text:
# The Symmetric Group

Part Of: Algebra sequence
Followup To
Content Summary: 1800 words, 18 min read

On Permutations

In the last few posts, we have discussed algebraic structures whose sets contain objects (e.g., numbers). Now, let’s consider structures over a set of functions, whose binary operation is function composition.

Definition 1. Consider two functions $f$ and $g$. We will denote function composition of $g(f(x))$ as $f \bullet g$. We will use this notation instead of the more common $g \circ f$.  Both represent the idea “apply $f$, then $g$“.

Consider $G = (\left\{ f(x) = 2x, g(x) = x+1, h(x) = 2x + 2, i(x) = 2x+1 \right\}, \bullet)$

Is this a group? Let’s check closure:

• $g \bullet f = 2(x+1) = h \in G$
• $f \bullet g = (2x)+1 = i \in G$
• $f \bullet h = 2(2x)+2 \not\in G$

Closure is violated. $G$ isn’t even a magma! Adding $j(x) = 4x+2$ to the underlying set exacerbates the problem: then both $f \bullet j \not\in G$ and $g \bullet j \not\in G$.

So it is hard to establish closure under function composition. Can it be done?

Yes. Composition exhibits closure on sets of permutation functions. Recall that a permutation is simply a bijection: it re-arrange a collection of things. For example, here are the six possible bijections over a set of three elements.

Definition 2. The symmetric group $S_n$ denotes composition over a set of all bijections (permutations) over some set of $n$ objects. The symmetric group is then of order $n!$.

The underlying set of $S_{3}$ is the set of all permutations over a 3-element set. It is of order $3! = 3 \times 2 \times 1 = 6$.

This graphical representation of permutations is rather unwieldy. Let’s switch to a different notation system!

Notation 3: Two Line Notation. We can use two lines to denote each permutation mapping $\phi$. The top row represents the original elements $x$, the bottom represents where each element has been relocated $\phi(x)$.

Two line notation is sometimes represented as an array, with the top row as matrix row, and bottom denoting matrix column. Then the identity matrix represents the identity permutation.

Definition 4. A cycle is a sequence of morphisms that forms a closed loop. An n-cycle is a cycle of length n. A 1-cycle does nothing. A 2-cycle is given the special name transposition. $S_3$ has two permutations with 3-cycles: can you find them?

Theorem 5. Cycle Decomposition Theorem. Every permutation can be decomposed into disjoint cycles. Put differently, a node cannot participate in more than one cycle. If it did, its parent cycles would merge.

Notation 6: Cycle Notation. Since permutations always decompose into cycles, we can represent them as $(A_1\ A_2\ \ldots\ A_n)$, pronounced “$A_1$ goes to $A_2$ goes to …”.

Cycle starting element does not matter: $(A\ C\ B) = (C\ B\ A) = (B\ A\ C)$.

The Cycle Algorithm

It is difficult to tell visually the outcome of permutation composition. Let’s design an algorithm to do it for us!

Algorithm 7: Cycle Algorithm. To compose two permutation functions $a(S)$ and $b(S)$, take each element $s \in S$ and follow its arrows until you find the set of disjoint cycles. More formally, compose these functions $x$ times until you get $(a \bullet b)^x(s) = s$.

Here’s a simple example from $S_{3} = ( \left\{ 0, 1, 2, 3, 4, 5 \right\}, \bullet )$.

Make sense? Good! Let’s try a more complicated example from $S_4$.

A couple observations are in order.

• 1-cycles (e.g., $(A)$) can be omitted: their inclusion does not affect algorithm results.
• Disjoint cycles commute: $(B\ D),(A\ C) = (A\ C),(B\ D)$. Contrast this with composition, which does not commute $(B\ D) \bullet (A\ C) \neq (A\ C) \bullet (B\ D)$.

Now, let’s return to $S_{3}$ for a moment, with its set of six permutation functions. Is this group closed? We can just check every possible composition:

From the Cayley table on the right, we see immediately that $S_{3}$ is closed (no new colors) and non-Abelian (not diagonal-symmetric).

But there is something much more interesting in this table. You have seen it before. Remember the dihedral group $D_{3}$? It is isomorphic to $S_{3}$!

If you go back to the original permutation pictures, this begins to make sense. Permutations $1$ and $2$ resemble rotations/cycles; $3$, $4$, and $5$ perform reflections/flips.

Generators & Presentations

In Theorem 7, we learned that permutations decompose into cycles. Let’s dig deeper into this idea.

Theorem 8. Every n-cycle can be decomposed into some combination of 2-cycles. In other words, cycles are built from transpositions.

The group $S_{3} = \left\{ 0, 1, 2, 3, 4, 5 \right\}, \bullet)$ has three transpositions $3 = (A\ B)$, $4 = (B\ C)$, and $5 = (A\ C)$.

Transpositions are important because they are generators: every permutation can be generated by them. For example, $1 = 3 \bullet 4$. In fact, we can lay claim to an even stronger fact:

Theorem 9. Every permutation can be generated by adjacent transpositions. Every permutation $S_{n} = \langle (1 2), (2 3), \ldots, (n-1\ n) \rangle$.

By the isomorphism $S_3 \cong D_3$ , we can generate our “dihedral-looking” Cayley graph by selecting generators $1=(A\ B\ C)$ and $3 = (B\ C)$.

But we can use Theorem 9 to produce another, equally valid Cayley diagram. There are two adjacent transpositions in $S_{3}$: $3$ and $4$. All other permutations can be written in terms of these two generators:

• $0 = 3 \bullet 3$.
• $1 = 3 \bullet 4$.
• $2 = 4 \bullet 3$.
• $5 = 3 \bullet 4 \bullet 3$

This allows us to generate a transposition-based Cayley diagram. Here are the dihedral and transposition Cayley diagrams, side by side:

We can confirm the validity of the transposition diagram by returning to our multiplication table: $1 \bullet 3 = 5$ means a green arrow $1 \rightarrow 5$.

Note that the transposition diagram is not equivalent cyclic group $C_6$, because arrows in the latter are monochrome and unidirectional.

We’re not quite done! We can also rename our set elements to employ generator-dependent names, by “moving clockwise”:

We could just as easily have “moved counterclockwise”, with names like $3 \mapsto r$, $1 \mapsto rl$. And we can confirm by inspection that, in fact, $rl = lrlr$ etc.

