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# Proof problem, help!

1. Mar 20, 2005

### Cyannaca

I would really appreciate if anyone could help me with this problem.

F is a group homomorphism from G= (E, *) to H= (F,#).
If , for all x e E , x*x e ker(f).
Show that for all x e E, f(x-1)=f(x)

Now I don't know how I should start the proof. Also, I would like to know if I can assume that Ker(f) is equal to the identity element.

2. Mar 20, 2005

### AKG

If $x*x \in Ker(f)$ then this means precisely that $f(x*x) = 1_F$ where $1_F$ is the identity element of H. You want to show something like:

$$f(x*y) = f(x)#f(y)$$

and

$$f(x^{-1}) = (f(x))^{-1}$$

In fact, you probably already have these proved somewhere. If you do, then use them and the rest should be very simple. Actually, I don't remember the definitions, but I have a feeling that the first thing (that f(x*y) = f(x)#f(y)) might just be a property of homomorphisms. The second property follows from the first at least in the case where x*x is in Ker(f), and probably in general too.

3. Mar 21, 2005

### matt grime

You've got F as a homomorphism, and the underlying set of H (at least that's what I presume H=(F,#) means. Usually there is no reason to distinguish the group from its elements like that. Also, if you're not going to tex it then it's common to indicate inverses by putting the minus 1 in an exponent, since x-1 doesn't really make sense for a group. You may not assume ker(f) is the identity element, since that is saying f is an injection which is not given.

From the definition y in ker f iff f(y)=e conclude something about f(xx), and use AKG's stuff above which is the (usual) definitoin of what a homomorphism is.

And remember showing that y is the inverse of x is showing exactly that xy=yx=e.

4. Mar 21, 2005

### AKG

Matt, I believe he/she meant f (lower case) to be the homomorphism, and F to be the set underlying group H.

5. Mar 21, 2005

### robert Ihnot

Is that what the guy is talking about, F(x-1) is actually F(x^(-1))? Just written differently? No wonder the problem look mysterious! This is a completely simple problem now that it is written so it can be read! If F(x*x) goes to 1, then x is its own inverse, since by group property there can be only a unique inverse.

You gotta be careful about how you write the inverse if you use a calculator, you know!

Last edited: Mar 21, 2005


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## College Algebra (10th Edition)

Published by Pearson

# Chapter 3 - Section 3.3 - Properties of Function - 3.3 Assess Your Understanding: 2

#### Answer

The slope is $1$.

#### Work Step by Step

The slope of the line through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $m=\dfrac{y_2-y_1}{x_2-x_1}$ Use the formula above to obtain: $m=\dfrac{3-8}{-2-3}=\dfrac{-5}{-5}=1$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

format: plain


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# Worded Problem, Probability Density, quantum mechanics

1. May 18, 2013

### Cogswell

1. The problem statement, all variables and given/known data

The needle on a broken car speedometer is free to swing and bounces off perfectly off the pins at either end, so that if you gave it a flick it's equally likely to come to rest at 0 and $\pi$

What is the probability density, $p(x)$?

2. Relevant equations

$$\int^{\pi}_{0} p(x) dx = \int^{\pi}_{0} |\Psi (x,t)|^2 dx = 1$$

3. The attempt at a solution

This may sound really dumb but I was thinking that since that it's equally likely to be from anywhere between 0 and pi, then it'll just be a straight horizontal line, p(x) = 1.
But since it has to be normalised from 0 to pi, then p(x) will just be 1/pi.

But that didn't seem right because it had to drop off to 0 at x=0 and x=pi so I was thinking of a sine curve, and maybe p(x) = sin(x)

How does that work though? It says it's equally likely to land anywhere from 0 to pi, but with a sine curve it's obvious that it'll have a way higher chance of landing in the middle...

#### Attached Files:

• ###### IMG_20130519_130041.jpg
File size:
24.6 KB
Views:
179
2. May 18, 2013

### Simon Bridge

Why do you need $p(x=0)=0$ and $p(x=\pi)=0$ ?

Note: if p(x) is a probability density function, then $$P(a<x<b)=\int_a^bp(x)dx$$

3. May 18, 2013

### Cogswell

Oh sorry I was thinking of a gaussian curve and how it drops off to zero at infinity.

So I was right about $p(x) = \dfrac{1}{\pi}$??

Then if I evaluate the integral $\displaystyle P(a < x < b) = \int^{\pi}_{0} \dfrac{1}{\pi} dx = 1$

If I evaluate it from $\displaystyle \theta$ to $d \theta$ I'll always get the same number, no matter what theta is: $\displaystyle \int^{\theta + d \theta}_{\theta} \dfrac{1}{\pi} dx = Constant$

It's the only function that I can think of that has the same probability no matter what theta is.

4. May 18, 2013

### Simon Bridge

OK you are having a bit of trouble having confidence in your deductions...

The top-hat distribution from x=0 to x=L would be: $p(x)=\frac{1}{L}( h(x)-h(x-L) )$ where $h(x)$ is the Heaviside step. Notice that the requirement $\int_{-infty}^\infty p(x)dx$ means that$P(0<x<L)=1$ and so does not need the points x=0 and x=L to be included?

I can write that out:$$p(x)=\left \{ \begin{array}{ll} \frac{1}{L} & : \; 0<x<L\\ 0 & : \; x\leq 0\\ 0 & : \; x\geq L \end{array} \right .$$
... but does it matter to the statistics if we do include 0 and L in the first inequalities?
(note: if you look up the Heaviside function, you'll see different people use different values at the limits.)

The question says that the probability of finding the needle in any arbitrarily small region inside 0≤x≤π is a constant - so far so good - and, intuitively, you'd figure that the probability of finding the needle exactly at x=0 or x=π would be zero. Does the top-hat distribution give you that?

Incidentally: since you mentioned QM, you could check to see if $\psi(x)=\frac{1}{\sqrt{L}}( h(x)-h(x-L) )$ is a solution to the Scrödinger equation ;)

Last edited: May 18, 2013
5. May 19, 2013

### vela

Staff Emeritus
Is the problem statement exactly as it was given to you? I ask because as you typed it, it doesn't say that the needle is likely to be anywhere between 0 and $\pi$; it said it's equally likely to be 0 or $\pi$.

6. May 19, 2013

### Cogswell

I'm pretty sure I typed it out exactly as the book had it.
I took a photo of the original problem (In the book Quantum Mechanics Second Edition by David J Griffiths)

It's near the very start of the book so I'm presuming it doesn't require that much knowledge about quantum mechanics.

Sorry I don't know what a top-hat distribution is, nor what a Heaviside step. (I quickly googled it a top-hat distribution and I see that it would be the shape I need for equal probability.)

I get that:

$$p(x)=\left \{ \begin{array}{ll} \frac{1}{L} & : \; 0<x<L\\ 0 & : \; x\leq 0\\ 0 & : \; x\geq L \end{array} \right .$$

where I presume L is $\pi$ in this case.

I don't know if the top hat distribution would give me 0 for x=0 and x=pi, we haven't needed to study it for this course.
From the diagrams it seems like it'll give me 1/pi for when x=0 and x=pi, because it looks like the infinite square well. But does it really matter what x=0 and x=pi are?
I guess the speedometer needle angle can never be zero because there's a pin there that stops it, and same with the angle pi.
But I don't quite get the top hat equation is. I google it and can't seem to find a general form of it.

7. May 19, 2013

### vela

Staff Emeritus
Oh, I see you left out a couple of words when you typed up the statement that completely change the meaning.

8. May 19, 2013

### Simon Bridge

... you didn;t type it out the way your attached picture tells it - but never mind, I followed the attachment and not what you wrote.

But you do know how to compute a probability from a probability density function don't you?

Anyway - you have seen it before - probably just not under that name.
Unless... where abouts in your education are you at right now?

How did you compute that?

You mean p(x)? - um - no it's pretty much the opposite of a square well ... very little like an infinite one. Did you try plotting it?

That was my question to you ;)
Note: you are the one who brought it up.

<puzzled> I gave you a general form of it - it's the piece-wise function with the L in it. Really generally, a uniform pdf : a<x<b is $$p(x) = \frac{1}{b-a}[h(x-a)-h(x-b)]$$ where h(x) is the Heaviside step function.

I googled for "top hat probability density function" and came up with lots of examples, i.e.
http://www.jhu.edu/virtlab/course-info/ei/notes/uncertainty_notes.pdf (p4)
http://www.pha.jhu.edu/~neufeld/numerical/lecturenotes5.pdf (slide 104)
...

It's one of the first distributions you get taught studying probability - the discrete version is the distribution obeyed by dice. It's one of the simplest around: try "uniform distribution":
http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)

That help?

But you don't need to know the name ... just the maths.
It is looking almost like you don't understand what a probability density function does.

Last edited: May 19, 2013
9. May 19, 2013

### Cogswell

Oh sorry I thought the hint part wasn't that important and that the all I needed in the question was the main part.

I'm guessing it's just the area under the curve of the probability density function, $\displaystyle P(a \le x \le b) = \int_{a}^{b} p(x)dx$

I don't think I've seen it before, We've only studied the basics of quantum mechanics.
I'm studyng at Otago University, New Zealand.
Second year physics (PHSI231), doing a bit of classical, quantum and thermal physics.

I didn't really compute it, I just looked at the pictures of what a top-hat distribution looked like.

Oh yea, like the opposite of an infinite square well.
How would you plot it, something like this?
Code (Text):

______
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|     |
|     |
_____|     |_____

I'm guessing since the integral I used to calculate it $\displaystyle \int_{0}^{\pi}p(x) dx = 1$ then it would have to include 0 and pi, because they are the bounds of the integral and so p(x) = 1/pi and also p(0) = 1/pi

Ah that Wikipedia page helped me heaps, thanks. It didn't show up when I googled stuff but I guess I just didn't know what to search.

So then the probability density would just be the top hat distribution (I don't really understand what a Heaviside step function is, and the wikipedia article on it is quite confusing for me):
$$p(x) = \frac{1}{\pi}[h(x)-h(x-\pi)]$$

$$p(x)=\left \{ \begin{array}{ll} \frac{1}{\pi} & : \; 0 \le x \le \pi \\ 0 & : \; x < 0\\ 0 & : \; x > \pi \end{array} \right .$$

I presume this is what I want? It satisfies all the conditions,

$$\displaystyle \int_{0}^{\pi} p(x) dx = 1$$

$$\displaystyle P(0 \le \theta < d \theta \le \pi) = \int_{\theta}^{\theta + d \theta} p(x) dx = \text{A Constant}$$

Haha sorry, I'm not really 'with it' when it comes to understanding physics concepts. I'm more of a Math Major. I do get that the area under the probability density function from a to b tells us that's the probability of finding the particle within that region. And that the wave function squared gives us a probability density function.

