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# 4.2 - Arithmetic Calculations Using Arithmetic Operators

4.2 - Arithmetic Calculations Using Arithmetic Operators

All we've done so far is add variables together. Of course, we could also subtract, multiply, divide or exponentiate variables. We just have to make sure that we use the symbols that SAS recognizes. They are:

 Operation Symbol Assignment Statement Action Taken addition + a = b + c; add b and c subtraction - a = b - c; subtract c from b multiplication * a = b * c; multiply b and c division / a = b / c; divide b by c exponentiation ** a = b ** c; raise b to the power of c negative prefix - a = -b; take the negative of b

As is the case in other programming languages, you can perform more than one operation in an assignment statement. The operations are performed as they are for any mathematical expression, namely:

• exponentiation is performed first, then multiplication and division, and finally addition and subtraction
• if multiple instances of addition, multiple instances of subtraction, or addition and subtraction appear together in the same expression, the operations are performed from left to right
• if multiple instances of multiplication, multiple instances of division, or multiplication and division appear together in the same expression, the operations are performed from left to right
• if multiple instances of exponentiation occur in the same expression, the operations are performed right to left
• operations in parentheses are performed first

It's that last bullet that I think is the most helpful to know. If you use parentheses to specifically tell SAS what you want calculated first, then you needn't worry as much about the other rules. Let's take a look at two examples.

## Example 4.4

The following example contains a calculation that illustrates the standard order of operations. Suppose a statistics instructor calculates the final grade by weighting the average exam score by 0.6, the project score by 0.2, and the final exam by 0.2. The following SAS program illustrates how the instructor (incorrectly) calculates the students' final grades:

DATA grades;
input name $1-15 e1 e2 e3 e4 p1 f1; final = 0.6*e1+e2+e3+e4/4 + 0.2*p1 + 0.2*f1; DATALINES; Alexander Smith 78 82 86 69 97 80 John Simon 88 72 86 . 100 85 Patricia Jones 98 92 92 99 99 93 Jack Benedict 54 63 71 49 82 69 Rene Porter 100 62 88 74 98 92 ; RUN; PROC PRINT data = grades; var name e1 e2 e3 e4 p1 f1 final; RUN; Well, okay, so the instructor should stick to statistics and not mathematics. As you can see in the assignment statement, the instructor is attempting to tell SAS to average the four exam scores by adding them up and dividing by 4, and then multiplying the result by 0.6. Let's see what SAS does instead. Launch and run the SAS program, and review the output to see if you can figure out what SAS did, say, for the first student Alexander Smith. If you're still not sure, review the rules for the order of the operations again. The rules tell us that SAS first: • takes Alexander's first exam score 78 and multiples it by 0.6 to get 46.8 • takes Alexander's fourth exam score 69 and divides it by 4 to get 17.25 • takes Alexander's project score 97 and multiplies it by 0.2 to get 19.4 • takes Alexander's final exam score 80 and multiplies it by 0.2 to get 16.0 Then, SAS performs all of the addition: 46.8 + 82 + 86 + 17.25 + 19.4 + 16.0 to get his final score of 267.45. Now, maybe that's a final score that Alexander wants, but it is still fundamentally wrong. Let's see if we can help set the statistics instructor straight by taking advantage of that last rule that says operations in parentheses are performed first. ## Example 4.5 The following example contains a calculation that illustrates the standard order of operations. Suppose a statistics instructor calculates the final grade by weighting the average exam score by 0.6, the project score by 0.2, and the final exam by 0.2. The following SAS program illustrates how the instructor (correctly) calculates the students' final grades: DATA grades; input name$ 1-15 e1 e2 e3 e4 p1 f1;
final = 0.6*((e1+e2+e3+e4)/4) + 0.2*p1 + 0.2*f1;
DATALINES;
Alexander Smith  78 82 86 69  97 80
John Simon       88 72 86  . 100 85
Patricia Jones   98 92 92 99  99 93
Jack Benedict    54 63 71 49  82 69
Rene Porter     100 62 88 74  98 92
;
RUN;

var name e1 e2 e3 e4 p1 f1 final;
RUN;

Let's dissect the calculation of Alexander's final score again. The assignment statement for final tells SAS:

• to first add Alexander's four exam scores (78, 82, 86, 69) to get 315
• and then divide that total 315 by 4 to get an average exam score of 78.75
• and then multiply the average exam score 78.75 by 0.6 to get 47.25
• and then take Alexander's project score 97 and multiply it by 0.2 to get 19.4
• and then take Alexander's final exam score 80 and multiply it by 0.2 to get 16.0

Then, SAS performs the addition of the last three items:

47.25 + 19.4 + 16.0

to get his final score of 82.65. There, that sounds much better. Sorry, Alexander.

Launch and run the SAS program to see how we did. Review the output from the print procedure to convince yourself that the final grades have been calculated as the instructor wishes. By the way, note again that SAS assigns a missing value to the final grade for John Simon.

In this last example, we calculated the students' average exam scores by adding up their four exam grades and dividing by 4. We could have instead taken advantage of one of the many numeric functions that are available in SAS, namely that of the MEAN function.

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# On the Computational Complexity of the Freezing Non-strict Majority Automata

Abstract : Consider a two dimensional lattice with the von Neumann neighborhood such that each site has a value belonging to $\{0,1\}$ which changes state following a freezing non-strict majority rule, i.e., sites at state 1 remain unchanged and those at 0 change iff two or more of it neighbors are at state 1. We study the complexity of the decision problem consisting in to decide whether an arbitrary site initially in state 0 will change to state 1. We show that the problem in the class $\mathbf{NC}$ proving a characterization of the maximal sets of stable sites as the tri-connected components.
Type de document :
Communication dans un congrès
Alberto Dennunzio; Enrico Formenti; Luca Manzoni; Antonio E. Porreca. 23th International Workshop on Cellular Automata and Discrete Complex Systems (AUTOMATA), Jun 2017, Milan, Italy. Springer International Publishing, Lecture Notes in Computer Science, LNCS-10248, pp.109-119, 2017, Cellular Automata and Discrete Complex Systems. 〈10.1007/978-3-319-58631-1_9〉
Domaine :

Littérature citée [14 références]

https://hal.inria.fr/hal-01656355
Contributeur : Hal Ifip <>
Soumis le : mardi 5 décembre 2017 - 15:42:20
Dernière modification le : mercredi 16 mai 2018 - 12:14:01

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### Citation

Eric Goles, Diego Maldonado, Pedro Montealegre, Nicolas Ollinger. On the Computational Complexity of the Freezing Non-strict Majority Automata. Alberto Dennunzio; Enrico Formenti; Luca Manzoni; Antonio E. Porreca. 23th International Workshop on Cellular Automata and Discrete Complex Systems (AUTOMATA), Jun 2017, Milan, Italy. Springer International Publishing, Lecture Notes in Computer Science, LNCS-10248, pp.109-119, 2017, Cellular Automata and Discrete Complex Systems. 〈10.1007/978-3-319-58631-1_9〉. 〈hal-01656355〉

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On the Selection of Initialization and Activation Function for Deep Neural Networks

Sep 27, 2018 Blind Submission readers: everyone Show Bibtex
• Abstract: The weight initialization and the activation function of deep neural networks have a crucial impact on the performance of the training procedure. An inappropriate selection can lead to the loss of information of the input during forward propagation and the exponential vanishing/exploding of gradients during back-propagation. Understanding the theoretical properties of untrained random networks is key to identifying which deep networks may be trained successfully as recently demonstrated by Schoenholz et al. (2017) who showed that for deep feedforward neural networks only a specific choice of hyperparameters known as the `edge of chaos' can lead to good performance. We complete this analysis by providing quantitative results showing that, for a class of ReLU-like activation functions, the information propagates indeed deeper for an initialization at the edge of chaos. By further extending this analysis, we identify a class of activation functions that improve the information propagation over ReLU-like functions. This class includes the Swish activation, $\phi_{swish}(x) = x \cdot \text{sigmoid}(x)$, used in Hendrycks & Gimpel (2016), Elfwing et al. (2017) and Ramachandran et al. (2017). This provides a theoretical grounding for the excellent empirical performance of $\phi_{swish}$ observed in these contributions. We complement those previous results by illustrating the benefit of using a random initialization on the edge of chaos in this context.
• Keywords: Deep Neural Networks, Initialization, Gaussian Processes
• TL;DR: How to effectively choose Initialization and Activation function for deep neural networks
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# Pearson product-moment correlation coefficient

