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# Confusion regarding one formulation of the Axiom of Choice.

One formulation of the Axiom of Choice is:

The Cartesian product of non-empty sets is always non-empty.

Cartesian product is defined as making "every possible pair" between elements of two sets. The implication is that through some aritrarily selected algorithm, we can select each element one-by-one, and pair it with another element (from another set). Does the Axiom of Choice imply that without assuming such an axiom, no such algorithm can exist? What if the elements are ordered? If I have a linearly ordered set, and I state that I my method of selection of elements is I will move from the lowest element to the highest element, am I still using the Axiom of Choice somehow? Is choosing even the lowest element in a set an exercise of the Axiom of Choice?

Thanks

• Well, in the cartesian product of a huge ammount (!) of sets, how do you "choose" an element from each set to form an element in the cartesian product and show it isn't empty? Well, there you go: that's AC ! – DonAntonio Mar 31 '14 at 12:42
• A total order isn't enough. (How do you choose the lowest positive real number? Or even the lowest real number?) To make that work, you need a well-ordering (usually taken to be a well-ordering of the discrete union of all the sets of which you're taking the product) to guarantee that there is always a lowest element. And thus you have a hint to the equivalence of AC to the well-ordering principle. – Arthur Mar 31 '14 at 12:45
• @DonAntonio- I don't get it. If all the sets are ordered, why can't you just choose the owedt element from all, and hence show that the cartesian product is non-empty? – algebraically_speaking Mar 31 '14 at 12:45
• The well ordering of any set is equivalent with AC, @algebraically_speaking. I'm afraid we're going around a circle here... – DonAntonio Mar 31 '14 at 12:48
• The Axiom of Choice can be used to show that every set is well-orderable, and from the reasoning you mentioned, the converse is readily proved. That's what Don is saying--he's not saying that without AC, no set is well-orderable (and there are many cartesian products that are non-empty without AC, too). – Cameron Buie Mar 31 '14 at 12:54

Cartesian product of two sets is used to "make every possible pair". And that is true even without the axiom of choice. If you take the product of any two non-empty sets then the product of them is non-empty.

But what about Cartesian product of three sets? Are these pairs? Triplets? If these are pairs, what coordinate is the additional one, is it $$((a,b),c)$$ or $$(a,(b,c))$$? Well, these questions don't matter because we have a natural identification between all three options, triplets and the two type of pairs. So instead we can define redefine the Cartesian product of $$A$$ and $$B$$ as the set of functions from $$\{0,1\}$$ into $$A\cup B$$ such that $$f(0)\in A$$ and $$f(1)\in B$$. It is not hard to see why this gives us the same structure as ordered pairs.

The thing about this approach is that now the product of three sets can be thought of as functions from $$\{0,1,2\}$$ rather than pairs of pairs and so on. Whenever we add more sets, we just increase the index set.

Another wonderful benefit of this definition is its extendability to infinite products. If you insist that products are just pairs of pairs of pairs, and so on. What objects do you have when you take the product of infinitely many sets? It is a nice exercise to see that such object would have to be ill-founded. So we cannot have that in the presence of regularity (and we don't want to have that in general).

On the other hand, with the proposed definition as functions taking the product of infinitely many sets is easy. We have an index set $$I$$ and a family of sets $$A_i$$ and we just take functions such that $$f(i)\in A_i$$. Does that look familiar? It should. It is in fact the definition of a choice function.

So it turns out that the Cartesian product of sets is actually the set of choice functions from them. The axiom of choice assures that every family of non-empty sets has a choice function, and therefore the Cartesian product of any family of non-empty sets is non-empty. But as I said, for finite families of sets we can prove that we don't need the axiom of choice for choosing. That is, the Cartesian product of finitely many sets (however large they might be, just not empty) is non-empty. In particular $$A\times B$$ is never empty if neither set is empty.

You also made several other mistakes.

1. Without the axiom of choice it is not true that every set is ordered. Not well-ordered, and not linearly ordered. It is true that there are ordered sets, both well-ordered and linearly ordered, but it is possible for sets which cannot be linearly ordered to exist.

2. Not every linear order has a least element. And not every set which can be linearly ordered can necessarily be well-ordered.

3. Sets are not ordered sets. Ordered sets are sets which have a fixed structure endowed on them. If I just give you $$\{a,b\}$$, you can either order it as $$a or $$b. But seeing how I haven't given you any information on what $$a$$ and $$b$$ are exactly, you have no preference for one way over the other. Therefore if you want this set to be ordered you have to choose an ordering.

It is possible that there is a countable family of pairs like that, that are so indiscernible from one another, that it is impossible to endow them all with orders at the same time. You can of course order each pair individually, and therefore you can order finitely many of them. But you cannot choose an order for each and every one of them at the same time without an appeal to the axiom of choice.

Regardless to that, if you are given structured sets. That is, sets which come along with a fixed structure, then it is possible to choose from each, under certain circumstances (e.g. the structures are well-orders, or there is some fixed element in each of the sets, etc.)

So if I would have given you $$\{a,b\}$$ and $$a; for each of the aforementioned pairs, then of course you can choose the minimal element in each pair.

But the point is that it is not necessary that every set can be ordered, and even if it is ordered, it is usually the case that there are many different (even if isomorphic) orders on each set, and to choose one for each set would be an appeal to the axiom of choice.

• I don't see how the case would differ if the number of sets is infinite. Say the number of sets is $\aleph$. Then why can't we select the lowest element from each set? Is it because we can;t "access" all $\aleph$ sets? – algebraically_speaking Mar 31 '14 at 12:48
• @algebraically_speaking, you're again assuming we can well-order all those sets. See the comments above. – DonAntonio Mar 31 '14 at 12:49
• @algebraically_speaking: Things don't exist just because you want them to. We need to prove their existence. Even if all sets can be well ordered, you may still have to access an infinite amount of information, which you cannot "just do". If a set can be well ordered to begin with, and it is not a singleton then there are many ways to well order it. How do you choose which order? Of course if the orders are given then you are right, assuming they are well orders. – Asaf Karagila Mar 31 '14 at 12:53
• @algebraically_speaking: I edited my answer (now that I sit in front of a keyboard). – Asaf Karagila Mar 31 '14 at 15:20
• Thanks Asaf for your detailed answer. – algebraically_speaking Apr 2 '14 at 2:33

Let me clear up a few definitions before answering.

We say a set $$X$$ is indexed by a set $$I$$ if there is a function from $$I$$ onto $$X$$--we typically denote this function by something like $$i\mapsto X_i,$$ which allows us to write $$X=\{X_i\}_{i\in I}$$ or $$X=\{X_i\mid i\in I\}.$$ Every set can be indexed by itself, of course.

Given an indexed set of sets $$\{X_i\}_{i\in I}$$ we may define the Cartesian product of the sets $$X_i$$ by $$\prod_{i\in I}X_i:=\left\{f:I\to\bigcup_{i\in I}X_i\mid \forall i\in I,f(i)\in X_i\right\}.$$ That is, the elements of a Cartesian product are functions from the index set into the union of all the sets, such that each $$i$$th element is a member of the $$i$$th set in the collection. This is the same idea as what is going on in the finite case, only the "tuples" aren't necessarily ordered (or even orderable, as far as we care).

To see why these are "the same," take any two sets $$A,B.$$ We can readily index the set $$\{A,B\}$$ with the set $$\{1,2\},$$ by $$1\mapsto A,2\mapsto B$$--that is, we can rewrite $$A=X_1,B=X_2$$ for the indexed notation. So, on the one hand, the elements of $$\prod_{i\in\{1,2\}}X_i$$ are functions on $$\{1,2\}$$ such that $$1$$ is sent to an element of $$X_1=A,$$ and $$2$$ is sent to an element of $$X_2=B.$$ But such functions are effectively ordered pairs whose first entry is an element of $$A$$ and whose second entry is an element of $$B,$$ meaning that this product is effectively the same as $$A\times B$$ in the sense you may be more familiar with. Note that the indexing makes a difference--if we'd chosen the indexing $$1\mapsto B,2\mapsto A,$$ then it would be "more natural" to say that the Cartesian product (in the sense defined above) was the same as $$B\times A$$ in the sense that you're used to.

