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# Resources tagged with: Mathematical reasoning & proof

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### There are 172 results

Broad Topics > Mathematical Thinking > Mathematical reasoning & proof

### Proof Sorter - Geometric Series

##### Age 16 to 18 Challenge Level:

Can you correctly order the steps in the proof of the formula for the sum of a geometric series?

### Proof Sorter - Sum of an AP

##### Age 16 to 18 Challenge Level:

Use this interactivity to sort out the steps of the proof of the formula for the sum of an arithmetic series. The 'thermometer' will tell you how you are doing

### Direct Logic

##### Age 16 to 18 Challenge Level:

Can you work through these direct proofs, using our interactive proof sorters?

### Proof Sorter - Quadratic Equation

##### Age 14 to 18 Challenge Level:

This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.

### Generally Geometric

##### Age 16 to 18 Challenge Level:

Generalise the sum of a GP by using derivatives to make the coefficients into powers of the natural numbers.

### Middle Man

##### Age 16 to 18 Challenge Level:

Mark a point P inside a closed curve. Is it always possible to find two points that lie on the curve, such that P is the mid point of the line joining these two points?

### Yih or Luk Tsut K'i or Three Men's Morris

##### Age 11 to 18 Challenge Level:

Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .

### Napoleon's Hat

##### Age 16 to 18 Challenge Level:

Three equilateral triangles ABC, AYX and XZB are drawn with the point X a moveable point on AB. The points P, Q and R are the centres of the three triangles. What can you say about triangle PQR?

### Proof: A Brief Historical Survey

##### Age 14 to 18

If you think that mathematical proof is really clearcut and universal then you should read this article.

### Dodgy Proofs

##### Age 16 to 18 Challenge Level:

These proofs are wrong. Can you see why?

### Tree Graphs

##### Age 16 to 18 Challenge Level:

A connected graph is a graph in which we can get from any vertex to any other by travelling along the edges. A tree is a connected graph with no closed circuits (or loops. Prove that every tree has. . . .

### Proof of Pick's Theorem

##### Age 16 to 18 Challenge Level:

Follow the hints and prove Pick's Theorem.

### Iffy Logic

##### Age 14 to 18 Challenge Level:

Can you rearrange the cards to make a series of correct mathematical statements?

### Proofs with Pictures

##### Age 14 to 18

Some diagrammatic 'proofs' of algebraic identities and inequalities.

### Euclid's Algorithm II

##### Age 16 to 18

We continue the discussion given in Euclid's Algorithm I, and here we shall discover when an equation of the form ax+by=c has no solutions, and when it has infinitely many solutions.

### Thousand Words

##### Age 16 to 18 Challenge Level:

Here the diagram says it all. Can you find the diagram?

### Multiplication Square

##### Age 14 to 16 Challenge Level:

Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?

### Sperner's Lemma

##### Age 16 to 18

An article about the strategy for playing The Triangle Game which appears on the NRICH site. It contains a simple lemma about labelling a grid of equilateral triangles within a triangular frame.

### To Prove or Not to Prove

##### Age 14 to 18

A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.

### Target Six

##### Age 16 to 18 Challenge Level:

Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.

### Dalmatians

##### Age 14 to 18 Challenge Level:

Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.

### Pair Squares

##### Age 16 to 18 Challenge Level:

The sum of any two of the numbers 2, 34 and 47 is a perfect square. Choose three square numbers and find sets of three integers with this property. Generalise to four integers.

### Impossible Triangles?

##### Age 16 to 18 Challenge Level:

Which of these triangular jigsaws are impossible to finish?

### Plus or Minus

##### Age 16 to 18 Challenge Level:

Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.

### Sprouts Explained

##### Age 7 to 18

This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .

### Janine's Conjecture

##### Age 14 to 16 Challenge Level:

Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .

### Mind Your Ps and Qs

##### Age 16 to 18 Short Challenge Level:

Sort these mathematical propositions into a series of 8 correct statements.

### Picture Story

##### Age 14 to 16 Challenge Level:

Can you see how this picture illustrates the formula for the sum of the first six cube numbers?

### There's a Limit

##### Age 14 to 18 Challenge Level:

Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?

### Exhaustion

##### Age 16 to 18 Challenge Level:

Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

### Polynomial Relations

##### Age 16 to 18 Challenge Level:

Given any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials. Try this one.

### Integral Inequality

##### Age 16 to 18 Challenge Level:

An inequality involving integrals of squares of functions.

##### Age 16 to 18 Challenge Level:

Find all positive integers a and b for which the two equations: x^2-ax+b = 0 and x^2-bx+a = 0 both have positive integer solutions.

### Square Mean

##### Age 14 to 16 Challenge Level:

Is the mean of the squares of two numbers greater than, or less than, the square of their means?

### Mechanical Integration

##### Age 16 to 18 Challenge Level:

To find the integral of a polynomial, evaluate it at some special points and add multiples of these values.

### Contrary Logic

##### Age 16 to 18 Challenge Level:

Can you invert the logic to prove these statements?

### Notty Logic

##### Age 16 to 18 Challenge Level:

Have a go at being mathematically negative, by negating these statements.

### Some Circuits in Graph or Network Theory

##### Age 14 to 18

Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.

### Rational Roots

##### Age 16 to 18 Challenge Level:

Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.

### Without Calculus

##### Age 16 to 18 Challenge Level:

Given that u>0 and v>0 find the smallest possible value of 1/u + 1/v given that u + v = 5 by different methods.

##### Age 16 to 18 Short Challenge Level:

Can you work out where the blue-and-red brick roads end?

### Tetra Inequalities

##### Age 16 to 18 Challenge Level:

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.

##### Age 14 to 18 Challenge Level:

Which of these roads will satisfy a Munchkin builder?

### On the Importance of Pedantry

##### Age 16 to 18

A introduction to how patterns can be deceiving, and what is and is not a proof.

### Breaking the Equation ' Empirical Argument = Proof '

##### Age 7 to 18

This article stems from research on the teaching of proof and offers guidance on how to move learners from focussing on experimental arguments to mathematical arguments and deductive reasoning.

### Magic Squares II

##### Age 14 to 18

An article which gives an account of some properties of magic squares.

### Advent Calendar 2011 - Secondary

##### Age 11 to 18 Challenge Level:

Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.

### An Introduction to Number Theory

##### Age 16 to 18

An introduction to some beautiful results of Number Theory (a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions)

### A Long Time at the Till

##### Age 14 to 18 Challenge Level:

Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?

### Calculating with Cosines

##### Age 14 to 18 Challenge Level:

If I tell you two sides of a right-angled triangle, you can easily work out the third. But what if the angle between the two sides is not a right angle?


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# Magnetic force under change of reference: Do Maxwell's equations hold?

Suppose I have a uniform magnetic field through all of space $$\textbf{B}(x,y,z)=\hat{\textbf{z}}$$ and a charge $$q$$ moving at a velocity of $$v\hat{\textbf{x}}$$. In this frame of reference, a magnetic force will act on $$q$$, pushing it up (or down, but let's assume up). Suppose instead I change my frame of reference so that $$q$$ is stationary at the origin. In this frame of reference, there is no force to act on $$q$$ and it shouldn't move. Naturally, this seems inconsistent.

The normal explanation given for this from what I've seen is that the magnetic field changes into an electric field upon a change of reference so that there is still a force in the second frame of reference and the results agree. I have some problems with this though:

This doesn't explain the source of the $$\textbf{E}$$ field in the second frame. Where does it come from? According to Maxwell's equations, $$\textbf{E}$$ fields can only be produced by time-varying magnetic fields or charges. There are no charges (apart from $$q$$) and no time-varying magnetic fields in either frame of reference - so Maxwell's equations seem to predict a zero electric field in either case. Am I wrong to say that Maxwell's equations must hold in all inertial frames of reference? How would both Maxwell's equations be satisfied and the same results be obtained upon transformation to the second frame of reference?

Let's just ignore Maxwell's equations even. Here's another reason it seems inconsistent: What if we started our analysis in the second frame of reference instead, where $$q$$ is stationary? In this frame, we don't expect $$q$$ to move upwards. Then presumably, after transforming the $$\textbf{E}$$ and $$\textbf{B}$$ fields into the first reference frame, you should still see the same result (no upwards movement). So explaining the problem as "Lorentz transforms make $$\textbf{B}$$ into $$\textbf{E}$$" doesn't seem to cut it, you could just as well do the analysis backwards and get contradictory results.

Maybe you could argue that due to the ultimate symmetry of the field, there is no way to distinguish any movement of the charge so that it's meaningless to talk about this. This just feels like a cop-out to me. It would also seem to be easily evaded, since what if I just defined $$\textbf{B}$$ to be zero outside of $$(-\infty,\infty)\times[-1,1]$$? In this case, any upward movement of the charge would be noticeable, and we still have the same problem that we started with.

EDIT: Regarding the original source of $$\textbf{B}$$, let's just say that it's created by either a large solenoid or two permanent magnets so that $$\textbf{B}$$ is approximately uniform in the region of interest. A finite $$\textbf{B}$$ field like this will create induced $$\textbf{E}$$ fields at the fringes when it moves, but this shouldn't affect the local uniform region.

