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# 1911 Encyclopædia Britannica/Swimming

SWIMMING (from "swim," A.S. swimman, the root being common in Teutonic languages), the action of self-support and self-propulsion on or in water; though used by analogy of inanimate objects, the term is generally connected with animal progression and specially with the art of self-propulsion on water as practised by man. Natation (the synonym derived from Lat. natare) is one of the most useful of the physical acquirements of man. There have been cases in which beginners have demonstrated some ability in the art upon their first immersion in deep water, but generally speaking it is an art which has to be acquired. For many years Great Britain held the supremacy in this particular form of athletics, but conti- nental, Australian and American swimmers have so much improved and have developed such speedy strokes, that the claim can no longer be maintained. English swimmers have, however, the satisfaction of knowing that in a great measure through them has come about the very great interest which is now taken in the teaching of swimming throughout the world, and more particularly on the continent of Europe, where they have made frequent tours and given instructive displays of swimming, life-saving (see Drowning), and water polo (q.v.); the latter a water game entirely British in its origin.

The teaching of swimming has been taken up in schools, and where the work is well done it is customary to use a form of land drill so as to impress upon the pupils some idea of the motions which have to be made in order to progress through the water. This drill is the preliminary practice to the teaching of the breast stroke. This stroke is about the most useful of all the known forms of swimming, more particularly when any one is thrown overboard in clothes; and though speed swimmers look upon it as obsolete, it is undoubtedly the best for a long-distance swim, such as across the English Channel, or other similar feats. A knowledge of it, as well as of the back stroke, is essential to the effective saving of life.

When learning the breast stroke, the first thing to avoid is undue haste and rapidity in the movements. It is this fault, probably born of nervousness, which causes many to aver that though eager to do so, they have never been able to learn to swim. Rapid action of the arms only exhausts the learner, whose breathing then becomes hurried and irregular, and as a consequence he fails to preserve the buoyancy necessary for carrying him along the surface. When starting for the first stroke the beginner should draw the elbows nearly to the side, at the same time bringing up the forearm and hands to the front of the chest with the palms of the hands downwards near to the surface of the water, the fingers being extended and closed and the forefingers and thumbs nearly touching. The hands are then pushed forward in front of the body to the full extent of the arms, the palms of the hands are turned slightly outwards, and the arms swept round until in a right angle with the shoulders, when the elbows are dropped and the hands come up in fiont of the chest for the next stroke. The arms should not be kept rigid, but allowed to work gracefully. As the arms are swept backwards the legs are drawn up, the knees being turned out- ward to the right and left and the heels nearly touching. The legs are then kicked outward and swept round as the arms are being pushed forward to their fullest extent, a " flip " being given with each of the feet, which must be kept loose at the ankles and in the same position as when standing. All beginners have the great fault of trying to make the limbs too rigid, thereby causing stiff- ness and possibly cramp. Another difficulty with them is the question of breathing, but if the learner will remember to inhale when making each backward sweep of the arms, much of the difficulty usually experienced at the start will be overcome. Expiration should be carried out during the other portion of each stroke. The important thing is to keep the body as level along the surface as possible, and at the same time get regular and natural breathing. The holding of the breath for two or three strokes will exhaust the beginner more than anything else.

A knowledge of the back stroke can easily be acquired by those who are able to swim on the breast, for the leg action is very similar and the principles relating to the use of the arms are almost the same. The arms, instead of being moved through the water, are lifted in the air and carried out to beyond the head with the palms upwards. The palms are then slightly turned and the arms swept round. Just as this action is being made the legs are drawn up as in the breast stroke, the body being allowed to travel on with the force of the kick as the arms are extended beyond the head. The great difficulty that a back swimmer has to contend with in open water is that of steering, and the best way to overcome it is to take an object for a guide before starting and hold the head slightly to the side so as to steer by it.

At one time the side stroke was the great racing stroke; the body being placed on the side, the upper arm worked from the head to the upper side of the body, the lower arm taken down- wards through the water to the underside of the body and a scissor-like kick made with the legs; but this has now been generally given up in favour of the over-arm, trudgen and crawl strokes.

In the over-arm stroke the body is usually turned on the right side. . At the start the lower arm is pulled downwards towards the hips, the fingers being kept closed and the hand flat, so as to present a large surface to the water. When the stroke is finished the hand is turned quickly palm upwards, so that together with the lower part of the arm it cuts the water sideways, the arm being almost bent double. Then, as it is shot forward, the hand is' gradually turned from palm upwards to palm downwards, until, when it arrives at its position beyond the head, it is ready for the next stroke. The recovery and the pull ought to be effected as quickly as possible. The upper arm stroke is started when the downward stroke of the under or right arm is finished. It is started in front of the forehead, the arm being slightly bent and the fingers pointing downwards. The hand is pulled past the face and chest with the arm bent at right angles and swept back in front of the body, the arm gradu- ally straightening as it leaves the water opposite the hip. When the hand is opposite the hip it should be brought quickly out of the water and sent forward for the next stroke. When the upper arm is opposite the shoulder in its pull through the water the legs are kicked wide apart and closed again at the moment when the hand leaves the water. The kick is completed and the legs straightened before the left hand is replaced ready for the next stroke. As the legs are opened the upper leg is kicked forward with the knee slightly bent, and the foot kept in its ordinary position. The lower leg is bent double until the heel approaches the thigh, which is brought backwards slightly. In the actual kick the upper leg is sent forward, and as it is straightened vigorously the under leg from the knee downward come/forward to meet it with a vicious kick; the swirl of the feet/Shd closing of the legs drives the body forward. This is what has come to be known in Great Britain as the " Northern Kick," by reason of its first being introduced by Lancashire swimmers.

The trudgen stroke, more commonly known as the trudgeon stroke, and on the continent of Europe as Spanish swimming, was first made prominent in England in 1873 by a swimmer named J. Trudgen, who stated that he had acquired a knowledge of it while in South America. It was, however, known to Clias, a writer on swimming, who described it in 1825 as " The Thrust." Trudgen's speed was so great for his time that swimmers quickly copied his style, and it is from this stroke that the crawl stroke has been developed. When swimming Trudgen kept on the chest and lifted the upper part of his body at each stroke out of the water, and at each swing of the arms pulled himself forward, a considerable swirl of the water occurring as each movement was finished. The arms were brought forward sideways, each completing a circle on each side of the body, and the head kept completely above water. Those who copied Trudgen soon found it was less laborious and equally as fast to use a double over-arm stroke with the head and chest well down, and thus have the body supported by the water, using the ordinary over- arm leg kick. At first it was considered a stroke only useful for short distances and for water polo where speed is essential, but the idea was quickly dispelled, and several men, as well as women, have swum as far as fifteen miles with this stroke.

The crawl stroke is, like the trudgen, an adaptation from native swimmers. It was not generally known in Great Britain until 1902, when Mr Richard Cavill came from Australia to compete in the English championships, but it is said to be common with natives of the South Sea Islands, and from there introduced into Australia about the year 1900. From thence it came to Europe, and there Mr C. M. Daniels, the American amateur champion, made so excellent a study of it that he not only so greatly increased his own pace as to be able to win the English championship, and beat the world's record for a hundred yards, but also introduced various improvements upon it. This stroke is distinct from any other form of swimming: the legs from the knee upwards are kept in line with the body and almost closed; there is no opening of the legs or drawing up of the knees as for the breast, back and side strokes. The swimmer lies flat upon -his breast on the surface, the lower part of the legs from the knee downward are alternately lifted above the surface up to the middle of the calf and then they are struck down upon the water with the instep with all force possible. This striking is done from an upward to a downward direction, one leg at a time. The arms are used somewhat similarly as in the trudgen stroke, they are bent at the elbows, dipped in just beyond the head and drawn smartly backwards till they come out of the water at the hips. The right arm is dipped in when the left foot strikes downward and vice versa. The result of this movement is that when one or the other of the limbs is pulling or propelling the body through the water at the same moment another limb is being recovered for the next stroke, most of the limbs are recovered through the air, fewer dead or retarding points are produced than in any other stroke, and less resistance is caused in the line of progress. In performing any other stroke most of the limbs are recovered through the water.

One of the most useful accomplishments for a swimmer is that of floating, but curiously enough many of them cannot acquire a knowledge of it. It is purely a matter of buoyancy, and requires constant practice before one can become perfect in it. In learning to float the beginner experiences great difficulty in overcoming the tendency of the legs to sink, and if after frequent trials they are still found to sink he should get some one to hold them up or else place them on the steps or behind the rail of the bath, and thus assisted learn to balance the body on the surface. Before doing so he should completely fill his lungs, spread his legs wide, and then lie backwards with the arms extended in a line with the body and beyond the head, with the .palms upwards, care being taken to throw as much weight beyond the head as possible. Furthermore he must lie perfectly still and take care not to hollow the back or raise the abdomen above water. One may sink for an instant, but if the breath be held the lips will come above the surface, when easy breathing may be indulged in. Only the face, chest and toes should appear above the surface of the water. Jf the feet still have a tendency to sink after they have been gently released from the step or rail, more weight should be thrown beyond the head by turning it well back and lifting the hands out of the water, which will raise the feet. A knowledge of floating is of good service to those attempting to save life and is also essential to those desirous of making a study of the many tricks and scientific feats which are performed by swimmers.

Speed swimming records are so frequently altered, that students had best obtain the Amateur Swimming Association's Annual Handbook, in which are detailed the accepted records up to date. The improvement in speed has been most remarkable. In 1877 the mile amateur record was 29 m. 25 ½ secs.; and that stood until 1892. The record in 1907 was 24 m. 42 ${\displaystyle {\tfrac {2}{5}}}$ sees, made by Mr D. Billington. The hundred yards record has been similarly reduced. In 1878 it was 1 m. 16 ¾ secs.; in 1888 it had been lowered by Mr J. Nuttall to 1 m. 6 ¼ sees.; and :n 1907 Mr C. M. Daniels, of America, created a world's record of 55 ${\displaystyle {\tfrac {2}{5}}}$ secs. The records over intermediate distances have also been considerably lowered and many long-distance swimming records have from time to time been created. One of the most remarkable of these long-distance swims is the race which is known as the " Swim through London," from Richmond lock and weir to Blackfriars, which was instituted in 1907 and won by Mr J. A. Jarvis of Leicester, in 3 hours 24 minutes 6 ${\displaystyle {\tfrac {4}{5}}}$ sees. In this event 34 started, and 21 finished the distance, which goes to show that much attention is being devoted to long- distance trials; in this event Miss Lilian M. Smith finished fourteenth. Much interest has centred in attempts to swim across the English Channel; Captain Webb, D. Dalton and F. Cavill, all claim to have done it, but only the swim of Captain Webb has been accepted as genuine. The first recorded attempt was made on the 24th of August 1872 by J. B. Johnson, who started from Dover, but remained in the water only 65 minutes. It was on the 12th of August 1875 that Captain Matthew Webb made his first attempt. He started from Dover and remained in the water 6 hours 49 m., when the weather became too rough for him to continue. It is estimated that he was about 13J m. across when he had to give up. On the 24th-25th of August 1875, he swam across the English Channel, diving from the Admiralty Pier, Dover, and touching Calais sands, France, after swimming for 21 hours 45 m. It is the greatest swim ever recorded, and at the time of the accomplishment created a great sensation in England. Since this great achievement, numerous unsuccessful attempts have been made, the best being those of Montague Holbein, Jabez Wolff and T. W. Burgess, and their efforts created an interest in long-distance swimming in all parts of the world, which has resulted in the accomplishment of trials and tests once thought impossible.