Now, where is it that the Axiom of Choice comes into play in ensuring that arbitrary Cartesian products of non-empty sets are non-empty? Well, let's start with an arbitrary non-empty indexed set of non-empty sets $$\{X_i\}_{i\in I}.$$ As it stands, we don't know anything about the set $$I$$ or the sets $$X_i,$$ except that they are non-empty. They may not necessarily be orderable at all so far as we are assuming. Even if we were to add the assumption of orderability, then we would still have to choose an ordering for each set $$X_i$$ (and for $$I,$$ if we like, though it isn't really necessary), and unless $$X_i$$ is a singleton, there will be more than one way to order $$X_i$$ if it can be ordered at all. Even if we assume that each set $$X_i$$ is already ordered, it won't be enough! After all, not every ordered set has a least element, as mentioned before, and though we could always "fix" a given order of that nature by pulling an element out of the usual order, and calling it the least element under a new order, we may have to choose which element to make the least element for every $$X_i$$! So, we would have to make the assumption that each $$X_i$$ is ordered in such a way that it has a least element, or your argument doesn't work. This is a fairly tall assumption, and (it turns out) requires the Axiom of Choice to prove that it is even possible! After all, even if we know that $$X_i$$ admits such an order for each $$i\in I,$$ we still have to choose an order for each such $$X_i,$$ and in general unless $$X_i$$ is a singleton, there will be more than one way to do this!

Let me outline (most of) the proof of the equivalence of some common formulations of the Axiom of Choice.

The following are equivalent:

• The Cartesian product of (a non-empty indexed set of) non-empty sets is non-empty.
• For every non-empty set $$\mathcal A$$ of pairwise-disjoint non-empty sets, there exists a set (which we'll call a "choice set" for $$\mathcal A$$) containing exactly one element from each element of $$\mathcal A.$$
• Every non-empty set of non-empty sets admits a choice function).
• Every set admits a well-order.

For the first implication (Cartesian product implies choice set), recall that every set can be indexed, so taking the Cartesian product of the elements of $$\mathcal A,$$ and picking out an element $$f$$ from this Cartesian product, the range of $$f$$ can be shown to be a choice set for $$\mathcal A.$$

For the second implication (choice set implies choice function), start with a non-empty set $$X$$ of non-empty sets, then note that $$\mathcal A:=\bigl\{\{A\}\times A\mid A\in X\bigr\}$$ is a non-empty set of pairwise disjoint non-empty sets. Furthermore, a choice set for $$\mathcal A$$ can be shown to be a choice function for $$X.$$

Proving the third implication (choice function implies well-orderability) is the most non-trivial of the results, and I won't get into it here, but I wanted to mention it.

The fourth implication (well-orderability implies Cartesian product), start with a non-empty indexed set of non-empty sets $$\{X_i\}_{i\in I}.$$ Since $$\bigcup_{i\in I}X_i$$ is a set, then it is well-orderable, so pick any well-ordering for this union. Now, by definition of well-ordering, we have a least element for each $$X_i,$$ so we can show that $$\prod_{i\in I}X_i$$ is non-empty, letting $$f:I\to\bigcup_{i\in I}X_i$$ be defined by letting $$f(i)$$ be the least element of $$X_i$$ for each $$i\in I.$$

Alternately, we can leave out the well-orderability, and prove that choice function implies Cartesian product. Indeed, suppose we have a non-empty indexed set of non-empty sets $$\{X_i\}_{i\in I}.$$ Let $$g$$ be a choice function for $$\{X_i\}_{i\in I},$$ meaning that $$g:\{X_i\}_{i\in I}\to\bigcup_{i\in I}X_i$$ is such that $$g(X_i)\in X_i$$ for all $$i\in I.$$ Defining $$f(i)=g(X_i)$$ for each $$i\in I$$ makes $$f$$ an element of $$\prod_{i\in I}X_i,$$ as desired.

• When I finally get behind an actual keyboard and plan on revise my answer, you make this plan almost irrelevant. :-) – Asaf Karagila Mar 31 '14 at 14:49
• @Caneron Buie- What I understood is that when there are multiple ways of doing something, and we don't have a clear choice of which way to choose, we use the AC to accomplish the task using an arbitrary way. Can this be stretched further? Say there are multiple ways of reaching a destination. And the ways can be ordered in different ways- comfort, length, etc. The selection of one particular order, or an arbitrary way, requires AC? Everyday choices, in some sense (I know I'm digressing from set theory) requires AC? – algebraically_speaking Apr 1 '14 at 4:02
• Say there is only one way to order a set (assume for a moment). Then either choosing that order and selecting its lowest/highest element, or choosing an altogether different element still requires AC? – algebraically_speaking Apr 1 '14 at 4:03
• @algebraically_speaking: That's not quite true. If we need to make any finite number of arbitrary choices, we can, with no need to use AC. So, if there are multiple ways of accomplishing a task, then the set of ways of accomplishing said task is non-empty, and we simply pick an arbitrary way from that set of all ways, with no need for AC. For example, if we know (somehow) that a given set can be ordered, we can just pick any ordering of that set to do the job. However, if we don't know enough about the set to conclude that it is orderable, then we can let Choice do it for us. – Cameron Buie Apr 1 '14 at 15:03
• I think Russell put it rather well: AC would be needed to select one sock from each of infinitely-many pairs of socks (assumed indistinguishable), but not from infinitely-many pairs of shoes (as we could just pick the right shoe from each pair, for example). – Cameron Buie Apr 1 '14 at 15:13


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# Math Help - This really is urgent (1 question)

1. ## This really is urgent (1 question)

okay , I have no idea how to work this out.

This is the question.....

There is a class who are having a competition, they have to guess the right combination of numbers.

The combination is ten numbers long, however the only numbers used are 1 2 and 3

The numbers are drawn from a giant basket, which contains 1 thousand no.1 balls ... 1 thousand no.2 balls and 1 thousand no.3 balls.. (stupid i know ahaha). NOTE : when a number is drawn it is put back in the basket with the other balls SO THERE IS ALWAYS THE SAME CHANCE OF DRAWING 1 2 OR 3 AND THE BALLS WILL NEVER RUN OUT

What is the total number of possible combinations?

examples of possible combinations

3131223132
1111111111
2121212121
3122212132

You get the idea......

the actual answer i need is

What is the total number of possible combinations

2. $3^{10}$

3. Originally Posted by Plato
$3^{10}$
Thank you so much for the fast reply, but I do not understand you.. I was looking for more of a number like 1846 or something

what does 3^{10} mean?

4. Go to bed.

5. Originally Posted by UsainBolt
$3^{10}$ means $3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$.

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# Relation between radio waves and photons generated by a classical current

Several questions have been posted on Physics SE regarding the relationship between photons and electromagnetic waves, and several good answers have been given. Some of those questions are listed below, but I didn't find any that requested a mathematically explicit analysis of what happens — in terms of photons — when an oscillating current generates an electromagnetic wave with a macroscopic wavelength, such as a radio wave.

I'm attempting to fill that gap by posting this question-and-answer.

I haven't found an equally explicit / equally narrated analysis anywhere else, but less-explicit / less-narrated references include:

• Itzykson and Zuber, Quantum Field Theory, section 4-1: "Quantized electromagnetic field interacting with a classical source";

• Cohen-Tannoudji, Dupont-Roc, and Grynberg, Atom-Photon Interactions, exercise 17: "Equivalence between a quantum field in a coherent state and an external field", and also exercise 9.

Some related posts, from newest to oldest:

Does a photon require an EM field to exist?

How to interpret a wavepacket in quantum field theory: is it one particle or a superposition of many?

Can the equation $$E = h\nu$$ be used not only for light, but for radio waves? (Since they are all part of the electromagnetic spectrum)

Electromagnetic waves and photons

What exactly is meant by the wavelength of a photon?

Why call it a particle and not a wave pulse?

Is double slit interference due to EM/de Broglie waves? And how does this relate to quantum mechanical waves?

What is the physical nature of electromagnetic waves?

Relation between Wave equation of light and photon wave function?

Sequence of E and B field in radio waves and in single photons

Photon Quantum Field proportional to Electromagnetic Field?

Light Waves and Light Photons gedanken Experiment

Do photons occupy space?

How is the classical EM field modeled in quantum mechanics?

Are coherent states of light 'classical' or 'quantum'?