Using the original clockwise notation, one presentation of $S_3$ becomes:

$S_3 = \langle l, r \mid r^2 = e, l^2 = e, (lr)^3 = e \rangle$

Towards Alternating Groups

Any given permutation can be written as a product of permutation. Consider, for example, the above equalities

• $1 = rl = lrlr = (lr)^3rl$. These have 2, 4, and 8 permutations, respectively.
• $5 = lrl = lrl^3 = l^3r^3l^3$. These have 3, 5, and 9 permutations, respectively.

Did you notice any patterns in the above lists? All expressions for $1$ require an even number of transpositions, and all expressions of $5$ require an odd number. In other words, the parity (evenness or oddness) of a given permutation doesn’t seem to be changing. In fact, this observation generalizes:

Theorem 10. For any given permutation, the parity of its transpositions is unique.

Thus, we can classify permutations by their parity. Let’s do this for $S_3$:

• $0, 1, 2$ are even permutations.
• $3, 4, 5$ of odd permutations.

Theorem 11. Exactly half of $S_n$ are even permutations, and they form a group called the alternating group $A_n$. Just as $\lvert S_n \rvert = n!$, $A_n$ has $\dfrac{n!}{2}$ elements.

Why don’t odd permutations form a group? For one thing, it doesn’t contain the identity permutation, which is always even.

Let’s examine $A^3 = (\left\{ 0, 1, 2 \right\}, \bullet )$ in more detail. Does it remind you of anything?

It is isomorphic to the cyclic group $C_3$ ..!

We have so far identified the following isomorphisms: $S_3 \cong D_3$ and $A_3 \cong C_3$. Is it also true that e.g., $S_4 \cong D_4$ and $A_4 \cong C_4$?

No! Recall that the $\lvert D_n \rvert=2n$ and $\lvert C_n \rvert = n$Only $n \neq 3$, these sets are not even potentially isomorphic.  For example:

• $\lvert D_4 \rvert=2 \times 4 = 8 \neq 24 = 4! = \lvert S_4 \rvert$.
• $\lvert C_4 \rvert= 4 \neq 12 = \frac{24}{2} = \lvert A_4 \rvert$.

For these larger values of $n$, the symmetric group is much larger than dihedral and cyclic groups.

Applications

Why do symmetric & alternating groups matter? Let me give two answers.

Perhaps you have seen the quadratic equation, the generic solution to quadratic polynomials $ax^2 + bx + c = 0$.

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Analogous formulae exist for cubic (degree-3) and quartic (degree-4) polynomials. 18th century mathematics was consumed by the theory of equations: mathematicians attempting to solve quintic polynomials (degree-5). Ultimately, this quest proved to be misguided: there is no general solution to quintic polynomials.

Why should degree-5 polynomials admit no solution? As we will see when we get to Galois Theory, it has to do with the properties of symmetric group $S_5$.

A second reason to pay attention to symmetric groups comes from the classification theorem of finite groups. Mathematicians have spent decades exploring the entire universe of finite groups, finding arcane creatures such as the monster group, which may or may not explain features of quantum gravity.

One way to think about group space is by the following periodic table:

Image courtesy of Samuel Prime

Crucially, in this diverse landscape, the symmetric group plays a unique role:

Theorem 12: Cayley’s Theorem. Every finite group is a subgroup of the subgroup $S_n$, for some sufficiently large $n$.

For historical reasons, subgroups of the symmetric group are usually called permutation groups.

Until next time.

Wrapping Up

Takeaways:

• The symmetric group $S_n$ is set of all bijections (permutations) over some set of $n$ objects, closed under function composition.
• Permutations can be decomposed into disjoint cycles: cycle notation uses this fact to provide an algorithm to solve for arbitrary compositions.
• All permutations (and hence, the symmetric group) can be generated by adjacent transpositions. This allows us to construct a presentation of the symmetric group.
• Permutations have unique parity: thus we can classify permutations as even or odd. The group of even presentations is called the alternating group $A_n$.
• It can be shown that $S_3 \cong D_3$ and $A_3 cong C_3$. However, for larger $n$, the symmetric and alternating group are much larger than cyclic and dihedral groups.

The best way to learn math is through practice! If you want to internalize this material, I encourage you to work out for yourself the Cayley table & Cayley diagram for $S_4$. 🙂

Related Resources

For a more traditional approach to the subject, these Harvard lectures are a good resource.


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# Measurement of azimuthal correlations of D mesons and charged particles in pp collisions at $\sqrt{s}=7$ TeV and p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV

Abstract : The azimuthal correlations of D mesons and charged particles were measured with the ALICE detector in pp collisions at $\sqrt{s}=7$ TeV and p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV at the Large Hadron Collider. D$^0$, D$^+$, and D$^{*+}$ mesons with transverse momentum $30.3$ Gev/$c$. The properties of the correlation peak induced by the jet containing the D meson, described in terms of the yield of charged particles in the peak and peak width, are compatible within uncertainties between the two collision systems, and described by Monte-Carlo simulations based on the PYTHIA and POWHEG event generators.
Type de document :
Article dans une revue
European Physical Journal C: Particles and Fields, Springer Verlag (Germany), 2017, 77 (245), 〈10.1140/epjc/s10052-017-4779-8〉
Domaine :

http://hal.in2p3.fr/in2p3-01321824
Contributeur : Dominique Girod <>
Soumis le : jeudi 26 mai 2016 - 12:17:32
Dernière modification le : mardi 19 juin 2018 - 01:19:14

### Citation

J. Adam, R. Vernet, A. Baldisseri, H. Borel, J. Castillo Castellanos, et al.. Measurement of azimuthal correlations of D mesons and charged particles in pp collisions at $\sqrt{s}=7$ TeV and p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV. European Physical Journal C: Particles and Fields, Springer Verlag (Germany), 2017, 77 (245), 〈10.1140/epjc/s10052-017-4779-8〉. 〈in2p3-01321824〉

### Métriques

Consultations de la notice

format: plain


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# Tarski-Knaster theorem

###### Theorem 1 (LATTICE-THEORETICAL FIXPOINT THEOREM).

Let

1. (i)

$\mathfrak{A}=(A,\ \leq)$  be a complete lattice,

2. (ii)

$f$ be an increasing function on $A$ to $A,$

3. (iii)

$P$ the set of all fixpoints of $f.$

Then the set $P$ is not empty and the system $(P,\ \leq)$ is a complete lattice; in particular we have

 $\cup P=\cup E_{x}[f(x)\geq x]\in P$

and

 $\cap P=\cap E_{x}[f(x)\leq x]\in P.$

This article follows, with clarification, Tarski’s original 1955 article [6], which may not be available to some readers.

