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# Cayley graphs, trees, Schreier’s theorem Part I (actually just groups)

6 Mar

So that’s a big agenda, but hey, dream big, right?  I’m giving a talk in our geometry, topology, and dynamics graduate seminar tomorrow on Schreier’s theorem, so I thought I’d try to write a blog post or three about it.   And then maybe blog about some cranberry cookies I made tonight after preparing my talk.

First, we’ve got to talk about the fundamental object of abstract algebra, pretty much the first thing you learn about as a math major: groups.  Abstractly, a group is a collection of objects that have some type of operation between them, where the operation satisfies a good number of properties: doing the operation to any two objects (or elements) results in another element in the group (this is called closure); there’s an identity element so that if you operate it with any other element, you get that other element back; the operation should be associative which means you can move parentheses around; and each element should have an inverse that takes you back to the identity.  Concretely, here are some groups:

• The integers, with addition as your group operation.  Adding any two integers gives you an integer (closure); adding 0 to any number gives you back, so 0 is the identity; addition is associative; and x+ (-x) = 0, so every element has an inverse.
• Symmetries of a square.  This means all the ways you can turn, flip, and change a square.  The group operation is doing one symmetry, and then another (like flipping, and then turning).  What’s interesting about this group is that it can be generated by just two different symmetries: rotating to the right by 90 degrees, and flipping vertically.   You can convince yourself of this by taking a piece of paper and numbering the corners 1, 2, 3, 4 and figuring out how many ways there are total to get the square back again.  Or you can try drawing a non-symmetric smiley face in Paint!

Symmetries of a square. Red guy is totally random

• Another is $\mathbb{Z}/ n\mathbb{Z}$, which is also written as $\mathbb{Z}_n$.  This can be read as the integers modulo n, which means every time you hit n, you jump back to 0.  It’s exactly how we count the hours up to 11, and then 12, and then we’re back at 1 again.  So our way of telling time $\mathbb{Z}_{12}$.  If I tell you that it’s 10 o’clock now, and ask you what time it’ll be in 70 hours, and you answer 8 o’clock, you just did modular arithmetic.  So you figured out that the closest multiple of 12 to 70 was 60, and then you knew we had to add 10 to our time, but since you’re not in not-the-US you knew the answer couldn’t be 20 o’clock, and you got 8 o’clock.  Good for you, you just did some group theory!

Flava Flav gets group theory

Okay, but I’m interested in geometric objects and pictures we can draw besides these little smiley face squares.  So how can I encode the data in our examples (the integers, the integers mod n, and the reflections of a square) into a geometric object?  I’ll use a diagram that computer scientists use a lot, called a graph.

Building our graph is like when we built our curve complex last time we did math.  We’ve got vertices, each of which represents a group element.  To connect two group elements with edges, we’ve got to figure out a generating set for our group.  This is a subset of the group elements, where if you operate with them over and over again, you end up with all of the group.  For instance, I can use {1,-1} as my generating set for the integers: I can add 1 to itself n times to get any positive integer n, and similarly for the negative integers.  And we used the rotation r and the vertical flip s in our smiley face picture above to generate the symmetries of a square.

So to connect two vertices, we draw an edge if we can operate by one of our generating set to get from one vertex to another.  So the picture of the integers looks like:

Here, we use little arrows for the vertices, and lines for the edges.  I very obviously didn’t make this picture.

This picture is from wikipedia!  It’s the picture for those rotations of the square.  Wikipedia used a letter ‘F’ instead of a smiley face.

So we can make these pictures for any group given some generating set (aside: generating sets aren’t unique.  For instance, you can generate the integers with $\pm 2, \pm 3$ instead of using 1, -1.  Different generating sets give you different pictures).  This picture is called the Cayley graph of the group with respect to the generating set S

A graph without any loops/cycles/ways to get from one vertex back to itself is called a tree.  Boom, we’ve got two vocabulary words from the title down!  In Part II, I’ll actually explain what Schreier’s theorem is and maybe give some idea as to its proof.

### 13 Responses to “Cayley graphs, trees, Schreier’s theorem Part I (actually just groups)”

1. - April 4, 2013

[…] theory.  From my last math post you know what a graph is: a collection of vertices with some edges connecting them.  We’ll […]

2. - July 23, 2013

[…] we weren’t together.  I decided to talk about graph theory, which I’ve introduced in this previous post.  I basically gave this blog post as my presentation, only with a powerpoint and being dressed up […]

3. - November 16, 2013

[…] (which you probably haven’t), check out the “groups” section in my post about Cayley graphs, which was really just about groups.  I just need you to know what the group of integers is […]

4. - February 25, 2014

[…] you don’t know what a group is, check out my quick intro post for some examples.  Wikipedia also has a significantly more exhaustive […]

5. - March 2, 2015

[…] algebra and topology- “group theory” is the study of groups, which we’ve seen a few times before, and “geometric” means that we’ll be looking at shapes.  Geometric […]

6. - April 26, 2015

[…] Last summer, I participated in an awesome research program which ended up with a good handful of original research projects all having to do with random groups.  Random group theory is rather modern- the first definition of a random group came about in the 1980s (just like me!)  It’s a very rich field with lots and lots of open problems, since you can dive into most properties of groups and ask if they apply to random groups. [Here’s the post defining groups if you forgot.] […]

7. - May 13, 2015

[…] you remember or know what a group is.  A pree is something on the way to being a group, but not quite: it’s a set with partial […]

8. - May 29, 2015

[…] talked before about what a group is– a set of elements with some operation that takes two elements to another one (like addition […]

9. - June 25, 2015

[…] that a group is a collection of elements paired with some kind of operation between them (the integers with […]

10. - July 30, 2015

[…] be talking about a property of groups, so brush up from a previous blog post or wikipedia.  First we need to define a  (total) ordering on a group: a binary […]

11. - September 24, 2015

[…] amount + an additive error term).  We say a group G is quasi-isometric to a space X if its Cayley graph is quasi-isometric to X.  Remember, a Cayley graph is a picture you can draw from a group if you […]

12. - November 14, 2015

[…] I say “group” when I’m obviously referring to a space, and then I mean the Cayley graph of that group (which changes depending on generating set, but if it’s a finite generating set […]

13. - November 8, 2016

[…] is the image of another under one of the reflections, you get something that looks a lot like the Cayley graph of the reflection triangle group.  The only difference is that each edge needs to be two edges […]


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# Integral of a non negative function over an infinite interval. [duplicate]

If $f(x) \geq 0$ and for every $x \in [0, \infty)$, $f(x)$ is Riemann integrable on $[0,n]$ for all n. If $\int_{0}^{\infty}f(x)dx < \infty$ then is it true that $\lim_{x \rightarrow \infty}f(x) = 0$. If yes then how?

## marked as duplicate by user147263, Jonas Meyer real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 14 '15 at 0:01

• No. $f$ might be "spiky" about each integer, with the height of each spike $1$; but, such that the widths of the bases of the spikes tend to zero sufficiently fast that their areas have finite sum. – David Mitra Jun 13 '15 at 19:38

Consider something like $$f(x) = \begin{cases} 1 & n< x < n+2^{-n}, \quad n \in \mathbb{N} \setminus \{0\} \\ 0 & \text{else} \end{cases}.$$ Then $$\int_0^{\infty} f(x) \, dx = \sum_{n=1}^{\infty} (n+2^{-n}-n) = 1,$$ but $f(x)$ is equal to $1$ for a set of $x$ including arbitrarily large reals.

To elaborate on David Mitra's comment:

Define $T(x) = \begin{cases}1-|x| & \text{if} \ |x| < 1 \\ 0 & \text{if} \ |x| \ge 1\end{cases}$ and let $f(x) = \displaystyle\sum_{n = 1}^{\infty}T\left(\dfrac{x-n}{n^2}\right)$.

Note that $T(x)$ is a piecewise linear function which connects $(-1,0)$ to $(0,1)$ to $(1,0)$. This spike has width $2$ (centered at $0$) and height $1$, so the area underneath $T(x)$ is $1$.

Then, $T\left(\dfrac{x-n}{n^2}\right)$ is $T(x)$ squished horizontally by a factor of $n^2$ and then translated to the right by $n$. Hence, $T\left(\dfrac{x-n}{n^2}\right)$ is a spike which has width $\dfrac{2}{n^2}$ (centered at $n$) and height $1$, so the area underneath $T\left(\dfrac{x-n}{n^2}\right)$ is $\dfrac{1}{n^2}$. Also, each of the $T\left(\dfrac{x-n}{n^2}\right)$'s are non-overlapping.

Hence, $\displaystyle\int_{0}^{\infty}f(x)\,dx = \displaystyle\int_{0}^{\infty}\sum_{n = 1}^{\infty}T\left(\dfrac{x-n}{n^2}\right)\,dx = \sum_{n = 1}^{\infty}\int_{0}^{\infty}T\left(\dfrac{x-n}{n^2}\right)\,dx = \sum_{n = 1}^{\infty}\dfrac{1}{n^2} < \infty$.

However, $f(n) = 1$ for each integer $n$, so $\displaystyle\lim_{x \to \infty}f(x) \neq 0$.

Note: We could also have a function, like $f(x) = \displaystyle\sum_{n = 1}^{\infty}nT\left(\dfrac{x-n}{n^3}\right)$, for which $\displaystyle\int_{0}^{\infty}f(x)\,dx < \infty$ but $f(x)$ is unbounded as $x \to \infty$.

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# The $$\pi$$s Go Full Circle

by Russell Euler (Northwest Missouri State University) and Jawad Sadek (Northwest Missouri State University)

This article originally appeared in:
Mathematics Magazine
February, 1999

Subject classification(s): Analysis | Real Analysis | Metric Spaces
Applicable Course(s): 4.9 Geometry | 4.11 Advanced Calc I, II, & Real Analysis | 4.17 Numerical Analysis

Explores notion of unit circle in various metrics

A pdf copy of the article can be viewed by clicking below. Since the copy is a faithful reproduction of the actual journal pages, the article may not begin at the top of the first page.

To open this file please click here.

These pdf files are furnished by JSTOR.

Classroom Capsules would not be possible without the contribution of JSTOR.

JSTOR provides online access to pdf copies of 512 journals, including all three print journals of the Mathematical Association of America: The American Mathematical Monthly, College Mathematics Journal, and Mathematics Magazine. We are grateful for JSTOR's cooperation in providing the pdf pages that we are using for Classroom Capsules.


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# How to get nearest 5 points mean with Nearest Neighbours?

In a dataset of longitude, latitude and price (of houses) I'm using sklearn's KNearestRegressor to get the 5 nearest neighbors mean price for each point. The problem is I want to do this for the whole dataset and each point is taking its own price into consideration since KNearestRegressor is a supervised algorithm, and I'm predicting the training set. How can I get the 5 nearest neighbors mean price for each point, not taking into consideration that point's price?

