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# A Geometrical Understanding of Matrices

## Beyond systems of equations

I took linear algebra in college, but the main thing I remember from the course was solving systems of linear equations by converting matrices to row echelon form. I did well in the class, but I had little intuition for matrices as geometric objects and almost none of the material stuck. In graduate school, I have discovered that having such a geometrical intuition for matrices—and for linear algebra more broadly—makes many concepts easier to understand, chunk, and remember.

For example, computing the determinant of a matrix is tedious. But if we think of the determinant of a matrix as the signed scale factor representing how much a matrix transforms the volume of an $n$-cube into an $m$-dimensional parallelepiped, it is straightforward to see why a matrix with a determinant of $0$ is singular. It is because the matrix is collapsing space along at least one dimension.

The goal of this post is to lay the ground work for understanding matrices as geometric objects. I focus on matrices because they effect how I think of vectors, vector spaces, the determinant, null spaces, spans, ranks, inverse matrices, singular values, and so on. In my mind, every one of these concepts is easier to understand with a strong, geometrical understanding of matrices. For simplicity, I will focus on real-valued matrices and vectors, and I error on the side of intuitive examples rather than generalizations.

## Matrices as linear transformations

In linear algebra, we are interested in equations of the following form:

Where $\textbf{b} \in \mathbb{R}^{m}$, $A \in \mathbb{R}^{m \times n}$, and $\textbf{x} \in \mathbb{R}^{n}$. One way to think about this equation is that $A$ represents a system of $m$ linear equations, each with $n$ variables, and $\textbf{x}$ represents a solution to this system. But there is another way to think of the matrix $A$, which is as a linear function $f$ from $\mathbb{R}^n$ to $\mathbb{R}^m$:

In my mind, the easiest way to see how matrices are linear transformations is to observe that the columns of a matrix $A$ represent where the standard basis vectors in $\mathbb{R}^n$ map to in $\mathbb{R}^m$. Let’s look at an example. Recall that the standard basis vectors in $\mathbb{R}^3$ are:

By definition, this means that every vector in $\mathbb{R}^3$ is a linear combination of this set of basis vectors:

Now let’s see what happens when we matrix multiply one of these standard basis vectors, say $\textbf{e}_2$, by a transformation matrix $A$:

In words, the second column of $A$ tells us where the second basis vector in $\mathbb{R}^3$ maps to in $\mathbb{R}^2$. If we horizontally stack the standard basis vectors into a $3 \times 3$ matrix, we can see where each basis vector maps to in $\mathbb{R}^2$ with a single matrix multiplication:

Now here’s the cool thing. We can express any transformed $2$-vector as a linear combination of $f(\textbf{e}_1)$, $f(\textbf{e}_2)$, and $f(\textbf{e}_3)$, where the coefficients are the components of the untransformed $3$-vector. For example, if $\textbf{x} = \begin{bmatrix} 1.2 & 1.5 & -2 \end{bmatrix}^{\top}$, then:

It’s worth staring at these equations for a few minutes. In my mind, seeing Equations \ref{eq1} and \ref{eq2} together helped me understand why matrix multiplication is defined as it is. When we perform matrix multiplication, we are projecting a vector or vectors into a new space defined by the columns of the transformation matrix. And $f(\textbf{x})$ is just a linear combination of the columns of $A$, where the coefficients are the components of $\textbf{x}$.

In my mind, changing how we see the equation $\textbf{b} = A \textbf{x}$ is not trivial. In the textbook Numerical Linear Algebra (Trefethen & Bau III, 1997), the authors claim that seeing matrices this way is “essential for a proper understanding of the algorithms of numerical linear algebra.”

This linearity is what allows us to say concretely that any linear transformation, $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, can be represented as a linear combination of the transformed basis vectors of $\textbf{x}$, which can be encoded as a matrix-vector multiplication:

See the Appendix for a proof of the linearity of matrix transformations.

## Visualizing matrix transformations

The linearity of matrix transformations can be visualized beautifully. For ease of visualization, let’s only consider $2 \times 2$ matrices, which represent linear transformations from $\mathbb{R}^2$ to $\mathbb{R}^2$. For example, consider the following matrix transformation $A$ of a vector $\textbf{x} = \begin{bmatrix} 1 & 2 \end{bmatrix}^{\top}$:

We can visualize two important properties of this operation (Figure $1$). First, the columns of $A$ represent where the standard basis vectors in $\mathbb{R}^2$ land in this transformed vector space.

Second, in the new vector space, the vector $f(\textbf{x})$ is a linear combination of these transformed vectors. This is particularly useful to visualize by coloring unit squares and seeing how they are transformed. You can imagine that not just a vector is transformed by a matrix but that all of $\mathbb{R}^n$ is transformed an $m \times n$ matrix. Note that the determinant of the matrix is also visualized here. It is $2$.

## Types of linear transformations

Now that we have some intuition for how matrix transformations represent linear functions, we might want to ask: what kind of transformations can we perform with matrices? A consequence of the linearity of matrix transformations is that we can only modify a vector space in particular ways such as by rotating, reflecting, scaling, shearing, and projecting. Intuitively, the hallmark of a linear transformation is that evenly spaced points before the transformation are evenly spaced after the transformation (Figure $2\text{a}$).

This makes sense. Linear functions are polynomials of degree one or less, meaning variables change at fixed rates. Conversely, what we cannot represent with a linear transformation is anything that would deform the space in such a way that evenly spaced points in $\mathbb{R}^n$ are unevenly spaced in $\mathbb{R}^m$ (Figure $2\text{b}$).

Now let’s look at some specific types of matrix transformations. To keep things simple, we’ll only consider square $n \times n$ matrices. This is by no means an exhaustive or formal look at every type of matrix. The idea is to see how our geometrical understanding so far relates to familiar types of matrices.

### Diagonal matrices

In my opinion, a diagonal matrix is the easiest matrix to understand once you understand that the columns of a matrix $A$ represent where the standard basis vectors map to in a new vector space defined by $A$. For example, consider an arbitrary $n \times n$ diagonal matrix with scalar values $\alpha_1, \alpha_2, \dots, \alpha_n$ along its diagonal. If we think of the columns of the matrix as defining a new vector subspace, it is clear that each component in a transformed vector $\textbf{x}$ is only modified by one value on the diagonal:

This means a diagonal matrix is constrained in how it can modify a set of vectors (Figure $3$). For example, it can stretch the $i$-th component of a vector with $\alpha_i > 1$, or it can reflect the $i$-th component with $\alpha_i < 0$. You cannot shear a vector subspace with a diagonal matrix.

Note that the identity matrix is a diagonal matrix where $\forall i, \alpha_i = 1$, meaning the standard basis vectors are not changed. It has a determinant of $1$ because it does not modify a vector subspace.

### Shear matrices

A shear matrix is so-named because you can imagine the unit square shearing. We can achieve shear with an off-diagonal element:

Without the off-diagonal element $\beta$, this transformation would just stretch $x_1$ by $\alpha_1$. The off-diagonal element means that a transformed vector’s $1$-st component is a linear combination of $\beta$ and $\alpha_1$. This results in a unit square that is sheared because one dimension is now a function of two dimensions (Figure $4$).

The matrix visualized in Figure $1$ is a concrete example of a shear matrix.

### Orthogonal matrix

An orthogonal matrix $A$ is a square matrix whose columns and rows are orthogonal unit vectors. Let $\textbf{a}_i$ represent the $i$-th column vector of $A$. Recall that two vectors are orthogonal if the dot product between them is $0$. Geometrically, this means that if you were to project one vector onto another, it would turn into a point rather than a line (Figure $5$).

So the columns $\textbf{a}_i^{\top} \textbf{a}_j = 0$ for all $i \neq j$. And since the columns are unit vectors, then $\textbf{a}_i^{\top} \textbf{a}_i = 1$. An immediate consequence of this is that if a matrix $A \in \mathbb{R}^{n \times n}$ is orthogonal, then:

If this is not obvious, write out the matrices explicitly as column and row vectors:

But what is the geometric intuition for orthogonal matrices? In words, an orthogonal matrix can rotate or flip a vector space, but it will not stretch it or compress it. But we can make our thinking more rigorous. The key is to observe a few key properties of orthogonal matrices:

#### 1. Angles are preserved

Given a set of vectors, $\{\textbf{v}_1, \textbf{v}_2, \dots \textbf{v}_k \}$, which we can represent as columns in a matrix $V$, an orthogonal matrix $A$ will preserve all the angles between pairs of vectors in this set after the transformation $AV$. To see this, let $\textbf{v}_i \cdot \textbf{v}_j$ denote the dot product between vectors $\textbf{v}_i$ and $\textbf{v}_j$. Then,

In words, all angles between pairs of vectors are preserved.

#### 2. Lengths are preserved

Given our same set of vectors $V$, an orthogonal matrix $A$ will preserve all lengths of the vectors $\textbf{v} \in V$ after the transformation $AV$. This proof is just a small modification of the previous proof:

In words, the length of any vector in $V$ is preserved after transformation.