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# Approximate biprojectivity and biflatness of some algebras over certain semigroups

Document Type: Original Article

Authors

1 Department of Mathematics, University of Tabriz, Tabriz, Iran.

2 Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran.

Abstract

We investigate (bounded) approximate biprojectivity of $l^1(S)$ for uniformly locally finite inverse semigroups. As a consequence, we show that when $S=\mathcal{M}(G, I)$ is the Brandt inverse semigroup, then $l^1(S)$ is (boundedly) approximately biprojective if and only if $G$ is amenable. Moreover, we study biflatness and (bounded) approximate biprojectivity of the measure algebra $M(S)$ of a topological Brandt semigroup.

Keywords

### References

O. Yu. Aristov, On approximation of at Banach modules by free modules, Sb. Math. 196 (2005), no. 11-12, 1553--1583

Y. Choi, Biatness of 1-semilattice algebras, Semigroup Forum 75 (2007), no. 2, 253--271.

H. G. Dales, F. Ghahramani and A. Yu. Helemskii, The amenability of measure algebras, J. London Math. Soc. (2) 66 (2002), no. 1, 213--226.

F. Ghahramani and R. J. Loy, Generalized notions of amenability, J. Funct. Anal. 208 (2004), no. 1, 229--260.

F. Ghahramani and Y.Zhang, Pseudo-amenable and pseudo-contractible Banach algebras, Math. Proc. Camb. Phil. Soc. 142 (2007), no. 1, 111--123.

A. R. Medghalchi and H. Pourmahmood-Aghababa, Figa-Talamanca-Herz algebras for restricted inverse semigroups and Clifford semigroups, J. Math. Anal. Appl. 395 (2012), no. 2, 473--485.

H. Pourmahmood-Aghababa, Approximately biprojective Banach algebras and nilpotent ideals, Bull. Aust. Math. Soc. 87 (2013), no. 1, 158--173.

P. Ramsden, Biatness of semigroup algebras, Semigroup Forum 79 (2009), no. 3, 515--530.

V. Runde, Biatness and biprojectivity of the Fourier algebra, Arch. Math. (Basel) 92 (2009), no. 5, 525--530.

H. Samea, Topological Brandt semigroups, Semigroup Forum 86 (2013), no. 2, 404--412.

G. L. Sleijpen, Convolution measure algebras on semigroups, PhD thesis, Katholieke Universiteit, The Netherlands 1976.

Y. Zhang, Nilpotent ideals in class of Banach algebras, Proc. Amer. Math. Soc. 127 (1999), no. 11, 3237--3242.

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# Dynamic model of an inverted pendulum (part 3)

This page is part of a serie of articles on how to write the model of an inverted pendulum. We strongly recommand to read the previous pages for a better understanding.

## Dynamics

The fundamental principle of dynamics can be apply to our system. For the trolley (body 1):

$$\begin{array}{r c l} m_1.{a_1}^x = m_1.\ddot{x_1}&=& | \vec{F_t} | + \lambda_1 \\ m_1.{a_1}^y = m_1.\ddot{y_1}&=& 0 + \lambda_2 \\ I_1.\ddot{\Theta_1} &=& 0 + \lambda_3 \ \end{array}$$

For the pendulum (body 2):

$$\begin{array}{r c l} m_2.{a_2}^x = m_2.\ddot{x_2} &=& 0 + \lambda_4 \\ m_2.{a_2}^y = m_2.\ddot{y_2} &=& | \vec{F_g} | + \lambda_5 = -g.m_2 + \lambda_5\\ I_2.\ddot{\Theta_2} &=& 0 + \lambda_6 \ \end{array}$$

Where $$\lambda_1$$, $$\lambda_2$$, $$\lambda_3$$, $$\lambda_4$$, $$\lambda_5$$ and $$\lambda_6$$ are the Lagrange's coefficients. These coefficients represent the interaction between bodies. In the present system, this is the interaction between the pendulum and the trolley.


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# If p and q are positive integers, is 21^p/630^q a terminating

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If p and q are positive integers, is 21^p/630^q a terminating  [#permalink]

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31 Dec 2017, 23:02
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If p and q are positive integers, is $$\frac{(21)^p}{(630)^q}$$ a terminating decimal

(1) p < 2q
(2) p > q

Source: Experts Global

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Re: If p and q are positive integers, is 21^p/630^q a terminating  [#permalink]

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01 Jan 2018, 01:20
pushpitkc wrote:
If p and q are positive integers, is $$\frac{(21)^p}{(630)^q}$$ a terminating decimal

(1) p < 2q
(2) p > q

Source: Experts Global

We'll simplify the expression so we understand what we need to do.
This is a Precise approach.

We'll extract all common factors between 21 and 630, creating an easier-to-understand expression.
$$\frac{(21)^p}{(630)^q}=\frac{(21)^p}{(21*30)^q}=\frac{(21)^{p-q}}{(30)^q}=\frac{(3*7)^{p-q}}{(3*10)^q}=\frac{(3)^{p-2q}*(7)^{p-q}}{(10)^q}$$
A terminating decimal is an integer divided by some power of 10.
Therefore, for the above to be a terminating decimal, $$(3)^{p-2q}*(7)^{p-q}$$ must be an integer.
As 3 and 7 are both prime numbers then this occurs only when p-2q and p-q are both positive. This happens only when p>2q.
Looking at our statements (1) lets us answer the question with a NO and (2) does not let us answer the question.

*Note 1 - a terminating decimal is basically one long integer with the decimal point somewhere in the middle.
Therefore you can multiply it by a power of 10 to create an integer. This also means that every terminating decimal is an integer divided by a power of 10.
*Note 2 - A number divided by another number is an integer only if the denominator has the same prime factors as the numerator.
As this can never happen for powers of 3 and 7, they must both be in the numerator meaning that p-2q and p-q need to be positive.
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Re: If p and q are positive integers, is 21^p/630^q a terminating &nbs [#permalink] 01 Jan 2018, 01:20
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Reference needed for projection from a linear subspace (projecting cone)

Hello,

I want to find a good reference on projecting cone, which is described below: Let $X$ be an irreducible subvariety of a projective space $\mathbb{P}^n$ (over $\mathbb{C}$ or in general an algebraic closed field of characteristic zero). Let $L$ be a linear subspace of $\mathbb{P}^n$ so that $L\cap X=\emptyset$. Then for an irreducible subvariety $Z\subset X$, we define $C_L(Z)$ to be the union of all lines joining a point $x\in Z$ and $y\in L$. I want to know answers to questions such as: i) degree of $C_L(Z)$, ii) whether $C_L(Z)$ intersects properly with a given subvariety $Y\subset X$, ...

Actually I know answer for ii) for generic linear subspace from a paper of Joel Roberts "Chow's moving lemma", but expect that in a recent reference there will be more discussions about other related results.

Can anyone help me with this? Thank you.

-

This is the portion concerning projecting cones of my explanation as posted in math.stackexchange regarding the degree of a projective variety. I think this answers exactly what you are asking for, including references within the text listed at the end. In particular, you can find the detailed construction of projecting cones at the beginning of part B. and just before the middle of C. The degree of it and the variety it comes from is the subject of the whole discussion (my notation is $C(X_k, L_r)$ for your $C_L(X)$, with $k:=\dim X$, $r=\dim L$). Whether it intersects properly a given subvariety $Y\subset X$ is true for generically chosen center $L_r$ (disjoint with X) by iterated application of Bertini's theorem to the sections $L_{r+1}\cap Y$, i.e., for $L_r$ in a dense Zariski-open subset of the Grassmannian $\mathbb{Gr}(r,\mathbb{CP}^{n})$, cf. reference at the end of point E. below.

Projecting Cones and Degree of a Variety: All the following definitions of degree are proved to be equivalent to each other so one can pick any of them as starting definition and get the rest (and more as mentioned in the link above) as interesting theorems.

• A. If $X$ is a hypersurface of $\mathbb{CP}^n$, i.e. $\dim X=n-1$, by [Hartshorne, Exercise I.2.8] it is given by the zero locus, $X=Z(f)$, of an irreducible homogeneous polynomial $f\in S_d$ of algebraic degree $d$, i.e. any monomial summand $a_{k_0\dots k_n}x_0^{k_0}\cdots x_n^{k_n}$ in $f$ has degree $d=\sum_i k_i$, where all such monomials generate the abelian group $S_d$, all of which make the ring of polynomials a graded ring $\mathbb{C}[x_0,\dots,x_n]=\bigoplus_{d=0}^\infty S_d$. So any such $f$ has a canonical associated (algebraic) degree, so

Degree of a hypersurface as the algebraic degree of its defining homogeneous polynomial: $$\deg X_{n-1}=\deg Z(f):=\deg(f)=d,\;\;\; f\in S_d$$

• B. If $k:=\dim X_k< n-1$, let $L_r\cong\mathbb{CP}^r$ be a generic (i.e. in general position) linear variety (linear projective vector subspace) of $\mathbb{CP}^n$ of dimension $r\leq n-k-1$. The projecting cone, $C(X_k, L_r)$, of $X_k$ from "vertex" $L_r$ is defined to be the joint locus of the subspaces $L_{r+1}$ that join the given $L_r$ with each point of $X_k$ (this generalizes the intuitive "cone" of lines obtained by projecting from a point). By [Beltrametti et al., sec.3.4.5] the projecting cone is also a, possibly reducible, algebraic variety of dimension $r+k+1$ (which justifies the upper bound of r at the beginning). For a generic $L_{n-k-2}$ the projecting cone of $X_k$ is thus a, possibly reducible, hypersurface as in A. above (if it is irreducible it is the case A. if it is reducible then its defining zero locus polynomial is reducible but has nevertheless well defined degree). Call

