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# Arnold's cat map

Picture showing how the linear map stretches the unit square and how its pieces are rearranged when the modulo operation is performed. The lines with the arrows show the direction of the contracting and expanding eigenspaces

In mathematics, Arnold's cat map is a chaotic map from the torus into itself, named after Vladimir Arnold, who demonstrated its effects in the 1960s using an image of a cat, hence the name.[1]

Thinking of the torus $\mathbb{T}^2$ as the quotient space $\mathbb{R}^2/\mathbb{Z}^2$ Arnold's cat map is the transformation $\Gamma : \mathbb{T}^2 \to \mathbb{T}^2$ given by the formula

$\Gamma \, : \, (x,y) \to (2x+y,x+y) \bmod 1.$

Equivalently, in matrix notation, this is

$\Gamma \left( \begin{bmatrix} x \\ y \end{bmatrix} \right) = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \bmod 1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \bmod 1.$

That is, with a unit size equal to the width of the square image, the image is sheared one unit up, then one unit to the right, and all that lies outside that unit square is shifted back by the unit until it's within the square.

## The discrete cat map

From order to chaos and back. Sample mapping on a picture of 150x150 pixels. The numbers shows the iteration step. After 300 iterations arriving at the original image
Sample mapping on a picture of a pair of cherries. The image is 74 pixels wide, and takes 114 iterations to be restored, although it appears upside-down at the halfway point.

It is possible to define a discrete analogue of the cat map. One of this map's features is that image being apparently randomized by the transformation but returning to its original state after a number of steps. As can be seen in the picture to the right, the original image of the cat is sheared and then wrapped around in the first iteration of the transformation. After some iterations, the resulting image appears rather random or disordered, yet after further iterations the image appears to have further order—ghost-like images of the cat, multiple smaller copies arranged in a repeating structure and even upside-down copies of the original image—and ultimately returns to the original image.

The discrete cat map describes the phase space flow corresponding to the discrete dynamics of a bead hopping from site qt (0 ≤ qt < N) to site qt+1 on a circular ring with circumference N, according to the second order equation:

$q_{t+1} - 3q_{t} + q_{t-1} = 0 \mod N$

Defining the momentum variable pt = qt - qt-1, the above second order dynamics can be re-written as a mapping of the square 0 ≤ q, p < N (the phase space of the discrete dynamical system) onto itself:

$q_{t+1} = 2q_{t} + p_{t} \mod N$
$p_{t+1} = q_{t} + p_{t} \mod N$

This Arnold cat mapping shows mixing behavior typical for chaotic systems. However, since the transformation has a determinant equal to unity, it is area-preserving and therefore invertible the inverse transformation being:

$q_{t-1} = q_{t} - p_{t} \mod N$
$p_{t-1} = -q_{t} + 2p_{t} \mod N$

For real variables q and p, it is common to set N = 1. In that case a mapping of the unit square with periodic boundary conditions onto itself results.

When N is set to an integer value, the position and momentum variables can be restricted to integers and the mapping becomes a mapping of a toroidial square grid of points onto itself. Such an integer cat map is commonly used to demonstrate mixing behavior with Poincaré recurrence utilising digital images. The number of iterations needed to restore the image can be shown never to exceed 3N.[4]

For an image, the relationship between iterations could be expressed as follows:

$\begin{array}{rrcl} n=0: \quad & T^0 (x,y) &= & \mbox{Input Image}(x,y) \\ n=1: \quad & T^1 (x,y) &= & T^0 \left( \bmod(2x+y, N), \bmod(x+y, N) \right) \\ & &\vdots \\ n=k: \quad & T^k (x,y) &= & T^{k-1} \left( \bmod(2x+y, N), \bmod(x+y, N) \right) \\ & &\vdots \\ n=m: \quad & \mbox{Output Image}(x,y) &=& T^m (x,y) \end{array}$


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# getCompSensitivity

Complementary sensitivity function from generalized model of control system

## Description

example

T = getCompSensitivity(CL,location) returns the complementary sensitivity measured at the specified location for a generalized model of a control system.

example

T = getCompSensitivity(CL,location,opening) specifies additional loop openings for the complementary sensitivity function calculation. Use an opening, for example, to calculate the complementary sensitivity function of an inner loop, with the outer loop open.

If opening and location list the same point, the software opens the loop after adding the disturbance signal at the point.

## Examples

collapse all

Compute the complementary sensitivity at the plant output, X, of the control system of the following illustration.

Create a model of the system by specifying and connecting a numeric LTI plant model G, a tunable controller C, and the AnalysisPoint block X. Use the AnalysisPoint block to mark the location where you assess the complementary sensitivity, which in this example is the plant output.

G = tf([1],[1 5]);
C = tunablePID('C','p');
C.Kp.Value = 3;
X = AnalysisPoint('X');
CL = feedback(G*C,X);

CL is a genss model that represents the closed-loop response of the control system from r to y. Examine the Control Design Blocks of the model.

CL.Blocks
ans = struct with fields:
C: [1x1 tunablePID]
X: [1x1 AnalysisPoint]

The model's blocks include the AnalysisPoint block, X, that identifies the analysis-point location.

Calculate the complementary sensitivity, T, at X.

T = getCompSensitivity(CL,'X')
T =

Generalized continuous-time state-space model with 1 outputs, 1 inputs, 1 states, and the following blocks:
C: Tunable PID controller, 1 occurrences.
X: Analysis point, 1 channels, 1 occurrences.

Type "ss(T)" to see the current value, "get(T)" to see all properties, and "T.Blocks" to interact with the blocks.

getCompSensitivity preserves the Control Design Blocks of CL, and returns a genss model. To get a numeric model, you can convert T to transfer-function form, using the current value of the tunable block.

Tnum = tf(T)
Tnum =

From input "X" to output "X":
-3
-----
s + 8

Continuous-time transfer function.

In the multiloop system of the following illustration, calculate the inner-loop sensitivity at the output of G2, with the outer loop open.

Create a model of the system by specifying and connecting the numeric plant models, tunable controllers, and AnalysisPoint blocks. G1 and G2 are plant models, C1 and C2 are tunable controllers, and X1 and X2 are AnalysisPoint blocks that mark potential loop-opening locations.

G1 = tf(10,[1 10]);
G2 = tf([1 2],[1 0.2 10]);
C1 = tunablePID('C','pi');
C2 = tunableGain('G',1);
X1 = AnalysisPoint('X1');
X2 = AnalysisPoint('X2');
CL = feedback(G1*feedback(G2*C2,X2)*C1,X1);

Calculate the complementary sensitivity, $T$, at X2, with the outer loop open at X1. Specifying X1 as the third input argument tells getCompSensitivity to open the loop at that location.

T = getCompSensitivity(CL,'X2','X1');
tf(T)
ans =

From input "X2" to output "X2":
-s - 2
----------------
s^2 + 1.2 s + 12

Continuous-time transfer function.

## Input Arguments

collapse all

Model of a control system, specified as a generalized state-space model (genss).

Locations at which you can perform sensitivity analysis or open loops are marked by AnalysisPoint blocks in CL. Use getPoints(CL) to get the list of such locations.

Location at which you calculate the complementary sensitivity function, specified as a character vector or cell array of character vectors. To extract the complementary sensitivity function at multiple locations, use a cell array of character vectors.

Each specified location must match an analysis point in CL. Analysis points are marked using AnalysisPoint blocks. To get the list of available analysis points in CL, use getPoints(CL).

Example: 'u' or {'u','y'}

Additional loop opening used to calculate the complementary sensitivity function, specified as a character vector or cell array of character vectors. To open the loop at multiple locations, use a cell array of character vectors.

Each specified opening must match an analysis point in CL. Analysis points are marked using AnalysisPoint blocks. To get the list of available analysis points in CL, use getPoints(CL).

Use an opening, for example, to calculate the complementary sensitivity function of an inner loop, with the outer loop open.

If opening and location list the same point, the software opens the loop after adding the disturbance signal at the point.

Example: 'y_outer' or {'y_outer','y_outer2'}

## Output Arguments

collapse all

Complementary sensitivity function of the control system, T, measured at location, returned as a generalized state-space model (genss).

• If location specifies a single analysis point, then T is a SISO genss model.

• If location is a vector signal, or specifies multiple analysis points, then T is a MIMO genss model.

collapse all

### Complementary Sensitivity

The complementary sensitivity function, T, at a point is the closed-loop transfer function around the feedback loop measured at the specified location. It is related to the open-loop transfer function, L, and the sensitivity function, S, at the same point as follows:

$T=\frac{L}{1-L}=S-1.$

Use getLoopTransfer and getSensitivity to compute L and S.

Consider the following model:

The complementary sensitivity, T, at y is defined as the transfer function from dy to y.

Observe that, in contrast to the sensitivity function, the disturbance, dy, is added after the measurement, y.

$\begin{array}{l}y=-GK\left(y+dy\right)\\ \to y=-GKy-GKdy\\ \to \left(I+GK\right)y=-GKdy\\ \to y=\underset{T}{\underbrace{-{\left(I+GK\right)}^{-1}GK}}dy.\end{array}$

Here, I is an identity matrix of the same size as GK. The complementary sensitivity transfer function at y is equal to -1 times the closed-loop transfer function from r to y.