#### 3. Area is preserved

Note that the determinant of an orthogonal matrix is either $1$ or $-1$. This is because,

Since the determinant represents the signed factor that the area of an $n$-cube is multiplied by when being transformed by a matrix, a determinant of $1$ or $-1$ means the cube is only rotated or reflected.

To summarize the previous three points: angles, lengths, and areas of a vector space transformed by an orthogonal matrix are all preserved. In my mind, this is a beautiful result. We have a geometric notion of an orthogonal matrix as a function that can only rotate or reflect a vector space, but we can combine this intuition with formal mathematical ideas such as orthogonality of vectors, the definition of the dot product, and the determinant to say something that is mathematically precise.

## Composing linear functions

Since $f(g(x))$ is linear if both $f$ and $g$ are linear, we can compose matrix transformations using matrix multiplications. In fact, because of singular value decomposition (SVD), we make a very strong claim: any matrix $A \in \mathbb{R}^{m \times n}$ has a singular value decomposition $U \Sigma V^{\top}$ where $U \in \mathbb{R}^{m \times m}$ and $V \in \mathbb{R}^{n \times n}$ are orthogonal matrices and $\Sigma \in \mathbb{R}^{m \times n}$ is diagonal. Intuitively, this means we can decompose any matrix transformation into simpler transformations such as unitary transformations and diagonal transformations. This falls out of the fact that the image of the unit sphere under any matrix transformation is a hyperellipse.

I want to save SVD for a future post, but let’s look at a simple example of matrix multiplication as function composition from (Paeth, 1986), who claims but does not prove that any $2 \times 2$ rotation matrix can be decomposed into three shear matrices. Following the example in that paper, consider decomposing a rotation matrix $R$ and three shear matrices $M_{i \in \{1,2,3\}}$:

Using a little trigonometry, we can solve for $\alpha = -\tan(\frac{\theta}{2})$, $\beta = \sin(\theta)$, and $\gamma = -\tan(\frac{\theta}{2})$. See the Appendix for a derivation. We can sanity check our work visualizing the matrix transformations for a specific $\theta$, say $\theta = \frac{\pi}{3}$ (Figure $6$).

I think this is a really beautiful way to think about matrix multiplication, and it also drives home the utility of a geometrical understanding. For example, image shearing and then rotating a unit square. Now imagine applying the rotation matrix first and then shearing the rotated matrix—be sure to shear with respect to the standard basis vectors, not along an axis parallel with the rotated matrix! You won’t get the same resultant matrix. (It may help to sketch this out.) What this means that matrix multiplication is not necessarily commutative, and this is a property you can visualize rather than just remember.

## Properties of matrices

With a geometrical understanding of matrices as linear transformations, many concepts in linear algebra are easier to appreciate. Here are just a concepts that are, in my mind, easier to understand geometrically.

### Invertibility

Thinking of a matrix as a geometric transformation or projection, it should be clear that a rectangular matrix cannot be invertible. This is because a rectangular matrix projects vectors into either a higher- or lower-dimensional space. Imagine you found an apple lying on the ground beneath an apple tree. Can you tell me how high it was before it fell to the ground?

### Rank

The rank of a matrix $A$ is the maximal number of linearly independent columns of $A$. Again, if we think of the columns of a matrix as defining where the standard basis vectors land after the transformation, then a low-rank matrix creates linear dependencies between standard basis vectors. The linear dependence between columns means that any projection will map a higher-dimensional vector space to a lower-dimensional vector space.

### Null space

The null space is the set of all vectors that land on the origin or become null after a transformation. For example, if a $3 \times 3$ matrix has rank $2$, that means it projects all the points in a 3-dimensional space onto a plane. This means there are a line of vectors that all land on the origin. If this is hard to visualize, imagine flattening a cardboard box in which the origin is in the center of the box. As you flatten the box, there must a line of vectors that all collapse to the origin.

### Spectral norm

The spectral norm computes the maximum singular value $\sigma_1$ of a matrix:

Since a matrix transforms a vector, we can think of the spectral norm as measuring the maximum amount that a matrix can “stretch” a vector. Many methods and proofs in numerical and randomized linear algebra rely on this norm. For example, algorithms in (Mahoney, 2016) use the spectral norm to bound randomized approximations of matrix multiplication.

### Determinant

The determinant has the property that it is multiplicative. That is:

You can easily find formal proofs of this, but let’s convince ourselves that this is at least plausible through geometrical intuition. Consider a scenario in which $A$ and $B$ are defined as follows:

Our geometrical understanding of matrices suggests that matrix $A$ dilates $\textbf{e}_1$ by $3$ and matrix $B$ dilates $\textbf{e}_1$ by $2$. Can you see why the determiant is multiplicative (Figure $7$)?

The determinant of $A$ is $3$, while the determinant of $B$ is $2$. And what is the determinant of $AB$? It is $6$ because it dilates a unit cube by $6$.

Another cool property of the determinant that is easy to visualize is its sign. The sign of the determinant is positive if the matrix does not change the chirality of the space and negative if it does. This is, I think, easy to visualize (Figure $8$).

Imagine a chiral object—that is, an object that cannot be mapped to its mirror image by rotations and translations alone, for example a hand. A matrix transformation on that chiral object that changes its chirality changes the sign of the determinant.

## Conclusion

While thinking about these ideas, I came across this lecture by Amritanshu Prasad, who argues at the end of the video for a geometrical understanding:

Most of us think fairly geometrically to start with, and algebra comes as part of training. Also, a lot of mathematics comes from trying to model problems in physics where we really need to think geometrically. So in that sense, a definition that is geometric has much more relationship to the physical world around us and also perhaps gives us a better intuition about the object we’re studying.

Mathematics is a contact sport, and I have no illusions that a geometrical understanding that is not paired with rigorous, numerical understanding is sufficient. It is impossible to deeply understand a mathematical idea if you cannot formalize it. That said, I have found that my ability to “think in linear algebra” has been strengthened by a better geometrical understanding of the material.

## Appendix

### 1. Proof that matrices are linear transformations

To prove that a function $f$ is linear, we need to show two properties:

1. $f(a) + f(b) = f(a + b)$
2. $f(ca) = c f(a)$

Let $A$ be an $m \times n$ vector, so the $i$-th column in $A$, $\textbf{a}_i$, is an $m$-vector and there are $n$ such vectors. And let $\textbf{v}$ and $\textbf{u}$ be two $n$-vectors. Then we can express a function $f$ as the linear combination of the columns of $A$ where the coefficients in are the scalar components in $\textbf{v}$ and $\textbf{u}$:

Proving both properties is straightforward with this representation because we only need to rely on properties of scalar addition and multiplication.

### 2. Derivation for rotation matrix decomposition

To solve for $\alpha$, $\beta$, and $\gamma$ in terms of $\theta$, let’s first multiply our matrices:

We are given that a $2 \times 2$ rotation matrix is equivalent to these three shear matrices:

Now let’s solve. Immediately, we have:

We can use this to solve for $\alpha$:

Finally, let’s solve for $\gamma$:

1. Trefethen, L. N., & Bau III, D. (1997). Numerical linear algebra (Vol. 50). Siam.
2. Paeth, A. W. (1986). A fast algorithm for general raster rotation. Graphics Interface, 86(5).
3. Mahoney, M. W. (2016). Lecture notes on randomized linear algebra. ArXiv Preprint ArXiv:1608.04481.


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Problem: Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.b. 100.0 mL of 0.10 M C2H5NH2 (Kb = 5.6 x 10 -4) titrated by 0.20 M HNO3

🤓 Based on our data, we think this question is relevant for Professor Eroy-Reveles' class at UCSC.

FREE Expert Solution

At halfway point,

Using the Henderson Hasselbalch equation:

$\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\mathbf{log}\left(\frac{{10}^{-14}}{{K}_{b}}\right)\mathbf{=}$10.75

10 mmol

At the equivalence point, 10 mmol of HNO3 is added. The strong acid will convert the weak base completely to its conjugate acid:

50 mL

0.067 M

$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\left[{C}_{2}{H}_{5}{\mathrm{NH}}_{2}\right]\left[{H}_{3}{O}^{+}\right]}{{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{NH}}_{\mathbf{3}}}}$

Determine the value of Ka:

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{5}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}\mathbf{=}$1.786×10-11

Substituting the equilibrium concentrations:

${\mathbf{1}}{\mathbf{.}}{\mathbf{78}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{11}}{\mathbf{=}}\frac{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{067}\mathbf{-}\mathbf{x}}$

Problem Details

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.

b. 100.0 mL of 0.10 M C2H5NH2 (Kb = 5.6 x 10 -4) titrated by 0.20 M HNO3

format: plain


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Flat metrics on $n$-toruses, their systoles, and the “shortest vector problem”

Apologies if this is too basic, but I haven't been able to find any info about the question. Is there anything known about the moduli space of flat metrics on an $n$-torus (i.e., $(S^1)^n$)? Information about the moduli space of lattices in families of Lie groups other than $\mathbb{R}^n$ would be of interest as well. (But I want to stress I am curious about families of Lie groups with the dimension going to infinity rather than a fixed-dimensional example.)

