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Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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17 Jul 2010, 04:36
7
2
DenisSh wrote:
Bunuel wrote:
gurpreetsingh's solution is correct.
Is \(\frac{x}{3}+\frac{3}{x}>2\)? --> is \(\frac{(x-3)^2}{x}>0\)? Answer: C.
How did you get \(\frac{(x-3)^2}{x}>0\)?
Step 1: \(\frac{x}{3}+\frac{3}{x}>2\)? Step 2 (multiply by 3x): \(x^2+9>6x\) Step 3 (move 6x to the left side): \(x^2-6x+9>0\) Step 4 (convert to the compact form): \((x-3)^2>0\)
Please explain...
This is the most common error when solve inequalities. I keep writing this over and over again:
Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign.
So you CANNOT multiply \(\frac{x}{3}+\frac{3}{x}>2\) by \(3x\) since you don't know the sign of \(x\).
What you CAN DO is: \(\frac{x}{3}+\frac{3}{x}>2\) --> \(\frac{x}{3}+\frac{3}{x}-2>\) --> common denominator is \(3x\) --> \(\frac{x^2+9-6x}{3x}>0\) --> multiply be 3 --> \(\frac{x^2+9-6x}{x}>0\) --> \(\frac{(x-3)^2}{x}>0\).
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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16 Jul 2010, 07:42
4
DenisSh wrote:
Is \(\frac{x}{3} + \frac{3}{x} > 2\)?
(1) \(x < 3\)
(2) \(x > 1\)
Please outline your approach!
gurpreetsingh's solution is correct.
Is \(\frac{x}{3}+\frac{3}{x}>2\)? --> is \(\frac{(x-3)^2}{x}>0\)? Now, nominator is non-negative, thus the fraction to be positive nominator must not be zero (thus it'll be positive) and denominator mut be positive --> \(x\neq{3}\) and \(x>0\).
Statement (1) satisfies the first requirement and statement (2) satisfies the second requirement, so taken together they are sufficient.
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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19 Jul 2010, 09:18
1
ulm wrote:
(x-3)^2/x >0
(x-3)^2 is always >0, therefore we just need x to be >0. The correct answer is (0;3) and (3;+inf) It's (B), because if we consider (C), we will loose answers. Try 10, 11, 12 and so on. It fits.
Generally, unknown (or expression with unknown) in even power is NOT always positive, it's non-negative. Not knowing this is the cause of many mistakes on GMAT.
So, \((x-3)^2\geq{0}\), because if \(x=3\), then \((x-3)^2=0\) and \(\frac{(x-3)^2}{x}\) also equals to zero (and not more than zero). We need statement (1) to exclude the possibility of \(x\) being 3 by saying that \(x<3\). That's why the answer to this question is C, not B.
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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19 Jul 2010, 09:48
ulm wrote:
(x-3)^2/x >0
(x-3)^2 is always >0, therefore we just need x to be >0. The correct answer is (0;3) and (3;+inf) It's (B), because if we consider (C), we will loose answers. Try 10, 11, 12 and so on. It fits.
ulm wrote:
I'm completely agree that x not equal 3 (as i wrote earlier;)
But we miss all positive x's that a > 3, don't we?
I'm not sure exactly what you're asking here. You said that the answer to this question should be B: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Statement (2) is: \(x>1\) --> denominator is positive, nominator is also positive EXCEPT for one value of \(x\): when \(x=3>1\), then \(\frac{(x-3)^2}{x}=0\) (not more than zero). Hence we have TWO different answers to the question "is \(\frac{(x-3)^2}{x}>0\)": one is NO for \(x=3>1\) and another is YES for all other values of \(x>1\). Two different answers = not sufficient.
Schools: Chicago Booth 2013, Ross, Duke , Kellogg , Stanford, Haas
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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21 Jul 2010, 09:41
Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...
1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient
2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient
When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C_________________
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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21 Jul 2010, 10:10
Michmax3 wrote:
Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...
1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient
2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient
When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C
Unfortunately your approach is not correct.
First of all if \(x=2\)--> \(\frac{x}{3} + \frac{3}{x}=\frac{13}{6} > 2\), so you made an error in calculations (\(\frac{2}{3} + \frac{3}{2}=\frac{13}{6}\neq{\frac{5}{6}}\)). Again \(\frac{x}{3} + \frac{3}{x}>2\) is true for ANY value of \(x\) but 3, for which \(\frac{x}{3} + \frac{3}{x}=2\).
Next: you say that "When combined, you know x has to be 2". Not so, as we are not told that \(x\) is an integer, hence \(x<3\) and \(x>1\) does not mean that \(x=2\), it can be 2.5 or 1.777, basically ANY number from 1 to 3, not inclusive.
Schools: Chicago Booth 2013, Ross, Duke , Kellogg , Stanford, Haas
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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21 Jul 2010, 10:30
Bunuel wrote:
Michmax3 wrote:
Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...
1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient
2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient
When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C
Unfortunately your approach is not correct.
First of all if \(x=2\)--> \(\frac{x}{3} + \frac{3}{x}=\frac{13}{6} > 2\), so made you an error in calculations (\(\frac{2}{3} + \frac{3}{2}=\frac{13}{6}\neq{\frac{5}{6}}\)). Again \(\frac{x}{3} + \frac{3}{x}>2\) is true for ANY value of \(x\) but 3, for which \(\frac{x}{3} + \frac{3}{x}=2\).
Next: you say that "When combined, you know x has to be 2". Not so, as we are not told that \(x\) is an integer, hence \(x<3\) and \(x>1\) does not mean that \(x=2\), it can be 2.5 or 1.777, basically ANY number from 1 to 3, not inclusive.
Hope it's clear.
Yes, then I guess I got lucky and need a lot more practice with these. Thanks for clarifying_________________
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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05 Oct 2010, 22:25
Bunuel wrote:
DenisSh wrote:
Bunuel wrote:
gurpreetsingh's solution is correct.
Is \(\frac{x}{3}+\frac{3}{x}>2\)? --> is \(\frac{(x-3)^2}{x}>0\)? Answer: C.
How did you get \(\frac{(x-3)^2}{x}>0\)?
Step 1: \(\frac{x}{3}+\frac{3}{x}>2\)? Step 2 (multiply by 3x): \(x^2+9>6x\) Step 3 (move 6x to the left side): \(x^2-6x+9>0\) Step 4 (convert to the compact form): \((x-3)^2>0\)
Please explain...
This is the most common error when solve inequalities. I keep writing this over and over again:
Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.
So you CAN NOT multiply \(\frac{x}{3}+\frac{3}{x}>2\) by \(3x\) since you don't know the sign of \(x\).
Wheat you CAN DO is: \(\frac{x}{3}+\frac{3}{x}>2\) --> \(\frac{x}{3}+\frac{3}{x}-2>\) --> common denominator is \(3x\) --> \(\frac{x^2+9-6x}{3x}>0\) --> multiply be 3 --> \(\frac{x^2+9-6x}{x}>0\) --> \(\frac{(x-3)^2}{x}>0\).
Hope it helps.
thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.
Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit_________________
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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06 Oct 2010, 00:23
hirendhanak wrote:
thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.
Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit
Consider the inequality \(\frac{x^2-12}{x}>-1\)
Lets say I multiply both sides by 7x without considering the signs of the variable, what happens ?
\(x^2-12>-x\) \(x^2+x-12>0\) \((x+4)(x-3)>0\)
Which is true whenever x>3 (both terms positive) or when x<-4 (both terms negative)
But since we haven't kept the Sign of x in mind when we multiplied in step 1, the solution is wrong.
For eg. Take x=-1 which according to us is not a solution. It is easy to see ((-1)^2-12)/(-1)=11>-1. So it should be a solution Similarly take x=-6 which according to us is a solution, but ((-6)^2-12)/-6=-4<-1. So it should not be a solution_________________
Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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06 Oct 2010, 04:09
hirendhanak wrote:
thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.
Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit
Consider a simple inequality \(4>3\) and some variable \(x\).
Now, you can't multiply (or divide) both parts of this inequality by \(x\) and write: \(4x>3x\), because if \(x=1>0\) then yes \(4*1>3*1\) but if \(x=-1\) then \(4*(-1)=-4<3*(-1)=-3\). Similarly, you can not divide an inequality by \(x\) not knowing its sign.
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Re: Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1[#permalink]
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08 Jun 2011, 03:52
is (x/3+3/x) > 2? (1) x < 3 (2) x > 1
This ds has been discussed thoroughly at http://gmatclub.com/forum/tricky-inequality-problem-97331.html. It inequality simplified there as [(x - 3)^2]/3 > 0. if i do not simplify then i can i solved it as: (1) if x = 2 then the (x/3+3/x) > 2 but if x = negative the (x/3+3/x) > 2 is not true. so Insufficient. (2) if x = 2 then (x/3+3/x) > 2 but if x = 1 then (x/3+3/x) > 2 is true. so insufficient.
for C x =2 and (x/3+3/x) > 2.
so why i will simplify as i am getting direct answer?_________________ |
As Stefan Wagner notes, the decision boundary for a logistic classifier is linear. (The classifier needs the inputs to be linearly separable.) I wanted to expand on the math for this in case it's not obvious.
The decision boundary is the set of x such that$${1 \over {1 + e^{-{\theta \cdot x}}}} = 0.5$$
A little bit of algebra shows that this is equivalent to$${1 = e^{-{\theta \cdot x}}}$$
and, taking the natural log of both sides,
$$0 = -\theta \cdot x = -\sum\limits_{i=0}^{n} \theta_i x_i$$
so the decision boundary is linear.
The reason the decision boundary for a neural network is not linear is because there are
two layers of sigmoid functions in the neural network: one in each of the output nodes plus an additional sigmoid function to combine and threshold the results of each output node. |
I'm trying to derive an HJB from a discrete time setting. At some point, I am left with
$$ \lim_{\Delta\to 0} \frac{v(c_{t+\Delta}, u_{t+\Delta}, t+\Delta) - v(c_{t}, u_{t}, t)}{\Delta}$$
and am not sure what to do. If $\Delta$ was only in one argument, this would be a partial differential. My hunch is that this is the total derivative in $t$, but I dont know how to show that.
How do I proceed with the expression above? |
There is a similar Phys.SE question here, but I still didn't get the idea. The problem is:
Write down the equations for one-dimensional motion of an ideal fluid in terms of the variables $a$ and $t$, where $a$ (called a Lagrangian variable) is the $x$ coordinate of a fluid particle at some instant $t=t_0$.
$x$ is clearly a function of $a$ and $t$, and so $x=x(a,t).$ One of the equations of motion is the equation of continuity: $$\frac{\partial\rho}{\partial t}+\nabla\cdot\rho\mathbf v= 0$$
I can't derive the solution, which was supposed to be $$\rho\frac{\partial x}{\partial a}=\rho_0$$
Where $ρ_0$ is the density at a at $t_0$.
This is what I've done so far: I have that $\frac{d\rho}{dt}=0$, and (I suppose) $$d\rho=\frac{\partial\rho}{\partial t}dt+\frac{\partial\rho}{\partial a}da$$
as well as
$$dx=\frac{\partial x}{\partial t}dt+\frac{\partial x}{\partial a}da$$
Aparently the main problem for me is that I don't know how to get to $\rho_0$... |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
"If problem $Y$ can be reduced to problem $X$ in polynomial time, we denote this by $Y \leq_p X$."
If $X$ is one of already known NP-complete problem then we can say that $Y$ is NP-complete.
No, that's backwards. Very informally, $Y\leq_{\mathrm{p}} X$ means that $Y$ is no harder to solve than $X$.
Still informally, but a bit more precisely, $Y\leq_{\mathrm{p}} X$ means that we can solve $Y$ by doing a little [a polynomial amount of] work and then solving $X$. A problem $L$ being
NP-hard means that we can solve every problem in NP by doing a little work and then solving $L$. Finally, $L$ is NP-complete if it's NP-hard and is, itself, in NP.
So, if $Y$ is
NP-complete, and $Y\leq_{\mathrm{p}} X$ then we can solve every problem in NP by doing a little work to convert to $Y$, then doing a little more work to convert to $X$, then solving $X$. This allows us to conclude that $X$ is NP-hard. If $X$ is also in NP, then it is NP-complete.
On the other hand, if $X$ is
NP-complete and $Y\leq_{\mathrm{P}}X$, all we can conclude is that $Y$ is in NP. Informally, we've been told that $Y$ is no harder than some NP-complete problem. $Y$ could be NP-complete, or it could be really easy, or something between the two.
From my understanding, $p$ should be always polynomial and this cannot be an exponential function such as $2^n$ or $3^n$.
It's not that "$p$ should always be polynomial": p doesn't denote a function; rather, it stands for the word "polynomial". (If you look carefully, you'll notice that I've typeset it as $\leq_{\mathrm{p}}$, denoting a type of reduction that's called "p", rather than $\leq_p$, which would be a type of reduction parameterized by some mathematical object $p$.)
So your remark is a bit like defining $A\to_{\mathrm{T}}B$ to mean that you can get from city $A$ to city $B$ by train and then saying, "From my understanding, T should always be a train and cannot be another means of transport, such as a car or bus." Well, yes: it's a train because we defined it to be a train. It makes complete sense to ask about cars and buses and functions that aren't polynomials, but "p" means "polynomial" because that's what we decided it would mean.
However, my question is if $Y$ or $X$ already has a lower bound that is not polynomial, which is $2^n$ or $3^n$ then can I still say that $Y$ is NP-complete?
This is a different question again. The "p" in $\leq_p$ refers to a polynomial amount of work done translating between two problems. That concept is independent of the difficulty of solving the problems themselves. For example, SAT and 3-colourability are both quite difficult to solve (nobody knows an efficient algorithm for either) but it's quite easy to translate between them (we make undergraduates do it as an exercise, and most of them survive).
As it stands, we don't have any super-polynomial lower bounds on the complexity of solving any
NP-complete problem on a deterministic Turing machine. That would prove that P$\,\neq\,$NP which would be... quite a big deal.
In other words, does lower bounds for $X$ and $Y$ matter?
NP-hardness is a lower bound: it says that the problem cannot be (much) easier than any problem in NP. More generally, if $Y\leq_{\mathrm{p}}X$ then a lower bound on $Y$ does translate to a lower bound on $X$. This is because the lower bound on $Y$ says "Every possible way of solving $X$ requires at least this much computational resources", and "any possible way" includes "translate to $X$ and then solve $X$."
In general, a lower bound on $X$ doesn't tell you anything. We just know that $Y$ is at most as hard as something which is at least "this hard", but that doesn't say anything, in the same way that "$x<y$ and $y>3$" doesn't restrict the value of $x$ at all. |
It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced.
Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit.
@Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form.
A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts
I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it.
Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis.
Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)?
No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet.
@MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it.
Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow.
@QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary.
@Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer.
@QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits...
@QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right.
OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ...
So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study?
> I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago
In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a...
@MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really.
When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.?
@tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...)
@MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences). |
Definitions. Suppose $X$ is a topological space.
$w(X)=\min\{|\mathcal B|:\mathcal B$ is a base for $X\}+\omega$ $e(X)=\sup\{|D|:D\subseteq X$ is closed and discrete$\}+\omega$ $K(X)$ is the collection of all compact subsets of $X$ $\mathbb R ^*=\beta\mathbb R \setminus \mathbb R$
The collection of all open subsets of $\mathbb R$ has cardinality $\mathfrak c$. The canonical basis for $\beta \mathbb R$ consists of sets of the form
$$\{p\in\beta\mathbb R :(\exists L\in p)(L\subseteq U)\}$$ where $U$ is open in $\mathbb R$. Thus $w(\beta\mathbb R) \leq \mathfrak c$ ( I suspect equal).
Now I appeal to a theorem, which may or may not be a deep result (I haven't read the proof) in cardinal invariants.
Theorem. If $X$ is compact $T_1$ then $|K(X)|\leq 2^{e(X)\cdot w(X)}$.
Note that $e(\mathbb R ^*)=\omega$ (really finite, but in the definition we require that it be at least $\omega$): If $D$ is an infinite discrete closed subspace of $\mathbb R^*$, then $D$ contains a copy of $\beta\omega$, contradicting $D$ discrete.