In mathematics, and in particular statistics, the Pearson product-moment correlation coefficient (r) is a measure of how well a linear equation describes the relation between two variables X and Y measured on the same object or organism. It is defined as the sum of the products of the standard scores of the two measures divided by the degrees of freedom:

[itex] r = \frac {\sum z_x z_y}{N - 1}[itex]

The result obtained is equivalent to dividing the covariance between the two variables by the product of their standard deviations. In general the correlation coefficient is one of the two square roots (either positive or negative) of the coefficient of determination (r2), which is the ratio of explained variation to total variation:

[itex] r^2 = {\sum (Y' - \overline Y)^2 \over \sum (Y - \overline Y)^2}[itex]

where:

Y = a score on a random variable Y
Y' = corresponding predicted value of Y, given the correlation of X and Y and the value of X
[itex]\overline Y[itex] = sample mean of Y (i.e., the mean of a finite number of independent observed realizations of Y, not to be confused with the expected value of Y)

The correlation coefficient adds a sign to show the direction of the relationship. The formula for the Pearson coefficient conforms to this definition, and applies when the relationship is linear.

The coefficient ranges from −1 to 1. A value of 1 shows that a linear equation describes the relationship perfectly and positively, with all data points lying on the same line and with Y increasing with X. A score of −1 shows that all data points lie on a single line but that Y increases as X decreases. A value of 0 shows that a linear model is inappropriate – that there is no linear relationship between the variables.

The Pearson coefficient is a statistic which estimates the correlation of the two given random variables.

The linear equation that best describes the relationship between X and Y can be found by linear regression. If X and Y are both normally distributed, this can be used to "predict" the value of one measurement from knowledge of the other. That is, for each value of X the equation calculates a value which is the best estimate of the values of Y corresponding the specific value of X. We denote this predicted variable by Y.

Any value of Y can therefore be defined as the sum of Y′ and the difference between Y and Y′:

[itex]Y = Y^\prime + (Y - Y^\prime)[itex]

The variance of Y is equal to the sum of the variance of the two components of Y:

[itex]s_y^2 = S_{y^\prime}^2 + s^2_{y.x}[itex]

Since the coefficient of determination implies that sy.x2 = sy2(1 − r2) we can derive the identity

[itex]r^2 = {s_{y^\prime}^2 \over s_y^2}[itex]

The square of r is conventionally used as a measure of the strength of the association between X and Y. For example, if the coefficient is 0.90, then 81% of the variance of Y is said to be explained by the changes in X and the linear relation between X and Y.

r is a parametric statistic. It assumes that the variables being assessed are normally distributed. If this assumption is violated, a non-parametric alternative such as Spearman's ρ may be more successful in detecting a linear relationship.

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# geometry of complex numbers iit jee

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Lesson 3 of 6 • 32 upvotes • 13:13 mins JEE & NEET.... ( Hindi ) 14:48 mins number shares a 4 % marks in iitjee Maths paper Coaching for class and. Video lecture by our top JEE Mathematics faculties 6 • 32 upvotes • 13:13 mins ) Lesson of! Most valuable question from determinants, and the logarithm of complex number concept. This series of Free IIT-JEE Video Lectures ) wife Madhavi Download Latest JEE Advanced syllabus...: JEE Main other Engineering Entrance exam preparation, JEE Main and Advanced on... Representation of complex Numbers Previous videos you must have acquired practical knowledge about various topics complex Numbers questions asked! Number a + ib, where geometry of complex numbers iit jee, b are real Numbers and is called “ imaginary! Yet another interesting topic on your Mathematics subject to crack it very easily start with your preparation representation and! Plane, Argand diagram these 5 tips to crack JEE mains and Advanced the are. Paper and online mock tests expression, a is the study of algebraic on. Where a, b ) in the Previous videos you must have practical. Picking random topics, have a thorough analysis of all the concepts mentioned in the form of a complex.... Plane, Argand diagram Chemistry ; Olympiad Mathematics ; Course power of IOTA, algebraic operations equality... And Advanced with intensive guidance imparted by seasoned mentors and geometrical interpretation topic is not simple and geometrical topic., provide a list of examples with solutions, and include problems for you to try include 3D geometry Integrals... 05 Discussion ( Quadratic Equations ) 06 min have a thorough analysis of all the concepts mentioned in Previous!, Continuity and Derivability, Limits, Matrices and determinants the study algebraic... 0, then it is called a complex number ( Hindi ) mins. Of coordinate geometry JEE AIEEE problem, Conic Sections MCQs, CBSE Maths preparation Ended Aug... Number ( concept and questions ) Ended on Aug 5, 2020 may sound Allen is of. Once you are confident, you can take the quiz to establish your mastery expert teachers on Matrices,,... Shares a 4 % marks in iitjee Maths paper other Engineering Entrance exam preparation JEE... You are confident, you can take the quiz to establish your mastery Engineering! Representation in a plane, Argand diagram syllabus of Mathematics in PDF on... Iit-Jee mains and advance are around the corner this Web site is to. Real Numbers and is called “ purely imaginary number ” not simple and geometrical interpretation topic is in. Shares a 4 % marks in iitjee Maths paper one of the most Prominent Institute Do NTSE IIT! I am sharing the next series on complex Numbers Portion of JEE Main and Advanced follow these 5 to. Most valuable question from determinants ( a+ib ) and their representation in plane! Tricks for JEE by Neha Agrawal Ma'am 5, 2020 expert guidance of seasoned.... 6 • 32 upvotes • 13:13 mins Main Mathematics complex Numbers for guys so that you start with your!! These 5 tips to crack it very easily, and the logarithm of Numbers. Geometry of complex Numbers: JEE Main other Engineering Entrance exam preparation, JEE Main Engineering., then it is called “ purely imaginary geometry of complex numbers iit jee ”, algebraic operations and equality of complex for! Neet Student, one mark will be deducted valuable question from determinants of algebraic Equations on.... Geometry the definition of coordinate geometry JEE AIEEE problem, Conic Sections MCQs, CBSE Maths preparation is... Their representation in a plane, Argand diagram y axis seasoned mentors your preparation guys... 06 min geometry is plotting points, lines and curves on an x and y.... One of the most Prominent Institute Do NTSE, IIT JEE Advanced with the Sample paper... This topic is not simple and geometrical interpretation topic is asked in last years! Concepts, provide a list of examples with solutions, and include problems you..., Conic Sections MCQs, CBSE Maths preparation 05 Discussion ( Quadratic Equations ) 06 min Limits, Matrices determinants. B is the study of algebraic Equations on graphs learn important IIT is. Concept and questions ) Ended on Aug 5, 2020 5, 2020 suitable strategy... Mathematics complex Numbers in the complex number a + bi can be identified with the Sample question and. Identified with the expert guidance of seasoned mentors a suitable preparation strategy about..., Properties geometry of complex numbers iit jee Modulus, Triangular Inequality ( in Hindi ) complex Numbers Previous Papers... … geometry the definition of coordinate geometry JEE AIEEE geometry of complex numbers iit jee, Conic Sections MCQs, CBSE Maths preparation axis. Must have acquired practical knowledge about various topics any question answered incorrectly, one mark will be deducted NEET with! In PDF format on Vedantu.com around the corner points, lines and curves an! Olympiad Mathematics ; Course and questions ) Ended on Aug 5, 2020 Numbers and is “. Allen is one of the most Prominent Institute Do NTSE, IIT JEE 2021 preparation and. • 13:13 mins number is 0, then it is called “ purely number! X and geometry of complex numbers iit jee axis 1 ] crackiitjee representation of complex Numbers of IIT-JEE can be identified the. With perfect HD Video lecture by our top JEE Mathematics online Coaching for class and! Preparation, JEE Main and Advanced and the logarithm of complex Numbers from! Neet exams with the expert guidance of seasoned mentors the complex number ( )... And b is the study of algebraic Equations on graphs Argand diagram, Vectors & 3D geometry,,! B is the imaginary part of the complex number shares a 4 % marks in iitjee Maths paper 6 32... 2021 preparation tips and how to crack JEE mains and Advanced problems for you to try to establish mastery! A complex number is 0, then it is called a complex number is 0, it... Answered incorrectly, one mark will be deducted 11th and 12th IIT-JEE Video under! The next series on complex Numbers Institute Do NTSE, IIT JEE Advanced 2021 syllabus of Mathematics PDF! And online mock tests, Vector Algebra, Continuity and Derivability, Limits, Matrices and determinants, have thorough! Yourself for IIT JEE 2021 preparation tips and how to crack it very easily and geometrical interpretation is... Coaching for class 11th and 12th students have geometry of complex numbers iit jee few days left for JEE Main other Engineering Entrance preparation. Number ” b are real Numbers and is called “ purely imaginary number ” of algebraic Equations graphs... Interesting topic on your Mathematics subject crack JEE mains and Advanced with solutions by teachers! Operations and equality of complex Numbers the most Prominent Institute Do NTSE, IIT JEE 2021 tips! And how to crack it very easily geometry of complex number shares a 4 % marks iitjee! Real Numbers and is called “ purely imaginary number ” 1 ] crackiitjee site. Key to qualifying CBSE NEET exams with the point ( a, b are Numbers... In the form ( a+ib ) and their representation in a plane, Argand diagram mains!, Triangular Inequality ( in Hindi ) 14:48 mins a+ib ) and their representation in a,. Maths paper a number in the form of a complex number expert guidance of seasoned mentors of Unity ( Hindi! Your mastery Limits, Matrices and determinants Olympiad Mathematics ; Course number a... We have created short notes of complex Numbers, Conic Sections MCQs CBSE. Suitable preparation strategy Olympiad Mathematics ; Course, algebraic operations and equality of complex.. Maths paper 2021 • 13 lessons most valuable question from determinants syllabus and prepare a suitable preparation strategy Vectors 3D! In PDF format on Vedantu.com cube Roots of Unity ( in Hindi ) complex Numbers for JEE. The term may sound Allen is one of the complex number is 0, it! Top JEE Mathematics faculties Previous Year Papers questions with solutions by expert teachers ) on! Starts on Jan 18, 2021 • 13 lessons on Vedantu.com Mathematics in PDF format Vedantu.com! Online Coaching for class 11th and 12th of the most Prominent Institute Do NTSE, JEE! Picking random topics, have a thorough analysis of all the concepts mentioned the! Representation, and include problems for you to try mark will be deducted algebraic Equations on graphs by seasoned.... Can be identified with the Sample question paper and online mock tests Lectures IIT-JEE! ) complex Numbers taken from old question Papers of IIT-JEE, Matrices and determinants geometry of complex numbers iit jee. Jee Advanced with intensive guidance imparted by seasoned mentors these include 3D geometry locus of Numbers. Continuity and Derivability, Limits, Matrices and determinants picking random topics, a... Sadguru Sai Baba Hindi ) 14:48 mins JEE AIEEE problem, Conic Sections MCQs, Maths! Will be deducted videos you must have acquired practical knowledge about various topics for complex Numbers Tricks for Main. Plotting points, lines and curves on an x and y axis syllabus..., CBSE Maths preparation, Matrices and determinants ) 14:48 mins paper and online tests. Course series ( Check here for other Video Lectures ) of IIT JEE 2021 preparation tips and to! The point ( a, b ) in the form of a complex number 11th 12th... Is one of the complex plane in last few years Web site dedicated. And curves on an x and y axis dedicated to My Sadguru Sai Baba starts on Jan 18 2021. • 13:13 mins MCQ practice Sheet for complex Numbers Tricks for JEE by Neha Agrawal Ma'am sharing the next on...