Ultimately, I'm interested in any information about the systoles of flat tori (i.e., the length of the shortest non-contractible loop). I do know the Gromov systolic inequality but was wondering what more is known (esp. since the Gromov inequality is straightforward in the flat case.)

I am asking (and the question has the computational-complexity tag) because I've been reading about the "shortest vector problem"*, and it struck me that it is equivalent to finding the systole of a flat torus, and so I was curious about any understanding of the problem from this perspective. (I guess this equivalence should predict that there can't be too much understanding of flat metrics on tori since the shortest vector problem is NP hard.)

*The shortest vector problem is: Given the generators of a Euclidean lattice, find the shortest (WRT the Euclidean norm say) element in the lattice.

• The "systolic geometry of flat tori" is exactly the same a the "geometry of numbers" and you will find the info you want more readily if you look under that heading. I like the book by Gruber and Lekkerkerker. Try that as starting point. – alvarezpaiva Aug 21 '17 at 7:47

The moduli space of $n$-tori (of volume 1) is the locally symmetric space (orbifold) $SO(n)\backslash SL(n,\mathbb{R})/SL(n,\mathbb{Z})$. The symmetric space is also the space of positive-definite symmetric matrices of determinant 1, and $SL(n,\mathbb{Z})$ acts on these as the symmetric square representation (i.e. by reparameterizing quadratic forms). Various things are known about its spine (the well-rounded retract), and the systole is known to be a topological Morse function.

But really when you are asking about systoles of tori, you are asking about lattice sphere packings, of which much is known. Given a systole of length $l$, there exists a ball at any point in the torus whose radius is $l/2$, and whose interior is embedded, and conversely. Hence lattices with maximal sphere packings achieve the maximal systole for volume 1. In a given dimension, this is equivalent to determining Hermite's constant. The optimal lattice packings are known in dimensions $\leq 8$ and $24$. Check out the encyclopedic

Conway, J.H.; Sloane, N.J.A., Sphere packings, lattices and groups. With additional contributions by E. Bannai, R. E. Borcherds, J. Leech, S. P. Norton, A. M. Odlyzko, R. A. Parker, L. Queen and B. B. Venkov., Grundlehren der Mathematischen Wissenschaften. 290. New York, NY: Springer. lxxiv, 703 p. (1999). ZBL0915.52003.

Gruber and Lekkerkerker is out of date. Some of the best work on the subject is by

Banaszczyk, W. Inequalities for convex bodies and polar reciprocal lattices in R^n. II. Application of K -convexity. Discrete Comput. Geom. 16 (1996), no. 3, 305-311.

or for a more elementary treatment

Bangert, Victor; Katz, Mikhail. Stable systolic inequalities and cohomology products. Dedicated to the memory of Jürgen K. Moser. Comm. Pure Appl. Math. 56 (2003), no. 7, 979-997.


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# LaTeX issue with citation [duplicate]

Hi all I am working on a citation for my whitepaper I used https://www.mybib.com/ to generate my .bib file. However I can't get the citation to show

Here is the code in my LaTeX file

 \documentclass{article}

\usepackage{biblatex}

\begin{document}
Someone must have found a simple solution  \cite{references}.
\printbibliography

\end{document}

Here is my References.bib file

 @misc{references,
month = {09},
title = {bibliographies - How do I add "citations" at the end of the document as done here?},
urldate = {2020-05-01},
year = {2016},
organization = {TeX - LaTeX Stack Exchange}
}

Here is the output of the code

I am using ubuntu, texmaker and all files are in the same directory

Running F11 bibtex in TexMaker I get the following error

Process started

 This is BibTeX, Version 0.99d (TeX Live 2017/Debian) The top-level auxiliary file: Using References-bib.aux I found no \citation commands---while reading file Using References-bib.aux I found no \bibdata command---while reading file Using References-bib.aux I found no \bibstyle command---while reading file Using References-bib.aux (There were 3 error messages)

Process exited with error(s)

I'm not sure if I am missing anything else THANKS ALL!!!


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0

# SLW-1 Modeling of Radiative Heat Transfer in Nonisothermal Nonhomogeneous Gas Mixtures With Soot

[+] Author and Article Information

Department of Mechanical Engineering,  Brigham Young University, Provo, UT 84604-4201

Denis Lemonnier

Institut Pprime,  CNRS-ENSMA-University of Poitiers, UPR 3346, 86961 Futuroscope, France

Brent W. Webb1

Department of Mechanical Engineering,  Brigham Young University, Provo, UT 84604-4201 e-mail: webb@byu.edu

1

Corresponding author.

J. Heat Transfer 133(10), 102701 (Aug 23, 2011) (9 pages) doi:10.1115/1.4003903 History: Received August 27, 2010; Revised March 10, 2011; Published August 23, 2011; Online August 23, 2011

## Abstract

The spectral line weighted-sum-of-gray-gases (SLW) model consisting only of a single gray gas and of one clear gas is developed as an efficient spectral method for modeling radiation transfer in gaseous medium. The model is applied here in prediction of radiative transfer in nonisothermal and nonhomogeneous gas mixtures with nongray soot. The absorption spectrum of the gas mixture and soot particles is treated as a spectrum of a single effective gas, reducing the problem to the simplest case of the SLW model with a single gray gas and a clear gas. Good accuracy can be achieved by the optimal choice of the model’s gray gas absorption coefficient and its weight by application of the absorption-line blackbody distribution functions of individual species in the mixture calculated with a high-resolution spectral database. The SLW-1 model is validated by comparison with benchmark solutions using the line-by-line method, the SLW method with a large number of gray gases, and the SNB model.

<>

## Figures

Figure 1

The SLW-1 spectral model

Figure 2

The total net radiative flux F(x) and the total divergence of the net radiative flux Q(x) for Example 1

Figure 3

Definition of the soot ALBDF, Fsoot(C,Tg,Tb,fv)

Figure 4

The SLW spectral model of soot and the SLW-1 spectral model of soot

Figure 5

Soot ALBDF Fsoot(C) at 1000 K for different soot volume fractions, and the total forward radiative flux F+(x) for Example 2

Figure 6

The SLW-1 model reference approach in nonisothermal and nonhomogeneous medium

Figure 7

Total divergence of the net radiative flux for Example 3

Figure 8

Total divergence of the net radiative flux for Example 4

Figure 9

Total divergence of the net radiative flux from Example 5

Figure 10

Total divergence of the net radiative flux from Example 6

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# Functional Equation (Probably needs a CAS)

1. Jun 1, 2009

### ritwik06

1. The problem statement, all variables and given/known data
If f'(x)>0 for all real positive x, where f:R+ ---> R and
f(x)+(1/x)=f-1(1/(f(x))),

f-1(1/(f(x)))>0 for all x>0. Find all the possible values of (i) f(2),(ii) f'(2) and (iii) Limit (x f(x)) as x ----->0 .

3. The attempt at a solution
Guessing from the last part, I intuitively, took f(x) = k/x. Using this, I get a quadratic in k which yields that k might be $$\frac{1 (+/-) \sqrt{5} }{2}$$. But this was just a guess. And there might still be other functions which satisfy that equation. I wonder if they could be manually solved for.
All that I need from you guys is to check out if all the possible solutions of this equation can be solved manually or not. If not, please let me know how can I get solutions to this equation using a Computer Algebra System. Actually, I have both Matlab R2008a and Mathematica 7 but I dont know how to use them to solve functional equations involving inverses. If anyone of you could provide me the syntax for these CAS on how to solve functional equations, I shall be very grateful.
Thanks.
Ritwik

2. Jun 5, 2009

### tiny-tim

Hi Ritwik! Thanks for the PM!

Yes, f(x) = (1 ± √5)/2x certainly works …

I think you've done very well to get that!

Sorry, but I haven't a clue how to get any other solution

(and I don't know any computing)


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# Periodic Sequence Factoring

This topic is 3723 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I have a sequence of characters. The characters are from an infinite alphabet with the only available operations being equality comparison and hashing. I know that the sequence of characters is periodic. That is, it is of the form:
Seq = ABkC

A = a1 ... an
B = b1 ... bm
C = c1 ... cl
Given the sequence, how can I compute k, l, m and n quickly? (The characters are assembly instructions, and I need to factor them out using a for loop).

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This certainly stretches my programming capability, but in an attempt to help I've found some good ideas here, if you haven't already seen it.

What you are saying, if I am not mistaken, is that these are T-periodic, of the form:

f(x+T) = f(x)

f(a+n) = f(a)f(b+m) = f(b)f(c+l) = f(c)

such that
Seq = f(a)*f(b)k*f(c)

and you are solving for n, m and l. In the absence of any generating function, the sequence for each would have to be subjected to 'pattern matching' or 'property matching', no? Are the character values represented by integers? It may be possible to optimally generate values for n, m and l (intelligently guess them) until the above equations are satisfied. Worst case, assuming they are indeed T-periodic, you could just start with n=2, for example, and iterate your sequence until f(a+n)=f(a). Another possibility is to take f(a)0 and search for the first re-occurence of it, for which there are excellent search algorithms available, and the distance becomes n (I have used this with pseudorandom number generators to verify/test claims about the period).