## 1 Definitions

The symbol $\emptyset$ denotes the empty set. The symbol $\forall$ is shorthand for ”for all”.

###### Definition 1.

A poset (partially ordered set) $\mathfrak{A}=(A,\ \leq)$ consists of a nonempty set $A$ and a relation

 $\leq:\ A\times A\longrightarrow\{\text{False},\ \text{True}\}$

that is reflexive ($a\leq a$ $\forall a\in A$), antisymmetric ($\forall\ a,\ b\in A,$ if $a\leq b$ and $b\leq a,$ then $a=b$). and transitive  ($\forall\ a,\ b,\ c\in A,$ if $a\leq b$ and $b\leq c,$ then $a\leq c$).

###### Definition 2.

If $\mathfrak{A}=(A,\ \leq)$ is a poset and $B$ is a nonempty subset of $A,$ then $\sup(B)$ is an element of $A,$ if it exists, with these properties:

1. 1.

$\forall\ b\in B,~{}b\leq\sup(B)$, (in shorthand, $B\leq\sup(B)$).

2. 2.

If $a\in A$ and $B\leq a,$ then $\sup(B)\leq a.$

Similarly $\inf(B)$ is an element of $A,$ if it exists, with these properties:

1. (3)

$\inf(B)\leq B,$

2. (4)

If $a\in A$ and $a\leq B,$ then $a\leq\inf(B).$

###### Theorem 2.

If $\sup(B)$ exists, then it is unique; if $\inf(B)$ exists, it is unique.

###### Definition 3.

Let $\mathfrak{A}=(A,\ \leq)$ be a poset. Then $\mathfrak{A}$ is a if $\forall\ a,\ b\in A,$ the set $\{a,\ b\}$ has a $\sup$ and an $\inf.$ We write

 $a\cup b=\sup(\{a,\ b\})\quad\text{and}\quad a\cap b=\inf(\{a,\ b\}).$

As an exercise, prove the following:

###### Theorem 3.

Let $\mathfrak{A}=(A,\ \leq)$ be a lattice. Then

1. 1.

$a\cup b=b\cup a,\qquad a\cap b=b\cap a$

2. 2.

$(a\cup b)\cup c=a\cup(b\cup c),\qquad(a\cap b)\cap c=a\cap(b\cap c)$

3. 3.

$a\cup(b\cap c)\leq(a\cup b)\cap(a\cup c).\qquad(a\cap b)\cup(a\cap c)\leq a% \cap(b\cup c).$

###### Definition 4.

A poset $\mathfrak{A}=(A,\ \leq)$ is a complete lattice if for each nonempty $B\subseteq A,$ both $\sup(B)$ and $\inf(B)$ exist.

In particular, this is the case when $B=\{a,\ b\},$ so a complete lattice is a lattice.

## 2 The Tarski-Knaster Theorem

Statement of the theorem, as it is in [6]:

###### Theorem 4 (LATTICE-THEORETICAL FIXPOINT THEOREM).

Let

1. (i)

$\mathfrak{A}=(A,\ \leq)$  be a complete lattice,

2. (ii)

$f$ be an increasing function on $A$ to $A,$

3. (iii)

$P$ the set of all fixpoints of $f.$

Then the set $P$ is not empty and the system $(P,\ \leq)$ is a complete lattice; in particular we have

 $\cup P=\cup E_{x}[f(x)\geq x]\in P$

and

 $\cap P=\cap E_{x}[f(x)\leq x]\in P.$

The symbols $\cup P$ and $\cap P$ are what I am calling $\sup(P)$ and $\inf(C).$ Now let me restate the theorem:

###### Theorem 5.

Let $\mathfrak{A}=(A,\ \leq)$ be a complete lattice. Let $f$ be an increasing function on $A$ to $A,$ that is, whenever $a\leq b,$ then $f(a)\leq f(b).$ Let $P$ be the set of all fixed points of $f:$

 $P=\{\ a\in A\ |\ f(a)=a\ \}.$

Then

1. 1.

$P\neq\emptyset$.

2. 2.

The system $\mathfrak{P}=(P,$ $\leq)$ is a complete lattice.

3. 3.

Set $B=\{\ b\in A\ |\ b\leq f(b)\ \}$. Then $B\neq\emptyset$ and $\sup(B)=\sup(P)\in P.$

4. 4.

Set $C=\{\ c\in A\ |\ f(c)\leq c\ \}$. Then $C\neq\emptyset\$and $\inf(C)=\inf(P)\in P.$

###### Lemma 1.

Let $\mathfrak{A}=(A,\ \leq)$ be a complete lattice and $\emptyset\neq A_{0}\subseteq A.$ Let

 $D=\{d\in A\ |\ d\leq A_{0}\},\qquad E=\{e\in A\ |\ A_{0}\leq e\}.$

($d\leq A_{o}$ means $d\leq a_{0}$ $\forall a_{0}\in A_{0}.$) Then both $B$ and $C$ are complete lattices, with $\leq$  inherited from that of $A.$

###### Proof (of Lemma).

We’ll do the case of $D;$ that for $E$ is similar (dual). First, $D\neq\emptyset$ because $\inf(A)\in D.$ Suppose $\emptyset\neq G\subseteq D.$ Obviously

 $\inf(G)\leq g\leq A_{0}$

for any $g\in G,$ so $\inf(G)\in D.$ As $g\leq A_{0}$ for all $g\in G,$ we have $\sup(G)\leq A_{0}$ by (2) in the definition of $\sup.$ So $\sup(G)\in D.$ As $G$ was any nonempty subset of $D,$ these statements say that $D$ is a complete lattice. ∎

