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Location of Repository

## Magnetic Catalysis and Quantum Hall Ferromagnetism in Weakly Coupled Graphene

### Abstract

We study the realization in a model of graphene of the phenomenon whereby the tendency of gauge-field mediated interactions to break chiral symmetry spontaneously is greatly enhanced in an external magnetic field. We prove that, in the weak coupling limit, and where the electron-electron interaction satisfies certain mild conditions, the ground state of charge neutral graphene in an external magnetic field is a quantum Hall ferromagnet which spontaneously breaks the emergent U(4) symmetry to U(2)XU(2). We argue that, due to a residual CP symmetry, the quantum Hall ferromagnet order parameter is given exactly by the leading order in perturbation theory. On the other hand, the chiral condensate which is the order parameter for chiral symmetry breaking generically obtains contributions at all orders. We compute the leading correction to the chiral condensate. We argue that the ensuing fermion spectrum resembles that of massive fermions with a vanishing U(4)-valued chemical potential. We discuss the realization of parity and charge conjugation symmetries and argue that, in the context of our model, the charge neutral quantum Hall state in graphene is a bulk insulator, with vanishing longitudinal conductivity due to a charge gap and Hall conductivity vanishing due to a residual discrete particle-hole symmetry.Comment: 35 page

Topics: High Energy Physics - Theory, Condensed Matter - Mesoscale and Nanoscale Physics
Year: 2011
DOI identifier: 10.1007/JHEP07(2011)037
OAI identifier: oai:arXiv.org:1104.4714


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# Mean and Variance of a Set of Sequences

I am trying to find the probability distribution (like mean and variance) of a set of sequences which have different length. I came into different measurement models that quantify the distance between difference sequences, but I cannot find something on the probability distribution of sequences. Is there anyway to extract the parameters for the distribution of a set of sequences?

There are a series of descriptive statistics of a set of sequences, which you can easily obtain for instance with the TraMineR R package (You find a lot of documentation on http://mephisto.unige.ch/traminer).

Here are a few examples:

• the relative frequency of each distinct sequence pattern;
• the successive cross-sectional state distributions at the successive positions;
• the sequence of cross-sectional entropies (i.e., the diversity of the state at the successive positions)
• longitudinal characteristics for each sequence such as longitudinal entropy, complexity index, ...

With TraMineR you can also compute pairwise dissimilarities, and based on these dissimilarities:

• the medoid (sequence with smallest sum of distances to all other sequences)
• other representative sequences, such as sequence with densest neighborhood.
• discrepancy measure (pseudo variance)

All this is indeed for categorical sequences, for which the mean does not make sense.

format: plain


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# Arc Length of $\ln\left(x\right)$

Calculus Level 5

Let $f(n)$ denote the arc length of the curve $y=\ln\left(x\right)$ in the interval $x\in\left [1,n\right ]$. Find $\lim_{n\to\infty}\big(n-f(n)\big).$ If your answer is of the form $\ln\left(\sqrt{a}-b\right)+\sqrt{c}-d$, where $a$, $b$, $c$, and $d$ are non-negative integers and $a$ and $c$ are not perfect squares, find $a+b+c+d$.

×


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# All Bicycle Cards

Well, the last puzzle I made was solved in about an hour, during which time I was busily creating more hints.

So this time I'm going to be less forthcoming with the giveaways and see what you cryptography nerds can come up with all on your own.

Here ya go:

• Here's a list of missing cards if it helps anyone: $$\begin{array}{|c|c|c|c|c|} 2 & H & D & C & S \\ 3 & & D & C & \\ 4 & & D & C & S \\ 5 & H & D & C & \\ 6 & H & D & C & \\ 7 & & D & C & S \\ 8 & & D & C & \\ 9 & H & & & \\ T & & D & & S \\ J & & D & C & \\ Q & & D & C & \\ K & H & D & & \\ A & & D & & \end{array}$$ Also interesting, there's only one diamond, a 9. Might have missed something here, they all start running together after a while... – Guest Feb 6 '18 at 21:59
• Looks like you made it an hour this time. – internet_user Feb 6 '18 at 22:25

The decoded message ...

... can be found at the bottom of this post. :)

First, let's transcribe the card layout, so that we can tackle it more easily:

        9S QH JH JS JH 9C AH JS JH 9C
TC TC TC 9C 9S 3S 5S JH KS 9C
8S KH 9C 5S JH 8S 5S 3H JH 9C
7H AS 9C 9S QH JH 9C 6S 8S JS
3H 4H 9D 9C 9S QH 8S KS JH 9C
6S QH 8S 9C QS AS 4H JH JS KS
9S AH AS 4H 9C 8H 7H AS AH JS
3S KC 9C 9S QH 8S KS JH 9C 6S
QH 8S 9C 4H 8S AS AC 9S 9C QS
AS 4H JH JS KS 9S AH AS 4H 9C
8H 7H AS AH JS 3S KC 9C AH AS
4H 9C 9S QH 8S KS JH 9C 6S QH
8S 9C 4H 8S AS AC 9S 9C TH AS
8S 6S 9C 6S QH AH 9S 9C 8H 7H
AS AH JS 3S 9C 7H KS QC

Frequency analysis tells us ...

... that there are only 24 different cards, distributed like this:

      25 × 9C              8 × AH 4H           3 × TC 8H 5S
12 × JH AS 8S        6 × KS 6S           2 × QS KC 3H
11 × 9S              5 × 7H              1 × TH QC KH 9D
9 × QH JS           4 × 3S

The most frequent card by far is the Nine of Clubs.