Complementary sensitivity at multiple locations, for example, u and y, is defined as the MIMO transfer function from the disturbances to measurements:

$T=\left[\begin{array}{cc}{T}_{du\to u}& {T}_{dy\to u}\\ {T}_{du\to y}& {T}_{dy\to y}\end{array}\right].$

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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Period and Frequency

## Horizontal distance traveled before y values repeat; number of complete waves in 2pi.

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Frequency and Period of Sinusoidal Functions

The transformation rules about horizontal stretching and shrinking directly apply to sine and cosine graphs.  If a sine graph is horizontally stretched by a factor of 12\begin{align*}\frac{1}{2}\end{align*} that is the same as a horizontal compression by a factor of 2.

How does the equation change when a sine or cosine graph is stretched by a factor of 3?

#### Watch This

http://www.youtube.com/watch?v=qJ-oUV7xL3w James Sousa: Amplitude and Period of Sine and Cosine

#### Guidance

The general equation for a sinusoidal function is:

f(x)=±asin(b(x+c))+d\begin{align*}f(x)={\pm} a \cdot \sin (b(x+c))+d\end{align*}

The ±\begin{align*}\pm\end{align*} controls the reflection across the x\begin{align*}x\end{align*}-axis.  The coefficient a\begin{align*}a\end{align*} controls the amplitude.  The constant d\begin{align*}d\end{align*} controls the vertical shift.  Here you will see that the coefficient b\begin{align*}b\end{align*} controls the horizontal stretch.

Horizontal stretch is measured for sinusoidal functions as their periods.  This is why this function family is also called the periodic function family.  The period of a sinusoid is the length of a complete cycle.  For basic sine and cosine functions, the period is 2π\begin{align*}2 \pi\end{align*}.  This length can be measured in multiple ways.  In word problems and in other tricky circumstances, it may be most useful to measure from peak to peak.

The ability to measure the period of a function in multiple ways allows different equations to model an identical graph.  In the image above, the top red line would represent a regular cosine wave.  The center red line would represent a regular sine wave with a horizontal shift.  The bottom red line would represent a negative cosine wave with a horizontal shift.  This flexibility in perspective means that many of the examples, guided practice and practice problems may have multiple solutions.  For now, try to always choose the function that has a period starting at x=0\begin{align*}x=0\end{align*}

Frequency is a different way of measuring horizontal stretch.  For sound, frequency is known as pitch.  With sinusoidal functions, frequency is the number of cycles that occur in 2π\begin{align*}2 \pi\end{align*}.  A shorter period means more cycles can fit in 2π\begin{align*}2 \pi\end{align*} and thus a higher frequency.  Period and frequency are inversely related by the equation:

period=2πfrequency\begin{align*}\text{period}={\frac{2 \pi}{\text{frequency}}}\end{align*}

The equation of a basic sine function is f(x)=sinx\begin{align*}f(x)={\sin} x\end{align*}.  In this case b\begin{align*}b\end{align*}, the frequency, is equal to 1 which means one cycle occurs in 2π\begin{align*}2 \pi\end{align*}.  This relationship is a common mistake in graphing sinusoidal functions.  Students find b=12\begin{align*}b={\frac{1}{2}}\end{align*} and then mistakenly conclude that the period is 12\begin{align*}\frac{1}{2}\end{align*} when it is in fact stretched to 4π\begin{align*}4 \pi\end{align*}.

Example A

Rank each wave by period from shortest to longest.

Solution:

The red wave has the shortest period.

The green and black waves have equal periods.  A common mistake is to see that the green wave has greater amplitude and confuse that with greater periods.

The blue wave has the longest period.

Example B

Identify the amplitude, vertical shift, period and frequency of the following function.  Then graph the function.

f(x)=2sin(x3)+1\begin{align*}f(x)=2 \sin \left(\frac{x}{3}\right)+1\end{align*}

Solution:  a=2,b=13,d=1\begin{align*}a=2, b=\frac{1}{3}, d=1\end{align*}. Since b=13\begin{align*}b=\frac{1}{3}\end{align*} (frequency), then the period must be 6π\begin{align*}6 \pi\end{align*}.

Often the most challenging part of graphing periodic functions is labeling the axes.  Since the period is 6π\begin{align*}6 \pi\end{align*}, start by drawing the sinusoidal axis shifted appropriately.  Then divide the 6π\begin{align*}6 \pi\end{align*} into four parts so that the 5 guiding points of the sine graph can be plotted with the amplitude and reflection in mind.  The very last thing to do is to draw and extend the curve.  Many students try to draw the curve too early and end up having to redo their work.

Example C

A measuring stick on a dock measures high tide to be 18 feet and low tide to be 6 feet.  It takes about 6 hours for the tide to switch between low and high tides.  Determine a graphical and algebraic model for the tides knowing that at t=0\begin{align*}t=0\end{align*} there is a high tide.

Solution:  Usually the best course of action for word problems is to identify information, plot points, sketch and then finally come up with an equation.

From the given information you can deduce the following points.  Notice how the sinusoidal axis can be assumed to be the average of the high and low tides.

 Time (hours) Water level (feet) 0 18 6 6 12 18 3 18+62=12\begin{align*}\frac{18+6}{2}=12\end{align*} 9 12

By plotting those points and filling in the sinusoidal axis you can observe a cosine graph.

The amplitude is 6 so a=6\begin{align*}a=6\end{align*}.  There is no vertical reflection.  Since the period is 12 you can determine the frequency b\begin{align*}b\end{align*}:

12=2πbb=π6\begin{align*}12={\frac{2 \pi}{b}} \rightarrow b={\frac{\pi}{6}}\end{align*}

The vertical shift is 12 so d=12\begin{align*}d=12\end{align*}.   Thus you have all the pieces to make an algebraic model:

f(x)=+6cos(π6x)+12\begin{align*}f(x)=+6 \cdot \cos \left(\frac{\pi}{6}x\right)+12\end{align*}

Concept Problem Revisited

If a sine graph is horizontally stretched by a factor of 3 then the general equation has b=13\begin{align*}b={\frac{1}{3}}\end{align*}.  This is because b\begin{align*}b\end{align*} is the frequency and counts the number (or fraction) of a period that fits in a normal period of 2π\begin{align*}2 \pi\end{align*}.  Graphically, the sine wave will make a complete cycle in 6π\begin{align*}6 \pi\end{align*} units like Example B.

#### Vocabulary

Period is the distance it takes for a repeating function to make one complete cycle.

Frequency is the number of cycles a function makes in a set amount of time or distance on the x\begin{align*}x\end{align*} axis.  For sine and cosine graphs this distance is 2π\begin{align*}2 \pi\end{align*}

#### Guided Practice

1. A fish is caught in a water wheel by the side of a river.  Initially the fish is 2 feet below the surface of the water.  Twenty seconds later the fish is 14 feet in the air at the top of the water wheel.  Model the fish’s height in a graph and an equation.

2. Graph the following function: g(x)=cos(8x)+2\begin{align*}g(x)=-\cos (8x)+2\end{align*}.

3. Given the following graph, identify the amplitude, period, and frequency and create an algebraic model.

1. Use logic to identify five key points.  Use those key points to come up with a sketch.  Use the sketch to identify information for the equation.

 Time (seconds) Fish height (feet) 0 -2 20 14 40 -2 10 −2+142=6\begin{align*}\frac{-2+14}{2}=6\end{align*} 30 6

The amplitude is 8 so a=8\begin{align*}a=8\end{align*}.  The function looks like a negative cosine graph.  The vertical shift is d=6\begin{align*}d=6\end{align*} and the period is 40.

40f(x)=2πbb=π20=8cos(π20x)+6

Notice how the labeling on the graph is extremely deliberate.  On both the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} axes, only the most important intervals are labeled.  This keeps the sketch accurate, evenly spaced on your paper and easy to read.

2. The labeling is the most important and challenging part of this problem.  The amplitude is 1.  The shape is a negative cosine.  The vertical shift is 2.  The period is 2π8=π4\begin{align*}\frac{2 \pi}{8}={\frac{\pi}{4}}\end{align*}.  Working with this small period may be challenging at first, but remember that halving fractions is as simple as doubling the denominator.

3. The amplitude is 3.  The shape is a negative cosine.  The period is 5π2\begin{align*}\frac{5 \pi}{2}\end{align*} which implies that b=45\begin{align*}b={\frac{4}{5}}\end{align*}.  The vertical shift is 1. f(x)=3cos(45x)+1\begin{align*}f(x)=-3 \cdot \cos \left(\frac{4}{5}x\right)+1\end{align*}.

#### Practice

Find the frequency and period of each function below.

1. f(x)=sin(4x)+1\begin{align*}f(x)={\sin (4x)+1}\end{align*}

2. g(x)=3cos(2x)\begin{align*}g(x)=-3 \cos (2x)\end{align*}

3. \begin{align*}h(x)={\cos} \left(\frac{1}{2}x\right)+2\end{align*}

4. \begin{align*}k(x)=-2 \sin \left(\frac{3}{4}x\right)+1\end{align*}

5. \begin{align*}j(x)=4 \cos (3x)-1\end{align*}

Graph each of the following functions.

6. \begin{align*}f(x)=3 \sin (2x)+1\end{align*}

7. \begin{align*}g(x)=2.5 \cos (\pi x)-4\end{align*}

8. \begin{align*}h(x)=-\sin (4x)-3\end{align*}

9. \begin{align*}k(x)={\frac{1}{2}} \cos (2x)\end{align*}

10. \begin{align*}j(x)=-2 \sin \left(\frac{3}{4}x \right)-1\end{align*}

Create an algebraic model for each of the following graphs.

