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# Fixing orientation of connected smooth manifold in $\mathbb{R}^n$ by a single chart

I am studying on Zorich, Mathematical Analysis II, 1st ed. pag. 174-175. After having properly explained how orientations (equivalence classes) are defined for smooth k-dimensional surfaces in $$\mathbb {R} ^ n$$ that can be described with a single map, move on to the more general case by defining the meanings of:

1. consistent charts,
2. orienting atlases,
3. equivalence classes for orienting atlases (possible orientations of the surface).

Having done this, he states without proof that a connected smooth k-dimensional surface can only have two possible orientations. From this statement he immediately deduces that in order to fix an orientation on a surface of this type it is not necessary to exhibit an entire atlas of consistent charts, but it is sufficient to exhibit a single chart.

I was trying to prove why, but I can't. I assumed, by absurdity, that I had two atlases of different orientation, made of pairwise consistent charts, containing a common chart $$\varphi_1$$:

$$A_1=\{\varphi_1,\varphi_2,...,\varphi_m,...\}$$ $$A_2=\{\varphi_1,\varphi'_2,...,\varphi'_m,...\}$$

but from here I can't get to any absurdity. Can anyone help me please?

• Can you include the definition of orientation used. Aug 18, 2020 at 11:31
• A surface is oriented when an equivalence class in the set of all orienting atlases is fixed. The equivalence relation on the set of atlases is defined by saying that two atlases are equivalent if their union is an orienting atlas. An orienting atlas is an atlas in which all charts are pairwise consistent. Two charts are consistent if their domains of action have empty intersection or if in the non-empty intersection the mutual transitions between one chart to the other have positive Jacobian. Aug 18, 2020 at 19:30
• The distinction between $\phi$ and $\varphi$ is too subtle; $A_1$ and $A_2$ looked identical to my eyes until I looked at the source. So I changed the notation to make the distinction more visually obvious. Aug 20, 2020 at 12:02

I'm going to use the terminology "manifold" instead of "surface", because "surface" usually means 2-dimensional.

Let me use the notation $$M$$ for the manifold in question.

You have to somehow make use of the hypothesis that the manifold $$M$$ is connected. Since manifolds are locally path connected, you can use the theorem that a connected, locally path connected space is path connected.

Consider the common chart $$\varphi_1 : U_1 \to \mathbb R^k$$ in $$A_1 \cap A_2$$, and fix a base point $$p \in U_1$$.

Now I'll prove directly that any chart in $$A_1$$ and any chart in $$A_2$$ are consistent at any point of their overlap.

Consider any $$x \in M$$, and pick charts $$\phi_I : U_I \to \mathbb R^k$$ in $$A_1$$ and $$\varphi'_J : U'_J \to \mathbb R^k$$ in $$A_2$$, such that $$x \in U_I \cap U'_J$$. We have to show that $$\varphi_I$$ and $$\varphi'_J$$ are consistent at the point $$x$$.

Using path connectivity of the manifold $$M$$, choose a continuous path $$\gamma : [0,1]$$ such that $$\gamma(0)=p$$ and $$\gamma(1)=x$$. Since the sets $$\{U_i \cap U'_j\}_{i,j}$$ cover $$M$$, their inverse images $$\{\gamma^{-1}(U_i \cap U'_j)\}_{i,j}$$ cover $$[0,1]$$. Applying the Lebesgue Number Lemma, we can choose an integer $$N \ge 1$$, and decompose $$[0,1]$$ into subintervals $$I_m = [\frac{m-1}{N},\frac{m}{N}]$$, $$m=1,\ldots,N$$, so that $$\gamma(I_m)$$ is a subset of one of the intersections $$U_{i(m)} \cap U'_{j(m)}$$.

We know that $$\varphi_{i(1)}$$ and $$\varphi'_{j(1)}$$ are both consistent with each other at $$\gamma(0)=p$$, because both are consistent with $$\varphi_1$$. Consider the path $$\gamma \mid I_1$$ and let $$t \in I_1 = [0,1/N]$$ vary from $$0$$ to $$1/N$$. As $$t$$ varies, the determinant of the derivative of the overlap map of the two charts $$\varphi_{i(1)}$$ and $$\varphi'_{j(1)}$$ varies continuously, it is nonzero everywhere, and it is positive at $$t=0$$, hence it is positive at $$t=1/N$$. This proves that $$\varphi_{i(1)}$$ and $$\varphi'_{j(1)}$$ are consistent at $$\gamma(1/N)$$.

Now we do an induction proof: assuming by induction that $$\varphi_{i(m)}$$ and $$\varphi'_{j(m)}$$ are consistent at $$\gamma(m/N)$$, we prove that $$\varphi_{i(m+1)}$$ and $$\varphi'_{j(m+1)}$$ are consistent at $$\gamma((m+1)/N)$$. Since $$\varphi_{i(m)}$$ and $$\varphi_{i(m+1)}$$ are consistent at $$\gamma(m/N)$$, and since $$\varphi'_{j(m)}$$ and $$\varphi'_{j(m+1)}$$ are consistent at $$\gamma(m/N)$$, it follows that $$\varphi_{i(m+1)}$$ and $$\varphi'_{j(m+1)}$$ are consistent at $$\gamma(m/N)$$. Now the proof continues as in the previous paragraph, using continuity of the determinant of the derivative of the overlap map of the two charts $$\varphi_{i(m+1)}$$ and $$\varphi'_{j(m+1)}$$ at $$\gamma(t)$$, as $$t \in I_{m+1}$$ varies from $$m/N$$ to $$(m+1)/N$$, and the consistency of those charts at $$\gamma(m/N)$$, to deduce consistency at $$\gamma((m+1)/N)$$. This completes the induction step.

To complete the proof, we have shown that $$\varphi_{i(N)}$$ and $$\varphi'_{j(N)}$$ are consistent at $$\gamma(N/N)=x$$. We also know that $$\varphi_I$$ is consistent with $$\varphi_{i(N)}$$, and $$\varphi'_J$$ is consistent with $$\varphi'_{j(N)}$$ at $$x$$. Therefore, $$\varphi_I$$ and $$\varphi'_J$$ are consistent at $$x$$.