Let's make some assumptions:

The cards represent a cipher. The reading direction is the usual western one: left to right, top to bottom. The cipher is a monoalphabetic substitution cipher, because the 26 letters can easily be represented with 24 cards.

The most frequent symbol in English texts without punctuation or spaces is usually E. It is space if spaces are also encoded. Here, 9C is never followed by another 9C, and its occurrence spreads out evenly. So let's assume that 9C is a space, which gives this pretty believable layout of words:

xxxxx xxx xxx xxxxx xx xxxxxx xx xxx xxxxxx xxxxx xxx xxxxxxxxxx xxxxxxx xxxxx xxx xxxxx xxxxxxxxxx xxxxxxx xxx xxxxx xxx xxxxx xxxx xxxx xxxxxx xxx

Let's rearrange the input according to these assumptions:

9S QH JH JS JH    AH JS JH    TC TC TC
9S 3S 5S JH KS    8S KH    5S JH 8S 5S 3H JH
7H AS    9S QH JH    6S 8S JS 3H 4H 9D
9S QH 8S KS JH    6S QH 8S
QS AS 4H JH JS KS 9S AH AS 4H
8H 7H AS AH JS 3S KC
9S QH 8S KS JH    6S QH 8S    4H 8S AS AC 9S
QS AS 4H JH JS KS 9S AH AS 4H
8H 7H AS AH JS 3S KC    AH AS 4H
9S QH 8S KS JH    6S QH 8S
4H 8S AS AC 9S    TH AS 8S 6S
6S QH AH 9S    8H 7H AS AH JS 3S    7H KS QC

There are several repeated words, for example 9S QH 8S KS JH, 6S QH 8S, QS AS 4H JH JS KS 9S AH AS 4H and 8H 7H AS AH JS 3S (KC). The additional (KC) could be an s, because it is a plural or a present-tense verb in third person singular, or it could be punctuation. TC TC TC stands out as a word with three identical letters. Aaa!

The previous frequency analysis doesn't make any letter stand out as a prime candidate for E. We could try several smaller words, for example the frequent the and and. Let's focus on QS AS 4H JH JS KS 9S AH AS 4H, which has AS and AHrepeated. This pattern can be found in many words, the most frequent being understand, particular, conversion, despatches, congestion, intolerant and convulsion. Then the list trickles out into the likes of Hanoverian, Manchurian and Talmudical.

The first one, understand, is already a hit and we get:

There are  • • •  types of people in the world: those who understand binary, those who don't understand binary, and those who don't know what binary is.

That's a variation of a popular joke. The punctuation was placed using common rules. But what should the three TCs represent? They must all be the same character, and the joke suggests digits. Maybe 111, because someone who uses unary numbers doesn't know what binary is?

Let's try to find some structure.

The alphabet used for decoding can be arranged according to the suit and rank of each card:

           A 2 3 4 5 6 7 8 9 T J Q K

C    '               _ 1   . ,
D                    :
H    A   L D     I B   K E H F
S    N   Y   P W   O T   R U S

Here, the _ marks the space character; blank entries aren't used in the encoded message. Unfortunately, that doesn't look very regular.

But ...

... if we arrange the ranks of cards in alphabetical order (Ace, Eight, Five, Four, Jack, King, Nine, Queen, Seven, Six, Ten, Three, Two, the table looks like this:

           A 8 5 4 J K 9 Q 7 6 T 3 2

C    '         , _ .     1
D                :
H    A B   D E F   H I   K L
S    N O P   R S T U   W   Y

And that does look familiar: It's the ASCII table, starting from the apostrophe and with the hyphen used as space:
           A 8 5 4 J K 9 Q 7 6 T 3 2

C    ' ( ) * + , - . / 0 1 2 3
D    4 5 6 7 8 9 : ; < = > ? @
H    A B C D E F G H I J K L M
S    N O P Q R S T U V W X Y Z

And indeed, the missing digits are 111.

So, here goes:

There are 111 types of people in the world: those who understand binary, those who don't understand binary, and those who don't know what binary is.

• I considered replying that this is not the correct answer, but decided that that would have been cruel. Very well done! Did you notice my (one and only) hint, the title? All Bicycle Cards, for alphabetizing the suits/values. – Chowzen Feb 7 '18 at 14:46
• Can someone tell me why it is 111 and not 11 (which is 3 in binary)? – Albino Feb 7 '18 at 14:48
• No, I didn't see the hint. I found out about the alhpabetical ordering after seeing that all the Hearts were letters from A to M and all the Spades were letters from N to Z and that the pairs EF and RS were in corresponding columns. – M Oehm Feb 7 '18 at 15:00
• @Albino This is from the answer: "Maybe 111, because someone who uses unary numbers doesn't know what binary is?" – Chowzen Feb 7 '18 at 15:02


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# How do these conditional probability tables work (resources)?