11.

12.

13.

14. At time 0 it is high tide and the water at a certain location is 10 feet high.  At low tide 6 hours later, the water is 2 feet high.  Given that tides can be modeled by sinusoidal functions, find a graph that models this scenario.

15. Find the equation that models the scenario in the previous problem.

### Vocabulary Language: English

Amplitude

Amplitude

The amplitude of a wave is one-half of the difference between the minimum and maximum values of the wave, it can be related to the radius of a circle.
sinusoidal function

sinusoidal function

A sinusoidal function is a sine or cosine wave.
sinusoidal functions

sinusoidal functions

A sinusoidal function is a sine or cosine wave.
Vertical shift

Vertical shift

A vertical shift is the result of adding a constant term to the value of a function. A positive term results in an upward shift, and a negative term in a downward shift.


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## A fact about multivariate polynomials over ZZ

Let $f(x_1,\ldots,x_n) \in \mathbb{Z}[x_1,\ldots,x_n]$, and let $p$ be a prime. Prove that $f(x_1,\ldots,x_n)^p = f(x_1^p,\ldots,x_n^p)$ mod $p$.

Remember that $\alpha^p \equiv \alpha$ mod $p$ for all $\alpha$. Say $f = \sum_i c_i \prod_j x_j^{e_{i,j}}$.

Then $f(x_1,\ldots,x_n)^p = (\sum_i c_i \prod_j x_j^{e_{i,j}})^p$ $= \sum_i c_i^p \prod_j x_j^{e_{i,j}p}$ $= \sum_i c_i \prod_j (x_j^p)^{e_{i,j}}$ $= f(x_1^p,\ldots,x_n^p)$ as desired.


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# XeLaTeX with fontspec not recognizing ANY fonts on a windows platform (with miktex) [closed]

Getting the message: font-not-found see fontspec documentation. I'm using cls file abbreviated at the bottom of this page whose origin is latextemplates.com/template/friggeri-resume-cv. I've tried all sorts of fonts from system fonts and tex fonts. Nothing seems to work! I would like to get fonts like in http://www.tug.dk/FontCatalogue/ to work, but I would be happy if I can just get one font to work!

\ProvidesClass{friggeri-cv}[2012/04/30 CV class]
\NeedsTeXFormat{LaTeX2e}

\DeclareOption{print}{\def\@cv@print{}}
\DeclareOption*{%
\PassOptionsToClass{\CurrentOption}{article}%
}
\ProcessOptions\relax

%%%%%%%%%%
% Colors %
%%%%%%%%%%

\RequirePackage{xcolor}

\definecolor{white}{RGB}{255,255,255}
...

%%%%%%%%%
% Fonts %
%%%%%%%%%
\RequirePackage{fontspec}
\RequirePackage[math-style=TeX,vargreek-shape=unicode]{unicode-math}
\RequirePackage[T1]{fontenc}

\RequirePackage{libertine}

\newfontfamily\bodyfont[]{\rm}
\newfontfamily\thinfont[]{\biolinum}

\defaultfontfeatures{Mapping=tex-text}
\setmainfont[Mapping=tex-text, Color=textcolor]{Tex Gyre Heros}

%\setmathfont{XITS Math}

%%%%%%%%%%
%%%%%%%%%%

\RequirePackage{tikz}

\newcommand{\rolefont}{%
\fontsize{14pt}{24pt}\selectfont%
\thinfont%
\color{white}%
}

\begin{tikzpicture}[remember picture,overlay]
\node [rectangle, fill=fillheader, anchor=north, minimum width=\paperwidth, minimum    height=4cm] (box) at (current page.north){};
\node [anchor=center] (name) at (box) {%
{\thinfont #1}{\bodyfont  #2}
};
\node [anchor=north] at (name.south) {%
\thinfont #3%
};
\end{tikzpicture}
\vspace{2.5cm}
\vspace{-2\parskip}
}

...
%%%%%%%%%%%%%%%%
% Other tweaks %
%%%%%%%%%%%%%%%%

• Without an MWE -- minimum working example; a compilable document that (i) starts with \documentclass and ends with \end{document} and (ii) generates the problem behavior you're looking to fix -- it's not going to be likely that anyone will be able to offer a diagnosis, let alone a solution. – Mico Dec 19 '13 at 16:51
• Don't test such things with complicated templates. Start with a small document which loads only fontspec and calls \setmainfont{Arial}. The line with \RequirePackage[T1]{fontenc} is wrong in your template. Delete it or move it before fontspec. – Ulrike Fischer Dec 19 '13 at 16:51


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# Understanding the steps in factoring $x^2+5xy+4y^2+5x+23y-6$

I understand the first two steps in factoring this expression, but the third step is unclear to me:

\begin{align}x^2+5xy+4y^2+5x+23y-6 &= x^2+(5y+5)x+4y^2+23y-6 \\ &= x^2+(5y+5)x+(4y-1)(y+6) \\ &= (x+(4y-1))(x+(y+6)) \\ &= (x+4y-1)(x+y+6).\end{align}

Why does the $$5y+5$$ disappear?

How can I learn to see such patterns and come up with this myself? I see that $$5y +5 = (4y -1) + (y+6)$$. Does this observation help me in this case? Usually when factoring by grouping the goal is to get one factor multiple times and then turn addition into multiplication.

In the same way, I don't understand the second step in here:

\begin{align}(x^2-5x)^2-18x^2+90x-144 &= (x^2-5x)^2-18(x^2-5x)-144 \\ &= (x^2-5x+6)(x^2-5x-24) \\ &= (x-2)(x-3)(x+3)(x-8) \\ &= (x-2)(x-3)(x-8)(x+3).\end{align}

I have a feeling that the same thing is happening and I don't understand why I can do that. I see that $$6\times 24 = 144$$, and that these constants have to be at the end of the factored terms, but how does this 'enhancing' of the content of the parenthesis work? Also, the first term was squared and then the square disappears.

• In the future, please take the time to enter all of the important parts of your question as text instead of pasting in pictures of them. Your question should be comprehensible with images disabled. Also, images are neither searchable nor accessible to screen readers. MathJax is used for formatting mathematical expressions on this site; you can find a quick reference and tutorial here. I’ve take the liberty of editing this question so that you can see how to format aligned equations. – amd Jan 29 at 19:29

You are right not to understand, as some of these transformations are rabbits pulled from a hat.

In the first case, you can begin by factoring the quadratic part

$$x^2+5xy+4y^2,$$ giving $$(x-4y)(x+y).$$

Then we decompose the linear part as a linear combination of the above factors,

$$5x+23y=\lambda(x-4y)+\mu(x+y)$$ the solution of which is $$\lambda=6,\mu=-1$$.

Now we write the tentative factorization

$$(x-4y-1)(x+y+6)$$ and by some magic, the constant term fits. (If it didn't, the factorization is not possible.)

In the second case, you may observe that $$-18x^2+90x$$ is a multiple of $$x^2-5x$$ that appears in the first term, and by setting $$t:=x^2-5$$ you can factor in terms of $$t$$, then factor the two factors:

$$t^2-18t-144=(t-24)(t+6).$$

But if the polynomial had not been presented in this "hinting" form, and just as

$$x^4-10x^3+7x^2+90x-144,$$

you would have been on your own. In case you suspected rational roots, the Rational Root Theorem would have been your best friend here.

Otherwise, the standard resolution procedure for the roots of a quartic is to depress the polynomial, i.e. shift the unknown so that the cubic term cancels out. Here, by substituting $$x:=t+\dfrac{10}4$$, the equation turns to

$$t^4-\frac{61t^2}2+\frac{121}{16}=0,$$ which, by luck, is biquadratic.

Begin with $$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2,\tag1$$ where LHS says that $$r_1,r_2$$ are the roots of a quadratic polynomial, and RHS gives the coefficients of the polynomial in terms of the roots (see also Vieta's formulae).

In your equation, the coefficient multiplying $$x$$ is the SUM (or better, opposite of the sum) and the last coefficient is the PRODUCT of the roots.
You already noticed that $$5y +5 = (4y -1) + (y+6),$$ or $$-(5y +5) = -(4y -1) +(- (y+6)).$$

Therefore, in $$(1)$$ is $$r_1=-(4y -1)$$ and $$r_2=- (y+6).$$


text:
linearlyNormalEmbedding -- get the linearly normal embedding

Synopsis

• Usage:
linearlyNormalEmbedding X
• Inputs:
• Outputs:
• an isomorphism from X to a linearly normal variety, whose inverse is a linear projection

Description

 i1 : K = ZZ/333331; i2 : X = PP_K^(1,7); -- rational normal curve of degree 7 o2 : ProjectiveVariety, curve in PP^7 i3 : time f = linearlyNormalEmbedding X; -- used 0.00946185 seconds o3 : MultirationalMap (automorphism of X) i4 : Y = (rationalMap {for i to 3 list random(1,ring ambient X)}) X; -- an isomorphic projection of X in PP^3 o4 : ProjectiveVariety, curve in PP^3 i5 : time g = linearlyNormalEmbedding Y; -- used 0.536626 seconds o5 : MultirationalMap (birational map from Y to curve in PP^7) i6 : assert(isIsomorphism g) i7 : describe g o7 = multi-rational map consisting of one single rational map source variety: curve in PP^3 cut out by 6 hypersurfaces of degree 4 target variety: curve in PP^7 cut out by 21 hypersurfaces of degree 2 base locus: empty subscheme of PP^3 dominance: true multidegree: {7, 7} degree: 1 degree sequence (map 1/1): [3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4] coefficient ring: K

Caveat

This is an experimental function.