• Why does $U_i\cap U_j'$ cover $M$? Aug 20, 2020 at 12:58
• $M=\cup _i U_i= \cup_j U'_j =\cup_i U_i \bigcap \cup_j U'_j= \cup_{i,j}U_i \cap U_j'$ an Aug 20, 2020 at 13:10
• I rewrote the notation a bit, regarding the sets $U_i \cap U'_j$ and how they cover $M$. Aug 20, 2020 at 14:22
• I also reread the induction step, and it was a bit incomplete, so I filled it out somewhat. Aug 20, 2020 at 14:27
• Ok, I understood this beautiful proof. There is another gap that I would fill: could you give me a reference to study the fact that "a connected, locally path connected space is path connected", please? Aug 20, 2020 at 14:50

Let $$M$$ be your $$k$$-dimensional surface oreinted with respect to the chart $$\{ \varphi_i\}_i$$, $$\varphi_i : \mathbb R^k\rightarrow U_i \subset_{open } M$$. $$\exists \ \omega\in \Omega^k(M)$$ such that $$\omega$$ is non-vanishing at every point. This is possible since $$M$$ is orientable. $$\varphi_i^*\omega=g_i \lambda$$ where $$\lambda=dx_1\wedge dx_2\wedge \dots dx_n$$ and $$g_i:\mathbb R^k \rightarrow \mathbb R$$ is a non-vanishing smooth function. Since the charts are consistent, either all $$g_i$$'s are positive or all negative. Assume that all the $$g_i$$'s are positive.

Now you have the charts $$\{ \varphi_1, \varphi_j'\}_j$$ As before we get $$\varphi^*_1 \omega =g_1\lambda$$ and $${\varphi'}_j^*\omega=h_j \lambda$$. By the same logic as above, we get either $$\{g_1, h_j \}_j$$ are all positive functions or all negative. But since $$g_1$$ is positive, we get all $$h_j$$'s are positive. Thus you get the same orientation.

• I think (but I'm not sure) that you are using the theory of differential forms, theory that I have not yet studied. I'm following rigorously Zorich book, and the chapter on differential forms is successive. Aug 20, 2020 at 13:04
• Ah I see. In that case, you will prefer Prof Lee's answer. But do come back to this once you have read differential forms. I studied orientation after differential forms/ homology and I guess that reflects in my answer. Aug 20, 2020 at 13:09


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# Reflecting linear velocity off surface easy, angular not so much...

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Hi I'm sure many of you know how to reflect an objects linear velocity off of a surface... It involves getting the dot product of the normal and the velocity, and then subtracting from the velocity a vector in the direction of the normal with size (dot product * 2)... This is all well and good, but what about when you have a rotating object that rotates into a surface? is there an easy way to kind of 'reflect' the angular velocity? The angular velocity is stored as a vector. The magnitude being the amount of radians to rotate and the direction being the axis around which to rotate. So, given the point on the surface, the surface normal, and the angular velocity, how would one get the new angular velocity? Thanks for taking the time to read this... Any help appreciated [edited by - nexius on August 16, 2003 12:47:46 AM]

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this is a lot more complicated than just velocity reflection. If you use proper dynamics, with forces and impulses, it involves the inertia tensor.

All I can give you is the equation for calculating the impulse generated at the point of contact.

Ra = (Pa - O)Va = V + (W x Ra)Impulse = [ -(1 + e) * (Va . Na) ] / [ 1/Mass + Na . (InvInertia * (Ra x Na)) x Ra]Force  = Impulse / dtTorque = (Ra x Impulse) / dtV'' = Force  * (1 / Mass)W'' = Torque * InvInertiaV += V'' * dtW += W'' * dtO += V * dtA += W * dt

V : linear velocity of the box
W : angular velocity ofd the box
O : centre of the box
A : angle of the box

Mass : mass of the box
InvInvertia : inverse inertia of the box
Pa : Point of contact
Na : normal at point of contact
e : elasticity of the box (how much it ''bounces'', in range [0, 1])
Va : velocity at the point of contact
Ra : position of the point of contact relative to the centre of the box
dt : Physics timestep
Force : Force of the collision
Torque : Amount of torque generated by the collision
V'' : linear acceleration of the object
W'' : rotational acceleration of the object.

the inertia tensor for a box of extents (or called half-size) (dx, dy, dz) is

x2 = (dx * dx * 4.0f);y2 = (dy * dy * 4.0f);z2 = (dz * dz * 4.0f);ix = (y2 + z2) / 12.0f;iy = (x2 + z2) / 12.0f;iz = (x2 + y2) / 12.0f;    | ix  0  0 |I = |  0 iy  0 |    |  0  0 iz |

the inverse inertia matrix is the transpose of the inertia matrix

I'' = Inertia.Transpose()

the body''s inverse inertia, as the body rotates in the world becomes then

InvInertia = Orientation * (I'' * 1/Mass) * Orientation.Transpose()

so, I'' and I inertia tensors are constant, and depends only on the shape of the object. On top of that, the InvInertia also depends on the current orientation of the object and the mass of the object.

all those equations work also for body vs. body collision, not just body vs. environment, but you need to extend the impulse calculation a little, to take into account the other object''s properties.

Also, one more thing. I tried using the angular velocity, as you did, with taking the angular velocity direction as an axis of rotation, and the magnitude as the rotation angle, but couldn''t get a decent result. I had to revert to a special equation that converts the rotational velocity into a quaternion, and does some euler integration on quaternions, then use that quaternion and turn it into a matrix to get the orientation matrix. That works. Anyway, try it.

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I''m struggling to understand how that works though.

The equation for impulse you gave appears to work out as a scalar, and yet it looks like you''re doing a cross product between it and Ra when calculating the torque, which confuses me...