I have tried sklearn's unsupervised NearestNeighbours but it only outputs the indexes of the nearest neighbors, not the mean price.

## 1 Answer

First of all you train/ fit a model on the training set, may be cross validate using a training set. But it is not meant to predict itself. That's what test set is for.

You can achieve your objective by predicting each observation in a Leave One Out manner (also used for cross validation):

For every observation (i) in your dataset, fit a KNN regressor using all data except (i), evaluate your prediction for (i).


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### A Thorough Treatment of Highly-Efficient NTRU Instantiations

Julien Duman, Kathrin Hövelmanns, Eike Kiltz, Vadim Lyubashevsky, Gregor Seiler, and Dominique Unruh

##### Abstract

Cryptography based on the hardness of lattice problems over polynomial rings currently provides the most practical solution for public key encryption in the quantum era. The first encryption scheme utilizing properties of polynomial rings was NTRU (ANTS '98), but in the recent decade, most research has focused on constructing schemes based on the hardness of the somewhat related Ring/Module-LWE problem. Indeed, 14 out of the 17 encryption schemes based on the hardness of lattice problems in polynomial rings submitted to the first round of the NIST standardization process used some version of Ring/Module-LWE, with the other three being based on NTRU. The preference for using Ring/Module-LWE is due to the fact that this problem is at least as hard as NTRU, is more flexible in the algebraic structure due to the fact that no polynomial division is necessary, and that the decryption error is independent of the message. And indeed, the practical NTRU encryption schemes in the literature generally lag their Ring/Module-LWE counterparts in either compactness or speed, or both. In this paper, we put the efficiency of NTRU-based schemes on equal (even slightly better, actually) footing with their Ring/Module-LWE counterparts. We provide several instantiations and transformations, with security given in the ROM and the QROM, that detach the decryption error from the message, thus eliminating the adversary's power to have any effect on it, which ultimately allows us to decrease parameter sizes. The resulting schemes are on par, compactness-wise, with their counterparts based on Ring/Module-LWE. Performance-wise, the NTRU schemes instantiated in this paper over NTT-friendly rings of the form $Z_q[X]/(X^d-X^{d/2}+1)$ are the fastest of all public key encryption schemes, whether quantum-safe or not. When compared to the NIST finalist NTRU-HRSS-701, our scheme is $15\%$ more compact and has a $15$X improvement in the round-trip time of ephemeral key exchange, with key generation being $35$X faster, encapsulation being $6$X faster, and decapsulation enjoying a $9$X speedup.

Available format(s)
Category
Public-key cryptography
Publication info
Preprint. MINOR revision.
Keywords
Lattice CryptographyNTRU
Contact author(s)
Julien Duman @ rub de
kathrin @ hoevelmanns net
eike kiltz @ rub de
gseiler @ inf ethz ch
unruh @ ut ee
History
Short URL
https://ia.cr/2021/1352

CC BY

BibTeX

@misc{cryptoeprint:2021/1352,
author = {Julien Duman and Kathrin Hövelmanns and Eike Kiltz and Vadim Lyubashevsky and Gregor Seiler and Dominique Unruh},
title = {A Thorough Treatment of Highly-Efficient NTRU Instantiations},
howpublished = {Cryptology ePrint Archive, Paper 2021/1352},
year = {2021},
note = {\url{https://eprint.iacr.org/2021/1352}},
url = {https://eprint.iacr.org/2021/1352}
}

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


text:
Minimal relative Schubert modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:53:03Z http://mathoverflow.net/feeds/question/91786 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91786/minimal-relative-schubert-modules Minimal relative Schubert modules Christopher Drupieski 2012-03-21T03:21:56Z 2012-03-22T16:02:19Z
I am trying to better understand the definition of certain objects called minimal relative Schubert modules. My primary reference is Chapters 1 and 2 of Wilberd van der Kallen's Lectures on Frobenius splittings and B-modules.

Here is the setup: $G$ is a reductive algebraic group over an algebraically closed field $k$, $B$ is a Borel subgroup of $G$, $T$ is a maximal torus of $B$, $X(T)$ is the character group of $T$, and $W = N_G(T)/C_G(T)$ is the associated Weyl group. In van der Kallen's notes, the Borel subgroup is assumed to correspond to the set of positive roots of $T$ in $G$.

Given an element $w \in W$, one has the Bruhat cell $(BwB)/B \subset G/B$, and its closure in $G/B$ is the Schubert variety $X_w$. We also have the boundary $\partial X_w$.

Given a $B$-module $M$, one has an associated $G$-vector bundle $\mathcal{L}(M)$ over $G/B$. We can restrict the vector bundle to the Schubert varietiy $X_w$, and then consider the space $H^0(X_w,\mathcal{L}(M))$ of global sections. van der Kallen denotes this space by $H_w(M)$, and calls it a dual Joseph module.

It follows from Remark 2.2.3 and Corollary 2.2.7 in van der Kallen's notes that the functor $H_w(-)$ can be realized as a composition of induction (and restriction) functors $\text{ind}_B^{P_s}(-)$, where $P_s$ is a minimal parabolic subgroup of $G$. Specifically, if $w = s_{\alpha_1} s_{\alpha_2} \ldots s_{\alpha_t}$ is a reduced expression for $w$, then

$H_w(-) = \text{ind}_B^{P_1} \circ \text{ind}_B^{P_2} \circ \cdots \circ \text{ind}_B^{P_t}(-)$,

where $P_i = P_{\alpha_i}$ is the minimal parabolic corresponding to the simple root $\alpha_i$.

Now to the definitions I am grappling with: Let $\mu \in X(T)$, and let $w \in W$ such that $-w\mu \in X(T)_+$ (i.e., $w\mu$ is an antidominant weight). Then the dual Joseph module $P(\mu)$ is defined by $P(\mu) = H_{w^{-1}}(w\mu) = H^0(X_{w^{-1}},\mathcal{L}(w\mu))$. Assume now that $w$ was chosen to be of minimal length with the property that $-w\mu \in X(T)_+$. Then the minimal relative Schubert module $Q(\mu)$ is defined by

$Q(\mu) = \text{ker}(\text{res}: H^0(X_{w^{-1}},\mathcal{L}(w\mu)) \rightarrow H^0(\partial X_{w^{-1}},\mathcal{L}(w\mu)))$.

So $Q(\mu) \subset P(\mu)$.

Is there a way to understand $Q(\mu)$ in terms of the homological properties of the induction functors $\text{ind}_B^{P_i}(-)$ and their evaluation maps $\varepsilon:\text{ind}_B^{P_i}(M) \rightarrow M$ ?

I am not familiar or comfortable at this point with the geometric notions involved in the above definitions, but am much more comfortable with the homological properties of the induction functors, since these can be defined much more algebraically (say, as given in Chapter I.3 of Jantzen's Representations of Algebraic Groups).
 http://mathoverflow.net/questions/91786/minimal-relative-schubert-modules/91841#91841 Answer by Chuck Hague for Minimal relative Schubert modules Chuck Hague 2012-03-21T17:52:57Z 2012-03-22T16:02:19Z
This is not really an answer to your question -- more of a long comment, I suppose -- but there is a somewhat intuitive way to understand what these modules look like, representation-theoretically speaking. It's easier to describe what the duals to these modules look like, so let me do that. First, for any weight $\mu$, let $\mu^+$ denote the unique dominant element in the Weyl group orbit of $-\mu$ and let $V(\mu^+)$ denote the Weyl module for $G$ of highest weight $\mu^+$ (i.e., $V(\mu^+) = H^0(-w_0\mu^+)^*$). Then the Joseph module $P(\mu)^*$ is just the $B$-submodule of $V(\mu^+)$ generated by any nonzero weight vector of weight $-\mu$. (Dually, this now describes the surjection $H^0(-w_0\mu^+) \twoheadrightarrow P(\mu)$).

Now set $$\lbrace \mu_1, \ldots, \mu_r \rbrace := \lbrace s \mu : s \in W \textrm{ is a simple reflection and } s \mu < \mu \rbrace .$$ Let $I(\mu)$ be the $B$-submodule of $V(\mu^+)$ generated by $P(\mu_i)^*$, $1 \leq i \leq r$. Equivalently, $I(\mu)$ is the $B$-submodule of $V(\mu^+)$ generated by nonzero weight vectors of weights $-\mu_1, \ldots, -\mu_r$. Then $Q(\mu)^*$ fits into an exact sequence $$0 \to I(\mu) \to P(\mu)^* \to Q(\mu)^* \to 0 .$$ Remark that we can also describe $I(\mu)$ as the submodule of $V(\mu^+)$ generated by all Joseph modules properly contained in $P(\mu)^*$.

As for the evaluation map $\varepsilon$, it can be described as follows. Let $V \subseteq V(\mu^+)$ denote the highest weight subspace of weight $\mu^+$. Then $\varepsilon : P(\mu) \twoheadrightarrow k_{-\mu^+}$ is dual to the inclusion $V \hookrightarrow P(\mu)^*$ of the highest weight subspace. (Remark also that $V = P(-\mu^+)^*$).


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0

# What is the integral of the quantity of the derivative with respect to x of the function f times another function of x defined as g subtracted by g prime times f divided by g squared with respect to x?

Mrkbh

Lvl 1
2010-11-05 07:52:02

âˆ« [f'(x)g(x) - g'(x)f(x)]/g(x)2 dx = f(x)/g(x) + C

C is the constant of integration.

Wiki User

2010-11-05 07:52:02
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# Time reversal symmetry for Zeeman fields

For a magnetic field along $\hat{x}$ the hamiltonian is given by: $H_{z}=V_x(c^{\dagger}_{\uparrow}c_{\downarrow}+c^{\dagger}_{\downarrow}c_{\uparrow})$.

If we follow the usual prescription of changing the spins: $\uparrow\quad \rightarrow \quad\downarrow$ and vice-versa, how do we see that the above hamiltonian breaks TR symmetry?

## 1 Answer

Time reversal sends $c_\uparrow\rightarrow c_\downarrow$ and $c_\downarrow\rightarrow -c_\uparrow$. With this transformation, it should be clear the term is time odd.

See, e.g., here for more information. The key thing to realize is that while time reversal must send $k\rightarrow -k$ and $+z\rightarrow -z$, that still leaves the possibility of gaining an arbitrary phase under time reversal. To determine this phase, you demand that time reversal also sends $+x$ spins to $-x$ spins, and $+y$ spins to $-y$ spins. With these constraints, you can find the phase factors above.