10. May 19, 2013

### Simon Bridge

That makes sense.

At level 2 you are expected top bring all your learning to bear on the course and not just the stuff specifically mentioned in the lectures. You are expected to have a core knowledge base to build on. What has happened here is that a prior familiarity with probability and statistics (part of mathematics), that you do not possess, has been expected of you.

In New Zealand - you get introduced to probability and statistics as part of the core math course at secondary school ... you will have met probabilities with dice by year 9.
You will have learned that the probability of a particular outcome is (by definition) the number of ways to get that outcome divided by the number of ways to get anything else.

When you specialize - you take a calculus or a statistics track in maths, but the core curriculum for both tracks includes basic probability and statistics - so you should have seen at least the normal distribution, and, by extension, continuous probability distributions, before you left secondary school. The continuous uniform distribution may not have been taught in the calculus track - that would depend on the school.

Otago does not require a statistics paper at stage 1 as a prerequisite for stage 2 QM as the paper does not require an understanding of stats above year 12 - but it is advisable if you did not take the stats track at secondary level.

So it is possible that you have done no probability or statistics since graduating secondary school.
If this is the case - then you should add a statistics paper to your course.

You say you have "got" the basic math for probability density functions but you had trouble with basic questions involving that knowledge... you "guess"ed about how to get a probability from a density function for eg, and were confused about what was part of physics and what part of math, you appeared to have trouble picturing the shapes of functions, and you don't seem to recall secondary math

I suspect you are used to the "applying formulas" part of math - when what you need is the "understanding" part.

Last edited: May 19, 2013
11. May 19, 2013

### Simon Bridge

Seeing you are at level 2 physics, then you probably have not met the Heaviside function in so many words yet - another function you will see a lot of but have yet to meet will be the Dirac delta function $\delta(x)$ - though you have probably seen the Kronecker delta $\delta_{jk}$.

The Heaviside function is simply a way of writing a piece-wise function on one line ...
h(x)=0: x<0 and h(x)=1: x>0 ... what happens at x=0 depends on who you ask.
I usually put h(0)=0, some people put h(0)=1/2 or 1. For the purposes above it doesn't matter.

You seem to have got it in the end though.
It shouldn't be this hard for you - this is math, not physics, and you say you are "math oriented".

12. May 20, 2013

### Cogswell

I remember being introduced to basic probability with coins and dice but not as much statistics. We only did like the mean, median and mode sort of stuff in High School.
I specialised in calculus, not statistics and never saw a normal distribution in High School. Right now I am taking a first year statistics paper, and we're only getting into the basics of a normal distribution.

You seem to know quite a lot about NZ and Otago university, which is weird. Not many people know much about New Zealand.

I am still quite confused about understanding physics concepts so that's probably why it seems like I'm lost.
Starting from the top, the whole concept of a particle not being in a particular location is new and surprising to me. I always thought Laplace's[/PLAIN] [Broken] Demon was sorta true and that if you measure the precise location and momentum of every particle you can predict the past present and future. Then they introduce this whole thing called Heisenberg's uncertainty principle to me and then everything's just changed.
And then there's the whole thing of a potential and a stray particle. Do they mean a particle as in a proton (Hydrogen ion)?

I'm not confused about shapes of functions (at least not continuous ones). In calculus we've only ever dealt with continuous functions and so the top hat distribution is still a bit weird for me.

I guess the NZ High School education system isn't that great at preparing you for university. I got straight into second year physics because of my 'strong' mathematics background (and I took a basic first year university maths paper and got A+).

Nope, never heard of the Heaviside function, but that (how you explained it) makes it so much simpler. I don't know why the Wikipedia article on it is so much more confusing to understand.

Last edited by a moderator: May 6, 2017
13. May 20, 2013

### Simon Bridge

This is basically what I'd come to suspect.

More than you'd suspect - Kiwis have been getting around and seem to have a pretty good international reputation.

I am very familiar with NZ education system, having works in it for decades across all the levels.

From what you'g got now, you should be able to cope with the maths of continuous probability distributions - bear in mind that p(x) can be any function at all so long as the total area under the graph can be normalized.

It's something everyone has to go through ... you get used to it.

The definition of a particle is deliberately left vague at this stage - just think of it as building upon what you already know about classical particles: a purely abstract idea which has application in the real world.
A good RL example of a particle is an electron.
The idea applies to any object whose internal structure and size does not affect it's dynamics.

The top-hat distribution is continuous in the sense that you saw in calc class.

Oh OK - that would explain a lot then. People who get the accelerated courses tend to struggle.

The stage 1 papers mostly concentrate on going over everything you were supposed to get by the end of secondary school ... makes sure everyone is on the same page. This is really great for solidifying what you already know and fills in a lot of gaps where you may have forgotten something or the teachers ran out of time or something.

You need a good crash course then - you should try to get hold of NZQA primer books - the sorts of things that prepare you for external exams. The NZQA website has past exam papers and exemplars that can make a useful study guide.
http://www.nzqa.govt.nz/ncea/assessment/search.do?query=mathematics&view=exams&level=03

Wikipedia math articles tend to be marginal in usefulness because they have been written by mathematicians who care more about being mathematically rigorous than being easy for a novice to understand.

Don't let the names of things throw you - focus on what the thing actually is.

Last edited by a moderator: May 6, 2017
14. May 23, 2013

### Cogswell

I'm trying to get A+ in this paper and it seems like it's near impossible! I have 2 weeks before my exams and I'm studying this at least 3 hours a day for this and reading the books and going through the questions but I'm still lacking a lot of base knowledge.
I think the best thing is to go through more questions so I understand how to do them.

Okay, so I have answered the first part of the question, which is:

$$p(\theta)=\left \{ \begin{array}{ll} \frac{1}{\pi} & \text{for} \; 0 \le \theta \le \pi \\ 0 & \text{for} \; \theta < 0\\ 0 & \text{for} \; \theta > \pi \end{array} \right .$$

The second part asks "Compute $< \theta >$, $< \theta^2 >$, and $< \sigma >$, for this distribution"
Exactly as is, no hints or anything.

So I know that $| \Psi (x,t) |^2 = p(x)$

Therefore: $\Psi (\theta , t)=\left \{ \begin{array}{ll} \frac{1}{\sqrt{\pi}} & \text{for} \; 0 \le \theta \le \pi \\ 0 & \text{for} \; \theta < 0\\ 0 & \text{for} \; \theta > \pi \end{array} \right .$

The expectation value is just:

$$< \theta > = \displaystyle \int^{\pi}_{0} \Psi (\theta , t)^* (\theta) \Psi (\theta , t) d \theta$$

$$< \theta > = \displaystyle \int^{\pi}_{0} \frac{1}{\sqrt{\pi}} ( \theta) \frac{1}{\sqrt{\pi}} d \theta$$

$$< \theta > = \displaystyle \frac{1}{\pi} \int^{\pi}_{0} \theta d \theta$$

$$< \theta > = \displaystyle \frac{1}{\pi} \left[ \dfrac{\theta ^2}{2} \right]^{\pi}_{0}$$

$$< \theta > = \dfrac{\pi}{2}$$

This seems correct, because that's the center of probability distribution.

And then, for the expectation value of $\theta ^2$

$$< \theta ^2 > = \displaystyle \int^{\pi}_{0} \frac{1}{\sqrt{\pi}} ( \theta ^2) \frac{1}{\sqrt{\pi}} d \theta$$

$$< \theta ^2 > = \dfrac{\pi ^2}{3}$$

And I know that $\sigma = \sqrt{< \theta ^2> - < \theta >^2}$ so I can just easil work ou sigma from that.

Is that right?

The last question then asks for me to compute $< \sin ( \theta )>$, $< \cos ( \theta )>$ and $< \cos ^2 ( \theta )$.
It should be fairly simple right, sort of like above, and for the $\cos^2$ I just use the trig identity to simplify it so I can integrate it?

15. May 25, 2013

### Simon Bridge

Doing good so far.
Just a niggle - in the very first line of math, the limits of the integral should be all space ... you got specific in the next line.

You are on the right track - notice that the sine and cosine parts are y and x components of the needle's arc?
For full marks you may want to comment on if <sinA> bears any relation to the <A> etc.
There is nothing wrong with just looking up such simple integrals.

16. May 25, 2013

### Cogswell

What do you mean, so I should integrate from negative infinity to infinity? But then how would I write the integral, since the wave function is a piecewise function? Something like this?

$$\displaystyle \int_{- \infty}^{0} 0^2 \theta d \theta + \int_{0}^{\pi} \dfrac{\theta}{\sqrt{\pi}^2}d \theta + \int_{\pi}^{\infty} 0^2 \theta d \theta$$

I think the sine and cosine things relate to the next question in the book:

So for this one, the probability density would be like a cosine curve?

Code (Text):

|
+
/|\
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-----|----|----|----x
-2r   |    2r
y

(Sorry for the horrible cosine curve drawn by the code tags, but you know what a cosine curve looks like)

So to normalise it, I'd need:

$$\displaystyle \int^{2r}_{-2r} Acos(\dfrac{\pi x}{2r}) dx = 1$$

Although I don't know what to do with the hint...

17. May 25, 2013

### Simon Bridge

The first line is fine the way you wrote it - but the limits should, technically, be all space. The second line is fine to follow.

After all, you started from the definition of the expectation value ... then you observe that some sub-areas are zero.

But, like I said, it's just a niggle.

That hint is a little bit like how you convert between polar and rectangular coordinates.
You know $p(\theta)$ of finding the needle between $\theta$ and $\theta+d\theta$ ... can you relate that range to x and x+dx?

The arclength in that range, for eg, $rd\theta$ for instance.

18. May 25, 2013

### Cogswell

Oh right I get it, because I wrote $\Psi (x,t)$ in the first integral and the wave function is always from negative infinity to infinity. And then I changed to the specific piece of the wave function which had a different boundary, limited from 0 to pi.

So the x position is related to theta by:

$x = \cos \left(\dfrac{\pi \theta}{2r} \right)$ for $0 \le \theta \le \pi$

$dx = - \dfrac{\pi}{2r} \sin \left(\dfrac{\pi \theta}{2r} \right) d \theta$

What do I need to do with these now?

Integrate?

$\displaystyle \int_{x}^{x + dx} A \cos \left(\dfrac{\pi \theta}{2r} \right) dx$

??

19. May 25, 2013

### Simon Bridge

In fact $\Psi(x,t)$ was overkill - you are only worried about the space part.

Note: everything inside a trig function should have the same dimensions as angle.

What you are trying to do is relate $\psi(x)$ with what you know about $\psi(\theta)$.
I'd start by sketching it out.

20. May 26, 2013

### Cogswell

Oh right I think the 2r part is supposed to be outside of the cosine function.