I've taken some probability classes before, but it's been a while. I'm looking for (1) some resource -- website, paper, software, book -- etc on conditional probabilities. Something which accessible to a junior/senior level undergrad. Specifically, I'm interested in the relationships described by the following example:

Suppose that the nodes $$A$$ and $$B$$ are independent, and the node $$C$$ depends on both (I understand this may be an example of what's called a directed acyclic graph, if that's important).
The type of dependencies associated with the nodes I described above rely on truth tables. For example, $$A$$ and $$B$$'s probability distributions:

$$D(A) = \begin{array}{c|c} \text{True} & \text{False} \\ \hline 0.3 & 0.7 \end{array}, \quad D(B) = \begin{array}{c|c} \text{True} & \text{False} \\ \hline 0.01 & 0.99 \end{array}$$

And $$C$$'s (in percentages)

Then, without observable evidence, the probability that $$C$$ is FALSE* is $$\approx 0.9688$$.
How does one arrive at that value?
Furthermore, how would I generalize this behavior to a larger graph, with possibly more complicated stuff like $$A$$ and $$B$$ not being independent.

• I´m not sure about the topic here. But I´ve calculated that the probability C is true is $\approx 1-0.9688=0.0312$ – callculus Sep 28 '18 at 14:00
• Have you checked your solution? – callculus Sep 28 '18 at 14:20
• Yep, sorry about that, I meant $P(C)$ is false. So you're right. – Zduff Sep 28 '18 at 14:30
• I have no recommendation for liturature. I can only show how to calculate $P(\overline C)$. – callculus Sep 28 '18 at 14:33
• Yep, that's all I need, if you don't mind. – Zduff Sep 28 '18 at 14:39

To calculate the probability that $$C$$ is false we calculate all combinations of $$A$$ und $$B$$ which lead to $$\overline C$$.

$$\overline C$$ is the complementary event of $$C$$.

$$P(\overline C)=P(A)\cdot P(B|A)\cdot P(\overline C|A\cap B)+P(\overline A)\cdot P(B|\overline A)\cdot P(\overline C|\overline A\cap B)+P( A)\cdot P(\overline B| A)\cdot P(\overline C|A\cap \overline B)+P(\overline A)\cdot P(\overline B|\overline A)\cdot P(\overline C|\overline A\cap \overline B)$$

The events are: $$A:=$$Event A is true, $$B:=$$Event B is true,$$\overline A:=$$Event A is false, $$\overline B:=$$Event B is false,$$C:=$$Event C is true, $$\overline C:=$$Event C is false

$$A$$ and $$B$$ are independent. Therefore $$P(B|A)=P(B)$$. This is the same for all combinations with the complementary events

$$P(\overline C)=P(A)\cdot P(B)\cdot P(\overline C|A\cap B)+P(\overline A)\cdot P(B)\cdot P(\overline C|\overline A\cap B)+P( A)\cdot P(\overline B)\cdot P(\overline C|A\cap \overline B)+P(\overline A)\cdot P(\overline B)\cdot P(\overline C|\overline A\cap \overline B)$$

And the conditional probability $$P(\overline C|A\cap B)=100 \%=1$$ is given in the table. This is the last row and third column. Similar for the combinations of the events A, B. Therefore

$$P(\overline C)=0.7\cdot 0.99\cdot 0.999+0.99\cdot 0.3\cdot 0.9+0.01\cdot 0.7\cdot 0.88+0.3\cdot 0.01\cdot 1 \approx 0.9688$$

And $$P(C)=1-P(\overline C)$$. This result can be also obtained by using the equivalent formula which is used to calculate $$P(\overline C)$$.


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# Harmonic series (mathematics)

infinite series of the reciprocals of the positive integers

In mathematics, the harmonic series is the divergent infinite series:

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots }$

Divergent means that as you add more terms the sum never stops getting bigger. It does not go towards a single finite value.

Infinite means that you can always add another term. There is no final term to the series.

Its name comes from the idea of harmonics in music: the wavelengths of the overtones of a vibrating string are 1/2, 1/3, 1/4, etc., of the string's fundamental wavelength. Apart from the first term, every term of the series is the harmonic mean of the terms either side of it. The phrase harmonic mean also comes from music.

## History

The fact that the harmonic series diverges was first proven in the 14th century by Nicole Oresme,[1] but was forgotten. Proofs were given in the 17th century by Pietro Mengoli,[2] Johann Bernoulli,[3] and Jacob Bernoulli.[4][5]

Harmonic sequences have been used by architects. In the Baroque period architects used them in the proportions of floor plans, elevations, and in the relationships between architectural details of churches and palaces.[6]

## Divergence

There are several well-known proofs of the divergence of the harmonic series. A few of them are given below.

### Comparison test

One way to prove divergence is to compare the harmonic series with another divergent series, where each denominator is replaced with the next-largest power of two:

{\displaystyle {\begin{aligned}&{}1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}+\cdots \\[12pt]\geq {}&1+{\frac {1}{2}}+{\frac {1}{\color {red}{\mathbf {4} }}}+{\frac {1}{4}}+{\frac {1}{\color {red}{\mathbf {8} }}}+{\frac {1}{\color {red}{\mathbf {8} }}}+{\frac {1}{\color {red}{\mathbf {8} }}}+{\frac {1}{8}}+{\frac {1}{\color {red}{\mathbf {16} }}}+\cdots \end{aligned}}}

Each term of the harmonic series is greater than or equal to the corresponding term of the second series, and therefore the sum of the harmonic series must be greater than or equal to the sum of the second series. However, the sum of the second series is infinite:

{\displaystyle {\begin{aligned}&{}1+\left({\frac {1}{2}}\right)+\left({\frac {1}{4}}\!+\!{\frac {1}{4}}\right)+\left({\frac {1}{8}}\!+\!{\frac {1}{8}}\!+\!{\frac {1}{8}}\!+\!{\frac {1}{8}}\right)+\left({\frac {1}{16}}\!+\!\cdots \!+\!{\frac {1}{16}}\right)+\cdots \\[12pt]={}&1+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+\cdots =\infty \end{aligned}}}

It follows (by the comparison test) that the sum of the harmonic series must be infinite as well. More precisely, the comparison above proves that

${\displaystyle \sum _{n=1}^{2^{k}}{\frac {1}{n}}\geq 1+{\frac {k}{2}}}$

for every positive integer k.