--random

[Edited by - random_thinker on October 2, 2007 4:25:16 AM]

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Quote:
 Original post by ToohrVykI have a sequence of characters. The characters are from an infinite alphabet with the only available operations being equality comparison and hashing. I know that the sequence of characters is periodic. That is, it is of the form:Seq = ABkCA = a1 ... anB = b1 ... bmC = c1 ... clGiven the sequence, how can I compute k, l, m and n quickly?(The characters are assembly instructions, and I need to factor them out using a for loop).

I think all you need here is the ability to detect cycles. Googling around a little for "cycle detection algorithm" yielded this which looks promising. Knuth's Art of Computer Programming has a cycle detection algorithm in it, which IIRC is similar if not identical to the one described at that URL.

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Whether or not these algorithms are useful depends upon the state of the data. For example, if you have a large quantity of stored sequence data then a specialized algorithm is certainly advantageous. But if you have yet generated the data, a just as quick method must be to monitor the values as they are generated and capture when they repeat.

For integers this is a straight-forward process but for float or double values you would have to apply some sort of signal processing (smoothing) algorithm.

--random

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Was an optimum algorithm found?

--random

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I am interested in this thread because my own pseudo-random period detection code is providing some interesting results. I suppose that the interesting thing about a prng algorithm is that it is a periodic function by design but there should be no detectable mathematical pattern within cycles, if that makes sense.

My own detection code follows values as they are generated and locates the first to re-occur and then identifies the first three consecutive reoccurences as the beginning of the next cycle. Whether this is strictly correct I have not been able to verify. Interestingly, for std::rand() the results of ten separately seeded runs results in the following:

// One set of 10 independently seeded runs, each through one cycle...Test period of pseudorandom generator...Warning:  First duplicate value occurs at n = 1038947364, continuing...Warning:  First duplicate value occurs at n = 160672077, continuing...ID            : C++ Standard Library rand()Min Period    : 96387719Max Period    : 1732857588Ave Period    : 6.91864e+08No of cycles  : 10Elapsed mins  : 2.04767// Another set of 10 independently seeded runs, each through one cycle...Test period of pseudorandom generator...Warning:  First duplicate value occurs at n = 597879981, continuing...Warning:  First duplicate value occurs at n = 1493682969, continuing...Warning:  First duplicate value occurs at n = 133528437, continuing...Warning:  First duplicate value occurs at n = 1195410326, continuing...Warning:  First duplicate value occurs at n = 1192369109, continuing...ID            : C++ Standard Library rand()Min Period    : 115934013Max Period    : 1493682971Ave Period    : 6.46274e+08No of cycles  : 10Elapsed mins  : 1.91167

Either I am doing something wrong or this is another reason to beware using std::rand(). The rarely available literature on it indicates that it should have a period of 232-1. The actual period is all over the place, and there is a high probability that numbers will be repeated within a cycle.

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text:
# orthogonal matrix with negative determinant

Question: Assume that $A$ is a $n\times n$ orthogonal matrix such that $det(A)<0$. Prove that there exists a nonzero vector $x\in\mathbb{R}^n$ such that $Ax=-x$.

Since $A$ is orthogonal and $det(A)<0$, its determinant is $-1$.

• Nonzero vector, of course. Do you mean $Ax = -x$? Otherwise $-I$ in odd dimension is counterexample. – Robert Israel May 1 '18 at 18:50
• @RobertIsrael Yes, I meant $Ax=-x$ and nonzero vector, Thank you! – user138017 May 1 '18 at 20:41
$$\det(A+I)=\det(A)\det(A^{-1}+I)=-\det(A^\top+I)=-\det(A+I)$$


text:
Semi: Prefix For Half

Geometry Level 2

True or False?

$$\quad$$ If the perimeter of a circle is $$\pi$$, then the perimeter of a semicircle with the same radius is $$\dfrac\pi2$$.

×


text:
Summation notation part 2

By Murray Bourne, 20 Feb 2006

Explanation

This is Part 2 of a (lengthy) exchange between a reader and myself regarding a particularly troubling summation notation question recently.

See the beginning of it here:

Summation Notation (part 1)

Continuing...

Hi again sir

Crazily, would it be ...

???

and the same for ...

???

Thanks for your help sir and I can't stop saying sorry because I know it must frustrating - I didn't know that for even a variable value, the same would apply when writing down the last few terms as natural numbers (specifically in this case).

Thanks again sir and sorry once again 🙁

Yousuf.

OK, Yay!! It is now correct!! Your first terms and last terms are now 100% perfect.

Now, please read the whole page again - especially the bits where I tried to convince you there is not 5 terms just because there is a 5 at the top, There can be hundreds of terms!

Then, when you are sure you have a handle on the expansion, write out the first few terms, the "..." and the last few terms for:

(This one never got answered properly before)

This one is new:

Hi again sir

Right, so drawing on everything done so far.

When using sigma series, the quantity at the top denotes the terminating term and the quantity at the bottom of sigma denotes the initiating term.

So for

This would be ...

The starting term is x + 1 and the terminating quantity is 5. They follow the pattern of natural numbers in that we +1 to the previous term after a.

Also,

Would give ...

This is because the starting term is 3 and the terminating quantity is n (which in essence goes on forever). They follow the pattern of natural numbers from 3 onwards by adding +1 to the previous term after a.

Is that OK sir?

Thanks again for your help sir! 🙂

Yousuf.

First one is excellent.

Second one - no, it doesn't go on forever. It goes up to the value of some "n" which we can define, or find if we have more information. It will only go on forever if the top number is infinity (in which case we would have "+ ..." at the end of our expansion. You correctly do not have that at the end).

When writing them out, please include the first few terms (like you did in both of them above, then the "...", then the last few terms before the final number (like you did in the first one but not the second one.

(1) Do the second one again and include the last few numbers before n. (Why? So I am convinced you know what is happening.)

(2) New one (also include the first few terms and the "..." and the last few terms before "2n"):

Hi again sir

Sorry about the wording of the second one sir but please correct me if I'm wrong on the reowrding because I actually meant that the pattern continues until it reaches the required quantity n unless I'm wrong sir.

So it'd be ...

The second one I think should be ...

???

But sir, apparantly, the textbook shows that we should ideally write the above as. ..

and

But I can see why we're using the "+...+" before the last few terms to gesticulate that the pattern proceeds in the pattern shown in the first few terms, but I'm just commenting on what the book defines in its way sir.

Thanks again sir, I really appreciate your help 🙂

Yousuf.

Where does 9, 10 and 11 come from?? What is 1 less than n (this will be the second last term)? What is 2 less than n (this will be the 3rd last term)??

The first one should be:

You only actually need 3 terms to indicate what is going on, not 6 as your textbook says. It doesn't hurt to have 6, you just don't need any more once the pattern is clear. And you have to show the last 3 terms, as I keep saying.

Try this again:

And here's another new one:

Show the first 3 or 4 terms, the "..." and the last 3 or 4 terms. There is no 9, 10 or 11 involved.

Hi again sir

Sorry sir, I have to apologise here: the ...+(n-2)+(n-1)+n part is new to me at the moment because thus far, the textbook has required only to show the first three/4 terms, then use the +...+ notation and then end it with a general rule. So like the textbook writes the following example as...

That's what I've been writing down thus far and I wasn't aware that the way you corrected me on the first one of those examples was actually the better way to write it and it's just my personal opinion to say that I believe the textbook should echo how you've written it because surely at A-level, the textbook should be showing this clearly, but they've just helped students take one small step from GCSE and in all honestly, the textbook doesn't even show anything like the way you wrote this one

a

nd from now on I would prefer to take this route.

I wrote this

because I thought we use +...+ to convey the pattern, and then terminate using the last 3 terms: the 2 from the last terms to gesticulate the pattern and end it with the general rule. I can now see why it's important to corroborate the lead-up of the general rule.

But, as you say sir, would the proper written manner be, for the fraction one ...

???

So in essence, we're just reversing as we approach the general rule (where we +1 all the time using natural numbers and when approaching the general rule, we go backwards until we reach the general rule - not quite sure how I can reword that sir!!)

Anyway, would this one be ...

???

??

Thanks sir 🙂

Yousuf.

Yay - all the ones you just did are good - the fraction one, the "2n" one (9, 10 and 11 will only be involved in those positions if n=6, and we don't know that to be true) and the "y" one.

So now, try this one:

Hi again sir

Sorry for the late reply sir.

This one looks a little unsettling sir, particularly the ending, so I'll plan my way through. We substitute r = n+3 into r and then add +1 to each consecutive term.

TERM 1 TERM2 TERM 3 TERM 4

Right, the last part is a little confusing, so this is what I've done. I've stated that the terminating term is 2n and we subtract our way using reversed natural numbers until we reach 2n isolated.

I couldn't correctly figure out if n + 3 would be involved in the foregoing, so I get ...

???

Thanks again sir,

Yousuf

Perfect! (I've noticed the more question marks you have at the end of your answer, the more correct it is ^_^)

Now, I'm not sure whether you recognised that last summation question, but in fact, it is part of the question you originally started with (right at the top of this page).

I hope you have a better idea what that question is asking now.

Do you think you can answer it?

