text:
# 4.6: Piecewise-defined Functions - Mathematics

In certain situations a numerical relationship may follow one pattern of behavior for a while and then exhibit a different kind of behavior. In a situation such as this, it is helpful to use what is known as a piecewise defined function - a function that is defined in pieces.

In the above example of a piecewise defined function, we see that the (y) values for the negative values of (x) are defined differently than the (y) values for the positive values of (x)

Sometimes we are given a graph and need to write a piecewise description of the function it describes.

The piecewise function pictured above could be described as follows:

Exercises 4.6
Sketch a graph for each of the piecewise functions described below.

Graph the piecewise-defined function shown below :

What are the domain and range ? Over what intervals is the function increasing or decreasing ?

Sketch the graph of y  =  4x + 11 for values of x between -10 and -2.

We can consider the following points to sketch the graph of y  =  4x + 11 :

*  y = 4x + 11 is a linear equation. Then, its graph will be a straight line.

*  y = 4x + 11 is in slope intercept form y = mx + b.

we get a positive slope m = 4.

So, the graph of y = 4x + 11 is a rising line.

Sketch the graph of y = x 2  - 1 for values of x between -2 and 2.

We can consider the following points to sketch the graph of y = x 2  - 1 :

*  y = x 2  - 1 is a quadratic equation. Then, its graph will be a parabola.

*  The sign of x 2   in y = x 2  - 1 is positive. So, the graph will be a open upward parabola.

*  We can write y = x 2  - 1 in vertex form as shown below.

we get the vertex (h, k)  =  (0, -1)

So, the graph of y = x 2  - 1 is a open upward parabola with the vertex (0, -1).

Sketch the graph of y  =  x + 1 for values of x between 2 and 10.

We can consider the following points to sketch the graph of y  =  x + 1 :

*  y = x + 1 is a linear equation. Then, its graph will be a straight line.

*  y = x + 1 is in slope intercept form y = mx + b.

we get a positive slope m = 1.

So, the graph of y = x + 1 is a rising line.

To determine the range, calculate the y-values that correspond to the minimum and maximum x-values on the graph.

For this graph, these values occur at the endpoints of the domain of the piecewise function,

Evaluate y = 4x + 11 for x = -10 :

Evaluate y = x + 1 for x = 10 :

Increasing and Decreasing Intervals :

The function is increasing when

The function is decreasing when

The following notes and exercises include piecewise examples, graphing steps, and problems written in function notation.

Click to select (large) image.

Then, right click to view or copy to desktop.

Webcomic #223 - "Peace-wise Function" (1-12-16)

• Finding f(x)
• Graphing functions
• Plotting points
• Piecewise (Split) Functions
• Continuous function
• Absolute values

#94 Math Masterpiece - Alfred Hitchock presents his 39 steps function (7-12-13)

"542: The Saddest Function" by Chris Burke (12/01/10) (Click lower right corner to view)

Read about "Maud Muller" by John Greenleaf Whittier (the source of the quote: "For of all sad words of tongue or pen, the saddest are these: 'What might've been!'")

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## Piecewise Functions

In mathematics, a piecewise-defined function (also called a piecewise function or a hybrid function) is a function which is defined by multiple sub-functions, each sub-function applying to a certain interval of the main function's domain (a sub-domain). Piecewise is actually a way of expressing the function, rather than a characteristic of the function itself, but with additional qualification, it can describe the nature of the function. For example, a piecewise polynomial function is a function that is a polynomial on each of its sub-domains, but possibly a different one on each.

The word piecewise is also used to describe any property of a piecewise-defined function that holds for each piece but not necessarily hold for the whole domain of the function. A function is piecewise differentiable or piecewise continuously differentiable if each piece is differentiable throughout its subdomain, even though the whole function may not be differentiable at the points between the pieces. In convex analysis, the notion of a derivative may be replaced by that of the subderivative for piecewise functions. Although the "pieces" in a piecewise definition need not be intervals, a function is not called "piecewise linear" or "piecewise continuous" or "piecewise differentiable" unless the pieces are intervals. 1

Maple is powerful software for exploring piecewise functions, and for analyzing, exploring, visualizing and solving virtually any mathematical problem. Student pricing available.

## 4.6: Piecewise-defined Functions - Mathematics

We can create new functions from existing ones in several ways.

When one creates a new function from existing functions in a "piecewise-defined" way, one breaks apart some domain into two or more disjoint pieces, using different functions to calculate the output for each $x$-value, where the function used is based upon the piece into which that particular $x$-value falls.

A simple example of a piecewise defined function is the absolute value function, $|x|$.

When $x ge 0$, the absolute value function doesn't really do anything -- it returns its input unchanged. So, for this piece $|x| = x$.

However, when $x lt 0$, the absolute value changes the sign of its input. Multiplying a value by negative one has the same effect -- so we can say that for this second piece $|x| = -x$.

Thus, we summarize how to calculate the output with the following "piecewise definition": $|x| = left< egin x & , & x ge 0 -x & , & x lt 0 end ight.$

Graphing a piecewise function can be accomplished by simply graphing the functions found in the respective "pieces", limiting the points drawn for each piece to the $x$-values that satisfy the appropriate condition.

For example, given the function $f(x) = left< egin sqrt <25-x^2>& , & x le 3 2x-5 & , & x gt 3 end ight.$

We find its graph by first graphing $y = sqrt<25-x^2>$ (a semi-circle with center at the origin and radius 5) and $y=2x-5$ (a line of slope 2 with $y$-intercept at $(0,-5)$), as shown at left below. Then, we discard those points that don't match the conditions provided (i.e., the dashed parts of the left graph below), to arrive at the graph of the piecewise-defined function on the right.

Note, that we use a filled-in point at $(3,4)$ to suggest that the output of the function at $x=3$ is determined by the semi-circular piece -- since the condition on this piece is true when $x=3$. Likewise, to indicate that the point where $x=3$ should be excluded from the linear piece (given the strict inequality), we place an open circle at $(3,1)$.

## 6.2: A Toy Rocket and a Drone (15 minutes)

### Activity

In this activity, students examine graphs of functions, identify and describe their key features, and connect these features to the situations represented. These key features include the horizontal and vertical intercepts, maximums and minimums, and intervals where a function is increasing or decreasing (or where a graph has a positive or a negative slope).

Neither the features nor the terms are likely new to students. The idea of intercepts was introduced in middle school and further developed in earlier units. Graphical features such as maximums and minimums have been considered intuitively in various cases. They are simply more precisely defined in here. In a later activity, students will distinguish between a maximum of a graph and the maximum of a function.

### Launch

Give students about 5 minutes of quiet work time. Follow with a whole-class discussion.

A toy rocket and a drone were launched at the same time.

Here are the graphs that represent the heights of two objects as a function of time since they were launched.

Height is measured in meters above the ground and time is measured in seconds since launch.

Expand Image

Description:
2 graphs. Horizontal axis, 0 to 7, time in seconds. Vertical axis, 0 to 50 by 10’s, height in meters. Graph R is parabolic and opens down. It starts at 0 comma 25. Its vertex is at 2 comma 45. It ends at 5 comma 0. Graph D is piecewise linear. It starts at 0 comma 0. Increases until 2 comma 20, horizontal until 5 comma 20, then decreases to 7 comma 0.

Analyze the graphs and describe—as precisely as you can—what was happening with each object. Your descriptions should be complete and precise enough that someone who is not looking at the graph could visualize how the objects were behaving.

### Student Response

For access, consult one of our IM Certified Partners.

### Anticipated Misconceptions

When analyzing the graphs and describing what is happening with each object, some students may mistakenly think that the horizontal axis represents horizontal distance, neglecting to notice that it represents time. They may then describe how the objects were moving vertically as they traveled horizontally, rather than with respect to the number of seconds since they took off. Encourage these students to check the label of each axis and revisit their descriptions.

### Activity Synthesis

Select a few students to share their description of the graphs, and have each student describe the motion of one flying object.

Display a blank coordinate plane for all to see. As each student shares their response, sketch a graph to match what is being described.

Expand Image

For any gaps in their description, make assumptions and sketch accordingly. (For example, if a student states that the toy rocket reaches a height of 45 meters after 2 seconds but does not state its starting height, start the curve at ((0,0)) , ((0,40)) , or any other point besides ((0,25)) .) If requested, allow students to refine their descriptions and adjust the sketch accordingly.

Next, invite other students to share their response to the last question. On the graphs, highlight the features students noted. (See Student Response for an example.) Use the terms vertical intercepts, horizontal intercepts, maximum, and minimum to refer to those features and label them on the graphs.

• A point on the graph that is as high as or higher than all other points is called a maximum of the graph (or a relative maximum, because its height is viewed in relation to other points shown on the graph).
• A point on the graph that is as low as or lower than all other points is called a minimum of the graph (or a relative minimum).

A graph could have more than one relative maximum or minimum. For instance, the points ((2,(D(2))) and ((5,D(5)) ) are both relative maximums, and ((0, D(0))) and ((7,D(7))) are both relative minimums.

If no students mentioned the intervals in which each function was increasing, staying constant, or decreasing, draw their attention to these features on the graphs and label them as such.

## Overview

#### Description

For courses in precalculus.

This package includes MyLab Math.

Ties concepts together using a functions approach

The Concepts Through Functions Series introduces functions at the start of each text, and maintains a continuous theme by introducing/developing a new function in every chapter.

Known for their ability to connect with today’s students, acclaimed authors Sullivan and Sullivan focus on the fundamentals — preparing for class, practice with homework, and reviewing key concepts — encouraging students to master basic skills and develop the conceptual understanding needed for this and future courses. Graphing utility coverage is optional, and can be included at the discretion of each instructor based on course needs.

Reach every student by pairing this text with MyLab Math

MyLab™ Math is the teaching and learning platform that empowers you to reach every student. By combining trusted author content with digital tools and a flexible platform, MyLab personalizes the learning experience and improves results for each student. Learn more about MyLab Math.

## Piecewise-defined function

On this page you can get various actions with a piecewise-defined function, as well as for most services - get the detailed solution.