Ways to use linearlyNormalEmbedding :

• "linearlyNormalEmbedding(EmbeddedProjectiveVariety)"

For the programmer

The object linearlyNormalEmbedding is .


text:
This is an old version, view current version.

21.2 Reparameterizations

Reparameterizations may be implemented directly using the transformed parameters block or just in the model block.

Beta and Dirichlet Priors

The beta and Dirichlet distributions may both be reparameterized from a vector of counts to use a mean and total count.

Beta Distribution

For example, the Beta distribution is parameterized by two positive count parameters $$\alpha, \beta > 0$$. The following example illustrates a hierarchical Stan model with a vector of parameters theta are drawn i.i.d. for a Beta distribution whose parameters are themselves drawn from a hyperprior distribution.

parameters {
real<lower = 0> alpha;
real<lower = 0> beta;
...
model {
alpha ~ ...
beta ~ ...
for (n in 1:N)
theta[n] ~ beta(alpha, beta);
...

It is often more natural to specify hyperpriors in terms of transformed parameters. In the case of the Beta, the obvious choice for reparameterization is in terms of a mean parameter $\phi = \alpha / (\alpha + \beta)$ and total count parameter $\lambda = \alpha + \beta.$ Following @[GelmanEtAl:2013, Chapter 5] the mean gets a uniform prior and the count parameter a Pareto prior with $$p(\lambda) \propto \lambda^{-2.5}$$.

parameters {
real<lower=0,upper=1> phi;
real<lower=0.1> lambda;
...
transformed parameters {
real<lower=0> alpha = lambda * phi;
real<lower=0> beta = lambda * (1 - phi);
...
model {
phi ~ beta(1, 1); // uniform on phi, could drop
lambda ~ pareto(0.1, 1.5);
for (n in 1:N)
theta[n] ~ beta(alpha, beta);
...

The new parameters, phi and lambda, are declared in the parameters block and the parameters for the Beta distribution, alpha and beta, are declared and defined in the transformed parameters block. And If their values are not of interest, they could instead be defined as local variables in the model as follows.

model {
real alpha = lambda * phi
real beta = lambda * (1 - phi);
...
for (n in 1:N)
theta[n] ~ beta(alpha, beta);
...
}

With vectorization, this could be expressed more compactly and efficiently as follows.

model {
theta ~ beta(lambda * phi, lambda * (1 - phi));
...
}

If the variables alpha and beta are of interest, they can be defined in the transformed parameter block and then used in the model.

Jacobians not Necessary

Because the transformed parameters are being used, rather than given a distribution, there is no need to apply a Jacobian adjustment for the transform. For example, in the beta distribution example, alpha and beta have the correct posterior distribution.

Dirichlet Priors

The same thing can be done with a Dirichlet, replacing the mean for the Beta, which is a probability value, with a simplex. Assume there are $$K > 0$$ dimensions being considered ($$K=1$$ is trivial and $$K=2$$ reduces to the beta distribution case). The traditional prior is

parameters {
vector[K] alpha;
simplex[K] theta[N];
...
model {
alpha ~ ...;
for (n in 1:N)
theta[n] ~ dirichlet(alpha);
}

This provides essentially $$K$$ degrees of freedom, one for each dimension of alpha, and it is not obvious how to specify a reasonable prior for alpha.

An alternative coding is to use the mean, which is a simplex, and a total count.

parameters {
simplex[K] phi;
real<lower=0> kappa;
simplex[K] theta[N];
...
transformed parameters {
vector[K] alpha = kappa * phi;
...
}
model {
phi ~ ...;
kappa ~ ...;
for (n in 1:N)
theta[n] ~ dirichlet(alpha);

Now it is much easier to formulate priors, because phi is the expected value of theta and kappa (minus K) is the strength of the prior mean measured in number of prior observations.

Transforming Unconstrained Priors: Probit and Logit

If the variable $$u$$ has a $$\mathsf{Uniform}(0, 1)$$ distribution, then $$\mbox{logit}(u)$$ is distributed as $$\mathsf{Logistic}(0, 1)$$. This is because inverse logit is the cumulative distribution function (cdf) for the logistic distribution, so that the logit function itself is the inverse cdf and thus maps a uniform draw in $$(0, 1)$$ to a logistically-distributed quantity.

Things work the same way for the probit case: if $$u$$ has a $$\mathsf{Uniform}(0, 1)$$ distribution, then $$\Phi^{-1}(u)$$ has a $$\mathsf{normal}(0, 1)$$ distribution. The other way around, if $$v$$ has a $$\mathsf{normal}(0, 1)$$ distribution, then $$\Phi(v)$$ has a $$\mathsf{Uniform}(0, 1)$$ distribution.

In order to use the probit and logistic as priors on variables constrained to $$(0, 1)$$, create an unconstrained variable and transform it appropriately. For comparison, the following Stan program fragment declares a $$(0, 1)$$-constrained parameter theta and gives it a beta prior, then uses it as a parameter in a distribution (here using foo as a placeholder).

parameters {
real<lower = 0, upper = 1> theta;
...
model {
theta ~ beta(a, b);
...
y ~ foo(theta);
...

If the variables a and b are one, then this imposes a uniform distribution theta. If a and b are both less than one, then the density on theta has a U shape, whereas if they are both greater than one, the density of theta has an inverted-U or more bell-like shape.

Roughly the same result can be achieved with unbounded parameters that are probit or inverse-logit-transformed. For example,

parameters {
real theta_raw;
...
transformed parameters {
real<lower = 0, upper = 1> theta = inv_logit(theta_raw);
...
model {
theta_raw ~ logistic(mu, sigma);
...
y ~ foo(theta);
...

In this model, an unconstrained parameter theta_raw gets a logistic prior, and then the transformed parameter theta is defined to be the inverse logit of theta_raw. In this parameterization, inv_logit(mu) is the mean of the implied prior on theta. The prior distribution on theta will be flat if sigma is one and mu is zero, and will be U-shaped if sigma is larger than one and bell shaped if sigma is less than one.

When moving from a variable in $$(0, 1)$$ to a simplex, the same trick may be performed using the softmax function, which is a multinomial generalization of the inverse logit function. First, consider a simplex parameter with a Dirichlet prior.

parameters {
simplex[K] theta;
...
model {
theta ~ dirichlet(a);
...
y ~ foo(theta);

Now a is a vector with K rows, but it has the same shape properties as the pair a and b for a beta; the beta distribution is just the distribution of the first component of a Dirichlet with parameter vector $$[a b]^{\top}$$. To formulate an unconstrained prior, the exact same strategy works as for the beta.

parameters {
vector[K] theta_raw;
...
transformed parameters {
simplex[K] theta = softmax(theta_raw);
...
model {
theta_raw ~ multi_normal_cholesky(mu, L_Sigma);

The multivariate normal is used for convenience and efficiency with its Cholesky-factor parameterization. Now the mean is controlled by softmax(mu), but we have additional control of covariance through L_Sigma at the expense of having on the order of $$K^2$$ parameters in the prior rather than order $$K$$. If no covariance is desired, the number of parameters can be reduced back to $$K$$ using a vectorized normal distribution as follows.

  theta_raw ~ normal(mu, sigma);

where either or both of mu and sigma can be vectors.


text:
# COURSE YEAR 2016

Module - I : May 08, 2016 – May 20, 2016, Roger Wiegand and Sylvia Wiegand
Topics: Rings, Fields, The characteristic of a field, Review of polynomial rings, Factoring polynomials, Extension fields, The subring gene-rated by a subset, The subfield generated by a subset, Construction of some extension fields, Stem fields, Algebraic and transcendental elements, Transcendental numbers
Brief review of basic notions (groups, rings, and fields): (also download from here)

• Characteristic of a field, "Freshman's Dream": $\left(a+ b\right)^p$ = $a^p$ + $b^p$
• Division Algorithm (with proof); $x-c \mid f(x)$ $\iff$ $f \left(a \right)$= 0
• This includes items 1.1-1.4, 1.5, 1.6, 1.7, 1.9, 1.11 $F[x]$ is a PID; precise definition of UFD. If degree $f\left(x\right)\le$ 3 then $f\left(x \right)$ is irreducible $\iff f \left(x \right)$ has no roots; fails for degree 4