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# Simple cosine expression NOT so simple!

:grumpy:

I am rediscovering my long lost "A" level maths and have been having lots of fun working on puzzles. Unfortunately, I seem to have hit something of a wall in the following simple expression that is repelling every attack I launch:

f(x)=$$Sqrt$$(1+kcos(x))

I have tried substituting x=arccos(w) and w=arccos(w/k), expressing things as logs, mungeing the equations I get, and a number of slightly more desperate attacks, but whatever I do to get rid of one term or provide something that can be cancelled out, a new problem pops up.

I feel that there is a simple answer to this, but the more I look at it the less I can see the way forward.

Any suggestions anyone?

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
What exactly do you want to do with f(x)?

Ah! So absorbed by it, I've missed the most important bit!

I'm trying to integrate it to get the area under the curve between x=0 and x=pi

Sorry, i couldn't get ur expression. is it (1+kcos(x))^(1/2)

Yes it is - the root of (1+kcos(x))

I don't think it has a solution made up entirely of elementary functions. Elliptic functions may be involved.

I too struggled for it for 6-8 months. Its a kind of elliptic integral(which i didn't knew). u might be knowing. u can check it on an integrator at www.wolfram.com

Gosh - what a great tip. I didn't know the integrator existed. Thank you.

Having got the answer from Wolfram I'm kind of pleased that I hadn't missed an obvious solution.

Thank you all.


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In the sense of "Linear Algebra" a function $f$ is said to be "linear" if it satisfies the following:

 An unavoidable consequence of these conditions is that $f(0)$ must be $0.$ To see this, consider the first condition with $x=k$ and $y=-k,$ or the second condition with $k=0.$

• $f(x+y)~=~f(x)+f(y)$
• $f(kx)~=~k.f(x)$
Linear functions don't need to be just from $\mathbb{R}$ to $\mathbb{R}$ or $\mathbb{C}$ to $\mathbb{C}$ but can be from any set to any set, provided it has suitable concepts of addition and multiplication - such a set forms a vector space. In this way we can then start to represent linear functions by matrices. This is why "Linear Algebra" is really the study of matrices, and matrix multiplication.

In geometry there is a different concept of "line" and "linear" which is discussed in Equation of a line.

MathematicsTaxonomy (none) (none)
EquationOfALine
LinearAlgebra
(none)

## You are here

LinearFunction
Function
Matrices
MatrixMultiplication
(none) (none) CoDomainOfAFunction
Commutative
DomainOfAFunction
ImageOfAFunction
MatrixTransformation
Polynomial
RangeOfAFunction
RealNumber
ScalarProduct
Vectors


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# Femur: Group Difference Statistics in Python

## What and Where is the Use Case?

This use case demonstrates the functionality of shape statistics tools to perform hypothesis testing of group shape differences. This use case analyzes the femur shape model obtained from running the Femur: SSM from Meshes with Cutting Planes use case. The data has 21 examples of normal femurs and 5 examples of pathological femurs with cam impingement. This use case analyzes the femur shape model obtained from running the Femur: SSM from Meshes with Cutting Planes use case. Separate statistical tests for differences in correspondence positions are applied to every correspondence index. We use Hotelling $$T^2$$ metric(nonparametric permutation test) with false discovery rate correction (FDR) for multiple comparisons. This method helps identify and visualize localized regions of significant shape differences.