• Derivative of a piecewise
• Plot a graph
• Curve sketching
• Defined integral
• Indefined integral of similar functions
• Limit of piecewises
• Fourier series (In common there are piecewises for calculating a series in the examples)
• Taylor series

The above examples also contain:

• the modulus or absolute value: absolute(x) or |x|
• square roots sqrt(x),
cubic roots cbrt(x)
• trigonometric functions:
sinus sin(x), cosine cos(x), tangent tan(x), cotangent ctan(x)
• exponential functions and exponents exp(x)
• inverse trigonometric functions:
arcsine asin(x), arccosine acos(x), arctangent atan(x), arccotangent acot(x)
• natural logarithms ln(x),
decimal logarithms log(x)
• hyperbolic functions:
hyperbolic sine sh(x), hyperbolic cosine ch(x), hyperbolic tangent and cotangent tanh(x), ctanh(x)
• inverse hyperbolic functions:
hyperbolic arcsine asinh(x), hyperbolic arccosinus acosh(x), hyperbolic arctangent atanh(x), hyperbolic arccotangent acoth(x)
• other trigonometry and hyperbolic functions:
secant sec(x), cosecant csc(x), arcsecant asec(x), arccosecant acsc(x), hyperbolic secant sech(x), hyperbolic cosecant csch(x), hyperbolic arcsecant asech(x), hyperbolic arccosecant acsch(x)
• rounding functions:
round down floor(x), round up ceiling(x)
• the sign of a number:
sign(x)
• for probability theory:
the error function erf(x) (integral of probability), Laplace function laplace(x)
• Factorial of x:
x! or factorial(x)
• Gamma function gamma(x)
• Lambert's function LambertW(x)

#### The insertion rules

The following operations can be performed

2*x - multiplication 3/x - division x^2 - squaring x^3 - cubing x^5 - raising to the power x + 7 - addition x - 6 - subtraction Real numbers insert as 7.5, no 7,5

## Lesson 6

This lesson has two aims. The first aim is to prompt students to write an equation of the form (y = a oldcdot b^x) to represent an exponential function without a context from limited information.

A second optional activity encourages students to look more closely at how different equivalent expressions can be written to highlight different aspects of a quantity. For example, we can write an expression that shows the growth factor of a bacteria population each week, or one that shows the daily growth factor.

In the first activity, to find (a) and (b) , students need to reason abstractly and make use of structure. Calculating the growth or decay factor, the (b) in (a oldcdot b^x) , from the coordinates of two points requires students to understand that exponential functions change by equal factors over equal intervals (MP7). In addition, because the first activity uses the Information Gap routine, students first must decide what information they need to solve the problem and why they need it. Obtaining useful information may take multiple rounds of questioning (MP1) and the use of increasingly precise language (MP6).


text:
# The value of lim x → 1 x 3 − 1 x − 1

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 2, Problem 1P
To determine

Expert Solution

## Answer to Problem 1P

The limit of the function is 23_

### Explanation of Solution

Result used:

Difference of squares formula: (a2b2)=(a+b)(ab)

Difference of cubes formula: (a3b3)=(ab)(a2+ab+b2)

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then limxaf(x)=f(a).

Calculation:

Obtain the limit of the function f(x)=x31x1 as x approaches 1 as follows.

Consider the function f(x)=x31x1.

Take the limit of the function as x approaches 1.

limx1f(x)=limx1x31x1=limx1(x13)1(x12)1=limx1(x26)1(x36)1=limx1(x16)21(x16)31

Let x6=t. This implies that, x=t6. Then t approaches 1 as x approaches 1.

On substitution,

limx1x31x1=limt1((t6)16)21((t6)16)31=limt1(t)21(t)31

Apply the difference of squares and cubes formulae in the numerator and the denominator as follows,

limx1x31x1=limt1(t+1)(t1)(t1)(t2+t+1)=limt1(t+1)(t2+t+1)=((1)+1)((1)2+(1)+1)[by direct substitution]=23

Thus, the limit of the function is 23_.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!

format: plain


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# Problem with Sales Regression Residuals

I'm trying to build a model for the ticket sales for different sporting events over a period of 30 days before the game to the day of the game. The problem that I'm having is that I can't seem to fit the right model to it. I've tried negative binomial, linear, poisson, but nothing seems to fit right. A problem that tends to occur is I get a residual plot that looks like the one below. Does anybody have any suggestions for things that I can try?

Residuals for the negative binomial:

• Ticket sales are probably all nonnegative, and they may be integer, so you should expect residuals that are skewed (simulate negbin data, fit them with the correct model and look at the residuals). That does not need to mean that your models are incorrect. – Stephan Kolassa Jun 26 '14 at 16:10
• You might also try plotting the residuals against each of the predictors. It might give you some idea where there is homogeneity. Doing so could suggest other models additions to the model, or where transformations are necessary. – Chris Jun 26 '14 at 16:31
• Just to give you an idea, I added what my residual plot was for the negative binomial. – user3704120 Jun 26 '14 at 18:08
• What are SINGLE.SALES? Are you modeling the number of tickets sold on a given day, the amount of sales in \$s, or something else? Also, how is it that your negative binomial model predicts$\approx -40\$? There is something wrong with that. – gung Jun 26 '14 at 18:15
• Sorry, single sales is just differentiating it from group sales or season tickets. It's number of tickets. – user3704120 Jun 26 '14 at 18:19


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# Fit statistics¶

## Introduction¶

This page describes the fit statistics used in gammapy. These fit statistics are used by datasets to perform model fitting and parameter estimation.

Fit statistics in gammapy are all log-likelihood functions normalized like chi-squares, i.e. if $$L$$ is the likelihood function used, they follow the expression $$2 \times log L$$.

All functions compute per-bin statistics. If you want the summed statistics for all bins, call sum on the output array yourself.

## Cash : Poisson data with background model¶

The number of counts, $$n$$, is a Poisson random variable of mean value $$\mu_{\mathrm{sig}} + \mu_{\mathrm{bkg}}$$. The former is the expected number of counts from the source (the signal), the latter is the number of expected background counts, which is supposed to be known. We can write the likelihood $$L$$ and applying the expression above, we obtain the following formula for the Cash fit statistic:

$C = 2 \times \left(\mu_{\mathrm{sig}} + \mu_{\mathrm{bkg}} - n \times log (\mu_{\mathrm{sig}} + \mu_{\mathrm{bkg}}) \right)$

The Cash statistic is implemented in cash and is used as a stat function by the MapDataset and the SpectrumDataset.

### Example¶

Here’s an example for the cash statistic:

>>> from gammapy.stats import cash
>>> data = [3, 5, 9]
>>> model = [3.3, 6.8, 9.2]
>>> cash(data, model)
array([ -0.56353481,  -5.56922612, -21.54566271])
>>> cash(data, model).sum()
-27.678423645645118

## WStat : Poisson data with background measurement¶

In the absence of a reliable background model, it is possible to use a second measurement containing only background to estimate it.

In the OFF region, which contains background only, the number of counts $$n_{\mathrm{off}}$$ is a Poisson random variable of mean value $$\mu_{\mathrm{bkg}}$$ In the ON region which contains signal and background contribution, the number of counts, $$n_{\mathrm{on}}$$, is a Poisson random variable of mean value $$\mu_{\mathrm{sig}} + \alpha \mu_{\mathrm{bkg}}$$, where $$\alpha$$ is the ratio of the ON and OFF region acceptances.

It is possible define a likelihood function and marginalize it over the unknown $$\mu_{\mathrm{bkg}}$$ to obtain $$\mu_{\mathrm{sig}}$$. This yields the so-called WStat or ON-OFF statistics which is traditionally used for ON-OFF measurements in ground based gamma-ray astronomy.

The WStat fit statistics is given by the following formula:

$\begin{split}W = 2 \big(\mu_{\mathrm{sig}} + (1 + \alpha)\mu_{\mathrm{bkg}} - n_{\mathrm{on}} - n_{\mathrm{off}} - & n_{\mathrm{on}} (\log{(\mu_{\mathrm{sig}} + \alpha \mu_{\mathrm{bkg}}) - \log{(n_{\mathrm{on}})}})\\ -& n_{\mathrm{off}} (\log{(\mu_{\mathrm{bkg}})} - \log{(n_{\mathrm{off}})})\big)\end{split}$

To see how to derive it see the wstat derivation.

The WStat statistic is implemented in wstat and is used as a stat function by the MapDatasetOnOff and the SpectrumDatasetOnOff.

### Caveat¶

• Since WStat takes into account background estimation uncertainties and makes no assumption such as a background model, it usually gives larger statistical uncertainties on the fitted parameters. If a background model exists, to properly compare with parameters estimated using the Cash statistics, one should include some systematic uncertainty on the background model.

• Note also that at very low counts, WStat is known to result in biased estimates. This can be an issue when studying the high energy behaviour of faint sources. When performing spectral fits with WStat, it is recommended to randomize observations and check whether the resulting fitted parameters distributions are consistent with the input values.

### Example¶

The following table gives an overview over values that WStat takes in different scenarios

>>> from gammapy.stats import wstat
>>> from astropy.table import Table
>>> table = Table()
>>> table['mu_sig'] = [0.1, 0.1, 1.4, 0.2, 0.1, 5.2, 6.2, 4.1, 6.4, 4.9, 10.2,
...                    16.9, 102.5]
>>> table['n_on'] = [0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 10, 20, 100]
>>> table['n_off'] = [0, 1, 1, 10 , 10, 0, 5, 5, 20, 40, 2, 70, 10]
>>> table['alpha'] = [0.01, 0.01, 0.5, 0.1 , 0.2, 0.2, 0.2, 0.01, 0.4, 0.4,
...                   0.2, 0.1, 0.6]
>>> table['wstat'] = wstat(n_on=table['n_on'],
...                        n_off=table['n_off'],
...                        alpha=table['alpha'],
...                        mu_sig=table['mu_sig'])
>>> table['wstat'].format = '.3f'
>>> table.pprint()
mu_sig n_on n_off alpha wstat
------ ---- ----- ----- ------
0.1    0     0  0.01  0.200
0.1    0     1  0.01  0.220
1.4    0     1   0.5  3.611
0.2    0    10   0.1  2.306
0.1    0    10   0.2  3.846
5.2    5     0   0.2  0.008
6.2    5     5   0.2  0.736
4.1    5     5  0.01  0.163
6.4    5    20   0.4  7.125
4.9    5    40   0.4 14.578
10.2   10     2   0.2  0.034
16.9   20    70   0.1  0.656
102.5  100    10   0.6  0.663

### Notes¶

All above formulae are equivalent to what is given on the XSpec manual statistics page with the substitutions:

$\begin{split}\mu_{\mathrm{sig}} = t_s \cdot m_i \\ \mu_{\mathrm{bkg}} = t_b \cdot m_b \\ \alpha = t_s / t_b \\\end{split}$


text:
# Checking screen sizes for responsive site

I wrote this piece of code to make any breakpoint to the users choice:

@mixin breakpoint($media) {$max-width: 0;
@if($media == 'phones') {$max-width: $phone-medium; } @else if ($media == 'monitors') {
$max-width:$screen-medium;

//etc etc etc

} @else if($media > 0) {$max-width: $media; } @media only screen and (max-width:$max-width) {
@content;
}
}

This function comes with these variables:

//Phone width sizes
$phone-small: 480px;$phone-medium: 768px;

//Screen widths
$screen-small: 1024px;$screen-medium: 1280px;
\$screen-big: 1920px;

This can be used by typing:

@include breakpoint('phones') {
....
}

or

@include breakpoint(450px) {
....
}

Is there any better way to do this?