• Is there any reference for this? I thought the procedure only required flipping the spins. (and k -> -k in momentum space) – Arnab Barman Ray May 28 '18 at 16:30


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# The Electron

Submitted by jwmoore on Wed, 01/12/2011 - 01:28

Near the middle of the nineteenth century the English chemist and physicist Michael FaradayThe electric charge carried by one mole of electrons, 9.648 670 x 104 C mol-1; abbreviated F. (1791 to 1867) established a connection between electricity and chemical reactions. He already knew that an electric current flowing into certain molten compounds through metalAn element characterized by a glossy surface, high thermal and electrical conductivity, malleability, and ductility. plates called electrodes could cause reactions to occur. Samples of different elements would deposit on the electrodes. Faraday found that the same quantity of electric charge was required to produce 1 mol of any element whose valence was 1. Twice that quantity of charge would deposit 1 mol of an element whose valence was 2, and so on. Electric charge is measured in units called coulombs, abbreviated C. One coulomb is the quantity of charge which corresponds to a current of one ampere flowing for one second. It was found that 96500 C of charge was required to deposit on an electrode l mol of an element whose valence is l. Faraday’s experiments strongly suggested that electricity, like matterAnything that occupies space and has mass; contrasted with energy., consists of very small indivisible particles. The name electronA negatively charged, sub-atomic particle with charge of 1.602 x 10-19 coulombs and mass of9.109 x 1023 kilograms; electrons have both wave and particle properties; electrons occupy most of the volume of an atom but represent only a tiny fraction of an atom's mass. was given to these particles, and an electric current came to be thought of as a flow of electrons from one place to another. When such a current flows into a chemical compoundA substance made up of two or more elements and having those elements present in definite proportions; a compound can be decomposed into two or more different substances., one electron is required for each atomThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. of a univalent element deposited on an electrode, two electrons for each atom of an element whose valence is 2, and so on. Thus an electric charge of 96500 C corresponds to 1 mol of indivisible electric particles (electrons). The relationship between electricity and atomic structure was further clarified by experiments involving cathodeThe electrode in an electrochemical cell where reduction occurs; the negatively charged electrode in a vacuum tube.-ray tubes in the 1890s. A cathode-ray tube can be made by pumping most of the air or other gasA state of matter in which a substance occupies the full volume of its container and changes shape to match the shape of the container. In a gas the distance between particles is much greater than the diameters of the particles themselves; hence the distances between particles can change as necessary so that the matter uniformly occupies its container. out of a glassA solid material that does not have the long-range order of a crystal lattice; an amorphous solid. A glass melts over a range of temperatures instead of having the definite melting temperature characteristic of crystalline solids. tube and applying a high voltage to two metal electrodes inside. If ZnS or some other phosphor is placed on the glass at the end of the tube opposite the negatively charged electrode (cathode), the ZnS emits light. This indicates that some kind of rays are streaming away from the cathode. When passed between the poles of a magnet, these cathode raysElectrons emitted from the surface of a cathode in a vacuum tube. behave the same way as the β particles described earlier. The fact that they were very small electrically charged particles led the English physicist J. J. Thomson (1856 to 1940) to identify them with the electrons of Faraday’s experiments. Thus cathode rays are a beam of electrons which come out of the solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. metal of the cathode. They behave exactly the same way no matter what the electrode is made of or what gas is in the tube. These observations allow one to conclude that electrons must be constituents of all matter. In addition to being deflected by a magnet, the electron beam in a cathode-ray tube can be attracted toward a positively charged metal plate or repelled from a negative plate. By adjusting such electrodes to exactly cancel the deflection produced by a magnet of known strength, Thomson was able to determine that the ratio of charge to massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. for an electron is 1.76 × 108 C/g. This is a rather large ratio. Either each electron has a very large charge, or each has a very small mass. We can see which by using Faraday’s result that there are 96 500 C mol–1 of electrons

$\frac{\text{96 500 C mol}^{-\text{1}}}{\text{1}\text{.76 }\times \text{ 10}^{\text{8}}\text{ C g}^{-\text{1}}}=\text{5}\text{.48 }\times \text{ 10}^{-\text{4}}\text{ g mol}^{-\text{1}}$

Thus the molar massThe mass of a mole of substance; the same as molecular weight for molecular substances. of an electron is 5.48 × 10–4 g mol–1, and if we think of the electron as an “atom“(or indivisible particle) of electricity, its atomic weightThe average mass of the naturally occurring isotopes of an element, taking into account the different natural abundances of the isotopes. Expressed relative to the value of exactly 12 for carbon-12; also called atomic mass. would be 0.000548—only $\tfrac{1}{1837}$ that of hydrogen, the lightest element known. In 1909 the American physicist Robert A. Millikan (1863 to 1953) was able to determine the charge on an electron independently of its mass. His value of 1.6 × 10–19 C can be combined with Thomson’s charge-to-mass ratio to give an independent check on the molar mass for the electron

$\frac{\text{1}\text{.60 }\times \text{ 10}^{-\text{19}}\text{ C}}{\text{1}\text{.76 }\times \text{ 10}^{\text{8}}\text{ C g}^{-\text{1}}}\text{ }\times \text{ 6}\text{.022 }\times \text{ 10}^{\text{23}}\text{ mol}^{-\text{1}}=\text{5}\text{.47 }\times \text{ 10}^{-\text{4}}\text{ g mol}^{-\text{1}}$

thus confirming that the electron has much less mass than the lightest atom. (The quantity 1.6 × 10–19 C is often represented by the symbol e. Thus the charge on a single electron is –e = –1.6 × 10–19 C. The minus sign indicates that the electron is a negatively charged particle.)


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USF Home > College of Arts and Sciences > Department of Mathematics & Statistics

Mathematics & Statistics

# Analysis (Leader: Prof. Boris Shekhtman )

## Friday, February 17, 2006

Title
Speaker

Time
Place

The Ramanujan Entire Function
University of Central Florida
5:00pm-6:00pm
PHY 120

Abstract

Ramanujan was a self-educated college dropout who did some of the best mathematics of the twentieth century. He extensively worked on the $$F(z)=1+\sum_{n=1}^\infty\frac{(-z)^nq^{n^2}}{(1-q)(1-q^2)\dotsc(1-q^n)},$$ which we refer to as the Ramanujan entire function. We demonstrate the significance of this function in number theory and analysis and give a new interpretation of the statement $$1+\sum_{n=1}^\infty\frac{z^nq^{n^2}}{(1-q)(1-q^2)\dotsc(1-q^n)} =\prod_{n=1}^\infty\left(1+\frac{zq^{2n-1}}{1-c_1q^n-c_2q^{2n}-\dotsb}\right)$$ in Ramanujan's lost notebook.

The coefficients $$c_1,c_2,\dotsc$$ turned out to have very interesting patterns and many open problems will be mentioned.

## Friday, February 10, 2006

Title
Speaker
Time
Place

Example of non uniquely minimal projection in $$L_p$$, Part II
Lesław Skrzypek
5:00pm-6:00pm
PHY 120

## Friday, February 3, 2006

Title
Speaker
Time
Place

Example of non uniquely minimal projection in $$L_p$$
Lesław Skrzypek
5:00pm-6:00pm
PHY 120


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# Force and momentum of water striking a water wheel?

Imagine water striking a water wheel. We can find the torque applied by calculating the force of the water and multiplying by the distance from the center.

To calculate force, we can use the formula F =$\ \frac{dp}{dt}$ .

However, I am confused about change in momentum. More specifically, when water hits the wheel, what will be the change in momentum? How can this be calculated?

To make the question easier, let's assume that a large drop of a certain mass m hits a stationary water wheel. Do we consider the water drop's final velocity to be zero at the point of impact? If not, how can we conceptually and mathematically determine the change in momentum, without experimenting?

Now, how about if the water wheel is already rotating. How can we conceptually imagine what the change in momentum would be (I assume lower than when the water wheel is stationary), and how can we mathematically calculate this if given real values?

It is possible that there are other factors, such as static friction of water wheel or mass of water wheel, however, I am not sure what is needed to be accounted for, so if anyone can improve this problem, please comment of edit.

Thanks for any help.

• volumes have been written on the dynamic analysis of waterwheels and turbines like this. A good starting point is wikipedia. suggest also that you migrate this to the engineering stack exchange, lots of experts there. – niels nielsen Aug 11 '18 at 5:33

The change in momentum is simply the difference between the momentum of the system before the event and the momentum of the system afterwards. Don't overthink it =)

The "correct" change in momentum depends on the exact shape of the water wheel, so your mileage may vary. However, a very reasonable first pass would be to assume the water "sticks" to the waterwheel. In this case, we can assume that the momentum of the water after the collision is exactly equal to the mass of the water multiplied by the tangental velocity of the wheel at the point where the water stuck.

However, given that you have a wheel rotating around a fixed point, it is easier to think in angular momentum. Angular momentum is also conserved, and the math is simpler. If you try to do this with just momentum, you'll have to consider how the forces are transmitted to the axel of the waterwheel. It'll yield the same answer, but using just linear momentum will require more math.

From this, we can put together a rough sketch of how a water wheel starts spinning. Start with the wheel at a standstill. One drop falls onto the wheel. Before the drop falls, the angular momentum of the drop is its mass multiplied by its downward velocity multiplied by its distance from the wheel (this is the definition of the angular momentum of a point mass), and the momentum of the waterwheel is zero. After the drop falls and "sticks" to the wheel, the momentum of the entire system is the angular velocity of the wheel after the collision multiplied by the moment of inertia of the waterdrop-waterwheel system. We can solve for this to see that the waterdroop-waterwheel system will start to turn ever so slightly, because of the collision.

As we have more collisions, the wheel starts to speed up. As the wheel speeds up, we start to find that the angular velocity of the water droplet before the collision and the angular velocity of the droplet after the collision start to get closer and closer together. This means the droplet will impart less and less angular momentum into the waterwheel. At some point, the wheel spins as fast as the droplets passing by it, and the wheel receives no angular momentum from the droplet, so it never accelerates faster.

Of course, this is only one effect. In a waterwheel like you have drawn, with no buckets, this sticking would only be momentary before the water rolls off. If there are buckets, we also have to consider that the water keeps acting on the waterwheel until it reaches the bottom. There are many ways this can be modeled, but energy may be the best way to do it. You can model the loss of gravitational potential energy as the bucket of water reaches the bottom of the wheel, and then look at how much kinetic energy is lost when you let the water go.

In a more complicated system, the water might splash or divert off at an angle. In such cases, you would need to take care to make sure it is modeled correctly.