###### Proof (of Theorem).

$B\neq\emptyset$ because $\inf(A)\in B.$ Hence $b_{1}=\sup(B)$ exists. If $b\in B,$ then $b\leq b_{1},$ which implies

 $f(b)\leq f(b_{1}).$

because $f$ is increasing., so

 $b\leq f(b)\leq f(b_{1}).$

This is true $\forall$ $b\in B.$ Hence

 $b_{1}=\sup(B)\leq f(b_{1}).$

Therefore $b_{1}\in B.$ Because $f$ is increasing, this implies

 $f(b_{1})\leq f(f(b_{1})),$

from which $f(b_{1})\in B.$ Hence $f(b_{1})\leq\sup(B)=b_{1}.$ From

 $b_{1}\leq f(b_{1})\leq b_{1}$

we have $f(b_{1})=b_{1},$ that is, $b_{1}\in P.$ This settles (1): $P\neq\emptyset,$ and incidentally proves $\sup(B)\in P.\vskip 3.0pt plus 1.0pt minus 1.0pt$

From this, $\sup(B)\leq\sup(P).$ But obviously $P\subseteq B,$ so $\sup(P)\leq\sup(B).$ Therefore $\sup(P)=\sup(B),$ completing the proof of (3). Statement (4) is proved similarly (or dually).

Next we prove (2). Let $\emptyset\neq P_{0}\subseteq P.$ Set

 $Q=\{\ q\in A\ |\ q\leq P_{0}\ \}.$

By the Lemma, $Q$ is a complete lattice. If $q\in Q,$ then $\forall\ p_{0}\in P_{0},$ we have $q\leq p_{0},$ hence

 $f(q)\leq f(p_{0})=p_{0}\ ,$

that is, $f(q)\in Q.$ In other words,

 $f:\ Q\longrightarrow Q.$

By what we have already proved for $A,$ applied to $Q,$ we have $\sup(Q)\in Q,$ and $\sup(Q)$ is a fixed point of $f.$ That is, $\sup(Q)\in P.$ But obviously $\sup(Q)=\inf(P_{0}),$ so $\inf(P_{0})\in P$ Similarly $\sup(P_{0})\in P,$ completing the proof of (2): $P$ is a complete lattice. ∎

This theorem was proved by A. Tarski [6]. A special case of this theorem (for lattices of sets) appeared in a paper of B. Knaster [3]. Kind of converse of this theorem was proved by Anne C. Davis [1]: If every order-preserving function $f\colon L\to L$ has a fixed point, then $L$ is a complete lattice.

## References

• 1 Anne C. Davis, A characterization of complete lattices., Pac. J. Math. 5 (1955), 311–319.
• 2 Thomas Forster, Logic, induction and sets, Cambridge University Press, Cambridge, 2003.
• 3 B. Knaster, Un théorème sur les fonctions d’ensembles., Annales Soc. Polonaise 6 (1928), 133–134.
• 4 M. Kolibiar, A. Legéň, T. Šalát, and Š. Znám, Algebra a príbuzné disciplíny, Alfa, Bratislava, 1992 (Slovak).
• 5 J. B. Nation: http://bigcheese.math.sc.edu/ mcnulty/alglatvar/Notes on lattice theory
• 6 Alfred Tarski, A lattice-theoretical fixpoint theorem and its applications., Pac. J. Math. 5 (1955), 285–309.
• 7 Wikipedia’s entry on http://en.wikipedia.org/wiki/Knaster-Tarski_theoremKnaster-Tarski theorem
 Title Tarski-Knaster theorem Canonical name TarskiKnasterTheorem Date of creation 2013-03-22 15:30:19 Last modified on 2013-03-22 15:30:19 Owner kompik (10588) Last modified by kompik (10588) Numerical id 18 Author kompik (10588) Entry type Theorem Classification msc 06B99 Synonym Tarski theorem Synonym Knaster-Tarski theorem Related topic lattice Related topic completelattice Related topic fixedpoint Related topic ProofOfSchroederBernsteinTheoremUsingTarskiKnasterTheorem Related topic Lattice Related topic CompleteLattice


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Homework Help: Division question

1. Oct 24, 2004

roger

HI

please can someone help me with this quick question on division :

If I have to do 3 divided by 4
____
4 |3

4 doesn't go into 3 so I put a zero on top .

When it gets to 4 into 2(from 30-4x7), it doesn't go, so why don't I put a zero on top but instead put a zero beside the 2 to make it 20 ?

thankyou

roger

2. Oct 24, 2004

dav2008

I have no idea what you're doing.

If you're doing 3 divided by 4 using long division:

http://home.comcast.net/~iberiaforums/division.GIF [Broken]

Last edited by a moderator: May 1, 2017
3. Oct 24, 2004

arildno

roger:
Do you want to perform 3:4 or 4:3?

4. Oct 24, 2004

roger

DEAR ARILDNO,

I AM TRYING TO DO 3 DIVIDED BY 4 LONG DIVISION

BUT WHY DONT I PUT A ZERO ON TOP FOR 4 INTO 2 ?

ROGER

5. Oct 24, 2004

dav2008

You never put a 0 on top.

What you do in the first step is put a decimal point on top.

6. Oct 24, 2004

roger

when you do 7x4 =28

and then 30 minus 28 = 2.

from here

4 doesnt go into 2 so why don't I put a zero on top ?

This is what I want to know ....

thanx

roger

7. Oct 25, 2004

arildno

Now, since to write down the long-division algoritm is a bit difficult (and besides, we do it differently in Norway from the US), I will show you the RATIONALE behind the long division technique instead; long division is simply a condensed version of what I'll present.