So something like this: $x = r \cos( \theta)$
And so $\theta = \arccos \left( \dfrac{x}{r} \right)$

So when the angle is zero, the needle's x position is r away, and when the angle is 90 degrees then the needle's x position is at 0.

And then $dx = -r \sin (\theta) d \theta$

Now do I replace $\theta$ with x for the probability density?

$p(\theta)=\left \{ \begin{array}{ll} \frac{1}{r} & : \; 0< \theta <r \\ 0 & : \; \theta \leq 0\\ 0 & : \; \theta \geq r \end{array} \right .$

Am I headed in the right direction?


text:
# Proof of Pythagorean identity of Secant and Tangent functions

In fact, the Pythagorean identity of secant and tan functions is proved mathematically in trigonometry by the geometrical method. Now, let us learn how to derive the proof of the Pythagorean identity for the secant and tangent functions.

### Write the relation between sides of the triangle

Draw a right triangle. It is considered as $\Delta BAC$ and its angle is represented by theta.

In this triangle, $\overline{BC}$, $\overline{AB}$ and $\overline{AC}$ are the opposite side, adjacent side and the hypotenuse respectively, and their respective lengths are written as $BC$, $AB$ and $AC$ mathematically.

As per the Pythagorean theorem, write the relation between the three sides of the right triangle in mathematical form.

${AC}^2 = {AB}^2+{BC}^2$

Remember this equation in your mind and it will be used for deriving the Pythagorean identity of secant and tan functions.

### Express the sides in trigonometric functions

According to the right triangle $BAC$, let us try to write the lengths of the sides in terms of the secant and tan functions. Firstly, we have to express the secant and tan functions in their ratio form for doing it.

$\sec{\theta} \,=\, \dfrac{AC}{AB}$

$\implies$ $AC = AB \times \sec{\theta}$

$\tan{\theta} \,=\, \dfrac{BC}{AB}$

$\implies$ $BC = AB \times \tan{\theta}$

### Express the equation in trigonometric form

We have already derived the lengths of the opposite site and hypotenuse in trigonometric functions in the above step.

1. $AC = AB \times \sec{\theta}$
2. $BC = AB \times \tan{\theta}$

Now, let us substitute them in the equation, which represents the relationship between then lengths of the sides of the right triangle.

${AC}^2 = {AB}^2+{BC}^2$

$\implies$ $(AB \times \sec{\theta})^2 = {AB}^2+(AB \times \tan{\theta})^2$

The terms in both sides of the equation can be simplified by the power rule of a product.

$\implies$ $(AB)^2 \times (\sec{\theta})^2$ $\,=\,$ ${AB}^2+(AB)^2 \times (\tan{\theta})^2$

$\implies$ ${AB}^2 \times \sec^2{\theta}$ $\,=\,$ ${AB}^2+{AB}^2 \times \tan^2{\theta}$

$\implies$ ${AB}^2 \times \sec^2{\theta}$ $\,=\,$ ${AB}^2 \times 1+{AB}^2 \times \tan^2{\theta}$

$AB^2$ is a common factor in the expression of the right hand side of the equation. So, it can be taken out common from them by the factorization.

$\implies$ ${AB}^2 \times \sec^2{\theta}$ $\,=\,$ ${AB}^2 \times (1+\tan^2{\theta})$

$\implies$ $\sec^2{\theta}$ $\,=\,$ $\dfrac{{AB}^2 \times (1+\tan^2{\theta})}{{AB}^2}$

$\implies$ $\sec^2{\theta}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{{AB}^2} \times (1+\tan^2{\theta})}{\cancel{{AB}^2}}$

$\implies$ $\sec^2{\theta}$ $\,=\,$ $1+\tan^2{\theta}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\sec^2{\theta}-\tan^2{\theta} \,=\, 1$

Therefore, it is proved that the subtraction of tan squared of an angle from the square of secant of angle is equal to one. It is called as the Pythagorean identity of secant and tan functions.

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text:
# Bootstrap and permutation tests of independence for point processes

Abstract : Motivated by a neuroscience question about synchrony detection in spike train analysis, we deal with the independence testing problem for point processes. We introduce non-parametric test statistics, which are rescaled general $U$-statistics, whose corresponding critical values are constructed from bootstrap and randomization/permutation approaches, making as few assumptions as possible on the underlying distribution of the point processes. We derive general consistency results for the bootstrap and for the permutation w.r.t. to Wasserstein's metric, which induce weak convergence as well as convergence of second order moments. The obtained bootstrap or permutation independence tests are thus proved to be asymptotically of the prescribed size, and to be consistent against any reasonable alternative. A simulation study is performed to illustrate the derived theoretical results, and to compare the performance of our new tests with existing ones in the neuroscientific literature.
Keywords :
Type de document :
Article dans une revue
The Annals of Statistics, Institute of Mathematical Statistics, 2015, 43 (6), pp.2537-2564. 〈10.1214/15-AOS1351〉
Domaine :

https://hal.archives-ouvertes.fr/hal-01001984
Contributeur : Mélisande Albert <>
Soumis le : mercredi 27 mai 2015 - 14:49:07
Dernière modification le : samedi 24 mars 2018 - 01:55:11
Document(s) archivé(s) le : mardi 15 septembre 2015 - 07:00:45

### Fichiers

Testindpp_Hal_v4.pdf
Fichiers produits par l'(les) auteur(s)

### Citation

Mélisande Albert, Yann Bouret, Magalie Fromont, Patricia Reynaud-Bouret. Bootstrap and permutation tests of independence for point processes. The Annals of Statistics, Institute of Mathematical Statistics, 2015, 43 (6), pp.2537-2564. 〈10.1214/15-AOS1351〉. 〈hal-01001984v4〉

### Métriques

Consultations de la notice

## 671

Téléchargements de fichiers


text:
Alexey V. BORISOV, Alexander V. SHAPOVALOV and  Andrey Yu. TRIFONOV
Theoretical Physics Department,
Tomsk State University,
36 Lenin ave.,
Tomsk,
634050, RUSSIA
and
Mathematical Physics Laboratory of the High Mathematics
and Mathematical Physics Department,
Tomsk Polytechnic University,
30 Lenin ave.,
Tomsk,
634050, RUSSIA
E-mail: shpv@phys.tsu.ru, trifonov@mph.phtd.tpu.edu.ru

Transverse Evolution Operator and Symmetry Operators for the Gross-Pitaevskii Equation in Semiclassical Approximation

Abstract:
The $n$-dimensional Gross-Pitaevskii equation (GPE) with an attractive self-action reads

\left(-i\hbar\partial_{t}+{\cal H}( \hat{\vec p},\vec x,t)-g^{2}|\Psi(\vec x,t,\hbar)|^{2}\right)\Psi(\vec x,t,\hbar)=0,

where $\vec x \in {\mathbb R}^{n}$, $t\in{\mathbb R}^{1}$, $\hat{\vec p}=-i\hbar\nabla_{\vec x}$, $g$ is a real nonlinearity parameter, $\hbar$ is  a small asymptotical parameter, $\hbar\to 0$; ${\cal H}( \hat{\vec p},\vec x,t)$ is a linear operator quadratic in $\hat{\vec p}$ and Weyl ordered in $\hat{\vec p}$ and $\vec x$. The GPE (1) is one of the basic model equations in the theory of Bose-Einstein condensate (BEC). Localized solutions of the GPE describe the condensate in external electro-magnetic fields including fields of magnetic traps. In the context of the complex WKB-Maslov method  a class of one-soliton trajectory concentrated functions (OSTCF) is introduced. Functions of this class are soliton-like fast-oscillating wave packets concentrated in a neighborhood of a trajectory in an effective phase space. A solution of the Cauchy problem is presented for the Eq. (1) in the class of OSTCF in semiclassical approximation. The evolution operator acting on the variables transversal on a wave packet vector is derived to obtain the leading term of the asymptotic solution in the class of OSTCF. A class of symmetry operators for the GPE is constructed in semiclassical approximation using the evolution operator. The three-dimensional GPE with the oscillator external field is considered as an illustration and the collapse problem is discussed.

The work has been  supported in part by the Grant for Support of Russian Scientific Schools 1743.2003.2, and the Grant of the President of the Russian Federation YD-246.2003.02.

See abstract in PDF.


text:
Venn diagram problem

paulmdrdo

Active member
Find a Venn diagram containing three sets A, B, C such that $A\cap B\,\neq\,\emptyset$, $B\cap C\,\neq\,\emptyset$,and $C\cap A\,\neq\,\emptyset$, but $A\cap B\cap C\,=\,\emptyset$.

Plato

Well-known member
MHB Math Helper
Find a Venn diagram containing three sets A, B, C such that $A\cap B\,\neq\,\emptyset$, $B\cap C\,\neq\,\emptyset$,and $C\cap A\,\neq\,\emptyset$, but $A\cap B\cap C\,=\,\emptyset$.

View attachment 1127
What you have is not $A\cap C\ne \emptyset$

Mathelogician

Member
You don't have to use perfect circles in Venn diagrams. Try more complex figures!
Also you don't have to show the intersections with shadows only! You can try figures containing objects.

soroban

Well-known member
Hello, paulmdrdo!

Find a Venn diagram containing three sets $$A, B, C$$ such that:
$$A\cap B\,\neq\,\emptyset,\;B\cap C\,\neq\, \emptyset,\,\text{and }C\cap A\,\neq\,\emptyset$$
but $$A\cap B\cap C\,=\,\emptyset$$

View attachment 1127
No.
You have: $$C \cap A \,=\,\emptyset$$

The diagram might look like this:

Code:
              *               *
*   *           *   *
*       *       *       *
*           *   *           *
*       A       *       B       *
*               *:::*               *
*               *:::::::*               *
*               *:::*               *
*               *               *
*     *-----*---*-----*     *
*   |:::*       *:::|   *
* |:*           *:| *
*               *
|       C       |
|               |
|               |
|               |
*---------------*


text:
# What is the average power consumed/cycle in ideal capacitor.

The average power consumed/cycle in an ideal capacitor is 0.

The average power consumed in an ideal capacitor is given as based on the instantaneous power which is supplied to the capacitor:

pc = iv= (im cos ωt)(vm sin ωt)
pc = imvm (cos ωt sin ωt)

$p_{c}=\frac{i_{m}v_{m}}{2}sin\;2\omega t$

We know that sin ωt = 0

Therefore, average power = 0


text:
# GATE Papers >> EEE >> 2007 >> Question No 56

Question No. 56

The linear operation L(x) is defined by the cross product L(x) = bXx, where b=[0 1 0]T and x=[x1 x2 x3]T are three dimensional vectors. The 3×3 matrix M of this operations satisfies $\mathrm{L}\left(\mathrm{x}\right)=\mathrm{M}\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]$

Then the eigenvalues of M are


text:
# Cardinal characteristics of the ideal of $\sigma$-continuity of the Pawlikowski function

A function $$f:X\to Y$$ between topological spaces is called $$\sigma$$-continuous if there exists a countable cover $$\mathcal C$$ of $$X$$ such that for every $$C\in\mathcal C$$ the restriction $$f{\restriction}C$$ is continuous.