• Thanks for taking the time to answer! I have a better idea of the situation now. However, as the water wheel speeds up, would the angular momentum added to the wheel simply be the difference between the the water's tangential velocity and the wheel's tangential velocity (at point of impact), multiplied by the water mass and the distance? I think it would make sense because when the water falls as fast as the wheel rotates, the velocity difference is 0, thus the angular momentum added would also be 0 (as you described). – F16Falcon Aug 11 '18 at 21:56
• @f16falcon you have the idea correct. I phrased it in general terms of conservation of angular momentum, but if you do the math, you get exactly what you say. – Cort Ammon Aug 11 '18 at 22:33


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# Enable Latex in Hugo

I was able to add $$\LaTeX$$ support to my hugo blog finally. Suprisingly adding it is not difficult with hugo blogs. All I needed to do is add katex support. This might look different depending on the theam that you are using. Now I am able to include some mathematical expressions using  or \$ depending on whether I want the expression to be inline or display mode. For example, the following is Holder's inequality in display mode:

$\sum_{i=1}^n x_iy_i \le \left(\sum_{i=1}^n x_i^p \right)^{1/p} \left(\sum_{i=1}^n y_i^q \right)^{1/q}$

which is fantastic. The only problem I am seeing is that some special characters in $$\LaTeX$$ are special characters in Markdown as well. For example, _ means something in both languages. My work around at the moment is to escape the underscores until I find a more convenient solution.


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# Making algorithms display after the paragraph and split them in pages

I am using algorithmicx to write algorithms. I want to be able to write algorithms so that they appear right after the paragraph they have been written and possibly split them from the end of one page to the starting of the next page if the space is not sufficient.

How to do so?

-
Please add a MWE. – Marco Daniel Sep 4 '11 at 1:13

To break down your post, you want:

1. To have an algorithm appear immediately after a paragraph in which it was discussed; and
2. To have the algorithm split across pages (if possible).

In order to fulfill both requests, you cannot place your algorithmicx environment within an algorithm environment, since the latter is a float. And, floats move around. Moreover, floats do not break over pages. So, you'll have to make your own environment that accounts for an algorithm caption and number. Here is a very basic example of what that would entail:

\documentclass{article}
\usepackage{algorithmicx}% http://ctan.org/pkg/algorithmicx
\usepackage{lipsum}% http://ctan.org/pkg/lipsum
\usepackage{xifthen}% http://ctan.org/pkg/xifthen
\usepackage{needspace}% http://ctan.org/pkg/needspace
\usepackage{hyperref}% http://ctan.org/pkg/hyperref

% ================ ALGORITHM ENVIRONMENT ================
\newcounter{myalg}% Algorithm counter
\newenvironment{myalg}[1][]%
{% \begin{myalg}[#1]
\needspace{2\baselineskip}% At least 2\baselineskip required, otherwise break
\noindent \rule{\linewidth}{1pt} \endgraf% Top rule
\refstepcounter{myalg}% For correct reference of algorithm
\centering \textsc{Algorithm}~\themyalg%
\ifthenelse{\isempty{#1}}{}{:\ #1}% Typeset name (if provided)
}{% \end{myalg}
\noindent \rule{\linewidth}{1pt}% Bottom rule
}%

\begin{document}

\lipsum[1-4]

Here is Algorithm~\ref{first-alg}:

\begin{myalg}[My first algorithm]
\label{first-alg}
\begin{algorithmic}[1]
\State This is some code
\State This is some more code
\State Here is some code
\State This is some code
\State This is some more code
\State Here is some code
\State This is some code
\State This is some more code
\State Here is some code
\State This is some code
\State This is some more code
\State Here is some code
\end{algorithmic}
\end{myalg}

That was Algorithm~\ref{first-alg}.

\lipsum[4-6]

\end{document}

In the above example, the myalg environment does not float and replaces the original algorithm environment. As an optional argument, myalg will typeset a caption. The xifthen package provides a conditional clause to check for a caption or not. The needspace package conditions on the amount of space left on the page. In order to not be stuck with a rule or caption at the page bottom, needspace will issue a \break if there is less then 2\baselineskip left on the page. The hyperref package was also included to show you that the hyperlinks work properly, while the lipsum package provides dummy text.

If you're into a more vibrant array of decorations for your algorithms, then the mdframed package allows easy breaking of decorative frames across pages:

This can be extended to include items into a "List of Algorithms" as well.

-


text:
# Lucas–Lehmer primality test

This article is about the Lucas–Lehmer test that only applies to Mersenne numbers. For the Lucas–Lehmer test that applies to a natural number n, see Lucas primality test. For the Lucas–Lehmer–Riesel test, see Lucas–Lehmer–Riesel test.

In mathematics, the Lucas–Lehmer test (LLT) is a primality test for Mersenne numbers. The test was originally developed by Édouard Lucas in 1856[1] and subsequently improved by Lucas in 1878 and Derrick Henry Lehmer in the 1930s.

## The test

The Lucas–Lehmer test works as follows. Let Mp = 2p − 1 be the Mersenne number to test with p an odd prime. The primality of p can be efficiently checked with a simple algorithm like trial division since p is exponentially smaller than Mp. Define a sequence ${\displaystyle \{s_{i}\}}$ for all i ≥ 0 by

${\displaystyle s_{i}={\begin{cases}4&{\text{if }}i=0;\\s_{i-1}^{2}-2&{\text{otherwise.}}\end{cases}}}$

The first few terms of this sequence are 4, 14, 194, 37634, ... (sequence A003010 in the OEIS). Then Mp is prime if and only if

${\displaystyle s_{p-2}\equiv 0{\pmod {M_{p}}}.}$

The number sp − 2 mod Mp is called the Lucas–Lehmer residue of p. (Some authors equivalently set s1 = 4 and test sp−1 mod Mp). In pseudocode, the test might be written as

// Determine if Mp = 2p − 1 is prime for p > 2
Lucas–Lehmer(p)
var s = 4
var M = 2p − 1
repeat p − 2 times:
s = ((s × s) − 2) mod M
if s == 0 return PRIME else return COMPOSITE

Performing the mod M at each iteration ensures that all intermediate results are at most p bits (otherwise the number of bits would double each iteration). The same strategy is used in modular exponentiation.

## Time complexity

In the algorithm as written above, there are two expensive operations during each iteration: the multiplication s × s, and the mod M operation. The mod M operation can be made particularly efficient on standard binary computers by observing that

${\displaystyle k\equiv (k\,{\bmod {\,}}2^{n})+\lfloor k/2^{n}\rfloor {\pmod {2^{n}-1}}.}$

This says that the least significant n bits of k plus the remaining bits of k are equivalent to k modulo 2n−1. This equivalence can be used repeatedly until at most n bits remain. In this way, the remainder after dividing k by the Mersenne number 2n−1 is computed without using division. For example,

 916 mod 25−1 = 11100101002 mod 25−1 = ((916 mod 25) + int(916 ÷ 25)) mod 25−1 = (101002 + 111002) mod 25−1 = 1100002 mod 25−1 = (100002 + 12) mod 25−1 = 100012 mod 25−1 = 100012 = 17.

Moreover, since s × s will never exceed M2 < 22p, this simple technique converges in at most 1 p-bit addition (and possibly a carry from the pth bit to the 1st bit), which can be done in linear time. This algorithm has a small exceptional case. It will produce 2n−1 for a multiple of the modulus rather than the correct value of 0. However, this case is easy to detect and correct.

With the modulus out of the way, the asymptotic complexity of the algorithm only depends on the multiplication algorithm used to square s at each step. The simple "grade-school" algorithm for multiplication requires O(p2) bit-level or word-level operations to square a p-bit number. Since this happens O(p) times, the total time complexity is O(p3). A more efficient multiplication algorithm is the Schönhage–Strassen algorithm, which is based on the Fast Fourier transform. It only requires O(p log p log log p) time to square a p-bit number. This reduces the complexity to O(p2 log p log log p) or Õ(p2).[2] Currently the most efficient known multiplication algorithm, Fürer's algorithm, only needs ${\displaystyle p\log p\ 2^{O(\log ^{*}p)}}$ time to multiply two p-bit numbers.

By comparison, the most efficient randomized primality test for general integers, the Miller–Rabin primality test, requires O(k n2 log n log log n) bit operations using FFT multiplication for an n-digit number, where k is the number of iterations and is related to the error rate. For constant k, this is in the same complexity class as the Lucas-Lehmer test. In practice however, the cost of doing many iterations and other differences leads to worse performance for Miller–Rabin. The most efficient deterministic primality test for any n-digit number, the AKS primality test, requires Õ(n6) bit operations in its best known variant and is dramatically slower in practice.

## Examples

The Mersenne number M3 = 7 is prime. The Lucas–Lehmer test verifies this as follows. Initially s is set to 4 and then is updated 3−2 = 1 time:

• s ← ((4 × 4) − 2) mod 7 = 0.

Since the final value of s is 0, the conclusion is that M3 is prime.

On the other hand, M11 = 2047 = 23 × 89 is not prime. Again, s is set to 4 but is now updated 11−2 = 9 times:

• s ← ((4 × 4) − 2) mod 2047 = 14
• s ← ((14 × 14) − 2) mod 2047 = 194
• s ← ((194 × 194) − 2) mod 2047 = 788
• s ← ((788 × 788) − 2) mod 2047 = 701
• s ← ((701 × 701) − 2) mod 2047 = 119
• s ← ((119 × 119) − 2) mod 2047 = 1877
• s ← ((1877 × 1877) − 2) mod 2047 = 240
• s ← ((240 × 240) − 2) mod 2047 = 282
• s ← ((282 × 282) − 2) mod 2047 = 1736

Since the final value of s is not 0, the conclusion is that M11 = 2047 is not prime. Although M11 = 2047 has nontrivial factors, the Lucas–Lehmer test gives no indication about what they might be.

## Proof of correctness

The proof of correctness for this test presented here is simpler than the original proof given by Lehmer. Recall the definition

${\displaystyle s_{i}={\begin{cases}4&{\text{if }}i=0;\\s_{i-1}^{2}-2&{\text{otherwise.}}\end{cases}}}$

The goal is to show that Mp is prime iff ${\displaystyle s_{p-2}\equiv 0{\pmod {M_{p}}}.}$

The sequence ${\displaystyle {\langle }s_{i}{\rangle }}$ is a recurrence relation with a closed-form solution. Let ${\displaystyle \omega =2+{\sqrt {3}}}$ and ${\displaystyle {\bar {\omega }}=2-{\sqrt {3}}}$. It then follows by induction that ${\displaystyle s_{i}=\omega ^{2^{i}}+{\bar {\omega }}^{2^{i}}}$ for all i:

${\displaystyle s_{0}=\omega ^{2^{0}}+{\bar {\omega }}^{2^{0}}=\left(2+{\sqrt {3}}\right)+\left(2-{\sqrt {3}}\right)=4}$

and

{\displaystyle {\begin{aligned}s_{n}&=s_{n-1}^{2}-2\\&=\left(\omega ^{2^{n-1}}+{\bar {\omega }}^{2^{n-1}}\right)^{2}-2\\&=\omega ^{2^{n}}+{\bar {\omega }}^{2^{n}}+2(\omega {\bar {\omega }})^{2^{n-1}}-2\\&=\omega ^{2^{n}}+{\bar {\omega }}^{2^{n}}.\end{aligned}}}

The last step uses ${\displaystyle \omega {\bar {\omega }}=\left(2+{\sqrt {3}}\right)\left(2-{\sqrt {3}}\right)=1.}$ This closed form is used in both the proof of sufficiency and the proof of necessity.

