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2.2: Derived Units

In addition to using some alternative basic units for length, time, mass and electric charge, we make use of many derived units in the SI system and a few are listed in Table $$\PageIndex{1}$$. Some derived units are sufficiently notorious so that they are named after famous scientists and represented by specific symbols. For example, the unit of kinematic viscosity is the stokes (St), after the British mathematician, Sir George G. Stokes (Rouse and Ince)7, while the equally important molecular and thermal diffusivities are known only by their generic names and represented by a variety of symbols. The key point to remember concerning units is that the basic units represented in Table $$2.1.1$$ are sufficient to describe all physical phenomena, while the alternate units illustrated Table $$2.1.2$$ and the derived units listed in Table $$\PageIndex{1}$$ are used as a matter of convenience.

Table $$\PageIndex{1}$$: Derived SI Units
Physical Quantity Unit (Symbol) Definition
force newton (N) $$kg$$ $$m/s^{ 2}$$
energy joule (J) $$kg$$ $$m^{2}/s^{ 2}$$
power watt (W) J/s
electrical potential volt (V) J/(A s)
electric resistance ohm ($$\Omega$$) V/A
frequency hertz (Hz) cycle/second
pressure pascal (Pa) $${ N/m}^{ 2}$$
kinematic viscosity stokes (St) $${ cm}^{ 2} /s$$
thermal diffusivity square meter/second $${ m}^{ 2} /s$$
molecular diffusivity square meter/second $${ m}^{ 2} /s$$

While the existence of derived units is simply a matter of convenience, this convenience sometimes leads to confusion. This is particularly true in the case of Newton’s second law which can be stated as

$\left\{\begin{array}{c} {\text{ force acting}} \\ {\text{ on a body}} \end{array}\right\}=\left\{\begin{array}{c} {\text{ time rate of change}} \\ {\text{ of linear momentum}} \\ {\text{ of the body}} \end{array}\right\} \label{2-5}$

In terms of mathematical symbols, we express this axiom as

$\mathbf{f}=\frac{d}{dt} \left(m\mathbf{v}\right) \label{2-6}$

Here we adopt a nomenclature in which a lower case, boldface Roman font is used to represent vectors such as the force and the velocity. Force and velocity are quantities that have both magnitude and direction and we need a special notation to remind us of these characteristics.

Let us now think about the use of Equation \ref{2-6} to calculate the force required to accelerate a mass of 7 kg at a rate of 13 $$m/s^2$$. From Equation \ref{2-6} we determine the magnitude of this force to be

$f=(7 \ kg )(13 \ m/s^{2})= 91 \ kg \ m/s^{2} \label{2-7}$

where $$f$$ is used to represent the magnitude of the vector $$\mathbf{f}$$. Note that the force is expressed in terms of three of the four fundamental standards of measure, i.e., mass, length and time. There is no real need to go beyond Equation \ref{2-7} in our description of force; however, our intuitive knowledge of force is rather different from our knowledge of mass, length and time. Consider for example, pushing against a wall with a “force” of 91 $$kg$$ $$m/s^2$$. This is simply not a satisfactory description of the event. What we want here is a unit that describes the physical nature of the event, and we obtain this unit by defining a unit of force as

$1 \text{ newton} = \left(1 \ kg \right)\times \left(1 \ m/s^{2} \right) \label{2-8}$

When pushing against the wall with a force of 91 $$kg$$ $$m/s^2$$ we feel comfortable describing the event as

\begin{align}\text{ force} &= 91 \ kg \ m/s^{2} \nonumber\\[4pt] &= \left(91 \ kg \ m/s^{2} \right) \times \left(1\right) \nonumber\\[4pt] &= \left(91 \ kg \ m/s^{2} \right) \times \left(\frac{1 \text{ newton}}{ kg \ m/s^{2}} \right) \nonumber\\[4pt] &= 91 \text{ newtons} \label{9} \end{align}

Here we have arranged Equation \ref{2-8} in the form

$1=\frac{\text{ newton}}{kg \ m/s^{2}} =\frac{ N}{\ kg \ m/s^{2}} \label{10}$

and multiplied the quantity 91 $$kg$$ $$m/s^2$$ by one in order to effect the change in units. Note that in our definition of the unit of force we have made use of a one-to-one correspondence represented by Equation \ref{2-8}. This is a characteristic of the S.I. system and it is certainly one of its attractive features. One must keep in mind that Equation \ref{2-8} is nothing more than a definition and if one wished it could be replaced by the alternate definition given by8

$1 \text{ euler} = \left(17.07 \ kg\right) \times \left(1 \ m/s^{2} \right) \label{11}$

However, there is nothing to be gained from this second definition and it is clearly less attractive than that given by Equation \ref{2-8}.

In the British system of units the one-to-one correspondence is often lost and confusion results. In the British system we choose our standards of mass, length and time as

Table $$\PageIndex{2}$$

 mass $$lb_m$$ length ft time s

and we define the pound-force according to

$1 \ lb_f =\left( 1 \ lb_m \right) \times \left(32.17 \ ft/s^{2} \right) \label{2-12}$

This definition was chosen so that mass and force would be numerically equivalent when the mass was acted upon by the earth’s gravitational field. While this may be a convenience under certain circumstances, the definition of a unit of force given by Equation \ref{2-12} is certainly less attractive than that given by Equation \ref{2-8}.

In summary, we note that there are only four standards needed to assign a numerical values to all observables, and the choice of standard is arbitrary, i.e. the standard of length could be a foot, an inch or a centimeter. Once the standard is chosen, i.e. the meter is the standard of length, other alternate units can be constructed such as those listed in Table $$2.1.2$$. In addition to a variety of alternate units for mass, length and time, we construct for our own convenience a series of derived units. Some of the derived units for the SI system are given in Table $$\PageIndex{1}$$. Finally, we find it convenient to tabulate conversion factors for the derived units for the various different systems of units and some of these are listed in Table $$\PageIndex{3}$$. An interesting history of the SI system is available from the Bureau International des Poids et Mesures9.

 Length Mass 1 inch (in) = 2.54 centimeter (cm) 1 pound mass ($$lb_m$$) = 453.6 gram (g) 1 angstrom $$(\mathop{\mathrm{ A}}\limits^{o}) = 10^{-8}$$ centimeter (cm) 1 g = $$10^{-3}$$ kilogram (kg) 1 foot (ft) = 0.3048 meter (m) 1 ton (short) = 2000 $$lb_m$$ 1 mile (mi) = 1.609 kilometer (km) 1 ton (long) = 2240 $$lb_m$$ 1 yard (yd) = 0.914 meter (m) 1 nanometer (nm) = $$10^{-9}$$ meter (m) Time 1 minute (min) = 60 second (s) Force 1 hour (h) = 60 minute (min) pound-force ($$lb_f$$) = 32.17 ft $$lb_m$$ $$s^{-2}$$ 1 day = 24 hours (h) dyne (dyn) = g cm $$s^{-2}$$ 1 newton (N) = $$10^{5}$$ dyne (dyn) Density 1 dyne (dyn) $$= 2.248 \times 10^{-6}$$ pound force ($$lb_f$$) 1 $$g/cm^{3} = 10^{3}$$ $$kg/m^{3}$$ 1 poundal = $$3.108 \times 10^{-2}$$ $$lb_f$$ 1 $$lb_m/ft^{3} = 16.018$$ $$kg/m^3$$ 1 pound-force ($$lb_f$$) = 4.448 newton (N) 1 $$lb_m/$$gal (US) = $$119.83 \ kg/m^{3}$$ Volume Pressure 1 $$in^3$$ = 16.39 $$cm^3$$ 1 atmosphere (atm) = 14.7 $$lb_f$$ /in$$^2$$ 1 $$ft^{3} = 2.83 \times 10^{-2}$$ $$m^{3}$$ 1 $$lb_f/in^{ 2} = 6.89 \times 10^{ 3}$$ $$N/m^{ 2}$$ 1 gallon (US) (gal) = 231 $$in^3$$ 1 atm = $$1.013 \times 10^{ 6}$$ dyne/$$cm^{ 2}$$ 1 quart (liquid) (qt) = 0.25 gal (US) 1 pascal (Pa) = 1 N/m$$^{ 2}$$ 1 barrel = 31.5 gal (US) 1 atmosphere (atm) = 1.01325 bar 1 gal (US) = 0.003785 $$m^{3}$$ 1 liter (L) = $$10^3$$ $$cm^3$$ Power 1 horsepower (hp) = 745.7 watt Area 1 hp = 42.6 Btu/min 1 $$in^2$$ = 6.452 $$cm^2$$ 1 watt = $$9.51 \times 10^{ -4}$$ Btu/s 1 $$ft^{2} =9.29 \times 10^{-2}$$ $$m^{2}$$ 1 ft $$lb_f/s$$ = 1.356 watt (W) 1 acre $$= 4.35 \times 10^{4}$$ $$ft^{ 2}$$ Temperature 1 C = 1.8 F Energy K = C + 273.16 1 calorie (cal) = 4.186 joule (J) R = F + 459.60 1 British Thermal Unit (Btu) = 252 cal F = 1.8C + 32 1 erg = $$10^{-7}$$ joule (J) 1 Btu = 1055 watt sec Flow Rate 1 ft $$lb_f$$ = 1.356 joule (J) 1 gal (US)/min = gpm = $$6.309 \times 10^{-5}$$ $$m^{ 3}/s$$ 1 $$ft^{ 3}$$/min = $$4.719 \times 10^{-5}$$ $$m^{ 3}/s$$ Viscosity 1 poise (P) = 1 g/cm s Kinematic Viscosity 1 poise (P) = 0.10 N $$s/m^2$$ 1 $$m^{ 2}/s$$ = $$3.875 \times 10^{ 4}$$ $$ft^{ 2}$$/hr 1 poise (P) = $$6.72\times 10^{-2}$$ $$lb_m$$/ft s 1 $$cm^{ 2}/s$$ = $$10^{ -4}$$ $$m^{ 2}/s$$ 1 centipoise (cP) = $$10^{-3}$$ kg/m s 1 stokes (St) = 1 $$cm^{2}/s$$


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Opened 7 years ago

Closed 7 years ago

# [[BibRef]] macro produce error

Reported by: Owned by: anonymous Roman Mohr normal TracBibPlugin normal 0.10

### Description

When I tried to render the citations using [[BibRef]] per the example, I got the following error:

Error: Macro BibRef(None) failed
expected string or buffer

Here's an example of my bibtex entry, just in case it's relevant:

@article{Quir94,
Adsnote = {Provided by the SAO/NASA Astrophysics Data System},
Author = {{Quirk}, J.~J.},
Date-Modified = {2010-05-04 10:41:10 -0400},
Doi = {10.1002/fld.1650180603},
Journal = {International Journal for Numerical Methods in Fluids},
Month = mar,
Pages = {555-574},
Title = {{A contribution to the great Riemann solver debate}},
Volume = 18,
Year = 1994,
Bdsk-Url-1 = {http://dx.doi.org/10.1002/fld.1650180603}}

and this is how I reference it :

[[BibAdd(source:trunk/hydro.bib)]]

Let's cite [[BibCite(Quir94)]].