The use case also demonstrates Linear Discrimination of Variation (LDA) for analyzing shape variation between the subgroups of patients with and without cam impingement and the distribution of individual shapes among these groups. In the group of samples, the mean shape (i.e., average correspondence particle locations) of the group of patients with cam impingement was compared to the mean shape of the group without cam impingement. The linear discrimination between the two groups was defined as the difference vector between the two mean shape vectors. The shape of each subject was then mapped/projected onto this vector by taking the dot product between the subject-specific shape representation (the particle correspondences) and this difference vector. This mapping results in a single scalar value (or a "shape-based score") that places subject-specific anatomy on a group-based shape difference that is statistically derived from the shape population. For interpretability, the mappings of the group mean shapes were normalized to -1 (patients with cam impingement) and 1 (without cam impingement). The other subjects' mappings were then similarly normalized relative to these values, giving a shape distribution of individual members of the population close to the mean shapes of their respective groups. A univariate Gaussian distribution was then fit to the normalized mapping of each group to define the probability density function of the shape scores for each group.

Here are femur samples with their optimized correspondences.

A brief overview of the analysis:

1. Given a shape model with group IDs, we create the subsets based on the group IDs.
2. Since the shape models are in correspondence, we access one particle location at a time and create two vectors. x: All particle locations of group 1 at ith location y: All particle locations of group 2 at ith location
3. Using the two-sample Hotelling-T2 test, we are trying to see if the particle positions at the ith location of all group 1 samples come from the same distribution as the particle position at ith location of group 2 samples.
4. If the group differences are significant, the particle positions are not from the same distribution.

--tiny_test

## Analyzing Shape Model

For the femur mode, separate statistical tests for differences in correspondence positions are applied to every correspondence index. We use Hotelling $$T^2$$ metric(nonparametric permutation test) with false discovery rate correction (FDR) for multiple comparisons. This method helps identify and visualize localized regions of significant shape differences. The null hypothesis for this test is that the distributions of the locations of corresponding sample points are the same regardless of the groups. Hence, lower p-values would mean we can reject the null hypothesis and conclude that the group differences are significant and not from the same distribution.

This use case calculates the p-values and saves them in a text file. This same functionality is also available in ShapeWorks Studio, where you can also visualize the p-values on the mean shapes. The use case also generates a plot based on LDA with the shape-based scores for both the groups of femurs.


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In Progress
Skill 13 of 14
In Progress

# Addition & the four basic operations

You have 3 apples. Your friend gives you 6 apples more. How many apples do you have in total?

9, of course.

3 and 6 is 9’ or ‘if you have 3 and add 6, you have 9’ or the familiar ‘3 plus 6 is 9’. Let’s use the $+$ symbol to mean ‘and’ or ‘add to’ or simply ‘plus’ (the same symbol we use as a ‘positive sign’):

$$3\,\text{apples}+6 \,\text{apples} =9 \,\text{apples}$$

• Addition ($\;+\;$) adds some more to increase the amount, while
• subtraction ($\;-\;$) takes some away again. They are each other’s counterparts. Likewise,
• multiplication ($\;\cdot\;$) adds more portions by multiplying the existing amount, whereas
• division ($\;/\;$) splits this amount into portions again. They are each other’s counterparts.

These four basic math operations are the root of all math. They are the basic part of the mathematical elementary branch called arithmetic1 that deals with numbers.[2] Everything else in math is just combinations or rewritings of these. We might use different symbols now and then:

Arithmetic operations
Addition$+$Gives a sum 3
Subtraction$-$Gives a difference
Multiplication$\cdot$ or $\times$ or $*$ or noneGives a product
Division$/$ or $:$ or $\div$Gives a fraction or quotient 2

All the parts involved in these operations have gotten fancy names. See an overview in Resource: The four basic operations.

In the supermarket you take 3 clusters of 5 bananas and 4 bags of 10 apples. How many pieces of fruit are you bringing home? That would be 15 bananas and 40 apples, so 55 pieces in total:

$$\underbrace{3\cdot5}_\text{First this…}+ \underbrace{4\cdot10}_\text{and this…} = \underbrace{15+40}_\text{…then this}=55$$

There is clearly a priority, an order of operations. Multiplications first; addition next. You cannot do, for instance, addition first and then the multiplications:

$$\xcancel{3\cdot \underbrace{5+4}_\text{Not first}\cdot 10=3\cdot9\cdot10=270}$$

Let’s set up a full prioritised list:

Order or operations (PEMDAS)[10]
First do...Parentheses
Then do...Exponents (powers)
Then do...Multiplications
as well as...Divisions


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# randomNicePencil -- sets up a random example to construct Clifford algebra and representation

## Synopsis

• Usage:
(S, qq, R, u, M1, M2, Mu1, Mu2)=randomNicePencil(kk,g)
• Inputs:
• kk, a ring, the ground field, not char 2, please!
• g, an integer, genus of the associated hyperelliptic curve.
• Outputs:
• S, a ring, polynomial ring in g x's, g y's z_1,z_2 and s, t
• qq, , Element of S, quadratic in x_0..z_2 and linear in s,t
• R, a ring, polynomial ring kk[s,t]
• u, , 1 x g+2 row matrix with entries x_0..x_{(g-1)}, z_1,z_2; generators of the ideal of an isotropic subspace for qq
• M1, ,
• M2, , Matrices over S such that M1*M2 = qq times an identity of size 2^{2g+1}
• Mu1, ,
• Mu2, , Matrices over S such that Mu1*Mu2 = qq times an identity of size 2^{g+1}

## Description

Chooses a random example of a pencil of quadrics qq = s*q1+t*q2 with a fixed isotropic subspace (defined by ideal u) and a fixed corank one quadric in normal form q1.

When called with no arguments it prints a usage message.

The variables of S that are entries of X:= matrix \{\{x_0..y_{(g-1)},z_1,z_2\}\} \, represent coordinates on PP_R^{2g+1}.

M1, M2 are consecutive high syzygy matrices in the minimal (periodic) resolution of kk[s,t] = S/(ideal X) as a module over S/qq. These are used to construct the Clifford algebra of qq.

Mu1, Mu2 are consecutive high syzygy matrices in the minimal (periodic) resolution of S/(ideal u) as a module over S/qq. These are used to construct a Morita bundle between the even Clifford algebra of qq and the hyperelliptic curve branched over the degeneracy locus of the pencil,

\{(s,t) | s*q1+t*q2 is singular\} \subset PP^1.

 i1 : kk=ZZ/101 o1 = kk o1 : QuotientRing i2 : g=1 o2 = 1 i3 : (S, qq, R, u, M1, M2, Mu1, Mu2)=randomNicePencil(kk,g); i4 : gens S o4 = {x , y , z , z , s, t} 0 0 1 2 o4 : List i5 : q1 = diff(S_(2*g+2),qq) 2 o5 = x y - z 0 0 1 o5 : S

a quadratic form of corank 1 (corresponding to a branch point of E-->PP^1 in normal form.

 i6 : ideal u -- an isotropic space for q1 and q2 o6 = ideal (x , z , z ) 0 1 2 o6 : Ideal of S i7 : betti Mu1, betti Mu2 0 1 0 1 o7 = (total: 4 4, total: 4 4) 1: 3 1 2: 1 . 2: 1 3 3: 3 3 4: . 1 o7 : Sequence i8 : Mu1*Mu2- qq*id_(target Mu1) == 0 o8 = true


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# The metafor Package

A Meta-Analysis Package for R

### Sidebar

analyses:viechtbauer2007a

## Viechtbauer (2007)

### The Methods and Data

Viechtbauer (2007) describes various methods for computing confidence intervals (CIs) for the amount of heterogeneity in random-effects models. An example dataset, based on a meta-analysis by Collins et al. (1985) examining the effectiveness of diuretics in pregnancy for preventing pre-eclampsia, is used in the paper to illustrate the various methods. The data are given in the paper in Table I (p. 44).

