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# ln|x| vs. ln(x)? When is the ln antiderivative marked as an absolute value?

One of the answers to the problems I'm doing had straight lines: $$\ln|y^2-25|$$

versus another problem's just now: $$\ln(1+e^r)$$

I know this is probably to do with the absolute value. Is the absolute value marking necessary because #1 was the antiderivative of a squared variable expression that could be either positive or negative (and had to be positive because, well, natural log) and the second was positive by default?

Sorry if this is me asking and answering my own question; I'd just love to get confirmation in case I'm wrong.

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Unless $r$ is complex, $e^r > 0$. –  vonbrand May 7 at 21:56

This is a rather subtle matter.

First thing is that to make sure we are talking sense, we should talk about antiderivatives on a specific interval. For example, we should not just say "$F(x)$ is an antiderivative of $f(x)$", but something like "$F(x)$ is an antiderivative of $f(x)$ for $1<x<2$".

Now if $x$ takes positive values then the derivative of $\ln x$ is $x^{-1}$. So we can say

$\ln x$ is an antiderivative of $x^{-1}$ for $x>0$.

On the other hand, if $x$ takes negative values then the derivative of $\ln(-x)$ is $x^{-1}$: you can check this by differentiation. So we can say

$\ln(-x)$ is an antiderivative of $x^{-1}$ for $x<0$.

Now if $x$ is positive then $x=|x|$, and if $x$ is negative then $-x=|x|$, so both our logarithm expressions above can be written as $\ln|x|$. We can therefore say

$\ln|x|$ is an antiderivative of $x^{-1}$ on any interval consisting of positive or negative $x$ values.

But. . . the matter starts to get seriously dodgy if we try to include both positive and negative $x$ values at the same time: because then to get an interval of $x$ values we would also have to include $x=0$, and neither $\ln|x|$ nor $x^{-1}$ makes sense if $x=0$.

In my view, writing the antiderivative as $\ln|x|$ is a neat way of summarising two results in one, but it carries a serious risk of disguising what is really going on. So my preference, if I have an integral giving $\ln(\hbox{something})$, is to work out whether the "something" is negative or positive, and put in a minus sign, or not, as appropriate.

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Thanks @JW, interesting link. I was aware of the issue of having different constants each side of $0$. Was also thinking of asking about $\int_{-1}^1 x^{-1}\,dx$. But I felt that both of these could tend to obscure my main point. –  David May 2 at 10:14
Very nice exposition of the situation (+1). I tried to describe this same situation in this answer. –  robjohn May 2 at 21:32
(For example, we should not just say "F(x) is an antiderivative of f(x)", but something like "F(x) is an antiderivative of f(x) for 1<x<2") - This is a very important point that most people ignore it. I used to ignore this too. (+1) –  math.n00b May 9 at 23:09

Yes you are right $1+e^x$ is positive everywhere, so putting an absolute value would be redundant, but $y^2-25$ is negative if $|y|\lt 5$ so we need the absolute value.

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I wish I could mark two answers and/or upvote this one too. Thank you very much, that succinctly makes sense of the difference between the problems I was working with! –  Calculistening helplessly May 2 at 8:31
No worries, happy to help, btw you can upvote more than one answer I believe. –  ellya May 2 at 8:37
More precisely, $\:y^2-25\:$ is negative if $\: \left|\hspace{.03 in}y\right|<5 \;$. $\;\;\;\;$ –  Ricky Demer May 2 at 9:23
thankyou, that is what I meant, i will edit. –  ellya May 2 at 9:25

logarithm is a function only defined on domain $(0,\infty)$, so it make no sense to input negative value.

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Well it's defined on $\mathbb C\backslash \{0\}$. –  Ruslan May 2 at 8:23
Well, if complex variable theory is involved, as long as we take care of the branches for logarithm we can input negative number. But I think the author is confused with the real number case. –  Poor Math Guy May 2 at 8:52
@Ruslan: usually a branch cut is also removed so that the function is differentiable on an open domain. –  robjohn May 2 at 21:34
Thanks for your note, but I think the author is confused with the real number case. So I just make the comment based on real logarithm. –  Poor Math Guy May 3 at 3:27

Just to make explicit what some were saying in the comments, you don't need much sophistication with complex numbers to make sense of this without the absolute value sign. You really only need to know the polar form of complex numbers. We can make use of the fact that we can represent a complex number $a+bi$ as $r e^{i\theta}$, by Euler's formula. The exponential gives a natural definition of $\log$: Given $a+bi=r e^{i\theta}$... $$\log(a+bi)=\log(r e^{i \theta})=\log(e^{\log(r)+i\theta})=\log(r)+i\theta$$

To make this a function in the usual sense, we can restrict $r\ge 0$ and $-\pi\le \theta <\pi$.

Then, $\log(x)$ where $x$ is real is defined as $\log x$ when $x>0$, and $\log |x|-\pi i$ when $x<0$. Differentiating with respect to $x$ in both cases leaves zero imaginary component and the expected result, so then you truly do have $\int \frac{1}{x} dx=\log(x)$.

Also note that when you move from an antiderivative to a definite integral, any constants in your function are subtracted away, so you still don't get a complex result.

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First of all let us assume that you are dealing with real valued functions on real variables. Then logarithm you are using comes with base Eulers constant e.whose value lies between 2 and 3 $loga=b \rightarrow e^{b}=a$ then naturally logarithm is defined in a domain which is set of positive real numbers. Secondly we cannot define inverse function unless the given function is bijective. If you are using complex valued function I think yopu need not put absolute value sign your function (log) will be a multiple valued function


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# Proof of a fact about traces

I'm following the open courseware content on Machine Learning from Stanford University. In the lecture notes, it is given that

$$\Delta_A \ tr(ABA^TC) = CAB + C^TAB^T$$

which I tried but couldn't prove easily. It is not required to follow the course content but I just wondered and wanted to learn its proof. Any suggestions?

Update: $A$, $B$, and $C$ are matrices and $\Delta_A$ is the gradient operation on matrix $A$.

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Dear Ismail: it's not necessary to open with a salutation or have a closing like "kind wishes". I guess $A$, $B$, $C$ are square matrices of the same size; could you say what is meant by $\Delta_A$ please? Also, a direct link to the lecture notes would be very nice. –  Todd Trimble Feb 27 '11 at 16:56
Dear Todd: Thanks for you advice. I will take your remarks about how to ask a question into consideration. The course notes are here: stanford.edu/class/cs229/notes/cs229-notes1.pdf Page 9, Eq. 3. –  İsmail Arı Feb 27 '11 at 17:07
I don't see the reason for getting a -1 for the question. Is that because I couldn't prove it easily? If so, why are there 8 current upvotes for the answer post. If I know the reason, I would update my questioning to mathoverflow community accordingly. –  İsmail Arı Apr 10 '11 at 23:23
I didn't downvote the question, but I can see a good reason for doing so, namely, the carelessness with which the question was presented. No indication of what $A$, $B$, $C$, and $\Delta_A$ were, not editing the question to include this information when the defect was brought to your attention - in fact, I've just decided to downvote the question. –  Gerry Myerson Apr 11 '11 at 2:18
Thanks for the warning. I updated the question. –  İsmail Arı Apr 11 '11 at 8:27

I guess $\Delta_A$ denotes the derivative with respect to the elements of the matrix $A$ (more conventionally denoted by $\partial_{A}$).
To evaluate the derivative with respect to $A_{ij}$, write out the trace in terms of components and then use $\partial_{A_{ij}} A_{mn} = \delta_{im} \delta_{jn}$, $$\partial_{A_{ij}} \text{tr}(A B A^T C) = \partial_{A_{ij}}\sum_{mnkl} A_{mn} B_{nk} A_{lk} C_{lm}= \sum_{kl} B_{jk}A_{lk} C_{li} + \sum_{mn} A_{mn} B_{nj} C_{im}$$ $$= ( C^T A B^T+ C A B )_{ij}.$$ This is the component-wise version of your identity.
Note to the comment of Todd Trimble: the matrices $A,B$, and $C$ do not have to be necessarily square matrices. Their dimension just has to "match" ($A \in \mathbb{R}^{m \times n}$, $B\in \mathbb{R}^{n \times n}$, $C \in \mathbb{R}^{m \times m}$, with $m$ and $n$ arbitrary integers).