• "This doesn't explain the source of the $E$ field in the second frame." - You haven't given a source for the $B$ field in the first frame, either! Why is the $B$ field allowed to magically exist but the $E$ field is not? – ACuriousMind Dec 1 '18 at 16:57
• Ok let's say it's created by two infinitely long permanent magnets. Or just consider the limit as the length of the two permanent magnets approaches infinity. All my argument requires is symmetry of the magnetic field along the x-axis so that there are no induced E fields in either frame of reference. – hddh Dec 1 '18 at 17:02
• In fact, I suspect the argument would even just work with one solenoid that is large enough so that $\textbf{B}$ is approximately uniform in the area of interest. Any induced $\textbf{E}$ fields due to $\textbf{B}$ changing at the fringes of a moving region shouldn't affect the local region of the approx. uniform $\textbf{B}$ field. – hddh Dec 1 '18 at 17:06
• Note that Maxwells equations are differential equations, and you can add a constant $E$ or $B$ field to any solution and it will still be a solution. Thus, Maxwells equations do not predict $E=0$ as you claim, they simply predict $E$ being constant across space and time. – Eddy Dec 1 '18 at 17:18
• if you have a magnet the magnet is moving in the reference frame in which the charge is at rest, thus generating an electric field – Wolphram jonny Dec 1 '18 at 17:23

You claim that the absence of charges or time-varying $$\mathbf B$$ fields implies that the $$\mathbf E$$ field is zero, but this is only true if we add the boundary condition that $$\mathbf E=0$$ at spatial infinity. The existence of a constant, uniform $$\mathbf E$$ field is perfectly consistent with Maxwell's equations, in precisely the same way as your initial $$\mathbf B$$ field is.

In any real situation, of course, your $$\mathbf B$$ field will be generated by sources of some description, and will not be constant or spatially uniform. The observer in the particle's rest frame will therefore see a time-varying $$\mathbf B$$ field and an induced $$\mathbf E$$ field. Maxwell's equations are preserved, and there is no contradiction between the observers.

You need to look into the math of this particular problem. We restrict ourselves to the classical interpretation. Let us have an inertial coordinate system $$O\hat{x}\hat{y}\hat{z}$$ and in it - a constant magnetic field $$\vec{B}$$ directed along the vertical axis $$O\vec{z}$$. You suggest that the magnetic field is described by the vertical basis unit vector $$\vec{B} = \hat{z}$$.

Furthermore, you have a particle $$Q$$ of charge $$q$$ and mass $$m=1$$ (for simplicity) moving in the magnetic field with initial velocity $$\vec{v}(0) = v_0\, \hat{x}$$. By $$\, \vec{r} = \vec{r}(t) = x(t)\, \hat{x} + y(t) \, \hat{y} + z(t) \, \hat{z}$$ denote the position vector of the particle $$Q$$ at time $$t$$.

The Lorentz law of motion states that the particle's position $$\vec{r} = \vec{r}(t)$$ changes with time $$t$$ according to the second order system of differential equations \begin{align} &\frac{d^2 \vec{r}}{dt^2}\, = \, q\frac{d \vec{r}}{dt} \times \vec{B}\, = \, q\frac{d \vec{r}}{dt} \times \hat{z}\\ & \frac{d \vec{r}}{dt}(0)\, =\, v_0 \,\hat{x}\\ & \vec{r}(0) = \vec{r}_0 \end{align} Now, observe that according to Lorentz's law, when a particle's velocity is zero, i.e. $$\frac{d\vec{r}}{dt} = \vec{0}$$, then the Lorentz force $$q\frac{d \vec{r}}{dt} \times \vec{B} = \vec{0}$$ so the said particle is stationary, that is, no force acts on it.

Next, you switch from the original inertial coordinate system $$O\hat{x}\hat{y}\hat{z}$$ to a non-inertial coordinate system $$Q\hat{x}\hat{y}\hat{z}$$ with origin at the moving charged particle $$Q$$ and coordinate axes, let's say, always staying parallel to the axes of $$O\hat{x}\hat{y}\hat{z}$$. Denote the position of any object in the non-inertial system $$Q\hat{x}\hat{y}\hat{z}$$ by $$\vec{R} = \vec{R}(t) = X(t)\, \hat{x} + Y(t) \, \hat{y} + Z(t) \, \hat{z}$$. Then the change of coordinates is given by the vector formula $$\vec{r} = \vec{R} + \vec{a}$$ where $$\vec{a} = \vec{a}(t)$$ is the position of the particle $$Q$$ at time $$t$$ with respect to $$O\hat{x}\hat{y}\hat{z}$$. Then $$\vec{a} = \vec{a}(t)$$ satisfies the equations $$\frac{d^2\vec{a}}{dt^2} = q \frac{da}{dt} \times \vec{B}$$ Consequently, we would like to represent Lorentz' s law in the new coordinate system $$Q\hat{x}\hat{y}\hat{z}$$.,

First, by differentiating the change of coordinates $$\vec{r} = \vec{R} + \vec{a}$$ with respect to time $$t$$, find that $$\frac{d\vec{r}}{dt} = \frac{d\vec{R}}{dt} + \frac{d\vec{a}}{dt}$$ and by differentiating one more time obtain $$\frac{d^2\vec{r}}{dt^2} = \frac{d^2\vec{R}}{dt^2} + \frac{d^2\vec{a}}{dt^2}$$ Second, the Lorentz's law yields \begin{align} \vec{0} &= \frac{d^2 \vec{r}}{dt^2} \, - \, q\frac{d \vec{r}}{dt} \times \vec{B} = \frac{d^2\vec{R}}{dt^2} + \frac{d^2\vec{a}}{dt^2} \, -\, q\left(\frac{d\vec{R}}{dt} + \frac{d\vec{a}}{dt}\right)\times \vec{B} = \\ &= \frac{d^2\vec{R}}{dt^2} + \frac{d^2\vec{a}}{dt^2} \, -\, q\frac{d\vec{R}}{dt} \times \vec{B} \, - \, q\frac{d\vec{a}}{dt}\times \vec{B} = \\ &= \frac{d^2\vec{R}}{dt^2} \, -\, q\frac{d\vec{R}}{dt} \times \vec{B} \, + \,\left(\frac{d^2\vec{a}}{dt^2} \, - \, q\frac{d\vec{a}}{dt}\times \vec{B} \right) =\\ & = \frac{d^2\vec{R}}{dt^2} \, -\, q\frac{d\vec{R}}{dt} \times \vec{B} \, + \, \vec{0} = \\ & = \frac{d^2\vec{R}}{dt^2} \, -\, q\frac{d\vec{R}}{dt} \times \vec{B} \end{align} Thus, the Lorentz's law in the non-inertial reference frame $$Q\hat{x}\hat{y}\hat{z}$$ is $$\frac{d^2\vec{R}}{dt^2} \, = \, q\frac{d\vec{R}}{dt} \times \vec{B}$$ which is exactly the same as the Lorentz's law in the inertial reference frame $$O\hat{x}\hat{y}\hat{z}$$. In the non-inertial frame though, the particle $$Q$$ is stationary, i.e. its velocity is zero. And as observed, whenever the velocity $$\frac{d\vec{R}}{dt} = \vec{0}$$, the Lorentz force of the form $$q\frac{d\vec{R}}{dt} \times \vec{B} = \vec{0}$$, and so the trajectory of $$Q$$ should be a fixed point (more precisely, the origin of the coordinate system) in the non-inertial system. As you can see, there is no evidence of a an electric field in the non-inertial coordinate system.

Now, if you introduce a non-inertial coordinate system attached at $$Q$$, which twists and rotates around $$Q$$, then things change substantially. Then, the fictitious forces of the non-inertial system could manifest as an additional term to the the magnetic field force and could play the role of an electric field in the Lorentz's law $$\frac{d^2\vec{r}}{dt^2} = q\vec{E}(\vec{r}, t) + q\frac{d\vec{r}}{dt}\times \vec{B}(\vec{r}, t)$$

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## Bulletin of the Belgian Mathematical Society - Simon Stevin

### Coexistence of Unbounded Solutions and Periodic Solutions of a Class of Planar Systems with Asymmetric Nonlinearities

#### Abstract

In this paper we will prove the coexistence of unbounded solutions and periodic solutions for a class of planar systems with asymmetric nonlinearities \begin{eqnarray*}\label{abstract} \left \{ \begin{array}{lll} u'=v-\alpha u^{+}+\beta u^{-} \\ v'=-\mu u^{+}+\gamma u^{-}-g(u)+p(t), \end{array} \right. \end{eqnarray*} where $g(u)$ is continuous and bounded, $p(t)$ is a continuous $2\pi$-periodic function and $\alpha, \beta\in \mathbb{R}, \mu, \gamma$ are positive constants.

#### Article information

Source
Bull. Belg. Math. Soc. Simon Stevin, Volume 17, Number 4 (2010), 577-591.

Dates
First available in Project Euclid: 24 November 2010

https://projecteuclid.org/euclid.bbms/1290608188

Mathematical Reviews number (MathSciNet)
MR2778438

Zentralblatt MATH identifier
1213.34062

Subjects
Primary: 34C11: Growth, boundedness 34C25: Periodic solutions

#### Citation

Liu, Qihuai; Sun, Xiying; Qian, Dingbian. Coexistence of Unbounded Solutions and Periodic Solutions of a Class of Planar Systems with Asymmetric Nonlinearities. Bull. Belg. Math. Soc. Simon Stevin 17 (2010), no. 4, 577--591. https://projecteuclid.org/euclid.bbms/1290608188


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SEARCH HOME
 Math Central Quandaries & Queries
 Question from laurie, a student: if shaded parts of a circle are 4/7 and the other is 1/3 what fraction is left

Hi Laurie,

I am going to solve a similar problem.

if shaded parts of a circle are $\large \frac25$ and the other is $\large \frac38$ what fraction is left

First I need to assume that the two shaded parts don't overlap.

The fraction of the circle that is shaded is the sum of $\large \frac25$ and $\large \frac38.$ To add the fractions I need a common denominator. The denominators are $5$ and $8$ so one common denominator is $5 \times 8 = 40.$ Now I need equivalent fractions to $\large \frac25$ and $\large \frac38$ both with denominators of $40.$

$\frac25 = \frac{2 \times 8}{5 \times 8} = \frac{16}{40}$

and

$\frac38 = \frac{3 \times 5}{8 \times 5} = \frac{15}{40}.$

Thus

$\frac25 + \frac38 = \frac{16}{40} + \frac{15}{40} = \frac{16 + 15}{40} = \frac{31}{40}.$

The fraction of the circle that is not shaded is then

$1 - \frac{31}{40} = \frac{40}{40} - \frac{31}{40} = \frac{9}{4}.$

Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.


