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# Interpretation of Orbital Magnetic Moment

The wavefunctions for atomic orbitals have always been described to me one of two ways:

1. As a "smeared out" electron standing wave with integer number circumference of de Broglie wavelengths

2. As a "probability cloud" wherein a point particle electron is likely to be observed.

While those 2 interpretations have never seemed compatible to me, one question seems incompatible with BOTH of these interpretations:

If an electron does not orbit classically, i.e. is not an accelerating point charge in a circular orbit, how does it create a magnetic field of magnitude equivalent to a current carrying loop of orbital circumference?

In other words, why does an electron that is not moving classically appear to produce a magnetic field one would expect from classical movement?

I'm having difficulty understanding how either a "smeared out" or "probability distribution" interpretation of electron orbitals under QM accounts for the non-spin related magnetic moment of an electron, just as if that electron were a point charge in a circular orbit, which QM emphatically refutes?

Actually, in a sense, there does exist something which moves. It's not something which can be measured to move, but rather part of quantum mechanical description, evolution of which much resembles classical electron motion: the phase of wavefunction.

See this java demonstration of hydrogen atom orbital. If you select "Complex Orbitals (phys.)" and then choose a state with $m\ne0$, then you'll see how the colours "rotate". This shows that the phase of the wavefunction does rotate around the $z$ axis. In a sense, this can be said to create the orbital magnetic moment.

But please note, I'll repeat: this is not an observable rotation by itself! Moreover, the angular velocity of phase is actually not fixed - it can be shifted by arbitrary constant (provided it's the same for all the states). But it does show how the wavefunction evolves in time. If you try to measure anything, all you can get is the magnitude squared of the wavefunction (for a large enough number of identical experiments). Still, the phase does play a role in creation of superpositions of states and finally in allowing the probability densities change in time, while for stationary states the probability densities are constant in time, as you already know.

EDIT in response to comment:

First, the wavefunction is not real in the sense $\psi\not\in\mathbb R$. It's complex, while all the directly measurable quantities must be real.

Second, despite in eigenstate the atom is stationary, we can make linear combinations of eigenstates with the same direction of magnetic moment such, that the result will be a wave packet measurably rotating in a definite direction.

For a simpler example of translational motion instead of rotational, consider a plain wave. Here's how it looks (blue real part, purple imaginary, yellow square of magnitude):

Now let's make up a Gaussian wave packet, like in this demonstration. It'll have the same peak state, but additionally some more states with higher and lower phase velocities (with smaller amplitudes). Here's its wave function (blue real part, purple imaginary):

Now this packet has measurable motion. Here's its square of magnitude, which is the probability density by Born rule:

You see that for a particle eigenstate, which is a plane wave, you can't measure its motion. But OTOH, it's the limit of infinitely delocalized wave packet, and for a localized wave packet you can measure its motion (in probabilistic sense, of course).

The same is for rotational motion, it just has some more technical features, which aren't relevant for understanding here.

• I feel like I'm 90% there but hanging up on this: If the wavefunction is not something physical in and of itself and merely the mathematical tool we use to describe the system we're looking at, I have trouble seeing how the change of phase in the WF can result in the physical property of magnetic moment - in other words the WF describing electron orbitals certainly seems VERY REAL and produces measurable consequences in atomic bonding behavior, magnetic moment, etc however we interpret it as non-real. Why? – JPattarini Jan 5 '15 at 16:06
• @JamesPattarini I've updated my answer to address your comment. – Ruslan Jan 5 '15 at 17:26
• @JamesPattarini: The wavefunction (in at least some sense) is real, as it can alter the results of measurements. However, if the wavefunction is multiplied by $e^{ix}$, where x is some real, the probabilities of the outcomes for every measurement will be the same. There exists a good analogy with symmetries in classical mechanics: the following two sentences describe exactly equivalent two physical systems. 1) Particle A moves with velocity $1m/s$ and B with $2m/s$ to the left. 2) Particle A moves with velocity $5m/s$ and B with $6m/s$ to the left. – kristjan Jan 5 '15 at 17:53
• @Ruslan Serious thanks for the detailed explanation, and the visuals help with the conceptualization. If you have any insight on physics.stackexchange.com/questions/156746/… it would be greatly appreciated. – JPattarini Jan 8 '15 at 17:11

It's neither. Those descriptions are just mental imagery for beginners. You need not waste any time on them. All of reality obeys the laws of quantum mechanics. What we call classical physics is just an approximation of quantum mechanics under certain circumstances.

This is important: classical physics derives from QM and NOT the other way round.

You have to stop thinking in terms of shiny classical billiard balls jumping around in some weird way and you have to start thinking about a complex field like object, that, when looked at from 30,000 feet, can produce distributions of expectation values that may look like particle tracks or electromagnetic waves.

For low energies the interactions of the different components of this field, i.e. those that stand for electrically charged fermionic states ("electrons") and those of uncharged bosonic states ("photons") can, on average, reproduce the usual phenomenology for electrons interacting with electromagnetic fields like Lorentz forces.

That is only a small fraction of what the QFT model can reproduce, though. It can also calculate magnetic moments with high precision, explain the hyperfine structure of atomic transition, explain matter and anti-matter and describe its production in e.g. pair-production processes and then some. Taken together, the most complete model that we have, describes almost every aspect of elementary particle physics and motivates the behavior of a wide variety of nuclear phenomena.

• Thanks for the detailed response - I thought I was getting away from the particle view by embracing the idea of a standing electron wave, but I hit a mental brick wall when I try to envision how a non-local object produces magnetic moment that seems to be best described by a discrete object moving classically in a circular path. It seems such an object should radiate EM radiation which is exactly why the Bohr model was abandoned - in other words I'd gladly embrace a different way of thinking about electrons than point particles but the (non-spin) magnetic moment calculation seems to demand it – JPattarini Dec 29 '14 at 14:32
• @JamesPattarini: The standing wave model is not quite sufficient to make the theory self-consistent. For one thing it does not cover all the properties of a quantum field. There is really no way getting around the definition of something fundamentally new (i.e. the quantum field), which in one approximation (large numbers of particles) has wave-like and i another (weak measurements) has particle track like properties. The good news is that once we understood that, the early ontological questions fell away completely. The bad news is that this new object is mathematically hard to deal with. – CuriousOne Dec 29 '14 at 15:53
• I think my hangup is on the "realness" of the wavefunction, since it's been beaten into me that it's a mathematical tool and not something real in and of itself, if picturing the 3D Schrodinger solutions for orbitals is not "really" what's going on then what CAN I picture in my mind's eye, or put another way: considering how perfectly the orbital picture describes physical atomic bonding behavior, why don't we consider the orbitals "real?" I imagine it's because treating the wavefunction as real in other scenarios seems nonsensical? – JPattarini Jan 5 '15 at 16:14

Although the standing wave is indeed a stationary state, don't make the mistake of thinking this means the electron is not moving. As you say, we know the electron is moving because there is a very real electric current and magnetic moment. So the electron is moving and its wavefunction is not moving.

The confusion comes from our everyday experience, where inertia and dynamics always come together in one package, which we call "motion". But with quantum mechanics we must separate these ideas. What I mean is, a quantum wave can contain motion (inertia) without itself being in motion (dynamic).

[By the way, of the two viewpoints you note, the probability cloud viewpoint is a much weaker view. I mean, from the wave function you can calculate the probability cloud, but the reverse is not possible.]

• Many thanks for the quick answer - for clarification, when you say "the electron is moving, but the wavefunction is not moving" how best to picture what the electron is "actually" DOING in this case? I ask because the whole issue with the Bohr model of imagining a tiny electron ball wizzing around a nucleus was abandoned because such an electron should radiate EM radiation - QM fixed that (I thought) by doing away with the idea of classical electron motion. So saying that the electron IS moving leaves me asking "in what manner should I conceive of the electron moving?" – JPattarini Dec 29 '14 at 13:54
• I find it hard to visualize, too. But really you have to abandon any hope that your visual picture will involve a particle in a well-defined place. I like to think of the electron standing wave as a smeared electron where it is simultaneously on all sides of the nucleus. It doesn't radiate since the charge and current patterns are steady. But even this view is limited. Another way to look at it is that the electron is a point particle but it is superposed on all sides of the nucleus---each superposed copy does radiate but the radiation is destructively interfered in the superposition. – Nanite Dec 31 '14 at 15:27

The standing wave describes the relative electron-nucleus motion. The electron itself moves since the atomic center of mass (a plane wave) involves the electron coordinates too.


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# Dot product - Area Calculations

###### No one has explained this idea yet.
Cite as: Dot product - Area Calculations. Brilliant.org. Retrieved from https://brilliant.org/wiki/dot-product-area-calculations/
×

format: plain


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# Regression model

1. Jun 5, 2016

### squenshl

1. The problem statement, all variables and given/known data
Suppose that we observe $i=1,2,\ldots,n$ independent observations that can be modeled as follows:
$$Y_i = i(\theta+\epsilon_i) \quad \text{where} \; \; \epsilon_i \sim N(0,\sigma^2).$$

1. Write the above as a regression model, $E(Y) = X\theta$, $\text{Cov}(Y) = \sigma^2W$ for matrices $X$ and $W$.

2. Show that $X^TW^{-1}X = n$.

3. Show that the least squares estimate for $\theta$ is given by
$$\hat{\theta} = \frac{1}{n}\sum_{i=1}^{n} \frac{Y_i}{i}.$$

Consider the following transformation: $Z_i = \frac{Y_i}{i}.$

4. Show the transformed model can be written as a regression model,
$$E(Z) = 1_n\theta, \quad \text{Cov}(Z) = \sigma^2I_n$$
where $1_n$ is a column vector of $1$s and $I_n$ is an identity matrix of dimension $n$.

5. Show that the least squares estimate from this model is exactly the same as the solution from part c).

2. Relevant equations

3. The attempt at a solution

$$X = \begin{bmatrix} 1 \\ 2 \\ \vdots \\ n \end{bmatrix}$$ but not sure on $W$. Once I can get this I can pretty much do the rest.

2. Jun 6, 2016

### squenshl

Never mind I got it but as usual your help was appreciated!!!!!!


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# Number of Ramsey ultrafilters

Hereafter, "ultrafilter" is intended to mean "nonprinicpal ultrafilter on the set, $\omega$, of natural numbers."

An ultrafilter $U$ is Ramsey if for every partition $P$ of $\omega$ into sets not in $U$, there is a set $a\in U$ such that $\forall p\in P, \vert{a\cap p}\vert = 1$.