Discussion of irreducibility in $\mathbb {Z}[X]$ vs in $\mathbb{Q}[X]$: (also download from here)
• Testing for irreducibility in $\mathbb{Z}[X]$ by checking irreducibility in $\mathbb{F}_p[X]$; proof of why it works, pitfalls of misinterpreting, why $p$ must not divide leading coefficient. Explained what Milne means by "factors non-trivially" (not quite the same, over $\mathbb{Z}$, as being reducible). Examples, of a polynomial that can be shown to be irreducible by going mod 2 and of another polynomial, $X^4-10X^2+1$ in $\mathbb{Z}[X]$, that factors non-trivially modulo every prime, even though it is irreducible (proof of irreducibility assigned in Problem Set 1).
• Finding a polynomial in $\mathbb{Z}[X]$ with $\sqrt 2+\sqrt 3$ as a root (the polynomial above, coincidentally).
• Gauss' Lemma 1.13, proved with a more rigorous version of Milne's proof. (Choose appropriate positive rational multiples of $g(x)$ and $h(x)$ so as to minimize their integral product. then show that this minimal product must be 1.)
• Vector spaces: Discussed how $F[X]$, and extension fields $K$ of $F$ are vector spaces over $F$.
• Proposition 1.8: proved that for $R$ a PID, gcd's exist and can be written as a combination of the generators. Discussed the Euclidean algorithm method of finding $d=\gcd(a,b)$ and r,s with $ra +sb=d$.
• Stated and proved one case of the $\frac{2}{3}$ Lemma ("two-out of three lemma"): If $a,b,c,d\in R$, a commutative ring, with $d\ne 0$ and $a \pm b=c$, and two of a, b, c are multiples of d, then so is the third.
• Discussed extension fields, notation, as on page 13 of Milne, including $F$-homomorphism'', and examples.
• We added the fact, which we proved later, that, if $\alpha$ and $\beta$ are roots of the same irreducible polynomial there exists an $F$--homomorphism from $F\big[ \alpha \big]$ to $F \big[ \beta \big]$
• Proved Proposition 1.20 Multiplication of degrees, part 2. (Will go over part 1 in lecture 4 today.)
• Proved Lemma 1.2.3: For $F\subseteq R$, where $F$ is a field and $R$ is an integral domain that has finite dimension over $F$ as a vector space, $R$ is a field.
• Discussed constructing an extension field where an irreducible polynomial has a root. Proved it is a field.
• Discussed examples such as building the complex numbers from the reals.
• Proved the finiteness Theorem: For $F \subseteq E \subseteq K$ fields, these are equivalent:
1. $E/F$ is finite (vector space dimension).
2. $E/F$ is finitely generated as a field extension and is algebraic over $F$.
3. There exist algebraic elements $\alpha_1$, $\alpha_2$, $\dots$, $\alpha_n$ of $E$, such that $E=F(\alpha_1, \alpha_2,$ $\dots, \alpha_n)$.
• Reviewed the Finite-algebraic Theorem from Lecture 5: For $F$ subseteq $E$ fields, these are equivalent:
1. $E/F$ is finite (i.e. finite vector space dimension).
2. $E/F$ is finitely generated as a field extension and $E$ is algebraic over $F$.
3. There exist algebraic elements $\alpha_1$, $\alpha_2$, $\dots$, $\alpha_n$ of $E$, such that $E=F(\alpha_1, \alpha_2, \dots,$ $\alpha_n)$.
• Stated and proved the Corollary: Let $K/F$ be a field extension and let $\overline F$ be the set of elements of $K$ that are algebraic over $F$. Then $\overline F$ is a subfield of $K$ with $F \subset \overline F \subset K$. The field $\overline F$ is called the algebraic closure of $F \in K$.
• Discussed the set of algebraic numbers in the complex numbers $\mathbb {C}$ over the rational numbers $\mathbb {Q}$.
• Showed that for $\alpha=\sqrt 2 + \sqrt 3$, $\mathbb{Q}(\alpha)= \mathbb{Q} (\sqrt2, \sqrt 3)$. Later showed that $\sqrt 3 \notin \mathbb{Q} (\sqrt2)$ and consequently that $\big[ \mathbb{Q}(\alpha): \mathbb{Q} \big]=4$ (By Eisenstein $x^2-2$, and $x^2-3$ are irreducible over $\mathbb{Q}$ and so $\sqrt 2, \sqrt 3 \notin \mathbb {Q}$).
• Discussed an example of the Louisville numbers (with big gaps in the decimal expansion): $\sum_{i=1}^\infty 1/10^{i!}$. These are transcendental and it is easier to show they are transcendental than other transcendental numbers (see book).
• Discussed countable and uncountable cardinalities.
• Stated and proved the Theorem : For $K/F$ a field extension with $F$ countable, the set of elements of $K$ that are algebraic over $F$ is countable.
• To prove this, stated and outlined/picture-proved $N$ lemmas:
For $X, X_n, Y$ sets
1. $X,Y$ countable $\implies X \times Y$ countable;
2. $X$ countable and an injection takes $Y$ into X $\implies Y$ is countable;
3. $X$ countable and a surjection takes $X \to Y \implies Y$ countable;
4. $Y$ a countable union of countable sets $X_n \implies Y$ is countable;
5. $F$ a countable field and $V$ a vector space over $F$ with a countable basis $\implies V$ is countable;
6. $\vdots$
7. $N$. If $X$ is countable and $Y \to X$ is countable-to-one, then $Y$ is countable.
• Corollary: In the complex numbers $\mathbb{C}$, the subfield of elements that are algebraic over the rationals $\mathbb{Q}$ is countable. Elements of this set are called algebraic numbers.
• Proved part $1$ of Proposition~1.20 Multiplication of degree (part 2 was proved in Lecture 5).
• Proved the following tw statements:
Proposition: (algebraic extensions are transitive): If $F \subseteq K \subseteq L$ are fields with $K/F$ algebraic and $L/K$ algebraic, then $L/F$ is algebraic.
Proposition: If $F\subseteq K \subseteq L$ are fields and $L=F[\alpha]$ where alpha is algebraic, then the minimal polynomial for $\alpha$ over $K$ divides the minimal polynomial for $\alpha$ over $F$.
• Worked two problems:
• Show if $[L:F]=p$, a prime integer, then there are no fields $K$ properly between $F$ and $L$.
• If $F \subseteq L$ is algebraic and $T$ is an integral domain with $F \subseteq T \subseteq L$, prove that $T$ is a field.
Another problem the students might like to try for extra practice is Exercise~1-5 in Milne.


text:
# Does the proof of the well-ordering principle not commit the circular reasoning fallacy?

It almost certainly doesn't, since it's an old and definitely a well-tested theorem, but I don't understand why.

Begging the question, or circular reasoning, as defined in my textbook (of sorts), is:

...the author of a proof uses in his argument the fact that he is supposed to prove.

The well-ordering principle, on the other hand, is defined as:

If S is a nonempty subset of N then there is m ∈ S such that m ≤ x for all x ∈ S. That is, S has a smallest or least element.

Then, as a proof for the well-ordering principle, they write:

We will use contraposition to prove the theorem. That is, by assuming that S has no smallest element we will prove that S = ∅. We will prove that n ∉ S for all n ∈ N. We do this by induction on n. Since S has no smallest element, we have 1 ∉ S. Assume that we have proved that 1, 2, · · · , n ∉ S. We will show that n + 1 ∉ S. If n + 1 ∈ S then n + 1 would be the smallest element of S since 1, 2, 3, · · · , n ∉ S, and this contradicts the assumption that S has no smallest element. Thus, we must have n + 1 ∉ S. Hence, by the principle of mathematical induction, n ∉ S for all n ∈ N. But this leads to S = ∅. This establishes a proof of the theorem

The well-ordering principle is trying to prove that every non-empty set has a smallest element, so why does its proof assume that n + 1, the only element in the set S, would be the smallest one?

I do get that n + 1 would be the smallest element if 1, 2, 3 . . . n do not belong to the set S (obviously), but does that assumption not make use of the well-ordering principle, which it's trying to prove?

(I'm a sophomore undergrad, so, if possible, relatively simple explanations would be appreciated.)

• I think that almost all the $k\in S$ in your post should be $k\notin S$. – Asaf Karagila Nov 7 '17 at 21:39
• @AsafKaragila Do you mean n? Also, yes, it should be. I copy-pasted from my e-textbook, but it seems it wasn't as perfect as I'd thought it would be. I'll fix it now. – Demetre Saghliani Nov 7 '17 at 21:41
• I used $k$ since in some places it has $1$ or $n+1$ and so on. – Asaf Karagila Nov 7 '17 at 21:41
• Two sentences before the one you put in boldface, it says "Assume that we have proved that $1,2,\cdots,n\notin S$." Given this assumption, if $n+1$ is in $S$, then $n+1$ is the smallest element of $S$, because all the smaller natural numbers $1,2,\cdots,n$ have been assumed to not be in $S$. (And $n+1$ would not necessarily be the only element in $S$. If it were the only element, then it would be the smallest even more obviously.) – Andreas Blass Nov 7 '17 at 21:52
• If a set has only one element, how can any element be bigger? – fleablood Nov 7 '17 at 21:56

The proof isn't circular --- it does assume that $\mathbb{N}\subseteq\mathbb{N}$ satisfies the well-ordering principle though, which may be your source of confusion. This assumption is fine though, as $\mathbb{N}$ vacuously has a least element (we define it that way!). I'll assume that $\mathbb{N} = \mathbb{Z}_{\geq 1}$. If you want $\mathbb{N}_0 = \mathbb{Z}_{\geq 0}$, the change should be minor.

An (informal) sketch of the proof is as follows.

We know that $\mathbb{N}$ satisfies the well-ordering principle, and has least element $1\in \mathbb{N}$. Now, consider $S\subseteq\mathbb{N}$, an arbitrary subset.

1. Assume $S$ doesn't satisfy the well-ordering princple. This must mean that $1\not\in S$ (otherwise $1$ would be the minimum element)
2. Now, via induction, we can show that if $n\not\in S$, then $n+1\not\in S$ (as otherwise $n+1$ would be the minimum element). So, we have that $1\not\in S$, then $2\not\in S$, etc.