## 1 Answer

Personally, I think it's better to make the breaks according to the content of the site, rather than to a fixed size.

This is even more true nowadays with docens of different screen sizes for each type of device.

You should make sure the site will always look good by making the breaks work depending on the design, and no the other way around.

• do you mean that i could remove the pre defined variables (such as 1920 etc) and keep it with the user input only? (So they have to enter a screensize themself) This is made for my own framework, so its not just for myself – Bas Sep 17 '14 at 20:12
• Yes, that way the layout can vary depending on the design of the site and the screen sizes where it "breaks". – Alexis Sep 18 '14 at 1:02
• Hmm, i'm not really sure, just about the way it had been written. Is there anything you can point out on that? Maybe about using 'max-width'? Maybe that i can use multiple selectors? – Bas Sep 18 '14 at 5:50
• What do you think about posting an example where we could test it as "users" and see how it works? – Alexis Sep 18 '14 at 12:55
• I already did, see where i typed This can be used by typing: – Bas Sep 18 '14 at 13:30


text:
# Period of motion of an object dropped through the Earth

1. Oct 25, 2016

### kubaanglin

1. The problem statement, all variables and given/known data
An object of mass $m$ moves in a smooth, straight tunnel dug between two points on the Earth’s surface. Show that the object moves with simple harmonic motion, $a = - ω^2 x$. Find the period of this motion. You can assume that the Earth’s density is uniform.
2. Relevant equations
$$F=G\frac{Mm}{r^2}$$
$$V_{sphere}=\frac{4}{3}πr^3$$
$$ρ = \frac{M}{V}$$
3. The attempt at a solution
I have been staring at this problem for the past 30 minutes. I would be very appreciative if someone could give me a hint on how to start this. I intuitively understand why the object would experience simple harmonic motion, I just can't figure out how to express it mathematically, especially since the hole does not have to go through the center of Earth.

Last edited: Oct 25, 2016
2. Oct 25, 2016

### Simon Bridge

For SHM, $F\propto r$ when r is the displacement from an equilibrium point.
Notice that the two points on the surface of the earth do not have to be diametrically opposite each other.
However - you have $F\propto r^{-2}$ for the case that r>R. But that is not the case here.

How does M vary with r when r<R (R=radius of the earth)?

3. Oct 25, 2016

### kubaanglin

$M=\frac{4}{3}πr^3ρ=(\frac{4}{3}πr^3)(\frac{M_{Earth}}{\frac{4}{3}πR^3})=\frac{r^3M_{Earth}}{R^3}$
$a=G\frac{M}{r^2}=G\frac{\frac{r^3M_{Earth}}{R^3}}{r^2}=G\frac{rM_{Earth}}{R^3}$
Is this correct so far?

4. Oct 26, 2016

### SammyS

Staff Emeritus
That is the magnitude of radial acceleration if the object were not constrained to move in the tunnel.

Multiply by mass, m, to find the magnitude of the gravitational force. What component of that force is in the direction of the tunnel?

5. Oct 26, 2016

### kubaanglin

$F = G\frac{rmM_{Earth}}{R^3}\cosθ$ where $θ$ is the angle between $r$ and the direction of motion. $\cosθ = \frac{x?}{r}$.

6. Oct 26, 2016

### haruspex

If you fall into a vertical hole, what is the angle between your direction of motion and the line joining you to the centre of the Earth?

7. Oct 26, 2016

### kubaanglin

The angle would be $0°$ so then $\cos(0°)=1$. That leaves me with $F = G\frac{rmM_{Earth}}{R^3}$. How would that account for the component of the force if the tunnel was not through the center of the Earth?

8. Oct 26, 2016

### haruspex

I'm sorry - I misread the question.
So let x be the distance from the midpoint of the tunnel. Express theta as a function of x and r.

9. Oct 26, 2016

### kubaanglin

So $\cos(θ)=\frac{x}{r}$ and then $F = G\frac{mMx}{R^3}$?

10. Oct 26, 2016

### haruspex

Yes.

11. Oct 26, 2016

### kubaanglin

Okay, so then
$a=ω^2x=G\frac{Mx}{R^3}$
$ω=\sqrt{\frac{GM}{R^3}}=\frac{2πr}{T}$
$T=\frac{2πr}{\sqrt{\frac{GM}{R^3}}}$
Is this correct?

Last edited: Oct 26, 2016
12. Oct 26, 2016

### haruspex

What is r there? Check the dimensional consistency.

13. Oct 26, 2016

### kubaanglin

Oh, $ω=\frac{2π}{T}\neq\frac{2πr}{T}$
Then this should be the answer:
$$T=\frac{2π}{\sqrt{\frac{GM}{R^3}}}$$
$$T=\frac{2π}{\sqrt{\frac{(6.67⋅10^{-11})(5.96⋅10^{24})}{(6.37⋅10^6)^3}}}=5066 s=84.44 min$$

Last edited: Oct 26, 2016
14. Oct 26, 2016

### haruspex

Looks right.

15. Oct 26, 2016

### kubaanglin

Thank you so much for your help!


text:
# Talk:Latin square

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## Latin squares and mathematical puzzles

I added a paragraph on a mathematical puzzle that illustrates a generalized version of Latin-square orthogonality. Cullinane 17:33, 14 August 2005 (UTC)

## Explanation of reverts

User:Patrick made quite a few changes which I reverted. The main error is in the interprettation of the square as a multiplication table. In this case the rows and columns are assumed indexed by the elements of the quasigroup. It is not the first row and first column that form the index. Reduced Latin squares correspond to quasigroups with an identity element, aka loops. For quasigroups without identity, the multiplication table is not reduced. Also, it is certainly NOT true that all the quasigroups on 4 elements are groups; the discussion of squares of order 4 is inaccurate. --Zero 01:07, 22 November 2005 (UTC)

I made some clarifications and corrections:
There are 576 Latin squares of size 4. Each can be interpreted as a multiplication table of a loop (i.e., a quasigroup with an identity element) where the indexes of the first row and first column both are the identity, hence the element in the first row and column is the identity, and the first element in each row is the row index, and the first element in each column is the column index.
Rearranging the 2nd to 4th row such that the order of the row indexes is the same as that of the column indexes, the number of possibilities reduces by a factor 6 to 96. Considering that the order of the elements is only a matter of notation of the table, the number of possibilities reduces again by a factor 6 to 16. Thus for a given set {a,b,c,d} we get 16 loops, which all turn out to be groups. There are 4 versions of the Klein four-group, differing only by the choice of which of the 4 elements is the identity. Apart from that there is only one, corresponding to the fact that the permutations of the other three elements form the automorphism group of this group, i.e., all three play the same role. The remaining 12 are all versions of the cyclic group of order 4, differing by the choice of which of the 4 elements is the identity, and which of the other 3 has order 2. Apart from that there is only one, corresponding to the fact that interchanging the remaining two elements generate the automorphism group of this group.
Frankly I don't see the point of any of the above. --Zero 11:41, 22 November 2005 (UTC)
The article only gives two 4x4 examples. My text discusses all Latin squares and is obviously more comprehensive. If you have criticism, that should be much more specific (errors, unclarities, etc.).--Patrick 12:23, 22 November 2005 (UTC)
This article is primarily about the combinatorial objects called Latin squares, not so much about algebraic structures related to them. From the combinatorial point of view, there are only two types of Latin square of order 4 since the others are all paratopic to one of those. So the existing list is complete (for order 5, also). Latin square do not have row and column indices (other than row and column numbers like every matrix has); rather, those indices are what must be added to a Latin square to make it into a multiplication table. Most of what you wrote above is about loops and groups and I don't think it would improve this page. --203.113.233.38 22:53, 22 November 2005 (UTC)
But it looks like the algebraic structures are not so different from "combinatorial objects", it is a way of explaining how 576 cases reduce to 2. If you like you can present the reasoning without (quasi)group terminology.--Patrick 00:30, 23 November 2005 (UTC)
The equivalence class operations are correctly defined in the article without any reference to algebraic structures or index rows/columns. That is the correct approach in an article on Latin squares, imo. Given those definitions, a demonstration of what happens for n=4 ought to use those definitions and not go off to another domain. --Zero 09:26, 23 November 2005 (UTC)

To see how many equivalence classes of each type there are is trivial for n=4. There are 4 reduced squares - two (A,B) with 2 in the (2,2) position and one each (C,D) with 3 or 4 in the (2,2) position. Every square is isotopic to one of A,B,C,D since every square can be reduced by sorting the rows and columns.

$A =\begin{matrix} 1&2&3&4\\2&1&4&3\\3&4&1&2\\4&3&2&1\end{matrix}\quad B =\begin{matrix} 1&2&3&4\\2&1&4&3\\3&4&2&1\\4&3&1&2\end{matrix}\quad C =\begin{matrix} 1&2&3&4\\2&3&4&1\\3&4&1&2\\4&1&2&3\end{matrix}\quad D =\begin{matrix} 1&2&3&4\\2&4&1&3\\3&1&4&2\\4&3&2&1\end{matrix}$

B is isotopic to C (swap rows 2,3, columns 2,3, and symbols 2,3). C is isotopic to D (swap rows 3,4, columns 3,4, and symbols 3,4). Next note that A contains six 2x2 Latin subsquares but B,C,D contain only four. Since all the operations that define paratopism preserve Latin subsquares, A lies in a different main class from B, C, D. Therefore there are two isotopy classes and two main classes. --Zero 09:26, 23 November 2005 (UTC)

I made a separate article, Small Latin squares and quasigroups. There seems to be an error in the above: of A and B, one is isotopic to C and D. Swapping the first two columns and the last two rows of A does not preserve the reduced form.--Patrick 10:31, 23 November 2005 (UTC)
Also, three have one Latin subsquare, and one has 6. The reasoning can still be used.--Patrick 10:54, 23 November 2005 (UTC)
Sorry, it is now fixed, I hope. --Zero 11:42, 23 November 2005 (UTC)
Of the 576 Latin squares, 288 are solutions of the 2 ×2 version of Sudoku, sometimes called Shi Doku [1]. With first row abcd and first column acbd this reduces to 2.
That could be mentioned in the existing section on Sudoku squares. --Zero 11:41, 22 November 2005 (UTC)
What do you think?--Patrick 03:01, 22 November 2005 (UTC)

## Name

Doesn't the name Latin Square originate from the ?first known example (from Roman times), namely the word square Sator Arepo Tenet Opera Rotas? Ben Finn 16:39, 2 April 2006 (UTC)

No, that is not a Latin square in the meaning of this article. For example, it has no "T" in the second row. As far as I have heard, the earliest known Latin squares appeared in Arabic manuscripts around the 12th century; this needs to be researched and put in the article. The name "Latin square" was coined by Euler in his paper of 1782 with a rather weak justification: he used Latin letters as the symbols. McKay 07:19, 17 May 2006 (UTC)

I believe that the term "Latin square" originates with Cayley (can't remember where I read that though), who was referring to Euler's work of 1782. Euler was using orthogonal Latin squares, one written with Latin letters and the other with Greek letters, so when Cayley wanted to talk about only one of the pair, he referred to the "Latin" square of Euler. Wcherowi (talk) 21:32, 10 August 2011 (UTC)

Donald Knuth has found references back to the 14th century for orthogonal latin squares. See [[2]]. —Preceding unsigned comment added by Neilerdwien (talkcontribs) 05:56, 5 December 2007 (UTC)

## Isotopy classes of order 7

The correct value is 564, not 563. See [3] for the history of this error. This is accepted by Colbourn and Dinitz. --McKay 07:29, 12 April 2006 (UTC)

## References

Where are the references for this article?