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Find the largest possible number of distinct integer values $$\{x_1, x_2, \ldots x_n\}$$, such that for a fixed reducible degree four polynomial with integer coefficients, $$|f(x_i)|$$ is prime for all $$i$$.

Details and assumptions

A polynomial with integer coefficients is called reducible if it is a product of two non-constant polynomials with integer coefficients.

×


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Existence results for partial neutral functional differential equations with unbounded delay. (English) Zbl 0915.35110
The authors prove existence of mild solutions of the Cauchy problem $${d\over dt} (x(t)+ F(t, x_t))= Ax(t)+ G(t, x_t),\quad t\ge\sigma,\quad x(\sigma)= \varphi\in \Omega,\tag 1$$ where $\Omega$ is an open subset of the phase space, $F,G: [0,a]\times \Omega\mapsto X$ are continuous functions, $0\le\sigma< a$ and $A$ is the infinitesimal generator of an analytic semigroup $T(.)$ of bounded linear operators on $X$. In the case $\sigma= 0$ the existence of strong solutions of the Cauchy problem (1) is proved.
Reviewer: D.Bainov (Sofia)

##### MSC:
 35R10 Partial functional-differential equations 34K40 Neutral functional-differential equations
Full Text:
##### References:
 [1] Ananjevskii, I. M.; Kolmanovskii, V. B.: Stabilization of some nonlinear hereditary mechanical systems. Nonlinear anal. 15, 101-114 (1990) [2] Corduneanu, C.; Lakshmikantham, V.: Equations with unbounded delay. Nonlinear anal. 4, 831-877 (1980) · Zbl 0449.34048 [3] Da Prato, G.; Lunardi, A.: Solvability on the real line of a class of linear Volterra integrodifferential equations of parabolic type. Ann. mat. Pura appl. (4) 150, 67-117 (1988) · Zbl 0646.45013 [4] Dhakne, M. B.; Pachpatte, B. G.: On a general class of abstract functional integrodifferential equations. Indian J. Pure appl. Math. 19, 728-746 (1988) · Zbl 0663.45009 [5] Fattorini, H. O.: Second order linear differential equations in Banach spaces. North-holland mathematics studies 108 (1985) · Zbl 0564.34063 [6] Goldstein, J.: Semigroups of linear operators and applications. (1985) · Zbl 0592.47034 [7] Gopalsamy, K.: Stability and oscillations in delay differential equations of population dynamics. (1992) · Zbl 0752.34039 [8] Gopalsamy, K.; Weng, P. -X.: On the stability of a neutral integro-partial differential equation. Bull. inst. Math. acad. Sinica 20, 267-284 (1992) · Zbl 0759.45007 [9] Grabosch, A.; Moustakas, U.: A semigroup approach to retarded differential equations. Lecture notes in math. 1184 (1986) [10] Györi, I.; Ladas, G.: Oscillation theory of delay differential equations. (1991) · Zbl 0780.34048 [11] Hale, J.; Kato, J.: Phase space for retarded equations with infinite delay. Funkcial. ekvac. 21 (1978) · Zbl 0383.34055 [12] Hale, J.: Asymptotic behavior of dissipative systems. (1988) · Zbl 0642.58013 [13] Hale, J.; Lunel, S. M. Verduyn: Introduction to functional differential equations. (1993) · Zbl 0787.34002 [14] Henriquez, H. R.: Periodic solutions of quasi-linear partial functional differential equations with unbounded delay. Funkcial. ekvac. 37, 329-343 (1994) · Zbl 0814.35141 [15] Henriquez, H. R.: Approximation of abstract functional differential equations with unbounded delay. Indian J. Pure appl. Math. 27, 357-386 (1996) · Zbl 0853.34072 [16] Henriquez, H. R.: Regularity of solutions of abstract retarded functional differential equations with unbounded delay. Nonlinear anal. 28, 513-531 (1997) [17] Hille, E.; Phillips, R. S.: Functional analysis and semi-groups. Amer. math. Soc. colloq. Publ. 31 (1957) · Zbl 0078.10004 [18] Hino, Y.; Murakami, S.; Naito, T.: Functional differential equations with infinite delay. Lecture notes in math. 1473 (1991) · Zbl 0732.34051 [19] Kwon, W. H.; Lee, G. W.; Kim, S. W.: Performance improvement using time delays in multivariable controller design. Internat. J. Control 52, 1455-1473 (1990) · Zbl 0708.93024 [20] Kartsatos, A. G.; Parrott, M. E.: The weak solution of a functional differential equation in a general Banach space. J. differential equations 75, 219-232 (1988) · Zbl 0666.34076 [21] Komura, J.: Differentiability of nonlinear semigroups. J. math. Soc. Japan 21, 375-402 (1969) · Zbl 0193.11004 [22] Liang, J.; Xiao, T.: Functional differential equations with infinite delay in Banach spaces. Internat. J. Math. math. Sci. 14, 497-508 (1991) · Zbl 0743.34082 [23] Marle, C. -M.: Mesures et probabilités. (1974) [24] Mcbride, A. C.: Semigroups of linear operators: an introduction. Pitman res. Notes math. Ser. 156 (1987) · Zbl 0635.47035 [25] Nagel, R.: One-parameter semigroups of positive operators. Lecture notes in math. 1184 (1986) · Zbl 0585.47030 [26] Parrott, M. E.: Linearized stability and irreducibility for a functional differential equation. SIAM J. Math. anal. 23, 649-661 (1992) · Zbl 0763.34058 [27] Pazy, A.: Semigroups of linear operators and applications to partial differential equations. (1983) · Zbl 0516.47023 [28] Petzeltova, H.: Solution semigroup and invariant manifolds for functional equations with infinite delay. Math. bohemica 118, 175-192 (1993) [29] Sadovskii, B. N.: On a fixed point principle. Funct. anal. Appl. 1, 74-76 (1967) · Zbl 0165.49102 [30] Salamon, D.: Neutral functional differential equations and semigroups of operators. Lecture notes in control and inform. Sci. 54 (1983) · Zbl 0518.93035 [31] Travis, C. C.; Webb, G. F.: Second order differential equations in Banach spaces. (1978) · Zbl 0455.34044 [32] Travis, C. C.; Webb, G. F.: Existence and stability for partial functional differential equations. Trans. amer. Math. soc. 200, 395-418 (1974) · Zbl 0299.35085 [33] Wiener, J.; Debnath, L.: The Fourier method for partial differential equations with piecewise continuous delay. Contemporary mathematics 129 (1992) · Zbl 0826.35131


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Contents 1 Encoding 1.1 Support with LaTeX and pdfLaTeX 1.2 Support with XeLaTeX or LuaLaTeX 1.3 Invalid/Unsupported Characters 2 Line Endings 3 Related Help Articles Encoding Overleaf uses the UTF-8 encoding for all text files. UTF-8 is the most widely used character encoding on the web today. You can use it to represent any unicode character, which includes an enormous variety of letters, numbers and symbols, including Greek letters and letters with accents. UTF-8 supersedes many older encodings, such as latin1, latin9, which are often found in LaTeX files and templates. TeX and LaTeX predate UTF-8 by several decades, so LaTeX support for UTF-8 is somewhat inconsistent and depends on which TeX engine you are using. Support with LaTeX and pdfLaTeX If you are using pdfLaTeX (i.e., running LaTeX using the pdfTeX engine), which is the default on Overleaf, you can typeset most accented letters and some symbols directly if you include this \usepackage command in the preamble of your document: \usepackage[utf8]{inputenc} This should be the only inputenc line in your preamble, so it should replace any that use other encodings. However, tex does not know how to typeset all UTF-8 characters. You may get an error like this, if it does not know how to typeset your character: Package inputenc Error: Unicode char \u8:��� not set up for use with LaTeX. You can sometimes work around this by using a different inputenc, utf8x, with an x. \usepackage[utf8x]{inputenc} The utf8x input encoding sets up more unicode characters for use with LaTeX. However, it is no longer maintained, and its use is discouraged. You should use utf8 unless you are certain you need utf8x. If that does not help, you may need to use TeX's built-in support for accents, e.g. by writing {\"e} instead of writing ë directly. You can find out more about special characters and accents in the LaTeX Wikibook. Alternatively, you can use a more modern TeX engine, such as XeLaTeX or LuaTeX. Support with XeLaTeX or LuaLaTeX If you are using one of these more modern LaTeX engines, you can use much more of UTF-8 directly. You can choose the engine on Overleaf by clicking on the gear icon in the top right of your project. Invalid/Unsupported Characters For technical reasons, Overleaf cannot store files that contain NUL characters or files that contain characters from outside of Unicode's Basic Multilingual Plane(BMP). The BMP contains only the first 65,536 unicode code points. Some unicode mathematical symbols and some symbols from eastern languages are outside of the BMP. Non-BMP mathematical symbols sometimes arise when copying from PDFs for web pages. If you try to upload a file that contains unsupported characters, Overleaf will respond with a "file contains invalid characters and can't be imported" error. The best workaround is to replace the non-BMP characters with equivalent LaTeX commands. Detexify is a useful tool for finding the LaTeX command that corresponds to a given symbol. Line Endings Overleaf uses unix-style newline endings, \n (LF), rather than Microsoft Windows-style line endings \r\n (CRLF). If you are using the git interface to edit your projects on Overleaf, then you can use your system's local line endings and have git translate them to unix line endings for you by enabling git's autocrlf setting. Related Help Articles How do I use letters with accents (e.g. in French or Portuguese)?


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# Verifying the Friis Transmission Equation with FEM–BEM Coupling

February 24, 2023

It’s possible to measure the far-field response of any scatterer or antenna with the Far-Field Domain feature of the RF Module, an add-on product to the COMSOL Multiphysics® software. For taking it a step further, COMSOL Multiphysics can replicate experimental setups to a greater extent, e.g., antenna gain measurement or emission or immunity testing of a circuit board for addressing EMI/EMC issues, thanks to the finite element method–boundary element method (FEM–BEM) coupling. This coupling can also be used to find wave propagation in the surrounding infinite free space in microstrip patch antenna models. In this blog post, we’ll take a deep dive into modeling transmitter and receiver antennas for line-of-sight (LoS) communication, one of the applications of FEM–BEM coupling.