Amplitude of an electromagnetic wave containing a single photon

Radio waves and frequency of photon

Reconciling refraction with particle theory and wave theory

Properties of the photon: Electric and Magnetic field components

In QED, the electromagnetic (EM) field and the charged matter are both quantum entities. This answer uses a semiclassical model instead, with a quantum field coupled to a prescribed classical current. This is an exactly solvable model inspired by QED. As a further simplification, the quantum field will be a scalar field instead of the EM field. By analogy, the quanta of this scalar field will be called "photons".

In the context of the free (non-interacting) quantum EM field, the word "photon" is typically used to mean a quantum of energy, and that's how I'm using the word here. The current will be active only during a finite time interval, and I'll only apply the word "photon" at the times when the current is not active, so that the meaning of "quantum of energy" is unambiguous.

To help limit the length of this post, familiarity with introductory QFT is assumed. The notation will be similar to that used in chapter 2 of Peskin and Schroeder's An Introduction to Quantum Field Theory.

The model and its exact solution

The Heisenberg picture will be used, so the state-vector is independent of time, but its physical significance still changes in time because the observables do. The equation of motion in the Heisenberg picture is $$$$\partial_\mu\partial^\mu\phi(t,\mathbf{x}) = J(t,\mathbf{x}) \tag{1}$$$$ where $$\phi$$ is the quantum field and where $$J$$ is a prescribed function that will be called the "current" by analogy with the EM case. The equal-time commutation relation for the quantum scalar field is $$$$\big[\phi(t,\mathbf{x}),\,\dot\phi(t,\mathbf{y})\big]=i\delta^3(\mathbf{x}-\mathbf{y}). \tag{2}$$$$ The quantum field $$\phi(t,\mathbf{x})$$ is the local observable corresponding to field-amplitude measurements.

Equations (1)-(2) can be solved exactly. The solution is $$$$\phi(t,\mathbf{x}) = \phi_0(t,\mathbf{x})+\phi_J(t,\mathbf{x}) \tag{3}$$$$ where:

• $$\phi_J$$ is a real-valued solution to (1), which commutes with everything;

• $$\phi_0$$ is an operator-valued solution to the $$J=0$$ version of (1) that satisfies the commutation relation (2).

From now on, suppose the current is non-zero only within the finite time interval $$0: $$$$J(t,\mathbf{x})=0 \hskip1cm \text{ except for }0 and choose $$$$\phi_J(t,\mathbf{x})=0 \hskip1cm \text{ for }t\leq 0. \tag{5}$$$$ These conditions are all satisfied by $$$$\phi_0(t,\mathbf{x}) =\int\frac{d^3p}{(2\pi)^3}\ e^{i\mathbf{p}\cdot \mathbf{x}}\, \frac{e^{-i\omega t}a_0(\mathbf{p})+e^{i\omega t}a^\dagger(-\mathbf{p})}{ \sqrt{2\omega}} \tag{6}$$$$ and $$$$\phi_J(t,\mathbf{x}) = \int ds\ \theta(t-s) \int\frac{d^3p}{(2\pi)^3}\ e^{i\mathbf{p}\cdot \mathbf{x}}\, i\,\frac{e^{-i\omega (t-s)}-e^{i\omega (t-s)}}{2\omega}\, \tilde J(s,\mathbf{p}) \tag{7}$$$$ with $$$$\omega\equiv \sqrt{\mathbf{p}^2} \hskip2cm \tilde J(s,\mathbf{p}) \equiv \int d^3x\ e^{-i\mathbf{p}\cdot \mathbf{x}}\,J(s,\mathbf{x}), \tag{8}$$$$ and where the operators $$a_0(\mathbf{p})$$ and their adjoints satisfy $$$$\big[a_0(\mathbf{p}),\,a_0^\dagger(\mathbf{p}')\big]=(2\pi)^3\delta^3(\mathbf{p}-\mathbf{p}'). \tag{9}$$$$ The operators $$a_0$$ and $$a_0^\dagger$$ are just a basic set of operators in terms of which everything else in the operator algebra may be expressed. Define a state-vector $$|0\rangle$$ by the conditions $$$$a_0(\mathbf{p})\,|0\rangle = 0 \hskip2cm \langle 0|0\rangle = 1 \tag{10}$$$$ for all $$\mathbf{p}$$, and suppose that the state of the system is the one represented by $$|0\rangle$$. The Heisenberg picture is being used here, so the state-vector has no time-dependence, but the physical state that it represents still changes in time because the observables do.

The rest of this answer addresses the interpretation of the state-vector (10) both for $$t<0$$ and for $$t>T$$, first in terms of photons and then as it relates to radio waves.

The interpretation in terms of photons

Equation (5) says that for $$t<0$$ we have the familiar free scalar field, and then we recognize the state defined by (10) as the vacuum state — the state of lowest energy, with no photons. This, of course, was the motive for choosing the state (10).

The question is what happens at $$t > T$$ in the aftermath of the temporary current $$J$$. For these times, equation (4) says that the factor $$\theta(t-s)$$ may be omitted in equation (7), because it is already enforced by the current itself. Therefore, for these late times, the solution (3) may be written $$$$\phi(t,\mathbf{x}) =\int\frac{d^3p}{(2\pi)^3}\ e^{i\mathbf{p}\cdot \mathbf{x}}\, \frac{e^{-i\omega t}a(\mathbf{p})+e^{i\omega t}a^\dagger(-\mathbf{p})}{ \sqrt{2\omega}} \tag{11}$$$$ with $$$$a(\mathbf{p}) \equiv a_0(\mathbf{p}) + a_J(\mathbf{p}) \hskip2cm a_J(\mathbf{p}) \equiv \frac{i}{\sqrt{2\omega}}\int ds\ e^{i\omega s}\tilde J(s,\mathbf{p}). \tag{12}$$$$ The complex-valued function $$a_J$$ encodes the effect of the current.

Before we can interpret the state (10) in terms of photons at times $$t>T$$, we need to determine which operators represent photon creation/annihilation operators at these times. The Hamiltonian associated with the equation of motion (1) is $$$$H(t) = \int d^3x\ \left(\frac{\dot\phi^2(t,\mathbf{x}) +(\nabla\phi(t,\mathbf{x}))^2}{2}-\phi(t,\mathbf{x}) J(t,\mathbf{x})\right). \tag{13}$$$$ Equations (9) and (12) imply $$$$\big[a(\mathbf{p}),\,a^\dagger(\mathbf{p}')\big] =(2\pi)^3\delta^3(\mathbf{p}-\mathbf{p}'). \tag{14}$$$$ Any any time for which $$J=0$$, equations (11) and (13)-(14) imply \begin{align} H(t)= \int\frac{d^3p}{(2\pi)^3}\ \omega\,a^\dagger(\mathbf{p})a(\mathbf{p}) +h(t) \tag{15} \end{align} where $$h(t)$$ is a real-valued function that doesn't affect this analysis. Whenever $$J=0$$, these equations all have the same form as they do in the free-field case (where $$J$$ is zero for all times). Based on this, we can interpret $$a(\mathbf{p})$$ and its adjoint as the operators that annihiliate and create (respectively) a photon with the indicated momentum at times $$t>T$$. The justification for this interpretation is identical to the corresponding justification for $$a_0$$ at times $$t<0$$.

Now that we know which operators create and annihilate photons at $$t>T$$, we can interpret the state $$|0\rangle$$ at these times. Equations (10) and (12) imply $$$$a(\mathbf{p})\,|0\rangle = a_J(\mathbf{p})\,|0\rangle, \tag{16}$$$$ which is the defining equation of a multi-mode coherent state. The state $$|0\rangle$$ was chosen because it represents the vacuum state for $$t<0$$, but equation (16) says that it is no longer the vacuum state with respect to observables at $$t>T$$. The vacuum state at $$t>T$$ is represented instead by the state-vector $$|T\rangle$$ that satisfies $$$$a(\mathbf{p})\,|T\rangle = 0. \tag{17}$$$$ Equation (14) implies that the relationship between the coherent state (16) and the vacuum state (17) is $$$$|0\rangle \propto \exp\big(A^\dagger \big)\,|T\rangle =|T\rangle + A^\dagger|T\rangle + \frac{1}{2!}(A^\dagger)^2|T\rangle + \frac{1}{3!}(A^\dagger)^3|T\rangle +\cdots \tag{18}$$$$ with $$$$A^\dagger \equiv \int\frac{d^3p}{(2\pi)^3}\ a_J(\mathbf{p}) a^\dagger(\mathbf{p}). \tag{19}$$$$ In words, the state at times $$t>T$$ is a special superposition of different numbers of identical photons, all with this same profile described by the complex-valued function $$a_J(\mathbf{p})$$.

The interpretation as a radio wave

At any time $$t$$, equations (3)-(10) imply $$$$\langle 0|\phi(t,\mathbf{x})|0\rangle=\phi_J(t,\mathbf{x}) \tag{20}$$$$ and $$$$\langle 0|\phi(t,\mathbf{x})\phi(t,\mathbf{y})|0\rangle - \langle 0|\phi(t,\mathbf{x})|0\rangle\, \langle 0|\phi(t,\mathbf{y})|0\rangle = \langle 0|\phi_0(t,\mathbf{x})\phi_0(t,\mathbf{y})|0\rangle. \tag{21}$$$$ Equation (20) says that the expectation value of the quantum field behaves just like a classical wave generated by the current $$J(t,\mathbf{x})$$. Equation (21) says that the fluctuations in the outcomes of field-amplitude measurements are just as small as they would be in the vacuum. If the current $$J$$ is large enough, so that the expectation value (20) is large enough, then the square root of (21) will be negligible compared to (20). In this case, we have a classical wave for all practical purposes. By choosing the oscillation frequency of the current, we can make it a radio wave.

Altogether, this shows that if we start with the vacuum at time $$t<0$$ and turn on a current during the interval $$0, then the state at times $$t>T$$ is a coherent state of photons, and the same state can also be interpreted as an effectively classical wave.

Classical superposition versus quantum superposition

Note that a classical superposition of two such effectively-classical waves is obtained by adding the corresponding single-photon profiles in the exponent of equation (18), like this: $$$$\exp\big(A_1^\dagger + A_2^\dagger\big)\,|T\rangle. \tag{22}$$$$ This follows from the fact that such a superposition is produced by a classical current of the form $$J=J_1+J_2$$, where $$J_1$$ and $$J_2$$ may be localized in different regions of space (for example). In contrast, a quantum superposition of two effectively-classical waves has the form $$$$\exp\big(A_1^\dagger\big)\,|T\rangle + \exp\big(A_2^\dagger\big)\,|T\rangle. \tag{23}$$$$ In this state, equation (21) doesn't hold; fluctuations in field-amplitude measurement-outcomes are typically just as large as the expectation value, so a quantum superposition of two effectively-classical waves not like a classical wave at all.


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Browse Questions

# A bucket filled with water is tied to a rope of length 0.5 m and is rotated in a circular path in vertical plane. The least velocity it should have at the lowest point of circle so that water does not spill is. $(g=10ms^{-2})$

$\begin {array} {1 1} (a)\;\sqrt 5 m/s & \quad (b)\;\sqrt {10} m/s \\ (c)\;\sqrt 5 m/s & \quad (d)\;2 \sqrt 5 \end {array}$

$(c)\;\sqrt 5 m/s$


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# The value of Kc for rxn below 2 H2S(g) <----> 2H2(g) + S2 (g) find value of [H2S] ?

## The value of Kc for rxn below 2 H2S(g) <----> 2H2(g) + S2 (g) is 2.2 x 10^-4 at 1400 K. A sample of [H2S] = 5.00 M is heated to 1400 K in a sealed vessel. After chemical equilibrium, find value of [H2S] No H2 or S2 was present original sample.

Apr 11, 2018

The equilibrium concentration of ${H}_{2} S$ is $4.78 M$

#### Explanation:

${K}_{c} = 2.2 \times {10}^{-} 4 = \frac{{\left[{H}_{2}\right]}^{2} \left[{S}_{2}\right]}{{\left[{H}_{2} S\right]}^{2}}$

Let

${n}_{{H}_{2} S}^{0}$ = the initial concentration of ${H}_{2} S = 5 M$

Assume that $x$ molar concentration of ${H}_{2} S$ decomposes at equilibrium. At equilibrium, the concentration of ${H}_{2} S$ will be ${n}_{{H}_{2} S}^{0} - x$.

This will produce $x$ molar concentration of ${H}_{2}$ and $\frac{x}{2}$ molar concentration of ${S}_{2}$. Thus at equilibrium we will have

${K}_{c} = \frac{{x}^{2} \left(\frac{x}{2}\right)}{{\left({n}_{{H}_{2} S}^{0} - x\right)}^{2}}$

Because the equilibrium constant is small, we will make the approximation that ${n}_{{H}_{2} S}^{0} - x \approx {n}_{{H}_{2} S}^{0}$. This allows us to write

${K}_{c} \approx \frac{{x}^{3}}{2 {\left({n}_{{H}_{2} S}^{0}\right)}^{2}}$

Solving for $x$ we have

$x = {\left[2 {K}_{c} {\left({n}_{{H}_{2} S}^{0}\right)}^{2}\right]}^{\frac{1}{3}} = {\left[2 \cdot 2.2 \times {10}^{-} 4 {\left(5\right)}^{2}\right]}^{\frac{1}{3}} \approx 0.22$

This would give the final concentration of ${H}_{2} S$ as

${n}_{{H}_{2} S}^{0} - x = 5 - 0.22 = 4.78 M$

As a check of our approximation we can calculate the equilibrium constant as

$\frac{{0.22}^{2} \left(\frac{0.22}{2}\right)}{{\left(4.78\right)}^{2}} \approx 2.3 \times {10}^{-} 4$

which is less than a 5% error from the given value.


text:
# Characteristically simple locally compact abelian groups

Say a topological group $G$ is topologically characteristically simple if there does not exist a closed subgroup $1 < K < G$ such that $K$ is invariant under all automorphisms of $G$ (here `automorphisms' are required to preserve the topology as well as the group structure).

Q1: Has someone written down a classification of locally compact second-countable abelian groups that are topologically characteristically simple?

I tried to derive a classification myself; the case I am having difficulty with is when $G$ is torsion-free and the group $pG$ of $p$-th powers in $G$ is a proper dense subgroup of $G$. (Another possible division of this case is between those $G$ with a dense divisible subgroup, and those with no divisible subgroup, but I don't know where to go from there.)

I know one such group for each $p$: let $G$ be the group of all functions from $\mathbb{N}$ to $\mathbb{Q}_p$ under pointwise addition such that all but finitely many values are in $\mathbb{Z}_p$, and topologise it so that the group of functions from $\mathbb{N}$ to $\mathbb{Z}_p$ is an open subgroup with the compact-open topology. This example is topologically characteristically simple with a dense divisible subgroup (the finitely supported functions).

Q2: Are there any other locally compact second-countable abelian groups $G$ such that $G$ is torsion-free, $pG$ is a proper dense subgroup (edit: and $x^{p^n} \rightarrow 1$ for all $x \in G$)? If so, are any of them topologically characteristically simple?

• In the first part of Q2 there's a trivial answer, e.g. the direct product of your $G$ (I met this group a couple of times, I don't remember where of if I used it anywhere) and the group of rationals, so probably only the TCS assumption makes the question nontrivial. An equivalent variant is to ask about classifying TCS locally compact groups with an open subgroup isomorphic to $\mathbf{Z}_p^\mathbf{N}$. – YCor Apr 24 '16 at 22:37
• Ah yes, I meant to exclude trivial examples like that (e.g. by also requiring $G$ to be locally elliptic). Indeed it is the TCS condition that gives some hope of a classification. It looks like all examples will have a compact open subgroup $U$ isomorphic to $\mathbb{Z}^{\mathbb{N}}_p$ with $G/U$ isomorphic to a direct sum of Prüfer $p$-groups, but there could be many ways of putting $U$ and $G/U$ together. – Colin Reid Apr 25 '16 at 6:11
• Or at least assume that $G$ is $p$-elliptic (i.e., is a $\mathbf{Z}_p$-module, or, equivalently but apparently more intrinsically, that $p^nx\to 0$ when $n\to\infty$ for all $x$). – YCor Apr 25 '16 at 6:35


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# Geometry of Molecules

Molecular geometry, also known as the molecular structure, is the three-dimensional structure or arrangement of atoms in a molecule. Understanding the molecular structure of a compound can help determine the polarity, reactivity, phase of matter, color, magnetism, as well as the biological activity.

## Introduction

To determine the shapes of molecules, we must become acquainted with the Lewis electron dot structure. Although the Lewis theory does not determine the shapes of molecules, it is the first step in predicting shapes of molecules. The Lewis structure helps us identify the bond pairs and the lone pairs. Then, with the Lewis structure, we apply the valence-shell electron-pair repulsion (VSPER) theory to determine the molecular geometry and the electron-group geometry.

To identify and have a complete description of the three-dimensional shape of a molecule, we need to know also learn about state the bond angle as well. Lewis Electron Dot Structures play crucial role in determining the geometry of molecules because it helps us identify the valence electrons. To learn how to draw a Lewis electron dot structure click the link above.

### Valence-Shell Electron-Pair Repulsion Theory

Now that we have a background in the Lewis electron dot structure we can use it to locate the the valence electrons of the center atom. The valence-shell electron-pair repulsion (VSEPR) theory states that electron pairs repel each other whether or not they are in bond pairs or in lone pairs. Thus, electron pairs will spread themselves as far from each other as possible to minimize repulsion. VSEPR focuses not only on electron pairs, but it also focus on electron groups as a whole. An electron group can be an electron pair, a lone pair, a single unpaired electron, a double bond or a triple bond on the center atom. Using the VSEPR theory, the electron bond pairs and lone pairs on the center atom will help us predict the shape of a molecule.

The shape of a molecule is determined by the location of the nuclei and its electrons. The electrons and the nuclei settle into positions that minimize repulsion and maximize attraction. Thus, the molecule's shape reflects its equilibrium state in which it has the lowest possible energy in the system. Although VSEPR theory predicts the distribution of the electrons, we have to take in consideration of the actual determinant of the molecular shape. We separate this into two categories, the electron-group geometry and the molecular geometry.

Electron-group geometry is determined by the number of electron groups.

Number of electron groups Name of electron group geometry
2 linear
3 trigonal-planar
4 tetrahedral
5 trigonal-bipyramidal
6 octahedral

Molecular geometry, on the other hand, depends on not only on the number of electron groups, but also on the number of lone pairs. When the electron groups are all bond pairs, they are named exactly like the electron-group geometry. See the chart below for more information on how they are named depending on the number of lone pairs the molecule has.

### VSEPR Notation

As stated above, molecular geometry and electron-group geometry are the same when there are no lone pairs. The VSEPR notation for these molecules are AXn. "A" represents the central atom and n represents the number of bonds with the central atom. When lone pairs are present, the letter Ex is added. The x represents the number of lone pairs present in the molecule. For example, a molecule with two bond pairs and two lone pairs would have this notation: AX2E2.

Geometry of Molecules Chart
Number of Electron Groups Electron-Group Geometry Number of Lone Pairs VSEPR Notation Molecular Geometry Ideal Bond Angles Examples
2 linear 1 AX2 180° BeH2
3 trigonal-planar 0 AX3 120° CO32-
1 AX2E 120° O3
4 tetrahedral 0 AX4

Tetrahedral

109.5° S042-
1 AX3E 109.5° H3O+
2 AX2E2 109.5° H2O
5 trigonal-bipyramidal 0 AX5 90°, 120° PF5
1 AX4Eb 90°, 120° TeCl4
2 AX3E2 90° ClF3
3 AX2E3 180° I3-
6 octahedral 0 AX6

octahedral

90° PF6-
1 AX5E 90° SbCl52-
2 AX4E2 90°

ICl4-

Example $$\PageIndex{1}$$:

Lets try determining the geometric structures of H2O and CO2. So starting off by drawing the Lewis structure:

H2O:

Water has four electron groups so it falls under tetrahedral for the electron-group geometry. The four electron groups are the 2 single bonds to Hydrogen and the 2 lone pairs of Oxygen. Since water has two lone pairs it's molecular shape is bent. According to the VSEPR theory, the electrons want to minimize repulsion, so as a result, the lone pairs are adjacent from each other.

CO2:

Carbon dioxide has two electron groups and no lone pairs. Carbon dioxide is therefore linear in electron-group geometry and in molecular geometry. The shape of CO2 is linear because there are no lone pairs affecting the orientation of the molecule. Therefore, the linear orientation minimizes the repulsion forces.

### Molecules with More than One Central Atom

The VSEPR theory not only applies to one central atom, but it applies to molecules with more than one central atom. We take in account the geometric distribution of the terminal atoms around each central atom. For the final description, we combine the separate description of each atom. In other words, we take long chain molecules and break it down into pieces. Each piece will form a particular shape. Follow the example provided below:

Butane is C4H10. C-C-C-C is the simplified structural formula where the Hydrogens (not shown) are implied to have single bonds to Carbon. You can view a better structural formula of butane at http://en.wikipedia.org/wiki/File:Butane-2D-flat.png
If we break down each Carbon, the central atoms, into pieces, we can determine the relative shape of each section. Let's start with the leftmost side. We see that C has three single bonds to 2 Hydrogens and one single bond to Carbon. That means that we have 4 electron groups. By checking the geometry of molecules chart above, we have a tetrahedral shape. Now, we move on to the next Carbon. This Carbon has 2 single bonds to 2 Carbons and 2 single bonds to 2 Hydrogens. Again, we have 4 electron groups which result in a tetrahedral. Continuing this trend, we have another tetrahedral with single bonds attached to Hydrogen and Carbon atoms. As for the rightmost Carbon, we also have a tetrahedral where Carbon binds with one Carbon and 3 Hydrogens.

Let me recap. We took a look at butane provided by the wonderful wikipedia link. We, then, broke the molecule into parts. We did this by looking at a particular central atom. In this case, we have 4 central atoms, all Carbon. By breaking the molecule into 4 parts (each part looks at 1 of the 4 Carbons), we determine how many electron groups there are and find out the shapes.

We aren't done, yet! We need to determine if there are any lone pairs because we only looked at bonds. Remember that electron groups include lone pairs! Butane doesn't have any lone pairs. Hence, we have 4 tetrahedrals. Now, what are we going to do with 4 tetrahedrals? Well, we want to optimize the bond angle of each central atom attached to each other. This is due to the electrons that are shared are more likely to repel each other. With 4 tetrahedrals, the shape of the molecule looks like this: http://en.wikipedia.org/wiki/File:Butane-3D-balls.png. That means that if we look back at every individual tetrahedral, we match the central Carbon with the Carbon it's bonded to.

### Bond Angles

Bond angles also contribute to the shape of a molecule. Bond angles are the angles between adjacent lines representing bonds. The bond angle can help differentiate between linear, trigonal planar, tetraheral, trigonal-bipyramidal, and octahedral. The ideal bond angles are the angles that demonstrate the maximum angle where it would minimize repulsion, thus verifying the VSEPR theory.

Essentially, bond angles is telling us that electrons don't like to be near each other. Electrons are negative. Two negatives don't attract. Let's create an analogy. Generally, a negative person is seen as bad or mean and you don't want to talk to a negative person. One negative person is bad enough, but if you have two put together...that's just horrible. The two negative people will be mean towards each other and they won't like each other. So, they will be far away from each other. We can apply this idea to electrons. Electrons are alike in charge and will repel each other. The farthest way they can get away from each other is through angles. Now, let's refer back to tetrahedrals. Why is it that 90 degrees does not work? Well, if we draw out a tetrahedral on a 2-D plane, then we get 90 degrees. However, we live in a 3-D world. To visualize this, think about movies. Movies in 3D pop out at us. Before, we see movies that are just on the screen and that's good. What's better? 3D or 2D? For bond angles, 3D is better. Therefore, tetrahedrals have a bond angle of 109.5 degrees. How scientists got that number was through experiments, but we don't need to know too much detail because that is not described in the textbook or lecture.

Using the example above, we would add that H2O has a bond angle of 109.5° and CO2 would have a bond angle of 180°.

### Steps Used to Find the Shape of the Molecule

To sum up there are four simple steps to apply the VSEPR theory.

1. Draw the Lewis Structure.
2. Count the number of electron groups and identify them as bond pairs of electron groups or lone pairs of electrons. Remember electron groups include not only bonds, but also lone pairs!
3. Name the electron-group geometry. (State whether it is linear, trigonal-planar, tetrahedral, trigonal-bipyramidal, or octahedral.)
4. Looking at the positions of other atomic nuclei around the central determine the molecular geometry. (See how many lone pairs there are.)

## Dipole Moments

A molecule is polar when the electrons are not distributed equally and the molecule has two poles. The more electronegative end of the molecule is the negative end and the less electronegative end is the positive end. A common example is HCl. Using the capital sigma + or - as a symbol to show the the positive end and the negative end we can draw the net dipole. So sigma + would be on the hydrogen atom and sigma - would be on the Chlorine atom. Using the cross bow arrow shown below we can show that it has a net dipole. The net dipole is the measurable, which is called the dipole moment. Dipole moment is equal to the product of the partial charge and the distance. The equation for dipole moment is as follows.

$\mu = \delta \times d$

with

• µ = dipole moment (debye)
• δ = partial charge (C)
• d = distance (m)

The units for dipole is expressed in debye which is also known as Coulombs x meter (C x m)

Example of a Dipole

The cross base arrow demonstrates the net dipole.

On the cross-base arrow, the cross represents the positive charge and the arrow represents the negative charge.

Here's another way to determine dipole moments. We need to comprehend electronegativity which is abbreviated EN. What is EN? Well, EN is how much an element really wants an electron. Think about basketball and how two players pass the ball to each other. Each player represent an element and the ball represents the electron. Let's say one player is a ball hog. The player that is the ball hog is more electronegative because he or she wants the ball more.

Here is a link that has all the EN listed: http://www.green-planet-solar-energy...electroneg.gif

What if we are not given EN? Luckily, there is a trend in the periodic table for EN. From bottom to the top, EN will increase. From left to right, EN will increase. The most electronegative element is Flourine with 4.0.

Now, we are ready to apply EN to determine whether or not molecules are polar. We look back at the picture of H2O above. The EN is given. What do we do with all the EN? We compare the EN between each bond. Oxygen has a greater EN than Hydrogen. Therefore, we can draw a cross bow arrow towards Oxygen. We have two arrows because Oxygen is bonded to two Hydrogens. Since both arrows point toward Oxygen, we can say that there is a net EN. We added the arrows that point to Oxygen and we end up with a new, bigger arrow. This is examplified in the picture above. If arrows are drawn away from each other like <--- and --->, then we are more likely to have no net EN because the molecule is symmetrical. Refer back to the Lewis dot diagram of CO2. The shape is linear and the EN arrows point towards Oxygen. The arrows are opposite of each other and have the same EN difference. Therefore, we have no net charge and the molecule is non-polar.

#### Summary of Dipole Moments

To recap, when a molecule is polar it means that the electron is not distributed evenly and there is a difference in the electronegativity of the atoms. If a molecule is polar, it means that it had a net dipole which results in having a dipole moment.

### Determining Polarity

Is it polar? There are three ways to go about determining whether a molecule is polar or not.

A. If the molecule has a net dipole, then it is polar.
B. If the structure is symmetric, then it is non-polar
C. There are three rules to this part:
1. When there are no lone pairs on the center atom, then the molecule is non-polar
2. If it is linear or square planar, then it is non-polar. (This rule is more important than rule 1, so it overrules it because it has lone pairs.)
3. If it has different terminal atoms, then it is polar. (This rule overrules rule 1 and 2 because it is more important.)

## References

1. Petrucci, Ralph H., William S. Harwood, F. Geoffrey Herring, & Jeffry D. Madura, General Chemistry, Principles and Modern Appplications Ninth Edition, Upper Saddle River, New Jersey
2. Tetrahedrality” and the Relationship between Collective Structure and Radial Distribution Functions in Liquid Water P. E. Mason and J. W. Brady J. Phys. Chem. B;2007
3. Inverted geometries at carbon Kenneth B. Wiberg Acc. Chem. Res.; 1984
4. "Molecular Geometries." Chemistry Foundations and Applications. Volume 3. Farmington, MI:Lagowski, J.J., 2004.

## Problems

### Part I

Draw the Lewis Structure and name the shape of each compound. Also determine the polarity and whether or not it has a dipole moment.

1. HClO3
2. SO3
3. PCl4
4. C2H4
5. SnCl3-

### Part II

Name the shape and determine whether they are polar or non-polar.

## Solutions

### Part I

1.

• Total # of electrons: 1+(3x6)+7=26
• electron group geometry: tetrahedral
• molecular: trigonal pyramidal
• ideal angle: 109.5°
• polar, has a dipole moment

2.

• Total # of electrons: (3x6)+6=24
• electronic group geometry: trigonal planar
• molecular geometry: trigonal planar
• ideal angle: 120°
• polar, has a dipole moment

3.

• Total # of electrons: (4x4)+5=19
• electronic group geometry: trigonal-bi-pyramidal
• molecular geometry: seesaw
• ideal angle: 90°, 120°
• polar, has a dipole moment

4.

• Total # of electrons: (1x4)+(4x2)=12
• electronic group geometry: trigonal planar
• molecular geometry: trigonal planar
• ideal angle: 120°
• non-polar, does not have a dipole moment

5.

• Total # of electrons: (7x3)+4=26
• electronic group geometry: tetrahedral
• molecular geometry: trigonal pyramidal
• ideal angle: 109.5°
• polar, has a dipole moment.

### Part II

1. electron group geometry: octahedral

molecular geometry: square planar

non polar because it is symmetrical

2. electron group geometry: octahedral

molecular geometry: square planar

polar because it is not symmetrical


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# How to solve a limit using L'Hôpital's rule

This is my limit:

$$\lim_{x \to 1} \left[(1-x)\tan\left(\frac{\pi x}{2}\right)\right]$$

My mind is probably playing games on me right now but can you help me?

-
$(1-x)tan\frac{\pi x}{2}=\frac{tan\frac{\pi x}{2}}{\frac{1}{1-x}}$ which is infinity over infinity case. You can apply L'hospital rule from there. –  user45765 Aug 16 '14 at 13:50

You can write the limit as $$\lim_{x\to 1}\frac{\tan{\pi x/2}}{1/(1-x)}$$ Which is $\frac{\infty}{\infty}$, so you can apply L'Hopital's from there.

EDIT: Technically $$\lim_{x\to 1}\frac{1}{1-x} = \text{DNE}$$ So we have to check the limit as $x$ approaches $1$ from both sides, in which case $$\lim_{x\to 1^+}\frac{1}{1-x} = -\infty$$ $$\lim_{x\to 1^-}\frac{1}{1-x} = \infty$$ Which still results in an $\frac{\infty}{\infty}$ case.

-
Yeah that's it thanks. I've been studying calculus for a bit and I can't solve this simple problem because of fatigue. Appreciate it. –  user3601507 Aug 16 '14 at 13:54
You're welcome, I have those days as well :) –  Vishwa Iyer Aug 16 '14 at 13:55

We can make the answer more easy like this:

$$\lim_{x\to1}(1-x)\tan\frac{\pi x}{2}=\lim_{x\to1}\frac{(1-x)\sin\frac{\pi x}{2}}{\cos \frac{\pi x}{2}}=\lim_{x\to1}\frac{(1-x)}{\cos \frac{\pi x}{2}}=\lim_{x\to1}\frac{-1}{-\frac\pi2\sin\frac{\pi x}{2}}=\frac2\pi$$

-

$\lim \limits_{x \to 1} (1-x)\tan\frac{\pi x}{2} = \lim \limits_{x \to 1} \dfrac{1-x}{\cot\frac{\pi x}{2}} ~\stackrel{L'hop}{=}~ \lim \limits_{x \to 1} \dfrac{-1}{-\csc^2\frac{\pi x}{2}\times \dfrac{\pi}{2}} = \dfrac{2}{\pi}$

-

$$\lim_{x \to 1} (1-x) \tan \frac{\pi x}{2}=\lim_{x \to 1} (1-x) \frac{\sin \frac{\pi x}{2}}{ \cos \frac{\pi x}{2}}=\lim_{x \to 1} \frac{1-x}{\cos \frac{\pi x}{2}} \cdot \lim_{x \to 1}\sin \frac{\pi x}{2}=\lim_{x \to 1} \frac{1-x}{\cos \frac{\pi x}{2}} \cdot 1\overset{DLH}{=} \lim_{x \to 1} \frac{-1}{\left (-\frac{\pi}{2} \sin \frac{\pi x}{2} \right )}=\frac{2}{\pi}$$

-
You forgot that $-1$ from the beginning :) –  Vishwa Iyer Aug 16 '14 at 13:57
@VishwaIyer I edited my post :) –  evinda Aug 16 '14 at 13:59

Setting $1-x=h\iff x=1-h$

$$\lim_{x\to1}(1-x)\tan\frac{\pi x}2=\lim_{h\to0}h\tan\left(\frac\pi2-\frac{\pi h}2\right)$$

$$=\lim_{h\to0}h\cot\dfrac{\pi h}2$$

$$=\lim_{h\to0}\cos\frac{\pi h}2\cdot\dfrac1{\lim_{h\to0}\dfrac{\sin\dfrac{\pi h}2}{\dfrac{\pi h}2}}\cdot\dfrac2\pi$$

$$=\cos0\cdot\frac11\cdot\frac2\pi$$

-


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# Is there a name for this family of sequences?

The sequence $${\displaystyle{M_n:=2^{p_n}-1}}$$, where $${\displaystyle{n\gt0}}$$ and $${p_n}$$ is the $${\displaystyle{n}^{th}}$$ prime number, is commonly known as the Mersenne numbers (not to be confused with the Mersenne primes, which also requires the number itself to be prime). They have the property that no member of this sequence is divisible by $${2}$$ (all Mersenne numbers are also odd numbers). In general, the sequences $${\displaystyle{a_{m,n}:=p_m^{p_n}-p_{m-1}\#}}$$, where $${\displaystyle{m,n>0}}$$ and $${\displaystyle{p_m\#:=\prod_{i=1}^mp_i}}$$ is the $${\displaystyle{m^{th}}}$$ primorial number, contain no members divisible by the first $${\displaystyle{m}}$$ primes. The Mersenne numbers are a special case where $${\displaystyle{m=1}}$$.

Do these sequences have a name? Do any important properties of Mersenne numbers besides the indivisibility by the first $${\displaystyle{m}}$$ primes generalize to these related sequences?

• .... and a property of all primes (other than $2$ itself). – David G. Stork Aug 24 at 2:55
• @hardmath I did not mean that all odd numbers are Mersenne numbers; I just meant that the Mersenne numbers a subset of the odd numbers (i.e. oddness is a property, but not the defining characteristic, of Mersenne numbers). This was to highlight the fact that the generalized sequences are not divisible by the first m primes. I was wondering if any of the important properties of the Mersenne numbers might also apply to these related sequences. – Evan Bailey Aug 24 at 3:15
• @DavidG.Stork I was referring to Mersenne numbers in general, not just Mersenne primes. I apologize if this was not clear in the post; I can edit it if that would be helpful. – Evan Bailey Aug 24 at 3:19
• The elements of the sequence $2^{p_n}-1$ are called "Mersenne numbers with prime exponent" , there is no more specific name. The sequences you mentioned in the later part are constructions of numbers having no small prime factor and thus having a "good" chance to be prime. Note that the factors of $2^p-1$ ($p$ prime) are even more restricted : They have to be of the form $2kp+1$ with positive integer $k$. – Peter Aug 24 at 12:41
• I think this is an interesting generalization of Mersenne numbers, and I did not not find these sequences described as such in places on the Web like Chris Caldwell's Prime Pages. Some simple factorizations can illustrate whether divisors have properties similar to what @Peter describes. – hardmath Aug 25 at 13:50

Let's broaden the family of sequences somewhat. Let $$b,c$$ be integers, with $$b\gt 1$$, and consider the integer sequence for $$k = 0,1,2,\ldots$$:

$$s_k = b^k - c$$

The case you ask about is when $$b=p$$ is prime and $$c = (p-1)\#$$ is a primorial number, and you restricted attention to the subsequence of prime exponents $$k$$.

Let's call these exact powers with a fixed offset. Negative as well as positive values of the offset constant $$c$$ are allowed, so these sequences include as subsequences Fermat numbers as well as Mersenne numbers. Some less well-known sequences in this family are the Cunningham numbers:

$$b^n \pm 1$$

and the so-called Crandall primes (after Richard Crandall's U.S. Patent 5,159,632, although there is prior art by Bender and Castagnoli):

$$2^q - c \;\;\text{ for small odd } c$$

The latter were studied as providing a richer supply of primes (than the few dozen Mersenne primes) for use as prime moduli of elliptic curve cryptosystems.

The key feature of these sequences is that they result from iteration of a first-degree univariate polynomial:

$$s_{k+1} = b s_k + (b-1)c$$

This leads to treatment by arithmetic dynamics. I'll illustrate some of the ideas by taking the example $$b=3$$ and $$c=2$$ of your constructions.


text:
I'm trying to install memcached and having no luck getting the site to run in local development using OS X 10.9 Mavericks. The homepage loads, but that's it. Everywhere else, I get this error:

Fatal error: Maximum execution time of 180 seconds exceeded in /Users/me/Sites/mysite/modules/field/modules/field_sql_storage/field_sql_storage.module on line 352

I get the error at the same point, no matter what the max_execution_time is set to.

I'd love to get this working, but I'd be content just to uninstall the whole mess. I can't seem to get the site working again even after I attempted that.

• Here's the memcached info from phpinfo()


text:
Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# A pendulum clock gives correct time at

Question:

A pendulum clock gives correct time at $20^{\circ} \mathrm{C}$ at a place where, $g=9.800 \mathrm{~m} / s^{2}$. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where $\mathrm{g}=9.788 \mathrm{~m} / \mathrm{s}^{2}$. At what temperature will it give correct time? Coefficient of linear expansion of steel $=12^{\times 10^{-6} /{ }^{\circ} \mathrm{C}}$.

Solution:


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# XAS $M_{4,5}$ partial excitations

asked by BODRY TEGOMO CHIOGO (2021/09/21 18:20)

Dear Quanty developpers,I was inspired by the tutorial for the calculation of partial excitation at the $L_{2,3}$ edge of NiO for writing a script for XAS at the $M_{4,5}$ edge of cerium with configuration-interaction. I would like to look at excitations into the $f^0$, $f^1$ and $f^2$ configurations but there is some problems.

Here are the restrictiions I used

T4f3dx = ConjugateTranspose(T3d4fx)
T4f3dy = ConjugateTranspose(T3d4fy)
T4f3dz = ConjugateTranspose(T3d4fz)
T4f3dl = ConjugateTranspose(T3d4fl)
T4f3dr = ConjugateTranspose(T3d4fr)

TXASf0x = Clone(T3d4fx)
TXASf1x = Clone(T3d4fx)
TXASf2x = Clone(T3d4fx)
TXASf0y = Clone(T3d4fy)
TXASf1y = Clone(T3d4fy)
TXASf2y = Clone(T3d4fy)
TXASf0z = Clone(T3d4fz)
TXASf1z = Clone(T3d4fz)
TXASf2z = Clone(T3d4fz)

TXASf0x.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",1,1}}
TXASf1x.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",2,2}}
TXASf2x.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",3,3}}

TXASf0y.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",1,1}}
TXASf1y.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",2,2}}
TXASf2y.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",3,3}}

TXASf0z.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",1,1}}
TXASf1z.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",2,2}}
TXASf2z.Restrictions = {NF,NB,{"0000000000 11111111111111 00000000000000",3,3}}

All these these transitions operators give the same spectrum instead of partial excitations into the $f^0$, $f^1$ and $f^2$ configurations as I expected. I do not understand where is the mistake.

Bodry

A note further important information to know is that the spectrum is calculated with restrictions set in the following way

NConfigurations = 3
CalculationRestrictions = {NF, NB, {'0000000000 00000000000000 11111111111111', 14 - (NConfigurations - 1), 14}}
XAS_spectrum = CreateSpectra(H_f,{TXASf0z,TXASf1z,TXASf2z,T3d4fz},psiList,{{"Emin",-20},{"Emax",30},{"NE",3500},{"Gamma",0.1},{'restrictions', CalculationRestrictions}})

, 2021/09/22 09:59

Dear Bodry,

As discussed by e-mail, but here as an answer to document also for others. If you include restrictions in your calculation there are two ways you can do this, either by adding them to operators, or by giving the restrictions as parameters to the functions that calculate the eigenstates or the spectra.

These two methods can not be used simultaneously and one method will override the other.

The input that gets what you want is:

NConfigurations = 3
CalculationRestrictions = {NF, NB, {'0000000000 00000000000000 11111111111111', 14 - (NConfigurations - 1), 14}}
H_f.Restrictions = CalculationRestrictions
XAS_spectrum = CreateSpectra(H_f,{TXASf0z,TXASf1z,TXASf2z,T3d4fz},psiList,{{"Emin",-20},{"Emax",30},{"NE",3500},{"Gamma",0.1}})

Some background information: Originally I implemented the restrictions as properties of operators. This however leads to problems, such as how to deal with two operators that one sums with different restrictions. As there is (I did not find) a good solution to this question the current code does not inherent restrictions if one acts on an operator to the result. As such, using restrictions as a property of operators is extremely dangerous coding. (easy to make errors that are hard to find). To make this better we added the restrictions to the functions. The call

XAS_spectrum = CreateSpectra(H_f,{TXASf0z,TXASf1z,TXASf2z,T3d4fz},psiList,{{"Emin",-20},{"Emax",30},{"NE",3500},{"Gamma",0.1},{'restrictions', CalculationRestrictions}})

is similar to the call

H_f.Restrictions = CalculationRestrictions
TXASf0z.Restrictions = CalculationRestrictions
TXASf1z.Restrictions = CalculationRestrictions
TXASf2z.Restrictions = CalculationRestrictions
T3d4fz.Restrictions = CalculationRestrictions
XAS_spectrum = CreateSpectra(H_f,{TXASf0z,TXASf1z,TXASf2z,T3d4fz},psiList,{{"Emin",-20},{"Emax",30},{"NE",3500},{"Gamma",0.1},{'restrictions', CalculationRestrictions}})

with the difference that the later changes the original operators and the former does not.

Best wishes, Maurits


text:
# Recursion in Chess AI

I am currently in the process of completely rewriting an older Chess project and have written the main recurring function in the AI:

//Finds and applies the best move for the ai.
function aiTurn() {
//Recursive function. recurBoard is the board array passed to the function.
//Colour is either -1 or 1, -1 means it's calculating for black and 1, white.
//depth is simply how deep the recursion has gone.
var recur = function(recurBoard, colour, depth) {
//Stops recurring when max depth is reached. Max is defined in index.html.
if (depth === max) return getEvaluation(recurBoard);
//Setting a baseline for the biggest value of all deeper moves.
var mVal = -10000*colour;
//Cycling through the entire board array.
for (var i = 0; i < 64; i++) {
//Defines unit as the numeric representation of the current board position
//(EG. -5 is a black queen, 1 is a white pawn and 6 is a white king).
var unit = recurBoard[i];
//Checks to see if the unit is a pawn
//(currently only have programmed pawn movement for the ai).
if (unit*colour != 1) continue;
//Makes an array of all the moves the specific unit can make
//(formated like this: [[9,17],[9,25]]
//This would be the moves of a pawn on square 9).
var moves = getMoves(unit, i, recurBoard);
//Loops through all the moves.
for (var j = 0; j < moves.length; j++) {
//Defines movToUnt as the unit the current move will move on top of.
var movToUnt = recurBoard[moves[j][1]];
//If moving on to a king simply return a massive advantage.
if (movToUnt === -6*colour) return 9000*colour;
//Recurs. move(from,to,board) returns a board after a given move.
var val = recur(move(moves[j][0], moves[j][1], recurBoard), -colour, depth+1);
//Checks to see if the value of the move is bigger than that of the current best move
if (val*colour > mVal*colour) mVal = val;
}
}
//Returns the value of the best move.
return mVal;
}

//This next part does pretty much the same thing as the for loop above, but stores the values.
var aiMoves = []
for (var i = 0; i < 64; i++) {
var u = board[i];
if (-u*playerSide != 1) continue;
var uMoves = getMoves(u, i, board);
for(var j = 0; j < uMoves.length; j++) {
if (board[uMoves[j][1]] === 6*playerSide) {         //Shouldn't be necessary.
uMoves[j][2] = -9000*playerSide;                //
aiMoves[aiMoves.length] = uMoves[j];            //
continue;                                       //
}                                                   //
uMoves[j][2] = recur(move(uMoves[j][0], uMoves[j][1], board), playerSide, 0);
aiMoves[aiMoves.length] = uMoves[j];
}
}
return aiMoves;
}

This properly recurs up to a given depth, but this gives the Chess AI the infamous "horizon effect" where it will make stupid decisions because it doesn't calculate far enough. This can be fixed simply by telling it to not stop recurring if it just killed a piece in its simulation.

• First step. Put your comments on line above where they are appropriate , not way out at the end of the line. Seriously, the code is so hard to review right now, that you may not even get a meaningful review. You should strive to keep your lines of code <80 characters per line. Even break up your comments across lines as necessary to keep within this limit. – Mike Brant Sep 3 '16 at 3:31
• Alright, thanks for the feedback. I've done that now. – Lasse Møldrup Sep 3 '16 at 11:41

While this is a reasonable initial attempt, this is not exactly what a modern chess engine would do. In fact, even if you would code the remaining pieces (you only have pawn now), the engine wouldn't be strong.

Before that, we should clear up some chess terminology:

• We don't say simulation, we say search
• The term AI is okay but even a better term would be engine

There're some very serious issues with your implementation:

• You will need to account for transposition. This can be done by hashing.
• You should take account of horizon effect by Quiescene Search.
• You should really consider alpha-beta search or the related nega-max search.
• In your code, you're just trying to brute force all moves in the deep-first fashion, this is absolutely unacceptable for chess programming. The search space will blow your code off. I repeat, you can't search every single position in chess, it's pointless.
• What kind of move representation are you using? This looks like a simple array to me, maybe you'd consider a faster implementation?
• Why is moving a pawn to the king is an advantage?

Depends on what you want to achieve, if you just want to code something that moves, it's fine. Unfortunately, the current code is nowhere close to anything what a modern chess engine is expected. As an experienced engine developer, I'd rate the code zero. It has lots improvement opportunities.

• Well I am very much a novice and really want to learn from what I write so i don't want to copy someone else's solution. It doesn't have to be particularly good as long as I learn from it. However I definitely see your point and will be doing some research on the things you listed. – Lasse Møldrup Sep 4 '16 at 16:55
• My move representation is indeed as simple as an array [from, to], what would be a faster solution? Moving a pawn to the king is an advantage as it wins you the game and it should also stop searching, which is why it returns. – Lasse Møldrup Sep 4 '16 at 16:56
• @LasseMøldrup why does moving a pawn to the King wins a game? We need to checkmate to win. – SmallChess Sep 4 '16 at 17:07
• Any reference to learn how to create such an engine ? – Steve Chamaillard Sep 4 '16 at 19:47
• @SteveChamaillard chessprogramming.wikispaces.com – SmallChess Sep 5 '16 at 2:24


text:
# Embedding covers of manifolds

I am considering $k$-fold covers of smooth manifolds (with smooth covering maps).

Let $f:M^m\to N^m$ be a smooth finite covering map.

--

The following implication is not true: $M$ can be embedded into $\mathbb{R}^n$ $\Rightarrow$ $N$ can be embedded into $\mathbb{R}^n$.

The easiest counterexample is $M=\mathbb{S}^{n-1}$ and $N=RP^{n-1}$.

--

But what about the converse of the implication:

$N$ can be embedded into $\mathbb{R}^n$ $\Rightarrow$ $M$ can be embedded into $\mathbb{R}^n$.

Intuitively, this makes sense since if $N$ can be embedded, we should be able to embed $M$ which is obtained by "cutting, unfolding and gluing several copies of $N$" and is therefore "less complicated".

Or is there an obvious counterexample I'm missing?

NB: I'm of course aware of the Whitney Embedding Theorem. The point is, however, to derive something better than $n=2m$ for $M$ based on the fact that I know something about the embeddability of $N$.

 It appears that adding the hypothesis of "closed" doesn't save us -- Section Three of (http://arxiv.org/abs/0810.2346) gives a few examples of closed three-manifolds (of Nil and Solv geometry) that embed in $S^4$ and which have (infinitely many) finite covers that do not embed in $S^4$.