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Question

# The following question consists of two statements, one labelled as “Assertion (A)” and the other labelled as “Reason (R)”. You are to examine these two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason (R) is the correct explanation for the given Assertion (A). Select your answer to these items using the codes given below and then select the correct option.Codes:(A) Both A and R are individually true and R is the correct explanation of A(B) Both A and R are individually true and R is not the correct explanation of A(C) A is true but R is false(D) A is false but R is trueAssertion (A): Let $f:[0,\infty )\to [0,\infty ],$be a function defined by $y=f\left( x \right)={{x}^{2}}$, then $\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=1$Reason (R): $\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1$(a) A(b) B(c) C(d) D

Hint: Differentiate the given function with respect to $x$ and with respect to $y$ twice.

A: Given $f:[0,\infty )\to [0,\infty ]$be a function defined by $y=f\left( x \right)={{x}^{2}}$, then we have to check the value of $\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)$.
Now, we take the given function,
$y={{x}^{2}}....\left( i \right)$
Now, we differentiate it with respect to $x$.
$\left[ \text{Also}\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]$
Therefore, $\dfrac{dy}{dx}=2x.....\left( ii \right)$
Again differentiating with respect to $x$,
We get, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2....\left( iii \right)$
As we have found that, $\dfrac{dy}{dx}=2x$
By taking reciprocal on both sides,
We get, $\dfrac{dx}{dy}=\dfrac{1}{2x}....\left( iv \right)$
Now, we differentiate with respect to $y$.
We get, $\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{d}{dy}\left( \dfrac{1}{2x} \right)$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}\dfrac{dx}{dy}$
Now, we put the value of $\dfrac{dx}{dy}$.
We get, $\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}.\dfrac{1}{2x}$
Hence, $\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{4{{x}^{3}}}....\left( v \right)$
Multiplying equation $\left( iii \right)$and $\left( v \right)$,
We get, $\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=2.\left( \dfrac{-1}{4{{x}^{3}}} \right)$
Therefore, $\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=\dfrac{-1}{2{{x}^{3}}}$.
Hence, given Assertion (A) is wrong.
R: Here we have to check whether $\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1$or not.
We know that any quantity when multiplied by its reciprocal gives a result as $1$.
That is $a\times \dfrac{1}{a}=1$.
Now, we put $a=\dfrac{dy}{dx}$.
We get, $\dfrac{dy}{dx}\times \dfrac{1}{\dfrac{dy}{dx}}=1$
Or, $\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1$
To verify it further, we multiply the equation $\left( ii \right)$and $\left( iv \right)$.
$\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=2x.\dfrac{1}{2x}$
Therefore, $\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1$ [Hence Proved]
Hence, Reason (R) is correct.
Therefore, option (d) is correct that is A is false and R is true

Note: Some students misunderstand that $\dfrac{{{d}^{2}}x}{d{{y}^{2}}}$and $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ are reciprocal of each other like $\dfrac{dx}{dy}$and $\dfrac{dy}{dx}$, but they are not as proved by above result also.


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# Im really stuck on this basic formula

z = pc

what do i need to do with p and c figures in order to get z this is prob really dumb but im so stuck please help.

format: plain


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# Space Elevator Problem Set

## Summary

Active Time
45 minutes to 1 hour
Total Project Time
45 minutes to 1 hour
Key Concepts
Forces, gravity, tension, circular motion, space exploration
Credits
Sabine De Brabandere, PhD, Science Buddies

## Introduction

Space elevators zipping people and materials up into space might seem like a very futuristic and improbable idea, but is it that difficult? This activity will guide you through the mathematics. Try it out and see what is possible with materials that can be produced with current technology.

This activity is not recommended for use as a science fair project. Good science fair projects have a stronger focus on controlling variables, taking accurate measurements, and analyzing data. To find a science fair project that is just right for you, browse our library of over 1,200 Science Fair Project Ideas or use the Topic Selection Wizard to get a personalized project recommendation.

## Materials

• Pen and paper
• Calculator
• Access to a computer program like Microsoft® Excel® that allows you to graph equations, or a graphing calculator

## Background Information

A space elevator is a transportation system that would be anchored to a celestial body and extend into space. Its purpose would be to move cargo through space without the use of spaceships. A space elevator would require much less fuel to transport cargo and might even be able to operate solely on solar power.

The most frequently proposed version suggests anchoring a cable to Earth at the equator. In this proposal, the elevator rotates with Earth, so it is stationary relative to Earth. The gravitational force generated by Earth's mass and the centrifugal force created by its rotation keep the elevator cable taut.

The gravitational force is strongest closer to Earth. As you move up along the cable, the gravitational force decreases and the centrifugal force increases. At geostationary orbit, these two forces are equal in strength. At this height, objects experience weightlessness and the cable experiences its maximal tension. The practical limitation on this idea is that the cable may break under this tension. Currently available mass-producible materials are not strong enough for this purpose.

But what if the elevator was anchored to a less-massive celestial body? With a reduced gravitational force, would a space elevator be attainable? This activity will guide you through the mathematics of a space elevator anchored to Earth's moon and extending toward Earth. In this proposal, the cable is kept taut by the competing forces of gravity created by the Moon and gravity created by Earth.

At a certain point along the cable (a Lagrange point), these two forces cancel each other, and one achieves weightlessness. This is the point along the cable where the tension is at its greatest. Would currently available materials be strong enough to sustain this tension? Try the activity to find out!

## Prerequisites

Knowledge of the following concepts is required:

• Newton's laws of motion
• Newton's law of universal gravitation
• Circular motion
• Algebra

## Instructions

Teachers: A PDF of this activity and an answer key are available for classroom use.
1. In our calculations, we will use a coordinate system where the origin is the center of Earth, and the x-axis extends from the center of Earth to the center of the Moon. In this reference frame, the variable x represents the distance from the center of Earth to any point on the line extending to the center of the Moon. This reference frame will rotate with the Earth-Moon system.

We will study what happens at point P at a distance x from the center of Earth. We will use D to indicate the distance from the center of Earth to the center of the Moon, M for the mass of Earth, and m for the mass of the Moon.

Quantity Variables Value
The Moon's Radius r 1,737 km
Distance Between the Center of Earth
and the Center of the Moon
D 384,400 km
Earth's Mass M 5.972 × 1024 kg
The Moon's Mass m 7.348 × 1022 kg

2. A mass placed at a point P between Earth and the Moon will feel the gravitational force generated by both these masses. Use Newton's law of universal gravitation to write a formula for the gravitational force felt by a point mass mtest placed at point P. Use the same sign convention for force as you did for the x-axis (the direction from the center of Earth toward the Moon is positive).

3. Consider the Earth-Moon system as a system of two point masses. Where, along the x-axis, is the center of mass of the Earth-Moon system located?

4. Gravity makes the Earth-Moon frame rotate around its center of mass, as shown in the figure (the red cross in the figure indicates the location of the center of mass of the system). The angular speed is $\omega = \sqrt{\frac{G\left(M +m \right)}{D^3}}$ radians per second, or approximately one full rotation per 27.3 days. Write the formula for the centrifugal force created by this rotation and felt by a point mass mtest placed at point P in the rotating reference frame.

5. Use your knowledge of how forces add up to find the formula for the gravitational and centrifugal forces combined.

6. Instead of a test mass mtest placed at point P, calculate this force (gravitational and centrifugal forces combined) on a tiny section of the cable with a length Δx placed at point P. Assume the cable has a fixed cross-sectional area A, and is made of a material with density ρ.

Hint: Find the formula for the mass of this tiny cylindrical piece of cable, and then substitute this mass in the formula for the force due to the gravitational and centrifugal forces combined.

7. Make a free-body diagram for this tiny section of cable. Make sure to include the gravitational forces of the Moon and Earth, the centrifugal force on the tiny section, and the tension force on each end of the tiny section of cable.

Hint: is the tension on each end of the cable the same or does the tension change along the cable?

8. Since in the rotating reference frame, the piece of cable does not move along the x-axis, there is no net force along the x-direction. In the formula, this becomes:

Where:

$F\left(x\right)$ represents the gravitational and centrifugal forces combined.

$T\left(x\right)$ represents the tension in the cable at the point P with x-coordinate x.

This tells you how the tension changes along the length of the cable. The beginning of the section feels a tension $T\left(x\right),$ the end feels a tension $T\left(x + \Delta x\right).$ Because $\Delta x$ is small, we can write this as $T\left(x\right) + \Delta T\left(x\right)$ where $\Delta T\left(x\right)$ represents the change in tension in the cable at point P.

Use this information to show that $\Delta T\left(x\right)$ must be due to the gravitational and centrifugal forces.

9. Combine the formulas found in steps 6 and 8 to get a formula for the change in tension at each position along the cable (the change of tension as a function of $x$ ).

10. With calculus (using integration), this equation leads to an equation that shows how the tension in the cable varies as a function of x.
11. Since the formula you found in step 9 only provides the change in tension, integration gives an equation for the tension $T\left(x\right)$ including an unknown constant. In order to solve for that constant, we need to know the tension at one point along the cable. We can use the point at the free end of the cable (the point farthest from the Moon and closest to Earth) because the tension at that point is known. What is the tension at that point on the cable?

12. Define the endpoint of the cable as a distance h above the center of Earth.
The equation for $T\left(x\right)$ then becomes:
This seems like a huge equation. We can make it more readable by writing it in the following way: where

Use a computer program like Excel or a graphing calculator to plot how $\varepsilon \left(x\right)$ changes with $x$ , with $x$ between [0, D] or [0, 384,400 km].

13. In order for a cable to remain taut, the tension in the entire cable must always be positive. If the tension in the cable is negative at any point, the cable will collapse. In other words, think about how you can pull on a rope to make it taut, but you cannot push on a rope to make it taut—the rope must always be in positive tension when it is taut.

Use the formula $T\left(x\right)$ and the graph $\varepsilon \left(x\right)$ to estimate the limitations this restriction sets on the maximal value for h. In other words, how far must the elevator reach so that $\varepsilon \left(h\right) - \varepsilon \left(x\right) \geq 0$ for all $x$ -values of points along the physical length of the cable.