### Sufficiency

The goal is to show that ${\displaystyle s_{p-2}\equiv 0{\pmod {M_{p}}}}$ implies that ${\displaystyle M_{p}}$ is prime. What follows is a straightforward proof exploiting elementary group theory given by J. W. Bruce[3] as related by Jason Wojciechowski.[4]

Suppose ${\displaystyle s_{p-2}\equiv 0{\pmod {M_{p}}}.}$ Then

${\displaystyle \omega ^{2^{p-2}}+{\bar {\omega }}^{2^{p-2}}=kM_{p}}$

for some integer k, so

${\displaystyle \omega ^{2^{p-2}}=kM_{p}-{\bar {\omega }}^{2^{p-2}}.}$

Multiplying by ${\displaystyle \omega ^{2^{p-2}}}$ gives

${\displaystyle \left(\omega ^{2^{p-2}}\right)^{2}=kM_{p}\omega ^{2^{p-2}}-(\omega {\bar {\omega }})^{2^{p-2}}.}$

Thus,

${\displaystyle \omega ^{2^{p-1}}=kM_{p}\omega ^{2^{p-2}}-1.\qquad \qquad (1)}$

For a contradiction, suppose Mp is composite, and let q be the smallest prime factor of Mp. Mersenne numbers are odd, so q > 2. Informally,[note 1] let ${\displaystyle \mathbb {Z} _{q}}$ be the integers modulo q, and let ${\displaystyle X=\left\{a+b{\sqrt {3}}\mid a,b\in \mathbb {Z} _{q}\right\}.}$ Multiplication in ${\displaystyle X}$ is defined as

${\displaystyle \left(a+{\sqrt {3}}b\right)\left(c+{\sqrt {3}}d\right)=[(ac+3bd)\,{\bmod {\,}}q]+{\sqrt {3}}[(ad+bc)\,{\bmod {\,}}q].}$

Clearly, this multiplication is closed, i.e. the product of numbers from X is itself in X. The size of X is denoted by ${\displaystyle |X|.}$

Since q > 2, it follows that ${\displaystyle \omega }$ and ${\displaystyle {\bar {\omega }}}$ are in X.[note 2] The subset of elements in X with inverses forms a group, which is denoted by X* and has size ${\displaystyle |X^{*}|.}$ One element in X that does not have an inverse is 0, so ${\displaystyle |X^{*}|\leq |X|-1=q^{2}-1.}$

Now ${\displaystyle M_{p}\equiv 0{\pmod {q}}}$ and ${\displaystyle \omega \in X}$, so

${\displaystyle kM_{p}\omega ^{2^{p-2}}=0}$

in X. Then by equation (1),

${\displaystyle \omega ^{2^{p-1}}=-1}$

in X, and squaring both sides gives

${\displaystyle \omega ^{2^{p}}=1.}$

Thus ${\displaystyle \omega }$ lies in X* and has inverse ${\displaystyle \omega ^{2^{p}-1}.}$ Furthermore, the order of ${\displaystyle \omega }$ divides ${\displaystyle 2^{p}.}$ However ${\displaystyle \omega ^{2^{p-1}}\neq 1}$, so the order does not divide ${\displaystyle 2^{p-1}.}$ Thus, the order is exactly ${\displaystyle 2^{p}.}$

The order of an element is at most the order (size) of the group, so

${\displaystyle 2^{p}\leq |X^{*}|\leq q^{2}-1

But q is the smallest prime factor of the composite ${\displaystyle M_{p}}$, so

${\displaystyle q^{2}\leq M_{p}=2^{p}-1.}$

This yields the contradiction ${\displaystyle 2^{p}<2^{p}-1}$. Therefore, ${\displaystyle M_{p}}$ is prime.

### Necessity

In the other direction, the goal is to show that the primality of ${\displaystyle M_{p}}$ implies ${\displaystyle s_{p-2}\equiv 0{\pmod {M_{p}}}}$. The following simplified proof is by Öystein J. Rödseth.[5]

Since ${\displaystyle 2^{p}-1\equiv 7{\pmod {12}}}$ for odd ${\displaystyle p>1}$, it follows from properties of the Legendre symbol that ${\displaystyle (3|M_{p})=-1.}$ This means that 3 is a quadratic nonresidue modulo ${\displaystyle M_{p}.}$ By Euler's criterion, this is equivalent to

${\displaystyle 3^{\frac {M_{p}-1}{2}}\equiv -1{\pmod {M_{p}}}.}$

In contrast, 2 is a quadratic residue modulo ${\displaystyle M_{p}}$ since ${\displaystyle 2^{p}\equiv 1{\pmod {M_{p}}}}$ and so ${\displaystyle 2\equiv 2^{p+1}=\left(2^{\frac {p+1}{2}}\right)^{2}{\pmod {M_{p}}}.}$ This time, Euler's criterion gives

${\displaystyle 2^{\frac {M_{p}-1}{2}}\equiv 1{\pmod {M_{p}}}.}$

Combining these two equivalence relations yields

${\displaystyle 24^{\frac {M_{p}-1}{2}}\equiv \left(2^{\frac {M_{p}-1}{2}}\right)^{3}\left(3^{\frac {M_{p}-1}{2}}\right)\equiv (1)^{3}(-1)\equiv -1{\pmod {M_{p}}}.}$

Let ${\displaystyle \sigma =2{\sqrt {3}}}$, and define X as before as the ring ${\displaystyle X=\{a+b{\sqrt {3}}\mid a,b\in \mathbb {Z} _{M_{p}}\}.}$ Then in the ring X, it follows that

{\displaystyle {\begin{aligned}(6+\sigma )^{M_{p}}&=6^{M_{p}}+\left(2^{M_{p}}\right)\left({\sqrt {3}}^{M_{p}}\right)\\&=6+2\left(3^{\frac {M_{p}-1}{2}}\right){\sqrt {3}}\\&=6+2(-1){\sqrt {3}}\\&=6-\sigma ,\end{aligned}}}

where the first equality uses the Binomial Theorem in a finite field, which is

${\displaystyle (x+y)^{M_{p}}\equiv x^{M_{p}}+y^{M_{p}}{\pmod {M_{p}}}}$,

and the second equality uses Fermat's little theorem, which is

${\displaystyle a^{M_{p}}\equiv a{\pmod {M_{p}}}}$

for any integer a. The value of ${\displaystyle \sigma }$ was chosen so that ${\displaystyle \omega ={\frac {(6+\sigma )^{2}}{24}}.}$ Consequently, this can be used to compute ${\displaystyle \omega ^{\frac {M_{p}+1}{2}}}$ in the ring X as

{\displaystyle {\begin{aligned}\omega ^{\frac {M_{p}+1}{2}}&={\frac {(6+\sigma )^{M_{p}+1}}{24^{\frac {M_{p}+1}{2}}}}\\&={\frac {(6+\sigma )(6+\sigma )^{M_{p}}}{24\cdot 24^{\frac {M_{p}-1}{2}}}}\\&={\frac {(6+\sigma )(6-\sigma )}{-24}}\\&=-1.\end{aligned}}}

All that remains is to multiply both sides of this equation by ${\displaystyle {\bar {\omega }}^{\frac {M_{p}+1}{4}}}$ and use ${\displaystyle \omega {\bar {\omega }}=1}$, which gives

{\displaystyle {\begin{aligned}\omega ^{\frac {M_{p}+1}{2}}{\bar {\omega }}^{\frac {M_{p}+1}{4}}&=-{\bar {\omega }}^{\frac {M_{p}+1}{4}}\\\omega ^{\frac {M_{p}+1}{4}}+{\bar {\omega }}^{\frac {M_{p}+1}{4}}&=0\\\omega ^{\frac {2^{p}-1+1}{4}}+{\bar {\omega }}^{\frac {2^{p}-1+1}{4}}&=0\\\omega ^{2^{p-2}}+{\bar {\omega }}^{2^{p-2}}&=0\\s_{p-2}&=0.\end{aligned}}}

Since ${\displaystyle s_{p-2}}$ is 0 in X, it is also 0 modulo ${\displaystyle M_{p}.}$

## Applications

The Lucas–Lehmer test is the primality test used by the Great Internet Mersenne Prime Search to locate large primes. This search has been successful in locating many of the largest primes known to date.[6] The test is considered valuable because it can provably test a large set of very large numbers for primality within an affordable amount of time. In contrast, the equivalently fast Pépin's test for any Fermat number can only be used on a much smaller set of very large numbers before reaching computational limits.

## Notes

1. ^ Formally, let ${\displaystyle \mathbb {Z} _{q}=\mathbb {Z} /q\mathbb {Z} }$ and ${\displaystyle X=\mathbb {Z} _{q}[T]/\langle T^{2}-3\rangle }$.
2. ^ Formally, ${\displaystyle \omega +\langle T^{2}-3\rangle }$ and ${\displaystyle {\bar {\omega }}+\langle T^{2}-3\rangle }$ are in X. By abuse of language ${\displaystyle \omega }$ and ${\displaystyle {\bar {\omega }}}$ are identified with their images in X under the natural ring homomorphism from ${\displaystyle \mathbb {Z} [{\sqrt {3}}]}$ to X which sends ${\displaystyle {\sqrt {3}}}$ to T.

## References

1. ^ The Largest Known Prime by Year: A Brief History
2. ^ Colquitt, W. N.; Welsh, L., Jr. (1991), "A New Mersenne Prime", Mathematics of Computation, 56 (194): 867–870, doi:10.2307/2008415, The use of the FFT speeds up the asymptotic time for the Lucas–Lehmer test for Mp from O(p3) to O(p2 log p log log p) bit operations.
3. ^ J. W. Bruce (1993). "A Really Trivial Proof of the Lucas–Lehmer Test". The American Mathematical Monthly. 100 (4): 370–371. doi:10.2307/2324959.
4. ^ Jason Wojciechowski. Mersenne Primes, An Introduction and Overview. 2003.
5. ^ Öystein J. Rödseth. A note on primality tests for N = h · 2n − 1. Department of Mathematics, University of Bergen.
6. ^


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# inputenc when we write in english

I will submit a paper in mathematic to a journal. I write it in english.