Degree of a subvariety as the degree of the generic projecting cones which are hypersurfaces: $$\deg X_k:=\max\limits_{L\in\mathbb{Gr}(n-k-2,\mathbb{CP}^n)}\{\deg C(X_k, L)\},\;\;\text{where } C(X_k, L_{n-k-2})=Z(g)\,\vert\, g\in S_p$$

where $L$ is an element of the Grasssmannian of the required dimension. The degree of a variety $X_r\subset\mathbb{CP}^n$ is thus defined to be the degree of the generic hypersurface-projecting-cone; this is proved to be well-defined as this max deg is constant for a dense Zariski-open subset of the Grassmannian, cf. [Harris, Exercise 18.2]. For example if $L_0$ is a generic point in $\mathbb{CP}^3$ and $X_1$ a spatial algebraic curve, for each point of $X_1$ there is only one line joining it with $L_0$. Moving along all such points of the curve we obtain a cone swept by the joining lines with the fixed $L_0$, cone which is an algebraic surface, thus the zero locus of an homogeneous polynomial in projective space. So we are calling the degree of the spatial curve the degree of its projecting cone generic surface polynomial. Note that a curve in projective space is generically given by the intersection of two surfaces of possibly different degrees, $X_1=Z(h_1,h_2)\subset\mathbb{CP}^3$, so it has no canonical unique polynomial degree as is the case for plane curves. It is also important to remember that any nonsingular algebraic curve (thus Riemann surface) is isomorphic to a smooth spatial curve in $\mathbb{CP}^3$ [Hartshorne, Cor.IV.3.6][Shafarevich vol. I, sec.5.4 Cor.2] and birational to a plane curve with at most node singularities [Hartshorne, Cor.IV.3.11]. Note also that the first definition A. above is necessary, since the projecting cone of a hypersurface cannot be defined due to the constraint $r\leq n-k-1$. So what we have done is defining, for any lower dimensional variety, associated hypersurfaces which have generically well-defined polynomial degree.

• C. The "vertex" $L_r$ of a generic projecting cone $C(X_k, L_r)$ of a variety $X_k$ is given by $n-r$ linearly independent linear equations: $L_r=Z(h_1,\dots,h_{n-r})$ where $h_i$ are linear forms which define hyperplanes $H_i=Z(h_i)\cong\mathbb{CP}^{n-1}$ within $\mathbb{CP}^n$, so that $L_r=\bigcap_{i=1}^{n-r} H_i$. Projecting cones take their name from the fact that they define a generalized projection of a variety to a linear subspace (e.g. projecting from a point a spatial curve into a plane): the projection [Shafarevich vol. I, sec.4.4 Ex.1], with center or vertex $L_r$, is the rational map $\pi_{L_r}(x):=[h_1(x):\dots :h_{n-r}(x)]$ which is a regular morphism on the Zariski-open set $\mathbb{CP}^n\setminus L_r$. Therefore its restriction to any variety disjoint from the vertex, $\pi_{L_r}\vert_{X_k}:X_k\rightarrow\mathbb{CP}^{n-r-1}$ is a regular map of it to a projective subspace. Take any linear variety disjoint from $L_r$ as representative, i.e. $\mathbb{CP}^{n-r-1}\cong L'_{n-r-1}\subset\mathbb{CP}^n$ such that $L_r\cap L'_{n-r-1}=\varnothing$, which is always possible by [Beltrametti et al., Th.3.3.8] (since $\dim L_r\cap L'_{n-r-1}\geq r+(n-r-1)-n=-1$ so they do not intersect necessarily). Now for every point $x\in\mathbb{CP}^n\setminus L_r$, in particular $X_k$, there is a unique $L''_{r+1}$ passing through the vertex $L_r$ and $x$ by elementary dimension counting. The locus of all these generators $L''_{r+1}$ is just the generic projecting cone $C(X_k,L_r)$ for generic center $L_r$!. Each generator intersects $L'_{n-r-1}$ in a unique point (solution of a system of $n-(r+1)+n-(n-r-1)=n$ linear equations) which corresponds to $\pi_{L_r}(x)$ through the isomorphism with $\mathbb{CP}^{n-r-1}$. Therefore, given a generic linear subspace $L_r$ we can regularly (rationally if $L_r\cap X_k\neq\varnothing$) project any variety $X_k\subseteq\mathbb{CP}^n$ to a lower dimensional generic linear subspace $L'_{n-r-1}\cong\mathbb{CP}^{n-r-1}$ by intersecting the projecting cone with it, $C(X_k,L_r)\cap L'_{n-r-1}$, and calling $\pi_{L_r}(X_k)\subset \mathbb{CP}^{n-r-1}$ the projection of $X_k$ from $L_r$ to $L'_{n-r-1}$. The case $r=0$ is the classical projection from a point into a hyperplane (like our spatial curve projected to a plane curve). Therefore for any $X_k$, projecting from a generic center $L_{n-k-2}$, we obtain a, possibly reducible, variety $\bar{X}_k$ in $\mathbb{CP}^{k+1}$ as projection; since this comes from the intersection of the hypersurface $C(X_k,L_{n-k-2})$ with a linear variety $L'_{k+1}$, by the projective dimension theorem [Hartshorne, Th.I.7.2] every of its irreducible components has dimension $\geq (n-1)+(k+1)-n=k$. In fact, if $\dim X_k\geq 2$ by repeated application of Bertini's theorem [Hartshorne, Th.II.8.18], any such intersection is generically not only smooth but connected and thus irreducible, thus any generic such projection is a hypersurface $\bar{X}_k\subset\mathbb{CP}^{k+1}$ (generically reducible for $X_1$ a curve) and so it has a defining zero locus irreducible homogeneous polynomial $q\in\mathbb{C}[x_0,...,x_{k+1}]$ with well-defined algebraic degree (if $X_1$ is a curve then each of its irreducible components after intersecting will be points solution of a reducible polynomial). This is equivalent to B. since the intersection of a generic projecting cone hypersurface with a generic linear variety has the same polynomial degree as the cone (solve as many variables as possible from the linear system defining the linear variety and substitute in the homogeneous polynomial of the cone; each of its equal-degree monomials produce new monomials in less variables but of the same degree as the original, so one gets a new homogeneous polynomial in less variables, i.e. a hypersurface in a lower-dimensional projective space). This hints a geometric construction for the theorem of the birational equivalence of any projective algebraic set of dimension $k$ with a hypersurface in $\mathbb{CP}^{k+1}$, cf. [Beltrametti et al., sec.2.6.11] and [Hartshorne, Prop.I.4.9].

Degree of a $k$-subvariety of $\mathbb{CP}^{n}$ as the polynomial degree of the, possibly reducible, hypersurface obtained by generically projecting to $\mathbb{CP}^{k+1}$, i.e. intersecting the projecting cone with a suitable generic linear subspace: $$\deg X_k:=\deg \pi_{L}(X_k)=\deg C(X_k,L_{n-k-2})\cap L'_{k+1},$$ for generic $L\in\mathbb{Gr}(n-k-2,\mathbb{CP}^n)$ and $L'\in\mathbb{Gr}(n-k-2,\mathbb{CP}^n)$.

...

• E. Following [Beltrametti et al., Prop.3.4.8] let us go back to B. or C. above, our projection of $X_k$ into a hypersurface of $\mathbb{CP}^{k+1}$ via generic center $L_{n-k-2}\subset\mathbb{CP}^{n}$. Take a generic line $l_1\subset\mathbb{CP}^{n}$ not contained in $L_{n-k-2}$, so that the linear space $\operatorname{Join}(L_{n-k-2}, l_1)=\langle L_{n-k-2}, l_1 \rangle$ is a generic $L'_{n-k}$ because this is just a projecting cone, thus having dimension $(n-k-2)+(1)+1$ (cf. beginning of B. above). It is clear that any such generic linear $(n-k)$-space can be obtained in this way by generically decomposing it into a line and a linear $n-k-2$-subspace contained in it. Now, the intersection of a $k$-variety with a generic hyperplane has irreducible components of dimension $k-1$ [Shafarevich vol. I, sec.6.2], thus $X_k\cap L'_{n-k}$ consists generically of a finite number of points. This number of points is constant in a dense Zariski-open subset of the Grassmannian $\mathbb{Gr}(n-k,\mathbb{CP}^{n})$, cf. [Harris, Ex.18.2]. This can be readily proved by noticing that it is the number of points in the generic fiber of $\pi_{\Lambda}$ with center a generic hyperplane $\Lambda\subset L'_{n-k}$ seen in D. above. To see this, note that our generic $L'_{n-k}$ can be thought as a projecting cone with center $\mathbb{CP}^{n-k-1}\cong\Lambda\subset L'_{n-k}$, and the fiber of $\pi_{\Lambda}$, which is finite by D., is by construction the intersection of the projecting cone with the variety. Therefore $\#\, (X_k\cap L'_{n-k})=\deg \pi_{\Lambda}$, showing equivalence with all the previous notions. It is worth mentioning that many (most) classical treatments define degree as this finite number of points of a generic intersection with a linear space of dimension the codimension of the variety. An independent proof is [Mumford, Th.5.1] where it is shown that generic linear $(n-k)$-subspaces meeting transversaly our variety $X_k$, do so in a common number of points: the degree. (This relates to definitions in differential topology of intersections of submanifolds meeting properly, i.e. $T_pX_k\cap T_pL'_{n-k}={0}$ and $T_pX_k\oplus T_pL'_{n-k}=T_p\mathbb{CP}^n$).

Degree of a subvariety as the number of points of intersection (generically transversal) with a generic codimensional linear variety (i.e., $\dim L=n-\dim X$): $$\deg X_k := \# (X_k\cap L_{n-k})= \# (X_k\cap C(X_k, \Lambda_{n-k-1})),$$ for generic $L\in\mathbb{Gr}(n-k,\mathbb{CP}^n)$, $\Lambda\in\mathbb{Gr}(n-k-1,\mathbb{CP}^n)$.

[Beltrametti; Carletti; Gallarati; Bragadin] - Lectures on Curves, Surfaces and Projective Varieties: A Classical View of Algebraic Geometry, (EMS 2009).

[Harris] - Algebraic Geometry: A First Course, (Springer 1992).

[Hartshorne] - Algebraic Geometry, (Springer 1977).

[Mumford] - Algebraic Geometry I, Complex Projective Varieties, (Springer 1976).

[Shafarevich] - Basic Algebraic Geometry I: Varieties in Projective Space, (Springer 1994).