Hint: The surface of the Moon is at $x = (D – r) \approx \left(384,400\: km – 1,737\: km\right),$ so the cable reaches from h to a height of 382,663 km. On your graph, you can find that $\varepsilon \left(surface \: of \: the \: Moon\right) \approx 4.$

14. The elevator will also fail if the cable breaks, or if at any point along the line, the stress it experiences is larger than the breaking stress. $\left(Stress\: S\: = \frac{Tension\: T}{Cross\: section\: A}\right).$
Use your graph of $\varepsilon \left(x\right)$ to find the point along the cable where the tension, and thus the stress, is the largest. (Assume a constant cross-section).

Hint: $T \left(x\right)$ will be maximal where $\varepsilon \left(x\right)$ is minimal.

Note: The Background section states that the tension is maximal at the point where the gravitational force of Earth cancels the gravitational force of the Moon (also called the Lagrange point). This is an alternative way to calculate the value for $x$ where $T\left(x\right)$ is maximal.

15. We can now concentrate on the point on the cable at the distance above Earth found in question 14 as that is the point along the cable where the tension is maximal.
Verify that at that point $\varepsilon \left(x\right) = 1.6.$

16. The cable will fail if the stress at that point is larger than the breaking stress of the cable material. Write the formula for the tension on the cable at that point and convert this formula to one for the stress on the cable at that point.

17. Rearrange your answer from question 16 to solve for specific stress $\left(Specific\: stress = \frac{Stress}{Density}\right),$ then look at the right side of the equation. Does the specific stress of the cable at the point where it is at its most depend on the material of which it is made?

18. Specific strength is defined as $\frac{Breaking\: Strength}{Density}$ or $\frac{S_{max}}{\rho}.$ Materials with a larger specific strength are preferred as cable material.

Use what you found in step 17 to explain why scientists use a constraint on specific stress of the material and not a constraint on the stress or the density of the material. Then state what this constraint could be.

19. Table 1 lists a few materials, together with their breaking strength, density, and specific strength.

Table 1
Characteristics of Some Cable Materials
Material Breaking Strength(*)
[ N/m² ]
Density
[ kg/m³ ]
Specific Strength
[Nm/kg]
Steel 5 × 108 80006.25 × 104
Titanium alloy 1.25 × 109 4800 2.6 × 105
Carbon fiber 4.3 × 109 1750 2.5 × 106
Carbon nanotube 6 × 1010 ~1000 6 × 107

* Note: Most metals—like steel, aluminum, and titanium alloy—are ductile, meaning they will stretch permanently before breaking (like bending a paper clip; it permanently deforms before it breaks). The yield strength is the maximum amount of stress that a material can bear without showing irreversible deformation. Typically, metal structures are designed to stay well below the yield strength, not the breaking strength. However, breaking strength and yield strength are usually the same order of magnitude, so this is a good approximation.

Calculate the specific stress experienced by the cable at the point where tension is maximal. Then, use the information from the table to evaluate if it exceeds the breaking strength of each material. Use the value $\varepsilon \left(h\right) = 4,$ the smallest value for $\varepsilon \left(h\right)$ obtained in step 13, for your calculations.

20. Make a conclusion: What materials could be used to make a space elevator stretching from the Moon toward Earth?

## What Happened?

The calculations explore a space elevator extending from the Moon. The calculations should have shown that the cable needs to extend to at least 76,880 km from the center of Earth for Earth's gravitational force to keep the cable taut. If you arrived at a different answer, check your calculations against our answer key.

The calculations show that for the cable to be able to support its own weight, it must be made from a material that is light and strong. Carbon fiber, as well as carbon nanotubes, are strong enough and, at the same time, light enough to support a cable with a constant cross-section.

## Digging Deeper

In the calculations, we assumed a cable with a constant cross-section. More-advanced designs use a tapered cable, one that is thin at the ends where tension is low and becomes thicker as it nears the place of maximal tension. Other designs use a combination of these scenarios. Although these designs yield more-attainable results, the general conclusion holds: at the time this activity was written, mass-producible materials like carbon fiber can hold up a space elevator anchored on the Moon, but are not strong enough to hold up a space elevator anchored on Earth.

## For Further Exploration

• Go through the calculations for a space elevator anchored on Earth. You can neglect any gravity generated by the Moon in this approach and set the tension at both ends of the cable to 0. What materials are strong enough to hold up this fixed cross-section cable?
• What would be the constraints of a space elevator anchored to Mars or another planet?
• What happens if you add a counterweight (a large point mass) at the end of an Earth-anchored or Mars-anchored cable?

### Activities

STEM Activity
1
2
3
4
5
3 reviews
Do words like "general relativity", "gravity well", and "space-time continuum" sound scary? Don't worry, you don't have to be Albert Einstein to understand them! Try this fun activity to learn about these concepts and black holes, using some common household materials. Read more

### Careers

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Astronomers think big! They want to understand the entire universe—the nature of the Sun, Moon, planets, stars, galaxies, and everything in between. An astronomer's work can be pure science—gathering and analyzing data from instruments and creating theories about the nature of cosmic objects—or the work can be applied to practical problems in space flight and navigation, or satellite communications. Read more
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Mathematicians are part of an ancient tradition of searching for patterns, conjecturing, and figuring out truths based on rigorous deduction. Some mathematicians focus on purely theoretical problems, with no obvious or immediate applications, except to advance our understanding of mathematics, while others focus on applied mathematics, where they try to solve problems in economics, business, science, physics, or engineering. Read more

## Reviews

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# PAT 1140. Look-and-say Sequence (20) – 甲级

Look-and-say sequence is a sequence of integers as the following:
D, D1, D111, D113, D11231, D112213111, …
where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, the 2nd number means that there is one D in the 1st number, and hence it is D1; the 2nd number consists of one D (corresponding to D1) and one 1 (corresponding to 11), therefore the 3rd number is D111; or since the 4th number is D113, it consists of one D, two 1’s, and one 3, so the next number must be D11231. This definition works for D = 1 as well. Now you are supposed to calculate the Nth number in a look-and-say sequence of a given digit D.
Input Specification:
Each input file contains one test case, which gives D (in [0, 9]) and a positive integer N (<=40), separated by a space.
Output Specification:
Print in a line the Nth number in a look-and-say sequence of D.
Sample Input:
1 8
Sample Output:
1123123111

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❤ 点击这里 -> 订阅PAT甲级乙级、蓝桥杯、GPLT天梯赛、LeetCode题解离线版


text:
- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnlilne Submission ㆍMy Manuscript - For Reviewers - For Editors
 Conformal deformation on a semi-Riemannian manifold (I) Bull. Korean Math. Soc. 2001 Vol. 38, No. 2, 223-230 Yoon-Tae Jung and Soo-Young Lee Chosun University, Chosun University Abstract : In this paper, we considered the uniqueness of positive time-solution to equation $\Box_{g} u(t,x) - c_n u(t,x) + c_n u(t,x)^{n+3 \over n-1 } = 0,$ where $c_n = {n-1 \over 4n}$ and $\Box_{g}$ is the d'Alembertian for a Lorentzian warped manifold $M = [a, \infty) \times_f N$. Keywords : conformal deformation, Lorentzian warped manifold, scalar curvature MSC numbers : 53C21, 53C50, 58C35, 58J05 Downloads: Full-text PDF


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A Appendix

# Functional A Posteriori Error Estimates for Elliptic Problems in Exterior Domains

## Abstract

This paper is concerned with the derivation of computable and guaranteed upper bounds of the difference between the exact and the approximate solution of an exterior domain boundary value problem for a linear elliptic equation. Our analysis is based upon purely functional argumentation and does not attract specific properties of an approximation method. Therefore, the estimates derived in the paper at hand are applicable to any approximate solution that belongs to the corresponding energy space. Such estimates (also called error majorants of the functional type) have been derived earlier for problems in bounded domains of (see [2, 3]).
Key Words A posteriori error estimates of functional type, elliptic boundary value problems in exterior domains
AMS MSC-Classifications 65 N 15

## 1 Introduction

The main focus of our investigations is to suggest a method of deriving guaranteed and computable upper bounds of the difference between the exact solution of an elliptic exterior domain boundary value problem and any approximation from the corresponding energy space. We discuss the method with the paradigm of the prototypical elliptic problem

 −divA∇u =f in Ω,\makebox[0.0pt][] (1.1) u|γ =g on γ:=∂Ω.\makebox[0.0pt][] (1.2)

We assume that with is an exterior domain, i.e. is compact, with Lipschitz continuous boundary (see Figure 1).

Throughout this paper we will use the weighted Lebesgue function spaces

 L2s(Ω):={φ∣ρsφ∈L2(Ω)},s∈R.

Here and denotes the radius vector. is a Hilbert space equipped with the scalar product

 ⟨φ,ψ⟩s,Ω:=⟨ρsφ,ρsψ⟩Ω:=∫Ωρ2sφψd% λ,

where and belong to and is Lebesgue’s measure. We denote the corresponding norms by . If then coincides with the usual Lebesgue space . For the sake of simplicity we keep the same notation for spaces of vector-valued functions. Moreover, we introduce the weighted Sobolev space

 H1−1(Ω):={φ∈L2−1(Ω)∣∇φ∈L2(Ω)},

which is a Hilbert space as well with respect to the scalar product

 Missing or unrecognized delimiter for \left

By we denote the closure of , the space of compactly supported smooth test functions, in the norm of . Whenever we consider Sobolev spaces of bounded domains we use the usual unweighted -scalar products and -norms.

For dimensions the solution theory for the problem (1.1)-(1.2) is based on the weighted Poincare/Friedrich estimate (see Corollary 16 (i) and Remark 17 of the appendix)

 Missing dimension or its units for \hskip (1.3)

the Lax-Milgram theorem and, if needed, an adequate extension operator for the boundary data. Let be some function in satisfying the boundary condition (1.2). The weak solution of (1.1)-(1.2) is then defined by the variational formulation

 ⟨A∇u,∇w⟩Ω =⟨f,w⟩Ω∀w∈∘H1−1(Ω).\makebox[0.0pt][] (1.4)

By (1.3) the left hand side of (1.4) is a strongly coercitive sesqui-linear form over provided that the real-matrix-valued function is measurable, bounded a.e., symmetric and uniformly strongly elliptic, i.e.