1. What is meant by "division" in this case?
Ordinarily, "division" is meant to be that process that rewrites a number, given as a FRACTION, into the equivalent DECIMAL REPRESENTATION of that number (which, as it happens, is a particular TYPE of fractional representation).
2. Let's look at the number given by fraction 3/4

We want to write 3/4 in its decimal representation, that is to find digits $$a_{i}$$ between 0 and 9, so that we have:
$$\frac{3}{4}=0.a_{1}a_{2}a_{3}....$$
Where the notation $$0.a_{1}a_{2}a_{3}....$$ MEANS:
$$0.a_{1}a_{2}a_{3}....\equiv\frac{a_{1}}{10}+\frac{a_{2}}{100}+\frac{a_{3}}{1000}++$$

3. Let's start!
a) We note that 3<4, so our first step is to multiply 3/4 with an appropriate representation of the number "1":
$$\frac{3}{4}=1*\frac{3}{4}=\frac{10}{10}*\frac{3}{4}=\frac{1}{10}*(\frac{30}{4})$$
Henceforth, we will work with the expression included in the parenthesis.
b)
We note that 30>4, and the greatest multiple of 4 which is less than 30, is 4*7=28.
Hence, we write 30=4*7+2
We therefore have the equality:
$$\frac{30}{4}=\frac{4*7+2}{4}$$
c)We now use the fact that we can split up a sum in the numerator into a sum of fractions:
$$\frac{4*7+2}{4}=\frac{4*7}{4}+\frac{2}{4}=7+\frac{2}{4}$$
The last step follows since 4 is a common factor in both the numerator and denominator in the first fraction.
d) Hence we have shown:
$$\frac{3}{4}=\frac{1}{10}*(7+\frac{2}{4})$$
This can be rewritten as:
$$\frac{3}{4}=\frac{7}{10}+\frac{1}{10}(\frac{2}{4})=0.7+\frac{1}{10}(\frac{2}{4})$$
e) We will now work with the parenthesized 2/4.
Since 2<4, we multiply 2/4 by an appropriate version of 1:
$$\frac{2}{4}=1*\frac{2}{4}=\frac{10}{10}*\frac{2}{4}=\frac{1}{10}*(\frac{20}{4})$$
f) We note that 20=5*4, so we have:
$$\frac{2}{4}=\frac{1}{10}*(\frac{5*4}{4})=\frac{1}{10}*(5)=\frac{5}{10}$$
g) We now look back at the equation in d):
$$\frac{3}{4}=0.7+\frac{1}{10}(\frac{2}{4})$$
With the result from f), we have:
$$\frac{3}{4}=0.7+\frac{1}{10}(\frac{5}{10})=0.7+\frac{5}{100}=0.7+0.05=0.75$$

And that's our result..

Last edited: Oct 25, 2004
8. Oct 25, 2004

HallsofIvy

I'm confused about your question "WHY DONT I PUT A ZERO ON TOP FOR 4 INTO 2 ?" because I certainly WOULD put a 0 on top!

4 goes into 3 0 times so I would put a 0 before the decimal point and continue:

__0.
4) 3.0

Now, 4 divides into 30 (ignore the decimal point now- we've already taken care of it) 7 times: 4*7= 28

__0.7
4) 3.00
_28_
20
and now 4 divides into 20 5 times, evenly

__0.75_
4) 3.000
_2 8__
20
20

Of course, you don't HAVE to put the "0 on top" because 0, after all, means NOTHING! 0.75 is the same as .75 although I think you will find that "0.75" is preferred in "formal" writing.

9. Oct 25, 2004

roger

I'm confused about your question "WHY DONT I PUT A ZERO ON TOP FOR 4 INTO 2 ?" because I certainly WOULD put a 0 on top!

4 goes into 3 0 times so I would put a 0 before the decimal point and continue:

__0.
4) 3.0

Now, 4 divides into 30 (ignore the decimal point now- we've already taken care of it) 7 times: 4*7= 28

__0.7
4) 3.00
_28_
20
and now 4 divides into 20 5 times, evenly
this is where my concern was...
I thought it was 4 divides into 2 NOT 20.
So is that to say because the decimal point has been put down, it is really the calculation of 30 divided by 4 ?
Because if it had been 2 instead of 20, my question was if 4 does not go into 2, then why do I not put a zero on top to indicate this .And then I would have brought down a zero to make it 20 and proceed from there......
If I ever have to bring down a zero, where does it come from ?
__0.75_
4) 3.000
_2 8__
20
20

Of course, you don't HAVE to put the "0 on top" because 0, after all, means NOTHING! 0.75 is the same as .75 although I think you will find that "0.75" is preferred in "formal" writing

10. Oct 26, 2004

HallsofIvy

"Because if it had been 2 instead of 20, my question was if 4 does not go into 2, then why do I not put a zero on top to indicate this "

And my answer was you certainly can "put a zero on top to indicate this."

You don't HAVE to since ".75" is commonly understood as "0.75" although, in my opinion, 0.75 is better.

11. Oct 26, 2004

roger

It still doesn't answer my query because if I had done it my way, I would have got an answer of 0.705

This is wrong it should be 0.75 but I'm referring to the zero after the 7 not before the decimal point ?

Roger

12. Oct 26, 2004

faust9

Ok, I don't understand why you WANT to put the zero on the top?

Lets walk through an easy example the way you want to do it and the correct way:

116/4

___2
4)116

2*4=8

11-8=3

Now 3/4=0

___20
4)116

bring down the 6

36/4=9

___209
4)116

Does that look like the correct answer? No. The reason you don't put the 0 on the top is that you don't redivide the remainder from one step to the next.

Good luck.

13. Oct 27, 2004

dav2008

Your question is like asking "why don't you add a zero when you divide 30/5 and get 6, not 60"

You just don't. That's how it's done.


text:
# Solve exponential equation using the mean value theorem

Solve $3^x + 6^x = 5^x + 4^x$ using Lagrange's mean value theorem's.

I believe all I have to do is apply Lagrange's theorem on some function on a given interval, but I know neither the function to apply on nor the interval.

So, please, help me if you know how.

You are trying to solve: $6^x-5^x=4^x-3^x$. The mean value theorem says that $6^x-5^x=xa^{x-1}$ for some $a\in [5,6]$, and that $4^x-3^x=xb^{x-1}$ for some $b\in [3,4]$.
So, given that you known $a\in [5,6]$ and $b\in [3,4]$, when is it possible that:
$$xa^{x-1}=xb^{x-1}?$$


text:
# Energy of neutral pion in proton's rest mass frame

Hello, my problem is as follows
For the decay $p\rightarrow e^+\pi ^0$ show that the energy of the pion in the proton’s rest
frame (i.e., the lab frame) is

$E_{\pi}=\frac{m_p^2+m_{\pi}^2-m_e^2}{2m_p}$
I've tried finding the invariant mass of the positron and pion as follows

$M^2=(E_e+E_{\pi})^2-(\mathbf{p_e}+\mathbf{p_{\pi}})\\$
$=E_e^2+E_{\pi}^2+2E_eE_{\pi}-p_e^2-p_{\pi}^2-2\mathbf{p_ep_{\pi}}\\$
$=m_e^2+m_{\pi}^2-2(E_eE_{\pi}+\mathbf{p_ep_{\pi}})\\$
$=m_e^2+m_{\pi}^2-2(E_eE_{\pi}+p_ep_{\pi}\textrm{cos}\theta)\\$
And this is presumably equal to the mass of the proton so

$m_e^2+m_{\pi}^2-2(E_eE_{\pi}+p_ep_{\pi}\textrm{cos}\theta)=m_p^2$
At this point I cannot see how to get any closer to the given answer. Any help with how to proceed from here, or advice on where I may have already gone wrong, would be greatly appreciated.

