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# Fastest Rolling Shape?

The following questions occurred to me. This is not research mathematics, just idle curiosity. Apologies if it is inappropriate.

1. Suppose you have a fixed volume V of maleable material, perhaps clay. The goal is to form it into a shape S (convex or nonconvex) that would roll down an inclined plane as fast as possible. The plane is tilted at θ with respect to the horizontal. The race track that quantifies "as fast as possible" is of length L. The shape S must be entirely behind a starting plane orthogonal to the inclined plane, and its race is finished when it is entirely ahead of a finishing plane, L distant, again orthogonal to the inclined plane. So if S is a disk of radius r, its center of gravity will have to travel a total distance of L + 2r to complete the race. S is released at rest, and rolls under gravity. Assume that the materials of S and of the inclined plane have a sufficient coefficient of friction μ such that there is no slippage, just pure rolling.
The stipulation that S start behind and end ahead of the orthogonal planes suggests that the optimal shape is an arbitrarily thin (and therefore arbitrarily long) cylinder, to minimize r. But this does not accord with (my) physical intuition. What am I missing here?

2. A variation on the same problem replaces the inclined plane with an inclined half-cylinder of radius R, like a rain gutter. It may be necessary to assume a relation between R and the volume V of material to make this problem reasonable (so that S fits in the gutter). But certainly this version's answer could not be an arbitrarily long cylinder!

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I'd like to retag this as "classical mechanics", but such a tag doesn't exist and I couldn't find a way to create it. – Victor Protsak May 25 '10 at 23:18
@Victor Protsak: Done. – Andrey Rekalo May 25 '10 at 23:33
Joseph, can you, please, clarify whether you allow for non-convex S ? If so and if, moreover, S can "stick out of the gutter", a variant of the "pair of wheels" construction may solve problem 2 as well, but I don't want to rewrite the answer for nothing. – Victor Protsak May 26 '10 at 6:00
The problem with one less dimension seems interesting as well: the fixed volume imposes greater restriction, and one can expect the disc to be the optimal shape (though it seems not obvious to prove). – Benoît Kloeckner May 26 '10 at 10:33
Another remark: I guess you assume that the shape stays in contact with the plane, but this is not physically the case for all shapes and length. I guess any non-round cylindrical shape will jump after a long enough run. – Benoît Kloeckner May 26 '10 at 10:36

There is one thing that's way more important than the precise definition of how the race starts and ends: the moment of intertia of $S$, which determines how much energy is wasted on the rotation. Let us assume that $S$ is cylindrically symmetrical with mass $m$, radius $R$, moment of intertia $I$, and let $k=I/mR^2$. Then from the conservation of energy,

$$\frac{I\omega^2}{2}+\frac{mv^2}{2}=mgx\sin\theta,$$

and using that $v=\omega R,$ we get that

$$\frac{dx}{dt}=v=\sqrt{\frac{2g\sin\theta}{k+1}}x^{1/2}.$$

This already makes it clear that in order to minimize the rolling time, $k$ must be as small as possible, but this can also be quantified by solving the separable ODE and finding that

$$T=\sqrt{\frac{2(k+1)d}{g\sin\theta}}\geq \sqrt{\frac{2L}{g\sin\theta}},$$

where $d=L+2R$ is the distance traveled. The conclusion, in the cylindrically symmetrical case of radius $R$, is that $S$ should be as close as possible to the infinitely thin heavy rod with a disk-shaped thin "flap" of radius $R$ attached, so that $k$ is just slightly over $1$. In a practical sense, imagine a wheel pair with very light rims and a very heavy axle. [I like to think of this situation as "anti-flywheel problem": we want to minimize rotation energy accumulated by $S$, so contrary to the usual flywheel, the wheel is light and the axle is heavy.] If your "malleable material" has uniform density and $R$ may vary, one sequence of mathematical solutions approaching the time bound consists of solid cylinders with radii $r_i$ shrinking to 0 and growing widths $\ell_i$, with two $w$-thin disks of radius $R_i$ attached, where

$$\pi(\ell_i r_{i}^2+2w R_i^2)=V, \quad r_i=o(R_i), \quad R_i\to 0.$$

[I've only put in an explicit sequence because of a clarification request in the comments; I actually think that this level of detail hinders understanding]. It's also clear that the cylindrical symmetry restriction is irrelevant.

The second problem is more difficult: I don't have much to say on it other than that, by the same token, a solid ball is better than a disk/cylinder even if it could be fitted in the "gutter".

P.S.: "Physical" intuition may be deceptive: many people assume that a ball will roll according to the same law of motion as a cylinder or an ideal point mass. This actually happened to some of my mathematically oriented classmates in a physics lab assignment, who tried to make up data in order to avoid doing tedious experiments.

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Is the cylindrical symmetry really irrelevant? For a fixed width the heavy rod is the best solution. But for a fixed maximum radius, your wheel pair with heavy axle is better. Since the inf lies in a non-compact direction, there may be multiple, not necessary the same methods of finding a minimizing sequence. In fact, it is probably also possible to construct a one-parameter family of shapes such that a stationary point (relative to the parameter) is achieved. – Willie Wong May 25 '10 at 23:35
Since there was no width restriction in problem 1, the answer is yes, cylindrical symmetry is irrelevant. The point is that you want to waste as little energy as possible on rotation. However, you've made a good point. Re-reading my answer, I see that I was a bit sloppy with the assumptions: when I said "heavy rod", I wasn't yet thinking of uniform density. Let me mull over it a bit more. – Victor Protsak May 26 '10 at 0:03
What does $I$ stand for in your formula, $k=I/mR^2$? – Gerry Myerson May 26 '10 at 0:17
I've put in some clarifications, but it's supposed to be a fun problem, you can work out the details yourselves! – Victor Protsak May 26 '10 at 1:15
Joseph, suppose you just take a cylinder of uniform density. If you make the radius smaller, the moment of inertia gets smaller, but the cylinder also has a smaller perimeter, so it has to rotate more to get to the bottom. These two effects cancel out -- it doesn't matter how large you make the cylinder, 1/3 of the energy goes into spinning the cylinder and 2/3 into rolling it. If the body has to be convex and circularly symmetric, the best shape is presumably a thin cylinder with two circular pyramids glued onto either side. Without the convexity restriction, Victor's analysis works – Peter Shor May 26 '10 at 21:03

I am going to post a bunch of assumptions which will impose further constraints on the 2nd problem. I think these assumptions maybe natural in view of the thought experiment.

1. Without loss of generality, we can assume that the gutter has width 1, so the width of the gutter is parametrized by $[0,1]$.
2. The cross-section of the gutter itself is parametrized by some convex function $g: [0,1] \to \mathbb{R}$ with $g(0) = g(1)$.
3. The object is not allowed to stick out of the gutter side-ways (else we can just take a heavy rod that rests on the top of the gutter...).
4. The object is assumed to have uniform density and total mass 1.
5. The object is obtained by rotation (since it is to rotate) around an axis.

First, by energy conservation. similar to described in Victor Protsak's answer, to minimize the total time of travel it suffices to minimize $$\mathcal{E} = (L + 2 R_0)( 1 + I / R_1^2 )$$ where $R_0$ is the maximum radius of the rotating body and $R_1$ is radius at which it is rolling, in other words the maximum distance from the axis to a point on the surface of the object that is in contact with the gutter. (Note that it is impossible for a ball of radius 1 to roll inside a gutter of radius 1 without slipping!!! We'll just conveniently forget frictional dissipation for our computations.)

Now, by the definition of the angular momentum and simple re-arrangement argument, it is clear that the cross-section of the object can be obtained as a function over its axis. More precisely, in a cylindrical coordinate system where $z$ is along the axis of rotation, the object $$S = \{ r \leq f(z) \}$$ (If there are any "holes" near the axis, by moving the mass inward closer to the axis it reduces both $I$ and $R$.)

It remains to write $I$ and $R$ as a functional of $f$. Working in the cylindrical coordinate still, the volume of $f$ is $$V(f) = \pi \int_0^1 f(z)^2 dz$$ By assumption this is a constant number.

The moment of inertia is $$I(f) = \int_0^1 \int_0^{f(z)} 2\pi r^3 \mu dr dz = \frac{\mu\pi}{2}\int_0^1 f(z)^4 dz$$ where $\mu = 1 / V$ the uniform mass density.

Lastly, $R_0 = \sup f$ by definition. But $R_1$ is a bit tricky to compute. A bit of thought tells you that $R_1 = \sup_{P} f(z)$, where $P$ is the set of points satisfying $f(z) + g(z) = \sup (f+g)$.

Observe that $R_1 \leq R_0$.

To summarize, the task then becomes: given a function $g(z)$ and constants $\alpha,\beta,V$, minimize the functional $$\mathcal{E} = (\alpha + \|f\|_\infty)(\beta + \frac{\|f\|^4_4}{R_1(f;g)^2} )$$ under the constraint that $\|f\|_2^2 = V$, where $\|\cdot\|_p$ denotes the Lebesgue $p$-norm over $[0,1]$.

If we impose the constraint that $R_1 = R_0$, then the problem reduces to the first one which Victor examined, where the solution is a Rolling cylinder. I rather suspect that the cylinder is also the solution in the general case. (I am pretty sure that to minimize $L^4$ while fixing $L^2$ on a compact interval one needs to make the function as flat as possible.)

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# What is probablity of getting 7 in a dice

Dear student,

The given question seems to be incomplete as 7 is not there on dice

possible outcomeswhen dice is roll= 1,2,3,4,5,6

so, probablity of getting 7 is =$\frac{0}{6}=0$

Regards

• 0
What are you looking for?


text:
Geometry

# Tangent - Sum and Difference Formulas

Which of the following is equal to

$\frac{ \tan 43 ^ \circ - \tan 14 ^ \circ } { 1 + \tan 43 ^ \circ \times \tan 14 ^ \circ } ?$

Which of the following is equal to

$\frac{ \tan 19 ^ \circ + \tan 26 ^ \circ } { 1 - \tan 19 ^ \circ \times \tan 26 ^ \circ } ?$

If $\tan x=\frac{1}{14}$, the value of $\tan \left(x+\frac{\pi}{4}\right)$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. What is the value of $a+b$?

If $\tan N ^ \circ = \tan 12 ^ \circ + \sec 12 ^ \circ$, where $0 \leq N \leq 90$, what is the value of $N$?

Which of the following expressions is equal to $\tan 2 \theta$?

×


text:
Candidate definitions for “1-braided 2-category”?

Recall that a braided monoidal category is a category $\mathcal C$, a functor $\otimes: \mathcal C \times \mathcal C \to \mathcal C$ satisfying some properties, and a natural isomorphism $b_{V,W}: V\otimes W \to W\otimes V$ satisfying some properties. Recall also that a monoidal category (just $\mathcal C,\otimes$ and their properties) is the same as a one-element 2-category: the objects of $\mathcal C$ become the morphisms, and the monoidal structure becomes composition.

Thus, is there a natural definition of "1-braided 2-category"? I'm calling it "1-braided" because the braiding acts on 1-morphisms (as opposed to "braided monoidal 2-category", where the braiding acts on the 0-morphisms).

I realize, of course, that if $V,W$ are morphisms of a 2-category so that $V\circ W$ is defined, then generally $W\circ V$ is not defined, so a priori asking for any relationship $V\circ W \cong W\circ V$ doesn't make sense. On the other hand, consider Aaron Lauda's categorification of $U\_q(\mathfrak{sl}\_2)$. It is a 2-category, but different hom-sets can be more-or-less identified.

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While clearly distinct, I'll add for completeness Ben's previous question which is related: mathoverflow.net/questions/4790/… –  David Jordan Jan 31 '10 at 22:55

In the Baez-Dolan periodic table a braided monoidal category is just a 2-monoidal category (that is a 3-category with one object and one 1-morphism). If you just think that a braiding means that the structure is inherently 3-dimensional, then you might just want to think about a 1-monoidal 2-category (that is a 3-category with one object).

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# Mogulus Goes Generic: Now Called LiveStream

If you want to convey that you’re a generic vendor of live-streaming services, then sure, why not change your name to LiveStream? Just forgive us if, having known your company for two years with the quirky if ambiguously pronounced name Mogulus, we have a little trouble reconciling something so plain.

Anyways, LiveStream, as Mogulus now wants to be known, is refreshing its free service to be more simple and accessible, and letting it be known that it’s signed up more than 1,000 pro accounts — which start at $350 per month and include features such as monetization, analytics and HD — since launching in December. The company has 250,000 total producer accounts. “We’ve been viewed as a bit too complicated,” LiveStream CEO Max Haot told us Monday. “Now you can sign up and go live in a single click.” But more than competitive positioning, LiveStream plans to succeed by simply building a more stable business than other live streamers like Ustream and Justin.tv. The company has adopted aggressive copyright policies, said Haot, including immediate takedowns of copyrighted streams for rights holders like the NBA, and limiting accounts to 50 concurrent viewers until they have been verified. That’s allowed it to secure enough advertising from Google (s goog) and ScanScout to cover the cost of bandwidth for the free service. The paid service is already profitable. And the company has some cash on reserve from its$12.7 million in funding from Gannett and angel investors.

It’s quite possible that the live-streaming market will become a lot bigger than it is today, and nobody will remember the days of trying to remember just how to say “Mogulus” (maw-gyew-lus? mo-guh-lus?). And hey, nobody had been using the URL livestream.com for a while now. But if this doesn’t pan out, I’m going to blame the company’s failure on a rebranding that is just so tragically boring.


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# Some errors when i try to connect Beaglebone with MATLAB

`

bbb = beaglebone
Warning: Cannot determine version number of the board.
In beaglebone/getBoardRevision (line 1693)
In beaglebone (line 276)
Warning: Cannot read the board serial number.
In beaglebone/getBoardSerialNumber (line 1704)
In beaglebone (line 278)
Index exceeds the number of array elements (0).
Error in beaglebone/getLEDConfiguration (line 1391)
trigger = ret{1};
Error in beaglebone/getAvailablePeripherals (line 1563)
obj.LED.(id).Trigger = obj.getLEDConfiguration(name);
Error in beaglebone (line 289)
obj = getAvailablePeripherals(obj);
`


text:
# Important Notations in Options Formulas

In the tutorial on risk management applications of options, we will look at how various options strategies can be used for managing risk in a company. Before we start let’s review the key notations used in options formulas:

Time 0 refers to the time when an options strategy is initiated

Time T refers to the time the option will expire

T – 0 is the time to expiry

t is the current time

S refers to the spot price. So, S0 is the price at time 0. ST is the price at time T.

X is the exercise or strike price

c is the call price. Small c is for European calls and capital C is for American calls.

p is the put price. Small p is for European puts and capital P is for American puts.

r is the risk-free rate

p is the profit

When we measure profit in an option transaction, we will consider only the final value minus the initial pay out (premium). We will not take into consideration the risk involved or the time value of money.

# R Programming Bundle: 25% OFF

Get our R Programming - Data Science for Finance Bundle for just $29$39.
Get it now for just \$29


text:
Question

# Calculate the volume of gas liberated at the anode at STP during the electrolysis of a $$CuSO_{4}$$ solution by a current of $$1\ A$$ passed for $$16$$ minutes and $$5$$ seconds.

A
224 mL
B
56 mL
C
112 mL
D
448 mL

Solution

## The correct option is A $$56\ mL$$Given:-Current, $$I=1A$$Time, $$t$$=$$16$$ min $$5$$ sec= $$965$$ seconds$$n=2$$$$Cu^{+2}+2e^-\longrightarrow Cu$$$$4OH^{-1}\longrightarrow O_{2} + 2H_{2}O +4e^{-1}$$$$W=\cfrac {I\times t\times M}{nF}$$$$\Rightarrow W= \cfrac {1\times 965\times 63.5}{2\times 96500}$$$$\Rightarrow W= 0.3175g$$$$2\times63.5g\longrightarrow 22.4\ litre O_{2\ }$$$$\Rightarrow 0.3175g \longrightarrow \cfrac {0.3175}{63.5\times2}\times 22.4$$Volume of gas liberated during electrolysis = $$56\ ml$$Chemistry

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# What would have been the Maxwell's equations if there were magnetic monopoles?

by roshan2004
Tags: equations, magnetic, maxwell, monopoles
 P: 139 I am specifically talking about differential forms of Maxwell's equations here- I think ( Tell me if I am incorrect here....) since the divergence of the magnetic field is zero, we have to say it is incorrect for my above problem, so the equation should have been the divergence of magnetic field is magnetic charge density divided by what ? ( Is it the permeability of free space ?) Also, I think ( not too sure) the faraday's law: $\frac{-\partial B}{\partial t}$ is also incorerct here- but I am not sure why is it and what should have been the equation for the above condition ?


text:
# How do you simplify -4(n-7)-4(-8n+6)?

Mar 12, 2017

$28 n + 4$
$- 4 n + 28$
The expression simplifies to $- 4 n + 28 - 4 \left(- 8 n + 6\right)$, and by distributing the $- 4$ into the final parenthesis, we get $+ 32 n - 24$.
Thus, $- 4 n + 28 + 32 n - 24 = 28 n + 4$.


text:
# LHO - coupled oscillators

by Brewer
Tags: coupled, oscillators
 P: 230 1. The problem statement, all variables and given/known data The question then goes on to say: Decompose the resulting oscillation as a superposition of symmetric and antisymmetric mode oscillations. Hence give A and B in terms of C 2. Relevant equations 3. The attempt at a solution Well as of yet I'm not sure I fully understand the question. I think all its asking me to do is to combine the two waves to give one equation that describes the resultant wave. As I understand the theory behind it, $$\omega_A$$ is 180 degrees out of phase with $$\omega_S$$. Does this mean that the superposition of the waves cancels each other out? Or does the oscillation of the coupling spring contribute to this? As far as I can gleam as well, A = C and B = -C. Is this right?
 Engineering Sci Advisor Thanks P: 6,041 In the general case, the phases of the two waves are completely independent. The general solution is x1s = Acos(ws t + ps) = x2s x1a = Bcos (wa t + pa) = -x2a where ps and pa are two independent phase angles. In the question, they have given you the particular solution that matches the initial conditions they specified, and that has ps = pa = 0. Your "A = C and B = -C" is not quite right. At time t = 0 you have x1 = x1s + x1a = A + B = C and x2 = x2s + x2a = A - B = 0. The waves don't cancel out because they have different frequencies. The formulas given in the question show that the coupling spring affects the frequency of the antisymmetric mode, but not the symmetric one. That makes sense - in the symmetric mode the coupling spring does not change length so its stiffness is irrelevant, but in the antisymmetric mode it does change length and its stiffness affects the frequency of the mode.
 P: 230 So, I'm not meant to write it as one function like Asinx + Bcosx = Xcos(x+w)? I'm still not 100% sure what's going on with this question and what its asking me. Is the x1 = x1s + x1a = A + B (and simililar for x2) part the first part of the question? Then that is used to give A and B in terms of C?
Engineering
Thanks
P: 6,041

## LHO - coupled oscillators

As far as I can see, all they want you to do is write the motion as

x1 = A cos w1 t + B cos w2 t
x2 = A cos w1 t - B cos w2 t

and use the conditions at t = 0 to find A and B in terms of C.

I agree it seems a strange question. They seem to have done all the hard stuff for you.
 P: 230 So I've done that and gotten that A=B=C/2. Does that sound about right?
Engineering


text:
# A+B for Input-Output Practice (I)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 119606    Accepted Submission(s): 65275

Problem Description
Too easy?! Of course! I specially designed the problem for acm beginners.
You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim.

Input
The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.

Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

Sample Input
1 5 10 20

Sample Output
6 30

Author
lcy

Recommend
JGShining   |   We have carefully selected several similar problems for you:  1001 1000 1008 1108 1040

Statistic | Submit | Discuss | Note

#include<stdio.h>
int main() {
int i,j,num;
while(scanf("%d %d",&i,&j)!=EOF)
printf("%d\n",i+j);
return 0;
}

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# Start a new discussion

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## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeAug 27th 2018

I have spelled out the proof (here) of the claim from

• Michel Dubois-Violette, Ivan Todorov, Exceptional quantum geometry and particle physics II (arXiv:1808.08110)

that the stabilizer in the automorphism group $F_4$ of the octonionic Albert algebra “of a 4d Minkowski subspace” happens to be the exact gauge group of the standard model of particle physics.

• CommentRowNumber2.
• CommentAuthorRichard Williamson
• CommentTimeAug 27th 2018
• (edited Aug 27th 2018)

Tweaking to use the new theorem environment syntax, just to check that it works in this case. [Edit: seems to have worked fine.]

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeAug 27th 2018

All right, thanks. Now I am so used to the old syntax… :-)

Is it okay if we keep using both dialects, or should I get into the habit of adopting the new one?

• CommentRowNumber4.
• CommentAuthorRichard Williamson
• CommentTimeAug 27th 2018
• (edited Aug 27th 2018)

No problem at all (for me at least!) using the old form, use whatever you prefer :-). I just wished to check that the new syntax worked OK on this example.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeDec 2nd 2019

added pointer to today’s

• Ivan Todorov, Exceptional quantum algebra for the standard model of particle physics (arXiv:1911.13124)
• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeDec 26th 2019
• (edited Dec 26th 2019)

added pointer to today’s


text:
# 0.2 Practice tests (1-4) and final exams  (Page 20/36)

 Page 20 / 36

62 . What would be your decision, if you were using α = 0.01?

## 10.3: comparing two independent population proportions

Use the information to answer the next six exercises. You want to know if proportion of homes with cable television service differs between Community A and Community B. To test this, you draw a random sample of 100 for each and record whether they have cable service.

63 . What are the null and alternative hypotheses for this study

64 . If 65 households in Community A have cable service, and 78 households in community B, what is the pooled proportion?

65 . At α = 0.03, will you reject the null hypothesis? What is your conclusion? 65 households in Community A have cable service, and 78 households in community B. 100 households in each community were surveyed.

66 . Using an alpha value of 0.01, would you reject the null hypothesis? What is your conclusion? 65 households in Community A have cable service, and 78 households in community B. 100 households in each community were surveyed.

## 10.4: matched or paired samples

Use the following information to answer the next five exercises. You are interested in whether a particular exercise program helps people lose weight. You conduct a study in which you weigh the participants at the start of the study, and again at the conclusion, after they have participated in the exercise program for six months. You compare the results using a matched-pairs t-test, in which the data is {weight at conclusion – weight at start}. You believe that, on average, the participants will have lost weight after six months on the exercise program.

67 . What are the null and alternative hypotheses for this study?

68 . Calculate the test statistic, assuming that ${\overline{x}}_{d}$ = –5, s d = 6, and n = 30 (pairs).

69 . What are the degrees of freedom for this statistic?

70 . Using α = 0.05, what is your decision regarding the effectiveness of this program in causing weight loss? What is the conclusion?

71 . What would it mean if the t -statistic had been 4.56, and what would have been your decision in that case?

## 11.1: facts about the chi-square distribution

72 . What is the mean and standard deviation for a chi-square distribution with 20 degrees of freedom?

## 11.2: goodness-of-fit test

Use the following information to answer the next four exercises. Nationally, about 66 percent of high school graduates enroll in higher education. You perform a chi-square goodness of fit test to see if this same proportion applies to your high school’s most recent graduating class of 200. Your null hypothesis is that the national distribution also applies to your high school.

73 . What are the expected numbers of students from your high school graduating class enrolled and not enrolled in higher education?

74 . Fill out the rest of this table.

Observed ( O ) Expected ( E ) O E ( O E )2 $\frac{{\left(O-E\right)}^{2}}{z}$
Enrolled 145
Not enrolled 55

75 . What are the degrees of freedom for this chi-square test?

76 . What is the chi-square test statistic and the p -value. At the 5% significance level, what do you conclude?

77 . For a chi-square distribution with 92 degrees of freedom, the curve _____________.

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text:
# Math Help - Columns width under multicolumn

1. ## Columns width under multicolumn

Good afternoon everybody,

I am a begineer in Latex and I have some trouble setting column's width in a Latex document. I have double columns under which I would like the columns to be split evenly (50% of the double column each) but I juste don't understant how to get this result. Here is a sample of what I get. (hope my explanation of what I want is clear enough).

$
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
& \multicolumn{2}{c|}{Linear Elements} & \multicolumn{2}{c|}{Quad Elements} & \multicolumn{2}{c|}{Cubic Elements} \\ \cline{2-7}
& 0,\Omega & 1,\Omega & 0,\Omega & 1,\Omega & 0,\Omega & 1,\Omega \\ \hline
\end{tabular}$

2. Originally Posted by asfd
Good afternoon everybody,

I am a begineer in Latex and I have some trouble setting column's width in a Latex document. I have double columns under which I would like the columns to be split evenly (50% of the double column each) but I juste don't understant how to get this result. Here is a sample of what I get. (hope my explanation of what I want is clear enough).

$
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
& \multicolumn{2}{c|}{Linear Elements} & \multicolumn{2}{c|}{Quad Elements} & \multicolumn{2}{c|}{Cubic Elements} \\ \cline{2-7}
& 0,\Omega & 1,\Omega & 0,\Omega & 1,\Omega & 0,\Omega & 1,\Omega \\ \hline
\end{tabular}$

This may be of use to you.

3. Originally Posted by asfd
Good afternoon everybody,

I am a begineer in Latex and I have some trouble setting column's width in a Latex document. I have double columns under which I would like the columns to be split evenly (50% of the double column each) but I juste don't understant how to get this result. Here is a sample of what I get. (hope my explanation of what I want is clear enough).

$
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
& \multicolumn{2}{c|}{Linear Elements} & \multicolumn{2}{c|}{Quad Elements} & \multicolumn{2}{c|}{Cubic Elements} \\ \cline{2-7}
& 0,\Omega & 1,\Omega & 0,\Omega & 1,\Omega & 0,\Omega & 1,\Omega \\ \hline
\end{tabular}$

The LaTeX tabular environment really isn't very helpful if you want to control widths of columns. The only way I know to get columns of equal width is to replace the "c" declaration in the table specification by something like "p{3em}". That will give a column of the specified width, but it will not be centred. However, you can then centre each individual table entry by padding it with flexible space in the form of an \hfil before and after it.
Code:
\begin{tabular}
{|c|p{3em}|p{3em}|p{3em}|p{3em}|p{3em}|p{3em}|}
\hline
& \multicolumn{2}{c|}{Linear Elements}
& \multicolumn{2}{c|}{Cubic Elements} \\
\cline{2-7}
&\hfil $0,\Omega$\hfil &\hfil $1,\Omega$\hfil
&\hfil $0,\Omega$\hfil &\hfil $1,\Omega$\hfil
&\hfil $0,\Omega$\hfil &\hfil $1,\Omega$\hfil \\
\hline
\end{tabular}
$\begin{tabular}{|c|p{3em}|p{3em}|p{3em}|p{3em}|p{3 em}|p{3em}|}
\hline
& \multicolumn{2}{c|}{Linear Elements} & \multicolumn{2}{c|}{Quad Elements} & \multicolumn{2}{c|}{Cubic Elements} \\ \cline{2-7}
&\hfil 0,\Omega\hfil &\hfil 1,\Omega\hfil &\hfil 0,\Omega\hfil &\hfil 1,\Omega\hfil &\hfil 0,\Omega\hfil &\hfil 1,\Omega\hfil \\ \hline
\end{tabular}$

Laborious and inelegant, but it should give you the desired result.

4. Originally Posted by Opalg
Laborious and inelegant, but it should give you the desired result.
Thanks a lot! Your last sentence pretty much says it all.


text:
# [SOLVED]how fast is the area of triangle formed

#### karush

##### Well-known member
9 A Ladder 13ft long is leaning against the side of a building.
If the foot of the ladder is pulled away from the building at a constant rate of 8in per second how fast is the area of triangle formed by the ladder, the building and the ground changing (in feet squared per second) at the instant when the top of the ladder is 12 feet above the ground

View attachment 1520

$$A=\frac{1}{2}BH=\frac{1}{2}xy$$

$$\frac{d}{dt}A=\frac{d}{dt} \left(\frac{1}{2} xy \right) \Rightarrow \frac{dA}{dt}=\frac{1}{2}x\frac{dy}{dt}+\frac{1}{2}y\frac{dx}{dt}$$

since $$\frac{dy}{dt}\text{ at y }= 12\text { is }\frac{-5 ft}{12 ft}\cdot \frac {2ft}{3 sec} = -\frac {5 ft}{18 sec}$$

then substituting

$$\frac{dA}{dt} = \frac{119}{36}\frac{ft^2}{sec}$$

hopefully

#### Ackbach

##### Indicium Physicus
Staff member
Looks good to me!


text:
# Pseudo arclength continuation

This is one of the various continuation methods implemented in BifurcationKit.jl. It is set by the option PALC(tangent = Bordered()) or PALC(tangent = Secant()) in continuation. See also PALC.

For solving

$$$\mathbb R^n\ni F(x,p) = 0 \quad\tag{E}$$$

using a Newton algorithm, we miss an equation. The simplest way is to select an hyperplane in the space $\mathbb R^n\times \mathbb R$ passing through $(x_0,p_0)$:

$$$N(x, p) = \frac{\theta}{n} \langle x - x_0, dx_0\rangle + (1 - \theta)\cdot(p - p_0)\cdot dp_0 - ds = 0\tag{N}$$$

with $\theta\in[0,1]$ and where $ds$ is the pseudo arclength (see [Keller]).

Parameter θ

The parameter θ in the struct ContinuationPar is very important. It should be tuned for the continuation to work properly especially in the case of large problems where the $\langle x - x_0, dx_0\rangle$ component in the constraint might be favored too much. Also, large θs favour p as the corresponding term in the constraint $N$ involves the term $1-θ$.

## Predictor

The possible predictors are listed in Predictors - Correctors.

## Corrector

The corrector is the newton algorithm for finding the roots $(x,p)$ of

$$$\begin{bmatrix} F(x,p) \\ N(x,p)\end{bmatrix} = 0\tag{PALC}$$$

## Linear Algebra

Let us discuss more about the norm and dot product. First, the option normC continuation specifies the norm used to evaluate the residual in the following way: $max(normC(F(x,p)), |N(x,p)|)<tol$. It is thus used as a stopping criterion for the corrector. The dot product (resp. norm) used in $N$ and in the (iterative) linear solvers is LinearAlgebra.dot (resp. LinearAlgebra.norm). It can be changed by importing these functions and redefining it. Note that by default, the $\mathcal L^2$ norm is used. These details are important because of the constraint $N$ which incorporates the factor length. For some custom composite type implementing a Vector space, the dot product could already incorporates the length factor in which case you should either redefine the dot product or change $\theta$.

The linear solver for the linear problem associated to (PALC) is set by the option linearAlgo in continuation: it is one of Bordered linear solvers (BLS).

## Dot product

In the constraint $N$ above, the scalar product is in fact saved in BifurcationKit.jl as dotp(x,y) -> dot(x,y)/length(y). This is used in the bordered linear solvers associated to PALC. If you want to use your own dot product, you can pass

dotPALC = BK.DotTheta(mydot),

to continuation. Additionally, you may want to provide the linear operator P such that mydot(x,y) = dot(x, A*y), especially if you intend too use the linear solver MatrixBLS. We refer to BifurcationKit.DotTheta for more details.

## Step size control

Each time the corrector fails, the step size $ds$ is halved. This has the disadvantage of having lost Newton iterations (which costs time) and imposing small steps (which can be slow as well). To prevent this, the step size is controlled internally with the idea of having a constant number of Newton iterations per point. This is in part controlled by the aggressiveness factor a in ContinuationPar.

### References

• Keller

Keller, Herbert B. Lectures on Numerical Methods in Bifurcation Problems. Springer, 1988


text:
## This is an unpublished test document with $LaTeX - Code$ in it. For example, $$\sqrt{3x-1}+(1+x)^2$$ is an squareroot.

• $\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$


text:
Dirichlet's unit theorem

Dirichlet's unit theorem In mathematics, Dirichlet's unit theorem is a basic result in algebraic number theory due to Peter Gustav Lejeune Dirichlet.[1] It determines the rank of the group of units in the ring OK of algebraic integers of a number field K. The regulator is a positive real number that determines how "dense" the units are.

The statement is that the group of units is finitely generated and has rank (maximal number of multiplicatively independent elements) equal to r = r1 + r2 − 1 where r1 is the number of real embeddings and r2 the number of conjugate pairs of complex embeddings of K. This characterisation of r1 and r2 is based on the idea that there will be as many ways to embed K in the complex number field as the degree {displaystyle n=[K:mathbb {Q} ]} ; these will either be into the real numbers, or pairs of embeddings related by complex conjugation, so that n = r1 + 2r2.

Note that if K is Galois over {Anzeigestil mathbb {Q} } then either r1 = 0 or r2 = 0.

Other ways of determining r1 and r2 are use the primitive element theorem to write {displaystyle K=mathbb {Q} (Alpha )} , and then r1 is the number of conjugates of α that are real, 2r2 the number that are complex; mit anderen Worten, if f is the minimal polynomial of α over {Anzeigestil mathbb {Q} } , then r1 is the number of real roots and 2r2 is the number of non-real complex roots of f (which come in complex conjugate pairs); write the tensor product of fields {displaystyle Kotimes _{mathbb {Q} }mathbb {R} } as a product of fields, there being r1 copies of {Anzeigestil mathbb {R} } and r2 copies of {Anzeigestil mathbb {C} } .

Als Beispiel, if K is a quadratic field, the rank is 1 if it is a real quadratic field, und 0 if an imaginary quadratic field. The theory for real quadratic fields is essentially the theory of Pell's equation.

The rank is positive for all number fields besides {Anzeigestil mathbb {Q} } and imaginary quadratic fields, which have rank 0. The 'size' of the units is measured in general by a determinant called the regulator. In principle a basis for the units can be effectively computed; in practice the calculations are quite involved when n is large.

The torsion in the group of units is the set of all roots of unity of K, which form a finite cyclic group. For a number field with at least one real embedding the torsion must therefore be only {1,−1}. There are number fields, for example most imaginary quadratic fields, having no real embeddings which also have {1,−1} for the torsion of its unit group.

Totally real fields are special with respect to units. If L/K is a finite extension of number fields with degree greater than 1 and the units groups for the integers of L and K have the same rank then K is totally real and L is a totally complex quadratic extension. The converse holds too. (An example is K equal to the rationals and L equal to an imaginary quadratic field; both have unit rank 0.) The theorem not only applies to the maximal order OK but to any order O ⊂ OK.[2] There is a generalisation of the unit theorem by Helmut Hasse (and later Claude Chevalley) to describe the structure of the group of S-units, determining the rank of the unit group in localizations of rings of integers. Ebenfalls, the Galois module structure of {Anzeigestil mathbb {Q} oplus O_{K,S}omal _{mathbb {Z} }mathbb {Q} } has been determined.[3] Inhalt 1 The regulator 1.1 Beispiele 2 Higher regulators 3 Stark regulator 4 p-adic regulator 5 Siehe auch 6 Anmerkungen 7 References The regulator Suppose that K is a number field and {displaystyle u_{1},Punkte ,u_{r}} are a set of generators for the unit group of K modulo roots of unity. There will be r + 1 Archimedean places of K, either real or complex. Zum {displaystyle uin K} , write {Anzeigestil u^{(1)},Punkte ,u^{(r+1)}} for the different embeddings into {Anzeigestil mathbb {R} } oder {Anzeigestil mathbb {C} } and set Nj to 1 oder 2 if the corresponding embedding is real or complex respectively. Then the r × (r + 1) Matrix {Anzeigestil links(N_{j}Protokoll übrig|u_{ich}^{(j)}Rechts|Rechts)_{i=1,dots ,r,;j=1,dots ,r+1}} has the property that the sum of any row is zero (because all units have norm 1, and the log of the norm is the sum of the entries in a row). This implies that the absolute value R of the determinant of the submatrix formed by deleting one column is independent of the column. The number R is called the regulator of the algebraic number field (it does not depend on the choice of generators ui). It measures the "Dichte" of the units: if the regulator is small, this means that there are "lots" of units.

The regulator has the following geometric interpretation. The map taking a unit u to the vector with entries {textstyle N_{j}Protokoll übrig|u^{(j)}Rechts|} has an image in the r-dimensional subspace of {Anzeigestil mathbb {R} ^{r+1}} consisting of all vectors whose entries have sum 0, and by Dirichlet's unit theorem the image is a lattice in this subspace. The volume of a fundamental domain of this lattice is {Anzeigestil R{quadrat {r+1}}} .

The regulator of an algebraic number field of degree greater than 2 is usually quite cumbersome to calculate, though there are now computer algebra packages that can do it in many cases. It is usually much easier to calculate the product hR of the class number h and the regulator using the class number formula, and the main difficulty in calculating the class number of an algebraic number field is usually the calculation of the regulator.


text:
### Symmetry, Integrability and Geometry: Methods and Applications (SIGMA)

SIGMA 12 (2016), 080, 29 pages      arXiv:1603.09335      https://doi.org/10.3842/SIGMA.2016.080

### Möbius Invariants of Shapes and Images

Stephen Marsland a and Robert I. McLachlan b
a) School of Engineering and Advanced Technology, Massey University, Palmerston North, New Zealand
b) Institute of Fundamental Sciences, Massey University, Palmerston North, New Zealand

Received April 01, 2016, in final form August 08, 2016; Published online August 11, 2016

Abstract
Identifying when different images are of the same object despite changes caused by imaging technologies, or processes such as growth, has many applications in fields such as computer vision and biological image analysis. One approach to this problem is to identify the group of possible transformations of the object and to find invariants to the action of that group, meaning that the object has the same values of the invariants despite the action of the group. In this paper we study the invariants of planar shapes and images under the Möbius group $\mathrm{PSL}(2,\mathbb{C})$, which arises in the conformal camera model of vision and may also correspond to neurological aspects of vision, such as grouping of lines and circles. We survey properties of invariants that are important in applications, and the known Möbius invariants, and then develop an algorithm by which shapes can be recognised that is Möbius- and reparametrization-invariant, numerically stable, and robust to noise. We demonstrate the efficacy of this new invariant approach on sets of curves, and then develop a Möbius-invariant signature of grey-scale images.

Key words: invariant; invariant signature; Möbius group; shape; image.

pdf (1957 kb)   tex (1548 kb)

References

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text:
Corpus ID: 8320935

# Tractable Epistemic Reasoning with Functional Fluents, Static Causal Laws and Postdiction

@article{Eppe2014TractableER,
title={Tractable Epistemic Reasoning with Functional Fluents, Static Causal Laws and Postdiction},
author={Manfred Eppe},
journal={ArXiv},
year={2014},
volume={abs/1403.0034}
}
Manfred EppeUniversity of Bremen, Germanymeppe@informatik.uni-bremen.de1. INTRODUCTIONEpistemic reasoning about action and change is a crucial requirement for systems that deal with incom-plete knowledge in the presence of abnormalities, unobservable processes, human-computer interactionand other real-world considerations. A particular type of epistemic inference which is required for di-agnosis tasks in an epistemic context is postiction (Eppe, 2013). Postdiction determines the conditionof an… Expand

#### References

SHOWING 1-10 OF 31 REFERENCES
An Epistemic Event Calculus for ASP-based Reasoning About Knowledge of the Past, Present and Future
• Computer Science, Mathematics
• LPAR
• 2013
We present a generalisation of the Event Calculus, specified in classical logic and implemented in ASP, that facilitates reasoning about non-binary-valued fluents in domains with non-deterministic,Expand
A Causal Theory of Ramifications and Qualifications
• Mathematics, Computer Science
• IJCAI
• 1995
This paper proposes a simple fixpoint condition defining the possible next states after performing an action, which guarantees minimal change between states, but also enforces the requirement that changes be "caused". Expand
Formalizing sensing actions A transition function based approach
• Mathematics, Computer Science
• Artif. Intell.
• 2001
A high-level action description language that allows specification of sensing actions and their effects in its domain description and allows queries with conditional plans is developed and provably correct translations of domain description in the language to axioms in first-order logic are given. Expand
Approximate Epistemic Planning with Postdiction as Answer-Set Programming
• Computer Science
• LPNMR
• 2013
A history-based approximation of the Possible Worlds Semantics $\mathcal{PWS}$ for reasoning about knowledge and action is proposed and its integration in a cognitive robotics framework for high-level control in a smart home is presented. Expand
Reasoning and planning with sensing actions, incomplete information, and static causal laws using answer set programming
• Computer Science
• Theory and Practice of Logic Programming
• 2007
This work extends the 0-approximation of sensing actions and incomplete information in Son and Baral (2001) to action theories with static causal laws and proves its soundness with respect to the possible world semantics and presents an answer set programming based conditional planner, called ASCP, that is capable of generating both conformant plans and conditional plans. Expand
A Formal Theory of Knowledge and Action
Abstract : Most work on planning and problem solving within the field of artificial intelligence assumes that the agent has complete knowledge of all relevant aspects of the problem domain andExpand
A logic-based calculus of events
• Computer Science, Mathematics
• New Generation Computing
• 2009
We outline an approach for reasoning about events and time within a logic programming framework. The notion of event is taken to be more primitive than that of time and both are representedExpand
Contingent Planning via Heuristic Forward Search witn Implicit Belief States
• Mathematics, Computer Science
• ICAPS
• 2005
This paper proposes the targeted use of an additional relaxation, mapping the relaxed contingent problem into a relaxed conformant problem, and shows that the resulting planning system, Contingent-FF, is highly competitive with the state-of-the-art contingent planners POND and MBP. Expand
Computational Complexity of Planning and Approximate Planning in Presence of Incompleteness
• Mathematics, Computer Science
• IJCAI
• 1999
It is shown that the planning problem in presence of incompleteness is indeed harder: it belongs to the next level of complexity hierarchy (in precise terms, it is Σ2P-complete), and under certain conditions, one of these approximations makes the problem NP-complete. Expand
Nonmonotonic causal theories
• Mathematics, Computer Science
• Artif. Intell.
• 2004
The nonmonotonic causal logic defined in this paper can be used to represent properties of actions, including actions with conditional and indirect effects, nondeterministic actions, and concurrentlyExpand


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text:
# Factor spaces of measurable spaces

Consider a measurable space $(\Omega, \mathcal{F})$. Given a partition $\mathcal{P}$ of $\Omega$ into measurable sets we consider the map $\pi\colon \Omega\to \mathcal{P}$ associating to each $\omega\in \Omega$ the atom $P\in \mathcal{P}$ that contains it. In the space $\mathcal{P}$ we consider the largest $\sigma$-algebra in which $\pi$ is measurable, that is, $A$ is a measurable subset of $\mathcal{P}$ if and only if $\pi^{-1}(A)$ is measurable.

I was reading Rokhlin's book "On the fundamental ideas of measure theory", and he says that ( see chapter 1, No. 2, page 5) the map $\pi$ constructed above satisfies the following property: If $B\subset \Omega$ is measurable then $\pi(B)$ is measurable.

If I understood right this is equivalent to say that $$\bigcup_{ P\in \mathcal{P},P\cap B\neq\emptyset,} P$$ is measurable. Since the partition $\mathcal{P}$ and the set $B$ may not be countable, I don't see why he assertion is true.

Can someone help me to clarify this fact or suggest some reference on elementary properties of factor spaces?

• The smallest $\sigma$-algebra on $\mathcal{P}$ that makes $\pi$ measurable is $\{\mathcal{P},\emptyset\}$. You are talking about the largest $\sigma$-algebra that makes $\pi$ measurable. – Michael Greinecker Dec 6 '16 at 17:33
• Yes, you are right!!! – Didi Dec 6 '16 at 20:13
• I have edited... – Didi Dec 6 '16 at 20:38

The general statement as written in the question is false. For example, consider the projection $\pi\colon[0,1]\times[0,1]\to[0,1]$, both $[0,1]\times[0,1]$ and $[0,1]$ with their Borel $\sigma$-algebras. There is a Borel subset of $[0,1]\times[0,1]$ whose $\pi$-image in $[0,1]$ is not Borel. To make the statement true, some assumptions must be added. For example, if $(\Omega,\mathcal{F})$ is a standard Borel space (i.e. isomorphic to $[0,1]$ with its Borel $\sigma$-algebra) and $B\in\mathcal{F}$ then $\pi(B)$ is measurable with respect to every measure on $\mathcal{P}$. A good reference is Fremlin's Measure Theory, chapters 42 and 43.
• Rokhlin does not work with Borel spaces. This is the paper where he introduces the notion of Lebesgue spaces. Here, the setting is abstract measure spaces and the $\sigma$-algebras are complete. – coudy Dec 3 '16 at 18:54


text:
# Complementability of finite dimensional subspaces

Suppose $$X$$ is a separable infinite dimensional Banach space, $$E$$ a finite dimensional subspace which is $$c$$-complemented in $$X$$. Is the following true?