If we assume the CH, one can construct a Ramsey ultrafilter by enumerating all paritions $\langle P_\alpha : \alpha<\omega_1\rangle$, and iteratively choosing infinite generating sets, $X_{\alpha+1}\subseteq X_\alpha$ such that either $X_{\alpha+1}\subset p$ for some $p\in P_\alpha$ or $\vert X_{\alpha+1}\cap p\vert \leq 1$ for all $p\in P_\alpha$. Because each limit stage $\beta$ is only countable we can find a set $X_\beta$ that is almost contained in $X_\alpha$ for each $\alpha<\beta$.

In fact, the construction above can be used to produce $\aleph_1$ distinct Ramsey ultrafilters: for a suitable $P_0$, there are $\aleph_1$ choices for $X_1$ that lead to distinct ultrafilters.

On the other hand, I believe that there are at least $2^\frak{c}$ ultrafilters: pick any almost disjoint family $\cal{A}$ of size $\frak{c}$, then for each $S\in \mathscr{P}(A)$ let $U_S$ be any ultrafilter containing $\{A\in S\}\cup\{\omega\setminus A: A\in \cal{A}\setminus S\}$.

My question is: when Ramsey ultrafilters exist, how many exist? Specifically I'm interested in the situation under the (G)CH.

Can the construction above be observed to produce the maximum number of ultrafilters? Or at least more than $\aleph_1$?

The Ramsey ultrafilters on $\omega$ are precisely those that are Rudin-Keisler minimal in $\beta\omega \setminus \omega$, and under CH there are $2^{2^\omega}$ of them. See Theorems 4.5 and 5.5 in W.W. Comfort, Ultrafilters: some old and new results, Bull. Amer. Math. Soc. 83 (1977), 417-455, available as a PDF from here.
It is consistent with ZFC that there are "few" Ramsey ultrafilters. For example, in the model of ZFC obtained by iterating $\omega_2$-many Sacks reals with countable support forcing over a model of ZFC+CH the continuum = $\omega_2 = 2^{\omega_1}$ and every Ramsey ultrafilter is $\omega_1$-generated. Thus there are only continuum-many of them.


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Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 6 Answer Key Functions.

Complete these exercises to review skills you will need for this chapter.

Evaluate each expression for the given value of x.

Question 1.
2x + 3 for x = 3 ____
2x + 3 for x = 3 Given
2(3) + 3 Substitute the given value of x for x
6 + 3 Multiply

Question 2.
-4x + 7 for x = -1 _____
To evaluate an expression for the given value of x, let us substitute the value of x in the expression.

Evaluating -4x + 7 for x = -1
-4x + 7 = -4(-1) + 7 Substitute the value of x.
= 4 + 7 Simplify.
= 11

Question 3.
1.5x – 2.5 for x = 3 _____
1.5x – 2.5 for x = 3 Given
1.5(3) – 2.5 Substitute the given value of x for x
4.5 – 2.5 Multiply
2 Subtract

Question 4.
0.4x + 6.1 for x = -5 ______
0.4x + 6.1 for x = -5 Given
0.4(-5) + 6.1 Substitute the given value of x for x
-2 + 6.1 Multiply

Question 5.
$$\frac{2}{3}$$x – 12 for x = 18 _____
$$\frac{2}{3}$$x – 12 for x = 18 Given
$$\frac{2}{3}$$(18) – 12 Substitute the given value of x for x
12 – 12 Multiply
0 subtract

Question 6.
–$$\frac{5}{8}$$x + 10 for x = -8 ______
To evaluate an expression for the given value of x, let us substitute the value of x in the expression.

Evaluating –$$\frac{5}{8}$$x + 10 for x = -8
–$$\frac{5}{8}$$x + 10 = –$$\frac{5}{8}$$(-8) + 10 Substitute the value of x.
= $$\frac{40}{8}$$ + 10 Simplify.
= 5 + 10
= 15

Define the variables for each situation. Then write an equation.

Question 7.
Jana’s age plus 5 equals her sister’s age.
j = Janas age Define the variables used in the situation
s = Jana’s sister age

s = j + 5 Identify the operation involved. Plus indicated addition

Question 8.
Andrew’s class has 3 more students than Lauren’s class.
A = Number in Andrew’s class Define the variables used in the situation
L = Number in Lauren’s class

A = L + 3 Identify the operation involved. more indicated addition

Question 9.
The bank is 50 feet shorter than the firehouse.
Given the statement The bank is 50 feet shorter than the firehouse’ the variables can be defined by letting x be the height of the bank; and y be the height of the firehouse.

Since x represents the height of the bank that is 50 feet shorter that the firehouse represented as variable y, we
can equate x to y – 50 that gives an equation
x = y – 50

Question 10.
The pencils were divided into 6 groups of 2.
g = number in pencils in groups Define the variables used in the situation
n = number of groups

$$\frac{12}{n}$$ = g Identify the operation involved. divided indicated division

Visualize Vocabulary
Use the ✓ words to complete the diagram. You can put more than one word in each section of the diagram.

Understand Vocabulary

Complete the sentences using the preview words.

Question 1.
A rule that assigns exactly one output to each input is a ___.
A rule that assigns exactly one output to each input is a function.

Explanation:
A function is defined as the relationship in which the input value is paired with exactly one output value.
By definition, a rule that assigns exactly one output to each input is a function.
Thus, the blank part of the statement is function

Question 2.
The value that is put into a function is the ___________
The value that is put into a function is the input

Explanation:
In a function, we have two types of values the input and the output values. Input value is defined as the independent variable that is being put into a function. It affects the output because output is the dependent variable or the result in a function.

By definition, the value that is put into a function is input

Thus, the blank part of the statement is input.

Question 3.
The result after applying the function machine’s rule is the ____.


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# Cayley graph on $D_{2n}$ and $\mathbb Z_n$

How we can make Cayley graph on $D_{2n}$ and $\mathbb Z_n$? What can be S in $Cay(D_{2n},S)$ and $Cay(\mathbb Z_n ,S)$, Please write one example. Thanks for advise.

• Well...what have you tried? – Elchanan Solomon Sep 21 '13 at 13:00

Here's an example of a Cayley graph for the group $\mathbb{Z}_{10}$:

The vertices are the underlying set of the group $\mathbb{Z}_{10}$, and we draw edges between $g$ and $g+h$ whenever $g,h \in S$ where $S=\{\pm 1, \pm 2\}$.

Here's an example of a Cayley graph for the dihedral group $D_{10}$:

The vertices are the underlying set of the dihedral group with presentation $$D=\langle f,r | f^2=e, r^5=e, rf=fr^{-1} \rangle,$$ (we can think of $f$ as "flip" and $r$ as "rotation") and we draw edges between $g$ and $g+h$ whenever $g,h \in S$ where $S=\{f,r,r^{-1}\}$.

• thanks, but can you give me an example for 3-hypergraph cayley on $D_{2n}$? – mehranian Sep 21 '13 at 13:31
• A 3-uniform Cayley hypergraph for $D_{2n}$, that'd be more difficult to draw. I'm not even sure of the definition in this case, but I expect if we take one hyperedge $\{g_1,g_2,g_3\}$, then generate the remaining edges as $\{hg_1,hg_2,hg_3\}$ for all $h \in D_{2n}$, this will give an example. – Rebecca J. Stones Sep 21 '13 at 13:36

Considering the definition of Cayley graph done by @Owen, you can do it as follows:

For $\mathbb Z_n=\langle a\rangle=\{1,a,a^2,\cdots a^{n-1}\}$:

And for $D_{2n}$, I did it for $D_6=\{1,a,a^2,b,ab,a^2b\}$ of order $6$ instead:

• Nothing like paper and pencil (or pen)! +1 – Namaste Sep 21 '13 at 13:49
• @amWhy: Pen,Amy. :-) – mrs Sep 21 '13 at 13:58

Recalling the definition of Cayley graph of a group $G$, remember that the vertex set is the set $G$ and two vertices, $g, h\in G$, are connected by an edge if there exists some $a\in S$ such that $g^{-1}h\in S$. So this means that $S\subset G$. Now in principal you could let $S$ be any subset of $G$ but usually (always?) you want to choose $S$ to be a generating set for $G$ because this ensures that the resulting graph is connected. So in paticular, since groups have many different generating sets, the resulting graphs will be different. So there is no such thing as $\textbf{the}$ Cayley graph of a group.

Just to give one example. If $G=\{e, g_1, ..., g_n\}$, note $|G|=n+1$, and if we let $S=G\setminus\{e\}$. Then Cayley($G, S)$= $K_{n+1}$, the complete graph on $n+1$ vertices.

However, in general you do not do this. Instead, you try to use the smallest $S$ possible. In particular for the case $\mathbb{Z}_n$ you can choose $S=\{1\}$. In this case you will get that the Cayley graph is a loop with $n$ points on it.

• thanks, but you can give me an example of 3-hypergraph cayley on $D_{2n}$? – mehranian Sep 21 '13 at 14:05


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Convergence of new inertial proximal methods for DC programming

Abstract : We present iterative methods for finding the critical points and/or the minima of extended real valued functions of the form $\phi = \psi+ g-h$, where $\psi$ is a differentiable function and g and h are convex, proper, and lower semicontinuous. The underlying idea relies upon the discretization of a first order dissipative dynamical system which allows us to preserve the local feature and to obtain some convergence results. The main theorems not only recover known convergence results in this field but also provide a theoretical basis for the development of new iterative methods.
Keywords :
Type de document :
Article dans une revue
SIAM Journal on Optimization, Society for Industrial and Applied Mathematics, 2008, 19 (1), pp.397-413. 〈10.1137/060655183〉

https://hal.univ-antilles.fr/hal-00779987
Contributeur : Pamphile Isch <>
Soumis le : mardi 22 janvier 2013 - 19:46:03
Dernière modification le : mercredi 18 juillet 2018 - 20:11:26

Citation

Abdellatif Moudafi. Convergence of new inertial proximal methods for DC programming. SIAM Journal on Optimization, Society for Industrial and Applied Mathematics, 2008, 19 (1), pp.397-413. 〈10.1137/060655183〉. 〈hal-00779987〉

Métriques

Consultations de la notice


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# Unfolding Polyhedra

I'm interested in learning more on unfolding polyhedra.

1. Are there any known algorithms that unfold polyhedra into nets? I'm interested in writing code on this in either MATLAB, Python, or C#.
2. On this page, Joseph O'Rourke goes into several unfoldings of a cube. What resources are there for studying the differences (or equivalencies) between the unfoldings?
-

## Counting the unfoldings

If you consider your polyhedron as a graph, with one vertex per face, and two vertices connected if the corresponding faces are adjacent, if your polyhedron has no vertex such that the sum of surrounding angles is $2\pi$ then I would say the number of unfoldings corresponds to the number of covering trees of this graph quotiented by the symmetries of your polyhedron. This question on counting the number of covering trees of a graph provides some insight on this problem and therefore on the potential number of unfoldings you may get.

