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www.whyslopes.com || Fit Browser Window || Site Map - 1200 items.

# Mathematics and Logic - Skill and Concept Development

Questions: Will these ends and values motivate? Will smaller & more steps in site lessons and lesson ideas build skills and confidence?
Should we emphasize how ideas & methods depend on earlier ones? Does concept & skill mastery need to be checked to be believed? What is a Variable?

30 pages en Francais || Parents - Help Your Child or Teen Learn
Online Volumes: 1 Elements of Reason || 2 Three Skills For Algebra || 3 Why Slopes Light Calculus Preview or Intro plus Hard Calculus Proofs, decimal-based.
More Lessons &Lesson Ideas: Arithmetic & No. Theory || Time & Date Matters || Algebra Starter Lessons || Geometry - maps, plans, diagrams, complex numbers, trig., & vectors || More Algebra || More Calculus || DC Electric Circuits || 1995-2011 Site Title: Appetizers and Lessons for Mathematics and Reason

Mathematics Concept & Skill Development Lecture Series: Webvideo consolidation of site lessons and lesson ideas in preparation. Price to be determined.

Bright Students: Top universities want you. While many have high fees: many will lower them, many will provide funds, many have more scholarships than students. Postage is cheap. Apply and ask how much help is available. Caution: some programs are rewarding. Others lead nowhere. After acceptance, it may be easy or not to switch.

For students of reason in society, science and technology: Pattern Based Reason describes origins, benefits and limits of rule- and pattern-based thought and actions. Not all is certain. We may strive for objectivity, but not reach it. Postscripts offer a story-telling view of learning: [ A ] [ B ] [ C ] [ D ] to suggest how we share theories and practices.

#### Site's Best Lessons

##### For Logic

These online chapters may amuse while leading to greater precision and comprehension in reading and writing at home, in school, at work and in mathematics.

1 versus 2-way implication rules - A different starting point - Writing or introducting the 1-way implication rule IF B THEN A as A IF B may emphasize the difference between it or the latter, and the 2-way implication A IF and ONLY IF B.

Deductive Chains of Reason - See which implications can and cannot be used together to arrive at more implications or conclusions,

Mathematical Induction - a light romantic view that becomes serious.

Responsibility Arguments - his, hers or no one's

Islands and Divisions of Knowledge - a model for many arts and disciplines including mathematics course design. Site Theme: Different entry points may be easier or harder for knowledge mastery.

##### For Arithmetic

Deciml Place Value - funny ways to read multidigit decimals forwards and backwards in groups of 3 or 6, US-CDN, UK-German and Metric SI style.

Decimals for Tutors - lean how to explain or justify operations. Long division of polynomials is easier for student who master long division with decimals.

Primes Factors - Efficient fraction skills and later studies of polynomials depend on this.

Fractions + Ratios - See how raising terms to obtain equivalent fractions leads to methods for addition, comparison, subtraction, multiplication and division of fractions.

Arithmetic with units - Skills of value in daily life and in the further study of rates, proportionality constants and computations in science & technology.

##### For Algebra

What is a Variable? - this entertaining oral & geometric view may be before and besides more formal definitions - is the view mathematically correct?

Formula Evaluation - Seeing and showing how to do and record steps or intermediate results of multistep methods allows the steps or results to be seen and checked as done or later; and will improve both marks and skill. The format here allows the domino effects of care and the domino effects of mistakes to be seen. It also emphasizes a proper use of the equal sign.

Solve Linear Eqns with & then without fractional operations on line segments - meet an visual introduction and learn how to present do and record steps in a way that demonstrate skill; learn how to check answers, set the stage for solving word problems by by learning how to solve systems of equations in essentially one unknown, set the stage for solving triangular and general systems of equations algebraically.

Function notation for Computation Rules - another way of looking at formulas. Does a computation rule, and any rule equivalent to it, define a function?

Axioms [some] as equivalent Computation Rule view - another way for understanding and explaining axioms.

Using Formulas Backwards - Most rules, formulas and relations may be used forwards and backwards. Talking about it should lead everyone to expect a backward use alone or plural, after mastery of forward use. Proportionality relations may be use backward first to find a proportionality constant before being used forwards and backwards to solve a problem.

www.whyslopes.com >> Geometry - maps plans trigonometry vectors >> 8 Unit-Circle Trigonometry >> 17B sine-cosine Angle Sum Formulas via cis Next: [17C sine and cosine double triple angle formulas.] Previous: [17A The complex-number valued trig function cis.]   [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21][22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45]

### A2. Trigonometric Angle Sum Formulas

From

$\mbox{ cis}( A) = \cos(A)+ i\sin( A) = a + bi$

and

$\mbox{ cis}(B) = \cos(B)+ i\sin(B) = c + id$

we have

\begin{eqnarray*} && \cos(A+B) + i \sin(A+B) \\ &=&\mbox{ cis}(A+B) \\ &=& \mbox{ cis}(A)\cdot\mbox{ cis}(B) \\ &=& (a+ib)( c+ id) \\ &=&( ac - bd)+ i( bc+ ad) \\ & = &{ \cos(A)\cos(B) -\sin(A)\sin(B) } \\ && \quad + i\cdot [\sin(A)\cos(B)+ \cos(A)\sin(B)] \end{eqnarray*}

#### Conclusion

$\cos(A+B) = \cos(A)\cos(B) -\sin(A)\sin(B)$

$\sin(A+B) = \sin(A)\cos(B)+ \cos(A)\sin(B)$

The verification or derivation of many trigonometric identities can be reduced to algebraic manipulations involving

$\mbox{ cis}(\theta) = \cos(\theta) + i \sin(\theta) = \exp( i \theta)$

www.whyslopes.com >> Geometry - maps plans trigonometry vectors >> 8 Unit-Circle Trigonometry >> 17B sine-cosine Angle Sum Formulas via cis Next: [17C sine and cosine double triple angle formulas.] Previous: [17A The complex-number valued trig function cis.]   [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21][22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45]

Road Safety Messages for All: When walking on a road, when is it safer to be on the side allowing one to see oncoming traffic?

### Site Reviews

1996 - Magellan, the McKinley Internet Directory:

Mathphobics, this site may ease your fears of the subject, perhaps even help you enjoy it. The tone of the little lessons and "appetizers" on math and logic is unintimidating, sometimes funny and very clear. There are a number of different angles offered, and you do not need to follow any linear lesson plan. Just pick and peck. The site also offers some reflections on teaching, so that teachers can not only use the site as part of their lesson, but also learn from it.

2000 - Waterboro Public Library, home schooling section:

CRITICAL THINKING AND LOGIC ... Articles and sections on topics such as how (and why) to learn mathematics in school; pattern-based reason; finding a number; solving linear equations; painless theorem proving; algebra and beyond; and complex numbers, trigonometry, and vectors. Also section on helping your child learn ... . Lots more!

2001 - Math Forum News Letter 14,

... new sections on Complex Numbers and the Distributive Law for Complex Numbers offer a short way to reach and explain: trigonometry, the Pythagorean theorem,trig formulas for dot- and cross-products, the cosine law,a converse to the Pythagorean Theorem

2002 - NSDL Scout Report for Mathematics, Engineering, Technology -- Volume 1, Number 8

Math resources for both students and teachers are given on this site, spanning the general topics of arithmetic, logic, algebra, calculus, complex numbers, and Euclidean geometry. Lessons and how-tos with clear descriptions of many important concepts provide a good foundation for high school and college level mathematics. There are sample problems that can help students prepare for exams, or teachers can make their own assignments based on the problems. Everything presented on the site is not only educational, but interesting as well. There is certainly plenty of material; however, it is somewhat poorly organized. This does not take away from the quality of the information, though.
... section Solving Linear Equations ... offers lesson ideas for teaching linear equations in high school or college. The approach uses stick diagrams to solve linear equations because they "provide a concrete or visual context for many of the rules or patterns for solving equations, a context that may develop equation solving skills and confidence." The idea is to build up student confidence in problem solving before presenting any formal algebraic statement of the rule and patterns for solving equations. ...

##### For Geometry

Maps + Plans Use - Measurement use maps, plans and diagrams drawn to scale.

Euclidean Geometry - See how chains of reason appears in and besides geometric constructions.

Coordinates - Use them not only for locating points in the plane or space.

Complex Numbers - Learn how rectangular and polar coordinates may be used for adding, multiplying and reflecting points in the plane, in a manner known since the 1840s for representing and demystifying "imaginary" numbers, and in a manner that provides a quicker, mathematically correct, path for defining "circular" trigonometric functions for all angles, not just acute ones, and easily obtaining their properties. Students of vectors in the plane may appreciate the complex number development of trig-formulas for dot- and cross-products. Lines-Slopes [I] - Take I & take II respectively assumes no knowledge and some knowledge of the tangent function in trigonometry.

What is Similarity - another view of using maps, plans and diagrams drawn to scale in the plane and space. May buildings in space are similar by design.

##### For Calculus

Why study slopes - this fall 1983 calculus appetizer shone in many classes at the start of calculus. It could also be given after the intro of slopes to introduce function maxima and minima at the ends of closed intervals.

Why factor polynomials - this 1995-96 lesson introduces calculus skills and concepts. It may also may be given to introduce further function maxima and minima both inside and at the ends of closed intervals.

Check Arith. Skills - too many calculus and precalculus students do not have strong arithmetic and computation skills. The exercises here check them while numerically hinting at equivalent computation rules.


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# Key Properties of the Normal Distribution

A random variable is said to have the normal distribution (Gaussian curve) if its values make a smooth curve that assumes a “bell shape.” A normal variable has a mean “μ,” pronounced as “mu,” and a standard deviation “σ,” pronounced as “sigma.” All normal distributions have a distinguishable bell shape regardless of the mean, variance, and standard deviation.

A normal distribution has certain properties that make it a useful tool in the world of finance.