 ∃cA>0∀ξ∈RN∀x∈ΩcA|ξ|2≤A(x)ξ⋅ξ.\makebox[0.0pt][] (1.5)

If then by the Cauchy-Scharz inequality the right hand side of (1.4) is a linear and continuous functional over . Thus, under these assumptions the problem (1.4) is uniquely solvable in by Lax-Milgram’s theorem.

If one can apply the same arguments with the difference that (1.3) has to be modified. For and, for example, we have by Corollary 16 (iii) and Remark 17

 ||φ||−1,Ω≤2∣∣∣∣φ′∣∣∣∣Ω∀φ∈∘H1−1(Ω).\makebox[0.0pt][] (1.6)

Hence, we get the same solution theory with tiny restrictions on , which easily can be removed by a translation. For the singularities are stronger and additionally we have to utilize logarithmic terms. By Corollary 16 (ii) and Remark 17 we have for domains , such that the complement contains the unit ball,

 Missing dimension or its units for \hskip (1.7)

where

 H1−1,ln(Ω):={φ∣φ/(rlnr),∇φ∈L2(Ω)}

is a Hilbert space equipped with the natural scalar product

 (φ,ψ)↦⟨φ/(rlnr),ψ/(rlnr)⟩Ω+⟨∇φ,∇ψ⟩Ω

and again denotes the closure of in the norm of . Consequently, we obtain for all with and all in satisfying the boundary condition (1.2) a unique solution belonging to .

We summarize the results in the following

###### Theorem 1

Let as well as and satisfying the boundary condition (1.2). Then the exterior boundary value problem (1.1)-(1.2) is uniquely weakly solvable in . The solution operator is continuous.

From the above discussion, it is clear that for the existence of weak solutions in suitable spaces can also be proved.

###### Remark 2

The boundary data and its extension can be described in more detail. In the bounded domain case it is well known that there exists a bounded linear trace operator and a corresponding bounded linear extension operator (right inverse) mapping to and vice verse. Hence, by restriction we get a bounded linear trace operator

 τγ:H1−1(Ω)→H1/2(γ)

and by extension and applying an obvious cutting technique we obtain a bounded linear extension operator

 E:H1/2(γ)→H1−1(Ω)

for our exterior domain , which even maps to functions with (arbitrarily thin) compact support. As in the bounded domain case, is a right inverse of . Then we may specify and as well as our variational formulation for : Find , such that

 Missing or unrecognized delimiter for \right

Finally, we introduce

 D(Ω):={φ∈L2(Ω)∣divφ∈L21(Ω)},

which is a Hilbert space with respect to the canonical scalar product

 Missing or unrecognized delimiter for \left

## 2 Upper bounds for the deviation from the exact solution in dimensions N≥3

Let be an approximation of , where is assumed just to belong to since the boundary condition may not be satisfied exactly. Our goal is to obtain upper bounds for the difference between and in terms of the norm

 Missing or unrecognized delimiter for \left

We use (1.4) and get for all

 ⟨A∇(u−v),∇w⟩Ω =⟨f,w⟩Ω−⟨A∇v,∇w⟩Ω.\makebox[0.0pt][] (2.1)

Before we proceed we note two useful results.

###### Theorem 3

Let be as above. Moreover, let be a linear and continuous functional over and , such that for all

 ⟨A∇(u−v),∇w⟩Ω=Φ(w)≤cΦ||∇w||A,Ω

holds. Then

 ||∇(u−v)||A,Ω ≤cΦ+2||∇(^u−^v)||A,Ω\makebox[0.0pt][] (2.2) for all ^u,^v∈H1−1(Ω), for which ^u−^v coincides with u−v on the boundary γ. If additionally u−v belongs to ∘H1−1(Ω) then ||∇(u−v)||A,Ω ≤cΦ.\makebox[0.0pt][] (2.3)

Proof We consider

 w:=u−v−(^u−^v)∈∘H1−1(Ω).

Using Cauchy-Schwarz’ inequality we obtain

 ||∇w||2A,Ω Missing or unrecognized delimiter for \left ≤(cΦ+||∇(^u−^v)||A,Ω)||∇w||A,Ω

and thus . By the triangle inequality we get (2.2). (2.3) is trivial since we can set , i.e. .

We may be more specific using the trace and extension operators from Remark 2.

###### Corollary 4

Let the assumptions of Theorem 3 be satisfied. Then

 Missing dimension or its units for \hskip

Here is the constant in the inequality

 Missing dimension or its units for \hskip (2.4)

Proof Setting and as well as using (2.4) proves the inequalities. We note that (2.3) follows directly from the corollary as well.

In the subsequent sections we introduce and discuss some different functionals and corresponding constants .

### 2.1 First estimate

For any and any we have

 Missing or unrecognized delimiter for \left =0.\makebox[0.0pt][] (2.5)

Combining (2.1) and (2.5) we obtain for all and all

 ⟨A∇(u−v),∇w⟩Ω Missing or unrecognized delimiter for \left (2.6)

By Cauchy-Schwarz’ inequality, (1.3) with and (1.5) we estimate the right hand side of (2.6) as follows:

 ∣∣⟨f+divy,w⟩Ω∣∣ Missing dimension or its units for \hskip (2.7) ≤cN√cA||f+divy||1,Ω||∇w||A,Ω ∣∣⟨y−A∇v,∇w⟩Ω∣∣ ≤||y−A∇v||A−1,Ω||∇w||A,Ω\makebox[0.0pt][] (2.8)

By Corollary 4 we arrive at the following result.

###### Proposition 5

Let be as in Theorem 3. Then

 \makebox[0.0pt][]||∇(u−v)||A,Ω≤cN√cA||f+divy||1,Ω+||y−A∇v||A−1,Ω+2cγ∣∣∣∣g−τγv∣∣∣∣H1/2(γ), (2.9)

where is an arbitrary vector field in .

###### Remark 6

If satisfies the prescribed boundary condition, then (2.9) implies

 \makebox[0.0pt][]||∇(u−v)||A,Ω≤cN√cA||f+divy||1,Ω+||y−A∇v||A−1,Ω. (2.10)

The estimates (2.9) and (2.10)) show that deviations from exact solutions of exterior boundary value problems have the same structure as for problems in bounded domains, namely they contain weighted residuals of basic relations with weights given by constants in the corresponding embedding inequalities.

### 2.2 Second estimate

Assume that is decomposed into two subdomains and with interface (see Figure 1) and that the fields exactly satisfy the relation

 divy+f=0in Ωe.\makebox[0.0pt][] (2.11)

In particular, this situation may arise if the source term has compact support and is represented (in the exterior domain ) as a linear combination of solenoidal fields having proper decay at infinity. In this case, the estimate of Proposition 5 turns trivially to

 Missing dimension or its units for \hskip (2.12)

which holds for all additionally satisfying (2.11), where the weight constant is

 \makebox[0.0pt][]co:=cN(1+||r||∞,Ωi)√cA, (2.13)

which follows directly from

 Missing dimension or its units for \hskip

But we also may derive another estimate. We rewrite (2.7) and use Cauchy-Schwarz’ inequality in

 Missing or unrecognized delimiter for \left (2.14)

and estimate

 ||w||Ωi≤cΩi||∇w||Ωi≤cΩi√cA||∇w||A,Ω.\makebox[0.0pt][] (2.15)

Here denotes a Poincare/Friedrich constant associated with the bounded domain , i.e. the best constant of the inequality

 ||φ||Ωi≤cΩi||∇φ||Ωi∀φ∈{ψ∈H1(Ωi)∣τ∂Ωiψ|γ=0 on γ},

where denotes the trace operator. In this case, we have again (2.12) but now with the (optional) weight constant

 \makebox[0.0pt][]co:=cΩi√cA. (2.16)

We note that the constant (2.13) may also be achieved by (2.7) and the argument (2.14) if we replace the estimate (2.15) by

 ||w||Ωi ≤(1+|r|∞,Ωi)||w||−1,Ωi≤(1+|r|∞,Ωi)||w||−1,Ω≤cN√cA(1+|r|∞,Ωi)||∇w||A,Ω.

We summarize and get our second a posteriori error estimate.

###### Proposition 7

For all with (2.11) we have

 ||∇(u−v)||A,Ω≤co||f+divy||Ωi+||y−A∇v||A−1,Ω+2cγ∣∣∣∣g−τγv∣∣∣∣H1/2(γ),

where is defined either by (2.13) or by (2.16).

###### Remark 8

In general, the number will be smaller and thus provides a better bound than . On the other hand, the number is an easily computable upper bound for the best possible constant .

### 2.3 Third estimate

Let and be the restrictions of some to and , respectively. Assuming and but not necessarily we use the equations

 Missing or unrecognized delimiter for \left =⟨τn,Γyi,τΓw⟩Γ,\makebox[0.0pt][] (2.17) Missing or unrecognized delimiter for \left =−⟨τn,Γye,τΓw⟩Γ,\makebox[0.0pt][] (2.18)

which hold for all and in the sense of the traces and respectively . At this point we assume that the interface is Lipschitz (in order to guarantee that the traces are well defined). By we denote the duality product of and . We recall that the normal traces and possess weak surface divergences in as well. If , then in and in . Hence, in this case adding (2.17) and (2.18) we obtain by (2.5)

 Missing or unrecognized delimiter for \right

for all . Therefore, we get

 τn,Γyi=τn,Γye

for all since is surjective.

On our way to find like in (2.6) we now insert (2.17), (2.18) instead of (2.5) into (2.1) and obtain

 ⟨A∇(u−v),∇w⟩Ω =⟨f+divyi,w⟩Ωi+⟨f+divye,w⟩Ωe\makebox[0.0pt][] (2.19) Missing or unrecognized delimiter for \right

The third term of will be estimated by (2.8) and for the last term we may use the continuity of the trace operator in combination with a Poincare/Friedrich estimate, i.e.

 ||τΓφ||H1/2(Γ)≤cΓ||∇φ||A,Ω∀φ∈∘H1−1(Ω),\makebox[0.0pt][] (2.20)

and obtain

 ∣∣⟨τn,Γye−τn,Γyi,τγw⟩Γ∣∣ ≤||τn,Γye−τn,Γyi||H−1/2(Γ)||τΓw||H1/2(Γ)\makebox[0.0pt][] (2.21) ≤cΓ||τn,Γye−τn,Γyi||H−1/2(Γ)||∇w||A,Ω.

To estimate the second term of we again use (1.3) and (1.5) and obtain

 ∣∣⟨f+divye,w⟩Ωe∣∣ Missing dimension or its units for \hskip (2.22) Missing dimension or its units for \hskip

Considering the first (and last) term of we have once more at least two options as in section 2.2 to obtain the estimate

 ∣∣⟨f+divyi,w⟩Ωi∣∣ ≤co||f+divyi||Ωi||∇w||A,Ω\makebox[0.0pt][] (2.23)

with defined either by (2.13) or (2.16).

Finally with (2.19) and (2.8), (2.21), (2.22), (2.23) we get by Corollary 4 the third estimate.

###### Proposition 9

For all with and we have

 \makebox[0.0pt][]||∇(u−v)||A,Ω ≤co||f+divyi||Ωi+cN√cA||f+divye||1,Ωe+||y−A∇v||A−1,Ω (2.24) +cΓ||τn,Γye−τn,Γyi||H−1/2(Γ)+2cγ∣∣∣∣g−τγv∣∣∣∣H1/2(γ)

with from Proposition 7. The right hand side of (2.24) vanishes if and only if coincides with and with .

###### Remark 10

There are many ways to deduce (2.20). We just mention that can be considered as a trace of a function defined in or or even of a function, which is just defined in a small neighborhood of . Thus, we may adjust the constant according to our needs.

###### Remark 11

This estimate suggests even a solution method: We construct approximations using locally supported trial functions in , e.g. FEM, and utilize global approximations properly behaving at infinity for . These two types of approximations are usually difficult to meet together exactly on the artificial boundary . However, Proposition 9 shows that this is not required because we can use instead the penalty term with known penalty factor . In addition, we have one more parameter, the ‘radius’ of the interface . Since is artificial and arbitrary we can use this parameter in the algorithm in order to obtain better results.

###### Remark 12

At this point we shall note that all our estimates are sharp, which easily can be seen by setting and .

###### Remark 13

In Propositions 5, 7, 9 we can always replace the last summand of the right hand side by or using Theorem 3 and Corollary 4.

## 3 Upper bounds in dimension N=2

Of course, Theorem 3 holds for as well and the modifications on the estimates depend just on the Poincare/Friedrich estimate and thus they are obvious using the proper Cauchy-Schwarz inequality. We achieve

###### Proposition 14

Let , such that contains the unit ball.

• For all

 ||∇(u−v)||A,Ω≤2√cA||rlnr(f+divy)||Ω+||y−A∇v||A−1,Ω+2cγ∣∣∣∣g−τγv∣∣∣∣H1/2(γ).
• For all with in

 ||∇(u−v)||A,Ω≤co||f+divy||Ωi+||y−A∇v||A−1,Ω+2cγ∣∣∣∣g−τγv∣∣∣∣H1/2(γ),

where .

• For all with and

 ||∇(u−v)||A,Ω Missing dimension or its units for \hskip +cΓ||τn,Γye−τn,Γyi||H−1/2(Γ)+2cγ∣∣∣∣g−τγv∣∣∣∣H1/2(γ).

Analogously, Remarks 8, 10, 11, 12, 13 hold.

## Appendix A Appendix

### a.1 Lower bounds for the error

We note by a standard variational argument

 ||∇(u−v)||2A,Ω Missing or unrecognized delimiter for \left

Thus, we obtain for all the estimate

 ||∇(u−v)||2A,Ω ≥2⟨A∇(u−v),∇w⟩Ω−||∇w||2A,Ω Missing or unrecognized delimiter for \left

which is sharp since one can put . But to exclude the unknown exact solution from the right hand side we need since then by (1.4)

 ||∇(u−v)||2A,Ω ≥2⟨f,w⟩Ω−⟨A∇(2v+w),∇w⟩Ω.\makebox[0.0pt][] (A.1)

But this estimate is no longer sharp because we can not put anymore. In fact, with and we get for

 Missing or unrecognized delimiter for \right

Hence, we obtain the estimate

 ||∇(u−v)||2A,Ω ≥2⟨f,w⟩Ω−⟨A∇(2v+w),∇w⟩Ω+2⟨τn,γA∇u,τγw⟩γ

for all , which is sharp and coincides with (A.1) if . But the unknown exact solution still appears on the right-hand side, i.e. the normal trace of on . Furthermore, if then (A.1) can not be sharp.

### a.2 Poincare type estimates for exterior domains

We introduce the radial derivative , where . Furthermore, and denote the open ball and sphere of radius centered at the origin in , respectively. We will use the ideas of [4, Lemma 4.1] and [1, Poincare’s estimate III, p. 57] with some minor useful modifications.

###### Lemma 15

Let , , be a domain and . For all the following Poincare estimates hold:

• If then

 (2β+N−2)∣∣∣∣rβ−1u∣∣∣∣Ω≤2∣∣∣∣rβ∂ru∣∣∣∣Ω.
• Let . If or then

 Missing dimension or its units for \hskip
• If then

 |2β−1|∣∣∣∣(1+r)β−1u∣∣∣∣Ω≤2∣∣∣∣(1+r)β∂ru∣∣∣∣Ω+|2min{0,2β−1}|1/2|u(0)|,

where will be extended by zero to .

For the estimates derived in this paper it suffices to set . In this particular case, the above lemma implies

###### Corollary 16

Let , , be a domain. For all the following Poincare estimates hold:

• If then

 Missing dimension or its units for \hskip
• If and then

 ||u/(rlnr)||Ω≤2||∂ru||Ω≤2||∇u||Ω.
• If then

 Missing dimension or its units for \hskip

Hence, if we have

 ||u||−1,Ω≤||u/(1+r)||Ω≤2||∂ru||Ω≤2∣∣∣∣u′∣∣∣∣Ω.
###### Remark 17

Of course, by continuity all these estimates extend to appropriate weighted -Sobolev spaces.

Proof Let , , be a domain and . By partial integration we get for all and

 2∫Ω∖Bεrαu∂rudλ =∫Ω∖Bεrα∂r|u|2dλ =−(α+N−1)∫Ω∖Bεrα−1|u|2dλ−εα∫Sε|u|2dσ.

Thus, for all and

 ∣∣∣∣rβ∂ru+γrβ−1u∣∣∣∣2Ω∖Bε =∣∣∣∣rβ∂ru∣∣∣∣2Ω∖Bε+|γ|2∣∣∣∣rβ−1u∣∣∣∣2Ω∖Bε+2γ⟨rβ∂ru,rβ−1u⟩Ω∖Bε=∫Ω∖Bεrαu∂rudλ =∣∣∣∣rβ∂ru∣∣∣∣2Ω∖Bε+γ(γ−2β−N+2)∣∣∣∣rβ−1u∣∣∣∣2Ω∖Bε−γε2β−1∫Sε|u|2dσ.

Now the left hand side of this equality converges by the monotone convergence theorem. Since , if and only if , and the right hand side converges for by Lebesgue’s dominated convergence theorem in . Hence, for we obtain

 ∣∣∣∣rβ∂ru+γrβ−1u∣∣∣∣2Ω=∣∣∣∣rβ∂ru∣∣∣∣2Ω+γ(γ−2β−N+2)∣∣∣∣rβ−1u∣∣∣∣2Ω.

Choosing we finally get by the triangle inequality

 γ∣∣∣∣rβ−1u∣∣∣∣Ω≤2∣∣∣∣rβ∂ru∣∣∣∣Ω.

Since we are especially interested in the case this estimate is only applicable in dimensions .

For we proceed as follows: For all we have

 2∫R±(1+r)αu∂rudλ=±2∫R±(1±t)αu(t)u(t)′


text:
# Find the Standard Form 4x-3y+3=0

4x-3y+3=0
The standard form of a linear equation is Ax+By=C.
Subtract 3 from both sides of the equation.
4x-3y=-3
Find the Standard Form 4x-3y+3=0

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text:
# To find last two digits of $2^{100}$

To find last two digits of $2^{100}$

I have just learned about modular arithmetic and I wanted to solve this problem. I only know about equivalence classes and about $a=b \pmod n$. I have also learned about multiplication and addition of classes. Can someone explain to me step by step on how to apply modular arithmetic to this problem?

Thanks

• – lab bhattacharjee Jul 31 '16 at 7:52
• @labbhattacharjee i need to know why and how of method and not the mechanical procedure – Gathdi Jul 31 '16 at 7:53
• Gathdi, just try it out. Experiment what happens with exponents smaller than $100$. You will find a period. It is not very short, but not excessively long either. After you see what happens, you can then take a look at it, and see what modular arithmetic is necessary to justify the result. – Jyrki Lahtonen Jul 31 '16 at 7:57
• @JyrkiLahtonen, +1 – Andreas Caranti Jul 31 '16 at 11:17
• @JyrkiLahtonen What do you mean by period? – Gathdi Jul 31 '16 at 14:58

Method 1(using fast power algorithm):

$2^0 \equiv 1 \pmod {100}$

$2^1 \equiv 2 \pmod {100}$

$2^2 \equiv 4 \pmod {100}$

$2^4 \equiv 16 \pmod {100}$

$2^8 \equiv 56 \pmod {100}$

$2^{16} \equiv 36 \pmod {100}$

$2^{32} \equiv 96 \pmod {100}$

$2^{64} \equiv 16 \pmod {100}$

Since $100 = 4 + 32 + 64$

$2^{100} \equiv 2^{4} 2^{32} 2^{64}\equiv 16 \times 96 \times 16 \equiv 76 \pmod {100}$

Method 2(using minimal cycle to improve):

Since $2^{22} \equiv 4 \pmod {100}$, $2^{100} \equiv 2^{22 \times 4 + 12} \equiv 4^4 \times 2^{12} \equiv 2^{20} \equiv 16 \times 96 \equiv 76 \pmod{100}$

Last but not least: Since $2^{20} \equiv 76 \pmod {100}$, $76 \times 76 \equiv 76 \pmod{100}$, $2^{100} = 2^{20 \times 5} \equiv 76^5 \equiv 76 \pmod{100}$ by induction.

$$2^{100} \equiv 4^{50} \equiv 0 \pmod {4}$$ $$2^{100} \equiv 1024^{10} \equiv (-1)^{10} \equiv 1 \pmod {25}$$

So what number less than $100$ is $0 \pmod 4$ and $1 \pmod {25}$ ?

$2^{100}\bmod{100}=$

$2^{10\cdot10}\bmod{100}=$

$(2^{10})^{10}\bmod{100}=$

$1024^{10}\bmod{100}=$

$24^{10}\bmod{100}=$

$24^{3\cdot3+1}\bmod{100}=$

$(24^{3})^{3}\cdot24\bmod{100}=$

$13824^{3}\cdot24\bmod{100}=$

$24^{3}\cdot24\bmod{100}=$

$13824\cdot24\bmod{100}=$

$24\cdot24\bmod{100}=$

$576\bmod{100}=$

$76\pmod{100}$

• $2^{3 \cdot 3 + 1}$ should probably be $24^\cdots$ – mathreadler Jul 31 '16 at 11:20
• @mathreadler: Thank you, added the missing digit :) – barak manos Jul 31 '16 at 13:11

Use the Chinese remainder theorem: $$\mathbf Z/100\mathbf Z\simeq \mathbf Z/4\mathbf Z\times\mathbf Z/25\mathbf Z.$$

Now $2^{100}\equiv 0\mod 4$, and $\;2^{100}=(2^{20})^5\equiv 1^5=1\mod 25\;$ by Euler's theorem.

So we have to solve the system of linear congruences: $\;\begin{cases} x\equiv 0\mod4,\\x\equiv 1\mod 25.\end{cases}$

From the Bézout's relation $\;25-6\cdot 4=1$, we deduce the solutions: $$x\equiv 0\cdot 24-1\cdot 6\cdot 4=-24\equiv 76\mod 100.$$

$2^{10}=1024$

$2^{100}=1024^{10}$

$2^{11} \equiv48$

$2^{12}\equiv96\equiv-2^2$

$2^{100}\equiv2^{12*8+2^4}\equiv-{2^2}^8*2^4$

$=-2^{20}=-2^2*256=-24\equiv76 \pmod {100}$

another method

By euler's function, $2^{\phi(25)}=2^{20}\equiv1 \pmod{25}$

$2^{100}\equiv1,26,51,76 \pmod{100}$

$2^{100}$ is multiple of 4, so $76$.

${\rm mod}\ 25\!:\ \ \ \ \color{#0a0}{2^{\large 10}} = 1024\equiv \color{#0a0}{\bf -1}\,\Rightarrow$ $\,\color{#c00}{2^{\large 90}}\equiv (\color{#0a0}{2^{\large 10}})^{\large 9}\equiv (\color{#0a0}{\bf -1})^{\large 9}\equiv \color{#c00}{\bf -1}$

${\rm mod}\ 100\!:\,\ 2^{\large 100} = 2^{\large 10} \color{#c00}{2^{\large 90}} = 2^{\large 10}(\color{#c00}{\bf -1}+25n) \equiv \color{#c00}{\bf -}2^{10}\!+0\equiv -1024\equiv -24$

• All mental arithmetic, no use of Euler/Fermat or CRT etc (but of course you should learn those methods because they will be essential for more general problems). – Bill Dubuque Jul 31 '16 at 14:54


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Public Group

# '...' in C++.

This topic is 4496 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

int printf(const char *buffer, ...); // Question.

How do you make use of the '...' in functions? Printf uses it... Any websites with info? Thanks!

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In this particular example, the ellipsis means there are optional, follow-on arguments that can be passed.

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You don't, because "..." is not C++. Using the dread ellipses construct "..." in C++ is punishable by death or just a beating handed to you by Fruny or Zahlman. If you want something like that you'll probably have to use streams.

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Quote:
 How do you make use of the '...' in functions? Printf uses it...

you try not to. this construct is rarley seen in c++.

but look up this, if you want to

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To be a little more informative: A function declared this way is called a "variadic" function. To make them work you need to do some work with some rather nasty macros (although it's not a very complicated process). Basically what happens is that all the things in that are passed get tossed onto the stack in order, regardless of their types or sizes, and it's up to the function to figure out how many things are there and what their types and sizes are. Since there isn't a reasonable way to do this in general, usually what these functions do is accept some information in the first argument (such as a printf-style 'format string') that is then interpreted to figure out what needs to go on.

As I'm sure you can imagine, this is tricky to get right, and it also can cause all kinds of undefined behaviour if the format string doesn't properly match the provided data (not to mention the annoyance of having to put it in in the first place, for those functions that *don't* format a string but just want extra arguments). It also just plain doesn't work with lots of C++ types - basically anything that isn't a POD type, because copy constructors aren't going to be invoked properly. Good luck dealing with references, too.

So yeah, as a rule of thumb, just don't freakin' do it, or even think about it. Set up some kind of streaming interface instead. If you do need printf-style formatted I/O (e.g. because you need "compound strings" for localization, where the order of outputted things might vary by language, and you don't want to have to recompile/change code to support it), you should take a look at boost::format.

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This seems relevant (as found in several signatures):

"Basically whenever you invoke the dread ellipses construct you leave the happy world of type safety." -- SiCrane

It should not be used in c++, and only very carefully in c.

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Now, if you are still really curious, here's an example:

void logMessage( const char* szMessage, ... )
{
static I8 szBuffer[4097];
va_list list;
va_start( list, szMessage );
vsprintf( szBuffer, szMessage, list );

if (LogManager::InstancePtr())
LogManager::Instance().logMessage (szBuffer);

va_end( list );
}

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Now now, the ellipses construct does have a place; so does goto. You just don't want to use it unless you're really, really sure you need it. In short, someone posting in 'For Beginners' with a question of the form 'How do you...?' is probably a good candidate for passing a linked list instead. :)

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Quote:
 Original post by King of MenNow now, the ellipses construct does have a place; so does goto. You just don't want to use it unless you're really, really sure you need it. In short, someone posting in 'For Beginners' with a question of the form 'How do you...?' is probably a good candidate for passing a linked list instead. :)

Huh?

Oh no no no, I'm not a beginner, definately not.

I have a templated doubly linked list with id's and iterators, and it's pretty good :).

Then, I have SDL/OpenGL graphics classes to handle sprites, text, stuff. I have an egine class, and event class, and so much more.

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I just love being able to contribute to a thread without ever having posted in it. However...
Quote:
 Original post by ZahlmanIt also just plain doesn't work with lots of C++ types - basically anything that isn't a POD type, because copy constructors aren't going to be invoked properly.

Actually, what happens on most compilers is the the copy construction of the argument proceeds correctly. It's just that the destructor isn't called. So not only are you leaving the happy world of type safety, you also have a potential memory leak every time you call a function with the ellipses construct, because you may have accidently passed a class type to it. Though I have noticed on one occassion, that had horrible function pointer typecasting involved, the opposite happening: the copy constructor wasn't called and the destructor was called. However, I'm not sure that the ellipses had anything to do with that, since the code was basically a walking undefined behavior.

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### 12.5 The tegprog environment and the tegrun macro

The texgraph.sty package also provides the environment:

\begin{tegprog}[<options>]{name}
<code TeXgraph>
\end{tegprog}

That environment saves the program <name>.teg. That program is created to be executed by the command tegrun, the parameters that will be used will be in the global variable param (list) of the program. That program also has a macro Return(string), that is writing the string in an output file. That file will be automatically included by the command tegrun.

Possible options are:

• file: that option shows that the inside of the environment is a whole TeXgraph source file, if not, this is only a command.
• commandchars: with that option, the environment may contain calls to TEX commands if the backslash $\$ is replaced by # before the commands name, ex: #command{...}. If this command contains macros that must not be developped, they will have to be preceeded by \noexpand.
• preload = < {"<file1>";"<file2>";...} >: loads one or several packages before creating the graphic, eg: preload={"papiers.mod";"draw2d.mod"}.

Once saved, a program can be executed in a TEX document with the command:

\tegrun{name}{param1 param2 ...}, it is saving the parameters in the file <name>.prm, ask TeXgraph to execute the program <name>.teg, and include the result file <name>.res. Here is an example:

\begin{tegprog}{PrintPgcd}
a:=param[1], b:=param[2],
if a<b then Echange(a,b) fi,
Return("\begin{tabular}{|c|c|c|}\par\hline{}a&b&r\tabularnewline\hline"),
r:=b,
while r>0 do
r:=mod(a,b),
Return(Concat(a,"&",b,"&",r,"\tabularnewline\hline")),
a:=b, b:=r
od,
Return("\end{tabular}")
\end{tegprog}
\newcommand{\PrintPgcd}[2]{\tegrun{PrintPgcd}{#1 #2}}%

In that example, the program PrintPgcd.teg is created, it calculates the gcd of two integers $a$ and $b$ by giving all the steps of the Euclid algorithm in the form of a tabular. The parameters list is in the variable param1 . The macro Return write in the output file whose name is PrintPgcd.res.

Then a two arguments macro called PrintPgcd is created. That macro calls the command \tegrun{PrintPgcd}{#1 #2}, that command write the two arguments in the parameter file PrintPgcd.prm, and ask TeXgraph to execute the program PrintPgcd.teg, and finally includes the file PrintPgcd.res.

The execution of \PrintPgcd{456}{166} gives

 a b r 456 166 124 166 124 42 124 42 40 42 40 2 40 2 0
.


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# Tag Info

## Hot answers tagged images

30

Try to keep as much of your answer in text as possible; obviously, some diagrams cannot be replicated on the site and need to be posted as images and it is certainly appropriate to do so. However, these diagrams should be split off from the rest of the text so that as much as possible remains in plain text. There's a number of reasons to do this, but the ...