### Line-of-Sight Communication

To understand LoS communication, it’s important to know how the power is received by the antenna. This is given by the well-known Friis transmission formula (in dB) as follows:

\text {P}_\text{r} = \text{P}_\text{t} + \text{G}_\text{t} (\theta_t, \phi_\text{t}) + \text {G}_\text{r} (\theta_\text{r}, \phi_\text{r}) + \text {Path Loss} \quad \quad \quad (1)

Here, \text {P}_\text{t} is transmitted power; \text{P}_\text{r} is received power; \text{G}_\text{t} (\theta_t, \phi_\text{t}) and \text {G}_\text{r} (\theta_\text{r}, \phi_\text{r}) are the gain of the transmitted and received antenna, respectively; and \text {Path Loss} (in dB) is given by

{\lambda} is the operating wavelength and d is the distance between the antennas. This distance typically represents the shortest distance between the antennas in the Fresnel zone. (This zone is an elliptical region formed between the antennas and does not contain any physical obstruction that may interfere with the signal traveling.)

Figure 1. Depiction of the typical LoS communication link.

Now, let’s rearrange Eq. 1 so that, with simulation, we can compare the right-hand side of the new equation (Eq. 3) with an S-parameter obtained from simulation. This equation is as follows:

\text{P}_\text{r}\text{ -P}_\text{t} = \text{G}_\text{t} (\theta_\text{t} , \phi_\text{t}) + \text{G}_\text{r}(\theta_\text{r} ,\phi_\text{r}) + \text {Path Loss} \quad \quad \quad (3)

Here, the left-hand side represents S21 (in dB).

### Modeling Transmitter and Receiver Antennas

From the Friis formula, it’s well understood that \text {P}_\text{r} can be maximized by minimizing the path loss. The path loss increases as operating frequencies and/or the distance between transmitter and receiver antenna increases. Figure 2 demonstrates this, with the transmitter and receiver modeled in the FEM domain and coupled with a predefined FEM–BEM coupling. Note that the transmitter is a microstrip patch antenna, with the Lumped Port 1 voltage excitation on in the Electromagnetic Waves, Frequency Domain 3: Tx and Rx Antennas (emw3) interface (also known as the emw3 interface). The receiver is a Yagi–Uda antenna, with the Lumped Port 2 excitation off in the emw3 interface.

Figure 2. The emw3 and embe interfaces are coupled with the Electric Field Coupling multiphysics interface.

You can also visualize the wave propagation from the transmitter to the receiver with the help of the Animation feature (shown below).

The propagation of EM waves from the transmitter to the receiver. This simulation uses the FEM–BEM coupling feature.

In the Friis formula, far-field antenna gain for both antennas is required. These are computed using the Far-Field Domain node in the:

• Electromagnetic Waves, Frequency Domain: For Yagi–Uda Antenna (Rx) (emw) interface — emw interface for short
• Electromagnetic Waves, Frequency Domain 2: For Microstrip Patch Antenna (Tx) (emw2) interface — emw2 interface for short

You can see such interfaces in the image below.

Figure 3. The emw and emw2 interfaces are used to find far-field gains of the Yagi–Uda and microstrip patch antennas.

For comparison, variables are created to formulate the Friis formula (Figure 4). The separation between the antennas is increased, and the received power in terms of the S-parameter is measured (Figure 5).

Figure 4. The far-field expressions are used to formulate the Friis equation. Here, lda0 and dist represent the operating free space wavelength and the distance between the Rx and Tx antennas, respectively. To access far-field gain at a particular coordinate defined via the phi angle parameter, a predefined  at3() function is used.

Figure 5. Comparison of the Friis formula with received power for varied distance.

Furthermore, \text {P}_\text{r} can be increased by aligning the main lobes of the antenna radiation pattern. Let’s verify this by rotating the receiver antenna at an azimuth angle. This angle variation could be from -90° to 90°. In other words, we are mimicking the antenna gain measurement setup. From Figure 6 it can be seen that the power received reaches its maximum at 0°. This configuration corresponds to both antennas having maximum gain along the x-axis.

Figure 6. Comparison of the Friis formula with received power for a varied angle.

### Opening Doors for EMI/EMC Testing

The beam envelope method available with the Wave Optics Module overcomes the barrier of wavelength-comparable geometry size in nonscattering EM problems, and it’s well suited for guided media. However, it is also possible to use a FEM–BEM coupling for modeling scattering EM problems without having to handle meshing requirements or geometry size restriction. One such application can be to set up an EMI/EMC test bench. For example, to perform the emission test for the RE102 military standard (up to 18 GHz frequency), the distance between the device under test (DUT) and the antenna is 1 meter. For 18 GHz of signal, a 1-meter distance is in order of 60 times wavelength, and modeling such a vast space through FEM is computationally very expensive. Instead of modeling the DUT and antenna in single FEM, we can segregate them into two FEM domains (with, of course, comparable wavelength size) and couple them with BEM, as illustrated in Figure 7. The detected power at the antenna can be a measure of radiated EM signal strength from the DUT.

Figure 7. An illustration of an EMI/EMC test bench setup for performing emission analysis.

### Concluding Thoughts

FEM–BEM coupling has opened doors for extensive EM simulations, which were restricted due to meshing requirements and computational resource limitations. Verifying the Friis transmission equation makes it more reliable in simulation cases where users are interested in emission and immunity testing in EMI/EMC analysis for their DUTs.

### Next Steps

Try the Verifying Friis Transmission Equation with FEM–BEM Coupling tutorial model yourself by clicking the button below, which will take you to the Application Gallery:

(Tip: Want extra modeling practice? Check out our FEM–BEM Coupling of a Microstrip Patch Antenna tutorial model.)


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# zbMATH — the first resource for mathematics

Neighbor sum distinguishing index. (English) Zbl 1272.05047
Summary: We consider proper edge colorings of a graph $$G$$ using colors of the set $$\{1,\dots ,k\}$$. Such a coloring is called neighbor sum distinguishing if for any pair of adjacent vertices $$x$$ and $$y$$ the sum of colors taken on the edges incident to $$x$$ is different from the sum of colors taken on the edges incident to $$y$$. The smallest value of $$k$$ in such a coloring of $$G$$ is denoted by $$\mathrm{ndi}_\Sigma (G)$$. In the paper we conjecture that for any connected graph $$G\neq C_5$$ of order $$n\geq 3$$ we have $$\mathrm{ndi}_\Sigma (G)\leq\Delta (G)+2$$. We prove this conjecture for several classes of graphs. We also show that $$\mathrm{ndi}_\Sigma (G)\leq 7\Delta (G)/2$$ for any graph $$G$$ with $$\Delta (G)\geq 2$$ and $$\mathrm{ndi}_\Sigma (G)\leq 8$$ if $$G$$ is cubic.

##### MSC:
 05C15 Coloring of graphs and hypergraphs
Full Text:
##### References:
 [1] Akbari, S.; Bidkhori, H.; Nosrati, N., R-strong edge colorings of graphs, Discrete. Math., 306, 3005-3010, (2006) · Zbl 1112.05035 [2] Bondy J.A., Murty U.S.R.: Graph Theory with Applications. Macmillan, London (1976) · Zbl 1226.05083 [3] Balister, PN.; Győri, E.; Lehel, L.; Schelp, R.H., Adjacent vertex distinguishing edge-colorings, SIAM. J. Discrete. Math., 21, 237-250, (2007) · Zbl 1189.05056 [4] Dénes, J., Keedwell, A.D.: Transversal and Complete Mappings, In: Latin Squares. New Developments in the Theory and Applications, Elsevier, Amsterdam (1991) · Zbl 1075.05034 [5] Edwards, K.; Horňák, M.; Woźniak, M., On the neighbour-distinguishing index of a graph, Graphs. Combin., 22, 341-350, (2006) · Zbl 1107.05032 [6] Hatami, H., Is a bound on the adjacent vertex distinguishing edge chromatic number, J. Combin. Theory. Ser B., 95, 246-256, (2005) · Zbl 1075.05034 [7] Mahéo, M., Saclé, J-F.: Some results on (Σ, $$p$$, $$g$$)-valuation of connected graphs, Rapport de Recherche 1497, Université de Paris-Sud, Centre d’Orsay (2008) · Zbl 1189.05056 [8] Zhang, Z.; Liu, L.; Wang, J., Adjacent strong edge coloring of graph, Appl. Math. Lett., 15, 623-626, (2005) · Zbl 1008.05050
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Network functions

## Recommended Posts

cherryhouse    100
I can't think of a good way of sending commands from my client to my server. I have one idea, but it isn't a very good way of going about this.. heres an example: The client sends a message(a string) to the server that reads "messagebox" and the server has a recv function like: recv(sd, buf, sizeof(buf), 0); if(buf == "messagebox") { messagebox(...); }else{ cout<<"Unknown command"; } I am looking for a nice way of creating commands to send over the network without the commands being if(string = cmd) type statements. Any insight on how I could do a command system not based on text and execution on hte client side?

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Bob Janova    769
switch(code){ ...}

Make the first four bytes (or whatever) of each message an integer that defines the action.

Also, if this is TCP, remember you need to mark where in the stream to split into messages; it doesn't guarantee that what you send in one call to send() will arrive in one call to recv(). If you're using UDP you can ignore this (but remember your data may not arrive at all).

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Woltan    122
I am no network expert,
but why dont you use a message system like windows for example

void Message(UINT messageid, int param1 int param2)

like this you dont have to use strings, which would increase performance.

once again i am complete noob, but maybe maybe i could help you ; )
cherio Woltan

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__filip    218
If all you want is a system not based on text comparison then you can just use predefined constants instead of strings.