## Computing one unfolding

If what you want is solely to compute one unfolding, then my strategy would be to use a breadth first traversal of the graph corresponding to your polyhedron to determine which adjacent faces should be kept glued in the unfolding. This technique is commonly used in algorithms for mesh parametrization in graphics, for texture mapping for instance. Roughly :

• take a random face $f_0$ and flatten in on the plane ;
• mark this face as done ;
• for each neighbouring face $f_i$ adjacent to $f_0$ :
• add $(f_0,f_i)$ to a queue ;
• mark $f_i$ as done ;
• while the queue is not empty :
• extract $(f_0,f_i)$ from the queue ;
• unfold $f_i$ rigidly next to $f_0$ ;
• for each neighbouring face $f_j$ of $f_i$ which is not marked as done :
• add $(f_i,f_j)$ to the queue ;
• mark $f_j$ as done.

Note however that this algorithm does not care on whether several unfolded faces overlap. I believe an example of a polyhedron which cannot be unfolded in a single piece without overlaps may be found, however I can't exhibit any which is compact. I believe this is related to surfaces with negative Gaussian curvature everywhere.

-
The existence of convex polyhedra without non-overlapping unfoldings is an open problem. For non-convex (even star-shaped) polyhedra it's relatively straightforward to exhibit: imagine, for instance, a cube with a 'spike' added to the center of each face. –  Steven Stadnicki Nov 14 '12 at 20:27


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# Arithmetic Progression - Understanding Question

My question is related to Arithmetic progression. I would solve it myself but I need some help understanding the question.

Q.The second, 31st and the last terms of an A.P are $$\frac{31}{4},\frac{1}{2}$$ and $$\frac{-13}{2}$$ respectively. Find the number of terms.

How can I use the $$Tn=a+(n-1)d$$ formula here, I need some explanation of this question.

-Thanks.

-
The second term is $31/4 = a + d$, the $31-$th term is $1/2 = a + (31-1)d$, solve these two equations for $a$ and $d$. then use $-13/2 = a + (n-1)d$ to solve for $n$. – Santosh Linkha Feb 18 '13 at 18:37

$a+d=\frac{31}{4}$, $a+30d=\frac12$, and $a+(l-1)d=-\frac{13}{2}$. Find $l$.

-

The algebraic approach described by Jacob Black is the right one. For fun, let us see how someone innocent of algebra might tackle things.

From the $2$nd term to the $31$-th (which takes $29$ steps), we go down by $\dfrac{31}{4}-\dfrac{1}{2}=\dfrac{31}{4}-\dfrac{2}{4}=\dfrac{29}{4}$.

So we go down by $\dfrac{1}{4}$ each time.

From the $31$-th term to the end, we went down by $\dfrac{1}{2}-\left(\dfrac{-13}{2}\right)=\dfrac{14}{2}=\dfrac{28}{4}$. So it took $28$ steps. Thus the total number of terms is $31+28=59$.

-


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# A strong-coupling analysis of two-dimensional O(N) sigma models with N<=2 on square, triangular and honeycomb lattices

Campostrini, Massimo and Pelissetto, Andrea and Rossi, Paolo and Vicari, Ettore (1996) A strong-coupling analysis of two-dimensional O(N) sigma models with N<=2 on square, triangular and honeycomb lattices. Physical review. B, Condensed matter and materials physics, 54 . pp. 7301-7317. ISSN 1550-235X

Full text not available from this repository.

## Abstract

The critical behavior of two-dimensional ${\rm O}(N)$ $\sigma$ models with $N\leq 2$ on the square, triangular, and honeycomb lattices is investigated by an analysis of the strong-coupling expansion of the two-point fundamental Green's function $G(x)$, calculated up to 21st order on the square lattice, 15th order on the triangular lattice, and 30th order on the honeycomb lattice. For $N<2$ the critical behavior is of power-law type, and the exponents $\gamma$ and $\nu$ extracted from our strong-coupling analysis confirm exact results derived assuming universality with solvable solid-on-solid models. At $N=2$, i.e., for the 2-$d$ XY model, the results from all lattices considered are consistent with the Kosterlitz-Thouless exponential approach to criticality, characterized by an exponent $\sigma=1/2$, and with universality. The value $\sigma=1/2$ is confirmed within an uncertainty of few per cent. The prediction $\eta=1/4$ is also roughly verified. For various values of $N\leq 2$, we determine some ratios of amplitudes concerning the two-point function $G(x)$ in the critical limit of the symmetric phase. This analysis shows that the low-momentum behavior of $G(x)$ in the critical region is essentially Gaussian at all values of $N\leq 2$. New exact results for the long-distance behavior of $G(x)$ when $N=1$ (Ising model in the strong-coupling phase) confirm this statement.

Item Type: Article Imported from arXiv Area02 - Scienze fisiche > FIS/02 - Fisica teorica, modelli e metodi matematici Dipartimenti (from 2013) > DIPARTIMENTO DI FISICA dott.ssa Sandra Faita 11 Feb 2014 22:37 11 Feb 2014 22:37 http://eprints.adm.unipi.it/id/eprint/1880

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# Robustly dealing with malformed Unicode files

I'm writing a script that deals reads UTF-8-encoded XML files and writes parts of those files into a tempfile for further processing.

Sometimes, the input files will have a few malformed characters. If it's just "a few" occurrences, I want those glossed over, but if the file is majorly corrupted, I want to see an error message. I have a solution for this working right now, but I have this nagging feeling that this isn't a clean way to do this.

This creates an example file with malformed data:

f = open("test.xml","w+b")
f.write("\x3c\x61\x3e\x31\x3c\x2f\x61\x3e\x0a\x3c\x62\x3e\x32\xc3\x0a\xbc\x3c\x2f\x62\x3e\x0a")
f.close()

Yes, I know it's probably a stray linebreak that's causing my "malformed utf-8" problem, but let's not make that the issue here.

Now, the following is a simplified version I have in place to deal with malformed utf-8 data up to a certain limit:

import codecs

sourcefile = open("test.xml", "r")

# First line always XML declaration, no special treatment

i = 1
unicodeerror = 0

while line:
try:
pos = sourcefile.tell()
except UnicodeDecodeError as e:
print "WARNING: Encoding error in line " + str(i+1)
sourcefile.seek(pos, 0)
unicodeerror += 1
i += 1

if unicodeerror > 5:
raise StandardError('ERROR: Too many Unicode errors')

As you can see, I read the input file as a byte string, and then read each line individually, doing the UTF-8-Decode within the try-except-clause. That way, I can catch the UnicodeDecodeError and count it. Only after counting it, reread the line, this time ignoring possible mistakes.

However, it "feels wrong" to simply read the file like this instead of doing a codecs.open(FILE, MODE, "utf-8"). With the codecs method, however, I see no way to reread the line with a different error handling directive after I hit the UnicodeDecodeException.

Am I being too cautious? Is this maybe the best / only reliable way to do this? OR is there a much easier solution (which can be achieved within the standard library!)?

The answer to your question depends on the version of python being used, and exactly what behavior you want. In Python 3.1 and up, the surrogateescape error handler can adeptly handle corrupted files that are otherwise encoded with UTF-8, and even write them back out again. Just use codecs.open(FILE, MODE, "utf-8", "surrogateescape"). If you want to know when invalid bytes were encountered, you can either scan for the relevant surrogate characters (those in the range U+DC80..U+DCFF), or compare to values you get from the replace error handler.

Through Python 2.7, it appears that the surrogateescape error handler is unavailable. You could contrast the results between the ignore and replace error handlers, but there is no convenient way to process the file in Unicode and write it out byte for byte.

In either version, you can theoretically write your own error handler that performs similarly to surrogateescape, ignore or replace, and counts the number of problems it sees. I'm uncertain whether this count can be scoped to a single file, however, as it appears an error handler only receives knowledge of the substring in question when an error is encountered.

On to a code review. Your code feels almost clunky to me. There's a stray import codecs, a lot of repetition (I can't fathom why re-reads the line with an error) and it doesn't try to leverage iterators. Perhaps you're on a particularly old version of Python (no, you use as in the except line), or perhaps you just haven't worked with iterators much. Probably most of this was from trying to extract the core issue, but I can only review it as is. I apologize for any errors in my code that follows; I haven't tested it, but I am willing to update it.

I would be strongly tempted to replace most of your code with something like this, and that's before trying to leverage anything I talked about above:

unicodeerror = 0
for i, line in enumerate(open("test.xml", "r")):
try:
line.decode("utf-8")
except UnicodeDecodeError:
print "WARNING: Encoding error in line", i+1
unicodeerror += 1

if unicodeerror > 5:
raise StandardError("ERROR: Too many Unicode errors")

If you need to do more, such as actually process the line, then it's reasonable to save the results of the decode, and to try again with an error handler in the except clause:

    try:
text = line.decode("utf-8")
except UnicodeDecodeError:
# error tracking omitted
text = line.decode("utf-8", "ignore")
process(text)

If you wanted to integrate a comparison between two error handlers, it would replace the exception, so you could do something like this:

import codecs
unicodeerror = 0
for i, (ignored, replaced) in enumerate(zip(
codecs.open("test.xml", "r", "utf-8", "ignore"),
codecs.open("test.xml", "r", "utf-8", "replace"))):
if ignored != replaced:
print "WARNING: Encoding error in line", i+1
unicodeerror += 1
if unicodeeror > 5:
raise StandardError("ERROR: Too many Unicode errors")

And I'll leave a custom error handler totally up to you. If it works, it would probably be a lot more efficient than this approach, but there's something to be said for the simplicity of the visible code.

As a side comment, while I found this question interesting enough to delve into, the more I did so, the more it felt like a question about the right way to do this than a request to review your code. Keep in mind that it's much more on topic here to review code and make suggestions on how to improve it. So this particular approach isn't necessarily a great fit for codereview.SE.

• Thanks for the competent and patient reply! To address your last issue first: I was unsure whether that went here or ... where else? SE central? I don't want to clutter the Code Review space with problem that don't belong here, so sorry for that. – Thor Sep 10 '14 at 16:32
• I went the readline-way only because when I first wrote this, I still used the strict error handler of the codecs.open method, and I didn't see a way of rereading the line with another error handler after catching the Unicode exception (yes, I do need the data in that line). I do like your last suggestion, using two file handles with contrasting error handlers. This has indeed improved the readability of my code loads, so thanks for that! – Thor Sep 10 '14 at 16:50
• Thanks for the interesting scenario! Thinking later I realized that with large files it may be important to change zip to itertools.izip, so keep that in mind too. – Michael Urman Sep 10 '14 at 23:31


text:
# Tag Info

33

The initial and final permutation have no influence on security (they are unkeyed and can be undone by anybody). The usual explanation is that they make implementation easier in some contexts, namely a hardware circuit which receives data over a 8-bit bus: it can accumulate the bits into eight shift registers, which is more efficient (in terms of circuit ...