For any $$\varepsilon>0$$, one can find $$x\notin E$$ such that $$\mathrm{span}\{E,x\}$$ is $$(c+\varepsilon)$$-complemented in $$X$$.

This is obviously true in Hilbert spaces, but not sure in general Banach spaces. Would reflexivity make a difference?

• This is not true because, if we use this condition repeatedly, we would get that there exist subspaces in $X$ with arbitrarily large dimension which are $1+\epsilon$-complemented. This is known to be false in a very strong sense, see G. Pisier, Acta Math. 151 (1983), no. 3-4, 181–208; or his book "Factorization of Linear Operators", 1986, Chapter 10, Corollary 10.8. Pisier mentions that (in 1986) it was unknown whether such spaces can be reflexive, and I do not know whether this question was solved. Sep 3, 2019 at 5:00
• It is a nice elementary exercise to prove this is true if $X$ is an $L_p$ space, $1\le p < \infty$. I think I will assign this problem to the students in my functional analysis class. No partial credit for doing the $p=2$ case. Sep 3, 2019 at 20:16


text:
Lemma 72.7.9. Let $S$ be a scheme. Given morphisms $f : X \to Y$, $g : Y \to Z$ of algebraic spaces over $S$ we have $g_{big} \circ f_{big} = (g \circ f)_{big}$.

Proof. This follows from the simple description of pushforward and pullback for the functors on the big sites from Lemma 72.7.8. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# R igraph manual pages

Use this if you are using igraph from R

 layout_with_gem {igraph} R Documentation

## The GEM layout algorithm

### Description

Place vertices on the plane using the GEM force-directed layout algorithm.

### Usage

layout_with_gem(
graph,
coords = NULL,
maxiter = 40 * vcount(graph)^2,
temp.max = vcount(graph),
temp.min = 1/10,
temp.init = sqrt(vcount(graph))
)

with_gem(...)

### Arguments

 graph The input graph. Edge directions are ignored. coords If not NULL, then the starting coordinates should be given here, in a two or three column matrix, depending on the dim argument. maxiter The maximum number of iterations to perform. Updating a single vertex counts as an iteration. A reasonable default is 40 * n * n, where n is the number of vertices. The original paper suggests 4 * n * n, but this usually only works if the other parameters are set up carefully. temp.max The maximum allowed local temperature. A reasonable default is the number of vertices. temp.min The global temperature at which the algorithm terminates (even before reaching maxiter iterations). A reasonable default is 1/10. temp.init Initial local temperature of all vertices. A reasonable default is the square root of the number of vertices. ... Passed to layout_with_gem.

### Details

See the referenced paper below for the details of the algorithm.

### Value

A numeric matrix with two columns, and as many rows as the number of vertices.

### Author(s)

Gabor Csardi csardi.gabor@gmail.com

### References

Arne Frick, Andreas Ludwig, Heiko Mehldau: A Fast Adaptive Layout Algorithm for Undirected Graphs, Proc. Graph Drawing 1994, LNCS 894, pp. 388-403, 1995.

layout_with_fr, plot.igraph, tkplot

Other graph layouts: add_layout_(), component_wise(), layout_as_bipartite(), layout_as_star(), layout_as_tree(), layout_in_circle(), layout_nicely(), layout_on_grid(), layout_on_sphere(), layout_randomly(), layout_with_dh(), layout_with_fr(), layout_with_graphopt(), layout_with_kk(), layout_with_lgl(), layout_with_mds(), layout_with_sugiyama(), layout_(), merge_coords(), norm_coords(), normalize()

### Examples

set.seed(42)
g <- make_ring(10)
plot(g, layout=layout_with_gem)

[Package igraph version 1.3.4 Index]


text:
## Friday, March 11, 2016

3/9/2016

### Section 4.11.1

The rowSums formula should read
$\mathbf{A1}=\sum_{i}^{n}\mathbf{A}_{*i},$
and the colSums formula should read
$\mathbf{A}^{\top}\mathbf{1}=\sum_{i}^{m}\mathbf{A}_{i*}$
Note: it is a little bit confusing. Following R semantics, the rowSums() method means "row-wise sum" which is the same as "sum of columns"; and vice versa, the colSums() method means "column-wise sum", which can be computed as the sum of matrix rows.

### Section 6.1

The dimensions of matrix $\mathbf{V}$ are $\mathbf{V}\in\mathbb{R}^{n\times k}$.
$\boldsymbol{a}_{pca}=\mathbf{V}^{\top}\left(\boldsymbol{a}-\boldsymbol{\mu}\right).\label{eq:to-pca}$
Formula (6.2) - (6.3) should read
$\begin{eqnarray} \boldsymbol{a} & = & \left(\mathbf{V}^{\top}\right)^{\text{-}1}\boldsymbol{a}_{pca}+\boldsymbol{\mu} \\ & = & \mathbf{V}\boldsymbol{a}_{pca}+\boldsymbol{\mu}.\label{eq:from-pca} \end{eqnarray}$

3/12/2016

### Section 8.3

In Step (1): Setup working directories and acquire data, the URL of the Wikipedia XML dump has since changed. The third command of Step (1) should read:

