mode_dec: Cycles to the previous mode.
5. mode_inc: Cycles to the next mode.
6. power_btn: Toggles the camera between idle/operating modes.
7. reset: Master reset signal.
8. sensor_data[]: A bus of 16 bits that contains the raw value for an imaginary 1x1 sensor.
9. shutter_btn: The main signal to take a picture.
10. shutter_dec: Decreases the shutter speed when in S or M mode.
11. shutter_inc: Increases the shutter speed when in S or M mode.
#### Output
There are just two outputs.
1. output_data[]: A bus of 16 bits that contains the post-processed output data.
2. output_data_valid: Signal that indicates top-level device that the bus output_data is valid and that the FSM has finished with the current task, and returning to the previous operating mode state.
### Layout
I used Cadence Encounter ® to produce the standard cell layout. The layout of the placed and routed design is shown in the image below. The final size of the total design is 52.3 μm in width and 48.45 μm in height. All the inputs signal pins are placed on the left and output signal pins are placed on the right.
Gaps in the standard-cell layout is filled with filler of size 1, 2, 4, and 8 to ensure design validations are passed.
### Testing Procedure
To ensure that the layout is functioning exactly as it is designed, the same test-bench used for project 1 and 2 is used here. There exists assertions placed throughout the test-bench to ensure that data flow and appropriate states are correct.
Lastly, I visually inspect the waveform in ModelSim using the output of previous project waveforms as reference (see next section).
#### Output Waveform
The image below shows the simulation waveform after layout using SDF. Note that some test signals are optimized away by ModelSim, however the result (sequences of states and output values) are still correct.
For reference, the image below shows the waveform of the synthesized FSM (mapped) from Cadence RTL. The output is identical.
For even further reference, the image below shows the waveform of the unmapped FSM. The output is still identical.
#### Test Files
The following is the test bench .sv file is used test the FSM. Linked version also available on GitHub here.
/* This is a modified version of the test_fsm.sv from project 1 and 2
* CHANGES INCLUDE:
* - Added timescale specification at the top
* - Commented out asserts that acces DUT.ST_... since those are reduced to bits
*/
timescale 1ps/1ps
module test_fsm();
logic clk;
logic reset;
logic power_btn;
logic mode_inc;
logic mode_dec;
logic fstop_inc;
logic fstop_dec;
logic shutter_inc;
logic shutter_dec;
logic shutter_btn;
logic [15:0] sensor_data;
logic [15:0] fsm_output;
logic fsm_output_valid;
// For simulation debugging
logic [799:0] current_test;
fsm DUT(
.output_data(fsm_output),
.output_data_valid(fsm_output_valid),
.*
);
// Clock generator
always #2 clk = ~clk;
initial begin
// Initial values
clk = 1;
reset = 0;
power_btn = 0;
## 2. Domino Logic
### Logic Function
All nodes are charged to VDD during the pre-charge phase, the only time that the input node to the inverter will discharge (go to 0) is that is that AB(C+D) is true.
$Out=A\wedge B\wedge(C\vee D)$
## 5. FPGA Logic
We first use a K-map to find the logic function from the truth table. Also for the sake of simplicity, I will reassign A3, A2 A1, A0 to $DCBA$. As shown in the drawing below, the sum of products can be deducted:
$Z=DC+C\bar{A}+\bar{D}\bar{C}BA$
The FPGA only has 3-bit LUT so we need to split this function into smaller functions such that each smaller function only takes in 3 inputs.
$Z=\overbrace{\underbrace{DC+C\bar{A}}_E+\underbrace{\bar{D}\bar{C}B}_FA}^{Z}$
As seen in the expression above, 3 logic blocks are required. Each block’s 3-LUT will be populated based on the above expressions:
$E=DC+C\bar A,\quad F=\bar D\bar C B,\quad Z=E+FA$
The completed PLB in the FPGA looks like this:
Note that because everything is combinational logic, the 2MUX at the end of each PLB is set to 1` to not use the flip-flops.
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## TGCD2 - Trending GCD (Hard)
no tags
This problem is a harder version of TRENDGCD.
Given $n$ and $m$, compute
$$S(n, m) = \sum_{i=1}^n \sum_{j=1}^m ij \cdot f(\gcd(i,j)),$$
where $f(n) = (\mu(n))^2 n$ and $\mu(n)$ is the Möbius function, that is, $f(n) = n$ if $n$ is square-free and $0$ otherwise. Especially, $f(1)=1$.
### Input
The first line contains an integer $T$, indicating the number of test cases.
Each of the next $T$ lines contains two positive integers $n$ and $m$.
### Output
For each test case, print $S(n, m)$ modulo $10^9+7$ in a single line.
### Example
Input:
542 1835 120 25123456789 987654321233333333333 233333333333
Output:
306395630128819897063534355737203
### Constraints
There are 6 test files.
Test #0: $1 \leq T \leq 10000$, $1 \leq n, m \leq 10^7$.
Test #1: $1 \leq T \leq 200$, $1 \leq n, m \leq 10^8$.
Test #2: $1 \leq T \leq 40$, $1 \leq n, m \leq 10^9$.