This proof, proposed by Nicole Oresme in around 1350, is considered to be a high point of medieval mathematics. It is still a standard proof taught in mathematics classes today.

### Integral test

Illustration of the integral test.

It is possible to prove that the harmonic series diverges by comparing its sum with an improper integral. Consider the arrangement of rectangles shown in the figure to the right. Each rectangle is 1 unit wide and 1/n units high, so the total area of the infinite number of rectangles is the sum of the harmonic series:

${\displaystyle {\begin{array}{c}{\text{area of}}\\{\text{rectangles}}\end{array}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots }$

The total area under the curve y = 1/x from 1 to infinity is given by a divergent improper integral:

${\displaystyle {\begin{array}{c}{\text{area under}}\\{\text{curve}}\end{array}}=\int _{1}^{\infty }{\frac {1}{x}}\,dx=\infty .}$

Since this area is entirely contained within the rectangles, the total area of the rectangles must be infinite as well. This proves that

${\displaystyle \sum _{n=1}^{k}{\frac {1}{n}}>\int _{1}^{k+1}{\frac {1}{x}}\,dx=\ln(k+1).}$

The generalization of this argument is known as the integral test.

## Rate of divergence

The harmonic series diverges very slowly. For example, the sum of the first 1043 terms is less than 100.[7] This is because the partial sums of the series have logarithmic growth. In particular,

${\displaystyle \sum _{n=1}^{k}{\frac {1}{n}}=\ln k+\gamma +\varepsilon _{k}\leq (\ln k)+1}$

where γ is the Euler–Mascheroni constant and εk ~ 1/2k which approaches 0 as k goes to infinity. Leonhard Euler proved both this and also that the sum which includes only the reciprocals of primes also diverges, that is:

${\displaystyle \sum _{p{\text{ prime }}}{\frac {1}{p}}={\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{11}}+{\frac {1}{13}}+{\frac {1}{17}}+\cdots =\infty .}$

## Partial sums

The first thirty harmonic numbers
n Partial sum of the harmonic series, Hn
expressed as a fraction decimal relative size
1 1 ~1

2 3 /2 ~1.5

3 11 /6 ~1.83333

4 25 /12 ~2.08333

5 137 /60 ~2.28333

6 49 /20 ~2.45

7 363 /140 ~2.59286

8 761 /280 ~2.71786

9 7129 /2520 ~2.82897

10 7381 /2520 ~2.92897

11 83711 /27720 ~3.01988

12 86021 /27720 ~3.10321

13 1145993 /360360 ~3.18013

14 1171733 /360360 ~3.25156

15 1195757 /360360 ~3.31823

16 2436559 /720720 ~3.38073

17 42142223 /12252240 ~3.43955

18 14274301 /4084080 ~3.49511

19 275295799 /77597520 ~3.54774

20 55835135 /15519504 ~3.59774

21 18858053 /5173168 ~3.64536

22 19093197 /5173168 ~3.69081

23 444316699 /118982864 ~3.73429

24 1347822955 /356948592 ~3.77596

25 34052522467 /8923714800 ~3.81596

26 34395742267 /8923714800 ~3.85442

27 312536252003 /80313433200 ~3.89146

28 315404588903 /80313433200 ~3.92717

29 9227046511387 /2329089562800 ~3.96165

30 9304682830147 /2329089562800 ~3.99499

The finite partial sums of the diverging harmonic series,

${\displaystyle H_{n}=\sum _{k=1}^{n}{\frac {1}{k}},}$

are called harmonic numbers.

The difference between Hn and ln n converges to the Euler–Mascheroni constant. The difference between any two harmonic numbers is never an integer. No harmonic numbers are integers, except for H1 = 1.[8]:p. 24[9]:Thm. 1

## Related series

### Alternating harmonic series

The first fourteen partial sums of the alternating harmonic series (black line segments) shown converging to the natural logarithm of 2 (red line).

The series

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-\cdots }$

is known as the alternating harmonic series. This series converges by the alternating series test. In particular, the sum is equal to the natural logarithm of 2:

${\displaystyle 1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-\cdots =\ln 2.}$

The alternating harmonic series, while conditionally convergent, is not absolutely convergent: if the terms in the series are systematically rearranged, in general the sum becomes different and, dependent on the rearrangement, possibly even infinite.

The alternating harmonic series formula is a special case of the Mercator series, the Taylor series for the natural logarithm.

A related series can be derived from the Taylor series for the arctangent:

${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+\cdots ={\frac {\pi }{4}}.}$

This is known as the Leibniz series.

### General harmonic series

The general harmonic series is of the form

${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{an+b}},}$

where a ≠ 0 and b are real numbers, and b/a is not zero or a negative integer.

By the limit comparison test with the harmonic series, all general harmonic series also diverge.