1. Its shorthand notation is $$X \sim N(\mu , \sigma^2)$$. This is read as “the random variable X has a normal distribution with mean μ and variance σ2”.
2. It has a symmetric shape: it can be cut into two halves that are mirror images of each other; as such, skewness = 0.
3. Kurtosis = 3. Note that kurtosis is a measure of flatness and excess kurtosis is measured relative to 3, the “normal kurtosis.”
4. The mean, mode, and median are all equal and lie directly in the middle of the distribution.
5. If $$X \sim N(\mu_x , \sigma^2_x)$$ and $$Y\sim N(\mu_y , \sigma^2_y)$$, then the two variables combined also have a normal distribution.
6. The standard deviation measures the distance from the mean to the point of inflection (the point where the curve changes an “upside-down-bowl” shape to a “right-side-up-bowl” shape).
7. Probabilities follow the empirical rule: first, about 68% of the total values lie within 1 standard deviation of the mean. In addition, 95% of the values lie within 2 standard deviations of the mean. Lastly, about 99.7% lie within 3 standard deviations of the mean.

## Area Under the Normal Distribution Curve

To determine the probability that a random variable X lies between two points, ‘a’ and ‘b’:

$$P\left( a < X < b \right) =\int _{ a }^{ b }{ f\left( x \right)dx }$$

The normal distribution is very important in statistical analysis, especially because of the central limit theorem. The theorem asserts that any distribution becomes normally distributed when the number of variables is sufficiently large. For instance, the binomial distribution tends to “change” into the normal distribution with mean nθ and variance nθ(1 – θ).

Explain the key properties of the normal distribution.

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#### Sentence Examples

• On the centric hypothesis two formulae are possible: (1) due to H.E.
• It remains, therefore, to consider Erlenmeyer's formula and those derived from the centric hypothesis.
• The centric formula proposed by Bamberger represents naphthalene as formed by the fusion of two benzene rings, this indicates that it is a monocyclic composed of ten atoms of carbon.
• When, as in the formation of naphthalene tetrachloride, for example, the one ring becomes saturated, the other might be expected to assume the normal centric form and become relatively inactive.
• According to Armstrong, anthracene behaves unsymmetrically towards substituents, and hence one lateral ring differs from the other; he represents the molecule as consisting of one centric ring, the remaining medial and lateral ring being ethenoid.


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# Introduction to Quotient Rule with example

0
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While using differentiation, two functions connected by different operators can be differentiated keeping in mind the different operators used. For example, a positive operator is used within two functions. Both the functions can be easily differentiated using the positive operator. The same thing happens with a negative operator as well. But it is not possible in the case of multiplication and division. For multiplication, each of the functions is differentiated separately while keeping the other function constant. But, in the case of division, the process gets more complicated. In this article, we would check out what the quotient rule in differentiation is all about.

Contents

## Introduction to Quotient Rule

Let us assume that there are two functions f(x) and g(x). Here the functions are differentiable, which means the derivative of both the functions prevails. Thus using the quotient rule, we can find the quotient of the given functions. According to the quotient rule, the derivative of the complete function is the derivative of the numerator times the denominator minus the derivative of the denominator times the numerator which is totally divided using the square of the denominator. Moreover, the quotient rule is quite similar to the product rule, just that we would put negative instead of positive sign. Another important thing is to keep in mind that we need to square the denominator. The formula looks as follows:

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H’(x) = d/dxf(x)/g(x) = [f’(x)g(x) – f(x)g’(x)]/[g(x)]2

Let us find whether the function is true or not.

H(x) = f(x)/g(x) {given}

Thus, using product rule, we get,

f’(x) = H’(x)g(x) + g’(x)H(x)

By putting the value of H(x),

f’(x) = H’(x)g(x) + g’(x)f(x)/g(x)

H’(x) = [f’(x) – g’(x)f(x)/g(x)]/g(x)

H’(x) = [f’(x)g(x) – g’(x)f(x)]/[g(x)]2 (Thus Proved)

## Example of Quotient Rule

Let us find out the quotient rule utilizing an example. We need to differentiate y=cos x / x2.

So, we can take u = cos x and v = x2

Therefore, the derivative of these two functions would be:

du / dx = – sin x and dv / dx = 2x.

Thus using the formula, we can write:

dy / dx = [v (du / dx) – u (dv / dx)] / [v2]

Thus by putting the values, we get:

dy / dx = [x2 * (- sin x) – cos x * 2 x] / [x2]2

So, by taking out the common factor ( – x) from the above equation, we get:

dy / dx = [ – x( x sin x + 2 cos x)] / x4

dy / dx = [ – ( x sin x + 2 cos x)] / x3

Thus this is the required derivative that we need to find out.

## Conclusion

From the above equation, we now know what a quotient rule is regarding the concept of calculus. It is a bit different from the product rule but more or less the formula to find the derivatives. Sometimes by working around with the quotient rule, individuals simply use the complete function to apply within the formula. This is not at all perfect. Firstly, we need to simplify the function, and then move forward with the application process. By utilizing simplification, the function becomes easy to understand and becomes simple and clear to derive.

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Cuemath has been helping students with their math lessons online. There are finer concepts regarding calculus, trigonometry, algebra, etc., which the teachers at Cuemath are eager to help students with. Moreover, there are test samples, and exams are also conducted online to strengthen the root level of each student. Book an online class at Cuemath and aid yourself with a profound knowledge of the subject. Cuemath classes are always there to provide help to students regarding different fields of mathematics.

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CodeForces - 1029A

You are given a string t consisting of n lowercase Latin letters and an integer number k.

Let’s define a substring of some string s with indices from l to r as s[l…r].

Your task is to construct such string s of minimum possible length that there are exactly k positions i such that s[i…i+n−1]=t. In other words, your task is to construct such string s of minimum possible length that there are exactly k substrings of s equal to t.

It is guaranteed that the answer is always unique.

Input
The first line of the input contains two integers n and k (1≤n,k≤50) — the length of the string t and the number of substrings.

The second line of the input contains the string t consisting of exactly n lowercase Latin letters.

Output
Print such string s of minimum possible length that there are exactly k substrings of s equal to t.

It is guaranteed that the answer is always unique.

Examples
Input
3 4
aba
Output
ababababa
Input
3 2
cat
Output
catcat

#include<iostream>
#include<string>
using namespace std;
int main(){
int n,m;
string str;
cin>>n>>m>>str;
int pos=0;
for(int i=0;i<n;i++)
if(str.substr(0,i)==str.substr(n-i,i)) pos=i;
cout<<str;
for(int i=1;i<m;i++)
cout<<str.substr(pos);
cout<<endl;
return 0;
}

10-29 31


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# The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral.

Solution:

Let the common ratio between the angles be $x$. Therefore, the angles will be $3 x, 5 x, 9 x$, and $13 x$ respectively.

As the sum of all interior angles of a quadrilateral is $360^{\circ}$,

$\therefore 3 x+5 x+9 x+13 x=360^{\circ}$

$30 x=360^{\circ}$

$x=12^{\circ}$

Hence, the angles are

$3 x=3 \times 12=36^{\circ}$

$5 x=5 \times 12=60^{\circ}$

$9 x=9 \times 12=108^{\circ}$

$13 x=13 \times 12=156^{\circ}$

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# GMAT Question of the Day (Jun 3): Geometry and Critical Reasoning

- Jun 3, 02:00 AM   Comments [0]

Math (DS) Water evaporates at the rate of two liters per hour per one square meter of surface. How long will it take to evaporate 30 liters of water from a swimming pool full of water with vertical walls and a flat bottom that holds 56...

# GMAT Question of the Day (Jun 2): Special Operations and Critical Reasoning

- Jun 2, 02:00 AM   Comments [0]

Math (PS) If , what is the last digit of ? (A) 8 (B) 6 (C) 4 (D) 2 (E) 1 OA and Explanation GMAT Daily Deals Manhattan GMAT is releasing 2 new books: Foundations of GMAT Verbal and Advanced Quant. Pre-order your copy today! Advanced Quant and Foundations of Verbal To welcome...

# GMAT Question of the Day (May 27): Algebra and Critical Reasoning

- May 27, 02:00 AM   Comments [0]

Math (DS) If is a positive integer, is \sqrt{x} < 2.5x - 5[/latex] ? 1. $x < 3$ 2. $x$ is a prime number OA and Explanation GMAT Daily Deals GMAT Club members receive 10% off of Veritas Prep GMAT Courses and MBA Admission Consulting. Sign up! Stacy Blackman Consulting offers...

# GMAT Question of the Day (May 26): Word Problem and Critical Reasoning

- May 26, 02:00 AM   Comments [0]

Math (PS) Two cities, Kensington MD and Reston VA are 30 km apart. From both of these cities, simultaneously, two hikers start their journeys towards each other. They are walking at a constant speed of 5 km/hour each. Simultaneously, a fly leaves the city of Kensington....

# Manhattan GMAT Releases New Books in Quant and Verbal

- May 23, 21:00 PM   Comments [2]

Manhattan GMAT is readying two new strategy supplements to expand their existing catalog of prep materials: Foundations of GMAT Verbal and Advanced GMAT Quant. Inspired by the success of Manhattan GMAT’s Foundations of GMAT Math strategy supplement, these two new books are intended to...

# GMAT Question of the Day (May 20): Arithmetic and Critical Reasoning

- May 20, 02:00 AM   Comments [0]

Math (PS) Which of the following integers is a divisor of ? (A) 15 (B) 14 (C) 13 (D) 7 (E) 2 OA and Explanation GMAT Daily Deals Stacy Blackman Consulting offers a comprehensive MBA consulting service. Learn more and get a free profile evaluation from on GMAT Club. Learn more! To welcome Precision...