Impulse = [ -(1 + e) * (Va . Na) ] / [ 1/Mass + Na . (InvInertia * (Ra x Na)) x Ra]

Could you explain where I went wrong?

Thanks

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Is it just in the same direction as the normal?

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sorry, yeah. the impulse vector if Impulse * Normal.

doing rotational stuff increase the complexity of the physics system by quite a lot. For example, if you do collision detection using swept boxes, due to a rotation between frames, the boxes could become entangled, which is a pain.

And you have to deal with inertia, and converting rotational velocity to a rotation matrix, and solving multiple contacts becomes a necessity (or you''d get things like a box endlessly tip-toeing on a plane, as the point of contact switches from one corner of the box to another, due to contact impulses making the box rotate). Quake and most first-person-shooters, don''t bother with rotations. The volumes are all axis aligned, and grenades and projectiles are just particles or small axis-aligned boxes.

The plus point with using rotations and a full-on physics system, you can use proper fricitonal forces, ragdolls and joints. It brings a lot of freedom to the engine, but it''s alot of work (and contact solving can be computationaly expensive).

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# Tag in S Tags in Struts2

2022-12-31   ES

Because the reference format required by IEEE is different from what I usually use, I have studied how to get it for a while, record it, and I am afraid of forgetting.

1. First of the nextIEEE’s thesis template, after entering the official website, click to downloadieeeconf.zip, the one inside is the template of Latex.Click here to download

2. After the template, the reference reference for adding IEEE needs to be used.BibTex‘s reference format, in Baidu’s academic or Google academic, after searching articles, click to reference, there will be

The reference format ofBibTex, as shown below:
Click the lower left cornerBibTex, the following description will be displayed:

@article{Wolfram2010Vox,
title={Vox Populi: The Public Searching of the Web},
author={Wolfram, Dietmar and Spink, Amanda and Jansen, Bernard J and Saracevic, Tefko},
journal={Journal of the Association for Information Science & Technology},
volume={52},
number={12},
pages={1073-1074},
year={2010},
}

1. and then create a name calledmylib.bibfile, copy the above description in

After

2. , open our downloaded template, pull to the bottom, and change the original code to the following form:
%\begin{thebibliography}{99}
\bibliographystyle{IEEEtran}
\bibliography{IEEEabrv,mylib}

%\end{thebibliography}

is the original\begin{thebibliography}{99}and\end{thebibliography}Note, plus

\bibliographystyle{IEEEtran}
\bibliography{IEEEabrv,mylib}

These two sentences, in the second sentencemylibis just what we just createdmylib.bibfile name name, then we can call inside

The reference ofBitTeXformat, the specific reference method is as follows:
Add to where you need to quote:

\cite{Wolfram2010Vox}

== Explain: == in\cite{}In parentheses, we are inmylib.bibThe first line of references added in the file, such as:@article{Wolfram2010Vox,
As long as the content of the Copy in parentheses can be referenced by the corresponding literature.
Use an example to explain:

This is a test operation, cite a article\cite{Wolfram2010Vox}

%\begin{thebibliography}{99}
\bibliographystyle{IEEEtran}
\bibliography{IEEEabrv,mylib}

%\end{thebibliography}

1. The above preparations are done well, and there is also a very critical operation, which is the method of compilation, otherwise it is still unsuccessful:
• Compilation: pdflatex -> bibtex -> pdflateX -> PDFlateX

Compile your .tex files with pdflateX, which generates a. AUX file, which tells Bibtex to use those applications;
Compiled .bib files with bibtex;
Compile your.tex files with pdflateX again. At this time, the reference is included in the document, but the number cited at this time may be incorrect;
Finally compile your.tex files with pdflateX. If everything goes well, this is normal.

After the above operation, you can see the desired reference effect. The reference part is automatically generated. It only needs to be added to the reference. It is particularly convenient. I also needed to add references by myself.
Last effect is shown in the figure:

source


text:
# NCERT Solutions For Class 12 Chemistry Chapter 16

## NCERT Solutions Class 12 Chemistry Chemistry in Everyday Life

NCERT solutions class 12 chemistry chapter 16 chemistry in everyday life is a crucial section in class 12 chemistry examination. The NCERT solutions class 12 chemistry chapter 16 chemistry in everyday life is provided here to help 12th standard students understand the concepts of the chapter in a better way. The solutions are prepared by subject experts according to the latest syllabus of the Central Board of Secondary Education. Students must practice the NCERT Solutions for class 12 chemistry chapter 16 regularly to excel in their board exam. Check the NCERT Solutions for class 12 chemistry chapter 16 pdf given below.

Question 1:

Why should we distinguish drugs in different ways?

The reason for classification of drugs are as follows:
(i) On the basis of pharmacological effect:
This classification is useful for doctors. It provides a whole range of drugs for the classification of drugs for various diseases.
(ii) On the basis of drug action:
This is based on the action of a drug on a particular biochemical process.
Thus this classification is important.
(iii) On the basis of chemical structure:

The range of drugs sharing common structural features and having similar pharmacological activity.
(iv) On the basis of molecular targets:

Some drugs have the same mechanism of action on targets. This classification is useful in such cases.

Question 2:
Explain drug targets or target molecules used in medicinal chemistry.

Drug targets are the key molecule that is responsible for certain metabolic pathways that can cause particular diseases. Proteins, nucleic acids, carbohydrates, and lipids are drug targets.

Chemical agents that are used to block these target molecules by fusing with the active sites of key molecules are called drugs.

Question 3:

Give some macromolecules that are chosen as drug targets.

Carbohydrates, lipids, proteins, and nucleic acids are the macromolecules that are chosen as drug targets

Question 4:

Medicine should not be taken without consulting a doctor. Why is it like that?

Medicines should not be taken without consulting a doctor because it can bind to more than one receptor site. Thus it can be harmful to some receptor sites. Medicines, when taken in higher doses, can cause harmful effects. So medicines can be poisonous.

Question 5:
What is chemotherapy?