Combining everything we have $|K(\mathbb R ^*)|\leq 2^{\omega\cdot \mathfrak c}=2^\mathfrak c$ (whereas $|\mathcal P (\mathbb R^*)|=2^{2^\mathfrak c}$).
Questions:
1) Is my calculation correct?
2) Is there an easier way to get my final conclusion? I feel there must be, but maybe not. We know $\beta\mathbb R$ has a countable dense subset, but I didn't see how to use that.
Edit: I might be an idiot. Why does the following not work: $\beta\mathbb R$ has a countable dense subset $\mathbb Q$, and we can uniquely identify an open subset of $\beta \mathbb R$ by its intersection with $\mathbb Q$. Thus there are only $2^\omega$ open subsets of $\beta\mathbb R$, whence there are also only $2^\omega$ closed subsets of $\beta\mathbb R$... clearly false. |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
(a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular?
(b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?
(c) Let $A$ be a $4\times 4$ matrix and let\[\mathbf{v}=\begin{bmatrix}1 \\2 \\3 \\4\end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}4 \\3 \\2 \\1\end{bmatrix}.\]Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular?
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
Prove that the matrix\[A=\begin{bmatrix}0 & 1\\-1& 0\end{bmatrix}\]is diagonalizable.Prove, however, that $A$ cannot be diagonalized by a real nonsingular matrix.That is, there is no real nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
An $n\times n$ matrix $A$ is called nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$.Using the definition of a nonsingular matrix, prove the following statements.
(a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular.
(b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then:
The matrix $B$ is nonsingular.
The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.)
Let $A$ be an $n\times (n-1)$ matrix and let $\mathbf{b}$ be an $(n-1)$-dimensional vector.Then the product $A\mathbf{b}$ is an $n$-dimensional vector.Set the $n\times n$ matrix $B=[A_1, A_2, \dots, A_{n-1}, A\mathbf{b}]$, where $A_i$ is the $i$-th column vector of $A$.
Prove that $B$ is a singular matrix for any choice of $\mathbf{b}$.
For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix. |
The Hamiltonian of helium can be expressed as the sum of two hydrogen Hamiltonians and that of the Coulomb interaction of two electrons.
$$\hat H = \hat H_1 + \hat H_2 + \hat H_{1,2}.$$
The wave function for parahelium (spin = 0) is
$\psi(1,2) = \psi_S(r_1, r_2)\dot \xi_A(s_1, s_2)$ with the first being a symmetric spatial function and the second being an antisymmetric one.
We can separate this into the normalized function
$\psi_S(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2)+ \psi_1(r_2)(\psi_2(r_1)]=\psi_S(r_2,r_1)$
For orthohelium the functions look like this:
$\psi(1,2) = \psi_A(r_1, r_2)\dot \xi_S(s_1, s_2)$ $\psi_A(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2) - \psi_1(r_2)(\psi_2(r_1)]=-\psi_A(r_2,r_1)$
Show the ground state of helium is parahelium. The hint is what happens to the wavefunction.
OK, so I start with that the Hamiltonian of the given wave function(s)
$$H_1 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_1^2} = E_1 \psi$$ $$H_2 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_2^2} = E_2 \psi$$ $$H_{1,2} = -\frac{e^2}{4\pi\epsilon_0 r_{1,2}}$$
OK, I was trying to get a handle on how to get started with this. So I wanted to check if what I have above is "allowed" -- that is, is the second derivative (the nabla, really) of the psi functions treatable this way, since they all have two variables (really two position vectors) in them? Basically this is all about how to set up the initial differentials I would solve.
EDIT: One thing I thought of doing was this (for $H_1$):
$H_1 = -\frac{\hbar^2}{2m}(\frac{\partial \psi}{\partial^2 r_1^2}+\frac{\partial \psi}{\partial^2 r_2^2})=E_1(\psi_1(r_2)+\psi_1(r_1))$
but again I don't know if that's kosher. |
Illinois Journal of Mathematics Illinois J. Math. Volume 51, Number 3 (2007), 951-976. Well-posedness for equations of Benjamin-Ono type Abstract
The Cauchy problem $u_t - |D|^{\alpha}u_x + uu_x=0$ in $(-T,T) \times \mathbb{R}$, $u(0)=u_0$, is studied for $1< \alpha <2$. Using suitable spaces of Bourgain type, local well-posedness for initial data $u_0 \in H^s(\mathbb{R}) \cap \dot{H}^{-\omega}(\mathbb{R})$ for any $s > -\tfrac{3}{4}(\alpha-1)$, $\omega:=1/\alpha-1/2$ is shown. This includes existence, uniqueness, persistence, and analytic dependence on the initial data. These results are sharp with respect to the low frequency condition in the sense that if $\omega<1/\alpha-1/2$, then the flow map is not $C^2$ due to the counterexamples previously known. By using a conservation law, these results are extended to global well-posedness in $H^s(\mathbb{R}) \cap \dot{H}^{-\omega}(\mathbb{R})$ for $s \geq 0$, $\omega=1/\alpha-1/2$, and real valued initial data.
Article information Source Illinois J. Math., Volume 51, Number 3 (2007), 951-976. Dates First available in Project Euclid: 13 November 2009 Permanent link to this document https://projecteuclid.org/euclid.ijm/1258131113 Digital Object Identifier doi:10.1215/ijm/1258131113 Mathematical Reviews number (MathSciNet) MR2379733 Zentralblatt MATH identifier 1215.35136 Subjects Primary: 35Q53: KdV-like equations (Korteweg-de Vries) [See also 37K10] Secondary: 35B30: Dependence of solutions on initial and boundary data, parameters [See also 37Cxx] 76B03: Existence, uniqueness, and regularity theory [See also 35Q35] 76B15: Water waves, gravity waves; dispersion and scattering, nonlinear interaction [See also 35Q30] Citation
Herr, Sebastian. Well-posedness for equations of Benjamin-Ono type. Illinois J. Math. 51 (2007), no. 3, 951--976. doi:10.1215/ijm/1258131113. https://projecteuclid.org/euclid.ijm/1258131113 |
Tagged: abelian group
Abelian Group Problems and Solutions.
The other popular topics in Group Theory are:
Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
Add to solve later
(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 497
Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.
Also determine whether the statement is true if $G$ is a non-abelian group.Add to solve later
Problem 434
Let $R$ be a ring with $1$.
A nonzero $R$-module $M$ is called irreducible if $0$ and $M$ are the only submodules of $M$. (It is also called a simple module.) (a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator.
Add to solve later
(b) Determine all the irreducible $\Z$-modules. Problem 420
In this post, we study the
Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Add to solve later
Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$. Problem 343
Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of $G$. |
In this paper (A simple provably secure key exchange by Ding et al.), I am trying to understand the correctness (which is given on page number 8) of the key exchange technique based on LWE.
To understand the correctness of the scheme, there is a very important lemma (on page number 6). The lemma is as follows:
Lemma 2. Let $q>8$ be an odd integer, the function $E$ defined above is a robust extractor with respect to $S$ with error tolerance $\frac{q}{4}-2$.
I'm confused by the proof of Lemma 2:
How does the author derive the condition that Lemma 2 is true for $q\gt 8$? How does $|y+ \sigma$ $\frac{(q-1)}{2} \mod q| \le \frac{q}{4}+1$? At the end of the proof of Lemma 2, the author writes that
Our robust extractor enjoys a very nice property which says that for uniformly random $x\in$$ \mathbb{Z} _q$, $E(x, \sigma$) is uniform in $\{0,1\}$ even conditioned on $\sigma$, where $\sigma \leftarrow S(x)$.
This property becomes Lemma 3 of this paper.
Can anyone answer the above three questions and thus the proof of Lemma 2 and Lemma 3? Thanks! |
So, we had a very brief introduction to the Nyquist-Shannon sampling theorem (the discrete time version).
While discussing this, we have seen that multiplying a discrete time signal $x[n]$ by an impulse train $p[n]$, we get a sampled signal $x_p[n]$, and if certain conditions are fulfilled, then it is possible to recover the original signal.
In the frequency domain, what it means (informally) is that we are taking $X(e^{j\omega})$ and shifting it by integer multiples of $2\pi/N$ (i.e the sampling frequency) and "adding" all of this to the plot of $X(e^{j\omega})$ giving us the plot of $X_p(e^{j\omega})$ (in other words we have replicas of $X(e^{j\omega})$)
After looking on the internet and doing a couple of drawings, I've seen that one important condition to recover the initial signal $x[n]$ is that $\omega_s > 2\cdot \omega_m$, where $\omega_s$ is the sampling frequency and $\omega_m$ is the frequency of the original signal (i.e the frequency of $x[n]$).
However, the teacher, in his lecture only mentioned the following condition:
"If the original signal satisfies the following condition: $$X(e^{j\omega}) = 0, \text{ for } \frac{\pi}{N} < |\omega| \leq \pi$$ then the sampled signal $x_p[n]$ contains all the information about the original signal and we can easily recover $x[n]$ from $x_p[n]$".
This makes a little sense to me, as I understand that if $X(e^{j\omega}) \neq 0 $ for all $\omega$, then we do, indeed, have overlapping (i.e aliasing) and it is not possible to recover the signal. But I do not understand why $X(e^{j\omega})$ has to be $0$ specifically for $\frac{\pi}{N} < |\omega| \leq \pi$.
Also, how come the teacher did not speak of the other condition (namely that $\omega_s > 2\cdot \omega_m$) ? Has he forgotten it, or is there something I am missing ? |
So, I am having a lot of troubles visualizing the following system.
The point of suspension of a plane simple pendulum of mass m and length l is constrained to move along a horizontal track and is connected to a point on the circumference of a uniform flywheel of mass M and radius a through a massless connecting rod also of length a, as shown in the figure. The flywheel rotates about a center fixed on the track. Find the hamiltonian for the combined system and determine Hamilton's equation of motion.
Here is my idea so first of all I would construct the Lagrangian and then from it I will get the hamiltonian, so for this I need to determine the kinetic energy and the potential energy.
I am having troubles constructing the cartesian coordinates of this thing which the only thing that I need I know how to do everything else.
Is the following correct
$x = l\sin(\theta) - a(\omega)t$? where I measure the angle of the regular m pendulum with respect to vertical and the other one with respect to horizontal ? and $\omega$ is how fast the flywheel rotates. |
DISCLAIMER - what follows is the simplified "analysis" that leads to the expression you were asking about. This is NOT in fact the correct way to treat this problem, as was pointed out in the comments.
When a blade with area $A$ moves through the air at a certain velocity $v$ and at an angle $\theta$, it "cuts through" an volume of air given by $V = A\sin\theta v$ every second - the projection of the area of the blade multiplied by the distance it moves in unit time.
This body of air is deflected downwards, and the velocity it attains is $v_d = v\sin\theta$. Now we can compute the mass of the body of air from the density and volume: $m = \rho V = \rho A \sin\theta v$. To accelerate a certain mass per unit time downwards, we need a force $F\Delta t = m\Delta v$. It follows that
$$\begin{align}F &= m\Delta v \\&= \left(\rho A \sin\theta v\right)\left( v\sin\theta\right)\\&=\rho A v^2 \sin^2\theta\end{align}$$
Now all you have to do is convert the linear velocity to a rotating velocity, noting that not every point on the blade will travel at the same speed when it's going around in a circle.
Note - lots of simplifying assumptions went into the above. The real physics of a blade moving through air has to take account of the fact that it's not only the air right in front of the blade that is moved... but the general expression (showing the linear relationship with density and area, the quadratic relationship with velocity, and the dependence on the square of the attack angle) looks just like the one you quoted. Of course when $\theta$ (or $\phi$, in the expression you quoted) gets too large, the air flow will "stall" and the force will get smaller, not larger. That shows the limitation of this simplistic approach (which only works over a limited range of angles and velocities, and with all the other simplifications already mentioned). |
I am trying to understand how to combine uncertainties when they are dependent and independent from each other.
Using this formula :
$$\delta z = \sqrt {\Biggl(\dfrac{\partial f}{\partial x} \delta x\Biggr)^2+\Biggl(\dfrac{\partial f}{\partial y} \delta y\Biggr)^2+2\Biggl(\dfrac{\partial f}{\partial x}\cdot \dfrac{\partial f}{\partial y}\Biggr)\text{cov}(x,y)}$$
Intuitively, if the covariance between the two is zero, the last term will disappear and the equation just becomes the square root of sums for combining uncertainty.
My question is does $\delta z$ then need to be divided by 2 to get the final uncertainty value.(i.e. divide by $N$) |
I would argue that at least when discussing linear models (like AR models), adjusted $R^2$ and AIC are not that different.
Consider the question of whether $X_2$ should be included in$$y=\underset{(n\times K_1)}{X_1}\beta_1+\underset{(n\times K_2)}{X_2}\beta_2+\epsilon$$This is equivalent to comparing the models\begin{eqnarray*}\mathcal{M}_1&:&y=X_1\beta_1+u\\\mathcal{M}_2&:&y=X_1\beta_1+X_2\beta_2+u,\end{eqnarray*}where $E(u|X_1,X_2)=0$. We say that $\mathcal{M}_2$ is the
true model if $\beta_2\neq0$.Notice that $\mathcal{M}_1\subset\mathcal{M}_2$. The models are thus nested.A model selection procedure $\widehat{\mathcal{M}}$ is a data-dependent rule that selects the most plausible of several models.
We say $\widehat{\mathcal{M}}$ is
consistent if \begin{eqnarray*}\lim_{n\rightarrow\infty}P\bigl(\widehat{\mathcal{M}}=\mathcal{M}_1|\mathcal{M}_1\bigr)&=&1\\\lim_{n\rightarrow\infty}P\bigl(\widehat{\mathcal{M}}=\mathcal{M}_2|\mathcal{M}_2\bigr)&=&1\end{eqnarray*}
Consider adjusted $R^2$. That is, choose $\mathcal{M}_1$ if $\bar{R}^2_1>\bar{R}^2_2$. As $\bar{R}^2$ is monotonically decreasing in $s^2$, this procedure is equivalent to minimizing $s^2$. In turn, this is equivalent to minimizing $\log(s^2)$. For sufficiently large $n$, the latter can be written as\begin{eqnarray*}\log(s^2)&=&\log\left(\widehat{\sigma}^2\frac{n}{n-K}\right) \\&=&\log(\widehat{\sigma}^2)+\log\left(1+\frac{K}{n-K}\right) \\&\approx&\log(\widehat{\sigma}^2)+\frac{K}{n-K} \\&\approx&\log(\widehat{\sigma}^2)+\frac{K}{n},\end{eqnarray*}where $\widehat{\sigma}^2$ is the ML estimator of the error variance. Model selection based on $\bar{R}^2$ is therefore asymptotically equivalent to choosing the model with the smallest$\log(\widehat{\sigma}^2)+K/n$.This procedure is inconsistent.
Proposition:$$\lim_{n\rightarrow\infty}P\bigl(\bar{R}^2_1>\bar{R}^2_2|\mathcal{M}_1\bigr)<1$$
Proof:\begin{eqnarray*}P\bigl(\bar{R}^2_1>\bar{R}^2_2|\mathcal{M}_1\bigr)&\approx&P\bigl(\log(s^2_1)<\log(s^2_2)|\mathcal{M}_1\bigr) \\&=&P\bigl(n\log(s^2_1)<n\log(s^2_2)|\mathcal{M}_1\bigr) \\&\approx&P(n\log(\widehat{\sigma}^2_1)+K_1<n\log(\widehat{\sigma}^2_2)+K_1+K_2|\mathcal{M}_1) \\&=&P(n[\log(\widehat{\sigma}^2_1)-\log(\widehat{\sigma}^2_2)]<K_2|\mathcal{M}_1) \\&\rightarrow&P(\chi^2_{K_2}<K_2) \\&<&1,\end{eqnarray*}where the 2nd-to-last line follows because the statistic is the LR statistic in the linear regression case that follows an asymptotic $\chi^2_{K_2}$ null distribution.QED
Now consider Akaike's criterion,$$AIC=\log(\widehat{\sigma}^2)+2\frac{K}{n}$$Thus, the AIC also trades off the reduction of the SSR implied by additional regressors against the "penalty term," which points in the opposite direction. Thus, choose $\mathcal{M}_1$ if$AIC_1<AIC_2$, else select $\mathcal{M}_2$.