curl http://dumps.wikimedia.org/enwiki/latest/enwiki-latest-pages-articles10.xml-p002336425p003046511.bz2 -o $WORK_DIR/wikixml/enwiki-latest-pages-articles.xml.bz2  We keep the Kindle edition updated with the Errata. If you have ever bought the print version from Amazon, the Kindle version is free via Amazon MatchBook. If you already have a Kindle version, you should be able just to reload it with the updated one. #### 8 comments: 1. Hi, Firstly I must say that your book is excellent. Thank you very much for writing the book that leads us to develop many distributed mathematical algorithms. I have a problem for the first example of the book. Page 19., Example 2.4 Simulating regression data with a small noise When I declare mxData object the scala interpreter launches that error: error: recursive value mxData needs type Can you explain me why? 1. Hello, Firstly, thanks for such an excellent book. Same problem here, I execute your code from your repository everything works well. Problem is when I try to execute the code from the book. I got the following error: Error:(26, 32) value checkpoint is not a member of org.apache.mahout.math.Matrix val drmXB = (1 cbind drmX).checkpoint() ^ Why code from your repository and the book are so different? Cheers. 2. Xavier -- thanks! I am not quite sure what you mean by "different". Unless it is a trivial one-, two- or maybe three-line example (which we still verified for errors on actual execution) all code is copied verbatim from the book's github code. Of course, many code examples are just fragments being illustrated. Omission of various boilerplate or irrelevant code (import, Scala unit test setup etc. etc.) is customary for printed examples that focus on a specific point being illustrated. That's why we publish the whole code as well so you could refer to it for other details if something doesn't seem to be working quite right. Specifically toward your problem: I am guessing you are referring to Example 2.3, which again is found in its entirety here: https://github.com/andrewpalumbo/mahout-samsara-book/blob/master/myMahoutApp/src/main/scala/myMahoutApp/LinearRegression.scala In example 2.3, the value drmXB (as its prefix implies) is of distributed matrix type -- DRMLike[?]. However, the compilation error you are getting implies that it is not a distributed matrix type, but rather an in-memory type (o.a.m.math.Matrix). The checkpoint() operation is a method on distributed matrices only (it initiates lazy optimization barrier). Chap. 4 goes over these specifics in more details. You must be doing something different -- my guess is in that line you are trying to run, drmX is not really a DRM but rather an in-memory Matrix type, which obviously does not have distributed optimizer contracts like checkpoint, hence the error you see. But I don't know what exactly you may be doing differently, since I do not see the rest of your code. Hope that helps. Best, -DL 3. Aykut, thank you -- and thank you for your question. This code is not meant to run in a shell, but rather in the application. The source code for this particular example is here: https://github.com/andrewpalumbo/mahout-samsara-book/blob/master/myMahoutApp/src/test/scala/myMahoutApp/MyAppSuite.scala. You should be able to compile it with maven and run tests, which pass. The book implies in the very chap. 2 that you download and compile corresponding version of Mahout locally, as the examples reuse test artifacts of Mahout which may not have been published to Maven Central for any given release. In fact, I always rely on my own recompilation rather than on binary releases (as i often work with a snapshot rather than a release). Even with Spark. :) Feel free to check out the example code and follow steps in chap2 to compile Mahout, make sure that you are compiling the same version the examples use (or adjust example dependencies to the version you compile). Please let me know if you have further difficulties. 4. Thanks problem solved function I was looking for it's in LinearRegression [1], sorry I didn't notice. Everything works! [1] https://github.com/andrewpalumbo/mahout-samsara-book/blob/master/myMahoutApp/src/main/scala/myMahoutApp/LinearRegression.scala 2. Thanks Dmitriy for your reply. I have created a replica from your code repository and it is working very well. Kind regards. 3. hello Sir, Firstly thanks for your excellent book.I have a problem for the example of parallel matrix multiplication.My code is here : package myMahoutApp.mthread /** * Created by chomon on 11/14/16. */ import org.apache.log4j.{BasicConfigurator, Level} import org.scalatest.{FunSuite, Matchers} import org.apache.mahout.math._ import scalabindings._ import RLikeOps._ import org.apache.mahout.logging._ import scala.concurrent.duration.Duration import scala.concurrent.{Await, Future} class MThreadSuite extends FunSuite with Matchers { BasicConfigurator.configure() private[mthread] final implicit val log = getLog(classOf[MThreadSuite]) setLogLevel(Level.DEBUG) test("mthread-mmul") { val m = 5000 val n = 300 val s = 350 val mxA = Matrices.symmetricUniformView(m, s, 1234).cloned val mxB = Matrices.symmetricUniformView(s, n, 1323).cloned // Just to warm up mxA %*% mxB MMul.mmulParA(mxA, mxB) val ntimes = 30 val controlMsStart = System.currentTimeMillis() val mxControlC = mxA %*% mxB for (i ← 1 until ntimes) mxA %*% mxB val controlMs = System.currentTimeMillis() - controlMsStart val cMsStart = System.currentTimeMillis() val mxC = MMul.mmulParA(mxA, mxB) for (i ← 1 until ntimes) MMul.mmulParA(mxA, mxB) val cMs = System.currentTimeMillis() - cMsStart debug(f"control:${controlMs / ntimes.toDouble}%.2f ms.")
debug(f"mthread: ${cMs / ntimes.toDouble}%.2f ms.") trace(s"mxControlC:$mxControlC")
trace(s"mxC:\$mxC")

(mxControlC - mxC).norm should be < 1e-5

def mmulParA(mxA: Matrix, mxB: Matrix): Matrix = {
val result = if (mxA.getFlavor.isDense)
mxA.like(mxA.nrow, mxB.ncol)
else if (mxB.getFlavor.isDense)
mxB.like(mxA.nrow, mxB.ncol)
else mxA.like(mxA.nrow, mxB.ncol)

val nsplits = Runtime.getRuntime.availableProcessors() min mxA.nrow
val ranges = createSplits(mxA.nrow, nsplits)
val blocks = ranges.map { r ⇒
Future {
r → (mxA(r, ::) %*% mxB)
}
}

Await.result(Future.fold(blocks)(result) {
case (result, (r, block)) ⇒
result(r, ::) := block
result
}, Duration.Inf)
}
}

def createSplits(nrow: Int, nsplits: Int):
TraversableOnce[Range] = {
val step = nrow / nsplits
val slack = nrow % nsplits
((0 until slack * (step + 1) by (step + 1)) ++

(slack * (step + 1) to nrow by step))
.sliding(2).map(s => s(0) until s(1))

}
}

Although I gave mmulParA method,it appears "cannot resolve symbol mmulParA".
How can I solve?

1. Cherry:

Thank you for all your questions. This is great that you are showing up the interest to learn.

(1) all the examples given in the book are available on Andrew's github here: https://github.com/andrewpalumbo/mahout-samsara-book. Just clone it, it all compiles and runs. Try to figure what is different.

(2) for strictly mahout related questions (rather than strictly book related -- i know you asked a few), please be sure to ask on mahout mailing list (user, or dev, doesn't matter much). The information how to subscribe is here: https://mahout.apache.org/general/mailing-lists,-irc-and-archives.html. It is ok to refer to the book or book examples, we all have copies, we all will find the reference if you make one.

Being Mahout's PMC (emeritus), we at Mahout have certain rules we abide by. One of them is to steer all questions to the mailing list. We don't answer direct project-related questions (usually), instead we ask people to go to the list and ask the question there. The reasons for this are: (1) you get better served because there are much more people there that can help you and that may know the answer to your particular question. (2) Other people learn from the same question as well as all answers are archived and searchable. This, to a certain degree, reduces the burden of answering the same question multiple times. (3) The project activity is measured, among other things, by the user list activity too. So the project gains every time you ask on its list.

If you can do that, that'd be great and we all at Mahout would be very thankful to you for doing this.

-Dmitriy


text:
# Homogeneous Flows, Moduli Spaces, and Arithmetic: Proceedings of the Clay Mathematics Institute Summer School, June 11--July 6, 2007, Centro Di Recerca Matematica Ennio de Giorgi, Pisa, Italy

American Mathematical Society for the Clay Mathematics Institute, Cambridge, MA, 2010 - Mathematics - 438 pages
This book contains a wealth of material concerning two very active and interconnected directions of current research at the interface of dynamics, number theory and geometry. Examples of the dynamics considered are the action of subgroups of $\mathrm{SL}(n,\mathbb{R})$ on the space of unit volume lattices in $\mathbb{R}^n$ and the action of $\mathrm{SL}(2,\mathbb{R})$ or its subgroups on moduli spaces of flat structures with prescribed singularities on a surface of genus $\ge 2$. Topics covered include the following: (a) Unipotent flows: non-divergence, the classification of invariant measures, equidistribution, orbit closures. (b) Actions of higher rank diagonalizable groups and their invariant measures, including entropy theory for such actions. (c) Interval exchange maps and their connections to translation surfaces, ergodicity and mixing of the Teichmuller geodesic flow, dynamics of rational billiards. (d) Application of homogeneous flows to arithmetic, including applications to the distribution of values of indefinite quadratic forms at integral points, metric Diophantine approximation, simultaneous Diophantine approximations, counting of integral and rational points on homogeneous varieties. (e) Eigenfunctions of the Laplacian, entropy of quantum limits, and arithmetic quantum unique ergodicity. (f) Connections between equidistribution and automorphic forms and their $L$-functions. The text includes comprehensive introductions to the state-of-the-art in these important areas and several surveys of more advanced topics, including complete proofs of many of the fundamental theorems on the subject. It is intended for graduate students and researchers wishing to study these fields either for their own sake or as tools to be applied in a variety of fields such as arithmetic, Diophantine approximations, billiards, etc.

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