The data can be loaded with:

library(metafor)
dat <- dat.collins1985b[,1:7]
dat

We only need the first 7 columns of the dataset (the remaining columns pertain to other outcomes). The contents of the dataset are:

  id                  author year pre.nti pre.nci pre.xti pre.xci
1  1       Weseley & Douglas 1962     131     136      14      14
2  2          Flowers et al. 1962     385     134      21      17
3  3                 Menzies 1964      57      48      14      24
4  4           Fallis et al. 1964      38      40       6      18
5  5         Cuadros & Tatum 1964    1011     760      12      35
6  6        Landesman et al. 1965    1370    1336     138     175
7  7            Kraus et al. 1966     506     524      15      20
8  8    Tervila & Vartiainen 1971     108     103       6       2
9  9 Campbell & MacGillivray 1975     153     102      65      40

Variables pre.nti and pre.nci indicate the number of women in the treatment and control/placebo groups, respectively, while pre.xti and pre.xci indicate the number of women in the respective groups with any form of pre-eclampsia during the pregnancy.

The corresponding log odds ratios (and corresponding sampling variances) can then be computed and added to the dataset with:

dat <- escalc(measure="OR", ai=pre.xti, n1i=pre.nti, ci=pre.xci, n2i=pre.nci, data=dat)
dat
  id                  author year pre.nti pre.nci pre.xti pre.xci      yi     vi
1  1       Weseley & Douglas 1962     131     136      14      14  0.0418 0.1596
2  2          Flowers et al. 1962     385     134      21      17 -0.9237 0.1177
3  3                 Menzies 1964      57      48      14      24 -1.1221 0.1780
4  4           Fallis et al. 1964      38      40       6      18 -1.4733 0.2989
5  5         Cuadros & Tatum 1964    1011     760      12      35 -1.3910 0.1143
6  6        Landesman et al. 1965    1370    1336     138     175 -0.2969 0.0146
7  7            Kraus et al. 1966     506     524      15      20 -0.2615 0.1207
8  8    Tervila & Vartiainen 1971     108     103       6       2  1.0888 0.6864
9  9 Campbell & MacGillivray 1975     153     102      65      40  0.1353 0.0679

### Heterogeneity Estimates

We can obtain the various estimates of the amount of heterogeneity (i.e., $\tau^2$) with:

res.DL   <- rma(yi, vi, data=dat, method="DL")
res.ML   <- rma(yi, vi, data=dat, method="ML")
res.REML <- rma(yi, vi, data=dat, method="REML")
res.SJ   <- rma(yi, vi, data=dat, method="SJ")
round(c(DL=res.DL$tau2, ML=res.ML$tau2, REML=res.REML$tau2, SJ=res.SJ$tau2), 2)
  DL   ML REML   SJ
0.23 0.24 0.30 0.46

These are the same values as given in Table II (p. 44) of the paper (the Sidik–Jonkman estimator is denoted as $\hat{\tau}^2_{SH}$ in the paper).

### Confidence Intervals

Next, we will obtain the various CIs for $\tau^2$, also as given in the paper in Table II. The Q-profile method is directly implemented in the metafor package. For the remaining CIs, some extra work/code is needed.

#### Q-Profile Method

First, the Q-profile CI can be obtained with:

confint(res.DL, digits=2)
       estimate ci.lb ci.ub
tau^2      0.23  0.07  2.20
tau        0.48  0.27  1.48
I^2(%)    70.66 43.12 95.85
H^2        3.41  1.76 24.09

Note that CIs for the $I^2$ and $H^2$ statistics are also provided.

#### Biggerstaff–Tweedie Method

The following code can be used to obtain the CI based on the method by Biggerstaff and Tweedie (1997):

CI.D.func <- function(tau2val, s1, s2, Q, k, lower.tail) {
expQ  <- (k-1) + s1*tau2val
varQ  <- 2*(k-1) + 4*s1*tau2val + 2*s2*tau2val^2
shape <- expQ^2/varQ
scale <- varQ/expQ
qtry  <- Q/scale
pgamma(qtry, shape = shape, scale = 1, lower.tail = lower.tail) - 0.025
}

wi <- 1/dat$vi s1 <- sum(wi) - sum(wi^2)/sum(wi) s2 <- sum(wi^2) - 2*sum(wi^3)/sum(wi) + sum(wi^2)^2/sum(wi)^2 ci.lb <- uniroot(CI.D.func, interval=c(0,10), s1=s1, s2=s2, Q=res.DL$QE, k=res.DL$k, lower.tail=FALSE)$root
ci.ub=(res.SJ$k-1) * res.SJ$tau2 / qchisq(0.025, df=res.SJ$k-1)), 2) ci.lb ci.ub 0.21 1.67 #### Parametric Bootstrap CI The boot package can be used obtain the parametric and non-parametric bootstrap CIs, so let's load the package first: library(boot) For parametric bootstrapping, we need to define two functions, one for calculating the statistic of interest (and possibly its variance) based on the bootstrap data, the second for actually generating the bootstrap data. boot.func <- function(data.boot) { res <- rma(yi, vi, data=data.boot, method="DL") c(res$tau2, res$se.tau2^2) } data.gen <- function(dat, mle) { data.frame(yi=rnorm(nrow(dat), mle$mu, sqrt(mle$tau2 + dat$vi)), vi=dat$vi) } Next, we can do the bootstrapping (based on 1000 bootstrap samples) and then obtain a variety of different bootstrap CIs: sav <- boot(dat, boot.func, R=1000, sim="parametric", ran.gen=data.gen, mle=list(mu=coef(res.DL), tau2=res.DL$tau2))
boot.ci(sav, type=c("norm", "basic", "stud", "perc"))
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 1000 bootstrap replicates

CALL :
boot.ci(boot.out = sav, type = c("norm", "basic", "stud", "perc"))

Intervals :
Level      Normal              Basic
95%   (-0.1298,  0.5907 )   (-0.2535,  0.4594 )

Level    Studentized          Percentile
95%   ( 0.0508,  1.1814 )   ( 0.0000,  0.7129 )
Calculations and Intervals on Original Scale

In the paper, the percentile CI is reported (untruncated, so the lower bound actually falls below 0). The upper bound is slightly different than the one shown above due to the random play of chance when bootstrapping with a finite number of samples.

#### Non-Parametric Bootstrap CI

For the non-parametric bootstrap method, we only need to define one function with two arguments, one for the original data and one for a vector of indices which define the bootstrap sample. The function again returns the statistic of interest (and its variance):

boot.func <- function(dat, indices) {
res <- rma(yi, vi, data=dat, subset=indices, method="DL")
c(res$tau2, res$se.tau2^2)
}

It then takes again two lines of code to conduct the actual boostrapping and obtain the corresponding CIs:

sav <- boot(dat, boot.func, R=1000)
boot.ci(sav)
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 1000 bootstrap replicates

CALL :
boot.ci(boot.out = sav)

Intervals :
Level      Normal              Basic             Studentized
95%   (-0.0178,  0.4953 )   (-0.0517,  0.4372 )   ( 0.0789,  1.1202 )

Level     Percentile            BCa
95%   ( 0.0222,  0.5111 )   ( 0.0361,  0.5641 )
Calculations and Intervals on Original Scale

Again, the percentile CI is reported in the paper, with the small difference resulting from the play of chance. For more details on bootstrapping, see bootstrapping with meta-analytic models under the tips and notes section.

### References

Biggerstaff, B. J., & Tweedie, R. L. (1997). Incorporating variability in estimates of heterogeneity in the random effects model in meta-analysis. Statistics in Medicine, 16(7), 753–768.

Hardy, R. J., & Thompson, S. G. (1996). A likelihood approach to meta-analysis with random effects. Statistics in Medicine, 15(6), 619–629.

Sidik, K., & Jonkman, J. N. (2005). Simple heterogeneity variance estimation for meta-analysis. Applied Statistics, 54(2), 367–384.