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If r=0.345, s=(0.345)^2, and t=root(0.345), which of the

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If r=0.345, s=(0.345)^2, and t=root(0.345), which of the [#permalink]

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If $$r=0.345$$, $$s=0.345^2$$, and $$t=\sqrt{0.345}$$, which of the following is the correct ordering of $$r$$, $$s$$, and $$t$$ ?

A. r < s < t
B. r < t < s
C. s < t < r
D. s < r < t
E. t < r < s
[Reveal] Spoiler: OA

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If r=0.345, s=(0.345)^2, and t=root(0.345), which of the [#permalink]

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08 Jun 2015, 13:22
If $$r=0.345$$, $$s=0.345^2$$, and $$t=\sqrt{0.345}$$, which of the following is the correct ordering of $$r$$, $$s$$, and $$t$$ ?

A. r < s < t
B. r < t < s
C. s < t < r
D. s < r < t
E. t < r < s

If $$x$$ is in the range $$0<x<1$$ then: $$0<x^2<x<\sqrt{x}<1$$.

Consider the simpler example: if $$x=\frac{1}{4}$$ then $$0<\frac{1}{16}<\frac{1}{4}<\frac{1}{2}<1$$.

So, $$s<r<t$$.

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Re: If r=0.345, s=(0.345)^2, and t=root(0.345), which of the [#permalink]

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10 Jun 2015, 20:24
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Hi All,

This prompt showcases some great Number Property rules that you're likely to face (in some way or another) on the Official GMAT.

Regardless of the actual numbers involved, IF you're dealing with a positive fraction (meaning a value that is greater than 0 and less than 1), then there are some rules to remember...

1) Squaring that fraction will make the result SMALLER
2) Square-rooting that fraction will make the result BIGGER

Here are some additional Number Property rules for NEGATIVE fractions (meaning a value that is less than 0 and greater than -1)....

1) Squaring that fraction will make the result BIGGER (since squaring a negative creates a positive) AND the result will be CLOSER to 0.
2) Cubing that fraction will make the result BIGGER (since the result will be closer to 0).

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** SVP Joined: 08 Jul 2010 Posts: 1920 Location: India GMAT: INSIGHT WE: Education (Education) Re: If r=0.345, s=(0.345)^2, and t=root(0.345), which of the [#permalink] Show Tags 20 Jun 2015, 19:53 girikorat wrote: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of the following is the correct ordering of r, s, and t ? A. r < s < t B. r < t < s C. s < t < r D. s < r < t E. t < r < s CONCEPT:For solving any such question, One must know that 1) If 0<x<1, then the higher powers of x result in smaller numbers and lower powers of x result in a bigger number 2) If x>1, then the higher powers of x result in Bigger numbers and lower powers of x result in a Smaller number Here since 0.345 is between 0 and 1 therefore the number with the highest power i.e. s must be least and the number with the Lowest power i.e. t must be Biggest number therefore s < r < t Answer: Option [Reveal] Spoiler: D _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html 22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 10747 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: If r=0.345, s=(0.345)^2, and t=root(0.345), which of the [#permalink] Show Tags 07 Jul 2015, 21:06 Hi All, This question is based on a couple of important Number Property rules regarding positive fractions (between 0 and 1): 1) When 'squaring' a positive fraction, the result gets SMALLER (e.g.. (1/2)^2 = 1/4). Since 'square-rooting' is the opposite of 'squaring'.... 2) When 'square-rooting' a positive fraction, the result gets BIGGER. While these rules are not tested that often on the Official GMAT, it's likely that you'll be tested on either or both at least once on Test Day. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: If r=0.345, s=(0.345)^2, and t=root(0.345), which of the [#permalink]

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Bunuel wrote:
If $$r=0.345$$, $$s=0.345^2$$, and $$t=\sqrt{0.345}$$, which of the following is the correct ordering of $$r$$, $$s$$, and $$t$$ ?

A. r < s < t
B. r < t < s
C. s < t < r
D. s < r < t
E. t < r < s

Let $$r = 0.36$$ ; $$s = 0.36^2$$ & $$t = \sqrt{0.36}$$

So, $$r = 0.36 , s ≈ 0.13$$ & $$t = 0.60$$

Thus, $$s < r < t = 0.13 < 0.36 < 0.60$$

Hence, answer will be (D) $$s < r < t$$
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Re: If r=0.345, s=(0.345)^2, and t=root(0.345), which of the [#permalink]

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05 Oct 2017, 09:14
Bunuel wrote:
If $$r=0.345$$, $$s=0.345^2$$, and $$t=\sqrt{0.345}$$, which of the following is the correct ordering of $$r$$, $$s$$, and $$t$$ ?

A. r < s < t
B. r < t < s
C. s < t < r
D. s < r < t
E. t < r < s

Since r is a positive decimal between 0 and 1, s = r^2 and t = √r. We see that when r is raised to a power > 1, its value decreases, and when it is raised to a power < 1, its value increases (recall that square root means ½ power). Thus, the correct ordering is s < r < t.

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Re: If r=0.345, s=(0.345)^2, and t=root(0.345), which of the   [#permalink] 05 Oct 2017, 09:14
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Archive

Archive for November, 2011

Edge detection: The Convolution Approach

The first approaches were performed with simple $3 \times 3$ kernels. For example, Faler came up with the following four simple masks that emulate differentiation:
$\begin{pmatrix} -1 & 0 & 1\\ -1 & 0 & 1\\ -1 & 0 & 1 \end{pmatrix}\quad \begin{pmatrix} 1 & 1 & 1\\ 0 & 0 & 0 \\ -1 & -1 & -1 \end{pmatrix}\quad \begin{pmatrix} -1 & -1 & -1 \\ -1 & 8 & -1 \\ -1 & -1 & -1 \end{pmatrix}\quad \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}$


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# Shipment of 50 light bulbs contains [0 defective light bulbs. In order t0 test the quality of the shipment; quality

###### Question:

shipment of 50 light bulbs contains [0 defective light bulbs. In order t0 test the quality of the shipment; quality control engineer selects lighc bulbs rndom Answer the following questions using the concepts from Chapter only . (a) How many different samples are possible? What is the probability the engineer will select exactly two delective light bulbs? (e) The shipment will be rejected ifat least one of the light bulbs in the sample delective. What is the probability the sample will be rejected?

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Adding in Rows

List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?

Magic Sums and Products

How to build your own magic squares.

Nine, Ten and One

Weekly Problem 3 - 2013
Can you find the value of t in these equations?

No Matter

Stage: 3 Short Challenge Level:

Think of a number, double it, then add $3$. Multiply your answer by $4$ and from this take away $5$. Now also take away the number you first thought of.

No matter what your first number was, your answer will be a multiple of which number?