Thanks

DEvens
$m_e^2+m_{\pi}^2-2(E_eE_{\pi}+p_ep_{\pi}\textrm{cos}\theta)=m_p^2$


text:
# Group-level random effect correlated with within-group independent variable

Dear StackExchange community,

# Question:

According to, e.g., Wikipedia, random effects models assume "that the individual specific effects are uncorrelated with the independent variables."

I understand that to mean that, in the context of a shared frailty model (see below), a control variable that varies on the group level must not be correlated with the group level frailty. But:

Question: are variables that vary within groups allowed to be correlated with the group-level frailty?

Example: imagine that

• one is modeling the time for siblings to reach their terminal educational degree,
• the control variables include dummy variables for each kind of terminal degree (BSc, MSc, etc.),
• the shared frailty at the sibling level represents the sibling-level speed, i.e., tendency to finish degrees faster

Imagine also, that people who run through their education at a faster pace also tend to continue into higher levels of education. Faster families will be overrepresented among higher degrees. Therefore, the researcher wants to correct for the sibling-level speed. But at the same time, the degree dummies will be correlated with the frailty.

Would the shared frailty model be valid in such a case? Would the model correctly estimate the associations between types of terminal degrees and the the time to completion?

# Background:

I would like to estimate a shared frailty model (see these lecture notes; the following draws on this document). It is a a random effects survival analysis model where unobserved heterogeneity between groups are modelled as a so-called shared random effect common to each group. It takes the form:

$h(t_{ij})=h_0(t)\exp(\beta' \mathbf x_{ij} + \varphi'w_i)$,

where $h(t_{ij})$ is the hazard of individual $i$ in group $j$ at time $t$ given the baseline hazard, $h_0$, the control variables for each individual, $\mathbf x_{ij}$, and the shared frailty for group $i$, $w_i$.

The $w_i$ are assumed to be drawn from an independent sample from a distribution with mean 0 and variance 1.

I hope my question makes sense and I look forward to hear your answers.


text:
Schmidt Decomposable Products of Projections

Schmidt Decomposable Products of Projections We characterize operators $$T=PQ$$ T = P Q (P, Q orthogonal projections in a Hilbert space $${\mathcal {H}}$$ H ) which have a singular value decomposition. A spatial characterizations is given: this condition occurs if and only if there exist orthonormal bases $$\{\psi _n\}$$ { ψ n } of R(P) and $$\{\xi _n\}$$ { ξ n } of R(Q) such that $$\langle \xi _n,\psi _m\rangle =0$$ ⟨ ξ n , ψ m ⟩ = 0 if $$n\ne m$$ n ≠ m . Also it is shown that this is equivalent to $$A=P-Q$$ A = P - Q being diagonalizable. Several examples are studied, relating Toeplitz, Hankel and Wiener–Hopf operators to this condition. We also examine the relationship with the differential geometry of the Grassmann manifold of underlying the Hilbert space: if $$T=PQ$$ T = P Q has a singular value decomposition, then the generic parts of P and Q are joined by a minimal geodesic with diagonalizable exponent. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Integral Equations and Operator Theory Springer Journals

Schmidt Decomposable Products of Projections

, Volume 89 (4) – Oct 11, 2017
24 pages

/lp/springer_journal/schmidt-decomposable-products-of-projections-04JMR7wHXy
Publisher
Springer Journals
Subject
Mathematics; Analysis
ISSN
0378-620X
eISSN
1420-8989
D.O.I.
10.1007/s00020-017-2402-x
Publisher site
See Article on Publisher Site

Abstract

We characterize operators $$T=PQ$$ T = P Q (P, Q orthogonal projections in a Hilbert space $${\mathcal {H}}$$ H ) which have a singular value decomposition. A spatial characterizations is given: this condition occurs if and only if there exist orthonormal bases $$\{\psi _n\}$$ { ψ n } of R(P) and $$\{\xi _n\}$$ { ξ n } of R(Q) such that $$\langle \xi _n,\psi _m\rangle =0$$ ⟨ ξ n , ψ m ⟩ = 0 if $$n\ne m$$ n ≠ m . Also it is shown that this is equivalent to $$A=P-Q$$ A = P - Q being diagonalizable. Several examples are studied, relating Toeplitz, Hankel and Wiener–Hopf operators to this condition. We also examine the relationship with the differential geometry of the Grassmann manifold of underlying the Hilbert space: if $$T=PQ$$ T = P Q has a singular value decomposition, then the generic parts of P and Q are joined by a minimal geodesic with diagonalizable exponent.

Journal

Integral Equations and Operator TheorySpringer Journals

Published: Oct 11, 2017

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text:
# Electric field

Electric field lines emanating from a point positive electric charge suspended over an infinite sheet of conducting material.

An electric field is a force field that surrounds electric charges that attracts or repels other electric charges.[1] [2] Mathematically the electric field is a vector field that associates to each point in space the force, called the Coulomb force, that would be experienced per unit of charge, by an infinitesimal test charge at that point.[3] Electric fields are created by electric charges, and by time-varying magnetic fields. Electric fields are important in many areas of physics, and are exploited practically in electrical devices and electronics technology, on a microscopic scale, the electric field is responsible for the attractive force between the atomic nucleus and electrons that holds atoms together, and the forces between atoms that cause chemical bonding. The electric field and the magnetic field together form the electromagnetic force, one of the four fundamental forces of nature.