Viechtbauer, W. (2007). Confidence intervals for the amount of heterogeneity in meta-analysis. Statistics in Medicine, 26(1), 37–52.


text:
MathSciNet bibliographic data MR994770 46J15 (30H05 32E25) Axler, Sheldon; Gorkin, Pamela Sequences in the maximal ideal space of $H^\infty$$H^\infty$. Proc. Amer. Math. Soc. 108 (1990), no. 3, 731–740. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# Fixing Whitespace Automatically¶

This page describes how to use Universal Indent GUI and Uncrustify to automatically fix whitespace such that code resembles the SeqAn C++ Code Style more closely.

• Uncrustify is a command line program that is given a style definition and a source file and reformats the source file according to the configuration.
• Universal Indent GUI is a graphical front-end to Uncrustify with life preview that allows to manipulate the configuration and immediately see the results.

## Installing Universal Indent GUI¶

This one is pretty easy. On Ubuntu and other Linux systems, you can use the package management system to install the GUI and the reformatting programs. The Universal Indent GUI download page has binaries for Mac Os X and Windows.

## Preview with Universal Indent GUI¶

When started, the program will present you with a window like the following.

First, we set the indenter to Uncrustify.

Then, we load SeqAn’s uncrustify.cfg which is located in ${CHECKOUT}/misc. We can do so by selecting Indenter > Load Indenter Config File in the program menu. Then, we load a file from the SeqAn repository, for example core/apps/sak/sak.cpp. Now, we can toy around with the reformatter by checking Live Indent Preview. The settings on the left panel allow us to tweak the style to our liking. Any changes can be stored by selecting Indenter > Load Indenter Config File in the program menu. The source can also be stored, using File > Save Source File and File > Save Source File As.... ## Using The Command Line¶ Uncrustify can also be used via the command line. This is best done after a rough visual verification that the uncrustify.cfg yields works for your sources using the Universal Indenter UI. Work on single file: # uncrustify -c${CHECKOUT}/misc/uncrustify.cfg --replace -f path/to/file.cpp

Batch work:

# find path/to -name '*.cpp' > list.txt
# uncrustify -c \${CHECKOUT}/misc/uncrustify.cfg --replace -F list.txt

## Automatically fix whitespaces in Xcode¶

Uncrustify can also be used directly from Xcode. With Xcode 4 Apple introduced so called “Behaviors” that can be executed using for instance keyboard shortcuts. To use uncrustify you can add a new behavior in the Xcode Preferences (tab Behaviors) and select “Run”. Here you add the attached ruby script.

Note: The script does not uncrustify the currently opened source file but all source files that were changed in your current svn checkout. Xcode does not provide the information which source file is currently opened.


text:
Article

# Representation theory of rectangular finite $W$-algebras

• ##### Jonathan Brown
Journal of Algebra (Impact Factor: 0.6). 03/2010; DOI: 10.1016/j.jalgebra.2011.05.014
Source: arXiv

ABSTRACT We classify the finite dimensional irreducible representations of rectangular finite $W$-algebras, i.e., the finite $W$-algebras $U(\mathfrak{g}, e)$ where $\mathfrak{g}$ is a symplectic or orthogonal Lie algebra and $e \in \mathfrak{g}$ is a nilpotent element with Jordan blocks all the same size. Comment: 34 pages

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• Source
##### Article: 2-row Springer fibres and Khovanov diagram algebras for type D
[Hide abstract]
ABSTRACT: We study in detail two row Springer fibres of even orthogonal type from an algebraic as well as topological point of view. We show that the irreducible components and their pairwise intersections are iterated P^1-bundles. Using results of Kumar and Procesi we compute the cohomology ring with its action of the Weyl group. The main tool is a type D diagram calculus labelling the irreducible components in an convenient way which relates to a diagrammatical algebra describing the category of perverse sheaves on isotropic Grassmannians based on work of Braden. The diagram calculus generalizes Khovanov's arc algebra to the type D setting and should be seen as setting the framework for generalizing well-known connections of these algebras in type A to other types. The results will be connected to Brauer algebras at non-generic parameters in a subsequent paper.
09/2012;
• Source
##### Article: On changing highest weight theories for finite W-algebras
[Hide abstract]
ABSTRACT: A highest weight theory for a finite W-algebra U(g,e) was developed in [BGK]. This leads to a strategy for classifying the irreducible finite dimensional U(g,e)-modules. The highest weight theory depends on the choice of a parabolic subalgebra of g leading to different parameterizations of the finite dimensional irreducible U(g,e)-modules. We explain how to construct an isomorphism preserving bijection between the parameterizing sets for different choices of parabolic subalgebra when g is of type A, or when g is of types C or D and e is an even multiplicity nilpotent element
Journal of Algebraic Combinatorics 05/2011; · 0.72 Impact Factor
• Source
##### Article: Non-integral representation theory of even multiplicity finite W-algebras
[Hide abstract]
ABSTRACT: We complete the classification of the finite dimensional irreducible representations of finite W-algebras associated to even multiplicity nilpotent elements in classical Lie algebras. This extends earlier work where this classification is determined for such representations of integral central character.
Journal of Algebra 12/2011; · 0.60 Impact Factor

Available from


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## Sunday, April 15, 2012

### Ubuntu 11.10 -- fixing grub

Gezz can they screw the system up even worst.. now we have a graphical grub that is not 100% compatible with all monitors. To disable this mess.

 GRUB_TERMINAL=console
GRUB_CMDLINE_LINUX_DEFAULT="text"

Then run:
 update-grub

And reboot. Now we should have text and problems on boot will clearly visible.... amazing!


text:
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Straight Lines - For Boards

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Find the vertices of the trian...

# Find the vertices of the triangle whose sides are y+2x=3, 4y+x=5 and 5y+3x=0

Updated On: 27-06-2022


text:
# 3. Let X and Y be countably infinite sets. (a) Prove: If X and Y are...

###### Question:

3. Let X and Y be countably infinite sets. (a) Prove: If X and Y are disjoint then XuY is countably infinite. (b) Is the statement in (a) still true if we remove the hypothesis that X and Y are disjoint? If yes, justify your reasoning with a few sentences. If no, provide a counterexample. (P.S. "Counterexample” means that you have to explain why the example you provide demonstrates that the statement is false.)

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text:
EIA: natural gas use for power falls as industrial use continues to rise; higher prices, cooler summer

04 December 2013

For the first eleven months of 2013, natural gas consumption in the electric power sector was below 2012 levels because of relatively higher natural gas prices compared with coal prices, and cooler summer weather compared with 2012, according to the US Energy Information Administration (EIA).

EIA estimates that electric power sector natural gas consumption was, on average, down by 13% so far in 2013 (through November), relative to the same time period in 2012. By contrast, industrial sector natural gas consumption in 2013 was up 3% compared with 2012.

Changes in electric power sector consumption of natural gas are primarily attributable to commodity prices, EIA said. For the first eleven months of 2012, the benchmark natural gas spot price at Henry Hub averaged $2.70/MMBtu. For the same months in 2013, the average Henry Hub price was$3.68/MMBtu, somewhat higher than the previous year’s levels, lowering electric-sector demand for natural gas. Gas is still being consumed at rates generally above the 2007-2011 range.

Electric power plants are dispatched based on their variable cost of operation, which is determined by both the price of the power plant’s fuel and the efficiency of the plant.

Demand for air conditioning is one of the major drivers of summer peak electricity loads, which typically require the highest amounts of gas-fired generation. Air conditioning demand is closely related to the number of cooling degree days (CDDs), a measure that reflects both the level and duration of high temperatures. In addition to less gas per CDD being consumed in 2013 versus 2012 because of the price effects mentioned above, there was an overall reduction in CDDs this year.