31

There are a variety of reasons why AES is more widely used: AES is a standard. AES has been vetted by cryptanalysts more extensively than Camellia. As a result, we can have greater confidence in the security of AES than in Camellia. Therefore, on the merits, there may be good reasons to choose AES over Camellia. AES is a government standard (FIPS). ...

26

For practical purposes, 128-bit keys are sufficient to ensure security. The larger key sizes exist mostly to satisfy some US military regulations which call for the existence of several distinct "security levels", regardless of whether breaking the lowest level is already far beyond existing technology. The larger key sizes imply some CPU overhead (+20% for ...

24

The really simple explanation for the difference between the two is this: ECB (electronic code book) is basically raw cipher. For each block of input, you encrypt the block and get some output. The problem with this transform is that any resident properties of the plaintext might well show up in the ciphertext – possibly not as clearly – that's what blocks ...

24

Applied Cryptography is book which is becoming, say, not-so-recent. NSA has quite a lot of budget, but not an infinite amount, and there are other organization, in particular big private corporation, which also have impressive means. Google or Apple, for instance, are companies with R&D activity in the area of cryptography, and who are able to ...

21

Assume that 1 evaluation of {DES, AES} takes 10 operations, and we can perform $10^{15}$ operations per second. Trivially, that means we can evaluate $10^{14}$, or about $2^{46.5}$ {DES, AES} encryptions per second. This is a simplistic view: we are ignoring here the cost of testing whether we found the correct key, and the key schedule cost. So on our ...

18

As a bonus feature, AES has hardware support in Intel processors which implement the AES instruction set, with AMD support coming soon in their Bulldozer based processors. The AES instructions set consists of six instructions. Four instructions, namely AESENC, AESENCLAST, AESDEC, AESDECLAST, are provided for data encryption and decryption (the ...

17

People found MARS to be clunky and overly complex, leading to more effort for implementation and optimization, and also a less clear overall security picture. Assessments of "security" are, in fact, extremely subjective, because they rely on speculations about unknown future cryptanalytic attack, empiric traditions (e.g. "more rounds" = "more security"), ...

16

A known-plaintext attack (i.e. knowing a pair of corresponding plaintext and ciphertext) always allows a brute-force attack on a cipher: Simply try all keys, decrypt the ciphertext and see if it matches the plaintext. This always works for every cipher, and will give you the matching key. (For very short plaintext-ciphertext pairs, you might get multiple ...

14

I'm just curious to know why the 128-bit version become the standard[.] That question is easy to respond. In the section Minimum Acceptability Requirements of Request for Candidate Algorithm Nominations for the AES, it says: The candidate algorithm shall be capable of supporting key-block combinations with sizes of 128-128, 192-128, and 256-128 ...

13

What you're looking for can be done using existing schemes for format preserving encryption (FPE). In general, FPE schemes convert an existing strong algorithm like AES into a block cipher that operates on a set of any size. For instance, FPE can encrypt 15 digit integers to other 15 digit integers (eg for credit card numbers, one of the common reasons for ...

13

Basically it's analysis of a cryptographic cypher by the means of finding a relationship between the difference in the input data and the output data. Ideally, the slightest difference in input data (cleartext), even a single bit, should produce a completely different cypthertext. However, if the cypher is not well-designed, a correlation between the two ...

13

The actual encryption algorithm is almost the same between all variants of AES. They all take a 128-bit block and apply a sequence of identical "rounds", each of which consists of some linear and non-linear shuffling steps. Between the rounds, a round key is applied (by XOR), also before the first and after the last round. The differences are: The longer ...

13

Well, the exact reason for an IV varies a bit between different modes that use IV. At a high level, what the IV does is act as a randomizer, so that each encrypted message appears to be encrypted to a random pattern, even if those messages are similar. In general, IVs disguise when you encrypt the same message twice (and more generally, when two messages ...

13

The reason why you see that is because Camellia is the highest-preference cipher in NSS (Chrome and Firefox). Servers that support Camellia and use the client-preferred cipher suite will use Camellia. NSS's rationale for this ordering is: National ciphers such as Camellia are listed before international ciphers such as AES and RC4 to allow servers ...

12

If you look closely at the definition of authenticated encryption modes, you will see they all are, actually, the combination of symmetric encryption and a MAC. Using traditional encryption and an independent MAC has a few tricky points, none of them being unsolvable: The encryption mode will use a key, and the MAC will also use a key; using the same key ...

12

Let a "block cipher" be defined with a fixed S-box $S$ (i.e. a permutation of some space) and a key $K$ (same size than a block), such that the encryption of a block $M$ is $C = S[P\oplus K]$. Everybody knows $S$ and can apply and invert it (that's a "S-box", not a "key" -- if the S-box is "key dependent" then the S-box is itself a block cipher in its own ...

12

There is only one main difference between PKCS#5 and PKCS#7 padding is the block size. PKCS#5 padding is only defined for 8-byte block sizes. PKCS#7 padding would work for any block size from 2 to 255 bytes. This is the definition of PKCS#5 padding (6.2): The padding string PS shall consist of 8 - (||M|| mod 8) octets all having value 8 - (||M|| mod ...

11

This approach, at a high level, is actually fairly common; many stream ciphers operate on this very principle. For instance, Salsa20 uses what is effectively a hash function (a PRF) to convert a secret input (that includes a counter) into the keystream which is XORed with the plaintext. However, this kind of function can be much faster than a secure ...

11

ECB and CBC are only about encryption. Most situations which call for encryption also need, at some point, integrity checks (ignoring the threat of active attackers is a common mistake). There are combined modes which do encryption and integrity simultaneously; see EAX and GCM (see also OCB, but this one has a few lingering patent issues; assuming that ...

11

Many cryptographic algorithms are expressed as iterative algorithms. E.g., when encrypting a message with a block cipher in CBC mode, each message "block" is first XORed with the previous encrypted block, and the result of the XOR is then encrypted. The first block has no "previous block" hence we must supply a conventional alternate "zero-th block" which we ...

11

Short version: It is quite likely that a large proportion of the keys have fixed points, but I don't have any idea on how to find them. Long version: A stochastic argument There are $2^{128}!$ permuations of 128-bit blocks, and of these, $!2^{128}$ (this is the subfactorial) are fixpoint-free. It is known that $\lim_{N\to\infty}\frac{!N}{N!} = \frac 1e ... 11 First of all, the avalanche effect is a desirable effect: it means that a very small change in the input will lead to a very big change in the output. A security algorithm that doesn't provide this avalanche effect can lead to an easy statistical analysis: if the change of one bit from the input leads to the change of only one bit of the output, then it's ... 11 Having taken The Design of Rijndael from the library just yesterday, I had a look on this problem, too. Fixee wrote in a comment: However, my question is not so much about security implications, but rather "how does omissions of MixColumns make the inverse cipher similar to the cipher?" and "how does this help in implementing the cipher?" The ... 11 Well, to start off with, IVs have different security properties than keys. With keys (as you are well aware), you need to hide them from anyone in the middle; if someone did learn your keys, then he could read all your traffic. IVs are not like this; instead, we don't mind if someone in the middle learns what the IV is; as long as he doesn't know the key, ... 11 A block cipher is an invertible transformation that maps an$n$bit block of bits to an$n$bit block of bits, under the control of a key (and where$n=128$in the case of AES) Now, we most often need to do things other than mapping blocks of$n$bits; how we do that is using the block cipher within a Mode of Operation. A mode of operation is just a way to ... 10 Well, as far as we know, the mode you suggest should be secure. Now, to be honest, AES256 versus your mode isn't quite a fair comparison; your mode gives somewhat less theoretical security; if you encrypt a known$2^n$block message, the key can be recovered with$2^{256-n}$effort; however, this observation doesn't really affect the practical security. ... 10 If a block cipher is linear with respect to some field, then, given a few known plaintext-ciphertext pairs, it is possible to recover the key using a simple Gaussian elimination. This clearly contradicts the security properties one expects from a secure block cipher. 10 The security of that approach is equivalent to that of normal CBC. Your scheme with first plaintext block$IV^\prime$is clearly identical to normal CBC with$IV=AES(IV^\prime)\$. Since a block cipher is a permutation over a block, a uniformly random first plaintext block will lead to a uniformly random IV for normal CBC. A ciphertext produced with your ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# The European Physical Journal C

## List of Papers (Total 3,650)

#### Anomaly-free models for flavour anomalies

We explore the constraints imposed by the cancellation of triangle anomalies on models in which the flavour anomalies reported by LHCb and other experiments are due to an extra U(1)$^\prime$ gauge boson Z$^\prime$. We assume universal and rational U(1)$^\prime$ charges for the first two generations of left-handed quarks and of right-handed up-type quarks but allow...

#### A precise determination of the top-quark pole mass

The Principle of Maximum Conformality (PMC) provides a systematic way to eliminate the renormalization scheme and renormalization scale uncertainties for high-energy processes. We have observed that by applying PMC scale setting, one obtains comprehensive and self-consistent pQCD predictions for the top-quark pair total cross section and the top-quark pair forward–backward...

#### Regge trajectories for heavy quarkonia from the quadratic form of the spinless Salpeter-type equation

In this paper, we present one new form of the Regge trajectories for heavy quarkonia which is obtained from the quadratic form of the spinless Salpeter-type equation (QSSE) by employing the Bohr-Sommerfeld quantization approach. The obtained Regge trajectories take the parameterized form $M^2={\beta }({c_l}l+{\pi }n_r+c_0)^{2/3}+c_1$, which are different from the present Regge...

#### The NNLO QCD soft function for 1-jettiness

We calculate the soft function for the global event variable 1-jettiness at next-to-next-to-leading order (NNLO) in QCD. We focus specifically on the non-Abelian contribution, which, unlike the Abelian part, is not determined by the next-to-leading order result. The calculation uses the known general forms for the emission of one and two soft partons and is performed using a...

#### Bouncing universe of entropy-corrected Friedmann equations

In this paper, we investigate the possibility of obtaining bouncing-oscillating solution in modified Friedmann equations with logarithmic entropy corrected, $S=\frac{A}{4G}+\alpha \ln \frac{A}{4G}+\beta \frac{4G}{A}$, for positive, negative and zero values of $(\alpha ,\beta )$ pre-factors and all kinds of curved universes. The results are argued using the dynamical system...

#### Status of quarkonia-like negative and positive parity states in a relativistic confinement scheme

Properties of quarkonia-like states in the charm and bottom sector have been studied in the frame work of relativistic Dirac formalism with a linear confinement potential. We have computed the mass spectroscopy and decay properties (vector decay constant and leptonic decay width) of several quarkonia-like states. The present study is also intended to identify some of the...