Can I delete the following code

 \usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}

Or I should replace it by another codes?

I ask this question, because some times when I want to open my tex file, I couldn't open it and I get this message. But when I change the folder of my tex file, it will be opened without any problem.

Thank you!

• you should keep T1 and use whatever input encoding you use: utf8 in preference to latin1, or if you only use ascii input then you can drop inputenc – David Carlisle Jan 1 '18 at 12:17
• the simplest thing is to write exactly what you wrote, unless you have a specific reason to change it. – David Carlisle Jan 1 '18 at 12:24
• Thank you: that is I use the following code? :\usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} because some times I find problems with 'inputenc' when I write in french – Student Jan 1 '18 at 12:32
• that is a completely separate issue, using non-ascii caracters in file names, it is simplest just not to do that. There is absolutely no way anyone could have guessed that was the issue from your original question. that is why questions should always have a complete test document, and if they are about an error message have the complete error (preferably as text from the log file, not as an image) – David Carlisle Jan 1 '18 at 12:41
• your image also shows you are using \bf don't do that! – David Carlisle Jan 1 '18 at 12:42

If your document really is saved in latin1 (ISO-8859-1) encoding then it is perfectly OK to leave

 \usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}

however latin1 is mostly regarded as a "legacy" encoding, and most operating systems and editors will (or should) default to UTF-8 (Unicode) encoding these days. If using UTF-8 then the lines should be

 \usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

If you can guarantee that you only use ASCII input then ASCII, UTF-8 and latin1 are the same encoding in that range so it doesn't really matter, you could delete the inputenc line but it is probably better to still specify the encoding your editor is using (which most likely is UTF-8 as noted above).

You should keep the T1 fontenc line unless you have specific reason to remove it.

As David stated utf-8 is preferable to latin1, to get this you should write

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

\usepackage[latin1]{inputenc}

\usepackage[T1]{fontenc}


text:
# learning/teaching approach to rigorous math with the goal of improving

I will state this now: yes, this is a subjective question. But I feel the answers people give may benefit students.

I want to get better at doing non trivial proofs. Real analysis is standard material some students have to learn. I have, broadly speaking, two choices for a textbook: 1) something like "baby Rudin", where I may struggle for long periods of time on problems, often having little direction and no progress, or 2) something easier, where smaller theorems are developed in the book and where there are less sophisticated jumps of reasoning (demonstrated, required) in the proofs.

What are the preferences of educators/students for using one over the other? Do you have reason to believe one is more successful at actually helping students improve? How do you determine success?

My beliefs: I tend to think the easier book may help me develop my own techniques (I could be wrong). I see more examples and I get more confidence in doing basic things. But more advanced material will probably be less accessible, so I know that at some point I will have to just deal with the sort of terseness that occurs in Rudin. Anyway, I can keep going back and forth about this.

-

If you haven't had much experience with writing proofs, I would suggest starting with something like Spivak's Calculus or, in fact, some abstract algebra, which most students find more accessible than analysis.

-
I have had exposure to proofs. So I'd want something with similar content to Rudin (basic topology and other more advanced material, maybe this). The problem comes when I try to prove his theorems. I don't feel like I'm making any progress whatsoever. – fancypants May 5 '13 at 19:24
Rudin has a hallowed reputation, but most students need a good teacher with it. Try looking at amazon.com/Introduction-Analysis-4th-William-Wade/dp/0132296381 or amazon.com/…. – Ted Shifrin May 5 '13 at 20:02

Both have their place. It depends in part on the taste of the reader. I personally would go for the "harder" text, complementing it with Wikipedia, random lecture notes, and sites like this one. It the "harder" is too hard for you, start with the "easy" one but if you are really interested you will have to tackle the "hard" one (and much, much harder stuff) pretty soon.

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"Mathematics is no more computation than typing is literature." - John Allen Paulos

There are two things that alienate people from math: 1. mindless computation. 2. obscure proofs. While both are important, the emphasis must be on discovery, exploration, and understanding. Oliver Heavenside once said, "I don't reject my food because I don't understand digestion," when defending his use of certain mathematical ideas he didn't fully understand. Exploration is a prerequisite to proofs. To extend the analogy; How can we determine what is happening in digestion if we are unaware the stomach exists? Few things help people understand better than Geometry, Symmetry, and Patterns. Rigorous math requires motivation which for me occurs when the context and importance are apparent. I agree with your assessment that the "easier" book is better since it will likely provide more opportunity to contextualize and thereby motivate you in some direction that may be far removed from the proofs you would learn in the "more advanced" book.

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text:
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# $$\S1.2$$ Structure and Formulations of Physical Theories

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Document type:

1. An Overview of Classical Concepts
Introduction;Thought experiments with waves;Thought experiments with bullets;
Waves versus particles.

2. Structure and Formulations of Physical Theories
Structure of physical theories; Formulations of classical mechanics;Thermodynamics and
statistical mechanics.

### Structure of Physical Theories

Classical theory of a physical system consists of components listed below.

1. States of the physical system; `Co-ordinates
2. Dynamical Variables
3. Laws of Motion
4. Forces, Interactions

At first we shall briefly explain each item in the table and later discuss them by means  of examples of different physical systems.

Physical states
By state of a physical system one means ways of specifying complete information about the system.

Dynamical variables
The dynamical variables of a classical system are functions of state of the system and can be computed when the state has been specified.

Laws of motion
Not only we are interested in knowing about a system at  a given time, we also want to know how the system changes with time. In order to describe behaviour of a system under time evolution one needs to know the laws of motion. Several different forms of the laws of motion are available for mechanical systems.

1. Newton's laws
2. Lagrangian equations of motion
3. Hamiltons equations
4. Poisson bracket formalism

When applicable, all the above formalism are equivalent.

Interactions
A classical description is completed by specifying the forces of the interactions of the system. It should be remarked that while the laws of motion are general and are applicable to a wide variety of physical systems, the nature of forces or the explicit form of interactions  differs from system to system. The interactions are specified by by giving explicit expressions for forces acting on the bodies in the system.

Classical Systems
Some examples of classical systems of interest are

1. System of point particles moving in a force field.
2. One or more point particle moving on a surface of a sphere.
3. Rigid body
4. Vibrating string or a spring
5. Electromagnetic waves
6. Charged particles interacting with  electromagnetic fields

System of point particles

You are all familiar with the newtonian mechanics from your school days. A complete specification of state of a particle requires three position coordinates and three velocities. The dynamical variables are experimentally measurable quantities such as energy, momentum and angular momentum. They are functions of coordinates and velocities.

The equations of motion are given by the second law. For a system of point particles, the behaviour of each particle is governed by the equation of motion
$$m_\alpha \DD[\vec{r}_\alpha]{t} = \vec{F}_\alpha \Label{EQ01}$$
Note that the \EOM {}  are a set of second order  differential equations in time. Therefore one needs to know the values of position and velocities at a time in order $t_0$ to be able to predict the state of the system at a later time. One also needs to have information about all the forces acting on the particle. The Newton's laws require that the equations of motion be set up using the Cartesian coordinates to describe the particle. For a system consisting of several particles one needs to knows all the forces, including the forces of constraint. In order to set up equations of motion in a non Cartesian system of coordinates one has to start from the Cartesian system and take into account of the constraints. In general finding solution may require a change variables from Cartesian coordinates to a new set of coordinates. Thus, for example, for a bead sliding on a sphere one should change from Cartesian coordinates to polar coordinates.

Waves

The state of a vibrating string is described completely by specifying the displacement and velocity of the string at each point. The vibrations are also governed by the Newton’s Laws which can be used to derive the wave equation giving the propagation of waves in a medium.

The systems consisting of charged particles interacting with electromagnetic fields are very important. These are governed by the Maxwell’s equations and the Lorentz force equation. The state is described by specifying position and momenta of the charged particles and the electromagnetic fields, or the scalar and vector potentials, at all points in the space.

### Formulations of Classical Mechanics

The Newtonian formulation has limitations which make it unsuitable for description of several physical systems. Many different formalisms exist which generalise the Newtonian formalism. We mention a few of these here which are useful for systems with a few degrees of freedom. Apart from Newtonian mechanics, other formulations of mechanics are

1. Lagrangian formulation
2. Hamiltonian and Poisson brackets
3. Hamilton Jacobi formulation

Each of the above formalism will be described briefly.

Lagrangian form of classical dynamics

In the Lagrangian approach the state of a system is described by a set of generalized coordinates and velocities. The generalized coordinates are not restricted to be Cartesian. They are a set of independent variables $$q_k$$ needed to specify the system completely. The knowledge of these variables $$q_k$$, called generalized coordinates, and their time derivatives allows us to compute all dynamical variables of the system. The dynamical laws or the equations of motion are given in terms of a single function of generalized coordinates and momenta, ${\cal L}(q,\dot{q}, t)$, called Lagrangian of the system. Knowing the Lagrangian, the equations of motion are given by
$$\dd{t}\left( \pp[ \cal L]{\dot{q_k} } \right) - \pp[\cal L] {q_k} = 0, \qquad k=1,2,\ldots \label{A:EQ01}$$
The Lagrangian formalism offers distinct advantages over the Newtonian formalism.

Hamiltonian form of classical dynamics
In the Hamiltonian approach to the classical mechanics, the state of a system at time $t$ is described by giving the values of generalised coordinates and momenta $q_k, p_k, (k=1,\ldots,n)$ at that time. The canonical momentum $p_k$ is defined as derivative of the Lagrangian of the system  w.r.t.  the generalised velocity $\dot{q}_k$:
$$p_k = \pp[L]{\dot{q}_k} \Label{EQ02}.$$

The interaction is specified by giving Hamiltonian $H(q,p)$ which determines the equations of motion. The equations of motion in the Hamiltonian approach take the form
$$\dot{q}_k = \pp[H]{p_k}, \qquad \dot{p}_k=-\pp[H]{q_k}, \qquad k=1,\ldots,n. \Label{EQ03}$$

Poisson bracket formalism

For two functions $F(q,p), G(q,p)$ of canonical variables, the Poisson bracket $[F,G]_\text{PB}$ is defined as
$$[F,G]_\text{PB} = \sum_k \left( \pp[F]{q_k} \pp[G]{p_k} - \pp[F]{p_k} \pp[G]{q_k} \right). \Label{EQ04}$$
The Hamilton's equations,\eqref{EQ03}, written in terms of Poisson brackets assume the form
$$\dot{q}_k = [q_k,H]_\text{PB}, \qquad \dot{p}_k=[p_k,H]_\text{PB}.\Label{EQ05}$$
In general the time evolution of any dynamical variable is given by
$$\dd[F]{t} = [F,H]_\text{PB}. \Label{EQ06}$$
The classical mechanics has been formulated in several different ways. We mention the Newtonian, the Lagrangian, the Hamiltonian and the Poisson bracket formulations. The Hamiltonian form of mechanics turns out to be the most convenient and suitable for  making a transition to quantum mechanics; the Schrodinger and Heisenberg formulations of quantum mechanics requiring an understanding of the Hamiltonian and Poisson bracket formulations. Frequently it is asked if Lagrangian formulation has a role in the quantum theory ? The answer is in affirmative and the Lagrangian plays  an essential role in the Feynman path integral approach to quantum mechanics.