The article looks really useful for my purposes. I have found vague allusions to some of these concepts in other resources, and discovered others on my own, but this is the only resource I have found that clearly and succinctly explains all of these concepts in one place (with better notation and terminology than I have found elsewhere).

However, for me to refer to this material for my project, I need to know where it came from, so I need references. In particular, I need to know the origin of the orthogonal array representation, and the table of latin square equivalence classes.

I will try to find a journal article that introduces these concepts, and if I do I will try to remember to come back and post those references here.DonkeyKong the mathematician (in training) 04:26, 16 May 2006 (UTC)

It seems that the paper ([4]) that McKay mentioned above is a suitable reference for most of the information in this article.
I'm new to Wiki editing, so I don't know how to put references into an article. If someone else could put that reference in there that would be good. Thanks.
You can download a citation from [5] (i don't know if that's helpful)
DonkeyKong the mathematician (in training) 06:19, 16 May 2006 (UTC)

## Question

If all Latin squares can be reduced, does this not mean that every quasigroup is a loop? —The preceding unsigned comment was added by 149.135.97.128 (talk) 11:08, 10 January 2007 (UTC).

No. The operations which can be used to reduce a Latin square (permutation of the rows and columns) change the quasigroup to a different quasigroup. These operations are called isotopies, so the conclusion is that every quasigroup is isotopic to a loop. It is not true that every quasigroup is a loop. McKay 02:40, 11 January 2007 (UTC)

## Latin Squares and projective \ affine planes

I might be missing something, but shouldn't the article include a note (or a section) about the relation of Latin squares to projective (or affine) planes? Denes and Keedwell show a proof of this in "Latin Squares and Their Applications" (although I don't think the proof is theirs).

Gal Diskin —The preceding unsigned comment was added by 85.65.215.115 (talk) 17:23, 11 January 2007 (UTC).

## Orthogonal Latin Squares

This note is a reminder for me and others that it might be worthwhile to include something about Orthogonal latin squares, as detailed e.g in [6]. The problem of constructing such matrixes and especially the impossible 6x6 matrix problem has been popularized in a number of books and articles and might make a good read. Also, in the section about error correcting codes, I committed the sin of writing "orthogonal latin squares" without defining what it means. :-] EverGreg 15:34, 13 May 2007 (UTC)

## Problem with mention of orthogonal Latin squares

The article states, in its first reference to orthogonal Latin squares:

"Sets of Latin squares that are orthogonal to each other have found an application as error correcting codes in situations where communication is disturbed by more types of noise than simple white noise, such as when attempting to transmit broadband internet over powerlines. [1] [2] [3]"

Unfortunately, the word orthogonal in this sense has *not* been defined previously, which makes this mention of the word orthogonal inappropriate. Further, the word "orthogonal" *has* been used previously (unnecessarily, in my opinion) to discuss the simple ordinary representation of a Latin square as a matrix, which the article somehow feels the need to refer to as an "orthogonal array".

It is certainly a good idea to discuss orthogonal Latin squares in this article, but this must be done in such a way that the concept is defined and the terminology is clear before it starts to be used without definition.

I hope this problem can be fixed.Daqu (talk) 22:44, 20 February 2009 (UTC)

## Applications for experimental design

I came here looking for how Latin squares are used in experimental design. It's mentioned in the intro, but I wonder if an expert could add a paragraph under applications. I'm at a conference and a lot of people have mentioned how they used Latin squares for randomizing their experiments, and I don't know what they mean, so I thought that might be a good thing to have here. Thanks 209.255.241.2 (talk) 16:54, 25 March 2010 (UTC)

You've got a good point. Information on this is rather fragmented on wikipedia. OK starting points are Blocking (statistics), Randomized block design and especially Hyper-Graeco-Latin square design. The bigger picture may perhaps be found in Design of experiments. None of these are the readable introductions they should have been, but hopefully they work as pointers to further reading in the literature. The gist of it is that when you do an experiment where say four variables can each take two values (high/low) you want to try every possible combination but there may not be time or money to do that. So you want to get as much info with as few experiments as possible. First you ofcourse avoid doing the same experiment twice, but you also want to avoid that two of your variables change in lockstep in each experiment, because then you wouldn't have been able to tell their effects apart. Following a latin square is one way to design a good set of experiments. EverGreg (talk) 18:17, 25 March 2010 (UTC)
I'll start adding some references on Latin squares here. When I'm done, I'll update the article. Kiefer.Wolfowitz (talk) 20:22, 22 May 2010 (UTC)

## Merger proposal

An anonymous user proposed a merger. It seems straightforward to me to put Graeco-Latin square as a chapter of Latin square. In addition: As the previous discussions on this page shows, it's a problem that orthogonal latin squares are not properly defined here, they seem to tie in closely with the text. The Graeco-Latin article seems to go a long way in defining it, so this could be a good start, though the text in the Graeco-Latin article is less accessible than that of Latin square. (That is, I still don't know what "a orthogonal latin square" is, and barely what two mutually orthogonal ones are.) EverGreg (talk) 22:10, 5 April 2010 (UTC)

Don't merge There are enough subtopics of Latin square to need different articles. A basic Latin square is essentially about one factor in addition to rows or columns, a Graeco-Latin square about two or more. As for orthogonal Latin squares, they have to come in comparison of pairs: there must be at least one cell (and so implicitly exactly one out of the n2 cells) which has any particular combination of the n posible values of each of the two symbols being considered.--Rumping (talk) 16:41, 22 July 2010 (UTC)
Due to the fact that no agreement has been reached — and only two comments — in eight months, I am removing the merge templates from each page. Supertouch (talk) 00:31, 4 November 2010 (UTC)

## Error in example

The examples for the two latin squares of order 5 at the end of the text are from the same main class. Correct representatives for the two main classes are given at http://cs.anu.edu.au/~bdm/data/latin.html . In one of the main classes, all ordered pairs of numbers must be at most once within the square, none of the currently given squares have this property.Random rings (talk) 12:45, 4 June 2010 (UTC)

## Orthogonal array representation (refers to non-existing figure)

It says" the orthogonal array representation of the first Latin square displayed above is:" But this example refers to a square with numbers which is not displayed in the article.(maybe it has been edited out at some time) Should we change this to (1,1,A), (1,2,B) and so on? Or put back the original figure. The whole expression "the first latin square displayed above" is also a bit vague. (figure number nn might be better?)Bj norge (talk) 19:00, 14 April 2011 (UTC)

## Error Correction Codes Example Unclear

Questions: 1. There are four frequencies and four time slots. What then is the meaning of 'the letter A would then be picked up as: 12 12 123 124'? 2. 'But the 1,2 case is the only one that yields a sequence matching a letter in the above table...'. Should it read 'sequence' instead of 'case'? 3. 'Again, we are able to infer from the table of encodings that it must have been the letter A being transmitted.' Is it too much to spell out *why* we are able to infer this? — Preceding unsigned comment added by Polariseke (talkcontribs) 15:54, 24 February 2012 (UTC)


text:
## Dual solutions in convex stochastic optimization

This paper studies duality and optimality conditions for general convex stochastic optimization problems. The main result gives sufficient conditions for the absence of a duality gap and the existence of dual solutions in a locally convex space of random variables. It implies, in particular, the necessity of scenario-wise optimality conditions that are behind many fundamental … Read more

## Solving large-scale unit-commitment problems using dual dynamic programming and open-source solvers

The astonishing dimensions and complexity of power systems render them impossible to be managed without the help of cutting-edge software. Due to a lack of scalable, reliable and well documented free and open-source solutions, system operators, regulators, and government agencies often rely on proprietary software to provide them information that ultimately will be used to … Read more

## Metrizing Fairness

We study supervised learning problems for predicting properties of individuals who belong to one of two demographic groups, and we seek predictors that are fair according to statistical parity. This means that the distributions of the predictions within the two groups should be close with respect to the Kolmogorov distance, and fairness is achieved by … Read more

## Multi-fidelity robust controller design with gradient sampling

Robust controllers that stabilize dynamical systems even under disturbances and noise are often formulated as solutions of nonsmooth, nonconvex optimization problems. While methods such as gradient sampling can handle the nonconvexity and nonsmoothness, the costs of evaluating the objective function may be substantial, making robust control challenging for dynamical systems with high-dimensional state spaces. In … Read more

## DiversiTree: A New Method to Efficiently Compute Diverse Sets of Near-Optimal Solutions to Mixed-Integer Optimization Problems

While most methods for solving mixed-integer optimization problems compute a single optimal solution, a diverse set of near-optimal solutions can often be more useful. We present a new method for finding a set of diverse solutions by emphasizing diversity within the search for near-optimal solutions. Specifically, within a branch-and-bound framework, we investigate parameterized node selection … Read more

## A Gauss-Newton-based Decomposition Algorithm for Nonlinear Mixed-Integer Optimal Control Problems

For the fast approximate solution of Mixed-Integer Non-Linear Programs (MINLPs) arising in the context of Mixed-Integer Optimal Control Problems (MIOCPs) a decomposition algorithm exists that solves a sequence of three comparatively less hard subproblems to determine an approximate MINLP solution. In this work, we propose a problem formulation for the second algorithm stage that is … Read more

## A new sufficient condition for non-convex sparse recovery via weighted $\ell_r\!-\!\ell_1$ minimization

In this letter, we discuss the reconstruction of sparse signals from undersampled data, which belongs to the core content of compressed sensing. A new sufficient condition in terms of the restricted isometry constant (RIC) and restricted orthogonality constant (ROC) is first established for the performance guarantee of recently proposed non-convex weighted $\ell_r-\ell_1$ minimization in recovering … Read more