# GMAT Question of the Day (May 19): Arithmetic and Critical Reasoning

- May 19, 02:00 AM   Comments [0]

Math (DS) , , and are integers. Is ? 1. 2. ( and ) OA and Explanation GMAT Daily Deals GMAT Club members receive 10% off of Veritas Prep GMAT Courses and MBA Admission Consulting. Sign up! Prepare for the GMAT fast! Take...

# GMAT Question of the Day (May 13): Algebra and Critical Reasoning

- May 13, 02:00 AM   Comments [0]

Math (DS) If is a positive integer, is 1. 2. OA and Explanation GMAT Daily Deals The GMAT Unlocked by Kaplan addresses some of the most frequently asked questions about the test and shares a bit about how CATs work. Get your free copy now! Need to...

# GMAT Question of the Day (May 12): Algebra and Critical Reasoning

- May 12, 02:00 AM   Comments [0]


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# Possible link between the electromagnetic field and gravitation- Was Nikola Tesla onto something?

The main philosophical question that is at the root of the fluid model presented here is the following: what is the ontological meaning of the four potential $\mathbf{A}$? James Clerk Maxwell used the concept of four-potential in his original formulation of Maxwell’s equation (later reformulated by Heaviside in their canonical vector form). However, to this day there is no conclusive agreement on what the entity “really” is. As the magnetic field is assumed to be divergence-free, it is a fact of vector calculus that there exists a vector potential that generates the magnetic field. But this is merely a mathematical theorem, it has no ontological added value in it. To me, the whole idea of describing fluids of space-time was to interpret the four-potential as a velocity field of the fluid. This is an ontological approach which helps me to find analogies and if one wants to interpret is as a potential instead of velocity, by all means do so. But the power of analogy is very useful when one tries to understand nature. So my working hypothesis is still (despite of the fact that the scientific community does not generally accept it) that the four-potential is to be interpreted as a velocity of an aether in 4-dimensional space-time.

As I explained earlier, the most important properties of fluids are the divergence, vorticity and viscosity (which are interrelated). I have already shown that the divergence of the fluid is directly linked to the Lorentz-gauge and that the vorticity of the fluid is essentially the electromagnetic field tensor. What occurred to me recently was the concept of “curvature”. Now curvature is important, as the general relativity describes that mass and energy alter the curvature of the space (i.e. the metric tensor differs from the Kronecker delta). Curvature is a nice heuristic concept more generally as well, because for example we can describe the flow of heat using the heat equation based on the curvature of the temperature field. Similarly, electrostatics and Newtonian gravity are based essentially on Poisson’s equation where charge or mass generate curvature on the potential. So I started thinking about curvature of the velocity field of aether.

So what is curvature? As a first approximation we can understand it through the second derivative or the Laplacian. In differential geometry we have concepts like the Riemann tensor, Ricci tensor and so forth. However, as I want to keep things as simple as possible, I start of with the Laplacian approach, which actually takes me quite far as you shall see. Let us then consider a 3-dimensional space and take the following vector calculus identity

$\nabla \times \nabla \times \mathbf{A}=\nabla (\nabla \cdot \mathbf{A})-\nabla\cdot \nabla\mathbf{A}$

This is a beautiful identity as it binds together the “curvature” and divergence of the vector field, which together manifest themselves as the curl of the curl of the vector field. What I realized, was that this actually means that the curl of the curl is basically a commutator, i.e. it measures how much the divergence and gradient do not commute locally. Not to mention that this is actually quite close to the definition of the Riemann tensor!

Now I shall the construct a curvature concept for the aether fluid. This is straightforward, but it is rather complicated in 4-dimensions, because the curl operator has to be generalized as I’ve done in my earlier posts. Nevertheless, the end result is a rank 3 tensor, which has 64 components.

I began my interest in this curl of the curl concept also because of some experiments that have to do with rotating magnetic fields. Nikola Tesla was one of the pioneers in the field. Apparently rotating magnetic fields can produce quite interesting effects which have to do with gravity. See for example the following article by the European Space Agency

http://www.esa.int/Our_Activities/Preparing_for_the_Future/GSP/Towards_a_new_test_of_general_relativity

Obviously it is also quite interesting that the rotation of the magnetic field happens to be linked to the curl of curl of the vector potential! So curvature, rotating magnetic fields and gravitation are all linked together in some mysterious way.

Let us then find what is the “curvature” of the “aether” field. What we already have is the curl of the velocity field, which I obtained using the sort of exterior/Grassmann algebra:

$\nabla^*\times \mathbf{A}=\begin{bmatrix} 0&-\frac{i}{c}\frac{\partial \phi}{\partial x}-\frac{i}{c}\frac{\partial A_1}{\partial t}&-\frac{i}{c}\frac{\partial \phi}{\partial y}-\frac{i}{c}\frac{\partial A_2}{\partial t}&-\frac{i}{c}\frac{\partial \phi}{\partial z}-\frac{i}{c}\frac{\partial A_3}{\partial t}\\ \frac{i}{c}\frac{\partial A_1}{\partial t}+\frac{i}{c}\frac{\partial \phi}{\partial x}&0&\frac{\partial A_1}{\partial y}-\frac{\partial A_2}{\partial x}&\frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial x}\\ \frac{i}{c}\frac{\partial A_2}{\partial t}+\frac{i}{c}\frac{\partial \phi}{\partial y}&\frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y}&0&\frac{\partial A_2}{\partial z}-\frac{\partial A_3}{\partial y}\\ \frac{i}{c}\frac{\partial A_3}{\partial t}+\frac{i}{c}\frac{\partial \phi}{\partial z}&\frac{\partial A_3}{\partial x}-\frac{\partial A_1}{\partial z}&\frac{\partial A_3}{\partial y}-\frac{\partial A_2}{\partial z}&0\\ \end{bmatrix}$

We can state this in a more simple notation using the symbols for electric and magnetic field

$\mathbf{F}=\nabla^*\times \mathbf{A}=\begin{bmatrix}0 & iE_x & iE_y & iE_z\\ -iE_x & 0 & -B_z & B_y \\ -iE_y & B_z & 0 &-B_x \\ -iE_z & -B_y & B_x & 0\\ \end{bmatrix}$

How to then apply the curl to a second rank tensor? We can do it by proceeding to each column vector separately. First we take the column vector and form a gradient tensor out of it. The first column vector is

$\mathbf{F}^1=\begin{bmatrix} 0 \\ -iE_x \\ -iE_y \\ -iE_z \\ \end{bmatrix}$

So that the gradient tensor is

$\nabla ^* \mathbf{F}^1=\begin{bmatrix} 0 & 0 &0 &0 \\ \frac{1}{c}\frac{\partial E_x}{\partial t} & -i\frac{\partial E_x}{\partial x} & -i\frac{\partial E_x}{\partial y} & -i\frac{\partial E_x}{\partial z} \\ \frac{1}{c}\frac{\partial E_y}{\partial t} & -i\frac{\partial E_y}{\partial x} & -i\frac{\partial E_y}{\partial y} & -i\frac{\partial E_y}{\partial z} \\ \frac{1}{c}\frac{\partial E_z}{\partial t} & -i\frac{\partial E_z}{\partial x} & -i\frac{\partial E_z}{\partial y} & -i\frac{\partial E_z}{\partial z} \\ \end{bmatrix}$

One should note btw that the diagonal contains the divergence of the electric field.

In similar manner, one has the second column vector,

$\mathbf{F}^2=\begin{bmatrix} iE_x \\ 0 \\ B_z \\ -B_y \\ \end{bmatrix}$

What one needs to do then is to take the transpose of the each gradient tensors and take the difference of the gradient tensor and its transpose. The result is a rank 3 tensor which contains just so much information that it needs another posting…


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# Would a strong diamagnetic material exhibit inductance?

Erm. This question got shot-down in electronics.stackexchange.com and somebody recommended I raise it on here, so ...

Do metals with the property of strong diamagnetism also exhibit inductance? Would a fluctuating magnetic field induce a weaker current in a strongly diamagnetic metal as compared against a ferromagnetic/paramagnetic metal?

I remember reading metal-detectors are based upon the property of inductance. So if a metal detector were to hover over a silver ingot/plate, would it succeed or fail?

-
From what kind of metals are transformer windings made? Are that metals ferro/diamagnetic? Is there any metal which is not basically diamagnetic? –  Georg Oct 17 '11 at 8:56
Georg, Good call - as far as i know transformer windings are usually copper/silver i.e. strongly diamagnetic. Thanks. –  Everyone Oct 17 '11 at 9:16
Most diamagnets have a very small magnetic susceptibility, of the order of $-10^{-5}$. Thus I would not expect any significant effect of their susceptibility on the induced eddy currents.
One exception are the superconductors. In some sense they can be considered like perfect diamagnets ($\chi = -1$). If you apply a field on a superconductor you will induce a significant current on its surface, enough to screen the field inside the superconductor (this is the Meissner effect).


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# How it works¶

The pemcoupling package provides command-line tools for computing and post-processing coupling functions. In summary, it performs the following steps:

1. preprocess data from PEM sensors and DARM
2. compute coupling functions for each sensor during each injection
3. estimates DARM ambients for every coupling function
4. aggregate data across multiple injections to produce a composite coupling function and estimated ambient for each sensor
5. export data in the form of:
1. CSV files containing single-injection coupling functions (in physical units and in counts) and estimated ambients
2. plots of single-injection coupling functions (in physical units and in counts)
3. spectrum plots showing sensor and DARM ASDs during and before injections, super-imposed with a estimated ambients
4. CSV files containing composite coupling functions (in physical units and in counts) and estimated ambients
5. plots of composite coupling functions (in physical units and in counts) and estimated ambients
6. multi-plots of composite coupling functions and estimated ambients, labeling data by injection.