### p-series

A generalization of the harmonic series is the p-series (or hyperharmonic series), defined as

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{p}}}}$

for any real number p. When p = 1, the p-series is the harmonic series, which diverges. Either the integral test or the Cauchy condensation test shows that the p-series converges for all p > 1 (in which case it is called the over-harmonic series) and diverges for all p ≤ 1. If p > 1 then the sum of the p-series is ζ(p), i.e., the Riemann zeta function evaluated at p.

The problem of finding the sum for p = 2 is called the Basel problem; Leonhard Euler showed it is π2/6. The value of the sum for p = 3 is called Apéry's constant, since Roger Apéry proved that it is an irrational number.

### ln-series

Related to the p-series is the ln-series, defined as

${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{n(\ln n)^{p}}}}$

for any positive real number p. This can be shown by the integral test to diverge for p ≤ 1 but converge for all p > 1.

### φ-series

For any convex, real-valued function φ such that

${\displaystyle \limsup _{u\to 0^{+}}{\frac {\varphi \left({\frac {u}{2}}\right)}{\varphi (u)}}<{\frac {1}{2}},}$

the series

${\displaystyle \sum _{n=1}^{\infty }\varphi \left({\frac {1}{n}}\right)}$

is convergent.[source?]

### Random harmonic series

The random harmonic series

${\displaystyle \sum _{n=1}^{\infty }{\frac {s_{n}}{n}},}$

where the sn are independent, identically distributed random variables taking the values +1 and −1 with equal probability 1/2, is a well-known example in probability theory for a series of random variables that converges with probability 1. The fact of this convergence is an easy consequence of either the Kolmogorov three-series theorem or of the closely related Kolmogorov maximal inequality. Byron Schmuland of the University of Alberta further examined[10] the properties of the random harmonic series, and showed that the convergent series is a random variable with some interesting properties. In particular, the probability density function of this random variable evaluated at +2 or at −2 takes on the value 0.124999999999999999999999999999999999999999764..., differing from 1/8 by less than 10−42. Schmuland's paper explains why this probability is so close to, but not exactly, 1/8. The exact value of this probability is given by the infinite cosine product integral C2[11] divided by π.

### Depleted harmonic series

The depleted harmonic series where all of the terms in which the digit 9 appears anywhere in the denominator are removed can be shown to converge and its value is less than 80.[12] In fact, when all the terms containing any particular string of digits (in any base) are removed the series converges.

## Applications

The harmonic series can be counterintuitive. This is because it is a divergent series even though the terms of the series get smaller and go towards zero. The divergence of the harmonic series is the source of some paradoxes.

• The "worm on the rubber band".[13] Suppose that a worm crawls along an infinitely-elastic one-meter rubber band at the same time as the rubber band is uniformly stretched. If the worm travels 1 centimeter per minute and the band stretches 1 meter per minute, will the worm ever reach the end of the rubber band? The answer, counterintuitively, is "yes", for after n minutes, the ratio of the distance travelled by the worm to the total length of the rubber band is
${\displaystyle {\frac {1}{100}}\sum _{k=1}^{n}{\frac {1}{k}}.}$

Because the series gets arbitrarily large as n becomes larger, eventually this ratio must exceed 1, which implies that the worm reaches the end of the rubber band. However, the value of n at which this occurs must be extremely large: approximately e100, a number exceeding 1043 minutes (1037 years). Although the harmonic series does diverge, it does so very slowly.

• The Jeep problem asks how much total fuel is required for a car with a limited fuel-carrying capacity to cross a desert leaving fuel drops along the route. The distance the car can go with a given amount of fuel is related to the partial sums of the harmonic series, which grow logarithmically. And so the fuel required increases exponentially with the desired distance.

The block-stacking problem: blocks aligned according to the harmonic series bridges cleavages of any width.
• The block-stacking problem: given a collection of identical dominoes, it is possible to stack them at the edge of a table so that they hang over the edge of the table without falling. The counterintuitive result is that they can be stacked in a way that makes the overhang as large as you want. That is, provided there are enough dominoes.[13][14]
• A swimmer that goes faster each time they touch the wall of the pool. The swimmer starts crossing a 10-meter pool at a speed of 2 m/s, and with every crossing, another 2 m/s is added to the speed. In theory, the swimmer's speed is unlimited, but the number of pool crosses needed to get to that speed becomes very large; for instance, to get to the speed of light (ignoring special relativity), the swimmer needs to cross the pool 150 million times. Contrary to this large number, the time needed to reach a given speed depends on the sum of the series at any given number of pool crosses:
${\displaystyle {\frac {10}{2}}\sum _{k=1}^{n}{\frac {1}{k}}.}$

Calculating the sum shows that the time required to get to the speed of light is only 97 seconds.