It can be seen that the $AIC$ is also inconsistent by continuing the above proof in line three with $P(n\log(\widehat{\sigma}^2_1)+2K_1<n\log(\widehat{\sigma}^2_2)+2(K_1+K_2)|\mathcal{M}_1)$. The adjusted $R^2$ and the $AIC$ thus choose the "large" model $\mathcal{M}_2$ with positive probability, even if $\mathcal{M}_1$ is the true model.
As the penalty for complexity in AIC is a little larger than for adjusted $R^2$, it may be less prone to overselect, though. And it has other nice properties (minimizing the KL divergence to the true model if that is not in the set of models considered) that are not addressed in my post. |
I'm using
clf = svm.SVC(kernel='linear') on a data set with only two classes $y \in \{-1, +1\}$ and the feature values of all samples are normalized between 0 and 1. We know from various studies (Guyon et al. 2003, Chang et al. 2008) that the magnitude of a feature weight in
clf.coef_ is indicative of the importance of the feature for the classification.
My question is how the sign (positive or negative) of the feature weight can be interpreted, when all the features values are positive?
Given that all my feature values are normalized between 0 and 1, I think that a negative feature weight contributes to a negative classification ($y=-1$), while a positive feature weight contributes to a positive classification ($y=+1$).
clf.coef_ is the coefficients in the primal problem. Looking at the formulation of a hard-margin primal optimization problem, using a linear kernel:
$$ \text{Minimize } \frac{1}{2} \mathbf{w}^T \mathbf{w} \\ \text{Subject to } y_i (\mathbf{w}^T\mathbf{x}_i - b) \geq 1, \quad i = 1, \ldots, N $$
where $y_i \in \{-1, +1\}$ is the class of sample $i$, $\mathbf{x}_i$ is the feature representation of sample $i$ and $\mathbf{w}$ is the feature weights, which I presume is what
clf.coef_ returns. $b$ is just an offset of the max-margin hyperplane. From the constraint $y_i (\mathbf{w}\mathbf{x}_i - b) \geq 1$, we can thus see that
If $y_i = -1$ Then the dot product $\mathbf{w}\mathbf{x}_i$ has to be negative. Since all feature values in $\mathbf{x}$ are positive, then a negative feature weight contributes to a negative classification. If $y_i = +1$ Then the dot product $\mathbf{w}\mathbf{x}_i$ has to be positive. Similarly, a positive feature weight then contributes to a positive classification.
A similar analysis can made for the soft-margin case.
Unfortunately, I haven't been able to find any academic sources for this interpretation of the sign of the feature weights. Perhaps the case of a binary classification with only positive features isn't general enough to deserve a study. My questions are thus:
Is my analysis and subsequent conclusion correct, that a negative feature weight contributes to a negative classification, and a positive feature weight contributes to a positive classification? Does anyone know of any academic sources that I can cite to strengthen my claim?
A comment in this post seems to support my claim, but no motivation or references are provided.
Update
I implemented some code to test my hypothesis. Using 9 features such that
features 1, 2, 3 occur more frequently in the negative class features 4, 5, 6 occur as frequently in both classes features 5, 6, 7 occur more frequently in the positive class
Code:
from __future__ import divisionimport numpy as npfrom sklearn import svmN = 1000 # number of samplesn_features = 9n_iter = 100# features 1, 2, 3 occur more in the negative class# features 4, 5, 6 occur as frequently in both classes# features 7, 8, 9 occur more frequently in the positve classmean_neg = [0.90, 0.80, 0.70, 0.20, 0.50, 0.70, 0.20, 0.10, 0.05]mean_pos = [0.05, 0.10, 0.20, 0.20, 0.50, 0.70, 0.70, 0.80, 0.90]cov = np.eye(n_features)*0.04 # std = 0.2 <=> var = std**2 = 0.04weights_avg = np.zeros(n_features)accuracy_avg = 0for _ in range(n_iter): # generate data X_neg = np.random.multivariate_normal(mean_neg, cov, N//2) X_pos = np.random.multivariate_normal(mean_pos, cov, N//2) # only values between 0 and 1 X_neg[X_neg < 0] = 0 X_neg[X_neg > 1] = 1 X_pos[X_pos < 0] = 0 X_pos[X_pos > 1] = 1 X = np.concatenate((X_neg, X_pos), axis=0) # training data y = np.ones(N) # labels y[:N//2] = -1 # first N/2 samples are from the negative class clf = svm.SVC(kernel='linear') clf.fit(X, y) # testing on training data, just to see if the SVM learns y_pred = clf.predict(X) accuracy_avg += np.sum(y_pred==y) / N weights_avg += clf.coef_[0] # coef_ has shape [n_class-1, n_features]accuracy_avg /= n_iterweights_avg /= n_iterprint("Average accuracy on training data: {}".format(accuracy_avg))print("Average linear SVM weights:")print(weights_avg)
Running this gave the following output
Average accuracy: 0.99999Average linear SVM weights:[-1.3517 -1.0827 -0.7347 0.0058 0.0237 -0.0354 0.7371 1.0735 1.3423]
And we see that the sign of the feature weight indeed is indicative of which class a feature contributes to, and that the magnitude also tells us how informative the feature is for the classification! |
Electrostatic Potential and Capacitance Combination of Capacitors and Energy Stored in a Capacitor When capacitors one connected in series all plates have same charge in magnitude. Potential differences \tt V_{1}:V_{2}:V_{3}=\frac{1}{C_{1}}:\frac{1}{C_{2}}:\frac{1}{C_{3}} Equivalent capacitance \tt \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} When ‘n’ identical capacitors one one connected in series \tt C_{s}=\frac{C}{n} Energy of capacitors \tt U_{1}:U_{2}:U_{3}=\frac{1}{C_{1}}:\frac{1}{C_{2}}:\frac{1}{C_{3}} Total energy of the combination U = U 1+ U 2+ U 3 In parallel combination of capacitors potential is same Total charge is Q = Q 1+ Q 2+ Q 3+ - - - - - The Ratio of charges in parallel combination Q 1: Q 2: Q 3= C 1: C 2: C 3 Equivalent capacitance in parallel C = C 1+ C 2+ C 3+ - - - - - “n” identical capacitors in parallel Cρ = nc. Energy of capacitors U 1: U 2: U 3= C 1: C 2: C 3 Total Energy of combination in parallel U = U 1+ U 2+ U 3 Two dielectrics K 1, K 2in parallel between capacitors then \tt C_{eff}=\frac{\varepsilon_{0}A}{d}\left(\frac{K_{1}+K_{2}}{2}\right) Two dielectrics K 1, K 2in series between capacitors the \tt C_{eff}=\frac{\varepsilon_{0}A}{d}\left(\frac{2K_{1}K_{2}}{K_{1}+K_{2}}\right) Two capacitors C 1C 2charged to V 1V 2potentials are connected in parallel then common parallel \tt V=\frac{Q_{1}+Q_{2}}{C_{1}+C_{2}}=\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}} Loss of Energy = \tt \frac{1}{2}\ \frac{C_{1}C_{2}}{C_{1}+C_{2}}\ \left(V_{1}-V_{2}\right)^2 Capacity of spherical capacitor \tt C=4\pi \varepsilon_{0}\ \frac{ab}{b-a}, (a, b = Radius) Capacity of cylindrical capacitor \tt C=\frac{2\pi\varepsilon_{0}l}{\log_{e}{\frac{b}{a}}}
Quantity When capacitor is fully charged with air between the two plates When the dielectric slab is introduced without the battery When the dielectric slab is introduced with the Battery Charge Q 0 Q 0 Q 0 Capacity C 0 KC 0 KC 0 P.D between the two plates V 0 \tt \frac{V_{0}}{K} V 0 Intensity of electric field E 0 \tt \frac{E_{0}}{K} E 0 Enegy stored U 0 \tt \frac{U_{0}}{K} KU 0 If "n" indentical charged liquid drops are combined to form bigdrop
Quantity For ecah charged small drop For the Brg drop Radius r R=n 1/3r Charge q Q = nq Capacity c C = n 1/3c Potential v V' = n 2/3V Energy u U' = n 5/3U Surface charge density σ σ' = n 1/3σ
View the Topic in this video From 01:36 To 18:00
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1. Work done in charging a capacitor gets stored in the capacitor in the form of its electric potential energy and it is given by
U = \frac{1}{2}CV^{2} = \frac{1}{2}QV = \frac{1}{2}\frac{Q^{2}}{C}
2. Energy Density
u = \frac{1}{2}\varepsilon_{0}E^{2} |
For the function of any given value, we have to determine the closest estimation value of a function and it is given by the Linear approximation Formula. The other name for this mathematical concept is tangent line approximation or approximate tangent value of a function.
This is the process to find the equation of a line or closest approximation of a given function for the given value. It is used in Trigonometry or optics. This is an infinitely small like where the curve is almost straight and imitates the function. For the real-valued function F (x), the linear approximation value is given as below –
\[\LARGE f(x) = f(x_{0}+f'(x_{0}(x-x_{0})+R_2\]
where R2 is the remainder term. The linear approximation, then, is given by
\[\LARGE f(x)\approx f(x_{0}+f'(x_{0}(x-x_{0})\]
Where,
f(x 0) is the value of f(x) at x = x 0. f'(x 0) is the derivative value of f(x) at x = x 0.
The approximation value is almost equal to the tangent line at a given point a?You could also opt for Euler’s method to find the solution for linear approximation values. In the end, only the closeness of the tangent line matters and tangent values can be calculated with specific formulas in mathematics. The other popular name for linear approximation is linearization.
So, how can you calculate the tangent at a given point a. There are two main things to remember here. First is the slope of a line and second is the particular point (a, b) line passes through. In this case, the equation of the line could be given as –
\[\LARGE Y – b = m(X-a)\]
Here, values of b and m will not be given automatically but you have to calculate it yourself only with formula derivatives.
Interpolation is a popular statistical tool in mathematics that is used to calculate the estimated values between two points. Here, we will discuss the formula for the concept. Interpolation is also used in science, businesses, or many other fields too. Every time when you have to predict values between data points, linear interpolation formula is helpful.
Take an example of Tomato plant planted by a curious gardener. He wanted to know how much the plant has grown on the third day? If it is 6mm tall on the third day then its height should increase in a linear pattern in coming days too. It means the growth chart should create a straight line when plotting data on the graph. Here, the linear Approximation formula comes in handy to solve the issue.The Linear Interpolation Formula in mathematics is given as below.
\[\large y=y_{1}+\frac{\left(x-x_{1}\right)\left(y-y_{1}\right)}{x_{2}-x_{1}}\]
Where,
x 1 and y 1 are the first coordinates x 2 and y 2 are the second coordinates x is the point to perform the interpolation y is the interpolated value.
Linear regression is the highly common and predictive analysis technique used by the mathematicians or scientists. In this case, there are two variables, one is taken as the explanatory variable, and the other is taken as the dependent variable. With the help of a linear regression model, you can always relate the weights of individuals with heights.
There are a variety of linear regression techniques available in the mathematical world. These are -simple linear regression. Multiple linear regression, Logistic regression, Ordinal regression, Multinomial regression, Discriminant analysis etc. The Formula for linear regression equation in mathematics is given by:
\[\large y=a+bx\]
a and b are given by the following formulas:
\[\large b\left(slope\right)=\frac{n\sum xy-\left(\sum x\right)\left(\sum y\right)}{n\sum x^{2}-\left(\sum x\right)^{2}}\]
\[\large a\left(intercept\right)=\frac{n\sum y-b\left(\sum x\right)}{n}\]
Where,
x and y are two variables on regression line. b = Slope of the line. a = y-intercept of the line. x = Values of first data set. y = Values of second data set. |
Inaccuracy of TAN near 75° in rad mode
06-30-2015, 03:18 PM (This post was last modified: 07-01-2015 02:36 AM by Marcio.)
Post: #1
Inaccuracy of TAN near 75° in rad mode
Hello,
It seems the 34s is having troubles getting the 10th digit correct when it evaluates the tangent of angles near 75 degrees in radians mode. I know this looks kinda silly but I thought it would be worth reporting.
\(TAN\left(73\frac{\pi}{180}\right) = 3.27085261848\leftarrow ...53\)
\(TAN\left(75\frac{\pi}{180}\right) = 3.73205080757\leftarrow ...48\)
\(TAN\left(77\frac{\pi}{180}\right) = 4.33147587428\leftarrow...17\)
The left arrow indicates what the last 2 digits should be.
Both sine and cosine are getting all digits right.
Thanks.
Marcio
CORRECTION: All answers given by the 34s in this thread are correct. The calculators I used to compare the results, those being the 50g, Prime and 35s, output inaccurate results for two main reasons, the first was the limited precision (12 digits) and the second was the numerical inaccuracy of tan as the angles approache 90 degrees. Read on for details!
06-30-2015, 06:22 PM (This post was last modified: 06-30-2015 06:31 PM by Gerald H.)
Post: #2
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
Confirmed for 73° for 3.3 3718.
A truly amazing result for tan(73*pi/180) as the HP 95LX, 100LX & 200LX give the same wrong answer, also in the following decimal digits!
ie ...848... in fact seems correct.
WP 34S exonerated!
06-30-2015, 06:34 PM
Post: #3
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
What is your gold standard, Marcio?
06-30-2015, 06:59 PM
Post: #4
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
OK, have now tested 32SII, 42S, 50G - all return the 853 answer.
The 3 organizers, TI-86, voyager 200, TI-30X Pro - all say 848, which is definitely the correct answer - not to mention the correct WP 34S!
06-30-2015, 07:03 PM
Post: #5
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
06-30-2015, 07:04 PM
Post: #6
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
My gold standard is the HP LX organiser series - I don't believe a false result has been found & they've been around donkey's years.
If you can get one I recommend a 200LX.
06-30-2015, 07:15 PM (This post was last modified: 06-30-2015 07:22 PM by Marcio.)
Post: #7
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
Well, yes. I should have checked Wolfram Alpha before opening this thread.
tan (73)
A bit surprised both the Prime and 50g gave incorrect answers.
OK then, if it is OK with you, I am gonna ask the moderators to delete all these which have been nothing but a futile exercise.
06-30-2015, 07:18 PM
Post: #8
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
The 853 answer is delivered only by the HP calculators using legacy software.
Have a look at how the HP models I mention cluster in this list:
http://www.rskey.org/~mwsebastian/miscprj/results.htm
as do the organisers, but in a different position.
06-30-2015, 07:20 PM (This post was last modified: 06-30-2015 07:21 PM by Gerald H.)
Post: #9
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
You certainly should not have consulted Wolfram Alpha - the thread is very interesting & useful.
I congratulate you on finding this error & on making it public.
The thread is most definitely useful.
06-30-2015, 07:26 PM
Post: #10
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
You shouldn't delete the thread - others should know Prime & 50G etc give SAME wrong answer!
06-30-2015, 07:33 PM (This post was last modified: 06-30-2015 07:35 PM by Marcio.)
Post: #11
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
06-30-2015, 07:37 PM (This post was last modified: 06-30-2015 07:57 PM by Gerald H.)
Post: #12
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
Not only Hosoda.
There's a library for the 50G called Long Float that delivers hundreds of decimal places & makes the same error!!!
Horror!!!
Edit: But read on, Hosoda & Long Float & HP will be justified!
06-30-2015, 07:39 PM
Post: #13
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
(06-30-2015 07:15 PM)Marcio Wrote: Well, yes. I should have checked Wolfram Alpha before opening this thread.
My '34S (in double-precision mode) differs from Wolfram Alpha - at least on tan (73*pi/180) - only starting in the 32nd decimal place:
WP-34S:3.270852618484140865308856257305407
.... W/A: 3.2708526184841408653088562573054107771059....
Jake
06-30-2015, 07:41 PM
Post: #14
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
My previous posting contradicts my posting #8 ("only").
Perhaps a hand calculation is the only way to settle the matter?
For me, Aribas returned 848..... so I'm pretty sure it's correct.
Maybe someone else can confirm the result?
06-30-2015, 07:41 PM (This post was last modified: 06-30-2015 07:56 PM by Dieter.)
Post: #15
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
(06-30-2015 03:18 PM)Marcio Wrote: It seems the 34s is having troubles getting the 10th dec number correct when it evaluates the tangent of angles near 75 degrees in radians mode. I know this looks kinda silly but I thought it would be worth reporting.
1. The 34s is fine.
A bit of simple calculus explains it all.
2. tan(75°) = 3,7320 50807 56887 72935 27446 34150 58723 ...
And that's exactly what the 34s returns.
3. 75° = 1,3089 96938 99574 71826 92768 07636 64595 ... rad
And that's exactly what the 34s returns.
However, on a simple 12-digit calculator, 75° is rounded to 1,3089 96939 00 rad. Due to roundoff in the last digit, the value in radians may be off by 5 units in the 13th digit. Since the derivative of tan x at this point is approx. 15, this translates to a potential error of 7 units in the last (12th) digit:
The exact argument (75*pi/180) is about halfway between 1,3089 96938 99 and ...900.
If it happens to get rounded down, you get tan(1,3089 96938 99) = 3,73205080748
If it happens to get (correctly) rounded up, you get tan(1,3089 96939 00) = 3,73205080763
You see that a change in 1 ULP of the argument makes the tanget change by 15 ULP (!), just as expected.
Since the argument always has a potential roundoff error of 0,5 ULP, it may and will be up to 7 ULP off.
Now try this with 89,9° and the last four (!) digits will be off.
Try 90° and the result becomes completely useless. ;-)
So the problem is the limited accuracy of the manually calculated argument. 75 degress is not exactly 1,30899693900 radians, so the tangent is inaccurate. On the other hand the 34s with its internal 39-digit precision gets it right. And so does the 35s or the 50G if you calculate tan(75°) in degrees mode: the internal calculations are done with three additional digits so the error does not show up. However, manually converting 75° to radians introduces a very slight error that is large enough to throw the tangent off in the last one or two digits. And that's exactly what you see here. The 34s returns a correct 12-digit result because even in SP mode it uses 16 digit precision and the error only affects the (not displayed) 15th digit.
Dieter
06-30-2015, 07:49 PM
Post: #16
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
OK, thanks for the detailed explanation.
I should also point out that the problem only rises (more significantly) when the angle is between 73 and 77 degrees. There must be a reason for that.
06-30-2015, 08:06 PM (This post was last modified: 06-30-2015 08:08 PM by Dieter.)
Post: #17
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
(06-30-2015 03:18 PM)Marcio Wrote: CORRECTION: All answers given by the 34s in this thread are correct. The answers given by the calculatorss I used to compare the results, those being the 50g, Prime and 35s, are not!
Right, the 34s results are correct.
But the 50g and 35s are fine as well. Their results are also correct.
The error is in the argument manually calculated by the user. It is rounded to 12 digits, and so these calculators return the exact result for this rounded argument.
The 34s would show the same error, but it uses at least 16 digits of which only 12 are displayed.
Dieter
06-30-2015, 08:11 PM (This post was last modified: 06-30-2015 08:12 PM by Marcio.)
Post: #18
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
(06-30-2015 08:06 PM)Dieter Wrote: Right, the 34s results are correct.
In this particular case, the 73 degrees, say it was rounded to 9 digits instead?
06-30-2015, 08:28 PM
Post: #19
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
(06-30-2015 07:49 PM)Marcio Wrote: I should also point out that the problem only rises (more significantly) when the angle is between 73 and 77 degrees. There must be a reason for that.
No, there is nothing special about 73, 75 or 77 degrees. The error gets larger as the angle approaches 90° and, here more important, if the angle in radians happens to round not very well to 12 digits.
Example:
tan 75,2° = 3,7848 48088 37
75,2° = 1,3124 87597 49973... radians, which rounds nicely to 12 or even 11 digits.
tan(1,3124 87597 50) = 3,7848 48088 37
q.e.d.
Dieter
06-30-2015, 08:45 PM (This post was last modified: 06-30-2015 08:53 PM by Dieter.)
Post: #20
RE: [WP 34s] Innacuracy of TAN near 75° in rad mode
(06-30-2015 08:11 PM)Marcio Wrote: In this particular case, the 73 degrees, say it was rounded to 9 digits instead?
?!? - sorry, I don't understand what you mean here. But the 73° case shows the roundoff problem very nicely:
73° = 1,27409035395586... rad
So the 12-digit result is almost halfway between 1,27409035395 and ...96.
tan(1,27409035395) = 3,27085261842
tan(1,27409035396) = 3,27085261853
The latter is what you get on a 12-digit calculator where 73° "equals" 1,27409035396.
The calculator should return tan(1,27409035395586...) = 3,27085261848
But all it can calculate is tan(1,27409035395) = 3,27085261842 or tan(1,27409035396) = 3,27085261853
The error in the result is the error in the argument (4,14E–12) times the tan derivative at this point (approx. 11,7), which yields 4,8E–11 or 5 units in the 12th digit – as you can see here: ...1848 vs. ...1853. Changing the argument by the smallest possible delta (1 ULP) makes the tangent change by 11–12 ULP (cf. ...1842 vs. ...1853).
Dieter
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Tagged: symmetric matrix Problem 572
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold). The time limit was 55 minutes. Problem 7. Let $A=\begin{bmatrix} -3 & -4\\ 8& 9 \end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix} -1 \\ 2 \end{bmatrix}$. (a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$. (b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$. Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular. Problem 9. Determine whether each of the following sentences is true or false. (a) There is a $3\times 3$ homogeneous system that has exactly three solutions. (b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric. (c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$. (d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
Add to solve later
(e) The vectors \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\] are linearly independent. Problem 564
Let $A$ and $B$ be $n\times n$ skew-symmetric matrices. Namely $A^{\trans}=-A$ and $B^{\trans}=-B$.
(a) Prove that $A+B$ is skew-symmetric. (b) Prove that $cA$ is skew-symmetric for any scalar $c$. (c) Let $P$ be an $m\times n$ matrix. Prove that $P^{\trans}AP$ is skew-symmetric. (d) Suppose that $A$ is real skew-symmetric. Prove that $iA$ is an Hermitian matrix. (e) Prove that if $AB=-BA$, then $AB$ is a skew-symmetric matrix. (f) Let $\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\mathbf{v}^{\trans}A\mathbf{v}=0$.
Add to solve later
(g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\mathbf{v}=\mathbf{0}$ for some vector $\mathbf{v}\in \R^n$. Then prove that $A\mathbf{v}=\mathbf{0}$. Problem 556
Let $\mathbf{v}$ be a nonzero vector in $\R^n$.
Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$. Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by \[A=I-a\mathbf{v}\mathbf{v}^{\trans},\] where $I$ is the $n\times n$ identity matrix.
Prove that $A$ is a symmetric matrix and $AA=I$.
Conclude that the inverse matrix is $A^{-1}=A$. Problem 538 (a) Suppose that $A$ is an $n\times n$ real symmetric positive definite matrix. Prove that \[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}\] defines an inner product on the vector space $\R^n$. (b) Let $A$ be an $n\times n$ real matrix. Suppose that \[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}\] defines an inner product on the vector space $\R^n$.
Prove that $A$ is symmetric and positive definite.Add to solve later
Problem 457
Let $A$ be a real symmetric $n\times n$ matrix with $0$ as a simple eigenvalue (that is, the algebraic multiplicity of the eigenvalue $0$ is $1$), and let us fix a vector $\mathbf{v}\in \R^n$.
(a) Prove that for sufficiently small positive real $\epsilon$, the equation \[A\mathbf{x}+\epsilon\mathbf{x}=\mathbf{v}\] has a unique solution $\mathbf{x}=\mathbf{x}(\epsilon) \in \R^n$. (b) Evaluate \[\lim_{\epsilon \to 0^+} \epsilon \mathbf{x}(\epsilon)\] in terms of $\mathbf{v}$, the eigenvectors of $A$, and the inner product $\langle\, ,\,\rangle$ on $\R^n$. ( University of California, Berkeley, Linear Algebra Qualifying Exam) Problem 396
A real symmetric $n \times n$ matrix $A$ is called
positive definite if \[\mathbf{x}^{\trans}A\mathbf{x}>0\] for all nonzero vectors $\mathbf{x}$ in $\R^n$. (a) Prove that the eigenvalues of a real symmetric positive-definite matrix $A$ are all positive.
Add to solve later
(b) Prove that if eigenvalues of a real symmetric matrix $A$ are all positive, then $A$ is positive-definite. Problem 385
Let
\[A=\begin{bmatrix} 2 & -1 & -1 \\ -1 &2 &-1 \\ -1 & -1 & 2 \end{bmatrix}.\] Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$. That is, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$. |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
Which countable discrete groups (apart from the infinite amenable ones) admit uncountably many mutually non-conjugate free ergodic probability measure preserving actions that are all mutually orbit equivalent?
We consider infinite countable groups and ask about the
this property: admitting uncountably many mutually non-isomorphis free ergodic probability measure preserving (pmp) actions that are all mutually orbit equivalent.
You ask which groups satisfy
this property. I do not know exactly. I think no one knows. But it is a big class of groups, maybe all groups which do not satisfy property (T).
In Corollary 2.6 of
A converse to Dye’s theorem Hjorth proves that property (T) groups do not satisfy this property.
On the other hand, amenable groups do satisfy
this property. Indeed, any amenable group has uncountably many mutually non-isomorphic pmp actions (say, by entropy theory), but these are all orbit equivalent, by the celebrated Ornstein-Weiss theorem.
To see that there are many non-amenable groups satisfying
this property, fix a pmp action of a non-amenable group $\Gamma_1$ on $X$ and uncountably many pmp actions of an amenable group $\Gamma_2$ on $Y_\alpha$. Then the non-amenable group $\Gamma=\Gamma_1\times \Gamma_2$ will satisfy this property, as witness by its actions on $X\times Y_\alpha$.
If you want to study
this property further, I suggest you to start with the following two things: try to show that if a group has a quotient that satisfies this property then the group satisfies this property, and try to show that this property is a measure equivalence invariant. |
I am looking at a spin 1/2 particle in a magnetic field. This has Hamiltonian $$H=-\mu s\cdot B_0$$ For simplicity, assume $B_0=B_0\hat z$ so $H=-\mu B_0$. I then apply a perturbative magnetic field such that $$V'=-\mu B_1 s_x$$ First I wanted to compute $E^{(1)}$ $$E^{(1)}_n=\langle\psi_n^{(0)}|-\mu B_1s_x|\psi_n^{(0)}\rangle=\mp \mu B_1 \hbar/2$$ Now I am looking to find the first order correction to the ground state wavefunction. I know that this is given as $$\psi^{(1)}_n=\sum_{n\neq n'} \psi^{(0)}_{n'}\frac{\langle\psi_{n'}^{(0)}|-\mu B_1s_x|\psi_{n}^{(0)}\rangle}{E_n^{(0)}-E_{n'}^{(0)}}$$ I am confused as to how to treat the summation. The only term I would get is if $n=n'$, but that would be degerate. So I am thinking that this first order correction is 0. Is this correct?
Spin1/2 particle
Ususally, in this kind of Hamiltonian, people uses $s=s_z$, where
$$s=s_z=\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right].$$
Then, your unperturbed hamiltonian $H_0$ is: $$H_0=-\mu s\cdot B_0 = -\mu \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right]B_{0,z}. $$
Then the eigen vectors of energy are:
$$|\psi^0_+\rangle=\left[ \begin{array}{c} 1 \\ 0\end{array} \right],$$ $$|\psi^0_-\rangle=\left[ \begin{array}{c} 0\\ 1 \end{array} \right].$$
Perturbation solution
Then you want to compute $|\psi_+\rangle$ and $|\psi_-\rangle$ for the perturbed Hamiltonian $H=H_0-\mu B_1 s_x$, where $$s_x=\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array} \right].$$
As you said, you have to compute the following quantities (note I use $+,-$ instead of $n=0,1$. Which became:
$$\psi^{(1)}_+=\sum_{n\neq +} \psi^{(0)}_{n'}\frac{\langle\psi_{n'}^{(0)}|-\mu B_1s_x|\psi_{+}^{(0)}\rangle}{E_+^{(0)}-E_{n'}^{(0)}}=\psi^{(0)}_{-}\frac{\langle\psi_{-}^{(0)}|-\mu B_1s_x|\psi_{+}^{(0)}\rangle}{E_+^{(0)}-E_{-}^{(0)}}$$
$$\psi^{(1)}_-=\sum_{n\neq -} \psi^{(0)}_{n'}\frac{\langle\psi_{n'}^{(0)}|-\mu B_1s_x|\psi_{-}^{(0)}\rangle}{E_-^{(0)}-E_{n'}^{(0)}}=\psi^{(0)}_{+}\frac{\langle\psi_{+}^{(0)}|-\mu B_1s_x|\psi_{-}^{(0)}\rangle}{E_-^{(0)}-E_{+}^{(0)}}$$
Put here vectors and matrices we just found and let me know if you get zero. |
Wave energy converters in coastal structures: verschil tussen versies
(→Application for wave energy converters)
(→Application for wave energy converters)
Regel 66: Regel 66:
<p>
<p>
−
For
+
For waves in
+ + + +
<br><div style="text-align: center;">
<div style="float: right">(4)</div>
<div style="float: right">(4)</div>
<math>P_{w1} = \frac{\rho g^2}{64 \pi} H_{m0}^2 T_e</math>
<math>P_{w1} = \frac{\rho g^2}{64 \pi} H_{m0}^2 T_e</math>
Versie van 3 sep 2012 om 10:23 Introduction
Fig 1: Construction of a coastal structure.
Coastal works along European coasts are composed of very diverse structures. Many coastal structures are ageing and facing problems of stability, sustainability and erosion. Moreover climate change and especially sea level rise represent a new danger for them. Coastal dykes in Europe will indeed be exposed to waves with heights that are greater than the dykes were designed to withstand, in particular all the structures built in shallow water where the depth imposes the maximal amplitude because of wave breaking.
This necessary adaptation will be costly but will provide an opportunity to integrate converters of sustainable energy in the new maritime structures along the coasts and in particular in harbours. This initiative will contribute to the reduction of the greenhouse effect. Produced energy can be directly used for the energy consumption in harbour area and will reduce the carbon footprint of harbours by feeding the docked ships with green energy. Nowadays these ships use their motors to produce electricity power on board even if they are docked. Integration of wave energy converters (WEC) in coastal structures will favour the emergence of the new concept of future harbours with zero emissions.
Inhoud Wave energy and wave energy flux
For regular water waves, the time-mean wave energy density E per unit horizontal area on the water surface (J/m²) is the sum of kinetic and potential energy density per unit horizontal area. The potential energy density is equal to the kinetic energy
[1] both contributing half to the time-mean wave energy density E that is proportional to the wave height squared according to linear wave theory [1]:
(1)
[math]E= \frac{1}{8} \rho g H^2[/math]
g is the gravity and [math]H[/math] the wave height of regular water waves. As the waves propagate, their energy is transported. The energy transport velocity is the group velocity. As a result, the time-mean wave energy flux per unit crest length (W/m) perpendicular to the wave propagation direction, is equal to
[1]:
(2)
[math] P= Ec_{g}[/math]
with [math]c_{g}[/math] the group velocity (m/s). Due to the dispersion relation for water waves under the action of gravity, the group velocity depends on the wavelength λ (m), or equivalently, on the wave period T (s). Further, the dispersion relation is a function of the water depth h (m). As a result, the group velocity behaves differently in the limits of deep and shallow water, and at intermediate depths:
[math](\frac{\lambda}{20} \lt h \lt \frac{\lambda}{2})[/math]
Application for wave energy convertersFor regular waves in deep water:
[math]c_{g} = \frac{gT}{4\pi} [/math] and [math]P_{w1} = \frac{\rho g^2}{32 \pi} H^2 T[/math]
The time-mean wave energy flux per unit crest length is used as one of the main criteria to choose a site for wave energy converters.
For real seas, whose waves are random in height, period (and direction), the spectral parameters have to be used like the spectral estimate of significant wave height [math]H_{m0} [/math] based on zero-order moment of the spectral function [math]H_{m0} = 4 \sqrt{m_0} [/math]
[math]P_{w1} = \frac{\rho g^2}{64 \pi} H_{m0}^2 T_e[/math]
If local data are available ([math]H_{m0}^2 [/math], T) for a sea state through in-situ wave buoys for example, satellite data or numerical modelling, the last equation giving wave energy flux [math]P_{w1}[/math] gives a first estimation. Averaged over a season or a year, it represents the maximal energetic resource that can be theoretically extracted from wave energy. If the directional spectrum of sea state variance F (f,[math]\theta[/math]) is known with f the wave frequency (Hz) and [math]\theta[/math] the wave direction (rad), a more accurate formulation is used:
[math]P_{w2} = \rho g\int\int c_{g}(f,h)F(f,\theta) dfd \theta[/math]
Fig 2: Time-mean wave energy flux along
West European coasts
[2] .
It can be shown easily that equation (4) can be reduced to (3) with the hypothesis of regular waves in deep water. The directional spectrum is deduced from directional wave buoys, SAR images or advanced spectral wind-wave models, known as third-generation models, such as WAM, WAVEWATCH III, TOMAWAC or SWAN. These models solve the spectral action balance equation without any a priori restrictions on the spectrum for the evolution of wave growth.
From TOMAWAC model, the near shore wave atlas ANEMOC along the coasts of Europe and France based on the numerical modelling of wave climate over 25 years has been produced
[3]. Using equation (4), the time-mean wave energy flux along West European coasts is obtained (see Fig. 2). This equation (4) still presents some limits like the definition of the bounds of the integration. Moreover, the objective to get data on the wave energy near coastal structures in shallow or intermediate water requires the use of numerical models that are able to represent the physical processes of wave propagation like the refraction, shoaling, dissipation by bottom friction or by wave breaking, interactions with tides and diffraction by islands.
The wave energy flux is therefore calculated usually for water depth superior to 20 m. This maximal energetic resource calculated in deep water will be limited in the coastal zone:
at low tide by wave breaking; at high tide in storm event when the wave height exceeds the maximal operating conditions; by screen effect due to the presence of capes, spits, reefs, islands,...
Technologies
According to the International Energy Agency (IEA), more than hundred systems of wave energy conversion are in development in the world. Among them, many can be integrated in coastal structures. Evaluations based on objective criteria are necessary in order to sort theses systems and to determine the most promising solutions.
Criteria are in particular:
the converter efficiency : the aim is to estimate the energy produced by the converter. The efficiency gives an estimate of the number of kWh that is produced by the machine but not the cost. the converter survivability : the capacity of the converter to survive in extreme conditions. The survivability gives an estimate of the cost considering that the weaker are the extreme efforts in comparison with the mean effort, the smaller is the cost.
Unfortunately, few data are available in literature. In order to determine the characteristics of the different wave energy technologies, it is necessary to class them first in four main families
[2].
An interesting result is that the maximum average wave power that a point absorber can absorb [math]P_{abs} [/math](W) from the waves does not depend on its dimensions
[4]. It is theoretically possible to absorb a lot of energy with only a small buoy. It can be shown that for a body with a vertical axis of symmetry (but otherwise arbitrary geometry) oscillating in heave the capture (or absorption) width [math]L_{max}[/math](m) is as follows [4]:
[math]L_{max} = \frac{P_{abs}}{P_{w}} = \frac{\lambda}{2\pi}[/math] or [math]1 = \frac{P_{abs}}{P_{w}} \frac{2\pi}{\lambda}[/math]
Fig 4: Upper limit of mean wave power
absorption for a heaving point absorber.
where [math]{P_{w}}[/math] is the wave energy flux per unit crest length (W/m). An optimally damped buoy responds however efficiently to a relatively narrow band of wave periods.
Babarit et Hals propose
[5] to derive that upper limit for the mean annual power in irregular waves at some typical locations where one could be interested in putting some wave energy devices. The mean annual power absorption tends to increase linearly with the wave power resource. Overall, one can say that for a typical site whose resource is between 20-30 kW/m, the upper limit of mean wave power absorption is about 1 MW for a heaving WEC with a capture width between 30-50 m.
In order to complete these theoretical results and to describe the efficiency of the WEC in practical situations, the capture width ratio [math]\eta[/math] is also usually introduced. It is defined as the ratio between the absorbed power and the available wave power resource per meter of wave front times a relevant dimension B [m].
[math]\eta = \frac{P_{abs}}{P_{w}B} [/math]
The choice of the dimension B will depend on the working principle of the WEC. Most of the time, it should be chosen as the width of the device, but in some cases another dimension is more relevant. Estimations of this ratio [math]\eta[/math] are given
[5]: 33 % for OWC, 13 % for overtopping devices, 9-29 % for heaving buoys, 20-41 % for pitching devices. For energy converted to electricity, one must take into account moreover the energy losses in other components of the system.
Civil engineering
Never forget that the energy conversion is only a secondary function for the coastal structure. The primary function of the coastal structure is still protection. It is necessary to verify whether integration of WEC modifies performance criteria of overtopping and stability and to assess the consequences for the construction cost.
Integration of WEC in coastal structures will always be easier for a new structure than for an existing one. In the latter case, it requires some knowledge on the existing coastal structures. Solutions differ according to sea state but also to type of structures (rubble mound breakwater, caisson breakwaters with typically vertical sides). Some types of WEC are more appropriate with some types of coastal structures.
Fig 5: Several OWC (Oscillating water column) configurations (by Wavegen – Voith Hydro).
Environmental impact
Wave absorption if it is significant will change hydrodynamics along the structure. If there is mobile bottom in front of the structure, a sand deposit can occur. Ecosystems can also be altered by change of hydrodynamics and but acoustic noise generated by the machines.
Fig 6: Finistere area and locations of
the six sites (google map).
Study case: Finistere area
Finistere area is an interesting study case because it is located in the far west of Brittany peninsula and receives in consequence the largest wave energy flux along the French coasts (see Fig.2). This area with a very ragged coast gathers moreover many commercial ports, fishing ports, yachting ports. The area produces a weak part of its consumption and is located far from electricity power plants. There are therefore needs for renewable energies that are produced locally. This issue is important in particular in islands. The production of electricity by wave energy will have seasonal variations. Wave energy flux is indeed larger in winter than in summer. The consumption has peaks in winter due to heating of buildings but the consumption in summer is also strong due to the arrival of tourists.
Six sites are selected (see figure 7) for a preliminary study of wave energy flux and capacity of integration of wave energy converters. The wave energy flux is expected to be in the range of 1 – 10 kW/m. The length of each breakwater exceeds 200 meters. The wave power along each structure is therefore estimated between 200 kW and 2 MW. Note that there exist much longer coastal structures like for example Cherbourg (France) with a length of 6 kilometres.
(1) Roscoff (300 meters) (2) Molène (200 meters) (3) Le Conquet (200 meters) (4) Esquibien (300 meters) (5) Saint-Guénolé (200 meters) (6) Lesconil (200 meters) Fig.7: Finistere area, the six coastal structures and their length (google map).
Wave power flux along the structure depends on local parameters: bottom depth that fronts the structure toe, the presence of caps, the direction of waves and the orientation of the coastal structure. See figure 8 for the statistics of wave directions measured by a wave buoy located at the Pierres Noires Lighthouse. These measurements show that structures well-oriented to West waves should be chosen in priority. Peaks of consumption occur often with low temperatures in winter coming with winds from East- North-East directions. Structures well-oriented to East waves could therefore be also interesting even if the mean production is weak.
Fig 8: Wave measurements at the Pierres Noires Lighthouse.
Conclusion
Wave energy converters (WEC) in coastal structures can be considered as a land renewable energy. The expected energy can be compared with the energy of land wind farms but not with offshore wind farms whose number and power are much larger. As a land system, the maintenance will be easy. Except the energy production, the advantages of such systems are :
a “zero emission” port industrial tourism test of WEC for future offshore installations.
Acknowledgement
This work is in progress in the frame of the national project EMACOP funded by the French Ministry of Ecology, Sustainable Development and Energy.
See also Waves Wave transformation Groynes Seawall Seawalls and revetments Coastal defense techniques Wave energy converters Shore protection, coast protection and sea defence methods Overtopping resistant dikes
References Mei C.C. (1989) The applied dynamics of ocean surface waves. Advanced series on ocean engineering. World Scientific Publishing Ltd Mattarolo G., Benoit M., Lafon F. (2009), Wave energy resource off the French coasts: the ANEMOC database applied to the energy yield evaluation of Wave Energy, 10th European Wave and Tidal Energy Conference Series (EWTEC’2009), Uppsala (Sweden) Benoit M. and Lafon F. (2004) : A nearshore wave atlas along the coasts of France based on the numerical modeling of wave climate over 25 years, 29th International Conference on Coastal Engineering (ICCE’2004), Lisbonne (Portugal), 714-726. De O. Falcão A. F. (2010) Wave energy utilization: A review of the technologies. Renewable and Sustainable Energy Reviews, Volume 14, Issue 3, April 2010, Pages 899–918. Babarit A. and Hals J. (2011) On the maximum and actual capture width ratio of wave energy converters – 11th European Wave and Tidal Energy Conference Series (EWTEC’2011) – Southampton (U-K). |
Of course, you must use the fact that for any point, there is a sequence of rationals and another sequence of irrationals which converge to that point.
Here is the conventional limit definition :
Given $f : D \to \mathbb R$ and $x \in D$, suppose there is a number $L$ such that for every $\epsilon > 0$ there is a $\delta > 0$ such that if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$. Then, $\lim_{t \to x}f(t)$ is said to exist and equal $L$.
Using a usual trick, we bring this down to convergence of sequences, and then get a sequential definition :
Given $f : D \to \mathbb R$ and $x \in D$, if there exists a number $L$ such that for every sequence $x_n$ converging to $x$, the sequence $f(x_n)$ converges to $L$. Then $\lim_{t \to x}f(t)$ is said to exist and equal $L$.
Now, if the limit
does not exist, then we must negate the above statement:
Given $f : D \to \mathbb R$ and $x \in D$, if
for all numbers $L$, there exists a sequence $x_n$ converging to $x$ such that $f(x_n)$ does not converge to $L$, then the limit $\lim_{x \to a} f(x)$ is said to not exist.
Now, we are in shape to attack the problem : all we need to do, is given any $a \neq 3$ and a candidate $L$ for the limit, find a sequence of points which converge to $a$ such that the function values don't go to $L$. To do this, we use the piecewise definition of our function.
Let $a \neq 3$. Suppose that $L$ is a candidate limit.
EDITED : At this stage, we want a sequence of numbers which does converge to $a$, but whose function values don't converge to $L$.
The answer to your question below is this : The pieces of $f$,
when treated as functions from the real line to itself, are continuous, and this fact can be used to prove that $f$ is not continuous at any point, other than $3$.
First, let $p_n$ is a sequence of rationals converging to $a$. We claim that $f(p_n) \to a-1$.
This follows from the fact that $p_n$ lies inside one piece of $f$, so we may use continuity of the function which defines $f$ on that piece. But if you want to argue by basics, then : $f(p_n) = p_n - 1$ from the fact that $p_n$ is rational. Now, since $p_n \to a$ and (the constant sequence)$-1 \to -1$, we can add limits to get $p_n -1 \to a-1$, and therefore $f(p_n) \to a-1$.
By uniqueness of limits, if $L \neq a-1$, then $p_n$ serves as a candidate for the sequence of values not converging to $L$, since $p_n \to a$ but $f(p_n) \not\to L$.
On the other hand, if $q_n$ is a sequence of irrationals converging to $a$, then we may repeat the above argument for the irrational piece to get $f(q_n) \to 5-a$. So $q_n$ now serves as a candidate for the sequence of values not converging to $L$, whenever $5-a \neq L$.
Now, if $a-1 \neq 5-a$,
which happens precisely when $a \neq 3$, we see that for any $L$, of course $L$ can't be equal to both $a-1$ and $5-a$, so take the case which it does not equal, and that sequence works.
Hence, the limit of the given function does not exist, for any $a \neq 3$. Of course, if $f$ has to be continuous at $a$, then $\lim_{x\to a} f(x)$ has to
exist and equal $f(a)$, but the limit doesn't exist if $a \neq 3$, so $f$ is not continuous at any such point. |
Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!! |
also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling
@heather well, there's a spectrum
so, there's things like New Journal of Physics and Physical Review X
which are the open-access branch of existing academic-society publishers
As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di...
Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago
> A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service”
for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty
> for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing)
@0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals.
@BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work...
@BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions.
Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley.
I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea.
@EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results...
Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town...
@EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit. |
Double side band suppressed carrier.
In standard AM modulation, two side bands and carrier are transmitted, this type of modulation is called DSB-FC but since carrier does not any information, it is wastage of power to transmit it, if we remove carrier from DSB-FC, we get DSB-FC.
Generation of DSB-SC :
To suppress carrier, we have to multiply modulating signal with carrier signal.
Therefore, equation of DSB-SC is
$V_c \ (t) = V_m \ V_c \ sin w_{mt}, \ sin \ w_c \ t$
The device for achieving product of modulation and carrier signal is product modulator.
DSB-FC using balanced modulator :
It consist of two AM modulator arranged in a balanced configuration, so as to suppress the carrier, hence the name balanced modulator.
In the above circuit, the JFFT $Q_1$ and $Q_2$ operated in non linear region of its characteristics, the RF carrier signal is applied to the gates of both JFET in phase, due to center tap transformer at input the 180 degree act of phase AF signal is applied to both gates JFET, input applied to $Q_1$ $V_1 + V_2$ however, that of $Q_2$ $V_1 - V_2$. the modulated o/p current of the FET's are combined by the center tap, primary grinding of the output transformer.
The current $I d_1$ and $I d_2$ of $Q_1$ and $Q_2$ are flowing in opposite direction in the primary winding of output transform, hence the resultant primary current is subtraction of $I d_1$ and $I d_2$, if the two FET's are perfectly balanced then the carrier frequency signal will be completely suppressed, the o/p voltage of output transformer proportioned to primary current and it contain two side band and unwind component which can be removed by tuning of output transformer. thus output balanced modulator contain only two side bands.
modulating signal $m(t) = v_m \ (os / 2 \pi f_m \ t)$, carrier signal = $v_c \ cos \ (2 \pi f_{c t})$
Equation g OSBSC wave :
s(+) = m (+) c (+)
$s \ (+) = v_m \ v_c \ cos \ (2 \pi f_m +) \ cos \ (2 \pi f_c +)$
$= \frac{vm.vc}{2} \ cos \ [2 \pi (f_c + f_m ) + ] + \frac{vm.vc}{2} [ 2 \ \pi (f_c f_m) +) $
DSBSC modulated wave $F_LSB \rightarrow f_c - f_m , f_{VSB} \rightarrow f_2 + f_m$
Power calculation $(P_t) = P_{USB} + P_{USB} = \frac{vm^2 vc^2}{4R}$ |
On the existence of weak solutions of an unsteady
p-Laplace thermistor system with strictly monotone electrical conductivities
Department of Mathematics, Humboldt University Berlin, Unter den Linden 6,10099 Berlin, Germany
$\Omega\subset\mathbb{R}^n$
$n=2$
$n=3$
$\text{(1)}\quad \nabla\cdot \boldsymbol{J}=0,\qquad \text{(2)}\quad \frac{\partial u}{\partial t}+\nabla\cdot\boldsymbol{q}=f(x,t,u,\nabla\varphi)\;\text{ in }\; \Omega\times\,]\,0,T\,[\,,$
$p$
$\varphi$
$u=$
$\varphi=$
$(\varphi,u)$
$\varphi$
$u$ Mathematics Subject Classification:35K92, 35Q79, 80A20. Citation:Joachim Naumann. On the existence of weak solutions of an unsteady p-Laplace thermistor system with strictly monotone electrical conductivities. Discrete & Continuous Dynamical Systems - S, 2017, 10 (4) : 837-852. doi: 10.3934/dcdss.2017042
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Almost everywhere convergence of the gradients of solutions to elliptic and parabolic equations,
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L. Boccardo and F. Murat,
Almost everywhere convergence of the gradients of solutions to elliptic and parabolic equations,
[3] [4]
N. Bourbaki,
[5]
H. Brezis,
[6] [7]
M. Bulíček, A. Glitzky and M. Liero, Systems describing electrothermal effects with
[8] [9] [10] [11] [12] [13]
A. Fischer, P. Pahner, B. Lüssem, K. Leo, R. Scholz, T. Koprucki, J. Fuhrmann, K. Gärtner and A. Glitzky,
Self-heating, bistability,
[14]
A. Glitzky and M. Liero, Analysis of
[15] [16] [17] [18]
M. Liero, T. Koprucki, A. Fischer, R. Scholz and A. Glitzky,
[19]
P. Lindqvist,
[20]
J. -L. Lions,
[21] [22]
J. Naumann,
On the existence of weak solutions to the equations of non-stationary motion of heat-conducting incompressible viscous fluids,
[23] [24] [25]
M. P. Shaw, V. V. Mitin, E. Schöll and H. L. Gubin,
The Physics of Instabilities in Solid State Electron Devices, Plenum Press, New York, 1992.
doi: 10.1007/978-1-4899-2344-8.