Hi again sir

Actually, I didn't take notice of it! I'll try and see if I can solve it sir.

The question is

which is saying that the starting term of the sequence of natural numbers (r) is n + 3 and the terminating term is 2n.

I think it's saying that the sum of the series of natural numbers, where the first term is n + 3 and the terminating term is 2n, is 312.

So... if I can draw back from the previous work, I did ...

And also

I'm still not sure what the notation is to solve this equation. DO I do this sir: do I take the first three terms and the last four terms thus ...

!!!!!!!!!!!!!! That can't be right!!!

But lets see if we can use AP notation involving

We know that from the series

, the first term is r = n + 3, so this gives ...

using

Also,

But the last term is a variable so

so

Sorry sir if it's confusing, but that's as far as I can go.

Sorry sir and thanks again, 🙂

Yousuf

(1) There are 2 formulas for the sum of an AP. What are they?

(2) For both formulas, you need to know the number of terms in your AP. So how many are there?

BTW, it's a shame they've used "n" in the sum sign - it will be confusing later. Let's make a slight change now, and then explain why later.

Trust me - it will be easier using "i" from now on. We will find the value of "i" then just change it back to "n" at the end.

Hi again sir

Oh yea! I was thinking about the sum formulas!!

(1) The formula for the sum of an AP is given by...

and also

(2) "Find the number of terms in the given sequence"

We know that ...

a = (i + 3)

u_n = 2i

d = 1 (common difference is +1)

We can use ...

I can see why there is the confusion (n = 2n)

Is that right sir? There are (i - 2) number of terms ?? (+ !!) That can't be right can it?

Thank you sir 🙂

Yousuf

Yes, correct. The number of terms is i-2.

You are extremely close, now to the final answer!!

Use the second sum formula (it's less messy).

(a) You know the number of terms - plug it in (i-2)

(b) You know the first term - plug it in (i+3)

(c) You know the final term - plug it in (2i)

Now go back and read the original question again (top of this page) and figure out what you need to do with the 312.

You should end up with a value of i at the end.

^_^

Thank you so much for replying back sir 🙂

Righto, we use

We know that

• n = i - 2

• a = i + 3

• l = 2i

Making the denominators common with for i:

Is that right sir?

Because I'm suggesting now that we should solve for i (which is n) ...

Changing the notation to a more easier format ...

Multiplying by - 2/5 ...

So in essence, ??

That's not right is it sir??

Yousuf.

Up to this line you were correct:

Next should be:

Keep going...

So we expand the brackets first to get ...

Then ...

We know that

So ...

Multiply by 2 throughout ...

So ...

Then ...

We need to solve this quadratic (sir, is it true we are using only the positive solution to this because the negative solution will be senseless?) ... two factors of 210 are 14 and 15 and we get -1 (coefficient of i) by doing 14-15 and since both numbers are + and -, we know that positive x negative = negative (as required for -210), so ... (actually I cheated here using GeoGebra!, but we should be able to do this using graphics calculators in the tests) ...

Therefore,

We change i back to n and get ...

So, the roots are and

Because we need the positive solution, the solution to the problem is n = 15.

We did it - yay!!! 😀 Many thanks to you sir for helping me in this sir !!

Thank you very much 🙂

Yousuf.

You're welcome, Yousuf!

Of course, you should substitute your answer into the original summation expression to ensure you are correct. this also helps you to understand what you have done.

And that's the end!

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# How can I calculate the buoyant force of a boat?

$F = V \times \frac{F}{V} = V \times g \times \frac{m}{V}$


text:
Viktor - 9 months ago 42
C++ Question

# How to determine, how many number of digits 3 is in variable

I have a variable which contains this numbers

int n = 6396339;

I need to determine how many digits 3 is in variable.
So far i trued like this:

int n = 6396339, counter = 0;

while (n > 0)
{
if ((n % 10) % 3 == 0) {
counter++;
}
n /= 10;
}
cout << counter << endl;

(n % 10) % 3 == 0

is true for any digit that is divisible by three, i.e. 0, 3, 6 and 9. Just check whether the digit is equal to 3:
(n % 10) == 3


text:
YSC, Title 7.  Election Chapter 4.  Voter Registration 401.  Place of registering and voting. 402.  Formality of registration required. 403.  Application for registration; affidavit. 404.  Registration of applicant. 405.  Entry of a voter's name in the general state register. 406.  Reregistration. 407.  When registration not required.      401.  Place of registering and voting.      (a)  Every person who has reached the age of 18 years, or who will have reached the age of 18 years on or before the date of the next election, and who has resided in the Federated States of Micronesia for nine months and in the election district of registration three months preceding the date of registration and who, except for the requirement of registration, is otherwise entitled to vote may register to vote in the election district in which he resides.      (b)  The Election Commissioner shall designate such place or places within the State where registration of voters may be made.      (c)  No person shall register to vote or vote as an elector of any other precinct than that in which he resides; provided, that where there is a mistake in placing the name of the voters on the list of voters of a precinct in which he does not actually reside, such voter shall nevertheless be allowed to vote therein, if otherwise qualified, and the member of the board of election of the particular election district and precinct where such voter has voted shall notify the Election Commissioner of the error in order that the name of such voter may be placed on the next succeeding list of voters of the precinct where he actually resides.      (d)  If any person resides in more than one precinct, he may choose the precinct in which he will register, but he shall register as an elector of one precinct only. Source:  YSL 2-81 1, modified. Cross-reference:  Yap Const., Art. IV, Sect. 2, provides that:  "The Legislature shall prescribe a minimum period of residence and the method of voting at elections, and shall provide for voter registration, . . ."      402.  Formality of registration required.      No person shall be entitled to vote in any state election or to be listed upon any general state register, or upon any precinct list, who fails to register with the formalities and subject to the restrictions and qualifications required by this title. Source:  YSL 2-81 1, modified.      403.  Application for registration; affidavit.      Any person qualified to and desiring to register as a voter in any election district, may present himself at any time during business hours to any of the members of the election board (herein empowered and authorized to administer oaths and take acknowledgments) or persons authorized by law to administer oaths, then and there to be examined under oath as to his qualification as an elector.  Each applicant shall make and subscribe to an application in substantially the following form: AFFIDAVIT ON APPLICATION FOR REGISTRATION Federated States of Micronesia         ss Yap State 1.     My full name is                                                                              . 2.     I was born at                                              and my date of birth is                               . 3.     My age is                                                      . 4.     I am now registering in                                        village                                          municipality. 5.     I am a citizen and resident of the Federated States of Micronesia. 6.     I was naturalized as a citizen of the Federated States of Micronesia at                            State on the                    day of                                              , 19             . 7.     I have resided in the Federated States of Micronesia (or Yap State) not less than nine months, and in Election District No. (or Representative District No.)                          not less than three months, immediately preceding this date on which I now offer to register, to wit, the                           day of                                                  , 19                . 8.     I am not currently under parole, probation, or sentence for any felony for which I have been convicted by any court within the Trust Territory, Federated States of Micronesia or any court within the jurisdiction of any foreign country for an offense in such a foreign country which would be a felony under Yap State Law. 9.     I am not currently under a judgment of mental incompetency or insanity. 10.     I solemnly swear that the foregoing statements are true, so help me God. Subscribed and sworn to before me this day                 of                               , 19       .      The applicant shall strike out allegations that are inapplicable, and shall swear to the truth of the allegations in his application.  In any case where the person who administers the oath shall so desire or believe the same to be expedient, he may demand that the applicant produce further information, a witness or witnesses to further substantiate the allegations of this application. Source:  YSL 2-81 1, modified.      404.  Registration of applicant.      Every affidavit on application for registration shall be submitted to the persons authorized to examine the qualifications of electors, not less than seven days before an election. Source:  YSL 2-81 1, as amended by YSL 5-10 1, modified.      405.  Entry of a voter's name in the general state register.      (a)  If the person authorized to receive an affidavit of application for registration is satisfied that the applicant is entitled to be registered as a voter, he shall number the affidavit consecutively as approved by him, and shall transmit the affidavit to the Election Commissioner.  The Election Commissioner shall thereupon enter or cause to be entered in the general state register the following facts:      (1)  Number of affidavit;      (2)  Date of registration;      (3)  Name of applicant in full;      (4)  Age;      (5)  If naturalized, the date of such naturalization;      (6)  Residence; and      (7)  Any other information which the Election Commissioner may deem necessary.      (b)  The Election Commissioner shall also forthwith enter or cause to be entered the name so registered in its proper place in the general alphabetical index, together with a reference to the page on which the registration appears.  A voter having once been registered shall not be required to register again for any succeeding election, except in case of change of name or residence or intervening disqualification; provided, that in the event the voting records are destroyed or lost, the Election Commissioner may require the re-registration of voters.      (c)  The Election Commissioner shall file the accepted affidavits in consecutive numbers and keep the same in some convenient place so as to be open to public inspection.   Source:  YSL 2-81 1, modified.      406.  Reregistration.      Any voter who changes his residence from one voting precinct to another, or who changes his name, after registration in any general state register, may register again in such general state register under the proper voting precinct or the proper name and the Election Commissioner or his authorized representative shall cancel the former registration by drawing one or more lines through the name of such voter as previously registered and enter or cause to be entered his own signature and the date of such cancellation with ink on the same line; provided, that no such registration shall be allowed on account of any change of residence or name made within seven days before an election. Source:  YSL 2-81 1, as amended by YSL 5-10 3, modified.      407.  When registration not required.      No registration in person shall be required of a full-time student at any institution of learning outside the State, but such person shall make and subscribe to an affidavit, substantially similar to  the form set forth in section 403 of this chapter and as the Election Commissioner may prescribe, to establish fully such person's right to vote.  Any duly qualified elector may challenge the acceptance of the voted ballot at the time of casting of the ballot under the provisions provided by law. Source:  YSL 2-81 1, modified.
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text:
# Potential in uniform fields

First in a book I found that the electric field strength $\mathrm{E}$ of a uniform electric field from a charged plate is equal to $\frac{q}{2\times \epsilon_0 \times S}$ where $q$ is the charge of the plate, $\epsilon_0$ is the electrical permitivitty and $S$ is the area of the plate.