## A level-set-based topology optimization strategy using radial basis functions and a Hilbertian velocity extension

This work addresses the structural compliance minimization problem through a level-set-based strategy that rests upon radial basis functions with compact support combined with Hilbertian velocity extensions. A consistent augmented Lagrangian scheme is adopted to handle the volume constraint. The linear elasticity model and the variational problem associated with the computation of the velocity field are … Read more

## Two Benders-type Branch-and-Cut Algorithms for Capacitated Facility Location with Single-Sourcing

We consider mixed 0-1 integer programs of the form $\min\{cx+hy: Ax+By \geq b, x \in \{0,1\}^n,y \in \{0,1\}^p \times \R^{N-p}_+\}$ in which the subproblem arising when $x$ is fixed has special structure. One such example is the capacitated facility location problem with single-sourcing in which the linear programming relaxation of the subproblem is a transportation … Read more

## The Null Space Property of the Weighted $\ell_r-\ell_1$ Minimization

The null space property (NSP), which relies merely on the null space of the sensing matrix column space, has drawn numerous interests in sparse signal recovery. This article studies NSP of the weighted $\ell_r-\ell_1$ minimization. Several versions of NSP of the weighted $\ell_r-\ell_1$ minimization including the weighted $\ell_r-\ell_1$ NSP, the weighted $\ell_r-\ell_1$ stable NSP, the … Read more


text:
#include "AliBaseESDTask.h"
#include "AliFMDEventInspector.h"
#include "AliAODCentralMult.h"
This graph shows which files directly or indirectly include this file:

Go to the source code of this file.


text:
# Estimating models

The logitr() function estimates multinomial logit (MNL) and mixed logit (MXL) models using “Preference” space or “Willingness-to-pay (WTP)” space utility parameterizations. The basic usage for estimating models is as follows:

model <- logitr(
data,
choiceName,
obsIDName,
parNames,
priceName = NULL,
randPars = NULL,
randPrice = NULL,
modelSpace = "pref",
weightsName = NULL,
options = list()
)

## Data format

The data argument must be provided a data.frame arranged in “long” or “tidy” format [@Wickham2014]. Each row should be an alternative from a choice observation. The choice observations do not have to be symmetric (i.e. each choice observation could have a different number of alternatives). The data must also include variables for each of the following arguments in the logitr() function:

• choiceName: A dummy variable that identifies which alternative was chosen (1 is chosen, 0 is not chosen). Only one alternative should be chosen per choice observation.
• obsIDName: A sequence of repeated numbers that identifies each unique choice observation For example, if the first three choice observations had 2 alternatives each, then the first 6 rows of the obsID variable would be 1, 1, 2, 2, 3, 3.
• parNames: The names of the variables that will be used as model covariates. For WTP space models, the price variable should not be included in parNames as it is provided separately with the priceName argument.

The “Data Formatting and Encoding” vignette has more details about the required data format.

## Preference space models

The logitr() function assumes that the deterministic part of the utility function, $$v_{j}$$, is linear in parameters, i.e.:

$$$u_{j} = v_{j} + \varepsilon_{j} \hspace{0.5in} \varepsilon_{j} \sim \textrm{Gumbel}\left(0,\frac{\pi^2}{6}\right) \label{eq:utilityComponents}$$$

$$$v_{j} = \mathbf{β}' \mathrm{\mathbf{x}}_{j} + \alpha p_{j} \hspace{0.5in} \label{eq:utilityPreferenceObserved}$$$

where $$\mathbf{β}$$ is the vector of coefficients for non-price attributes $$\mathrm{\mathbf{x}}_{j}$$, $$\alpha$$ is the coefficient for price $$p_{j}$$. Accordingly, each parameter in the parNames argument is an additive part of $$v_{j}$$.

For example, if the observed utility for yogurt purchases was $$v_{j} = \beta x_{j} + \alpha p_{j}$$, where $$p_{j}$$ is “price” and $$x_{j}$$ is “brand”, then the parNames argument should be c("price", "brand"). This model can be estimated as follows using the yogurt data frame:

mnl_pref <- logitr(
data       = yogurt,
choiceName = "choice",
obsIDName  = "obsID",
parNames   = c("price", "brand"))
)

Use the summary() function to view a summary of the estimated model results. In the following summary, three coefficients are shown for the “brand” attribute, with "dannon" set as the reference level:

summary(mnl_pref)
#> =================================================
#> MODEL SUMMARY:
#>
#> Model Space:    Preference
#> Model Run:          1 of 1
#> Iterations:             21
#> Elapsed Time:  0h:0m:0.13s
#> Exit Status:             3
#> Weights Used?:       FALSE
#> robust?              FALSE
#>
#> Model Coefficients:
#>               Estimate StdError    tStat pVal signif
#> price        -0.366543 0.024365 -15.0439    0    ***
#> feat          0.491433 0.120061   4.0932    0    ***
#> brandhiland  -3.715428 0.145416 -25.5504    0    ***
#> brandweight  -0.641128 0.054498 -11.7643    0    ***
#> brandyoplait  0.734496 0.080642   9.1082    0    ***
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Model Fit Values:
#>
#> Log-Likelihood:         -2656.8878799
#> Null Log-Likelihood:    -3343.7419990
#> AIC:                     5323.7758000
#> BIC:                     5352.7168000
#> McFadden R2:                0.2054148
#> Number of Observations:  2412.0000000

## WTP space models

For models estimated in the WTP space, the modelSpace argument should be set to "wtp". In addition, the parNames argument should not contain the name of the column for the price attribute because there is no estimated “price” parameter in WTP space models. Instead, a “scale” parameter given by $$\lambda$$ will be estimated, and the column for “price” should be provided separately using the priceName argument.

For example, if the observed utility was $$v_{j} = \lambda \left(\omega x_{j} - p_{j}\right)$$, where $$p_{j}$$ is “price” and $$x_{j}$$ is “brand”, then the parNames argument should be "brand" and the priceName argument should be "price". This model can be estimated as follows using the yogurt data frame:

mnl_wtp <- logitr(
data       = yogurt,
choiceName = "choice",
obsIDName  = "obsID",
parNames   = "brand",
priceName  = "price",
modelSpace = "wtp",
options    = list(numMultiStarts = 10)
)

Since WTP space models are non-convex, it is recommended to use a multi-start search to run the optimization loop multiple times to search for different local minima. This was implemented in the above example by passing a numMultiStarts argument inside a list to the options argument (other options can be passed in the same list).

In the WTP space model, the estimated “brand” coefficients have units of dollars and can be interpreted as the average willingness to pay for each brand relative to the “Dannon” brand, all else being equal.

## Mixed logit models

To estimate a mixed logit model, use the randPars argument in the logitr() function to denote which parameters will be modeled with a distribution. The current package version supports normal ("n") and log-normal ("ln") distributions.

For example, assume the observed utility for yogurts was $$v_{j} = \beta x_{j} + \alpha p_{j}$$, where $$p_{j}$$ is “price” and $$x_{j}$$ is “brand”. To model "brand" parameter, $$\beta$$, with a normal or log-normal distribution, set randPars = c(size = "n") or randPars = c(size = "ln"):

mxl_pref <- logitr(
data       = yogurt,
choiceName = "choice",
obsIDName  = "obsID",
parNames   = c("price", "brand"),
randPars   = c(brand = "n"),
options    = list(numMultiStarts = 10)
)

Since mixed logit models are non-convex, it is again recommended to use a multi-start search to run the optimization loop multiple times to search for different local minima. Note that mixed logit models can take a long time to estimate, so setting large number for numMultiStarts could take hours or longer to complete.

## View results

Use the summary() function to print a summary of the results from an estimated model. The function will print the following based on the model settings:

• For a single model run, it prints some summary information, including the model space (Preference or WTP), log-likelihood value at the solution, and a summary table of the model coefficients.
• For MXL models, the function also prints a summary of the random parameters.
• If you set keepAllRuns = TRUE in the options argument, summary() will print a summary of all the multistart runs followed by a summary of the best model (as determined by the largest log-likelihood value).

Use statusCodes() to print a description of each status code from the nloptr optimization routine.

## Computing and comparing WTP

For models in the preference space, a summary table of the computed WTP based on the estimated preference space parameters can be obtained with the wtp() function. For example, the computed WTP from the previously estimated fixed parameter model can be obtained with the following command:

wtp(mnl_pref, priceName = "price")
#>                Estimate StdError    tStat  pVal signif
#> lambda         0.366543 0.024384  15.0320 0e+00    ***
#> feat           1.340724 0.359525   3.7292 2e-04    ***
#> brandhiland  -10.136406 0.582781 -17.3932 0e+00    ***
#> brandweight   -1.749121 0.182274  -9.5961 0e+00    ***
#> brandyoplait   2.003848 0.143285  13.9851 0e+00    ***

The wtp() function divides the non-price parameters by the negative of the price parameter. Standard errors are estimated using the Krinsky and Robb parametric bootstrapping method [@Krinsky1986]. Similarly, the wtpCompare() function can be used to compare the WTP from a WTP space model with that computed from an equivalent preference space model:

wtpCompare(mnl_pref, mnl_wtp, priceName = "price")
#>                      pref           wtp  difference
#> lambda           0.366543     0.3665854  0.00004238
#> feat             1.340724     1.3405708 -0.00015315
#> brandhiland    -10.136406   -10.1357346  0.00067139
#> brandweight     -1.749121    -1.7490629  0.00005807
#> brandyoplait     2.003848     2.0038178 -0.00003018
#> logLik       -2656.887880 -2656.8878779  0.00000194

## Making predictions

Estimated models can be used to predict expected choices and choice probabilities for a set (or multiple sets) of alternatives based on an estimated model. As an example, consider predicting choice probabilities for two of the choice observations from the yogurt dataset:

alts <- subset(
yogurt, obsID %in% c(42, 13),
select = c('obsID', 'choice', 'price', 'feat', 'brand')
)

alts
#>     obsID choice price feat   brand
#> 49     13      0   8.1    0  dannon
#> 50     13      0   5.0    0  hiland
#> 51     13      1   8.6    0  weight
#> 52     13      0  10.8    0 yoplait
#> 165    42      1   6.3    0  dannon
#> 166    42      0   6.1    1  hiland
#> 167    42      0   7.9    0  weight
#> 168    42      0  11.5    0 yoplait

In the example below, the expected choice probabilities for these two sets of alternatives are computed using the fixed parameter mnl_pref model:

probs <- predictProbs(
model     = mnl_pref,
alts      = alts,
obsIDName = "obsID"
)

probs
#>     obsID  prob_mean   prob_low  prob_high
#> 49     13 0.43684871 0.41507538 0.45803979
#> 50     13 0.03313009 0.02620996 0.04185603
#> 51     13 0.19155738 0.17623217 0.20774895
#> 52     13 0.33846382 0.31891140 0.35928491
#> 165    42 0.60763975 0.57309149 0.64106191
#> 166    42 0.02602079 0.01841343 0.03671153
#> 167    42 0.17803594 0.16200334 0.19478044
#> 168    42 0.18830353 0.16779635 0.20987379

The resulting probs data frame contains the expected choice probabilities for each alternative. The low and high values show a 95% confidence interval, estimated using the Krinsky and Robb parametric bootstrapping method [@Krinsky1986]. You can change the CI level by setting alpha to a different value (e.g. a 90% CI is obtained with alpha = 0.05).