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E. Zeidler,
[1]
Genni Fragnelli, Dimitri Mugnai, Nikolaos S. Papageorgiou.
Robin problems for the
[2]
Wen Tan.
The regularity of pullback attractor for a non-autonomous
[3]
Everaldo S. de Medeiros, Jianfu Yang.
Asymptotic behavior of solutions to a perturbed p-Laplacian problem with Neumann condition.
[4]
Nikolaos S. Papageorgiou, Vicenţiu D. Rǎdulescu, Dušan D. Repovš.
Nodal solutions for the Robin
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VicenŢiu D. RǍdulescu, Somayeh Saiedinezhad.
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Zuodong Yang, Jing Mo, Subei Li.
Positive solutions of $p$-Laplacian equations with
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Leszek Gasiński.
Positive solutions for resonant boundary value problems with the scalar p-Laplacian and nonsmooth potential.
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A diffusion problem of Kirchhoff type involving the nonlocal fractional
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Shanming Ji, Yutian Li, Rui Huang, Xuejing Yin.
Singular periodic solutions for the
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On positive solutions for classes of p-Laplacian semipositone systems.
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$$
\frac{1}{\sum_{l\in \color{cyan}{L}} |\color{green}{\hat{y}}_l|} \sum_{l \in \color{cyan}{L}} |\color{green}{\hat{y}}_l| \phi(\color{magenta}{y}_l, \color{green}{\hat{y}}_l) $$ \(\color{cyan}{L}\) is the set of labels \(\color{green}{\hat{y}}\) is the true label \(\color{magenta}{y}\) is the predicted label \(\color{green}{\hat{y}}_l\) is all the true labels that have the label \(l\) \(|\color{green}{\hat{y}}_l|\) is the number of true labels that have the label \(l\) \(\phi(\color{magenta}{y}_l, \color{green}{\hat{y}}_l)\) computes the precision or recall for the true and predicted labels that have the label \(l\). To compute
precision, let \(\phi(A,B) = \frac{|A \cap B|}{|A|}\). To compute
recall, let \(\phi(A,B) = \frac{|A \cap B|}{|B|}\).
How is Weighted Precision and Recall Calculated?
Let’s break this apart a bit more.
Continue reading “Weighted Precision and Recall Equation” |
Let $C := A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2}$; this is a positive semi-definite matrix with the same trace as $(A+B)^2$. We show that the eigenvalues of $C$ are majorised by the eigenvalues of $(A+B)^2$, that is to say that the sum of the top $k$ eigenvalues of $C$ is at most the sum of the top $k$ eigenvalues of $(A+B)^2$ for any $k$. By the Schur concavity of $(\lambda_1,\dots,\lambda_n) \mapsto \lambda_1 \dots \lambda_n$, this gives the claimed determinantal inequality.
The sum of the top $k$ eigenvalues of $C$ can be written as$$ \hbox{tr}( C P_V )$$where $V$ is the $k$-dimensional space spanned by the top $k$ eigenvectors of $C$. This can be rearranged as$$ \hbox{tr}( (A+B) (A^{1/2} P_V A^{1/2} + B^{1/2} P_V B^{1/2}) ). \quad\quad (*)$$
We can conjugate $A+B$ to be a diagonal matrix $\hbox{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_1 \geq \dots \geq \lambda_n \geq 0$. In particular we have $A+B \leq \lambda_k I + D$ in the sense of positive definite matrices, where $D := \hbox{diag}(\lambda_1-\lambda_k, \dots, \lambda_{k-1}-\lambda_k, 0, \dots, 0)$. Using the fact that $\hbox{tr}(XZ) \leq \hbox{tr}(YZ)$ whenever $X,Y,Z$ are positive semi-definite with $X \leq Y$, we can bound (*) by
$$ \hbox{tr}( (\lambda_k I + D) (A^{1/2} P_V A^{1/2} + B^{1/2} P_V B^{1/2}) )$$
which rearranges as
$$ \lambda_k \hbox{tr}( (A+B) P_V ) + \hbox{tr}( P_V (A^{1/2} D A^{1/2} + B^{1/2} D B^{1/2}) ).$$Using $A+B \leq \lambda_k I + D$ for the first term and $P_V \leq I$ for the second term, this is bounded by$$ \lambda_k^2 \hbox{tr}( P_V ) + \lambda_k \hbox{tr}( D P_V ) + \hbox{tr}( A^{1/2} D A^{1/2} + B^{1/2} D B^{1/2} ).$$For the second term we use $P_V \leq 1$, and for the third term we use the cyclic property of trace to bound by$$ \lambda_k^2 \hbox{tr}( P_V ) + \lambda_k \hbox{tr}( D ) + \hbox{tr}( (A+B) D ).$$For the first term we write $\hbox{tr}(P_V) = k = \hbox{tr}(P_W)$, where $W$ is the span of the first $k$ basis vectors $e_1,\dots,e_k$. For the third term we use $A+B \leq \lambda_k I + D$ to bound the above by$$ \lambda_k^2 \hbox{tr}( P_W ) + \lambda_k \hbox{tr}( D ) + \hbox{tr}( (\lambda_k I + D) D ).$$Since $D = P_W D P_W$, we can collect terms to obtain$$ \hbox{tr}( P_W (\lambda_k I + D)^2 P_W ).$$But by construction, $P_W (\lambda_k I + D)^2 P_W = \hbox{diag}( \lambda_1^2, \dots, \lambda_k^2, 0, \dots, 0 )$, so we have bounded (*) by the sum of the top $k$ eigenvalues of $(A+B)^2$, as required. |
I would like to align a rather complicated formula, consisting ofan (outer) optimization problem with several (inner) constraint blocks, whichshould be aligned around equal signs. I have opted to use nested
aligns and got around to this:
\begin{equation} \begin{aligned} \min_{\vec{y}, \vec{v}}\ & f(\vec{y}) + g(\vec{v}) \\ \textrm{s.t. } & \begin{aligned} \dot{\vec{y}} & = f(\vec{y}, \vec{v}) \\ \vec{y}(0) & = \vec{y}_0 \\ \end{aligned} \end{aligned}\end{equation}
which yields the following result:
Unfortunately, the
s.t. is centered with respect to the constraint block, whereas I need both blocks to be aligned at their respective tops, so I need the
s.t. to come up.
How can I adjust the code to achieve this effect? |
By displaying the change of signal energy on a two dimensional time-frequency images based on time-frequency analysis, a new mathematical morphology method to distinguish target from the nonlinear time-frequency curve is presented.
A novel approach to detect abnormal transient vibration signal for complex mechanical system is presented,where alias-free exponential bilinear time-frequency transformation is adopted to avoid the frequency alias and information loss in the traditional bilinear distributions. Simultancously,the cross-terms are effectively deduced with higher time-frequency resolution,which is validated by a gearbox early fault detection.
The scattering function is one of most important mathematical method describing time-frequency dispersion channel performance, which embodies a concentrated reflection of the dispersion characteristics of the mobile radio channel in time and frequency domain, and reflects the angle dispersion distributed performance of the channel to a certain extent.
Aimed at all kinds of methods of time-frequency analysis currently,this paper emphasizes on introducing the HHT method ,which deeply depicts the relationship between time and frequency and gets the time-frequency-energy figure of the analysed signal using the integration of EMD and Hilbert transform.
Wavelet analysis is a new mathematical method and has acquired wide application in the field due to its fine character of making analysis according to multiple resolutions in time and frequency domains.
The characteristics of the urban heat island(UHI) intensity in Shanghai were studied and calculated using wavelet transformation, based on per hour difference in temperatures in 2000 observed from Davis automatic stations installed at the Urban and Suburb of Shanghai.
We present two-sided singular value estimates for a class of convolution-product operators related to time-frequency localization.
The Balian-Low theorem (BLT) is a key result in time-frequency analysis, originally stated by Balian and, independently, by Low, as: If a Gabor system $\{e^{2\pi imbt} \, g(t-na)\}_{m,n \in \mbox{\bf Z}}$
Gabor Time-Frequency Lattices and the Wexler-Raz Identity
Gabor time-frequency lattices are sets of functions of the form $g_{m \alpha , n \beta} (t) =e^{-2 \pi i \alpha m t}g(t-n \beta)$ generated from a given function $g(t)$ by discrete translations in time and frequency.
High-Order Orthonormal Scaling Functions and Wavelets Give Poor Time-Frequency Localization
Gabor time-frequency lattices are sets of functions of the form $g_{m \alpha , n \beta} (t) =e^{-2 \pi i \alpha m t}g(t-n \beta)$ generated from a given function $g(t)$ by discrete translations in time and frequency.
We present an explicit, straightforward construction of smooth integrable functions with prescribed gaps around the origin in both time and frequency domain.
From the analysis of the electric field structure the conclusion is drawn that the bulk of the AKR power is carried by the signal component fast variable in time and frequency (flickering component).
The spectrometer provides the programming of time and frequency parameters of pulse sequences and allows phase-sensitive detection of echo signals with their subsequent representation in digital form.
Experimental data are presented on time and frequency dependences of the reverberation level for bistatic transmission and reception at low acoustic frequencies.
The equipment is aimed at manipulating five sectionalizing posts and provided with both, telecontrol and telesignaling. Time-frequency system is employed for the channel. Telesignaling from the five sectionalizing posts works continually and automatically in cyclic order. Self-checking duel information codes are adopted for each operation, in which the direct code is being compared with its inverse counterpart. The logic design is suitable for use, the circuitry is simple, and the equipment is reliable and easy...
The equipment is aimed at manipulating five sectionalizing posts and provided with both, telecontrol and telesignaling. Time-frequency system is employed for the channel. Telesignaling from the five sectionalizing posts works continually and automatically in cyclic order. Self-checking duel information codes are adopted for each operation, in which the direct code is being compared with its inverse counterpart. The logic design is suitable for use, the circuitry is simple, and the equipment is reliable and easy in maintenance.
126 lateral geniculate neurones were examined on unanaesthetized immobili-zed cats.The temporal frequency tuning curves of single neurones were measuredby stimulating the cat's eye with a sinusoidally modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its surround.Theaverage discharging rate was used as an index to judge the sensitivity to different mo-dulating frequencies.By comparing the mean impulse rates responding to modula-ted light stimulations(modulated discharge...
126 lateral geniculate neurones were examined on unanaesthetized immobili-zed cats.The temporal frequency tuning curves of single neurones were measuredby stimulating the cat's eye with a sinusoidally modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its surround.Theaverage discharging rate was used as an index to judge the sensitivity to different mo-dulating frequencies.By comparing the mean impulse rates responding to modula-ted light stimulations(modulated discharge rate)with those responding to unmodulatedlight(unmodulated discharge rate),two opposite effects were observed.The majorityof the cells(93.7)% showed modulation-excitatory curves in which the modulateddischarge rates were higher than the unmodulated ones.A minority of the cells(6.3%)showed modulation-inhibitory curves in which the modulated dischargerates were lower than the unmodulated ones.The modulation-excitatory curvescould further be grouped into two subtypes,the“band-pass filters”and the“low-pass filters”.Similarly,the modulation-inhibitory curves could also be dividedinto two subtypes,the“band-rejection filters”and the“low-rejection filters”.Onboth sides of the temporal frequency tuning curve subsidiary sidebands oppositein direction to the main part of the curves could often be seen,i.e.inhibitorysidebands flanking the modunation-excitatory tuning curve and excitatorysidebands flanking the modulation-inhibitory tuning curve.Most of the tuningcurves have only one peak.of the 110 curves measured the peak loci show anormal distribution which centered at 7 Hz.The tuning curves due to stimulationof different locations of the periphery were almost the same.But the tuningproperties of the receptive field centers were different from those of the periphe-ries either in shape or in band-width.
This paper makes to model and predict the measured data in Variant Sample Time of Hydrogen, Cesium and Rubidium atomic clocks, mainly according to the principle of ARIMA models of time series analysis method with the Bias function theory of noise Variance of measuring statistics of atomic time/frequency and method of spline functions. Some preliminary useful predicting models and conclusions obtained in variant noise process have pretical sense in optimal predictions and monitoring of the behaviour of atomic...
This paper makes to model and predict the measured data in Variant Sample Time of Hydrogen, Cesium and Rubidium atomic clocks, mainly according to the principle of ARIMA models of time series analysis method with the Bias function theory of noise Variance of measuring statistics of atomic time/frequency and method of spline functions. Some preliminary useful predicting models and conclusions obtained in variant noise process have pretical sense in optimal predictions and monitoring of the behaviour of atomic time/frequency and in developing new optimal filting control method to improve the performances of atomic clocks. |
This is a writeup of my project for the Google Summer of Code 2017. The associated repository contains examples of estimating various models. In addition to this repository, I have collaborated in HamiltonianABC and its branches as part of the GSOC 2017.
This summer I have had the opportunity to participate in the Google Summer of Code program. My project was in the Julia language and the main goal was to implement Indirect Inference (A. A. Smith 1993; A. Smith 2008) to overcome the typically arising issues (such as intractable or costly to compute likelihoods) when estimating models using likelihood-based methods. Hamiltonian Monte Carlo was expected to result in a more efficient sampling process.
Under the mentorship of Tamás K. Papp, I completed a major revision of Bayesian estimation methods using Indirect Inference (II) and Hamiltonian Monte Carlo. I also got familiar with using git, opening issues, creating a repository among others.
Here I introduce the methods with a bit of context, and dicuss an example more extensively.
Usually when we face an intractable likelihood or a likelihood that would be extremely costly to calculate, we have the option to use an alternative auxiliary model to extract and estimate the parameters of interest. These alternative models should be easier to deal with. Drovandi et al. reviews a collection of parametric Bayesian Indirect Inference (pBII) methods, I focused on the parametric Bayesian Indirect Likelihood for the Full Data (pdBIL) method proposed by Gallant and McCulloch (2009). The pdBIL uses the likelihood of the auxiliary model as a substitute for the intractable likelihood. The pdBIL does not compare summary statistics, instead works in the following way:
First the data is generated, once we have the data, we can estimate the parameters of the auxiliary model. Then, the estimated parameters are put into the auxiliary likelihood with the observed/generated data. Afterwards we can use this likelihood in our chosen Bayesian method i.e. MCMC.
To summarize the method, first we have the parameter vector $\theta$ and the observed data y. We would like to calculate the likelihood of $\ell(\theta|y)$, but it is intractable or costly to compute. In this case, with pdBIL we have to find an auxiliary model (A) that we use to approximate the true likelihood in the following way:
Then we compute the MLE of the auxiliary likelihood under
x to get the parameters denoted by $\phi$. \
In the first stage of my project I coded two models from Drovandi et al. using pdBIL. After calculating the likelihood of the auxiliary model, I used a Random Walk Metropolis-Hastings MCMC to sample from the target distribution, resulting in Toy models. In this stage of the project, the methods I used were well-known. The purpose of the replication of the toy models from Drovandi et al. was to find out what issues we might face later on and to come up with a usable interface. This stage resulted in HamiltonianABC (collaboration with Tamás K. Papp).