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

This problem is taken from the UKMT Mathematical Challenges.
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# Chapter 1 - Trigonometric Functions - Section 1.2 Angle Relationships and Similar Triangles - 1.2 Exercises: 57

30 meters

#### Work Step by Step

1. It is similar triangles problem. We need to put shadow over shadow and set it equal to actual height over actual heght of the tree. $\frac{3}{45}=\frac{2}{t}$ 2. Then solve for t: $3t= 90$ $t=30$ 3. Don't forget the units 30 meters

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#### On some p-adic Galois representations and form class groups

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Let K be an imaginary quadratic field of discriminant dK$d_K$ with ring of integers OK$\mathcal {O}_K$. When K is different from Q(-1)$\mathbb {Q}(\sqrt {-1})$ and Q(-3)$\mathbb {Q}(\sqrt {-3})$, we consider a specific elliptic curve EJK$E_{J_K}$ with j-invariant j(OK)$j(\mathcal {O}_K)$ which is defined over Q(j(OK))$\mathbb {Q}(j(\mathcal {O}_K))$. In this paper, for each positive integer N we compare the extension field of Q$\mathbb {Q}$ generated by the coordinates of N-torsion points on EJK$E_{J_K}$ with the ray class field K(N)$K_{(N)}$ of K modulo NOK$N\mathcal {O}_K$. By using this result, we investigate the image of the p-adic Galois representation attached to EJK$E_{J_K}$ for a prime p, in terms of class field theory. Second, we construct the definite form class group of discriminant dK$d_K$ and level N which is isomorphic to Gal(K(N)/Q)$\mathrm{Gal}(K_{(N)}/\mathbb {Q})$.
Publisher
WILEY
Issue Date
2022-04
Language
English
Article Type
Article
Citation

MATHEMATIKA, v.68, no.2, pp.535 - 564

ISSN
0025-5793
DOI
10.1112/mtk.12141
URI
http://hdl.handle.net/10203/295856
Appears in Collection
MA-Journal Papers(저널논문)
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text: I. sp II. In a nutshell if you count them there are 16 total bonds. See the answer. Further Reading. Chapter 7, Problem 65QAP. The carbon-carbon triple bond in acetylene is the shortest (120 pm) and the strongest (965 kJ/mol) of the carbon-carbon bond types. * Triple bond contains 1 sigma bond and 2 pi bonds. Ethyne, … Another 2 atoms of carbon has double bond berween them. Still have questions? In which compound are all the carbon atoms sp 2 hybridized? I would agree with you that there are 6 sigma bonds in the molecule. 2 sigma bonds between H2 and C. 1 sigma bond between C and C. 1 sigma bond between C and H. 1 sigma bond between C and C. 3 sigma bonds between C and H3 Three sigma bonds are formed from each carbon atom for a total of six sigma bonds total in the molecule. No. Source(s): Professional chemist since 1987. Well sigma bonds are a single bond basically, and pi-bonds are double. Show transcribed image text. Relevance. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes! 0 3. hcbiochem. Answer. of double bonds in the compound. H = -311 kcal/ mol. The given molecule is, As we know that carbon has '4' valence electrons and hydrogen has '1' valence electron. arrow_forward. The Number Of Sigma And Pi Bonds In The Molecule Below Are Sigma A) 4 B) 6 C) 9 D) 4 E) 9 Pi 2 0. 5. of sigma and pi bonds can also be identified by noting the No. There are also 2 C-C sigma bonds using 4 electrons giving us a total of 16 electrons in sigma bonds. Since propyne has a triple bond, therefore it has minimum bond length. Each carbon to hydrogen bond contains one sigma bond, so in propyne you have 3 of those on the carbon atom 1, none on carbon atom 2, and one on carbon atom 3. Lv 7. There were many diagrams which had no annotations and were difficult to interpret. Sigma and Pi Bonds. (Total 1 mark) p - block element; Share It On Facebook Twitter Email. If you draw it out explicitly, you should find: Three "C"-"H" sigma bonds on the "CH"_3 Two "C"-"C" sigma bonds on … Give the number of sigma and pi bonds in the molecule in Question 58. check_circle Expert Solution. The number of sigma and pi-bonds are … The pi bond is the "second" bond of the double bonds between the carbon atoms and is shown as an elongated green lobe that extends both above and below the plane of the molecule. As per the definition, • A sigma bond is formed when the two unfilled atomic orbitals of atom overlaps with their internuclear axes. In a triple bond, there are 2 pi bonds (and 1 sigma bond). The sp 2 hybridized carbons now form three sigma bonds and one pi bond, as illustrated below. Where are the Sigma and Pi Bonds in the Compound C3H4? To help understand the relative stabilities of alkyne isomers, heats of … * Double bond contains 1 pi bond and 1 sigma bond. In this case there are 16 C−C sigma bonds and 10 C−H bonds. Physical Properties include nonpolar due to slight solubility in polar solvents and insoluble in water. 2 sigma and ... Identify the number of sigma $$(\sigma )$$ and pi $$(\pi )$$ bonds present in a molecule of... 15M.3.hl.TZ1.20a: Explain, on a molecular level, why ozone dissociates with radiation of a longer wavelength... 15M.3.hl.TZ1.20b: Nitrogen(II) oxide, NO, is a primary pollutant that depletes the ozone … Science Anatomy & Physiology Astronomy Astrophysics Biology Chemistry Earth Science Environmental Science Organic Chemistry Physics Math Algebra Calculus Geometry Prealgebra Precalculus Statistics Trigonometry Humanities English Grammar U.S. … Select one: a. seven σ and three π b. seven σ and two π c. five σ and five π d. five σ and three π e. five σ and five π f. a huge number of bonds What is the number of sigma ( σ ) and pi ( π ) bonds present in sulphuric acid molecule ? The pi-bonds however, do not encompass a great amount of energy even though the concentration is small within the molecule. Alkyne contain one triple bonded atom In triple bond 2 are pi bond and one sigma So, here 2 bond is +nt do you know chemistry yeah hotelcalifornia hotelcalifornia There would be a maximum of two pi-bonds in . Many candidates however omitted mention of orbitals when trying to describe the formation of sigma and pi bonds or to explain hybridization. Want to see this answer and more? Answer Save. Each triple bond contains one sigma bond, so thats the fifth one. Check out a sample textbook solution. And 3 pi bonds (1 on the C=C double bond and 2 on the C≡C triple bond) Note: people are saying C≡C triple bonds have 2 sigma bonds and 1 pi bonds, this is not true. c.iv. To calculate total number of sigma bonds present in a molecule is very important topic. Here is the simple structure And here is the structure made on the basis of valence bond theory you can refer to them to understand the structure in detail. Pi bonds are contained in double and triple bonds. Explanation: The molecule usually called 3-Hexyne which will have a triple bond in its structure as the name itself indicates the triple bond which will be generally termed as ‘yne’ according to nomenclature. How many sigma and pi bonds are in the whole molecule shown below? Additionally, there exist six carbon-hydrogen sigma bonds. A double bond (an "ene") is a sigma and a pi. The number of sigma bond, pi bond and lone pairs of electrons can be determined by Lewis structure. Four in total. Additional higher level (AHL) » Topic 14: Chemical bonding and structure » 14.1 Covalent bonding and electron domain and molecular geometries. : In C176H250, Y = 250, therefore A =[(3 x 250)/2] = 375 -2 = 373 single bonds. The number of nodal planes present in s × s antibonding orbitals is (a) 1 (b) 2 (c) 0 (d) 3. Sigma between C and H, Pi between each … How many sigma and pi bonds are there in propyne, CH 3 CCH? Question: For The Molecule (a) C2H2 And (b) CO3(-2)ii) Give The Total Number Of Sigma And Pi Bonds In The Molecule,iii) Predict The Type Of Hybrid Orbital Possibly Existed In The Central Atom(s). It seems not too simple. There are a number of isomers of C3H4 cyclopropene- 7 sigma bonds - one pi methyl acetylene- 6 sigma - 2pi propadiene- 6 sigma - 2 pi The sigma (sigma) bonds are simply the number of single bonds shown here: 6 xx (sp^2) "C"-"C" sigma (sigma) bonds 6 xx (sp^2) "C"-"H" sigma (sigma) bonds Then, when we incorporate the additional electrons that are delocalized throughout the ring, it is easiest to count the pi (pi) bonds when using the major resonance structure, where all the pi electrons are depicted as localized within pure … 7 sigma and 1 pi C. 6 sigma and 2 pi D. 5 sigma and 3 pi (Total 1 mark) 28. Sandeep J . Anyway, the simplest answer is as follows: The carbon suboxide is C3O2. To calculate sigma and pi bonds you need to have basic knowledge of structural formula.That is - * Single bond contain one sigma bond . Sigma:16. See solution. Acetylene is said to have three sigma bonds and two pi bonds. 1 point for correct hybridization (ii) indicate the total number of sigma (σ) bonds and the total number of pi (π) bonds in the molecule. These two perpendicular pairs of p orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds). The carbon-carbon triple bond in acetylene is the shortest (120 pm) and the strongest (965 kJ/mol) of the carbon-carbon bond types. Propyne is a three-carbon alkyne: "H"_3"C"-"C"-="C"-"H" "Prop" means 3, and "yne" means there is a triple bond. Lv 7. Any time you form a double or triple bond, the first two electrons are for the sigma bond and any additional … 0 0. These two perpendicular pairs of p orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds). … The remaining 2 electrons (18 - 16 = 2) make up the second bond between two of the carbons which is a pi bond (sidewise overlap of bonding orbitals --- not as strong as a sigma bond). Thus, each carbon atom in the ethene molecule participates in three sigma bonds and one … There are other … It is 1 sigma and 2 pi. A. Therefore, the total number of valence electrons in = 3(4) + 4(1) = 16. This plane contains the six atoms and all of the sigma bonds. It is H(H) Sigma C Pi C Pi C H (H) sigma diene. And 2xxC-N pi bonds. Therefore, the total number of pi bonds in a benzene molecule is 3. (a) 6σ , 2 π (b) 6 σ , 0 π (c) 2σ , 4 π (d) 2 σ , 2 π . 1 decade ago. Answer: (a) 1 Explanation: In an antibonding molecular orbital, most of the electron density is located away from the space between the nuclei, as a result of which there is a nodal plane (i.e, a plane at which the electron … Show 96 … Quick Calculation of Sigma and Pi Bonds Lovish Raheja Abstract The research ile is mainly aimed to provide chemists with a useful theoretical technique for the calculation of sigma and pi bonds even in the most complex molecule (except in the co-ordination molecule). Molecules with rings have additional sigma bonds, such as benzene rings, which have 6 C−C sigma bonds within the ring for 6 carbon atoms. 2 sigma and 2 pi B. arrow_back. According to Lewis-dot structure, there are 16 number of bonding electrons and 0 number of non-bonding electrons. See the answer. Want to see the full answer? This problem has been solved! Chris. Sigma bonds are formed by end on overlap, and the geometry doesn't allow for more than one of these to happen in the same overall bond. 1 decade ago. The hybrid orbitals are oriented in opposite directions, forming an angle of … sp 2 III. By counting the hydrogens as well. Expert Answer 100% (2 ratings) Previous question Next question Transcribed Image Text from this Question. 2 Answers. Syllabus sections. answered Apr 9, 2018 by BISHNU (32.0k points) selected Apr 14, 2018 by Vikash Kumar . This problem has been solved! Acetylene is said to have three sigma bonds and two pi bonds. 21. For the molecule (a) C2H2 and (b) CO3(-2) ii) Give the total number of Sigma and Pi bonds in the molecule, iii) Predict the type of hybrid orbital possibly existed in … 1 decade ago. There is sigma between each pair of atoms plus a pi between each pair of carbons. 1 Answer. the number of sigma ( σ ) and pi ( π ) bonds present in sulphuric acid molecule is ← … The number of sigma and pi-bonds are 6 and 2 respectively. Discussion and formation of a chart which shows the order of formation and number of sigma (σ) versus pi (π) bonds in single, double, and triple bonds. The combustion of Ethyne is a major contributor from CO 2, water, and the ethyne molecule?? The 2p z electrons of the carbon atoms now form a pi bond with each other. … (You probably want to draw the full Lewis Dot structure for each molecule to get accurate count) (Molecule A) (Molecule B) (Molecule C) CH2C=CH : Br: CH,CCH : Br: Propyne CHCCHC Propanone (Acetone) A) number of sigma bonds = pi= ----- B) number of sigma bonds = ----- pi= ----- C) number of sigma bonds = ----- pi= D) What … Which species does not have delocalized electrons? Two sigma and 2 pi bonds. (a) H2C =CH- CH3. Chapter 7, Problem 63QAP. It can form a total of three sigma bonds. Figure $$\PageIndex{3}$$: Sigma and pi … * See Solution *Response … The aromatic features alternating double bonds between carbon atoms. Favorite Answer. (Total 1 mark) 29. Question 2. A. Best answer. How many sigma (σ) bonds and pi (π) bonds are in the following molecule? Organic Chemistry . 0 votes . How many sigma and pi bonds are there in propyne, $${\text{C}}{{\text{H}}_{\text{3}}}{\text{CCH}}$$? Difference between sigma and pi bonds; Orbital overlap Always 8. In a double bond, there is 1 pi bond (and 1 sigma bond for all intents and purposes). The content of energy involved in the alkyne molecule contributes to this high amount of energy. The sp 2 hybridized orbital in the carbon atom is made up of a 2s electron , a 2p x electron, and a 2p y electron. H-C-=N formally possesses 1xxC-H and 1xxC-N sigma bonds. 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# Are insulators non-ohmic conductors?