## References

1. Oresme, Nicole (c. 1360). Quaestiones super Geometriam Euclidis [Questions concerning Euclid's Geometry].
2. Mengoli, Pietro (1650). "Praefatio [Preface]". Novae quadraturae arithmeticae, seu De additione fractionum [New arithmetic quadrature (i.e., integration), or On the addition of fractions]. Bologna: Giacomo Monti.
Let S denote the sum of the series. Group the terms of the series in triplets: S = 1 + (1/2 + 1/3 + 1/4) + (1/5 + 1/6 + 1/7) + (1/8 + 1/9 + 1/10) + … Since for x > 1, 1/x − 1 + 1/x + 1/x + 1 > 3/x, then S > 1 + 3/3 + 3/6 + 3/9 + … = 1 + 1 + 1/2 + 1/3 + … = 1 + S, which is false for any finite S. Therefore, the series diverges.
3. Bernoulli, Johann (1742). "Corollary III of De seriebus varia". Opera Omnia. Lausanne & Basel: Marc-Michel Bousquet & Co. vol. 4, p. 8.
4. Bernoulli, Jacob (1689). Propositiones arithmeticae de seriebus infinitis earumque summa finita [Arithmetical propositions about infinite series and their finite sums]. Basel: J. Conrad.
5. Bernoulli, Jacob (1713). Ars conjectandi, opus posthumum. Accedit Tractatus de seriebus infinitis [Theory of inference, posthumous work. With the Treatise on infinite series…]. Basel: Thurneysen. pp. 250–251.
From p. 250, prop. 16:
"XVI. Summa serei infinita harmonicè progressionalium, 1/1 + 1/2 + 1/3 + 1/4 + 1/5 &c. est infinita. Id primus deprehendit Frater:…"
[16. The sum of an infinite series of harmonic progression, 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + …, is infinite. My brother first discovered this…]
6. Hersey, George L. Architecture and Geometry in the Age of the Baroque. pp. 11–12, 37–51.
7. Sloane, N. J. A. (ed.). "Sequence A082912 (Sum of a(n) terms of harmonic series is > 10n)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
8. Julian Havil, Gamma: Exploring Euler’s Constant, Princeton University Press, 2009.
9. Thomas J. Osler, “Partial sums of series that cannot be an integer”, The Mathematical Gazette 96, November 2012, 515–519. https://www.jstor.org/stable/24496876?seq=1#page_scan_tab_contents
10. Schmuland, Byron (May 2003). "Random Harmonic Series" (PDF). American Mathematical Monthly. 110: 407–416. doi:10.2307/3647827.
11. Eric W. Weisstein, Infinite Cosine Product Integral at MathWorld.
12. Graham, Ronald; Knuth, Donald E.; Patashnik, Oren (1989), Concrete Mathematics (2nd ed.), Addison-Wesley, pp. 258–264, ISBN 978-0-201-55802-9
13. Sharp, R. T. (1954). "Problem 52: Overhanging dominoes" (PDF). Pi Mu Epsilon Journal. 1 (10): 411–412.


text:
# What are the auto-equivalences of the category of groups?

My question is motivated by Are the inner automorphisms the only ones that extend to every overgroup?

What are the auto-equivalences of the category of groups? What kind of structure do they form?

There are certainly some not-so-trivial examples (which are still trivial if one just focus on a single group), e.g. the functor send an abstract group G to the group G^op with the same underlying set with multiplication g,h ---> hg, where hg is the product of h and g in G. [The notation G^op will make sense if you think G as the category BG]

Suppose $F:\mathrm{Grp}\to\mathrm{Grp}$ is an equivalence. The object $\mathbb{Z}\in\mathrm{Grp}$ is a minimal generator (it is a generator, and no proper quotient is also a generator), and this property must be preserved by equivalences. Since there is a unique minimal generator, we can fix an isomorphism $\phi:\mathbb Z\to F(\mathbb Z)$. Now $F$ must preserve arbitrary coproducts, so for all cardinals $\kappa$, the isomorphism $\phi$ induces an isomorphism $\phi_\kappa:L_\kappa\to F(L_\kappa)$, where $L_\kappa$ is the free product of $\kappa$ copies of $\mathbb Z$. In particular, if $1$ is the trivial group, $\phi_0:1\to F(1)$ is an isomorphism.

Next pick a group $G\in\mathrm{Grp}$, and consider a free presentation $L_1\to L_0\to G\to1$, that is, an exact sequence with the $L\_i$ free. (For simplicity, we can take $L\_0=L(G)$ the free group on the underlying set of $G$, and $L_1$ to be the free group on the underlying subset the kernel of the obvious map $L_1\to G$; this eliminates choices) Since $F$ is an equivalence, we have another exact sequence $F(L_1)\to F(L_0)\to F(G)\to F(1)$. Fixing bases for $L_1$ and $L_0$ we can use $\phi$ to construct isomorphisms $L_i\to F(L_i)$ for both $i\in\{0,1\}$. Assuming we can prove the square commutes, one gets an isomorphism $\phi_G:G\to F(G)$—this should not be hard, I guess.

The usual arguments prove then in that case the assignment $G \mapsto \phi_G$ is a natural isomorphism between the identity functor of $\mathrm{Grp}$ and $F$.

• I think there are some subtleties in this approach. Let's say we've shown that F(Z) is abstractly isomorphic to Z. Since Z has a nontrivial automorphism, there are two isomorphisms ϕ we might choose. If we pick the wrong one, we won't be able to extend it to a natural isomorphism between the identity and F, for the reason I described in my answer. So there must be more work involved in constructing ϕ_G. – Reid Barton Dec 5 '09 at 6:00
• @Reid: automorphisms of the identity functor and autoequivalences of the category are different things, and your argument applies to the first. – Mariano Suárez-Álvarez Dec 5 '09 at 20:27
• What I am saying is that you are trying to construct an isomorphism between F and the identity, by extending an isomorphism between F(Z) and Z--but there are two isomorphisms of the latter type, and only one of the former, so you might go wrong! – Reid Barton Jan 6 '10 at 21:05
• (And there is only one isomorphism of the former type precisely because the identity functor has no automorphisms.) – Reid Barton Jan 6 '10 at 21:08
• It seems that if the isomorphism phi: Z -> F(Z) is the wrong one, then it wouldn't induce an isomorphism on free groups phi_k: L_k -> F(L_k). So doesn't that imply that the isomorphism Z -> F(Z) was the right one? – Tom Church Jan 7 '10 at 16:03

Just to make this a little more visible: for a proof that every autoequivalence of the category of groups is naturally isomorphic to the identity, see page 31 of Peter Freyd's book "Abelian Categories".