## Definition of an electric field

An electric charge, such as an electron, creates and is surrounded by an electric force field, this field will exert a force, either repulsive or attractive, on any other electric charge placed in its vicinity.[4] The electric field ${\displaystyle \mathbf {E(x)} }$ at a point ${\displaystyle \mathbf {x} }$ is defined as the (vector) force, ${\displaystyle \mathbf {F} }$, that would be exerted on a stationary test particle of unit positive charge placed at that point. The force on a particle of charge ${\displaystyle q}$ will be, given by:${\displaystyle \mathbf {F} =q\mathbf {E} }$. with SI units of newtons per coulomb (N⋅C−1) or, equivalently, volts per metre (V⋅m−1), which in terms of SI base units are kg⋅m⋅s−3⋅A−1. If the charge is moving, there is a second force field to consider: the magnetic field, the force on a test charge from both fields combined is called the Lorentz force.[5]

## Sources of electric field

### Causes and description

Electric fields are caused by electric charges, described by Gauss's law, [6] or varying magnetic fields, described by Faraday's law of induction.[7] Together, these laws are enough to define the behavior of the electric field as a function of charge repartition and magnetic field. However, since the magnetic field is described as a function of electric field, the equations of both fields are coupled and together form Maxwell's equations that describe both fields as a function of charges and currents.

In the special case of a steady state (stationary charges and currents), the Maxwell-Faraday inductive effect disappears, the resulting two equations (Gauss's law ${\displaystyle \nabla \cdot \mathbf {E} ={\frac {\rho }{\varepsilon _{0}}}}$ and Faraday's law with no induction term ${\displaystyle \nabla \times \mathbf {E} =0}$), taken together, are equivalent to Coulomb's law, written as ${\displaystyle {\boldsymbol {E}}({\boldsymbol {r}})={1 \over 4\pi \varepsilon _{0}}\int \rho ({\boldsymbol {r'}}){{\boldsymbol {r}}-{\boldsymbol {r'}} \over |{\boldsymbol {r}}-{\boldsymbol {r'}}|^{3}}d^{3}r'}$ for a charge density ${\displaystyle \mathbf {\rho } (\mathbf {r} )}$ (${\displaystyle \mathbf {r} }$ denotes the position in space).[8] Notice that ${\displaystyle \varepsilon _{0}}$, the permitivity of vacuum, must be substituted if charges are considered in non-empty media.

### Continuous vs. discrete charge representation

The equations of electromagnetism are best described in a continuous description. However, charges are sometimes best described as discrete points; for example, some models may describe electrons as point sources where charge density is infinite on an infinitesimal section of space.

A charge ${\displaystyle q}$ located at ${\displaystyle \mathbf {r_{0}} }$ can be described mathematically as a charge density ${\displaystyle \rho (\mathbf {r} )=q\delta (\mathbf {r-r_{0}} )}$, where the Dirac delta function (in three dimensions) is used. Conversely, a charge distribution can be approximated by many small point charges.

## Superposition principle

Electric fields satisfy the superposition principle, because Maxwell's equations are linear, as a result, if ${\displaystyle \mathbf {E} _{1}}$ and ${\displaystyle \mathbf {E} _{2}}$ are the electric fields resulting from distribution of charges ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}}$, a distribution of charges ${\displaystyle \rho _{1}+\rho _{2}}$ will create an electric field ${\displaystyle \mathbf {E} _{1}+\mathbf {E} _{2}}$; for instance, Coulomb's law is linear in charge density as well.

This principle is useful to calculate the field created by multiple point charges. If charges ${\displaystyle q_{1},q_{2},...,q_{n}}$ are stationary in space at ${\displaystyle \mathbf {r} _{1},\mathbf {r} _{2},...\mathbf {r} _{n}}$, in the absence of currents, the superposition principle proves that the resulting field is the sum of fields generated by each particle as described by Coulomb's law:

${\displaystyle \mathbf {E} (\mathbf {r} )=\sum _{i=1}^{N}\mathbf {E} _{i}(\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}q_{i}{\frac {\mathbf {r} -\mathbf {r} _{i}}{|\mathbf {r} -\mathbf {r} _{i}|^{3}}}}$

## Electrostatic fields

Illustration of the electric field surrounding a positive (red) and a negative (blue) charge
Experiment illustrating electric field lines. An electrode connected to an electrostatic induction machine is placed in an oil-filled container. Considering that oil is a dielectric medium, when there is current through the electrode, the particles arrange themselves so as to show the force lines of the electric field.

Electrostatic fields are E-fields which do not change with time, which happens when charges and currents are stationary; in that case, Coulomb's law fully describes the field.

### Electric potential

If a system is static, such that magnetic fields are not time-varying, then by Faraday's law, the electric field is curl-free; in this case, one can define an electric potential, that is, a function ${\displaystyle \Phi }$ such that ${\displaystyle \mathbf {E} =-\nabla \Phi }$.[9] This is analogous to the gravitational potential.

### Parallels between electrostatic and gravitational fields

Coulomb's law, which describes the interaction of electric charges:

${\displaystyle \mathbf {F} =q\left({\frac {Q}{4\pi \varepsilon _{0}}}{\frac {\mathbf {\hat {r}} }{|\mathbf {r} |^{2}}}\right)=q\mathbf {E} }$

is similar to Newton's law of universal gravitation:

${\displaystyle \mathbf {F} =m\left(-GM{\frac {\mathbf {\hat {r}} }{|\mathbf {r} |^{2}}}\right)=m\mathbf {g} }$

(where ${\displaystyle \mathbf {\hat {r}} =\mathbf {\frac {r}{|r|}} }$).

This suggests similarities between the electric field E and the gravitational field g, or their associated potentials. Mass is sometimes called "gravitational charge" because of that similarity.[citation needed]

Electrostatic and gravitational forces both are central, conservative and obey an inverse-square law.

### Uniform fields

A uniform field is one in which the electric field is constant at every point, it can be approximated by placing two conducting plates parallel to each other and maintaining a voltage (potential difference) between them; it is only an approximation because of boundary effects (near the edge of the planes, electric field is distorted because the plane does not continue). Assuming infinite planes, the magnitude of the electric field E is:

${\displaystyle E=-{\frac {\Delta \phi }{d}}}$

where Δϕ is the potential difference between the plates and d is the distance separating the plates. The negative sign arises as positive charges repel, so a positive charge will experience a force away from the positively charged plate, in the opposite direction to that in which the voltage increases; in micro- and nano-applications, for instance in relation to semiconductors, a typical magnitude of an electric field is in the order of 106 V⋅m−1, achieved by applying a voltage of the order of 1 volt between conductors spaced 1 µm apart.