### Thermodynamics and Statistical Mechanics

For systems   consisting of a large number of particles, such as gases, the classical mechanics, in the form used for point particles, , is not very useful. One needs to use statistical methods. While thermodynamics and statistical mechanics were successful in describing the behaviour of a large number of systems very closely, there were some notable disagreements with experiments.

When it comes to the structure of thermodynamics and statistical mechanics, they too have the same structure that is outlined above for mechanics. Except that most of the time we talk about equilibrium situations. Therefore, we do not quite discuss time evolution. A discussion of time evolution  of statistical systems falls under the non-equilibrium statistical mechanics and that is very hard subject.

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# A captious work problem: same paths but same forces? [closed]

A man jumps onto a chair. A man climbs onto a chair by putting a leg first and then the other.

In both cases, the work has been the same. TRUE or FALSE...?

Spoiler!: The path is the same, so the change in potential energy is the same. But Work equals potential energy only if there are not non-conservative forces and how can you tell if there are not? Or more simple, how do you know that the change in kinetic energy is the same in both cases?

• Think about what non-conservative forces might act on the man in each case. Are these forces likely to be at all significant compared to the work done against gravity? – tok3rat0r May 14 '15 at 22:25
• I could manage to finish with almost exactly the same velocity in the second case and that would give me 0 work. But if I jump, the change in kinetic energy would be more significant. That's what I'm thinking. – cla May 14 '15 at 22:35
• You said yourself: "The path is the same, so the change in potential energy is the same. But work equals potential energy [if there are no non-conservative forces]". So does the change in kinetic energy matter? In both cases, the man starts with zero velocity on the ground and ends with zero velocity on the chair. Does the change in speed used to get there make any difference (if we can neglect non-conservative forces like friction and air resistance)? – tok3rat0r May 14 '15 at 22:43
• I was thinking that when you jump you land on the chair with a non zero velocity. But maybe considering that was wrong. And my problem was not friction and air resistance, but the forces the man does, wich are completly dissipative. Maybe the aim of the problem was just about the path independence of weight force as I first thought, but I wanted to make sure it wasn't captious. Thanks for the discution! – cla May 14 '15 at 23:07
• (Of course, when jumping onto the chair the man first has to bend his legs to store energy in his muscles ready for the jump. This energy is turned into kinetic energy when he jumps, which is turned into potential energy as he rises to the level of the chair. One could argue that because the process of converting chemical energy into kinetic energy in the muscles is not 100% efficient, he uses slightly more energy when he jumps. But this would be a pretty small effect by any reasonable standard, and in any case it adds nothing to the 'mechanical work' as the term is properly defined) – tok3rat0r May 14 '15 at 23:27

## 3 Answers

The answer depends on how particular you want to be about it. Upon first examination you might say that in the Earth's conservative gravitational field, that the work must be the same, since the start and endpoints are the same and we have ignored any non-conservative forces (e.g. friction, air resistance, etc.). This is more than likely the response that would be expected on an assignment in most classes. One could argue however that if air resistances are taken into account, that more work is done in the jumping case, since one has to combat the higher resistive forces of the air.

I'll answer the question for completeness. As discussed in the comments, if you think about all the posible ways of dissipating energy, maybe you will doubt of the veracity of the assertion made in the exercice. But, the clue is not to think in if the teacher wanted to be "captious"... but that in this case all these ways of dissipating are really small effects and means nothing in comparison to the mechanical energy, and so it is irrelevant for the purpose of the exercice.

Work is the change in kinetic energy. In both cases, the man starts on the ground at rest and end on the chair at rest. In both cases, the net work is zero.

If you want to be more specific and describe the work due to his muscles, that positive work must exactly offset the negative work due to other sources. Gravity does negative work equal to minus the change in gravitational potential energy of the man-earth system, so it doesn't depend on the path. Air drag does some tiny amount of negative work which does depend on the path. That's probably pretty small either way, so it's probably okay to ignore it.

• Work $W=mgh$ in a gravitational field, and there is only Potential Energy stored, no Kinetic Energy, at the end of doing work against a conservative force. The net work done by the man is positive. A gravitational field is a conservative force, and work done against a gravitational field alone leaves no remnant of Kinetic Energy. The energy expended to move his muscles will at least equal the Potential Energy stored in the Gravitational Field. And yes, there will be non-conservative forces acting within the muscles, and on air, which will produce random Kinetic Energy/Thermal Energy. – Thomas Lee Abshier ND Jun 22 '18 at 2:38


text:
2 1 2
Chapter 6
shall see in Example 6.3 that there is an easier way to calculate the flux control coefficients, but
from the calculations above we can already draw the main conclusion that
( J
decreases while Cf
increases when x increases. Since
J(x)
is a monotonous function of
x
so must also In
J(x)
be a
monotonous function of In x and consequently
dJ
/(:,
8J' x
dlnJ'
1
8k{
J
dx J'
d\nx
must decrease monotonously to zero when x (or
k f
increases to infinity. Since CJ
*-
= 1
the
value of
Cf2
must correspondingly increase towards 1.
The conclusion is that the control shifts from the first reaction step to the second for increasing
activity of the first enzyme. The rate of change varies with the environmental conditions as was
seen for the case of increasing 5. For large
s
the concentration of the first enzyme (represented
by the parameter
k f
must be relatively large to change the relative size of C1, and C,2 since the
competitive inhibition of the first enzyme by
St
is less pronounced when ,v is large.
Having introduced the flux control coefficients and explained their significance by means of a
simple example we can now turn to a general unbranched pathway with I reaction steps:
S
S, S S2 .
......
S , , f t
P
(6.32)
There are 7 - 1
intermediates 5 ,.
...........
S
, , and
I
pathway reactions mediated by the enzymes
£ ,.
.........
In the steady state the rates of all the pathway reactions
r,
are equal and equal to the
flux
J
through the pathway. We wish to calculate the flux J of substrate of known concentration
that can be processed by the pathway, and we wish to investigate how the pathway architecture
can be improved to allow a larger flux to be processed.
The flux control coefficients C1
, are defined by
c / = £ L a /
J de,
(6.33)
In Note 6.3 it is shown that irrespective of the kinetics of the individual enzyme catalyzed
reactions the sum of the flux control coefficients is 1
for the unbranched pathway.
I C / = 1
(6.34)
i =
l
This result is named the
flux control summation theorem
N ote 6.3 Proof o f the flux-control sum m ation theorem
If in a thought experim ent w e increase all the enzym e activities sim ultaneously by the sam e fractional
am ount
a,
i.e.
.


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# RIntro / RIntroBrief.Rnw

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 \documentclass{article} \title{A Very Brief Introduction to R} \author{Sean Davis} \date{June 21, 2013} \begin{document} \maketitle \section{Vectors} <<>>= x = 1 x y = rep('a',10) y z = rnorm(15) z z<0 z[z<0] @ \section{Matrices} <<>>= m = matrix(rnorm(100),ncol=5) ncol(m) nrow(m) dim(m) summary(m) # first column of m m[,1] # first row of m m[1,] # get a big matrix from R data(volcano) dim(volcano) @ \section{Factors} Remember that factors are used for storing categorical variables. They often look like character vectors, but they do not have quotes'' when R prints them. Also, they behave more like numeric vectors. <<>>= cvec = rep(c('a','b'),each=10) cvec f = factor(cvec) f levels(f) # we can change the levels of the factor levels(f) = c('d','e') f as.character(f) @ \section{Data Frames} <<>>= df = data.frame(m,f) dim(df) head(df) colnames(df) df$X1 summary(df) @ \section{Plotting} This is where your curiosity and experimentation come in handy. R has many plotting capabilities and some trial and error may be necessary to get what you want from R graphics. <>= plot(df[,1],df[,2],col=df$f,pch=2,main='A simple plot') @ \section{Package installation} <>= source('http://bioconductor.org/biocLite.R') library(BiocInstaller) biocLite(c('genefilter','qvalue')) @ \subsection{Getting Help on a Package} <>= help(package='genefilter') @ \end{document}


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# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Example 5.3.2.4 (Constant Diagrams). Let $\operatorname{\mathcal{C}}$ be a category, let $X$ be a simplicial set, and let $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Set_{\Delta }}$ be the constant diagram taking the value $X$. Combining Remark 5.3.2.3 with Example 5.3.2.2, we obtain a canonical isomorphism of simplicial sets $\underset { \longrightarrow }{\mathrm{holim}}( \mathscr {F} ) \simeq \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \times X$. In particular, if $X = \Delta ^0$, then the projection map $\underset { \longrightarrow }{\mathrm{holim}}( \mathscr {F} ) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ is an isomorphism.


text:
Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# Which is considered a stronger lewis acid, BF3 or BCl3?

"BCl"_3 is the stronger Lewis acid.

We would expect "BF"_3 to be stronger, because "F" is more electronegative than "Cl".

Chemists explain this unexpected result by an electronic argument and a steric argument.

The electronic argument — backbonding

The boron atom in "BF"_3 is sp^2 hybridized, with a vacant 2p orbital.

The "F" atoms can also be sp^2 hybridised, with lone pairs in their 2p orbitals.

These "F" orbitals can overlap with the orbital on "B", thereby increasing the electron density on the boron atom and making it less acidic.

This effect is called backbonding, because electron density is leaving the more electronegative atom.

In "BCl"_3, the 3p orbitals on "Cl" are bigger than the 2p orbital on "B", so orbital overlap is less efficient, and backbonding is less important.

Hence, the greater backbonding in "BF"_3 makes it a weaker Lewis acid.

The steric argument — ligand close-packing (LCP)

The LCP model is based on the observation that the "X" atoms (ligands) in "AX"_n systems are always the same distance from each other.

For example, the distance between the "F" atoms in "BF"_3 and "BF"_4^- is 226 pm, despite the longer "B-F" distance in the tetrahedral structure.

(from alpha.chem.umb.edu)

It is as if the "F" atoms are closest-packed (like in a crystal), with the "F" atoms having a ligand radius of 113 pm.

When the "BF"_3 forms a Lewis complex, the "F" atoms remain close-packed, but the "B-F" bonds must become longer in the new tetrahedral geometry.