After the first stage, I worked through Betancourt (2017) and did a code revision for Tamás K. Papp’s DynamicHMC.jl which consisted of checking the code and its comparison with the paper. In addition to using the Hamiltonian Monte Carlo method, the usage of the forward mode automatic differentiation of the ForwardDiff package was the other main factor of this stage. The novelty of this project was to find a way to fit every component together in a way to get an efficient estimation out of it. The biggest issue was to define type-stable functions such that to accelerate the sampling process.
After the second stage, I coded economic models for the DynamicHMC.jl. The Stochastic Volatility model is one of them. In the following section, I will go through the set up.
The continuous-time version of the Ornstein-Ulenbeck Stochastic - volatiltiy model describes how the return at time t has mean zero and its volatility is governed by a continuous-time Ornstein-Ulenbeck process of its variance. The big fluctuation of the value of a financial product imply a varying volatility process. That is why we need stochastic elements in the model. As we can access data only in discrete time, it is natural to take the discretization of the model.
The discrete-time version of the Ornstein-Ulenbeck Stochastic - volatility model:
The discrete-time version was used as the data-generating process. Where yₜ denotes the logarithm of return, $x_{t}$ is the logarithm of variance, while $\epsilon_{t}$ and $\nu_{t}$ are unobserved noise terms.
For the auxiliary model, we used two regressions. The first regression was an AR(2) process on the first differences, the second was also an AR(2) process on the original variables in order to capture the levels.
""" lag_matrix(xs, ns, K = maximum(ns))Matrix with differently lagged xs."""function lag_matrix(xs, ns, K = maximum(ns)) M = Matrix{eltype(xs)}(length(xs)-K, maximum(ns)) for i ∈ ns M[:, i] = lag(xs, i, K) end Mend"first auxiliary regression y, X, meant to capture first differences"function yX1(zs, K) Δs = diff(zs) lag(Δs, 0, K), hcat(lag_matrix(Δs, 1:K, K), ones(eltype(zs), length(Δs)-K), lag(zs, 1, K+1))end"second auxiliary regression y, X, meant to capture levels"function yX2(zs, K) lag(zs, 0, K), hcat(ones(eltype(zs), length(zs)-K), lag_matrix(zs, 1:K, K))end
The AR(2) process of the first differences can be summarized by: \ Given a series Y, it is the first difference of the first difference. The so called “change in the change” of Y at time t. The second difference of a discrete function can be interpreted as the second derivative of a continuous function, which is the “acceleration” of the function at a point in time t. In this model, we want to capture the “acceleration” of the logarithm of return.
The AR(2) process of the original variables is needed to capture the effect of $\rho$. It turned out that the impact of ρ was rather weak in the AR(2) process of the first differences . That is why we need a second auxiliary model.
I will now describe the required steps for the estimation of the parameters of interest in the stochastic volatility model with the Dynamic Hamiltonian Monte Carlo method. First we need a callable Julia object which gives back the logdensity and the gradient in DiffResult type. After that, we write a function that computes the density, then we calculate its gradient using the ForwardDiff package in a wrapper function.
Required packages for the StochasticVolatility model:
using ArgCheckusing Distributionsusing Parametersusing DynamicHMCusing StatsBaseusing Base.Testusing ContinuousTransformationsusing DiffWrappersimport Distributions: Uniform, InverseGamma
struct StochasticVolatility{T, Prior_ρ, Prior_σ, Ttrans} "observed data" ys::Vector{T} "prior for ρ (persistence)" prior_ρ::Prior_ρ "prior for σ_v (volatility of volatility)" prior_σ::Prior_σ "χ^2 draws for simulation" ϵ::Vector{T} "Normal(0,1) draws for simulation" ν::Vector{T} "Transformations cached" transformation::Ttransend
After specifying the data generating function and a couple of facilitator and additional functions for the particular model (whole module can be found in
src folder), we can make the model structure callable, returning the log density. The logjac is needed because of the transformation we make on the parameters.
function (pp::StochasticVolatility)(θ) @unpack ys, prior_ρ, prior_σ, ν, ϵ, transformation = pp ρ, σ = transformation(θ) logprior = logpdf(prior_ρ, ρ) + logpdf(prior_σ, σ) N = length(ϵ) # Generating xs, which is the latent volatility process xs = simulate_stochastic(ρ, σ, ϵ, ν) Y_1, X_1 = yX1(xs, 2) β₁ = qrfact(X_1, Val{true}) \ Y_1 v₁ = mean(abs2, Y_1 - X_1*β₁) Y_2, X_2 = yX2(xs, 2) β₂ = qrfact(X_2, Val{true}) \ Y_2 v₂ = mean(abs2, Y_2 - X_2*β₂) # We work with first differences y₁, X₁ = yX1(ys, 2) log_likelihood1 = sum(logpdf.(Normal(0, √v₁), y₁ - X₁ * β₁)) y₂, X₂ = yX2(ys, 2) log_likelihood2 = sum(logpdf.(Normal(0, √v₂), y₂ - X₂ * β₂)) logprior + log_likelihood1 + log_likelihood2 + logjac(transformation, θ)end
We need the transformations because the parameters are in the proper subset of $\Re^{n}$, but we want to use $\Re^{n}$. The ContinuousTransformation package is used for the transformations. We save the transformations such that the callable object stays type-stable which makes the process faster.
$\nu$ and $\epsilon$ are random variables which we use after the transformation to simulate observation points. This way the simulated variables are continuous in the parameters and the posterior is differentiable.
Given the defined functions, we can now start the estimation and sampling process:
RNG = Base.Random.GLOBAL_RNG# true parameters and observed dataρ = 0.8σ = 0.6y = simulate_stochastic(ρ, σ, 10000)# setting up the modelmodel = StochasticVolatility(y, Uniform(-1, 1), InverseGamma(1, 1), 10000)# we start the estimation process from the true valuesθ₀ = inverse(model.transformation, (ρ, σ))# wrap for gradient calculationsfgw = ForwardGradientWrapper(model, θ₀)# samplingsample, tuned_sampler = NUTS_tune_and_mcmc(RNG, fgw, 5000; q = θ₀)
The following graphs show the results for the parameters:
Analysing the graphs above, we can say that the posterior values are in rather close to the true values. Also worth mentioning that the priors do not affect the posterior values.
1)
Difficult auxiliary model
We faced serious problems with this model. \ First of all, I coded the MLE of the finite component normal mixture model, which computes the means, variances and weights of the normals given the observed data and the desired number of mixtures. With the g-and-k quantile function, I experienced the so called “isolation”, which means that one observation point is an outlier getting weight 1, the other observed points get weight $\theta$, which results in variance equal to $\theta$. There are ways to disentangle the problem of isolation, but the parameters of interests still did not converge to the true values. There is work to be done with this model.
2)
Type-stability issues
To use the automatic differentiation method efficiently, I had to code the functions to be type-stable, otherwise the sampling functions would have taken hours to run. See the following example:
function simulate_stochastic(ρ, σ, ϵs, νs)N = length(ϵs)@argcheck N == length(νs)xs = Vector(N)for i in 1:N xs[i] = (i == 1) ? νs[1]*σ*(1 - ρ^2)^(-0.5) : (ρ*xs[i-1] + σ*νs[i])endxs + log.(ϵs) + 1.27end
function simulate_stochastic(ρ, σ, ϵs, νs) N = length(ϵs) @argcheck N == length(νs) x₀ = νs[1]*σ*(1 - ρ^2)^(-0.5) xs = Vector{typeof(x₀)}(N) for i in 1:N xs[i] = (i == 1) ? x₀ : (ρ*xs[i-1] + σ*νs[i]) end xs + log.(ϵs) + 1.27end
More involved models
Solving isolation in the three component normal mixture model
Updating shocks in every iteration
Optimization |
Let's say that I have $n$ people and $m \le n$ of this people are malicious. Then I run a process that chooses 6 people randomly and assigns each of them to one of two different groups $A$ and $B$ such that each group in the end has 3 people. What is the probability that there will be at least a malicious person in every group?
I'm not sure, but it might be something like this:
The number of ways we can select two groups with 3 people each is $${n\choose 3}\cdot{n-3\choose 3}\cdot{1\over 2}$$ and the number of ways that at least one group has no malicious people is (the use of principle of inclusion and exclusion) $$2\cdot {n-m\choose 3}\cdot {n-3\choose 3}-{n-m\choose 3}{n-m-3\choose 3}{1\over 2}$$ i.e. the number of ways A doesn't have malicious people+ the number of ways B doesn't have malicious people -the number of ways A and B doesn't have malicious people.
Not sure if this is the correct answer, but maybe my reasoning can help you nonetheless.
The total number of arrangements is: ${n \choose 6} \cdot { 6 \choose 3}$.
${n \choose 6}$ represents the total ways of choosing 6 people out of the n people
${ 6 \choose 3}$ represents the number of ways of splitting the people into two groups
Now I will (attempt) to find the number of event in which one of the groups has no malicious persons...which is the complement of what you require.
This requires 4 separate cases:
Case 1: None of the chosen 6 people are malicious
${n-m \choose 6}\cdot {6 \choose 3}$
Case 2: one person is malicious
$2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3}$
Case 3: Two persons are malicious and placed in the same group
$2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3}$
Case 4: Three persons are malicious and placed in the same group
$2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}$
So the final answer is: $$1-\dfrac{{n-m \choose 6}\cdot {6 \choose 3}}{{n-m \choose 6}\cdot {6 \choose 3} + 2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3} + 2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3} + 2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}}$$
You can also think of it like this: $n$ persons are split up in group $A$ containing $3$ persons, group $B$ containing $3$ persons and group $C$ containing $n-6$ persons. Now $m$ of the $n$ persons are chosen randomly and are labeled as "malicious". What is the probability that the groups $A$ and $B$ contain at least one labeled person? In that way of thinking my answer is best understood.
Let $A$ denote the event that a malicious person is in group $A$ and let $B$ denote the event that a malicious person is in group $B$. Then:$$P(A\cap B)=1-P(A^{\complement}\cup B^{\complement})=1-P(A^{\complement})-P(B^{\complement})+P(A^{\complement}\cap B^{\complement})=$$$$1-\frac{\binom30\binom{n-3}{m}}{\binom{n}{m}}-\frac{\binom30\binom{n-3}{m}}{\binom{n}{m}}+\frac{\binom60\binom{n-6}{m}}{\binom{n}{m}}=$$$$\frac{\binom{n}{m}-2\binom{n-3}{m}+\binom{n-6}{m}}{\binom{n}{m}}$$ |
Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
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(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. Problem 613
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective.
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(d) Determine the group structure of the kernel of $\phi$. Problem 612
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$. (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
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(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611
An $n\times n$ matrix $A$ is called
orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices.
Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 607
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.
Let \[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\] where \begin{align*} p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\ p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3. \end{align*} (a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$. (b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 606
Let $V$ be a vector space and $B$ be a basis for $V$.
Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$. Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form
\[\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) What is the dimension of $V$? (b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 605
Let $T:\R^2 \to \R^3$ be a linear transformation such that
\[T\left(\, \begin{bmatrix} 3 \\ 2 \end{bmatrix} \,\right) =\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \text{ and } T\left(\, \begin{bmatrix} 4\\ 3 \end{bmatrix} \,\right) =\begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix}.\] (a) Find the matrix representation of $T$ (with respect to the standard basis for $\R^2$). (b) Determine the rank and nullity of $T$.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 604
Let
\[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 603
Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the functions \[f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)\] in $C[-2\pi, 2\pi]$.
Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 601
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let \[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of $V$. (b) Find a basis of $W$. (c) Find the dimension of $W$.
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(The Ohio State University, Linear Algebra Midterm) Read solution |
Trident
Every day one sees politicians on TV assuring us that nuclear deterrence works because there no nuclear weapon has been exploded in anger since 1945. They clearly have no understanding of statistics.
With a few plausible assumptions, we can easily calculate that the time until the next bomb explodes could be as little as 20 years.
Be scared, very scared.
The first assumption is that bombs go off at random intervals. Since we have had only one so far (counting Hiroshima and Nagasaki as a single event), this can’t be verified. But given the large number of small influences that control when a bomb explodes (whether in war or by accident), it is the natural assumption to make. The assumption is given some credence by the observation that the intervals between wars are random [download pdf].
If the intervals between bombs are random, that implies that the distribution of the length of the intervals is exponential in shape, The nature of this distribution has already been explained in an earlier post about the random lengths of time for which a patient stays in an intensive care unit. If you haven’t come across an exponential distribution before, please look at that post before moving on.
All that we know is that 70 years have elapsed since the last bomb. so the interval until the next one must be greater than 70 years. The probability that a random interval is longer than 70 years can be found from the cumulative form of the exponential distribution.
If we denote the true mean interval between bombs as $\mu$ then the probability that an intervals is longer than 70 years is
\[ \text{Prob}\left( \text{interval > 70}\right)=\exp{\left(\frac{-70}{\mu_\mathrm{lo}}\right)} \]
We can get a lower 95% confidence limit (call it $\mu_\mathrm{lo}$) for the mean interval between bombs by the argument used in Lecture on Biostatistics, section 7.8 (page 108). If we imagine that $\mu_\mathrm{lo}$ were the true mean, we want it to be such that there is a 2.5% chance that we observe an interval that is greater than 70 years. That is, we want to solve
\[ \exp{\left(\frac{-70}{\mu_\mathrm{lo}}\right)} = 0.025\]
That’s easily solved by taking natural logs of both sides, giving
\[ \mu_\mathrm{lo} = \frac{-70}{\ln{\left(0.025\right)}}= 19.0\text{ years}\]
A similar argument leads to an upper confidence limit, $\mu_\mathrm{hi}$, for the mean interval between bombs, by solving
\[ \exp{\left(\frac{-70}{\mu_\mathrm{hi}}\right)} = 0.975\]
so \[ \mu_\mathrm{hi} = \frac{-70}{\ln{\left(0.975\right)}}= 2765\text{ years}\]
If the worst case were true, and the mean interval between bombs was 19 years. then the distribution of the time to the next bomb would have an exponential probability density function, $f(t)$,
\[ f(t) = \frac{1}{19} \exp{\left(\frac{-70}{19}\right)} \]
There would be a 50% chance that the waiting time until the next bomb would be less than the median of this distribution, =19 ln(0.5) = 13.2 years.
In summary, the observation that there has been no explosion for 70 years implies that the mean time until the next explosion lies (with 95% confidence) between 19 years and 2765 years. If it were 19 years, there would be a 50% chance that the waiting time to the next bomb could be less than 13.2 years. Thus there is no reason at all to think that nuclear deterrence works well enough to protect the world from incineration.
Another approach
My statistical colleague, the ace probabilist Alan Hawkes, suggested a slightly different approach to the problem,
via likelihood. The likelihood of a particular value of the interval between bombs is defined as the probability of making the observation(s), given a particular value of $\mu$. In this case, there is one observation, that the interval between bombs is more than 70 years. The likelihood, $L\left(\mu\right)$, of any specified value of $\mu$ is thus
\[L\left(\mu\right)=\text{Prob}\left( \text{interval > 70 | }\mu\right) = \exp{\left(\frac{-70}{\mu}\right)} \]
If we plot this function (graph on right) shows that it increases with $\mu$ continuously, so the maximum likelihood estimate of $\mu$ is infinity. An infinite wait until the next bomb is perfect deterrence.
But again we need confidence limits for this. Since the upper limit is infinite, the appropriate thing to calculate is a one-sided lower 95% confidence limit. This is found by solving
\[ \exp{\left(\frac{-70}{\mu_\mathrm{lo}}\right)} = 0.05\]
which gives
\[ \mu_\mathrm{lo} = \frac{-70}{\ln{\left(0.05\right)}}= 23.4\text{ years}\]
Summary
The first approach gives 95% confidence limits for the average time until we get incinerated as 19 years to 2765 years. The second approach gives the lower limit as 23.4 years. There is no important difference between the two methods of calculation. This shows that the bland assurances of politicians that “nuclear deterrence works” is not justified.
It is not the purpose of this post to predict when the next bomb will explode, but rather to point out that the available information tells us very little about that question. This seems important to me because it contradicts directly the frequent assurances that deterrence works.
The only consolation is that, since I’m now 79, it’s unlikely that I’ll live long enough to see the conflagration.