Contents:


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# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
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The Hausdorff measure of noncompactness of operators on the matrix domains of triangles in the spaces of strongly ${C}_{1}$ summable and bounded sequences. (English) Zbl 1244.47046

Let $\omega$ be the space of all complex sequences $x={\left({x}_{k}\right)}_{k=1}^{\infty }$, $A={\left({a}_{nk}\right)}_{n,k=1}^{\infty }$ be an infinite matrix of complex numbers and $X$ a subset of $\omega$. The set

${X}_{A}=\left\{x\in \omega :Ax=\sum _{k=1}^{\infty }{a}_{nk}{x}_{k}\in X\right\}$

is called the matrix domain of $A$ in $X$. Given any sets $X$ and $Y$ in $\omega$, let $\left(X,Y\right)$ denote the class of all matrices $A$ such that $X\subset {Y}_{A}$. Let $T={\left({t}_{nk}\right)}_{n,k=1}^{\infty }$ be a triangle, that is, ${t}_{nk}=0$ for $k>n$ and ${t}_{nn}\ne 0$ $\left(n=1,2,\cdots \right)$, let $e\in \omega$ be the sequence with ${e}_{k}=1$ for all $k$ and $1\le p<\infty$. The sets of strongly ${C}_{1}$-summable and bounded sequences

${w}_{0}^{p}=\left\{x\in \omega :\underset{n\to \infty }{lim}\left(\frac{1}{n}\sum _{k=1}^{n}{\left|{x}_{k}\right|}^{p}\right)=0\right\},$
${w}^{p}=\left\{x\in \omega :x-\xi ·e\in {w}_{0}^{p}\phantom{\rule{4.pt}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{some}\phantom{\rule{4.pt}{0ex}}\text{complex}\phantom{\rule{4.pt}{0ex}}\text{number}\phantom{\rule{4.pt}{0ex}}\xi \right\}$

and

${w}_{\infty }^{p}=\left\{x\in \omega :\underset{n}{sup}\left(\frac{1}{n}\sum _{k=1}^{n}{\left|{x}_{k}\right|}^{p}\right)<\infty \right\}$

were defined and studied by I. J. Maddox [“On Kuttner’s theorem”, J. Lond. Math. Soc. 43, 285–290 (1968; Zbl 0155.38802)].

In the paper under review, the authors apply the Hausdorff measure of noncompactness to characterize the classes of compact operators given by infinite matrices $A\in \left(X,Y\right)$, where $X$ is one of spaces ${\left({w}_{0}^{p}\right)}_{T},{\left({w}^{p}\right)}_{T}$ or ${\left({w}_{\infty }^{p}\right)}_{T}$ and $Y$ is the space ${c}_{0}$ of null sequences or the space $c$ of convergent sequences. Moreover, they give sufficient conditions for the compactness of operators $A\in \left(X,Y\right)$ when $X$ is again one of spaces ${\left({w}_{0}^{p}\right)}_{T},{\left({w}^{p}\right)}_{T}$ or ${\left({w}_{\infty }^{p}\right)}_{T}$ and the final $Y$ is the space ${l}_{\infty }$ of bounded sequences.

##### MSC:
 47H08 Measures of noncompactness and condensing mappings, $K$-set contractions, etc. 40C05 Matrix methods in summability 40J05 Summability in abstract structures
##### References:
 [1] Cooke, R. C.: Infinite matrices and sequence spaces, (1950) · Zbl 0040.02501 [2] Altay, B.; Başar, F.; Malkowsky, E.: Matrix transformations on some sequence spaces related to strong Cesàro summability and boundedness, Appl. math. Comput. 211, No. 2, 255-264 (2009) · Zbl 1171.40002 · doi:10.1016/j.amc.2009.01.062 [3] Başar, F.; Malkowsky, E.; Altay, B.: Matrix transformations on the matrix domains of triangles in the spaces of strongly C1- summable and bounded sequences, Publ. math. Debrecen 73, No. 1 – 2, 193-213 (2008) · Zbl 1164.46003 [4] Djolović, I.; Malkowsky, E.: A note on compact operators on matrix domains, J. math. Anal. appl. 340, No. 1, 291-303 (2008) · Zbl 1147.47002 · doi:10.1016/j.jmaa.2007.08.021 [5] Jarrah, A. M.; Malkowsky, E.: Ordinary, absolute and strong summability and matrix transformations, Filomat 17, 59-78 (2003) [6] Maddox, I. J.: On kuttner’s theorem, J. London math. Soc. 43, 285-290 (1968) · Zbl 0155.38802 · doi:10.1112/jlms/s1-43.1.285 [7] Malkowsky, E.; Rakočević, V.: The measure of noncompactness of linear operators between spaces of strongly C1-summable and bounded sequences, Acta math. Hungar. 89, No. 1 – 2, 29-45 (2000) · Zbl 0963.47038 · doi:10.1023/A:1026721323997 [8] Malkowsky, E.; Rakočević, V.: An introduction into the theory of sequence spaces and measures of noncompactness, zbornik radova, Zbornik radova, matematički institut SANU belgrade 9, No. 17, 143-234 (2000) · Zbl 0996.46006 [9] Malkowsky, E.; Rakočević, V.: On matrix domains of triangles, Appl. math. Comput. 189, No. 2, 1146-1163 (2007) · Zbl 1132.46011 · doi:10.1016/j.amc.2006.12.024 [10] Wilansky, A.: Summability through functional analysis, Summability through functional analysis 85 (1984) · Zbl 0531.40008


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Enter the final elastic potential energy (J), the spring constant (N/m), and the change in position (m) into the calculator to determine the Initial Elastic Potential Energy.

## Initial Elastic Potential Energy Formula

The following equation is used to calculate the Initial Elastic Potential Energy.

PEi = PEf – f*x

• Where PEi is the Initial Elastic Potential Energy (J)
• PEf is the final elastic potential energy (J)
• f is the spring constant (N/m)
• x is the change in position (m)

## How to Calculate Initial Elastic Potential Energy?

The following example problems outline the steps and information needed to calculate the Initial Elastic Potential Energy.

Example Problem #1

1. First, determine the final elastic potential energy (J). In this example, the final elastic potential energy (J) is determined to be 400 .
2. Next, determine the spring constant (N/m). For this problem, the spring constant (N/m) is measured to be 3 .
3. Next, determine the change in position (m). In this case, the change in position (m) is found to be 5.
4. Finally, calculate the Initial Elastic Potential Energy using the formula above:

PEi = PEf – f*x

Inserting the values from above and solving the equation yields:

PEi = 400- 3*5 = 375 (J)

Example Problem #2

Using the same method as above, first, we need to measure or determine the variables required by the equation. For this example problem, these are provided as follows:

final elastic potential energy (J) = 500

spring constant (N/m) = 4

change in position (m) = 3

Enter these given values into the calculator or above yields:

PEi = 500- 4*3 = 488 (J)


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# Product lattice

Could someone explain how to construct a product lattice, or point me to an explanation on the web?

-
I recommend that you read Davey and Priestley "Introduction to Lattices and Order". Specifically sections 1.26 and 2.14. –  Loop Space Jun 1 '10 at 12:42
I don't have a copy! :( And I need to multiply two lattices... –  user6473 Jun 1 '10 at 12:46

Please click through and read the discussion there. But to summarize, one choice is to form the product order, where you consider pairs $(a,b)$ and the orders are inherited in each coordinate separately. Another choice is to use the lexical order on pairs, but this only results in a lattice when there are certain completeness or linearity assumptions. The lexical product $K\times L$, however, amounts to replacing each node in $K$ with a copy of $L$, and this is often what is desired.


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# AP Microeconomics : AP Microeconomics

## Example Questions

1 2 5 6 7 8 9 10 11 13 Next →

### Example Question #92 : Competition

Patents, limiting the number of licenses available, and economies of scale are all examples of:

None of the other answers are correct.

barriers to entry

factors of production

sources of demand

sources of supply

barriers to entry

Explanation:

Patents, limiting the number of licenses available, and economies of scale can all hinder a firm's ability to enter the market. A patent prevents a firm from replicating a product that originated from another firm. A limited number of licenses can exclude firms who are unable to obtain licenses from entering the market. Economies of scale can prevent smaller firms from entering the market by making such an action cost-prohibitive.

### Example Question #91 : Perfectly Competitive Markets

Labor, capital, human capital, and natural resources are all examples of:

factors of production

None of the other answers are correct.

sources of demand

barriers to entry

sources of supply

factors of production

Explanation:

All of these are examples of factors of production, which are the inputs necessary for a production process.

### Example Question #93 : Competition

A consumer's indifference curves for two different goods will be a straight lines when:

The prices of the two goods are equal

Utility is maximized

Both goods are perfect substitutes

None of the other answers are correct.

Both goods are perfect compliments

Both goods are perfect substitutes

Explanation:

When both goods are perfect substitutes, the consumer is indifferent between the amount of consumption of the two goods. Thus, his/her indifference curves will be straight lines, because the rate of substitution between the two goods will be constant.

### Example Question #94 : Competition

Which of the following market structures is characterized by low barriers to entry, homogeneous products, and a large number of firms?

Oligopoly

Perfect Competition

Monopolistic Competition

Monopoly

Perfect Competition

Explanation:

In perfectly competitive markets, there are low barriers to entry (think of a food truck--all one needs in order to operate are a truck, fuel, cooking equipment, and a business license), homoegenous products (consider a farm--nearly all cucumbers are identical to one another), and a large number of firms (this is what makes the environment so competitive).

In monopolistic competition and oligopolies, products tend to be heterogeneous, and in a monopoly, there are extremely high barriers to entry.

### Example Question #123 : Ap Microeconomics

For which of the following market structures is demand always equal to marginal revenue?

Oligopoly

Monopoly

Perfect competition

Monopolistic competition

Perfect competition

Explanation:

The marginal revenue curve always has the same intercept on the price axis as the demand curve and twice the slope of the demand curve.

For market structures of oligopoly, monopoly, and monopolistic competition, the firm faces a demand curve that is downward sloping (indicated market power), so the marginal revenue curve would be steeper than the demand curve.