(Using simplified versions of the functions send and recv)

enum COMMAND{    NO_COMMAND,    CMD_MESSAGEBOX,    CMD_SOMECOMMAND,    CMD_SOMEOTHERCOMMAND,};void send_command(){   int cmd=CMD_MESSAGEBOX;   send(&cmd,sizeof(cmd));}void recv_command(){   int cmd=NO_COMMAND;   recv(&cmd,sizeof(cmd));   switch (cmd)   {   case CMD_MESSAGEBOX:      //Handle messagebox command      break;   case CMD_SOMECOMMAND:      //Handle somecommand command      break;   case CMD_SOMEOTHERCOMMAND:      //Handle someothercommand command      break;   default:      //Handle unknown command      break;   }}

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Typically you use a byte or two preceeding the data to tell the other end what kind of data it is. On the other end you would use a switch statement to differentiate between the label bytes.

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Anon Mike    1098
Depending on exactly how much control you need you might also want to look into Remote Procedure Calls (RPC). RPC's whole reason for existing is to be able to call functions on another computer in a pseudo-transparent manner. It's not really suitable for real-time guys but it fits your original example neatly.

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hplus0603    11356
The Etwork networking library makes it easy to send and receive commands. It takes care of the tokenization for you (i e, the TCP-is-a-stream-not-a-message problem).


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# Introduction

Poles and polars are a powerful geometrical concept that can be used to complete interesting proofs in often elegant ways. But what exactly are they?

Before we can define what poles and polars are, we must first familiarize ourselves with the concept of an inverse. If we start with a circle centered at $$O$$ with radius $$r$$ and a point $$P$$ which isn’t on the center of the circle, then the inverse of point $$P$$ is the point $$P'$$ such that $$O, P,$$ and $$P'$$ are collinear and $$OP\cdot OP' = r^2$$.

It’s important to note that the way we are currently thinking about lengths, the above definition would have two solutions in which $$P'$$ is reflected about $$O$$. But we can introduce directed lengths, which claims that the product of two collinear segments is positive if they point in the same direction, and negative if they point in opposite directions. Under that concept of lengths, the condition that $$O$$, $$P$$, and $$P'$$ are collinear and $$OP\cdot OP' = r^2$$ now has only one solution.

It’s also important to know how to construct the inverse. If we are given a point $$P$$ outside a circle centered at $$O$$, then we can construct a tangent from $$P$$ to the circle and drop an altitude from the point of tangency to the line $$\overline{OP}$$. The point where the altitude intersects $$\overline{OP}$$ will be the inverse of point $$P$$.

We can prove this by using similar triangles. If we call the point of tangency $$M$$, then we know that $$\Delta PMO$$ is a right triangle. The altitude of a right triangle splits that triangle into two similar triangles. Since $$\Delta MP'O$$ and $$\Delta PMO$$ are similar, we know that $$\frac{OP'}{OM}=\frac{OM}{OP}$$. $$M$$ is on the circle centered at $$O$$, so $$OM=r$$, and $$\frac{OP'}{r} = \frac{r}{OP}$$. Multiplying out the denominators gives $$OP\cdot OP' = r^2$$, which satisfies the definition of the inverse.

Now, we’ll define poles and polars: if we start with a circle centered at $$O$$ and a point $$P$$ which we will call a pole, then the polar of pole $$P$$ is the line perpendicular to $$OP$$ which passes through the inverse of $$P$$.

Again, we’ll start with a construction. If we start with a circle centered at $$O$$ and a pole $$P$$ that lies outside the circle, then we can draw the tangents from $$P$$ that intersect the circle at points $$E$$ and $$F$$. As it turns out, line $$EF$$ goes through the inverse of $$P$$, and is perpendicular to $$OP$$, so therefore it is the polar of pole $$P$$. Likewise, if we start with a chord $$EF$$, we can draw the tangents to the circle at points $$E$$ and $$F$$, and they will intersect at the pole $$P$$.

# La Hire’s Theorem

Now we can go on to prove La Hire’s Theorem, an important statement about poles and polars: if we have poles $$P$$ and $$Q$$ such that $$Q$$ lies on the polar of $$P$$, then $$P$$ must also lie on the polar of $$Q$$.

Proving this statement becomes easier when we break it down into simpler geometric statements. Saying $$Q$$ is on the polar of $$P$$ is equivalent to saying that $$\angle OP'Q$$ is a right angle. Likewise, to prove that $$P$$ is on the polar of $$Q$$, we just have to show that $$\angle OQ'P$$ is a right angle. We know that $$OP \cdot OP' = r^2$$ and $$OQ \cdot OQ' = r^2$$, so we also know that $$OP\cdot OP' = OQ \cdot OQ'$$. Therefore, $$\frac{OP}{OQ'}=\frac{OQ}{OP'}$$. Next, we can use SAS similarity to prove that $$\triangle OP'Q \sim \triangle OQ'P$$. We know $$\angle P'OQ \cong Q'OP$$ since they’re the same angle, and we know that the surrounding sides are proportional based on what we just showed. Therefore, the two triangles are similar, and so their angles are equal. We’re given that $$\angle OP'Q$$ is a right angle, so because of the similar triangles we’ve established, $$\angle OQ'P$$ must also be a right angle. Therefore, if $$Q$$ is on the polar of $$P$$, then $$P$$ is on the polar of $$Q$$.

## Example Problem

Here’s an example of a problem that’s made significantly easier by using the concept of poles and polars.

Problem: Given a triangle $$\triangle ABC$$, with an incircle centered at incenter $$I$$ which intersects $$BC$$, $$AC$$, and $$AB$$ at points $$D$$, $$E$$, and $$F$$ respectively, prove that if line $$EF$$ intersects $$BC$$ at point $$T$$, then $$\overline{IT}\perp\overline{AD}$$.

Proof: First, notice that $$\overline{EF}$$ is the polar of $$A$$, since we’ve already shown that polars are defined by the points of tangency from the pole. Therefore, if $$T$$ lies on $$\overline{EF}$$, then $$T$$ is on the polar of $$A$$. By La Hire’s, $$A$$ must also lie on the polar of $$T$$. We also know that $$D$$ must also be on the polar of $$T$$, since it is a point of tangency. Therefore, $$\overline{AD}$$ must be the polar of $$T$$, and so by the definition of a polar, it must be perpendicular to $$\overline{IT}$$.

## Corollary

Now we can go on to prove an extension of La Hire’s Theorem: three points $$P$$, $$Q$$, and $$R$$ are collinear if and only if their polars are concurrent.

There are two things we have to prove here. Since the statement is bidirectional, we have to prove both directions separately.

First, we will prove that if $$P$$, $$Q$$, and $$R$$ are collinear, then their polars are concurrent. Let’s say the the line that goes through all three is labelled $$\ell$$. Then we can construct the pole of polar $$\ell$$ and label it $$L$$.

Next, we can apply La Hire’s Theorem three times; if $$P$$ lies on the polar of $$L$$, then $$L$$ lies on the polar of $$P$$; if $$Q$$ lies on the polar of $$L$$, then $$L$$ lies on the polar of $$Q$$; if $$R$$ lies on the polar of $$L$$, then $$L$$ lies on the polar of $$R$$. Therefore, we’ve proved that $$L$$ lies on the polars of $$P$$, $$Q$$, and $$R$$, so therefore the three polars are concurrent.

Next, we’ll use a similar strategy to prove that if the polars of $$P$$, $$Q$$, and $$R$$ are concurrent, then they are collinear. Since the three lines are concurrent, call their intersection $$L$$, construct the polar of $$L$$ and call it $$\ell$$, and apply La Hire’s three times to show that $$P$$, $$Q$$, and $$R$$ all lie on the polar of $$L$$, $$\ell$$, and so they must be collinear.

# Pascal’s Theorem

Before we prove more theorems about poles and polars, we have to get familiar with Pascal’s Theorem. It states the following: given a hexagon $$ABCDEF$$ inscribed in a circle, then the intersections of opposite sides are collinear. To clarify, if we label $$AF\cap CD$$ as $$G$$, $$AB \cap DE$$ as $$H$$, and $$BC \cap EF$$ as $$I$$, then $$G$$, $$H$$, and $$I$$ are collinear. It is important to note that a hexagon does not have to be a convex geometric shape, but rather any collection of 6 points that are connected by 6 lines in a cycle.

## Menelaus’s Theorem

Before we prove Pascal’s Theorem, we must familiarize ourselves with Menelaus’s Theorem. This is a generalization of Ceva’s Theorem, but can be super useful in interesting situations. It states that given triangle $$\triangle ABC$$ and points $$X$$, $$Y$$, and $$Z$$ on sides $$BC$$, $$AC$$, and $$AB$$ respectively (with $$X$$ on the extension of side $$BC$$ rather than on the line segment itself), then $$X$$, $$Y$$, and $$Z$$ are collinear if and only if $$\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=-1$$. Note that the product is negative because since $$BX$$ and $$XC$$ run in opposite directions, dividing one by the other yields a negative result.

This is also a bidirectional statement so we must prove both directions separately. The first direction we’ll prove is that if the points are collinear, then the equation is true. We can drop altitudes to line $$XYZ$$ from $$A$$, $$B$$, and $$C$$, which meet the line at points $$D$$, $$E$$, and $$F$$ respectively. From here, we can find multiple similar triangles to get ratios that equal each other. $$\triangle ADZ \sim \triangle BEZ$$ because they share a right angle and vertical angles, so $$\frac{AZ}{ZB}=\frac{AD}{BE}$$. Likewise, $$\triangle XEB \sim \triangle XFC$$ because of two shared angles, so $$\frac{BX}{XC}=-\frac{BE}{CF}$$ (Note the negative sign to indicate that $$\frac{BX}{XC}$$ is negative). Finally, $$\triangle ADY \sim \triangle CFY$$ by angle-angle similarity, so $$\frac{CY}{YA}=\frac{CF}{AD}$$. Finally we can start considering the equation we are trying to prove: $\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=-1$

We’ve found that each of these factors is equal to another ratio, so we can substitute those in: $\frac{AD}{BE}\cdot\frac{-BE}{CF}\cdot\frac{CF}{AD}=-1$

And now that the factors in the numerators and denominators are the same except for a negative sign, the left side is equal to -1 and the theorem has been proved true.