-
Thank you very much for your very detailed answer, which is very helpful. –  nono Nov 5 '12 at 14:36
You are welcome! I always wanted to understand the classical constructions concerning projecting cones and its relation to degree of a variety. –  Javier Álvarez Nov 6 '12 at 8:52


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# Chapter 5 : Polynomials

Chapter 5 : Polynomials 5.1 Exponent Properties 5.2 Negative Exponents 5.3 Scientific Notation 5.4 Introduction to Polynomials 5.5 Multiplying Polynomials 5.6 Multiply Special Products 5.7 Divide Polynomials


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## On Convergent Sequences

So here's a question that I've been preoccupied with in the last couple of days.  Suppose I have a convergent (at the sum) sequence, if you like a geometric sequence, say $1/2, 1/4, 1/8, ... = 1$.  If one were to multiply the elements of the sequence by its corresponding index, would this still be a convergent sequence? In this case, I'm concerned with $1/2 * 1, 1/4 * 2, 1/8 * 3, ...$ and its convergence (at the sum).  I wish to explore whether there are particular circumstances in which this is indeed the case.

I'll leave the question open ended for a bit.  I have solved this via derivative considerations, I think (I do not mean it as a pun... I think I have solved this by considering derivatives of functions).

Categories:

## On Naturally Arising Differential Equations

So if you have been following the argument a bit, it turns out that

is the starting at time t = 1 transition probability propagator of a probability distribution, say $c_0(x)$ at t=0, in the interval x = 0 to 1.  A question that I tried to answer was how zeros are propagated via the propagator or at the probability distribution, which lead to theorems that I dubbed "Shadow Casting" because, under that context, it turns out that a zero, if found on the propagator, remained in place until infinity, and via the propagator it appears on the probability distribution we want to propagate as well (therefore casting a "shadow").  I hadn't thought of the following approach until recently, and I haven't worked it out completely, but it connects to the theory of Ordinary Differential Equations which seems interesting to me. Here's the argument:

Suppose we focus on $p(x,y,1)$ for the time being, and wish to find the zeros on the transition probability surface.  Thus we seek $p(x,y,1) = 0$ and suppose $y(x)$ is an implicit function of $x$. We have

Now let $\mathbf{P}_y(y)$ is a collection derived from $y(x)$, so that, for example,

and I think we have successfully created a link to ODEs.  To find the zeros on the surface (and other time surfaces of the propagator) we stick in the correct $\alpha$ and solve, using the familiar methods (solve the homogeneous equation and the particular solution via sin-cos-exponential solutions, variation of parameters, power series, etc.).

I'm working out the specificities, for example including the constraints we know on $f_n^*(x)$ or $\mathbf{P}_x(x)$.  Perhaps this approach will help us broaden the spectrum of differential equations we can solve, by linking via Shadow Casting.

It may seem basic, but I think there is some untapped power here.

Additionally, I have been working on clarifying some thoughts on polynomials that converge in area in the interval from 0 to 1, but all those details tend to be a bit drab and I keep having trouble focusing.  Nevertheless, there is a lot of clarity that I have been able to include, and it is now in the newest draft of "Compendium".  By the way, I renamed it. It is now "Compendium of Claims and Proofs on How Probability Distributions Transform".  There's still soooo much more to do.

Here it is! part-i-v28

## On Riding the Wave

So here it is, the pinnacle of my research effort thus far. I'll start by the definitions:

Definition 1. Let $f(x,y)$ and $g(x,y)$ be surfaces so that $f,g \colon [0,1] \times [0,1] \to \mathbb{R}$. The star operator $\star \colon [0,1]^2 \times [0,1]^2 \to [0,1]^2$ takes two surfaces and creates another in the following way:

with the central transformation being defined by $\diamond \colon [0,1]^2 \times [0,1]^2 \to [0,1]^2$

and the last transformation that takes $h(x,z) \rightsquigarrow h(x,y)$ we will call $j \colon [0,1]^2 \to [0,1]^2$. Thus

Definition 2. Define a continuous, bounded surface $p(x,y)$, with $p \colon [ 0, 1 ] \times [ 0, 1 ] \to \mathbb{R}^+ \cup \{0\}$,  and let $\int_0^1 p(x,y) \, dx = 1$ be true regardless of the value of $y$.    In other words, integrating such surface with respect to $x$ yields the uniform  probability distribution $u(y)$, $u \colon [0,1] \to \{1\}$. We will call this a strict Pasquali patch, and such is intimately related to probability notions.  With $p \colon [ 0, 1 ] \times [ 0, 1 ] \to \mathbb{R}$, we have a more general definition for a Pasquali patch.

Construction 1. Let $p(x,y) = \sum_{i = 1}^n f_i(x) \cdot g_i(y) = \mathbf{f}(x) \cdot \mathbf{g}(y)$, a function which consists of a finite sum of pairs of functions of $x$ and $y$. In the spirit of conciseness, we omit the transpose symbology, thusly understanding the first vector listed in the dot product as a row vector and the second vector as a column vector.  Then $p(x,y)$ is a Pasquali patch provided

and $F_n \neq 0$. Thus, we may choose $n-1$ arbitrary functions of $x$,  $n-1$ arbitrary functions of $y$, an $n$th function of $x$ so that $F_n \neq 0$, and

We may write the normalized version as:

and again observe that the unit contribution to the integral of the Pasquali patch is provided by $f_n^*(x)$, so that $F_n^* = 1$.

Claim 1. Pasquali patches constructed as by Construction 1 are closed.

Proof. To make the proof clear, let us relabel the normalized version of Construction 1 as

with $\int_0^1 \mathbf{P}_x(x) \, dx = 0$ so as to manipulate the equation more simply, and

with $F_n^* = 1$ and $H_n^* = 1$. Then

with $\alpha = \int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{Q}_y(y) \, dy$, $\beta = \int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{1} \, dy = 0$, and $\gamma = \int_0^1 f_n^*(1-y) \cdot \mathbf{Q}_y(y) \, dy$. We're not too concerned of the form of the resultant star product as much as its structure. Observe

can be folded back into function vectors $\mathbf{R}_x(x)$ and $\mathbf{R}_y(y)$. Thus the structure of Construction 1 functions is preserved when we multiply one by another, showing closure. Of course the property of Construction 1 being Pasquali patches means $r(x,y)$ is closed under that property, and so is a Pasquali patch also, as can be seen when we integrate across $x$:

and the unit contribution is given by $h_n^*(x)$. \qed

Claim 2. Pasquali patches constructed as by Construction 1 have powers:

with

and

Proof by Induction. First, using the formula observe

and the second power

which is exactly what we expect from the definition of Construction 1 and Claim 1. Next, let us assume that the formula works and

Let us examine $p_{k+1}(x,y) = p_k(x,y) \star p(x,y)$

term by term upon dotting. The first dot the first term is:

The first dot the last term is:

The last dot the first term is:

The last dot the last term is:

The middle dot the first term is:

Finally, the middle dot the last term vanishes:

Putting all this information together we get:

\qed

Claim 3. It follows that

and $\lvert \alpha \rvert < 1$, $\gamma$ bounded, both conditions necessary and sufficient to establish that such a limiting surface indeed exists (convergence criterion). Furthermore, we check that this is indeed a Pasquali patch.

Proof. To reach a steady state limit,

Next, the steady state limit must be solely functions of $x$, so the functions of $y$ must vanish at the limit. Thus, it follows that $\lvert \alpha \rvert < 1$. We have now established bounds on $\alpha$, which happen to be exactly the radius of convergence of the geometric series:

and

gives the desired result:

As a check, we integrate across $x$ to corroborate the definition of Pasquali patch:

Claim 4.

is the eigenfunction corresponding to eigenvalue $\lambda = 1$ of all Construction 1 functions, through each power independently.

Proof. An eigenfunction $e(x)$ has the property

where the eigenfunction's   corresponding eigenvalue is $\lambda$.  The claim is more ambitious, and we will show that $p_\infty(x) \star p_n(x,y) = 1 \cdot p_\infty(x)$ for any $n \in \mathbb{Z}^+$.  The left-hand side is

Observe the first term dotted with the middle and last term produce $\beta$ which annihilates the results, so that the only relevant term is the first dot the first:

The second term dot the first produces:

The second term and the second:

and the second by the last term gives

Factoring gives

The parenthetical part of this last formulation is equivalent to

within the bounds already established for $\alpha$, and the result of the star product is

as we wanted to show. \qed

Claim 5.

$A$ is a constant, is the eigenfunction corresponding to eigenvalue $\lambda = \alpha$ of

Proof. The eigenfunction equation is suggestive of what we must do to prove the claim: $e(x) \star p(x,y) = \lambda e(x)$. We must show that, starring the eigenfunction with $p(x,y)$, we obtain $\alpha$ times the eigenfunction.

Thus:

\qed

Part I v27

## On Proving Eigenvalue = 1 for Particular Surfaces

So, as you have read here, I've been saying for quite a few years now that (bounded) smooth surfaces on $\left[ 0, 1 \right]^2 \to \mathbb{R}$ have certain invariants when viewed from a particular perspective (eigenvalues, eigenfunctions).  There are particular surfaces which I dubbed Pasquali patches (for lack of a better word) and a particular construction, namely

with $F_2 = \int_0^1 f_2(x) \, dx \neq 0$ which are very special in that they have lots of properties which are interesting and closely tied to probability theory.  I have now proven for this particular construction that it possesses two very specific eigenvalues given a particular operator "star" $\star$... one of which is $\lambda_1 = 1$, regardless of function choices for $f_1(x), g_1(y)$ and almost arbitrary choice of $f_2(x)$, which, by requirement needs $F_2 \neq 0$.  This mimics well known probability mathematics (except in the surface realm) and operator theory/linear algebra. I think of this as a very proud accomplishment.

Thusly, I have revamped the relevant sections in Compendium full of new and juicy recharacterizations in order to be able to do just this... particularly Section 2, the definition of Pasquali patches and Section 11.6, Relevant Generalizations as Applied to Pasquali Patches (where I have included such a proof).  Section 2 is now a lot tighter than it used to be, and I'm trying really hard to go over everything to close all the loopholes I've left so that Compendium isn't just notes but an actual... Volume of Mathematics or Book or something.

I am very close to a full eigenvalue theory which I will apply to a generalization of the Pasquali patch formula above, giving a multiplicity of calculable eigenvalues for, not just Pasquali patches, but any sufficiently well behaved surface on $\left[ 0, 1 \right]^2 \to \mathbb{R}$.