### comment:1 Changed 7 years ago by Roman Mohr

Status: new → assigned

OK, FINALLY I found some time to look at this error. The funny thing is, that when the file is attached to the page, everything works perfect. If a file "source:trunk/somefile.bib" is loaded everything works, but if your file is loaded from the repository the macro can't load the file. I'm very curious what's responsible for that error.

### comment:2 Changed 7 years ago by Roman Mohr

Resolution: → fixed assigned → closed

(In [9945]) don't call close() on the repository. Fixes #8024

### comment:3 Changed 7 years ago by Roman Mohr

In http://trac.edgewall.org/ticket/4972#comment:5 it is expliticly said:

You should never call close() by yourself, unless you've created the Repository instance. ![...]

... but I did.

Fixed in r9945

### Modify Ticket

Change Properties

format: plain


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# From Trees to General Graphs: Counting Independent Sets up to the Tree Threshold

 COMPUTER SCIENCE/DISCRETE MATH I Topic: From Trees to General Graphs: Counting Independent Sets up to the Tree Threshold Speaker: Dror Weitz Affiliation: DIMACS Date: Monday, January 30 Time/Room: 11:15am - 12:15pm/S-101

We present a novel tree representation for the hard-core lattice gas model (weighted independent sets) on a general graph. We use this representation to show that for any graph of maximum degree D, the Gibbs measure is unique (the influence of any boundary condition decays with distance) provided that the activity parameter \lambda < \lambda_c, where \lambda_c is the critical activity for the regular tree of degree D. This resolves an open conjecture in statistical physics. Also, since \lambda_c is known, this extends the known uniqueness regime for many interesting graphs, including the square integer lattice Z^2. Our proof is algorithmic in nature, consisting of an elegant recursive procedure for calculating the probabilities that a given vertex is occupied. This procedure yields an efficient deterministic approximation scheme for counting independent sets of any graph of maximum degree D in the above regime of \lambda. This extends the regime of \lambda for which an efficient approximation scheme is known to exist, and includes the interesting case of \lambda=1 (all independent sets are equally weighted) and maximum degree D=5.


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## Basic College Mathematics (10th Edition)

Angles are Supplementary when their sum is equal to 180$^{\circ}$. In this case one can subtract 5$^{\circ}$ from 180$^{\circ}$ to find the supplement. 180$^{\circ}$-5$^{\circ}$=175$^{\circ}$


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Feeds:
Posts

## The free cocompletion I

Let $C$ be a (locally small) category. Recall that any such category naturally admits a Yoneda embedding

$\displaystyle Y : C \ni c \mapsto \text{Hom}(-, c) \in \widehat{C}$

into its presheaf category $\widehat{C} = [C^{op}, \text{Set}]$ (where we use $[C, D]$ to denote the category of functors $C \to D$). The Yoneda lemma asserts in particular that $Y$ is full and faithful, which justifies calling it an embedding.

When $C$ is in addition assumed to be small, the Yoneda embedding has the following elegant universal property.

Theorem: The Yoneda embedding $Y : C \to \widehat{C}$ exhibits $\widehat{C}$ as the free cocompletion of $C$ in the sense that for any cocomplete category $D$, the restriction functor

$\displaystyle Y^{\ast} : [\widehat{C}, D]_{\text{cocont}} \to [C, D]$

from the category of cocontinuous functors $\widehat{C} \to D$ to the category of functors $C \to D$ is an equivalence. In particular, any functor $C \to D$ extends (uniquely, up to natural isomorphism) to a cocontinuous functor $\widehat{C} \to D$, and all cocontinuous functors $\widehat{C} \to D$ arise this way (up to natural isomorphism).

Colimits should be thought of as a general notion of gluing, so the above should be understood as the claim that $\widehat{C}$ is the category obtained by “freely gluing together” the objects of $C$ in a way dictated by the morphisms. This intuition is important when trying to understand the definition of, among other things, a simplicial set. A simplicial set is by definition a presheaf on a certain category, the simplex category, and the universal property above says that this means simplicial sets are obtained by “freely gluing together” simplices.

In this post we’ll content ourselves with meandering towards a proof of the above result. In a subsequent post we’ll give a sampling of applications.

## How to invent intuitionistic logic

(This is an old post I never got around to finishing. It was originally going to have a second half about pointless topology; the interested reader can consult Vickers’ Topology via Logic on this theme.)

Standard presentations of propositional logic treat the Boolean operators “and,” “or,” and “not” as fundamental (e.g. these are the operators axiomatized by Boolean algebras). But from the point of view of category theory, arguably the most fundamental Boolean operator is “implies,” because it gives a collection of propositions the structure of a category, or more precisely a poset. We can endow the set of propositions with a morphism $p \to q$ whenever $p \Rightarrow q$, and no morphisms otherwise. Then the identity morphisms $\text{id}_p : p \to p$ simply reflect the fact that a proposition always implies itself, while composition of morphisms

$\displaystyle (p \Rightarrow q) \wedge (q \Rightarrow r) \to (p \Rightarrow r)$

is a familiar inference rule (hypothetical syllogism). Since it is possible to define “and,” “or,” and “not” in terms of “implies” in the Boolean setting, we might want to see what happens when we start from the perspective that propositional logic ought to be about certain posets and figure out how to recover the familiar operations from propositional logic by thinking about what their universal properties should be.

It turns out that when we do this, we don’t get ordinary propositional logic back in the sense that the posets we end up identifying are not just the Boolean algebras: instead we’ll get Heyting algebras, and the corresponding notion of logic we’ll get is intuitionistic logic.

## The type system of mathematics

Occasionally I see mathematical questions that seem “grammatically incorrect” in some sense.

Example. “Is $[0, 1]$ open or closed?”
Example. “Is $\{ 1, 2, 3 \}$ a group?”
Example. “What’s the Fourier series of $\sin x + \sin \pi x$?”

Here are some sillier examples.

Example. “Is a rectangle prime?”
Example. “Is $17 \in 3$?”
Example. “What’s the Fourier series of the empty set?”

What all of these examples have in common is that they are type errors: they are attempts to apply some mathematical process to a kind of mathematical object it was never intended to take as input. If you tried to write a program in some highly mathematical programming language to answer these questions, it (hopefully!) wouldn’t compile.

Mathematical objects are usually not explicitly thought of as having types in the same way that objects in a programming language with a type system has types. Ordinary mathematics is supposed to be formalizable within Zermelo-Fraenkel (ZF) set theory, possibly with the axiom of choice, and in ZF every mathematical object is constructed as a set. In that sense they all have the same type. (In particular, the question “is $17 \in 3$?” is perfectly meaningful in ZF! This is one reason not to like ZF as a foundation of mathematics.) However, I think that in practice mathematical objects are implicitly thought of as having types, and that this is a mental habit mathematicians pick up but don’t often talk about.

Instead of thinking in terms of set theory, thinking of mathematical objects as having types allows us to import various useful concepts into mathematics, such as the notions of type safety, typecasting, subtyping, and overloading, that help us make more precise what we mean by a mathematical sentence being “grammatically incorrect.” The rest of this post will be a leisurely discussion of these and other type-based concepts as applied to mathematics in general. There are various categorical ideas here, but for the sake of accessibility we will restrict them to parenthetical remarks.

## Constructions vs. specifications

What does it mean to define a mathematical object?

Roughly speaking, there are two strategies you could use. One is to construct that object from other objects. For example, $\mathbb{R}$ can be constructed from $\mathbb{Q}$ using Cauchy sequences or Dedekind cuts.

Another is to give a list of properties that uniquely specify the object. For example, $\mathbb{R}$ can be defined as the unique Dedekind-complete totally ordered field.

Let’s call these strategies construction and specification. Construction has the advantage of concreteness, and it also proves that the object exists. However, it has the disadvantage that constructions are not unique in general and you might not be working with the most useful construction at a given time; to prove things about an interesting mathematical object you might have to switch between various constructions (while also proving that they all construct the same object).

Relying excessively on construction also has a curious disadvantage when talking to students about the object in question: namely, they might ask you how to show that object $X$ satisfies property $P$, and you might respond “well, it’s obvious if you use construction $A$,” and they might respond “but I was taught that $X$ is [construction $B$]!” The possibility of multiple different constructions of the same mathematical object makes construction somewhat unsatisfying in terms of describing what you even mean when you say “object $X$.”

This is the appeal of specification. Although specification is more abstract and harder to grasp at first, it has the advantage you don’t have to use any particular construction of an object to prove things about it: instead, you should in principle be able to prove everything from the properties alone, since those uniquely specify the object. This often leads to cleaner proofs. Specifications also usually give a better motivation for looking at an object: it’s easier to make sense of the claim that we want a mathematical object with various properties than to make sense of the claim that we want a mathematical object which is constructed in some arbitrary-looking (to the uninitiated) fashion.

Below we’ll go through a few examples of mathematical objects and some constructions and specifications of them.

## Regular and effective monomorphisms and epimorphisms

Previously we observed that although monomorphisms tended to give expected generalizations of injective function in many categories, epimorphisms sometimes weren’t the expected generalization of surjective functions. We also discussed split epimorphisms, but where the definition of an epimorphism is too permissive to agree with the surjective morphisms in familiar concrete categories, the definition of a split epimorphism is too restrictive.

In this post we will discuss two other intermediate notions of epimorphism. (These all give dual notions of monomorphisms, but their epimorphic variants are more interesting as a possible solution to the above problem.) There are yet others, but these two appear to be the most relevant in the context of abelian categories.

## Z[sqrt{-3}] is the Eisenstein integers glued together at two points

Today’s post is a record of a very small observation from my time at PROMYS this summer. Below, by $\text{Spec } R$ I mean a commutative ring $R$ regarded as an object in the opposite category $\text{CRing}^{op}$.


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# Notes

Below are series of typset notes that I am making publically available. They are in the form of $\LaTeX$ PDFs or beamer slides. Descriptions are provided above the links. Any and all comments/criticisms are appreciated, and the best way to contact me is through my contact page.

#### Quantum Groups Notes

These are typeset lecture notes for a topics in algebra class on quantum groups B. Brubaker taught during the fall 2020 semester. It assumes no background with quantum groups, strong knowledge of abstract and linear algebra, and some familiarity with representation theory (in particular highest weight theory).

#### Notre Dame RTG Topology & Geometry: Closed Geodesics

These are typeset notes of a talk I particularly enjoyed at the Notre Dame RTG topology & Geometry conference in 2019. The talk was delivered by gradaute student Gabor Székelyhidi on closed geodesics.

#### UMN REU 2020 Background Theory Notes and Exercises

These are my notes for the UMN REU in 2020. They include background theory and a few series of short exercises some of which have full solutions or partial solutions.

#### Presentation Notes

This section houses my raw notes, scripts, and beamer slides for all the presentations and talks I have given.


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## Elementary Technical Mathematics

$$2\times 10^{20}.$$
By raising each factor to the exponent and multiplying the double exponents, we have $$\frac{(50,000)(640,000,000)}{(0.0004)^2}=\frac{32,000,000,000,000}{0.00000016}\\ =2\times 10^{12+8}=2\times 10^{20}.$$ which is in the scientific notation.


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# Is there an internet application to create ASCII art tables?

Exists an internet application that can create ASCII - art tables similar to this one?:

------------------------------------------------
|           | ColA            | ColB           |
------------------------------------------------
| Row1      | CellA1          | CellB1         |
------------------------------------------------
| Row2      | CellA2          | CellB2         |
------------------------------------------------
| Row3      | CellA3          | CellB3         |
------------------------------------------------

It does not need to make use of ASCII (it can be Unicode), I suggested ASCII in the feeling of ASCII art.

This would certainly serve for websites that do not sustain correct tables (like Stack Exchange).

0
2019-05-18 21:02:38
Source Share

Now there is: Format Text as Table.

I've been suggesting to create this energy for some time. I was in fact motivated by MySQL is command line energy and also the absence of tables on SO. So many thanks for advising me to make it.

And also many thanks to @Lipis for the Unicode char suggestion.

Below is an instance of the result:

+------+--------+--------+
|      | ColA   | ColB   |
+------+--------+--------+
| Row1 | CellA1 | CellB1 |
| Row2 | CellA2 | CellB2 |
| Row3 | CellA3 | CellB3 |
+------+--------+--------+

0
2019-05-31 01:04:30
Source


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Techniques for making tesselated shapes appear smoother?

What are some common techniques for making the edges on a tesselated shape less apparent without increasing the amount of tesselation which is actually carried out?

As an example, I followed along with this tutorial on tesselation shaders, and one can control the inner and outer tesselation levels. With relatively low levels of both, one can achieve a pretty good sphere. However, in order for the surface to actually appear smooth, a fairly bright light seems to be required.

This particular implementation uses triangular tesselation beginning from an icosahedron to approach a sphere.

My understanding is that performing the tesselation with quads would result in a smoother surface.

Here is a picture which illustrates how direct lighting makes it much harder to see the level of tesselation. On the top right of the sphere, where the light is hitting it strongly, no edges are visible, wheres on the bottom left, the edges can easily be seen. Perhaps in practice, simply applying a texture would mask many of the edges?

GL_SMOOTH interpolates the vertex attributes of the primitive (triangle), it looks like your normals are setup per face, rather than per vertex.

If using normal per face (i.e. a single normal for the triangle), increasing the number of triangles will produce a less faceted look as there is more granularity when interpolating the normals over the surface of a given surface area given the increase in the number of triangles within that area. It is recommended to use normals per vertex for smooth surfaces when feature edges are not important (i.e. Not for cubes!).

If using normal per vertex (i.e. a normal for each vertex, resulting in each triangle having 3 vertices), each 3 vertices of a face will be slightly different if the neighbouring triangles are not coplanar to the triangle in question (i.e. not a flat surface). This will aid in the smoothness of the shading.

Normals per vertex can be easily calculated by averaging the three normals of the three vertices defining a triangle, however your surface needs to be defined with shared vertices for this. The exception to this is if you have a sphere or ellipsoid, in which case you can calculate a more precise normal easily by unitising vertex - sphereOrigin rather than averaging the three normals of the face.

On a side note: Are you using two lights in the scene of the screen shot you posted or one? If two it looks like the normals are currently not correct as the shading does not appear correct since there is a well-lit area of the surface on the bottom left region. With a single light, either the front (top right) of the sphere would be lit, or the bottom left, but not both.

Also note that simply increasing the power of the light simply over exposes the surface colour rather than improving the smoothing. You lose the surface definition.

glShadeModel(GL_SMOOTH); - probably in PezInitialize - should do what you need.

• I'm fairly certain that is the default mode, so will not change anything in my picture. Also, I'm more interested in, for instance, what something like that does (which is just interpolate colors between vertices), rather than just a formula to fix the problem. Mar 28 '20 at 16:18


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# B Properties of antimatter?

Tags:
1. Dec 21, 2016

### hsdrop

do we know what the proportions of antimatter is compared to matter ?? like does gravity electromagnetism the strong and weak forces exc.....as well as other thing like does it react to it self the same way matter does to it self??

2. Dec 21, 2016

### Simon Bridge

In the observable Universe, matter vastly outnumbers antimatter ... but, within experiments, antimatter is always accompanied by an equal amount of matter.
Antimatter has the same interactions as regular matter, including the self-interaction ... however, there is this thing called the CP violation which may explain the dominance of matter over antimatter.

Thus: if you search using "CP violation" among your terms, you will find lots of information on the subects you are asking about.

3. Dec 22, 2016

### Orodruin

Staff Emeritus
It is worth mentioning that this is not due to a large asymmetry between matter and antimatter. In the early universe, there was almost equal amounts. What happened was that there was a slight excess of matter and when matter and antimatter annihilated away, this small excess was left as it had nothing to annihilate with. This small excess is all the matter we see today.

As far as we are aware, gravity, electromagnetism, and the strong force affect matter and antimatter in the same way. The weak force displays CP violation. It is worth noticing that the observed CP violation in the weak force is not sufficient to explain why there was slightly more matter in the early universe. This indicates that there should exist another source of CP violation that we do not know about (yet!). This is an open field of research.

4. Jan 30, 2018

### Kagrat

Go to this link and watch this video. It will make you understand more.

5. Feb 1, 2018

### ChrisVer

In general no, we don't. We only know that we love in a matter-dominated (dominated in the sense of matter-antimatter) world. Experiments, like PAMELA, aim in measuring such an asymmetry in the Universe. PAMELA aimed to do that by measuring the flux of antimatter in cosmic rays which would come from the residual antimatter from the Big Bang. As far as I know PAMELA did not observe any kind of primary cosmic ray antiproton result (https://arxiv.org/pdf/1801.10310.pdf Fig.13 and paragraph under it).

Yes most of the times, except for if there is CP-violation in the game. In the CP violation scenario, matter and antimatter interactions differ.
One nice example is shown in this beautiful plot from Babar:

The idea is that you have the production of $B_{\text{tag}}^0\bar{B}^0$ mesons... and you measure the $B_{\text{tag}}^0$ meson's evolution over time (compared to the $\bar{B}^0$)..The nice thing with the neutral B mesons is that there is a possible transition $B\rightarrow \bar{B} \rightarrow B^0 \rightarrow ...$ etc. Now you can measure how much time it takes the $B_\text{tag}^0$ to decay (and also see if it decays as a $B$ or $\bar{B}$) ($t$ or $t'$ respectively) and compare it to the time of decay of $\bar{B}^0$ ($\bar{t}$).
Then you have this variable $\Delta t = \bar{t} - t$ (blue markers) or $=\bar{t}-t'$ (red-markers) (so if the tag decays second, the time difference is negative).
The fact that the $B^0_\text{tag}$ and $\bar{B}_\text{tag}^0$ have so different $\Delta t$ (their asymmetry is shown in the bottom panel) is obviously due to different behavior between matter $B^0_\text{tag}$ and antimatter $\bar{B}^0_\text{tag}$

One day.... Until that one day, when axions are observed...


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# LaTeX/生成封面和标题

## 标准标题页面

 \documentclass{article}% use option titlepage to get the title on a page of its own. \usepackage{blindtext} \title{The Triangulation of Titling Data in Non-Linear Gaussian Fashion via $\rho$ Series\thanks{No procrastination}} \date{2017\\ December} \author{John Doe\\ Magic Department\thanks{I am no longer a member of this department}, Richard Miles University \and Richard Row, \LaTeX\ Academy} \begin{document} \maketitle \section{Introduction} \blindtext \end{document}
\thanks  命令会存储内容，并生成与标题对应的脚注。正如其命令名称所显示的，该命令可以用来对某人致谢。也可以在脚注中打印电子邮件地址或者类似的内容。

 \documentclass{scrbook} \setkomafont{author}{\scshape} \usepackage{blindtext} \title{How hard would it be to build a spaceship from scrap} \author{Carl Capybara\thanks{I never procrastinate} \and Walter Wombat} \subtitle{A closer look at the expenses} \subject{a funny paper} \begin{document} \maketitle \addchap{Introduction} \blindtext \end{document}

 \documentclass{memoir}% use option titlepage to get the title on a page of its own. \usepackage{blindtext} \title{The influence of colour on the floating velocity of rubber ducks} \author{Peter Piranha} \renewcommand{\maketitlehookb}{\centering You won't expect the results} \begin{document} \maketitle \chapter{Introduction} \blindtext \end{document}

## 自订标题页面

### 建立标题

Normally, the benefit of using LaTeX instead of traditional word processing programs is that LaTeX frees you to concentrate on content by handling margins, justification, and other typesetting concerns. On the other hand, if you want to write your own title format, it is exactly the opposite: you have to take care of everything--this time LaTeX will do nothing to help you. It can be challenging to create your own title format since LaTeX was not designed to be graphically interactive in the adjustment of layout. The process is similar to working with raw HTML with the added step that each time you want to see how your changes look, you have to re-compile the source. While this may seem like a major inconvenience, the benefit is that once the customized title format has been written, it serves as a template for all other documents that would use the title format you have just made. In other words, once you have a layout you like, you can use it for any other documents where you would like the same layout without any additional fiddling with layout.

First step: since you'll be working only on the first page of your document and you'll have to compile very often, you don't have to compile the whole document each time, you only need to take a look at the first page. That is why we'll first create a dummy document for preparing the title and then we'll simply include it within the existing big document we are writing. Call the dummy document test_title.tex and put the following code in it:

\documentclass[12pt,a4paper]{report}

\usepackage{graphicx}

\newcommand{\HRule}{\rule{\linewidth}{0.5mm}}

\begin{document}

\input{./title.tex}
\end{document}

It is meant to be compiled with pdflatex to create a PDF in output. It is a very basic document, but take care that it has the same settings of the document you are writing, so the output won't change when you include the title in your document. In this case (see the first line) the font size is set to 12pt and the paper size is an A4. The package graphicx is included to insert an image in the title. Then a command is defined called \HRule; it will just insert a horizontal line whose length is like the size of the paper and whose thickness is 0.5 mm. If you want you can change its settings in the definition. Finally the document starts and it simply includes the title.tex file, that must be placed in the same directory of our dummy file test_title.tex.

Now create the title.tex and write in it:

\begin{titlepage}

\end{titlepage}

all the things you want to put in the title must be inside the titlepage environment. Now if you compile test_title.tex you will see a preview of your title in the test_title.pdf file. Here is what you need to know to write your title:

Alignment
if you want to center some text just use \begin{center} ... \end{center}. If you want to align it differently you can use the environment flushright for right-alignment and flushleft for left alignment.
Images
the command for including images is the following (the example is for a small logo, but you can introduce any image of any size): \includegraphics[width=0.15\textwidth]{./logo}. There is no \begin{figure} as you usually do because you don't want it to be floating, you just want it there where you placed it. When handling it, remember that it is considered like a big box by the TeX engine.
Text size
If you want to change the size of some text just place it within brackets, {like this}, and you can use the following commands (in order of size): \Huge, \huge, \LARGE, \Large, \large, \small, \footnotesize, \tiny. So for example:
{\large this text is slightly bigger than normal}, this one is not

\normalsize is used to create text at the default size for the document.
New lines
you can force the start of a new line by \\. If you want to add more vertical space you don't need to use several new-line commands, just insert some vertical space. For example, this way \\[1cm] you start a new line after having left 1 cm of empty space.

you can insert the date of the current day with the command \today. If you do not wish to insert any date, keep it blank e.g. \date{}
Filling the page
the command \vfill keeps on adding empty spaces until the page is full. If you put it in the page, you are sure that all the following text will be placed at the bottom of the page.

### 实例

All these tips might have made you confused. Then, here is a practical example. Get the test_title.tex described above and here is an example of a title.tex. On the right you can see the output after you compile test_title.tex in PDF:

 \begin{titlepage} \begin{center} % Upper part of the page \includegraphics[width=0.15\textwidth]{./logo}\\[1cm] \textsc{\LARGE University of Beer}\\[1.5cm] \textsc{\Large Final year project}\\[0.5cm] % Title \HRule \\[0.4cm] { \huge \bfseries Lager brewing techniques}\\[0.4cm] \HRule \\[1.5cm] % Author and supervisor \begin{minipage}{0.4\textwidth} \begin{flushleft} \large \emph{Author:}\\ John \textsc{Smith} \end{flushleft} \end{minipage} \begin{minipage}{0.4\textwidth} \begin{flushright} \large \emph{Supervisor:} \\ Dr.~Mark \textsc{Brown} \end{flushright} \end{minipage} \vfill % Bottom of the page {\large \today} \end{center} \end{titlepage}

The picture is from a file called logo.png that is in the same directory of both title.tex and test_title.tex. Since I wanted to insert both the author and supervisor names properly aligned I used a trick: I created two small minipages, one on left and one on the right. Their width is a bit less than half of page width (as you can see, they are exactly 40% of the text width). Within the minipages I have used different alignments. Using \vfill I could write the date exactly at the bottom of the page.

As you can see, the code looks "dirtier" than standard LaTeX source because you have to take care of the output as well. If you start changing font's output it will get more confused, but you can do it: it's only for the title and your complicated code will be isolated from all the rest within its own file title.tex.

### 整合标题页

Assuming that your title page is now contained in a file named title.tex, it must be placed in the same directory as the main document. In order to integrate it, the input command must be used by placing \input{./title.tex} at the top of the document. Don't forget to add the commands \usepackage{graphicx} and \newcommand{\HRule}{\rule{\linewidth}{0.5mm}} in the preamble section as well.

For example, the top section of your document would look like:

...
\usepackage{graphicx}

\newcommand{\HRule}{\rule{\linewidth}{0.5mm}}

\begin{document}

\input{./title.tex}
\tableofcontents
...

# 更多标题页面的例子

The titlepages package presents many different styles for title pages.

TeX.SE has a collection of titlepages.

Another small collection can be found on Github.


text:
# Assume that radon measurements are normally distributed

#### palpulla

##### New Member
with mean 4.1 and standard deviation 0.2.

-Find the median of the measurement?

-What is the probability the radon measurements exceed 4.5 picocuries?


text:
# Topics in Abstract Algebra/Commutative algebra

The set of all prime ideals in a commutative ring $A$ is called the spectrum of $A$ and denoted by $\operatorname{Spec}(A)$. (The motivation for the term comes from the theory of a commutative Banach algebra.)

## Spec A

The set of all nilpotent elements in $A$ forms an ideal called the nilradical of $A$. Given any ideal $\mathfrak a$, the pre-image of the nilradical of $A$ is an ideal called the radical of $\mathfrak a$ and denoted by $\sqrt{\mathfrak a}$. Explicitly, $x \in \sqrt{\mathfrak a}$ if and only if $x^n \in \mathfrak a$ for some $n$.

Proposition A.14. Let $\mathfrak i, \mathfrak j \triangleleft A$.

• (i) $\sqrt{\mathfrak i^n} = \sqrt \mathfrak i$
• (ii) $\sqrt{\mathfrak i \mathfrak j} = \sqrt{\mathfrak i \cap \mathfrak j} = \sqrt \mathfrak i \cap \sqrt \mathfrak j$

Proof. Routine. $\square$

Exercise. A ring has only one prime ideal if and only if its nilradical is maximal.

Exercise. Every prime ideal in a finite ring is maximal.

Proposition A.2. Let $A \ne 0$ be a ring. If every principal ideal in $A$ is prime, then $A$ is a field.
Proof. Let $0 \ne x \in A$. Since $x^2$ is in $(x^2)$, which is prime, $x \in (x^2)$. Thus, we can write $x = ax^2$. Since $(0)$ is prime, $A$ is a domain. Hence, $1 = ax$. $\square$

Lemma. Let $\mathfrak p \triangleleft A$. Then $\mathfrak p$ is prime if and only if

$\mathfrak p \subsetneq \mathfrak a \triangleleft A, \mathfrak p \subsetneq \mathfrak b \triangleleft A$ implies $\mathfrak a \mathfrak b \not\subset \mathfrak p$

Proof. ($\Rightarrow$) Clear. ($\Leftarrow$) Let $\overline{x}$ be the image of $x \in A$ in $A / \mathfrak p$. Suppose $\overline{a}$ is a zero-divisor; that is, $\overline{a}\overline{b} = 0$ for some $b \in A \backslash \mathfrak p$. Let $\mathfrak a = (a, \mathfrak p)$, and $\mathfrak b = (b, \mathfrak p)$. Since $\mathfrak a \mathfrak b = ab + \mathfrak p \subset \mathfrak p$, and $\mathfrak b$ is strictly larger than $\mathfrak p$, by the hypothesis, $\mathfrak a \subset \mathfrak p$. That is, $\overline{a} = 0$. $\square$

Theorem A.11 (multiplicative avoidance). Let $S \subset A$ be a multiplicative system. If $\mathfrak a \triangleleft A$ is disjoint from $S$, then there exists a prime ideal $\mathfrak p \supset \mathfrak a$ that is maximal among ideals disjoint from $S$.
Proof. Let $\mathfrak m$ be a maximal element in the set of all ideals disjoint from $S$. Let $\mathfrak a$ and $\mathfrak b$ be ideals strictly larger than $\mathfrak m$. Since $\mathfrak m$ is maximal, we find $a \in \mathfrak a \cap S$ and $b \in \mathfrak b \cap S$. By the definition of $S$, $ab \in S$; thus, $\mathfrak a \mathfrak b \not \subset \mathfrak m$. By the lemma, $\mathfrak m$ is prime then. $\square$

Note that the theorem applies in particular when $S$ contains only 1.

Exercise. A domain A is a principal ideal domain if every prime ideal is principal.

A Goldman domain is a domain whose field of fractions $K$ is finitely generated as an algebra. When $A$ is a Goldman domain, K always has the form $A[f^{-1}]$. Indeed, if $K = A[s_1^{-1}, ..., s_n^{-1}]$, let $s = s_1...s_n$. Then $K = A[s^{-1}]$.

Lemma. Let $A$ be a domain with the field of fractions $K$, and $0 \ne f \in A$. Then $K = A[f^{-1}]$ if and only if every nonzero prime ideal of $A$ contains $f$.
Proof. ($\Leftarrow$) Let $0 \ne x \in A$, and $S = \{ f^n | n \ge 0 \}$. If $(x)$ is disjoint from $S$, then, by the lemma, there is a prime ideal disjoint from $S$, contradicting the hypothesis. Thus, $(x)$ contains some power of $f$, say, $yx = f^n$. Then $yx$ and so $x$ are invertible in $A[f^{-1}].$ ($\Rightarrow$) If $\mathfrak p$ is a nonzero prime ideal, it contains a nonzero element, say, $s$. Then we can write: $1 / s = a / f^n$, or $f^n = as \in \mathfrak p$; thus, $f \in \mathfrak p$. $\square$

A prime ideal $\mathfrak{p} \in \operatorname{Spec}(A)$ is called a Goldman ideal if $A/\mathfrak p$ is a Goldman domain.

Theorem A.21. Let $A$ be a ring and $\mathfrak a \triangleleft A$. Then $\sqrt \mathfrak a$ is the intersection of all minimal Goldman ideals of A containing $\mathfrak a$
Proof. By the ideal correspondence, it suffices to prove the case $\mathfrak a = \sqrt \mathfrak a = 0$. Let $0 \ne f \in A$. Let $S = \{ f^n | n \ge 0 \}$. Since $f$ is not nilpotent (or it will be in $\sqrt{(0)}$), by multiplicative avoidance, there is some prime ideal $\mathfrak g$ not containing $f$. It remains to show it is a Goldman ideal. But if $\mathfrak p \triangleleft A / \mathfrak g$ is a nonzero prime, then $f \in \mathfrak p$ since $\mathfrak p$ collapses to zero if it is disjoint from $S$. By Lemma, the field of fractions of $A / \mathfrak g$ is obtained by inverting $f$ and so $\mathfrak g$ is a Goldman ideal. Hence, the intersection of all Goldman ideals reduces to zero. $\square$

In some rings, Goldman ideals are maximal; this will be discussed in the next section. On the other hand,

Lemma. Let $\mathfrak a \triangleleft A$. Then $\mathfrak a$ is a Goldman ideal if and only if it is the contraction of a maximal ideal in $A[X]$.

Theorem. The following are equivalent.

1. For any $\mathfrak a \triangleleft A$, $\mathfrak a$ is the intersection of all maximal ideals containing $\mathfrak a$.
2. Every Goldman ideal is maximal.
3. Every maximal ideal in $A[X]$ contracts to a maximal ideal in $A$.

Proof. Clear. $\square$

A ring satisfying the equivalent conditions in the theorem is called a Hilbert-Jacobson ring.

Lemma. Let $A \subset B$ be domains such that $B$ is algebraic and of finite type over $A$. Then $A$ is a Goldman domain if and only if $B$ is a Goldman domain.
Proof. Let $K \subset L$ be the fields of fractions of $A$ and $B$, respectively. $\square$

Theorem A.19. Let $A$ be a Hilbert-Jacobson ring. Then $A[X]$ is a Hilbert-Jacobson ring.
Proof. Let $\mathfrak q \triangleleft A[X]$ be a Goldman ideal, and $\mathfrak p = \mathfrak q \cap A$. It follows from Lemma something that $A / \mathfrak p$ is a Goldman domain since it is contained in a $A[X] / \mathfrak q$, a Goldman domain. Since $A$ is a Hilbert-Jacobson ring, $\mathfrak p$ is maximal and so $A / \mathfrak p$ is a field and so $A[X] / \mathfrak q$ is a field; that is, $\mathfrak q$ is maximal. $\square$

 To do: Explain why $A / \mathfrak p$ is a field (or point to a location where it can be understood why it is so...).

Theorem A.5 (prime avoidance). Let $\mathfrak p_1, ..., \mathfrak p_r \triangleleft A$ be ideals, at most two of which are not prime, and $\mathfrak a \triangleleft A$. If $\mathfrak a \subset \bigcup_1^r \mathfrak p_i$, then $\mathfrak a \subset \mathfrak p_i$ for some $\mathfrak i$.
Proof. We shall induct on $r$ to find $a \in \mathfrak a$ that is in no $\mathfrak p_i$. The case $r=1$ being trivial, suppose we find $a \in \mathfrak a$ such that $a \not\in \mathfrak{p}_i$ for $i < r$. We assume $a \in \mathfrak{p}_r$; else, we're done. Moreover, if $\mathfrak{p}_i \subset \mathfrak{p}_r$ for some $i < r$, then the theorem applies without $\mathfrak{p}_i$ and so this case is done by by the inductive hypothesis. We thus assume $\mathfrak p_i \not\subset \mathfrak p_r$ for all $i < r$. Now, $\mathfrak a \mathfrak p_1 ... \mathfrak p_{r-1} \not\subset \mathfrak {p}_r$; if not, since $\mathfrak p_r$ is prime, one of the ideals in the left is contained in $\mathfrak p_r$, contradiction. Hence, there is $b$ in the left that is not in $\mathfrak {p}_r$. It follows that $a + b \not\in \mathfrak p_i$ for all $i \le r$. Finally, we remark that the argument works without assuming $\mathfrak p_1$ and $\mathfrak p_2$ are prime. (TODO: too sketchy.) The proof is thus complete. $\square$

An element p of a ring is a prime if $(p)$ is prime, and is an irreducible if $p = xy \Rightarrow$ either $x$ or $y$ is a unit..

We write $x | y$ if $(x) \ni y$, and say $x$ divides $y$. In a domain, a prime element is irreducible. (Suppose $x = yz$. Then either $x | y$ or $x | z$, say, the former. Then $sx = y$, and $sxz = x$. Canceling $x$ out we see $z$ is a unit.) The converse is false in general. We have however:

Proposition. Suppose: for every $x$ and $y$, $(x) \cap (y) = (xy)$ whenever (1) is the only principal ideal containing $(x, y)$. Then every irreducible is a prime.
Proof. Let $p$ be an irreducible, and suppose $p | xy$ and $p \not| y$. Since $(p) \cap (x) = (px)$ implies $px | xy$ and $p | y$, there is a $d$ such that $(1) \ne (d) \supset (p, x)$. But then $d | p$ and so $p | d$ ($p$ is an irreducible.) Thus, $p | x$. $\square$

Theorem A.16 (Chinese remainder theorem). Let $\mathfrak a_1, ..., \mathfrak a_n \triangleleft A$. If $\mathfrak a_j + \mathfrak a_i = (1)$, then

$\prod \mathfrak a_i \to A \to A / \mathfrak a_1 \times \cdots \times \mathfrak a_n \to 0$

is exact.

The Jacobson radical of a ring $A$ is the intersection of all maximal ideals.

Proposition A.6. $x \in A$ is in the Jacobson radical if and only if $1 - xy$ is a unit for every $y \in A$.
Proof. Let $x$ be in the Jacobson radical. If $1-xy$ is not a unit, it is in a maximal ideal $\mathfrak m$. But then we have: $1 = (1 - xy) + xy$, which is a sum of elements in $\mathfrak m$; thus, in $\mathfrak m$, contradiction. Conversely, suppose $x$ is not in the Jacobson radical; that is, it is not in some maximal ideal $\mathfrak m$. Then $(x, \mathfrak m)$ is an ideal containing $\mathfrak m$ but strictly larger. Thus, it contains $1$, and we can write: $1 = xy + z$ with $y \in A$ and $z \in \mathfrak m$. Then $1 - xy \in \mathfrak m$, and $\mathfrak m$ would cease to be proper, unless $1 - xy$ is a non-unit. $\square$

Note that the nilradical is contained in the Jacobson radical, and they coincide in particular if prime ideals are maximal (e.g., the ring is a principal ideal domain). Another instance of this is:

Exercise. In $A[X]$, the nilradical and the Jacobson radical coincide.

Theorem A.17 (Hopkins). Let A be a ring. Then the following are equivalent.

1. A is artinian
2. A is noetherian and every prime ideal is maximal.
3. $\operatorname{Spec}(A)$ is finite and discrete, and $A_\mathfrak m$ is noetherian for all maximal ideal $\mathfrak m$.

Proof. (1) $\Rightarrow$ (3): Let $\mathfrak p \triangleleft A$ be prime, and $x \in A / \mathfrak p$. Since $A / \mathfrak p$ is artinian (consider the short exact sequence), the descending sequence $(x^n)$ stabilizes eventually; i.e., $x^n = u x^{n+1}$ for some unit u. Since $A / \mathfrak p$ is a domain, $x$ is a unit then. Hence, $\mathfrak p$ is maximal and so $\operatorname{Spec}(A)$ is discrete. It remains to show that it is finite. Let $S$ be the set of all finite intersections of maximal ideals. Let $\mathfrak i \in S$ be its minimal element, which we have by (1). We write $\mathfrak i = \mathfrak m_1 \cap ... \cap \mathfrak m_n$. Let $\mathfrak m$ be an arbitrary maximal ideal. Then $\mathfrak m \cap \mathfrak i \in S$ and so $\mathfrak m \cap \mathfrak i = \mathfrak i$ by minimality. Thus, $\mathfrak m = \mathfrak m_i$ for some i. (3) $\Rightarrow$ (2): We only have to show $A$ is noetherian. $\square$

A ring is said to be local if it has only one maximal ideal.

Proposition A.17. Let $A$ be a nonzero ring. The following are equivalent.

1. $A$ is local.
2. For every $x \in A$, either $x$ or $1 - x$ is a unit.
3. The set of non-units is an ideal.

Proof. (1) $\Rightarrow$ (2): If $x$ is a non-unit, then $x$ is the Jacobson radical; thus, $1 - x$ is a unit by Proposition A.6. (2) $\Rightarrow$ (3): Let $x, y \in A$, and suppose $x$ is a non-unit. If $xy$ is a unit, then so are $x$ and $y$. Thus, $xy$ is a non-unit. Suppose $x,y$ are non-units; we show that $x+y$ is a non-unit by contradiction. If $x+y$ is a unit, then there exists a unit $a\in A$ such that $1=a(x+y)=ax+ay$. Thus either $ax$ or $1-ax=ay$ is a unit, whence either $x$ or $y$ is a unit, a contradiction. (3) $\Rightarrow$ (1): Let $\mathfrak i$ be the set of non-units. If $\mathfrak m \triangleleft A$ is maximal, it consists of nonunits; thus, $\mathfrak m \subset \mathfrak i$ where we have the equality by the maximality of $\mathfrak m$. $\square$

Example. If $p$ is a prime ideal, then $A_p$ is a local ring where $p$ is its unique maximal ideal.

Example. If $\sqrt \mathfrak i$ is maximal, then $A/\mathfrak i$ is a local ring. In particular, $A/\mathfrak m^n, (n \ge 1)$is local for any maximal ideal $\mathfrak m$.

Let $(A, \mathfrak m)$ be a local noetherian ring.

A. Lemma

• (i) Let $\mathfrak i$ be a proper ideal of $A$. If $M$ is a finite generated $\mathfrak i$-module, then $M = 0$.
• (ii) The intersection of all $\mathfrak{m}^k$ over $k \ge 1$ is trivial.

Proof: We prove (i) by the induction on the number of generators. Suppose $M$ cannot be generated by strictly less than $n$ generators, and suppose we have $x_1, ... x_n$ that generates $M$. Then, in particular,

$x_1 = a_1 x_1 + a_2 x_2 + ... + a_n x_n$ where $a_i$ are in $\mathfrak i$,

and thus

$(1 - a_1) x_1 = a_2 x_2 + ... + a_n x_n$

Since $a_1$ is not a unit, $1 - a_1$ is a unit; in fact, if $1 - a_1$ is not a unit, it belongs to a unique maximal ideal $\mathfrak{m}$, which contains every non-units, in particular, $a_1$, and thus $1 \in \mathfrak{m}$, which is nonsense. Thus we find that actually x_2, ..., x_n generates $M$; this contradicts the inductive hypothesis.$\square$

An ideal $\mathfrak q \triangleleft A$ is said to be primary if every zero-divisor in $A/\mathfrak q$ is nilpotent. Explicitly, this means that, whenever $xy \in \mathfrak q$ and $y \not\in \mathfrak q$, $x \in \sqrt{\mathfrak q}$. In particular, a prime ideal is primary.

Proposition. If $\mathfrak q$ is primary, then $\sqrt{\mathfrak q}$ is prime. Conversely, if $\sqrt{\mathfrak q}$ is maximal, then $\mathfrak q$ is primary.
Proof. The first part is clear. Conversely, if $\sqrt \mathfrak q$ is maximal, then $\mathfrak m = \sqrt \mathfrak q / \mathfrak q$ is a maximal ideal in $A/\mathfrak q$. It must be unique and so $A/\mathfrak q$ is local. In particular, a zero-divisor in $A/\mathfrak q$ is nonunit and so is contained in $\mathfrak m$; hence, nilpotent. $\square$

Exercise. $\sqrt{\mathfrak q}$ prime $\not\Rightarrow \mathfrak q$ primary.

Theorem A.8 (Primary decomposition). Let $A$ be a noetherian ring. If $\mathfrak i \triangleleft A$, then $\mathfrak i$ is a finite intersection of primary ideals.
Proof. Let $S$ be the set of all ideals that is not a finite intersection of primary ideals. We want to show $S$ is empty. Suppose not, and let $\mathfrak i$ be its maximal element. We can write $\mathfrak i$ as an intersection of two ideals strictly larger than $\mathfrak i$. Indeed, since $\mathfrak i$ is not prime by definition in particular, choose $x \not\in \mathfrak i$ and $y \not\in \mathfrak i$ such that $xy \in \mathfrak i$. As in the proof of Theorem A.3, we can write: $\mathfrak i = \mathfrak j (\mathfrak i + x)$ where $\mathfrak j$ is the set of all $a \in A$ such that $ax \in \mathfrak i$. By maximality, $\mathfrak j, \mathfrak i + x \not\in S$. Thus, they are finite intersections of primary ideals, but then so is $\mathfrak i$, contradiction. $\square$

Proposition. If $(0)$ is indecomposable, then the set of zero divisors is a union of minimal primes.

## Integral extension

Let $A \subset B$ be rings. If $b \in B$ is a root of a monic polynomial $f \in A[X]$, then $b$ is said to be integral over $A$. If every element of $B$ is integral over $A$, then we say $B$ is integral over $A$ or $B$ is an integral extension of $A$. More generally, we say a ring morphism $f: A \to B$ is integral if the image of $A$ is integral over $B$. By replacing $A$ with $f(A)$, it suffices to study the case $A \subset B$, and that's what we will below do.

Lemma A.9. Let $b \in B$. Then the following are equivalent.

1. $b$ is integral over $A$.
2. $A[b]$ is finite over $A$.
3. $A[b]$ is contained in an $A$-submodule of $B$ that is finite over $A$.

Proof. (1) means that we can write:

$b^{n+r} = -(b^{r+n-1} a_{n-1} + ... b^{r+1} a_1 + b^r a_0)$

Thus, $1, b, ..., b^{n-1}$ spans $A[b]$. Hence, (1) $\Rightarrow$ (2). Since (2) $\Rightarrow$ (3) vacuously, it remains to show (3) $\Rightarrow$ (1). Let $M_{/A[b]}$ be generated over $A$ by $x_1, ..., x_n$. Since $b x_i \in M$, we can write

$b x_i = \sum_{j=1}^n c_{ij} x_j$

where $c_{kj} \in A$. Denoting by $C$ the matrix $c_{ij}$, this means that $\det(bI-C)$ annihilates $M$. Hence, $\det(bI-C) = 0$ by (3). Noting $\det(bI-C)$ is a monic polynomial in $b$ we get (1). $\square$

The set of all elements in B that are integral over A is called the integral closure of A in B. By the lemma, the integral closure is a subring of $B$ containing $A$. (Proof: if $x$ and $y$ are integral elements, then $A[xy]$ and $A[x-y]$ are contained in $A[x, y]$, finite over $A$.) It is also clear that integrability is transitive; that is, if $C$ is integral over $B$ and $B$ is integral over $A$, then $C$ is integral over $A$.

Proposition. Let $f:A \to B$ be an integral extension where $B$ is a domain. Then

• (i) $A$ is a field if and only if $B$ is a field.
• (ii) Every nonzero ideal of $B$ has nonzero intersection with $A$.

Proof. (i) Suppose $B$ is a field, and let $x \in A$. Since $x^{-1} \in B$ and is integral over $A$, we can write:

$x^{-n} = -(a_{n-1}x^{-(n-1)} + ... + a_1 x^{-1} + a_0)$

Multiplying both sides by $x^{n-1}$ we see $x^{-1} \in A$. For the rest, let $0 \ne b \in B$. We have an integral equation:

$-a_0 = b^n + a_{n-1}b^{n-1} + ... + a_1 b = b(b^{n-1} + a_{n-1}b^{n-2} + ... + a_1)$.

Since $B$ is a domain, if $n$ is the minimal degree of a monic polynomial that annihilates $b$, then it must be that $a_0 \ne 0$. This shows that $bB \cap A \ne 0$, giving us (ii). Also, if $A$ is a field, then $a_0$ is invertible and so is $b$. $\square$

Theorem (Noether normalization). Let $A$ be a finitely generated $k$-algebra. Then we can find $z_1, ..., z_d$ such that

1. $A$ is integral over $k[z_1, ..., z_d]$.
2. $z_1, ..., z_d$ are algebraically independent over $k$.
3. $z_1, ..., z_d$ are a separating transcendence basis of the field of fractions $K$ of $A$ if $K$ is separable over $k$.

Exercise A.10 (Artin-Tate). Let $A \subset B \subset C$ be rings. Suppose $A$ is noetherian. If $C$ is finitely generated as an $A$-algebra and integral over $B$, then $B$ is finitely generated as an $A$-algebra.

Exercise. A ring morphism $f:A \to \Omega$ (where $\Omega$ is an algebraically closed field) extends to $F: A[b] \to \Omega$ (Answer: http://www.math.uiuc.edu/~r-ash/ComAlg/)

## Noetherian rings

Exercise. A ring is noetherian if and only if every prime ideal is finitely generated. (See T. Y. Lam and Manuel L. Reyes, A Prime Ideal Principle in Commutative Algebra for a systematic study of results of this type.)

The next theorem furnishes many examples of a noetherian ring.

Theorem A.7 (Hilbert basis). $A$ is a noetherian ring if and only if $A[T_1, ... T_n]$ is noetherian.
Proof. By induction it suffices to prove $A[T]$ is noetherian. Let $I \triangleleft A[T]$. Let $L_n$ be the set of all coefficients of polynomials of degree $\le n$ in $I$. Since $L_n \triangleleft A$, there exists $d$ such that

$L_0 \subset L_1 \subset L_2, ..., \subset L_d = L_{d+1} = ...$.

For each $0 \le n \le d$, choose finitely many elements $f_{1n}, f_{2n}, ... f_{m_nn}$ of $I$ whose coefficients $b_{1n}, ... b_{m_nn}$ generate $L_n$. Let $I'$ be an ideal generated by $f_{jn}$ for all $j, n$. We claim $I = I'$. It is clear that $I \subset I'$. We prove the opposite inclusion by induction on the degree of polynomials in $I$. Let $f \in I$, $a$ the leading coefficient of $f$ and $n$ the degree of $f$. Then $a \in L_n$. If $n \le d$, then

$a = a_1 b_{1n} + a_2 b_{2n} + ... + a_{m_n} b_{{m_n}n}$

In particular, if $g = a_1 f_{1n} + a_2 f_{2n} + ... + a_{m_n} f_{{m_n}n}$, then $f - g$ has degree strictly less than that of $f$ and so by the inductive hypothesis $f - g \in I'$. Since $g \in I'$, $f \in I'$ then. If $n \ge d$, then $a \in L_d$ and the same argument shows $f \in I'$. $\square$

Exercise. Let $A$ be the ring of continuous functions $f: [0, 1] \to [0, 1]$.$A$ is not noetherian.

Let $(A, \mathfrak m)$ be a noetherian local ring with $k = A/\mathfrak m$. Let $\mathfrak i \triangleleft A$. Then $\mathfrak i$ is called an ideal of definition if $A / \mathfrak i$ is artinian.

Theorem. $\dim_k (\mathfrak m / {\mathfrak m}^2) \ge \dim A$

The local ring $A$ is said to be regular if the equality holds in the above.

Theorem. Let $A$ be a noetherian ring. Then $\dim A[T_1, ..., T_n] = n + \dim A$.
Proof. By induction, it suffices to prove the case $n = 1$. $\square$

Theorem. Let $A$ be a finite-dimensional $k$-algebra. If $A$ is a domain with the field of fractions $K$, then $\dim A = \operatorname{trdeg}_k K$.
Proof. By the noether normalization lemma, $A$ is integral over $k[x_1, ..., x_n]$ where $x_1, ..., x_n$ are algebraically independent over $k$. Thus, $\dim A = \dim k[x_1, ..., x_n] = n$. On the other hand, $\operatorname{trdeg}_k K = n$. $\square$

Theorem. Let $A$ be a domain with (ACCP). Then $A$ is a UFD if and only if every prime ideal $\mathfrak p$ of height 1 is principal.
Proof. ($\Rightarrow$) By Theorem A.10, $\mathfrak p$ contains a prime element $x$. Then

$0 \subset (x) \subset \mathfrak p$

where the second inclusion must be equality since $\mathfrak p$ has height 1. ($\Leftarrow$) In light of Theorem A.10, it suffices to show that $A$ is a GCD domain. (TODO: complete the proof.) $\square$

Theorem. A regular local ring is a UFD.

Theorem A.10 (Krull's intersection theorem). Let $\mathfrak i \triangleleft A$ be a proper ideal. If $A$ is either a noetherian domain or a local ring, then $\bigcap_{n \ge 1} \mathfrak i^n = 0$.

Theorem A.15. Let $\mathfrak i \triangleleft A$. If $A$ is noetherian,

$\sqrt \mathfrak i^n \subset \mathfrak i \subset \sqrt \mathfrak i$ for some $n$.

In particular, the nilradical of $A$ is nilpotent.
Proof. It suffices to prove this when $\mathfrak i = 0$. Thus, the proof reduces to proving that the nilradical of A is nilpotent. Since $A$ is nilpotent, we have finitely many nilpotent elements $x_1, ..., x_n$ that spans $\sqrt{(0)}$. The power of any linear combination of them is then a sum of terms that contain the high power of some $x_j$ if we take the sufficiently high power. Thus, $\sqrt{(0)}$ is nilpotent. $\square$

Proposition A.8. If $A$ is noetherian, then $\hat A$ is noetherian.

Corollary. If $A$ is noetherian, then $A[[X]]$ is noetherian.

## Zariski topology

Given $\mathfrak a \triangleleft A$, let $\operatorname{V}(\mathfrak a) = \{ \mathfrak p \in \operatorname{Spec}(A) | \mathfrak p \supset \mathfrak a \}$. (Note that $\operatorname{V}(\mathfrak a) = \operatorname{V}(\sqrt \mathfrak a)$.) It is easy to see

$V(\mathfrak a) \cup V(\mathfrak b) = V(\mathfrak a \mathfrak b) = V(\mathfrak a \cap \mathfrak b)$, and $\cap_\alpha V(\mathfrak a_\alpha) = V((\mathfrak a_\alpha | \alpha))$.

It follows that the collection of the sets of the form $\operatorname{V}(\mathfrak a)$ includes the empty set and $\operatorname{Spec}(A)$ and is closed under intersection and finite union. In other words, we can define a topology for $\operatorname{Spec}(A)$ by declaring $\operatorname{Z}(\mathfrak i)$ to be closed sets. The resulting topology is called the Zariski topology. Let $X = \operatorname{Spec}(A)$, and write $X_f = X \backslash V ((f)) = \{ P \in X | P \ni f \}$.

Proposition A.16. We have:

• (i) $X_f$ is quasi-compact.
• (ii) $X_{fg}$ is canonically isomorphic to $\operatorname{Spec}(A[f^{-1}])_g$.

Proof. We have: $X_f \subset \bigcup_\alpha X_{f_\alpha} = X \backslash V((f_\alpha|\alpha)) \Leftrightarrow (f) \subset (f_\alpha|\alpha) \Leftrightarrow f \in (f_{\alpha_1}, ..., f_{\alpha_n})$. $\square$

Exercise. Let $A$ be a local ring. Then $\operatorname{Spec}(A)$ is connected.

Corollary. $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$ is a closed surjection.

Theorem A.12. If $A_m$ is noetherian for every maximal ideal $\mathfrak m$ and if $\{ \mathfrak m \in \operatorname{Max}(A) | x \in m \}$ is finite for each $x \in A$, then $A$ is noetherian.

## Integrally closed domain

Lemma A.8. In a GCD domain, if $(x, y) = 1 = (x, z)$, then $(x, yz) = 1$.

Proposition A.9. In a GCD domain, every irreducible element is prime.
Proof. Let $x$ be an irreducible, and suppose $x | yz$. Then $x | (x, yz)$. If $(x, yz) = 1$, $x$ is a unit, the case we tacitly ignore. Thus, by the lemma, $d = (x, y)$, say, is a nonunit. Since $x$ is irreducible, $x | d$ and so $x | y$. $\square$

In particular, in a polynomial ring that is a GCD domain, every irreducible polynomial is a prime element.

Theorem (undefined: ACC). Let A be a ring that satisfies the ascending chain conditions on principal ideals (example: noetherian ring). Then every x in $A$ is a finite product of irreducibles.

Theorem A.10. Let A be a domain. The following are equivalent.

1. Every nonzero nonunit element is a finite product of prime elements.
2. (Kaplansky) Every nonzero prime ideal contains a prime element.
3. $A$ is a GCD domain and has (ACC) on principal ideals.

Proof. (3) $\Rightarrow$ (2): Let $\mathfrak{p} \in \operatorname{Spec}(A)$. If $\mathfrak{p}$ is nonzero, it then contains a nonzero element x, which we factor into irreducibles: $x = p_1 ... p_n$. Then $p_j \in \mathfrak{p}$ for some $j$. Finally, irreducibles are prime since $A$ is a GCD domain. (2) $\Rightarrow$ (1): Let $S$ be the set of all products of prime elements. Clearly, $S$ satisfies the hypothesis of Theorem A.11 (i.e., closed under multiplication). Suppose, on the contrary, there is a nonzero nonunit $x$. It is easy to see that since $x \not\in S$, $(x)$ and $S$ are disjoint. Thus, by Theorem A.11, there is a prime ideal $\mathfrak p$ containing $x$ and disjoint from $S$. But, by (2), $\mathfrak p$ contains a prime element $y$; that is, $\mathfrak p$ intersects $S$, contradiction. (1) $\Rightarrow$ (3): By uniqueness of factorization, it is clear that $A$ is a GCD domain. $\square$

A domain satisfying the equivalent conditions in the theorem is called a unique factorization domain or a UFD for short.

Corollary. If $A$ is a UFD, then $A[X]$ is a UFD. If A is a principal ideal domain, then $A[[X]]$ is a UFD.

Theorem A.13 (Nagata criterion). Let A be a domain, and $S \subset A$ a multiplicatively closed subset generated by prime elements. Then $A$ is a UFD if and only if $S^{-1}A$ is a UFD.


text:
# Math Help - Mixing

1. ## Mixing

Consider H to be a interval and f: H->H be a cts fxn.

For nontrivial closed intervals, suppose that U and V are contained in H, we can find m such that V contained in f^{m}(U).

Show that the periodic pts are dense in H.

I need to show that every point of H is the limit of some
sequence of points in H. Not sure where to proceed.

Show that f is topologically transitive.

f:H->H is topologically transitive if for all open sets U,V (non empty) there exists x in U and n such that f^{n}(x) in V. ie f^{n}(U) intersection V nonempty. Doesn't that just follow from assumption?

2. Suppose p is a repelling ﬁxed point of a map f . Then there is some nbhd Nϵ (p) so that for any x not equal to p and x ∈ Nϵ (p), there is some n* so that that for all n ≥ n*, f^{n}(x) not in Nϵ(p).
Let d = ϵ. If k = n* we have

|f^{k}(x) -f^{k}(p)| = |f^{k}(x) -(p)| = >= ϵ = d

therefore has sensitive dependence.


text:
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text:
# If 0 < a < b, where a, b $\in\mathbb{R}$, determine $\lim \bigg(\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}}\bigg)$

If 0 < a < b, where a, b $\in\mathbb{R}$, determine $\lim \bigg(\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}}\bigg)$

The answer (from the back of the text) is $\lim \bigg(\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}}\bigg) = b$ but I have no idea how to get there. The course is Real Analysis 1, so its a course on proofs. This chapter is on limit theorems for sequences and series. The squeeze theorem might be helpful.

I can prove that $\lim \bigg(\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}}\bigg) \le b$ but I can't find a way to also prove that it is larger than b

Thank you!

-
Though it's obvious, you should indicate what variable you are taking the limit with respect to, at to what point. – David Mitra Dec 9 '12 at 0:25

Hint: Divide top and bottom by $b^n$.

At the top we get $b+a\left(\dfrac{a}{b}\right)^n$.

Remarks: $1$.) Presumably it has already been shown that if $|x|\lt 1$ then $\lim_{n\to\infty} x^n=0$. If not, let $|x|=1+t$. By the Binomial Theorem, or the Bernoulli Inequality, $(1+t)^n \ge 1+nt$. From this one can prove quickly, even in full $\epsilon$-$N$ style, that $\lim_{n\to \infty}\frac{1}{(1+t)^n}=0$.

$2$) Suppose one does not think of the standard trick that we used. Knowing the answer is then helpful, for then it is natural to look at $$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}-b.$$ After some simplification, this becomes $\dfrac{a^n-a^{n+1}}{a^n+b^n}$, and now what to do may be clearer.

-

HINT: Use the standard trick from first-year calculus, and divide top and bottom by the biggest thing in the denominator, $b^n$:

$$\lim_{n\to\infty}\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\lim_{n\to\infty}\frac{\dfrac{a^{n+1}}{b^n}+b}{\dfrac{a^n}{b^n}+1}\;.$$

There’s still some work to be done, but it’s easy work.

-


text:
# Three resistors of value 9Ω, 4Ω and 12Ω are connected in parallel. What will be the equivalent resistance of the circuit?

This question was previously asked in
RRC Group D Previous Paper 47 (Held On: 10 Dec 2018 Shift 1)
View all RRC Group D Papers >
1. 0.8
2. 2.2
3. 1.9
4. 2.6

Option 2 : 2.2
Free
Polity - I
9734
10 Questions 20 Marks 12 Mins

## Detailed Solution

The correct answer will be 2.2Ω.

Key Points

Concept:

When the resistances are connected in series, the equivalent resistance is given by

$${R_{eq}} = {R_1} + {R_2} + \ldots + {R_n}$$

The equivalent resistance is greater than the largest resistance in the series circuit

When the resistances are connected in parallel, the equivalent resistance is given by

$$\frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \ldots + \frac{1}{{{R_n}}}$$

The equivalent resistance is lesser than the smallest resistance in the series circuit

Calculation:

R1 = 9 ohm, R2 = 4 ohm,  R= 12

$$\frac{1}{{{R_{eq}}}} = \frac{1}{{9}} + \frac{1}{{4}} + \frac{1}{{12}}$$

Req = 2.2 Ω


text:
The largest permutation group without 2-cycles is $A_n$, which has size $n!/2$. I think the largest permutation group without 2-cycles or 3-cycles is much smaller, but I can't figure out if it should be polynomially smaller (eg. of size $n!/n^3$), or more dramatically smaller (eg. of size $(.5n)!$).

The largest group I could come up with is {$\phi(x_1,\dots,x_{n/2})$\{\phi(x_1,\dots,x_{n/2}) \circ \phi(x_{n/2+1},\dots,x_n) | \phi \in S_{n/2}$}S_{n/2}\}$, which has size $(.5n)!$.

EDIT: Posting this here since the answers below pointed me in the right direction, but ended up conjecturing something that was not quite correct. The group

$\{(g_1, g_2, \dots, g_{d-1}\!,\ g_1g_2\cdots g_{d-1}\!)\ |\ g_i \in S_{n/d}\} \le S_{n/d}^d \le S_n$

has no 2-cycles or 3-cycles, and has $(n/d)!^{d-1}$ elements. When $d = \log n$, this is $n!/n^{\Theta(n\log\log(n)/\log(n))}$, which is smaller than $n!/poly(n)$ but larger than $(cn)!$ for any $c<1$.

You can do a little bit better by using a wreath product instead of a direct product, and by tweaking $d$, but I think this is more or less optimal.

2 fixed a typo

The largest permutation group without 2-cycles is $A_n$, which has size $n!/2$. I think the largest permutation group without 2-cycles or 3-cycles is much smaller, but I can't figure out if it should be polynomially smaller (eg. of size $n!/n^3$), or more dramatically smaller (eg. of size $(.5n)!$).

The largest group I could come up with is {$\phi(x_1,\dots,x_{n/2}) \circ \phi(x_{n+1},\dots,x_n) phi(x_{n/2+1},\dots,x_n) | \phi \in S_{n/2}$}, which has size $(.5n)!$.

1

# Largest permutation group without 2-cycles or 3-cycles

The largest permutation group without 2-cycles is $A_n$, which has size $n!/2$. I think the largest permutation group without 2-cycles or 3-cycles is much smaller, but I can't figure out if it should be polynomially smaller (eg. of size $n!/n^3$), or more dramatically smaller (eg. of size $(.5n)!$).

The largest group I could come up with is {$\phi(x_1,\dots,x_{n/2}) \circ \phi(x_{n+1},\dots,x_n) | \phi \in S_{n/2}$}, which has size $(.5n)!$.


text:
f

how to redefine \begin{align*} ... \end{align*} to \begin{eqnarray*} .... \end{eqnarray*}

Hello;

I am not good with Latex syntax. I think I need to use
\renewenvironment, but I am not getting the syntax right.

Scientific word insists in generating displayed math using

\begin{align*} ... \end{align*}

and htlatex has a bug in that it does not generate bit mapped
images in this case, so that math looks ugly.

The fix that I found after trial and error is to replace the
above with

\begin{eqnarray*} ....  \end{eqnarray*}

So I assume I need to put something in the preamble to
tell it that. (I know how to make sure this is effective
only only htlatex is running and not pdflatex).

So I just need to know what the syntax is supposed to be
(I need to learn Latex programming more).

(ps. I first found this bug in Nov 2009, and to this day, it
is still not fixed, 3 years later.

http://12000.org/my_notes/on_tex4ht_displayed_math/note.htm

thank you,
--Nasser

 0
Nasser
2/11/2013 2:43:09 AM
comp.text.tex 39029 articles. 3 followers.

1 Replies
458 Views

Similar Articles

[PageSpeed] 44

FYI;

Please no need to look into this issue. I have
posted this question on another forum and will

thank you,
--Nasser

On 2/10/2013 8:43 PM, Nasser M. Abbasi wrote:
> Hello;
>
> I am not good with Latex syntax. I think I need to use
> \renewenvironment, but I am not getting the syntax right.
>
> Scientific word insists in generating displayed math using
>
>    \begin{align*} ... \end{align*}
>
> and htlatex has a bug in that it does not generate bit mapped
> images in this case, so that math looks ugly.
>
> The fix that I found after trial and error is to replace the
> above with
>
> \begin{eqnarray*} ....  \end{eqnarray*}
>
> So I assume I need to put something in the preamble to
> tell it that. (I know how to make sure this is effective
> only only htlatex is running and not pdflatex).
>
> So I just need to know what the syntax is supposed to be
> (I need to learn Latex programming more).
>
> (ps. I first found this bug in Nov 2009, and to this day, it
> is still not fixed, 3 years later.
>
> http://12000.org/my_notes/on_tex4ht_displayed_math/note.htm
>
> thank you,
> --Nasser
>

 0
Nasser
2/11/2013 6:40:31 AM

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joke: LaTeX Error: \Begin{document} ended by \end{document}.
\documentclass{article} \begin{document} \def\else{Ha! Ha!} \end{document} Hello! When running the example above, I get the following message: LaTeX Error: \Begin{document} ended by \end{document}. Well. OK. Why is this an error? Regards Ulrich "Ulrich Diez" <ulrich.diez@alumni.uni-tuebingen.de> writes: > \documentclass{article} > \begin{document} > \def\else{Ha! Ha!} > \end{document} > > Hello! > > When running the example above, I get the following message: > > LaTeX Error: \Begin{document} ended by \end{document}. > > Well. OK. Why is this an error? Because you broke LaTeX. Use \newcommand\else{Oops} and find out. Hell, you get bogus "Extra }" errors from fragile commands in moving arguments, without even breaking LaTeX itself. Some other attractive command names are \box, \output, \above, \count, \day, \if, \mark, \number, \over, \read, ... all of which are fundamental TeX commands, used internally. -- Donald Arseneau asnd@triumf.ca Donald Arseneau <asnd@triumf.ca> writes: > > Well. OK. Why is this an error? > Because you broke LaTeX. OK, I did understand the posting to be somewhat tongue-in-cheek (or "cheeky") but I didn't notice the "subject" or realize it was a straight-out joke. Sorry to be a rube. In my defence, I've been typing "less secure" in ernest today. -- Donald Arseneau ...

align, followed by text and an other align
Hello, I would like to write some formulas in the align environment, then some normal text and then again some formulas in the align environment. Is there a way to do so, so that all the formulas align in the same way? Regards, Passien Passien Begoen schrieb: > I would like to write some formulas in the align environment, then some > normal text and then again some formulas in the align environment. Is > there a way to do so, so that all the formulas align in the same way? Hello, try the \intertext command: \begin{align} A &= B \\ \intertext{text} C &= D \end{ali...

begin declare continue handler for sqlstate '42704' begin /* Ignore */ end@
I have a script that looks like this: I'm not getting the syntax for declaring a continue handler. I have: set path sysibm,sysfun,sysproc,sysibmadm@ begin declare continue handler for sqlstate '42704' begin /* Ignore */ end@ alter table can drop constraint can_fk2@ end Which results in this: begin declare continue handler for sqlstate '42704' begin /* Ignore */ end DB21034E The command was processed as an SQL statement because it was not a valid Command Line Processor command. During SQL processing it returned: SQL0104N An unexpected token "declare" was found following "begin ". Expected tokens may include: "JOIN <joined_table>". SQLSTATE=42601 Thanks. gimme_this_gimme_that@yahoo.com wrote: > I have a script that looks like this: > > I'm not getting the syntax for declaring a continue handler. > > I have: > > set path sysibm,sysfun,sysproc,sysibmadm@ > > begin declare continue handler for sqlstate '42704' > begin /* Ignore */ end@ > alter table can drop constraint can_fk2@ > end > > > Which results in this: > > begin declare continue handler for sqlstate '42704' begin /* Ignore */ > end > DB21034E The command was processed as an SQL statement because it was > not a > valid Command Line Processor command. During SQL processing it > returned: > SQL0104N An unex...

Posgres Adding braces at beginning and end of text (html) content
Hi All, I'm fairly new to both tcl and postgres and I'm having this peculiar problem. I'm storing raw html in a text field and I want users who know HTML to update the content in a textarea field. The problem is that postgres is adding braces to the beginning and ending of the content -- On creation and every time I update, using ncgi and tclobdc. ie I input:
Hello World

xyz
 .......... into the text area field, save: set page_content [ ncgi::value fieldname] database connect dbh $datasource$dbuser $dbpassword set sql "INS... Begin..End Sequence Hello all I have a window with a button-method to save my transaction which in turn calls 2 more methods e.g. Method SaveTrans() Class MyWind ...... Commands to save transaction then SELF:UpdateCustomer() SELF:UpdateAsset() RETURN NIL If any one should fail, I do not want to complete the others so a BEGIN..END SEQUENCE should work. My question is where is the best place to start the SEQUENCE and the best places to put the BREAK statements? Thanks Trevor Hi Trevor BEGIN SEQUENCE SELF:UpdateCustomer() SELF:UpdateAsset() RECOVER (eventu... The beginning of the end for MS SEATTLE - Microsoft Corp. said on Thursday that its quarterly profit and revenue rose thanks to strong sales of personal computers. The Redmond, Washington-based company reported a net profit of$2.69 billion, or 25 cents per share, for its fourth fiscal quarter ended June 30, compared with a profit of \$1.48 billion, or 14 cents per share, a year earlier. http://www.msnbc.msn.com/id/5489825/ DFS comments: Earnings are up only 82% from the same quarter a year ago?!?!? Horrible! They're dying for sure. It's only a matter of time. You all saw it coming, years ago. The end is nea...

valarray, why no begin() / end()?
Does any one know why there is no begin() / end() functions for valarray? -- Christopher Diggins http://www.cdiggins.com [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] "christopher diggins" <cdiggins@videotron.ca> writes: | Does any one know why there is no begin() / end() functions for valarray? Because they are thought of as "whole thingy" collections rather than sequences. You might argue that associative containers are not sequences, but the difference here is that, fun...

The Beginning of the End of the Internet?
FCC Policies That Damaged Media Now Threatening Internet REMARKS OF MICHAEL J. COPPS FEDERAL COMMUNICATIONS COMMISSIONER "THE BEGINNING OF THE END OF THE INTERNET? DISCRIMINATION, CLOSED NETWORKS, AND THE FUTURE OF CYBERSPACE" as presented to NEW AMERICA FOUNDATION WASHINGTON, DC OCTOBER 9, 2003 http://hraunfoss.fcc.gov/edocs_public/attachmatch/DOC-239800A1.doc Forwarded from the Private Wireless Forum for Mobile Communication Professionals http://groups.yahoo.com/group/PrivateWirelessForum ...

As most people know by now Hyperthreading, has a security flaw found by researchers working with the BSD projects. http://www.daemonology.net/hyperthreading-considered-harmful/ Another good article can be found here: http://www.eweek.com/article2/0,1759,1815954,00.asp After reading the eweek article, it seems that a few people are miffed at this discovery, one even suggests it might be easier to just take the machine. Another one suggest that it would best be solved in a software update, and for now to just turn off HyperThreading. We know that Intel is moving away from Netb...

latex eqnarray alignment
I have a long list of mathematical equations that I want to align. The alignment I want is given by \begin{array}{lcc} a& = & 1 b& = & 2 .......etc \end{array} However the list of equations is so long that it extends to more than one page and the 'array' environment does seem to wrap over more than one page. I can fix the page breaking problem by using the 'eqnarray*' environment: \begin{eqnrray*} a& = & 1 b& = & 2 .......etc \end{eqnarray*} but now the alignment is effectively {rcl} instead of the {lcc} that I want. Can anyone suggest a...

The beginning of the end for google?
The beginning of the end for google? ------------------------------------ Google has just 'innocently' agreed to skew search results from not showing up popular sites for media content to please Hollywood. This innocent move effectively puts 3rd parties in control of the google search results. The very same thing that killed all other search engines because their politics and revenue streams directly interfered with search results leading to search results that have no connection to end user expectations. If 3rd parties are now in charge of google's search res...

the beginning of the end for SCO
IBM Takes New Tack in SCO Legal Battle Steven J. Vaughan-Nichols Aug 17 2004 IBM is trying to knock out SCO's Linux intellectual property claims with a new tactic. [...] "The focus of IBM's motion is not on the Unix System V code that SCO claims IBM copied. It is instead on source code that IBM claims was developed independently, is not subject to copyright infringement claims, and so (says IBM) can freely be contributed to Linux." IBM does not say, however, that it actually had contributed any code to Linux. [Kelly Talcott - Kirkpatrick & Lockhart] This also means...

Temporaries
Sorry about this, but its driving me up the wall. First some code: typedef unsigned int size_t; struct AddOp { template<class I1, class I2> static inline float call(size_t i, const I1& i1, const I2& i2) { return i1[i] + i2[i]; } }; struct MultOp { template<class I1, class I2> static inline float call(size_t i, const I1& i1, const I2& i2) { return i1[i] * i2[i]; } }; class Expr { float data; public: Expr(float arg_data) : data(arg_data) {} float operator[](size_t i) const {return data;} }; template<class I1, class I2, class Op> class Expr3 { const I1& i1; const I2& i2; public: Expr3(const I1& ai1, const I2& ai2) : i1(ai1), i2(ai2) { } float operator[](size_t i) const { return Op::call(i, i1, i2); } template<class E> Expr3<Expr3, E, AddOp> operator+(const E& e) const { return Expr3<Expr3, E, AddOp>(*this, e); } }; struct Vec4 { Vec4(float all) : x(all), y(all), z(all), w(all) { } Vec4(float ax, float ay, float az, float aw = 1) : x(ax), y(ay), z(az), w(aw) { } float x, y, z, w; float operator[](size_t i) const { switch(i) { case 0: return x; case 1: return y; case 2: return z; case 3: return w; default: return 0xdeadbeef; } // HACKHACKHACK :-) // return (&x)[i] } const Vec4& operator=(float arg) { x = y = z = w = arg; return *this; } Expr3<Vec4, Expr, MultOp> operator*(float e) const { // **** r...

The Beginning of the End for Microsoft
Contrary to what most people believe, the Beginning of the End for Microsoft will NOT start by events unfolding in the United States but in the European Union, the most powerfull economic block in the world. The Beginning will be marked by the day that the European Commission announces that henceforth ALL personal computers sold in the EU must be sold seperately from the operating system. From that day on, Microsoft will lose at least half of all new sales to Linux and support for the Linux operating system by ISV's will increase markedly. Huge competition breaks out between Linux vendors...

beginning and ending of a month
Is there an easy way to determine that a date is withing a particular month of a year? On Jun 7, 9:21 am, william kossack <wskoss...@comcast.net> wrote: > Is there an easy way to determine that a date is withing a particular > month of a year? Welllll the easiest way is a simple if statement with the between dates... (using > < etc..); But it reallys depends on how robust/ automated you want your program to be... Example if you want to flag Oct2007 obs... (if you only want to output then use 'then output;') see below.. Remember you asked for the easiest way! no do...

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Q&A for professional and enthusiast programmers

The TeX Catalogue OnLine, Alphabetic Ctan Edition
Search - Home - Alpha - Brief - Topical - Hierarchical TeX Catalogue (Alphabetic Ctan Edition) Graham Williams A - B - C - D - E - F - G - H ...

Fluxions for Fun and Profit: Euler, Trapezoidal, Verlet, or Runge-Kutta?
First of all, just pretend I didn't bring up anything complicated. We're exposed to the effects of differential equations every day, whether ...

Fundamental Theorem of Arithmetic
Let us begin by noticing that, in a certain sense, there are two kinds of natural number : composite numbers and prime numbers. Composite numbers ...

A Line Charge: Electrostatic Potential and Field
The potential of a finite line charge is calculated. Approximations are performed and shown to correspond to known results, i.e. the infinite ...

Sharpe Ratio for Algorithmic Trading Performance Measurement
Discusses the Sharpe Ratio is applied to algorithmic trading strategies, as well as its limitations and drawbacks

Order Statistics 10/30
Order Statistics 10/30

What if we got the sign wrong on monetary policy?
I've been following with interest the rumblings of economists playing with an amazing idea what if we have the sign wrong on monetary policy? ...

2.3.2. What is artifact (single-point) calibration?
The operational definition of bias is that it is thedifference between values that would be assigned to an artifact by theclient laboratory and ...

Predictability and correlation
Today another little note that I discovered while teaching. Warning: this will only be of any interest at all to time-series finance academics. ...

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