Can insulators/ dielectrics be considered as non-ohmic conductors ? As they apparently breakdown when sufficiently large potential differences are applied across them.

Ohm's Law is an idealization. No material follows it exactly, even for small applied voltages. But if your voltage is small enough, the deviation from Ohm's Law will be so small as to be unmeasurable ... but it's still there.

What we usually mean in Ohm's Law is that the behavior is linear for the purpose of whatever the application is. Once we have enough deviation that our application is affected, then we say it's non-ohmic.

So there's no answer to your question. It depends on the application and the sensitivity of your equipment, and your tolerance to non-linearity. For example, if you apply 1,000,000 volts to a small piece of plastic you will say that the material is non-ohmic. But if your equipment (and what you care about) is limited to 1 V, you'd probably say that it's ohmic.

There are two related questions I answered: Does perfect insulation exist and Graph of Electrical Resistivity of Air vs Air Pressure

The crux of it is, that everything will conduct eventually, even nothing (look at the first link).

• To call capacitors (which are made with dielectrics) non-Ohmic conductors, can apply if you feed them with AC voltage. They will allow currents to pass, as they change polarization in sync with the change of voltage. This isn't a conduction current but a displacement current.
• Another case where insulators conduct is at breakdown. I would be careful to call them anything but an insulator during breakdown, as the chemical structure, the conduction mechanism and generally the physics behind it change a lot.
• Insulators will conduct very small amounts of current when they are in a non-breakdown situations, these currents are usually negligible. They can rise during AC operation at specific frequencies, as you don't have instant polarization in materials. Therefore complex permittivity and loss calculation in capacitors could be interesting to you.
• You did not answer my question. Oct 12 '15 at 16:18
• @Abhirikshma I didn't intend to. I left it open to you to conclude what you want to call it, as the term used isn't defined consistently (so that everyone thinks about the same thing). Oct 12 '15 at 17:25

As the other answers have stated, all materials are non-Ohmic to some degree. The main point of this answer is to elaborate some details and state one case of "severe non-Ohmicness", where the $I$-$V$-dependence is non-analytic around $V = 0$: The Zener current.

Most insulators are actually just bad Ohmic conductors (at room temperature and sufficiently low voltage).

Typical insulators are band insulators, which means they have a filled band and a band gap to the next higher band. When we turn on the oven (i.e. consider the material at non-zero temperature) particle-hole-pairs will be excited which carry the current. As the band gap is usually much larger ($\approx 1\,\mathrm{eV}$) than room temperature ($25\,\mathrm{meV}$), the number of particle-hole-pairs is small, and so they conduct badly (because there are few current carriers). The distinction semiconductor/insulator is not a sharp one, but just an arbitrary quantitative distinction. (For simplicity we ignore complicated insulators based on many-body effects such as Mott insulators).

But there is one exception: If the temperature is very low (compared to the bandgap) another, non-Ohmic effect dominates: Zener tunnelling. Here, the electric field induces tunnelling from the valence band into the conduction band. Typical for tunnelling effects it has a non-perturbative $e^{-1/V}$-dependency, so it can in no limit be described as Ohmic.