Test #3: $1 \leq T \leq 10$, $1 \leq n, m \leq 10^{10}$.
Test #4: $1 \leq T \leq 2$, $1 \leq n, m \leq 10^{11}$.
Test #5: $T = 1$, $1 \leq n, m \leq 235711131719$.
@Speed Addicts: My solution runs in 20.76s (total time). (approx 3.46s per file)
WARNING: The time limit may be somewhat strict.
Added by: liouzhou_101 Date: 2019-03-27 Time limit: 20s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: All Resource: TRENDGCD
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# Riemann Zeta Function integral
I was reading about the Riemann Zeta Function when they mentioned the contour integral $$\int_{+\infty}^{+\infty}\frac{(-x)^{s-1}}{e^x - 1} dx$$ where the path of integration "begins at $+\infty$, moves left down the positive real axis, circles the origin once in the positive (counterclockwise) direction, and returns up the positive real axis to $+\infty$." They specified that $(-x)^{s-1} = e^{(s-1)\log(-x)}$ so that $\log(-x)$ is real when x is a negative real number. Furthermore, $\log(-x)$ is undefined when x is a positive real number as a result of this choice of $\log$.
They proceeded to evaluate the integral by splitting it into $$\int_{+\infty}^{\delta}\frac{(-x)^{s-1}}{e^x - 1}dx + \int_{|x|=\delta}\frac{(-x)^{s-1}}{e^x - 1} dx + \int_{\delta}^{+\infty}\frac{(-x)^{s-1}}{e^x - 1} dx$$ I do not understand the justification for the path of the second integral; if $(-x)^{s-1}$ is undefined on the non-negative real axis, how can the path of integration cross the real axis at $x=\delta$, when $\delta$ is implicitly non-negative?
EDIT: I am aware that $\log(-x)$ is defined on the entire plane except for some ray from the origin. What I am specifically confused about is how the contour is allowed to cross the ray on which $\log(-x)$ is undefined.
EDIT: They describe the first and third integrals of the sum as being taken "slightly above" and "slightly below" the positive real axis as the function is undefined on the positive real axis.
## 2 Answers
Here is a picture that matches the description of the path of integration:
Note that we are taking the positive real axis as the branch cut for $\log(-x)$, and we have taken the branch of $\log(-x)$ whose argument goes from $-\pi$ to $\pi$.
The red and blue lines should be infinitesimally above and below the positive real axis.
There is more to what they specified than you've reproduced here. $\log(-x)$ is not necessarily undefined when the values are allowed to be complex. In fact, if $x = re^{i\theta}$, then $-x = re^{i(\theta + \pi)}$. Therefore we can define $\log(-x) = \log(r) + i(\theta + \pi)$. The problem is, $-x = re^{i(\theta + (2n+1)\pi)}$ as well, for any integer $n$. So which value of $\theta$ do we choose? No matter what the choice, at some place as you circle the origin, $\theta$ must suddenly jump in value to get back to the other side of its interval of definition. So we have to choose to place this discontinuity somewhere.
Since they need $\log(-x)$ defined on both the positive and negative real axis, and since they choose to circle $0$ on the upper side (counterclockwise from positive to negative circles above $0$), they probably chose to put the discontinuity on the negative imaginary axis. I.e., they defined $\log(-re^{i\theta}) = \log(r) + i\theta$ where $-\frac \pi 2 < \theta < \frac{3\pi} 2$.
So $(-x)^{s-1}$ is also defined everywhere except on the negative imaginary axis, and they circled the origin as they did exactly to avoid crossing that axis, even at its tip ($0$).
• Sorry if I was vague. I know that log(-x) is not necessarily undefined; however, in the text that I was reading they defined the branch of log in the integral specifically so that log(-x) is undefined when x is a positive real number. Furthermore, even if they did choose some other angle, the contour in the second integral is a circle about the origin, so some point of discontinuity occurs on the contour, which is precisely what I am confused about. I will edit the question to make this more clear. – arbitrary username Nov 24 '15 at 3:42
• They are a little sloppy with the notation, but if you read carefully the text you posted here, you see that they mean to only integrate on the upper half of the circle, passing from the positive real axis to the negative real axis. It doesn't even work to go all the way around, as that would leave them back on the positive axis, not the negative axis where the curve continues. And if as you claim that defined $\log(-x)$ to be undefined on the positive real axis, then the entire 3rd integral is undefined. Check again. The discontinuity is the negative imaginary axis. – Paul Sinclair Nov 24 '15 at 3:48
• The curve is supposed to return up the positive real axis. They require the first and third integrals to be taken "slightly above" and "slightly below" the positive real axis, which I take to mean they are integrating over the lines parameterized by f(t) = t ± i\delta, where \delta can be arbitrarily small. I understand your confusion about the contour, as that is precisely what I am confused about. – arbitrary username Nov 24 '15 at 3:56
• Now you are giving critical additional information: the integrals are not actually on the real line. Then yes, they can be undefined on the line itself. The contour around the origin is also off the line (note again, they describe it as only going halfway around, not all the way). so it never reaches the undefined place (though why it is even needed if they are not integrating along the real line, I don't know). – Paul Sinclair Nov 24 '15 at 4:22
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