Choice probabilities can also be predicted used WTP space models. For example, the predicted choice probabilities using the mnl_wtp model are nearly identical with those from the mnl_pref model:

probs <- predictProbs(
model     = mnl_wtp,
alts      = alts,
obsIDName = "obsID"
)

probs
#>     obsID  prob_mean   prob_low  prob_high
#> 49     13 0.43686166 0.41514663 0.45812349
#> 50     13 0.03312933 0.02651411 0.04212799
#> 51     13 0.19154885 0.17606783 0.20749348
#> 52     13 0.33846015 0.31828622 0.35897979
#> 165    42 0.60767236 0.57218829 0.64084438
#> 166    42 0.02601763 0.01836858 0.03677997
#> 167    42 0.17802398 0.16177482 0.19462268
#> 168    42 0.18828603 0.16728440 0.20955128

Similar to the predictProbs() function, the predictChoices() function can be used to predict choices using the results of an estimated model. In the example below, choices are predicted for the entire yogurt dataset, which was used to the estimate the mnl_pref model:

choices <- predictChoices(
model     = mnl_pref,
alts      = yogurt,
obsIDName = "obsID"
)

# Preview actual and predicted choices
#> 2.6     2      0              0
The resulting choices data frame contains the same alts data frame with an additional column, choice_predict, which contains the predicted choices. You can quickly compute the accuracy by dividing the number of correctly predicted choices by the total number of choices:
chosen <- subset(choices, choice == 1)
chosen$correct <- chosen$choice == chosen$choice_predict sum(chosen$correct) / nrow(chosen)
#> [1] 0.3681592


text:
## Solve x, y, z

Arc length of AB=60deg, Arc length of BC= 90deg

x=

y=

z=
• Anonymous commented

z = 105 degree y = 75 degree


text:
Canadian Mathematical Society www.cms.math.ca
 location:  Publications → journals → CMB
Abstract view

# On the Chow Groups of Supersingular Varieties

Published:2002-06-01
Printed: Jun 2002
• Najmuddin Fakhruddin
Features coming soon:
Citations   (via CrossRef) Tools: Search Google Scholar:
 Format: HTML LaTeX MathJax PDF PostScript

## Abstract

We compute the rational Chow groups of supersingular abelian varieties and some other related varieties, such as supersingular Fermat varieties and supersingular $K3$ surfaces. These computations are concordant with the conjectural relationship, for a smooth projective variety, between the structure of Chow groups and the coniveau filtration on the cohomology.
 MSC Classifications: 14C25 - Algebraic cycles 14K99 - None of the above, but in this section

© Canadian Mathematical Society, 2014 : https://cms.math.ca/


text:
# Temperature coefficient of resistivity

I came across a question recently:

A wire has a resistance of 2.1 ohm at 300 K, and 2.7 ohm at 373 K. What is the temperature coefficient of resistance of the metal?

I tried by simply putting the formula R2 = R1 (1 + A(T2 - T1)), and got the answer as 2/511.
But when I did -
R2 / R1 = Ro (1 + A(T2 - To) / Ro (1 + A(T1 - To)
taking To as 273 K, I was getting 2/457.

This got me wondering, I can take any value of To and get a different answer. Why is this so, since I am simply taking out the ratio here?

Temperature coefficient of resistance, $\alpha$, is most easily defined in terms of Celsius temperature, $\theta$. Thus$$R_\theta=R_0 (1 +\alpha \theta)$$ in which $R_0$ is the resistance at temperature 0°C and $R_\theta$ is the resistance at Celsius temperature $\theta$.
Note that $R_0$ has to be the resistance at temperature 0°C; it can't be the resistance at some other temperature of your choice.


text:
# Algebra Examples

Find the X and Y Intercepts
To find the x-intercept, substitute in for and solve for .
Rewrite the equation as .
Move to the right side of the equation by subtracting from both sides of the equation.
Take the root of both sides of the to eliminate the exponent on the left side.
The complete solution is the result of both the positive and negative portions of the solution.
First, use the positive value of the to find the first solution.
Next, use the negative value of the to find the second solution.
The complete solution is the result of both the positive and negative portions of the solution.
To find the y-intercept, substitute in for and solve for .
Solve the equation.
Simplify each term.
Remove parentheses around .
Raising to any positive power yields .
Subtract from to get .
These are the and intercepts of the equation .
x-intercept:
y-intercept:

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text:
# Mass Renormalization: Geometric Series of One-Particle Irreducible Diagrams

Pretty much everywhere I look it is stated that the full two point Green function (let's say for the Klein-Gordon field) is a geometric series in the one particle irreducible diagrams, ie. in momentum space,

$$G(k) = G_0(k)+G_0(k)\Sigma(k)G_0(k)+G_0(k)\Sigma(k)G_0(k)\Sigma(k)G_0(k) + \dots$$ $$= G_0(k)\big(1+\Sigma(k)G_0(k)+\Sigma(k)G_0(k)\Sigma(k)G_0(k)+\dots\big)$$

and that the sum of this is, using the geometric series formula, $$G(k)=\frac{G_0(k)}{1-\Sigma(k)G_0(k)}=\frac{1}{k^2+m^2-\Sigma(k)-i\epsilon}$$

(for instance on the first page on the top right hand corner of here, or the bottom/top of pages 56/57 here).

However, what is the justification for summing this geometric series in that way? It never seems to be justified and it doesn't seem that $$|\Sigma(k)G_0(k)|<1\,.$$ Even if I do renormalize the mass $$m$$ so that $$m_R^2 := m^2-\Sigma(k)$$ is finite, if the geometric series summation isn't justified the infinities won't cancel out and everything will still diverge. Is there a resummation going on implicitly? Is this step completely nonperturbative?

$$\mathbf{Edit\;1}$$: Basically as I see it, the situation is this: I ask what the amplitude for the propagator is (in momentum space), and you say $$\frac{G_0(k)}{1-\Sigma(k)G_0(k)}=\frac{1}{k^2+m_R^2-i\epsilon}$$ where $$m_R^2=m^2-\Sigma(k)$$ is the (finite) mass. Then I notice that you can calculate it approximately using the perturbation series given by the theory, which is $$G_0(k)\big(1+\Sigma(k)G_0(k)+\Sigma(k)G_0(k)\Sigma(k)G_0(k)+\dots\big)\,.$$ However I notice that $$\Sigma(k)G_0(k)>1$$ and I know that for $$x>1$$ the perturbation series given by $$1+x+x^2+\cdots$$ is not a good approximation to $$\frac{1}{1-x}\,,$$ at any order of the perturbation series. So it would seem that perturbation theory fails because it doesn't give a good approximation, at any order, for the propagator. So then I ask what is the justification for this whole thing in the first place? Are they basically just seeing that the perturbation series for the propagator is the same as the perturbation series for $$\frac{1}{1-x}\,,$$ but realizing that $$|x|>1$$ and then assuming that the actual propagator must be $$\frac{1}{1-x}\,?$$ Because this would seem to be a completely nonperturbative step.

$$\mathbf{Edit\;2}$$: I am going to do the actual computation which is confusing me, and if someone can point out where it is wrong (if it is) that would be a great help. By the way I did a Wick rotation previously, but I am not going to do that this time:

We have that $$\frac{G_0(k)}{1-\Sigma(k)G_0(K)}=\frac{1}{k^2-m_R^2}$$

where $$m_R$$ is the physical (and finite) mass, and where $$G_0(k)=\frac{1}{k^2-m^2}$$ where $$m$$ is the cut-off dependent bare mass. Rearranging, I get $$\Sigma(k)G_0(k)=1-G_0(k)(k^2-m_R^2)=1-\frac{k^2-m_R^2}{k^2-m^2}\,.$$ Let's look at the region where $$m_R^2 $$m^2\to\infty$$ as the cut-off is removed, so this is a very large region. Now it looks to me like that in the above expression for $$\Sigma(k)G_0(k)\,,$$ that the right hand side is ALWAYS greater than one, and even possibly near infinity for certain values of $$k$$ (or maybe it doesn't obtain values near infinity because $$k$$ has to be restricted to values below the cut-off, but this isn't really important). This would seem to make the whole geometric summation suspicious. Is there a mistake?

If I have a basic misunderstanding of how this works then I'd like to know, this is the most basic case of renormalization in QFT but I don't understand it.

• This things are hardly ever justified in the literature. I guess, the best a-posteriori justification is that it simply works: perturbative QFT gives sensible results on the $n$-loop level, but I hope somebody will answer this question to give a better one. Sep 22, 2016 at 18:32
• Theoretical physicists love to talk about asymptotic expansion in these cases. But I also hope that someone in the field answers with more details and clarifies the issue. Sep 22, 2016 at 18:39
• I am not sure you are aware that the two propagators $\frac{1}{k^2+m^2}$ and $\frac{1}{k^2-m^2}$ are not wick rotated part but actually are propagators in two different metric convention used in literature. They are written such that $k_0$ has poles $\sqrt {|k|^2+m^2}$ on real axis. The problem you are describing is just a result of the failed analysis you have done and would also appear with $\frac{1}{k^2+m^2}$. I will address this in further comments.
– ved
Sep 29, 2016 at 6:20
• Using two different metric convention is not going to change any physics and as such $\Sigma(k)$ will have the same behaviour. In the first case, you had $m_R^2=m^2-\Sigma(k)$ as the physical mass as a result bare mass has to go to $+\infty$ to cancel the $+\infty$ coming from $\Sigma(k)$ to get a finite result. In the second case, bare mass has to go to $-\infty$ to cancel the $+\infty$ coming from $\Sigma(k)$ in $-(m^2+\Sigma(k))$ as changing the metric convention won't change $\Sigma(k)$ behaviour.
– ved
Sep 29, 2016 at 6:32
• This would basically lead to $m_R^2<k^2<-\infty$, which does not make sense and you might want to reverse it to $-\infty<k^2<m_R^2$. If $k^2$ is always than physical mass then it would never appear as a pole in the propagator and you will never be able to generate physical mass by radiative correction as a result your whole calculation is for nothing and that's why I was treating k in high-energy limit in the answer.
– ved
Sep 29, 2016 at 6:39

You are asking a legitimate question. We can turn the whole situation on its head. Rather than beginning with the geometric series $$$$G(k) = G_0(k)+G_0(k)\Sigma(k)G_0(k)+G_0(k)\Sigma(k)G_0(k)\Sigma(k)G_0(k) + \dots, \quad (1)$$$$ and arriving at $$G(k)=\frac{G_0(k)}{1-\Sigma(k)G_0(k)}=\frac{1}{k^2+m^2-\Sigma(k)-i\epsilon} ,\quad (2)$$ we can go the other way around. Instead, we shall assert that eq. (2) is valid and call the dubious expansion of eq. (1) into question, and rightfully so (OP has called out the folly in assuming $$|\Sigma(k)G_0(k)|<1$$).

If you do your homework by poking around the QFT literature, you will uncover plenty of non-perturbative ways of deriving eq. (2) without resort to eq. (1). The old school perturbative QFT makes intuitive sense, but is a mathematical train wreck.

• Can you give a reference for a book that derives this in a non perturbative way? Dec 20, 2021 at 11:47
• @Suriya, try any book detailing Schwinger-Dyson Equation and its first order approximate solution. Mar 9 at 21:56

Perturbative expansions are typically divergent in field theory so I would venture that the full resumation of the 1 particle irreducible diagrams is actually divergent.

But this is no trouble if you just want to renormalize. Keep in mind that in practice you always renormalize at finite order in your expansion parameter so there really is no need to worry about the infinite resummation. The sum of the truncated geometric series is well defined, so in order to renormalize you define the counterterms such that your propagator is finite at the order on which you are doing your computations.

• OP's question was: how can be taking the sum of the complete geometrical series (which diverges, so the expression for the sum is not correct) justified in actual computations. I have no doubt that what you wrote is true, but it does not answer OP's question. Sep 23, 2016 at 6:30
• @SolenodonParadoxus I answered it in the sense that I have said that the complete series is (most likely) divergent so rigorously it is not justified to do it. When have you seen the resummation of the geometric series of all 1PI diagrams in an actual computation? Sep 23, 2016 at 9:57
• @AnarchistBirdsWorshipFungus Are you suggesting that if I set $m^2 = m_R^2+\Sigma(k)$ for some fixed $m_R\,,$ then the perturbation series, at some order, is a good approximation to $\frac{1}{k^2+m_R^2-i\epsilon}$? Because it doesn't seem to be and if not how is the summation justified?
– JLA
Sep 23, 2016 at 17:50
• JLA are you aware that even after renormalization, when you get finite contributions at every order in perturbation theory, and even when you are considering the coupling much smaller than one the radius of convergence of perturbative expansions is zero? this is the key of my point. The full summation of the perturbative expansion is (believed) to be divergent. Chech this physics.stackexchange.com/q/70411 . Therefore most likely it is never legitimate to sum all the diagrams, so don't worry about doing it Sep 30, 2016 at 13:47
• The full $\Sigma$ after summing all diagrams even adding all counterterms is (most likely) divergent. Sep 30, 2016 at 13:48

Your basic misunderstanding is that $|\Sigma(k)G_0(k)|>1$, which is not true and basically three principles are used at least to make sure that $|\Sigma(k)G_0(k)|<1$. First is very basic which is the renormalizability of the theory and second is local nature of counterterms. A third ingredient is not needed except in case of scalar particles which is termed as "naturalness" in the standard model.

The mass correction term $\Sigma(k)$ is a sum of infinitely many graphs contributing to the self energy of the particle which in general will have a linear divergence (like self-energy of electron) or quadratic divergence (like self energy of photon) in high-energy limit. In case of particles other than spin zero different regularization schemes can be used to make this behaviour only logarithmically divergent and a redefinition of mass parameter ( renormalization) absorbs this divergence.

This basically tells you that $\Sigma(k)$ behaves as $ln(k/m)$ with k can be replaced by a scale $\Lambda$ and you will have $|\Sigma(k)G_0(k)|$ behaviour as $ln(k)/k^2$ as $$G_0(k)=\frac{1}{k^2+m^2}$$

Now $ln(k)/k^2$ is always less than $1$ (as $ln(k)-k^2$ is always negative).

In case of scalar particles however you have $\Sigma(k)$ behaving as badly as quadratically divergent (not enough symmetries in the theory), for which you need to have a lot of cancellation involving $\Lambda^2$ term to smoothen out the resulting integrals to not diverge as badly as quadratic. There is however no reason to expect that such cancellations should "naturally" occur as a result you need to fine-tune the theory to obtain a proper cancellation in which you can have $m^2_{physical}<<\Lambda$. As a result of these cancellations, you will have a soften up behaviour of $\Sigma(k)$ which will not diverge as badly as $\Lambda^2$ and you will have $|\Sigma(k)G_0(k)|<1$ as required.

Fine-tuning problem however is not a physical solution and hence must be supplemented with extra symmetry of theory. A case of which is supersymmetry which supplies the symmetry behaviour of spin 1/2 superpartner to make the divergence logarithamic but that is another question.

I hope this explain all your concerns and try to get a grab of good qft book like Srednicki's qft.

• Apart from domain related issues, note that the rhs of your first equation is a well defined operator precisely if the geometric series with $Q=BA^{-1}$ converges. (take, for instance, $A=1$, $B=2$). Sep 23, 2016 at 6:39
• All right, see above.
– ved
Sep 23, 2016 at 9:51
• @ved I appreciate your response, can you see my edit to my question? Hopefully it's more clear now.
– JLA
Sep 23, 2016 at 18:11
• @JLA I have removed the unnecessary portions and made it as clear as I can.
– ved
Sep 24, 2016 at 7:40
• @ved Thank you for your response. Can you see the "Edit 2" in my post? I wrote it largely in response to your post.
– JLA
Sep 24, 2016 at 19:42

You may consider this as a selective diagramm summation and a definition of eventually divergent series as a finite function. This was known even in early Euler times how to sum such series when the formal expanstion parameter is not actually small. Read Hardy's book on this subject.

The true problem is not here. In Classical Electrodynamics one can calculate exactly the back reaction, and elas, it is a divergent addendum to the particle mass. It is because the back reaction (self-action) is a mainly a self-induction of a point-like charge. No external force can accelerate a charge due to this infinite selft-induction. Note, the self-action idea is an ansatz, not something unquestionable. That is why we have to discard the mass corrections in practically any theory: even finite, they are unnecessary because we had the observed mass in the original equations. Thus, discarding (mass renormalization).

In addition, the electron propagator is not gauge invariant.

Worse, in QED there always a practically unity probability to emit soft photons because the electron is permanently coupled to the photon oscillators. Your exercise does not take it into account; in QED it is taken into account later on. An electron is actually an infra-particle (google it).


text:
# Find bounded solutions of second order ODE system

Resulting from a Classical Mechanics problem, I have the following system of second order ODEs

\begin{align} \ddot{x} &= \frac{-kx}{(x^2+y^2)^{3/2}} \\ \ddot{y} &= \frac{-ky}{(x^2+y^2)^{3/2}} \end{align}

With $$k>0$$ a real constant. I want to find the initial conditions for which the solutions of the system are bounded, and the initial conditions for which the system passes through the origin.

I know the origin is an unstable equilibrium point of the system, but I have no idea how to continue from there. I guess converting into polar coordinates would help with it, but I have only seen examples of converting systems of first order equations, not second order.

Hint.

Calling $$p = (x(t),y(t))$$ and also

$$\cases{ T = \frac 12 ||\dot p||^2\\ U = -\frac{k}{||p||}}$$

we have the lagrangian

$$L = T - U = \frac 12 ||\dot p||^2+\frac{k}{||p||}$$

with the movement equations

$$\cases{ \ddot x(t) = -\frac{k x(t)}{\left(x(t)^2+y(t)^2\right)^{3/2}}\\ \ddot y(t) = -\frac{k y(t)}{\left(x(t)^2+y(t)^2\right)^{3/2}}}$$

The total energy is given by

$$E = T + U = \frac 12 ||\dot p||^2-\frac{k}{||p||}$$

NOTE

Making a change of coordinates

$$\cases{ x(t) = r(t)\cos(\theta(t))\\ y(t) = r(t)\sin(\theta(t)) }$$

on the former lagrangian we arrive at

$$L = \frac{1}{2} \left(r'(t)^2+r(t)^2 \theta '(t)^2-\frac{2 k}{r(t)}\right)$$

and now deriving the movement equations we arrive at

$$\left\{ \begin{array}{c} r''(t) = r(t) \theta '(t)^2-\frac{k}{r(t)^2} \\ 2 r(t)r'(t) \theta '(t)+r(t)^2 \theta ''(t)= (r(t)^2\theta'(t))' = 0 \\ \end{array} \right.$$

so we have

$$r(t)^2\theta'(t) = h_0$$

and substituting into the first movement equation we get

$$r''(t) = \frac{h_0^2}{r(t)^3}-\frac{k}{r(t)^2}$$

etc.

• I can't integrate the $E$ equation to know $\dot{p}$ because of the norm, so I should do a change of coordinates? Sep 14 '20 at 18:17
• For a given set of initial conditions, $E$ is constant all along the trajectory. For instance, for $p_0 = (x_0, y_0)$ and $\dot p_0 = (v_{x_0}, v_{y_0})$ we have $E_0 = \frac 12(v_{x_0}^2+ v_{y_0}^2)-\frac{k}{\sqrt{x_0^2+y_0^2}}$ Sep 14 '20 at 19:36
• so, going with the energetic argument, for certain values of the energy the trajectory will be bounded and for other it will not. Sep 14 '20 at 19:56
• Please. Have a look at ocw.mit.edu/courses/aeronautics-and-astronautics/… Sep 14 '20 at 20:05