Now we just have to prove the other direction, but this is a relatively easy proof by contradiction: let’s assume that three points $$X$$, $$Y$$, and $$Z$$ satisfy the equation without being collinear. Then there must exist a unique point $$X'$$ that also lies on $$BC$$ while being collinear with $$Y$$ and $$Z$$. Since we’ve already proved the other direction of Menelaus, we know that $$X'$$, $$Y$$, and $$Z$$ satisfy the equation. If both sets of points satisfy the equation, then it must be true that $$\frac{BX}{XC}=\frac{BX'}{X'C}$$ in order for both equations to simultaneously hold. And that is a contradiction, because under the assumption that $$X$$ and $$X'$$ are distinct points on $$BC$$, the two ratios cannot be equal.

## Proof of Pascal’s Theorem

First, we’ll define some more points: let $$CD \cap EF$$ be labelled $$U$$, and $$AB \cap EF$$ be labelled $$V$$, and $$AB \cap CD$$ be labelled $$W$$.

Now we can apply Menelaus with $$\triangle UVW$$ and various lines. First, if we use Menelaus with $$\triangle UVW$$ and line $$EHD$$, we get $\frac{VH}{WH}\cdot\frac{WD}{UD}\cdot\frac{UE}{VE}=-1$

Next, we can apply Menelaus with $$\triangle UVW$$ and line $$AGF$$ to get $\frac{VA}{WA}\cdot\frac{WG}{UG}\cdot\frac{UF}{VF}=-1$

Next, we’ll apply Menelaus with $$\triangle UVW$$ and line $$BIC$$ to get $\frac{VB}{WB}\cdot\frac{WC}{UC}\cdot\frac{UI}{VI}=-1$

Now we can multiply all three equations together: $\frac{VH}{WH}\cdot\frac{WD}{UD}\cdot\frac{UE}{VE}\cdot\frac{VA}{WA}\cdot\frac{WG}{UG}\cdot\frac{UF}{VF}\cdot\frac{VB}{WB}\cdot\frac{WC}{UC}\cdot\frac{UI}{VI}=-1$

That’s a lot of factors being multiplied, but a little bit of organizing helps us cancel out most of them: $\frac{WD \cdot WC}{WA \cdot WB} \cdot \frac{VA \cdot VB}{VE \cdot VF} \cdot \frac{UE \cdot UF}{UC \cdot UD} \cdot\frac{VH}{WH}\cdot\frac{WG}{UG}\cdot\frac{UI}{VI}=-1$

We can use power of a point here to cancel out the first three fractions. As a reminder, power of a point claims that for a fixed point $$P$$ and circle $$\omega$$, every chord that passes through $$P$$ and intersects $$\omega$$ at points $$E$$ and $$F$$ will have a constant product $$PE\cdot PF$$. In this case, it helps us show that $$WD\cdot WC = WA \cdot WB$$, and $$VA \cdot VB = VE \cdot VF$$, and $$UE \cdot UF = UC \cdot UD$$. Cancelling out these equal factors gives us $\frac{VH}{WH}\cdot\frac{WG}{UG}\cdot\frac{UI}{VI}=-1$

Notice that this is exactly what Menelaus says when applied to $$\triangle UVW$$ and line $$GHI$$. Since we’ve shown that Menelaus is bidirectional, this proves that $$G$$, $$H$$, and $$I$$ are collinear, and our proof of Pascal’s Theorem is complete.

Just to show that Pascal’s Theorem is true for all cyclic hexagons, including convex hexagons, here’s a diagram of the same proof drawn out for a convex hexagon.

Note that all the steps in the proof are the same and still hold for this configuration, except for one complication: line $$EHD$$ now has 3 points that lie outside the borders of the triangle. But that’s okay, because we can actually show that we can apply Menelaus to a set of three collinear points where every point lies outside the borders of the triangle. The proof is exactly the same: if we draw perpendiculars from points $$U$$, $$V$$, and $$W$$ to line $$EHD$$, we can use the resulting similar triangles as outlined in the proof of Menelaus above, which confirms that $$\frac{VH}{WH}\cdot\frac{WD}{UD}\cdot\frac{UE}{VE}=-1$$ and completes the proof.

# Fundamental Theorem of Poles and Polars

Now we can go on to prove another theorem about poles and polars: given a cyclic quadrilateral $$ABCD$$ inscribed in a circle, let $$AB \cap CD$$ be labelled $$F$$, and $$AD \cap BC$$ be labelled $$E$$, and $$AC \cap BD$$ be labelled $$K$$. Then $$EK$$ is the polar of $$F$$, $$KF$$ is the polar of $$E$$, and $$EF$$ is the polar of $$K$$.

We can apply Pascal’s Theorem here, but we must first establish how we interpret "lines" defined by a single point on a circle. For example, we know what line $$AB$$ looks like, but how would we interpret line $$AA$$? For the purposes of this article, this will represent the line tangent to the circle at point $$A$$. You can conceptualize it this way: if Pascal’s Theorem is true for any hexagon, then it is true for hexagons where $$A$$ and $$B$$ are extremely close together. When $$A$$ and $$B$$ are extremely close together, the line between them very closely resembles the tangent to the circle at point $$A$$. So if Pascal’s Theorem is true when line $$AB$$ is very similar to the tangent at $$A$$, then it should also be true when line $$AB$$ is exactly the tangent at $$A$$, which we’ll denote as line $$AA$$.

Now we can prove that $$EK$$ is the polar of $$F$$ by applying Pascal’s Theorem. If we look at hexagon $$ACCBDD$$, Pascal’s Theorem states that $$AC \cap BD$$, $$CC \cap DD$$, and $$CB \cap DA$$ are collinear, so $$K$$, $$E$$, and $$CC \cap DD$$ are collinear. We can also apply Pascal’s Theorem to hexagon $$CAADBB$$ to find that $$CA \cap DB$$, $$AA \cap BB$$, and $$AD \cap BC$$ are collinear, so $$K$$, $$E$$, and $$AA \cap BB$$ are collinear. Combining the two applications of Pascal’s Theorem shows that $$K$$, $$E$$, $$CC \cap DD$$, and $$AA \cap BB$$ are all collinear. Notice that $$CC \cap DD$$ is the pole of polar $$CD$$, since we’ve shown that poles can be constructed by the intersection of two tangents to a circle. And now we can apply La Hire’s: since $$F$$ lies on $$CD$$, which is the polar of $$CC \cap DD$$, then $$CC\cap DD$$ must lie on the polar of $$F$$. For the same reason, if $$F$$ lies on $$AB$$, which is the polar of $$AA \cap BB$$, then $$AA \cap BB$$ must lie on the polar of $$F$$. Therefore, the line containing $$E$$, $$K$$, $$CC \cap DD$$, and $$AA \cap BB$$ must be the polar of $$F$$, and therefore $$EK$$ is the polar of $$F$$.

Very similar proofs can show that $$KF$$ is the polar of $$E$$ and $$EF$$ is the polar of $$K$$, but we can skip the tedious work by making a clever observation: we’ve already proven that $$EK$$ is the polar of $$F$$, which means that in general, for any cyclic quadrilateral $$ABCD$$, the line between $$AD\cap BC$$ and $$AC \cap BD$$ will be the polar of $$AB \cap CD$$. This means that we can take the same diagram and change the cyclic quadrilateral we choose to consider. For example, in the given diagram, we can choose to consider quadrilateral $$ADCB$$ instead, and what we’ve already proved can be used to demonstrate that the line between $$AB \cap DC$$ and $$AD \cap DB$$ is the polar of $$AD \cap BC$$, so $$KF$$ is the polar of $$E$$. We can even choose to consider quadrilateral $$ACBD$$ — poles, polars, La Hire’s Theorem, and Pascal’s Theorem do not depend on polygons in question being convex, so we can let our quadrilateral intersect itself. Considering what we’ve proved in the context of this new quadrilateral shows that the line between $$AD \cap CB$$ and $$AB \cap CD$$ will be the polar of $$AC \cap BD$$, so $$EF$$ is the polar of $$K$$.

## Brocard’s Theorem

Brocard’s Theorem is a corollary of the Fundamental Theorem of Poles and Polars. It starts off with the same setup: a cyclic quadrilateral $$ABCD$$ with $$AD \cap BC$$ labelled as $$E$$, $$AB \cap CD$$ labelled as $$F$$, and $$AC \cap BD$$ labelled as $$K$$. Brocard’s Theorem then states that the orthocenter of triangle $$\triangle EFK$$ is the circumcenter of $$ABCD$$, which we’ll call $$O$$.

We can first apply the Fundamental Theorem of Poles and Polars that we have just proved, which shows that each side of the triangle is the polar of the opposite vertex. Next, we can look at the altitudes to each side. For example, if $$EF$$ is the polar of $$K$$, then line $$OK$$ must be perpendicular to $$EF$$. That means the altitude from $$K$$ must pass through $$O$$. For the same reason, the altitudes through $$E$$ and $$F$$ must also pass through $$O$$. Therefore, the orthocenter of triangle $$\triangle EFK$$ must be the circumcenter of $$ABCD$$, $$O$$.

## Example Problem

Here’s a problem that uses the theorems we’ve just discussed.

Problem: Given a semicircle with diameter $$AB$$ and points $$C$$ and $$D$$ on the semicircle, define $$E$$ as the intersection between the tangents to the semicircle at $$C$$ and $$D$$ (in other words, $$E = CC \cap DD$$, and define $$F$$ as $$AC \cap BD$$. Prove that $$EF \perp AB$$.