#### Fragmentation-fra ction ratio $$f_{\Xi _ b}/f_{\Lam bda _ b}$$fΞ b/fΛ b in b- and c- baryon de cays

We study the ratio of fragmentation fractions, $f_{\Xi _b}/f_{\Lambda _b}$, from the measurement of $\Xi _b^0\rightarrow \Xi _c^+\pi ^-$ and $\Lambda _b^0\rightarrow \Lambda _c^+\pi ^-$ with $\Xi _{c}^{+}/\Lambda _{c}^{+}\rightarrow p K^-\pi ^+$. With the branching fraction $\mathcal {B}(\Xi _c^+\rightarrow pK^-\pi ^+)=(2.2\pm 0.8)\%$ obtained under the U-spin symmetry...

#### Some dipole shower studies

Parton showers have become a standard component in the description of high-energy collisions. Nowadays most final-state ones are of the dipole character, wherein a pair of partons branches into three, with energy and momentum preserved inside this subsystem. For initial-state showers a dipole picture is also possible and commonly used, but the older global-recoil strategy remains...

#### Non-perturbative electroweak-scalegenesis on the test bench of dark matter detection

We revisit a recently proposed scale invariant extension of the standard model, in which the scalar bi-linear condensate in a strongly interacting hidden sector dynamically breaks scale symmetry, thereby triggering electroweak symmetry breaking. Relaxing the previously made assumption on $U(N_f)$ flavor symmetry we find that the presence of the would-be dark matter candidate...

#### Once more on Coulomb-nuclear interference

This is a critical reconsideration of the standard way of account for Coulomb-nuclear interference in the elastic scattering amplitude.

#### Torsion axial vector and Yvon-Takabayashi angle: zitterbewegung, chirality and all that

We consider propagating torsion as a completion of gravitation in order to describe the dynamics of curved-twisted space-times filled with Dirac spinorial fields; we discuss interesting relationships of the torsion axial vector and the curvature tensor with the Yvon-Takabayashi angle and the module of the spinor field, that is the two degrees of freedom of the spinor field itself...

#### Realistic simplified gaugino-higgsino models in the MSSM

We present simplified MSSM models for light neutralinos and charginos with realistic mass spectra and realistic gaugino-higgsino mixing, that can be used in experimental searches at the LHC. The formerly used naive approach of defining mass spectra and mixing matrix elements manually and independently of each other does not yield genuine MSSM benchmarks. We suggest the use of...

#### Hidden role of Maxwell superalgebras in the free differential algebras of D = 4 and D = 11 supergravity

The purpose of this paper is to show that the so-called Maxwell superalgebra in four dimensions, which naturally involves the presence of a nilpotent fermionic generator, can be interpreted as a hidden superalgebra underlying $\mathcal {N}=1, {\hbox {D}}=4$ supergravity extended to include a 2-form gauge potential associated to a 2-index antisymmetric tensor. In this scenario...

#### On the coverage of the pMSSM by simplified model results

We investigate to which extent the SUSY search results published by ATLAS and CMS in the context of simplified models actually cover the more realistic scenarios of a full model. Concretely, we work within the phenomenological MSSM (pMSSM) with 19 free parameters and compare the constraints obtained from SModelS v1.1.1 with those from the ATLAS pMSSM study in arXiv:1508.06608. We...

#### Disclosing $$D^*\bar{D}^*$$ molecular states in the $$B_c^- \rightarrow \pi ^- J/\psi \omega$$Bc-→π-J/ψω decay

We study the $B_c^- \rightarrow \pi ^- J/\psi \omega$ and $B_c^- \rightarrow \pi ^- D^* \bar{D}^*$ reactions and show that they are related by the presence of two resonances, the X(3940) and X(3930), that are of molecular nature and couple most strongly to $D^* \bar{D}^*$, but also to $J/\psi \omega$. Because of that, in the $J/\psi \omega$ mass distribution we find...

#### Radiative seesaw model and DAMPE excess from leptophilic gauge symmetry

In the light of the $e^{+}+e^{-}$ excess observed by DAMPE experiment, we propose an anomaly-free radiative seesaw model with an alternative leptophilic $U(1)_X$ gauge symmetry. In the model, only right-handed leptons are charged under $U(1)_X$ symmetry. The tiny Dirac neutrino masses are generated at one-loop level and charged leptons acquire masses though the type-I...

#### keV-Scale sterile neutrino sensitivity estimation with time-of-flight spectroscopy in KATRIN using self-consistent approximate Monte Carlo

We investigate the sensitivity of the Karlsruhe Tritium Neutrino Experiment (KATRIN) to keV-scale sterile neutrinos, which are promising dark matter candidates. Since the active-sterile mixing would lead to a second component in the tritium $\upbeta$-spectrum with a weak relative intensity of order $\sin ^2\theta \lesssim 10^{-6}$, additional experimental strategies are...

#### Running bumps from stealth bosons

For the ‘stealth bosons’ S, light boosted particles with a decay $S \rightarrow A A \rightarrow q \bar{q} q \bar{q}$ into four quarks and reconstructed as a single fat jet, the groomed jet mass has a strong correlation with groomed jet substructure variables. Consequently, the jet mass distribution is strongly affected by the jet substructure selection cuts when applied on the...

#### Singlet structure function $$F_1$$ in double-logarithmic approximation

The conventional ways to calculate the perturbative component of the DIS singlet structure function $F_1$ involve approaches based on BFKL which account for the single-logarithmic contributions accompanying the Born factor 1 / x. In contrast, we account for the double-logarithmic (DL) contributions unrelated to 1 / x and because of that they were disregarded as negligibly small...

#### Three waves for quantum gravity

Using effective field theoretical methods, we show that besides the already observed gravitational waves, quantum gravity predicts two further massive classical fields leading to two new massive waves. We set a limit on the masses of these new modes using data from the Eöt-Wash experiment. We point out that the existence of these new states is a model independent prediction of...


text:
# DelValle Institute 2014 Year-in-Review

See the maps below for a look at where we’ve been, and where our students came from in 2014.


text:
# Question #1c217

Mar 14, 2017

$\text{n-butane}$ DOES NOT generate any geometric isomers.......
And nor does $\text{isobutane}$, ${\text{H"_3"CCH(CH"_3")CH}}_{3}$.
However, $\text{2-butylene}$ can generate cis and trans isomers as shown:
These are GEOMETRIC isomers, because the $C - C$ connectivity is IDENTICAL for each isomer: $C 1$ connects to $C 2$.........to $C 4$. Their different geometry gives rise to different physical and chemical properties.


text:
## Algebraic constructions of K_{s,t}-free graphs

#### Boris BukhCarnegie-Mellon University

How many edges can an $n$-vertex graph have without containing $G$ as a subgraph? If $G$ is not bipartite, then the asymptotics is known, but for only a few bipartite graphs the humankind knows the answer. In all the resolved instances the construction is algebraic. It turns out that no real algebraic construction of extremal $K_{4,4}$-free graphs is possible.
In this talk, I will explain what that means, and sketch a new construction of extremal $K_{s,t}$-free graphs for $t$ much larger than $s$.

Back to Workshop IV: Finding Algebraic Structures in Extremal Combinatorial Configurations


text:
A calculator for solving the capacitance, reactance, and admittance of a capacitor.

(pF)
(Ω)
(S)

### Overview

This calculator is designed to compute for a capacitor's reactance and admittance given the capacitance value and the frequency. It can be also used the other way around, i.e., give the reactance value and it will give the capacitance. Note that the input capacitance must be in picofarads (pF) while the input frequency can be in Gigahertz (GHz) or Megahertz (MHz).

### Equation

$$X_{C} = \frac{1}{2 \pi f C}$$

Where:

$$X_{C}$$ = capacitive reactance

$$f$$ = frequency

$$C$$ = capacitance

### Application

A capacitor's opposition to alternating current is called its reactance. This reactance is dependent on the frequency of the alternating current (see equation above). For low frequencies, the capacitor acts like a resistor with a very high resistance. In fact, at direct currents where the frequency is zero, the capacitor acts like an open switch. Conversely, the capacitor acts like a resistor with a very low resistance at very high frequencies. This characteristic of the capacitor makes it suitable for use in tons of applications, particularly in filters, phase shifters, and couplers.


text:
Connection with golden ratio?

Consider the following problem:

Let $p\in\mathbb{Z}[x]$ be a polynomial with integer coefficient. Suppose that the leading coefficient is 1, all roots are real and in $(0, 3)$. Find all possible roots of $p$.

The roots 1 and 2 are abvious, but there are more possible roots: the roots of $x^2-3x+1$, $$\frac{3+\sqrt{5}}{2}\quad\text{ and }\quad\frac{3-\sqrt{5}}{2}.$$

Other possible roots I didn't find, but what is interesting (at least for me) I notice that the above non-integer roots have the form $$\frac{3+\sqrt{5}}{2} = 1+\phi\quad\quad\frac{3-\sqrt{5}}{2} = 2-\phi,$$ where $$\phi = \frac{\sqrt{5}+1}{2}$$ is the golden ratio constant.

1. Is the set of all possible roots $\{1, 1+\phi, 2-\phi, 2\}$ ?
2. (if the answer of question 1 is positive) Why the golden ratio constant appears in this problem ?
• I'm slightly confused. So is your question to find all real algebraic integers between 0 an 3? Yes $\phi$ is one of them, but not all of them have anything to do with $\phi$ – Mathmo123 Aug 5 '14 at 13:26
• @Mathmo123: Not just real algebraic number, for example, $\phi$ is not one of them, because it is the root of $p(x)=x^2-x-1$, but the other root of $p$ is $$\frac{1-\sqrt{5}}{2}\not\in (0,3)$$ – Mher Aug 5 '14 at 13:49
• I don't have time to chase this up right now, but I think the question is related to the concept of "transfinite diameter" of an interval. – Gerry Myerson Aug 6 '14 at 7:37
• @Mher: Alternatively,$$\frac{3+\sqrt{5}}{2}=\phi^2,\;\;\frac{3-\sqrt{5}}{2}=1/\phi^2$$ – Tito Piezas III Jun 18 '15 at 6:19

This answer is only for degree $2$.

There are only four pairs $(a,b)=(-2,1),(-3,1),(-3,2),(-4,4).$ (Maybe you want to eliminate $(-2,1),(-4,4)$, though.)