A typical example of a function (of the first Baire class) which is not $$\sigma$$-continuous is the Pawlikowski function $$P:(\omega+1)^\omega\to\omega^\omega$$ (which is defined as the countable power $$P=f^\omega$$ of a bijection $$f:\omega+1\to\omega$$).

Let $$\mathcal I_P$$ be the $$\sigma$$-ideal of subsets $$X$$ of the compact metrizable space $$(\omega+1)^\omega$$ such that $$P{\restriction}X$$ is $$\sigma$$-continuous.

I am interesting in evaluating the standard cardinal characteristics $$\mathrm{add}(\mathcal I_P)$$, $$\mathrm{cov}(\mathcal I_P)$$, $$\mathrm{non}(\mathcal I_P)$$, $$\mathrm{cof}(\mathcal I_P)$$ of the $$\sigma$$-ideal $$\mathcal I_P$$.

It seems that among these four cardinal characteristics only the covering number $$\mathrm{cov}(\mathcal I_P)$$ was studied in the literature. In particular, Cichon, Morayne, Pawlikowski and Solecki proved that $$\mathrm{cov}(\mathcal I_P)\ge\mathrm{cov}(\mathcal M)$$ where $$\mathcal M$$ is the $$\sigma$$-ideal of meager subset in $$\mathbb R$$. Steprans gave a combinatorial description of the ideal $$\mathcal I_P$$ and proved the consistency of the strict inequality $$\mathrm{cov}(\mathcal I_P)>\mathrm{cov}(\mathcal M)$$. Steprans also observed that for every $$A\in\mathcal I_P$$ the image $$P(A)$$ is a meager subset of $$\omega^\omega$$, which implies that $$\mathrm{cov}(\mathcal I_P)\ge\mathrm{cov}(\mathcal M)$$ and $$\mathrm{non}(\mathcal I_P)\le\mathrm{non}(\mathcal M)$$. On the other hand, it can be shown that $$\mathrm{non}(\mathcal I_P)\ge\mathfrak p$$.

It is well-known that the strict inequality $$\mathfrak p<\mathrm{non}(\mathcal M)$$ is consistent. In particular, according to Table 4 in the survey paper of Blass, this strict inequality holds in the random, Hechler, Laver, and Mathias forcing models.

Problem 1. Which of the inequalities $$\mathfrak p<\mathrm{non}(\mathcal I_P)$$ and $$\mathrm{non}(\mathcal I_P)<\mathrm{non}(\mathcal M)$$ is consistent?

Problem 2. What is the value of $$\mathrm{non}(\mathcal I_P)$$ (and other cardinal characteristics of the ideal $$\mathcal I_P)$$ in the random, Hechler, Laver, and Mathias forcing models?


text:
# Counting to 99 with Fingers

May 24, 2020 · 18:19 pm · Niche Idea

Consider the following:

HandValue
0
1
2
3
4
5
6
7
8
9

If each hand represents a digit in base $10$, then two hands can express anywhere from $0$ to $99$. For example, we can show $10$ by showing $1$ on the left hand and $0$ on the right hand:

LeftRight

Streaming direct thought dumps from Yuto Nishida.
Connect with me on LinkedIn! I'm currently listening to:


text:
# How to find drag of a cylinder?

How can I find a drag on the cylinder placed in uniform water flow, if I know a velocity field around it?

In the case of a potential flow, the resistance is zero (d'Alembert paradox). In the case of viscous flow, it is necessary to calculate the pressure distribution and the derivative of velocity along the normal. Calculate the integral $$F_D=\int _S{\left( \mu \frac {\partial u_t}{\partial n}n_y-pn_x\right)\mathrm dS}$$ The drag coeﬃcient is $$c_D=\frac {2F_D}{\rho U_0^2D}$$. Here $$\mu$$ is viscosity coefficient, $$\vec n=(n_x,n_y)$$ - normal vector, $$D$$ - diameter. The algorithm for solving the problem in the case of a laminar flow with the formation of Karman vortices, see on Solver for unsteady flow with the use of Mathematica FEM . Figure $$1$$ shows the velocity distribution at $$Re=100$$, the drag and lift coeﬃcients, and the pressure diﬀerence $$\Delta P$$ is deﬁned, with the front and end point of the cylinder.
Experimental data of the velocity distribution around the cylinder at $$Re=4216.8$$ @Anton Kovalyov


text:
# Using the Thermal Conductivity in the Design of Thermoplastic Composites

Scientists from Shanghai Second Polytechnic University and Dongua University, both in Shanghai, China, recently published work in the Journal of Applied Polymer Science concerning designing thermoplastic composites with improved thermal transport properties. The paper, which may be found here, characterizes the synergistic effect of alumina and graphene addition to the thermal properties, including thermal conductivity, of the thermoplastic polymer material under study.

Alumina, with a thermal conductivity of approximately 30 W/mK, is a relatively good thermal conductor compared to most ceramics. It is also chemically inert, which makes it an attractive material for thermal conductivity improvement of polymer materials. However, relatively large wt % loadings are typically needed to attain good thermal transport properties, resulting in a heavy material which may be unattractive for applications where low weight is important.

Graphene, by contrast, has an extremely high intrinsic thermal conductivity. While the initial work in Nano Letters measured a thermal conductivity of single-layer graphene approaching 5300 W/mK, most research since then has measured a considerably lower, though still extremely large, thermal conductivity ranging between 1500W/mK and 2500 W/mK. Graphene is highly sensitive to its environment owing to the sensitivity of its phonon transfer efficiency to disruption, so graphene supported on an amorphous substrate has a much lower thermal conductivity, on the order of 500 W/mK to 600 W/mK. As an allotrope of carbon with high aromaticity, graphene is fairly chemically inert. In addition to its chemical stability, it has greatly superior thermal conductivity relative to alumina, and is considerably lighter than alumina, which makes it an attractive candidate for designing thermoplastic composites with improved thermal conductivity.
Figure 1. C-Therm TCi Thermal Conductivity Analyzer.

Using a C-Therm TCi Thermal Conductivity Analyzer (Figure 1), the researchers were able to characterize their materials for thermal conductivity at a variety of graphene and alumina loadings, in addition to examining the thermal conductivity of both materials.
Figure 2. Effect of Graphene and Alumina loading on thermal conductivity of thermoplastic polymers.

One of the key results the researchers were able to obtain is shown in Figure 2. Here, improvement of thermal conductivity relative to the base polymer material is plotted for various thermoplastic composites. It can be seen that the greatest synergistic effect occurs at a loading of 64 wt% alumina, 1 wt% graphene, and the balance thermoplastic polymer. The researchers go on to detail the postulated mechanism of this synergistic effect. Graphs of thermal conductivity as a function of weight % loading of each constituent, SEM characterization data, and powder X-ray diffraction data are also presented.

For those who are interested in more information, a link to where the paper may be purchased from the Journal of Applied Polymer Science may be found here


text:
# Floral symmetry

[Left] Normal Streptocarpus flower (zygomorphic or mirror-symmetric), and [right] peloric (radially symmetric) flower on the same plant

Floral symmetry describes whether, and how, a flower, in particular its perianth, can be divided into two or more identical or mirror-image parts.

Uncommonly, flowers may have no axis of symmetry at all, typically because their parts are spirally arranged.

## Actinomorphic

Wurmbea stricta, its tepals in actinomorphic arrangement

Most flowers are actinomorphic ("star shaped", "radial"), meaning they can be divided into 3 or more identical sectors which are related to each other by rotation about the center of the flower. Typically, each sector might contain one tepal or one petal and one sepal and so on. It may or may not be possible to divide the flower into symmetrical halves by the same number of longitudinal planes passing through the axis: Oleander is an example of a flower without such mirror planes. Actinomorphic flowers are also called radially symmetrical or regular flowers. Other examples of actinomorphic flowers are the lily (Lilium, Liliaceae) and the buttercup (Ranunculus, Ranunculaceae).

## Zygomorphic

Satyrium carneum. Ground orchid with typical zygomorphic floral anatomy

Zygomorphic ("yoke shaped", "bilateral" – from the Greek ζυγόν, zygon, yoke, and μορφή, morphe, shape) flowers can be divided by only a single plane into two mirror-image halves, much like a yoke or a person's face. Examples are orchids and the flowers of most members of the Lamiales (e.g., Scrophulariaceae and Gesneriaceae). Some authors prefer the term monosymmetry or bilateral symmetry.[1] The asymmetry allows pollen to be deposited in specific locations on pollinating insects and this specificity can result in evolution of new species.[2]

Globally and within individual networks, zygomorphic flowers are a minority. Plants with zygomorphic flowers have smaller number of visitor species compared to those with actinomorphic flowers. Sub-networks of plants with zygomorphic flowers share greater connectance, greater asymmetry and lower coextinction robustness for both the plants and the visitor species. Plant taxa with zygomorphic flowers can have a greater risk of extinction due to pollinator decline.[3]

## Asymmetry

A few plant species have flowers lacking any symmetry, and therefore having a "handedness". Examples: Valeriana officinalis and Canna indica.[4]

## Differences

Actinomorphic flowers are a basal angiosperm character; zygomorphic flowers are a derived character that has evolved many times.[5]

Some familiar and seemingly actinomorphic so-called flowers, such as those of daisies and dandelions (Asteraceae), and most species of Protea, are actually clusters of tiny (not necessarily actinomorphic) flowers arranged into a roughly radially symmetric inflorescence of the form known as a head, capitulum, or pseudanthium.

## Peloria

Digitalis purpurea (common foxglove) displaying an aberrant peloric terminal flower and normal zygomorphic flowers

Peloria or a peloric flower is the aberration in which a plant that normally produces zygomorphic flowers produces actinomorphic flowers instead. This aberration can be developmental, or it can have a genetic basis: the CYCLOIDEA gene controls floral symmetry. Peloric Antirrhinum plants have been produced by knocking out this gene.[5] Many modern cultivars of Sinningia speciosa ("gloxinia") have been bred to have peloric flowers as they are larger and showier than the normally zygomorphic flowers of this species.