The first step is to use Pascal’s Theorem on hexagon $$ACCBDD$$ to show that $$AC \cap BD$$, $$CC \cap DD$$, and $$CB \cap DA$$ are collinear. If we call $$CB \cap DA$$ point $$G$$, then that means $$E$$, $$F$$, and $$G$$ are collinear. Let’s also label $$AB \cap CD$$ as $$H$$. By the Fundamental Theorem of Poles and Polars, line $$FG$$ is the polar of $$H$$. Since $$E$$, $$F$$, and $$G$$ are all collinear, we can also say that line $$EF$$ is the polar of $$H$$. Finally, by the definition of polars, line $$EF$$ must be perpendicular to the line connecting the center of the circle to $$H$$. Since line $$AB$$ goes through both the center of the circle and point $$H$$, then we have proven that $$EF$$ is perpendicular to $$AB$$.

# Citations

1. AoPS user Potla, 3-12-2012, "Poles and Polars - Another Useful Tool!",https://artofproblemsolving.com/community/c1642h1027724_poles_and_polars__another_useful_tool

2. Kin Y. Li, November 2006, "Pole and Polar", Mathematical Excalibur Volume 11 Number 4, https://www.math.ust.hk/excalibur/v11_n4.pdf

3. Tyler Zhu, 10-22-2015, "Poles and Polars", Irvington Math Club,https://tylerzhu.com/assets/handouts/poles_and_polars.pdf

4. Pascal’s Theorem. Brilliant.org. Retrieved January 18, 2021, fromhttps://brilliant.org/wiki/pascals-theorem/

5. Wellesley math Professor CONCEIVES formulas for health, wealth, and love. (2017, July 07). Retrieved April 01, 2021, from https://www.wellesley.edu/news/2017/stories /node/120511


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# Are there any practical implementation of a homomorphic hashing or signature scheme?

A homomorphic hash function is a function $H : A \to B$ between two sets with some algebraic structure $(A, *)$ and $(B, \star)$ such that

• $H$ is collision resistant, i.e. it is hard to find $x \neq y$ such that $H(x) = H(y)$ and
• $H$ is a homomorphism, i.e. $H(x * y) = H(x) \star H(y)$.

Are there any practical realizations of such a homomorphic hash function, or even a homomorphic signature scheme (i.e., where we can "add" valid signatures to get a signature of the "sum" of two messages)?

Even better, are there even any libraries implementing this?

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As far as I know, there is yet no practically efficient implementation of fully homomorphic encryption on the horizon. So the answer to your question would evidently be negative, at least for a good hashing scheme, IMHO. –  Mok-Kong Shen Feb 27 '13 at 11:00
FYI, I posted a question on Meta a while back about this type of question and whether we should allow them. Perhaps you would like to weigh in? –  mikeazo Feb 27 '13 at 12:31
sashank, I think you might need to specify more precisely exactly what you mean by homomorphic hashing. –  D.W. Feb 27 '13 at 15:45
@D.W., sashank: I edited the question to contain an explanation of what is searched here. –  Paŭlo Ebermann Feb 27 '13 at 20:16
You might want to add an additional constraint; the Identity function pedantically meets all the requirements listed; it is hard (impossible) to find $x \neq y$ with $I(x) = I(y)$, and for it homomorphic with any operation $\star$, that is, $I(A \star B) = I(A) \star I(B)$ –  poncho Feb 27 '13 at 22:20


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This paper is concerned with the statistics of the height of rise and full for continuous random processes. In particular, approximate methods are given for determining the probability density of the increment in a random continuous function as the function passes from one extremum to the next. Application of the general result is made to the case of processes with a Gaussian distribution. Numerical results are given for four special cases of stationary Gaussian processes. Computed results are found to agree well with available experimental data. The knowledge of such statistical information is of use in studies dealing with fatigue under random loadings.

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# Is there a set that is both a sigma algebra and a topology but not a powerset?

Is there a set that is both a sigma algebra, $\Sigma$, and a topology, $\tau$, but not a powerset, $\mathcal{P}$?

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What do you think? – Asaf Karagila Sep 24 '12 at 19:01
I think it is if and only if. i.e., powerset $\iff$ the set is both a sigma algebra and a topology. So the answer to my question is no. – torrho Sep 24 '12 at 19:43

This can be easily done on large enough infinite sets, for example consider $$\mathcal A=\{A\subseteq\mathbb R\mid A\subseteq\mathbb N\text{ or }\mathbb R\setminus A\subseteq\mathbb N\}.$$

Let us see that this is a topology and a $\sigma$-algebra:

1. Clearly $\mathbb R$ and $\varnothing$ are in this collection.
2. It is closed under countable intersections, suppose $A_n$ is a collection of sets in $\cal A$. If the intersection is empty then we are done, otherwise
• if there is at least one set which is a subset of $\mathbb N$ then obviously their intersection is a subset of $\mathbb N$ and therefore is in $\cal A$; otherwise
• all sets come are subsets of $\mathbb R\setminus B$, where $B\subseteq\mathbb N$, then by DeMorgan this is again true.
3. Closed under any sort of unions, take $\{A_i\mid i\in I\}$ any collection of sets from $\cal A$, and by similar (but dual) arguments to before the union is in $\cal A$.
4. Closure under complement is trivial.

It is quite obvious that $\cal A\neq P(\Bbb R)$ since the rationals (for example) are not in $\cal A$.

(Note that this is essentially equivalent to two copies of $\cal P(\Bbb N)$ though, but it's not quite that.)

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Isn't this simply $\mathcal A=\mathcal P(\mathbb N)\subset \mathcal P(\mathbb R)$? And how is $\mathbb R\in\mathcal A$? Or maybe I don't understand your definition right. But for me $\exists B\subset \mathbb N\colon A\cup B=\mathbb N\land A\cap B=\emptyset \lor A\subseteq \mathbb N$ is simply equivalent to $A\subseteq \mathbb N$. – Hagen von Eitzen Sep 24 '12 at 20:13
@Hagen $\varnothing$? – Asaf Karagila Sep 24 '12 at 20:28
Is this supposed to mean that the choice $B=\emptyset$ shows that $A=\mathbb R$ is an element of $\mathcal A$? $\emptyset\subseteq \mathbb N$ - check. $\mathbb R\cup \emptyset=\mathbb N$ - fail. – Hagen von Eitzen Sep 24 '12 at 20:31
@Hagen: I found the definition of $\mathcal{A}$ a bit hard to parse sensibly, but Asaf appears to mean the set of all $A\subseteq\Bbb R$ such that either $A\subseteq\Bbb N$, or $A\supseteq\Bbb R\setminus\Bbb N$; equivalently, $A\subseteq\Bbb N$ or $\Bbb R\setminus A\subseteq\Bbb N$. – Brian M. Scott Sep 24 '12 at 21:08
@Clive: There was a typo, it is now corrected. – Asaf Karagila Sep 25 '12 at 6:01

Yes. If $X$ is any set containing at least two elements then the collection $\{ X, \varnothing \}$ is both a $\sigma$-algebra and a topology on $X$ but is not the powerset.

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If you add the Hausdorff condition, then you can achieve the full power set result, as I explain below. Meanwhile, in general, here is a sense in which every such algebra/topology is in fact a power set.

Theorem. Every family $\cal{A}$ of sets that is both a $\sigma$-algebra and a topology on a set $X$ (or even just an algebra and a topology) is isomorphic, as a partial order under $\subseteq$, to a power set order. Specifically, $\cal{A}$ is isomorphic to the power set of the collection of minimal nonempty elements of $\cal{A}$, which form a partition of $X$.

Proof. Suppose that $\cal{A}$ is closed under complement and is a topology on $X$. Thus, $\cal{A}$ is closed under complement and arbitrary unions, and hence by de Morgan it is closed under arbitrary intersection. For each $x\in X$, let $[x]$ be the intersection of all $A\in\cal{A}$ with $x\in A$. Thus, $[x]\in \cal{A}$, and these are simply the minimal nonempty elements of $\cal{A}$. These partition $X$, and form the equivalence classes of the relation $x\sim y$ iff $x$ and $y$ are in all the same elements of $\cal{A}$. Every element of $\cal{A}$ is a union of the various classes $[x]$, and furthermore, every union of classes $[x]$ is in $\cal{A}$. So the partial order $\langle \cal{A},{\subseteq}\rangle$ is isomorphic to the power set of the classes $\langle P(X/\sim),\subseteq\rangle$, since every element $A\in \cal{A}$ breaks into the union of pieces of the partition $A=\bigcup_{a\in A}[a]$. QED

Let us now add a separation requirement to the topology to achieve the full power set result.

Theorem. Every family $\cal{A}$ of sets that is both an algebra and a Hausdorff topology on $X$ is the full power set of $X$.

Proof. If the topology is Hausdorff, or merely even $T_1$, then the relation $x\sim y$ I mention above is just $x=y$, and so $[x]=\{x\}$. So all the singletons are in $\cal{A}$, which is closed under arbitrary unions. So ${\cal A}=P(X)$. QED

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thank you for your reply. – torrho Sep 26 '12 at 4:32

Hint: a topology on $X$ that contains all singletons $\{x\}$ for $x \in X$ is the power set. Try a very simple $\sigma$-algebra on a finite set.

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# How far is the list of known primes known to be complete?

So there is always the search for the next "biggest known prime number". The last result that came out of GIMPS was $2^{74\,207\,281} - 1$, with over twenty million digits. Wikipedia also lists the twenty highest known prime numbers, only the four smallest on that list have fewer than three million digits.

For some while now, I have been wondering about the smaller prime numbers we haven't found. How far up is the list of known primes known to be complete? Since $500$ to $1000$ digit primes are considered safe for the RSA algorithm, I'd assume that it's well below that. How far along the number line have we checked that there are no more primes to be found? How fast is this boundary moving forward, currently? Have we, for instance, checked the primality of all numbers below $10^{100}$, or are we stuck somewhere south of $10^{20}$?

