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# Silver bromide

Silver bromide

chembox new
OtherNames = bromargyrite
bromyrite
silver(I) bromide
Section1 = Chembox Identifiers
CASNo = 7785-23-1

Section2 = Chembox Properties
Formula = AgBr
MolarMass = 187.772 g/mol
Appearance = Pale yellow solid
photosensitive
Density = 6.473 g/cm3, solid
MeltingPt = 432 °C
BoilingPt = >1502°C
Boiling_notes = decomposes
Solubility =14 μg/100 ml (20 °C)
SolubilityProduct = 5.4 × 10 -13
BandG

ElectronMobility = 4000 cm2/V-s
RefractIndex = 2.253

Section3 = Chembox Structure
CrystalStruct =
Coordination =
MolShape =

Section4 = Chembox Thermochemistry
DeltaHf = −99.5 kJ/mol
Entropy = 107.1 J.K−1.mol−1
HeatCapacity =270 J.kg–1.K–1

Section8 = Chembox Related
OtherAnions = Silver(I) fluoride
Silver chloride
Silver iodide
OtherCations = Copper(I) bromide
Mercury(I) bromide

Silver bromide (AgBr), a soft, pale-yellow, insoluble salt well known (along with other silver halides) for its unusual sensitivity to light. This property has allowed silver halides to become the basis of modern photographic materials. Greenwood, N.N., Earnshaw, A. (1984) "Chemistry of the Elements." New York: Permagon Press. pp. 1185–87.] AgBr is widely used in black-and-white photography film and is believed by some to have been used for faking the Shroud of Turin. Due to these photosensitive properties silver bromide is considered also an ionic semiconductor.cite journal
author=Hamilton, J.F.
title=Physical Properties of Silver Halide Microcrystals
journal=Photographic Science and Engineering
year=1974 | volume=18 | issue=5 | pages= 493–500
] The salt can be found naturally as the mineral bromargyrite (bromyrite).

Preparation

Although the compound can be found in mineral form, AgBr is typically prepared by the reaction of silver nitrate with an alkali bromide, typically potassium bromide:

:AgNO3(aq) + KBr(aq) → AgBr(s)+ KNO3(aq)

Although less convenient, the salt can be also prepared directly from its elements

Modern preparation of a simple, light-sensitive surface involves forming an emulsion of silver halide crystals in a gelatine, which is then coated on a film or other support. The crystals are formed by precipitation in a controlled environment to produce small, uniform crystals (typically < 1 μm in diameter and containing ~1012 Ag atoms) called grains.

Physical properties

Crystal structure

AgF, AgCl, and AgBr all have face-centered cubic (fcc) rock-salt (NaCl) lattice structure with the following lattice parameters:cite journal
author=Glaus, S. and Calzaferri, G.
title=The band structures of the silver halides AgF, AgCl, and AgBr:A comparative study
journal= Photochem. Photobiol. Sci
year=2003 | volume=2 | issue= | pages= 398–401 | doi=10.1039/b211678b
]

The larger halide ions are arranged in a cubic close-packing, while the smaller silver ions fill the octahedral gaps between them, giving a 6-coordinate structure where a silver ion Ag+ is surrounded by 6 Br- ions and vice-versa. The coordination geometry for AgBr in the NaCl structure is unexpected for Ag(I) which typically forms linear, trigonal (3-coordinated Ag) or tetrahedral (4-coordinated Ag) complexes.

Unlike the other silver halides, iodargyrite (AgI) contains a hexagonal zincite lattice structure.

olubility

The silver halides have a wide range of solubilities, noting that the solubility of AgF is about 6 x 107 times greater than that of AgI. These differences are attributed to the relative solvation enthalpies of the halide ions; the enthalpy of solvation of fluoride is anomalously large.Lide, David R. (ed). (2005)"Handbook of Chemistry and Physics", 86th Edition, The Chemical Rubber Publishing Co., Cleveland. ]

Photosensitivity

Although photographic processes have been in development since the mid 1800’s, there were no suitable theoretical explanations until 1938 with the publication of a paper by R.W. Gurney and N.F. Mott. (Proc. Roy. Soc. A164, 151-167 (1938)). This paper triggered a large amount of research in fields of solid-state chemistry and physics, as well more specifically in silver halide photosensitivity phenomena.

Moreover, further research into this mechanism revealed that the photographic properties of silver halides (in particular AgBr) were a result of deviations from an ideal crystal structure. Factors such as crystal growth, impurities, and surface defects all contribute to affect concentrations of point ionic defects and electronic traps, which subsequently affect the sensitivity to light and allow for the formation of a latent image.

;Frenkel defects and quadropolar deformationThe major defect in silver halides is the Frenkel defect, where silver ions are located interstitially (Agi+) in high concentration with their corresponding negatively-charged silver ion vacancies (Agv-). What is unique about AgBr Frenkel pairs is that the interstitial Agi+ are exceptionally mobile, and that its concentration in the layer below the grain surface (called the space charge layer) far exceeds that of the intrinsic bulk. cite journal
author=Slifkin, L. M.
title=The Physics of Lattice Defects in Silver Halides
journal=Crystal Lattice Defects and Amorphous Materials
year=1989 | volume=18 | issue= | pages= 81–96
] The formation energy of the Frenkel pair is low at 1.16 eV, and the migration activation energy is unusually low at 0.05 eV (compare to NaCl: 2.18 eV for the formation of a Schottky pair and 0.75 eV for cationic migration). These low energies result in large defect concentrations, which can reach near 1% near the melting point.

The cause of the low activation energy in silver bromide can be attributed the silver ions’ high quadrupolar polarizability; that is, it can easily deform from a sphere into an ellipsoid. This property, a result of the d9 electronic configuration of the silver ion, facilitates migration in both the silver ion and in silver ion vacancies, thus giving the unusually low migration energy (for Agv-: 0.29-0.33 eV, compared to 0.65 eV for NaCl).

Studies have demonstrated that the defect concentrations are strongly affected (up to several powers of 10) by crystal size. Most defects, such as interstitial silver ion concentration and surface kinks are inversely proportional to crystal size, although vacancy defects are directly proportional. This phenomenon is attributed to changes in the surface chemistry equilibrium, and thus affects each defect concentration differently.

Impurity concentrations can be controlled by crystal growth or direct addition of impurities to the crystal solutions. Although impurities in the silver bromide lattice are necessary to encourage Frenkel defect formation, studies by Hamilton have shown that after a particular concentration of impurities, the number of defects of interstitial silver ions and positive kinks reduce sharply by several orders of 10. After this point only silver ion vacancy defects, which actually increase in several orders of magnitude, are prominent.

;Electron traps and hole trapsWhen light is incident on the silver halide grain surface, a photoelectron is generated when a halide loses its electron to the conduction band: cite journal
author=Malinowski, J.
title=The Role of Holes in the Photographic Process
journal=The Journal of Photographic Science
year=1968 | volume=16 | issue=2 | pages= 57–62
]

:X + hν → X + e

After the electron is released, it will combine with an interstitial Agi+ to create a silver metal atom Agi0:

:e- + Agi+ → Agi0

Through the defects in the crystal the electron is able to reduce its energy and become trapped in the atom. The extent of grain boundaries and defects in the crystal affect the lifetime of the photoelectron, where crystals with a large concentration of defects will trap an electron much faster than a purer crystal.

When a photoelectron is mobilized a photohole h• is also formed, which, too, needs to be neutralized. The lifetime of a photohole, however, does not correlate with that of a photoelectron. This detail suggests a different trapping mechanism; Malinowski suggests that the hole traps may be related to defects as a result of impurities. Once trapped, the holes attract mobile, negatively-charged defects in the lattice: the interstitial silver vacancy Agv-:

:h• + Agv- <-> h.Agv

The formation of the h.Agv lowers its energy sufficiently to stabilize the complex and reduce the probability of ejection of the hole back into the valance band (the equilibrium constant for hole-complex in the interior of the crystal is estimated at 10-4.

Additional investigations on electron- and hole-trapping demonstrated that impurities also can be a significant trapping system. Consequently, interstitial silver ions may not be reduced. Therefore, these traps are actually oss mechanisms and are considered trapping inefficiencies. For example, atmospheric oxygen can interact with photoelectrons to form an O2- species, which can interact with a hole to reverse the complex and undergo recombination. Metal ion impurities such as copper(I), iron(II), and cadmium(II) have demonstrated hole-trapping in silver bromide.

;Crystal surface chemistry;Once the hole-complexes are formed, they diffuse to the surface of the grain as a result of the formed concentration gradient. Studies demonstrated that the lifetime of holes near the surface of the grain are much longer than those in the bulk and that these holes are in equilibrium with adsorbed bromine. The net effect is an equilibrium push at the surface to form more holes. Therefore, as the hole-complexes reach the surface, they disassociate:

:h.Agv- → h• + Agv- → Br → ½ Br2

By this reaction equilibrium, the hole-complexes are constantly consumed at the surface, which acts as a sink, until removed from the crystal. This mechanism provides the counterpart to the reduction of the interstitial Agi+ to Agi0, giving an overall equation of:

:AgBr -> Ag + ½ Br2

;Latent image formation and photographyNow that some of the theory has been presented, the actual mechanism of the photographic process can now be discussed. To summarize, as a photographic film is subjected to an image, photons incident on the grain produce electrons which interact to yield silver metal. More photons hitting a particular grain will produce a larger concentration of silver atoms, containing between 5 and 50 silver atoms (out of ~1012 atoms), depending on the sensitivity of the emulsion. The film now has a concentration gradient of silver atom specks based upon varying intensity light across its area, producing an invisible "latent image". While this process is occurring, bromine atoms are being produced at the surface of the crystal. To collect the bromine, a layer on top of the emulsion, called a sensitizer, acts as a bromine acceptor.