19

There's a meta.SO thread on this issue, though perhaps it's out of date. The advice there is essentially the same, though they suggest using repeated  's or spaces inside a code block to achieve the same effect. So I suppose that's better than my suggestion below in the sense that it doesn't depend on MathJax, but it still isn't automatic and nice. ...

19

It can be done in MathJax, but it is painful and a complete hack. Here is one way: \require{enclose} \def\uline#1#2{\enclose{updiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} \def\dline#1#2{\enclose{downdiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} % \def\place#1#2#3{\smash{\rlap{\hskip{#1em}\raise{#2em}{#3}}}} % \hskip 1em % \place{0}{12}{\bullet} ...

17

Yes. The advantages of replacing images with MathJax and MarkDown are: It provides searchable text; It is more compatible with accessibility features; It is easier to read; It provides MSE with a more uniform look-and-feel; It does not depend on servers and services beyond StackExchange's control. That said, please improve the post in every other aspect ...

17

My thought process: If it's TeX-ifiable, do so. If it's too long to texify and it's recent, link the OP to the LaTeX guide. If it's too long to texify and it's old, too bad. If it's too long to texify because it's simply a lazy "dumping" of a question onto SE, then vote to close.

16

It is sort of possible via the MathJAX extension HTML using commands of the form style{css}{math}. Basically, this style command will attach some extra CSS style attributes to the math elements and let the browser render them. However, how are they rendered depend on browser and even for a single browser, it takes a lot tweaking to make it looks right (...

16

I am the "commenter" that you're referring to in the last sentence. Before explaining anything, I would like to draw your attention to the benefits of having things properly typed. More efficient (math) searching We can easily find a bunch of questions with math expressions like "$n < 2^n$" with the help of Approach0. Figure 1: Logo of the math search ...

15

If it is a good explanation of the answer it is not noisy. Generally, I like to see animated gifs with three attributes: few (but not zero), slow, small (in memory size) I really don't think that we will have the problem of being overwhelmed by animated gifs.

15

Some decide instead of typing the actual text of the question, typesetting formulae with LATEX and using plotting software, it would be better for them to avoid the effort Okay, hold on. You seem to be under the impression that computer typesetting and plotting is easy. Something anyone who uses this site should have in their skillset. Do you really expect ...

15

Our standard procedure when we get any request for a special site plugin like this is to ask for two things: community support on meta evidence that it would make a significant improvement to post and/or community quality In other words, we'd like for you to edit this meta question to show us examples of several posts on the site that would benefit ...

13

«For the screen readers of visually impaired, for browsers that don't show images, for cases where Imgur is blocked, and for search engine indexes: please enter a good description.»

13

If you know Mathematica syntax you can use Wolfram Alpha for simple graphs. Then post the image here:

12

Answers are meant to be enlightening and useful, but not necessarily original. As long as the sources are properly attributed, I do not find any issues in posting an excerpt from a textbook as an answer. In fact, it's very likely that the exposition in the textbook is hard -if not impossible- to improve upon. So there may not be any point in rewriting the ...

12

In this answer I posted the following plot created in $\mathrm{\LaTeX}$, using the TikZ package. Or a contour such as this one

12

I think TikZ rendering would be a great asset to the site. Well-made pictures can be an enormous aid in learning, from diagrams for path integrals in complex analysis to this image illustrating the proof of the Cantor–Schröder–Bernstein Theorem in set theory. Commutative diagrams in particular are essential in areas such as algebra, algebraic geometry, ...

11

You can scale down images by appending l m or s to the filename, before .extension. After upload, the image below was named http://i.stack.imgur.com/2pBS0.png Here's the l version with URL http://i.stack.imgur.com/2pBS0l.png; the images scaled with l should not be wider than the allocated horizontal space for the display of question and answers. And ...

10

Just add hover text in parentheses: [1]: http://i.stack.imgur.com/..... (hover text) or in quotation marks: [1]: http://i.stack.imgur.com/..... "hover text" Like this: Works for links, too. There's a good reason to have alt-text and hover-text as separate attributes: one is intended for users who don't see the image, the other for those who do.

10

We have a special close reason for questions consisting only of a copy of an exercise from a textbook: This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where ...

10

People should be encouraged to avoid putting the main content of their question in an image. This makes the question harder to find using search and inaccessible to the visually impaired, which is less than ideal. Making something harder to find via search is especially unfortunate, as it means that others who might be interested in the same question won't ...

10

A quick Google search produced this handy run-down of the GIF file format, which includes this section on looping: Example (not in sample file): 21 FF 0B 4E 45 54 53 43 41 50 45 32 2E 30 03 01 05 00 00 The GIF89 specification allows for application-specific information to be embedded in the GIF file itself. This capability is not much used. About the ...

9

Disclaimer: I can't recall ever drawing as part of a question or answer on this site, but I do draw a lot of math (on a "Wacom" tablet) as part of my vocation. There are many (better) third-party solutions for this, including: MS Paint (or Mac/Linux equivalent). Export. Upload. This site, which was the first hit when I googled "draw". Sure, it ...

9

The answer to the question in the title is that It can't be done. MathJax interacts poorly with MarkDown (they overwrite each other's pieces). The way around it is to escape everything inside a MathJax environment and hide it from view of MarkDown. Therefore by design loading of pictures/links using MarkDown syntax inside a MathJax environment will not ...

9

The most you can do is provide a link to the image. Comments are not meant to take up a lot of space, and one thing that pictures tend to do is take up space. However, it's common enough that comments may contain the need to reference images in order to help clarify or expand on the point they are making. So simply linking to the images is perfectly fine. ...

9

I'm the wrong person to tell the whole story, but I know that you can resize images by using an appropriate HTML img attribute. Below I reuse an animation that is a small part of an answer of mine from main. First 200 pixels wide  Then resized <img src="http://i.stack.imgur.com/dIISR.gif" ...

8

Here is the meta.SO post about stack.imgur.com. Unfortunately it is not the case that this site preserves images indefinitely (see the comments on Jeff Atwood's answer); however, they will remain significantly longer than most images hosted on the imgur site. Hooray! I was mistaken earlier; it is the case that images uploaded via stack.imgur.com will be ...

8

Well, there are many ways to create a picture, and upload it here. Some of the more obvious alternatives are: (1) MS Paint, GIMP, or other paint programs (2) Illustrator, Corel, Powerpoint, Inkscape, or other vector drawing packages (3) TikZ or MetaPost or Asymptote, if you think that writing code is a good way to create pictures. (4) Mathematica, Maple, ...

7

You could use: Geogebra Mathematica Wolfram Alpha SageMath It depends on the depth of the question. But for the questions you are making, I guess Geogebra is enough.

Only top voted, non community-wiki answers of a minimum length are eligible


text:
Note that if â¦ ConvergenceinProbability RobertBaumgarth1 1MathematicsResearchUnit,FSTC,UniversityofLuxembourg,MaisonduNombre,6,AvenuedelaFonte,4364 Esch-sur-Alzette,Grand-DuchédeLuxembourg ð«ð-convergence ð«1-convergence a.s. convergence convergence in probability (stochastic convergence) Theorem 2.11 If X n âP X, then X n âd X. Convergence in probability essentially means that the probability that jX n Xjexceeds any prescribed, strictly positive value converges to zero. (a) We say that a sequence of random variables X. n (not neces-sarily deï¬ned on the same probability space) converges in probability to a real number c, and write X. i.p. Assume that X n âP X. Convergence in probability Deï¬nition 3. To convince ourselves that the convergence in probability does not 5.2. 2. We say V n converges weakly to V (writte Lecture 15. Proof: Let F n(x) and F(x) denote the distribution functions of X n and X, respectively. Convergence with probability 1 implies convergence in probability. 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text:
A Compact Routing Scheme and Approximate Distance Oracle for Power-Law Graphs

Compact routing addresses the tradeoff between table sizes and stretch, which is the worst-case ratio between the length of the path a packet is routed through by the scheme and the length of a shortest path from source to destination. We adapt the compact routing scheme by Thorup and Zwick to optimize it for power-law graphs. We analyze our adapted routing scheme based on the theory of unweighted random power-law graphs with fixed expected degree sequence by Aiello, Chung, and Lu. Our result is the first theoretical bound coupled to the parameter of the power-law graph model for a compact routing scheme. In particular, we prove that, for stretch 3, instead of routing tables with $\tilde{O}(n^{1/2})$ bits as in the general scheme by Thorup and Zwick, expected sizes of $O(n^{\gamma}\log n)$ bits are sufficient, and that all the routing tables can be constructed at once in expected time $O(n^{1+\gamma}\log n)$, with $\gamma=(\tau-2)/(2\tau-3)+\epsilon$, where $\tau$ in (2,3) is the power-law exponent and $\epsilon>0$ (which implies $\epsilon < \gamma < 1/3 + \epsilon$). Both bounds also hold with probability at least 1-1/n (independent of $\epsilon$). The routing scheme is a labeled scheme, requiring a stretch-5 handshaking step and using addresses and message headers with $O(\log n\log\log n)$ bits, with probability at least 1-o(1). We further demonstrate the effectiveness of our scheme by simulations on real-world graphs as well as synthetic power-law graphs. With the same techniques as for the compact routing scheme, we also adapt the approximate distance oracle by Thorup and Zwick for stretch 3 and obtain a new upper bound of expected $\tilde{O}(n^{1+\gamma})$ for space and preprocessing for random power-law graphs.

 PDF file

## Details

 Type TechReport Number MSR-TR-2009-84
> Publications > A Compact Routing Scheme and Approximate Distance Oracle for Power-Law Graphs


text:
# Weekly NDVI

To assess the condition of the vegetation for the purpose of rating fire danger in conterminous U.S., weekly images of Normalized Difference Vegetation Index (NDVI) are utilized. NDVI is derived using the following equation:

$$NDVI= \frac{Near Infrared-Red Visible Wavelength}{Near Infrared+Red Visible Wavelength}$$

NDVI ranges from -1.0 to 1.0, with features such as barren, sand, and snow resulting in low values (0.1 or less) and surfaces with sporadic vegetation such as grasslands and senescing vegetation registering moderate NDVI values (0.2 – 0.5). In contrast high and dense vegetation coverage on a surface gives high NDVI indexes (0.6 – 0.9). For the purposes of generating Fire Danger Forecast products, 1 km resolution, 7‑day Visible Infrared Imaging Radiometer Suite (VIIRS) NDVI composites are used (Figure 1).