3. Now, we've shown that every $n\geq 1$ has $n\not\in S$. So, $S = \emptyset$ is an empty subset of $\mathbb{N}$.

• Your observation that the proof does assume that $\Bbb{N} \subseteq \Bbb{N}$ satisfies the well-ordering principle is incorrect. The sketch quoted in the question and spelled out in more detail in your points 1, 2 and 3 provides a valid proof that any non-empty $S \subseteq \Bbb{N}$ has a minimum element that doesn't make any use of that assumption. – Rob Arthan Nov 7 '17 at 22:28
• @RobArthan The proof does require $\mathbb{N}$ to have a minimal element. I referred to this as "$\mathbb{N}\subseteq\mathbb{N}$ satisfying the well-ordering principle" as I suspected this was where OP was confused. $\mathbb{N}$ having a minimum element is equivalent to it satisfying the well-ordering principle, so I saw no harm in mentioning that. – Mark Nov 7 '17 at 23:01
• To have a minimal element is not the same as satisfying the well-ordering principle. Your answer is simply incorrect. – Andrés E. Caicedo Nov 8 '17 at 23:41

The proof you've provided has many typos. It should read:

We will use contraposition to prove the theorem. That is, by assuming that $S$ has no smallest element we will prove that $S = \emptyset$. We will prove that $n \not\in S$ for all $n\in\mathbb{N}$. We do this by induction on $n$. Since $S$ has no smallest element, we have $1 \not\in S$. Assume that we have proved that $1, 2,\ldots, n \not\in S$. We will show that $n + 1 \not\in S$. If $n + 1 \in S$ then $n + 1$ would be the smallest element of $S$ since $1, 2, 3, \ldots , n \not\in S$, and this contradicts the assumption that $S$ has no smallest element. Thus, we must have $n + 1 \not\in S$. Hence, by the principle of mathematical induction, $n \not\in S$ for all $n\in \mathbb{N}$. But this leads to $S=\emptyset$. This establishes a proof of the theorem.

Now, the bolded part is true because if $S$ is a set of natural numbers that doesn't contain $1,\ldots,n$ but does contain $n+1$, then $n+1$ would be the least element of $S$. This follows because $n+1$ is the smallest natural number that isn't $1,2,\ldots,n$.

The idea is as follows: Because $1$ is the smallest natural number, if $1\in S$, then $1$ has to be the smallest element. So $1\notin S$.

Now by induction, if $1,\ldots,n\notin S$, then it has to be the case that $n+1\notin S$ as well, since $n+1$ is the smallest number which is larger than all $1,\ldots,n$. So if $n+1\in S$, then by the assumption that $1,\ldots,n\notin S$, it has to be that $n+1$ is the smallest element of $S$, which is again a contradiction to the assumption on $S$.

By the induction principle, it follows that for every $n\in\Bbb N$, $n\notin S$. But that means that $S=\varnothing$.

In other words, there is no begging of the question here. We simply prove by induction, if $1,\ldots,n\notin S$, then $n+1\notin S$ either.

Either the way you have copied the proof is wrong, or your book is wrong, since almost all of you $\in$ should be replaced with $\notin$.

The argument assumes that $S$ has no smallest element. The aim, then is to show that $S$ must be empty, by showing that no natural number is in $S$. If $1$ is in $S$, then this contradicts the fact that $S$ has no smallest element, since $1$, being the smallest natural number would have to be the smallest element in $S$. Thus, $1\notin S$. Now, proceed by induction, assuming that for some fixed $n\geq 1$, $1,\ldots, n\notin S$. Then, if $n+1$ were in $S$, it would have to be the smallest element in $S$, since none of the ones that could be smaller, namely, $1,\ldots, n$ are in $S$. But then $S$ would have a smallest element, $n+1$, which would contradict the assumption that $S$ has no smallest element. Therefore, $n+1$ cannot be in $S$. Hence, by induction, no natural number is in $S$, so $S$ must be empty.

It looks to me like you have miscopied the proof or that there was a misprint in the proof itself. Obviously, it makes no sense to say that "Assume that we have proved 1, 2, ..., n $\in$ S" and then say "if n+1 $\in$ S the n+ 1 is the smallest number in S! I suspect rather that they meant to say "Assume that we have proved 1 2, ..., n $\notin$ S". If n+ 1$\in$ S while all numbers n and less are not in S, then certainly n+ 1 is the smallest number in S.

• They did. The copy-paste didn't quite carry it all over. I've fixed it, now. – Demetre Saghliani Nov 7 '17 at 21:51

Given a set $X$, $x$ is the least element of $X$ if i) $x\in X$ and ii) for all $y\in X; x\le y$.

Claim: if $S=\{n+1\}$ then $n+1$ is the least element of $S$.

Proof.

i) $n+1\in S$. That is true.

ii) for all $y\in S$ then $y=n+1$. And $y=n+1\le n+1$.

End of Proof.

One element is one element, and the author was taking the above as self evident

• Have you read the final edit (N. 6) of the post? "I do get that n + 1 would be the smallest element if 1, 2, 3 . . . n do not belong to the set S (obviously), but does that assumption not make use of the well-ordering principle, which it's trying to prove?" – Demetre Saghliani Nov 7 '17 at 22:04
• No, he is not assuming S has a lowest element. He has demonstrated that S has no elements that are smaller than n+1, then n+1 is smaller than all others. That's definition and he has now proven that n+1 is the least. – fleablood Nov 7 '17 at 22:11


text:
## Dalton's Law

Dalton's law gives the relation between the total pressure of a gas in a container and the partial pressure of its constituents.

The sum of all partial pressures in a mixture of gases is equal to the total pressure.

$p_t_o_t_a_l=p_1+p_2+p_3+{_\dots}+p_n$

Note that:

The gas mixture in a closed volume does not separate. In an air-to-helium mixture, the helium does not rise to the top of the container even though it is lighter than the other gases. This is because of the thermal speed of the particles there are inter-particle collisions in the gas as well as with the container walls and therefore always a mixing process, called "diffusion"


text:
Rigidity of the magic pentagram game

# Rigidity of the magic pentagram game

Amir Kalev Joint Center for Quantum Information and Computer Science, University of Maryland, College Park, MD 20742-2420, USA    Carl A. Miller Joint Center for Quantum Information and Computer Science, University of Maryland, College Park, MD 20742-2420, USA National Institute of Standards and Technology, Gaithersburg, MD 20899, USA
July 25, 2019
###### Abstract

A game is rigid if a near-optimal score guarantees, under the sole assumption of the validity of quantum mechanics, that the players are using an approximately unique quantum strategy. Rigidity has a vital role in quantum cryptography as it permits a strictly classical user to trust behavior in the quantum realm. This property can be traced back as far as 1998 (Mayers and Yao) and has been proved for multiple classes of games. In this paper we prove ridigity for the magic pentagram game, a simple binary constraint satisfaction game involving two players, five clauses and ten variables. We show that all near-optimal strategies for the pentagram game are approximately equivalent to a unique strategy involving real Pauli measurements on three maximally-entangled qubit pairs.

## I Introduction

Quantum rigidity is a strengthening of the guarantee that quantum behavior is taking place. It essentially ascertains that observing certain correlations in a system, for example, correlations that violate Bell inequalities, is sufficient by itself to determine the quantum state and the measurements used to obtain these correlations. This notion was expressed in the work of Mayers and Yao on “self-checking quantum sources” Mayers and Yao (1998) in 1998, and it can be traced back even earlier Popescu and Rohrlich (1992); Summers and Werner (1987). Rigidity is a central tool for quantum computational protocols that involve untrusted devices, since it allows a user to verify the internal workings of a device based only on its external behavior (see, e.g., Reichardt et al. (2013)).

Since its introduction the notion of rigidity has seen good deal of work, generally focused either on proving rigidity for particular classes of games, or proving that rigid games exist that self-test particular quantum states. Two-player games that are known to be rigid include the CHSH game Popescu and Rohrlich (1992); McKague et al. (2012), the magic square game Wu et al. (2016), the chained Bell inequalities Šupić et al. (2016), the Mayers-Yao criterion Mayers and Yao (1998); Magniez et al. (2006), Hardy’s test Rabelo et al. (2012), the Hadamard-graph coloring game Mančinska (2014), and various classes of binary games Miller and Shi (2013); Wang et al. (2016); Bamps and Pironio (2015). New results on rigid games add to the tools available for protocols based on untrusted devices.

In the current paper we prove that the magic pentagram game (see Figure 1) is rigid. This game is a natural one to study: in particular, it was originally proposed alongside the magic square game Mermin (1990), and it shares some of the same properties that make the magic square game useful in cryptography (in particular, it shares the property that an optimal strategy must yield a perfect shared key bit pair between two parties, which was exploited in Jain et al. (2017)). From a resource standpoint, it also offers an improvement over the magic square game: whereas the magic square game requires questions to self-test EPR pairs, we will prove that the magic pentagram game self-tests EPR pairs with questions. If we compare the number of bits of randomness needed to generate the questions set to the number EPR pairs tested, the magic square has a ratio of , while the magic pentagram game has a ratio of .

The optimal strategy for the magic pentagram game is shown in Figure 2. Our main result is summarized below, and proved formally in Propositions 9, 10, 12, and Corollary 11.

###### Theorem 1 (Informal).

Suppose that Alice and Bob have a strategy for the magic pentagram game that wins with probability . Then, after the application of a local isometry on Alice’s and Bob’s systems, the following statements hold.