Chemotherapy is the use of chemicals for therapeutic effects. Using chemicals for prevention, diagnosis and treatment of diseases are examples

Question 6:

Name the forces which are involved in holding the drugs to the active site of enzymes?

The forces responsible are
(1) Ionic bonding
(2) Hydrogen bonding
(3) Dipole − dipole interaction
(4) van der Waals force

Question 7:

Antacids and antiallergic drugs intervene with the function of the histamines, and why do they not intervene with each other?

Certain drugs affect particular receptors only. Antacids and antiallergic drugs do not intervene with each other because they work on different receptors. This is the reason why antacids and antiallergic drugs intervene with the function of histamines but not with each other.

Question 8:
Level of noradrenaline if low can cause depression. Name the kind of drugs used to solve this problem? Give the names of two drugs.

Anti-depressant drugs are used to reduce the effect of depression. These drugs
contain enzymes which catalyze the degradation of the noradrenaline, neurotransmitter. Therefore the neurotransmitter is metabolized slowly and can activate the receptor for a larger period of time.
Two anti-depressant drugs are:
1. Iproniazid
2. Phenelzine

Question 9:
Explain the term ‘broad spectrum antibiotics?

Antibiotics which are effective against a large range of gram-negative and gram-positive
bacteria are known as broad-spectrum antibiotics. Eg: Chloramphenicol

This is used for the treatment of acute fever, typhoid, meningitis, dysentery, pneumonia and some forms of urinary infections. Vancomycin and ofloxacin are the other two broad spectrum antibiotics. Amoxicillin and ampicillin −synthetically modified from penicillin− are also broad spectrum antibiotics.

Question 10:
How are antiseptics different from disinfectants? Give one example of each.

Antiseptics and disinfectants are really effective against micro-organisms. Antiseptics
are used for living tissues like cuts, wounds, diseased skin surfaces and ulcers,
while disinfectants used for objects such as floors, drainage system,
instruments, etc. Disinfectants are harmful to the living tissues.
Iodine is a strong antiseptic. Tincture of iodine is applied to wounds. 1 percent solution of phenol is used as a disinfectant.

Question 11:
Ranitidine and cimetidine better antacids than sodium hydrogen carbonate or aluminum hydroxide or magnesium. Explain

Magnesium hydroxide, sodium hydrogen carbonate, and aluminum
hydroxide are antacids which neutralize excess hydrochloric acid in the stomach.
The reason for the release of excess acid, however, remains untreated.
Cimetidine and ranitidine are good antacids as they control the cause of acidity. These
drugs avoid the interaction of histamine with the receptors present in the stomach walls and hence can decrease the amount of acid released by the stomach.

Question 12:
Which substance can be used as an antiseptic as well as a disinfectant?

The substance which can be used as an antiseptic, as well as a disinfectant, is phenol. 0.2 percent solution of phenol can be used as an antiseptic and for disinfectant 1 percent of the solution should be used.

Question 13:
What are the main constituents of dettol?

The main constituents of dettol are chloroxylenol and $$\alpha-terpineol$$.

Question 14:
What meant by tincture of iodine? Mention its use?

2-3 percent of iodine in alcohol-water mixture is reffered as tincture of iodine and it is mainly applied to wounds.

Question 15:
What are Food preservatives?

Chemicals which prevent microbial growth is referred to as Food preservatives. They reduce spoilage. Sugar, table salt, vegetable oil, salts of propanoic acid and sodium benzoate ($$C_{6}H_{3}COONa$$), are some food preservatives.

Question 16:
Why is aspartame used for cold foods and drinks only?

Aspartame is unstable at cooking temperature and hence their use is only limited to cold foods and drinks.

Question 17:
What are artificial sweetening agents? Give two examples.

Those chemicals that sweeten food are referred to as artificial sweetening agents. They do not add calories to our body and also do not harm the human body. Some artificial sweeteners are sucrolose, aspartame, alitame and saccharin.

Question 18:
Give the sweetening agent which is used in the preparation of sweets for diabetic patients.

Saccharin, aspartame and alitame are sweetening agents used in preparing sweets for diabetic patients.

Question 19:
What is the disadvantage in using alitame as an artificial sweetener?

Alitame is a high potency sweetener. It is difficult to control the sweetness of food while
using alitame as an artificial sweetener.

Question 20:
Why is synthetic detergents better than soap?

Synthetic detergents work both in soft water and hard water while soaps work in soft water. Soaps are not effective in hard water. Therefore, synthetic detergents are better than soaps.

Question 21 :
illustrate with examples the cationic detergent, anionic detergent and non-ionic detergent

Quaternary ammonium salts of acetates, chlorides, or bromides are usually called Cationic detergents.
Cationic detergents are namedso because of the reason that the cationic part of the above compound detergents has a extended hydrocarbon chain and a positive charge on the N atom.
eg : cetyltrimethylammonium bromide

(ii) Anionic detergents
following are the type of anionic detergents :
1. Sodium alkyl sulphates: such detergents are basically sodium salts of long chain alcohols. They are made by reacting such alcohols by concentrated sulphuric acid and later
reacting by sodium hydroxide. Suitable Examples of such detergents are sodium lauryl sulphate
($$C_{11}H_{23}CH_{2}OSO_{3}^{-}Na^{+}$$) and sodium stearyl sulphate($$C_{17}H_{35}CH_{2}OSO_{3}^{-}Na^{+}$$).

2.Sodium alkylbenzenesulphonates: such detergents are sodium salts of long chain
alkylbenzenesulphonic acids. They are synthesized by Friedel-Crafts alkylation of benzene
along with long chain alkyl halides or alkenes. The resultant product is initially reacted with
concentrated sulphuric acid and later reacted with sodium hydroxide. An example of anionic detergents is Sodium 4-(1-dodecy) benzenesulphonate (SDS)

(iii) Non-ionic detergents
molecules of such detergents have no ions. They are a good example of esters of alcohols which have high molecular mass. These are prepared by reacting stearic acid and polyethylene glycol.