## Electrodynamic fields

Electrodynamic fields are E-fields which do change with time, for instance when charges are in motion.

The electric field cannot be described independently of the magnetic field in that case. If A is the magnetic vector potential, defined so that ${\displaystyle \mathbf {B} =\nabla \times \mathbf {A} }$, one can still define an electric potential ${\displaystyle \Phi }$ such that:

${\displaystyle \mathbf {E} =-\nabla \Phi -{\frac {\partial \mathbf {A} }{\partial t}}}$

One can recover Faraday's law of induction by taking the curl of that equation

[10]
${\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial (\nabla \times \mathbf {A} )}{\partial t}}=-{\frac {\partial \mathbf {B} }{\partial t}}}$

which justifies, a posteriori, the previous form for E.

## Energy in the electric field

The total energy per unit volume stored by the electromagnetic field is[11]

${\displaystyle u_{EM}={\frac {\varepsilon }{2}}|\mathbf {E} |^{2}+{\frac {1}{2\mu }}|\mathbf {B} |^{2}}$

where ε is the permittivity of the medium in which the field exists, ${\displaystyle \mu }$ its magnetic permeability, and E and B are the electric and magnetic field vectors.

As E and B fields are coupled, it would be misleading to split this expression into "electric" and "magnetic" contributions. However, in the steady-state case, the fields are no longer coupled (see Maxwell's equations), it makes sense in that case to compute the electrostatic energy per unit volume:

${\displaystyle u_{ES}={\frac {1}{2}}\varepsilon |\mathbf {E} |^{2}\,,}$

The total energy U stored in the electric field in a given volume V is therefore

${\displaystyle U_{ES}={\frac {1}{2}}\varepsilon \int _{V}|\mathbf {E} |^{2}\,\mathrm {d} V\,,}$

## Further extensions

### Definitive equation of vector fields

In the presence of matter, it is helpful to extend the notion of the electric field into three vector fields:[12]

${\displaystyle \mathbf {D} =\varepsilon _{0}\mathbf {E} +\mathbf {P} \!}$

where P is the electric polarization – the volume density of electric dipole moments, and D is the electric displacement field. Since E and P are defined separately, this equation can be used to define D, the physical interpretation of D is not as clear as E (effectively the field applied to the material) or P (induced field due to the dipoles in the material), but still serves as a convenient mathematical simplification, since Maxwell's equations can be simplified in terms of free charges and currents.

### Constitutive relation

The E and D fields are related by the permittivity of the material, ε.[13][12]

For linear, homogeneous, isotropic materials E and D are proportional and constant throughout the region, there is no position dependence: For inhomogeneous materials, there is a position dependence throughout the material:

${\displaystyle \mathbf {D(r)} =\varepsilon \mathbf {E(r)} }$

For anisotropic materials the E and D fields are not parallel, and so E and D are related by the permittivity tensor (a 2nd order tensor field), in component form:

${\displaystyle D_{i}=\varepsilon _{ij}E_{j}}$

For non-linear media, E and D are not proportional. Materials can have varying extents of linearity, homogeneity and isotropy.

## References

1. ^ Purcell and Morin, Harvard University. (2013). Electricity and Magnetism, 820pages (3rd ed.). Cambridge University Press, New York. ISBN 978-1-107-01402-2.
2. ^ Browne, p 225: "... around every charge there is an aura that fills all space. This aura is the electric field due to the charge, the electric field is a vector field... and has a magnitude and direction."
3. ^ Richard Feynman (1970). The Feynman Lectures on Physics Vol II. Addison Wesley Longman. ISBN 978-0-201-02115-8.
4. ^ Purcell, p5-7: "... the building blocks of matter possess a quality called charge. Two important aspects of charge are conservation and quantitization. The electric force between two charges is given by Coulomb's law".
5. ^ Purcell, p278: "A charge that is moving parallel to a current of other charges experiences a force perpendicular to its own velocity." Additional discussion on p 237-238: Chapter section 5.2: Magnetic Forces.
6. ^ Purcell, p 25: "Gauss's Law: the flux of the electric field E through any closed surface... equals 1/e times the total charge enclosed by the surface."
7. ^ Purcell, p 356: "Faraday's Law of Induction."
8. ^ Purcell, p7: "... the interaction between electric charges at rest is described by Coulomb's Law: two stationary electric charges repel or attract each other with a force proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.
9. ^ gwrowe (8 October 2011). "Curl & Potential in Electrostatics". physicspages.com. Retrieved 21 January 2017.
10. ^ Huray, Paul G. (2009). Maxwell's Equations. Wiley-IEEE. p. 205. ISBN 0-470-54276-4.
11. ^ Introduction to Electrodynamics (3rd Edition), D.J. Griffiths, Pearson Education, Dorling Kindersley, 2007, ISBN 81-7758-293-3
12. ^ a b Electromagnetism (2nd Edition), I.S. Grant, W.R. Phillips, Manchester Physics, John Wiley & Sons, 2008, ISBN 978-0-471-92712-9
13. ^ Electricity and Modern Physics (2nd Edition), G.A.G. Bennet, Edward Arnold (UK), 1974, ISBN 0-7131-2459-8
*Purcell, Edward; Morin, David (2010). ELECTRICITY AND MAGNETISM (3rd ed.). Cambridge University Press, New York. ISBN 978-1-107-01402-2.
*Browne, Michael (2011). PHYSICS FOR ENGINEERING AND SCIENCE (2nd ed.). McGraw-Hill, Schaum, New York. ISBN 978-0-07-161399-6.


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##### 9.  A thin P-type silicon sample is unifo...

9.  A thin P-type silicon sample is uniformly illuminated with light which generates excess carriers.  The recombination rate is directly proportional to

(A) The minority carrier mobility

(B) The minority carrier recombination lifetime

(C) The majority carrier concentration

(D) The excess minority carrier concentration

Hint: