text:
You have the inflation restriction sequence $0\rightarrow H^1(K^{nr}/K,\mathbf{F}_p)\rightarrow H^1(K,\mathbf{F}_p)\rightarrow H^1(K^{nr},\mathbf{F}_p)^{\mathrm{Gal}(K^{nr}/K)}\rightarrow 0$ with the zero on the end because $\hat{\mathbf{Z}}$ has cohomological dimension one. The first term is $\mathbf{F}_p$ by evaluation on Frobenius. The second term can be understood via class field theory. Any continuous homomorphism $G_K\rightarrow\mathbf{F}_p$ factors through the maximal abelian pro-$p$ quotient of $G_K$, which is isomorphic to $\mathbf{Z}_p^{d+1}\times\mu$ where $d$ is the degree over $\mathbf{Q}_p$ and $\mu$ is the group of $p$-power roots of unity in $K$. Since $\mathrm{Hom}_{cts}(\mathbf{Z}_p^{d+1},\mathbf{F}_p)$ is just $d+1$ copies of $\mathrm{Hom}(\mathbf{F}_p,\mathbf{F}_p)$, its dimension is $d+1$. If $\mu$ is trivial, then $\mathrm{Hom}(\mu,\mathbf{F}_p)=0$. If $K$ contains a primitive $p$-th root of unity, then $\mathrm{Hom}(\mu,\mathbf{F}_p)=\mathrm{Hom}(\mu_p,\mathbf{F}_p)$ is $1$-dimensional. So the dimension of the middle piece is $d+1$ or $d+2$ according as $K$ does or doesn't contain a primitive $p$-th root of unity. So the dimension of the last term is either $d$ or $d+1$.
Well, you have the inflation restriction sequence $0\rightarrow H^1(K^{nr}/K,\mathbf{F}_p)\rightarrow H^1(K,\mathbf{F}_p)\rightarrow H^1(K^{nr},\mathbf{F}_p)^{\mathrm{Gal}(K^{nr}/K)}\rightarrow 0$ with the zero on the end because $\hat{\mathbf{Z}}$ has cohomological dimension one. The first term is $\mathbf{F}_p$ by evaluation on Frobenius. The second term can be understood via class field theory. Any continuous homomorphism $G_K\rightarrow\mathbf{F}_p$ factors through the maximal abelian pro-$p$ quotient of $G_K$, which is isomorphic to $\mathbf{Z}_p^{d+1}\times\mu$ where $d$ is the degree over $\mathbf{Q}_p$ and $\mu$ is the group of $p$-power roots of unity in $K$. Since $\mathrm{Hom}_{cts}(\mathbf{Z}_p^{d+1},\mathbf{F}_p)$ is just $d+1$ copies of $\mathrm{Hom}(\mathbf{F}_p,\mathbf{F}_p)$, its dimension is $d+1$. If $\mu$ is trivial, then $\mathrm{Hom}(\mu,\mathbf{F}_p)=0$. If $K$ contains a primitive $p$-th root of unity, then $\mathrm{Hom}(\mu,\mathbf{F}_p)=\mathrm{Hom}(\mu_p,\mathbf{F}_p)$ is $1$-dimensional. So the dimension of the middle piece is $d+1$ or $d+2$ according as $K$ does or doesn't contain a primitive $p$-th root of unity. So the dimension of the last term is either $d$ or $d+1$.


text:
zbMATH — the first resource for mathematics

Variations on the Banach-Stone theorem. (English) Zbl 1027.54021
The authors review some classical and recent contributions concerning extensions, generalizations and variants of the Banach-Stone theorem on determination of the topology of a compact space $$K$$ by the linear metric structure of $$C(K)$$. The paper consists of five sections: introduction (a short description of the content), classical results, isometries, algebra isomorphisms, vector lattice isomorphisms. The paper is based on the lectures delivered by the second author in a Summer Course of the University of Cantabria, held in Laredo in August 2001.

MSC:
 54C35 Function spaces in general topology 46E15 Banach spaces of continuous, differentiable or analytic functions 46E25 Rings and algebras of continuous, differentiable or analytic functions 54D30 Compactness
Full Text:

format: plain


text:
Posted on Categories:Hypothesis Testing, 假设检验, 数据科学代写, 统计代写, 统计代考

# 数据科学代写|假设检验代考Hypothesis Testing代考|STAT1070 Basic Tools for Judging Robustness

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## 数据科学代写|假设检验代考Hypothesis Testing代考|Basic Tools for Judging Robustness

There are three basic tools used to establish whether quantities such as measures of location and scale have good properties: qualitative robustness, quantitative robustness, and infinitesimal robustness. This section describes these tools in the context of location measures, but they are relevant to measures of scale, as will become evident. These tools not only provide formal methods for judging a particular measure, they can be used to help derive measures that are robust.

Before continuing, it helps to be more formal about what is meant by a measure of location. A quantity that characterizes a distribution, such as the population mean, is said to be a measure of location if it satisfies four conditions, and a fifth is sometimes added. To describe them, let $X$ be a random variable with distribution $F$, and let $\theta(X)$ be some descriptive measure of $F$. Then $\theta(X)$ is said to be a measure of location if for any constants $a$ and $b$,

The first condition is called location equivariance. It simply requires that if a constant $b$ is added to every possible value of $X$, a measure of location should be increased by the same amount. Let $E(X)$ denote the expected value of $X$. From basic principles, the population mean is location equivariant. That is, if $\theta(X)=E(X)=\mu$, then $\theta(X+b)=E(X+b)=\mu+b$. The first three conditions, taken together, imply that a measure of location should have a value within the range of possible values of $X$. The fourth condition is called scale equivariance. If the scale by which something is measured is altered by multiplying all possible values of $X$ by $a$, a measure of location should be altered by the same amount. In essence, results should be independent of the scale of measurement. As a simple example, if the typical height of a man is to be compared to the typical height of a woman, it should not matter whether the comparisons are made in inches or feet.

## 数据科学代写|假设检验代考Hypothesis Testing代考|Qualitative Robustness

To understand qualitative robustness, it helps to begin by considering any function $f(x)$, not necessarily a probability density function. Suppose it is desired to impose a restriction on this function so that it does not change drastically with small changes in $x$. One way of doing this is to insist that $f(x)$ be continuous. If, for example, $f(x)=0$ for $x \leq 1$, but $f(x)=10,000$ for any $x>1$, the function is not continuous, and if $x=1$, an arbitrarily small increase in $x$ results in a large increase in $f(x)$.

A similar idea can be used when judging a measure of location. This is accomplished by viewing parameters as functionals. In the present context, a functional is just a rule that maps every distribution into a real number. For example, the population mean can be written as
$$T(F)=E(X),$$
where the expected value of $X$ depends on $F$. The role of $F$ becomes more explicit if expectation is written in integral form, in which case this last equation becomes
$$T(F)=\int_{-\infty}^{\infty} x d F(x) .$$
If $X$ is discrete and the probability function corresponding to $F(x)$ is $f(x)$,
$$T(F)=\sum x f(x),$$
where the summation is over all possible values $x$ of $X$.
One advantage of viewing parameters as functionals is that the notion of continuity can be extended to them. Thus, if the goal is to have measures of location that are relatively unaffected by small shifts in $F$, a requirement that can be imposed is that when viewed as a functional, it is continuous. Parameters with this property are said to have qualitative robustness.

## 数据科学代写|假设检验代考Hypothesis Testing代考|Qualitative Robustness

$$T(F)=E(X),$$

$$T(F)=\int_{-\infty}^{\infty} x d F(x) .$$

$$T(F)=\sum x f(x),$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


text:
# Fermat's Little Theorem

• October 19th 2009, 04:07 PM
kyldn6
Fermat's Little Theorem
Prove that $12\ |\ n^2-1$ if g.c.d. (n,6) = 1.
• October 19th 2009, 07:32 PM
chisigma
If g.c.d. (n,6) = 1 that means that $2 \nmid n$ and $3 \nmid n$, i.e. n is and odd number non multiple of 3. But if $2 \nmid n$ and $3 \nmid n$ means also that $n-1$ and $n+1$ are both even numbers and 3 devides one of them. The conclusion is that...

$2\cdot 2\cdot 3 \mid (n+1)\cdot (n-1)$

Kind regards

$\chi$ $\sigma$
• October 19th 2009, 09:47 PM
tonio
Quote:

Originally Posted by chisigma
If g.c.d. (n,6) = 1 that means that $2 \nmid n$ and $3 \nmid n$,

i.e. n is and odd number non multiple of 3. But if $2 \nmid n$ and $3 \nmid n$ means also that $n-1$ and $n+1$ are both even numbers and 3 devides one of them. The conclusion is that...

$2\cdot 2\cdot 3 \mid (n+1)\cdot (n-1)$

Kind regards

$\chi$ $\sigma$

.


text:
# Hotelling's lemma

Hotelling's lemma is a result in microeconomics that relates the supply of a good to the maximum profit of the producer. It was first shown by Harold Hotelling, and is widely used in the theory of the firm.

Specifically, it states: The rate of an increase in maximized profits w.r.t. a price increase is equal to the net supply of the good. In other words, if the firm makes its choices to maximize profits, then the choices can be recovered from the knowledge of the maximum profit function.

## Formal Statement

Let ${\displaystyle p}$  denote a variable price, and ${\displaystyle w}$  be a constant cost of each input. Let ${\displaystyle x:{\mathbb {R} ^{+}}\rightarrow X}$  be a mapping from the price to a set of feasible input choices ${\displaystyle X\subset {\mathbb {R} ^{+}}}$ . Let ${\displaystyle f:{\mathbb {R} ^{+}}\rightarrow {\mathbb {R} ^{+}}}$  be the production function, and ${\displaystyle y(p)\triangleq f(x(p))}$  be the net supply.

The maximum profit can be written by

${\displaystyle \pi (p)=\max _{x}p\cdot y(p)-w\cdot x(p).}$

Then the lemma states that if the profit ${\displaystyle \pi }$  is differentiable at ${\displaystyle p}$ , the maximizing net supply is given by

${\displaystyle y^{*}(p)={\frac {d\pi (p)}{dp}}.}$

## Proof for Hotelling's lemma

The lemma is a corollary of the envelope theorem.