• Just as a start, here is First fifty million primes Jul 7, 2016 at 10:49
• @YuriyS I don't want the list. I want to know how long the (longest) list (we can currently make) is. Jul 7, 2016 at 10:51
• @barakmanos According to wiki (the list I linked in the question) it is $19249×2^{13\,018\,586} + 1$, which is just shy of $4$ million digits, and there are $11$ known Mersenne primes that are larger. Jul 7, 2016 at 10:55
• @Arthur, here is exactly your question considered Jul 7, 2016 at 10:57
• @joriki: I agree, but I can only delete mine, which would leave this thread kind of enigmatic, so if you have the privilege of doing it, then please go ahead, I will not bare a grudge for that :) Jul 7, 2016 at 11:05

The maximum of such list is far smaller than mentioned 500-digits. Due to the prime number theorem $\pi(x) \approx x/\log(x)$ so one could estimate that the list of prime numbers up to $x$ would require at the order of $x$ digits to represent.

So by using the sieve of Atkin the complexity is $O(x)$ for both time consumption and memory consumption. And all memory available is small enough to be feasible to traverse. This means that the memory available for storing such a list is what should be the limiting factor.

Basically this boils down to that the largest such list is as large as anybody have place (and need) for.

Now the total global storage is estimated to be at the order of $10^{21}$ bytes which means that the upper bound of such a prime number list is there. So there exist no complete list with primenumbers up to more than about twenty digits (and not even that since not all storage is devoted to store prime numbers).

• Excellent line of reasoning. If you can't store it, you can't (continue to) know it. Astonishing remindee of the power of exponents... Jul 8, 2016 at 1:28
• Suppose someone with extremely many huge hard disks kept a list of all prime numbers with at most 21 decimal digits. It would be extremely easy to come up with a continuation of that list. For example the first prime over $10^{21}$ (i.e. 22 digits) is $1{,}000{,}000{,}000{,}000{,}000{,}000{,}117$. It takes a few milliseconds to find and prove with standard math software (I used PARI/GP). Jul 8, 2016 at 8:26

This may be a somewhat unsatisfying answer, but no-one's really keeping a complete list of known primes (to the best of my knowledge). Moreover, it's fairly easy to come up with large primes, and it's fairly easy to "guarantee" (guarantee being a slippery term), that a given large number is prime.

The Miller-Rabin primality test is an algorithm that takes a number $n$, and a "certainty" parameter $m$, and (in layman's terms) if $n$ is prime, it will return "PRIME". If $n$ is composite, it will almost certainly (again, a slippery term) return "COMPOSITE", but there's a small probability $(\frac{1}{4^m})$ that it will return "PRIME".

However, by setting $m$ high enough, to, say, something greater than 40, then it essentially means the probability of a composite number being declared prime is smaller than you winning the jackpot in the lottery twice in one week. Thus, for almost ALL practical purposes, it suffices to work with primes that pass the Miller-Rabin test to a high degree of certainty. Henri Cohen famously called such numbers "Industrial Grade Primes".

If you're still interested in having "proof" that a number is prime, may I suggest reading up on prime certificates. I haven't ever personally come across a situation in which you'd prefer a certified prime to an industrial grade prime however.

Finally, as a quick example, Mathematica can generate very large primes easily. The Mathematica command "RandomPrime[{10^1000, 10^1001}]" generates a random 1000 digit prime in 0.40625 seconds on my five year old desktop machine. This should give you some indication as to why mathematicians generally don't keep long lists of all known primes.

• so RSA uses "Industrial Grade Primes" right ? Jul 7, 2016 at 11:31
• @skyking Best two I know of are: 1) the elliptic curve primality test, which is $O((\log(n))^6)$, on the condition that certain (probably true) conjectures are true, and 2) Miller's test, upon which the Miller-Rabin test is based, which is a deterministic test. It only works under the assumption of GRH, but it works in $O((\log(n))^4 \log^k((\log(n))^4)$, for some integer $k$. Jul 7, 2016 at 14:23
• "I haven't ever personally come across a situation in which you'd prefer a certified prime to an industrial grade prime"; one case might be if you're generating a prime for a long term DH group (e.g. IKE group 14) or EC curve (e.g. P256 or Curve25519); since such a group/curve might be used literally billions of times, it might make sense to go to the extra effort of doing the full certification. Jul 7, 2016 at 16:54
• Are there any Industrial Grade Primes that are known to be composite? Jul 7, 2016 at 18:10
• @MadMonty The AKS primality test is deterministic, its correctness is proven unconditionally, and a variant of it has running time roughly $O((\log n)^6)$. Jul 7, 2016 at 21:21

It seems that this link, provided by Yuriy S in the comments above more or less addresses my question. It states that prime gap searches have checked the primality (but not stored the primes) for all numbers up to about 20 digits (the exact bound is always changing, since the numbers are relatively small).

• Should've probably posted as an answer, since you got upvoted. Not that I mind though, glad I helped Jul 7, 2016 at 12:39
• I'm not familiar with the algorithms used by the maximal prime gap project; however I'd think that if you're searching for a prime gap $> k$, you wouldn't need to check the primality of each integer. Instead, if you're at the verified prime $p$, the next integer you check is not $p+2$, but instead $p+k$ (and if that's not prime, work your way backwards). If you verify that (say) $p+k-6$ is prime, you don't need to check the primality of the other integers in the range... Jul 7, 2016 at 17:01
• I wish someone know, just an order of magnitude, how many primes there are with less than 20 digits. this QA is remarkably unmathematical :) Everyone's just saying "there's a lot!" and "there's a hell of a lot!" Geesh! Jul 7, 2016 at 20:18
• @JoeBlow: according to the prime number theorem, there are approximately $10^{20} / \log 10^{20} \approx 2,171,472,409,516,259,138$ primes in that range. While this is just an approximation (and the lower dozen digits are almost certainly wrong), this is certainly within an order of magnitude. Jul 7, 2016 at 20:23
• @poncho People also counted the exact number of primes below $10^{20}$, it is $2{,}220{,}819{,}602{,}560{,}918{,}840$, see this Wikipedia subsection. That number divided by your number is about $1.02$ according to that table. This fraction tends to $1$ by the prime number theorem. Jul 8, 2016 at 8:59

Listing primes in order is a fairly trivial problem — in fact, I believe we have programs that can compute lists of primes faster than they can be written to disk, let alone displayed in any human readable format.

The thing is, there are a lot of primes. The entire internet put together probably couldn't store the list of all 20 digit primes.

But that's okay, because we don't need lists of primes — whenever we need primes, we can just generate them.


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# How to find linear speed

By John

Linear speed always measures the concrete distance traveled by a moving object. So, the linear speed measures in distance units per time units. For example meter per second. If an object is doing circular motion then the term linear is used as straightening out the arc traveled by the object alongside the circle. It results in a line of the same length. So, the common definition of speed as distance over time is correct. In this topic, we will discuss the linear speed formula with examples. Let us learn the concept!

Linear Speed Formula
Concept of Linear Speed
The linear speed of a point on a rotating object can be determined by its distance from the center of the rotation. The angular speed is measuring the angle that an object travels through in a certain amount of time. The angular speed will measure in units of radians per second i.e. radian per second.

It has 2\pi radians in a full circle. At a distance r i.e. radius from the center of the rotation. Then a point on the object has a linear speed equivalent to the angular speed multiplied by the distance r. Its measuring units are meters per second, meter per second.

The Formula for Linear Speed
We can compute Linear speed as,

s=dt
Where,

s Speed
d Distance
t Time
i.e.v=ω×r
v = linear speed (m per s)

Where,

v Linear speed
ω Angular speed
r The radius of the rotation
Solved Examples
Q.1: A power wheel begins and revolving at 10.0 revolutions per second. The diameter of the wheel is 4 m. What will be the linear speed of a point of the wheel on the surface, in meters per second?

Solution: First start to find the angular speed. The revolutions per second must be converted to radians per second. There are 2π radians in a full circle.

ω=10.0rev/s
r=42=2m
Using the formula

v=ω×r,
The linear speed of a point on the surface of the wheel is,

v = 125.6 m per s.

The linear speed of a point of the wheel is around 125.6 m per s.

Q.2: A sensor is connected under a car wheel that measures the linear speed. The sensor is 0.080 m from the center of rotation. At that point, the sensor reads that the linear speed of the wheel is around 8.00 m per s. So, if the radius of the wheel is 0.220 m, what is the linear speed on the outside edge of the wheel?

Solution: The linear speed is different at various distances from the center of rotation, but the angular speed is similar everywhere on the wheel. To solve this problem, the first step is to find the angular speed using the linear speed at the position of the sensor, 0.080 m.

The formula v = ω×r can be rearranged to solve for the angular speed:

v=ω×r


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# Nilpotent operator / Orthogonal projection

If you have a nilpotent operator $A \colon V \to V$, $V$ being $A$-cyclic (meaning that $V$ is generated by a single vector $v$ in $V$), is it true that the minimal polynomial of the Gram operator $P = A^{T} A$ is $p(t) = t^2 - t$, i.e., $P$ is the orthogonal projection?

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Do you have any normalizing condition on $A$ (for instance norm one)? Because otherwise if one $A$ satisfies your condition, so will a scalar multiple and you wouldn't get a projection. –  Martin Argerami May 29 '13 at 1:52
You're right, this isn't true under the given conditions. –  Br09 May 29 '13 at 2:08

If you grant Jordan canonical form, then in a suitable basis $A$ looks like $$\begin{bmatrix}0&1&0&\dots&0 \\ 0&0&1&\dots&0 \\ & & \ddots&\ddots \\ 0&0&\dots&0&1 \\ 0&0&\dots &0&0 \end{bmatrix}\,.$$ Then $$A^\top A = \begin{bmatrix}0 & \\ & 1 & \\ & & \ddots \\ & & & 1\end{bmatrix}\,.$$ So 'twould appear you're right.
But .... Note that once you bring the transpose into the game, things are no longer basis-independent. The basis that brings $A$ into this nice form is not likely to be orthonormal, and so, if you're bringing the dot product into the game, it really is not meaningful to describe this in terms of an orthogonal projection. The characteristic polynomial of $A$ is basis-independent, but the characteristic polynomial of $A^\top A$ is not.