If $x^2+ax+b=0\ (a,b\in\mathbb Z)$, then we have $$0\lt \frac{-a-\sqrt{a^2-4b}}{2}\lt 3,\ \ 0\lt\frac{-a+\sqrt{a^2-4b}}{2}\lt 3$$ $$\iff 0\lt -a-\sqrt{a^2-4b}\lt 6,\ \ 0\lt -a+\sqrt{a^2-4b}\lt 6\tag1$$ Since we have $0\lt -2a\lt 12\iff -6\lt a\lt 0$, $$(1)\iff a\lt -\sqrt{a^2-4b}\lt a+6,\ \ a\lt\sqrt{a^2-4b}\lt a+6$$ $$\iff a\lt -\sqrt{a^2-4b},\ \ \sqrt{a^2-4b}\lt a+6$$ $$\iff -a\gt \sqrt{a^2-4b},\ \ \sqrt{a^2-4b}\lt a+6$$ $$\iff (-a)^2\gt a^2-4b,\ \ a^2-4b\lt (a+6)^2\iff b\gt 0,\ \ b\gt -3a-9.$$

With $a^2-4b\ge 0$, we have $$-6\lt a\lt 0,\ \ b\gt 0,\ \ b\gt -3a-9,\ \ b\le \frac{a^2}{4}.$$

Considering these conditions on $ab$ plane will give you all possible pairs $(a,b)$, which are the followings :

$$(a,b)=(-2,1),(-3,1),(-3,2),(-4,4)$$

• OK, greate, but it is not supposed that $\deg p=2$. – Mher Aug 5 '14 at 13:10
• @Mher: Oh, my bad. I got it. – mathlove Aug 5 '14 at 14:25
• What about $(-2,1)$? Like $(-4,4)$ was the the repeated root at $2$ right? – snulty Aug 6 '14 at 7:21
• @snulty: Oh, you are right. Thanks. – mathlove Aug 6 '14 at 8:23

Polya proved that any interval of length less than 4 contains only finitely many sets of conjugate algebraic integers (in the language of the original question, only finitely many polynomials with integer coefficients, leading coefficient 1, all roots real and in any given interval of length less than 4). I'm not sure where Polya proved this. It might be in his paper Ueber Ganzwertige Ganze Funktionen, Rend Circ Mat Palermo 40 (1915) 1-16.

Raphael M Robinson, Algebraic equations with span less than 4, Math Comp 18 (1964) 547-559, MR 0169374, 29 #6624 (I believe the paper is freely available on the American Math Society website) found all the irreducible polynomials of degree between 2 and 6, integer coefficients, leading coefficient 1, all roots real, greatest root minus smallest ("span") less than 4, and also several such polynomials of degrees 7 and 8.

It was later proved by Capparelli et al that Robinson's list was complete through degree 8.

The only ones with span under 3 were the ones OP found with $\sqrt5$, and the polynomial $x^2-2$. But there's no way to translate $x^2-2$ so its roots are between $0$ and $3$ and stay algebraic integers.

The ones with higher degree all have span close to 4, so I'm convinced that OP has found all the ones with roots between $0$ and $3$, even though I haven't been able to find a paper that says exactly that.

It might be worth having a look at Capparelli et al, On the span of polynomials with integer coefficients, Math Comp 79 (2010) #270 967-981, MR 2600551, 2011b:12001, where Robinson's computations are extended to degree 14, just to see whether it gives any history, or cites Polya, or mentions span 3.


text:
# Gradient of an angle in terms of the vertices

Let $\theta(\vec p, \vec q, \vec r)$ be the angle theta between 3D real vectors $(\vec{q}-\vec{p})$ and $(\vec{r} - \vec{p})$. What is a simple expression of $\nabla \theta$ in terms of $\vec{p}$, $\vec{q}$, and $\vec{r}$ that is efficient for computation?

The context is that I'm writing a minimization function that operates on a deformable triangle mesh; I would like to use a harmonic angle potential to prevent significant deformation of the angles in the mesh. The potential energy function for this would be $E(\theta) = K (\theta - \theta_0)^2$ and the gradient in terms of $\theta$ is $\nabla E = 2 K (\theta - \theta_0)$ for a constant $K$ and an initial angle $\theta_0$. Finding the form of $\nabla E$ in terms of the vector values is quite trivial with the exception of finding the form of $\nabla \theta$ in terms of the vectors.

It's easy enough to pop this expression into Mathematica and get a monstrous form of the gradient out in terms of $(x_p, y_p, z_p, \; x_q, y_q, z_q, \; x_r, y_r, z_r)$; if I were only calculating this gradient over a few values this would be fine, but I'll be calculating this for hundreds of thousands of angles, and need the calculation to be very efficient. My intuition is that the derivative in terms of $\vec{q}$ or $\vec{r}$ should be orthogonal to $\vec{q} - \vec{p}$ or $\vec{r} - \vec{p}$, respectively, and that the derivative in terms of $\vec{p}$ should be parallel to the angle bisector. This doesn't particularly help me come up with something more efficient, however. I've found numerous resources that talk about this particular problem (here, for example, section on Angle Potential), usually interms of molecular dynamics simulation, but have been unable to find a simple form of the gradient.

I am also looking answer for the same question. So far I am done with below. I am referring to the equation in Gromacs manual for harmonic angle bond. $$V(\theta_{ijk})=\frac{1}{2}k_{ijk}^\theta (\theta_{ijk}-\theta_{0})^{2}$$ $$\theta_{ijk}=cos^{-1} \frac{\overrightarrow{r_{ij}}.\overrightarrow{r_{kj}}}{r_{ij}r_{kj}}$$ $$F_{i}=-\frac{\partial V(\theta)}{\partial r_{i}}$$ $$F_{i}= -k_{ijk}^\theta.(\theta_{ijk}-\theta_{0}).\frac{\partial \theta_{ijk}}{\partial r_i}$$ $$\frac{\partial \theta_{ijk}}{\partial r_i}=\frac{-1}{\sqrt{1-u^{2}}}.\frac{\partial u}{\partial r_i}$$ where $$u=\frac{\overrightarrow{r_{ij}}.\overrightarrow{r_{kj}}}{r_{ij}r_{kj}}$$ Now, $$\frac{\partial u}{\partial r_i}=\frac{1}{r_{ij}r_{kj}}(\overrightarrow{r_{ij}}.\frac{d\overrightarrow{r_{kj}}}{dr_i}+\frac{d\overrightarrow{r_{ij}}}{dr_i}.\overrightarrow{r_{kj}} ) ??$$ I am stuck in the last part...Anybody with better grip in maths can please point out the missing detail...

• This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. – Mike Pierce Apr 12 '15 at 3:39
• @mapierce271 It is the same question with different notation..I didnt want to create a duplicate entry...Also couldnt comment on the question because of low reputation, so I put the partial answer... – Sumith Yesudasan Apr 12 '15 at 4:47
• @mapierce271 I think this partial answers contributes significantly, so I think it's ok to leave this here. Maybe make it Community Wiki and encourage editing/collaboration? – Newb Apr 12 '15 at 5:23

This website gives an answer to the problem; see equations 4.108 - 4.110: http://www.mbnexplorer.com/documentation/3-energy-and-force-calculation/33-molecular-mechanics-potential

In case the link disappears, here are the equations, translated into the notation of the original question; note that these give the gradient, as the question asked, while the above website, which is about the molecular dynamics problem in particular, gives the force ($-\nabla$).

$$E(\vec{p}, \vec{q}, \vec{r}) = K\left( \theta(\vec{p}, \vec{q}, \vec{r}) - \theta_0\right)^2\\ \nabla E(\vec{p}, \vec{q}, \vec{r}) = \left( \frac{\partial E(\vec{p}, \vec{q}, \vec{r})}{\partial \vec{p}}, \frac{\partial E(\vec{p}, \vec{q}, \vec{r})}{\partial \vec{q}}, \frac{\partial E(\vec{p}, \vec{q}, \vec{r})}{\partial \vec{r}}\right)\\ \frac{\partial E(\vec{p}, \vec{q}, \vec{r})}{\partial \vec{p}} = -\frac{\partial E(\vec{p}, \vec{q}, \vec{r})}{\partial \vec{q}} - \frac{\partial E(\vec{p}, \vec{q}, \vec{r})}{\partial \vec{r}}\\ \frac{\partial E(\vec{p}, \vec{q}, \vec{r})}{\partial \vec{q}} = \frac{2K\left( \theta(\vec{p},\vec{q},\vec{r}) - \theta_0\right)}{||\vec{q}-\vec{p}||\,\sin\left(\theta(\vec{p},\vec{q},\vec{r})\right)} \left( \frac{\vec{r}-\vec{p}}{||\vec{r}-\vec{p}||} - \frac{\vec{q}-\vec{p}}{||\vec{q}-\vec{p}||}\cos\left(\theta(\vec{p},\vec{q},\vec{r})\right) \right)\\ \frac{\partial E(\vec{p}, \vec{q}, \vec{r})}{\partial \vec{r}} = \frac{2K\left( \theta(\vec{p},\vec{q},\vec{r}) - \theta_0\right)}{||\vec{r}-\vec{p}||\,\sin\left(\theta(\vec{p},\vec{q},\vec{r})\right)} \left( \frac{\vec{q}-\vec{p}}{||\vec{q}-\vec{p}||} - \frac{\vec{r}-\vec{p}}{||\vec{r}-\vec{p}||}\cos\left(\theta(\vec{p},\vec{q},\vec{r})\right) \right)$$

While I haven't derived the correct solution, I did plop the solution into a Mathematica instance and verified that it agreed with the numerical solution for a wide range of values.

This is a very late answer, but I just leave it here for others, since it seems these things are not readily available online:

$$\theta = acos (\frac{\vec r_{ij} \cdot \vec r_{kj}}{r_{ij} r_{kj}})$$

Let's look at the derivative with respect to $\vec r_i$:

$$\frac{\partial \theta}{\partial \vec r_i} = \frac{\partial \theta}{\partial r_{ij}} \frac{\partial r_{ij}}{\partial \vec r_{ij}} \frac{\partial \vec r_{ij}}{\partial \vec r_{i}} = \frac{-1}{\sqrt{1-(\frac{\vec r_{ij} \cdot \vec r_{kj}}{r_{ij} r_{kj}})^2}} (\frac{\partial \vec r_{ij}}{\partial r_{ij}} \cdot \frac{\vec r_{kj}}{r_{ij} r_{kj}} + \frac{-1}{r_{ij}^2} \frac{\vec r_{ij} \cdot \vec r_{kj}}{r_{kj}}) \frac{\partial r_{ij}}{\partial \vec r_{ij}} \frac{\partial \vec r_{ij}}{\partial \vec r_{i}}$$

Since $\vec r_{ij} = \vec r_i - \vec r_j$, the last derivative term is simply identity. Also, for $\vec r = (x,y,z)$, $r = \sqrt{x^2+y^2+z^2}$, so the second last derivative is:

$$\frac{\partial r}{\partial \vec r} = \frac{1}{2 \sqrt{x^2+y^2+z^2}}(2x, 2y, 2z) = \frac{\vec r}{r}$$

Thus: $$\frac{\partial \theta}{\partial \vec r_i} = \frac{- r_{ij} r_{kj}}{\sqrt{r_{ij}^2 r_{kj}^2 - (\vec r_{ij} \cdot \vec r_{kj})^2}} (\frac{\vec r_{kj}}{r_{ij} r_{kj}} - \frac{\vec r_{ij} \cdot \vec r_{kj}}{r_{ij}^2 r_{kj}} \frac{\vec r_{ij}}{r_{ij}}) = \frac {-1}{r_{ij}} \frac{r_{ij}^2 \vec r_{kj} - (\vec r_{ij} \cdot \vec r_{kj}) \vec r_{ij}} {\sqrt{r_{ij}^4 r_{kj}^2 - (\vec r_{ij} \cdot \vec r_{kj})^2 r_{ij}^2}}$$

The numerator is the vector triple product, so the final result is: $$\frac{norm(\vec r_{ij} \times (\vec r_{ij} \times \vec r_{kj}))}{r_{ij}}$$

You can apply the same idea to find the derivative with respect to $\vec r_k$ and $\vec r_j$. Also, this also works with dihedral angle formula, just with more things to take care of.