Specifically, the maximum profit can be rewritten as ${\displaystyle \pi (p,x^{*})=p\cdot f(x^{*}(p))-w\cdot x^{*}(p)}$  where ${\displaystyle x^{*}}$  is the maximizing input corresponding to ${\displaystyle y^{*}}$ . Due to the optimality, the first order condition gives

${\displaystyle {\frac {\partial \pi }{\partial x}}{\bigg |}_{x=x^{*}}=0.}$

(1)

By taking the derivative by ${\displaystyle p}$  at ${\displaystyle x^{*}}$ ,

${\displaystyle {\frac {d\pi }{dp}}={\frac {\partial \pi }{\partial x}}{\bigg |}_{x=x^{*}}{\frac {\partial x}{\partial p}}+{\frac {\partial \pi }{\partial p}}={\frac {\partial \pi }{\partial p}}=f(x^{*}(p))=y^{*}(p)}$

where the second equality is due to (1). QED

## Application of Hotelling's lemma

Consider the following example.[1] Let output ${\displaystyle y}$  have price ${\displaystyle p}$  and inputs ${\displaystyle x_{1}}$  and ${\displaystyle x_{2}}$  have prices ${\displaystyle w_{1}}$  and ${\displaystyle w_{2}}$ . Suppose the production function is ${\displaystyle y=x_{1}^{1/3}x_{2}^{1/3}}$ . The unmaximized profit function is ${\displaystyle \pi (p,w_{1},w_{2},x_{1},x_{2})=py-w_{1}x_{1}-w_{2}x_{2}}$ . From this can be derived the profit-maximizing choices of inputs and the maximized profit function, a function just of the input and output prices, which is

${\displaystyle \pi (p,w_{1},w_{2})={\frac {1}{27}}{\frac {p^{3}}{w_{1}w_{2}}}}$

Hotelling's Lemma says that from the maximized profit function we can find the profit-maximizing choices of output and input by taking partial derivatives:

${\displaystyle {\frac {\partial \pi (p,w_{1},w_{2})}{\partial p}}=y={\frac {1}{9}}{\frac {p^{2}}{w_{1}w_{2}}}}$

${\displaystyle {\frac {\partial \pi (p,w_{1},w_{2})}{\partial w_{1}}}=-x_{1}=-{\frac {1}{27}}{\frac {p^{3}}{w_{1}^{2}w_{2}}}}$

${\displaystyle {\frac {\partial \pi (p,w_{1},w_{2})}{\partial w_{2}}}=-x_{2}=-{\frac {1}{27}}{\frac {p^{3}}{w_{1}w_{2}^{2}}}}$

Note that Hotelling's lemma gives the net supplies, which are positive for outputs and negative for inputs, since profit rises with output prices and falls with input prices.

## Criticisms and empirical evidence

A number of criticisms have been made with regards to the use and application of Hotelling's lemma in empirical work.

C. Robert Taylor points out that the accuracy of Hotelling's lemma is dependent on the firm maximizing profits, meaning that it is producing profit maximizing output ${\displaystyle y^{*}}$  and cost minimizing input ${\displaystyle x^{*}}$ . If a firm is not producing at these optima, then Hotelling's lemma would not hold.[2]


text:
# Check if all non-zero elements in a matrix are connected

### Input:

A matrix containing integers in the range [0 - 9].

### Challenge:

Determine if all non-zero elements are connected to each other vertically and/or horizontally.

### Output:

A truthy value if all are connected, and a falsy value if there are non-zero elements/groups that aren't connected to other elements/groups.

### Test cases:

Test cases are separated by line. Test cases can be found in more convenient formats here (Kudos to Dada).

The following are all connected and should return a truthy value:

0
---
0 0
---
1 1 1
0 0 0
---
1 0 0
1 1 1
0 0 1
---
0 0 0 0 0 0
0 0 3 5 1 0
0 1 0 2 0 1
1 1 0 3 1 6
7 2 0 0 3 0
0 8 2 6 2 9
0 0 0 0 0 5

The following are all not-connected, and should return a falsy value:

0 1
1 0
---
1 1 1 0
0 0 0 2
0 0 0 5
---
0 0 5 2
1 2 0 0
5 3 2 1
5 7 3 2
---
1 2 3 0 0 5
1 5 3 0 1 1
9 0 0 4 2 1
9 9 9 0 1 4
0 1 0 1 0 0

This is , so the shortest submission in each language wins. Explanations are encouraged!

Inspired by this challenge.

• Perhaps the input should contain only ones and zeros (or truthys and falsys), as this is essentially about connected components. – NikoNyrh Jan 30 '18 at 14:15
• Can we take input as an 1d array and a number of columns? – ovs Jan 30 '18 at 15:05
• @ovs sure. I can't see that it should give you any advantages over other people that have already answered. – Stewie Griffin Jan 30 '18 at 17:06
• Related: how many zeroes do you need to change to make all nonzero elements connected – dylnan Jan 30 '18 at 17:34
• Related: count the number of components (but with diagonal entries adjacent). – Misha Lavrov Feb 5 '18 at 14:53

# Retina 0.8.2, 80 77 bytes

Td@1
11
_
+m^((.)*)(1|_)( |.*¶(?<-2>.)*(?(2)(?!)))(?!\3)[1_]
$1_$4_
^\D+$ Try it online! Edit: Saved 1 byte thanks to @FryAmTheEggman. Explanation: Td@1  Simplify to an array of @s and 1s. 11 _  Change one 1 to a _. +m^((.)*)(1|_)( |.*¶(?<-2>.)*(?(2)(?!)))(?!\3)[1_]$1_$4_  Flood fill from the _ to adjacent 1s. ^\D+$

Test whether there are any 1s left.

• @FryAmTheEggman Thanks, and you gave me an idea as to how to save another two bytes too! – Neil Jan 31 '18 at 9:33

# MATL, 7 bytes

4&1ZI2<

This gives a matrix containing all ones as truthy output, or a matrix containing at least a zero as falsy. Try it online!

You can also verify truthiness/falsiness adding an if-else branch at the footer; try it too!

### Explanation

4       % Push 4 (defines neighbourhood)
&       % Alternative input/output specification for next function
1ZI     % bwlabeln with 2 input arguments: first is a matrix (implicit input),
% second is a number (4). Nonzeros in the matrix are interpreted as
% "active" pixels. The function gives a matrix of the same size
% containing positive integer labels for the connected components in
% the input, considering 4-connectedness
2<      % Is each entry less than 2? (That would mean there is only one
% connected component). Implicit display

• OP's note: in case there's any doubt: the outputs are perfectly fine and adheres to the linked meta post. – Stewie Griffin Jan 30 '18 at 21:59
• It's blowing my mind that MATL/matlab considers an array of numbers to be truthy IFF it contains no zeros. mathworks.com/help/matlab/ref/if.html (earlier comment deleted) – Sparr Jan 31 '18 at 1:56
• @Sparr (Actually, it is iff it contains no zeros and is not empty.) I was also confused when I learned that any nonempty array is truthy in other languages – Luis Mendo Jan 31 '18 at 10:13

# Wolfram Language (Mathematica), 54 bytes

Saved 2 bytes thanks to user202729.

Max@MorphologicalComponents[#,CornerNeighbors->1<0]<2&

Try it online!

# JavaScript (ES6), 136 135 bytes

Returns a boolean.

m=>!/[1-9]/.test((g=(y,x=0)=>1/(n=(m[y]||0)[x])&&!z|n?(m[y++][x]=0,z=n)?g(y,x)&g(--y-1,x)&g(y,x+1)||g(y,x-1):g(m[y]?y:+!++x,x):m)(z=0))

### Test cases

let f =

m=>!/[1-9]/.test((g=(y,x=0)=>1/(n=(m[y]||0)[x])&&!z|n?(m[y++][x]=0,z=n)?g(y,x)&g(--y-1,x)&g(y,x+1)||g(y,x-1):g(m[y]?y:+!++x,x):m)(z=0))

console.log('[Truthy]')

console.log(f([
[0]
]))
console.log(f([
[0,0]
]))
console.log(f([
[1,1,1],
[0,0,0]
]))
console.log(f([
[1,0,0],
[1,1,1],
[0,0,1]
]))
console.log(f([
[0,0,0,0,0,0],
[0,0,3,5,1,0],
[0,1,0,2,0,1],
[1,1,0,3,1,6],
[7,2,0,0,3,0],
[0,8,2,6,2,9],
[0,0,0,0,0,5]
]))

console.log('[Falsy]')

console.log(f([
[0,1],
[1,0]
]))
console.log(f([
[1,1,1,0],
[0,0,0,2],
[0,0,0,5]
]))
console.log(f([
[0,0,5,2],
[1,2,0,0],
[5,3,2,1],
[5,7,3,2]
]))
console.log(f([
[1,2,3,0,0,5],
[1,5,3,0,1,1],
[9,0,0,4,2,1],
[9,9,9,0,1,4],
[0,1,0,1,0,0]
]))

### Commented

The recursive function g() first looks for a non-zero cell (as long as the globally-defined flag z is set to 0) and then starts flood-filling from there (as soon as z != 0).

m =>                               // given the input matrix m
!/[1-9]/.test((                  // test whether there's still a non-zero digit
g = (y, x = 0) =>              //   after recursive calls to g(), starting at (x=0,y=0):
1 / (n = (m[y] || 0)[x]) &&  //     n = current cell; if it is defined:
!z | n ? (                   //       if z is zero or n is non-zero:
m[y++][x] = 0,           //         we set the current cell to zero
z = n                    //         we set z to the current cell
) ?                        //         if z is non-zero:
g(y, x) &                //           flood-fill towards bottom
g(--y - 1, x) &          //           flood-fill towards top
g(y, x + 1) ||           //           flood-fill towards right
g(y, x - 1)              //           flood-fill towards left
:                          //         else:
g(m[y] ? y : +!++x, x)   //           look for a non-zero cell to start from
:                            //       else:
m                          //         return the matrix
)(z = 0)                       //   initial call to g() + initialization of z
)                                // end of test()

# C, 163 bytes

Thanks to @user202729 for saving two bytes!

*A,N,M;g(j){j>=0&j<N*M&&A[j]?A[j]=0,g(j+N),g(j%N?j-1:0),g(j-N),g(++j%N?j:0):0;}f(a,r,c)int*a;{A=a;N=c;M=r;for(c=r=a=0;c<N*M;A[c++]&&++r)A[c]&&!a++&&g(c);return!r;}

Loops through the matrix until it finds the first non-zero element. Then stops looping for a while and recursively sets every non-zero element connected to the found element to zero. Then loops through the rest of the matrix checking if every element is now zero.

Try it online!

Unrolled:

*A, N, M;

g(j)
{
j>=0 & j<N*M && A[j] ? A[j]=0, g(j+N), g(j%N ? j-1 : 0), g(j-N), g(++j%N ? j : 0) : 0;
}

f(a, r, c) int*a;
{
A = a;
N = c;
M = r;

for (c=r=a=0; c<N*M; A[c++] && ++r)
A[c] && !a++ && g(c);

return !r;
}

# Perl, 80797873 70 bytes

Includes +2 for 0a

Give the input matrix without spaces on STDIN (or in fact as rows separated by any kind of whitespace)

perl -0aE 's%.%$".join"",map chop,@F%seg;s%\b}|/%z%;y%w-z,-9%v-~/%?redo:say!/}/' 000000 003510 010201 110316 720030 082629 000005 ^D  Easier to read if put in a file: #!/usr/bin/perl -0a use 5.10.0; s%.%$".join"",map chop,@F%seg;s%\b}|/%z%;y%w-z,-9%v-~/%?redo:say!/}/

# APL (Dyalog Unicode), 36 22 bytesSBCS

~0∊∨.∧⍨⍣≡2>+/|↑∘.-⍨⍸×⎕

Try it online!

thanks @H.PWiz

⍸×⎕ coordinates of non-zero squares

2>+/|↑∘.-⍨ adjacency matrix of the neighbour graph

∨.∧⍨⍣≡ transitive closure

~0∊ is it all 1s?

# Jelly, 23 bytes

FJṁa@µ«Ḋoµ€ZUµ4¡ÐLFQL<3

Try it online!

## Explanation.

The program labels each morphological components with a different numbers, then check if there are less than 3 numbers. (including 0).

Consider a row in the matrix.