Weekly Relative Greenness:

Relative greenness (RG) (Figure 2) is calculated from weekly composites of the NDVI previously derived from Moderate Resolution Imaging Spectrometer (MODIS) NDVI data, and more recently from VIIRS NDVI composites. RG indicates how green each pixel currently is in relation to the range of historical NDVI observations for it. RG values are scaled from 0 to 100, with low values indicating the vegetation is at or near its minimum greenness. The purpose of using relative greenness in the Wildland Fire Potential Index model is to define the proportion of live and dead vegetation. Specifically, RG is derived using the following equation:

RG=(NDo−NDmn)(NDmx−NDmn)∙100

Where, NDo = highest observed NDVI value for the 1-week composite period, NDmn = historical minimum NDVI value for a given pixel, NDmx = historical maximum NDVI value for a given pixel


text:
# 17th International Conference on Ion Sources

15-20 October 2017
CERN
Europe/Zurich timezone

## Commissioning of the ECR Ion Source of the High Intensity Proton Injector of the Facility for Anti Proton and Ion Research at CEA-Saclay

18 Oct 2017, 14:30
30m
CERN

#### CERN

Centre international de Conférence Genève (CICG). http://www.cicg.ch/
Oral presentaion Production of high intensity ion beams

### Speaker

Olivier Tuske (CEA SACLAY / Irfu/DACM/LEDA)

### Description

The Irfu/SACM at Saclay is in charge of developing and building the ion source and the low energy line or the proton linac of the FAIR accelerator complex (Facility for Antiproton and Ion Research) located at GSI (Darmstadt) in Germany. The FAIR Facility will deliver stable and rare isotope beams covering a huge range of intensities and beam energies for experiments in the fields of atomic physics, plasma physics, nuclear physics, hadron physics, nuclear matter physics, material physics and biophysics. A significant part of the experimental program at FAIR is dedicated to antiproton physics that requires an ultimate number $7\times 10^{10}$ cooled pbar/h.

The high-intensity proton beam that is necessary for antiproton production will be delivered by a dedicated 75 mA/70 MeV proton linac. A 2.45 GHz microwave ion source will deliver a 100 mA H$^+$ beam pulsed at 4 Hz with an energy of 95 keV. A dual solenoids Low Energy Beam Transport line (LEBT) allows the injection of the proton beam into the radio frequency quadrupole (RFQ) within an acceptance of 0.3π mm.mrad (norm. rms). An electrostatic chopper system located between the second solenoid and the RFQ is used to cut the beam macro pulse from the source to inject 36 μs long beam pulses into the RFQ. At present time, a Ladder-RFQ are under construction at the University of Frankfurt.

This article reports the first beam measurements obtained since mid of 2016. Proton beams have been extracted from the ECR ion source and analyzed just after the extraction column on a dedicated diagnostic chamber. Emittance measurements, as well as extracted current and beam proportion analysis have been performed in different configurations of ion source parameters, such as magnetic field profile, RF power, gas injection or extraction voltage.

### Primary authors

Olivier Tuske (CEA SACLAY / Irfu/DACM/LEDA) Olivier Delferriere (CEA SACLAY Irfu/DACM/LEDA) Nicolas Chauvin (CEA SACLAY Irfu/DACM/LEDA) Yannick Gauthier (CEA SACLAY Irfu/DACM/LEDA)

### Presentation materials

 tuske olivier id293.pdf tuske olivier id293.pdf tuske olivier id293.pptx


text: Submit Abstract for AHAMC 2011 Send your abstract (in an editable format such as Tex or Latex (preferred format), plain text, MS word, but not in PDF format) to the conference email adress: ahamc@ualberta.ca before June 15, 2011, Copy and paste the following AMS-Latex codes between the two lines, then add your own information to prepare your abstract. Please do not use macros. Then submit your AMS-latex source code (not its produced PDF file) to ahamc@ualberta.ca. A sample abstract and its AMS-latex output are provided at the bottom of this page. %%%%%%%%%%%%%%%%%%%%latex stencil for title and abstract of AHAMC 2011%%%%%%%%%%%%%% \documentclass[12pt]{article} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsfonts} \begin{document} \bigskip %your title goes here \centerline{\textbf{         }} %speaker's name goes here. If your have more than one authors, %use $~^*$ to indicate the speaker \centerline{        } %Your email address goes here. Use \@ instead of @ \centerline{\texttt{     }} %Your mailing address goes here %use \\ to break your address into several lines {\centering \\} %Your abstract goes here \end{document} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%end of abstract%%%%%%%%%%%%%%%%%%%% The following is a sample abstract. Its AMS-Latex output is here \documentclass[12pt]{article} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsfonts} \begin{document} \medskip %your title goes here \centerline{\textbf{Matrix Extension with Symmetry and Applications to Multiwavelets}} %speaker's name goes here. If your have more than one authors, %use $~*$ to indicate the speaker \centerline{Bin Han$^*$ and Xiaosheng Zhuang} %Your email adress goes here. Use \@ instead of @ \centerline{\texttt{bhan\@ualberta.ca, johnsen99\@gmail.com}} %Your mailing address goes here %use \\ to break your address into several lines {\centering Department of Mathematical and Statistical Sciences,\\ University of Alberta, Edmonton, Alberta,\\ Canada T6G 2G1 \\} %Your abstract goes here In this talk, we will discuss the matrix extension with symmetry and its applications to the construction of multiwavelets and framelets. \end{document} %end of abstract


text:
# Announcement

The Graduate Student / Postdoc Seminar has moved!

The new page can be found here.

# Next Talk

Speaker: Paul Bourgade TBA December 1, 1:00 p.m. (pizza and drinks at 12:45 p.m.) WWH 1302

#### Abstract

tba

This seminar is meant to benefit young mathematicians, particularly graduate students and postdocs.
It aims to accomplish the following:
• provide a venue for talks that young mathematicians will understand
• expose students to areas of research at the Courant Institute
The research talks should be fairly introductory and accessible to students and non-specialists in the audience.

# Schedule Fall 2017

## October 6

Speaker: Robert Kohn Energy driven pattern formation Abstract Energy-driven pattern formation examines how energy minimization leads to the formation of defects and microstructure in a variety of physical systems. Examples include the wrinkling of elastic membranes, the formation of domains in a magnetic material, and the twinning produced by martensitic phase transformation. These systems can be described by Landau theories'' -- essentially, nonconvex variational problems regularized by higher-order singular perturbations. I will show in various examples -- some old, some new -- how one can identify the scaling law of the minimum energy, and how this sheds light on the underlying patterns.

## October 13

Speaker: Aaditya Rangan Covariate-corrected biclustering methods for gene-expression and GWAS data Abstract A common goal in data-analysis is to sift through a large matrix and detect any significant submatrices (i.e., biclusters) that have a low numerical rank. To give an example from genomics, one might imagine a data-matrix involving several genetic-measurements taken across many patients. In this context a ‘bicluster’ would correspond to a subset of genetic-measurents that are correlated across a subset of the patients. While some biclusters might extend across most (or all) of the patients, it is also possible for biclusters to involve only a small subset of patients. Detecting biclusters such as these provides a first step towards unraveling the physiological mechanisms underlying the heterogeneity within a patient population. In this talk I'll describe a simple algorithm for tackling this biclustering problem – i.e., for detecting low-rank submatrices from within a larger data-matrix. The basic method itself is very straightforward (c.f. Rangan 2012), and involves examining the 'loops' (i.e., 2-x-2 submatrices) within the data-set. Importantly, this method can easily be modified to account for many considerations which commonly arise in practice. For example, by counting loops slightly differently, we can correct for controls: finding biclusters that manifest only within a ‘case’-population without manifesting within a ‘control’-population. Moreover, this methodology can also correct for categorical- and continuous-covariates, as well as sparsity within the data. I'll illustrate these practical features with two examples; the first drawn from gene-expression analysis and the second drawn from a much larger genome-wide-association-study (GWAS). In addition to being quite practical, this loop-counting method exhibits a few interesting mathematical features, which I'll mention towards the second half of the talk. To begin with, I'll point out some of the well known phase-transitions for the biclustering problem (i.e., parameters which determine whether or not the problem is 'easy' or 'hard'). I'll also briefly explain why the loop-counting method is 'close to optimal' within a certain subset of local algorithms. Finally, I'll demonstrate that the loop-counting method actually outperforms what are called 'spectral-methods' (based on the singular-value-decomposition) near the computational phase-transition, and explain why this is the case.

## October 27

Speaker: Robert Young Quantifying simple connectivity: an introduction to the Dehn function Abstract Many theorems start by taking an existence theorem and asking How many?'' or How big?'' or How fast''. The best-known example may be the prime number theorem. Euclid proved that infinitely many primes exist, and the prime number theorem describes how quickly they grow. I'll discuss what happens when you apply the same idea to simple connectivity. In a simply-connected space, any closed curve is the boundary of some disc, but how big is that disc? And what can that tell you about the geometry of the space?

## November 10

Speaker: Edwin Gerber Untangling the annual cycle of the tropical tropopause layer with an idealized moist model Abstract What is the tropical tropopause layer? Why should I care if it has an annual cycle? These, and more climate science oriented questions, will be addressed in my seminar this Friday. But as not to leaving you in total suspense, the tropical tropopause layer plays a critical role in regulating the chemical composition of the stratosphere, with significant impacts on surface climate. It also provides a curious dynamical puzzle, exhibiting a distinct annual cycle despite the semi-annual solar forcing of the tropics. In this talk, I will introduce a new intermediate complexity atmospheric model that allows us to explore the key processes regulating the tropical tropopause.