Question 22:

Biodegradable detergents are detergents which are degraded by bacteria. They have straight hydrocarbon chains. For example: sodium lauryl sulphate

They have highly-branched hydrocarbon chains. For example:
sodium -4- (1, 3, 5, 7- tetra methyl octyl) benzene sulphonate

Question 23:

Explain why soaps do not work in hard water?

Long-chain fatty acids of sodium or potassium salts are present in soaps. Magnesium and calcium are contained in the hard water. Insoluble calcium or magnesium salts of fatty acids are formed when the ions displace sodium or potassium on dissolving soaps in hard water.

These insoluble salts separate as scum.

This is the reason why soaps do not work in hard water.

Question 24:
Which one should be used for finding the hardness of water, soaps or synthetic detergents?

Soaps will precipitate in hard water but it won’t get precipitated in soft water and hence it can be used for finding the hardness of water. Synthetic detergents, on the other hand, will not get precipitated both in hard water and soft water and cannot be used for finding the hardness of water.

Question 25:
Explain the cleansing action of soaps?

Soap molecules form micelles around an oil droplet (dirt) in such a way that the
hydrophobic parts of the stearate ions attach themselves to the oil droplet and the
hydrophilic parts project outside the oil droplet. Due to the polar nature of the hydrophilic
parts, the stearate ions (along with the dirt) are pulled into water, thereby removing the
dirt from the cloth.

Question 26:

You have water with dissolved calcium hydrogen carbonate. Will you use soap or synthetic detergents for cleaning clothes in this water? Explain

Usually, synthetic detergents are used for washing clothes. Soaps when dissolved in water containing calcium ions, these ions form insoluble salts which is of no use. Synthetic detergents are dissolved in water containing calcium ions, these ions form soluble salts that act as cleansing agents.

Question 27:
Label the hydrophilic and hydrophobic parts in the following compounds.


text:
Article

Full entry | PDF   (0.2 MB)
Keywords:
iteration digraph; height; Carmichael lambda function; fixed point; regular digraph
Summary:
A power digraph, denoted by $G(n,k)$, is a directed graph with $\mathbb Z_{n}=\{0,1,\dots ,n-1\}$ as the set of vertices and $E=\{(a,b)\colon a^{k}\equiv b\pmod n\}$ as the edge set. In this paper we extend the work done by Lawrence Somer and Michal Křížek: On a connection of number theory with graph theory, Czech. Math. J. 54 (2004), 465–485, and Lawrence Somer and Michal Křížek: Structure of digraphs associated with quadratic congruences with composite moduli, Discrete Math. 306 (2006), 2174–2185. The heights of the vertices and the components of $G(n,k)$ for $n\geq 1$ and $k\geq 2$ are determined. We also find an expression for the number of vertices at a specific height. Finally, we obtain necessary and sufficient conditions on $n$ such that each vertex of indegree $0$ of a certain subdigraph of $G(n,k)$ is at height $q\geq 1$.
References:
[1] Burton, D. M.: Elementary Number Theory. McGraw-Hill (2007).
[2] Carlip, W., Mincheva, M.: Symmetry of iteration graphs. Czech. Math. J. 58 (2008), 131-145. DOI 10.1007/s10587-008-0009-8 | MR 2402530 | Zbl 1174.05048
[3] Carmichael, R. D.: Note on a new number theory function. Am. Math. Soc. Bull. 16 (1910), 232-238. DOI 10.1090/S0002-9904-1910-01892-9 | MR 1558896
[4] Chartrand, G., Oellermann, O. R.: Applied and Algorithmic Graph Theory. International Series in Pure and Applied Mathematics McGraw-Hill (1993) \MR 1211413. MR 1211413
[5] Deo, N.: Graph theory with Application to Engineering and Computer Sciences. Prentice-Hall Series in Automatic Computation. Englewood Cliffs, N.J.: Prentice-Hall (1974). MR 0360322
[6] Ellson, J., Gansner, E., Koutsofios, L., North, S. C., Woodhull, G.: Graphviz--open source graph drawing tools. Mutzel, Petra (ed.) et al., Graph drawing. 9th international symposium, GD 2001, Vienna, Austria, September 23-26, 2001 Revised papers. Berlin: Springer. Lect. Notes Comput. Sci. 2265 (2002), 483-484. MR 1962414 | Zbl 1054.68583
[7] Husnine, S. M., Ahmad, U., Somer, L.: On symmetries of power digraphs. Util. Math. 85 (2011), 257-271. MR 2840802
[8] Kramer-Miller, J.: Structural properties of power digraphs modulo $n$. Proceedings of the 2009 Midstates Conference on Undergraduate Research in Computer Science and Mathematics, Oberlin, Ohio (2009), 40-49.
[9] Lucheta, C., Miller, E., Reiter, C.: Digraphs from powers modulo $p$. Fibonacci Q. 34 (1996), 226-239. MR 1390409 | Zbl 0855.05067
[10] Somer, L., Křížek, M.: On a connection of number theory with graph theory. Czech. Math. J. 54 (2004), 465-485. DOI 10.1023/B:CMAJ.0000042385.93571.58 | MR 2059267
[11] Somer, L., Křížek, M.: Structure of digraphs associated with quadratic congruences with composite moduli. Discrete Math. 306 (2006), 2174-2185. DOI 10.1016/j.disc.2005.12.026 | MR 2255611
[12] Somer, L., Křížek, M.: On semiregular digraphs of the congruence $x^{k}\equiv y \pmod n$. Commentat. Math. Univ. Carol. 48 (2007), 41-58. MR 2338828
[13] Somer, L., Křížek, M.: On symmetric digraphs of the congruence $x^{k}\equiv y \pmod n$. Discrete Math. 309 (2009), 1999-2009. DOI 10.1016/j.disc.2008.04.009 | MR 2510326
[14] Wilson, B.: Power digraphs modulo $n$. Fibonacci Q. 36 (1998), 229-239. MR 1627384 | Zbl 0936.05049
[15] MATLAB, The language of technical computing (version 7.0.0.19920 (R14)).