Charles Darwin explored peloria in Antirrhinum (snapdragon) while researching the inheritance of floral characteristics for his The Variation of Animals and Plants Under Domestication.[6] Later research, using Digitalis purpurea, showed that his results[7] were largely in line with Mendelian theory.[8]

## Symmetry groups

If we consider only those flowers which consist of a single flower, rather than a flower head or other form of inflorescence, we can categorize their symmetries into a relatively small number of two-dimensional symmetry groups. These groups are characterized by two types of symmetries: reflection (or mirror) symmetries, and rotational symmetries. Figures that are left invariant under reflections about a single axis have reflection symmetry, which is described by the cyclic group of order 2, ${\displaystyle C_{2}}$ (sometimes denoted ${\displaystyle Z_{2}}$). Figures that are left invariant under rotations by ${\displaystyle 2\pi /n}$ have a rotational symmetry belonging to the cyclic group of order ${\displaystyle n}$, ${\displaystyle C_{n}}$ (or ${\displaystyle Z_{n}}$). Many flowers that are invariant under rotations by ${\displaystyle 2\pi /n}$ are also invariant under reflections about ${\displaystyle n}$ distinct axes, the combination of these two symmetries forms the larger dihedral group of dimension ${\displaystyle n}$, ${\displaystyle D_{n}}$ (which has order ${\displaystyle 2n}$).

Flowers with bilateral symmetry, such as orchids have reflection symmetry about a single axis and no rotational symmetry, meaning that they are described simply by the reflection group ${\displaystyle C_{2}}$.

Monocots are identifiable by their trimerous petals, meaning that they are often invariant under rotations by ${\displaystyle 2\pi /3}$ and thus have rotational symmetry. Monocots that exhibit rotational symmetry but not mirror symmetry (for instance, if their petals exhibit chirality) are described by the cyclic group of order 3, ${\displaystyle C_{3}}$, and monocots with both rotational symmetry and reflection symmetry about 3 axes are described by the dihedral group of dimension 3, ${\displaystyle D_{3}}$.

Eudicots with tetramerous or pentamerous petals are often invariant under rotations by ${\displaystyle \pi /2}$ or ${\displaystyle 2\pi /5}$. Again, whether they also have mirror planes decides whether they belong to dihedral (${\displaystyle D_{4}}$ and ${\displaystyle D_{5}}$) or cyclic groups (${\displaystyle C_{4}}$ or ${\displaystyle C_{5}}$). Most eudiocots will have ${\displaystyle D_{4}}$ or ${\displaystyle D_{5}}$ symmetry, but, as was the case with monocots, those that exhibit chirality will only have cyclic symmetry of order the number of petals. For example, the individual petals of flowers in the genus Hypericum have no axis under which they are invariant under reflections, so their symmetry is described by ${\displaystyle C_{5}}$.

We can see the trend forming that, in general, the order of the cyclic group or dimension of the dihedral group that describes a flower's symmetry will correspond to the merosity of its petals. However, the sepals of some monocot flowers develop to replicate the petals, thus, superficially, certain monocots can appear to have rotational symmetry of order 6 and belong to either symmetry group ${\displaystyle D_{6}}$ or ${\displaystyle C_{6}}$. Some composite flowers may also have at least a superficial cyclical or dihedral symmetry. How exact this symmetry is depends on the structure of the head of the flower. Even in monocots and eudicots, flower symmetries are rarely perfect, as any imperfections in the petals will result in imperfect invariance under rotations or reflections.

## References

1. ^
2. ^ Zimmerman, Erin (26 October 2020). "Asymmetry allows divergent use of pollinators, but often represents a one-way evolutionary street". Botany ONE. Retrieved 30 October 2020.
3. ^ Yoder JB, Gomez G, Carlson CJ (2020). "Zygomorphic flowers have fewer potential pollinator species". Biology Letters. 16 (9): 20200307. doi:10.1098/rsbl.2020.0307. PMC 7532724. PMID 32871089.
4. ^ Weberling, Focko (1992). Morphology of Flowers and Inflorescences. Cambridge University Press. p. 19. ISBN 0-521-25134-6.
5. ^ a b Losos, J.B.; Mason, K.A.; Singer, S.R. Biology (8th ed.). New York: McGraw Hill.
6. ^ Darwin 1868, pp. 33–34
7. ^ Darwin 1868, p. 46
8. ^ Keeble, Frederick; Pellew, C; Jones, WN (1910). "The Inheritance of Peloria and Flower-Colour in Foxgloves (Digitalis purpurea)". New Phytologist. 9 (1–2): 68–77. doi:10.1111/j.1469-8137.1910.tb05554.x. JSTOR 2427515.


text:
# Logarithmic decrement of a lightly damped oscillator

1. ### kraigandrews

108
1. The problem statement, all variables and given/known data
The logarithmic decrement δ of a lightly damped oscillator is defined to be the natural logarithm of the ratio of successive maximum displacements (in the same direction) of a free damped oscillator. That is, δ = ln(An/An+1) where An is the maximum displacement of the n-th cycle. Derive the simple relationship between δ and Q.
Find the spring constant k and damping constant b of a damped oscillator with mass m, frequency of oscillation f and logarithmic decrement δ.
[Data: m = 4.0 kg; f = 0.9 Hz; δ = 0.029.]
First, the spring constant k...

Also, the damping constant b...

2. Relevant equations

$\beta$=b/(2m)
Q=$\omegao/(2\beta) 3. The attempt at a solution Given the diff eq: d2x/dt2+2[itex]\beta$(dx/dt)+$\omega$o2x=0

I can solve this to find x(t), however I feel this is irrelevant because no initial condition or boundary conditions are given, so I am kinda lost here as to where go or to start at for that matter. Any suggestions are greatly appreciated, Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. ### ideasrule

2,315
The initial conditions don't matter. Can you write the exponential term of the equation of motion in terms of Q? This term controls the amplitude of the oscillator, so the ratio between successive amplitudes is just the ratio between successive exponential terms.


text:
## Algebraic & Geometric Topology

### Homotopy groups of diagonal complements

#### Abstract

For $X$ a connected finite simplicial complex we consider $Δd(X,n)$, the space of configurations of $n$ ordered points of $X$ such that no $d + 1$ of them are equal, and $Bd(X,n)$, the analogous space of configurations of unordered points. These reduce to the standard configuration spaces of distinct points when $d = 1$. We describe the homotopy groups of $Δd(X,n)$ (resp. $Bd(X,n)$) in terms of the homotopy (resp. homology) groups of $X$ through a range which is generally sharp. It is noteworthy that the fundamental group of the configuration space $Bd(X,n)$ abelianizes as soon as we allow points to collide, ie $d ≥ 2$.

#### Article information

Source
Algebr. Geom. Topol., Volume 16, Number 5 (2016), 2949-2980.

Dates
Revised: 15 January 2016
Accepted: 7 February 2016
First available in Project Euclid: 16 November 2017

https://projecteuclid.org/euclid.agt/1510841235

Digital Object Identifier
doi:10.2140/agt.2016.16.2949

Mathematical Reviews number (MathSciNet)
MR3572355

Zentralblatt MATH identifier
1355.55012

Subjects
Primary: 55Q52: Homotopy groups of special spaces
Secondary: 55P10: Homotopy equivalences

#### Citation

Kallel, Sadok; Saihi, Ines. Homotopy groups of diagonal complements. Algebr. Geom. Topol. 16 (2016), no. 5, 2949--2980. doi:10.2140/agt.2016.16.2949. https://projecteuclid.org/euclid.agt/1510841235

#### References

• M,A Armstrong, The fundamental group of the orbit space of a discontinuous group, Proc. Cambridge Philos. Soc. 64 (1968) 299–301
• A Bj örner, M Las Vergnas, B Sturmfels, N White, G,M Ziegler, Oriented matroids, 2nd edition, Encyclopedia of Mathematics and its Applications 46, Cambridge Univ. Press (1999)
• A Bj örner, V Welker, The homology of “$k$–equal” manifolds and related partition lattices, Adv. Math. 110 (1995) 277–313
• P,V,M Blagojević, B Matschke, G,M Ziegler, A tight colored Tverberg theorem for maps to manifolds, Topology Appl. 158 (2011) 1445–1452
• P,V,M Blagojević, G,M Ziegler, Convex equipartitions via equivariant obstruction theory, Israel J. Math. 200 (2014) 49–77
• C-F B ödigheimer, I Madsen, Homotopy quotients of mapping spaces and their stable splitting, Quart. J. Math. Oxford Ser. 39 (1988) 401–409
• B Branman, I Kriz, A Pultr, A sequence of inclusions whose colimit is not a homotopy colimit, New York J. Math. 21 (2015) 333–338
• F Cohen, E,L Lusk, Configuration-like spaces and the Borsuk–Ulam theorem, Proc. Amer. Math. Soc. 56 (1976) 313–317
• N Dobrinskaya, V Turchin, Homology of non-$k$–overlapping discs, Homology Homotopy Appl. 17 (2015) 261–290
• C Eyral, Profondeur homotopique et conjecture de Grothendieck, Ann. Sci. École Norm. Sup. 33 (2000) 823–836
• E,D Farjoun, Cellular spaces, null spaces and homotopy localization, Lecture Notes in Mathematics 1622, Springer (1996)
• M,A Guest, A Kozlowski, K Yamaguchi, Stable splitting of the space of polynomials with roots of bounded multiplicity, J. Math. Kyoto Univ. 38 (1998) 351–366
• A Hatcher, Algebraic topology, Cambridge Univ. Press (2002)
• U Helmke, Topology of the moduli space for reachable linear dynamical systems: the complex case, Math. Systems Theory 19 (1986) 155–187
• T,D Imbo, C Shah Imbo, E,C,G Sudarshan, Identical particles, exotic statistics and braid groups, Phys. Lett. B 234 (1990) 103–107
• S Kallel, Spaces of particles on manifolds and generalized Poincaré dualities, Q. J. Math. 52 (2001) 45–70
• S Kallel, R Karoui, Symmetric joins and weighted barycenters, Adv. Nonlinear Stud. 11 (2011) 117–143
• S Kallel, W Taamallah, The geometry and fundamental group of permutation products and fat diagonals, Canad. J. Math. 65 (2013) 575–599
• M Khovanov, Real $K(\pi,1)$ arrangements from finite root systems, Math. Res. Lett. 3 (1996) 261–274
• B Kloeckner, The space of closed subgroups of $\mathbb R^n$ is stratified and simply connected, J. Topol. 2 (2009) 570–588
• K,H Ko, H,W Park, Characteristics of graph braid groups, Discrete Comput. Geom. 48 (2012) 915–963
• J,R Munkres, Elements of algebraic topology, Addison–Wesley Publishing Company, Menlo Park, CA (1984)
• M Nakaoka, Cohomology of symmetric products, J. Inst. Polytech. Osaka City Univ. Ser. A 8 (1957) 121–145
• S Smale, A Vietoris mapping theorem for homotopy, Proc. Amer. Math. Soc. 8 (1957) 604–610
• Q Sun, Configuration spaces of singular spaces, PhD thesis, University of Rochester, Ann Arbor, MI (2010) Available at \setbox0\makeatletter\@url http://search.proquest.com/docview/871891981 {\unhbox0
• D Tamaki, Cellular stratified spaces, I: Face categories and classifying spaces, preprint (2011)
• S Zanos, Méthodes de scindements homologiques en topologie et en géométrie, PhD thesis, Université Lille 1 (2009)


text:
Grammar Error beta
Weekly occurences of strange grammar questions
Home/Exclamation mark/

# Exclamation Mark

The exclamation mark, also sometimes referred to as the exclamation point in American English, is a punctuation mark usually used after an interjection or exclamation to indicate strong feelings or high volume (shouting), or to show emphasis, and often marks the end of a sentence, for example: Watch out! Similarly, a bare exclamation mark (with nothing before or after) is often used in warning signs.

License: CC BY-SA 3.0. Source: wikipedia (1)

Apostrophe

Colon

Comma

Dash and hyphen

Ellipsis

Period

Question mark

Quotation mark

Semicolon


text:
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# M20-23

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M20-23  [#permalink]

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16 Sep 2014, 00:08
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Question Stats:

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### HideShow timer Statistics

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

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Official Solution:

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in $$2(C_{3}^{2}) = 6$$ different ways.

Answer: B
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Re: M20-23  [#permalink]

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02 Nov 2014, 12:39
I solved using "slot" method. 2*3*1=6

Because the last digit must be 5, there is only one option, therefore, I placed a 1 in this slot. the next slot to the left, therefore, only has 3 possible options (2,3, or 4, since we've already placed the 5, and the numbers cannot repeat). The next slot only has 2 possible options. The product of these is 6.

Is there a simple way to solve this in x!/(y!z!) notation?

Thanks
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Re: M20-23  [#permalink]

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03 Nov 2014, 01:35
1
JackSparr0w wrote:
I solved using "slot" method. 2*3*1=6

Because the last digit must be 5, there is only one option, therefore, I placed a 1 in this slot. the next slot to the left, therefore, only has 3 possible options (2,3, or 4, since we've already placed the 5, and the numbers cannot repeat). The next slot only has 2 possible options. The product of these is 6.

Is there a simple way to solve this in x!/(y!z!) notation?

Thanks

Your approach is correct.

As for "x!/(y!z!) notation", the solution provided uses 3C2, which is 3!/(2!1!)...
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Re: M20-23  [#permalink]

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04 Nov 2014, 17:45
Thanks for the clarification. I'm not familiar with the "C" notation.
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M20-23  [#permalink]

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05 Nov 2014, 04:34
1
1
JackSparr0w wrote:
Thanks for the clarification. I'm not familiar with the "C" notation.

nCk is choosing k items out of n distinct items so that the order of the selection does not matter and it equals to $$\frac{n!}{k!(n-k)!}$$. For example, choosing 3 out of 5, 5C3, is $$\frac{n!}{k!(n-k)!}=\frac{5!}{3!(5-3)!}=10$$: we can choose 10 different 3-item groups out of 5 different items.

Check for more here: math-combinatorics-87345.html

Hope it helps.
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Re M20-23  [#permalink]

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20 Mar 2015, 10:38
I think this question is good and helpful.
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Re: M20-23  [#permalink]

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10 Oct 2015, 07:50
Bunuel, I believe your annotation of your original combination may be incorrect.

You stated 2C3, whereas later you mentioned 3C2. Its a typo I believe, or I could be misinterpreting the C sign, where n is at the top and k is at the bottom.

Thanks
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Re: M20-23  [#permalink]

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11 Oct 2015, 05:09
TaiwanTaiwan wrote:
Bunuel, I believe your annotation of your original combination may be incorrect.

You stated 2C3, whereas later you mentioned 3C2. Its a typo I believe, or I could be misinterpreting the C sign, where n is at the top and k is at the bottom.

Thanks

Where do I have 2C3?
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Re: M20-23  [#permalink]

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14 Oct 2017, 05:34
I am wondering why one would use the formula 2 * C2,3? Is that the same as 3!/(3-2)! = 6? (In general, n!/(k-n)!) Thanks!
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Re: M20-23  [#permalink]

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14 Oct 2017, 05:41
tmtsn wrote:
I am wondering why one would use the formula 2 * C2,3? Is that the same as 3!/(3-2)! = 6? (In general, n!/(k-n)!) Thanks!

3C2, $$C^2_3$$, $$C^3_2$$, all mean $$\frac{3!}{2!(3-2)!}$$
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M20-23  [#permalink]

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14 Oct 2017, 06:03
Bunuel wrote:
tmtsn wrote:
I am wondering why one would use the formula 2 * C2,3? Is that the same as 3!/(3-2)! = 6? (In general, n!/(k-n)!) Thanks!

3C2, $$C^2_3$$, $$C^3_2$$, all mean $$\frac{3!}{2!(3-2)!}$$

Thanks, but why use the combinations formula instead of permutations formula? Why 2 * 3C2 instead of 3!/(3-2)! which both have the answer 6?

Sorry if I'm being unclear..
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Re: M20-23  [#permalink]

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14 Oct 2017, 06:06
tmtsn wrote:
Bunuel wrote:
tmtsn wrote:
I am wondering why one would use the formula 2 * C2,3? Is that the same as 3!/(3-2)! = 6? (In general, n!/(k-n)!) Thanks!

3C2, $$C^2_3$$, $$C^3_2$$, all mean $$\frac{3!}{2!(3-2)!}$$

Thanks, but why use the combinations formula instead of permutations formula? Why 2 * 3C2 instead of 3!/(3-2)! which both have the answer 6?

Sorry if I'm being unclear..

Yes, we could use $$P^2_3=6$$, instead, which is the same as $$2*C^2_3$$.
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Re: M20-23  [#permalink]

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21 Oct 2017, 03:21
Bunuel wrote:
Official Solution:

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in $$2(C_{3}^{2}) = 6$$ different ways.

Answer: B

I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?
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Re: M20-23  [#permalink]

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21 Oct 2017, 03:43
aashishagarwal2 wrote:
Bunuel wrote:
Official Solution:

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in $$2(C_{3}^{2}) = 6$$ different ways.

Answer: B

I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?

We multiply 3C2 by 2, which takes care of all arrangements.
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M20-23  [#permalink]

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21 Oct 2017, 03:46
Ok, understood. Thanks a ton!

Posted from my mobile device
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Re: M20-23  [#permalink]

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21 Oct 2017, 10:53
Bunuel wrote:
aashishagarwal2 wrote:
Bunuel wrote:
Official Solution:

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in $$2(C_{3}^{2}) = 6$$ different ways.

Answer: B

I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?

We multiply 3C2 by 2, which takes care of all arrangements.

Hi Bunuel, I have a follow-up question to this. If multiplying by 2 takes care of all arrangements, is it like every permutation is equal to double of combination (provided no other constraints on permutation)? Like can we just calculate combination and 2x it?

Thanks for being so kind and and answering all our queries with such patience.
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Re: M20-23  [#permalink]

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21 Oct 2017, 11:09
1
aashishagarwal2 wrote:
Hi Bunuel, I have a follow-up question to this. If multiplying by 2 takes care of all arrangements, is it like every permutation is equal to double of combination (provided no other constraints on permutation)? Like can we just calculate combination and 2x it?

Thanks for being so kind and and answering all our queries with such patience.

No. We multiply here by 2! (2 factorial) because it's the number of arrangements of 2 distinct objects. If there were n distinct objects it would have been n!.
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Re: M20-23 &nbs [#permalink] 21 Oct 2017, 11:09
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text:
## Installation

### Need help with Mail::Sender

by Jim Fischer -
Number of replies: 6
There seems to be some problems with installing Mail::Sender. I am using the instructions for installing 2.12 with ubuntu 16.04:

http://webwork.maa.org/wiki/Installation_Manual_for_2.12_on_Ubuntu_16.04#.WG-bflzD42g

When I run check_modules.pl apache2 I get the output shown at bottom of this post.

I tried to follow these instructions:
sudo perl -MCPAN -e shell [sudo] password for wwadmin: <wwadmin password>

and accept any defaults. Then

cpan> install Mail::Sender

and accept any defaults. Now

 cpan> exit
However, the first question said something about defaults and I answered yes, when I think I should have answered No. Unfortunately, once this process has been screwed up it doesnt seem to be reversible. I have tried to reinstall and I have tried to uninstall Mail::Sender to start over. Neither methods have fixed my problems.

I really dont want to start over entirely and could use some help in figuring out what to do to fix this efficiently.

Here is the output from check_modules.pl apache2:

Checking your $PATH for executables required by WeBWorK...$PATH= /home/wwadmin/bin
/usr/local/sbin
/usr/local/bin
/usr/sbin
/usr/bin
/sbin
/bin
/usr/games
/usr/local/games
/snap/bin
/opt/webwork/webwork2/bin

curl found at /usr/bin/curl
mkdir found at /bin/mkdir
mv found at /bin/mv
mysql found at /usr/bin/mysql
tar found at /bin/tar
git found at /usr/bin/git
gzip found at /bin/gzip
latex found at /usr/bin/latex
pdflatex found at /usr/bin/pdflatex
dvipng found at /usr/bin/dvipng
ppmtopgm found at /usr/bin/ppmtopgm
pnmtops found at /usr/bin/pnmtops
pnmtopng found at /usr/bin/pnmtopng
pngtopnm found at /usr/bin/pngtopnm

Checking your @INC for modules required by WeBWorK...
@INC= /etc/perl
/usr/local/lib/x86_64-linux-gnu/perl/5.22.1
/usr/local/share/perl/5.22.1
/usr/lib/x86_64-linux-gnu/perl5/5.22
/usr/share/perl5
/usr/lib/x86_64-linux-gnu/perl/5.22
/usr/share/perl/5.22
/usr/local/lib/site_perl
/usr/lib/x86_64-linux-gnu/perl-base
.

Prototype mismatch: sub main::from_json: none vs ($@) at (eval 110) line 2. Prototype mismatch: sub main::to_json: none vs ($@) at (eval 110) line 2.
Mail::Sender is deprecated and you should look to Email::Sender instead at (eval 140) line 2.
wwadmin@mathdev:/usr/local/share/perl/5.22.1/Mail\$ Mail::Sender is deprecated and you should look to Email::Sender instead at (eval 140) line 2.
-bash: syntax error near unexpected token `('

### Re: Need help with Mail::Sender

by Jim Fischer -
Update...

Ive looked this over and it appears that the latest version of Mail::Sender is getting loaded. I'm not sure why there is the following line in the output shown above:

Mail::Sender is deprecated and you should look to Email::Sender instead at (eval 140) line 2.

However, Im starting to think that the webwork system and the modules are all good and that the email issue could be a problem on the mail server side that I am trying to use. I have asked my school's IT department to see if there is something on their end that is keeping WebWork from connecting to the school email server.

### Re: Need help with Mail::Sender

by Jim Fischer -
I put in a request for our IT department to see if there is something on their side, such as blocked port or ?

In the meantime, I will post what is happening in case someone here can figure out why the email is not working.

1. From Faculty to Student(s) no WW error is presented, the system says email is sent. However, no email is received.

2. From Student to Faculty, using email instructor button. WW times out and gives the following error:

Failed to open the mailer: connect() failed: Input/output error

Mail::Sender appears to be loading and I am using the same settings in site.conf as are being used in another server that is running WW with no email issues.

### Re: Need help with Mail::Sender

by Michael Gage -
It seems to me that IT might be able to look at logs at the smtp to see if the mailer connection request has been received.

### Re: Need help with Mail::Sender

by Jim Fischer -
Thanks Mike.

Okay, I made more progress. Our school is using Authentication with their smtp email server. They think the reason why the production server running WW is sending email while the development server is not is because they had setup an exception for the authentication on the production server.

Their desire is to have WebWork work with the authentication in place when communicating with the school smtp server.

I have been asked if WebWork can do this and if so is there anything I need to do differently say when entering things into site.conf.

Thanks in case you can help me clear this up. Keep in mind I really don't know much about this stuff (in case that isnt obvious already!)

### Re: Need help with Mail::Sender

by Michael Gage -
Hi Jim,

Here is a reply to my question to our local IT guy, Hoss Firooznia, about
what kind of authorization they were talking about.

"They're most likely using the auth mechanism that's part of the SMTP protocol. Here's how to use it in Perl:

http://search.cpan.org/~jenda/Mail-Sender/Sender.pm#Using_MailMsg_and_authentication

"

You can run grep on the webwork code and see where the smtp request is formulated -- probably in SendMail.pm.

The other thing they might be talking about would be TLS (transport layer security) which some

smtp servers use.  We have this turned of by default in site.conf because many smtp servers become

confused if WeBWorK sends them tls prompts, but that might not be the case in your situation.

I don't have time to trouble shoot this myself right now, the semester is starting, but let me know how far you and your IT people get and I'll try to help out in two weeks or so if the situation is not resolved.

Take care,

Mike


text:
# TransformedRegion, RegionPlot broken in 10.0.2.0?

After having had trouble applying TranslationTransform to a simple region like so

RegionPlot[TransformedRegion[Rectangle[],TranslationTransform[{2,0}]]]

I tried to find my error by researching Mathematica's online documentation, specifically ref/TransformedRegion. After having failed to find a glitch in my code, I just did an "Evaluate Notebook" from the "Evaluation" menu, with a discouraging result:

1. Error messages: RegionPlot::nnregion, RegionPlot::invplotreg, TransformedRegion::reg, DiscretizeRegion::regp, RegionDistance::regp, Volume::reg, Integrate::ilim, MinValue::elemc, Reduce::elemc, RegionBounds::reg, RegionDistance::regp (in that order).
2. Different RegionPlot-outputs, which, as comments showed, are due to wrong PlotRange (so, there is a workaround for this by manually changing that option).

My question is: Is my 10.0.2.0 installation damaged or might this be a problem with Mathematica itself?

I have the same results on two machines (Windows 7 and Windows 8.1, both 64bit).

## Examples

Before evaluating the documentation page

After

• A damaged installation, likely not. It's a miscalculated default PlotRegion, that's all. – rcollyer Jan 28 '15 at 13:40
• @rcollyer: You may be right, but the display (especially the closed boundaries) are highly misleading. In that state, the region plotting functionality is not really usable with TransformedRegion. – Jinxed Jan 28 '15 at 14:16
• @rcollyer: And the error messages resulting from processing the documentation page indicate, that there is something wrong here. – Jinxed Jan 28 '15 at 14:21
• I don't contest those conclusions, just the conclusion that it is a damaged installation. You can verify that it is the PlotRange by setting PlotRange -> {{-1, 0.6}, {0, 1.4}} for the first example and PlotRange -> {{-8, 15}, {-4, 8}} for the second. – rcollyer Jan 28 '15 at 14:47
• @rcollyer: I take it you have the same behavior (error messages, PlotArea-problem) when evaluating the online documentation reference page? – Jinxed Jan 28 '15 at 16:59

## Problem Confirmed

Wolfram has confirmed this to be a problem with Mathematica 10.0.2.0, not a damaged installation. There is no known workaround for now.

Transcript of the confirmation mail (dated 2015-01-28):

Hello -

http://support.wolfram.com


text:
Setup for multiple strains, multiple conditions, and replicates
1
1
Entering edit mode
7.3 years ago
montezuma ▴ 10

Hello all,

I have RNA-Seq data from an experiment with multiple strains and conditions and wanted to clarify the correct setup for DEG using the tuxedo pipeline.

The experiment is looking at two C. elegans strains: wild-type and mutant, both water treated or drug treated, and triplicate for each, so 12 samples total (2 x 2 x 3). What I have done:

1. I ran tophat for each sample, and then cufflinks separately for each sample.
2. To compare drug treatment to water treatment, I cuffmerged all cufflinks files from the wild-type strain and then ran cuffdiff from that output. I did the same process for the mutant strain.
3. This results in DEG for drug to water treatment in each strain separately, but I also want a comparison between strains. What would be the correct way of comparing the effect of drug treatment between the wild-type and mutant strain?

Thanks for the help, much appreciated!

cufflinks tuxedo rna-seq • 2.0k views
0
Entering edit mode
7.0 years ago

I have a similar setup and people work differently with these kinds of setup depending on the tool and workflow they use. There is no straight answer to this, after testing and searching, I can share with you what I do

What I do is use Tuxedo suite for most of the analysis.

1. Run Tophat individually on all samples and biological replicates (you can merge technical replicates before this step) which maps the reads.
2. Run Cuffquant which generates .cxb profiles and these can be used later via cuffdiff.
3. I completely skip the cuffmerge, because I am only interested in DEG not novel isoform discovery.
4. Run Cuffdiff here I combine all the replicates and conditions together. Replicates and same conditions are seperated by a comma (,) where as different conditions are seperated by space in a time series manner.

For the last step, you can run several scripts each time answering different question

WT1rep1,WT1rep2,WT2rep1,WT2rep2 Cond1rep1,Cond1rep2,Cond2rep1,Cond2rep2 -> will give you DEG b/w WT and mutant strain

WT1rep1,WT1rep2 WT2rep1,WT2rep2 -> will give you the differences between WT strains

WT1rep1,WT1rep2 Cond1rep1,Cond1rep2 and WT2rep1,WT2rep2 Cond2rep1,Cond2rep2, DEG's from these two should have a good overlap to say your strains and replicates are good in terms of reproducibility and experimentation.

Also this might be helpful

RNA seq time course data experiment design

HTH


text:
GARCH_VL - Long-run Volatility of the GARCH Model

Calculates the long-run average volatility for the given GARCH model.

Syntax

GARCH_VL(alphas, betas, innovation, v)

alphas are the parameters of the ARCH(p) component model (starting with the lowest lag).

betas are the parameters of the GARCH(q) component model (starting with the lowest lag).

innovation is the probability distribution model for the innovations/residuals (1=Gaussian (default), 2=t-Distribution, 3=GED).

value Description
1 Gaussian or Normal Distribution (default)
2 Student's t-Distribution
3 Generalized Error Distribution (GED)

v is the shape parameter (or degrees of freedom) of the innovations/residuals probability distribution function.

Remarks

1. The underlying model is described here.
2. The long-run variance of a GARCH process is defined as follow:
1. $$\sigma_{\infty}^2 \rightarrow V_L=\frac{\alpha_o}{1-\sum_{i=1}^{max(p,q)}\left(\alpha_i+\beta_i\right)}$$
3. The long-run variance is not affected by our choice of shock/innovation distribution.
4. The number of parameters in the input argument - alpha - determines the order of the ARCH component model.
5. The number of parameters in the input argument - beta - determines the order of the GARCH component model.

Examples

Example 1:

 1 2 3 4 5
A B
GARCH(1,1)
Mean -0.160
Alpha_0 0.608
Alpha_1 0.000
Beta_1 0.391

Formula Description (Result)
=GARCH_VL($B$3:$B$4,$B$5) GARCH long-run average volatility (0.999)


text:
Sakulpipatsin P.,Boelman E.C., Cauberg J.J.M.
Year:
2006
Bibliographic info:
27th AIVC and 4th Epic Conference "Technologies & sustainable policies for a radical decrease of the energy consumption in buildings", Lyon, France, 20-22 November 2006

In cold and moderate climates, improvements in building shell insulation and air-tightness imply a shiftin heating loads from transmission and infiltration towards ventilation. Heat recovery from the ventilation airflow plays an increasingly important role in minimising energy needs. Such heat recovery systems rely on the input of electric power (to drive fans, heat pumps, etc.) in order to recover thermal energy. Since electricity input is relatively small compared to the amounts of thermal energy recovered, such systems are efficient from an energy viewpoint. One important yet often overlooked aspect, however, is the difference in quality between the high-grade electricity input and the lower grade thermal energy recovered. This paper analyzes the effectiveness of heat recovery from ventilation airflows from the viewpoint of exergy. The results provide a common basis for evaluating different forms of energy (e.g. thermal and electric), considering their different abilities to produce work in relation to a given environment.


text:
# Rotating multiple points at once in 2D

I currently have an editor that creates shapes out of (X, Y) coordinates and then triangulate that to make up a shape of those points.

What will I have to do to rotate all of those points simultaneously? Say I click the screen in my editor, it locates the point where I've clicked and if I move the mouse up or down from that point it calculates rotation on X and Y axis depending on new position relevant to first position, say I move up 10 on the Y axis it rotates that way and the same way for X. Or simply, somehow to enter rotation degree: 90, 180, 270, 360, for example.

I use VertexPositionColor at the moment.

What are the best algorithms or methods that I can look at to rotate multiple points in 2D at once?

Also:

Since this is an editor I do now want to rotate it on the Matrix, so if I want to rotate the whole shape 180 degree that's the new "position" of all the points, so that's the new rotation = 0 for example. Later on I probably will use World Matrix rotation for this, but not now.

• What do you mean by "rotation on X axis"? In 2D, you can only rotate anything around a point, not around a line. – Martin Sojka Oct 25 '12 at 10:44
• @MartinSojka You have to specify the rotation axis in 2D as well. For example: Matrix worldMatrix = Matrix.CreateRotationY(angle); – Marton Oct 25 '12 at 10:50
• @Marton: What would a rotation by, say, 30 degrees around the Y axis in 2D exactly mean - where would the points end up? Remember that neither the points nor the axis can be located outside those two dimensions, neither pre nor post rotation. – Martin Sojka Oct 25 '12 at 10:52
• @Marton: There is no Z axis in 2D. And if you change just the X coordinate, that's either scaling or movement, or a combination thereof. – Martin Sojka Oct 25 '12 at 11:01
• I think you're overcomplicating things. For a rotation in 2D, you need a point (the point around which the rotation happens) and some way to specify the angle of rotation - examples would be an angle value, a complex number describing the rotation, or two directional vectors - one for the direction pre rotation, one for the direction post rotation. So far, I don't even see where your origin point is supposed to be located. On the one hand, you write something about clicking on the screen to locate the point; on the other hand, you bring Blender as an example, which doesn't do that. – Martin Sojka Oct 25 '12 at 12:07

You need to define a center for rotation... in this case the box center..

then when you detect the mouse down you store vector (A) from the center to the mouse coords... this is your initial vector.... ans should store your shape rotation as initial rotation.

then while you move the mouse and it continues being down you calculate the second vector (B) from the center to the current mouse coords...

now you can calculate the angle between to vectors... as you should know it can be calculated normalizing the two vectors and calcultaing the dot product... so you get the cosine... now you get the angle with an arccosine function...

this angle should be added to the initial rotation of your shape... ans set the new rotation for your shape.

so you only have to build your transform matrix for your shape...

the easiest way is to store your vertex coords relative to the center of the shape, if you don't store that way you should get that points relative to it...

Your transform matrix should similar to this...

// Supposing your coords are absolute
Matrix transform = Matrix.CreateTranslation( -Center )
* Matrix.CreateRotationZ(rotation)
* Matrix.CreateTranslation (Center );

// Supposing your coords are relative, center is (0,0)
Matrix transform = Matrix.CreateRotationZ(rotation)

You should realize that usually the right way is the easy way... you should use matrix transforms... I'd do something similar to this:

  class Bone2D {
Vector2 Translation;
float Rotation;
float Scale;
Bone Parent;
Matrix Local { get { return Matrix.CreateScale(scale)
* Matrix.CreateRotationZ(rotation)
* Matrix.CreateTranslation(translation); } }
Matrix Absolute { get{ return (Parent == null)
?  Local
: Parent.Absolute * Local; }}
}

class Shape : Bone2D {

Vector2[] RelativeCoords;  // Relative to your shape center
Vector2[] AbsoluteCoords {
get{
return RelativeCoords.Select(r => Vector2.Transform(r, Absolute)
.ToArray();
} }
}

This way the shape you show in your example, would have two bones...

1) the parent bone that translates the shape,

2) the shape that is a bone itself and can be rotated easily... or scaled or translated again relative to the parent bone of course... ;)

This editor is quite old... I have a new version... but the video shows what you can achieve doing it right... http://www.youtube.com/watch?feature=player_detailpage&v=K9R5v5Va4CI#t=28s

I hope this to be useful... ;)

We have the following points:

Origin O = (xO, yO), the origin (or centre) of the object and the origin point for our rotation.

Starting point S = (xS, yS), the point at which the mouse was pressed down and/or the mouse was when the rotation was initiated.

Current (mouse) point M = (xM, yM), the point the mouse is at right now during the rotational operation.

Your rotation can then be expressed as 2x2 matrix calculated from the unit vectors pointing from the origin to the starting point and to the current mouse point respectively.

DS = (xDS, yDS) = (S - O) / ||S - O||

DM = (xDM, yDM) = (M - O) / ||M - O||

Warning

We can calculate the inverse of the second matrix so easily because we know that the matrix is made out of an unit vector - and thus the determinant has the value of exactly 1. This is not a generic formula for calculating such.

The resulting matrix can again be expressed as a unit vector of direction, in case you ever need it:

DR = (xDM xDS + yDM yDS, - xDM yDS + yDM xDS)

The rotation matrix describes a rotation about the origin (0, 0) - not about your origin. For this, the points first have to be moved so that the object origin is the same as world origin, rotated, then moved back to their original position. In other words:

Protated = R x (Poriginal - O) + O

You can express the translations as 3x3 matrices if you first extend your 2D vectors by an anonymous dimension with the constant value of 1 for all of them; the three transformations can then be pre-calculated into one single matrix by simple matrix multiplication.

• An example in C#/XNA would be nice, not sure whether to (S - O) / (S-O) which is 0, or ((S-O) / S) - O or if it's just a variation of the first, still confused. – Deukalion Oct 26 '12 at 14:37
• @Deukalion ||v|| is the mathematical notation for "length of vector v". In other words, calculate (S-O) (a vector), sum the squares of its elements, then calculate the square root of the sum. – Martin Sojka Oct 26 '12 at 14:48


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## MATLAB For Beginners: 20-Minute Video Training Course. (Grab The Training Course and The Cheat Sheet for FREE)

In this article, you’ll learn the basics of MATLAB Programming in 5 minutes, including how to:

• Create variables, vectors, and matrices
• Use MATLAB functions and scripts
• Create loops and conditions
• Read and write text and Excel files

## Creating Variables, Vectors, and Matrices

### How to Define a Variable in MATLAB

To define a variable, write the name of the variable followed by “=” and the value of the variable:

numberVariable = 5; % variable containing a number
stringVariable = 'string'; % variable containing a string
• The “;” is used to prevent MATLAB from displaying the value of your variable in your workspace when you define it.
• The “%” is used to add a comment to your code.

Furthermore, MATLAB remembers what you wrote. If you want to access a command that you wrote earlier in your workspace, use the up arrow key to get your previous command.

### How to Create a Vector in MATLAB

In MATLAB, you can create row vectors and column vectors:

• Row vector: create a vector in MATLAB by putting numbers (or strings) next to each other and enclosing them with square brackets:
vector = [1 2 3 4 5]; % row vector
stringVector = [{'speed'} {'size'}]; % string row vector

This defines the following vectors:

$vector=\begin{pmatrix}1 &2 &3&4 &5\end{pmatrix}~~~~~~~~~~~~~~~~~~stringVector=\begin{pmatrix}\{'speed'\}& \{'size'\}\end{pmatrix}$

• Column vector: use a single quotation mark to define a column vector:
vector = [1 2 3 4 5]'; % column vector
stringVector = [{'speed'} {'size'}]'; % string column vector

$vector=\begin{pmatrix}1 \\2 \\3\\4 \\5\end{pmatrix}~~~~~~~~~~~~~~~~~~stringVector=\begin{pmatrix}\{'speed'\}\\\{'size'\}\end{pmatrix}$

As you can see, you need to use braces to separate strings in a vector; otherwise, MATLAB tries to concatenate them.

If you want to learn all the technical stuff on vectors, see MATLAB Vector Tutorial: Create, Add, Concatenate, and Extract

### How to Create a Matrix in MATLAB

You can create a matrix in MATLAB by writing numbers or strings next to each other to define the rows and using a semicolon to go to the next row:

 matrix = [1 2 3 ; 4 5 6 ; 7 8 9];
stringMatrix = [{'position'} {'speed'} ; {'size'} {'acceleration'}];

The result is the following matrices:

$matrix=\begin{pmatrix}1 &2 &3\\4 &5 &6 \\7 &8 &9\end{pmatrix}~~~~~~~~~~~~~~~~~~stringMatrix=\begin{pmatrix}\{'position'\}& \{'speed'\}\\\{'size'\} & \{'acceleration'\}\end{pmatrix}$

You can then multiply matrices as the following:

 A*B % classic matrix multiplication in MATLAB

## MATLAB: Function and Script

### Writing a Script

To write a new MATLAB script, follow these steps:

1. Go to the top left of your MATLAB windows, under the “HOME” tab
2. Click on “New Script“:
3. Save your script by using the keyword shortcut “CTRL+S” or by clicking “Save” at the top left of the MATLAB editor:

### Defining a MATLAB Function

Here’s the syntax to define a MATLAB function:

function [out1, out2, out3] = functionName(in1, in2)

Instructions

end
• Outputs: list every output of your function separated with commas in between square brackets.
• Inputs: list input separated by commas and enclosed by parenthesis.

If you want to know all the nitty-gritty details of MATLAB functions, see MATLAB Function: Vector, Matrix, Or Single Input/Output Functions.

## Creating Loops and Conditions

### MATLAB for Loop

You can define a for loop in MATLAB as the following:

for i = 1:n % n is the number of loops you want
Instructions; % what you want to do n times
end

For example, here’s how to create a for loop that counts to 10:

for i = 1:10
i
end

To see more examples of for loops, see MATLAB For Loop: Example of Simple, Backward, and Nested For Loops.

### While Loop

If you want to define the while loop in MATLAB, you can use the following syntax:

while condition
Instructions
end

For example, here’s how to create a wide loop that counts to 10:

i = 10;
while i~=10 % while i is not equal to 10
i=i+1
end

### If Else MATLAB

The basic if else MATLAB structure is:

if firstCondition
firstInstruction
elseif secondCondition
secondInstruction
else
otherInstructions
end

For example, if your condition is “A is not equal to B,” you’d have:

• A and B matrices:
if ~isequal(A, B)
instruction
end
• A and B numbers:
if A~=B
instruction
end

If you want to learn more complicated examples such as switch case structures, A less than or equal to B, or strings and logical operators, see If Else MATLAB: Conditional Statements if A Is Not Equal to B.

## MATLAB: Reading and Writing Files

### Text Files

• To read a text file, you need to open it with the fopen MATLAB command, read it using a while loop, and close it using the fclose MATLAB command:

fid = fopen(fileName);
while ~feof(fid)
textLineEntry = fgetl(fid);
end
fclose(fid);

• Using the same structure, we can employ the fprintf MATLAB command to write a new text file. For example, here’s how to write the word “Text” in a new text file named “textFileName.txt”:

fid = fopen('textFileName.txt', 'at');
fprintf(fid, 'Text');
fclose(fid);

To learn more about the fprintf MATLAB command, see fprintf in MATLAB: Text, Matrix, and Complex Number.

### Excel Files

• Reading an Excel file: an easy way to read an Excel file is to use the xlsread MATLAB command. Here’s the basic syntax for that command:

[numbers text textAndNumbers] = xlsread(excelFileName, sheetNumber);

If you want more explanations and examples of how to use this function, see Xlsread Tutorial: Extract Data from Excel in MATLAB.

• Writing into an Excel file: to write into an Excel file, use the following:

xlswrite(fileName, data, sheetNumber);

For example, if you want to write the following matrix into an Excel file:

$\begin{pmatrix} \{'firstWord'\}&\{'secondWord'\}\\\{'thirdWord'\} &\{'fouthWord'\} \end{pmatrix}$

You can do the following:

variable1 = 'firstWord';
variable2 = 'secondWord';
variable3 = 'thirdWord';
variable4 = 'fourthWord';
data = [{variable1} {variable2} ; {variable3} {variable4}];
fileName = 'excelFile.xlsx';
sheetNumber = 1;
xlswrite(fileName, data, sheetNumber);