## December 1

Speaker: Paul Bourgade TBA Abstract tba

# Contact Info

Marguerite Brownbrownml [at] cims [dot] nyu [dot] edu
Reza Gheissarireza [at] cims [dot] nyu [dot] edu
Benjamin McKennamckenna [at] cims [dot] nyu [dot] edu

## Previous semesters

### Spring 2011 schedule

Descriptions of earlier talks are here.

Department of Mathematics
Courant Institute of Mathematical Sciences
New York University
251 Mercer St.
New York, NY 10012


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# If the hypotenuse of an isosceles right triangle has length of 8, then

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If the hypotenuse of an isosceles right triangle has length of 8, then  [#permalink]

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18 Oct 2018, 00:53
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If the hypotenuse of an isosceles right triangle has length of 8, then the area of the triangle is

A. 4
B. 4√2
C. 8
D. 8√2
E. 16

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If the hypotenuse of an isosceles right triangle has length of 8, then  [#permalink]

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18 Oct 2018, 01:01
Bunuel wrote:
If the hypotenuse of an isosceles right triangle has length of 8, then the area of the triangle is

A. 4
B. 4√2
C. 8
D. 8√2
E. 16

An isosceles right triangle has angle 45-45-90, whose sides are in the ratio $$1:1:\sqrt{2}$$

As the hypotenuse of the isosceles triangle is 8, the other sides measure $$\frac{8}{\sqrt{2}}$$ = base = height

Therefore, the area for this triangle is $$\frac{1}{2}*\frac{8}{\sqrt{2}}*\frac{8}{\sqrt{2}} = \frac{64}{4} = 16$$(Option E)
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If the hypotenuse of an isosceles right triangle has length of 8, then &nbs [#permalink] 18 Oct 2018, 01:01
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# YuMi or Sawyer (or something else) for RL research?

We are planning to purchase a cobot for our research lab to run experiments in human-robot interactions. Our research would need cobot reinforcement learning in various modes of interaction with humans (and occasionally with other robots). We have so far narrowed down the options to Sawyer and ABB YuMi. YuMi has a relatively small payload limit, but that should be okay for us. We don't see that many robot RL papers that use ABB YuMi and that makes us wonder whether there are limitations/challenges/etc. when it comes to using YuMi for research. Also, in addition to Sawyer and Yumi, other suggestions are also appreciated.

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We don't see that many robot RL papers that use ABB YuMi and that makes us wonder whether there are limitations/challenges/etc. when it comes to using YuMi for research.

There was no support for EGM on YuMi officially. That has changed recently, with the introduction of RobotWare 6.09. While it is possible to achieve somewhat decent on-line performance using just RAPID (with fi something like kth-ros-pkg/yumi), EGM is really the future and in my opinion the only control interface that would make something like RL on actual hw interesting (in the case of ABB robots).

It's most likely not the only reason you don't see many papers, but for researchers interested in running experiments on actual hw (which is a requirement for quite some venues now) and/or looking to run RL training phases not in simulation but on real hw, not having a decent, real-time external control interface will make them look at other robots. And there are quite a few (just not humanoid style upper torsos with two arms).

ABB has released two packages that help with integrating EGM (into any application, not just ROS): ros-industrial/abb_libegm and its companion ros-industrial/abb_librws. These are officially supported by ABB and under active development.

Sawyer has a very nice ROS interface, which YuMi does not (yet) have. That is certainly a negative on the side of YuMi: it will take you somewhat longer to get started with YuMi. However, as @stevemacenski already wrote: Rethink is no more, and it's somewhat unclear (to me at least) what the future of the research side of Baxter and Sawyer is.

Community-wise I would expect there to be a bit more people using Sawyer with ROS than YuMi (Sawyer was released not that long before Rethink stopped operations), but EGM is used with ABB's other robots, so there is a decent number of people using that.

And just to avoid a false dilemma: there are quite a few other robot brands that have decent, real-time capable external motion control interfaces. You may have already evaluated those, and crossed them off your list, but just in case you may want to look at other options as well.

more

PS: I will reach out to ABB to ascertain the status of EGM on YuMi and update this answer.

Edit: I've updated the answer. EGM is officially supported on YuMi with the release of RobotWare 6.09.

( 2019-09-21 05:34:32 -0600 )edit

Ah, I had assumed if they had narrowed down to the 2 robot models that meant the asker had already validated they both meet the specifications required to do RL work. If the YuMi doesn't give you the interfaces you need then I'd look at other options than those 2, but I think choosing a Rethink robot is a poor long-term choice unless its the only option

( 2019-09-21 19:50:15 -0600 )edit

Well as I wrote, there is now support for the kind of interface I'd like to use with YuMis: EGM.

( 2019-09-22 05:02:56 -0600 )edit

If those are the two you're looking at, YuMi, hands down.

Rethink robotics shut down, so you'd be buying hardware with little if any support. There supposed to be revived in some fashion down the line but if you're making a lab purchase, going with who you know will support you is important if something breaks or you have questions.

more


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Tile detection on a 2D isometric layered diamond shaped map - What are the math?

What are the math for tile detection on a 2D isometric layered diamond shaped map ?

Here is an image of a layer (this is an image from an old folder, I didn't write the display yet for my current project, so it's incomplete but you get the idea) :

I have them staggered on top of each other. Tile detection could works with a loop but it would take too much time. The basic math for floor tiles doesn't work.

A layer is composed of a tiles(array) :

A tile_size * tile_size image representing a 3D square.

A long time ago I asked this question with an account I'v lost unfortunatly and I had one answer :

• Create an hollow ground

But I never understood what that was and when I type "tile detection hollow ground" on google, I got nothing.

I also found several solution I do not understand the concept :

• Ray scan on neightbours (I know the theory for ray scanning, but neightbours ??? All of that sounds very difficult for me)

• using color mask for a pixel perfect detection (no time to implement such a method)

So, was left the math. And here I'm stuck. I'm looking at my formula 'isometric to screen' conversion(for display), hoping a miracle ha ha :

int x = (i - j) * TILE_SIZE_HALF,
y = (i + j) * TILE_SIZE_QUARTER - (layer_id * TILE_SIZE_HALF);
//This one is totally working

Anyway, is there even a formula able to solve tile detection for such a map ? Am I missing something ?

What I'm looking for :

   //With given x & y (mouse coordinates on screen)

int i = ?,
j = ?,
layerID = current_layer_index,
//tile's index in tile array(layers are tile arrays) :
tileIndex = calculated from I & J;
//Where goes the offsets in these calculations ?

How am I gonna know which layer the tile is from ?

I decided to only edit the selected layer (current_selected_layer variable) and then check if the tile is empty. If yes, I'll select the "same tile" but the layer under the selected one and so on. Now the problem should be less complicated since it's like there is only one layer. What bother me are the map offsets.

EDIT : I didn't tought about the corner... This is not gonna work. If I don't find a solution, I'll maybe switch to classic 2D. Sounds very complicate to code tile selection with isometric map.

There is another problem I have no idea how to handle these :

(this is a layered map, the screen do not represent the final result)

End note: Pseudo code is welcome. Also, consider you have all informations about the map. Since I use a pseudo MVC code, I have access to all data everywhere (tile size, layers number, etc...). There is no restriction.

The isometric stuff is just a projection, based on that, you have to unproject your mouse coords.

the following code is in pseudo javascript:

var x = MOUSE_X,
y = MOUSE_Y,
retX = 0,
retY = 0;

retX = Math.round(x / TILE_WIDTH) * TILE_WIDTH;
retY = (Math.round(y / (TILE_HEIGHT/2)) - 1) * TILE_HEIGHT/2;

if ((Math.round(y / (TILE_HEIGHT/2)) - 1) % 2) {
retX -= (TILE_WIDTH/ 2);
}

In this case, mouse coords are unprojected from isometric ones, so you can, after this, request your tile by doing:

map[retx + MAP_WIDTH * rety] (1D Array)

map[retx][rety] (2D Array)

• Thanks for your answer. It's a layered map, I don't understand where are the layers in your pseudo-code. The screen you saw, there are several of this on top of each other. – Sanchez Tanguy May 17 '17 at 14:25
• This pseudo-code would work for one layered map with flat tiles (a classic one) but I don't think it's gonna work for mine. – Sanchez Tanguy May 17 '17 at 14:54
• oh, i see, sorry, somehow i thought you wanted to know how to do it in 2D. In your case, the easiest (and most common) approach is intersecting a ray from your camera to the pointer with the terrain cells. Read this stackoverflow.com/questions/2093096/implementing-ray-picking – PRDeving May 17 '17 at 16:31
• @SanchezTanguy Sure, why not? "Cameras" in games are just a fiction, a useful conceptual tool to encapsulate some algebra related to positioning content on the player's screen. They don't do anything magic that you can't reproduce in your own code. – DMGregory May 17 '17 at 19:16
• I'm pretty sure the 1d array's lookup algorithm is retx + WIDTH * rety – Bálint May 17 '17 at 19:29

Not efficient, but accurate on complex structures, is drawing all cubes with a different color and no shade(r) onto a black background into a framebufferobject and later testing which color a given pixel has. You should draw the objects without blending and without texture into the buffer.

Another way is, you ray-trace until you reach a cube or a limit.

First, I'll deal with the simpler problem with the edge and corner of the map you showed above. For that, you will only be concerned with the top of the tile when a person selects it, so if they click on the outer edge along the side or corner of the tile, than they are out of bounds and clicking off the map. It doesn't matter than they click the side of a tile. I would reject such a click as out of bounds.

As for layer selection, this is something I struggled with as well, and still do a little, but I have come up with several ideas on this. One idea I had was to store the highest layer on your map. Then when you click, you check to see if there is a tile at that location on the highest layer, so basically, take your normal flat selection, and offset it by the height of the layer; so, if your layers are all 16 pixels higher for each layer, than you would do your normal detection, except you would adjust the Y location by -16 * LAYER. You check the highest layer where the mouse clicked, if there is no tile at that height at that location, than check the next layer down until you end up at the ground level.

That was one idea anyhow. The other idea was to use mouse maps to determine if the tile you selected was the right one, of if you're clicking a tile next to it that is at a higher level.