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# 9.5: Precipitation Reactions

• Anonymous
• LibreTexts

##### Learning Objectives
• To identify a precipitation reaction and predict solubility.

A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. When a colorless solution of silver nitrate is mixed with a yellow-orange solution of potassium dichromate, a reddish precipitate of silver dichromate is produced.

$\ce{ AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + KNO_3(aq)} \label{4.2.1}$

This unbalanced equation has the general form of an exchange reaction:

$\ce{ AC + BD \rightarrow }\underset{insoluble}{\ce{AD}} + \ce{BC} \label{4.2.2}$

Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Precipitation reactions are used to isolate metals that have been extracted from their ores, and to recover precious metals for recycling.

Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of aqueous $$NaCl$$ solution is mixed with 500 mL of aqueous $$KBr$$ solution, the final solution has a volume of 1.00 L and contains $$\ce{Na^{+}(aq)}$$, $$\ce{Cl^{−}(aq)}$$, $$\ce{K^{+}(aq)}$$, and $$\ce{Br^{−}(aq)}$$. As you will see in (Figure $$\PageIndex{1}$$), none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other.

## Predicting Precipitation Reactions

Previously, we had discussed the prediction of double replacement reactions without considering solubility. Now that we have learned how to use solubility rules, we will revisit double replacement reactions with this new knowledge. The steps for predicting the result of a double replacement reaction are essentially the same as before, but with an additional step included for using the solubility rules.

We will go through the same examples as previously, but adding the additional staff of using solubility rules. When a reaction does occur, we will obtain the same result but with the additional information of which of the products is solid and which is aqueous. If the reaction does not occur, we are now able to determine that using the solubility rules when we were not able to do so earlier.

Predict what will happen when aqueous solutions of barium chloride and lithium sulfate are mixed.

Change the partners of the anions and cations on the reactant side to form new compounds (products):

Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains $$\ce{Ba^{2+}}$$, $$\ce{Cl^{−}}$$, $$\ce{Li^{+}}$$, and $$\ce{SO4^{2−}}$$ ions. The only possible exchange reaction is to form $$\ce{LiCl}$$ and $$\ce{BaSO4}$$.

Correct the formulas of the products based on the charges of the ions.

No need to correct the formula as both compounds already have their charges balanced.

$\ce{BaCl_2(aq) + Li_2SO_4(aq) \rightarrow BaSO_4 + LiCl} \nonumber$

Refer to the solubility rules table to determine insoluble products which will therefore form a precipitate.

$\ce{BaCl_2(aq) + Li_2SO_4(aq) \rightarrow BaSO_4(s) + LiCl(aq)} \nonumber$

Table 7.5.1 from the previous section shows that $$\ce{LiCl}$$ is soluble in water, but $$\ce{BaSO4}$$ is not soluble in water.

Balance the equation:

$\ce{BaCl_2(aq) + Li_2SO_4(aq) \rightarrow BaSO_4(s) + 2LiCl(aq)} \nonumber$

Although soluble barium salts are toxic, $$\ce{BaSO4}$$ is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine $$\ce{BaSO4}$$ particles in water.

##### Example $$\PageIndex{1}$$

Predict what will happen if aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.

Solution

Steps Example

Change the partners of the anions and cations on the reactant side to form new compounds (products).

Correct the formulas of the products based on the charges of the ions.

$$\ce{RbOH(aq) + CoCl2(aq) \rightarrow RbCl + Co(OH)_2}$$

Refer to the solubility rules table to determine insoluble products which will therefore form a precipitate.

$$\ce{RbOH(aq) + CoCl_2(aq) \rightarrow RbCl(aq) + Co(OH)_2(s)}$$
Balance the equation.

$$\ce{RbOH(aq) + CoCl_2(aq) \rightarrow RbCl(aq) + Co(OH)_2(s)}$$

##### Example $$\PageIndex{2}$$

Predict what will happen if aqueous solutions of strontium bromide and aluminum nitrate are mixed.

Solution

Steps Example

Change the partners of the anions and cations on the reactant side to form new compounds (products).

Correct the formulas of the products based on the charges of the ions.

$$\ce{SrBr_2(aq) + Al(NO_3)_3(aq) \rightarrow Sr(NO_3)_2 + AlBr_3}$$

Refer to the solubility rules table to determine insoluble products which will therefore form a precipitate.

$$\ce{SrBr_2(aq) + Al(NO_3)_3(aq) \rightarrow Sr(NO_3)_2(aq) + AlBr_3(aq)}$$

According to Table 7.5.1 from the previous section, both $$\ce{AlBr3}$$ (rule 4) and $$\ce{Sr(NO3)2}$$ (rule 2) are soluble.

If all possible products are soluble, then no net reaction will occur.

$$\ce{SrBr_2(aq) + Al(NO_3)_3(aq) \rightarrow}$$

NO REACTION

##### Exercise $$\PageIndex{2}$$

Using the information in Table 7.5.1 from the previous section, predict what will happen in each case involving strong electrolytes.

1. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.
2. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.
3. Solid sodium fluoride is added to an aqueous solution of ammonium formate.
4. Aqueous solutions of calcium bromide and cesium carbonate are mixed.
Fe(OH)2 precipitate is formed.
Hg3(PO4)2 precipitate is formed.
No Reaction.
CaCO3 is precipitate formed.

## Summary

In a precipitation reaction, a subclass of exchange reactions, an insoluble material (a precipitate) forms when two electrolyte solutions are mixed. To predict the product of a precipitation reaction, all species initially present in the solutions are identified, as are any combinations likely to produce an insoluble salt.


