In the multiple regression model, the goodness-of- t measure R-squared always increases (or remains the same) when an additional explanatory variable is added. Slides Prepared by JOHN S. We'll just use the term "regression analysis" for all these variations. While many statistical software packages can perform various types of nonparametric and robust regression. Orlov Chemistry Department, Oregon State University (1996) INTRODUCTION In modern science, regression analysis is a necessary part of virtually almost any data reduction process. wnarifin@usm. Note that the intercept will always be zero and so we could have used regression without an intercept to obtain the same regression coefficients (although the standard errors will be slightly different). A simple linear regression equation for this would be $$\hat{Price} = b_0 + b_1 * Mileage$$. Multiple regression is a flexible method of data analysis that may be appropriate whenever a quantitative variable (the dependent or criterion variable) is to be examined in relationship to any other factors (expressed as independent or predictor variables). i = be the value of. This PowerPoint is a workshop on multiple linear regression and analysis of covariance. Multiple Linear Regression Multiple linear regression attempts to model the relationship between two or more explanatory variables and a response variable by fitting a linear equation to observed data. Lecture 1 Introduction to Multi-level Models Regression Analysis? • Specification of predictor variables from multiple levels. The goal of a model is to get the smallest possible sum of squares and draw a line that comes closest to the data. When I started experimenting with machine learning, I wanted to come up with an application that would solve a real-world problem but would not be too complicated to implement. This statistics is for multiple linear regression technique. Once you have completed the test, click on 'Submit Answers' to get your results. | PowerPoint PPT presentation | free to view. Multiple Regression and Correlation Dr. Multiple regression generally explains the relationship between multiple independent or predictor variables and one dependent or criterion variable. Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression …\1. Objectives of Multiple Regression In selecting suitable applications of multiple regression, the researcher must consider three primary issues: 1. Secondary Data Analysis • Starting Off Right: Effects of Rurality on Parent‟s Involvement in Children‟s Early Learning (Sue Sheridan, PPO) – Data from the Early Childhood Longitudinal Study – Birth Cohort (ECLS-B) were used to examine the influence of setting on parental involvement in preschool and the effects of involvement on. Example: estimated coefficient of education equals 92. Dummy Variables Dummy Variables A dummy variable is a variable that takes on the value 1 or 0 Examples: male (= 1 if are male, 0 otherwise), south (= 1 if in the south, 0 otherwise), etc. Multiple regression analysis is almost the same as simple linear regression. “encoding model”. NASCAR Race Crashes SAS Program SAS Output. Multiple Regression Now, let’s move on to multiple regression. , the dependent variable) of a fictitious economy by using 2 independent/input variables:. Simple linear regression and multiple regression using least squares can be done in some spreadsheet applications and on some calculators. The example below demonstrates the use of the summary function on the two models created during this tutorial. One regressor should not be a linear function of another. This lesson explores the use of a regression analysis to answer. In the orange juice classification problem, Y can only take on two possible values: 0 or 1. As a data scientist, one must always explore multiple options for solving the same analysis or modeling task and choose the best for his/her particular problem. Understand and use bivariate and multiple linear regression analysis. In the following example, we will use multiple linear regression to predict the stock index price (i. patient, f. Chapter 9 Correlational Research Designs What are correlational research designs, and why are they used in behavioral research. Edward’s University Chapter 12 Simple Linear Regression Simple Linear Regression Model Least Squares Method Coefficient of Determination Model Assumptions Testing for Significance Using the Estimated Regression Equation for Estimation and Prediction Computer Solution Residual Analysis: Validating Model Assumptions Simple Linear Regression Model y = b0. Regression analysis helps in establishing a functional Relationship between two or more variables. Amaral November 21, 2017 Advanced Methods of Social Research (SOCI 420). Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression …\1. Introduction to Correlation and Regression Analysis. Simple regression analysis uses a single x variable for each dependent “y” variable. Multiple Linear Regression Model We consider the problem of regression when study variable depends on more than one explanatory or independent variables, called as multiple linear regression model. One will not encounter cases in a multiple regression path model where one could go through the same variable twice. The generic form of the linear regression model is y = x 1β 1 +x 2β 2 +. Multiple linear regression attempts to fit a regression line for a response variable using more than one explanatory variable. selection of the dependent and independent variables. The model states that the expected value of Y--in this case, the expected merit pay increase--equals β0 plus β1 times X. In ordinary samples we would expect 95% of cases to have standardized residuals within about +2 and -2. Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression …\1. If you click. y = c + ax c = constant a = slope. It can also handle multiple predictor variables. Multiple Linear Regression and Matrix Formulation Introduction I Regression analysis is a statistical technique used to describe relationships among variables. , the dependent variable) of a fictitious economy by using 2 independent/input variables:. This is a simplified tutorial with example codes in R. One type of analysis many practitioners struggle with is multiple regression analysis, particularly an analysis that aims to optimize a response by finding the best levels for different variables. Your print-out of the slides needs to be modified at one point. Use Multiple Regression to model the linear relationship between a continuous response and up to 12 continuous predictors and 1 categorical predictor. The simple proportional hazards model generalizes to a multiple regression model in much the same way as for linear and logistic regression. Posc/Uapp 816 Class 14 Multiple Regression With Categorical Data Page 3 1. Lecture 6:. Standard multiple regression is the same idea as simple linear regression, except now you have several independent variables predicting the dependent variable. If two variables are trending over time, a regression. Assessing Studies Based on Multiple Regression Part III. – The test is multiple linear regression isThe test is multiple linear regression is H 0 = β 1 = β 2 = … = β p = 00. Root MSE = s = our estimate of σ = 2. This is what regression analysis can do! For example, we'll see (via regression) that Fords. Regression analysis is a statistical process for estimating the relationships among variables. " In the main dialog box, input the dependent variable. txt) or view presentation slides online. Lecture 18: Multiple Logistic Regression Mulugeta Gebregziabher, Ph. Clearly, it is nothing but an extension of Simple linear regression. We expect to build a model that fits the data better than the simple linear regression model. Dummy variables are useful because they enable us to use a single regression equation to represent multiple groups. A multiple regression based model will use the data to build a function that predicts the outcome based on the independent variables. Technically, a regression analysis model is based on the sum of squares, which is a mathematical way to find the dispersion of data points. 8 Quantitative Business Analysis for Decision Making Multiple Linear Regression Analysis Outlines Multiple Regression Model Estimation Testing Significance of Predictors Multicollinearity Selection of Predictors Diagnostic Plots Multiple Regression. When you have more than two events, you ca n extend the binary logistic regression model, as described in Chapter 3. y = c + ax c = constant a = slope. Frank Wood, fwood@stat. Econometrics I: Multiple Regression: Inference - H. One of these variable is called predictor variable whose value is gathered through experiments. When you look at the output for this multiple regression, you see that the two predictor model does do significantly better than chance at predicting cyberloafing, F(2, 48) = 20. In each ex-ample, you will ﬁrst learn about the speciﬁc ingredi-ents required for the power or sample size computa-tion for the linear model being considered. It is also called the coefficient of determination, or the coefficient of multiple determination for multiple regression. Nonlinear regression The model is a nonlinear function of the parameters. University of Dayton. The program’s graph, regression, and correlation functions can respectively produce scatterplots, provide regression equation coefficients, and create correlation matrices. Assumptions of Multiple Regression This tutorial should be looked at in conjunction with the previous tutorial on Multiple Regression. When using multiple regression to estimate a relationship, there is always the possibility of correlation among the independent variables. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Further details and related logical operations can be found in the R documentation. Multiple Personality Disorder. A sound understanding of the multiple regression model will help you to understand these other applications. Switching Regression Models — Estimation (8) First obtain the expected values of the residuals that are truncated. We will also learn how to decide whether a group of explanatory variables can be excluded from the model. "Linear Regression and Modeling" is course 3 of 5 in the Statistics with R Coursera Specialization. In this paper, the risk factors for a disease of the eye (retinopathy of prematurity) are identi ed using logistic regression analysis. Multiple regression is a very advanced statistical too and it is extremely powerful when you are trying to develop a "model" for predicting a wide variety of outcomes. With only one independent variable, the regression line can be plotted neatly in two dimensions. However, performing a regression does not automatically give us a reliable relationship between the variables. xls Data Sample (showing only the variables to be used in analysis) Variables Used. Chapter 8: Multiple Choice Questions. Lecture 1 Introduction to Multi-level Models Regression Analysis? • Specification of predictor variables from multiple levels. Simple linear regression. Variable and Dummy Coded Region Variable 273. Chapter 3: Multiple regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) regression model to con-sider the possibility that there are additional explanatory factors that have a systematic ef-fect on the dependent variable. We are not going to go too far into multiple regression, it will only be a solid introduction. zeigler-hill. pdfs and a PowerPoint presentation (also found on the left side of the site), and have integrated SPSS procedures into the discussion. Regression analysis is a statistical process for estimating the relationships among variables. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Many people find this too complicated to understand. I also wanted to practice working with regression algorithms. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. MR&B3  is intended to offer a conceptually-oriented introduction to multiple regression (MR) and structural equation modeling (SEM), along with analyses that flow naturally from those methods. ppt - Free download as Powerpoint Presentation (. ppt from AA 1Multiple Regression (Part 3: Variables Selection) 1 Topic Outline Comparing Two Multiple Regression Models Partial F -Test Using Sequential Sum of. In the simplest case, we would use a 0,1 dummy variable where a person is given a value of 0 if they are in the control group or a 1 if they are in the treated group. We are not going to go too far into multiple regression, it will only be a solid introduction. – The test is multiple linear regression isThe test is multiple linear regression is H 0 = β 1 = β 2 = … = β p = 00. An Introduction to Logistic Regression: From Basic Concepts to Interpretation with Particular Attention to Nursing Domain ure" event (for example, death) during a follow-up period of observation. Wouldn’t it be great if there was a more accurate way to predict whether your prospect will buy rather than just taking an educated guess? Well, there is…if you have enough data on your previous prospects. This model is built using, for example, a set of real world data listing the outcome in various cases. Welcome to Part 1 of Regression & Classification - Simple Linear Regression: Step 1. Multiple Regression Multiple Proportional Hazards Regr (Cox model) For time dependent outcomes (ie time to death), we model the hazard rate, h , the event rate per unit time (for death, it is the mortality rate). How to Run a Multiple Regression in Excel. Wooldridge, Introductory Econometrics, 4th ed. If two variables are trending over time, a regression. Multiple regression is an extension of simple linear regression. Multiple linear regression has one y and two or more x variables. 8%, with 95% confidence. is a way of understanding the relationship between two variables. For instance, when we predict rent based on square feet alone that is simple linear regression. Stepwise multiple linear regression has proved to be an extremely useful computational technique in data analysis problems. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Background Multiple Linear Regression is widely used in academics and also in MR We can consider it the start of Multivariate Analysis, for our course Any idea what the following are: Multivariate analysis Multiple Linear Regression (MLR). The precision of a method (Table 4) is the extent to which the individual test results of multiple injections of a series of standards agree. For example, linear regression can be used to quantify the relative impacts of age, gender, and diet (the predictor variables) on height (the outcome variable). Regression assumes that variables have normal distributions. regress is useful when you simply need the output arguments of the function and when you want to repeat fitting a model multiple times in a loop. 3 Notes: PowerPoint xxx PDF. Suppose we have a cohort of. Lecture 24: Partial correlation, multiple regression, and correlation Ernesto F. Third, multiple regression offers our first glimpse into statistical models that use more than two quantitative. If you have the Excel desktop application, you can use the Open in Excel button to open your workbook and use either the Analysis ToolPak's Regression tool or statistical functions to perform a regression analysis there. Uses of Regression Analysis 1. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. 1 Introduction In this chapter we extend the simple linear regression model, and allow for any number of independent variables. You probably remember the concept of simple linear regression intuition from your high school years. Similar tests. Notes on Regression Model • It is VERY important to have theory before starting developing any regression model. Lecture 4: Multivariate Regression Model in Matrix Form In this lecture, we rewrite the multiple regression model in the matrix form. Partial r is just another way of standardizing the coefficient, along with beta coefficient (standardized regression coefficient)$^1$. selection of the dependent and independent variables. edu|http://www. Simple linear regression is a bivariate situation, that is, it involves two dimensions, one for the dependent variable Y and one for the independent variable x. ppt), PDF File (. Formula 13. This correlation may be pair-wise or multiple correlation. The linear regression model (LRM) The simple (or bivariate) LRM model is designed to study the relationship between a pair of variables that appear in a data set. Regression analysis is a very widely used statistical tool to establish a relationship model between two variables. , between an independent and a dependent variable or between two independent variables). There are two models of logistic regression, binary logistic regression and multinomial logistic regression. If the universe is only 14 billion years old, how can it be 92 billion light years wide? - Duration: 9:47. It is the correlation between the variable's values and the best predictions that can be computed linearly from the predictive variables. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. It can also be used to estimate the linear association between the predictors and reponses. pdfs and a PowerPoint presentation (also found on the left side of the site), and have integrated SPSS procedures into the discussion. Lecture 18: Multiple Logistic Regression - p. You can use it to predict values of the dependent variable, or if you're careful, you can use it for suggestions about which independent variables have a major effect on the dependent variable. Known also as curve fitting or line fitting because a regression analysis equation can be used in fitting a curve or line to. Intercept: the intercept in a multiple regression model is the mean for the response when. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. The data was split into three employment sectors Teaching, government and private industry Each sector showed a positive relationship Employer type was confounded with degree level Simpson’s Paradox In each of these examples, the bivariate analysis (cross-tabulation or correlation) gave misleading results Introducing another variable gave a. Here, the summary(OBJECT) function is a useful tool. +x K β K +ε where y is the dependent or. Upload and Share PowerPoint Presentations. Pantula David A. covariates for the. MV - Multiple Regression. The generic form of the linear regression model is y = x 1β 1 +x 2β 2 +. Typically points further than, say, three or four standard deviations from the mean are considered as "outliers". This post will show you examples of linear regression. Regression Analysis. patient, f. In this exercise we apply the same method to validating the Novy-Marx Quality and Value screener’s ability to predict the 5-year total return. Multiple Regression Analysis using SPSS Statistics Introduction. 01, with an R-square of. Briefly speaking, the goal of the multiple linear regression is to point out the relation between a dependent variable (explained, endogenous or resultative) and a great deal of independent variables (explanatory, factorial,. Comparing Multiple Regression Model Results against Historic Demand. However, it is critical to recognize that multiple regression is inherently a correlation technique and cannot. " In the main dialog box, input the dependent variable. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Rawlings Sastry G. Regression analysis is a statistical technique for estimating the relationship among variables which have reason and result relation. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. persistence of shocks will be infinite for nonstationary series • Spurious regressions. Multiple linear regression Situations frequently occur when we are interested in the dependency of a variable on several explanatory (independent) variables. If two variables are trending over time, a regression. RSM is a method used to locate the optimum value of the response and is one of the final stages of experimentation. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Suppose we have a cohort of. Cherry Blossoms WS: Test Review. ID: 305741 ID: 305741. Linear regression Linear dependence: constant rate of increase of one variable with respect to another (as opposed to, e. As a data scientist, one must always explore multiple options for solving the same analysis or modeling task and choose the best for his/her particular problem. 1 Introduction to Poisson Regression As usual, we start by introducing an example that will serve to illustrative regression models for count data. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. Multiple (Linear) Regression. This tutorial will explore how R can be used to perform multiple linear regression. Dummy variables are useful because they enable us to use a single regression equation to represent multiple groups. is a way of understanding the relationship between two variables. 1 Introduction In this chapter we extend the simple linear regression model, and allow for any number of independent variables. This article is a part of the guide:. Suppose that there is a cholesterol lowering drug that is tested through a clinical trial. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. We have been reviewing the relationship between correlation (r and r2) and regression (R, R2 and 1-R2) in class, through lectures, blog. Nonlinear Regression Functions Chapter 9. Regression Analysis. (3) Does the combination of predictors in this fitted multiple regression explain significant variation in the response? Statistics 621 Multiple Regression Practice Questions. Multiple Regression Analysis using SPSS Statistics Introduction. We now perform multiple linear regression to obtain the standardized regression coefficients shown in range J19:J21. Multiple linear regression Situations frequently occur when we are interested in the dependency of a variable on several explanatory (independent) variables. selection of the dependent and independent variables. Consider ﬁrst the case of a single binary predictor, where x = (1 if exposed to factor 0 if not;and y =. The output varies linearly based upon the input. Multiple regression uses the ordinary least squares solution (as does bi. The whole model F test (test of the useful of the model) tests whether the slopes on all variables in multiple regression are zero, i. Ryan McEwan and Julia Chapman. Simple linear regression involves a single independent variable. 12-1 Multiple Linear Regression Models • Many applications of regression analysis involve situations in which there are more than one regressor variable. Introduction to Linear Regression and Correlation Analysis Fall 2006 – Fundamentals of Business Statistics 2 Chapter Goals To understand the methods for displaying and describing relationship among variables. ppt from AA 1Multiple Regression (Part 3: Variables Selection) 1 Topic Outline Comparing Two Multiple Regression Models Partial F -Test Using Sequential Sum of. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. Chapter 8: Multiple Choice Questions. Here is image from Linear Regression, posted by Ralf Adam, on March 30, 2017, image size: 63kB, width: 1200, height: 1455, Least to Greatest, Where, Least Common. Regression with two or more predictors is called multiple regression Available in all statistical packages Just like correlation, if an explanatory variable is a significant predictor of the dependent variable, it doesn't imply that the explanatory variable is a cause of the dependent variable. It is discussed in Response Surface Methods. Hey I would like to make a scatter plot with p-value and r^2 included for a multiple linear regression. 0 Chapter 6: Multiple Linear Regression Topics Explanatory Modeling Predictive Modeling Example: Prices of Toyota Corolla ToyotaCorolla. Multiple regression is a very advanced statistical too and it is extremely powerful when you are trying to develop a "model" for predicting a wide variety of outcomes. Regression is the analysis of the relation between one variable and some other variable(s), assuming a linear relation. Also referred to as least squares regression and ordinary least squares (OLS). Regression when all explanatory variables are categorical is “analysis of variance”. 5 Correlation and Regression Simple regression 1. Notes on Regression Model • It is VERY important to have theory before starting developing any regression model. The Regression df is the number of independent variables in the model. We can have only two models or more than three models depending on research questions. For example, consider the cubic polynomial model which is a multiple linear regression model with three regressor variables. Multiple Regression 18. X2 1 or even interactions X1 X2. The following model is a multiple linear regression model with two predictor variables, and. Find PowerPoint Presentations and Slides using the power of XPowerPoint. This post will show you examples of linear regression. Please access that tutorial now, if you havent already. The basic command is "regression": "linear. Chapter 3: Multiple regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) regression model to con-sider the possibility that there are additional explanatory factors that have a systematic ef-fect on the dependent variable. Objectives. MULTIPLE LINEAR REGRESSION ANALYSIS USING MICROSOFT EXCEL by Michael L. Here is image from Linear Regression, posted by Ralf Adam, on March 30, 2017, image size: 63kB, width: 1200, height: 1455, Least to Greatest, Where, Least Common. Frank Wood, fwood@stat. 1 Introduction. " Coefficient table, bottom. Under Type of power analysis, choose ‘A priori…’, which will be used to identify the sample size required given the alpha level, power, number of predictors and effect size. View and Download PowerPoint Presentations on Spss Tutorial For Multiple Logistic Regression PPT. One of the applications of multiple linear regression models is Response Surface Methodology (RSM). State the multiple regression equation. For ordina l categorical variables, the drawback of the. Linear, Ridge Regression, and Principal Component Analysis Geometric Interpretation I Each column of X is a vector in an N-dimensional space (NOT the p-dimensional feature vector space). Linear Regression. The model is linear because it is linear in the parameters , and. We now perform multiple linear regression to obtain the standardized regression coefficients shown in range J19:J21. In R, multiple linear regression is only a small step away from simple linear regression. Regression Analysis. The basic command is “regression”: “linear. Multiple linear regression in R Dependent variable: Continuous (scale/interval/ratio) Independent variables: Continuous (scale/interval/ratio) or binary (e. y = c + ax c = constant a = slope. 8%, with 95% confidence. We can run regressions on multiple different DVs and compare the results for each DV. Multiple Regression Models • Advantages of multiple regression • Important preliminary analyses • Parts of a multiple regression model & interpretation • Differences between r, bivariate b, multivariate b & • Steps in examining & interpreting a full regression model Advantages of Multiple Regression Practical issues …. See the Handbook for information on these topics. Multiple Regression - Linearity. ppt - Free download as Powerpoint Presentation (. This correlation may be pair-wise or multiple correlation. Arial Franklin Gothic Book Perpetua Wingdings 2 Calibri Equity 1_Equity 2_Equity 3_Equity 4_Equity Microsoft Equation 3. Be sure to tackle the exercise and the quiz to get a good understanding. One type of analysis many practitioners struggle with is multiple regression analysis, particularly an analysis that aims to optimize a response by finding the best levels for different variables. 2% with 95% confidence. Lecture 18: Multiple Logistic Regression - p. The measured standard deviation can be subdivided into 3 categories: repeatability, intermediate precision and reproducibility (4, 5). Find PowerPoint Presentations and Slides using the power of XPowerPoint. By selecting the features like this and applying the linear regression algorithms you can do polynomial linear regression; Remember, feature scaling becomes even more important here; Instead of a conventional polynomial you could do variable ^(1/something) - i. MR&B3  is intended to offer a conceptually-oriented introduction to multiple regression (MR) and structural equation modeling (SEM), along with analyses that flow naturally from those methods. Introduction to Correlation and Regression Analysis. It is used to also to determine the overall fit of the model and the contribution of each of the predictors to the total variation. Uses of Regression Analysis 1. Variable Interactions 5. 11 LOGISTIC REGRESSION - INTERPRETING PARAMETERS 11 Logistic Regression - Interpreting Parameters Let us expand on the material in the last section, trying to make sure we understand the logistic regression model and can interpret Stata output. Figure 14 – Model Summary Output for Multiple Regression. Starting values of the estimated parameters are used and the likelihood that the sample came from a population with those parameters is computed. On the contrary, regression is used to fit a best line and estimate one variable on the basis of another variable. For simple linear regression (only one x), the Regression df is 1. Source: Hair et al. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. Predictions in neurophysiology. Nonlinear regression The model is a nonlinear function of the parameters. The only difference between simple linear regression and multiple regression is in the number of predictors (“x” variables) used in the regression. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. 12-1 Multiple Linear Regression Models • Many applications of regression analysis involve situations in which there are more than one regressor variable. specification of a statistical relationship, and 3. Upload and Share PowerPoint Presentations. Welcome to Part 1 of Regression & Classification - Simple Linear Regression: Step 1. Popular spreadsheet programs, such as Quattro Pro, Microsoft Excel,. biostatcourse. Teaching\stata\stata version 13 – SPRING 2015\stata v 13 first session. Under Type of power analysis, choose ‘A priori…’, which will be used to identify the sample size required given the alpha level, power, number of predictors and effect size. How to improve leadership skills in the workplace ppt for write my essay nursing best cover letter editing site uk , writing a scientific report buy a college essay online. Logistic regression is a statistical method for analyzing a dataset in which there are one or more independent variables that determine an outcome. Regression with categorical variables and one numerical X is often called “analysis of covariance”. It could be, for example, performance, education, or experience. The course starts with a discussion of the logic of the multivariate regression model and the central assumptions underlying the ordinary least squares approach. Also referred to as least squares regression and ordinary least squares (OLS). Linear regression for the advertising data Consider the advertising data shown on the next slide. The method of multiple regression sought to create the most closely related model. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Multiple regression is used to build a model that allows us to study this interplay. What to report? What a statistics program gives you: For a simple regression (one independent variable), statistics programs produce two estimates, a (the "constant term") and b (the "linear coefficient"), for the parameters α and β, respectively. For example: (x 1, Y 1). 2% with 95% confidence.


HP 11C real root finder [Newton Method]
01-09-2014, 11:59 PM
Post: #1
 Carlos CM (Mexico) Junior Member Posts: 35 Joined: Dec 2013
HP 11C real root finder [Newton Method]
LBL 0
RCL 1
GSB 1
RCL 1
GSB 2
/
CHS
RCL 1
+
STO 3
RCL 1
-
ABS
RCL 2
X>Y?
RTN
RCL 3
STO 1
GTO 0

LBL 1
f(x)=0 code

LBL 2
f'(x)=0 code

R1: old value (or initial value)
R2: Tolerance of error (0.001, 0.0001, 0.000000001 etc)

THE RESULT:
R3: new value (root)

EXAMPLE

Find the root of f(x)=x^3 - 3x^2-6x+8

rewritten as f(x)= [(x-3)x-6]x +8

f(x) code:

LBL1
ENTER
ENTER
ENTER
3
-
*
6
-
*
8
+
RTN

f'(x)= 3x(x-6)-6
f'(x) code

LBL 2
ENTER
ENTER
6
-
*
3
*
6
-
RTN

The roots are [-2, 1, 4]

give a initial value, for example: -3

-3 STO 1

give a tolerance of error, for example: 0.0001

0.0001 STO 2

BEGIN THE PROGRAM...

GSB 0

runnning...

when the error < TOL the program stop and display:

0.0001

you can find the root in the register 3

RCL 3

DISPLAY: -2.0001

try with initial value = 2

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Best Regards
01-10-2014, 12:57 AM
Post: #2
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Using storage arithmetic allows us to get rid of a few steps:

LBL 0
RCL 1
GSB 1
RCL 1
GSB 2
/
STO- 1
ABS
RCL 2
X<=Y?
GTO 0
RCL 1
RTN

You made a mistake calculating the 1st derivative.
This is the corrected program:

f'(x)= 3x^2 - 6x - 6 = (3x - 6)x - 6
f'(x) code

LBL 2
ENTER
ENTER
3
*
6
-
*
6
-
RTN

Kind regards
Thomas
01-10-2014, 05:28 PM
Post: #3
 Carlos CM (Mexico) Junior Member Posts: 35 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]

Best regards

Best Regards
01-12-2014, 08:31 AM
Post: #4
 Namir Senior Member Posts: 688 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Here is a version that requires coding f(x) only since it approximate f'(x) as:

f'(x) = (f(x+h) - f(x))/h

Where h = 0.001*(ABS(X)+1)

The new version uses registers R1 through R4.

Code:
LBL 0 RCL 1 ABS 1 + EEX 3 CHS * STO 3            # calculate and store increment h RCL 1 GSB 1 STO 4            # calculate and store f(x) RCL 1 RCL 3 + GSB 1            # calculate f(x+h) RCL 4 - 1/X RCL 4 * RCL 3 *                    # calculate diff = h *f(x)/(f(x+h) - f(x)) STO- 1 ABS RCL 2 X<=Y? GTO 0 RCL 1 RTN
01-12-2014, 01:26 PM
Post: #5
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
(01-12-2014 08:31 AM)Namir Wrote:  ...h = 0.001*(ABS(X)+1)

First of all, instead of multiplying with $$10^{-3}$$, dividing by $$10^{3}$$ is one step shorter. ;-)

This method for determining h will work in most cases, but not for very small arguments. Consider $$x = 10^{-4}$$ or even $$x = 10^{-40}$$. That's why I prefer $$h = x/10^4$$. On the 34s, the result can be easily rounded to 1 or 2 significant digits (RSD 1) to prevent slight roundoff errors. As usual, $$x=0$$ is handled as $$x=1$$.

Dieter
01-14-2014, 08:56 PM
Post: #6
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
If you only want to solve polynomials with real coefficients Bairstow's Method can be used.
In a nutshell: instead of a root a quadratic factor is found in each iteration. To find the roots of this factor the classic formula is used.

Example:

$$x^3-3x^2-6x+8=0$$

First enter the coefficients of the polynomial. Always start with register 9:

1 STO 9
-3 STO .0
-6 STO .1
8 STO .2

Then specify the order of the polynomial with a loop-control value. It defines the registers you used for the coefficients:

9.012 STO 6

And now give an initial guess. Probably {1, 1} will do in all cases:

1 STO 7
STO 8

Start the program B:

GSB B

Now the coefficients of the quadratic factor can be found in registers 7 and 8:

RCL 7
1.000000

RCL 8
-2.000000

The coefficient of the first term is always 1. Thus the factor is $$x^2+x-2$$.
Now we can solve this quadratic equation using program A:

1
RCL 7
RCL 8
GSB A

1.000000
X<>Y
-2.000000

Thus: $$x^2+x-2=(x-1)(x+2)$$.

What is left? Let's have a look at it:

RCL 6
9.010

RCL 9
1.000000

RCL .0
-4.000000

This is a linear factor. Thus we end up with the following factorization: $$x^3-3x^2-6x+8=(x-1)(x+2)(x-4)$$
Therefore the solutions are: {1, -2, 4}

Example from Bunuel66's solution to the crossed ladders problem

$$C^4-30C^3+700C^2-21,000C+157,500=0$$

Enter the coefficients of the polynomial:

1 STO 9
-30 STO .0
700 STO .1
21,000 STO .2
157,500 STO .3

Specify the loop-control value:

9.013 STO 6

Use the initial guess {1, 1}:

1 STO 7
STO 8

Start the program:

GSB B

The coefficients of the quadratic factor are in registers 7 and 8:

RCL 7
-35.178025

RCL 8
248.596895

Solve this quadratic equation using program A:

1
RCL 7
RCL 8
GSB A

25.384938
X<>Y
9.793087

RCL 6
9.011

RCL 9
1.000000

RCL .0
5.178025

RCL .1
633.555781

If we try to solve this quadratic equation we get an Error 0.
That's because we're trying to calculate the square root of a negative value.
However we can still get the desired result:

RCL 9
RCL .0
RCL .1
GSB B
Error 0

<-
CHS
626.852796

$$\sqrt{x}$$
25.037029

X<>Y
-2.589012

Thus the final list of solutions is:
• 25.384938
• 9.793087
• -2.589012$$\pm$$25.037029i

Cheers
Thomas
01-15-2014, 05:53 AM
Post: #7
 Namir Senior Member Posts: 688 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
(01-12-2014 01:26 PM)Dieter Wrote:
(01-12-2014 08:31 AM)Namir Wrote:  ...h = 0.001*(ABS(X)+1)

First of all, instead of multiplying with $$10^{-3}$$, dividing by $$10^{3}$$ is one step shorter. ;-)

This method for determining h will work in most cases, but not for very small arguments. Consider $$x = 10^{-4}$$ or even $$x = 10^{-40}$$. That's why I prefer $$h = x/10^4$$. On the 34s, the result can be easily rounded to 1 or 2 significant digits (RSD 1) to prevent slight roundoff errors. As usual, $$x=0$$ is handled as $$x=1$$.

Dieter

Sure, dividing by 1000 is a step shorter. Originially, I had learned to calculate h using an If statement:

Code:
If |x| >= 1 Then   h= x/100 else   h = 1/100 end if

Until I realized one day that .01*(|x|+1) does the job and eliminates the need for labels and GOTOs. I recently started using .001 instead of 0.01. Using the expression for h ensures that if x=0, h is not zero.

Namir
01-15-2014, 08:38 PM
Post: #8
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Namir, sorry that I was not able to express myself clearly. I wanted to point out that I do not think it's a good idea to use h = 0,01 or 0,001 for any value below 1. This may lead to significant errors since h may be much, much larger than x. For instance, if x = 1E-10, h = 1E-3 is not recommended. Here, h = 1E-13 should be better.

That's why I prefer to set h = 0,001 x or similar. With the only exception x=0 where x may be 0,001.

(01-15-2014 05:53 AM)Namir Wrote:  Until I realized one day that .01*(|x|+1) does the job and eliminates the need for labels and GOTOs. I recently started using .001 instead of 0.01. Using the expression for h ensures that if x=0, h is not zero.

This can be coded this way, for instance:
Code:
 X=0?  e^x  EEX   3   /
Look, no labels or gotos required. ;-)

Or even more elegant on the 34s:
Code:
 X=0?  INC X  SDR 3  RSD 1   ' optional

Dieter
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# Basic Taylor Polynomial Question involving e^(-x)^2

1. Nov 15, 2014

### RJLiberator

1. The problem statement, all variables and given/known data
Consider:

F(x) =$\int_0^x e^{-x^2} \, dx$
Find the Taylor polynomial p3(x) for the function F(x) centered at a = 0.

2. Relevant equations
Tabulated Taylor polynomial value for standard e^x

3. The attempt at a solution

I started out by using the tabulated value for Taylor polynomial e^x and replaced all x with -x^2.
Going to the degree of three I received:

e^(-x)^2 = 1 - (x^2)/1! + x^4/2! - x^6/3!

Is this all there is to it? The question simply states to find the Taylor Polynomial p3(x) for the function, nothing else. I don't need to find the sum or the remainder per the problems question. Correct?

If I were to go at the sum, would I integrate each value?

Thank you for any help, I think I am just getting used to the wording here more than anything.

2. Nov 15, 2014

### gopher_p

The Taylor polynomial $p_3$ of degree $3$ for $F$ centered at $0$ should look something like $$p_3(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\frac{F'''(0)}{6}x^3,$$ yes? If you remember the Fundamental Theorem of Calculus and what is says about computing $F'(x)$, it shouldn't be too hard to get the coefficients, right? So you can just do the problem "properly" without worrying whether your methods work.

I'm not saying your way won't work, though you'd theoretically only need the degree $2$ Taylor polynomial for $f(x)=e^{-x^2}$ to extract the degree $3$ polynomial for $F$. But the question you'd want to ask is whether the derivative of the degree $n$ Taylor polynomial of a function $g$ really is the degree $n-1$ Taylor polynomial for the function $g'$ and vice -versa; i.e. is the (anti)derivative of the Taylor polynomial the Taylor polynomial of the (anti)derivative? I don't think that question is too incredibly difficult to answer for those who understand what it is asking.

3. Nov 17, 2014

### RJLiberator

Hm. Bare with me. I am in the early process of understanding Taylor Series wordage:
So what you are saying is that I should take the derivative of e^(-x)^2 a few times to acquire the coefficients? If I do this I answer:
f(0) = 1
f'(0) = 1
f''(0) = 2
f'''(0) = 4
Which would mean my original equation is incorrect. I am not sure what you mean by doing the problem properly without worrying.

I guess my first question would be: Is my answer correct? All I did was input -x^2 to the e^x common taylor series polynomial expansion for the first 3 degrees.

Then my second question would be how to proceed with this problem 'properly' ? I will review the fundamental theorem of calculus.

Question 3: Out or curiosity, why would I need to only go to degree 2 for this problem? Doesn't it state to go to degree 3?

4. Nov 17, 2014

### Staff: Mentor

1. Yes, what you did looks fine.
2. Integrate the series you found.
3. If you have terms up to degree 2, when you integrate, what will be the highest degree you get?

5. Nov 17, 2014

### RJLiberator

Mark44, you are shedding light on this topic!

1. Thank you for your confirmation.
2. Hm, ok. I am going to mess around with a few things. I know my coefficients then, now I got to find where to plug them in and integrate.
3. Ah.. integrating would bring it to degree three.

6. Nov 17, 2014

### gopher_p

The way that I was suggesting you do the problem is as follows;

For $F(x)=\int_0^xe^{-t^2}\ \mathrm{d}t$,

$F(0)=\int_0^0e^{-t^2}\ \mathrm{d}t=0$

$F'(x)=e^{-x^2}\Rightarrow F'(0)=1$, where the derivative is given by the FTC.

$F''(x)=-2xe^{-x^2}\Rightarrow F''(0)=0$

$F'''(x)=-2e^{-x^2}+4x^2e^{-x^2}\Rightarrow F'''(0)=-2$

Then the Taylor polynomial of degree three centered at $a=0$ is given by $$p_3(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\frac{F'''(0)}{6}x^3=x-\frac{1}{3}x^3$$

So we're just (a) appealing to the definition of the Taylor polynomial and (b) computing the coefficients directly. I guess that's what I meant by "properly".

7. Nov 18, 2014

### RJLiberator

OK, so with your help, I've been working through this problem.

Question 1: So for F(0) you made the bounds 0 to 0. I don't understand why. I see that the answer is clearly 0 when the bounds are both 0 to 0, but why do this?

Other then this part, everything you wrote now makes complete sense to me.

Question 2: In my original solution where I just inputted -x^2 to the e^x tabulated value i had NOT integrated yet. Did I HAVE to integrate after I replaced those values? If so, the two answers match up nicely. I would imagine so, by the definition of the Taylor Series, correct?

8. Nov 18, 2014

### gopher_p

One of the "skills" that you learned in you preparation for calculus was that, given a formula for $F(x)$, you compute $F(a)$ by plugging in $a$ for $x$. So given $F(x)=\int_0^xe^{-t^2}\ \mathrm{d}t$, get $F(0)$ by plugging in $0$ for $x$; $F(0)=\int_0^0e^{-t^2}\ \mathrm{d}t$. This is why it is important to use a different variable of integration - $t$ here - inside the integral.

I'm confused by the question. It's clear that the degree two Taylor polynomial $\tilde p_2$ for $f(x)=e^{-x^2}$ is (a) not a degree three polynomial and (b) not a Taylor polynomial for the function $F$, right? So something must be done if we are to extract $p_3$ from $\tilde p_2$.

The real question, in my mind, is whether it is true that $p_3(x)=\int_0^x\tilde p_2(t)\ \mathrm{d}t$. In this case it is. It works here because the lower limit of integration, $0$, is the same as the center of the Taylor polynomials. You need to be careful in a more general setting.

9. Nov 18, 2014

### PeroK

You should note that where you wrote:

F(x) =$\int_0^x e^{-x^2} \, dx$

gopher_p wrote:

F(x) =$\int_0^x e^{-t^2} \, dt$

I think you need to understand the difference and why the second one correctly defines F as a function of x, with t as the dummy variable.

This may also be the reason why you did not understand the substitution $x = 0$

10. Nov 18, 2014

### RJLiberator

Guys, thanks for your help here.

You were a help to me and my classmates.

Also, thanks for pointing out my subtle mistakes that cause errors in conceptual understanding. I am starting to understand this problem due to this.

I see why the dummy variable matters.

As always, excellent help.


# Find the function $f(x)$ if $f(x+2f(y))=f(x)+y+f(y)$

Let $f:\mathbb{R}\to \mathbb{R}$ such $f(x)$ at $x=0$ continuous, and for any $x,y\in \mathbb{R}$ such $$f(x+2f(y))=f(x)+y+f(y)$$ Find $f(x)$.

Let $x=0,y=0$ then we have $$2f(2f(0))=f(0)++f(0)$$ Let $y=2f(0)$ then we have $$f(x+2f(2f(0))=f(x)+2f(0)+f(2f(0))$$ so we have $$f(x+4f(0))=f(x)+4f(0)$$ on the other hand we have $$f(x+4f(0))=f(x+2f(0))+0+f(0)=2f(0)+f(x)$$ so we have $f(0)=0$ Then I can't deal with this problem

• Maybe I am off base but what happens if you plug in $y=x$? Sep 23, 2016 at 5:29
• @CameronWilliams you get that $x+2f(x)$ is a fixed point. Sep 23, 2016 at 5:30
• Am I missing something or does $f(x)=x$ work? Sep 23, 2016 at 6:21
• @Mastrem It does, but presumably the question is to find all such functions. Sep 23, 2016 at 6:22
• @Mastrem it works, but no one says this equation has only one solution. You have to prove then that it's the only one. Sep 23, 2016 at 6:23

Substitute $x\mapsto y$ to get $$f(y+2f(y))=y+2f(y).$$ Now substitute $y\mapsto y+2f(y)$: $$f(x+2f(y+2f(y)))=f(x)+y+2f(y)+f(y+2f(y)).$$ Thus $$f(x+2y+4f(y))=f(x)+2y+4f(y).$$ Substituting $x\mapsto x+2y+2f(y)$, $$f(x+2y+4f(y))=f(x+2y+2f(y))+y+f(y).$$ Substituting $x\mapsto x+2y$, $$f(x+2y+2f(y))=f(x+2y)+y+f(y).$$ Combining the last 3 equations, $$f(x+2y)=f(x)+2f(y).$$ Thus $$f(x)=f(0+2x/2)=f(0)+2f(x/2)=2f(x/2),$$ so $$f(x+y)=f(x+2y/2)=f(x)+2f(y/2)=f(x)+f(y).$$ Since you are given that $f$ is continuous at $0$, it must be continuous everywhere. Now there is a standard argument to show that $f(x)=f(1)x$ (prove it on rationals and extend by continuity). Finally using the original equation, we see $f(1)=1$ or $-\frac12$, so the solutions are $f(x)=x$ or $f(x)=-x/2$.

Setting $x=y$ gives $f(x+2f(x))=x+2f(x)$, and so $f(y)=y$ whenever $y$ is of the form $x+2f(x)$ for some $x$. Note also that if $f(y)=y$, then for any $x$, $f(x+2y)=f(x)+2y$.

Now consider the function $g(x)=f(x)-x$. From the previous paragraph, we see that if $f(y)=y$ then $g(x+2y)=g(x)$. In particular, every number of the form $2x+4f(x)$ is a period of $g$. Let $P\subseteq\mathbb{R}$ be the group of periods of $g$, i.e. the set of $p$ such that $g(x)=g(x+p)$ for all $x$. There are now two cases.

The first case is that $P$ is not cyclic. Then $P$ is dense in $\mathbb{R}$, and so each coset of $P$ is dense in $\mathbb{R}$, and in particular $0$ is in the closure of each coset. But by definition of $P$, $g$ is constant on each coset of $P$, and so by continuity of $g$ at $0$ this constant value must be equal to $g(0)$. Thus $g(x)=g(0)$ for all $x$, and $g$ is constant everywhere. Thus $f(x)$ has the form $f(x)=x+c$ for some constant $c$. Plugging this into the functional equation gives $$x+2(y+c)+c=x+c+y+y+c,$$ so $c=0$. Thus $f(x)=x$.

The second case is that $P$ is cyclic, generated by some $p\in\mathbb{R}$ (possibly $0$). Then every number of the form $2x+4f(x)$ is an integer multiple of $p$. But $2x+4f(x)$ is continuous at $0$, so this means $2x+4f(x)$ is constant in a neighborhood of $0$. As you've shown, $f(0)=0$, so $2x+4f(x)=0$ for all $x$ in some neighborhood of $0$, so $f(x)=-x/2$ for all $x$ in some neighborhood of $0$.

Now suppose $\epsilon>0$ is such that $f(x)=-x/2$ whenever $|x|\leq\epsilon$. If $0\leq a\leq \epsilon$, the functional equation with $x=a$ and $y=-\epsilon$ then gives $$f(a+\epsilon)=-\frac{a+\epsilon}{2}.$$ That is, $f(x)=-x/2$ is also valid if $\epsilon\leq x\leq 2\epsilon$. Similarly, using $y=\epsilon$ gives that $f(x)=-x/2$ is also valid if $-2\epsilon\leq x\leq -\epsilon$.

That is, if $f(x)=-x/2$ whenever $|x|\leq\epsilon$, $f(x)=-x/2$ whenever $|x|\leq 2\epsilon$ as well. Repeating this argument over and over, we get that $f(x)=-x/2$ for all $x$.

Thus the only possibilities for $f$ are $f(x)=x$ and $f(x)=-x/2$, and you can easily check that both of these work.

• Nice try! Even if using machinery which .. outguns the problem! I downvoted by accident.(and I thought I had already cancelled it. Seems I hadn't!) I will make a trivial edit so that I'll be able to cancel it. P.S.: the part where you are arguing about the cosets of the group of periods is not very clear. Could you expand it a little? (what is the equivalence relation here? why does the continuity of $g$ implies that every coset accumulates at $0$ ?). Sep 23, 2016 at 18:35
• I've clarified that part a bit. Sep 23, 2016 at 19:12
• Thanks for your response. As I've already said: very nice ! +1 Sep 25, 2016 at 19:49


Geometry - Prove a right triangle resulting from three inscribed circles

Two half circles and a full circle fit inside a larger quarter circle as shown in the diagram. The centers of the two half circles are on the two sides of the quarter circle, respectively.

Prove that the triangle formed by the centers of three smaller circles, $$\triangle O_1O_2O_3$$, is a right triangle.

I was able to apply the Pythagorean formula to a few triangles involving the radii of the inscribed circles and proved that the ratios of the three radii are 1:2:3. Then, the centers of the three circles form a triangle with side-length ratios 3:4:5, hence, a right triangle.

On the other hand, I feel the proof may be an overkill, and evaluating the three radii explicitly may be unnecessary. There ought to be clean geometric solutions to show directly that the vertex $$O_3$$ is of a right angle, which I am not sure how to figure out.

• Suppose $OO_1$ and $OO_2$ are x and y axis. If you complete half and quarter circles, you will have a big circle containing two big circles, two medium and four little circles. the center of little circles are symmetric about x and y axis. since x and y axis are perpendicular then the lines connecting the centers of little circles are perpendicular that is triangle $O_1O_2O_3$ is right angle at $O_3$. – sirous Sep 14 '19 at 21:01
• Insightful. A good case of can’t-see-forest-staring at trees. Thanks – Quanto Sep 14 '19 at 21:10
• @sirous I finally read your comment. I think it would be clearer if you "complete" the half and quarter circles by reflecting the figure over one axis and then the other. But your construction is much more elegant than mine and would make a fine answer. – David K Sep 15 '19 at 0:30
• @sirous: Certainly, "the lines connecting the centers of the little circles are perpendicular", but how do you know that those lines contain $O_1$ and $O_2$? – Blue Sep 15 '19 at 3:51
• @Blue - critical observation. guess a simple geometric proof remains elusive – Quanto Sep 15 '19 at 14:42

Given the quarter circle with $$O$$ as the center of the arc, and given semicircles with centers $$O_1$$ and $$O_2$$ on the two straight sides of the quarter circle such that $$O_1$$ and $$O_2$$ both are tangent to the arc of the quarter circle and such that $$O_1$$ and $$O_2$$ are tangent to each other.

This is based on the figure in the question, except that we do not assume that $$O_1$$ is the midpoint of one side of the quarter circle.

Let the radii of the quarter circle, the semicircle about $$O_1$$, and the semicircle about $$O_2$$ be $$r,$$ $$r_1,$$ and $$r_2$$ respectively.

Let $$P$$ be the fourth vertex of the rectangle that has three of its vertices at $$O,$$ $$O_1,$$ and $$O_2.$$ The sides of this rectangle are $$OO_1 = r - r_1$$ and $$OO_2 = r - r_2.$$ The diagonals are $$OP = O_1O_2 = r_1 + r_2.$$

Let $$P_1$$ be the point of intersection of the semicircle about $$O_1$$ with $$O_1P$$ and $$P_2$$ be the point of intersection of the semicircle about $$O_2$$ with $$O_2P.$$ Then $$PP_1 = r - r_1 - r_2 = PP_2.$$ Moreover, the distance from $$P$$ to the arc of the quarter circle is $$r - (r_1 + r_2) = r - r_1 - r_2.$$

Then a circle of radius $$r - r_1 - r_2$$ about $$P$$ is tangent to the quarter circle and to the semicircles about $$O_1$$ and $$O_2.$$ Therefore $$P$$ is $$O_3$$ in the figure, the center of the circle inscribed between the two semicircles and the quarter circle.

Since $$O_3$$ is a vertex of the rectangle $$OO_1O_3O_2,$$ the triangle $$\triangle O_1O_3O_2$$ is a right triangle. $$\square$$

• +1 a nice generalization. – achille hui Sep 15 '19 at 16:27

Let $$R=6a$$ then $$r_1=3a$$

$$\triangle OO_1O_2: \;\;\;\; (R-r_2)^2+r_1^2 =(r_1+r_2)^2$$

So $$\boxed{r_2= 2a}$$

So $$\triangle O_3O_1O_2$$ is right iff $$(r_1+r_2)^2 = (r_1+r_3)^2+(r_2+r_3)^2$$

or $$r_1r_2 = r_3(r_1+r_2+r_3)$$

or $$6a^2 = r_3(5a+r_3)\iff r_3^2+5ar_3-6a^2 =0$$

So we have to prove: $$\boxed{\color{red}{r_3=a}}$$

If we put everything in coordinate system then we have: $$O(0,0)$$, $$O_1(3a,0)$$, $$O_2(0,4a)$$ and let $$O_3(m,n)$$, so $$\begin{eqnarray} m^2+n^2 &=& (6a-r_3)^2\;\;\;\;\;\; (OO_3 = R-r_3)\\ (m-3a)^2+n^2 &=& (3a+r_3)^2\;\;\;\;\;\; (O_1O_3 = r_1+r_3)\\ m^2+(n-4a)^2 &=& (2a+r_3)^2\;\;\;\;\;\; (O_2O_3 = r_2+r_3) \end{eqnarray}$$

If we substract 1.st and 2.nd equation we get $$m = 6a-3r_3$$ and if we substract 1.st and 3.rd equation we get $$n=6a-2r_3$$

Now put this again in 1.st equation and we get $$\boxed{\color{green}{r_3=a}}$$

Following is an answer based on circle inversion w/o computing the radius of circle centered at $$O_3$$.

Label the contact points as illustrated below. We are going to show $$\angle OYD = \angle OXC = 45^\circ$$ Form these, it is easy to deduce $$\begin{cases} O_2O_3 \parallel O_2D \parallel OX,\\ O_1O_3 \parallel O_1C \parallel OY\end{cases}$$ and hence $$\angle O_1O_3O_2 = 90^\circ$$

Choose a coordinate system so that $$O$$ is the origin and $$X = (1,0)$$, $$Y = (0,1)$$. The arc $$XEY$$ will be part of the unit circle. Under circle inversion with respect to the unit circle centered at $$O$$, we get what illustrated below:

• The red semicircle centered at $$O_1$$ get mapped to a ray $$y = 1$$ in the upper half plane.
• The green semicircle centered at $$O_2$$ get mapped to another semicircle centered at some point of $$y$$-axis and touching the red line. This means this inverted green semicircle has radius $$1$$.
As a result, the inverted image of point $$A$$ is $$A' = (0,3) \implies A = (0,\frac13)$$.

• The blue circle get mapped to a circle placed symmetrically between the inverted green semicircle and the arc $$XEY$$. As a result, the inverted image of point $$C$$ is $$C' = (1,1)$$.
This implies $$\angle OXC = \angle OXC' = 45^\circ$$.

For the other angle, translate the coordinate axis so that $$Y$$ is the new origin. The new coordinate of $$A$$ is now $$(0,-\frac23)$$. Under circle inversion with respect to the unit circle centered at $$Y$$, it now looks like:

• The arc $$XEY$$ get mapped to a ray on the line $$y = -\frac12$$.
• The green semicircle centered at $$O_2$$ get mapped to a ray on the line $$y = -\frac32$$.
• The red semicircle get mapped to an arc on a circle sandwiched between above two rays touching the $$y$$-axis. So its radius is $$\frac12$$.
• The blue circle get mapped to an circle also sandwiched between above two rays. This means the inverted image of $$D$$ is located at $$D'' = (\frac32,-\frac32)$$. As a result, $$\angle OYD = \angle OYD' = 45^\circ$$.

Let $$OO_1$$ and $$OO_2$$ are x and y axis respectively. Mirror the figure about y axis, then mirror the resulted figure about x axis. You get two big, two medium and four small circles indescribable in a large circle. If you connect the centers of four small circles you get a square. So angle at $$O_3$$ is $$90^o$$. Now let the radius of small circle is $$r_1=1$$ and that of large circle is $$r_2=6$$. Suppose the location of center of small circle is $$O_3(3, 4)$$ which satisfies:

$$3^2+4^2= [OO_3=6-1=5]^2$$

Then we can claim that the radius of medium circle is $$r_3=3-1=2$$ and those of big and large circles are $$r_4=4-1=3$$ and $$r_2=2\times 3=6$$ respectively that is all involved circle are mutually tangent.If radius of medium circle is $$r_3=2$$ that means that it's center locates on a line passing $$O_3$$ which has a distance=4 from x axis. similarly if the radius of big circle is $$r_4=3$$ that would mean that it's center locates on a line passing $$O_3$$ which has distance =3 from y axis.That is triangle $$O_1O_2O_3$$ is right triangle at $$O_3$$


# Find the probability that, on Tuesday, it does not rain in Greg's area and he does not get a quiz.

The probability of rain in Greg's area on Tuesday is $0.3$. The probability that Greg's teacher will give him a pop quiz in Tuesday is $0.2$. The events occur independently of each other. What is the probability of neither events occur?

My approach:

Probability of rain or quiz or both = $0.3+0.2= 0.5$

So, probability of neither= $1-0.5$ = $0.5$

Question: Actual probability of this problem is $0.7\cdot0.8$ = $0.56$. But, I don't understand what is the mistake in above approach?

• They are independent not mutually exclusive. For mutually exclusive events you have $Pr(A\text{or}B)=Pr(A)+Pr(B)$. That is not true of independent events. – JMoravitz Aug 23 '16 at 6:11
• Principle of inclusion-exclusion 2set case: $Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$ – JMoravitz Aug 23 '16 at 6:18

Note the following two things:

Principle of inclusion-exclusion (2-set case) $$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$$

Definition of independent events

The following are equivalent statements

• $A$ and $B$ are independent events

• $Pr(A\mid B) = Pr(A)$

• $Pr(B\mid A) = Pr(B)$

• $Pr(A\cap B)=Pr(A)\cdot Pr(B)$

For your problem: you know $Pr(A)=0.3, Pr(B)=0.2$ and they are independent, so

$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B) = 0.3+0.2-0.3\cdot 0.2 = 0.5-0.06 = 0.44$$

So the probability of at least one of the events occurring is $0.44$. The probability of no events occurring is then one minus that, i.e. $1-0.44=0.56$.

Alternatively, if $A$ and $B$ are independent, then $A^c$ and $B^c$ are also independent. We have then $Pr(A^c\cap B^c)=Pr(A^c)Pr(B^c)=(1-0.3)(1-0.2)=0.7\cdot 0.8 = 0.56$

Your approach double-counts some possibilities. The $0.3$ covers all possibility of rain - so, it includes both "rain and quiz" and "rain and no quiz". The $0.2$ covers all possibility of a quiz, so it includes both "rain and quiz" and "quiz and no rain". So $0.3 + 0.2$ covers "rain and quiz", "rain and no quiz", "rain and quiz" again, and "quiz and no rain". To cut out one of the countings of "rain and quiz", we need to subtract the probability of that event, which is $0.3 \cdot 0.2 = 0.06$. So the probability of "rain or quiz or both" is $0.2 + 0.3 - 0.06 = 0.44$. The probability of neither is then $1 - 0.44 = 0.56$.


# 4.4 The mean value theorem  (Page 2/7)

 Page 2 / 7

Verify that the function $f\left(x\right)=2{x}^{2}-8x+6$ defined over the interval $\left[1,3\right]$ satisfies the conditions of Rolle’s theorem. Find all points $c$ guaranteed by Rolle’s theorem.

$c=2$

## The mean value theorem and its meaning

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions $f$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem ( [link] ). The Mean Value Theorem states that if $f$ is continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right),$ then there exists a point $c\in \left(a,b\right)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$

## Mean value theorem

Let $f$ be continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right).$ Then, there exists at least one point $c\in \left(a,b\right)$ such that

$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

## Proof

The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$ Since the slope of that line is

$\frac{f\left(b\right)-f\left(a\right)}{b-a}$

and the line passes through the point $\left(a,f\left(a\right)\right),$ the equation of that line can be written as

$y=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right).$

Let $g\left(x\right)$ denote the vertical difference between the point $\left(x,f\left(x\right)\right)$ and the point $\left(x,y\right)$ on that line. Therefore,

$g\left(x\right)=f\left(x\right)-\left[\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)\right]\text{.}$

Since the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\left(a\right)=0=g\left(b\right).$ Since $f$ is a differentiable function over $\left(a,b\right),$ $g$ is also a differentiable function over $\left(a,b\right).$ Furthermore, since $f$ is continuous over $\left[a,b\right],$ $g$ is also continuous over $\left[a,b\right].$ Therefore, $g$ satisfies the criteria of Rolle’s theorem. Consequently, there exists a point $c\in \left(a,b\right)$ such that $g\prime \left(c\right)=0.$ Since

$g\prime \left(x\right)=f\prime \left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a},$

we see that

$g\prime \left(c\right)=f\prime \left(c\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

Since $g\prime \left(c\right)=0,$ we conclude that

$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

In the next example, we show how the Mean Value Theorem can be applied to the function $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right].$ The method is the same for other functions, although sometimes with more interesting consequences.

## Verifying that the mean value theorem applies

For $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right).$ Find these values $c$ guaranteed by the Mean Value Theorem.

We know that $f\left(x\right)=\sqrt{x}$ is continuous over $\left[0,9\right]$ and differentiable over $\left(0,9\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ ( [link] ). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative ${f}^{\prime }\left(x\right)=\frac{1}{\left(2\sqrt{x}\right)}.$ The slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ is given by

$\frac{f\left(9\right)-f\left(0\right)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.$

We want to find $c$ such that ${f}^{\prime }\left(c\right)=\frac{1}{3}.$ That is, we want to find $c$ such that

$\frac{1}{2\sqrt{c}}=\frac{1}{3}.$

Solving this equation for $c,$ we obtain $c=\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints.

find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one
state and prove L hospital rule
I want to know about hospital rule
Faysal
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were$48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach \$60 billion?
what is the derivative of x= Arc sin (x)^1/2
y^2 = arcsin(x)
Pitior
x = sin (y^2)
Pitior
differentiate implicitly
Pitior
then solve for dy/dx
Pitior
thank you it was very helpful
morfling
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.


# Is $x^2 \equiv -1 \pmod{365}$ solvable?

I know that a similar question already exists, but I have a different question to ask.

We want to examine if $x^2 \equiv -1 \pmod{365}$ has a solution.

My thought is: $365=5\cdot 73$. So,The congruence $x^2 \equiv -1 \pmod{365}$ has solution, if and only if, the congrueces $x^2 \equiv -1 \pmod 5$ and $x^2 \equiv -1 \pmod{73}$ has solutions. So, if we use Legendre's Symbol we have

• $x^2 \equiv -1 \pmod 5$ has solution $\iff (-1/5)=1$ (and with simple calculations, indeed)
• $x^2 \equiv -1 \pmod{71}$ has solution $\iff (-1/73)=1$

Now, can we conclude that the congruence $x^2 \equiv -1 \pmod{365}$ has solution?

And more general: If we have the congruence $x^2 \equiv a \pmod n$ with $n=p_1^{n_1}\cdots p_k^{n_k},\ \gcd(a,n)=1$, which is equivalent with the system $x^2 \equiv a {\pmod p_1^{n_1}},\ldots,x^2 \equiv a \pmod{p_k^{n_k}}$, can we conclude that the first has solution if and only if each one of $x^2\equiv a\pmod{p_i^{n_i}},\ \forall i=1,\ldots,k$ has solution?

Thank you.

• Change $71$ to $73$ and it's fine. – Robert Israel Feb 7 '17 at 16:12
• @RobertIsrael Thank you for your comment and your correction. – Chris Feb 7 '17 at 16:15
• Note the use of \pmod rather than \bmod in my edit to the question. – Michael Hardy Feb 7 '17 at 16:51
• @RobertIsrael Ok, thank you – Chris Feb 7 '17 at 16:58

We show the following result:

Let $n,m$ be coprime integers. Let $a \in \Bbb Z$. Then $a$ is a square modulo $nm$ iff $a$ is square mod $n$ and $a$ is square mod $m$.

Assume that there are integers $x_i$ such that $x_1^2 \equiv a \pmod m$ and $x_2^2 \equiv a \pmod n$ where $m,n$ are coprime. We show that $y^2 \equiv a \pmod {nm}$ has a solution (the converse is obvious).

We know that there is an integer $y$ such that $y \equiv x_1 \pmod m$ and $y \equiv x_2 \pmod n$, by the Chinese remainder theorem.

Then $y^2-a$ is a multiple of $m$ and $n$, so it is a multiple of $nm$ since $(n,m)=1$. Therefore $y^2 \equiv a \pmod {nm}$.

More generally, we have the following theorem (see Ireland Rosen, p. 50)

Let $a,n \in \Bbb N$ be coprime integers. Write $n=2^e p_1^{e_1} \cdots p_k^{e_k}$ as product of distinct prime powers. Then $a$ is a square modulo $n$ if and only if

• $a$ is a square modulo $p_i$ (for $1 \leq i \leq k$) ; this is equivalent to $a^{\dfrac{p_i-1}{2}} \equiv 1 \pmod{p_i}$

• $e>1 \implies a \equiv 1 \pmod{2^{2+r}}$ where $r = 1$ if $e \geq 3$ and $r=0$ if $e=2$.

• It might be worth explicitly mentioning the CRT here where it's used, but this is elsewise a good, clean answer. – Steven Stadnicki Feb 7 '17 at 16:20
• Thank you , now it's clear. – Chris Feb 7 '17 at 16:27
• @Chris : you're welcome. It is chapter 5 in Ireland, Rosen. – Watson Feb 7 '17 at 16:28
• Tell me please something else. If we have the congruence $x^2 \equiv 2 \bmod 118$ could we say that $118=2\cdot 59$ so it's in the form of $2p^m$ and $2^{\phi(118)/ \gcd(2,\phi(118))} \equiv -1 \bmod 118$ so there are no solutions (because it is not a quadratic residue mod 118)? – Chris Feb 7 '17 at 16:46
• @Chris : since $2$ is not a square mod $59$, it is not a square mod $118$, that's correct! – Watson Feb 7 '17 at 16:50

We have $365 = 5 \times 73$. The congruence becomes $x^2 = -1 \mod 5$ and $x^2 = -1 \mod 73$.

We have if $p = 1 \mod 4 \implies x^2 = -1 \mod p$ has exactly $2$ solutions.

Thus $x^2 = -1 \mod 5$ has solutions $x_0,x_1$ and $x^2 = -1 \mod 73$ has solutions $y_0,y_1$.

The original solutions satisfies either: $x = x_0 \mod 5, x = y_0 \mod 73$; $x = x_0 \mod 5, x = y_1 \mod 73$; $x = x_1 \mod 5, x = y_0 \mod 73$ ; $x = x_1 \mod 5, x = y_1 \mod 73$.

For each pair of congruence , $x$ is uniquely determined $\mod 365$ by the Chinese remainder theorem. Hence the original congruence has $4$ solutions.

If $\,m,n\,$ are coprime then, by CRT, solving an integer coefficient polynomial $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ and mod $\,n.\,$ By CRT, each combination of a root $\,r_i\bmod m\,$ and a root $\,s_j\bmod n\,$ corresponds to a unique root $\,t_{ij}\bmod mn,\,$ i.e.

$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{mn}&\overset{\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod m\\f(x)\equiv 0\pmod n\end{array} \\ &\iff& \begin{array}{}x\equiv r_1,\ldots,r_k\pmod m\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod n\end{array}\\ &\iff& \left\{ \begin{array}{}x\equiv r_i\pmod m\\x\equiv s_j\pmod n\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\ &\overset{\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\ \end{eqnarray}\qquad\qquad$$

Yes. Let a solution to $x^2+1 \equiv 0 \pmod {5}$ be $r_{1}$, and let a solution to $x^2+1 \equiv 0 \pmod {73}$ be $r_{2}$

Then note that by CRT, we have that there exists such $x$ that $$x \equiv r_{1} \pmod {5}$$ $$x \equiv r_{2} \pmod {73}$$ Exists. Then note that for such $x$, $$x^2+1 \equiv 0 \pmod {5}$$ $$x^2+1 \equiv 0 \pmod {73}$$ Which gives $$x^2+1 \equiv 0 \pmod {365}$$

• Seems alright now. – Wojowu Feb 7 '17 at 16:20

It is true because of the Chinese remainder theorem, which asserts the map \begin{align} \mathbf Z/n\mathbf Z&\longrightarrow \mathbf Z/p_1^{n_1}\mathbf Z\times\dotsm\times\mathbf Z/p_k^{n_k}\mathbf Z\\ x\bmod n&\longmapsto(x\bmod p_1^{n_1},\dots,x\bmod p_k^{n_k}) \end{align} is a ring isomorphism.


# Math Help - Cartesian equations

1. ## Cartesian equations

Given $x=2 \sin (nt+\frac{\pi}{3})$ and $y=4 \sin (nt+ \frac{\pi}{6})$, express $x$ and $y$ in terms of $\sin nt$ and $\cos nt$. Find the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.

I have expressed $x$ and $y$ in terms of $\sin nt$ and $\cos nt$ already.
$x=2 \sin (nt+\frac{\pi}{3})$
$x=2(\frac{1}{2} \sin nt+\sqrt{3} \cos nt)$
$x= \sin nt+ \sqrt{3} \cos nt$
and
$y=4 \sin (nt+ \frac{\pi}{6})$
$y=4(\frac{\sqrt{3}}{2} \cos nt+\frac{1}{2} \sin nt)$
$y= 2\sqrt{3} \cos nt +2 \sin nt$
Now my problem is how do i form the Cartesian equation in x and y as t varies?
I take n as constant and t varies? Can anyone explain?
Thanks

2. Originally Posted by arze
Given $x=2 \sin (nt+\frac{\pi}{3})$ and $y=4 \sin (nt+ \frac{\pi}{6})$, express $x$ and $y$ in terms of $\sin nt$ and $\cos nt$. Find the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.

I have expressed $x$ and $y$ in terms of $\sin nt$ and $\cos nt$ already.
$x=2 \sin (nt+\frac{\pi}{3})$
$x=2(\frac{1}{2} \sin nt+\sqrt{3} \cos nt)$
$x= \sin nt+ \sqrt{3} \cos nt$
and
$y=4 \sin (nt+ \frac{\pi}{6})$
$y=4(\frac{\sqrt{3}}{2} \cos nt+\frac{1}{2} \sin nt)$
$y= 2\sqrt{3} \cos nt +2 \sin nt$
Now my problem is how do i form the Cartesian equation in x and y as t varies?
I take n as constant and t varies? Can anyone explain?
Thanks
$x=2 \sin (nt+\frac{\pi}{3})$ and $y=4 \sin (nt+ \frac{\pi}{6})$

$x=2 \sin (nt+\frac{\pi}{6}+\frac{\pi}{6})$

let $\alpha=nt+\frac{\pi}{6}$

$x=2 \sin (\alpha+\frac{\pi}{6})$ and $y=4 \sin (\alpha)$

$\frac{x}{2}=\sin (\alpha)\cos(\frac{\pi}{6}) + \cos (\alpha)\sin(\frac{\pi}{6})$

and

$\frac{y}{4}=\sin (\alpha)$

use $\sin^2 (\alpha) + \cos^2(\alpha) = 1$ to eliminate $\alpha$ and get your equation

3. Hello, arze!

Given: . $\begin{array}{ccc}x &=&2 \sin \left(nt+\frac{\pi}{3}\right) \\ y &=& 4 \sin \left(nt+ \frac{\pi}{6}\right)\end{array}$

Express $x$ and $y$ in terms of $\sin nt$ and $\cos nt$.

Find the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.

I have expressed $x$ and $y$ in terms of $\sin nt$ and $\cos nt$ already.

. . $\begin{array}{ccccc}x\:=\: 2\sin\left(nt+\tfrac{\pi}{3}\right) \:=\:2\left(\tfrac{1}{2}\sin nt+\tfrac{\sqrt{3}}{2}\cos nt\right) &\Rightarrow& x\:=\: \sin nt+ \sqrt{3} \cos nt \\ \\[-4mm] y\:=\:4\sin\left(nt+ \tfrac{\pi}{6}\right) \:=\:4\left(\tfrac{\sqrt{3}}{2} \cos nt+\tfrac{1}{2} \sin nt\right) &\Rightarrow& y \:=\: 2\sqrt{3} \cos nt +2 \sin nt \end{array}$

. . . . . Good work!

Now my problem is how do i form the Cartesian equation in $x$ and $y$ as $t$ varies?

We have: . $\begin{array}{cccc}x &=& \sin nt + \sqrt{3}\cos nt & {\color{blue}(1)} \\ \\[-4mm] \dfrac{y}{2} &=& \sqrt{3}\sin nt + \cos nt & {\color{blue}(2)}\end{array}$

$\begin{array}{ccccc}\text{Square }{\color{blue}(1)}: & x^2 &=& \sin^2\!nt + 2\sqrt{3}\sin nt\cos nt + 3\cos^2\!nt \\
\text{Square }{\color{blue}(2)}: & \dfrac{y^2}{4} &=& 3\sin^2\!nt + 2\sqrt{3}\sin nt\cos nt + \cos^2\!nt \end{array}$

Add: . $x^2 + \frac{y^2}{4} \;=\;4\sin^2\!nt + 4\sqrt{3}\sin nt\cos nt + 4\cos^2\!nt$

. . . . $x^2 + \frac{y^2}{4} \;=\;4\underbrace{\left(\sin^2\!nt + \cos^2\!nt\right)}_{\text{This is 1}} + 2\sqrt{3}\underbrace{\left(2\sin nt\cos nt\right)}_{\text{This is }\sin2nt}$

. . . . $x^2 + \frac{y^2}{4} \;=\;4 + 2\sqrt{3}\sin(2nt)$

4. Originally Posted by Soroban
Hello, arze!

We have: . $\begin{array}{cccc}x &=& \sin nt + \sqrt{3}\cos nt & {\color{blue}(1)} \\ \\[-4mm] \dfrac{y}{2} &=& \sqrt{3}\sin nt + \cos nt & {\color{blue}(2)}\end{array}$

Add: . $x^2 + \frac{y^2}{4} \;=\;4\sin^2\!nt + 4\sqrt{3}\sin nt\cos nt + 4\cos^2\!nt$

. . . . $x^2 + \frac{y^2}{4} \;=\;4\underbrace{\left(\sin^2\!nt + \cos^2\!nt\right)}_{\text{This is 1}} + 2\sqrt{3}\underbrace{\left(2\sin nt\cos nt\right)}_{\text{This is }\sin2nt}$

. . . . $x^2 + \frac{y^2}{4} \;=\;4 + 2\sqrt{3}\sin(2nt)$
if you want to do it that way, solve (1) and (2) for sin and cos:

$2\sin nt = \frac{\sqrt3y}{2}-x$

$2\cos nt = \sqrt3x-\frac{y}{2}$

then use

$\sin^2nt + \cos^2nt = 1$

to get an equation in x and y

5. Hello, Tesla!

Absolutely right!

It's an old habit.

Whenever I see something like: . $\begin{array}{ccc}x &=& a\sin\theta \\ y &=& b\cos\theta\end{array}$

. . I tend to use the square-and-add approach to simplify.

I must learn to get out of my "box".


# Why can't $y=xe^x$ be solved for $x$?

I apologize for my mathematical ignorance regarding this, but could someone help me understand why it isn't possible to (symbolically) find an inverse function for $f(x)=xe^x$?

The most obvious (but presumably the most trivial) is that $f$ does not pass the "horizontal line test". However, if we restrict the domain to $x\geq-1$ this should not be a problem (derivative is positive for $x>-1$ so function is strictly increasing). So now my question becomes: "Why can't we find an inverse function for $f$ over the interval $[-1,\infty)$?"

Perhaps it is because $e^x$ is transcendental (not algebraic). However, we can find an inverse for $g(x) = e^x$, which is also transcendental. Is that because we're "cheating" by defining another transcendental function, namely $\ln(x)$, to be its inverse? In other words, would it be fundamentally no different to define a new function, call it $\text{lnx}(x)$ (if that's not already something else), to be the inverse of $xe^x$ over $[-1,\infty)$ and then say that $f$ has a "closed form" / "symbolic" / ??? inverse function $f^{-1}(x)=\text{lnx}(x)$ over the interval $[-1, \infty)$?

### SageMath source to generate plot

xs = (x,-5,2) ys = (y,-1,5) p1 = implicit_plot(x*exp(x)-y,xs,ys, color='blue', legend_label='y=x*e^x') p2 = implicit_plot(x-y,xs,ys, color='orange', linestyle='dashed', legend_label='y=x') p3 = implicit_plot(exp(x)-y,xs,ys, color='green', linestyle='dotted', legend_label='y=e^x') combined = p1 + p2 + p3 combined.axes_labels(['x', 'y']) combined.legend(True) combined.show(title='Transcendental Stuff', frame=True, axes=True, legend_loc='lower right')

• it can be solved by the following function $$\{\{x\to W(y)\}\}$$ the product Logarithmus by Mathematica – Dr. Sonnhard Graubner Mar 21 '18 at 15:49
• It can be solved, and an inverse function can be found in the given interval, just not analytically. – Thern Mar 21 '18 at 15:51
• Yes, you're right that "fundamentally" it would "be no different" to invent a label for the inverse function, which is guaranteed to exist because your function is injective. – symplectomorphic Mar 21 '18 at 15:51
• This question is so high quality, why aren't people upvoting this – vrugtehagel Mar 21 '18 at 15:54
• @vrugtehagel that is what I have about this site more than anything else. There's so much good content that goes unappreciated. This is an excellent question. I upvoted it. – user223391 Mar 21 '18 at 16:03

It can be solved by inventing new functions, but it cannot be solved in closed form using trigonometric, logarithmic or exponential etc. Read this: Chow, Timothy Y. (May 1999), "What is a Closed-Form Number?"

• I believe you, but I was trying to get a better answer to the why part of the question. – iX3 Mar 21 '18 at 16:02
• Read the linked article for the why part. – Shubhashish Mar 21 '18 at 16:06
• Wow, this paper talks about exactly what I wanted to know! (Did you just add that? I can't believe I didn't see it before.) BTW, it was hard to read on that site (not scalable, not searchable, etc.), but I found a link to a PDF version on the author's website here – iX3 Mar 21 '18 at 16:18
• Yes I have updated the link – Shubhashish Mar 21 '18 at 16:19

You can use the Lambert-$W$ function to solve it symbolically.

$y = xe^x$ gives $x = W(y)$.

You may run solve(x*exp(x)-y,x) on SymPy Live as an alternative to SageMath.

• Just for reference, here is a Wolfram Alpha example solving $xe^x=\pi$ – gt6989b Mar 21 '18 at 15:56
• Thank you for showing me the Lambert-W function (was just finding that in the answer to another question on here too). It seems that what determines whether it is possible to solve something in "closed form" has largely to do with what functions/expressions are allowed. e.g. $y = e^x$ cannot be "solved for $x$" unless something like $\ln$ is allowed. Similarly, the answer to my question depends on whether or not something like $W$ is acceptable. – iX3 Mar 21 '18 at 16:00
• @iX3 In the tag info of (closed-form), it's said that a "closed form expression" is anything written in terms of "known" functions. Though the Lambert-W function is not elementary, but I believe it qualifies a known function. – GNUSupporter 8964民主女神 地下教會 Mar 21 '18 at 16:09
• Hmm, solve(x*exp(x)-y,x) --> Traceback (most recent call last):...line 1055, in solve solution = _solve_system(f, symbols, **flags) ... line 158, in _lambert solns[i] = tmp.subs(u, rhs) AttributeError: 'dict' object has no attribute 'subs' I may need to go read & learn about SymPy to figure out what's wrong. (I think SageMath uses SymPy though, so in theory I should be able to access all of the same functionality through that.) – iX3 Mar 21 '18 at 18:25
• @iX3 I've edited my post in response to your comment. – GNUSupporter 8964民主女神 地下教會 Mar 21 '18 at 18:33

As indicated in the comments and in another answer, there is a special function that has been defined which can serve as an inverse here. However, this doesn't address your basic question, which is: why is this necessary?

Before I talk about that, let me say something about $\ln x$ and $e^x$. Those are both transcendental functions, and they're inverses of each other, but neither is really defined simply to be the inverse of the other. Both of these functions arise very naturally, each on their own. The natural log function is the integral of $x^{-1}$, for a suitably defined integral, and the exponential function is the solution of a differential equation modeling the simplest kind of constant relative growth. They end up being inverses of each other, and that's a cool set of facts to understand and wonder at.

Anyway, you can't use elementary functions to invert lots of things, such as $f(x)=xe^x$. This often happens when we mix different types of functions together. The functions $g(x)=x$ and $h(x)=e^x$ are perfectly invertible, and they are, respectively, polynomial and exponential. The function $f=gh$, on the other hand, is the product of a polynomial function with an exponential function. We expect that to be more complicated. It's not solvable, with the usual algebraic methods, because whatever technique we apply to simplify $e^x$ messes up the polynomial part. Similarly, if you try to solve $e^x(x^2-x)=k$, anything you do to simplify the polynomial part will just make an exponential mess.