1. The shared state is within Euclidean distance from a state of the form , where denotes a Bell state and denotes an arbitrary bipartite state. (Proposition 12.)

2. The post-measurement states under Alice’s and Bob’s measurements are approximated (up to ) by the corresponding post-measurement states from the strategy in Figure 2. (Propositions 910 and Corollary 11.)

Our proof is self-contained and borrows techniques from previous papers on rigidity McKague et al. (2012); McKague (2016); Wu et al. (2016). One of the challenges for the magic pentagram game is that the first player may associate two different measurements to a single observable — for example, in Figure 1, Alice may use a different measurement for vertex depending on whether the context is or . (This does not occur in the magic square game.) Our early technical work addresses this fact — see Propositions 56 and the discussion that follows.

The coefficients of the error terms for Theorem 1 are not given explicitly, and optimizing these coefficients is left as an open problem. (Tracing through the steps of the current proof might yield coefficients in the thousands.)

In the larger picture, the magic square game and the magic pentagram game are examples of binary constraint satisfaction games Cleve and Mittal (2014). Arkhipov Arkhipov (2012) proved that a certain natural subclass of binary constraint satisfaction problems — specifically, those that are based on XOR clauses where every variable is in exactly two clauses — are all in a precise sense reducible to the magic square game and the magic pentagram game. This suggests that our result is a step towards a full classification of winning quantum strategies within this class.

## Ii The magic pentagram game

The pentagram game is a binary constraint satisfaction game between two parties, Alice and Bob. Its rules can be defined, as its name suggests, on a pentagram hypergraph, see Fig. 1. The five hyperedges of the pentagram (the clauses or contexts) are labeled , and each contains four vertices. The hyperedges are each assigned a value: , and . The rules of the games are as follows:

• A context is chosen and a vertex is chosen (both uniformly at random). The context is given to Alice and the vertex is given to Bob.

• Alice assigns either or to each vertex in the context , and Bob assigns or to .

• Alice and Bob can communicate and agree on a strategy prior to the beginning of the game, but are not allowed to communicate once the game has begun.

The game is won if the following two conditions both hold:

• The product of the values returned by Alice is equal to the pre-assigned value .

• Alice and Bob return the same value for .

There is no classical strategy to win this game perfectly, as is easily verified. However, it can be won with probability 1 using quantum resources Mermin (1990, 1993). A winning strategy is schematically shown in Fig. 2, with and denoting the Pauli operators , and the identity operator, respectively. They share six qubits, three at Alice’s lab () and three at Bob’s (), prepared in the maximally entangled state

 |Φ+⟩⊗3=3⨂i=11√2(|0⟩Qi|0⟩Qi+3+|1⟩Qi|1⟩Qi+3), (1)

where are the eigenbasis of the Pauli operator. (When no confusion arises we drop the tensor product symbol and the subscript labels for Alice and Bob’s subsystems.) Upon receiving a hyperedge label , Alice measures the four Pauli observables associated with the four vertices of on her three qubits, and then assigns to each vertex the value she obtains for the corresponding observable. These observables are reflection operators (i.e., Hermitian operators having eigenvalues in ) such that observables of adjacent vertices (vertices that are connected by the same hyperedge) all commute and thus can be measured simultaneously. Bob measures the observable of his input vertex on his three qubit system and assigns a value to the vertex according to the outcome of his measurement. By construction of this strategy, the winning conditions for this game, as listed above, are fulfilled for every input value and .

We note that in this strategy any two non-adjacent observables anti-commute. (This will become important in later proofs.)

## Iii Strategies for the magic pentagram game

Our goal is to relate arbitrary strategies for the magic pentagram game to the strategy in Figure 2. The class of strategies that we study are captured in the following definition.

###### Definition 2.

A projective strategy for the magic pentagram game consists of the following data:

1. The shared state: Two finite-dimensional Hilbert spaces and , and a unit vector .

2. Alice’s measurements: For each , a projective measurement on , where varies over the set of all functions from to whose parity is equal to .

3. Bob’s measurements: For vertex , a projective measurement .

The functions obtained from these measurements specify the output values for Alice and Bob. Note that we could have allowed for the shared state to be mixed and for the measurements to be general positive-operator valued measures (POVMs). However standard techniques imply that any such strategy is a partial trace of one in the above form, so there is no generality lost.

Additionally, we make the following definition.

###### Definition 3.

A reflection is a Hermitian automorphism whose eigenvalues are contained in . A reflection strategy for the magic pentagram game consists of the following data:

1. The shared state: Two finite-dimensional Hilbert spaces and , and a linear map satisfying .

2. Alice’s reflections: Reflections

 {Rjv∣j∈{C,D,E,F,G},v∈j}

on such that the reflections that belong to any context all commute () and their product is equal to .

3. Bob’s reflections: Reflections on .

Note that any projective strategy can be converted into a reflection strategy, and vice versa, via the relations

 Rjv = ∑t(v)=0Mjt−∑t(v)=1Mjt, (2) Sv = N0v−N1v (3) L = ⟨ΦB|ψ⟩ (4)

where on .

The probability distribution obtained from a projective measurement on is given by , and the probability distribution obtained from a projective measurement on is given by . For any context and any vertex , the probability that Alice and Bob will assign different values to the vertex in a given reflection strategy is given by

 ∥∥ ∥∥(I+Rjv2)L(I−Sv2)∥∥ ∥∥22+∥∥ ∥∥(I−Rjv2)L(I+Sv2)∥∥ ∥∥22.

Thus the losing probability (that is, one minus the expected score) for the reflection strategy is given by

 plose = 120∑v∈j⎛⎝∥∥ ∥∥(I+Rjv2)L(I−Sv2)∥∥ ∥∥22 +∥∥ ∥∥(I−Rjv2)L(I+Sv2)∥∥ ∥∥22⎞⎠ = 120∑v∈j∥∥L−RjvLSv∥∥22 = 120∑v∈j∥∥RjvL−LSv∥∥22.

Thus we have the following.

###### Proposition 4.

Let be a reflection strategy for the magic pentagram game which achieves winning probability . Then, for any context and vertex ,

 ∥∥RjvL−LSv∥∥2 ≤ O(√ϵ), (5)

Next we prove a series of properties for near-optimal strategies, all of which are consequences of Proposition 4.

###### Proposition 5 (Changing contexts).

Let

 (L,{Rjv},{Sv})

be a reflection strategy with expected score . Let be a sequence of vertices and and be sequences of contexts such that for all . Then,

 ∥∥Rj1v1Rj2v2⋯RjnvnL−Rj′1v1Rj′2v2⋯Rj′nvnL∥∥2 ≤ O(n√ϵ).
###### Proof.

Applying Proposition 4 inductively, we find that and are both within Euclidean distance from . ∎

The next two propositions certify the relation between reflection operators in a strategy with expected score . For convenience, hereafter we refer to sequences of matrices satisfying as -approximate sequences.

###### Proposition 6 (Approximate commutativity).

Let be a reflection strategy with expected score . Let and be adjacent vertices, such that , and let be the other hyperedge which contains . Then,

 ∥∥RjvRj′wL−Rj′wRjvL∥∥2 ≤ O(√ϵ) (6) ∥LSwSv−LSvSw∥2 ≤ O(√ϵ). (7)
###### Proof.

The desired result follows easily by applications of Proposition 4. ∎

Each vertex has two reflection operators for Alice ( and , where ). It is helpful for some of the proofs that follow to single out one distinguished reflection operator for each vertex. We therefore make the following (arbitrary) assignments,

 R1:=RG1R6:=RE6R2:=RG2R7:=RF7R3:=RE3R8:=RD8R4:=RF4R9:=RD9R5:=RE5R10:=RC10. (8)
###### Proposition 7 (Approximate anti-commutativity).

Let be a reflection strategy with expected score , and let and be non-adjacent vertices (i.e., vertices that never occur in the same context). Then,

 ∥∥RjvRj′wL+Rj′wRjvL∥∥2 ≤ O(√ϵ) (9) ∥LSwSv+LSvSw∥2 ≤ O(√ϵ). (10)
###### Proof.

By Proposition 5, it suffices to prove these relations with replaced by . We give a proof for , which generalizes to cover all other cases by symmetry. The proof is inspired by the proof of rigidity for the magic square game Wu et al. (2016). Applying the rules for Alice’s measurements from Definition 3 and the foregoing propositions, we find that the following sequence is an -approximate sequence:

 R7R3L, (R4R9R6)(R6R8R5)L, R4R9R8R5L, R4(R1R10R8)R8R5L, R4R1R10R5L, R4R1(R2R7)L, −R3R7L,

and relation (10) follows similarly. ∎

The next proposition follows from Propositions 4, 6, and 7.

###### Proposition 8.

Let be a sequence of vertices and . Then,

 ∥∥Rj1v1⋯RjiviRji+1vi+1⋯RjnvnL −bRj1v1⋯Rji+1vi+1Rjivi⋯RjnvnL∥∥2 ≤ O(n√ϵ) ∥∥LSv1⋯SviSvi+1⋯Svn −bLSv1⋯Svi+1Svi⋯Svn∥∥2 ≤ O(n√ϵ),

where if are adjacent and if are non-adjacent.