Part I v26


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NYJC/2021/JC1/Promo/P1/Q10

Mrs Tan wants to build a wooden fence using vertical planks of equal width and thickness but different lengths. In her plan, the length of the first wooden plank is 2 metres and the length of the planks forms a geometric progression. The length of the 10 th plank is $1.5$ metres.
(i) If the cost of wooden plank is $\$ 18$per metre, show that the cost of the fence will never exceed$\$1200$.
Mrs Tan realised that her plan is not feasible and now wants to build her fence using several identical panels. Each panel comprises 10 vertical planks with identical dimensions to the first 10 planks in her original plan.

[4]

(ii) Find the total length, $T$ metres, of the planks to be used in each panel.

[2]

(iii) She hires a contractor to install the fence. The contractor misunderstands her instructions and uses 10 planks to construct a panel so that the lengths form an arithmetic progression with common difference $d$ metres. If the total length of the planks to be used for one panel is still $T$ metres and the length of the first plank is still 2 metres, find the value of $d$.

[2]

The contractor offers to paint the fence for Mrs Tan. He buys a 5-litre can of paint to do the paint job. To save costs, he fills the can to the 5-litre mark with turpentine to form a uniform mixture when the level of paint in the can falls to the 4-litre mark. He then repeats this process whenever the level of the mixture falls to the 4-litre mark.
(iv) State, in terms of $n$, an expression for the volume of paint remaining in the mixture after the $n$th refill.

[2]

(v) Find the minimum number of refills taken before the mixture is more than $80 \%$ turpentine.

[2]


text:
Determine whether there is a train collision

1. Aug 28, 2005

Two trains, one traveling at 60 miles/hr and the other at 80 miles/hr, are headed toward one another on a straight level track. When they are 2.0 miles apart, both engineers simultaneously see the other's train and apply their brakes. If the brakes decelerate each train at the rate of 3.0 ft/sec^2, determine whether there is a collision.

Ok, so is it correct to say that if $d > \frac{(v_{1}-v_{2})^{2}}{2a}$ there will be no collision, and if $d < \frac{(v_{1}-v_{2})^{2}}{2a}$ there will be a collision (d is distance, v is velocity, and a is acceleration). When I do this, I get that there will be a collision, but the correct answer is that there will be no collision. What am I doing wrong?

Thanks

2. Aug 28, 2005

StatusX

You need to use two equations and see how far each train gets, and then see whether the sum is greater or less than 2 miles. The reason you can't get away with one equation is because once the 60 mph train stops, it doesn't start going backwards. It stops accelerating, but the 80 mph train continues slowing down. Plus, if you were going to do it that way, you would use v1+v2 in your equation, and 6 ft/sec^2 as the acceleration, since both trains are accelerating.

3. Aug 29, 2005

would I use $x = x{0} + v_{x}_{0}t + \frac{1}{2}(v_{x}_{0} + v_{x})$?

4. Aug 29, 2005

StatusX

I think your last term should be 1/2 a t^2, and remember you need to find the t where each train stops and plug that in to get x. You don't need an x_0, you just need to know if the total distance the trains travel once they start braking is more or less than 2 miles.

5. Aug 30, 2005

So $v_{x} = v_{x}_{0} + a_{x}t$. I got $t = 20, t = 26.66$ for 60 mph and 80 mph respectively. After plugging these times in $x = v_{x}_{0}t + \frac{1}{2}a_{x}t^{2}$ I got $1066.1334 + 600 = 1666.1334 ft$ which is less than two miles. Is this correct?

6. Aug 30, 2005

StatusX

7. Aug 30, 2005

should i convert the acceleration from ft/sec to miles/hour? Also the time in seconds to time in minutes?

8. Aug 30, 2005

StatusX

You just need to make sure everything has the same units of length and time (eg, feet and seconds).

9. Aug 30, 2005


text:
• May 22, 2015, 09:17:37 AM

### Author Topic: "I'm never shopping THERE again!" Share your story!  (Read 1425902 times)

0 Members and 1 Guest are viewing this topic.

#### Cami

• Hero Member
• Posts: 1307
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3915 on: April 11, 2013, 12:22:25 PM »
Because she wasn't aware that it was meant to be a collectible and that writing on a receipt +bag over the comic would do so much damage?

The way she went at it with pushing for a 10 and actually writting it would be borderline for me.
Telling people about the survey is fine, joking about the buyers giving a ten is fine, actually writting it is a bit too much.

At one place I worked we were required by our district manager to specifically request top scores and to write it on the receipts. So it might have been the cashier being too pushy or it might have been a corporate requirement.

That doesn't make it okay to push for high scores, but it does mean that the wrong person can get blamed.
My dd worked retail and was required to point out the survey and required to ask for a 10. Failing to do so would get the employee a reprimand. Failing to do so five times would get the employee fired.

Which would be counterproductive for me, because I pretty much automatically downgrade anyone who asks for a 10.

Rob
You need to keep in mind that you're punishing the cashier for doing as they are required.

#### RebeccainGA

• Hero Member
• Posts: 1207
• formerly RebeccainAR
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3916 on: April 11, 2013, 12:29:02 PM »
I had one today. I use a CPAP machine, and the masks you use with them are a prescription item. I thought I had to go through the medical goods firm that my insurance company recommends - and they charge me, until I've met my deductible, aprox $400US for a mask kit. Exorbitant, but I was OK with it - thought that was my only option, service was reasonable, etc. Then I had an issue with something being billed incorrectly, and they got very snarky with me. So I started looking around. Found that I could just buy the masks, without the insurance, for less than 25% of what I was paying. Thought I'd check with the DME and see if they would price match - after all, I'd be saving myself $$and my insurance company$$$ -  they pretty much told me to pound sand. So I bought them from a wholesaler instead of through the insurance people.

Shame - we'd been buying $from the medical goods firm since DP got out of the hospital, and will continue to buy $ from someone - but I don't think it'll be that company, after the way they treated me. I mean, tell me it's not possible, that's one thing, but to be snarky over their mistake and then over a simple, routine request?? No thanks.

#### Sophie Jenkins

• Member
• Posts: 127
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3917 on: April 11, 2013, 12:49:03 PM »
And the thing is, I'm not going to give 10s across the board if the company doesn't actually deserve the 10s. 10s to me indicate that you went well above and beyond -- the equivalent of a server earning a 25% tip or better. 20% tip service, good to very good service, does not earn a 10, and I'm not going to be guilted in to giving a 10 because of poor corporate policy.

Rob

If what you're hoping to effect with your surveys is actual change within a company, you'd be far better off contacting corporate directly. Poor surveys don't send any message to corporate other than telling them who they need to punish for following their own ridiculous corporate requirements.

Nothing cashiers do lets them win- follow corporate guidelines and customers are unhappy and get them in trouble. Don't follow corporate guidelines to keep the customers happy and they get in trouble for not following guidelines.

Better to not fill out a survey at all than get someone in trouble for something they have no control over.

#### Carotte

• Hero Member
• Posts: 1384
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3918 on: April 11, 2013, 01:29:41 PM »
Well, I think my bank will soon get the boot, depending on how they are going to explain how it came to be that I'm making a transfer to myself now  .
Background: for the past ten years or so my mom has made a monthly deposit of 15€ every months, it's called "permanent transfer LASTNAME". This month (havent checked las month statement yet), I have a "permanet transfer LASTNAME" in my debit and one in my credit line.
Checked online, both are from/to my account, with my name (we have the same LASTNAME). And none from my mother.

Unless they appologize and bend backward while serving me maccarons I'm leaving, pronto. I was already thinking about since some of their charges are much higher than other banks.

They messed two peoples account (we do have the same lastname and bank, and my mothers has a right to handle mine if I'm away or unable but that's not a reason!), made her deposit vanish and created a new monthly deposit that I never agreed upon.
And it takes them 14 days to have the deposit appear in the depositee account? I might be wrong but I bet other banks are faster than that!

#### DottyG

• Super Hero!
• Posts: 18204
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3919 on: April 11, 2013, 01:37:03 PM »
I started a spin-off thread for the customer survey discussion.

#### Jocelyn

• Hero Member
• Posts: 3530
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3920 on: April 11, 2013, 09:05:50 PM »
I had one today. I use a CPAP machine, and the masks you use with them are a prescription item. I thought I had to go through the medical goods firm that my insurance company recommends - and they charge me, until I've met my deductible, aprox $400US for a mask kit. Exorbitant, but I was OK with it - thought that was my only option, service was reasonable, etc. Then I had an issue with something being billed incorrectly, and they got very snarky with me. So I started looking around. Found that I could just buy the masks, without the insurance, for less than 25% of what I was paying. Thought I'd check with the DME and see if they would price match - after all, I'd be saving myself $$and my insurance company$$$ -  they pretty much told me to pound sand. So I bought them from a wholesaler instead of through the insurance people.

Shame - we'd been buying $from the medical goods firm since DP got out of the hospital, and will continue to buy $ from someone - but I don't think it'll be that company, after the way they treated me. I mean, tell me it's not possible, that's one thing, but to be snarky over their mistake and then over a simple, routine request?? No thanks.

CPAP.com
A friend gave me the referral, and I've been happy with the things I've bought from them. Huge selection, and much cheaper than my previous provider.

#### Slartibartfast

• Super Hero!
• Posts: 12117
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3921 on: April 11, 2013, 11:42:13 PM »
This is kind of more like a product review  but I was really disappointed.  I was intrigued by the concept of getting our phone line via our DSL - right now we pay for local calls and DSL through our phone company and long-distance through a 3rd party.  The phone company won't give us a bundle discount because apparently local doesn't count for that, so we pay ~$40/month between the two phone bills (not counting our cell phones, which we use minimally). There are a few companies which offer devices that do this - MagicJack advertised the most, but I read some reviews and thought NetDUO sounded the best. ~$80 covered the gear and the first year, then it would be $30/YEAR after that (instead of$40/month).  Huge savings, right?

I've gone back and forth on this, but my conclusion is that the call quality varies between "slightly better than adequate" and "completely useless."  As in, I was all ready to call our phone company and cancel our regular landline, but the interference on the call was so bad the phone tree couldn't figure out what buttons I was pressing.  It seems like most of the time, the call will be reasonably clear on one end and staticky on the other.

As much as I'd love to save $450/year on my phone bill, I want my phone to actually be useful as a phone! I guess the next step is to add up how much time DH and I actually spend on our phones and see whether that$450/year would pay for us to have phones we could use for more than just emergencies . . .

#### jedikaiti

• Swiss Army Nerd
• Hero Member
• Posts: 3167
• A pie in the hand is worth two in the mail.
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3922 on: April 12, 2013, 12:04:23 AM »
A friend of mine bought a MagicJack to take to England when she moved, so she'd have a US number available for family & friends to call her. Works like a charm, I can't hear the difference.
What part of v_e = \sqrt{\frac{2GM}{r}} don't you understand? It's only rocket science!

"The problem with re-examining your brilliant ideas is that more often than not, you discover they are the intellectual equivalent of saying, 'Hold my beer and watch this!'" - Cindy Couture

#### Slartibartfast

• Super Hero!
• Posts: 12117
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3923 on: April 12, 2013, 01:20:22 AM »
Grr, just spent the last half an hour perusing my phone company's website.  On a whim, I decided to check the company-assigned email address they won't let me change and which I've never actually checked since I first became a customer in 2003.  Apparently sometime between when I signed up for "unlimited" DSL and now, they a) put a cap on bandwidth, and b) started sending me threatening emails that I was getting close to going over it.  To the email address I have never read and only marginally knew existed.  Oh, and the notice that they're raising my DSL bill by $3/month starting with my next bill? Yeah, that went to the phantom email address too. I would switch companies if I could - and since I'm not 100% tied to the idea of a landline, I might actually have the freedom to do so. Nothing makes you feel like a chump quite like seeing an advertised rate for new customers at less than half what you're getting charged for the same exact service #### Carotte • Hero Member • Posts: 1384 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3924 on: April 12, 2013, 05:49:49 AM » I would switch companies if I could - and since I'm not 100% tied to the idea of a landline, I might actually have the freedom to do so. Nothing makes you feel like a chump quite like seeing an advertised rate for new customers at less than half what you're getting charged for the same exact service Same goes for banks and what mine charges vs. what others have free... You do have to excavate the info as to what/when it is free, but I'm eligible for most anyway if i move my two accounts. #### JenJay • I'm a nonconformist who doesn't conform to the prevailing standards of nonconformity. • Super Hero! • Posts: 7127 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3925 on: April 12, 2013, 07:32:30 AM » I won't stop shopping at Whittards, but I'm seriously considering doing all my purchases online from now on. The last few times I've been to my local branch, the sales assistants have been quite "enthusiastic", to say the least, but today was utter nuts. I just popped in there to get another jar of my favourite instant coffee. Between picking up the jar and getting to the till (approximately 5 metres, if that), I was asked by two separate shop assistants if I needed any help. Fair enough. I held up the jar and said "Just this, thanks" to both of them, and they backed off. However, when I got to the till, the cashier: - asked me if I was sure I only wanted one jar. (Well, duh, if I'd wanted more, I would have picked up more. Surely?) - tried to persuade me into buying ground coffee instead of instant. (I don't have a coffee maker.) - tried to persuade me into buying a coffee maker / percolator. (I don't want a coffee maker. I am a heathen who likes instant!) - asked me if I'd considered buying any tea or hot chocolate products. (I came in for the coffee, dude.) - asked me if I'd had a chance to look at their china cups/mugs. (Seriously, I just came in for the coffee.) - asked me if I wanted to buy some single-serve-sachets of their new instant tea flavours - "only £1 today!". (There is a massive sign, and no, I don't.) - asked me if I wanted to be on their mailing list, and sounded quite put out when I said I already was. (Which I am.) - started on a clearly practised lecture on the evils of plastic bags. (At this point, I had to interrupt him, and tell him that I didn't need a bag, but that I was in a bit of a hurry.) After all this, he then finally got around to actually scanning my jar of coffee, and letting me pay, but I could tell that he wasn't happy that I hadn't taken any of his other suggestions. All in all, a quite uncomfortable experience, which lasted almost 15 minutes when it could have been done in 2. So yeah, I think I'm going to shop online from now on. Please consider contacting corporate with a generic complaint about the employees trying to up-sell you (I say generic because if the man you were dealing with was, in fact, doing what he was directed to do then I'd hate to see him get complained about personally). At my last job we had to treat customers like that and believe me, we hated it as much as they did. Unfortunately getting caught not offering "selling suggestions" was a big deal and would get you called into the manager's office for a lecture. We not only had secret shoppers but they'd send employees from other stores around to act like customers and check up on us. Finally after enough customers complained they put a stop to it. Oregon #### RebeccainGA • Hero Member • Posts: 1207 • formerly RebeccainAR ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3926 on: April 12, 2013, 08:32:02 AM » CPAP.com A friend gave me the referral, and I've been happy with the things I've bought from them. Huge selection, and much cheaper than my previous provider. Actually, that's who I went with! I've also now recommended them to my mother - they have been great so far. :-) #### artk2002 • Super Hero! • Posts: 13524 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3927 on: April 12, 2013, 02:40:59 PM » This is kind of more like a product review but I was really disappointed. I was intrigued by the concept of getting our phone line via our DSL - right now we pay for local calls and DSL through our phone company and long-distance through a 3rd party. The phone company won't give us a bundle discount because apparently local doesn't count for that, so we pay ~$40/month between the two phone bills (not counting our cell phones, which we use minimally).  There are a few companies which offer devices that do this - MagicJack advertised the most, but I read some reviews and thought NetDUO sounded the best.  ~$80 covered the gear and the first year, then it would be$30/YEAR after that (instead of $40/month). Huge savings, right? I've gone back and forth on this, but my conclusion is that the call quality varies between "slightly better than adequate" and "completely useless." As in, I was all ready to call our phone company and cancel our regular landline, but the interference on the call was so bad the phone tree couldn't figure out what buttons I was pressing. It seems like most of the time, the call will be reasonably clear on one end and staticky on the other. As much as I'd love to save$450/year on my phone bill, I want my phone to actually be useful as a phone!  I guess the next step is to add up how much time DH and I actually spend on our phones and see whether that \$450/year would pay for us to have phones we could use for more than just emergencies . . .

For work, I have Voice Over IP (VOIP), which is what you're doing here. Call quality stinks, especially if someone in the house looks at YouTube video or browses a web site. It also sends calls directly to voice mail; I just missed a call from by boss even though I was sitting right next to the phone the whole time. I knew about the call because the phone happily beeped telling me that I had a message.
Twenty years from now you will be more disappointed by the things that you didn't do than by the ones you did do. So throw off the bow lines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover. -Mark Twain

#### GSNW

• Member
• Posts: 576
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3928 on: April 12, 2013, 02:50:01 PM »
We ordered a new sink from Lowe's.  Our old one was junk and I really wanted a nice DEEP single-basin sink.  I found just the one I wanted (available special order only) and a good faucet to go with.  Ordered both together and verified that both could be picked up together at Lowe's when the sink came in.  Paid at time of order.

1.  Got a call two days later informing that the faucet was in stock, would I come get it?  No, I said, I don't want it until the sink comes in.  Verified again that I could pick up both together.

2.  Got an email informing that the sink would be shipped to the store that is about 25 miles from my home.  Reiterated that I needed it shipped to the store 2 miles from my home, yes, that's where I want it, that's where it needs to go.  Order had to be canceled and re-processed for this to happen.

3.  Got a call informing me that sink had arrived.  Yay!

4.  Drove to store to get sink the day after the phone call.  Advised by CS desk that sink was at the 25-mile away store.  No apology, the guy could not have cared less.  Basically... go get your sink and get out of my face.

5.  Drove to far away store in Saturday traffic.  Fun.  Arrived and caused mass confusion as they searched high and low for sink.  Turns out sink WAS AT THE STORE CLOSE TO MY HOUSE THAT I HAD ALREADY BEEN AT.

6.  Went back.  It took them an hour to find the sink, at which point I was hoping would be big enough to drown myself in.  Again, no apologies or "sorry you've been wasting your time all day..."

7.  Sink comes out.  No faucet.  Apparently since I didn't pick up the faucet when they asked, they returned it and refunded the cost on my card (which I should have noticed, but that's hardly the point).

I WILL NEVER GIVE THEM ANOTHER DIME, EVER.  This was in January.  I didn't even bother writing a letter, because literally nothing they could do short of a free lifetime charge account would make me want to deal with their employees again.

#### Outdoor Girl

• Super Hero!
• Posts: 15043
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3929 on: April 12, 2013, 02:55:12 PM »
The other issue with VOIP, at least here in Ontario, is that it doesn't necessarily register the address correctly when you dial 911.  A Bell landline call will automatically bring up the address for the 911 operators.  A Rogers Cable (in my area) VOIP call will bring up the address but most of the other smaller companies don't bring up the address right away.  There is an extra step involved and if the address isn't correct on the account, EMS could start heading to the wrong address before this is determined.

I still have a landline with Bell.  Mainly because I talked them down in price so it is comparable to switching to Rogers.  I play the two of them off each other all the time, trying to get the best deal.
After cleaning out my Dad's house, I have this advice:  If you haven't used it in a year, throw it out!!!!.
Ontario


text:
# Homework Help: Determine whether the sequences are increasing

1. Nov 4, 2007

### will_lansing

1. The problem statement, all variables and given/known data
Determine whether the sequences are increasing, decreasing, or not monotonic.
1) an= $$\frac{\sqrt{n+2}}{4n+2}$$
2) an=$$\frac{1}{4n+2}$$
3) an=$$\frac{cosn}{2^{n}}$$
4) an=$$\frac{n-2}{n+2}$$
2. Relevant equations

3. The attempt at a solution
I thought that the number 1 and 2 were decreasing because $$a_{n}$$ $$\geq$$ $$a_{n+1}$$ in both cases
#3 is increasing because $$a_{n}$$$$\leq$$ $$a_{n+1}$$
#4 is monotonic because if n=3 then its 0

for number #1 I compared it to $$a_{n+1}$$=$$\frac{\sqrt{n+3}}{4n+6}$$
and i found that it was smaller than an so it was decreasing
i did this procedure for the rest of them and i still get the wrong answer


text:
# Command - line conclusion from command history

So, I've considered background and also at Ctrl+R, yet they are not what I assumed I recognized. Exists a manner in which I can key in the start of a command, and also cycle via the suits in my background with some bash faster way?

# mysq(some shortcut key)

Gives me:

# mysqldump  --add-drop-table -e -q -n -C -u
(some shortcut key)
#  mysql -u ben.dauphinee -p

0
2019-05-13 05:00:20
Source Share

Pressing Ctrl+R will certainly open the reverse background search. Currently start keying your command, this will certainly offer the first suit. By pushing Ctrl+R once more (and also once more) you can cycle via the background.

mysq(Ctrl+R)

Would offer:

mysqldump  --add-drop-table -e -q -n -C -u

Ctrl+R once more:

mysql -u ben.dauphinee -p

0
2019-05-17 18:46:10
Source

You can make use of the readline regulates history-search-backward and also history-search-forward to browse in between commands lines starting with the prefix you've currently keyed in. Neither of these commands are bound to type in the default arrangement.

Zsh (zle) has comparable commands history-beginning-search-backward and also history-beginning-search-forward, additionally not bound to keys by default. There are additionally history-search-backward and also history-search-forward, which makes use of the first word of the existing command as the prefix to look no matter the arrow placement.

0
2019-05-17 18:32:53
Source


text:
Data

1. Title: Webs and $q$-Howe dualities in types $\mathbf{B}\mathbf{C}\mathbf{D}$
2. Authors: Antonio Sartori and Daniel Tubbenhauer
3. Status: To appear in Trans. Amer. Math. Soc. Last update: Tue, 3 Apr 2018 13:17:55 EST
5. LaTex Beamer presentation: Slides1, Slides2

Abstract

We define web categories describing intertwiners for the orthogonal and symplectic Lie algebras, and, in the quantized setup, for certain orthogonal and symplectic coideal subalgebras. They generalize the Brauer category, and allow us to prove quantum versions of some classical type $\mathbf{B}\mathbf{C}\mathbf{D}$ Howe dualities.

A few extra words

Consider the following question: Given some Lie algebra $\mathfrak{g}$, can one give a generator-relation presentation for the category of its finite-dimensional representations, or for some well-behaved subcategory?
Maybe the best-known instance of this is the case of the monoidal category generated by the vector representation of (quantum) $\mathfrak{sl}_2$. Its generator-relation presentation is known as the Temperley-Lieb category and goes back to work of Rumer-Teller-Weyl and Temperley-Lieb.
Several generalizations of this were found over the years. The most important one for our paper was seminal work of Cautis-Kamnitzer-Morrison who a generator-relation presentation of the monoidal category generated by (quantum) exterior powers of the vector representation of quantum $\mathfrak{gl}_n$.
Their crucial observation was that a classical tool from representation and invariant theory, known as skew Howe duality, can be quantized and used as a device to describe intertwiners of quantum $\mathfrak{gl}_n$. The diagrammatic presentation is provided by so-called (type) $\mathbf{A}$-web categories.
The idea which started this paper was to apply Cautis-Kamnitzer-Morrison's approach to types $\mathbf{B}\mathbf{C}\mathbf{D}$. However, quantization outside of type $\mathbf{A}$ turns out be be quite delicate. In particular, we cannot use the road map given by Cautis-Kamnitzer-Morrison since quantization of Howe's classical dualities in types $\mathbf{B}\mathbf{C}\mathbf{D}$ is not a straightforward affair.
To overcome this problem, we consider alternative quantizations in types $\mathbf{B}\mathbf{C}\mathbf{D}$ provided by so-called coideal subalgebras of quantum $\mathfrak{gl}_n$. (In short, these have “nicer” quantum factors than the quantum enveloping algebras, but worse “topological behaviour”.)
Our approach then goes as follows: In order to quantize Howe dualities in types $\mathbf{B}\mathbf{C}\mathbf{D}$, we define extended web categories, which we call cup- and dot-web categories, and prove that they act on the representation categories of the coideal subalgebras. (And they provide diagrammatic descriptions of these categories by playing the roles of “thickened” Brauer categories.) We will then show that these extended web categories can be used to recovering some versions of quantum Howe duality in types $\mathbf{B}\mathbf{C}\mathbf{D}$.
Note that our approach goes somehow the opposite way with respect to Cautis-Kamnitzer-Morrison's road map: instead of using quantum Howe duality to obtain a web calculus, we use our web categories to prove quantized Howe dualities. We define these cup- and dot-web categories in our paper. All the reader needs to know about them before looking into the paper is summarized in the Figure below.

Note that these can be seen as “extended type $\mathbf{A}$ webs”.


text:
1. ## Differentiating sin(x/y)

How do we differentiate something like this? Chain rule should be used here, so the first term will be cos(x/y), but then I don't understand how to break up this (x/y) function?

2. ## Re: Differentiating sin(x/y)

Oh, wait, I just looked up the answer. It seems like I should just use quotient rule?

3. ## Re: Differentiating sin(x/y)

First, the answer is yes. By the way, it is d/dx right? What you are differentiating WITH RESPECT TO is important.

Yes d/dx.

*deleted*

7. ## Re: Differentiating sin(x/y)

Are we assuming that y is a function of x? Or are you doing a partial derivative?

8. ## Re: Differentiating sin(x/y)

It's actually an expression in implicit differentiation.

9. ## Re: Differentiating sin(x/y)

In other words, y is a function of x.

So like you have suggested, you need to differentiate this function implicitly. How will you do this?

10. ## Re: Differentiating sin(x/y)

$\cos(\dfrac{x}{y}) \cdot \dfrac{y-xy'}{y^2}$ right?

11. ## Re: Differentiating sin(x/y)

That's correct for now. But remember, our aim is to solve for y' so that means your original expression needs to be an EQUATION.

What was the original equation?

12. ## Re: Differentiating sin(x/y)

Well, I just wanted to clear up this part , that's why I posted only it. I was confused whether I should apply the chain rule to the inside term x\y and how if so.

13. ## Re: Differentiating sin(x/y)

You already did. But like I said, you can not continue unless you have an EQUATION to start with.

14. ## Re: Differentiating sin(x/y)

$$\frac{d}{dx}\left[\sin(\dfrac{x}{y})\right]=\frac{d}{dx}\left[ x+y \right]$$

$$\cos(\dfrac{x}{y}) \cdot \dfrac{y-xy'}{y^2}=1+y'$$

$$\cos(\dfrac{x}{y})y - \cos(\dfrac{x}{y})xy' = y^2+y^2y'$$

$$\cos(\dfrac{x}{y})y - y^2 = \cos(\dfrac{x}{y})xy' + y^2y'$$

$$\cos(\dfrac{x}{y})y - y^2 = y' \left( \cos(\dfrac{x}{y})x + y^2 \right)$$

$$y'=\dfrac{\cos(\dfrac{x}{y})y - y^2}{\cos(\dfrac{x}{y})x + y^2 }$$

15. ## Re: Differentiating sin(x/y)

$\frac{dy}{dx} = \frac{-f_{x}}{f_{y}}$ where $f_{x}, f_{y}$ are the partial derivatives of x and y respectively. If you don't know what partial derivatives are, then for the x partial. treat y as though it is a constant. and similarly for the y partial, treat x as it is a constant.

I get $\frac{dy}{dx} = \frac{y}{x}$

Page 1 of 2 12 Last


text:
## College Algebra (6th Edition)

According to the definition of Scientific Notation given in the textbook, a number is written in scientific notation when it is expressed in the form $a \times 10^n$. $a$ can take any value between $-10$ and $10$, excluding $0$, while $n$ should be an integer and can take any integer value. In the given case, $a=40$ and $n=2$. While the value of $n$ fulfills the conditions, the absolute value of $a$ is greater than $10$ and does not fulfill the conditions. The proper Scientific Notation for $40\times10^2$ would be $(4\times10)\times10^2=4\times10^{1+2}=4\times10^3$.


text:
# Tag Info

## Hot answers tagged stochastic-processes

3

Set $$M_t := f(X_t) - \int_0^t Af(X_s) \, ds$$ and write $P_t f(x) := \mathbb{E}^x f(X_t)$ for the semigroup associated with $(X_t)_{t \geq 0}$. Using the definition of the generator, it is not difficult to see that $$\frac{d}{dt} (\mathbb{E}^x f(X_t)) = \frac{d}{dt} P_t f(x) = P_t Af(x).$$ By the fundamental theorem of calculus, this implies $$P_t ... 3 \nu \circ f^{-1} denotes the push-forward measure (or image measure) of f with respect to \nu, i.e.$$\nu_1(B) = (\nu \circ f^{-1})(B) = \int 1_B(f(x)) \, \nu(dx)$$for any Borel set B. This implies$$\int g(x) \, \nu_1(dx) = \int g(f(x)) \, \nu(dx)$$for any g \in L^1(\nu_1). 3 Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it. If we define$$f_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$then f_n \uparrow 1_{(0,\infty)}. Obviously, f_n is not differentiable at x=0 at x=1/n, but for the moment we forget about this (that's the non-rigorous part) and apply ... 3 You need to show independence of increments, i.e. if 0\le a<b<c<d then (N_1(d)+N_2(d)) - (N_1(c)+N_2(c)) is independent of (N_1(b)+N_2(b)) - (N_1(a)+N_2(a)), and similarly for more than two intervals. You can prove that by using independence of increments of each of the two processes separately plus independence of N_1 and N_2. You also ... 3 Hmm. I don't know how you can directly find A, B, C but I can suggest a way to go about the problem. Now that you have found the eigenvalues \mu_i, solve for the corresponding normalized eigenvectors e_i. Then$$P^n = E^{-1}\Lambda^n E$$where \Lambda is the diagonal matrix of eigenvalues and E = \sum_{i=0}^2 {e_ie_i^T}. Then you can read ... 2 To help hone your intuition, it might help to think about the following simple example, which shows that the Blumenthal law is using much more than just the continuity of Brownian motion. Let Z be a random variable with P(Z=1) = P(Z=-1) = 1/2, i.e. a coin flip. Set X_t = tZ, so that the process X_t just moves with constant speed \pm 1 depending ... 2 Why not just computing P^2, then interpolating? We have:$$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix},\qquad ...

2

The result is false in this generality. Fix some $z\in S$ and define $O\mu=\mu_c+\mu_d(S)\delta_z$, where $\mu_d$ and $\mu_c=\mu-\mu_d$ are the discrete and continuous parts of $\mu$, respectively. Then $O$ satisfies your conditions, but cannot possibly be given by a kernel, if there is a non-discrete probability measure $\nu$ on $S$. Why not? If ...

2

No. The disjoint events $\{B_n\}$ are not independent. They are mutually exclusive. Hint: Begin with the Law of Total Probability: $\mathsf P(A) = \sum\limits_{n=1}^\infty \mathsf P(A\mid B_n)\;\mathsf P(B_n)$ Yes. (But watch your shift key; you have a $P(X\mid y=y)$ typo. Additionally you need $X=x$ there not $X$ by itself).

2

For sake of compact notation, we let $\Delta N_i \mathop{:=} N_i(t+\Delta t)-N_i(t)$, denote the change in event count over a period $\Delta t$, and make note of the memoryless property of Poisson distribution's waiting times.   Meaning, $\mathsf P(\Delta N_i=n)$ depends only on $\Delta t, \lambda_i,$ and $n$; it is irrespective of $t$.   Indeed: ...