For perfectly competitive market structures, however, the firm faces a demand curve that is horizontal, with a slope of 0 (i.e. it is perfectly elastic). For a firm in perfect competitive market structure, therefore, the marginal revenue curve, which has the same intercept and twice the slope, results in the exact same curve as the demand curve.

### Example Question #1 : Marginal Revenue Product Of Labor Mrp

The following question will refer to this table:

If the wage paid to labor is $8 and the price of each unit of output is$2, how many units of labor will be hired?

one

five

four

two

three

four

Explanation:

A firm will hire labor until the marginal revenue product of labor is lower than the cost of labor/wage, which is $8. The marginal revenue product of labor is the marginal product of labor multiplied by the product's price. The marginal revenue of the fourth unit of labor is$10 (five units multiplied by $2) and the marginal revenue of the fifth unit of labor is$6 (three units multiplied by $2). Thus, the firm will hire four units of labor. ### Example Question #2 : Marginal Revenue Product Of Labor Mrp A firm is hiring labor in a perfectly competitive labor market. If the marginal revenue product of labor is greater than the wage rate, which of the following is true? Possible Answers: The firm should invest in more capital. The firm should hire more labor. The firm is maximizing its profit. The firm is losing money. None of the other answers. Correct answer: The firm should hire more labor. Explanation: A profit maximizing firm will hire labor until the marginal product of labor is greater than the wage rate. If the marginal product of labor is greater than the wage rate, then the firm should hire more labor until the two values are equal. ### Example Question #3 : Marginal Revenue Product Of Labor Mrp In a perfectly competitive labor market, an employer will continue hiring until ___________ . Possible Answers: wage equals marginal product of labor. wage equals the price of the good or service produced. wage equals marginal cost of labor. wage equals marginal revenue. wage equals marginal revenue product of labor. Correct answer: wage equals marginal revenue product of labor. Explanation: Like any market, the profit maximization level is the point at which marginal cost equals marginal revenue. In this case, the marginal cost of labor is the labor wage. Thus, the profit maximization point for the employer is the point at which wage equals the marginal revenue product of labor. ### Example Question #4 : Marginal Revenue Product Of Labor Mrp Refer to the following chart for this question: If each product is sold for$2, what is the marginal revenue product of labor for the fourth unit of labor?

$4$8

$10$12

$6 Correct answer:$8

Explanation:

The marginal revenue product of labor for the fourth unit of labor is its marginal product multiplied by the cost of the product. The marginal product of the fourth unit of labor is 4 (the difference between total production at four units of labor and three units of labor), and cost of the product is $2, so the marginal revenue product of labor for the fourth unit is$8.

### Example Question #5 : Marginal Revenue Product Of Labor Mrp

Refer to the following chart for this question:

If each product is sold for $2, and each unit of labor costs$10, how many units of labor will the firm hire?

one

three

five

two

four


text:
# What are the general solutions to a hard sphere collision? [duplicate]

Surely someone has found the solutions to the hard sphere collisions (in $$n$$ dimensions) of two bodies of mass $$m_1$$ and $$m_2$$, respectively--that is the resultant velocities (or momenta) of the two bodies after a hard sphere collision given $$\vec{p}_1$$ and $$\vec{p}_2$$. I have a solution (both for elastic and inelastic collisions), but I wonder whether or not it's correct.

FYI, here are my solutions for elastic, hard-sphere collisions:

$$\vec{p}_{1f} = \vec{p}_1 + 2\frac{\left(m_1\vec{p}_2 - m_2\vec{p}_1\right)\circ \hat{r}}{m_1 + m_2}\hat{r}, \\ \vec{p}_{2f} = \vec{p}_2 - 2\frac{\left(m_1\vec{p}_2 - m_2\vec{p}_1\right)\circ \hat{r}}{m_1 + m_2}\hat{r}.$$

wher $$\hat{r}$$ is in the direction of the "hard sphere" (i.e. the unit vector connecting the two centers of the two spheres). You can see that my formulation trivially conserved momentum--the question is whether or not it conserves energy as well.

If it helps, here is my full explanation (this is quite long and the link will expire in 30 days from 09/23/2014). For these purposes, you should skip to the section on n-dimensional cases (page 6). I wrote this as a reference to myself for doing physics simulations since I could not find any reference on hard sphere collisions (other than references setting up the problem which give no solutions).

Here is an abbreviated work on finding the final momenta from the elastic case:

Just as in the $$1$$D case, we add and subtract the same momentum from each particle--except now we are adding and subtracting a vector: $$\vec{p} = p\hat{r}$$, which means that $$|\vec{p}| = p$$ (such that $$\vec{p}_{1f} = \vec{p}_1 + p\hat{r}$$ and $$\vec{p}_{2f} = \vec{p}_2 - p\hat{r}$$). This gives the final energy as:

\begin{align} E_f = \frac{|\vec{p}_1 + p\hat{r}|^2}{2m_1} + \frac{|\vec{p_2} - p\hat{r}|^2} {2m_2} \end{align}

We now write the final energy in terms of the magnitude of $$p$$ and the dot product between $$\hat{r}$$ and $$\vec{p}_1$$ and likewise for $$\vec{p}_2$$. Note that if you choose the direction for $$\vec{p}$$ to act a priori (like in the hard-sphere case), the dot products $$\vec{p}_1\cdot \hat{r}$$ and $$\vec{p}_2\cdot\hat{r}$$ will be known:

\begin{align*} E_f = \frac{p_1^2 + p^2 + 2p\hat{r}\circ\vec{p}_1}{2m_1} + \frac{p_2^2 + p^2 - 2p\hat{r}\circ\vec{p}_2}{2m_2} \end{align*}

As usual the elastic case greatly simplifies:

\begin{align*} \require{cancel} \cancel{\frac{p_1^2}{2m_1}} + \cancel{\frac{p_2^2}{2m_1}}= \frac{\cancel{p_1^2} + p^2 + 2p\hat{r}\circ\vec{p}_1}{2m_1} + \frac{\cancel{p_2^2} + p^2 - 2p\hat{r}\circ\vec{p}_2}{2m_2} \\ \frac{m_1 + m_2}{2m_1m_2}p^2 - 2\frac{m_1 \vec{p}_2\circ\hat{r} - m_2\vec{p}_1\circ\hat{r}}{2m_1m_2}p = 0 \end{align*}

This leads to a trivial solution of $$p = 0$$ (no collision occurs) and:

\begin{align} \frac{m_1 + m_2}{2m_1m_2}p - 2\frac{m_1 \vec{p}_2\circ\hat{r} - m_2\vec{p}_1\circ\hat{r}}{2m_1m_2} = 0 \\ p = 2\frac{m_1 \vec{p}_2\circ\hat{r} - m_2\vec{p}_1\circ\hat{r}}{m_1 + m_2} \end{align}

Hence my solutions of:

$$\vec{p}_{1f} = \vec{p}_1 + 2\frac{\left(m_1\vec{p}_2 - m_2\vec{p}_1\right)\circ \hat{r}}{m_1 + m_2}\hat{r}, \\ \vec{p}_{2f} = \vec{p}_2 - 2\frac{\left(m_1\vec{p}_2 - m_2\vec{p}_1\right)\circ \hat{r}}{m_1 + m_2}\hat{r}.$$

## marked as duplicate by ja72, John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 20 '16 at 16:02

• What does $\vec p\circ \hat r$ mean here? Is it a dot product $\vec p\cdot\hat r$? – Ruslan Sep 22 '14 at 8:57
• Is it a 1D problem ? Otherwise you need an extra parameter, defining the impact distance. – Tom-Tom Sep 22 '14 at 10:23
• Hi Jared. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Sep 23 '14 at 6:39
• @Jared: Right, the reason I ask is partly to probe that your terminology agrees with my terminology:) If you are not interested in questions sensitive to the hard ball's finite size or internal degrees of freedom, then you may want to speak of point particles instead to simplify the question formulation altogether. – Qmechanic Sep 23 '14 at 19:39
• Possible duplicate of Elastic collision between two circles – ja72 May 20 '16 at 13:51

## Finding the missing equations

Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the problem.

The additional equations come from the fact that if you assume a frictionless impact*, then the impulse/force imparted to each object must be normal to the surface. This gives a constraint on the direction of the impulse. Along with the magnitude, this direction would give us n degrees of freedom. So without it, it gives us n-1: exactly the number we need.

## Solving for final velocities

Since we have spheres, the direction normal to the impact surface will be the vector that points from one COM to the other. Let's call this normal vector $\hat{j}$.

Now the velocities of balls A and B can be given as $V_{a1}$ and $V_{b1}$ before the collision and $V_{a2}$ and $V_{b2}$ after the collision.