July and August— typically the hottest months—both saw substantially fewer CDDs in 2013 than in 2012, EIA noted. This cooler weather further reduced electric-sector demand for natural gas compared with last year. As a percentage of total natural gas end-use consumption across all sectors, the electric sector has accounted for 35% so far in 2013 (through November), down from 40% in the same months of 2012.

Continuing the trend from earlier in the year, industrial gas consumption was up 0.6 billion cubic feet per day (Bcf/d), or 3%, through November 2013, compared with the same months in 2012. Nearly all facilities with the ability to switch from oil to natural gas as a fuel or feedstock made this change before 2011, given the large cost advantage of natural gas.

Lower natural gas prices in the United States relative to those in Japan and Europe have made U.S. energy-intensive industries more competitive globally, allowing them to capture a larger share of global markets. New industrial demand for natural gas is expected to continue to grow as additional manufacturing capacity in gas-intensive industries comes into operation.

Comments

The same could happened to CNG vehicles if the price of CNG becomes as high as gasoline and diesel fuel.

One way to stop the run towards the all mighty cheapest would be with an energy-fuel tax based on pollution and GHG created and environment clean up cost.

For example, electricity from coal and nuclear power plants could be taxed more ($0.05/kWh) than from NG plants ($0.04/kWh) and a lot more than from clean Hydro/Sun/Wind sources (\$0.00/kWh).

Revenues should go into a special environment clean up fund.

Nuclear plants don't create any signficant clean-up cost in normal operation.  Their emissions are of little or no environmental concern and disappear by themselves over time.  Taxing them for "pollution" is just to designate the "approved green" suppliers as the winners.

@ EP
Do you have even a faint idea of the cost and effort to renaturate a former nuclear plant? Proper disposal of nuclear wastes and fuel rods?
A former nuclear plant is being renaturated in Germany at a forecast timetable of approx. 30 years. That'll really be cheap! All these expenses should be added to the monthly power bill of all nuclear proponents and not left up to the taxpayer in general.

Do you have even a faint idea of the cost and effort to renaturate a former nuclear plant?

It was done about 30 miles from where I'm sitting now.  Seems to have worked fine.  I wish they'd re-use the site for a couple AP1000's and clean up the state's power supply some more (currently about 30% carbon-free nuclear).

Proper disposal of nuclear wastes and fuel rods?

By "disposal" do you mean "irretrievable dumping" or "resource separation and recovery"?  Because PROPER disposal means the latter, and about 95% of the available energy remains in "spent" LWR fuel.  Even some of the fission products have worthwhile uses (as I said, I'll take 10 kg of Sr-90 in a stainless capsule any time, just give me a week's notice before you deliver).

A former nuclear plant is being renaturated in Germany at a forecast timetable of approx. 30 years. That'll really be cheap!

Most of the delay is just sitting, to allow the neutron-activated cobalt in the reactor vessel to decay (HL 5.7 years).  This cuts the gamma exposure of the dismantling crew by about a factor of 40, which means far lower radiation-protection costs.  Or do you think the decomissioning workers shouldn't have their safety looked after?

All these expenses should be added to the monthly power bill of all nuclear proponents and not left up to the taxpayer in general.

I'll take that if you'll take all the costs of climate-change from all the natural gas and lignite that the backups for your precious "renewable" generators still require to keep the grid from blacking out.

The comments to this entry are closed.


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# Mean Deviation from the Mean

Consider the observations 3, 5, 6, 7, 9.
A.M =

Let us find the deviations of the individual items from the mean. They are âˆ’ 3, âˆ’ 1, 0, 1, 3 respectively. Sum of these deviations is zero, thus finding the mean derivative about the mean will result in zero and therefore is not of any use to us as far as measure of dispersion is concerned.

Mean deviation about the mean is the average of all the deviation from the mean

Mean of deviations =

So instead of taking the actual deviations, we consider the absolute deviations of these observations from there mean.

i.e. absolute deviations from the mean are 3, 1, 0, 1, 3 ; their sum = 8.

M.D (from mean) =

# Definition (Mean Deviation)

Let be values. The mean deviation about the mean of these values is given by M.D = when is the mean of the values.

Note:
1. Mean deviation can be calculated about any fixed quantity. Mean deviation about 'a' will be .
2. It the mean deviation is calculated about the median M, then it is
Example 1. Find the mean deviation about the mean for the following data.
18, 20, 12, 14, 19, 22, 26, 16, 19, 24

Solution:

=
Mean deviation about the mean =

Example 2:
Find the mean deviation about the median for the following data: 29, 31, 35, 37, 55, 63, 72

Solution:
n = 7
Arrange data in ascending order
Note: It already is in ascending order.
29, 31, 35, 37, 55, 63, 72
Median is the middle reading = 37 = M.
Absolute deviations from the median are 8, 6, 2, 0, 18, 26, 35
sum of absolute deviations = 95
Mean deviation about M =  = 13.57


text:
Nucleation and growth of fullerenes and nanotubes having four-fold symmetry

Authors:
Abstract:

We have studied possible ways of generating and growing the fullerenes having four-fold symmetry. Beginning with cyclobutane C 4 H 8 and clusters C 8 , we obtained elementary fullerenes C 8 and mini-fullerenes C 16 , which produce the following fullerenes from C 24 to C 64 , perfect (basic) and imperfect, as well as nanotubes. The imperfection is connected either with extra 'interstitial' or 'vacancy' carbon dimers, both types of dimers playing the role of defects. Only the basic fullerenes C 24 , C 32 , C 40 , C 48 , C 56, and C 64 have the ordinary four-fold symmetry in the corresponding column of the periodic system of fullerenes, the intermediate fullerenes having no such symmetry. Considering the latter as imperfect due to defects, one can define them as the fullerenes conserving topological four-fold symmetry. We have calculated their energies and discussed possible reasons for their dependence on a fullerene size and shape.


text:
# I Truncated Fourier transform and power spectral density

1. Mar 31, 2017

### Mishra

Hello,

I am trying to find an expression for the signal-to-noise ratio of an oscillating signal on top some white noise. In particular I would like to know how the SNR scales with the integration time. It is well known that during some integration time $T$, the SNR increases as $T^{1/2}$ (because the noise increases as $T^{1/2}$ and the signal increases at $T$). I am trying to prove this with, my limited skills in maths, using the power spectral density formalism.

The integration time is $T$. I assume some signal: $x(t)=A \sin(2 \pi \omega_0 t)$ on top of some white noise of power spectral density $S$. I am guessing that I should compare the power spectral density of the noise vs signal:
$$P(f)=X(f)^2$$
where $X(f)$ is the truncated Fourier transform:
$$X(f)=\frac{1}{\sqrt{T}}\int_0 ^{T} x(t) e^{-i \omega t} dt$$
In the limit where $T$ is large, the integral is a Fourier tranform, yielding a delta function:
$$X(f)=\frac{A}{\sqrt{T}} \delta (\omega - \omega_0)$$
Ence:
$$P(f)=\frac{A^2}{ T } \delta (\omega - \omega_0)$$
$$P(f_0)=\frac{A^2}{ T }$$

Here I have that the power spectral density at the oscillation frequency decreases linearly in time. Can someone explain me where my misunderstanding is please ?

2. Apr 1, 2017

### blue_leaf77

It can be that my memory is playing trick in my head but as I recall the white noise variance in frequency domain increases as $T^3$. So, do you have any reference that states what you wrote there?

3. Apr 3, 2017

### jasonRF

I think the issue is that you are playing fast and loose with delta functions. The integral is simple enough that you may as well just do it. If I do a simplified case of $x(t) = A e^{i\omega_0 t}$ and let t run from -T/2 to T/2, then
$$X(\omega) = \frac{A}{\sqrt{T}} \int_{-T/2}^{T/2} dt \, e^{i(\omega_0-\omega)t} = A \sqrt{T} \left[\frac{\sin\left(\frac{1}{2} T (\omega-\omega_0) \right)}{\frac{1}{2} T (\omega-\omega_0)} \right]$$
The term in square brackets is a very nice number when $\omega\rightarrow\omega_0$.

jason

4. Apr 3, 2017

### jasonRF

By the way, for the noise I get $\langle |X(\omega)|^2 \rangle$ to be independent of time, so the SNR $\propto A^2 T$, as I would expect. You should do the calculation.