«Ḋo   Given [1,2,3,0,3,2,1], return [1,2,3,0,2,1,1].
«     Minimize this list (element-wise) and...
Ḋ      its dequeue. (remove the first element)
So min([1,2,3,0,3,2,1],
[2,3,0,3,2,1]    (deque)
) =    [1,2,0,0,2,1,1].
o   Logical or - if the current value is 0, get the value in the input.
[1,2,0,0,2,1,1] (value)
or [1,2,3,0,3,2,1] (input)
=  [1,2,3,0,2,1,1]

Repeatedly apply this function for all row and columns in the matrix, in all orders, eventually all morphological components will have the same label.

µ«Ḋoµ€ZUµ4¡ÐL  Given a matrix with all distinct elements (except 0),
label two nonzero numbers the same if and only if they are in
the same morphological component.
µ«Ḋoµ          Apply the function above...
€           for €ach row in the matrix.

Z        Zip, transpose the matrix.
U       Upend, reverse all rows in the matrix.
Together, ZU rotates the matrix 90° clockwise.
4¡    Repeat 4 times. (after rotating 90° 4 times the matrix is in the
original orientation)
ÐL  Repeat until fixed.

And finally...

FJṁa@ ... FQL<3   Main link.
F                 Flatten.
J                Indices. Get [1,2,3,4,...]
ṁ               ṁold (reshape) the array of indices to have the same
shape as the input.
a@             Logical AND, with the order swapped. The zeroes in the input
mask out the array of indices.
...         Do whatever I described above.
F       Flatten again.
Q      uniQue the list.
L     the list of unique elements have Length...
<3   less than 3.

• Imaginary bounty if you can do it in linear time. I think it's not possible in Jelly, even ¦ takes O(n). – user202729 Jan 31 '18 at 7:20
• (without Python eval, of course) – user202729 Jan 31 '18 at 7:32

 \m->null.snd.until(null.fst)(\(f,e)->partition(\(b,p)->any(==1)[(b-d)^2+(p-q)^2|(d,q)<-f])e).splitAt 1.filter((/=0).(m!)).indices$m  extracted from Solve Hitori Puzzles indices m lists the (line,cell) locations of the input grid. filter((/=0).(m!)) filters out all locations with non-zero values. splitAt 1 partitions off the first member into a singleton list next to a rest list. any(==1)[(b-d)^2+(p-q)^2|(d,q)<-f] tells if (b,p) touches the frontier f. \(f,e)->partition(\(b,p)->touches(b,p)f)e splits off touchers from not[yet]touchers. until(null.fst)advanceFrontier repeats this until the frontier can advance no further. null.snd looks at the result whether all locations to be reached were indeed reached. Try it online! # Grime, 37 bytes C=[,0]&<e/\0{/e\0*0$e|CoF^0oX
eC|:\0

Prints 1 for match and 0 for no match. Try it online!

## Explanation

The nonterminal C matches any nonzero character that is connected to the first nonzero character of the matrix in the English reading order.

C=[,0]&<e/\0{/e\0*0$e|CoF^0oX C= A rectangle R matches C if [,0] it is a single character other than 0 & and < it is contained in a rectangle S which matches this pattern: e/\0{/e\0*0$e           R is the first nonzero character in the matrix:
e                        S has an edge of the matrix over its top row,
/0{/                    below that a rectangle of 0s, below that
e\0*0$e a row containing an edge, then any number of 0s, then R (the unescaped 0), then anything, then an edge. |CoF^0oX or R is next to another match of C: CoF S is a match of C (with fixed orientation) ^0 followed by R, oX possibly rotated by any multiple of 90 dergees.  Some explanation: e matches a rectangle of zero width or height that's part of the edge of the input matrix, and $ is a "wildcard" that matches anything. The expression e/\0{/e\0*0$e can be visualized as follows: +-e-e-e-e-e-e-e-+ | | | \0{ | | | +-----+-+-------+ e \0* |0|$   e
+-----+-+-------+

The expression CoX^0oX is actually parsed as ((CoF)0)oX; the oF and oX are postfix operators and concatenation of tokens means horizontal concatenation. The ^ gives juxtaposition a higher precedence then oX, so the rotation is applied to the entire sub-expression. The oF corrects the orientation of C after it is rotated by oX; otherwise it could match the first nonzero coordinate in a rotated English reading order.

eC|:\0
e       Match entire input against pattern:
:    a grid whose cells match
C      C
|     or
\0  literal 0.

This means that all nonzero characters must be connected to the first one. The grid specifier : is technically a postfix operator, but C|:\0 is syntactic sugar for (C|\0):.

# Java 8, 226 219 bytes

m->{int c=0,l=m[0].length,i=m.length*l;for(;i-->0;)if(m[i/l][i%l]>0){c++;f(m,i/l,i%l);}return c<2;}void f(int[][]m,int x,int y){try{if(m[x][y]>0){m[x][y]=0;f(m,x+1,y);f(m,x,y+1);f(m,x-1,y);f(m,x,y-1);}}finally{return;}}

Explanation:

Try it online.

m->{                   // Method with integer-matrix parameter and boolean return-type
int c=0,             //  Amount of non-zero islands, starting at 0
l=m[0].length,   //  Amount of columns in the matrix
i=m.length*l;    //  Index integer
for(;i-->0;)         //  Loop over the cells
if(m[i/l][i%l]>0){ //   If the current cell is not 0:
c++;           //    Increase the non-zero island counter by 1
f(m,i/l,i%l);} //    Separate method call to flood-fill the matrix
return c<2;}         //  Return true if 0 or 1 islands are found, false otherwise

void f(int[][]m,int x,int y){
// Separated method with matrix and cell input and no return-type
try{if(m[x][y]>0){   //  If the current cell is not 0:
m[x][y]=0;         //   Set it to 0
f(m,x+1,y);        //   Recursive call south
f(m,x,y+1);        //   Recursive call east
f(m,x-1,y);        //   Recursive call north
f(m,x,y-1);}       //   Recursive call west
}finally{return;}    //  Catch and swallow any ArrayIndexOutOfBoundsExceptions
//  (shorter than manual if-checks)

# Perl 5, 131 129 + 2 (-ap) = 133 bytes

push@a,[@F,0]}{push@a,[(0)x@F];$\=1;map{//;for$j(0..$#F){$b+=$a[$'][$j+$_]+$a[$'+$_][$j]for-1,1;$\*=$b||!$a[$'][$j];$b=0}}0..@a-2

Try it online!

# Python 2, 211163 150 bytes

m,w=input()
def f(i):a=m[i];m[i]=0;[f(x)for x in(i+1,i-1,i+w,i-w)if(x>=0==(i/w-x/w)*(i%w-x%w))*a*m[x:]]
f(m.index((filter(abs,m)or[0])[0]))<any(m)<1>q

Try it online!

Output is via exit code. Input is as an 1d list and the width of the matrix.

# JavaScript (Node.js), 115 102 bytes

A=>w=>(A.find(F=(u,i)=>u&&[-w,-1,1,w].map(v=>v*v>1|i%w+v>=0&i%w+v<w&&F(A[v+=i],v),A[i]=0)),!+A.join)

Try it online!

Receive input as (1d-array)(width). Rewrites all entries in one of the connected non-zero regions with 0 and checks whether the resultant array is a zero matrix. The matrix will become zero if there is zero or one connected non-zero region.

A=>                         // all rows are concatenated into an 1-d array
w=>                        // width of the original matrix
(
A.find(                  // find the first non-zero entry
F=(                     // helper function
u,                     // value of the entry
i                      // index of the entry
)=>
u                      // return 0 if value is 0 or out of range
&&[-w,-1,1,w].map(     // return an array otherwise, and find the neighbors
v=>                   // for each neighbor
v*v>1                // if y-coord change: let go anyway
|i%w+v>=0&i%w+v<w    // if x-coord change: let go only if on the same row
&&F(A[v+=i],v),      // recurse on the neighbor
A[i]=0                // before recursing change the current entry to 0
)
),                       // if all entries were 0, no change is made
!+A.join               // return whether all entries are now 0
)                         // if all entries are 0, then +A.join == 0

Tip: Use Array.prototype.find if you want to map through an array with a function that the return values can be ignored but to stop once the function is executed.

In this case, I want to map through the array to change only the first connected non-zero region found, but not the other regions (otherwise the array will become a zero matrix no matter what the original input is), so find` is used here to save some bytes.


text:
# 3.7 – Graphs of Quadratic Functions

## Objectives

• Determine the number of times that a given function crosses the x-axis.
• Find the vertex and x-intercepts of the graph of a quadratic function.
• Use the quadratic formula to find the vertex of a given function.
• Determine if a discriminant is positive, negative, or zero when given the graph of the function.
• Match a graph to its quadratic equation.

## Key Terms

• Average Rate of Change – A quotient that tells, on average, how some quantity changes over time.
• The ratio between the change in the dependent variable and the change in the independent variable.
• Instantaneous Slope – The slope of a line that is tangent to a curve at a point.
• It may be estimated by calculating the average slope between the point and a nearby point on the curve.
• The closer the nearby point, the better the estimation will be.
• In nonlinear functions, the average rate of change and the instantaneous slope are NOT the same thing.
• Vertex – Marks four features of a parabola:
• Its sharpest turn
• Its center
• Its line of symmetry
• The halfway point between its two x-intercepts

## Notes

Review
• Review
• Roots of a Polynomial
• The x-intercepts
• The places where the graph crosses, or touches, the x-axis
• The x-values that make y = 0
• Roots are also called Zeros (where the parabola crosses the x-axis)
• To find the zeros of a parabola, set the y equal to 0 and solve for x.
• Use any of the methods learned in this chapter to solve for x.
• Averages
• To find the average of 2 values, add the values together and divide by 2.
• Ex. 4 and 14
• Add 4 + 14:  18
• Divide by 2:  9
• The average of 4 and 14 is 9.
• When quadratic equations are graphed, they form parabolas.
• The vertex of a parabola is the point at which the direction of the function changes.
• The vertex of a parabola is also either the minimum or the maximum of the function, depending on whether the leading coefficient is positive or negative.
• Minimum: parabola opens upward
• Maximum: parabola opens downward

Finding the Vertex of a Parabola
• Option 1:  Finding the Vertex by Factoring & finding Roots
• 1st: find the x-intercepts (roots)
• 2nd: factor
• 3rd: set equal to zero & solve
• 4th: list the roots as coordinates (x,0) and (x,0)
• 5th: find the average of the x-values
• 6th: substitute the average into the original equation and find the y-value

• Example
• 1st & 2nd: $y=x^2-6x+5$ factors into (x – 1)(x – 5) = 0
• 3rd: x – 1 = 0 results in x = 1
• 3rd: x – 5 = 0 results in x = 5
• 4th: The roots are (1,0) and (5,0) since we know the roots are on the x-axis (where the y-value is equal to zero)
• 5th: find the average:

• 6th: substitute the x-value for the vertex into the original equation to find the y-value:

• Remember the signs of the equation and the signs of the binomial factors (x – #)(x + #) can help find the roots

• Option 2:  Finding the Vertex of a parabola with the Quadratic Formula
• Formula for finding the x-value of the vertex: $- \frac{b}{2a}$

* Imaginary numbers ARE complex roots

• Example 1:  $y=x^2-6x+5$
• a: 1
• b: -6
• c: 5
• Substitute:  $- \frac{-6}{(2)(1)}$
• Solve for the x-value:  3
• Substitute 3 for x and solve for y (see step 6 above)

• Example 2

Finding the Equation of a Parabola
• Parabolas are symmetrical curves, meaning that each side of the curve is an exact reflection of the other side.
• The line of symmetry (also called the axis of symmetry) that divides the parabola into two mirror images runs directly through its vertex, as shown in the image below.