Let's call the impulse that is imparted on ball A $J$. Then we have:

$$J \bullet\hat{j} =|J|$$ $$V_{a1} + \frac{J}{M_a}=V_{a2}$$ $$V_{b1} - \frac{J}{M_b}=V_{b2}$$

Then we can write the conservation of energy (multiplied by 2) as: $$M_a V_{a1} \bullet V_{a1} + M_b V_{b1} \bullet V_{b1} = M_a V_{a2} \bullet V_{a2} + M_b V_{b2} \bullet V_{b2}$$

Plugging in for our velocities after to collision gives:

$$M_a V_{a1} \bullet V_{a1} + M_b V_{b1} \bullet V_{b1} = M_a \left( V_{a1} + \frac{J}{M_a}\right) \bullet \left( V_{a1} + \frac{J}{M_a}\right) + M_b \left( V_{b1} - \frac{J}{M_b}\right) \bullet \left( V_{b1} - \frac{J}{M_b}\right)$$

$$M_a V_{a1} \bullet V_{a1} + M_b V_{b1} \bullet V_{b1} = M_a V_{a1}\bullet V_{a1} + 2 J\bullet V_{a1} + \frac{J\bullet J}{M_a} + M_b V_{b1}\bullet V_{b1} - 2 J\bullet V_{b1} + \frac{J\bullet J}{M_b}$$

$$0 = 2 J\bullet V_{a1} + \frac{J\bullet J}{M_a} - 2 J\bullet V_{b1} + \frac{J\bullet J}{M_b}$$

dividing by the magnitude of $J$ yields:

$$0 = 2 \hat{j} \bullet (V_{a1}-V_{b1}) + |J|\left(\frac1{M_a} + \frac1{M_b}\right)$$

$$|J| = 2 \frac{\hat{j} \bullet (V_{a1}-V_{b1})}{\frac1{M_a} + \frac1{M_b}}$$

### Final Velocities

$$V_{a2}=V_{a1}+2 \frac{\hat{j} \bullet (V_{a1}-V_{b1})}{M_a\left(\frac1{M_a} + \frac1{M_b}\right)}$$ $$V_{b2}=V_{b1}+2 \frac{\hat{j} \bullet (V_{b1}-V_{a1})}{M_b\left(\frac1{M_a} + \frac1{M_b}\right)}$$

### Finding Direction of Impulse

Let's call the vector pointing from the center of ball B to the center of ball A $R$. We know that before the impact $R=R_0+R't$ where $R_0$ is the difference in the balls' initial positions $P_a-P_b$ and $R'$ is the difference in the balls' initial velocities $V_{a1}-V_{b1}$.

At the moment of impact: $$\hat{j}=\frac{R}{|R|}$$ and $$|R|=r_a+r_b$$ Where $r_a$ and $r_b$ are the radii of the balls.

We can transform this second equation to find t:

$$R\bullet R=(r_a+r_b)^2$$ $$R'\bullet R't^2+2R_0\bullet R't+R_0\bullet R_0-(r_a+r_b)^2=0$$ $$t=-\frac{R_0\bullet R'+\sqrt{(R_0\bullet R')^2-R'\bullet R'(R_0\bullet R_0-(r_a+r_b)^2)}}{R'\bullet R'}$$

If this time is negative or imaginary, then the collision will not occur. Otherwise $R$ at the moment of impact can be found by plugging in t:

$$R=R_0-R'\frac{R_0\bullet R'+\sqrt{(R_0\bullet R')^2-R'\bullet R'(R_0\bullet R_0-(r_a+r_b)^2)}}{R'\bullet R'}$$

### Wrap Up

So given initial positions $P_a$ and $P_b$, initial velocities $V_{a1}$ and $V_{b1}$, and ball radii $r_a$ and $r_b$ the final velocities $V_{a2}$ and $V_{b2}$ can be found be following the following procedure:

Calculate the relative positions and velocities: $$R_0=P_a-P_b$$ $$R'=V_{a1}-V_{b1}$$

Calculate impact time: $$t=-\frac{R_0\bullet R'+\sqrt{(R_0\bullet R')^2-R'\bullet R'(R_0\bullet R_0-(r_a+r_b)^2)}}{R'\bullet R'}$$ If t is negative or imaginary, then no collision occurs, and the initial velocities are the final velocities. Otherwise:

Calculate the impulse direction: $$\hat{j}=\frac{R_0+R't}{|R_0+R't|}$$ Calculate the impulse: $$J=2 \hat{j}\frac{\hat{j} \bullet R'}{\frac1{M_a} + \frac1{M_b}}$$ Calculate the final velocities: $$V_{a2} = V_{a1} + \frac{J}{M_a}$$ $$V_{b2} = V_{b1} - \frac{J}{M_b}$$

*The collision must be frictionless if we're not considering moments of inertia as any friction would apply torque to the ball which would transfer energy and rotational inertia to the ball. The problem becomes slightly more complicated in that case as one has to assume how the friction will cause the balls to interact, which is not trivial as many of the simple assumptions conflict with energy conservation.

• Thank you! I used your solution in a pool simulator, worked perfectly! :) – PaulisDontExcludeMe Oct 1 '18 at 2:38

Here is a general figure of an hard spheres collision drawn in the center of mass of the mass $m_2$ before the collision. The black dot is attached to this frame.

To solve the problem, you need to observe

• Conservation of energy: $m_1v_1^2=m_1(v'_1)^2+m_2(v'_2)^2$.
• Conservation of momentum: $m_1\vec v_1=m_1\vec v'_1+m_2\vec v'_2$
• Conservation of torque with respect to the center of mass of $m_2$: $m_1v_1b=m_1v'_1b'$.

I have drawn a 2D problem, so you have four unknowns (two for $\vec v'_1$ and two for $\vec v'_2$). And there are four equations (one for energy and torque, two for momentum). So the problem is solved.

In three dimensions, you have $6$ unkowns, energy conservation provides one equation, momentum conservation provides $3$ equations and torque conservation provides two.

• You need the conservation of angular momentum because it provides the last equation you need to solve it all. For example, if $m_2=\infty$ you have $\alpha=\alpha'$ but this is no more true if $m_2$ is finite. – Tom-Tom Sep 23 '14 at 9:29
• I actually solved the system of equations I have given using your own method. The conservation of angular momentum gives $\vec r_1\times(\vec p_1+p\hat r)+\vec r_2\times(\vec p_2-p\hat r)=\vec r_1\times\vec p_1+\vec r_2\times\vec p_2$. This simplifies into $\hat r\times(\vec r_1-\vec r_2)=0$, which matches your definition of $\hat r$ (that you did not justify). – Tom-Tom Sep 24 '14 at 14:26
• I think my confusion (adding to everyone else's) is that any force that acts along the vector connecting two point masses will conserve angular momentum. The key here is that the interaction happens at a distance. If it were simply two point particles and the distance at which the collision occurred were zero, then angular momentum doesn't really come into play. I.e. it's not determinate what the final state would be because there are many possible outcomes that would all conserve angular momentum--this is not the case when the interaction happens at a distance. – Jared Sep 25 '14 at 5:41
• Your remark the distance between the centers is really nice. – Tom-Tom Sep 25 '14 at 7:30
• You left out rotational energy in conservation of energy 1/2 I $\omega^2$. So, you need 4 angular velocities and 8 angles to describe rotation rates and axes of two bodies before and after. – eshaya Sep 26 '14 at 20:37

This thread is a little old, but I thought I would chip in in case anyone comes looking for this answer. In short, Tom-Tom is exactly correct in his analysis of the problem. However, I am going to break this down in steps for a full 3D case and try to explain some of the points there was contention about (e.g. angular momentum).

First of all, here are the assumptions we are working with:

• Hard sphere collision is a technical term meaning the following: 1) collisions are elastic (which means kinetic energy is conserved); 2) the spheres have a finite radius (as someone else has pointed out, if they don't this is a point-like collision.)
• While one could work with a simulation in which we accounted for rigid body rotation, I will assume that the "spin" is zero. Spin adds a good deal of complexity to the computation.

I'm not going to actually solve each of the terms (it gets pretty messy), but I am going to outline how the process goes. For completeness, I'm not going to assume that we know the bodies will collide. We shall see as part of the analysis that this determination falls out pretty easily.

Here is the procedure:

1) First move to the rest frame of one of the particles. Now we are comparing a point and a 3D line, which greatly simplifies the geometry, rather than two 3D lines.

2) Recall the fundamental 3D geometry proof that a line and a point in 3D space can always be contained in a single plane. We can, in principle, do a coordinate transformation (rotation in $\theta$ and $\phi$ in spherical coordinates around the particle at the origin).

3) Good, now we have a line and a point in a single plane. It should be obvious that another rotation (in polar angle) can ensure we have the moving particle moving the x-axis only. This reproduces a picture like the first figure Tom-Tom posted. Now we have reduced our 3D problem to the 2D problem Tom-Tom gave a method for!

4) Collision detection: how do we know if they hit? We need to check 2 things: 1) is $R_1 + R_2 > b$, where $R_i$ are the radii and $b$ is the impact parameter; 2) we need to check that the moving particle gets within $\sqrt{(R_1 + R_2)^2 - b^2}$ of the x-axis. That's just the Pythagorean theorem for the triangle the spheres make at contact. If both criteria are met, we have a collision.

5) Now we're all setup, and we work from Tom-Tom's solution. I won't work the rest out, but I will make a couple of notes to clear up confusion.

• Angular momentum: Contrary to Jared's comment above, we cannot in general pick a coordinate system where the angular momentum is zero. Angular momentum will only be zero in two special cases: 1) the particles are at rest with respect to each other; 2) the moving particle is headed directly at the stationary particle. Why? $\vec{L} = \vec{r}\times\vec{p}$. If the particles are at rest, $\vec{p} = 0 \to L = 0$ (obviously we don't have a collision in this case either). If the impact parameter is zero, that means $\vec{r}$ and $\vec{p}$ are lined up with each other. Recall, $|\vec{L}| = |\vec{r}||\vec{p}|Sin(\theta)$, where $\theta$ is the angle between the two vectors. In the $b = 0$ case, $\theta = 0 \to |\vec{L}| = 0$. In all other cases, angular momentum will be non-zero, no matter what coordinate system you choose!

• The radii are important! They will enter into the angular momentum equation. To solve the angular momentum equation, first note that $\vec{r_{2}}' = \vec{v_2}' t$, therefore the angle between momentum and the radial vector will be zero giving the angular momentum of the second body as zero (as Tom-Tom's equation has indicated). They are not parallel for the first sphere. Staring at the geometry for a few minutes should convince you, rather, that $\frac{\vec{v_1}'}{t} = \vec{r_1}' + \vec{\alpha}$, where $\alpha$ is the vector pointing from the center of body 1 to the center of body 2. This vector has magnitude $R_1 + R_2$. We can then use the Pythagorean theorem to break $\alpha$ into components (note: the y component is just b). The x component will be $\sqrt{(R_1 + R_2)^2 - b^2}$. We use the vector addition above to eliminate $\vec{r_i}'$ for $\vec{v_i}'$. Note: t is arbitrary and can be set to 1. Finally, carry out the angular momentum cross product. You will see that the $\sqrt{(R_1 + R_2)^2 - b^2}$ term survives, and will therefore enter into your final equations!