My first question is how to find the potential of a point in the uniform field (is there a formula using the electric field strength and may be some distance to the plate) or at least the voltage between two points (again formula using the electric fields strength and the distance between the two points).

Now we are talking about a capacitor. In some sources it is said that one plate has negative electric potential (for example $V$) and the other $-V$. But in other sources one plate is said to have a zero potential and the other to have nonzero. Are these both true? In my opinion the negative plate would make negative potential in the positive plate and vice versa (we can look at the uniform field of the capacitor as made of from the two uniform fields created by the negative and the positive plate). I'm really confused.

Also I want to ask whether there is a formula for finding the electric potential of a point in the capacitor because for the voltage between two points it is $E\times d$ where $d$ is the distance between them. Finally, I have found that the electric potential in the midpoint of the capacitor (i.e. if the distance between the plates is $d$, the distances between this point and the plates is $\frac{d}{2}$).

The relationship between electric field and electric potential is just that the electric field is (minus) the gradient of the potential. Thus in the case of a uniform field extending from a uniformly charged plate (let's call it along the z-axis, with the late in the x,y plane) $$E_z = - \frac{dV}{dz}$$.
This of course means that $$V(z) = - \int E_z\ dz = E_z z + V_0,$$ where $E_z$ is the constant field and $V_0$ is an arbitrary constant. What this means is that you can set the zeropoint of the voltage to be anything you like, it is only its gradient that matters. The potential difference simply depends on the difference in $z$ coordinates, $\Delta z$. ie. $V_1 = E_z z_1 + V_0$ and $V_2 = E_z z_2 + V_0$, so the potential difference $V_2- V_1 = E_z \Delta z$.
Therefore the potential at some distance $z$ between the plates of an ideal, parallel-plate capacitor with uniform electric field $E$ and the positively charged plate is at $z=0$ is $E z + V_0$. You can choose $V_0$ to be anything you like, but whatever you choose, the potential difference between the plates $(E d + V_0) - V_0$ is exactly the same.
Thus when you ask what is the potential at a point midway between the plates, all you can really say is what is the potential difference between that point and one of the plates, in this case $Ed/2$.
• The potential difference is the charge on the plates divided by the capacitance $=Ed$. You can set the zeropoint of the potential to whatever is convenient. – ProfRob Apr 3 '15 at 23:43


text:
# Proof Clarification type decomposition of von Neumann algebras

I'm reading a proof of the type decomposition theorem for von Neumann algebras (into a direct sum of 3 von Neumann algebras which are in turn of type 1,2,3)

One constructs maximal projections $p^1,p^2,p^3\in M$ so that $p^i$ is the largest central projection with $p^iM$ being of type $i$. These three projections are easily seen to be pairwise orthogonal. Now it remains to show that $1=p^1+p^2+p^3$. One does that by showing $q=1-p^1-p^2-p^3$ is the null projection.

The author goes on to show that $qM$ does not contain any non-zero finite projections(so $qM$ can't be of types 1,2) and that it can't be of type 3 either. Why does that imply that $q=0$ ?

I can't give a reference since the source is a lecture I attend.

I cannot comment without seeing your notes. One usually defines "type III" as "not type I nor II". This gives, by the maximality of $p^1$ and $p^2$, $p^3=1-p^1-p^2$. Because if $p^3$ is not $1$, then there would be a type I or II part, contradicting maximality.


text:
# Surface production equipment

Series Development Geology Reference Manual Methods in Exploration Production engineering methods Surface production equipment James Jennings Web page AAPG Store
Wellhead in Northeastern BCbarrels condensate, Canada The wellhead is a dual completion. Two different zones have been completed however production from the zones remain separate from the zones to the wellhead.

There are three major components of surface production equipment:

• Separators and heater treaters
• Tank batteries and meter facilities

Production engineers often design all equipment on the lease. After the oil and gas leaves the lease, pipeline or facilities engineers take over.

## Contents

The wellhead is the equipment at the surface that provides support for the tubulars inside the well, a pressure seal between the tubulars, and a means of controlling production from the well. Typically, the wellhead consists of a casing head for each casing string, a tubing head, and a Christmas tree. For each string of pipe in the well, casing, or tubing, some means of support and pressure sealing must be provided. This is the function of the casing and tubing heads. The Christmas tree provides the necessary valving and chokes to control the production from a well capable of flowing. For a well that is being pumped, the Christmas tree is replaced by wellhead equipment that accommodates the pumping operation.

Figure 1 shows a typical wellhead for a flowing well. Notice that a choke is provided to control the rate of production from the well in addition to the tubing wing valve, which provides for a complete shut-off of the production. The choke can be either fixed or variable in size. The choke is nothing more than a small orifice, usually from 1/8 to 3/4 in. in diameter, that restricts the flow rate.

Figure 1 Typical wellhead for a flowing well with a single-wing, single-completion threaded manifold.

Other valves are present on the side of the wellhead. These are called casing valves and they provide access to the various annulii between casing strings and tubing. Normally, the wellhead is fitted with pressure gauges for monitoring pressure within the different annulii and in the tubing.

The flow rate from either an oil well or a gas well can be easily estimated from the wellhead pressure if the wellhead pressure is at least twice the flowline pressure. For an oil well, the Gilbert equation is commonly used:

$q={\frac {P_{{{\rm {tf}}}}S^{{2}}}{600{\sqrt {R}}}}$

where

• q = gross liquid flow rate (bblbarrels/day)
• Ptf = flowing tubing head pressure (psia)
• R = gas to liquid ratio (MSCF/bblbarrels)
• S = choke size (1/64 in.)

For a gas well, the following equation is used:

$q=390{\frac {d^{{2}}P_{{{\rm {tf}}}}}{{\sqrt {GT}}}}$

where

• q = gas flow rate (MSCF/day)
• Ptf = flowing tubing head pressure (psia)
• d = choke size (in.)
• G = gas specific gravity
• T = wellhead temperature (°R)

## Separators and heater treaters

Once liquids are brought to the surface, the oil, gas, and water must be separated for ease of measurement and transportation. A separator is a vessel used to separate liquid from gas. In some cases, the liquid may be additionally separated into individual oil and water streams. A separator is commonly given any of the following names:

• Gun barrel
• Free water knock out
• Knock out
• Trap
• Scrubber
• Stage separator

A heater treater is simply a separator that is designed to separate primarily oil from water. Heating of the mixture normally speeds up and improves the separation process.

Several physical processes are commonly used in the separation process:

• Gravity settling
• Centrifugal force
• Impingement
• Electrostatic precipitation
• Filtration
• Heat

The design of a particular separator depends on the nature of the flow stream to be separated. For a gas well, the separator usually separates a small amount of liquid from the gas. In an oil well, the separation may involve a small amount of gas for the amount of liquid. In general, the well stream separator must separate the mostly liquid fluids from the mostly gas fluids. In addition, it must separate liquid hydrocarbon from liquid water and remove most of the entrained liquid mist from the gas.

To accomplish the separation, the separator is usually designed to control and dissipate the well stream flowing energy. Once gas and liquid velocities are slow enough, gravity causes the liquid to settle and the gas to rise. The size of the vessel must be such that adequate time is allowed for this settling to occur before the fluid leaves the separator. If water is to be separated from oil, then the liquid residence time depends on the volume of the fluid being handled and the specific gravity of the two liquids. Many times, a mist extractor composed of vanes, mesh pads, or a cyclonic passage is used to remove residual liquid droplets from the gas stream.

There are three types of separators: vertical (Figure 2), horizontal (Figure 3), and spherical (Figure 4). Horizontal separators are found in both the single tube and double tube design. Advantages of the vertical separator include

• Good for predominantly liquid streams
• Can handle producing stream surges without carryover
• Occupies little space (small footprint)
• Easily cleaned of sand and mud

Advantages of the horizontal separator are

• Good for predominantly gas streams
• Easy to fabricate, ship, and install
• Low profile

And for the spherical separator, the advantages are

• Good for high pressure gas wells
• Compact, small size

Figure 5 shows a typical vertical heater treater. Notice that a heater treater is simply a separator in which a firetube has been placed to heat the liquid mixture as it enters the vessel. This heating of the oil and water mixture reduces the viscosity and promotes the separation of the two phases. The fuel for the firetube is usually taken from the gas produced from the well.