• To find the equation of a parabola, set up the binomial factors as the opposite values of the roots (zeros)
• When a is positive, the root is (x – a).
• When a is negative, the root is (x + a).
• Ex. Roots: x = 6 and x = 2 will be the factors: (x – 6) and (x – 2). Use FOIL to find the equation: $x^2-8x+12$


text:
# Sign of potential energy of SHO

1. Dec 27, 2013

### devang2

Suppose the particle is at distance x from mean position and moving away . The standard formula for calculating potential U is U=-w here wis the work w=kxdx when the particle is moving away .On integration U=1/2Kx^2 . When the particle is moving towards mean position w=-kxdx on integration U=-1/2kx^2 thus the sign changes with direction Is it correct

2. Dec 27, 2013

### devang2

sorry i omitted in my previous post that work is calculated by dot product formula

3. Dec 27, 2013

### Tanya Sharma

Suppose there is a spring with its left end attached to a wall and right end having a block of mass m .Now we pull it slowly towards right ,then the work done by external agent is stored as potential energy.

W = ∫ F.dx
= ∫kxdxcos0 (since force is parallel to the displacement )
= [(1/2)kx2]0x
= (1/2)kx2

So , U = (1/2)kx2

4. Dec 28, 2013

### devang2

Thank you for the reply . So you mean that whether the spring is stretched or compressed the potential energy is always positive.While calculating the potential of particle executing simple harmonic oscillations the energy is always positive . I have used basic relation dU=-w to calculate the potential energy which gives opposite signs of potential energy when the particle is moving away or towards the mean position . Please take some trouble to point out where i have gone wrong while using the above formula

5. Dec 28, 2013

### Tanya Sharma

See...understand one thing dU = -dW,if dW is work done by the conservative force,spring force in our spring block example . But dU = dW if dW is work done by external agent .

Do you agree with this ?

6. Dec 28, 2013

### Philip Wood

This is correct. The reason is that a system's potential energy is the work it can do because of configuration (i.e. the relative positions of its parts). So if a system DOES work dW, it loses potential energy dU.

Suppose we consider force and displacement components in the x direction, that is the direction which tends to stretch the spring. If the spring is already extended, Fx will be negative, so for positive dx, dW = Fx dx will be negative and dU will be positive.
If the spring is compressed Fx will be positive, and if we compress it further, dx will be negative, so dW = Fx dx will be negative and dU will (again) be positive.

7. Dec 29, 2013

### devang2

Thank you for the reply .AS you know for oscillator force is always -kx and always directed towards mean position . i calculate the work dw by applying dot product of force and displacement. While dot product formula you know scalar values of force and displacement are used . When the particle is moving away from mean position angle is 180 because force and displacement are in opposite direction hence dw is negative but when particle is moving towards mean position force and displacement are in the same direction hence angle is 0 therefore according to product dw is positive. I shall be very thankful if you point out where i have gone wrong

8. Dec 29, 2013

### Staff: Mentor

The potential energy is at a maximum when the displacement is at a maximum, and it's zero at the mean position. So it's not at all surprising that the sign of $dW$ is different when the displacement is increasing and when the displacement is decreasing; in one case we're putting energy into the spring and in the other case we're taking energy out of it.

9. Dec 29, 2013

### Philip Wood

My method also uses dot product. If two vectors, A and B, have no y or z components, then it is easy to show that A.B = AxBx. The components, Ax and Bx, are scalar, but can be positive or negative.

Suggest you read my post carefully again, with this in mind. Or, if you prefer, use magnitudes (scalar and non-negative) of vectors, and put in the cosine value, +1 or -1, separately as you follow my argument. You'll find it gives the same result as my method each time. The argument shows that the more you deform the spring - either by extending it or compressing it - the more energy it stores. If you repeat the argument, changing the sign of dx, you'll find that if you reduce the deformation - either the extension or the compression - the spring stores less energy.

What I found confusing in your last post - and it may have some bearing on your own difficulty - is what you mean by 'displacement'. Are you using it to mean displacement of the mass from its equilibrium position (that's the usual meaning) or are you using it to mean the incremental displacement, dx?

Last edited: Dec 29, 2013
10. Dec 30, 2013

### devang2

Thank you for the trouble you have taken for the clarification . By displacement i mean the distance of the mass of the particle from the mean position executing harmonic oscillation.
suppose spring is stretched by certain distance , the work involved is w(!) and when the spring is released to restore the equilibrium original position work involved is W(2) because spring is in original state no net work is done therefore w(!)+w(2)=0 which leads to the conclusion the sign of work during stretching and decompressing are opposite .Similarly when the particle executing harmonic oscillation is moving away from mean position it is like stretching and when it is moving towards mean position it is like decompressing hence sign of work when it is moving away from mean position is negative and when it moves towards mean position it is positive which finally leads to the conclusion that sign of potential energy when particle is moving away is positive and it is negative when motion of particle is towards mean position I shall be very thankful if you let me know if i am right or wrog

11. Dec 30, 2013

### Philip Wood

Yes, that's right. The work done BY the spring is negative when it is being compressed or extended, and positive when being decompressed or de-extended. But, since I can see you like precision, I'll point out that your last sentence isn't quite correct...

What you should say is that the sign of the CHANGE in potential energy when particle is moving away is positive and it is negative when the motion of the particle is towards the mean position. [The potential energy itself is always positive (or zero) if we adopt the usual convention of assigning zero to the potential energy when the spring is undeformed.]

12. Dec 30, 2013

### vanhees71

I don't understand the problem. It's pretty simple. The potential of a force (if it exists!) is defined by
$$\vec{F}(\vec{r})=-\vec{\nabla} V(\vec{r}).$$
Given that, for the isotropic harmonic oscillator,
$$\vec{F}(\vec{r})=- m \omega^2 \vec{r},$$
you find by simple integration or taking the appropriate line integral (e.g., a straight line from the origin to $\vec{r}$) you get
$$V(\vec{r})=\frac{m \omega^2}{2} \vec{r}^2.$$
You easily check that indeed this is the potential, i.e., that $\vec{F}$ is indeed a conservative field.

Of course it also satisfies the local condition
$$\vec{\nabla} \times \vec{F}=0,$$
and since it is defined and analytic everywhere in $\mathbb{R}^3$ there must exist a potential.

13. Dec 31, 2013

### devang2

thank you for the analytical reply. My problem is very basic and simple ,it is about the sign of potential which changes sign with change in direction of motion of the particle executing harmonic oscillation though magnitude remains same .There is change in sign if potential is calculated by applying the basic formula ,dU=-W u is the potential and w is the work which i calcule by using dot product. Please calculate the potential by using this method , you will will see that sign of potential is positive when particle is moving away from mean position while it is negative when particle is moving towards mean position.Please pay attention to the sign. I shall be very thankful if you let me know your views about the change in sign .All the text books assign positive sign yo the potential disregarding the change in direction of motion .You know when mass is shifted to infinity from earth against gravity potential is positive but it is negative when it is brought from infinity moving in the direction of gravity thus it is seen that there is change in sign with the direction so same thing is applicable to motion of particle executing harmonic scillation because it is continusuosly changing direction of motion

14. Dec 31, 2013

### Philip Wood

You're still making the mistake that I pointed out in my last post! It's not the potential energy, but the CHANGE in potential energy which is positive when the particle is moving away from the mean position and negative when the particle is moving towards the mean position. The potential energy itself is always positive (if we assign zero to it at the mean position, when the spring is unstretched). It is more positive the more the spring is stretched.

Say if you're still not clear.

[PS: You've started to write 'potential' instead of 'potential energy'. Stick with potential energy!]

Last edited: Dec 31, 2013
15. Jan 1, 2014

### devang2

Thank you very much indeed for the clarification and pointing out ther mistake from which it can be concluded that work is negative when the particle is moving away from mean posiytion whereas it is positive whan the particle is moving towards mean position .

16. Jan 1, 2014

### Philip Wood

That's right: work done by the spring (or other body responsible for the restoring force) is negative when the particle moves away from the mean position, so the spring gains potential energy (dU = -dW). If we stretch the spring by X, then potential energy stored is

$$U = \int_{x = 0}^{X}{dU} = -\int_{x =0}^{X}{dW} = -\int_{0}^{X}{-kx dx} =\frac{1}{2}k X^2$$

You won't find it done quite like this in the textbooks. I'm using dW to mean an increment of work done BY the spring, following - I hope - what you were doing. If you consider the work done ON the spring as it is stretched, you can dispense with the minus signs in the above derivation.

Last edited: Jan 1, 2014
17. Jan 2, 2014

### devang2

Your way of expressing makes it easy for me to understand the problem.If the particle is moving towards the mean position,applying the method as adopted when the particle is moving away from mean position potential energy of the particle in SHO is negative that is -1/2kx^2. I shall be very thankful if you can clarify it . My contention is that the potential energy changes sign with change in direction of motion

18. Jan 2, 2014

### vanhees71

No, it doesn't. It doesn't even depend on the velocity. The potential is $V(\vec{r})=k \vec{r}^2/2$ and thus independent on the direction of the position vector and doesn't depend on velocity at all. It's the force, which is
$$\vec{F}=-k \vec{r}$$
that changes it's direction with the position vector, not the potential!

19. Jan 2, 2014

### Philip Wood

deVang2. Same mistake AGAIN ! When the particle is moving TOWARDS the mean position, the CHANGE in potential energy is negative, so the spring is LOSING the potential energy that it had when it had a greater displacement. By the time it has got back to x = 0 it has lost ALL the potential energy. So at x = 0, it has zero potential energy, not a negative potential energy. Its potential energy never goes negative.

20. Jan 2, 2014

### Philip Wood

Same mistake AGAIN ! As the particle moves towards the mean position the CHANGE in potential energy is negative, so the spring is losing the potential energy which it had at displacement X. When it has got back to the mean position it has lost ALL the potential energy. Its potential energy is now zero. The potential energy is never negative.


text:
# A note on homogeneous Sobolev spaces of fractional order

created by brasco on 20 Apr 2018
modified on 23 Jun 2018

[BibTeX]

Submitted Paper

Inserted: 20 apr 2018
Last Updated: 23 jun 2018

Pages: 33
Year: 2018

Abstract:

We consider a homogeneous fractional Sobolev space obtained by completion of the space of smooth test functions, with respect to a Sobolev--Slobodecki\u{\i} norm. We compare it to the fractional Sobolev space obtained by the $K-$method in real interpolation theory. We show that the two spaces do not always coincide and give some sufficient conditions on the open sets for this to happen. We also highlight some unnatural behaviors of the interpolation space. The treatment is as self-contained as possible.


text:
# networkx.algorithms.tree.mst.minimum_spanning_tree¶

minimum_spanning_tree(G, weight='weight', algorithm='kruskal', ignore_nan=False)[source]

Returns a minimum spanning tree or forest on an undirected graph G.

Parameters
• G (undirected graph) – An undirected graph. If G is connected, then the algorithm finds a spanning tree. Otherwise, a spanning forest is found.

• weight (str) – Data key to use for edge weights.

• algorithm (string) – The algorithm to use when finding a minimum spanning tree. Valid choices are ‘kruskal’, ‘prim’, or ‘boruvka’. The default is ‘kruskal’.

• ignore_nan (bool (default: False)) – If a NaN is found as an edge weight normally an exception is raised. If ignore_nan is True then that edge is ignored instead.