So as long as your temperature is sufficiently high and your voltage sufficiently low (compared to the bandgap) your typical insulator will be an Ohmic conductor.

On breakdown, many materials will be damaged/transformed non-reversible, so the term "non-Ohmic insulator" might not be the best choice for that scenario.

Actually, all materials(non conducting) in this universe can be considered as non-ohmic conductors because on applying certain voltage, they break down because of the breakage of bonds in the lattice. But perfect insulators, unlike resistors, don't allow current to flow through them at all until this voltage is attained. Their V-I graph is as follows:

Actually, after breaking voltage, materials produce very large amounts of current because of availability of excess amounts of electrons due to breakage of bonds (that is how breaking voltage gets its name). The same goes for dielectrics but they tend to breakdown at a much lower voltage than insulators.

• This is wrong. Even "perfect" band insulators (at $T=0$) allow Zener tunneling which has a non-analytical $e^{-1/V}$ dependence. There is no perfectly insulating region. Oct 13 '15 at 23:28
• Yes, there are no perfect insulating regions. I was just trying to say that even hypothetical "perfect" insulators would conduct current after some voltage. Oct 14 '15 at 5:50


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# I need help solving this riddle! [closed]

You walk 30 feet and turn left, you walk another 30 feet and turn left, you continue to walk 30 feet and turn left, and you do this one more time.

Where are you?

Hint: The answer is not where you started, but you can use this to help get the answer

## closed as too broad by puzzledPig, QuantumTwinkie, JMP, athin, AProughMay 4 '18 at 16:36

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Welcome to Puzzling.SE! Is this a riddle you created yourself (and if not, could you provide attribution?) – puzzledPig May 4 '18 at 14:38
• Does "turn left" imply a 90 degree turn? If not, it depends on the angle(s) that you are turning. – David Starkey May 4 '18 at 16:53

Assuming your initial direction was north, you are now

30 feet north,30 feet west

because

"do this one more time" means do the "3 steps" one more time. So you walked 30 feet a total of 6 times, not 4 times.

You are

home. 30 feet is the distance between the bases on a baseball diamond.

• Isn't it around 80-90 feet? – Anastasiya V May 4 '18 at 14:46
• Darn. Should have looked that up. 30 feet isn't even used in Little League. – generalcrispy May 4 '18 at 14:51
• Matches my initial thoughts, but googling told me No. – AndyT May 4 '18 at 15:14

Using a baseball diamond you would be at 2nd base.

  START ->  h
1st 30 -> /
1   3  <- END
2nd 30 -> \ / <- 3rd 30
2

Doing this one more time:

            h
5th 30 -> / \ <- 4th 30
1   3 <- START
6th 30 -> \ /
END -> 2

Leaving you on second base

It's hard to say without any other reference, but you may be

30 feet south of the North Pole,

or

30 feet north of the South Pole.

• Why only N or S? it could be any direction, as it depends from the first direction – Kepotx May 4 '18 at 15:00

30 feet to the left of where you started

because

The "this" that you're told to do one more time is "turn left". So you've walked 3 sides of a square and then turned so that you're facing the same direction as you were originally

You are

30 feet to the left of where, and how, you started walking, facing original direction

Because

"you do this one more time." is the left turn, not the walk and left turn. So you take three walks and turn left four times. Assuming you start facing north: Walk North, Face West, Walk West, Face South, Walk South, Face East, Face North. The North/South walks cancel so you end up 30ft west facing North

you are walking on

3d shape, like pyramid.

shown as below

imagine that you start walking

from any corner point to the top 30 feet and so on (for simplicity you can assume every corner is 30 feet long) you will not end up where you start since you go up and down etc.

You are

Under the ground

because

I'll assume you're not doing this "one more time", but you "do this one"(turn left) "more time" (probably a lot more), wich means you kind of drill the ground and ends up under the ground


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## Week 2 – Math 10

This week I learned how negative exponents work and how they effect their base’s. Negative exponents unlike positive exponents don’t increase the numbers size. Normal exponents multiply a number over and over (ex. $3^3 = 3\cdot3\cdot3 = 27$) and negative exponents turn the number into a fraction (ex $3^{-3} = \frac{1}{27}$).

You can turn it into a normal number by following the next steps. I will use $5^{-4}$ for this example.

First, if the exponent is negative, then you turn it into a fraction of $\frac{x}{1}$.

Then you put the denominator on the bottom, therefor making the exponent positive.

Alternatively, I learned that if the negative was originally on the bottom, you would then move it to the top.

Then you find out what the product of the power is and put that underneath a 1, in this case $\frac{1}{625}$

I learned that if it were for example $5x^{-4}$, then only the $x$ and its exponent would be moved to the denominator as in $\frac {5}{x^4}$

If it were $(5x)^{-4}$, then both would move to the bottom.

if it were $(5x^{-4})^{-4}$ would equal $5x^{16}$ because you multiply the exponent.


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# A naive question about composition factor of a representation

Let $G$ be a Lie group, and $(\pi,V)$ is a continuous representation of $G$ which has finite composition series. A question I have which might be somehow naive is that: for any irreducible representation $(\sigma,W)$ of $G$, is it true that $(\sigma,W)$ occurs as one composition factor if and only if the set $Hom_G(V,W)$ is nonzero?

I have no idea how difficult or how easy this question might be, and any reference or answer is appreciated.

Edit: Thanks a lot for all of your answers, comments and examples. Now if $G$ is real reductive, $(\pi,V)$ is smooth admissible. Is there a way to determine all of the composition factors of $V$?

-
I'm not too familiar with the etiquette on MO myself, but since your original question was already answered, you might want to make a separate question about real reductive groups. –  Kimball Mar 21 '11 at 14:48

No. This holds only if $V$ is semisimple. Consider the case when $V$ has two composition factors. This means that one is an invariant subspace and the other is the quotient of $V$ by this invariant subspace. If there is also an invariant subspace isomorphic to the quotient then $V$ is the direct sum of these two representations and so $V$ is decomposable.
Bruce's answer is perfectly satisfactory, but you might want to see an explicit example. Let $B$ be the group of all upper triangular matrices in $GL_2({\mathbb R})$ and let $\pi:B\to GL_2({\mathbb R})$ be the inclusion map, which you might consider as a representation on $\mathbb R^2$. Let $\chi_1,\chi_2:B\to GL_1({\mathbb R})$ be the representations given by $$\chi_j\left(^{a_1}\ ^x_{a_2} \right)=a_j.$$ Then $\chi_1$ is a subrepresentation and $\chi_2$ is a quotient of $\pi$. So both are subquotients, but $Hom_B(\pi,\chi_1)$ is zero.


text:
# Inputenc: utf8 characters in cross-references

I am using the hack given here as accepted solution to refer to a description list item by name instead of number. It works nicely, except when one of the words in the name ends with a non-ascii character, in which case it kills the space after it (only in the reference). Why is that? Is there a way to modify the hack that avoids this problem? I am using pdflatex with hyperref and inputenc.

A MWE:

\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{hyperref}
\makeatletter
\let\orgdescriptionlabel\descriptionlabel
\renewcommand*{\descriptionlabel}[1]{%
\let\orglabel\label
\let\label\@gobble
\phantomsection
\edef\@currentlabel{#1}%
\let\label\orglabel
\orgdescriptionlabel{#1}%
}
\makeatother
\begin{document}
\begin{description}
\item[Arzelà and Volterra\label{av}] Reference: \ref{av}.
\item[Arzel{à} and Volterra\label{av2}] Reference: \ref{av2}.
\item[Arzel\a and Volterra\label{av3}] Reference: \ref{av3}.
\end{description}
\end{document}

An even worse problem arises with ß, which I don't need (try Weierstraß and Kummer).

-

Use \protected@edef\@currentlabel{#1}%`


text:
# What is a quasicrystal?

Posted by in math

This post was originally a talk given in Pizza Seminar on October 15, 2014. Most of the mathematical content is from Wikipedia and this paper by Niizeki.

## Abstract

The crystallographic restriction theorem says a finite-order automorphism of $$\mathbb{Z}^3$$ (in chemistry language, a symmetry of a crystal) has order 1, 2, 3, 4, or 6. But $$\mathbb{Z}^3$$ is small potatoes; so let’s go to $$\mathbb{Z}^n$$, do some Galois theory in cyclotomic fields, and get us some icosahedral symmetry! (Icosahedrite is a real thing, say serious scientists.) Pretty pictures will be provided, including demonstrations of self-similarity. Self-similarity is always cool, right?

## The crystallographic restriction theorem

A crystal is a regular grid (like $$\mathbb{Z}^3$$) of atoms. There is a remarkable theorem about what symmetries a crystal can admit:

Theorem (crystallographic restriction theorem)
If $$A \in \operatorname{Aut} \mathbb{Z}^3$$ has finite order, then the order is 1, 2, 3, 4, or 6.
Proof
Properties of $$A$$ imply properties of its eigenvalues:
$$A$$ eigenvalues
finite order on the unit circle $$S^1$$
real in complex conjugate pairs (so one is $$\pm 1$$)
integer their sum, $$\operatorname{tr} A$$, is an integer

So if $$e^{i\theta}$$ is an eigenvalue of $$A$$, then $\operatorname{tr} A = e^{i\theta} + e^{-i\theta} \pm 1 = 2 \cos \theta \pm 1.$ Therefore $$\cos \theta \in \frac{1}{2} \mathbb{Z}$$. The eigenvalues $$e^{i\theta}$$ satisfying this are of order 1, 2, 3, 4, and 6.

## X-ray crystallography: peeking inside real crystals

### Setup

Here’s how scientists figure out the internal structure of a crystal. To start, shine an x-ray on it and record the scattered radiation on film. For our purposes, an x-ray is a function $$f$$ satisfying $f(x+?) = e^{i k \cdot x} f(?)$ for some vector $$k$$ that depends on the ray direction and wavelength. This lets us calculate the factors by which $$f$$ changes from source $$x_1$$ to atom $$x$$ and from atom to film $$x_2$$.

The total change in $$f$$ is by a factor of $$e^{i[(k_1 - k_2) \cdot x - k_1 \cdot x_1 + k_2 \cdot x_2]} .$$ Integrating this over all atom positions $$x$$ in the crystal yields the total radiation bounced in the direction of $$k_2$$. So…

Upshot
The photo records (part of) the Fourier transform of the distribution of atoms.1

### A troublesome photo

Since rotational symmetries of a lattice are inherited by its Fourier transform, photos like the one below would seem to violate the crystallographic restriction theorem.

### Demo: Making waves

To get a picture of what kind of crystal we’re looking at, let’s approximate the inverse transform by adding waves at $$n$$ regularly spaced angles. Try it for 5, 8, 12, and 7.

You can animate these too; and Jason Davies has a WebGL adaptation of the animation.

## Slicing higher-dimensional lattices

### The problem: locating atoms

The last demo showed you the graph of $$f(x) = \displaystyle\sum_{i=1}^n \cos (k_i \cdot x)$$ in $$\mathbb{R}^2$$ for evenly spaced $$k_i$$ on the unit circle. This is all very nice and smooth, but we want to actually place atoms—discrete points—in a grid. Let’s put them in the hotspots—i.e. where $$f$$ hits some threshold.

To turn this into a discrete process, use the following trick: realize the waves $$\cos (k_i \cdot x)$$ as 2-D slices of independent waves in $$\mathbb{R}^n$$—then hotspots are where the slice passes near lattice points.

### The explicit recipe

Let $$f$$ be the restriction of the function $\mathbb{R}^n \ni x \mapsto \sum_{i=1}^n \cos (e_i \cdot x)$ to a 2-plane on which the basis vectors $$e_i$$ project to $$k_i$$. To choose the 2-plane, let $$A \in \operatorname{GL}_n \mathbb{Z}$$ act by cycling the basis vectors $$e_i$$. Diagonalizing over $$\mathbb{C}$$ gets you the matrix $\begin{pmatrix} \zeta & & & & \\ & \zeta^{-1} & & & \\ & & \zeta^2 & & \\ & & & \zeta^{-2} & \\ & & & & \ddots \end{pmatrix}$ where $$\zeta = e^{2 \pi i / n}$$. Over $$\mathbb{R}$$, the eigenvalues pair into 2×2 blocks, corresponding to 2-planes on which $$A$$ acts as rotation by a multiple of $$\frac{2\pi}{n}$$. Pick any one of them.

Upshot
• The 2-plane is a 2-dimensional subrepresentation $$V$$ of $$\langle A \rangle \curvearrowright \mathbb{R}^n$$.
• Atoms (hotspots) are points of $$V$$ that lie kinda near points of $$\mathbb{Z}^n$$.

### Demo: Atoms in 2-D quasicrystals

With this in hand, a computer can be convinced2 to find the atoms for you, drawing larger dots for better hotspots. Try it for several values of $$n$$; I suggest 5, 7, 8, 6, 2, 12, and 11.

Remark
An alternative method is to slice fundamental domains ($$n$$-cubes) with the 2-plane, which yields pictures like the Penrose tiling below. Jason Davies has an applet you can play with, based on an older one by Greg Egan.

## Self-similarity

At this point, we could go home—we have a description of quasicrystals from which we can produce both interactive computer renderings and theorems (like quasicrystals not actually having translational symmetry, due to $$V$$ containing no integer points besides 0).

Of course, there is more—the “quasi”-ness of quasicrystals allows a sort of self-similarity that honest crystals can’t have. To see it in action, go back to that atoms demo and try zooming in or out for a while.

To explain this self-similarity—the apparent indistinguishability of different zoom levels—we will need a bit of number theory.

### A number-theoretic model of quasicrystals

Theorem (crystallographic restriction in arbitrary dimensions)
$$\operatorname{GL}_m \mathbb{Z}$$ contains a matrix $$A$$ which has $$\zeta_n = e^{2\pi i/n}$$ as an eigenvalue (so it rotates some 2-plane by $$\frac{2 \pi}{n}$$) if and only if $$m \geq \varphi(n)$$, where $$\varphi$$ is Euler’s totient function.
Proof
The Galois conjugates of $$\zeta_n$$ must also be eigenvalues of $$A$$.
• Since $$\zeta_n^n = 1$$, they’re $$n$$th roots of 1.
• Since $$\zeta_n$$ generates all $$n$$th roots of 1, they’re primitive—i.e. $$\zeta_n^k$$ where $$k$$ and $$n$$ are coprime.

The number of such $$k$$ is $$\varphi(n)$$. To prove that $$\varphi(n)$$ works, observe that $\mathbb{Z}[\zeta_n] \hookrightarrow \mathbb{Q}(\zeta_n) \cong_{\mathbb{Q}\text{-vect}} \mathbb{Q}^{\varphi(n)} \subset \mathbb{R}^{\varphi(n)}$ as a discrete subgroup, since the field norms of the leftmost terms are in $$\mathbb{Z}$$.

Remark
If $$p$$ and $$q$$ are large enough primes, you can produce an integer matrix $$A$$ of order $$pq$$ and size $$\varphi(p) + \varphi(q) < \varphi(pq)$$, but it won’t rotate any 2-plane by $$\frac{2\pi}{pq}$$.

### A mathematical description of zooming

Here’s how we implement3 zooming: scale the 2-plane $$V$$ by some factor and the unseen $$m-2$$ dimensions by some other factor to preserve $$m$$-dimensional volume. That way the total visual density of dots remains about the same. Here’s how we interpret zooming and self-similarity:

• Zooming acts by scalars on $$V$$ and the other irreducible sub-representations of $$\langle A \rangle \curvearrowright \mathbb{R}^{\varphi(n)}$$, so zooming is a 1-parameter subgroup $$\gamma$$ in $$\operatorname{SAut}_A \mathbb{R}^{\varphi(n)}$$ (volume preserving automorphisms).
• Visually, transforming by anything in $$\operatorname{Aut} \mathbb{Z}[\zeta_n]$$ isn’t detectable.
• Self-similarity means that the image of $$\gamma$$ in $$\operatorname{SAut}_A \mathbb{R}^{\varphi(n)} / \operatorname{Aut} \mathbb{Z}[\zeta_n]$$ accumulates on itself (or is periodic).

The existence of self-similarity rests on one key fact from number theory.

Theorem (Dirichlet’s Unit Theorem)
Suppose a number field has $$r$$ embeddings in $$\mathbb{R}$$ and $$c$$ embeddings in $$\mathbb{C}$$. Then its group of units has rank $r + \frac{c}{2} - 1 .$
Corollary
$$\operatorname{SAut}_A \mathbb{R}^{\varphi(n)} / \operatorname{Aut} \mathbb{Z}[\zeta_n]$$ is compact.
Proof
Since automorphisms act by scalars on all irreducible subrepresentations,

\begin{align*} \operatorname{Aut}_A \mathbb{R}^{\varphi(n)} &\cong (\mathbb{C}^\times)^{\frac{\varphi(n)}{2}} \cong (\mathbb{R} \times S^1)^{\frac{\varphi(n)}{2}} \\ \operatorname{SAut}_A \mathbb{R}^{\varphi(n)} &\cong \mathbb{R}^{\frac{\varphi(n)}{2} - 1} \times (S^1)^{\frac{\varphi(n)}{2}} . \end{align*}

Since $$\mathbb{Z}[\zeta_n] = \mathcal{O}_{\mathbb{Q}(\zeta_n)}$$, Dirichlet’s Unit Theorem implies that $$\operatorname{Aut} \mathbb{Z}[\zeta_n]$$ contains a copy of $$\mathbb{Z}^{\frac{\varphi(n)}{2} - 1}$$. Then $$\operatorname{Aut} \mathbb{Z}[\zeta_n]$$, being discrete (it’s conjugate to a subgroup of $$\operatorname{SL}(\varphi(n), \mathbb{Z})$$), is cocompact.

Self-similarity now follows, having been reduced to the statement that a 1-parameter subgroup of a torus is recurrent.

Remark
The image of $$\gamma$$ is periodic when (and only when) $$\varphi(n) = 2$$—so when $$n$$ is 5, 8, 10, or 12. This makes the self-similarity exact—i.e. different zoom levels can look identical (not just suspiciously similar)—and allows one to build games like Richard Schwartz’s Lucy and Lily based on facts like $(2 + \sqrt{5})(2 - \sqrt{5}) = -1.$
Remark
Michael Rule’s collection of quasicrystals graphics includes an alternative description of zooming that sidesteps the number theory, instead using Shepard Tones to produce self-similarity.

## 3-D quasicrystals

### The n-dimensional viewpoint

A quasicrystal with irreducible symmetry $$G$$ is an irreducible-over-$$\mathbb{R}$$ subrepresentation $$V$$ of an irreducible-over-$$\mathbb{Z}$$ representation $$G \hookrightarrow \operatorname{SL}(m,\mathbb{Z})$$. The atoms are points of $$\mathbb{Z}^m$$ near $$V$$, and the Galois conjugates of $$V$$ (the other subrepresentations over $$\mathbb{R}$$) allow the possibility for self-similarity.

### The search for 3-D quasicrystals

To look for 3-dimensional quasicrystals, one first finds a list of the seven irreducible finite subgroups of $$\operatorname{SL}(3,\mathbb{R})$$. Most of these preserve an ordinary 3-D lattice—i.e. the representation can be defined over $$\mathbb{Z}$$. The two exceptions are the two icosahedral groups—the alternating group $$A_5$$ and the product $$A_5 \times \{\pm 1\}$$—so up to finite index subgroups, there’s only one irreducible 3-D quasicrystal, which you can explore in the following demo.

The standard representation of $$A_5$$ has one Galois conjugate, which can be found in a list of representations of $$A_5$$. The two representations can be described explicitly as follows: in the standard representation, $$A_5$$ acts by rotation on the icosahedron with vertices $$\pm(\sqrt{5}, 0)$$ and $$\pm(1, 2\zeta_5^k)$$ in $$\mathbb{R} \times \mathbb{C}$$. To obtain the other representation, merely conjugate the action on the set of vertices by the permutation induced by $$\zeta_5 \mapsto \zeta_5^2$$.

Oh, and—yes, they’re real.

## Footnotes

1. The photo loses phase information, the recovery of which is a nontrivial problem

2. For efficiency reasons, the demo actually uses the number-theoretic model from the section after.

3. Self-similarity is what makes zooming possible (i.e. not unbearably slow) in the atoms demo—the atom finder can get away with exploring a 2-dimensional slice instead of the entire $$m$$-dimensional space.


text:
### Soft Walks: Real-Time, Two-Ways Interaction between a Character and Loose Grounds

#### Eurographics 2021 (short)

Chloé Paliard [1], Eduardo Alvarado [2], Damien Rohmer [2], Marie-Paule Cani [2]

[1] LTCI, Télécom Paris, Institut Polytechnique de Paris, Palaiseau, France
[2] LIX, Ecole Polytechnique/CNRS, Institut Polytechnique de Paris, Palaiseau, France

## Abstract

When walking on loose terrains, possibly covered with vegetation, the ground and grass should deform, but the character’s gait should also change accordingly. We propose a method for modeling such two-ways interactions in real-time. We first complement a layered character model by a high-level controller, which uses position and angular velocity inputs to improve dynamic oscillations when walking on various slopes. Secondly, at a refined level, the feet are set to locally deform the ground and surrounding vegetation using efficient procedural functions, while the character’s response to such deformations is computed through adapted inverse kinematics. While simple to set up, our method is generic enough to adapt to any character morphology. Moreover, its ability to generate in real time, consistent gaits on a variety of loose grounds of arbitrary slope, possibly covered with grass, makes it an interesting solution to enhance films and games.

Code: [Github]
Paper: [HAL]

## Bibtex

@inproceedings{Paliard2021,
title = {{Soft Walks: Real-Time, Two-Ways Interaction between a Character and Loose Grounds}},
author = {Paliard, Chlo{\'{e}} and Alvarado, Eduardo and Rohmer, Damien and Cani, Marie-paule},
doi = {10.2312/egs.20211019},
booktitle = {Eurographics (Short)},
year = {2021},
month = {may},
url = {https://hal.inria.fr/hal-03200160/},
}


text:
# Caloric equation of state for the Van der Waals gas

I'm currently trying to reproduce a specific derivation of the caloric equation of state for the Van der Waals gas, which I saw a couple of months ago in a Thermodynamics lecture. I'm well aware of the fact that there are already multiple derivations on this site and online (see [1], [2], [3], [4], [5]), but none of them are what I'm looking for.

The things that I remember from the derivation:

1. The inner energy was a function of $$T$$ and $$v$$.
2. We used the fact that $$s$$ is a state variable.
3. After using the first law we expanded all differentials in some variables.
4. With this we obtained $$u(T)=f(T)-a/v$$ and then proceeded to find $$f(T)$$ by using the fact that $$C_V = (\delta Q/\partial T)_V = (\partial u/\partial T)_V$$.

What I've tried so far: Using the first law we obtain $$Tds=du+pdv$$. We can express $$du$$ and $$dv$$ in terms of $$T,p$$, which leads to the equation $$ds = \frac{1}{T}\left(\frac{\partial u}{\partial p}\Bigg|_{T} + p \frac{\partial v}{\partial p}\Bigg|_{T}\right)dp+\frac{1}{T}\left(\frac{\partial u}{\partial T}\Bigg|_{p}+ p\frac{\partial v}{\partial T}\Bigg|_{p}\right)dT.$$ Using the fact that $$s$$ is a state variable we get the relation $$-\frac{\partial v}{\partial p}\Bigg|_{T} = \frac{1}{T}\left(\frac{\partial u}{\partial T}\Bigg|_{p}+p \frac{\partial v}{\partial T}\Bigg|_{p}\right).$$ This simplifies the first equation a bit but isn't really helpful since it's impossible to acutally take the derivative of $$v$$ with respect to any variable due to the form of the Van der Waals equation.

Since I couldn't figure anything out with this I tried the following as another approach: \begin{align} ds &= \frac{1}{T} \left(\frac{\partial u}{\partial v}\Bigg|_{T}dv + \frac{\partial u}{\partial T}\Bigg|_{v}dT\right) + \frac{p}{T}dv \\ & = C_V \frac{dT}{T}+ \frac{1}{T}\left(\frac{\partial u}{\partial v}\Bigg|_{T}+p \right)dv.\end{align} This approach seems to similar to the one from [1], but I'm 100% sure that we didn't use the Helmholtz equation as suggested by @juanrga.

TL;DR Could someone help me derive $$u(T)=C_VT-a/v$$ (caloric equaiton of state) for the Van der Waals gas without using any fancy equations, but rather fundamentals like the fact that $$s$$ is a state variable, etc.

From the formal point of view, the root of the problem is trivially: to have $$U(V,T)$$ or any equivalent information, one has to integrate the pressure. Knowing $$P(T,V,N)$$, we know the partial derivative of the Helmholtz free energy $$F(T,V,N)$$ with respect to $$V$$ at fixed $$T,N$$. The integration of $$P = - \left. \frac{\partial{F}}{\partial{V}} \right|_{T,N}$$ can provide $$F$$ only within an arbitrary function of $$N$$ and $$T$$. Which is equivalent to know almost nothing about the thermal equation of state.
The real answer is that knowledge of the equation of state $$P=P(T,V,N)$$ alone does not allow to obtain the caloric equation of state. Some additional information must be added, for example, information about the constant volume specific heat as a function of (T,V,N). In any case, it is an independent additional information.


text:
Corpus ID: 236447852

# Inverted c-functions in thermal states

@inproceedings{Cartwright2021InvertedCI,
title={Inverted c-functions in thermal states},
author={Casey Cartwright and Matthias Kaminski},
year={2021}
}
• Published 2021
• Physics
We first compute the effect of a chiral anomaly, charge, and a magnetic field on the entanglement entropy in N = 4 Super-Yang-Mills theory at strong coupling via holography. Depending on the width of the entanglement strip the entanglement entropy probes energy scales from the ultraviolet to the infrared energy regime of this quantum field theory (QFT) prepared in a given state. From the entanglement entropy, we compute holographic c-functions and demonstrate an inverted c-theorem for them… Expand

#### References

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We propose the c-function as a new and accurate probe to detect the location of topological quantum critical points. As a direct application, we consider a holographic model which exhibits aExpand
c -theorem for anisotropic RG flows from holographic entanglement entropy
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We construct a $\mathcal{N}$-function for Lovelock theories of gravity, which yields a holographic c-function in domain-wall backgrounds, and seemingly generalizes the concept for black holeExpand
Charged magnetic brane solutions in AdS5 and the fate of the third law of thermodynamics
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text:
# Spin expectation values in Dirac theory

The wave function of a free Dirac particle moving with momentum $\vec{p}=(p,0,0)^T$ is given in the rest frame and the laboratory frame as

$$\Psi_r=N_r\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)e^{-\frac{i}{\hbar}mc^2t}\qquad\Psi_l = N_l\sqrt{\frac{E+mc^2}{2mc^2}}\left(\begin{array}{c}1\\0\\0\\\frac{pc}{E+mc^2}\end{array}\right)e^{-ip_\mu x^\mu/\hbar},$$

where $N_r$ and $N_l$ are normalization constants which are

$$N_i =\left\{\begin{array}{l}1/\sqrt{V},\quad i=r\\\sqrt{\frac{mc^2}{EV}},\quad i=l\end{array}\right.$$

and which I calculated using $\int\mathop{d^3x}\Psi_i^\dagger\Psi_i=1$.

Now, I would like to calculate the spin expectation values $\left<\sigma_z\right> = \frac{1}{N}\int\mathop{d^3x}\Psi^\dagger\Sigma_z\Psi$ in the rest and the laboratory frame, where $\Sigma_z = \frac{\hbar}{2}\left(\begin{array}{cc}\sigma_z & 0 \\ 0 & \sigma_z\end{array}\right)$ is the spin operator and $\sigma_z$ the third Pauli matrix. For the rest frame, I get

$$\left<\sigma_z\right>_r = \frac{\hbar\sqrt{V}}{2}\int\mathop{d^3x}(1,0,0,0)e^{i mc^2t/\hbar}\left(\begin{array}{cc}\sigma_z & 0 \\ 0 & \sigma_z\end{array}\right)\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)e^{-imc^2t/\hbar}=\frac{V^{3/2}\hbar}{2}$$

and analogously in the lab frame

$$\left<\sigma_z\right>_l = \frac{\hbar}{2}\sqrt{\frac{E}{mc^2}}V^{3/2}\frac{mc^2}{E+mc^2}.$$

I'm quite sure that these results cannot be true. Especially in the rest frame, I would expect an expectation value which is independent of the volume $V$. So it would be nice if someone could tell me how to do the calculation correctly.

• I think there might be some ambiguity with what you call „N“. One the one hand every Psi has its own N, but then in the normalization there should be a N^2 on the left-hand side. Similarly, I think there should be a N^2 in the definition of <sigma>. – Stephan Nov 16 '17 at 11:17
• @Stephan I edited my post because the line with the normalization constants was nonsense, as you said. In the definition of $<\sigma>$, you mean that it should be $1/N^2$ instead of $1/N$? But then the volumes don't vanish from the results. The definition in my post was taken from an exercise sheet I found on the internet. – MeMeansMe Nov 16 '17 at 11:35

Short answer: I believe your "$N$" in $\langle\sigma_z\rangle=\ldots$ refers to the normalization of the spinors, which is (since you've already normalized them) $N=1$.
Long answer: An expectation value is defined as $$\langle A \rangle = \frac{\langle \Psi |\hat A|\Psi\rangle}{\langle \Psi|\Psi\rangle}.$$ This means, for the spin in $z$-direction, we need to calculate $\langle \Psi |\hat \Sigma_z|\Psi\rangle$ and $\langle \Psi|\Psi\rangle$. Let's do the second one first: \begin{align*} \langle \Psi|\Psi\rangle &= \int\limits_V \text d^3x \Psi^\dagger \Psi\\ &= \frac{1}{V}\int\limits_V \text d^3x \,(1,0,0,0)\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\text e^{-\text i \frac{mc^2t}{\hbar}}e^{+\text i \frac{mc^2t}{\hbar}}\\ &= \frac{1}{V}\int\limits_V \text d^3x \, \cdot 1 = 1. \end{align*} Now onto the first term: \begin{align*} \langle \Psi |\hat \Sigma_z|\Psi\rangle &= \int\limits_V \text d^3x \Psi^\dagger \hat\Sigma_z\Psi\\ &= \frac{1}{V}\frac{\hbar}{2} \int\limits_V \text d^3x \, \underbrace{(1,0,0,0)\begin{pmatrix}1&&&\\&-1&&\\&&1&\\&&&-1\end{pmatrix}\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}_{1}\,\cdot 1\\ &= \frac{1}{V}\frac{\hbar}{2} \int\limits_V \text d^3x \,\cdot 1\\ &= \frac{V}{V}\frac{\hbar}{2} = \frac{\hbar}{2} \end{align*}
• This makes sense, thank you. In the lab frame, I get $\langle\sigma_z\rangle = \frac{\hbar (mc^2)^2}{E^2+Emc^2}$. Do you think this is right? – MeMeansMe Nov 16 '17 at 14:16
• I get $\langle\Sigma_z\rangle = \frac{\hbar}{2}\frac{mc^2}{E}$. – Stephan Nov 16 '17 at 15:24