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# Holomorphic Logarithms and Roots

In this post we’ll make sense of a holomorphic square root and logarithm. Wrote this up because I was surprised how hard it was to find a decent complete explanation.

Let ${f : U \rightarrow \mathbb C}$ be a holomorphic function. A holomorphic ${n}$th root of ${f}$ is a function ${g : U \rightarrow \mathbb C}$ such that ${f(z) = g(z)^n}$ for all ${z \in U}$. A logarithm of ${f}$ is a function ${g : U \rightarrow \mathbb C}$ such that ${f(z) = e^{g(z)}}$ for all ${z \in U}$. The main question we’ll try to figure out is: when do these exist? In particular, what if ${f = \mathrm{id}}$?

## 1. Motivation: Square Root of a Complex Number

To start us off, can we define ${\sqrt z}$ for any complex number ${z}$?

The first obvious problem that comes up is that there for any ${z}$, there are two numbers ${w}$ such that ${w^2 = z}$. How can we pick one to use? For our ordinary square root function, we had a notion of “positive”, and so we simply took the positive root.

Let’s expand on this: given ${ z = r \left( \cos\theta + i \sin\theta \right) }$ (here ${r \ge 0}$) we should take the root to be

$\displaystyle w = \sqrt{r} \left( \cos \alpha + i \sin \alpha \right).$

such that ${2\alpha \equiv \theta \pmod{2\pi}}$; there are two choices for ${\alpha \pmod{2\pi}}$, differing by ${\pi}$.

For complex numbers, we don’t have an obvious way to pick ${\alpha}$. Nonetheless, perhaps we can also get away with an arbitrary distinction: let’s see what happens if we just choose the ${\alpha}$ with ${-\frac{1}{2}\pi < \alpha \le \frac{1}{2}\pi}$.

Pictured below are some points (in red) and their images (in blue) under this “upper-half” square root. The condition on ${\alpha}$ means we are forcing the blue points to lie on the right-half plane.

Here, ${w_i^2 = z_i}$ for each ${i}$, and we are constraining the ${w_i}$ to lie in the right half of the complex plane. We see there is an obvious issue: there is a big discontinuity near the point ${z_5}$ and ${z_7}$! The nearby point ${w_6}$ has been mapped very far away. This discontinuity occurs since the points on the negative real axis are at the “boundary”. For example, given ${-4}$, we send it to ${-2i}$, but we have hit the boundary: in our interval ${-\frac{1}{2}\pi \le \alpha < \frac{1}{2}\pi}$, we are at the very left edge.

The negative real axis that we must not touch is is what we will later call a branch cut, but for now I call it a ray of death. It is a warning to the red points: if you cross this line, you will die! However, if we move the red circle just a little upwards (so that it misses the negative real axis) this issue is avoided entirely, and we get what seems to be a “nice” square root.

In fact, the ray of death is fairly arbitrary: it is the set of “boundary issues” that arose when we picked ${-\frac{1}{2}\pi < \alpha \le \frac{1}{2}\pi}$. Suppose we instead insisted on the interval ${0 \le \alpha < \pi}$; then the ray of death would be the positive real axis instead. The earlier circle we had now works just fine.

What we see is that picking a particular ${\alpha}$-interval leads to a different set of edge cases, and hence a different ray of death. The only thing these rays have in common is their starting point of zero. In other words, given a red circle and a restriction of ${\alpha}$, I can make a nice “square rooted” blue circle as long as the ray of death misses it.

So, what exactly is going on?

## 2. Square Roots of Holomorphic Functions

To get a picture of what’s happening, we would like to consider a more general problem: let ${f: U \rightarrow \mathbb C}$ be holomorphic. Then we want to decide whether there is a ${g : U \rightarrow \mathbb C}$ such that

$\displaystyle f(z) = g(z)^2.$

Our previous discussion when ${f = \mathrm{id}}$ tells us we cannot hope to achieve this for ${U = \mathbb C}$; there is a “half-ray” which causes problems. However, there are certainly functions ${f : \mathbb C \rightarrow \mathbb C}$ such that a ${g}$ exists. As a simplest example, ${f(z) = z^2}$ should definitely have a square root!

Now let’s see if we can fudge together a square root. Earlier, what we did was try to specify a rule to force one of the two choices at each point. This is unnecessarily strict. Perhaps we can do something like the following: start at a point in ${z_0 \in U}$, pick a square root ${w_0}$ of ${f(z_0)}$, and then try to “fudge” from there the square roots of the other points. What do I mean by fudge? Well, suppose ${z_1}$ is a point very close to ${z_0}$, and we want to pick a square root ${w_1}$ of ${f(z_1)}$. While there are two choices, we also would expect ${w_0}$ to be close to ${w_1}$. Unless we are highly unlucky, this should tells us which choice of ${w_1}$ to pick. (Stupid concrete example: if I have taken the square root ${-4.12i}$ of ${-17}$ and then ask you to continue this square root to ${-16}$, which sign should you pick for ${\pm 4i}$?)

There are two possible ways we could get unlucky in the scheme above: first, if ${w_0 = 0}$, then we’re sunk. But even if we avoid that, we have to worry that we are in a situation, where we run around a full loop in the complex plane, and then find that our continuous perturbation has left us in a different place than we started. For concreteness, consider the following situation, again with ${f = \mathrm{id}}$:

We started at the point ${z_0}$, with one of its square roots as ${w_0}$. We then wound a full red circle around the origin, only to find that at the end of it, the blue arc is at a different place where it started!

The interval construction from earlier doesn’t work either: no matter how we pick the interval for ${\alpha}$, any ray of death must hit our red circle. The problem somehow lies with the fact that we have enclosed the very special point ${0}$.

Nevertheless, we know that if we take ${f(z) = z^2}$, then we don’t run into any problems with our “make it up as you go” procedure. So, what exactly is going on?

## 3. Covering Projections

By now, if you have read the part of algebraic topology. this should all seem very strangely familiar. The “fudging” procedure exactly describes the idea of a lifting.

More precisely, recall that there is a covering projection

$\displaystyle (-)^2 : \mathbb C \setminus \{0\} \rightarrow \mathbb C \setminus \{0\}.$

Let ${V = \left\{ z \in U \mid f(z) \neq 0 \right\}}$. For ${z \in U \setminus V}$, we already have the square root ${g(z) = \sqrt{f(z)} = \sqrt 0 = 0}$. So the burden is completing ${g : V \rightarrow \mathbb C}$.

Then essentially, what we are trying to do is construct a lifting ${g}$ for the following diagram: Our map ${p}$ can be described as “winding around twice”. From algebraic topology, we now know that this lifting exists if and only if

$\displaystyle f_\ast(\pi_1(V)) \subseteq p_\ast(\pi_1(E))$

is a subset of the image of ${\pi_1(E)}$ by ${p}$. Since ${B}$ and ${E}$ are both punctured planes, we can identify them with ${S^1}$.

Ques 1

Show that the image under ${p}$ is exactly ${2\mathbb Z}$ once we identify ${\pi_1(B) = \mathbb Z}$.

That means that for any loop ${\gamma}$ in ${V}$, we need ${f \circ \gamma}$ to have an even winding number around ${0 \in B}$. This amounts to

$\displaystyle \frac{1}{2\pi} \oint_\gamma \frac{f'}{f} \; dz \in 2\mathbb Z$

since ${f}$ has no poles.

Replacing ${2}$ with ${n}$ and carrying over the discussion gives the first main result.

Theorem 2 (Existence of Holomorphic ${n}$th Roots)

Let ${f : U \rightarrow \mathbb C}$ be holomorphic. Then ${f}$ has a holomorphic ${n}$th root if and only if

$\displaystyle \frac{1}{2\pi i}\oint_\gamma \frac{f'}{f} \; dz \in n\mathbb Z$

for every contour ${\gamma}$ in ${U}$.

## 4. Complex Logarithms

The multivalued nature of the complex logarithm comes from the fact that

$\displaystyle \exp(z+2\pi i) = \exp(z).$

So if ${e^w = z}$, then any complex number ${w + 2\pi i k}$ is also a solution.

We can handle this in the same way as before: it amounts to a lifting of the following diagram. There is no longer a need to work with a separate ${V}$ since:

Ques 3

Show that if ${f}$ has any zeros then ${g}$ possibly can’t exist.

In fact, the map ${\exp : \mathbb C \rightarrow \mathbb C\setminus\{0\}}$ is a universal cover, since ${\mathbb C}$ is simply connected. Thus, ${p(\pi_1(\mathbb C))}$ is trivial. So in addition to being zero-free, ${f}$ cannot have any winding number around ${0 \in B}$ at all. In other words:

Theorem 4 (Existence of Logarithms)

Let ${f : U \rightarrow \mathbb C}$ be holomorphic. Then ${f}$ has a logarithm if and only if

$\displaystyle \frac{1}{2\pi i}\oint_\gamma \frac{f'}{f} \; dz = 0$

for every contour ${\gamma}$ in ${U}$.

## 5. Some Special Cases

The most common special case is

Corollary 5 (Nonvanishing Functions from Simply Connected Domains)

Let ${f : \Omega \rightarrow \mathbb C}$ be continuous, where ${\Omega}$ is simply connected. If ${f(z) \neq 0}$ for every ${z \in \Omega}$, then ${f}$ has both a logarithm and holomorphic ${n}$th root.