It's the blending of different types of functions, which are amenable to transforming with completely different tools, that makes such functions complicated. Other notorious examples include $\frac{\sin x}{x}$, and $e^{x^2}$. It's not always that they're hard to invert, but try doing integral calculus with such functions!

• Thank you; this makes a lot of sense, at least intuitively. Is there a more precise definition regarding the "blending of different types of functions" and does that always produce tough situations like this? – iX3 Mar 21 '18 at 16:10
• I'd say the different kinds are 1. polynomial/rational/radical (algebraic), 2. Trigonometric, 3. Inverse trigonometric, 4. Exponential, 5. Logarithmic. If it's done right, to allow for cancellation, you can blend 2 with 3 and 4 with 5 and still be able to work analytically. If you're very careful, you can blend any kinds, as long as there's a way to cancel stuff until you have something pure. I don't know of any real precise definitions and theory that address this "blending" though. – G Tony Jacobs Mar 21 '18 at 16:14

I am going to break your question into two separate questions:

1. Does an inverse function to $y=x \cdot e^x$ exist on the domain $[-1,+\infty)$?
2. Can we find that inverse function in the "standard list" of functions that everyone knows?

Question 1 has an easy answer: yes, that inverse function exists. Its existence is an application of the inverse function theorem which most students first encounter without proof in an ordinary calculus class, and which they then encounter with proof in some kind of advanced calculus class. And then if you're quick enough, you might even be able to attach a name to that function, although it appears that Lambert beat you to it as shown in the answer of @GNUSupporter.

Question 2 is more difficult to answer. First there is no "standard list" of functions that everyone knows. Nonetheless, perhaps that's just because we haven't tried hard enough to list all functions. Can we keep adding functions to the "standard list" until we don't have to add any more? For example, as suggested in the answer of @GNUSupporter, can we just add the "Lambert-W function" $x=W(y)$ to our list if we had never heard of it before, and then declare ourselves happy? Well, maybe, but then you would be perfectly justified to ask whether there is an inverse function to $z=yW(y)$ on some appropriately chosen domain......................

These kinds of questions are addressed (but not definitively answered) in a branch of mathematics called differential algebra which is a kind of broad generalization of differential equations.

• Thank you for the reference to differential algebra. I gather from your answer that the "answer" to my "why?" question is basically "Differential algebra deals with this question but does not provide a complete/general answer." Is that about right? – iX3 Mar 21 '18 at 16:12
• Yes, that's about right. Nonetheless, one can often answer specific questions, which is what's so nice about the link to Chow's paper provided in the answer of @ShubhashishChauhan. – Lee Mosher Mar 21 '18 at 17:05
• Although, to qualify my last comment, even Chow's paper does not provide a definitive answer to questions about Lambert's function, instead reducing them to applications of big unsolved conjectures. – Lee Mosher Mar 21 '18 at 17:13


# Inequalities with two absolute values with greater than symbol. Please tell me the proper way of doing $|4x-1|>|3x+2|$

I have finals in a couple of days. I need help with inequalities. I have spent around 15 hours trying my hand at inequalities and I am still trying to figure things out. In this case I need help with an inequality with two absolute values with greater than symbol.

The following is a picture of the inequality ($$|4x-1|>|3x+2|$$) and the procedure I tried, which is something I applied from an answer by Isaac to another question in this site, although I think I did something wrong because it looks like I got the answer wrong or incomplete.

Please tell me what I did wrong and how to do it the right way.

PS: Does this method also apply if there is a less than sign?

• Alternatively (and I'd say it's faster) you could note that $|a|>|b|$ is equivalent to $a^2>b^2$.
– dfnu
Nov 15, 2019 at 8:32
• Two years later I have completely forgotten how to do this. :D Nov 19, 2021 at 23:50

Here’s another way to look at it: $$|4x-1|>|3x+2| \iff (4x-1)^2-(3x+2)^2>0 \iff (7x+1)(x-3)>0$$

Now LHS has only two obvious zeros, and outside the interval with those end points, it has to be positive.

——-

P.S. When you have something like $$f(x)>0$$, where $$f$$ is continuous, it can change sign only when there are roots. So its roots effectively partition the real line into intervals which are either solutions or not. In the above case, two roots are $$-\frac17, 3$$, so we need to consider only the three intervals $$(-\infty, -\frac17), (-\frac17, 3)$$ and $$(3, \infty)$$. Obviously the LHS is positive for $$x\to \pm\infty$$ and negative for $$x=0$$, so the first and last intervals are solutions, the middle one isn’t.

BTW, whether the intervals to consider are open / closed depends on whether the inequality is strict or $$\geqslant$$.

• 7x+1>0 gives me (-1/7,+) which seems to be invalid. What is wrong? Nov 15, 2019 at 16:18
• @dfnu +macavity x-3 gives me (3,+), with a combined solution of (-1/7,+). Nov 15, 2019 at 16:56
• @freethinker36 If you need $A\cdot B>0$, you need $A, B$ to have the same sign. Not just $A$ or just $B$. Nov 15, 2019 at 16:58
• It would be great if you can elaborate. I'm learning, thus what may look evident, for me may not be evident. And tomorrow is my test and have to study two classes; already taking time of the second class to try to cement my understanding of inequalities. But I appreciate whatever guidance has been provided! Nov 15, 2019 at 17:05
• OK, I have added some explanations, do let me know if youre having more doubts after studying it. Nov 15, 2019 at 17:11

You're doing almost everything right; you're just misinterpreting the results at the end.

In particular, the idea is to combine intervals where the inequality holds. In the first region, $$\left(-\infty, -\frac23\right)$$, you arrived at $$x<3$$. It seems that you rejected this result, judging by the "$$\times$$" beside your work. This is where you went wrong.

Instead, think of it like this:

Within the region $$\left(-\infty, -\frac23\right)$$, which $$x$$ satisfy $$x<3$$?

All $$x$$ do within this region. Another way to see this is that you are effectively taking the intersection of the two: $$\left(-\infty, -\frac23\right) \cap (-\infty,3) = \left(-\infty, -\frac23\right)$$

The other two regions are handled the same way.

• My rationale for rejecting it was that 3>-2/3, so x<3 is not within the region of x<-2/3. Nov 15, 2019 at 6:33

I tried again, following the answer of Théophile, and I was able to reach the correct solution. Below is the pic of the procedure I used. Thanks!


# Are right continuous functions measurable?

Are right-continuous function from $\mathbb{R}^n$ to $\mathbb{R}$ necessarily semi-continuous? If not, are they necessarily Borel measurable? Is there a topological characterization of right-continuous functions (as there is of continuous ones)? Are CDFs of $n$-dimensional random vectors measurable?

Note: A function $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is right-continuous iff it is right-continuous at every point $x \in \mathbb{R}^n$. A function $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is right-continuous at $x \in \mathbb{R}^n$ iff given any infinite sequence of points in $\mathbb{R}^n$ $(y_0,y_1,\dots)$ that converges to $x$ from above (i.e. the sequence converges to $x$ in the usual, Euclidean sense and in addition every sequence element is greater than or equal to $x$ component-wise), the sequence $(f(y_0), f(y_1), \dots)$ converges to $f(x)$ in the usual sense.

• You might want to add the definition of right-continuity for a function defined on $\mathbb R^n$ when $n\gt1$.
– Did
Aug 2, 2012 at 8:28
• What are the definition of right-continuous function, and semi-continuous?
– Paul
Aug 2, 2012 at 8:44

The answer to the first question is no, even in the case $n=1$: The characteristic function of the half open interval $[0,1)$ is right continuous, but neither upper nor lower semicontinuous.

A right continuous function $\mathbb{R}\to\mathbb{R}$ is indeed Borel measurable. By definition, the inverse image $E$ of an open set has the property that for any $x\in E$, there is some $\delta>0$ so that $[x‚x+\delta)\subseteq E$. It follows that $E$ is a countable union of half open intervals, and hence is Borel measurable. I am not sure about the answer to this one when $n>1$ (the countable union argument no longer holds), but my guess is that right continuous functions are still measurable.

Topological characterization: If we write $\le$ for pointwise comparison on $\mathbb{R}^n$, we can make a one-sided topology by declaring a set $V\subseteq\mathbb{R}^n$ to be open if, for each $x\in V$, there is some $\delta>0$ so that $\{y\ge x\colon\lvert y-x\rvert<\delta\}\subseteq V$. Then the right continuous maps $\mathbb{R}^n\to\mathbb{R}$ are just the ones that are continuous from this topology to the usual topology on $\mathbb{R}$.

I am not too sure on the CDF question either. (I assume CDF stands for cumulative distribution function, in the sense of $F(x)=\mathrm{P}\{X\le x\}$, where $X$ is a random $n$-vector.) It might help that $F$ is not only right continuous, but also monotone. So the set $\{x\colon F(x)\le p\}$ has a particularly simple structure; I imagine it must be measurable, but right now I don't see a proof.

• Thanks, Harald, but it is in fact the case n>1 that i am interested in. See my clarification of the term "right-continuous" in a comment to my original post. Also, of the four questions i posed in my original post i am most interested in the last one concerning CDF. Aug 2, 2012 at 10:49
• @EvanAad: This topology appears to be the product of $n$ copies of the Sorgenfrey line. But I don't quite see why the open sets of this topology are Borel. It certainly has a basis of Borel sets, but I don't think it's second countable, so why does it follow that any union of such sets is also Borel? Aug 2, 2012 at 21:54
• @EvanAad: In fact, let $E$ be a non-measurable subset of $\mathbb{R}$ and let $A = \bigcup_{x \in E} [x,\infty)^2$. $A$ is open in our topology on $\mathbb{R}^2$ (the Sorgenfrey plane), but not Borel with respect to the usual topology, since its intersection with the diagonal line $y=-x$ is a skewed copy of $E$. One might think one could exploit this to make a right-continuous function which is not Borel, but I still don't see how. Aug 3, 2012 at 13:11
• @NateEldredge: Did you mean $\bigcup_{x\in E}[x,\infty)\times[-x,\infty)$? Aug 3, 2012 at 13:51
• @HaraldHanche-Olsen: Oops, yes, I did. Aug 3, 2012 at 14:25

Here is a proof that any function $F: \mathbb{R}^n \to \mathbb{R}$ which is right continuous is Borel measurable.

Define $F_n(x) = F\left(\frac{[nx]+1}{n}\right)$, where $[x]$ is the greatest integer smaller than $x$. Clearly $F_n$ are measurable since they are step functions.

From the right continuity of $F$ we get that $F_n\to F$, since $\frac{[nx]+1}{n} \downarrow x$.

Hence since $F$ is the pointwise limit of measurable functions, it is measurable too.

• Very Nice, Milind! Aug 24, 2017 at 13:58
• to make this a bit more solid, you would need to clarify that $n=1$ or properly define $[x]$ for an $n$-vector $x$. Oct 12, 2018 at 11:58