From this point, it's just messy algebra. We have 4 unknowns and 4 equations. Solve those equations, and you'll have the velocities. A small amount of extra work can tell you, e.g. exactly when the collision occurs, which is important for simulations and will tell you how far to translate the spheres with the new velocity after collision (to match simulation timestep).

The very last step, after you've done all that, is to remember to transform back to the system coordinates. To do this, just unroll everything backwards, flipping the signs of angles and translations.

Hope that clears things up for someone that's looking at this problem. This is a classic one of those problems that seems like it should be trivially easy, but turns out to be quite involved to solve in the general case.

UPDATE

So, this all got me thinking and I went back to re-examine the problem a bit. In short, the problem turns out to be much more subtle than I originally thought. Here are a couple of new comments:

• The reason you need angular momentum is that angular momentum must be conserved for the simulation to be physical. If you break angular momentum, you break rotational invariance of your system. If you're making a simple game engine, that might be OK, but if you're trying to do physics, that's definitely not allowed. Angular momentum conservation is what makes the radii important.

• I need to retract what I said about spin. Spin is necessary (as I will show below). However, Newtonian gravity cannot be the source of spin. To introduce spin requires having net torque. Recall that for any (conservative) potential of the form $V = V(r)$, the force due to a sphere is equivalent to the force due to a point at its center. Briefly, any torque that would be created on one side of the sphere is canceled by the other side, and only force along the radial line survives. Newtonian gravity is clearly a potential of this form (as is, e.g. Coulomb, Yukakawa, and almost any "real" potential we deal with.)

• So, I worked out a solution that takes into account linear momentum, angular momentum and kinetic energy. I'll only put the x-component of the velocity vector here (the others are simply related by plugging this back into the equations:

$v_{xf} = \frac{\left(\beta ^2-1\right) \sqrt{\frac{v_i^2 \left(\beta ^2 (\mu +2)^2-\mu ^2\right)}{\beta ^2-1}}+v_i \left(\beta ^2 (\mu +2)-1\right)}{\mu +1}$

If you're curious, the following code will get this result in Mathematica:

(* Setup vectors: No Q *)
vi = {v1x, 0, 0};
v1 = {v1xf, v1yf, 0};
v2 = {v2xf, v2yf, 0};
\[Alpha] = {-Sqrt[(R1 + R2)^2 - b^2], b, 0}; (* The vector that connects the two centers at impact *)

r1 = \[Alpha] +
v1 \[CapitalDelta]t; (* NOTE: \[CapitalDelta]t should drop out! *)

r2 = v2 \[CapitalDelta]t;

(* Physics *)
Eqns0 = Simplify[{
m1 (vi.vi) == m1 (v1.v1) + m2 (v2.v2)(*+Q m1(vi.vi)*),
m1 v1x == m1 v1xf + m2 v2xf,
0 == m1 v1yf + m2 v2yf,
m1 v1x b == (m1 (r1 \[Cross] v1) + m2 (r2 \[Cross] v2))[[3]]
} /. {m1 -> m, m2 -> \[Mu] m, b -> \[Beta] (R1 + R2)},
\[Beta] < 1 && \[Beta] >= 0 && m > 0 && \[Mu] > 0 && R1 > 0 &&
R2]; Eqns0 // MatrixForm

Eqns1 = Eqns0 /. Solve[Eqns0[[2]], v2xf] //
Flatten; Eqns1 // MatrixForm
Eqns2 = Eqns1 /. Solve[Eqns1[[3]], v2yf] //
Flatten; Eqns2 // MatrixForm
(Eqns2[[1]] /. Solve[Eqns2[[4]], v1yf] // Flatten)[[1]]
FullSimplify[
Solve[%, v1xf], \[Beta] < 1 && \[Beta] >= 0 && m > 0 && \[Mu] > 0 &&
R1 > 0 && R2] // Flatten

I have put everything in dimensionless units to make the result more transparent. Specifically, $\mu$ is the ratio $m_1/m_2$, $\beta$ is the ratio $b/(R_1+R_2)$, where $b$ is the impact parameter. Note that the dimensionless parameters allow you to quickly verify that all of the terms on the right hand side do in fact have dimensions of velocity.

But wait! This is where things get interesting. Note that the term in the radical can become negative for real situations, which leads to complex velocities. This can't be allowed. We can plot $\beta$ vs $\mu$ to show the "allowed" region:

This is clearly a problem. It is certainly physically possible for $\beta$ to take any value between 0 and 1, but we get imaginary velocities if $\beta$ is too large compared to $\mu$.

In short, the resolution is that (classical) hard sphere scattering cannot be elastic in general (though it can be in special cases). I'll explain in a moment why that is the case, but first, I'll give a solution that "works." Here, I introduce another dimensionless parameter $Q$, which is the ratio of energy loss over initial kinetic energy. The modified solution is:

$v_{xf} = \frac{\left(\beta ^2-1\right) \sqrt{\frac{v_1^2 \left(\beta ^2 (\mu +2)^2+\mu (\mu (Q-1)+Q)\right)}{\beta ^2-1}}-v_i \left(\beta ^2 (\mu +2)-1\right)}{\mu +1}$

We can now plot $Q$ vs. $\beta$ for varying values of $\mu$:

It's important to note that this only provides a bound on $Q$. It does not tell us what value of $Q$ we should use. In other words, the problem is no longer fully constrained. So, what's going on here and how can we resolve it?

Here's the bottom line: hard sphere scattering is not physical! To have a true hard sphere (i.e. impenetrable) we would require the force at the surface of the spheres to be infinite, i.e. the force law would be a Dirac delta function. The impulse experienced, then, would be: $\Delta p = F \Delta t = (\infty)(0)$. This is undefined. It’s not surprising, then that we might get some strange physics out of hard sphere scattering. Moreover, it’s not surprising the problem would be unconstrained.

We can say a little more though. Up till now, I assumed that we could use the old standby: $\vec{F} = - \nabla V(r)$. But, this rule is only valid for conservative forces. Recall that a conservative force is a path independent force. In other words, if you start at point $A$ and take any path you like returning to point $A$ the potential remains the same. This is clearly not the case for hard sphere scattering. Any path that doesn’t involve a collision is going to be dramatically different than any path that does. In fact, this is a general rule for any truly localized force (note: Yukawa potential is a “localized” force, but in reality, it just has exponential fall-off, so it’s not a localized force in this sense).

In the case of a non-conservative force, we can no longer use $\vec{F} = - \nabla V(r)$. Rather, we must use the Helmholtz decomposition, which states that any force vector field can be decomposed as: $\vec{F} = - \nabla V(r) + \nabla \times \vec{A}$. That is, we must introduce a vector potential.

This gets us back to the notion of spin. A quick calculation will verify that the vector potential will generically introduce a net torque in the scattering process, which will produce spin (of course, for very special $\vec{A}$ this will not happen).

The real problem now is that $\vec{A}$ is completely unconstrained by the problem. It is associated with $Q$, and therefore the internal degrees of freedom of the sphere. But, we have assumed from the beginning that the sphere is a “black box” that we don’t specify physics for. This is an unphysical assumption, and it turns out that it matters. The obvious internal degrees of freedom to account for are the spin vector and the temperature (it’s a fairly natural assumption that collisions between real bodies would effect the heat content of the bodies). The argument above suggests that at least some of that energy should go into spin energy, but probably not all of it. Just how much should go into each of these degrees of freedom depends on the specifics of internal physics.

I think that’s about as far as we can go with the problem. If we want to fully specify the problem we need to create something like an “equation of state” (by analogy) for the spheres which determines how the energy gets distributed in bodies. A reasonable approximation might be to just specify a ratio of temperature to spin, though physical intuition suggests that higher impact parameters should transfer more spin and less temperature, so perhaps this should be a function of $\beta$. A suggestion off the top of my head would be to assume spin as power series in $\beta$, where the series constants could be experimentally determined or estimated by more detailed theoretical models of the internal physics of the bodies.

If you want a physical model of scattering, you need to specify the physics of the bodies. One clean solution that makes sense for a lot of cases (e.g. actual macroscopic rigid bodies like pool balls) would be to use Yukawa scattering. The Yukawa potential is conservative, and was specifically designed to approximate the effects of electrodynamic screening that occurs in macroscopic, (mostly) electrically neutral bodies.

• I appreciate your answer--my question was to find solutions (to check against mine). I didn't use angular momentum at all and I don't think it's necessary (albeit the canonical approach). I used conservation of linear momentum and only looked at solutions which trivially conserved this by adding opposite and equal momentum vectors in the direction of the vector connecting the centers (this is the only place that the radii come into play; I don't think the impact parameter affects this--just changes when a collision occurs and thus $\hat{r}$). See my edited question (soon). – Jared Mar 29 '16 at 23:59
• I am aware of the "usual" way to set up the equations--but I have never found (and still cannot find) solved solutions. My approach seems to be much simpler--I'm mainly wondering whether or not it's valid. This approach also correctly gives the solutions in the 1D case without having to go to the center of mass frame. Again to reiterate, I'm not concerned with spin at all (that indeed would require some conservation of angular momentum and, most likely, how spin is transferred when two bodies collide). – Jared Mar 30 '16 at 0:14
• ...and as far as spin goes and being artificial (in the classical sense), I would again refer to my simulation of two bodies colliding with initially no spin (but a non-zero total angular momentum) and then exhibiting spin after the collision (this is undergoing Newtonian gravity and with my inelastic results...and a large amount of fine tuning). Newtonian gravity, in this case, is what provides the centripetal force necessary for spin (rigid bodies require convenient molecular forces). – Jared Mar 30 '16 at 2:20
• Please see my (substantial) updates to my original post... – Lowell Apr 5 '16 at 2:30
• I'm not going to lie, I don't fully understand what you did. However there are some things which do not make sense to me. Namely $\mu = \frac{M_1}{M_2}$. Your graph of $Q$ with respect to $\beta$, it seems to me, should be symmetric about $\mu = 1$. There should be a lot of symmetry here--especially when talking about the $Q$ value. It shouldn't matter which particle we label as $1$ and $2$--you should get the same $Q$ value under change of indexes, shouldn't you? This would mean we would expect the same curve for $\mu = .1$ as we do for $\mu = 10$. – Jared Apr 6 '16 at 0:24


text:
# How fixed air, gas and fixed weights in rigid airships is computed in this book?