2

Define an array as a sequence of random variables on a probability space $(\Omega, \mathcal{F},P)$. Introduce doubly infinite arrays of random variables $X_{n,j},~\mathcal{F}_{n,j}$ for $j,n\geq 1$, and set sub $\sigma -$algebras of a $\sigma -$algebra. Adapting the array to the filtration, then $\{X_{n,j}\}$ is a MDA if the relation ...

1

We know the distribution of the vector $V:=(X(7)-X(5), X(5)-X(1), X(1))$, since the increments are independent. Defining the function $f\colon (x,y,z)\mapsto (x+y+z)(y+z)z$, the wanted expectation is $\mathbb E\left[f(V)\right]$.

1

First, $N_1(0) + N_2(0) = 0 + 0 = 0$. Second, we need to show $\{N_1(t) + N_2(t)\}$ has independent increments, namely, for $0 < t_1 < t_2 < \cdots < t_k$, $N_1(t_1) + N_2(t_1), (N_1(t_2) + N_2(t_2)) - (N_1(t_1) + N_2(t_1)), \ldots, (N_1(t_k) + N_2(t_k)) - (N_1(t_{k - 1}) + N_2(t_{k - 1}))$ are independent, which follows from that $\{N_1(t)\}$ ...

1

The only requirement is that the random variables $X_t$ are defined on the same probability space.

1

Stochastic process is a collection of random variables — nothing more and nothing less. That means your collection of iid standard normal random variables is an example of stochastic process that even has a special name — it's called a white noise.

1

Hint: You know that for $n=0$ we get $$A + B = I,$$ and for $n=1$ we get $$A + B/3 = P.$$ So you can express $A$ and $B$ in terms of $I$ and $P$...

1

Some related summary comments on ergodic Markov chains. (There are several fleas hopping around triangles elsewhere on this site, each according to the same transition matrix, but with varying questions about the associated process; search 'Markov flea vertex'.) With help from @Did, essentially by manipulating difference equations, I believe you have ...

1

The closed communicating class might be the whole set. For the second part, you will need an infinite state-space.

1

In some sense, there is an $x$ on the right-hand side because the equality $$\mathbb{E}_x(Y \circ \theta_s \mid \mathcal{F}_s^+)(\omega) = \mathbb{E}_{B_s(\omega)}(Y)$$ holds only $\mathbb{P}_x$-almost surely. For a function $f: \mathbb{R} \to \mathbb{R}$ the Markov property reads $$\mathbb{E}_x (f(B_{t+s}) \mid \mathcal{F}_s^+) = \mathbb{E}_{B_s} f(B_{t}), ... 1 Since P is symmetric, we know from the spectral theorem that P has real eigenvalues and is diagonalizable. Since P is a doubly stochastic matrix, the stationary distribution is uniform, i.e.$$\pi=\left(\frac16,\frac16,\frac16,\frac16,\frac16,\frac16\right). Since $P$ has a stationary distribution, we know that $1$ is an eigenvalue. In particular, ...

1

That is mistaken. Let's look at the probability distribution of $X_2$ given $X_0=6$. \begin{align} & \Pr(X_2 = 6\mid X_0=6) \\[10pt] = {} & \Pr\Big( \underbrace{(X_1 = 1\ \&\ X_2 = 6)\text{ or }(X_1=2\ \&\ X_2=6)\text{ or }\cdots}_{\text{five disjuncts}}\mid X_0=6\Big) \\[10pt] = {} & \underbrace{\Pr\Big(X_1 = 1\ \&\ X_2 = 6 \mid ...

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text: The vector rotations of the plane (i.e. centred in the origin), are derived analytically (by coordinates) as linear applications of determinant $$1$$, which makes it possible to characterise them integrally


text:
Chord algorithm - how to map nodes to keys? [closed]

I'm trying to understand Chord algorithm (for Peer-to-Peer systems) but I can't understand how it maps the documents to the nodes.

I have one example where we have:

• keys with 4 bits
• 5 active nodes with the following identifiers: 1, 5, 7, 10, 13

After mapping the documents to the nodes, we have:

Key 0 - Node 1

Key 2 - Node 5

Key 8 - Node 10

Key 11 - Node 13

Key 15 - Node 1

If somebody could help me with the steps that lead to this results I'd be truly grateful.

• Welcome to CS.SE! What research have you done? What resources have you read? Chord is described in a bunch of places. – D.W. Jan 17 '17 at 0:51

NodeID = KeyID % N


text:
Now showing items 33521-33540 of 56284

• #### Probabilistic analysis in surface water hydrology: Open drainage channels and detention basins

(1996)
A reliability-based method for analyzing open drainage channels and detention basins using the full marginal distributions of uncertain parameters together with correlation coefficients will be presented in this work. A ...
• #### Probabilistic analysis of the plastic collapse of random media

(1997)
Two new, efficient probabilistic methods are presented for evaluating the reliability of an elastic/perfectly plastic continuous medium exhibiting randomly varying yield strength. Both the Tresca and Mohr-Coulomb yield ...
• #### Probabilistic analysis of the stability of imperfection sensitive arch and shell structures

(1996)
The stability of arch and shell structures with random imperfections subjected to random loading is investigated. Arches are analyzed under different types of transverse loads while axial and/or pressure loading are ...
• #### Probabilistic Determination of Thermal Conductivity and Cyclic Behavior of Nanocomposites via Multi-Phase Homogenization

(2013-09-16)
A novel multiscale approach is introduced for determining the thermal conductivity of polymer nanocomposites (PNCs) reinforced with single-walled carbon nanotubes (SWCNTs), which accounts for their intrinsic uncertainties ...
• #### Probabilistic Fragility of Interdependent Urban Systems Subjected to Seismic Hazards

(2012)
Urban service networks have come under increased pressure due to expansion of urban population, decrease of capital investment, growing interdependence, and man-made and natural hazards. This thesis introduces a simulation-based ...
• #### Probabilistic Models for Genetic and Genomic Data with Missing Information

(2013-09-16)
Genetic and genomic data often contain unobservable or missing information. Applications of probabilistic models such as mixture models and hidden Markov models (HMMs) have been widely used since the 1960s to make inference ...
• #### Probabilistic phenomena in random combinatorial problems

(2003)
This is an experimental investigation of three combinatorial problems. I examined the average-case complexity of random 3-SAT and of 3-Colorability of random graphs, and the satisfiability of random 1-3-HornSAT. All these ...
• #### Probabilistically Bounded Soft Sphere Detection for MIMO-OFDM Receivers: Algorithm and System Architecture

(2009-10-01)
Iterative soft detection and channel decoding for MIMO OFDM downlink receivers is studied in this work. Proposed inner soft sphere detection employs a variable upper bound for number of candidates per transmit antenna and ...
• #### The Probability of a Gene Tree Topology within a Phylogenetic Network with Applications to Hybridization Detection

(4/19/2012)
• #### Probing and Manipulating Ultracold Fermi Superfluids

(2012)
Ultracold Fermi gas is an exciting field benefiting from atomic physics, optical physics and condensed matter physics. It covers many aspects of quantum mechanics. Here I introduce some of my work during my graduate study. ...
• #### Probing color coherence effects in pp collisions at √s = 7TeV

(2014)
A study of color coherence effects in pp collisions at a center-of-mass energy of 7 TeV is presented. The data used in the analysis were collected in 2010 with the CMS detector at the LHC and correspond to an integrated ...
• #### Probing depths of low energy electrons in metals

(1992)
Spin-polarized electron energy-loss spectroscopy has been used to investigate the probing depth of low energy ($\sim$30 eV) electrons in metals. A beam of spin-polarized electrons is directed at the surface of the sample ...
• #### Probing lipid membrane electrostatics

(2009)
The electrostatic properties of lipid bilayer membranes play a significant role in many biological processes. Atomic force microscopy (AFM) is highly sensitive to membrane surface potential in electrolyte solutions. With ...
• #### Probing molecular adsorption and mechanics at the atomic scale: The Nanocar family of molecules

(2008)
Molecular machines, typically thought to be only the fanciful imaginings of speculative fiction, have taken great strides in recent years towards real-world viability and usefulness. Under variable temperature scanning ...
• #### Probing the Association of Triblock Copolymers with Supported Lipid Membranes using Microcantilevers

(2014)
Pluronics are a class of amphiphilic triblock copolymers that are known to interact with cellular membranes in interesting ways. The solubility of these triblock copolymers in free lipid membranes can be altered with ...
• #### Probing the Surface-Vacuum Interface with Spin-Sensitive Metastable Atom Deexcitation, Electron Capture and Electron Emission Spectroscopies

(1993)
Spin-Polarized Metastable Atom Deexcitation (SPMDS) and Electron Capture (ECS) Spectroscopies probe the exponential tails of electronic wavefunctions extending from the surface into the vacuum, and are consequently extremely ...
• #### the problem building an educated community

(1965-09-16)
• #### Problem generator system in Engineering Design Tutor

(2002)
The Internet has provided an ideal environment for engineering education. This thesis describes the development of a new problem generator used in the Engineering Design Tutor (EDT) system to provide multi-step problem ...
• #### The Problem of Creation

(1965-03-04)

(1928-01)


text:
## Intersections of automorphism fixed subgroups in the free group of rank three.(English)Zbl 1054.20009

Let $$F_n$$ be a free group of finite rank $$n$$. A subgroup $$H$$ of $$F_n$$ is called ‘1-auto-fixed’ when there exists an automorphism $$\varphi$$ of $$F_n$$ such that $$H=\text{fix\,}\varphi=\{x\in F_n:\varphi x=x\}$$. By definition an ‘auto-fixed’ subgroup of $$F_n$$ is an arbitrary intersection of 1-auto-fixed subgroups.
It is shown that the intersection $$H\cap K$$ of each 1-auto-fixed subgroup $$H$$ with any finitely generated subgroup $$K$$ of $$F_n$$ has rank not exceeding $$n$$. Thus the author generalises the celebrated theorem of Bestvina-Handel showing that a 1-auto-fixed subgroup of $$F_n$$ has rank at most $$n$$.
It is also shown that in the case of the group $$F_3$$ any auto-fixed subgroup is in fact 1-auto-fixed. A similar assertion is valid in the case of the group $$F_2$$ and remains open for the groups $$F_n$$ for $$n\geq 4$$.

### MSC:

 20E05 Free nonabelian groups 20F28 Automorphism groups of groups 20E36 Automorphisms of infinite groups 20E07 Subgroup theorems; subgroup growth

### Keywords:

free groups; automorphisms; fixed subgroups
Full Text:

### References:

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