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22
Jun 10

## Reaching escape velocity

Sample once from the Uniform(0,1) distribution. Call the resulting value $x$. Multiply this result by some constant $c$. Repeat the process, this time sampling from Uniform(0, $x*c$). What happens when the multiplier is 2? How big does the multiplier have to be to force divergence. Try it and see:

iters = 200
locations = rep(0,iters)
top = 1
multiplier = 2
for(i in 1:iters) {
locations[i] = runif(1,0,top)

top = locations[i] * multiplier
}

windows()
plot(locations[1:i],1:i,pch=20,col="blue",xlim=c(0,max(locations)),ylim=c(0,iters),xlab="Location",ylab="Iteration")

# Optional save as movie, not a good idea for more than a few hundred iterations. I warned you!
# library("animation")
# saveMovie(for (i in 1:iters) plot(locations[1:i],1:i,pch=20,col="blue",xlim=c(0,max(locations)),ylim=c(0,iters),xlab="Location",ylab="Iteration"),loop=1,interval=.1)


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# Simulated Annealing InitialPoints option

I have a cost function that depends on 4 parameters: CostFunction[{tx,ty,tz,ox}].

I used NMinimize[] function with SimulatedAnnealing method to determine its minimum. I looked for an example that uses the InitialPoints option but I did not find any. For this reason I am posting my question.

My code is like this (I cannot post the CostFunction[]):

  NMinimize[{
CostFunction[{tx, ty, tz, ox}],
tx >= 0 && tx <= a, ty >= 0 && ty <= b, tz >= 0 && tz <=c, ox >=0 && ox <= d},
{tx,ty,tz,ox},
Method -> {"SimulatedAnnealing", "PerturbationScale" -> 3, "SearchPoints" -> 10}]

The above code works correctly but when I add the "InitialPoints"->{0,0,0,0}, I receive the following error:

NMinimize::parchange: "Inappropriate parameter: \!$$\"InitialPoints\"$$ -> {0,0,0,0}, changed to Automatic"

Another option is not very clear "RandomSeed" ?

-

Borrowing the function from the documentation in the optimization tutorial

Clear[a, f];
a = Reverse /@ Distribute[{{-32, -16, 0, 16, 32}, {-32, -16, 0, 16, 32}},  List];
f = 1/(0.002 + Plus @@ MapIndexed[1/(#2[[1]] + Plus @@ (({x, y} - #1)^6)) &, a]);

NMinimize[f, {{x, -50, 50}, {y, -50, 50}}, Method -> {"SimulatedAnnealing",
"InitialPoints" -> Flatten[Table[{i, j}, {i, -45, 45, 5}, {j, -45, 45, 5}], 1]}]
{0.998004, {x -> -31.9783, y -> -31.9783}}

I think the problem with your use of InitialPoints is that you have to give it a list of points, and not a single point! Randomseed is used in many functions to specify a starting point for the random number generator so that all runs of the algorithm use the same set of random numbers (this may be desired for repeatability).

-
Ok! Thanks, in fact I am developping a program to align each frame from a video with a model. In ieration K, I minimize the costfunction[] to get the optimal rigid transformation between model and video frame, in the next iteration k+1, I want to inject the transformation given in the iteration k as initial point in order to get rapidly the convergence in the next iteration (k+1). Thanks ! – phdstudent Oct 24 '13 at 2:04
Have you looked at ImageAlign or ImageCorrespondingPoints. These are likely to do a better job than a direct optimization. – bill s Oct 24 '13 at 2:33
Thanks for your response, But I need precisily to do direct optimization, because in my case, I can not take a corresponding points. Thanks a lot – phdstudent Oct 24 '13 at 2:58
Please, What is the exactly role of the option InitialPoints. Going back to my exemple above, I pick InitialPoints like this Flatten[Table[{i, j, k, l}, {i, -500, 500, 50}, {j, -500, 500, 50}, {k, -90, 90, 20}, {l, -90, 90, 20}], 3]; The algorithm become very slow. I think that, it uses the InitialPoints as a starting points and from each points it look for an optimum, for this reason it is slow! And what is the relation between the option SearchPoints and InitialPoints? is the algorithm takes the SearchPoints from the IntialPoints? – phdstudent Oct 24 '13 at 17:35
Simulated annealing works by starting at a large number of initial guesses and then moving in (some kind of mostly) downhill direction until it converges. You want to start at a large variety of initial guesses in order not to miss any minima. – bill s Oct 24 '13 at 17:38


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## put image on a *.tex

kile
2008-04-19
2012-07-03
• debianizate - 2008-04-19

I am trying to put an image an a .tex file with kile, i am search in the gui but i don't find a button to put an image. I search in the help ( debian lenny ): it says:

figure

\begin{figure}[placement]

body of the figure

\caption{figure title}
\end{figure}

Figures are objects that are not part of the normal text, and are usually "floated" to a convenient place, like the top of a page. Figures will not be split between two pages.
The optional argument [placement] determines where LaTeX will try to place your figure. There are four places where LaTeX can possibly put a float:
h (Here) - at the position in the text where the figure environment appears.
t (Top) - at the top of a text page.
b (Bottom) - at the bottom of a text page.
p (Page of floats) - on a separate float page, which is a page containing no text, only floats.
The standard report and article classes use the default placement tbp.
The body of the figure is made up of whatever text, LaTeX commands, etc. you wish. The \caption command allows you to title your figure.

I can't unsderstand the help, maybe the help clearly tells to me how to do that but I don't understand, please anybody can help me.

• debianizate - 2008-04-19

i can put the picture.

• dvase - 2008-04-19

Hello,
You will need to include a statement telling LaTeX the filename of the picture to include.