Finally, let’s return to the question of ${f = \mathrm{id}}$ from the very beginning. What’s the best domain ${U}$ such that we can define ${\sqrt{-} : U \rightarrow \mathbb C}$? Clearly ${U = \mathbb C}$ cannot be made to work, but we can do almost as well. For note that the only zero of ${f = \mathrm{id}}$ is at the origin. Thus if we want to make a logarithm exist, all we have to do is make an incision in the complex plane that renders it impossible to make a loop around the origin. The usual choice is to delete negative half of the real axis, our very first ray of death; we call this a branch cut, with branch point at ${0 \in \mathbb C}$ (the point which we cannot circle around). This gives

Theorem 6 (Branch Cut Functions)

There exist holomorphic functions

\displaystyle \begin{aligned} \log &: \mathbb C \setminus (-\infty, 0] \rightarrow \mathbb C \\ \sqrt[n]{-} &: \mathbb C \setminus (-\infty, 0] \rightarrow \mathbb C \end{aligned}

satisfying the obvious properties.

There are many possible choices of such functions (${n}$ choices for the ${n}$th root and infinitely many for ${\log}$); a choice of such a function is called a branch. So this is what is meant by a “branch” of a logarithm.

The principal branch is the “canonical” branch, analogous to the way we arbitrarily pick the positive branch to define ${\sqrt{-} : \mathbb R_{\ge 0} \rightarrow \mathbb R_{\ge 0}}$. For ${\log}$, we take the ${w}$ such that ${e^w = z}$ and the imaginary part of ${w}$ lies in ${(-\pi, \pi]}$ (since we can shift by integer multiples of ${2\pi i}$). Often, authors will write ${\text{Log } z}$ to emphasize this choice.

Example 7

Let ${U}$ be the complex plane minus the real interval ${[0,1]}$. Then the function ${U \rightarrow \mathbb C}$ by ${z \mapsto z(z-1)}$ has a holomorphic square root.

Corollary 8

A holomorphic function ${f : U \rightarrow \mathbb C}$ has a holomorphic ${n}$th root for all ${n \ge 1}$ if and only if it has a holomorphic logarithm.

# Algebraic Topology Functors

This will be old news to anyone who does algebraic topology, but oddly enough I can’t seem to find it all written in one place anywhere, and in particular I can’t find the bit about ${\mathsf{hPairTop}}$ at all.

In algebraic topology you (for example) associate every topological space ${X}$ with a group, like ${\pi_1(X, x_0)}$ or ${H_5(X)}$. All of these operations turn out to be functors. This isn’t surprising, because as far as I’m concerned the definition of a functor is “any time you take one type of object and naturally make another object”.

The surprise is that these objects also respect homotopy in a nice way; proving this is a fair amount of the “setup” work in algebraic topology.

## 1. Homology, ${H_n : \mathsf{hTop} \rightarrow \mathsf{Grp}}$

Note that ${H_5}$ is a functor

$\displaystyle H_5 : \mathsf{Top} \rightarrow \mathsf{Grp}$

i.e. to every space ${X}$ we can associate a group ${H_5(X)}$. (Of course, replace ${5}$ by integer of your choice.) Recall that:

Definition 1

Two maps ${f, g : X \rightarrow Y}$ are homotopy equivalent if there exists a homotopy between them.

Thus for a map we can take its homotopy class ${[f]}$ (the equivalence class under this relationship). This has the nice property that ${[f \circ g] = [f] \circ [g]}$ and so on.

Definition 2

Two spaces ${X}$ and ${Y}$ are homotopic if there exists a pair of maps ${f : X \rightarrow Y}$ and ${g : Y \rightarrow X}$ such that ${[f \circ g] = [\mathrm{id}_X]}$ and ${[g \circ f] = [\mathrm{id}_Y]}$.

In light of this, we can define

Definition 3

The category ${\mathsf{hTop}}$ is defined as follows:

• The objects are topological spaces ${X}$.
• The morphisms ${X \rightarrow Y}$ are homotopy classes of continuous maps ${X \rightarrow Y}$.

Remark 4

Composition is well-defined since ${[f \circ g] = [f] \circ [g]}$. Two spaces are isomorphic in ${\mathsf{hTop}}$ if they are homotopic.

Remark 5

As you might guess this “quotient” construction is called a quotient category.

Then the big result is that:

Theorem 6

The induced map ${f_\sharp = H_n(f)}$ of a map ${f: X \rightarrow Y}$ depends only on the homotopy class of ${f}$. Thus ${H_n}$ is a functor

$\displaystyle H_n : \mathsf{hTop} \rightarrow \mathsf{Grp}.$

The proof of this is geometric, using the so-called prism operators. In any case, as with all functors we deduce

Corollary 7

${H_n(X) \cong H_n(Y)}$ if ${X}$ and ${Y}$ are homotopic.

In particular, the contractible spaces are those spaces ${X}$ which are homotopy equivalent to a point. In which case, ${H_n(X) = 0}$ for all ${n \ge 1}$.

## 2. Relative Homology, ${H_n : \mathsf{hPairTop} \rightarrow \mathsf{Grp}}$

In fact, we also defined homology groups

$\displaystyle H_n(X,A)$

for ${A \subseteq X}$. We will now show this is functorial too.

Definition 8

Let ${\varnothing \neq A \subset X}$ and ${\varnothing \neq B \subset X}$ be subspaces, and consider a map ${f : X \rightarrow Y}$. If ${f(A) \subseteq B}$ we write

$\displaystyle f : (X,A) \rightarrow (Y,B).$

We say ${f}$ is a map of pairs, between the pairs ${(X,A)}$ and ${(Y,B)}$.

Definition 9

We say that ${f,g : (X,A) \rightarrow (Y,B)}$ are pair-homotopic if they are “homotopic through maps of pairs”.

More formally, a pair-homotopy ${f, g : (X,A) \rightarrow (Y,B)}$ is a map ${F : [0,1] \times X \rightarrow Y}$, which we’ll write as ${F_t(X)}$, such that ${F}$ is a homotopy of the maps ${f,g : X \rightarrow Y}$ and each ${F_t}$ is itself a map of pairs.

Thus, we naturally arrive at two categories:

• ${\mathsf{PairTop}}$, the category of pairs of toplogical spaces, and
• ${\mathsf{hPairTop}}$, the same category except with maps only equivalent up to homotopy.

Definition 10

As before, we say pairs ${(X,A)}$ and ${(Y,B)}$ are pair-homotopy equivalent if they are isomorphic in ${\mathsf{hPairTop}}$. An isomorphism of ${\mathsf{hPairTop}}$ is a pair-homotopy equivalence.

Then, the prism operators now let us derive

Theorem 11

We have a functor

$\displaystyle H_n : \mathsf{hPairTop} \rightarrow \mathsf{Grp}.$

The usual corollaries apply.

Now, we want an analog of contractible spaces for our pairs: i.e. pairs of spaces ${(X,A)}$ such that ${H_n(X,A) = 0}$ for ${n \ge 1}$. The correct definition is:

Definition 12

Let ${A \subset X}$. We say that ${A}$ is a deformation retract of ${X}$ if there is a map of pairs ${r : (X, A) \rightarrow (A, A)}$ which is a pair homotopy equivalence.

Example 13 (Examples of Deformation Retracts)

1. If a single point ${p}$ is a deformation retract of a space ${X}$, then ${X}$ is contractible, since the retraction ${r : X \rightarrow \{\ast\}}$ (when viewed as a map ${X \rightarrow X}$) is homotopic to the identity map ${\mathrm{id}_X : X \rightarrow X}$.
2. The punctured disk ${D^2 \setminus \{0\}}$ deformation retracts onto its boundary ${S^1}$.
3. More generally, ${D^{n} \setminus \{0\}}$ deformation retracts onto its boundary ${S^{n-1}}$.
4. Similarly, ${\mathbb R^n \setminus \{0\}}$ deformation retracts onto a sphere ${S^{n-1}}$.

Of course in this situation we have that

$\displaystyle H_n(X,A) \cong H_n(A,A) = 0.$

## 3. Homotopy, ${\pi_1 : \mathsf{hTop}_\ast \rightarrow \mathsf{Grp}}$

As a special case of the above, we define

Definition 14

The category ${\mathsf{Top}_\ast}$ is defined as follows:

• The objects are pairs ${(X, x_0)}$ of spaces ${X}$ with a distinguished basepoint ${x_0}$. We call these pointed spaces.
• The morphisms are maps ${f : (X, x_0) \rightarrow (Y, y_0)}$, meaning ${f}$ is continuous and ${f(x_0) = y_0}$.

Now again we mod out:

Definition 15

Two maps ${f , g : (X, x_0) \rightarrow (Y, y_0)}$ of pointed spaces are homotopic if there is a homotopy between them which also fixes the basepoints. We can then, in the same way as before, define the quotient category ${\mathsf{hTop}_\ast}$.

And lo and behold:

Theorem 16

We have a functor

$\displaystyle \pi_1 : \mathsf{hTop}_\ast \rightarrow \mathsf{Grp}.$

Same corollaries as before.


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Acceleration G and Braking G

A lap of Mazda Raceway Laguna Seca, showing speed (black), acceleration G (green) and braking G (red). The acceleration and braking G channels are derived from the longitudinal acceleration data (blue).

Two channels can be generated by dissecting the longitudinal acceleration channel into its positive and negative components. The first channel, which we will call Acceleration G, shows only the values for positive longitudinal acceleration – when the bike is what we commonly refer to as accelerating. The second channel we’ll call Braking G, and shows only the values for negative acceleration – when the bike is decelerating, or braking. Here, the braking forces will displayed as positive values rather than negative. The real value of these new channels is that it is much easier to visualize braking and acceleration as separate, positive channels, rather than looking only at longitudinal G. Additionally, separating them into two distinct channels makes it easier to use the data in other math channels.

These channels are calculated as:

$\text{Acceleration G = max (Longitudinal G, 0)}$

$\text{Braking G = max (-Longitudinal G, 0)}$

Here, the max () function refers to the maximum value. Your software should have this function; if not, the channels can be generated using some math and the abs (absolute value) function, or some creative addition and subtraction.

The attached chart shows a lap of Mazda Raceway Laguna Seca with the longitudinal acceleration broken down into these components.  Speed is shown in black. As we have discussed before, it is difficult to determine much information at a glance from the longitudinal acceleration (blue) trace on its own; this is mostly because it has both positive and negative components, and many software packages do not clearly show the zero axis. Split into its two components for acceleration and braking (green and red, respectively), it is much easier to see at a glance how much the rider is accelerating out of each corner or braking into each corner.

Now it is much easier to see maximum values as well as how quickly those values are reached. For braking G, it is also useful to see how braking tapers off through the braking zone and as the rider arcs into the corner. These traits give the acceleration G and braking G channels their characteristic shapes, and we can also use additional math channels to put values on that information and better compare data from lap to lap.