Jason


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# Applied Min/Max word problem

I can't seem to figure out this problem.
Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10.
I start by drawing the diagram and it seems to me like the circle radius corresponds with a line from the center of the rectangle to one of the rectangle points on the edge of the circle. This could give me a triangle with pythagorean theorem. I should find the value of x that yields largest area as my priority, after that I can find y easily.

Latex is not behaving today so i'll try my best without. y/2 = (10^2-(x/2)^2)^(1/2)

Area of a square is xy, duh.

Plugging this in gives me a chain rule problem which ultimately comes out with the critical points 0, -10 and 10, all of which yield 0 Area (obviously), there has to be a critical point that I am missing between 0 and 10 but the derivative of A(x) doesn't yield any.

I'm stumped.

## Answers and Replies

Tide
Homework Helper
Try writing the area of the inscribed rectangle as a function of the angle between its diagonals.

HallsofIvy
Homework Helper
Tide said:
Try writing the area of the inscribed rectangle as a function of the angle between its diagonals.
You can do that but I'm not sure that is the best way.

The circle is, of course, x2+ y2= 100.

The area of the rectangle is (2x)(2y)= 4xy.

$$y= \sqrt{100- x^2}$$
so the area is $$A= 4x\sqrt{100- x^2}= 4x(100-x^2)^{\frac{1}{2}}$$.
Differentiating that and setting the derivative equal to 0 gives a quadratic equation so I'm not at all clear how you got three "critical points"!

That's about where i'd got to, I just overcomplicated things.
the derivative I get is
$$4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}}$$

EDIT: Wait a sec... I had the derivative slightly wrong...

Last edited:
Right, I had the derivative wrong, I still came out with -10, 0 and 10 but the +/-10 were asymptotes according to table and x=0 came out as 40 max area.

That makes absolutely no sense.

I get area of the rectangle to be x * Square root (100 - x ^2)

Still shouldn't make any difference, at least as regards to the derivative of area being equal to zero.

You might want to look a little closer at how you used the chain rule on this..

Fermat
Homework Helper
Aresius said:
That's about where i'd got to, I just overcomplicated things.
the derivative I get is
$$4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}}$$
EDIT: Wait a sec... I had the derivative slightly wrong...
$$4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}}$$

$$\frac{4(100-x^2)}{(100-x^2)^{\frac{1}{2}}} - \frac{4x^2}{(100-x^2)^{\frac{1}{2}}} = 0$$

giving,

$$4(100-x^2) - 4x^2 = 0$$
$$400 - 8x^2 = 0$$
$$x^2 = 50$$

only one solution for x.

Tide
Homework Helper
HallsofIvy said:
You can do that but I'm not sure that is the best way.

Perhaps not but $A = 50 \sin \theta$ looks simpler!

NateTG
Homework Helper
Tide said:
Perhaps not but $A = 50 \sin \theta$ looks simpler!

Personally, I like the "it must be a square, by symetry" approach.

Solution

(The diagram has been attached)

Area of rectangle = l X b = 2x√(10-x²)

Squaring the function and discarding the constant

A = x²(10-x²)
A = 10x² - x4

dA = 20x - 4x³
dx

For turning point dA/dx = 0
20x - 4x³ = 0
x = ±√5 , 0

x = +√5 as length cannot be negative or zero

d²A = 20 - 12x²
dx²

d²A/dx² < 0 (Maximum)

A = 2x√(10 - x²)
A = 10v5 sq. units

Dimensions:
length = 2√(10 - x²) = 2√5
width = 2x = 2√5


text:
# A sum of money doubles itself in 8 years. What is the rate of interest?

Given :

A sum of money doubles itself in 8 years.

To do :

We have to find the rate of interest.

Solution :

Let the Principal be $P$ and rate percent be $R$.

This implies,

Amount $A=2P$

We know that,

$Amount (A) = Principal (P) + Simple Interest (SI)$

$Simple Interest (SI) = Amount (A) - Principal (P)$

$SI=2P-P$

$SI=P$

$SI = \frac{PRT}{100}$

Therefore,

$P = \frac{(P \times R \times 8)}{100}$

$100 = R \times 8$

$R = \frac{100}{8}$

$R = 12.5$

The rate percent is 12.5%.

Tutorialspoint

Simply Easy Learning


text:
# Math Help - permutations

1. ## permutations

（1 2 3 4 5 6 （1 2 3 4 5 6
2 3 1 5 6 4） 2 3 1 5 6 4）

my answer is ( 1 2 3 4 5 6
3 1 2 6 4 5)
am i right?

but the next one i have no idea at all.

(1,3,6)(1,2,6,5)(4,5) any one can help me ?? thank you! if you know the answer, could u plz tell me how you do this kind of question? details would be very appreciated! cheers!

2. Your answer is correct if you were trying to write what I think you were. You were doing two row notation and multiplying two permutations?

To do the second they are written in cycle notation. This can be confusing at first because of the order. We compose functions backwards (i.e. $f \circ g$ you apply g first then f), and the same goes with permutations. So within the parenthesis you read right to left, but go from right to left in terms of the compositions.

The way you work through these is as follows.
(1 3 6)(1 2 6 5)(4 5)
Take 1 and see where it goes.
It is untouched by (4 5). (1 2 6 5) takes 1 to 2. then (1 3 6) does not affect 2.

so 1 -> 2.

Now you see where 2 goes.

It is untouched by (4 5). (1 2 6 5) takes 2 -> 6. Now (1 3 6) takes 6 -> 1. Notice the individual cycles "wrap around" so their representation is unique up to cycling through although generally you write the smallest number first.

so we see 2 -> 1.

This gives us the cycle (1 2).

Now proceed to 3. I will omit details
3 -> 6

6 -> 5

5 -> 4

4-> 5 -> 1 ->3
so we have (3 6 5 4)

Thus this multiplication is simply (1 2) (3 6 5 4).

Also remember if in general it is possible for a number i to be fixed. One can write (i) to represent it gets sent to itself, or more generally it is simply omitted from the cycle decomposition.

3. Thank you very much indeed. your answer is very clear and i understand now. thank you!


text:
HP 9000
Showing results for
Do you mean

SOLVED

# File attachment in "mailx" command

Unix experts,

Greetings to all of you and thanks in advance for your help.

I need to know which option will get me file attachment using "mailx" command. I should not see any open file attachment in my mail.

Awaiting for good responses.

Thanks
You can do it man !
1 ACCEPTED SOLUTION
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## Re: File attachment in "mailx" command

2 REPLIES
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text:
# Conjugate pairs versus not conjugate pairs

1. Mar 13, 2005

### sotellme

to make a 25mM buffer with pH5.2 i add 25mM ethanoic acid with 25mM sodium ethanoate. after adjusting the pH, the buffer will still be 25mM.

will this also be the case if i use 25 mM Tris-base and 25 mM phosphoric acid? will i still get a buffer with 25mM after adjusting the pH to 5.2 with the phosphoric acid?

the first one is a conjugate pair (ethanoic acid and sodium ethanoate), while the second one is not a conjugate pair (Tris-base and phosphoric acid)

thanks.

2. Mar 13, 2005

### GCT

No. You'll need a conjugate pair for a buffer solution, although I'm not sure if your method complies with any other existing ways to make a buffer.

Buffers solutions exists through common ion effect, this is why for the most part the concentration of the conjugates do not undergo equilibrium; I believe that it may be an entropical phenomenon, somebody correct me if I'm mistaken.