A minimal example:

\documentclass{article}

\usepackage{graphicx}

\begin{document}

\begin{figure}[h]
\includegraphics{Your_Picture}
\end{figure}

\end{document}

Kile has a GUI that helps with this sort of thing, it can be found under "LaTeX -> Image Insertion"


text:
All Questions

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Uniqueness of equivalent martingale measure in Black Scholes-Model

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Pricing options under restricted domain

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Wholesale credit risk management

Trying to read up on "Wholesale credit risk ", can't find any useful references, why the emphasis on wholesale? - Any help greatly appreciated.
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Hedging bond with CDS of different maturity

Say I buy a 10-year bond with a notional of 100k. To hedge my credit risk entirely I could buy a 10-year CDS, also on a notional of 100k. Now, if there are only 5-year CDS trading and no 10-year CDS, ...
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I'm working in the commodity market and I've to price Swing Options with MATLAB, preferably with finite element. Has anyone already priced these kind of derivatives? I'm thinking about using the ...
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Data on banks’ leverage

Does someone know free resources to estimate the leverage of the banking and financial sector at an aggregate level? In particular I would be interested in something like Federal Reserve’s Flow of ...
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Testing the validity of a factor model for stock returns

Consider the following m regression equation system: $$r^i = X^i \beta^i + \epsilon^i \;\;\; \text{for} \;i=1,2,3,..,n$$ where $r^i$ is a $(T\times 1)$ vector of the T observations of the dependent ...
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garchOxFit in R-oxo file does not match

Could someone please help me with trying to get the Ox interface to work in R. I get the following errors as output: This version may be used for academic research and teaching only Link error: ...
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Data on margin volumes?

I came across a Financial Times article today that said "Peaks in margin trading have been a precursor to bear runs in the past, notably in March 2000 and July 2007." I'm curious if anyone here would ...


text:
+0

# Help CPhil!!!

+8
355
4
+273

A bag contains 16 congruent blocks, 4 of which are blue and 12 of which are red. Two blocks are randomly pulled from the bag at the same time. What is the probability that both blocks are blue? Express your answer as a common fraction.

Jan 20, 2022

#1
+2602
+1

Because both blocks are pulled at the same time, there is a $${1 \over 4} * {3 \over 16} = {3 \over 64}$$ chance that both balls will be blue. Correct me if I am wrong.

Jan 20, 2022
#2
+1

A bag contains 16 congruent blocks, 4 of which are blue and 12 of which are red. Two blocks are randomly pulled from the bag at the same time. What is the probability that both blocks are blue? Express your answer as a common fraction.

Arbitrarily call either block the first one.

The chance that the first block is blue is 4 / 16.

For both blocks to be blue, the first one had to be blue.

Therefore, 3 of the remaining 15 blocks are blue.

The chance that the second block is blue is 3 / 15.

For both things to happen, multiply their chances.

( 4 / 16 ) • ( 3 / 15 )  =  12 / 240  =>  this reduces to 1 / 20.

.

Guest Jan 20, 2022
#4
+2602
+1

Ah...

BuilderBoi  Jan 20, 2022
#3
+273
+9

nevermeind solved it myself 20 minits ago

its 1/20

Jan 20, 2022


text:
# Is there a formal way to show that $X \cap Y \subseteq X$.

The question is in the title. It is trivial that $X \cap Y \subseteq X$. Because $X \cap Y$ only contains elements that are both in $X$ and in $Y$. So every element in $X \cap Y$ is also an element of $X$ and so $X \cap Y \subseteq X$ is true.
Is there a more formal way to proof this?

• Your proof is fine and sufficienlty formal. Good job (: – Stefan Mesken Jul 25 '16 at 22:42

Use the definition of subset ($A\subseteq B$ if $x\in A$ implies $x\in B$) and the definition of intersection.
Let $x\in X\cap Y$. Then $x\in X$ (and $x\in Y$) by the definition of intersection.


text:
# Why is time represented on the vertical axis in space-time diagrams?

One thing that bugs me when learning and explaining special relativity concepts is that the time axis on space-time charts always on the vertical axis. While almost everywhere else time is going on the horizontal axis.

Moreover space-time coordinates often written in the order of $(t, x, y, z)$. And in case of 1 dimensions $(t, x)$. Now when we speak of the event $(A, B)$, $A$ goes on the vertical axis $B$ on the horizontal and not vice versa as it usually goes on the coordinate charts.

So I always need to twist my mind a bit when working with it.

When did we started putting time on the vertical axis and what's the reason behind it?


text:
# proportional

1. ### What to do with proportion response variable

Hey! I am so confused as to what to do with my data. I have 5 types of communication signal A,B,C,D,E recorded. For 26 individuals, I have counted the total number of signals they produced, and converted the number of A,B,C,D,E from the total into mean proportions eg. mean proportion of total...


text:
# Differences

This shows you the differences between two versions of the page.

 — preprints:2009:ttp09-14 [2016/03/17 11:03] (current) Line 1: Line 1: + ====== TTP09-14 CKM ELEMENTS FROM SQUARK-GLUINO LOOPS ====== + + +   I present results for the finite renormalization of the Cabibbo-Kobayashi-Maskawa ( + CKM) + matrix induced by gluino--squark diagrams in the MSSM with non-minimal + sources of flavour violation. Subsequently I derive bounds on the + flavour--off--diagonal elements of the squark mass matrices by requiring that the + radiative corrections to + the CKM elements do not exceed the experimental values. The constraints on + the associated dimensionless quantities $\delta^{d\,LR}_{ij}$, $j>i$, are + stronger than the bounds from flavour-changing neutral current (FCNC) + processes if gluino and squarks are heavier than 500$\,\gev$. The + results imply that it is still possible to generate all observed flavour + violation from the soft supersymmetry-breaking terms without conflicting + with present-day data on FCNC processes. Therefore we reappraise the idea that onl + y $Y^q_{33}$ + is different from zero at tree-level and all other Yukawa couplings are generated + radiatively. This model solves the SUSY flavour as well as the SUSY CP problem. The + presented results are published in \cite{Crivellin:2008mq}. + + |**Andreas Crivellin**  | + |**     To appear in the proceedings of 44th Rencontres de Moriond EW 2009: Electroweak Interactions and Unified Theories, La Thuile, Italy, 7-14 Mar 2009. **  | + | {{preprints:2009:ttp09-14.pdf|PDF}} {{preprints:2009:ttp09-14.ps|PostScript}} [[http://arxiv.org/abs/0905.3130|arXiv]]   | + | |