Your functor –op is naturally isomorphic to the identity, via the natural transformation G → Gop sending g to g-1. (If you asked about monoids, not groups, then –op would be a nontrivial autoequivalence.)

I'm pretty sure the category of groups has no nontrivial autoequivalences—more precisely the groupoid of autoequivalences is contractible. Certainly the identity autoequivalence has no natural automorphisms, because the only possible components of a natural automorphism on the object Z are +1 and -1, and by naturality the latter would have to extent to the "automorphism" sending x to x-1 on every group, which isn't a group homomorphism in general.

I don't know quite how to prove that every autoequivalence of the category of groups is naturally isomorphic to the identity. I would start with some easier examples, like the category of sets or the category of pointed sets, and then maybe try to use facts about cogroup objects in the category of groups to reduce the problem for groups to the case of pointed sets.

• This is on page 31 of Peter Freyd's "Abelian Categories" book. – Gonçalo Marques Dec 4 '09 at 19:44
• You should make that an answer to the question! – Reid Barton Dec 5 '09 at 3:19


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# Math Help - combinations

1. ## combinations

Suppose that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked for defects.
a. How many different samples can be chosen?
b. How many samples will contain at least one defective board?
c. What is the probability that a randomly chosen sample of five contains at least one defective board?

A) $\binom{40}{5} = \frac{40!}{5!(35!)} = 658\;008$ it didn't state something along the lines "{1,2,3,4,5} is different from {1,4,3,2,5}" so I am assuming its not a permutation question

B) $\binom{5}{1}+\binom{5}{2}+\binom{5}{3} = 25$

C) $P(E) = \frac{N(E)}{N(S)} = \frac{25}{658\;008}$

2. ## Re: combinations

Hello, Jonroberts74!

Suppose that three computer boards in a production run of forty are defective.
A sample of five is to be selected to be checked for defects.
a. How many different samples can be chosen?
b. How many samples will contain at least one defective board?
c. What is the probability that a randomly chosen sample of five contains at least one defective board?

$(a)\;\binom{40}{5} = \frac{40!}{5!(35!)} = 658\,008$ . Correct!

$(b)\;\binom{5}{1}+\binom{5}{2}+\binom{5}{3} = 25$ . This is incorrect.

We have: . $\begin{Bmatrix}\text{3 def.} & (D) \\ \text{37 good} & (G)\end{Bmatrix}$

This is what you want: . $\displaystyle\begin{Bmatrix} \text{1D,4G} & \text{2D,3G} & \text{3D,2G} \\ {3\choose1}{37\choose4} & {3\choose2}{37\choose3} & {3\choose3}{37\choose2} \end{Bmatrix}$

3. ## Re: combinations

Originally Posted by Soroban
Hello, Jonroberts74!

We have: . $\begin{Bmatrix}\text{3 def.} & (D) \\ \text{37 good} & (G)\end{Bmatrix}$

This is what you want: . $\displaystyle\begin{Bmatrix} \text{1D,4G} & \text{2D,3G} & \text{3D,2G} \\ {3\choose1}{37\choose4} & {3\choose2}{37\choose3} & {3\choose3}{37\choose2} \end{Bmatrix}$
$\binom{3}{1}\binom{37}{4}+\binom{3}{2}\binom{37}{3 }+\binom{3}{3}\binom{37}{2} = 222\,111$

C) $P(E) = \frac{N(E)}{N(S)} = \frac{222\,111}{658\,008} \approx 33.76\%$


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# powerEpiInt.default0: Power Calculation Testing Interaction Effect for Cox... In powerSurvEpi: Power and Sample Size Calculation for Survival Analysis of Epidemiological Studies

## Description

Power calculation testing interaction effect for Cox proportional hazards regression with two covariates for Epidemiological Studies. Both covariates should be binary variables. The formula takes into account the correlation between the two covariates.

## Usage

 1 powerEpiInt.default0(n, theta, p, psi, G, rho2, alpha = 0.05)

## Arguments

 n total number of subjects. theta postulated hazard ratio. p proportion of subjects taking the value one for the covariate of interest. psi proportion of subjects died of the disease of interest. G a factor adjusting the sample size. The sample size needed to detect an effect of a prognostic factor with given error probabilities has to be multiplied by the factor G when an interaction of the same magnitude is to be detected. rho2 square of the correlation between the covariate of interest and the other covariate. alpha type I error rate.

## Details

This is an implementation of the power calculation formula derived by Schmoor et al. (2000) for the following Cox proportional hazards regression in the epidemiological studies:

h(t|x_1, x_2)=h_0(t)\exp(β_1 x_1+β_2 x_2 + γ (x_1 x_2)),

where both covariates X_1 and X_2 are binary variables.