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I also do not start work for another 1.5 months so I have a lot of free time right now (I am allocating this time to study for my GMAT). I am used to normally seeing positive results after putting in effort, but it seems as though my score is stuck beneath a ceiling. Since I have not started work, I am really short on money and already pushed my funds when purchasing the Manhattan Prep Books. I saw some posts with similar stories to mine and they recommended getting a tutor, but honestly, I can not afford that. I am really frustrated and know that I can do better but I am not sure how and would LOVE feedback and support. I have purchased the Manhattan Prep GMAT Books as well as the Official Guide Books (haven't quite looked at them yet. hehehe) and the Kaplan GMAT Premier 2016. I have taken 6 practice tests so far. KAPLAN: CAT 1: 590 IR2 Q41 V30 (April 8) ~~This was kind of my diagnostic test CAT 2: 650 IR4 Q44 V35 (April 10) At this point, I purchased the Manhattan Prep Books and went through all of them and the practice problems provided. KAPLAN: CAT 3: 640 IR7 Q44 V34 (May 4) CAT 4: 630 IR3 Q44 V33 (May 8) ~~I was really tired when I took this test and my mom kept interrupting my exam CAT 5: 660 IR6 Q46 V34 (May 10) Manhattan Prep: CAT 1: 650 IR 4.5 Q44 V35 (May 17) Analytical Section: I usually finish my essays with ease and have 5 minutes to spare even after checking my essay for errors. I have compared my essays to the ones in the Kaplan book and I think I should be getting 5/6 essays, but am not too sure. Considering paying for the GMAT Write. IR Section: When it comes to the IR section, either the questions seem super easy or extremely difficult. On the tests with the low IR scores, I had difficulty in simply trying to figure out what the questions were asking. I also often run out of time before I can finish the questions because I sometimes don't figure out how to answer the question until I have worked out some scenarios, which leads me to realize how to actually answer the question and then I spend time calculating the right answer. The fact is, I know how to answer the questions, but it is just taking me awhile to figure out how to answer them and to figure out what the question is asking. Quantitative Section: I almost always finish my quantitative section with about 10 to 15 minutes to spare. I have always been a fast test taker (idk if that's a good or bad thing. ahahahah) and usually finish sections early (did so for almost all standardized tests including SATs, all those random tests they gave us in high school, and even finals in college). It seems as though I at least one problem wrong in every category type. But I have noticed that I struggle more with Combinatorics, Consecutive Integers, Exponents and Roots, Divisibility and Primes (for Data Sufficiency), Geometry, Inequalities, and Percents. In response, I will review the problem and redo it later on and try to learn from the concept. I have reviewed those sections in the Manhattan Books and the Kaplan Books, but when the next question comes on the next test, I feel like I haven't made much progress. What should I do here? Sometimes I get questions wrong from simple mistakes but most of the time it's because I am just confused on how to do the problem and when I try to work out different scenarios, I get even more confused. At this point, I just make my best guess (I typically get these questions right) or I randomly select an answer when I really have no idea. When I review my tests, I got through the problem and try to understand the reason why I got it wrong and then I redo the problem later (so that I wouldn't have memorized the answer). When that same or a similar problem comes up on the next test, I can handle it with no problem, but I feel like I need to understand the methods more, but even when I apply the methods learned from the books, I am still confused during the test. Verbal Section: I also finish this section about 10-15 minutes early. For the sentence correction, I have noticed that I get confused with idioms easily. I am a native English speaker, but when I have choose between two phrases, I get confused. I have spent some time reading and memorizing idioms, but I feel like I can't memorize all of them and often forget them. oops! I have also noticed that I will get modifiers confused and add extra modifiers when they are not necessary. For Critical Reasoning, I often get half of them wrong (). I normally get stuck between the out-of-scope answer that is meant to throw you off and the actual answer. It's so frustraatttiinnggg!!! Okay, calming down now. Anyway, I also feel like for half of questions that I get wrong, I really have no idea what the author is saying. I separate the premises from the conclusion and try to look at only what is provided in the blurb but I feel like I often don't know where the author is going with something. For Reading Comprehension, I will get like one passage all correct and then like half or more wrong on the other passages. I know that you are supposed to separate your own knowledge from what is provided, but there are some passages that I really don't understand and get bored while reading. I try to tell myself to just read the passage and act as if I am going to explain it to someone later, but then I get confused on words and jargon. There are times when a detail question will ask me a question and it is based upon a word that I don't know. Even when reading the sentences around it and trying to pay attention to the tone and words used, I struggle with trying to figure out what the author is trying to convey. When I read the passages, I typically feel like I have a good grasp of what's going on. I create passage maps and write down topics, scope, and purpose on my scratch paper. I typically get main idea questions right but get confused on specific detail (I have difficulty finding all the details of the passage, especially if I can't understand what's going on, unless I paraphrase like each sentence and write it down [I find that when I do this, I have enough time to finish and actually do a bit better, but sometimes I still get the questions wrong bc I get confused]). I also get confused on inference questions for both RC and CR. I am not sure why, maybe because I am having difficulty finding the argument and trying to see what the author is assuming? I am not sure. I am planning on taking the test on June 24th and am pretty much free every weekday and when I am not busy serving at church, so any advice or tactics on how to improve my score are greatly appreciated! This is my first time posting and I have never really done this before, so please go all out with what you think is wrong with how I am studying, how I can improve, or any questions you may have. thanks so much! Last edited by maddiggity on Tue May 30, 2017 6:32 pm; edited 1 time in total maddiggity Newbie | Next Rank: 10 Posts Joined 18 May 2017 Posted: 4 messages Thu Jun 15, 2017 9:03 am I am also planning to apply in a couple of years, but I wanted to get my GMAT out of the way so that I don't have to worry about studying for it while I am working. ### GMAT/MBA Expert Rich.C@EMPOWERgmat.com Elite Legendary Member Joined 23 Jun 2013 Posted: 9324 messages Followed by: 478 members Upvotes: 2867 GMAT Score: 800 Thu Jun 15, 2017 9:28 am Hi maddiggity, GMAC has just announced that - starting July 11th - Test Takers can choose which the order in which they take the sections of the GMAT (from among 3 options). This could be significantly advantageous, especially for those Test Takers who tend to get too tired to perform at a high level later on in the Exam. As such, you might want to consider rescheduling your GMAT until later, so that you can have that option too. The extra study time might also be helpful for your preparations. GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com maddiggity Newbie | Next Rank: 10 Posts Joined 18 May 2017 Posted: 4 messages Tue May 30, 2017 6:20 pm Hi everyone. Today I took another Manhattan Prep and scored 620 (IR 3 Q42 V33). I was unable to study for a week or so. I am wondering if maybe that contributed to my lower score. I tried a different tactic of paying more attention to the questions and working on solutions a bit more and actually finished my quantitative section right on time instead of way ahead. I did, however, finish my verbal section 24 minutes early, even though I spent a lot more time outlining the passages and reading the questions. For the Critical Reasoning questions, I have been able to narrow down the answer choices to two potential answers, but on the ones that I got wrong, it was because I chose the wrong answer of the two potential answers. I feel stuck and frustrated. Please help! ### GMAT/MBA Expert DavidG@VeritasPrep Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 120 members Upvotes: 1153 GMAT Score: 770 Wed May 31, 2017 6:42 am maddiggity wrote: Hi everyone. Today I took another Manhattan Prep and scored 620 (IR 3 Q42 V33). I was unable to study for a week or so. I am wondering if maybe that contributed to my lower score. I tried a different tactic of paying more attention to the questions and working on solutions a bit more and actually finished my quantitative section right on time instead of way ahead. I did, however, finish my verbal section 24 minutes early, even though I spent a lot more time outlining the passages and reading the questions. For the Critical Reasoning questions, I have been able to narrow down the answer choices to two potential answers, but on the ones that I got wrong, it was because I chose the wrong answer of the two potential answers. I feel stuck and frustrated. Please help! There's no magic formula, but here's what I suggest: give yourself a little time before you take your next exam. You've taken 5 this month. That's not unreasonable, but when you're not seeing a positive trajectory, it's helpful to take a step back and see if you can tweak your approach. First, review the 7 tests you've taken again. Boil down the issues into 4-5 concrete strategic adjustments you can make before the next test. Maybe you can be better at picking easy numbers for tough algebra problems. Maybe you need to drill away on exponent problems, etc. Next, try to adjust your mindset. Once you get caught in the cycle of "I need to improve, why am I not improving, oh no, it's getting worse!" the struggles can become a self-fulfilling prophesy. Take the emotion out of it. This is a game that you're attempting to master, and, so long as you keep calmly assessing problem areas and addressing them, in time you'll get where you need to be. Don't read too much into each practice exam score. See it as an opportunity to see what's working and what isn't. If you find the frustration difficult to shake, try working in some mindfulness meditation, which research has shown to be beneficial for test-takers: https://www.theatlantic.com/health/archive/2013/05/study-meditation-improves-memory-attention/275564/ Once you've broken down the old tests, done some drilling, and practiced a bit of mindfulness, take an official test. Repeat the process. Then check back in and let us know where you are. _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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### GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
Joined
23 Jun 2013
Posted:
9324 messages
Followed by:
478 members
2867
GMAT Score:
800
Sat Jun 03, 2017 10:41 am

Based on everything that you've written, there are a variety of different issues to discuss:

1) You've been studying for about 2 months. Many Test Takers spend 3 months (or more) of consistent study time before they hit their 'peak' scores, so it's likely that you just have not put in enough time and effort yet to have scored higher. In that same way, you might naturally improve with time. However...
2) Many Test Takers who use a 'book heavy' study approach end up getting 'stuck' at a particular score level - and that appears to have happened to you as well. GMAC has publicly stated that the Official Score that you earn on Test Day is within +/- 30 points of actual ability. Assuming a similar 'swing' in how your CATs function, your 5 recent CAT score results show that you essentially performed the same each time (640 +/- a few points). You handle certain aspects of the Test consistently well, but you also make certain consistent mistakes. You appear to have 'self-taught' your way to this score level, but "your way" of approaching the GMAT is potentially keeping you from scoring any higher than that.
3) A 780+ is a score that over 99% of Test Takers will never hit, regardless of the number of time that they take the Official GMAT. As such, that goal isn't particularly realistic. Thankfully, no Business School actually requires applicants to score that high - so it's important to remember that the score that you "want" and the score that you "need" (to get into your first choice Business School) are NOT the same thing.
4) Finishing the Quant or Verbal sections early is NOT a good thing - it almost always means that you've 'rushed' through certain questions (and gotten them wrong) when you could have slowed down, taken more notes, done more work (or double-checked the work that you did do) and picked up some points.
5) Your CAT scores that you're a strong critical thinker, but you haven't focused enough on learning/practicing Tactics and learning the various built-in patterns that occur on the GMAT. All of that knowledge/activity will likely require that you invest in some new, non-book study resources.
6) You don't want to spend much money on the process. While that's a perfectly fine sentiment, you might look at all of this as more of an investment in your future. A few hundred dollars spent in the near term could potentially be worth millions of extra dollars to you later on.

1) What Schools are you planning to apply to?
2) When are you planning to apply?

GMAT assassins aren't born, they're made,
Rich

_________________
Contact Rich at Rich.C@empowergmat.com

maddiggity Newbie | Next Rank: 10 Posts
Joined
18 May 2017
Posted:
4 messages
Thu Jun 15, 2017 9:02 am
Thanks for the advice everyone. I took a couple days off of studying and just took some time to relax and take a break. I plan to take the test on June 24, so about a week and a half from now. In the meantime, I have been reviewing old tests and I went through all of the practice questions in the Official GMAT Guides for Quantitative and Verbal Review 2017. I am planning on taking a official practice test either this weekend or next week and will let you know about how it goes.

In terms of schools, I think that 720 is a bit more realistic at this point. I want to go to USC, but am also considering UCI, UCLA, Claremont, UC Berkeley, and Harvard.

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