I hope I am posting this in the right forum. My current problem is part aviation, part maths. I have been struggling with the following computation ( Burgess, Charles Paine. Airship Design. Honolulu, Hawaii: University Press of the Pacific, 2004. page 14. The pdf is available at http://www.xlta.org/library/burgess.pdf, note that the first edition of this book was written in 1927. The present snippet uses the typographical conventions from the pdf.)

Problem 1: Find the volume and horsepower of a rigid airship to carry a military load of 15,000 lbs.at 60 knots (101.3 ft. / sec.). for 60 hours, 85% of the total volume being filled with helium lifting .064 lb./ft.**3 (94 % pure) in the standard atmosphere. Since the hull is specified in the condition of the problem to be 85% full of gas, the weight of the gas is 85% D multiplied by the difference between the weight of air and the lift of gas per unit volume, and divided by the weight of air per unit volume. The total weight of air and gas is therefore given:

$$Air\; and\; gas = [ .15 + .85 ( .07635 - .064 ) / .07635 ] \times D = .288 D$$

While this formula is not an issue, the subsequent formula has been giving me headaches.

From existing data, the fixed weights exclusive of power plant, power cars, and fuel system equal -3D, and the crew, stores and ballast amount to .055 D. There remains for power plant, fuel, and military load

$$( 1 - .288 - .30 - .055) D = .357 D$$ Question is how did the author arrive at this result? Not matter how I approach solving the equation, I am unable to get the same result. Problem is that the result of this equation is used at a later stage. I also considered that 1 might be an i, but there was no mention of any constant i in the chapter this formula was taken from. I have no clue what I am missing here.

• I agree the "1" can easily be mistaken with a "I", but I also think this is a "1". I'm glad now we can use modern tools to write equation and avoid such ambiguity. – Manu H Jun 30 at 11:42
• A lower-case L "l" was often used as a substitute for the digit one "1" in early typography and typewriters. An off-topic anecdote: My father learned to type on a typewriter without a digit one key and this bit him hard when he was programming in the punch card days. He one spent spent 3 weeks chasing down a bug in a program because he'd used an "l" (ell) when he needed a "1" (one) and the compiler happily created a whole new variable for him. Lesson: always predeclare your variables and shun compilers that will create variables for you on the fly! – FreeMan Jun 30 at 13:57
• @FreeMan Thanks for your note on typography, that actually helps. Apart from dealing with all the math, the print is not great and although I've come across several "old" books with "strange" typography, trying to figure out variables in formulae makes it even harder. Luckily, pdf files can be magnified, another benefit of "the modern era" (not implying that everything is better now). – Sasquatch Jul 1 at 22:45
• I think there is a typo in the second quote which initially threw me: -3D should be .30 D. BTW: I've read some of the old fluid dynamics books of the Saturn 5 era before modern CFD. Excellent read. – tim Jul 1 at 23:56

• $$D$$ is the total displacement (the weight of the air displaced by the airship)
• $$0.288 \cdot D$$ is the weight of helium and air inside the air ship
• $$0.055 \cdot D$$ is the weight of crew, storage, and ballast
• $$x \cdot D$$ is the unknown weight of power plants, power cars, and the fuel system
• $$0.3 \cdot D$$ is anything else (fixed weights)

The total must be one:

$$.288 + .055 + .3 + x = 1$$

or equivalent, the total must be the displacement:

$$D \cdot (.288 + .055 + .3 + x) = 1 \cdot D$$

solve for x and you get:

$$x \cdot D = (1 - .288 - .055 - .3) \cdot D$$

or just

$$x = 1 - .288 - .055 - .3 = .357$$

• Could it be that there is a typo in your formula. I think it should be 0.055 instead of 0.55 (cf. edit(s) above). I tried to edit it, but could not save the changes due to minimum amount of characters required. Anyway, I entered the formula in the calculator and got the result. Still can't tell why I wasn't able to compute this formula in the first place, since it is quite simple, maybe too simple. – Sasquatch Jul 1 at 22:39
• Yes, indeed, thanks! – bogl Jul 1 at 22:47


text:
# Does the Standard Model have a Landau pole?

I have seen the statement that the Standard Model has a Landau pole, or at least it its believed that it does at $\sim 10^{34}$ GeV. Has this actually been proven (at least in perturbation theory, as in QED) or what kind of evidence is there to support this?

-
I'm not sure if it's the case, and it's possible that the answer is sensitive to measured constants (the Higgs mass, the top mass, alpha-strong). A well-known recent paper that runs relevant SM couplings up to the Planck scale is inspirehep.net/record/1116539 - it hints that the Higgs coupling doesn't blow up but actually almost vanishes at the Planck scale. –  Vibert Feb 19 '13 at 15:29
@Vibert: The $\lambda$ result in that paper is interesting, but the $U(1)$ coupling shows no signs of slowing down. I think right now I'd bet that the Standard Model has at least one Landau pole. –  user1504 Feb 20 '13 at 13:16


text:
# Introduction to set theory/Lecture 1

## Introduction

We will start the course by introducing Propositional Logic. Even though this is a set theory class and not a logic course, most of notations from the logic courses can be used in set theory. Furthermore, logic is important in various proofs we will encounter in this course.

## Notations

Here are the notations and what they mean:

Symbols Meaning
${\displaystyle \land }$ and (conjunction)
${\displaystyle \lor }$ or (nonexclusive disjunction)
${\displaystyle \lnot }$ not (negation)
${\displaystyle \to }$ if then/implies
${\displaystyle \leftrightarrow }$ if and only if

## Truth Table

Truth tables are used to analyze formulae of propositional logic.

### Example

Truth table for ${\displaystyle p\to (q\to p)}$

${\displaystyle p\,\!}$ ${\displaystyle q\,\!}$ ${\displaystyle q\to p}$ ${\displaystyle p\to (q\to p)}$
T T T T
T F T T
F T F T
F F T T

## Tautology

### Definition

A formula ${\displaystyle \theta \,\!}$ of propositional logic is a tautology if only T's occur in the ${\displaystyle \theta \,\!}$ column of the truth table.

### Examples

Truth table for ${\displaystyle (p\to \lnot p)\to \lnot p=\theta }$

${\displaystyle p\,\!}$ ${\displaystyle \lnot p}$ ${\displaystyle p\to \lnot p}$ ${\displaystyle \theta \,\!}$
T F F T
F T T T

Truth table for ${\displaystyle (p\to (q\leftrightarrow \lnot q))\to \lnot p=\theta }$

${\displaystyle p\,\!}$ ${\displaystyle q\,\!}$ ${\displaystyle \lnot q}$ ${\displaystyle q\leftrightarrow \lnot q}$ ${\displaystyle p\to (q\leftrightarrow \lnot q)}$ ${\displaystyle \lnot p}$ ${\displaystyle \theta \,\!}$
T T F F F F T
T F T F F F T
F T F F T T T
F F T F T T T

Truth table for ${\displaystyle (\lnot p\vee q)\to (p\to q)=\theta }$

${\displaystyle p\,\!}$ ${\displaystyle q\,\!}$ ${\displaystyle \lnot p}$ ${\displaystyle \lnot p\vee q}$ ${\displaystyle p\to q}$ ${\displaystyle \theta \,\!}$
T T F T T T
T F F F F T
F T T T T T
F F T T T T

## Tautological Equivalence

### Definition

The proposition formulas ${\displaystyle \varphi \,\!}$ and ${\displaystyle \theta \,\!}$ are tautologically equivalent if ${\displaystyle \varphi \leftrightarrow \theta }$ is a tautology.

### Examples

Contraposition: ${\displaystyle p\to q=\theta }$ is tautologically equivalent to ${\displaystyle \lnot q\to \lnot p=\varphi }$.

${\displaystyle p\,\!}$ ${\displaystyle q\,\!}$ ${\displaystyle \lnot q}$ ${\displaystyle \lnot p}$ ${\displaystyle \theta \,\!}$ ${\displaystyle \varphi \,\!}$ ${\displaystyle \theta \leftrightarrow \varphi }$
T T F F T T T
T F T F F F T
F T F T T T T
F F T T T T T

de Morgan's Law I: ${\displaystyle \lnot (p\lor q)=\theta }$ is tautologically equivalent to ${\displaystyle \lnot p\land \lnot q=\varphi }$.

${\displaystyle p\,\!}$ ${\displaystyle q\,\!}$ ${\displaystyle \lnot p}$ ${\displaystyle \lnot q}$ ${\displaystyle p\lor q}$ ${\displaystyle \theta \,\!}$ ${\displaystyle \varphi \,\!}$ ${\displaystyle \theta \leftrightarrow \varphi }$
T T F F T F F T
T F F T T F F T
F T T F T F F T
F F T T F T T T

de Morgan's Law II: ${\displaystyle \lnot (p\land q)=\theta }$ is tautologically equivalent to ${\displaystyle \lnot p\lor \lnot q=\varphi }$. Truth table for Assignment #1

## Related Resources

The materials in this course overlap with Introductory Discrete Mathematics for Computer Science, particularly Lesson 1.