Returns

G – A minimum spanning tree or forest.

Return type

NetworkX Graph

Examples

>>> G = nx.cycle_graph(4)
>>> T = nx.minimum_spanning_tree(G)
>>> sorted(T.edges(data=True))
[(0, 1, {}), (1, 2, {}), (2, 3, {})]

Notes

For Borůvka’s algorithm, each edge must have a weight attribute, and each edge weight must be distinct.

For the other algorithms, if the graph edges do not have a weight attribute a default weight of 1 will be used.

There may be more than one tree with the same minimum or maximum weight. See networkx.tree.recognition for more detailed definitions.

Isolated nodes with self-loops are in the tree as edgeless isolated nodes.


text:
# How to show that $f(x) = \cos^2(x)\sin(x)$ is symmetric about the line $x=\frac{1}{2} \pi$?

I really have trouble with making any exercises regarding point symmetry and line symmetry. For example:

Show that $f(x) = \cos^2(x)\sin(x)$ is line symmetrical in the line $x=\dfrac{1}{2} \pi$.

So I need to show $f(a-p) = f(a+p)$, but I have no idea how to. Even when I see the answers, I don't know. I just see them rewriting cosines and sines in different forms.

-

Note that $$\sin(\tfrac{\pi}{2}-x)=\cos(x)\qquad \text{ and }\qquad \cos(\tfrac{\pi}{2}-x)=\sin(x)$$ (see Wikipedia for example; you can also just draw a triangle with angles $x$, $\frac{\pi}{2}-x$, and $\frac{\pi}{2}$).
Using the above facts together with the facts $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$, we see that $$\sin(\tfrac{\pi}{2}+x)=\cos(x)\qquad\text{ and }\qquad \cos(\tfrac{\pi}{2}+x)=-\sin(x).$$ Thus, we have $$f(\tfrac{\pi}{2}-x)=\bigg[\cos(\tfrac{\pi}{2}-x)\bigg]^2\sin(\tfrac{\pi}{2}-x)=\sin^2(x)\cos(x)$$ and $$f(\tfrac{\pi}{2}+x)=\bigg[\cos(\tfrac{\pi}{2}+x)\bigg]^2\sin(\tfrac{\pi}{2}+x)=(-\sin^2(x))\cos(x)=\sin^2(x)\cos(x)$$ so that $$f(\tfrac{\pi}{2}-x)=f(\tfrac{\pi}{2}+x).$$
I suppose so; I certainly keep very few trig identities in memory, I just work it out when necessary. Sometimes drawing a triangle is the best way to go, but if you're comfortable with complex numbers and their relation with trigonometric functions, e.g. these identities and de Moivre's formula $$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx),$$ then that provides a very useful technique as well. – Zev Chonoles Apr 5 '13 at 6:28


text:
# Fermat's principle

The actual ray path between two points is the one for which the optical path length is stationary with respect to variations of the path. Can you explain me what is extremum path?

• Extremum means maximum or minimum. In the Fermat principle, it is usually (but not absolutely always) the path that takes the minimum time. Because the derivative (speed of change) in the extremum is zero, the path is stationary (no change due to small variations). – safesphere Sep 22 '17 at 5:36
• Take a look in my answer therein : Why one should follow Snell's law for shortest time? May be help you. – Frobenius Sep 22 '17 at 6:04

1. An global (local) extremum is a global (local) maximum or minimum, respectively.

2. Note that the principle of least action is more accurately a 'principle of stationary action', since we are looking for stationary/critical paths rather than minimum paths.

3. Similarly, Fermat's principle aka. the principle of least time is more accurately a 'principle of stationary time'.

As an example suppose you want to find the path that light takes in going from £A£ to $B$ as a result of a reflection.

There are an infinite number of possible paths and below I have shown three.

You will note that the tame taken for the light to travel from $A$ to $B$ depends on the path taken but one of the paths $AYZ$ takes a minimum (extremum) path.
Fermat says that that is the path taken by the light and it also happens to be the path such that the angle of incidence is equal to the angle of reflection.

Feynman devotes a whole chapter "Optics: The Principle of Least Time" on this topic.


text:
Modulo Multiplication is Closed

Theorem

Multiplication modulo $m$ is closed on the set of integers modulo $m$:

$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m \in \Z_m$.

Proof

From the definition of multiplication modulo $m$, we have:

$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$

By the Division Theorem:

$x y = q m + r$ where $0 \le r < m$

Therefore for all $0 \le r < m$:

$\eqclass {x y} m = \eqclass r m$

Therefore from the definition of integers modulo $m$:

$\eqclass {x y} m \in \Z_m$

$\blacksquare$


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are reading an older version of this FlexBook® textbook: CK-12 Geometry - Basic Go to the latest version.

6.2: Properties of Parallelograms

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Define a parallelogram.
• Understand the properties of a parallelogram
• Apply theorems about a parallelogram’s sides, angles and diagonals.

Review Queue

1. Draw a quadrilateral with one set of parallel sides.
2. Draw a quadrilateral with two sets of parallel sides.
3. Find the measure of the missing angles in the quadrilaterals below.

Know What? A college has a parallelogram-shaped courtyard between two buildings. The school wants to build two walkways on the diagonals of the parallelogram and a fountain where they intersect. The walkways are going to be 50 feet and 68 feet long. Where would the fountain be?

What is a Parallelogram?

Parallelogram: A quadrilateral with two pairs of parallel sides.

Notice that each pair of sides is marked parallel. Also, recall that two lines are parallel when they are perpendicular to the same line. Parallelograms have a lot of interesting properties.

Investigation 6-2: Properties of Parallelograms

Tools Needed: Paper, pencil, ruler, protractor

1. Draw a set of parallel lines by drawing a 3 inch line on either side of your ruler.
2. Rotate the ruler and repeat so you have a parallelogram. If you have colored pencils, outline the parallelogram in another color.
3. Measure the four interior angles of the parallelogram as well as the length of each side. What do you notice?
4. Draw the diagonals. Measure each and then measure the lengths from the point of intersection to each vertex.

To continue to explore the properties of a parallelogram, see the website: http://www.mathwarehouse.com/geometry/quadrilaterals/parallelograms/interactive-parallelogram.php

Opposite Sides Theorem: If a quadrilateral is a parallelogram, then the opposite sides are congruent.

If then

Opposite Angles Theorem: If a quadrilateral is a parallelogram, then the opposite angles are congruent.

If then

Consecutive Angles Theorem: If a quadrilateral is a parallelogram, then the consecutive angles are supplementary.

If then

$m \angle A + m \angle D = 180^\circ\\m \angle A + m \angle B = 180^\circ\\m \angle B + m \angle C = 180^\circ\\m \angle C + m \angle D = 180^\circ$

Parallelogram Diagonals Theorem: If a quadrilateral is a parallelogram, then the diagonals bisect each other.

If then

Proof of Opposite Sides Theorem

Given: $ABCD$ is a parallelogram with diagonal $\overline{BD}$

Prove: $\overline{AB} \cong \overline{DC},\overline{AD} \cong \overline{BC}$

Statement Reason
1. $ABCD$ is a parallelogram with diagonal $\overline{BD}$ Given
2. $\overline{AB} \| \overline{DC}, \overline{AD} \| \overline{BC}$ Definition of a parallelogram
3. $\angle ABD \cong \angle BDC, \angle ADB \cong \angle DBC$ Alternate Interior Angles Theorem
4. $\overline{DB} \cong \overline{DB}$ Reflexive PoC
5. $\triangle ABD \cong \triangle CDB$ ASA
6. $\overline{AB} \cong \overline{DC}, \overline{AD} \cong \overline{BC}$ CPCTC

The proof of the Opposite Angles Theorem is almost identical. For the last step, the angles are congruent by CPCTC.

Example 1: $ABCD$ is a parallelogram. If $m \angle A = 56^\circ$, find the measure of the other angles.

Solution: Draw a picture. When labeling the vertices, the letters are listed, in order, clockwise.

If $m \angle A = 56^\circ$, then $m \angle C = 56^\circ$ by the Opposite Angles Theorem.

$m \angle A + m \angle B & = 180^\circ \quad \text{by the Consecutive Angles Theorem.}\\56^\circ + m \angle B & = 180^\circ\\m \angle B & = 124^\circ \quad m \angle D = 124^\circ \quad \text{because it is opposite angles with} \ \angle B.$

Example 2: Algebra Connection Find the values of $x$ and $y$.

Solution: Opposite sides are congruent.

$6x - 7 & = 2x + 9 && y + 3 = 12\\4x & = 16 && \qquad y = 9\\x & = 4$

Diagonals in a Parallelogram

From the Parallelogram Diagonals Theorem, we know that the diagonals of a parallelogram bisect each other.

Example 3: Show that the diagonals of $FGHJ$ bisect each other.

Solution: Find the midpoint of each diagonal.

$&\text{Midpoint of} \ \overline{FH}: \qquad \left ( \frac{-4 + 6 }{2}, \frac{5 - 4}{2} \right ) = (1, 0.5)\\&\text{Midpoint of} \ \overline{GJ}: \qquad \left ( \frac{3 - 1}{2}, \frac{3 - 2}{2} \right ) = (1,0.5)$

Because they are the same point, the diagonals intersect at each other’s midpoint. This means they bisect each other.

This is one way to show a quadrilateral is a parallelogram.

Example 4: Algebra Connection $SAND$ is a parallelogram and $SY = 4x - 11$ and $YN = x + 10$. Solve for $x$.

Solution:

$SY & = YN\\4x - 11 & = x + 10\\3x & = 21\\x & = 7$

Know What? Revisited The diagonals bisect each other, so the fountain is going to be 34 feet from either endpoint on the 68 foot diagonal and 25 feet from either endpoint on the 50 foot diagonal.

Review Questions

• Questions 1-6 are similar to Examples 2 and 4.
• Questions 7-10 are similar to Example 1.
• Questions 11-23 are similar to Examples 2 and 4.
• Questions 24-27 are similar to Example 3.
• Questions 28 and 29 are similar to the proof of the Opposite Sides Theorem.
• Question 30 is a challenge. Use the properties of parallelograms.

$ABCD$ is a parallelogram. Fill in the blanks below.

1. If $AB = 6$, then $CD =$ ______.
2. If $AE = 4$, then $AC =$ ______.
3. If $m \angle ADC = 80^\circ, m \angle DAB$ = ______.
4. If $m \angle BAC = 45^\circ, m \angle ACD$ = ______.
5. If $m \angle CBD = 62^\circ, m \angle ADB$ = ______.
6. If $DB = 16$, then $DE$ = ______.
7. If $m \angle B = 72^\circ$ in parallelogram $ABCD$, find the other three angles.
8. If $m \angle S = 143^\circ$ in parallelogram $PQRS$, find the other three angles.
9. If $\overline{AB} \perp \overline{BC}$ in parallelogram $ABCD$, find the measure of all four angles.
10. If $m \angle F = x^\circ$ in parallelogram $EFGH$, find the other three angles.

For questions 11-19, find the measures of the variable(s). All the figures below are parallelograms.

Use the parallelogram $WAVE$ to find:

1. $m \angle AWE$
2. $m \angle ESV$
3. $m \angle WEA$
4. $m \angle AVW$

Find the point of intersection of the diagonals to see if $EFGH$ is a parallelogram.

1. $E(-1, 3), F(3, 4), G(5, -1), H(1, -2)$
2. $E(3, -2), F(7, 0), G(9, -4), H(5, -4)$
3. $E(-6, 3), F(2, 5), G(6, -3), H(-4, -5)$
4. $E(-2, -2), F(-4, -6), G(-6, -4), H(-4, 0)$

Fill in the blanks in the proofs below.

1. Opposite Angles Theorem

Given: $ABCD$ is a parallelogram with diagonal $\overline{BD}$

Prove: $\angle A \cong \angle C$

Statement Reason
1. Given
2. $\overline{AB} \| \overline{DC},\overline{AD} \| \overline{BC}$
3. Alternate Interior Angles Theorem
4. Reflexive PoC
5. $\triangle ABD \cong \triangle CDB$
6. $\angle A \cong \angle C$
1. Parallelogram Diagonals Theorem

Given: $ABCD$ is a parallelogram with diagonals $\overline{BD}$ and $\overline{AC}$

Prove: $\overline{AE} \cong \overline{EC}, \overline{DE} \cong \overline{EB}$

Statement Reason
1.
2. Definition of a parallelogram
3. Alternate Interior Angles Theorem
4. $\overline{AB} \cong \overline{DC}$
5.
6. $\overline{AE} \cong \overline{EC}, \overline{DE} \cong \overline{EB}$
1. Challenge Find $x, y^\circ,$ and $z^\circ$. (The two quadrilaterals with the same side are parallelograms.)

1. $3x+x+3x+x=360^\circ\!\\{\;}\qquad \qquad \quad \ 8x = 360^\circ\!\\{\;}\quad \ \qquad \quad \ \ \ \ x = 45^\circ$
2. $4x+2=90^\circ\!\\{\;}\quad \ 4x=88^\circ\!\\{\;}\qquad x=22^\circ$

8 , 9 , 10

Date Created:

Feb 22, 2012

Dec 11, 2014
Files can only be attached to the latest version of None


text:
# [tex-live] XeTeX and LaTeX-Beamer

Zdenek Wagner zdenek.wagner at gmail.com
Sun Jan 27 19:42:44 CET 2008

2008/1/27, Jonathan Kew <jonathan_kew at sil.org>:
> On 27 Jan 2008, at 3:55 pm, Ulrich Groh wrote:
>
> > How can I combine XeTeX and Latexbeamer?
>
> I'm not sure of the exact details, but I think you need to update the
> PGF and Beamer packages from CTAN, as the versions shipped on TeX
> Live 2007 had some incompatibilities with XeTeX.
>
PGF in TL 2007 does not recognise XeTeX but it is sufficient to add
the following definition to your preamble:
\def\pgfsysdriver{pgfsys-dvipdfm.def}

> The XeTeX mailing list (see http://tug.org/mailman/listinfo/XeTeX/)
> is probably the best place to find users who could give more
> information on this.
>
> Jonathan
>
>

--
Zdeněk Wagner
http://hroch486.icpf.cas.cz/wagner/
http://icebearsoft.euweb.cz


text:
## Introduction

In recent years, there has been a significant interest in 2D vdW materials due to the realization of magnetism in the 2D limit. Over just the past 5 years, various 2D vdW magnetic materials, such as CrI31, Cr2Ge2Te62, VSe23, MnSe24, and FexGeTe2 (x = 3–5)5,6 have been reported. These materials contain 3d magnetic elements in their 2D atomically thin crystals and thus exhibit intrinsic magnetic properties. Compared to bulk 3D materials, the 2D vdW magnetic materials provide additional controllability via strain, surface doping, gating, Moiré potential, etc. Therefore, the discovery of 2D vdW magnetic materials has provided an interesting route to explore new physical phenomena and design novel devices. The realization of magnetic 2D vdW materials is reminiscent of the possible existence of the heavy-fermion state in 2D materials.

The heavy-fermion materials containing rare-earth or actinide ions can be regarded as prototypes of lattice Kondo systems, where the interplay between the Ruderman–Kittel–Kasuya–Yoshida (RKKY) interaction and the Kondo effect determines the ground state. When the RKKY interaction dominates, local moments tend to form long-range magnetic order at low temperatures, usually the antiferromagnetic (AFM) phase. On the other hand, when the Kondo effect prevails, local magnetic moments are screened by itinerant conduction electrons leading to the paramagnetic (PM) Kondo resonance state (strong hybridization between f electrons and conduction electrons; fc hybridization). Due to the small energy scales of two competing interactions, the ground state can be easily tuned through a quantum phase transition by external perturbations, such as pressure, chemical doping, or magnetic field7,8,9,10,11,12,13,14,15,16,17. In the vicinity of QCP, anomalous phenomena often occur including unconventional superconductivity and non-Fermi liquid. If a heavy-fermion state can be realized in 2D vdW materials, it can provide a new platform to study QCP, magnetism, and superconductivity with additional controllability.

There have been several attempts to realize the heavy-fermion state in the 2D limit. The dimensionality tuning from 3D to 2D heavy-fermion state was firstly reported by using the epitaxially grown CeIn3/LaIn3 superlattice18. Several other artificial Kondo superlattices have been studied to investigate the interaction between two different phases, such as unconventional superconductivity and magnetic order, in the quasi-2D regime19,20,21,22,23,24. Recently, it was reported that the artificial 2D rare-earth free heavy-fermion state can be realized from a 1T/1H-TaS2 heterostructure25,26. The Kondo coupling between the localized moment in the 1T-TaS2 layer and the itinerant electrons in the 1H-TaS2 layer gives rise to a heavy-fermion state. Trilayer twisted graphene was also suggested as a vdW platform for the realization of heavy-fermion physics27. However, a genuine 2D heavy-fermion state within the monolayer limit has not yet been reported.

A recent high-throughput computational study proposed 1036 easily exfoliable compounds from experimentally known compounds28. Their calculations using vdW density functional theory were validated against random phase approximation calculation and experimental results in already well-known 2D materials. This large portfolio of materials enables us to investigate the novel 2D vdW exfoliable materials with desired properties. In order to explore the potential 2D vdW heavy-fermion candidates, we considered 32 lanthanide materials containing 4f orbitals, which is the essential ingredient of the intrinsic heavy-fermion state. We believe that this is a good starting point to explore new 2D vdW heavy-fermion materials.

In this study, we investigate possible 2D vdW heavy-fermion materials based on the reported exfoliable lanthanide materials28. By analyzing the electronic structure and the Kondo resonance state, Our result reveals that the CeSiI system is the most promising 2D vdW heavy-fermion candidate. The dynamical mean-field theory calculation combined with density functional theory (DFT+DMFT) shows that the electronic properties do not change upon exfoliation and the Kondo resonance state of the CeSiI monolayer can be sensitively tuned by strain and surface doping. Based on these findings, we predict the ground state of the CeSiI monolayer in response to the external perturbation by comparing it with other well-known Ce-based heavy-fermion materials.

## Results

### Screening from the database

We first briefly investigate the electronic structures for 32 layered lanthanide materials proposed to be exfoliable. Not only localized f electrons but also itinerant conduction electrons are essential to forming a heavy-fermion state. Thus, one can easily get an important clue to identify heavy-fermion materials by simply checking the conduction bands at the Fermi level (EF). Therefore, we perform DFT open-core calculations, which consider f electrons as core electrons, to examine the existence of non-f conduction bands at EF.

Figure 1 shows the density of states (DOS) at EF for 32 layered lanthanide materials obtained from DFT calculation within the open-core approach. Open symbols indicate La (f0) and Lu (f14) compounds. Since they have empty or fully occupied 4f shells, the lattice Kondo effect cannot be expected. On the other hand, the compounds denoted by square symbols also cannot be heavy-fermion compounds since they exhibit an insulating nature, indicating that there are no itinerant conduction bands to screen localized 4f electrons. Finally, we are left with only four compounds (CeSiI, ErHCl, TmI2, and YbI2) satisfying the minimal necessary conditions. The four remaining candidates can be again divided into two categories. The conduction bands of CeSiI at EF are mainly coming from the ligand atoms, while those of other compounds are coming from solely lanthanide 5d orbitals.

For comparison, α-Ce is a typical heavy-fermion system whose 4f electron is screened by its own 5d orbitals. In contrast, γ-Ce, which has an isostructure with a larger volume, shows weak 4f–5d hybridization yielding a small Kondo resonance peak29,30. Since ErHCl, TmI2, and YbI2 have ligand atoms between their lanthanide atoms, the distance between the lanthanide atoms of those compounds is even greater than that of γ-Ce (3.65 Å). Therefore, 4f electrons are hardly screened by neighboring lanthanide 5d orbitals or the Kondo cloud at each lanthanide site hardly sees each other to form the coherent heavy-fermion state.

Although CeI2 is not classified into exfoliable compounds from the previous study28, it has the same structure as TmI2 and YbI2. CeI2 has a magnetically ordered ground state and the previous photoemission spectroscopy (PES) study showed that the f1 peak is almost absent at EF, indicating 4f–5d hybridization in CeI2 is much smaller than α and γ-Ce cases30. Our DFT+DMFT calculation on CeI2 also verifies the absence of a Kondo peak near EF (see Supplementary Fig. 1). Considering these aspects, we conclude CeSiI is the most promising candidate for 2D vdW heavy-fermion systems among 32 exfoliable lanthanide compounds. Very recently, it was reported that bulk CeSiI has a magnetically ordered ground state31. However, it can be sensitively tuned by external perturbations, which will be discussed in later sections.

### Structural and electronic properties of CeSiI

The CeSiI vdW layer consists of a silicene layer, surrounding triangular Ce cage (orange dashed line), and outer iodine layers. The vdW gap can be defined as the iodine to iodine as depicted in Fig. 2a. The electronic structures calculated by DFT within the open-core approach are presented in Fig. 2b in terms of DOS (right) and band dispersion (left). Analysis of DOS and band dispersion leads to the following observation: (1) The iodine p state occurs primarily between −5 and −2 eV indicating that the iodine p state is fully occupied, and their contribution at EF is negligible. (2) The band dispersion near EF resembles that of pure silicene compound while the Dirac point at K point is moved to −2 eV in contrast to pure silicene, where the Dirac point is located at EF32. This indicates there is charge transfer from the Ce atom to the silicene layer. (3) Therefore, the bands crossing EF should be coming from Si px and py antibonding state. However, the contribution from Si at EF is quite small compared to the total DOS as shown in the right panel of Fig. 2b.

The most of contribution at EF is originated from the interstitial region. Figure 2c shows the calculated charge density distribution near EF (energy range −1.5 to 1.5 eV with respect to EF) for CeSiI on (110) plane (see the red thick line in the left lower panel in Fig. 2a). The electron density is delocalized all over the Ce cage (between Ce atoms and silicene layer), not localized at specific atoms. This feature agrees well with the Zintl–Klemm feature of [Ce3+ISi]e as suggested by the previous study33. Floating electrons on the silicene layer and the interstitial region can be the ingredient of a heavy-fermion state.

To investigate the hybridization between Ce 4f and conduction electrons (fc hybridization) properly, DFT+DMFT calculations are performed. The calculated spectral function of bulk and monolayer CeSiI are shown in Fig. 3a at 10 K. Unlike CeI2 case, the Kondo resonance peak is clearly observed near EF, indicating substantial hybridization between Ce 4f and conduction electrons. Therefore, CeSiI is located in the vicinity of the PM heavy-fermion state. The electronic structures of bulk and monolayer CeSiI are almost identical, from which similar chemical and physical properties are expected. The temperature-dependent evolution of the Kondo resonance peak also does not change upon exfoliation as shown in the inset of Fig. 3a, b. Therefore, the CeSiI monolayer can provide an ideal platform for exploring the heavy-fermion state in the 2D monolayer limit.

### Effect of strain and surface doping

Now we investigate the external perturbation effect on the Kondo resonance state of the CeSiI monolayer. Figure 4 shows the electronic structure of the CeSiI monolayer with biaxial strain from −4% to 4%. Upon compressive (tensile) strain, the volume of the Ce cage (orange dashed line in Fig. 2a, c) surrounding the silicene layer decreases (increases). Therefore, one can easily expect stronger fc hybridization under compressive strain due to the enhanced conduction electron density in the Ce cage. Figure 4a shows the spectral function of Ce 4f state at 10 K depending on the strain. The Kondo resonance peak is sensitively affected by the strain.

The T-dependent 4f spectral function at EF, A4f(ω = 0) is shown in Fig. 4b. The Kondo resonance peak is enhanced by the compressive strain for a wide temperature range, resulting in a higher Kondo temperature, TK energy scale. TK can be estimated from the logarithmic T dependence behavior of A4f(ω = 0) ~ ln(TK/T) (dashed line in Fig. 4b)34,35. The estimated TK for monolayer without strain is 120 K. TK slightly decreases to 110 K upon 4% tensile strain while it is strongly enhanced up to 200 K upon the compressive strain. The strain-induced fc hybridization change is also clearly observed in the momentum-resolved spectral function as shown in Fig. 4c–e. Not only the Kondo resonance state from j = 5/2 state at EF but also those from j = 7/2 state at 0.3 eV are clearly enhanced upon the compressive strain. The kink feature at 0.3 eV above EF associated with j = 7/2−c hybridization is very weak for 4% tensile strain but it becomes much clear and stronger for compressive strain (dashed circle in Fig. 4c). In general, the kink feature in the spectral function is related to the formation of the Kondo resonance and the fc hybridization in heavy-fermion systems36.

Another notable aspect is band-dependent fc hybridization. This feature is most clearly observed in the case of −4% compressive strain. At the one-third point along the K−Γ line (white arrow), two dispersive bands are hybridized with Ce 4f states. However, the hybridization strengths of the two bands are quite different. One band is strongly hybridized with the Ce 4f state resulting in a large hybridization gap of 50 meV while the hybridization gap from the other band is only 10 meV. This difference arises from the different orbital character of the conduction band. The left conduction band mainly comes from Ce $${d}_{{z}^{2}}$$, whereas the right band mainly comes from Si px+py orbitals (see Supplementary Fig. 2). The hybridization between Ce 4f and Si px+py states is much stronger and sensitively affected by the strain compared to the hybridization between Ce 4f and Ce $${d}_{{z}^{2}}$$ state.

Next, we investigate the surface doping effect on the Kondo resonance state. There are two Ce-I layers above and below the silicene layer within a monolayer as shown in Fig. 5a. By using the virtual crystal approximation, we control the number of electrons of the upper iodine layer, mimicking the surface doping situation. The calculated Ce 4f orbitals occupancy of each upper and lower Ce layer is presented in Fig. 5b. Regardless of surface doping, Ce 4f electron occupancy remains constant for both Ce layers. However, the Kondo resonance peak of the upper Ce layer is sensitively affected by surface doping while that of the lower Ce layer does not change as shown in Fig. 5c. The surface hole doping significantly enhances only the Kondo resonance state of the upper Ce layer. This suggests that by adjusting surface doping, the Kondo resonance peaks of the upper and lower layer are separately controlled within the monolayer. In addition, I substitution by Te, CeSiI1−xTex, would be considered an effective way to enhance the Kondo resonance of CeSiI monolayer.

### Phase diagram and ground state prediction

An important remaining question is how sensitive the ground state of the CeSiI monolayer can be controlled by strain or surface doping. To answer this question, we construct a phase diagram as shown in Fig. 6, by calculating the inverse DOS (1/A4f(ω = 0)) at EF (x-axis) and inverse quasiparticle lifetime Γ = −ZImΣ(0) (where Z−1 = m*/m = 1−∂ReΣ(ω)/∂ωω→0) (y-axis) based on PM DFT+DMFT calculations (at T ~ 46 K) and in comparison with well-known Ce-based heavy-fermion materials as shown in Fig. 6. Both quantities of each material are normalized by those of CeCoIn5 for better comparison. The materials having PM heavy-fermion ground states are denoted by circles while those having magnetically ordered ground states are denoted by diamonds, respectively. Note that CeCoIn5 is located in the close vicinity of the QCP and magnetically ordered CeRhIn5 undergoes a superconducting transition only above 2 GPa, indicating the proximity to the QCP. Interestingly, the well-known Ce-based materials are well classified with respect to the CeCoIn5 point. The typical PM heavy-fermion systems are located below the CeCoIn5 point and magnetically ordered systems are located above the CeCoIn5. The PM ground state of CeRhIn5 under 7 GPa is also well captured8,9,37.

Along with the well-known Ce-based compounds, CeSiI monolayers with and without external perturbation (strain, surface doping, and Si substitution) are also displayed with black and open squares in Fig. 6. For the surface doping case, only the Γ and 1/A4f(ω = 0) of upper layer Ce are shown in the figure because those of lower layer Ce do not change upon surface doping on the upper iodine layer. CeSiI monolayer is located above the CeCoIn5 point, indicating a magnetically ordered ground state and it is consistent with the recent experimental result on bulk CeSiI31. It is rather located closer to CeIn3 and CeCu2Ge2. Although CeRhIn5, CeIn3, and CeCu2Ge2 have magnetically ordered ground states, they have been intensively considered as AFM heavy-fermion systems since their ground state can be easily turned into PM heavy-fermion states by external pressure or chemical doping7,8,9,10,11,12,13,14,15,16,17. Therefore, CeSiI should also be located not far from the QCP, and its ground state can be easily tuned by external perturbations like CeIn3 and CeCu2Ge2.

As we have discussed above, compressive strain and surface hole doping strongly enhance the Kondo resonance peak of the CeSiI monolayer. With 4% compressive strain, it is located right below the CeCoIn5 indicating a possible quantum phase transition to PM heavy-fermion state. With 0.1 surface hole doping, it moves further below CeCoIn5 and is located near the well-known PM heavy-fermion materials CeIrIn5 and CeCu2Si2. Therefore, CeSiI monolayer is a promising 2D vdW heavy-fermion candidate and its ground state can be tuned by strain or surface doping (or I substitution by Te), providing alternative ways to explore the QCP, magnetism, and unconventional superconductivity.

## Discussion

The interesting points of this new type of heavy-fermion system are low dimensionality and frustration originating from its own triangular lattice. During the last couple of decades, the degree of quantum fluctuation has been studied in addition to the Kondo coupling and RRKY interaction. This quantum fluctuation can be enhanced by both lattice frustration and geometrical low-dimensionality. Together with the Kondo coupling strength, this new microscopic quantity (i.e., frustration) enables the construction of a 2D parameter space, zero-temperature global phase diagram38,39. So far, a complete understanding of this global phase diagram is still lacking.

By enhancing spatial dimensionality to reduce the quantum fluctuation, the cubic Ce3Pd20Si6 was studied40. One can tune Ce3Pd20Si6 from a small Fermi surface (FS) AFM state, through a large FS AFM phase, into a large FS PM state. Recently, it is reported that geometrically frustrated CePdAl hosts a stable quantum critical phase, possibly PM small FS phase. YbRh2Si241 and β-YbAlB442 also show similar quantum phase transitions. Although they do not have geometrical frustration, their quasi-2D crystal structure may enhance the quantum fluctuation and lead to a similar quantum phase transition trajectory on the global phase diagram.

However, an understanding of the microscopic processes remains missing. CeSiI monolayer can provide a new platform to study the global phase diagram. Due to its low dimensionality and lattice frustration, one can study the upper part of the global phase diagram, where the quantum fluctuation is strong. Besides, this 2D vdW heavy-fermion system has additional control parameters. In addition to magnetic field and pressure (uniform strain), uniaxial strain can be an external tuning parameter in the CeSiI monolayer. By inducing anisotropy via uniaxial strain, frustration can be controlled, providing a unique opportunity to explore the entire global phase diagram.

Another interesting aspect is the critical behavior of resistivity near this QCP with low dimensionality or frustration. In the epitaxially grown CeIn3/LaIn3 superlattice, the quantum phase transition within the 2D limit was observed by reducing the superlattice period. Near the QCP, the resistivity shows T-linear behavior deviating from the Fermi liquid behavior ρ(T) = ρ0 + ATα, where α = 2. This linear T-dependence (α = 1) is also in contrast to the resistivity behavior observed near the pressure-induced QCP in the bulk CeIn3 (α = 1.5)18. YbRh2Si2 also shows T-linear resistivity near the QCP41,43, while β-YbAlB4 recovers the Fermi liquid behavior (α = 2) at a very low temperature42. Therefore, it would be interesting to study the resistivity behavior in the CeSiI monolayer to further understand the strange metallic behavior in the global phase diagram.

Finally, the CeSiI monolayer is a new building block to study emergent phenomena in vdW heterostructures. Although 1T/1H-TaS2 heterostructure and twisted trilayer graphene can also be 2D heavy-fermion building blocks, delicate preparation is required since those heavy-fermion states are not intrinsic. It would be interesting to study vdW heterostructures which consist of CeSiI monolayer and other 2D vdW components, including magnetic materials. Although the epitaxially grown Kondo superlattice, such as CeCoIn5/CeIn3 and CeCoIn5/CeRhIn5, have been used to study the dimensionality effect and the interactions between heavy-fermion state (or superconducting state) and magnetism (bosonic excitations)22,23, CeSiI monolayer provides more degrees of freedom to make various interfaces in combination with other 2D vdW materials or substrates.

To summarize, we have investigated 2D vdW heavy-fermion systems from the experimentally known bulk compounds by using DFT+DMFT calculations. Among them, CeSiI is the most promising candidate. We have found that the Kondo coupling strength of CeSiI does not change upon exfoliation and can be easily controlled by strain and surface doping. Finally, we have predicted the ground state of the CeSiI monolayer in response to external perturbations. Although the CeSiI monolayer has a magnetically ordered ground state, we found that the compressive strain and surface hole doping strongly enhance the Kondo coupling strength, indicating a possible quantum phase transition to PM heavy-fermion state. Our result thus suggests that the CeSiI monolayer can be a genuine 2D vdW heavy-fermion system and provide a new playground to study the QCP and emergent phenomena in vdW heterostructures.

## Methods

### DFT and DFT+DMFT calculations

We first performed DFT open-core calculation using the Vienna ab-initio Simulation Package (VASP) for the screening process44. The experimental crystal structure of 32 lanthanide materials employed from the Inorganic Crystal Structure Database (ICSD) was used in the calculations. For structural relaxations of the strained CeSiI cases, DFT open-core calculations with VASP were also employed since DFT calculations with 4f orbitals significantly overestimate fc hybridization, resulting in a smaller lattice constant. DFT open-core calculation with vdW correction well reproduced the experimental crystal structure of CeSiI. The internal atomic positions were fully relaxed for biaxial strained case, while the CeSiI monolayer structure was just adopted for surface doping cases.

To study the fc hybridization properly, we employed a DFT+DMFT calculation as implemented in DFT+Embedded DMFT (eDMFT) Functional code45. DFT calculations were performed by using WIEN2k code46 and the correlation effect of Ce 4f orbitals is treated by a DMFT loop. The hybridization energy window from −10 to 10 eV with respect to the EF was chosen and U = 5 eV and J = 0.68 eV were used for all Ce-based compounds. The continuous-time quantum Monte Carlo (CTQMC) solver was basically adopted for all calculations (Fig. 6)47. The temperature evolution of the Kondo resonance peak on the real axis was analyzed by using vertex corrected one-crossing approximation (OCA) solver to avoid the ambiguity arising from analytic continuation, which is inevitable for the CTQMC solver (Figs. 3 and 4)45.


text:
# Continuous seminorms

If I have a locally convex vector space $S$ equipped with a countable family $(p_i)_{i \in I}$ of seminorms, is it correct that the topology remains unchanged if I add an extra seminorm $q$ satisfying : $\exists J \subset I$ ($J$ finite) $\exists c>0 | \forall x \in S$ $q(x) \leq \Sigma_J p_i(x)$ ?

• you can better express your question, introducing some notation and definition? honestly I do not understand. – Andrew Oct 4 '16 at 21:57

## 1 Answer

We start with the assumption that $S$ is equipped exactly with the topology induced by $(p_i)_{i\in I}$ - the smalled topology on $S$ which makes these seminorms continuous.

Since $q$ is a seminorm, it is continuous if and only if it is continuous at zero.

Take a sequence $x_n\to 0$. Then for all $j$, since $p_j$ are continuous, we have $p_j(x_n)\to p_j(0) = 0$ as $n\to \infty$.

Then, $$0\leq q(x_n) \leq \sum_{j\in J}p_j(x_n),$$

and using the sandwich theorem, we see that $q(x_n) \to 0 = q(0)$.

Consequently, adding $q$ in the original family of seminorms will not alter the topology.


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# Long Range Shooting Question

1. Jul 12, 2012

### mram10

I was speaking with a physics friend (BS) and he told me if I fired a bullet at a target 1000 yds away, while simultaneously dropping one the same distance as the computed drop(255in), they would strike the ground at the same time. I understand in a vacuum how that could work, however with the kinetic energy behind the fired bullet, along with the shape of the projectile and the viscosity of air, this seems like a different animal entirely.

Just to clarify the test:
- bullet fired from high powered rifle at a target 1000 yds away.
- barrel in parallel to the ground (not arcing path)
- both bullets are 300gr and alike

Thanks guys.

2. Jul 12, 2012

### tiny-tim

welcome to pf!

hi mram10! welcome to pf!

let's see …

we're only interested in the vertical ("y") component of motion

let the velocity be the vector v = (vx,vy) = (dx/dt,dy/dt)

i] if we assume that air resistance is proportional (and opposite) to velocity, A = -kv,

then the component of air resistance in the vertical direction is kvy, = kdy/dt

and so the vertical equation of motion is md2y/dt2 = mg - kdy/dt, and this is irrespective of whether there is any horizontal motion

ii] however, if air resistance is proportional to speed-squared, A = -k|v|v,

then the component of air resistance in the vertical direction is k|v|vy, = k|v|dy/dt = k√((dx/dt)2 + (dy/dt)2)dy/dt

which is obviously smallest if dx/dt = 0

in other words: if air resistance is proportional to the square of the speed, then it has a greater effect on reducing vertical motion if the bullet is shot than if it is dropped

(but if air resistance is proportional to the speed itself, then there's no difference)

3. Jul 12, 2012

### PaulS1950

The vertical velocity is due to the acceleration of gravity. If the rifle barrel is level with the ground (90 degrees to gravity) then both bullets will hit the ground at the same time.
The one fired from the rifle will trave a lot farther but it falls at the same speed. The bullet does not produce lift.

4. Jul 12, 2012

### rcgldr

There could be a difference if the vertical component of drag was different between a dropped bullet versus one where the spin keeps it horizontal. If the bullet's center of mass versus the center of drag allowed the dropped bullet to rotate towards a vertical positiion during the drop, then there could be less vertical drag on the dropped bullet, but I don't think this is being considered in these type of thought problems. You could change the comparason so that the dropped bullet had the same spin as the fired bullet to eliminate any pitch effects of the bullet. Other issues like curvature of the earth, or spinning bullet, cross wind, and magnus effect (could result in lift or sink) are usually not considered either.

Last edited: Jul 13, 2012
5. Jul 12, 2012

### AlephZero

Tiny Tim's analysis is a reasonable starting point if you ignore the shape of the bullet (or assume it is a non-spinning sphere). For any realistic speed the drag force will be roughly proportional to the square of the speed, not the speed.

Any additional forces caused by the bullet "flying" will tend to make it fall slower than a sphere, so all the effects "add up" rather than "cancel out".

A practical way to investigate this is look at the adjustments to the rifle sights, and figure out the how the aiming position changes for different ranges. That will tell you how far the bullet actually "falls" below a straight line of flight.

6. Jul 13, 2012

### mram10

Wow, most of this is over my head, but I am slowly understanding. Thank you for your time guys. I will run some ballistic tests to see the true drop from a 90deg to gravity barrel so I can compare the times. I know it drops 255" and takes 1.34sec to reach 1000yds, however, the barrel is aiming slightly upward (arc).

7. Jul 13, 2012

### rorix_bw

Good luck! This is the correct approach - testing by experiment.

As others have said,

The *theory* is that they will strike the ground at the same time as the gun only provides horizontal and not vertical movemet to the bullet. All of the vertical movement comes from gravity, which is the same on both bullets.

However, in practice many things might affect this. Don't expect the results to be *exactly* identical. Remember your friend is talking about a theoretical system devoid of practical effects.

BTW if you were in the army artillery they would have taught you what your friend said about the time being equal (at least in the UK) as they teach ballistics to gunners (or at least they did, when my friend was in).

8. Jul 13, 2012

### mram10

Not an artillary guy. I just build and shoot long range rifles. My 338 edge is my long range elk rifle and does a great job. I digress.. :)

According to my ballistic software it takes 1.34 seconds to travel 1000 yds and drop 255 from a parallel barrel. Now I just need to drop a 300gr bullet from 255 inches and see how long it takes. Anyone have an easy formula that can find the time to fall?

9. Jul 13, 2012

### Staff: Mentor

Sure...
General formula for something starting at rest and accelerating at a constant rate is $$s=\frac{1}{2}at^{2}$$ where a is the acceleration, t is the time, and s is the distance. Turn that around because we know s and want to know t, and we have $t=\sqrt{\frac{2s}{a}}$. Here s is 255 inches, a is 32 ft/sec2 or 384 in/sec2... So t is theoretically 1.15 seconds, a bit longer allowing for air resistance.

That's more different from 1.34 seconds than I expected to see when I started the calculation... Maybe your ballistic software knows about some interaction between a fast-moving bullet and the air that tends to keep the bullet up a bit longer? No way the numbers should be different in a vacuum.

10. Jul 13, 2012

### rorix_bw

Earlier you had the same numbers with a slightly elevated barrel.

Ballistics software I have seen takes altitude, air temperature and wind speed and direction into account. What parameters are you using?

A ball round will rotate in flight and could generate lift from that - not sure how that works.

11. Jul 14, 2012

### mram10

Parameters:
3000ft
40deg F
300gr
.758 Ballistic Coefficient
2810 feet per second

If the program takes into account a non-level barrel, then the time would be slightly less for a level barrel, correct(due to less effect from gravity)?

12. Jul 14, 2012

### Staff: Mentor

Upwards-pointing barrel keeps the bullet in the air a bit longer than a level barrel, downwards-pointing barrel keeps the bullet in the air for a bit less time. Level barrel is the one that should be compared with the dropped bullet.

13. Jul 14, 2012

### mikeph

I can't see how there can be any lift, or in fact any perpendicular forces due to the air resistance on a spherically symmetric spinning bullet.

14. Jul 14, 2012

### rorix_bw

It's possible the ballistic software is measuring the drop relative to the where the sights are aimed, and not where the barrel is pointed - these won't be in line, as the shooter "zeroes in" the gun (aligns the sights with the impact point) at a chosen range.

I found this, which describes the lift generated by a spinning ball.

http://www.grc.nasa.gov/WWW/k-12/airplane/beach.html

15. Jul 14, 2012

### Staff: Mentor

This requires an axis of rotation which is not aligned with the direction of movement - something you will never see in bullets. In table tennis, it is common to use this effect.

Is it possible to "zero in" the gun for a distance of 0? This should allow horizontal shots.

16. Jul 14, 2012

### mram10

I tried to zero it at 0 and the program wouldn't let me. I can try to do it for 1 yd and have it shoot 2" low since the scope sits 2" higher than the barrel center.

17. Jul 15, 2012

### rorix_bw

Except in a crosswind ... and some weird situations I don't remember.

18. Jul 15, 2012

### A.T.

If the spin keeps the orientation constant but the movement direction changes due to gravity, there could be a slight misalignment, and some lift could be produced. But I don't know if this is significant in reality.


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# Payback Period and NPV: Taxes and Straight-Line Depreciation Assume that United Technologies is e...

Payback Period and NPV: Taxes and Straight-Line Depreciation Assume that United Technologies is evaluating a proposal to change the company's manual design system to a computer-aided design (CAD) system. The proposed system is expected to save 10,000 design hours per year; an operating cost savings of S40 per hour. The annual cash expenditures of operating the CAD system are estimated to be $200,000. The CAD system requires an initial investment of$500,000. The estimated life of this system is five years with no salvage value. The tax rate is 35 percent, and United Technologies uses straight-line depreciation for tax purposes. United Technologies has a cost of capital of 20 percent Compute the annual after-tax cash flows related to the CAD project Payback period. () Net present value. (Round answer to the nearest whole number. Use a negative sign with your answer if appropriate.) \$ 0


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Home » Uncategorized » Is it zero?

# Is it zero?

If is an even continuous function defined on and all its midpoint Riemann sums are zero ( i.e   for every ), then is ?

Solution

Let be an absolutely summable sequence, and use it to define a function as

It is evident that is even. By the Weierstrass M-Test converges uniformly on to . Hence, is continuous, and using
the fact that we have that

Next, we make two observations:

• If for every then by the definition of the Riemann integral we have that

which yields .

• Using the identity

we obtain the following string of bi-implications for every

In order for all of the mid-point Riemann sums of to be zero, it is thus
necessary and sufficient that

(i) and
(ii)

For this reason we choose the sequence as follows

Finally, is continuous , even , non zero and its Riemann mid-points are zero. To see that is non zero we simply calculate ;


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### Iff

Prove that if n is a triangular number then 8n+1 is a square number. Prove, conversely, that if 8n+1 is a square number then n is a triangular number.

### Relative Powers

Weekly Problem 10 - 2007
The square of a number is 12 more than the number itself. The cube of the number is 9 times the number. What is the number?

### Nine Sum to Square

Weekly Problem 32 - 2007
One of these numbers is the largest of nine consecutive positive integers whose sum is a perfect square. Which one is it?

# Generating Triples

##### Stage: 4 Challenge Level:
Here is an extract from Charlie's table:

Where is the $5^2$?
Where is the $12^2$?
Where is the $13^2$?

Where might we find a similar set of related square numbers?