## Tank batteries and metering

Figure 6 Orifice meter for gas measurement.

Tanks must be provided to hold both oil and water for shipping or disposal. Usually, at least two oil tanks are used, one for shipping and one for filling. The volume of oil being shipped is sometimes determined by simply measuring the height of the fluid in the tank, or “strapping” the tank. Many of the more modern production facilities have lease automatic custody transfer (LACT) units installed. These stations continuously measure the flow into the shipping point and periodically sample the product being shipped so that oil gravity, temperature, pressure, and water content are known. The metering in this case is done with a positive displacement meter.

The measurement of gas is usually done with an orifice meter. Figure 6 shows such an installation. For a given meter installation, the gas flow rate depends on the pressure on both sides of the orifice and the temperature of the gas. These pressures and temperatures are normally recorded on a circular chart. This chart is later used to determine the total gas flow over a particular time period. Many installations today have digital flow calculating and recording devices installed at the meter. These installations have proven to be accurate, and they provide for the telemetering of the flow information from a remote installation.


text:
⚠Our tutors found the solution shown to be helpful for the problem you're searching for. We don't have the exact solution yet.

# Solution: What is the molecularity of each of the following elementary reactions? Write the rate law for the reaction.Cl2 (g) → 2 Cl (g)

###### Problem

What is the molecularity of each of the following elementary reactions? Write the rate law for the reaction.

Cl2 (g) → 2 Cl (g)


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### SecureNN: Efficient and Private Neural Network Training

Sameer Wagh, Divya Gupta, and Nishanth Chandran

##### Abstract

Neural Networks (NN) provide a powerful method for machine learning training and inference. To effectively train, it is desirable for multiple parties to combine their data -- however, doing so conflicts with data privacy. In this work, we provide novel three-party secure computation protocols for various NN building blocks such as matrix multiplication, convolutions, Rectified Linear Units, Maxpool, normalization and so on. This enables us to construct three-party secure protocols for training and inference of several NN architectures such that no single party learns any information about the data. Experimentally, we implement our system over Amazon EC2 servers in different settings. \\ Our work advances the state-of-the-art of secure computation for neural networks in three ways: \begin{enumerate} \item Scalability: We are the first work to provide neural network training on Convolutional Neural Networks (CNNs) that have an accuracy of $>99\%$ on the MNIST dataset; \item Performance: For secure inference, our system outperforms prior 2 and 3-server works (SecureML, MiniONN, Chameleon, Gazelle) by $6\times$-$113\times$ (with larger gains obtained in more complex networks). Our total execution times are $2-4\times$ faster than even just the online times of these works. For secure training, compared to the only prior work (SecureML) that considered a much smaller fully connected network, our protocols are $79\times$ and $7\times$ faster than their 2 and 3-server protocols. In the WAN setting, these improvements are more dramatic and we obtain an improvement of $553\times$! \item Security: Our protocols provide two kinds of security: full security (privacy and correctness) against one semi-honest corruption and the notion of privacy against one malicious corruption [Araki~\etal~CCS'16]. All prior works only provide semi-honest security and ours is the first system to provide any security against malicious adversaries for the secure computation of complex algorithms such as neural network inference and training. \end{enumerate} Our gains come from a significant improvement in communication through the elimination of expensive garbled circuits and oblivious transfer protocols.

Available format(s)
Category
Cryptographic protocols
Publication info
Published elsewhere. 19th Privacy Enhancing Technologies Symposium (PETS 2019)
Keywords
secure computationneural network traininginformation-theoretic security
Contact author(s)
nichandr @ microsoft com
t-digu @ microsoft com
snwagh @ gmail com
History
2019-03-08: revised
See all versions
Short URL
https://ia.cr/2018/442

CC BY

BibTeX

@misc{cryptoeprint:2018/442,
author = {Sameer Wagh and Divya Gupta and Nishanth Chandran},
title = {SecureNN: Efficient and Private Neural Network Training},
howpublished = {Cryptology ePrint Archive, Paper 2018/442},
year = {2018},
note = {\url{https://eprint.iacr.org/2018/442}},
url = {https://eprint.iacr.org/2018/442}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# How to create symbolic links in Windows? [closed]

I've tried Winbolic (Windows XP), but I can't get it to work the way I want.

Am I wrong here?

If I create a symbolic link within a folder in my %PATH%, the referenced directory should work as if it had actually been added to %PATH%, right?

-
–  Roger Pate May 24 '10 at 2:19
NOTE: Windows 7 supports sysmlinks. –  guillermooo May 24 '10 at 6:28
Duplicate question with better discussion at SuperUser: What are the various link types in Windows? How do I create them? –  mwolfe02 Sep 14 at 3:05

## closed as off topic by Nick T, Bennor McCarthy, ppeterka, mindas, cjstehnoFeb 28 at 14:34

Questions on Stack Overflow are expected to relate to programming within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

Have you looked at Junction? Also, Windows Vista introduced the mklink command line tool to create symbolic links. Scott Hanselman has a write-up on it at http://www.hanselman.com/blog/MoreOnVistaReparsePoints.aspx. If you're speaking solely about Windows XP, I've only ever used the Junction tool.

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works as required! Thanks. –  Jan Galinski May 10 '11 at 9:40
Looks like another option could be the /J switch on mklink (like "mklink /J <JUNCTION> <DIR>"). Note that it isn't a separate executable, it is a command in cmd so if you're calling from PowerShell, you'll have to shell out to cmd (i.e. "cmd /c mklink /J <JUNCTION> <DIR>") –  David Mohundro May 10 '11 at 19:14
Wow junction works great. It creates a real symbolic link so even Cygwin will follow it. –  styfle Sep 19 '11 at 19:20

the referenced directory should work as if it had actually been added to %PATH%, right

No, if you had C:\bin added to your %PATH% and inside c:\bin you created a Junction called c:\bin\anotherDir\ to point to c:\anotherDir\ so that you could run some.exe from the command line like:

> anotherDir\some.exe

This would not work since the %PATH% variable does not work with sub directories (symlinked or not)

you would need to create a hardlink directly to some.exe, so the hardlink c:\bin\some.exe points to c:\anotherDir\some.exe

So in summary I don't believe you can do exactly what you are asking. That said I do recommend Link Shell Extension for creating Junctions and hardlinks. It offers more capabilities than most junction tools in a very easy to use windows explorer extension

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Try Symlinker a GUI tool to create Symlink in Windows works on all windows Versions

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Can it create symlinks on Windows XP? –  Sebastian Godelet Dec 11 '12 at 8:20

Junction is a great utility but one thing to watch for is that junction points look exactly the same as ordinary folders in Windows Explorer.

Use 'junction.exe -s' on the command line to list all junctions in folders and sub-folders.

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I haven't heard of Winbolic but I'm not aware of anything built-in that will give you this functionality. Perhaps Winbolic just adds some functionality to explorer.exe to simulate symlinks, and this is why it's not working for you in other places?

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A few weeks ago I had the problem to make a symbolic link to a network share. This helped: http://www.pearlmagik.com/winbolic/ :

Winbolic Link creates special folders which serve as links to the contents of another folder. They are functionally similar to "symbolic links" or "symlinks" which unix users are familiar with.

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It should be noted that for links to a network share no actual symlink is created, but a Shell Folder: However, on the file system, the folder actually just contains two files that tell Windows to display the folder as though it is the target folder. Not all programs are aware of this type of link and may show as the content of the folder the two files instead of the content of the target folder. This is especially true for command line tools. –  mthomas Dec 1 '12 at 23:17

Have you tried linkd from Windows 200x Resource Kit? I've heard that it does for pre-Vista versions of Windows same thing that mklink does for Vista which is what you are looking for.

-

Here is an article that discusses every aspect of symlinks on windows:

http://www.shell-shocked.org/article.php?id=284

can you undownvote me now :)

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-1: broken link –  deft_code Apr 8 '11 at 4:58
Hopefully the site will come back up. –  joedevon Apr 11 '11 at 22:06
Unfortunately, it didn't. –  Rafael Almeida Aug 31 '12 at 17:59
And this is why any good answer should include the most relevant information from the website you link to. –  Thor84no Dec 7 '12 at 13:24
@Thor84no found it on archive.org see edit above. –  joedevon Dec 8 '12 at 19:30

Windows uses shortcuts as symlinks; there isn't any other way (that I know of) to create the equivalent of a symlink.

The file system does support hard links, but Windows doesn't actually provide a way to do it. I believe sysinternals provides a mechanism for hard links, though.

EDIT: as @David Mohundro pointed out (while I was typing this), Junction is the thing I was thinking of from the sysinternals guys.