It takes more energy to lengthen the short, strong "B-F" bonds than the longer, weaker "B-Cl" bonds.

Hence "BF"_3 is a weaker Lewis acid than "BCl"_3.


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# What's so special about the group axioms?

I've only just begun studying group theory (up to Lagrange) following on from vector spaces and I am still finding them almost frustratingly arbitrary. I'm not sure what exactly it is about the axioms that motivated them defining groups.

My textbook asks you to list 'common features' of vector spaces and then later defines a set of axioms for vector spaces under addition, scalar multiplication and both, noting that the axioms under addition form an Abelian group. So are groups just a generalisation of vector spaces under any binary operation?

My main problem is that the book notes that 'the axioms may seem rather arbitrary' and links groups with vector spaces but doesn't elaborate.

When introducing groups, you are tasked to complete Cayley tables for the symmetries of an equilateral triangle and square. Then, similarly to the delivery of vector spaces, notes that the tables have common properties (Closure, Identity, inverse and association) and defines a group as a set of elements under a binary operation that has these features.

So what's so important about these 4 properties? For example, if 1 or 2 the properties were excluded form the axioms, or we added an extra few properties as axioms how would that cripple the effectiveness of groups?

Are the group axioms ever difficult to work with or do they always work, forgive the crude Littlewood analogy, like a mathematical skeleton key?

What is it about these 4 properties that make groups such a powerful tool in mathematics and physics?

My best guess is that a group is the best way to express our sense of symmetry and what is symmetric mathematically, but I would prefer some elaboration.

• It is about defining things so we know what we are working with faster, generalising things that are equivalent and being able to prove these things generally. What textbook are you working with? It doesn't seem very good(from what little I have read, and I may be entirely wrong[about the textbook]).
– user142198
Feb 23 '15 at 0:17
• Note that objects obeying only a subset of the group axioms are studied: see for instance monoids, semigroups, etc. You are right that symmetries have group structure, and that the importance of symmetry in geometry and physics motivates the study of group theory. Feb 23 '15 at 0:18
• There is nothing special about the group axoims, other than that there are a lot of common systems that satisfy them. By studying them, we can learn about all the systems at once. Feb 23 '15 at 3:41
• @GridleyQuayle mathematicians before the 19th, indeed, the late 19th century, simply did not abstract examples into axiomatic systems in the way do today. This was a century in which there was constant dissension against the idea of a 4-dimensional space, let alone an $n$-dimensional; only visionaries like Grassmann could even conceive of an axiomatization as natural as that of vector spaces. Feb 23 '15 at 4:41
• @Shaun: The distinction between groups and inverse semigroups depending on whether they are distance-preserving or not in your first comment lacks a bit of context (I guess one finds it in the book you linked to).
– j.p.
Feb 23 '15 at 12:51

Yes, groups express symmetries of objects. If you want a symmetry to be a way to map an object into itself which preserves some property (say, location of vertices on a square, or distance between points in the plane,) and you want symmetries to be things that you can (a) chain together and (b) undo, then you've got a group. (As long as you also admit the trivial symmetry.) In this picture of symmetries as mappings, you don't really have to specify associativity, as it's a natural aspect of anything that looks like composition of functions (more precisely, of anything which fits into a category.)

You shouldn't think that groups are just generalizations of vector spaces. Anything that's a generalization from one case is a bad generalization! Groups are also generalizations of numbers (under addition and, in some cases, multiplication), symmetries of geometric objects (both discrete ones you mention and continuous ones,) and functions, under not only addition and multiplication but, critically, composition. An important historical example was the group of permutations of roots of a polynomial, which is used in the proof that quintic equations can't be solved in radicals and has led to huge areas of modern math.

As for loosening and strengthening some of the axioms: if you throw out inverses, you get monoids (semigroups, if you also throw out the identity.) These are extremely important objects in their own right, but they're too general to have a nice structure theory. The biggest problem is with quotients: surjective homomorphisms $G\to H$ naturally correspond to "normal subgroups" $K\subset G$, i.e. subgroups with $gKg^{-1}\subset K$ for all $g\in G$. There's no such simple way to characterize quotients $M\to N$ of monoids in terms of submonoids, because there's not necessarily a $g^{-1}$ for every $g$, so the most basic isomorphism theorems, which lay the foundation for everything you're likely to learn in elementary group theory, are no longer true. Thus we like to use groups instead of monoids when we can, and many actual things in the world come as groups, so we do so.

Strengthening the axioms does not "cripple the effectiveness" of groups so much as weakening them does. Various objects subject to axioms extending those of a group include abelian groups, vector spaces, modules, rings and algebras over rings, topological groups, Lie groups, and many more, and all of these are of great importance. But there are groups that are none of these things, (every group is in some sense topological, but we don't always want to think about the topology) so we start with just a plain group and strengthen the axioms in appropriate contexts.

• Regarding the third paragraph, algebras like groups and rings, whose congruences are determined by a single congruence class, are called ideal-determined. There is an interesting discussion of this and related properties in the paper referenced in this answer. Feb 23 '15 at 2:36
• @BillDubuque, thanks very much for the information! Feb 23 '15 at 4:35
• Great answer. My summary: the group axioms are sufficient to provide a rich structure but simple enough to have (very) wide applicability. Feb 23 '15 at 4:39
• More about loosening the axioms: If you discard associativity from the axioms you get loops, and if you in addition discard the identity, you get quasigroups. There's also a broad field dedicated to these concepts. But as someone who works in loop theory, I have to admit that life without associativity is tough!! Feb 23 '15 at 10:49
• In the last paragraph, when you say, "every group is in some sense topological", do you mean because you can equip the discrete topology to any group, or is there something deeper? Aug 28 '18 at 2:07

At some level, these features are arbitrary - mathematicians study many objects which have more or less axioms than a group. They study abelian groups (adding the axiom $ab=ba$), monoids (which don't need inverses), loops (which don't need to be associative), and other objects. Groups, from a purely mathematical point of view, are fairly well-behaved objects - we don't need to put parenthesis everywhere (associativity), we're allowed to make cancellations (inverses, identity), and there's no magical, "oops, multiplying those two gives something entirely different" (closure). At some level, groups are interesting as a pretty general object for which we get many interesting results (like Lagrange's theorem or the classification of simple groups) which don't hold when we start to take away axioms, and adding more axioms does necessarily give us a lot of extra theorems.

As far as applications go, groups really are good at capturing the notion of symmetry. A concept from category theory that is relevant is the group of automorphisms of an object. What does this mean? Well, to put it roughly, an automorphism is a function which completely preserves every quality that we care about - like the symmetries of a polygon (or polyhedron) might move all the points in the ambient space, but preserve the actual polygon. We might also think about how many physical systems are translation or rotation or reflection invariant (or even time-reversal-invariance in Newtonian physics), meaning that those functions preserve all the laws of motion. And it happens that the structure of automorphisms (with respect to composition) forms a group - it is closed, since if $f$ and $g$ preserve all the desired properties, then doing $f$ then $g$ must also preserve everything desired. Moreover, every automorphism must have an inverse (and this is in fact the definition of an automorphism), as if it did not, we could conclude that some property was lost by the morphism. Associativity follows from the fact that function composition is associative, and identity follows from the fact that the identity morphism, which changes nothing, is obviously an automorphism. Outside of group theory, which studies groups generally for the sake of studying groups, this tends to be why groups arise so much - you want to know about the symmetries of an object? Well, you're looking at a group! (And, indeed, all groups can be interpreted as a set of functions closed under composition - this is Cayley's theorem)

Don't think of a group as a abstracted vector space. When you drop commutativity, you end up with objects that behave very differently from vector spaces.

The heuristic that I use is that if an operation is not commutative, then the operation should be thought of as function composition. If it is commutative, then it should be thought of multiplication, like numbers. So if we restrict to abelian groups, it turns out that they actually do behave a lot like vector spaces. Indeed, abelian groups are modules over the ring $\mathbb Z$, and vector spaces are modules over a field $k$ (usually $\mathbb R$ or $\mathbb C$). I like to think of abelian groups sort of as counting numbers, but where you can consider torsion. A commutative ring, to me, is a collection of functions, where the multiplication operation is multiplication of the outputs (think of $k[x]$, the set of all polynomials with coefficients in a field $k$, or $\mathbb Z$). A non-commutative ring is a collection of functions, where the multiplication is function composition (think $M_{n\times n}$, the collection of all $n\times n$ matricies, or the collection of all linear transformations from a vector space of dimension $n$ with basis to itself).

So if groups are like composition of morphisms (eg. functions), but the morphisms have to be invertable, then we are thinking of a collection of isomorphisms (eg. bijections). Indeed, if we look at a category (eg. all vector spaces and linear transformations between them, or groups and group homomorphisms, or sets and functions), and we look at all of the isomorphisms of an object, then that will satisfy the axioms of a group. In fact, an alternative way of describing a group is as a category with one object, and all isomorphisms are isomorphisms. To me, this is the definition of a group, because it tells you exactly what a group is for. And what is an isomorphism? It is a way of transforming an object, such that you can always reverse that transformation. A symmetry.

To cure the feeling of arbitrariness, it might be good to have examples. So we already have $Aut(X)$, the collection of automorphisms on $X$ (isomorphisms from $X$ to itself), and from that we get the symmetric group $S_n$ (automorphisms (or permutations) of a set with $n$ elements), and $GL_n(k)$, which is $Aut(V)$, where $V$ is an $n$-dimenstional $k$-vector space. Then we have useful subsets of automorphisms, like $SL_n(k)$, invertible matrices with determinant 1, and $A_n$, the collection of even permutations on a set with $n$ elements. We also have for a field extension $K/k$ (think of taking a field k and adding in some elements to get a bigger field, like take $\mathbb R$ and add in the solution to the equation $x^2 + 1 = 0$ to get $\mathbb C$) we have the automorphisms that fix the underlying field $k$, $Aut_k(K)$, which comes up in Galois theory, where you are looking at permutations of roots of polynomial equations (like complex conjugation will swap the roots of the above polynomial while keeping $\mathbb R$ fixed).

I do sympathize with you. Group theory really shouldn't be taught first in a course in abstract algebra, because all the important examples of groups come when you've seen a bunch more theory, so the whole thing comes off as fairly unmotivated. If you read Visual Group Theory by Carter, you might be able to get some motivation, especially if you look at the last part on Galois theory. Galois theory is really powerful, and solves all sorts of interesting problems that were long unsolved before the invention of Galois Theory. Problems like the impossibility of squaring a circle or of trisecting an angle with straightedge and compass, and of a formula for solving quintic equations using only radicals (as in, a quadratic equation, but for degree 5 polynomials). This last one vitally uses properties of the group $S_5$.