## Iv Rigidity

In this section, we will use the following notation: will denote qubit registers (each with a fixed isomorphism to ). The linear maps denote the Hadamard maps , and the linear maps denote the Pauli operators. For any reflection on , and , let the map

 Ci(U):Qi⊗HA⊗HB→Qi⊗HA⊗HB (11)

denote the controlled operation . Note that these maps interact as follows:

 XiCi(U)Xi=Ci(U)U=UCi(U) (12) ZiCi(U)=Ci(−U)=Ci(U)Zi (13)

The next theorem asserts that some of the reflections in a near-optimal strategy for the magic pentagram game can be simulated by Pauli operators. Let

 X′1=R6X′4=S6X′2=R5X′5=S5X′3=R7X′6=S7Z′1=R10Z′4=S10Z′2=R9Z′5=S9Z′3=R8Z′6=S8, (14)

where the s are given in Eq. (8). These operators are chosen so that for (and similarly for ) the pairs , belong to non-adjacent vertices, while all the other pairs of operators belong to adjacent vertices. Thus the approximate commutativity conditions and anti-commutativity conditions are what one would expect for the corresponding Pauli operators. We note that the particular choice of the s and s here is not unique. The following results will hold for any choice of s and s as long as they satisfy the required approximate commutation relations.

###### Proposition 9.

Let be a reflection strategy with expected score . Then, there exists an isometry from to such that for all ,

 ∥∥XiΨAL−ΨAX′iL∥∥2 ≤ O(√ϵ) (15) ∥∥ZiΨAL−ΨAZ′iL∥∥2 ≤ O(√ϵ). (16)
###### Proof.

Our construction of the isometries follows previous papers on rigidity (e.g., McKague (2016)). For each define

 Ψi:HA→HA⊗Qi (17)

by

 Ψi(z) = (18)

Then, the following is an -approximate sequence:

 XiΨiL, Xi[Ci(X′i)]Hi[Ci(Z′i)](L⊗|+⟩), [Ci(X′i)]X′iXiHi[Ci(Z′i)](L⊗|+⟩), [Ci(X′i)]HiZiX′i[Ci(Z′i)](L⊗|+⟩), [Ci(X′i)]HiZi[Ci(−Z′i)]X′i(L⊗|+⟩), [Ci(X′i)]Hi[Ci(Z′i)]X′i(L⊗|+⟩), ΨiX′iL.

Thus,

 ∥∥XiΨiL−ΨiX′iL∥∥2 ≤ O(√ϵ).

Additionally, the following is an -approximate sequence:

 ZiΨiL, Zi[Ci(X′i)]Hi[Ci(Z′i)](L⊗|+⟩), [Ci(X′i)]ZiHi[Ci(Z′i)](L⊗|+⟩), [Ci(X′i)]HiXi[Ci(Z′i)](L⊗|+⟩), [Ci(X′i)]Hi[Ci(Z′i)]Z′iXi(L⊗|+⟩), [Ci(X′i)]Hi[Ci(Z′i)]Z′i(L⊗|+⟩), ΨiZ′iL.

Thus,

 ∥∥ZiΨiL−ΨiZ′iL∥∥2 ≤ O(√ϵ).

Also, if with , then by Proposition 6, the following is a -approximate sequence:

 X′kΨiL, X′k[Ci(X′i)]Hi[Ci(Z′i)](L⊗|+⟩), X′k[Ci(X′i)]HiL[Ci(Z′i+3)](I⊗|+⟩), X′k[Ci(X′i)]LHi[Ci(Z′i+3)](I⊗|+⟩), [Ci(X′i)]X′kLHi[Ci(Z′i+3)](I⊗|+⟩), [Ci(X′i)]X′kHi[Ci(Z′i)]L(I⊗|+⟩), [Ci(X′i)]Hi[Ci(Z′i)]X′kL(I⊗|+⟩), ΨiX′kL.

Therefore

 ∥∥X′kΨiL−ΨiX′kL∥∥2 ≤ O(√ϵ) (19)

and by similar reasoning,

 ∥∥Z′kΨiL−ΨiZ′kL∥∥2 ≤ O(√ϵ). (20)

Define by the same expression used to define , except with the operators replaced with :

 Φi(z) = [Ci(X′i+3)]Hi[Ci(Z′i+3)](z⊗|+⟩). (21)

Then, by Proposition 4. Let

 ΨA = Ψ1Ψ2Ψ3. (22)

Then, the following is an -approximate sequence:

 X2ΨAL, X2Ψ1Ψ2Ψ3L, Ψ1X2Ψ2Ψ3L, Ψ1X2Ψ2LΦ3, Ψ1Ψ2X′2LΦ3, Ψ1Ψ2X′2Ψ3L, Ψ1Ψ2Ψ3X′2L.

Therefore,

 ∥∥X2ΨAL−ΨAX′2L∥∥2 ≤ O(√ϵ). (23)

The desired result for follows by similar reasoning. ∎

Likewise, we have the following.

###### Proposition 10.

Let be a reflection strategy with expected score . Then, there exists an isometry from to such that for all ,

 ∥∥LΨ†BXi−LX′iΨ†B∥∥2 ≤ O(√ϵ) (24) ∥∥LΨ†BZi−LZ′iΨ†B∥∥2 ≤ O(√ϵ). (25)
###### Proof.

Define for by the same expression (18) that was used in the previous proof, and let . The desired result follows by the same reasoning that was used to prove Proposition 9. ∎

Note that Propositions 9 and 10 easily generalize to sequences of measurements — for example, the following is an -approximate sequence:

 X1X2ΨAL, (26) X1ΨAX′2L, (27) X1ΨALX′5, (28) ΨAX′1LX′5, (29) ΨAX′1X′2L. (30)

Applying this method inductively, we have the following corollary.

###### Corollary 11.

The isometries from Proposition 9 and 10 satisfy the following. For any sequence and corresponding sequence ,

 ∥∥M1⋯MnΨAL−ΨAM′1⋯M′nL∥∥2 ≤ O(n√ϵ).

For any sequence and corresponding sequence ,

 ∥∥LΨ†BNn⋯N1−LN′n⋯N′1Ψ†B∥∥2 ≤ O(n√ϵ).\qed

Finally, we prove the following proposition, which addresses the image of the under the isometry . For each , let

 ϕ+i:Qi→Qi+3 (31)

be defined by

 ϕ+i = ⎡⎢⎣1√2001√2⎤⎥⎦. (32)

(This is a matrix expression for an EPR pair.) Let

 ϕ−i = ⎡⎢⎣1√200−1√2⎤⎥⎦ (33) ψ+i = ⎡⎢⎣01√21√20⎤⎥⎦ (34) ψ−i = ⎡⎢⎣01√2−1√20⎤⎥⎦. (35)
###### Proposition 12.

Let be the operators from Propositions 9 and 10. Then, for some ,

 ∥∥L′⊗ϕ+1⊗ϕ+2⊗ϕ+3−ΨALΨ†B∥∥2 ≤ O(√ϵ). (36)
###### Proof.

Let . By the score assumption,

 ∥∥X′iLX′i+3−L∥∥2 ≤ O(√ϵ) (37) ∥∥Z′iLZ′i+3−L∥∥2 ≤ O(√ϵ), (38)

for , therefore by Propositions 9 and 10,

 ∥XiPXi+3−P∥2 ≤ O(√ϵ) (39) ∥ZiPZi+3−P∥2 ≤ O(√ϵ), (40)

Note that , while the other Bell states fail significantly to satisfy the same equalities:

 Xiϕ−iXi = −ϕ−i (41) Ziψ+iZi = −ψ+i (42) Ziψ−iZi = −ψ−i. (43)

Write

 P = ∑v1,v2,v3v1⊗v2⊗v3⊗Pv1,v2,v3, (44)

where varies over . Conditions (39) and (40) imply that all components except must have Euclidean norm less than . The desired result follows. ∎

## V Summary and Conclusions

Quantum rigidity allows a classical user to certify manipulations of quantum systems, thus enabling quantum cryptography in a scenario in which the user does not trust her quantum apparatus (device-independent quantum cryptography). In this paper we have expanded the toolbox for the device-independent setting by showing that the magic pentagram game is rigid. In particular, this means that it is possible to certify the existence of ebits using a game that consists of only questions.

In our style of proof we have reduced some of the arguments for rigidity to bare manipulations of sequences of operators (see the proofs in section IV). This style in particular allows us to cleanly handle conditions such as approximate commutativity and anti-commutativity. Such an approach could be useful for proving more general results.

A natural next step would be to try to parallelize our result (following Reichardt et al. (2013); Ostrev (2015); Ostrev and Vidick (2016); Coladangelo (2016); Natarajan and Vidick (2017); McKague (2016, 2017); Chao et al. (2016); Coudron and Natarajan (2016)) to show that parallel copies of the magic pentagram game can be used to certify a maximally entangled state of arbitrary size. Then, we could try to choose a small subset of the questions from the parallelized game and prove that that subset is adequate to achieve rigidity.

The magic pentagram game is an example of a binary constraint satisfaction XOR game in which every variable appears in exactly two contexts. This class of games was studied in Arkhipov (2012), and the author proved that any game in the class that exhibits pseudo-telepathy must in a sense contain either the magic square game or the magic pentagram game (as topological minors of its relational graph). An interesting further direction would be to explore further the consequences for our rigidity result (and Wu et al. (2016)) for the class from Arkhipov (2012).

###### Acknowledgements.
The authors would like to thank Cedric Lin for bringing Ref. Arkhipov (2012) to our attention, and Matthew McKague for helpful technical discussions about our proofs. AK is funded by the US Department of Defense.

## References

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