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The mklink command was introduced since Windows Vista. –  Dan Dascalescu Dec 3 '12 at 14:42


text:
# Mark 8:1-38

8  In those days the multitude being very great, and having nothing to eat, Jesus called his disciples unto him, and saith unto them, 2  I have compassion on the multitude, because they have now been with me three days, and have nothing to eat: 3  And if I send them away fasting to their own houses, they will faint by the way: for divers of them came from far. 4  And his disciples answered him, From whence can a man satisfy these men with bread here in the wilderness? 5  And he asked them, How many loaves have ye? And they said, Seven. 6  And he commanded the people to sit down on the ground: and he took the seven loaves, and gave thanks, and brake, and gave to his disciples to set before them; and they did set them before the people. 7  And they had a few small fishes: and he blessed, and commanded to set them also before them. 8  So they did eat, and were filled: and they took up of the broken meat that was left seven baskets. 9  And they that had eaten were about four thousand: and he sent them away. 10  And straightway he entered into a ship with his disciples, and came into the parts of Dalmanutha. 11  And the Pharisees came forth, and began to question with him, seeking of him a sign from heaven, tempting him. 12  And he sighed deeply in his spirit, and saith, Why doth this generation seek after a sign? verily I say unto you, There shall no sign be given unto this generation. 13  And he left them, and entering into the ship again departed to the other side. 14  Now the disciples had forgotten to take bread, neither had they in the ship with them more than one loaf. 15  And he charged them, saying, Take heed, beware of the leaven of the Pharisees, and of the leaven of Herod. 16  And they reasoned among themselves, saying, It is because we have no bread. 17  And when Jesus knew it, he saith unto them, Why reason ye, because ye have no bread? perceive ye not yet, neither understand? have ye your heart yet hardened? 18  Having eyes, see ye not? and having ears, hear ye not? and do ye not remember? 19  When I brake the five loaves among five thousand, how many baskets full of fragments took ye up? They say unto him, Twelve. 20  And when the seven among four thousand, how many baskets full of fragments took ye up? And they said, Seven. 21  And he said unto them, How is it that ye do not understand? 22  And he cometh to Bethsaida; and they bring a blind man unto him, and besought him to touch him. 23  And he took the blind man by the hand, and led him out of the town; and when he had spit on his eyes, and put his hands upon him, he asked him if he saw ought. 24  And he looked up, and said, I see men as trees, walking. 25  After that he put his hands again upon his eyes, and made him look up: and he was restored, and saw every man clearly. 26  And he sent him away to his house, saying, Neither go into the town, nor tell it to any in the town. 27  And Jesus went out, and his disciples, into the towns of Caesarea Philippi: and by the way he asked his disciples, saying unto them, Whom do men say that I am? 28  And they answered, John the Baptist: but some say, Elias; and others, One of the prophets. 29  And he saith unto them, But whom say ye that I am? And Peter answereth and saith unto him, Thou art the Christ. 30  And he charged them that they should tell no man of him. 31  And he began to teach them, that the Son of man must suffer many things, and be rejected of the elders, and of the chief priests, and scribes, and be killed, and after three days rise again. 32  And he spake that saying openly. And Peter took him, and began to rebuke him. 33  But when he had turned about and looked on his disciples, he rebuked Peter, saying, Get thee behind me, Satan: for thou savourest not the things that be of God, but the things that be of men. 34  And when he had called the people unto him with his disciples also, he said unto them, Whosoever will come after me, let him deny himself, and take up his cross, and follow me. 35  For whosoever will save his life shall lose it; but whosoever shall lose his life for my sake and the gospel’s, the same shall save it. 36  For what shall it profit a man, if he shall gain the whole world, and lose his own soul? 37  Or what shall a man give in exchange for his soul? 38  Whosoever therefore shall be ashamed of me and of my words in this adulterous and sinful generation; of him also shall the Son of man be ashamed, when he cometh in the glory of his Father with the holy angels.


text:
Creating pdf with inverted colors using pdflatex (like white on black etc)

Is there a simple way of a creating PDF with inverted colours?

A partial solution would be just changing font to white and making the background black.

Is there simple "full" solution of inverting whole PDF pages? (including colors in lstlistings, images, etc)

You can set a pagecolor with the background package and the page color using the color command. You will need to pass the values onto setlistings as well. Color inversion of images is perhaps not possible although you can manipulate them as well. What do you exactly want the images to look like?

\documentclass{article}
\usepackage{graphicx}
\usepackage{listings}
\lstnewenvironment{teX}[1][]
{\lstset{language=[LaTeX]TeX}\lstset{escapeinside={(*@}{@*)},
numbers=left,numberstyle=\normalsize,stepnumber=1,numbersep=5pt,
breaklines=true,
%firstnumber=last,
%frame=tblr,
framesep=5pt,
basicstyle=\normalsize\ttfamily,
showstringspaces=false,
keywordstyle=\ttfamily\color{orange},
%identifierstyle=\ttfamily,
stringstyle=\color{maroon},
rulecolor=\color{gray!10},
xleftmargin=0pt,
xrightmargin=0pt,
aboveskip=\medskipamount,
belowskip=\medskipamount,
backgroundcolor=\color{black}, #1
}}
{}
\usepackage{hyperref}
\usepackage{background}
\SetBgContents{}
\begin{document}
\pagecolor{black}
\color{green}
\begin{teX}
\documentclass{article}
\usepackage{graphicx}
\usepackage{listings}
\lstnewenvironment{teX}[1][]
\end{teX}
\end{document}

• I just would like to invert colors of whole pages. This can be done by generating pdf first and than including it and reverting colors. I mentioned about images, just because I don't want to change settings of each used object, but to have a kind of global switch. – Grzegorz Wierzowiecki Feb 26 '12 at 19:59
• @GrzegorzWierzowiecki You will then have to look at external tools, or add a switch as an option when you generate the pdf i.e., the pagecolor and color can be set to be activated via an if statement. Best tool externally is Acrobat Professional. – Yiannis Lazarides Feb 26 '12 at 20:07
• It's not available for Linux and it's 270$for commercial use, for Teachers is 120$. I will not pay 120\$, just for inverting colors in my pdf ;). – Grzegorz Wierzowiecki Feb 26 '12 at 20:14
• @GrzegorzWierzowiecki I would invest 2 minutes and get a trial version ; besides in Poland anything is possible;) – Yiannis Lazarides Feb 26 '12 at 20:22

A better solution (perhaps the best solution) is to use crop package with invert option.

• Not really: "This option doesn’t invert pictures, nor does it really swap text and paper color." – Martin Schröder Feb 27 '12 at 7:36


text:
Mathematics

# $\angle\, A$ is supplementary to $\angle\, B,\, \angle\, B$ is complementary to $\angle\, C$. If $\angle\, A\, =\, 118^\circ$, what is the measure of $\angle\, C$ ?

$28^\circ$

##### SOLUTION

$\angle\, B\, =\, 180^\circ\, -\, \angle\, A\, =\, 180^\circ\, -\, 118^\circ\, =\, 62^\circ$

$\angle\, C\, =\, 90^\circ\, -\, 62^\circ\, =\, 28^\circ$.

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text:
# CIRCUIT ANALYSIS: 8 Resistors, 1 IVS, 1 Differential Op-Amp - Find Vo

## Homework Statement

The circuit below is a differential amplifier driven by a bridge, find $V_0$.

http://img476.imageshack.us/img476/8827/chapter5problem48fi2.jpg [Broken]

## Homework Equations

KCL, v = i R, Ideal Op-Amp relationships

## The Attempt at a Solution

http://img259.imageshack.us/img259/8314/chapter5problem48part2fs0.jpg [Broken]

By the Ideal Op-Amp relationships, we know that there is 0 current at both input terminals of the Op-Amp.

KCL at $V_2$)

$$\frac{0.005\,-\,V_2}{40000}\,=\,\frac{V_2}{60000}\,+\,\frac{V_2\,-\,V_4}{20000}$$

$$18\,V_2\,-\,12\,V_4\,=\,0.003$$ < ------ Equation 1

KCL at $V_4$)

$$\frac{V_2\,-\,V_4}{20000}\,=\,\frac{V_4}{80000}$$

$$4\,V_2\,-\,5\,V_4\,=\,0$$ <----- Equation 2

Now, using equations 1 and 2, I get $V_2\,=\,0.003571\,V$ and $V_4\,=\,0.002857\,V$. Does that seem right?

By the relationships of an ideal Op-Amp, $V_3\,=\,V_4$, so $V_3\,=\,0.002857\,V$.

KCL at $V_1$)

$$\frac{0.005\,-\,V_1}{10000}\,=\,\frac{V_1}{30000}\,+\,\frac{V_1\,-\,V_3}{20000}$$

$$11\,V_1\,-\,3\,V_3\,=\,0.03$$ <----- Equation 3

KCL at $V_3$)

$$\frac{V-1\,-\,V_3}{20000}\,=\,\frac{V_3\,-\,V_0}{80000}$$

$$4\,V_1\,-\,3\,V_3\,+\,V_0\,=\,0$$ <----- Equation 4

Using equations 3 and 4, I get $V_0\,=\,0.004363\,V$.

$$V_0\,=\,4.363\,mV$$ <------ Is that right?

Last edited by a moderator:

Related Introductory Physics Homework Help News on Phys.org
Bump - Bump:)

Looks good. Quick question though, why don't you use linear algebra instead of messy substitutions?