Suppose we want to check if the hazard ratio of the interaction effect X_1 X_2=1 to X_1 X_2=0 is equal to 1 or is equal to \exp(γ)=θ. Given the type I error rate α for a two-sided test, the power required to detect a hazard ratio as small as \exp(γ)=θ is

power=Φ≤ft(-z_{1-α/2}+√{\frac{n}{G}[\log(θ)]^2 p (1-p) ψ (1-ρ^2)}\right),

where ψ is the proportion of subjects died of the disease of interest, and

ρ=corr(X_1, X_2)=(p_1-p_0)\times √{\frac{q(1-q)}{p(1-p)}},

and p=Pr(X_1=1), q=Pr(X_2=1), p_0=Pr(X_1=1|X_2=0), and p_1=Pr(X_1=1 | X_2=1), and

G=\frac{[(1-q)(1-p_0)p_0+q(1-p_1)p_1]^2}{(1-q)q (1-p_0)p_0 (1-p_1) p_1}.

If X_1 and X_2 are uncorrelated, we have p_0=p_1=p leading to 1/[(1-q)q]. For q=0.5, we have G=4.

## Value

The power of the test.

## References

Schmoor C., Sauerbrei W., and Schumacher M. (2000). Sample size considerations for the evaluation of prognostic factors in survival analysis. Statistics in Medicine. 19:441-452.

powerEpiInt.default1, powerEpiInt2
 1 2 3 4  # Example at the end of Section 4 of Schmoor et al. (2000). powerEpiInt.default0(n = 184, theta = 3, p = 0.61, psi = 139 / 184, G = 4.79177, rho2 = 0.015^2, alpha = 0.05)


text:
# How do you multiply xsqrt(10x)*7sqrt(15x)?

Mar 3, 2018

The result is $35 {x}^{2} \sqrt{6}$.

#### Explanation:

Multiplication: $\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$

Simplification: $\sqrt{{a}^{2}} = a$

For this problem, first, multiply the radicals (in blue) and their coefficients (in red) together:

$\textcolor{w h i t e}{=} \textcolor{red}{x} \textcolor{b l u e}{\sqrt{10 x}} \cdot \textcolor{red}{7} \textcolor{b l u e}{\sqrt{15 x}}$

$= \textcolor{red}{x} \cdot \textcolor{b l u e}{\sqrt{10 x}} \cdot \textcolor{red}{7} \cdot \textcolor{b l u e}{\sqrt{15 x}}$

$= \textcolor{red}{x} \cdot \textcolor{red}{7} \cdot \textcolor{b l u e}{\sqrt{10 x}} \cdot \textcolor{b l u e}{\sqrt{15 x}}$

$= \textcolor{red}{7 x} \cdot \textcolor{b l u e}{\sqrt{10 x}} \cdot \textcolor{b l u e}{\sqrt{15 x}}$

$= \textcolor{red}{7 x} \cdot \textcolor{b l u e}{\sqrt{10 x \cdot 15 x}}$

$= \textcolor{red}{7 x} \cdot \textcolor{b l u e}{\sqrt{10 \cdot 15 \cdot x \cdot x}}$

$= \textcolor{red}{7 x} \setminus \cdot \textcolor{b l u e}{\sqrt{150 \cdot x \cdot x}}$

$= \textcolor{red}{7 x} \cdot \textcolor{b l u e}{\sqrt{150 \cdot {x}^{2}}}$

Next, use the multiplication rule backwards:

$\textcolor{w h i t e}{=} \textcolor{red}{7 x} \cdot \textcolor{b l u e}{\sqrt{150 \cdot {x}^{2}}}$

$= \textcolor{red}{7 x} \cdot \textcolor{b l u e}{\sqrt{150}} \cdot \textcolor{b l u e}{\sqrt{{x}^{2}}}$

Now, use the simplification rule:

$\textcolor{w h i t e}{=} \textcolor{red}{7 x} \cdot \textcolor{b l u e}{\sqrt{150}} \cdot \textcolor{b l u e}{\sqrt{{x}^{2}}}$

$= \textcolor{red}{7 x} \cdot \textcolor{b l u e}{\sqrt{150}} \cdot \textcolor{red}{x}$

$= \textcolor{red}{7 x} \cdot \textcolor{red}{x} \cdot \textcolor{b l u e}{\sqrt{150}}$

$= \textcolor{red}{7 {x}^{2}} \cdot \textcolor{b l u e}{\sqrt{150}}$

Technically, this answer is correct, but it can be simplified further by factoring $150$ and then using the simplification rule backward again:

$\textcolor{w h i t e}{=} \textcolor{red}{7 {x}^{2}} \cdot \textcolor{b l u e}{\sqrt{150}}$

$= \textcolor{red}{7 {x}^{2}} \cdot \textcolor{b l u e}{\sqrt{6 \cdot 25}}$

$= \textcolor{red}{7 {x}^{2}} \cdot \textcolor{b l u e}{\sqrt{6}} \cdot \textcolor{b l u e}{\sqrt{25}}$

$= \textcolor{red}{7 {x}^{2}} \cdot \textcolor{b l u e}{\sqrt{6}} \cdot \textcolor{b l u e}{\sqrt{{5}^{2}}}$

$= \textcolor{red}{7 {x}^{2}} \cdot \textcolor{b l u e}{\sqrt{6}} \cdot \textcolor{red}{5}$

$= \textcolor{red}{7 {x}^{2}} \cdot \textcolor{red}{5} \cdot \textcolor{b l u e}{\sqrt{6}}$

$= \textcolor{red}{35 {x}^{2}} \cdot \textcolor{b l u e}{\sqrt{6}}$