During film development the latent image is intensified by addition of a chemical, typically hydroquinone, that selectivity reduces those grains which contain atoms of silver. The process, which is sensitive to temperature and concentration, will completely reduce grains to silver metal, intensifying the latent image on the order of 1010 to 1011. This step demonstrates the advantage and superiority of silver halides over other systems: the latent image, which takes only milliseconds to form and is invisible, is sufficient to produce a full image from it.

After development, the film is "fixed," during which the remaining silver salts are removed to prevent further reduction, leaving the "negative" of the film. The agent used is sodium thiosulphate, and reactions according to the following equation:

:AgX(s) + 2Na2S2O3(aq) -> Na3 [Ag(S2O3)2] (aq) + NaX(aq)

An indefinite number of positive prints can be generated from the negative by passing light through it and undergoing the same steps outlined above.

emiconductor properties

As silver bromide is heated within 100°C of its melting point, an Arrhenius plot of the ionic conductivity shows the value increasing and "upward-turning." Other physical properties such as elastic moduli, specific heat, and the electronic energy gap also increase, suggesting the crystal is approaching instability. This behavior, typical of a semi-conductor, is attributed to a temperature-dependence of Frenkel defect formation, and, when normalized against the concentration of Frenkel defects, the Arrhenius plot linearizes.

Chemistry

Silver bromide reacts readily with liquid ammonia to generate a variety of amine complexes:Leden, I., Persson, G., (1961). ACSAA4. "Acta Chem. Scand." 15: 607 - 614. ]

AgBr + nNH3 $longrightarrow$ Ag(NH3)21+:::::{AgBr(NH3)2}
{AgBr2(NH3)2}1-
{AgBr(NH3)}
{AgBr2(NH3)}1-

Silver bromide reacts with triphenylphosphine to give a tris(triphenylphosphine) product:Engelhardt L.M., Healy P.C., Patrick, V.A., and White, A.H. (1987). "Lewis-Base Adducts of Group-11 Metal(I) Compounds. XXX. 3:1 Complexes of Triphenylphosphine With Silver(I) Halides." Aust. J. Chem. 40 (11). 1873-1880.] it is something that makes the black and white pictures clear

::

References

ee also

*Photography
*Science of photography
*Silver chloride

Wikimedia Foundation. 2010.

### Look at other dictionaries:

• silver bromide — n. a yellow white, crystalline compound, AgBr, which becomes dark when exposed to light: the main salt used in photographic emulsions …   English World dictionary

• silver bromide — noun a bromide that darkens when exposed to light; used in making photographic emulsions • Hypernyms: ↑bromide • Substance Holonyms: ↑emulsion, ↑photographic emulsion * * * noun Etymology: silver ( …   Useful english dictionary

• silver bromide — sidabro bromidas statusas T sritis chemija formulė AgBr atitikmenys: angl. silver bromide rus. серебра бромид; серебро бромистое …   Chemijos terminų aiškinamasis žodynas

• silver bromide — Chem. a yellowish, water insoluble powder, AgBr, which darkens on exposure to light, produced by the reaction of silver nitrate with a bromide: used chiefly in the manufacture of photographic emulsions. [1875 80] * * * …   Universalium

• silver bromide — /sɪlvə ˈbroʊmaɪd/ (say silvuh brohmuyd) noun a light sensitive compound, AgBr, which is of basic importance in photography. It is obtained from the reaction of a bromide with a silver salt …   Australian English dictionary

• silver bromide — noun Date: 1869 a compound AgBr that is extremely sensitive to light and is much used for photographic materials …   New Collegiate Dictionary

• silver bromide — sil′ver bro′mide n. chem. a yellowish powder, AgBr, that darkens on exposure to light: used chiefly in photographic emulsions • Etymology: 1875–80 …   From formal English to slang

• Bromide (photography) — Bromide, in photography, refers to a type of photographic printing paper coated in an emulsion of silver bromide, or a photograph made thereof. Images are imprinted on the bromide paper via the gelatin silver process.It was invented in the 19th… …   Wikipedia

• Silver nitrate — Silver nitrate …   Wikipedia

• Bromide (disambiguation) — Bromide can refer to:In language: * Bromide (language), a figure of speech meaning a tranquilizing cliché.In chemistry: * Bromide, the anion of bromine, or any ionic salt containing bromide as the only anion, or (as a common name) any covalent… …   Wikipedia


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# Homework Help: Orthogonal Basis and Inner Products

1. Mar 3, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

3. The attempt at a solutionSince A is a vector in V and since the $A_i$ form a basis, we can write A as a linear combination of the $A_i$. We write $A = x_1 A_1 + ... + x_n A_n$. Thus, we have,

$<x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i>$. Because two orthogonal vectors, when multiplied via inner product together give the zero vector, we simplify,

$0 = x_i <A_i,A_i>$. Because we have presumed that the $A_i ≠ 0$, we cannot have $<A_i,A_i> = 0$, so we must have that the $x_i = 0$. So we have $<A,A_i> = <x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i> = x_i <A_i,A_i> = 0 <A_i,A_i> = <0,A_i> = 0. Which shows that A = 0. I don't know if I've done this correctly. I feel like it's not entirely convincing towards the end. Advice? 2. Mar 3, 2012 ### Deveno you're fine up to here. at this point, stop thinking about the "inner products", and start thinking about which linear combination of the Ai, A has to be. 3. Mar 3, 2012 ### HallsofIvy You can say more than that- since you are told that the basis is "orthonormal", you know that [itex]<A_i, A_i>= 1$

Once you have $x_i= 0$ for all x, you have immediately that $A= 0A_1+ 0A_2+ ...= 0$

4. Mar 3, 2012

### TranscendArcu

Okay. That makes sense. Thank you both.

This problem is related to the idea of the inner product. I don't think I need any help on the first parts, but I'd like to talk about the third part of the problem.

So, finding the length of a given vector given this inner product:
$<(x,y),(x,y)> = 5x^2 + y^2$.

Taking the length, we have

$|(x,y)| = \sqrt{5x^2 + y^2}$, which we define as equaling 1. Squaring both sides we find,

$5x^2 + y^2 = 1$. I think this is the equation of the circle, but I'm not sure. If it is, then my picture has y-intercepts at 1,-1 and x-intercepts at .5,-.5.

Is this correct?

format: plain


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#### You may also like

A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles?

### Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

When experimenting with numbers in equations it is often a good idea to try common numbers such as $0, 1, 2$ and $-1$. You might also try really big numbers such as a million or really small numbers such as $0.000001$.


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## anonymous 4 years ago An equilateral triangle with sides of length 6 is inscribed in a circle. What is the diameter of the circle?

1. anonymous

|dw:1337641995914:dw| because your dealing with a 30-60-90 triangle, you can solve that triangle for r: 3^2 = (3/2)^2 + r^2 solve for r, your radius.

2. anonymous

Okay! Thank you! :)

3. anonymous

yw... you want an answer verification?

4. anonymous

5. anonymous

ok.. tell me what u got an i'll tell you mine...

6. anonymous

Well, for the answer to the question, I got 6.9.. Is that right?

7. anonymous

i got 3(sqrt3)/2 or about 2.6

8. anonymous

hmm... wanna go over the work?

9. asnaseer

6.9 is the correct answer for the diameter

10. anonymous

ohh... hahah... thanks @asnaseer ..

11. asnaseer

yw :)

12. anonymous

sorry, I meant @octoberchick , your correct... :)

13. asnaseer

btw: I did it in a slightly different way: |dw:1337643034380:dw| bisect one of the angles of the equilateral triangle to get...

14. asnaseer

|dw:1337643091539:dw|

15. asnaseer

then, in the right angled triangle ABC:$\cos(30)=6/d$therefore:$d=6/\cos(30)$

16. anonymous

wow.... isn't it great having to see other methods? and this is still a 30-60-90 triangle so trig functions aren't really needed still. thanks... :)

17. asnaseer

true... :)

18. anonymous

and i think that is a lot more clear ^^^... :)


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## The footprint bound

Studying the set of common zeros of systems of polynomial equations is a fundamental problem in algebra and geometry. In this post we will look at estimating the cardinality of the set of common zeros, when we already know that it is finite. In fact, with some extra theory one can also determine whether the set of zeros is finite or not.

My main motivation here is to popularise the so-called footprint bound, especially among combinatorialists (combinators?).  As an application, we will see a nice “conceptual” proof of the so-called Alon-Füredi theorem, which is a result that in particular solves the following well-known geometric problem whose $3$-dimensional case appeared as Problem 6 in IMO 2007:

Let $S$ be a finite subset of a field $F$ and let $a \in S^n$. What is the minimum number of hyperplanes in $F^n$ that can cover all the points in $S^n$, while missing the point $a$?

(The answer is $n(|S| - 1)$.)

To be able to state the footprint bound we need some definitions. A monomial order is a total order on the set of all monomials in $F[x_1, \dots, x_n]$ which satisfies the property $u \leq v \implies uw \leq vw$ for any monomials $u, v, w$. For example, we can look at the graded lexicographic order in which we first compare the total degree of the monomials and if that’s equal then we compare them lexicographically, i.e., for $u = (u_1, \dots, u_n)$ and $v = (v_1, \dots, v_n)$ we have $\prod x_i^{u_i} > \prod x_i^{v_i}$ if $\sum u_i > \sum v_i$ or if $\sum u_i = \sum v_i$ and the left-most non-zero entry of the vector $u - v$ is positive (so for example, we will have $x_1^2 x_2^3 \geq x_1^2 x_2 x_3^2$).

Now given a polynomial $f$ in $F[x_1, \dots, x_n]$ and a monomial ordering $\leq$, we denote the largest monomial  appearing in $f$ w.r.t. “$\leq$” by $LM(f)$ (a.k.a. leading monomial). And if $J$ is an ideal of polynomials then we define $LM(J) = \{LM(f) : f \in J\}$. Given an ideal $J$, we are going to give an upper bound on the set $V(J) = \{x \in F^n : f(x) = 0~\forall f \in J\}$. For that, let $\Delta(J)$ denote the set of those monomials which are not contained in the (monomial) ideal $\langle LM(J) \rangle$ (these are also known as standard monomials). Then this set $\Delta(J)$ is known as the footprint of the ideal $J$.

Theorem 1 (Footprint bound [7], [5,§5.3]) If $\Delta(J)$ is finite, then we have $|V(J)| \leq |\Delta(J)|$. Moreover, the bound is sharp if the ideal $J$ is radical.
Proof. Exercise! (show that $\Delta(J)$ is a basis of the vector space $F[x_1, \dots, x_n]/J$.)

Of course, the usefulness of this bound (at least in combinatorial situations) depends on how easily we can calculate, or estimate, $|\Delta(J)|$. Say $J$ is generated by $g_1, \dots, g_r$ and let $\Delta(g_1, \dots, g_r) = \{x^u : x^u \not \in \langle LM(g_i) \rangle ~\forall i\}$. Then clearly $\Delta(J) \subseteq \Delta(g_1, \dots, g_r)$, and since we have a finite list of monomials to check, with right choice of $g_i$‘s it can be easier to compute $\Delta(g_1, \dots, g_r)$. Ideally, we would like $\Delta(J)$ to be equal to $\Delta(g_1, \dots, g_r)$, and in fact this does happen when $g_1, \dots, g_r$ forms a Gröbner basis of $J$ (if you know what a Gröbner basis is; but you don’t really need to for the rest of this post). Let’s consider an example. Let $n = 2$ and $J = \langle x^2 y^3 - y, x^3y - x, x^4 - y^3, y^4 - xy^2 \rangle \subset F[x, y]$, and consider the graded lexicographical order. Then the multiples of the leading monomials $x^2y^3, x^3y, x^4, y^4$ can be seen pictorially as follows:

The monomials which are multiples of these four monomials, i.e. the monomials in the ideal generated by these leading monomials, correspond to the integer points in the shaded region. Therefore, the elements of $\Delta(x^2 y^3 - y, x^3y - x, x^4 - y^3, y^4 - xy^2)$ corresponds to the points in the unshaded region, and hence $|\Delta(J)| \leq |\Delta(x^2 y^3 - y, x^3y - x, x^4 - y^3, y^4 - xy^2)| = 12$, which implies that these polynomials have at most $12$ common solutions (Wolfram Alpha tells me that there are in fact only $2$).

Say we are now interested in estimating the number of zeros of a polynomial $f \in F[x_1, \dots, x_n]$ contained in a finite grid $S = S_1 \times \cdots \times S_n$, where $S_i \subseteq F$, assuming that $f$ does not vanish on all points of $S$. This is precisely what the Alon-Füredi theorem is all about. Let $s_i = |S_i|$ and $g_i = \prod_{a \in S} (x_i - a)$. Then $g_i$ is a polynomial of degree $s_i$, and the points of the $S$ are the common solutions of the polynomials $g_1, \dots, g_n$. What we are interested in is $V(J)$ for $J = \langle f, g_1, \dots, g_n \rangle$. For that we will first assume that $f$ is in its reduced form, so that $LM(f)$ (with respect to any monomial ordering) is of the form $\prod x_i^{u_i}$ where $0 \leq u_i \leq s_i - 1$ for all $i$. We can always do this by reducing higher powers of $x_i$ by using the equation $g_i(x_i) = 0$ (the so-called grid reduction [3, Chapter 8]). Say $\deg (f) = d$ and $x^u = \prod x_i^{u_i}$ is the leading monomial of $f$, with $\sum u_i = d$. Then $\Delta(x^u, x_1^{s_i}, \dots, x_n^{s_n})$ is equal to the set of all monomials of the form $\prod x_i^{v_i}$ where there exists an index $j$ for which $v_j < u_j$. The number of such monomials is in fact equal to $\prod s_i - \prod (s_i - u_i)$ because $\prod s_i$ is the total number of reduced monomials, and $\prod (s_i - u_i)$ of them are those which are multiples of $x^u$. If we let $c_i = s_i - u_i$, then we have $1 \leq c_i \leq s_i$ and $\sum c_i = (\sum s_i) - d$, which directly implies the Alon-Füredi theorem:

Theorem 2 (Alon-Füredi [1, Theorem 5]) Let $f$ be a polynomial which does not vanish on the entire grid $S = S_1 \times \cdots \times S_n$, then $f$ does not vanish on at least $\min \{ \prod c_i : 1 \leq c_i \leq s_i ~\forall i,~ \sum c_i = (\sum |S_i|) - \deg f\}$ points of $S$.

Compared to the other proof of the Alon-Füredi theorem, I feel like this proof gives us a better understanding of where that bound comes from and “why” this theorem is true. Also, the same proof also gives the generalized version of the Alon-Füredi theorem [3, Theorem 9.1.2] (over fields). Moreover, one can also prove Alon’s combinatorial nullstellensatz using the footprint bound as follows.

Theorem 3 (Combinatorial Nullstellensatz) Let $f$ be such that $LM(x) = x^u$  with $u = (u_1, \dots, u_n)$ satisfying $u_i < |S_i|$ for all $i$. Then there exists an $a \in S$ for which $f(a) \neq 0$.
Proof. The size of the footprint of the ideal $I = \langle f, g_1, \dots, g_n \rangle$ is strictly less than $|S|$ since $u$, being the exponent of a leading monomial, is not contained in it. Therefore by the footprint bound we have $|V(I)| < |S|$, which implies that there is a point where $f$ does not vanish.

So, footprint bound is a nice algebraic result which has several interesting applications and extensions [2, 4, 6, 7, 8, 9, 10]. It is particularly useful for studying systems of polynomial equations over a finite field $\mathbb{F}_q$, where we can take the grid $S$ to be equal to $\mathbb{F}_q^n$. For further reading on this bound see the papers in the references below.

And now for the extra theory, here is a characterisation of systems of polynomial equations which have only finitely many common zeros (the so-called zero dimensional affine varieties over the algebraic closure of the field).

Theorem 4 ([5, Page 234]) Let $k$ be an algebraically closed field and let $V = V(I)$ be an affine variety in $k^n$ where $I$ is an ideal in $k[x_1, \dots, x_n]$. Then the following statements are equivalent:

1. $V$ is a finite set.
2. For each $i$, there exists some $m_i \geq 0$ such that $x_i^{m_i} \in \langle LM(I) \rangle$.
3. Let $G$ be a Gröbner basis for $I$. Then for each $i$, there exists some $m_i \geq 0$ such that $x_i^{m_i} = LM(g_i)$ for some $g_i \in G$.
4. The set $\Delta(I)$ is finite.
5. The $k$-vector space $k[x_1, \dots, x_n]/I$ is finite-dimensional.

References

[1] N. Alon and Z. Füredi. Covering the cube by affine hyperplanes. European J. Combin., 14(2):79–83, 1993.
[2] P. Beelen, M. Datta, and S. R. Ghorpade. Vanishing ideals of projective spaces over finite fields and a projective footprint bound. arXiv:1801.09139.
[3] A. Bishnoi, Some contributions to incidence geometry and polynomial method, PhD Thesis (2016), Ghent University.
[4] C. Carvalho. On the second Hamming weight of some Reed–Muller type codes. Finite Fields Appl., 24:88–94, 2013.
[5] D. Cox, J. Little, and D. O’Shea. Ideals, Varieties, and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative Algebra. Springer-Verlag New York, Inc., Secaucus, NJ, USA, 2007.
[6] O. Geil. On the second weight of generalized Reed-Muller codes. Designs, Codes and Cryptography, 48(3): 323-330, 2008.
[7] O. Geil and T. Høholdt. Footprints or generalized Bezout’s theorem. IEEE Trans- actions on Information Theory, 46(2):635–641, 2000.
[8] O. Geil and C. Thomsen. Weighted Reed–Muller codes revisited. Designs, Codes and Cryptography, 66(1):195–220, 2013.
[9] O. Geil and U. Martínez-Penãs. Bounding the number of common zeros of multivariate polynomials and their consecutive derivatives, arXiv:1707.01354.
[10] R. Pellikaan and X.-W. Wu. List decoding of q-ary Reed-Muller codes. IEEE Transactions on Information Theory, 50(4):679–682, 2004.


text:
# Simple Fractional Equations

Welcome to Class !!

In today’s Mathematics class, We will be looking at Simple Fractional Equations. We hope you enjoy the class!

##### SOLVING EQUATION EXPRESSIONS

WORD PROBLEMS

Worked Examples:

1. Find 1/4 of the positive difference between 29 & 11
2. The product of a certain number and 5 is equal to twice the number subtracted from 20. Find the number
3. The sum of 35 and a certain number is divided by 4 the result is equal to double the number. Find the number.

Solutions:

1. Positive Difference 29 – 11 = 18

1. Let the number be x

x X 5 = 20 – 2x

5x = 20 – 2x

5x + 2x = 20

7x = 20

x = 20/7 = 2

1. Let the number be n

sum of 35 and n = n + 35

EVALUATION

1. From 50 subtract the sum of 3 & 5 then divide the result by 6
2. The sum of 8 and a certain number is equal to the product of the number and 3 find the number

###### SOLVING EQUATION EXPRESSIONS WITH FRACTION

Always clear fractions before beginning to solve an equation: –

To clear fractions, multiply each term in the equation by the LCM of the denominations of the fractions.

Examples:

Solve the following

2x + 18 + 10 + 5x = 0

1. 2x =

Multiply by the LCM (14)

28x = 31x – 33

28x – 31x = -33

-3x = -33

x = 33/3 = 11

EVALUATION

Solve the following equations.

##### Furthermore, we can consider the word equations or expressions into:
• Sum & Differences
• Products
• Expressions with fractions & equations

SUM & DIFFERENCES

The sum of a set of numbers is a result obtained when the numbers are added together. The difference between two numbers is a result of subtracting one number from the other.

Worked Examples:

1. Find the sum of -2 & -3.4
2. Find the positive difference between 19 & 8
3. The difference between two numbers is 7. If the smaller number is 7 find the other.
4. The difference between -3 and a number is 8, find the two possible values for the number.
5. Find the three consecutive numbers whose sum is 63.

Solutions:

1. -2 + -3.4 = -5.4
2. 19 – 8 = +11
3. let the number be Y i.e. Y -7 = 7

i.e. Y = 7 + 7 = 14

1. Let M represent the number

M – (-3) = 8

m + 3 = 8

m = 8 – 3

m  = +5

also -3 – m = 8

-m = 8 + 3

-m = 11

m = -11

the possible values are +5 & -11

1. Consecutive numbers are 1, 2, 3,4,5,6, Consecutive odd numbers are

1, 3,5,7,9……….. consecutive even numbers are 2, 4, 6, 8,10……….

Representing in terms of X, we have 2X, 2X + 2, 2X + 4, 2X + 6, 2X + 8, 2X + 10…………

for consecutive even numbers, we have X, X + 2, X + 4, X + 6…….

for consecutive odd numbers, we have  X + 1, X + 2, X + 3, X + 4…

for consecutive numbers.

let the first number be x,

let the second number be x + 1

let the third number be x + 2

Therefore x + x + 1 + x + 2 = 63

3x + 3 = 63

3x = 63 – 3

3x = 60

x = 60 /3

= 20

The numbers are 20, 21, and 22.

EVALUATION

1. Find the sum of all odd numbers between 10 and 20
2. The sum of four consecutive odd numbers is 80 find the numbers
3. The difference between 2 numbers is 9, the largest number is 32 find the numbers.

PRODUCTS

The product of two or more numbers is the result obtained when the numbers are multiplied together.

Worked Examples:

1. Find the product of – 6, 0.7, &
2. The product of two numbers is 8. If one of the numbers is 1/4 find the other.
3. Find the product of the sum of -2 & 9 and the difference between -8 & -5.

Solutions

1. Products -6 x 0.7 x 20/3

-6 x 7/10 x 20/3 =

= -2 x 7 x 2 = -28

1. Let the number be x

$\frac{1}{4}$

X x = 8 multiply both sides by 4

x = 8 x 4 = 33

1. Sum = -2 + 9 = 7

Difference = -5-(-8) = -5 + 8 = 3

Products= 7 x 3 = 21

EVALUATION

1. The product of three numbers is 0.084 if two numbers are 0.7 & 0.2 find the third number
2. Find the product of the difference between 2 & 7 and the sum of 2 & 7
3. From 50 subtract the sum of 3 & 5 then divide the result by 6
4. The sum of 8 and a certain number is equal to the product of the number and 3 find the number

We have come to the end of this class. We do hope you enjoyed the class?

Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.

In our next class, we will be talking about Change Of Subject Formulae. We are very much eager to meet you there.

LEARN TO CODE IN 8 WEEKS. Pay Only ₦25000 To Join Class💃

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Don`t copy text!


text:
E. Convex Countour
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given an strictly convex polygon with n vertices. It is guaranteed that no three points are collinear. You would like to take a maximum non intersecting path on the polygon vertices that visits each point at most once.

More specifically your path can be represented as some sequence of distinct polygon vertices. Your path is the straight line segments between adjacent vertices in order. These segments are not allowed to touch or intersect each other except at the vertices in your sequence.

Given the polygon, print the maximum length non-intersecting path that visits each point at most once.

Input

The first line of input will contain a single integer n (2 ≤ n ≤ 2 500), the number of points.

The next n lines will contain two integers xi, yi (|xi|, |yi| ≤ 109), denoting the coordinates of the i-th vertex.

It is guaranteed that these points are listed in clockwise order.

Output

Print a single floating point number, representing the longest non-intersecting path that visits the vertices at most once.

40 00 11 11 0
3.4142135624


text:
Sam Buss ; Neil Thapen - DRAT and Propagation Redundancy Proofs Without New Variables

lmcs:5750 - Logical Methods in Computer Science, April 23, 2021, Volume 17, Issue 2 - https://doi.org/10.23638/LMCS-17(2:12)2021
DRAT and Propagation Redundancy Proofs Without New Variables

Authors: Sam Buss ; Neil Thapen

We study the complexity of a range of propositional proof systems which allow inference rules of the form: from a set of clauses $\Gamma$ derive the set of clauses $\Gamma \cup \{ C \}$ where, due to some syntactic condition, $\Gamma \cup \{ C \}$ is satisfiable if $\Gamma$ is, but where $\Gamma$ does not necessarily imply $C$. These inference rules include BC, RAT, SPR and PR (respectively short for blocked clauses, resolution asymmetric tautologies, subset propagation redundancy and propagation redundancy), which arose from work in satisfiability (SAT) solving. We introduce a new, more general rule SR (substitution redundancy). If the new clause $C$ is allowed to include new variables then the systems based on these rules are all equivalent to extended resolution. We focus on restricted systems that do not allow new variables. The systems with deletion, where we can delete a clause from our set at any time, are denoted DBC${}^-$, DRAT${}^-$, DSPR${}^-$, DPR${}^-$ and DSR${}^-$. The systems without deletion are BC${}^-$, RAT${}^-$, SPR${}^-$, PR${}^-$ and SR${}^-$. With deletion, we show that DRAT${}^-$, DSPR${}^-$ and DPR${}^-$ are equivalent. By earlier work of Kiesl, Rebola-Pardo and Heule, they are also equivalent to DBC${}^-$. Without deletion, we show that SPR${}^-$ can simulate PR${}^-$ provided only short clauses are inferred by SPR inferences. We also show that many of the well-known "hard" principles have small SPR${}^-$ refutations. These include the pigeonhole principle, bit pigeonhole principle, parity principle, Tseitin tautologies and clique-coloring tautologies. SPR${}^-$ can also handle or-fication and xor-ification, and lifting with an index gadget. Our final result is an exponential size lower bound for RAT${}^-$ refutations, giving exponential separations between RAT${}^-$ and both DRAT${}^-$ and SPR${}^-$.

Volume: Volume 17, Issue 2
Published on: April 23, 2021
Accepted on: February 20, 2021
Submitted on: September 7, 2019
Keywords: Computer Science - Logic in Computer Science,Mathematics - Logic,68T15 (Primary) 03B35, 03B05 (Secondary)


text:
# Why my Tikz picture doesn't rotate?

Here is my code in Tikzformat picture and I made many changes to rotate it, include using some package and changing in angle values but it still doesn't rotate, I don't know what's the problem.

\documentclass[border=10pt]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.8}
\usepackage{tikz, tikz-3dplot}
\begin{document}
\tdplotsetmaincoords{20}{45}
\begin{tikzpicture}[scale=1]
\tdplotsetrotatedcoords{20}{30}{90}
\begin{axis}[
axis lines=center,
axis on top,
xlabel={$x$}, ylabel={$y$}, zlabel={$z$},
view/h=110,
title={\tiny $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, Rotate Please},
]
surf,
z buffer=sort,
mesh/interior colormap={purplegreen}{color=(purple) color=(green)},
colormap={periodic}{%
color=(blue)
color=(yellow)
color=(orange)
color=(red)
color=(orange)
color=(yellow)
color=(blue)},
domain=-2:2, domain y=0:360,
samples=20, samples y=20,
variable=\u, variable y=\v,
point meta=u,
]
({2 * cos(v)},{3 * sin(v)},{u});
\end{axis}
\end{tikzpicture}

\end{document}

Also I don't want the numbers appear on axis', what should I make changes.

• You are using pgfplots! Use view={20}{45} instead! (NO, you cannot use tikz-3dplot's rotation command to rotate pgfplots's plot. They are simply indepedent.) – Symbol 1 Apr 21 '17 at 3:45
• Instead of 'view/h=110' ? – Nosrati Apr 21 '17 at 3:47
• Oops I mislooked that. So you did set the view. Then what you do want to achieve? which part of the picture should be rotated? – Symbol 1 Apr 21 '17 at 3:57
• I want to rotate whole of picture include cylinder and axis'. – Nosrati Apr 21 '17 at 4:02
• Then that is fairly impossible. The axis draws everything on the 2D plane! – Symbol 1 Apr 21 '17 at 4:10

Let us sit down and see what are the options:

First of all, NO, you cannot use \tdplotxxx to rotate axis. The former is from tikz-3dplot and the latter is from pgfplots. They are basically working independently. (off topic: I used to think tikz-3dplot>pgfplots; But now I know pgfplots>>tikz-3dplot.)

The following uses only pgfplots

• If you want to rotate just the data points, not axis, then there is a phase in pgfplots that allows you to modify data points. Search filter points in the manual.

• If you want to rotate the data points together with axis, then you can use view={12}{34} together with rotate=56 (the 2D rotation). The combination of these two should give you all possible 3D rotations.

• If you want to rotate just the axis, combine the previous two methods.

If you want to stick to tikz-3dplot, then you have to give up axis.

• I could't find filter points in my manual file: Tantau Till, TikZ and pgf, Manual for version 3.0.1a (2013). The combine view={12}{34} with rotate=56 gives me something as I wanted. Thank you. – Nosrati Apr 23 '17 at 17:54
• @MyGlasses I mean the manual of pgfplots. Sometimes I mistakenly assume that I am clear. – Symbol 1 Apr 23 '17 at 18:24
• I don't want the numbers appear on axis', what should I make changes in my file, please. – Nosrati Apr 23 '17 at 18:33
• Something like xticklabel={}. – Symbol 1 Apr 23 '17 at 18:40


text:
• Create Account

### #ActualJaey

Posted 18 February 2013 - 09:04 AM

In addition to all of the above points, the standard and most "cross-platform" way to include SDL is:

#include "SDL.h"

Rather than:

#include <SDL/SDL.h>

thanks a lot. that seems to have been the issue. Thanks again

### #1Jaey

Posted 18 February 2013 - 09:03 AM

In addition to all of the above points, the standard and most "cross-platform" way to include SDL is:

#include "SDL.h"

Rather than:

#include <SDL/SDL.h>

thanks man. that seems to have been the issue. Thanks again

PARTNERS


text:
← Back to mit.katmh.com

I attempted to take a shower between :55 and :05 but ended up joining this Zoom at :11 — not bad. I knew how to do the problems after watching lecture and this Organic Chemistry Tutor video yesterday, but this recitation was incredibly helpful in clarifying some concepts. The TA's really good!

• How do we convert to hexadecimal a bitstring whose length is not a multiple of 4? Since it's unsigned, we add 0s to the left side
• Decimal to hexadecimal: 0-9, 10=A, 11=B, 12=C, 13=D...

• I was so caught up in binary <> hex that I forgot that decimal <> hex is also a thing
• To sum hex numbers, convert to binary
• Restrictions, overflow, etc. only apply to signed numbers

Two's complement overview

• Different system, as opposed to binary/unsigned
• Most significant bit is negative; if there's a 1 in the most significant bit, we subtract that value (e.g. $-2^4$)
• To negate a number in two's complement: invert all the digits and add 1
• "Normal" subtraction doesn't work in two's complement; instead, invert the second number and add

Range and mod

• N-bit two's complement number: what's the range of values these N bits could represent? $[-2^{N-1}, 2^{N-1}-1]$
• Most negative: e.g. $N=4$: Leftmost bit is $-2^3$, so most negative number we can represent is $-2^3$, or $-2^{N-1}$
• Most positive will look like $0111...111$: $2^0 + 2^1 + ... + 2^{N-2}$ = sum of geometric series = $2^{N-1} - 1$

• Holy shit I did not realize this was a geometric sum
• How many numbers are in this range: Subtract ends of range and add 1: $2^N$
• Two's complement is modular. Regular mod: N bits can represent $[0, 2^N - 1]$. Here it's shifted over, but the range is the same size: $[-2^{N-1}, 2^{N-1} - 1]$

Overflows with two's complement operations

• Pos + neg: inside range, because most significant bits are 0 and 1
• Pos + pos: sometimes in range
• Neg + neg: sometimes in range
• We drop overflow because two's complement can only represent a certain range of numbers


text:
# What motivated Gauss' quadratic sums?

The proofs of quadratic reciprocity based on quadratic Gaussian sums involve sums of the form

$$\sum_{k=1}^{p-1}\left(\frac k p \right)\zeta^{ak}$$

Where $\left(\frac k p \right)$ is a Legendre symbol and $\zeta$ is a $p$-th root of unity. Where were these sums considered (if in the DA, which section) ? Can someone summarize what motivated these seemingly arbitrary expressions?

One answer is that Gauss sums are special cases of Lagrange resolvents; these in turn are expressions that Lagrange used for analysing the solution of equations by radicals.

Another source are Euler's and Lagrange's factorizations of the cyclotomic equations $4(x^p-1)/(x-1)$ in the form $Y^2 - p^*Z^2$, where $p^* = (-1)^{\frac{p-1}2}p$, for $p = 3, 5, 7$ in connection with problems on special cases of the quadratic reciprocity law. Both of them were unable to find the factorization for $p = 11$, but Gauss realized that the roots of the linear factors of the quadratic polynomials are roots of unity $\zeta^r$ and $\zeta^s$, respectively, where $r$ and $n$ denote the quadratic residues and nonresidues modulo $p$. Once you start writing down the polynomials $(X - \zeta)(X-\zeta^4)(X-\zeta^9)\cdots$ and look at the coefficient of the second largest power of $X$, you will discover quadratic Gauss sums.

Can someone summarize what motivated these seemingly arbitrary expressions?

They are far from arbitrary.

Because quadratic reciprocity is about whether a prime $p$ is a square modulo another prie $q$, it is natural to look for an actual square-root when it is (an idea that was already familiar to Euler). The use of quadratic Gauss sums is to construct the desired square-root (but possibly in a finite extension). With that explicit goal in mind, it is in fact not so hard to see patterns emerging when doing computations, especially if you have been thinking long and hard about Legendre's symbols and even if you don't know the answer beforehand. I have sometimes asked students to guess them for instance and they do succeed (occasionally).

But though the students do not know the result, they have a tremendous advantage: they are conducting the guessing in a course on finite extension of finite fields or algebraic number theory. That idea, i.e the idea of looking for the square-root in extensions (of the integers or of finite fields) and of studying the behavior of primes in such extensions, was absolutely groundbreaking.

• So is the sum in the OP a representation of a square root? – Conifold Apr 8 '15 at 0:04


text:
# Anomaly detection

### I don't define what is normal, but I know it when I see it

Working out when something you are seeing is not something you should have been expecting, in some sense which will not be made rigorous here, not yet.

Nuit Blanche has a roundup.

Special case: “trend detection”. e.g. Gnip Trend detection Niko12 looks fun, and comes with a cute explanation of how you might do this nonparametrically.


text:
# A geometric characterization of Rees algebras in categories without Choice

Before asking my question, a caveat: The category theorist in me would like me to ask this question in more generality, but I will restrict my scope since what I'm really after is some geometric intuition.

Let $R$ be a commutative ring. A $\mathbb Z$-filtered $R$-module is a diagram $\cdots \hookrightarrow X_{\leq -1} \hookrightarrow X_{\leq 0} \hookrightarrow X_{\leq 1} \hookrightarrow \dots$, where each $X_{\leq i}$ is an $R$-module and each map is an inclusion. I will abuse notation and denote by $X$ both the diagram and its direct limit (the union of the $X_{\leq i}$s). I am happy to assume that the inverse limit (the intersection of the $X_{\leq i}$s) is trivial. The associated graded of $X$ is the $\mathbb Z$-graded $R$-module $\operatorname{gr}(X) = \bigoplus (X_{\leq i}/X_{\leq i-1})$, where the piece $(X_{\leq i}/X_{\leq i-1})$ is put in grading $i$.

The Rees algebra of $X$ is the $R[\epsilon]$-module $$\operatorname{Rees}(X) = \sum_i \epsilon^i X_{\leq i}[\epsilon] \subseteq X[\epsilon^{\pm 1}].$$ I.e. you look at the $X[\epsilon^{\pm 1}] = X \otimes_R R[\epsilon,\epsilon^{-1}]$, and inside it you take the $R[\epsilon]$-module generated by all elements of the form $\epsilon^i x$ for $x\in X_{\leq i}$.

Then the Rees algebra interpolates between $X$ and $\operatorname{gr}(X)$ in the following sense: $$\operatorname{Rees}(X)/(\epsilon=1) = \operatorname{Rees}(X) \otimes_{R[\epsilon]} R[\epsilon]/(\epsilon-1) \cong X$$ $$\operatorname{Rees}(X)/(\epsilon=0) = \operatorname{Rees}(X) \otimes_{R[\epsilon]} R[\epsilon]/(\epsilon) \cong \operatorname{gr}(X)$$ Here the "$=$" signs are definitions and the "$\cong$" signs are canonical isomorphisms. The first of these should be completely obvious; the second is an important calculation.

The Rees algebra has the following geometric interpretation. The affine line is $\mathbb A^1 = \operatorname{spec}(R[\epsilon])$, and so any $R[\epsilon]$-module is a sheaf over $\mathbb A^1$. Consider the sheaf defined by $\operatorname{Rees}(X)$. The above isomorphisms say that $\operatorname{gr}(X)$ is the fiber of $\operatorname{Rees}(X)$ over $0 \in \mathbb A^1$, and (the union of) $X$ is the fiber over $1\in \mathbb A^1$.

Moreover, $\mathbb A^1$ has an action by the group $\mathbb G_m = \mathrm{GL}(1) = \operatorname{spec}(R[\lambda^{\pm 1}])$ given by $\epsilon \mapsto \lambda \epsilon$. This action extends to $\operatorname{Rees}(X)$, by restricting the action on $X[\epsilon^{\pm 1}]$. Thus:

$\operatorname{Rees}(X)$ is a $\mathbb G_m$-equivariant sheaf over $\mathbb A^1$.

Since $\mathbb G_m$ fixes $0 \in \mathbb A^1$, the fiber $\operatorname{gr}(X)$ of $\operatorname{Rees}(X)$ over $0$ inherits a $\mathbb G_m$-action. This is precisely the grading, wherein the $i$th graded piece is the weight space on which $\lambda \in \mathbb G_m$ acts by $\lambda^i$. I have been told that one should think of $X$-as-a-filtered-thing not as the fiber over $1$ but rather as the fiber over the generic point in $\mathbb A^1$.

Now, not every $\mathbb G_m$-equivariant sheaf over $\mathbb A^1$ arises as the Rees algebra of a filtered $R$-module. When $R$ is a field, I believe the correct statement is that:

Over a field, the $\mathbb G_m$-equivariant sheaves over $\mathbb A^1$ that arise as Rees algebras of filtered modules are precisely the $\mathbb G_m$-equivariant vector bundles.

Put another way, over a field, there always exist isomorphisms between $X$ and $\operatorname{gr}(X)$ (i.e. trivializations of the sheaf near $0$). The existence of such isomorphisms requires the axiom of choice, which says that every epimorphism splits. But this description doesn't make sense in generality. Hence:

Question: When $R$ is not a field, so that I do not necessarily have isomorphisms of $R$-modules $X \cong \operatorname{gr}(X)$, what is the geometric characterization that determines when a $\mathbb G_m$-equivariant sheaf on $\mathbb A^1$ is the Rees algebra of a filtered module?

• What seems obvious is that, if $M$ is the graded therefore filtered $R[\epsilon]$-module, then $X$ should be $M / \langle \epsilon-1\rangle$ and therefore filtered. Trying to get an isomorphism from there, I keep running into $\epsilon$ vs $\epsilon^{-1}$ issues that always bedevil me with these algebras. – Allen Knutson Nov 4 '12 at 1:39
• @Allen: I also occasionally get confused with $\epsilon$ versus $\epsilon^{-1}$. I think I have done everything correctly, of course, but if there is an error, please let me know! The idea is that the minimum $i$ for which $x\in X_{\leq i}$ is also the minimum $i$ for which $\epsilon^i x\in \mathrm{Rees}(X) = M$. – Theo Johnson-Freyd Nov 4 '12 at 3:25
• What in the world does any of this have to do with Choice? – Jason Starr Feb 2 '14 at 14:22
• @JasonStarr The "axiom of choice" is a property that some categories enjoy and other categories do not. Category $\mathcal C$ "satisfies the axiom of choice" if every epi in $\mathcal C$ splits. I generally work with a category of Sets that does satisfy this useful property; then Vect does as well. But $R$-mod for $R$ a non-semisimple ring does not; neither does Top. It was in that sense that I meant the reference to axiom of choice. – Theo Johnson-Freyd Feb 2 '14 at 18:08
• @TheoJohnson-Freyd: Okay, but what in the world does any of that have to do with your question? – Jason Starr Feb 2 '14 at 18:58

## 1 Answer

Here is a formulation: if $\mathcal{F}$ is quasi-coherent $\mathbf{G}_{m, R}$-equivariant $\mathcal{O}_{\mathbf{A}^1_R}$-module such that $\epsilon$ is a nonzero divisor on $\mathcal{F}$, then one gets a filtered module as in your question.

On the one hand, it is clear that the modules you've constructed have this property.

On the other hand, let $\mathcal{F}$ be such a quasi-coherent equivariant sheaf and set $M = \Gamma(\mathcal{F})$. This module has an action of $\mathbf{G}_{m, R}$ and hence comes with a weight decomposition $M = \bigoplus M_n$ (to actually prove this you need to do a little bit of work). Observe that $\epsilon : M_n \to M_{n + 1}$ as $\epsilon$ has weight $1$. If $\epsilon$ is a nonzero divisor these maps are injective and one sees that $M$ is the Rees module associated to $M_i \to M_{i + 1} \to \ldots$.


text:
## How to Pick Your Next Gig: Evaluating Startups - Part I

14 Aug 2017

This blog post is based on my experience interviewing with a range of tech companies during my junior year at Princeton. At the end of the process, I was fortunate to have the choice between a large public company, a then recently-minted unicorn, and a then-Series B startup with 100 employees based in San Francisco.

This piece attempts to codify how I went about making my decision, and in particular, lays out a methodology for how to evaluate startups. Though I’m writing primarily for recent college graduates, I’ve tried to make the discussion generally applicable to anyone looking to work for a small company.

Choosing a company to work for is an investment. While a venture capitalist might put in a large amount of money and a small amount of their time (i.e. in monthly board meetings) into a portfolio company, as an employee, the primary asset you invest is your time.

The returns you make can take a number of forms - career advancement, personal growth and fulfillment, and financial payoff. Making any decision comes with an opportunity cost - the foregone returns of another choice you could have made.

Finally, unlike a venture capitalist, as an employee, you commit to work at a single place, so you can only make one investment at any given point in time. So while Marc Andreessen might be okay with a 1-in-10 hit rate,1 you have to set the bar a little higher.

All of this is not to paint the decision as a paralyzingly difficult one, but to place it in its proper context - as a prospective employee, you are an investor, and you evaluate companies as much as they evaluate you.

The brand recognition of a company, the capital raised, the prestige of the investors, your friends' opinion - all of these might matter, but perhaps only as secondary signals of a company's future prospects. What, then, should you look for, and how much weight should you put on each factor?

Here's a laundry list of potential criteria that you might consider in evaluating a startup:

In the remainder of the post, I'll address each of these in turn, and provide a "star rating" to indicate how strongly you should consider each factor when making your decision.

I assume that you are interested in some combination of the following: personal growth, career growth, and financial upside. A convenient truth is that these three goals tend to be tightly correlated, and joining an early-stage, fast-growing startup with strong founders and talented employees is likely to satisfy all three.

This post is about how to identify such companies.

### The criteria

#### Current traction

One fact that makes evaluating startups as a prospective employee particularly difficult is that most key metrics are not public information. Statistics such as the number of monthly or daily active users (MAUs/DAUs), annual revenue, and months of runway are often not even known to current employees, let alone available on the internet.

Moreover, the founders and upper management are unlikely to share this information in the interview process, but if you do get the chance to speak with them, it is definitely worth a shot to ask.

Even if these numbers are known, they may not actually be the strongest signals. Revenue, for example, is often an irrelevant measure for early-stage, consumer-facing companies (e.g. Facebook in 2008), while number of clients may be too coarse a metric for early-stage, enterprise tech companies (for many years, Palantir only had one customer: the US government).

You also have to be careful about companies cherry picking statistics that paint them in a favorable light. A mobile app company may choose to reveal total downloads, but not monthly active users, or their shockingly high churn (i.e. uninstall) rate.

A famous example where these metrics sharply conflicted is Draw Something, a social drawing app, which, along with its parent company OMGPop, was acquired by Zynga in March 2012 for $180 million. Within two months of the purchase, daily active users had fallen by a third, from a peak of 15 million on the day of the sale to 10 million by early May. Draw Something relied on aggressive growth hacking, via close integration with Facebook, and sacrificed the opportunity to build a sustainable product for rapid growth. The result? One of the greatest "pops" of the social-local-mobile app era. #### Growth rate This one is tricky. Growth figures, especially when measured over only two data points (metrics today vs. metrics last year), are often hard to evaluate, unless absolute numbers are known as well. A representative from Facebook could have reported a 2150% growth in revenue in 2005. While Facebook was indeed growing incredibly fast at that time, that particular statistic is meaningless, as revenue was nearly zero in 2004. This is not just a straw man argument. Various startups that I've interviewed with have claimed that they "doubled in revenue since last year," or even that they've been "doubling in revenue every year" when the company has only been in existence for 3 years, without providing a clear estimate of current revenue. You should ask yourself why a company is choosing to share growth numbers, but not any absolute figures. It's likely because growth statistics are a lot more flattering to the company. But you should know: being a derivative of the yearly revenue (or total users) graph, growth figures will almost always contain less information. Of course, don't be pedantic. If a founder mentions that their service has over a million users, and is sporting 200% year-over-year growth, but won't give exact numbers, you probably have enough information to judge that the business is growing rapidly. Joining a company that is growing at breakneck pace is one of the smartest career decisions you can make early on. Such a startup will offer many opportunities to take charge and grow into leadership roles, and will attract many other intelligent and ambitious young people - colleagues who may become your business partners or fellow founders one day. Finally, as many have said, getting a win on your record early in your career is highly valuable (1, 2), and as I discuss in the last section, early hypergrowth is one of the strongest signals that a company will do very, very well in the future. Of course, true hypergrowth is rare, and identifying such companies when they're relatively small and unheard of is challenging. But your alternative is the even harder problem of trying to turn around a slow-growing company as employee #50. On the flip side, joining a startup that "fails" a year or two after you join is not as bad as you may think. Assuming the startup had clear potential and a high hiring bar, no one will hold it against you that the company didn't do as well as hoped. This attitude may not hold outside the San Francisco Bay Area, or the United States, but what is most threatened by spending time at a company that flounders is not your CV, but just that: your time. Staying at a company with no clear growth prospects for five years translates to five lost years you could have spent growing and learning. Note that this idea that a failed startup does not equal a black mark on your career holds for both the founders and the employees. That said, you are probably less likely to regret time spent working on a problem you deeply care about as a founder than as employee #50. #### Number of employees Cisco Systems has 70,000 employees today, and its stock price has hovered between$15 and $30 for the past sixteen years. When WhatsApp was acquired by Facebook for$19 billion in February 2014, it had 55 employees.

Evidently, headcount alone says little about a company's quality. But the number of employees, combined with the most recent valuation of the company, can give you some idea for how much money you stand to make if you join.

Here's the heuristic. Stocks generally vest over four years, and if you stay the four years, your ownership of the company will amount to the following:
\begin{aligned} \text{Fraction ownership } = \frac{1}{\text{(Number of employees)}^2} \end{aligned}

This can now be used to calculate your potential upside. If you join a 100-person startup valued at 500 million, and work there for four years, you'll be granted shares constituting about \begin{aligned} \text{Fraction ownership } &= \frac{1}{\text{(Number of employees)}^2} \\ &= \frac{1}{\text{(100)}^2} \\ &= 0.01\% \end{aligned} of the company. If the company is then acquired for5 billion, you will stand to make:
\begin{aligned} \text{Equity payoff } &= \text{Fraction ownership } * \text{Exit valuation} \\ &= 0.0001 * \text{\5,000,000,000} \\ &= \500,000 \end{aligned}
(Here I'm not considering the strike price of your stock options, i.e. how much you'll need to pay the company to exercise them.)

This equation, when shifted by one, predicts that a startup's first employee will own 25% of the company, a very reasonable estimate for the eventual ownership stake of a solo founder (1, 2). As more anecdotal evidence, it correctly predicted the equity I was offered at a few startups, interviewing for entry-level roles. Finally, it fulfills a key mathematical constraint that we might expect any equity distribution scheme to satisfy, namely that the total ownership of every employee in the company should sum to 1.2

Take this formula for what it is: a rough heuristic for your ownership stake in a startup. In particular, percent ownership alone says nothing about your potential upside. To estimate upside, you need to consider a span of possible trajectories that the company could take after you join. If you join a company that goes under, or is acquired for peanuts, it won't matter if you own 1% of it or 0.01% of it. 1% of 0 is still 0.

For more precise predictions, please see this calculator, which takes different inputs, but offers upside estimates for a range of possible company outcomes.

#### Strength of founders

Silicon Valley loves to lionize its founders. To be a successful founder is to be a statesman, a crusader, a messiah - the enlightened champion of a new way of life. In fact, early CEOs often paint themselves into the founding story of a company (1, 2) to buy further legitimacy: the idea being, to be the real deal, you have to be there from the beginning. And yet the Valley's most revered "founder-CEOs" (the most venerable job title in Silicon Valley) are also at times its most polarizing, its most reviled.

My goal in this section is to filter through the hype, and really evaluate whether founders matter. Do one or two people at the beginning of a startup's history, who do a tiny fraction of the total work that goes into building a massively successful company, really deserve that much credit? How disproportionate is a founder's influence on the eventual success and failure of the company? And what founder qualities and qualifications are necessary for, and, more interestingly, indicative of a startup's future success?

...

What kinds of past experience are signs of a great founder?

• Pivotal role in a very successful company - This is rare, but a very strong sign when true. Two great examples are Dustin Moskovitz, a co-founder of Facebook, and Adam D'Angelo, Facebook's first CTO, who each went on to start extremely well-run companies, Asana and Quora, respectively. Dustin and Adam combine front-line experience in a rocket ship, with a genuine thirst to build companies of their own - both were in some sense sidelined at circa 2008 Facebook, which only had room for one real leader at its helm. Asana emerged as an internal project at Facebook, and Adam left two years of equity at Facebook on the table to found Quora, strong circumstantial evidence that the problems they chose to tackle were relevant and valuable.

• The fourth-time founder, aka the serial entrepreneur - I have more mixed feelings about this one. Being a serial entrepreneur is considered a great resume item in some circles, and there are of course a few examples of wildly successful born-again founders - notably, Elon Musk, and even Steve Jobs, who founded NeXT Computer between 1980s Apple and iPod/iPhone-era Apple. Entrepreneurship is a difficult journey, and the greatest founders tend to treat their companies as their life's work. A long history of eight-figure acquisitions doesn't speak very well to a founder's dedication, and their ability to navigate difficult times. At the same time, an exceptional past result (e.g. a company sold for $500 million) is very strong evidence of caliber and ability to execute. • Domain expertise - This is valuable, but probably not strictly necessary. This Y Combinator-backed waste management company recently raised a$12 million Series A round, and includes a founder whose family has been in the waste recycling business for four generations. That sounds splendid to me, and is an attribute that investors often assign big plus points to. But being an outsider can sometimes constitute a great advantage as well. Elon Musk had no formal experience in rockets or electric cars when he started SpaceX and Tesla, two areas that seem to have extremely high barriers to entry. Moral: we probably overestimate how hard it is for smart people to immerse themselves in new, hard problems they care about.

Continue reading: Part II of Evaluating Startups

### Footnotes

1. The number one rule of startup investing is that nearly all of an investor's returns are concentrated in a few big winners. As of 2012, three-quarters of Y Combinator's \$10 billion portfolio value was concentrated in two companies: Dropbox and Airbnb. This, combined with the difficulty of identifying the biggest winners in their infancy, means that early-stage investors go out of their way to court a large number of companies with a small chance at astronomical success.

2. My heuristic for your expected ownership stake in a company, reproduced here:
\begin{aligned} \text{Fraction ownership } = \frac{1}{\text{(Number of employees)}^2} \end{aligned}
holds the following nice property that we might expect of any equity distribution scheme:
\begin{aligned} \text{Total ownership } &= \int_1^\infty \text{Fraction ownership} \\ &= \int_1^\infty \frac{1}{n^2} \text{ dn} \\ &= -\frac{1}{n} \bigg|_1^\infty \\ &= 1 \end{aligned}


text:
# aliquote

## < a quantity that can be divided into another a whole number of time />

While browsing GitHub, I noticed that the nvim-lua organization offers a “starter kit” init.lua file for Neovim users. It is a single Lua file, which is quite handy when you want to quickly bootstrap a sane config on external servers. I tested it in my own environment and it’s pretty good actually. If you are new to Neovim, and especially Lua config files, I recommend taking a look at how this file is organized. Except for a few packages (theme and snippets), and the default key mappings, you will likely find a working setup for your daily editing tasks. I like the idea of having a single file for the whole config, although I can understand why we may want to split our whole config in separate files, which is actually what I do.


text:
# URL - linebreak and page break

I'm using luatatex 1.0.4 and I already found some solutions for breaking urls. Nevertheless I have special cases for which I didn't find a solution. My sample code looks like:

\documentclass[a4paper,11pt]{article}
\usepackage{fontspec}
\usepackage[english, ngerman]{babel}
\usepackage{hyperref}
\def\UrlBreaks{\do\/\do-\do\_\do\%} % for breaking long URLS additonally
\begin{document}

I have the list of links collected in each sections - just as a regular text. They are itemize. Since the lists are quite long it can happen that an url is broken into two lines just when the page break occurs. As a consequnce the page number and the page header of the next page are clickable from the link. How to disable this behaviour?


text:
# #StackBounty: #machine-learning #time-series #neural-networks #data-leakage data leakage when scaling time series

### Bounty: 150

Suppose I want to forecast future values of $$y$$ past values of features $$x$$.
In this example I am using:

• the training set goes from $$t_0$$ to $$t_{15}$$
• values from $$x_{t_0}$$ to $$x_{t_{10}}$$ to forecast $$y_{t_{11}}$$
• values from $$x_{t_1}$$ to $$x_{t_{11}}$$ to forecast $$y_{t_{12}}$$
• and so on until I use $$x_{t_6}$$ to $$x_{t_{15}}$$ to forecast $$y_{t_{16}}$$

I scale the my feature $$x$$ using only data in the training set (up to $$t_{15}$$)

Nevertheless when I try to predict $$y_{t_{17}}$$ you can see from the picture below that I use some data points that have also been used for scaling.

Is this leakage?

Get this bounty!!!

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text:
## Precalculus (6th Edition) Blitzer

$26^\circ, 64^\circ$
Step 1. Given the model function $d=\frac{v^2_0}{16}\ sin\theta\ cos\theta$, with $v_0=90\ ft/sec$ and $d=200\ ft$, we have $\frac{90^2}{16}\ sin\theta\ cos\theta=200$ or $sin(2\theta)=0.7901$ Step 2. With $sin^{-1}(0.7901)\approx52^\circ$, the general solutions for the above equation are $2\theta=360^\circ k+52^\circ$ and $2\theta=360^\circ k+180^\circ-52^\circ$, which gives $\theta=180^\circ k+26^\circ$ and $\theta=180^\circ k+64^\circ$ where $k$ is an integer. Step 3. Within $(0^\circ,90^\circ)$, we have $\theta=26^\circ, 64^\circ$


text:
# Generate random string to match the goal string with minimum iteration and control statements

The following is my python code to find the number of iterations needed to generate random string iteratively to match the input string.

Here I have repeated a similar kind of for loop in both the functions getRandomString() and getMatchString(). I tried to avoid it but couldn't. I want this to be combined together into a single generateString() function with minimum iteration statements. Any suggestions to make it so or any other possible way to make the code better without complicating much?

The input is any string statements like "Hello, World! How Are Y@u doiNG?" and the output is the number of iterations needed to match the input string e.g 408. Your suggestions to improve the code will be helpful.

import random

goalString = input()
stringLen = len(goalString)

alphanum = '''abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ !@#&()–[{}]:;‘,'?/*1234567890'''
alphaLen = len(alphanum)

def getRandomString():
randomString = ''
for i in range(stringLen):
randomString = randomString + alphanum[random.randrange(alphaLen)]

return randomString

def getMatchString():
betterString = getRandomString()
newString = ''
iteration = 1 #included the iteration of generating randomString
while betterString != goalString:
charIndex = [i for i in range(stringLen) if betterString[i] == goalString[i]]
for i in range(stringLen):
if i in charIndex:
newString = newString + betterString[i]
else:
newString = newString + alphanum[random.randrange(alphaLen)]

betterString = newString
newString = ''
iteration = iteration + 1

return iteration

print(getMatchString())

Sample Input: "Hello, World! How Are Y@u doiNG?"

Sample Output: 408 ##May change depending on the random iteration##

• Please do NOT edit the question after an answer has been posted, especially the code. Everyone who sees the post should see what the first person that posted an answer saw. Please read What should I do after someone answers?. – pacmaninbw Jan 26 at 13:26
• You can ask follow up questions by linking to this question. – pacmaninbw Jan 26 at 13:30
• @pacmaninbw Just thought to optimize the original code so as to get a more refined code. Won't do that from now and add up the link questions to the original one as suggested by you. – Jerin06 Jan 26 at 13:45
• Please stop modifying the code, as you were asked by others. Protip: if you follow suggestion #1 on that list of options then you might be able to earn more reputation points. – Sᴀᴍ Onᴇᴌᴀ Jan 26 at 15:10

Since your result is only the number of iterations, you don't actually have to work with strings. Just keep track of how many characters are still wrong.

import random

goalString = input()

alphaLen = 82

def getMatchString():
wrong = len(goalString)
iteration = 0
while wrong:
wrong = sum(random.choices((0, 1), (1, alphaLen-1), k=wrong))
iteration = iteration + 1
return iteration or 1

print(getMatchString())

Or handle the characters separately and take the max number of iterations:

import random

goalString = input()

alphaLen = 82

def getMatchChar():
iteration = 1
while random.randrange(alphaLen):
iteration += 1
return iteration

def getMatchString():
return max(getMatchChar() for _ in goalString)

print(getMatchString())

A version that's perhaps slightly clearer:

import random

alphanum = '''abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ !@#&()–[{}]:;‘,'?/*1234567890'''

def getMatchChar(goalChar):
iteration = 1
while random.choice(alphanum) != goalChar:
iteration += 1
return iteration

def getMatchString(goalString):
return max(getMatchChar(goalChar) for goalChar in goalString)

goalString = input()
print(getMatchString(goalString))

• Liked the way how you approached solving the problem. Both options provide the right solution. I understood the first one which seems to be directly solving it. However, I couldn't get your idea behind the second option. Could you explain it a little bit? The second one seems to produce a different range of iteration results and you have taken maximum out of it. – Jerin06 Jan 27 at 5:42
• @Jerin06 In your way, you keep going until all characters match. That is, the slowest-matching character is what matters. The one that took the maximum time to match. That's what my second option does. It produces the same range of iteration results (in getMatchString, not in getMatchChar, of course). – superb rain Jan 27 at 11:26
• @Jerin06 Added yet another version that's like the second one but perhaps a little clearer (without being much longer or slower). – superb rain Jan 27 at 14:11
• Yes, this one is much easier for me to understand. You have simplified the code very well. The only thing I am not sure about here is how long the while loop runs here and whether that increases the time complexity compared to the previous codes. If there is no problem with time complexity, I prefer this last option added by you. – Jerin06 Jan 27 at 14:32
• @Jerin06 Compared to yours, mine decrease the time complexity, as I don't keep looking over and over again at already matched characters. – superb rain Jan 27 at 16:31

Find my suggestions below.

goalString = input()

Let the user know what to do, input accepts a string as an argument:

goalString = input('Type a string:')

alphanum = '''abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ !@#&()–[{}]:;‘,'?/*1234567890'''

If you need all printable characters you can use string.printable:

import string

alphabet = string.printable
# alphabet contains: 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!"#\$%&'()*+,-./:;<=>?@[\]^_{|}~

Or choose a combination of other constants in the string module.

Generating a random string

def getRandomString():
randomString = ''
for i in range(stringLen):
randomString = randomString + alphanum[random.randrange(alphaLen)]

return randomString

• Avoid using globals passing the length as an argument
• String concatenation can be done more efficiently with join
• Use random.choices to avoid the for-loop
• PEP 8: function names should be lowercase, with words separated by underscores. Variable names follow the same convention as function names.
def generate_random_string(n):
alphabet = string.printable
return ''.join(random.choices(alphabet, k=n))

The function to match the string creates many strings using concatenation. It can be done more efficiently by using a list of booleans (as @rperezsoto suggested).

def get_matching_string(s):
n = len(s)
matches = [False] * n
i = 0
while not all(matches):
matches = [matches[i] or s[i] == c for i, c in enumerate(generate_random_string(n))]
i += 1
return i

The key part is to use the operator or that "remembers" positive matches.

I want this to be combined together into a single generateString() function with minimum iteration statements

I would suggest keeping two separate functions as in your first solution. The code will be easier to test and reuse. Additionally, you can extract the logic to request the input and print the result from the function getMatchString with a main function:

def main():
s = input('Type a string:')
print(get_matching_string(s))

if __name__ == "__main__":
main()

• string.printable and random.choices make things easier here. Appreciate your effort in explaining them in detail. Yes, instead of focussing on reducing functions, I should create reusable functions. Good, you mentioned many helpful tips like PEP 8 for creating function names. Thanks! – Jerin06 Jan 26 at 14:51

It took me a while to understand the code but I hope that I understood it properly. Lets say you have the string "abcd"

   goal string      "abcd"
1st iteration    "ftcP" -> you save the 3rd position as found
2nd iteration    "aXlL" -> At this pint both 1st and 3rd positions have been found
3rd iteration    "...." -> Still 3rd and 1st
4th iteration    "UbDd" -> all positions have matched at least once.
Stop and return 4 as the result

If the previous example is what you want to achieve below is my code suggestion

from random import choice

pool = '''abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ !@#&()–[{}]:;‘,'?/*1234567890'''
GetNewChar = lambda : choice(pool)

def GetRandomString(stringLen):
characters = [GetNewChar() for i in range(stringLen)]
return ''.join(characters)

def IterationsToMatchString(goal_string):
IsGoalAchieved = [False for i in goal_string]
iterations = 0
while not all(IsGoalAchieved):
iterations += 1
test_string = GetRandomString(len(goal_string))
Matches = [g==t for g,t in zip(goal_string,test_string)]
for i,t in enumerate(IsGoalAchieved):
if t is False:
IsGoalAchieved[i] = Matches[i]
return iterations

example = input()
print(IterationsToMatchString(example))

`

regardless of which variable naming convention you use I suggest not using the same name for the function and the output unless you are writing a decorator function. In your "randomString" function you return a variable named "randomString" which makes the code difficult to understand and is likely to bug. I would suggest changing the name of the function to improve reading.

Also you abuse the global keyword. You can just pass arguments to the function. In this case your function randomString generates a random string of a fixed length. Therefore you can always call that function with the length as a parameter.

Also as you have seen I have substituted the usage of randrange by choice (but does the same). I prefer it because we are picking one item out of a certain sequence.I have used a lambda function to provide a more "readable" name that gives information about what is being done.

• Thanks for your suggestions. I will make the changes as mentioned by you to improve the code readability. I'll also go through your code and come back if I need any clarification. – Jerin06 Jan 26 at 10:36
• Does your code work? I couldn't get output from your code due to some error message. – Jerin06 Jan 26 at 11:16
• I forgot to "initialize" the iterations variable with "0" and also put an non neede argument in the GetNewChar. I have edited the post and it should work now – rperezsoto Jan 26 at 13:10
• Yes, it is working now. You have used some wonderful techniques. Very helpful for me to enhance my coding. Thanks! – Jerin06 Jan 26 at 13:22
• If you could please point me towards some reference for that I would love to learn more. I know that this code can get much better and I am aware that even within the answer the naming convention is not even consistent. And also feel free to edit the answer, I will be able to learn from the changes. – rperezsoto Jan 26 at 20:01


text:
Electric Flux Video Lessons

Concept

# Problem: The three small spheres shown in the figure (Figure 1) carry charges q1 = 4.35nC , q2 = -8.00nC , and q3= 2.25nC.Find the net electric flux through the closed surface S1 shown in cross-section in the figure.

###### FREE Expert Solution

Gauss law:

$\overline{){{\mathbf{\varphi }}}_{{\mathbf{E}}}{\mathbf{=}}\frac{{\mathbf{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$

97% (365 ratings)
###### Problem Details

The three small spheres shown in the figure (Figure 1) carry charges q1 = 4.35nC , q2 = -8.00nC , and q3= 2.25nC.

Find the net electric flux through the closed surface S1 shown in cross-section in the figure.


text:
## Jonask Group Title prove that the directrix of an ellipse is $x=\frac{a^2}{c}$ one year ago one year ago

this ellipse is the general ellipse with $\frac{ x^2 }{ a^2}+\frac{ y^2 }{ b^2 }=1,a>b$

i have the proof but i dont understand the ratio set up here can i post the link

$r=\frac{ a-c }{ d-a }=\frac{ a }{ d }$

\[a^2=b^2+c^2 i dont get this ratio .why it holds

6. experimentX

??|dw:1364055606757:dw||dw:1364055629069:dw|

7. experimentX

i haven't done conics for quite while, but this shouldn't be difficult.|dw:1364055735803:dw|

9. experimentX

what's r there?

its defined in the 1st link

11. experimentX

can't find it ... equation no?

24

13. experimentX

what is this 'c' again?

14. experimentX

this 'c' looks like foci.

15. experimentX

this is pretty ovbioius ... this ratio must be constant


text:
1. Engineering
2. Civil Engineering
3. draw the shear force and bending moment diagrams for the...

# Question: draw the shear force and bending moment diagrams for the...

###### Question details
Draw the shear force and bending moment diagrams for the beam shown. Determine the maximum shear force and the maximum bending moment.


text:
UGCNET-Nov2017-ii-47

760 views

Which of the following statements about ERP systems is true?

1. Most ERP software implementations fully achieve seamless integration
2. ERP software packages are themselves combinations of separate applications for manufacturing, materials, resource planning, general ledger, human resources, procurement and order entry
3. Integration of ERP systems can be achieved in only one way
4. An ERP package implemented uniformly throughout an enterprise is likely to contain very flexible connections to allow charges and software variations

edited
0
Yes option B is correct

 Some ERP software packages are themselves combinations of separate applications for manufacturing, materials resource planning, general ledger, human resources, procurement, and order entry. option b

Related questions

1 vote
1
2.1k views
Which speed up could be achieved according to Amdahl's Law for infinte number of processes if $5\%$ of a program is sequential and the remaining part is ideally parallel? Infinite $5$ $20$ $50$