Anyone younger than me would be advised to get off their backsides and do something about it, before you are destroyed by innumerate politicians.
Postscript
While talking about politicians and war it seems relevant to reproduce Peter Kennard’s powerful image of the Iraq war.
and with that, to quote the comment made by Tony Blair’s aide, Lance Price
It’s a bit like my feeling about priests doing the twelve stations of the cross. Politicians and priests masturbating at the expense of kids getting slaughtered (at a safe distance, of course). |
Let $\{X_n\}$ be a sequence of compact subsets of a metric space $M$ with $X_1\supset X_2\supset X_3\supset\dotsm$. Prove that if $U$ is an open set containing $\bigcap X_n$, then there exists $X_n\subset U$.
So suppose for contradiction that there doesn't exist $X_n\subset U$, i.e. every $X_n$ has a part outside $U$. Since $\bigcap X_n$ is a subset of an open set $U$, every element of $\bigcap X_n$ has an open ball within $U$. To use compactness, I want to find an open cover for $X_n$. But I don't see any open cover to use...
Edit: Okay, following the hint that someone posted (and deleted), I think I got it.
Compact sets are closed, so $X_n$ is closed for all $n$, so $X_n'$ is open for all $n$ (where $'$ denotes complement.) Note that $X_n$ has an open cover consisting of $U,X_{n+1}',X_{n+2}',\ldots$ since $X_{n+1}'\cup X_{n+2}'\cup\ldots = (X_{n+1}\cap X_{n+2}\cap \ldots)' = (\bigcap X_i)'$. Since $X_n$ is compact, there exists a finite subcover $U,X_{a_1}',X_{a_2}',\ldots,X_{a_k}'$ where $a_1<a_2<\ldots<a_k$. But this is a contradiction, since the part of $X_n$ which is the part of $X_{a_k}$ outside $U$ is not covered. |
Object A can move at 50km/h, wants to intercept object B (currently $15^{\circ}$, east of north from A) moving at 26km/h, $40^{\circ}$ east of north. What angle should A take to intercept B? AB is 20km apart
The provided answer looks like:
Choose x axis along 20km distance.
$26t \sin{(40-15)} = 50t \sin{\theta}$
$\theta = \sin^{-1}{\frac{11}{50}} = 12.7$
$15 + 12.7 = 27.7$
I took a different approach and used $\cos$ and got a different answer ... why is that?
$26t \cos{(40-15)} = 50t \cos{\theta}$ |
Bonds, swaps and swaptions are traded securities and their prices
are directly observable in the market. The bond price depends
crucially on the random fluctuation of the interest rates over the
term of bond’s life. Unlike bonds, interest rates themselves are not
“tradeable” securities.We only trade bonds and other fixed income
instruments that depend on interest rates.
Your Question
Firstly Vasicek (1977) proposed the stochastic process for the short rate $r_t$ under the physical measure to be governed by the Ornstein–Uhlenbeck process$$dr_t=\kappa(\theta -r_t)dt+\sigma \color{red}{dW_t}\tag 1$$where $dW_t$ is a White noise or the differential of the Wiener process. In economic time series, the white noise series is often thought of as representing innovations , or shocks . That is, $dW_t$ represents those aspects of the time series of interest which could not have been predicted in advance.
The process $(1)$ is sometimes called the elastic random walk or mean reversion process.The instantaneous drift $\kappa(\theta -r_t)$ represents the effect of pulling the process toward its long-term mean $\theta$ with magnitude proportional to the deviation of the process from the mean. The mean reversion assumption agrees with the
economic phenomenon that interest rates appear over time to be pulled back to some long-run average value. To explain the mean reversion phenomenon, we argue that when interest rates increase, the economy slows down and there is less demand for loans; this leads to the tendency for rates to fall.Indeed$$\mathbb{E}\left[ {{r}_{T}}|{{r}_{t}} \right]=\theta +({{r}_{t}}-\theta ){{e}^{-\kappa (T-t)}}$$ |
Electronic Communications in Probability Electron. Commun. Probab. Volume 3 (1998), paper no. 3, 21-27. Eventual Intersection for Sequences of Lévy Processes Abstract
Consider the events $\{F_n \cap \bigcup_{k=1}^{n-1} F_k = \emptyset\}$, $n \in N$, where $(F_n)_{n=1}^\infty$ is an i.i.d. sequence of stationary random subsets of a compact group $G$. A plausible conjecture is that these events will not occur infinitely often with positive probability if $P\{F_i \cap F_j \ne \emptyset \mid F_j\} \gt 0$ a.s. for $i \ne j$. We present a counterexample to show that this condition is not sufficient, and give one that is. The sufficient condition always holds when $F_n = \{X_t^n : 0 \le t \le T\}$ is the range of a Lévy process $X^n$ on the $d$-dimensional torus with uniformly distributed initial position and $P\{\exists 0 \le s, t \le T : X_s^i = X_t^j \} \gt 0$ for $i \ne j$. We also establish an analogous result for the sequence of graphs $\{(t,X_t^n) : 0 \le t \le T\}$.
Article information Source Electron. Commun. Probab., Volume 3 (1998), paper no. 3, 21-27. Dates Accepted: 13 April 1998 First available in Project Euclid: 2 March 2016 Permanent link to this document https://projecteuclid.org/euclid.ecp/1456935909 Digital Object Identifier doi:10.1214/ECP.v3-989 Mathematical Reviews number (MathSciNet) MR1625695 Zentralblatt MATH identifier 0903.60060 Subjects Primary: 60J30 Secondary: 60G17: Sample path properties 60G57: Random measures 60J45: Probabilistic potential theory [See also 31Cxx, 31D05] Rights This work is licensed under aCreative Commons Attribution 3.0 License. Citation
Evans, Steven; Peres, Yuval. Eventual Intersection for Sequences of Lévy Processes. Electron. Commun. Probab. 3 (1998), paper no. 3, 21--27. doi:10.1214/ECP.v3-989. https://projecteuclid.org/euclid.ecp/1456935909 |
Almost everyone has participated in survey at least once in their life. The meaning of survey could be anything when the same question is asked from multiple people, it is called the survey. The most common category is formal surveys where a set of questions are asked by scientists and researchers in the form of written questions, the user can answer these questions either verbally on in written as well.
With a deep analysis of answers, they could find out about a particular thing, product, or service. Here, the term comes statistical significance that is defined as the measure of probability not just due to the chance.
A statistically significant result can be attained when p-value is less than the significance level. Here, the final output is given in terms of alpha or also named as the Type 1 error. In simple terms, there is no formula given for the statistical significance but it can be calculated in different terms by multiple testing techniques like z testing, t testing etc.
Now, you can see that there are multiple surveys or tests are conducted everyday and not all of them are useful? So, what is that? A survey can be considered useful only if it has the statistical significance, a low probability value for the hypothesis is not true. In other words, a survey is called the statistically significant only if it has the high probability for a given hypothesis that is being set true.The formula and terminologies related to this formula is given as:
\[\large Z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\]
Where, x¯ is the sample mean, μ is the population mean, σ is the sample standard deviation, n is the sample size. |
Article ID 0020 April 2018
We applied fast Fourier transform techniques and Morlet wavelet transform on the time series data of coronal index, solar flare index, and galactic cosmic ray, for the period 1986–2008, in order to investigate the long- and mid-term periodicities including the Rieger ($\sim$130 to $\sim$190 days), quasi-period ($\sim$200 to $\sim$374 days), and quasi-biennial periodicities ($\sim$1.20 to $\sim$3.27 years) during the combined solar cycles 22–23. We emphasize the fact that a lesser number of periodicities are found in the range of low frequencies, while the higher frequencies show a greater number of periodicities. The rotation rates at the base of convection zone have periods for coronal index of $\sim$1.43 years and for solar flare index of $\sim$1.41 year, and galactic cosmic ray, $\sim$1.35 year, during combined solar cycles 22–23. In relation to these two solar parameters (coronal index and solar flare index), for the solar cycles 22–23, we found that galactic cosmic ray modulation at mid cut-offrigidity (${\rm Rc} = 2.43$GV) is anti-correlated with time-lag of few months.
Article ID 0021 April 2018
Thorne– $\dot{\rm Z}$ytkow objects (T$\dot{\rm Z}$Os), originally proposed by Thorne and $\dot{\rm Z}$ytkow, may form as a result of unstable mass transfer in a massive X-ray binary after a neutron star (NS) is engulfed in the envelope of its companion star. Using a rapid binary evolution program and the Monte Carlo method, we simulated the formation of T$\dot{\rm Z}$Os in close binary stars. The Galactic birth rate of T$\dot{\rm Z}$Os is about $1.5 \times 10^{−4}$ yr$^{−1}$. Their progenitors may be composed of a NS and a main-sequence star, a star in the Hertzsprung gap or a core-helium burning, or a naked helium star. The birth rates of T$\dot{\rm Z}$Os via the above different progenitors are $1.7 \times 10^{−5}$,$1.2 \times 10^{−4}$, $0.7\times 10^{−5}$, $0.6\times 10^{−5}$ yr$^{−1}$, respectively. These progenitors may be massive X-ray binaries. Wefound that the observational properties of three massive X-ray binaries (SMC X-1, Cen X-3 and LMC X-4) in which the companions of NSs may fill their Roche robes were consistent with those of their progenitors.
Article ID 0022 April 2018 Review
The effect of compressive viscosity, thermal conductivity and radiative heat-loss functions on the gravitational instability of infinitely extended homogeneous MHD plasma has been investigated. By taking in account these parameters we developed the six-order dispersion relation for magnetohydrodynamic (MHD)waves propagating in a homogeneous and isotropic plasma. The general dispersion relation has been developed from set of linearized basic equations and solved analytically to analyse the conditions of instability and instability of self-gravitating plasma embedded in a constant magnetic field. Our result shows that the presence of viscosity and thermal conductivity in a strong magnetic field substantially modifies the fundamental Jeans criterion of gravitational instability.
Article ID 0023 April 2018
In this paper, the Planck absolute entropy and the Bekenstein–Smarr formula of the rotating Banados–Teitelboim–Zanelli (BTZ) black hole are presented via a complex thermodynamical system contributed by its inner and outer horizons. The redefined entropy approaches zero as the temperature of the rotating BTZ black hole tends to absolute zero, satisfying the Nernst formulation of a black hole. Hence, it can be regarded as the Planck absolute entropy of the rotating BTZ black hole.
Article ID 0024 April 2018
By systematically searching the region of far infrared loops, we found a number of huge cavity-like dust structures at 60 $\mu$m and 100 $\mu$m IRIS maps.By checking these with AKARI maps (90 $\mu$m and 140 $\mu$m), twonew cavity-like structures (sizes $\sim 2.7 {\rm pc}\times 0.8$ pc and $\sim 1.8 {\rm pc} \times 1$ pc) located at R.A. $({\rm J2000}) = 14^{\rm h}41^{\rm m}23^{\rm s}$ and Dec. $ ({\rm J2000}) = −64^{\circ}04'17''$ and R.A. $({\rm J2000}) = 05^h05^m35^s$ and Dec. $({\rm J2000}) = −69^{\circ}35'25''$ wereselected for the study. The difference in the average dust color temperatures calculated using IRIS and AKARI maps of the cavity candidates were found to be $3.2 \pm 0.9$ K and $4.1 \pm 1.2$ K, respectively. Interestingly, the longer wavelength AKARI map gives larger values of dust color temperature than that of the shorter wavelength IRIS maps. Possible explanation of the results will be presented.
Article ID 0025 April 2018
In this paper we discuss our choice of a large unbiased sample used for the survey of red giant branch stars for finding Li-rich K giants, and the method used for identifying Li-rich candidates using low-resolution spectra. The sample has 2000 giants within a mass range of 0.8 to 3.0$M_{\odot}$. Sample stars were selected from the Hipparcos catalogue with colour (B–V) and luminosity (L/L$_{\odot}$) in such way that the sample covers RGB evolution from its base towards RGB tip passing through first dredge-up and luminosity bump. Low-resolution (R $\approx$ 2000, 3500, 5000) spectra were obtained for all sample stars. Using core strength ratios of lines at Li I 6707{\AA} and its adjacent line Ca I 6717{\AA} we successfully identified 15 K giants with A(Li) > 1.5 dex, which are defined as Li-rich K giants. The results demonstrate the usefulness of low-resolution spectra to measure Li abundance and identify Li-rich giants from a large sample of stars in relatively shorter time periods.
Article ID 0026 April 2018
The giant impact hypothesis is the dominant theory explaining the formation of our Moon. However, the inability to produce an isotopically similar Earth–Moon system with correct angular momentum has cast a shadow on its validity. Computer-generated impacts have been successful in producing virtual systems that possess many of the observed physical properties. However, addressing the isotopic similarities between the Earth and Moon coupled with correct angular momentum has proven to be challenging. Equilibration and evection resonance have been proposed as means of reconciling the models. In the summer of 2013, the Royal Society called a meeting solely to discuss the formation of the Moon. In this meeting, evection resonance and equilibration were both questioned as viable means of removing the deficiencies from giant impact models. Themain concerns were that models were multi-staged and too complex.We present here initial impact conditions that produce an isotopically similar Earth–Moon system with correct angular momentum. This is done in a single-staged simulation. The initial parameters are straightforward and the results evolve solely from the impact. This was accomplished by colliding two roughly half-Earth-sized impactors, rotating in approximately the same plane in a high-energy, off-centered impact, where both impactors spin into the collision.
Article ID 0027 April 2018
Effects of electron temperature on the propagation of electron acoustic solitary waves in plasma with stationary ions, cold and superthermal hot electrons is investigated in non-planar geometry employing reductive perturbation method. Modified Korteweg–de Vries equation is derived in the small amplitude approximation limit. The analytical and numerical calculations of the KdV equation reveal that the phase velocity of the electron acoustic waves increases as one goes from planar to non planar geometry. It is shown that the electrontemperature ratio changes the width and amplitude of the solitary waves and when electron temperature is not taken into account,our results completely agree with the results of Javidan & Pakzad (2012). It is found thatat small values of τ , solitary wave structures behave differently in cylindrical ($m = 1$), spherical ($m = 2$) and planar geometry ($m = 0$) but looks similar at large values of τ . These results may be useful to understand the solitary wave characteristics in laboratory and space environments where the plasma have multiple temperature electrons.
Current Issue
Volume 40 | Issue 5 October 2019
Since January 2016, the Journal of Astrophysics and Astronomy has moved to Continuous Article Publishing (CAP) mode. This means that each accepted article is being published immediately online with DOI and article citation ID with starting page number 1. Articles are also visible in Web of Science immediately. All these have helped shorten the publication time and have improved the visibility of the articles.
Click here for Editorial Note on CAP Mode |
Article ID 0028 June 2018
In this paper we establish a relation between direct radiations (generally called radiation factor) and reflected radiations (albedo) to show their effects on the existence and stability of non-collinear libration points in the elliptic restricted three-body problem taking into account the oblateness of smaller primary. It isdiscussed briefly when $\alpha = 0$ and $\sigma = 0$, the non-collinear libration points form an isosceles triangle with the primaries and as e increases the libration points $L_{4,5}$ move vertically downward ($\alpha$, $\sigma$ and $e$ represents the radiation factor, oblateness factor and eccentricity of the primaries respectively). If $\alpha = 0$ but $\sigma \neq 0$, the libration points slightly displaced to the right-side from its previous location and form scalene triangle with the primaries and go vertically downward as $e$ increases. If $\alpha \neq 0$ and $\sigma\neq 0$, the libration points $L_{4,5}$ form scalene triangle with the primaries and as e increases $L_{4,5}$ move downward and displaced to the left-side. Also, the libration points $L_{4,5}$ are stable for the critical mass parameter $\mu \leq \mu_c$.
Article ID 0029 June 2018
Regular monitoring the trajectory of asteroids to a future time is a necessity, because the variety of known probably unsafe near-Earth asteroids are increasing. The analysis is perform to avoid any incident or whether they would have a further future threat to the Earth or not. Recently a new Near Earth Asteroid (2017 SB20) has been observed to cross the Earth orbit. In view of this we obtain the trajectory of Asteroid in the circular restricted three body problem with radiation pressure and oblateness.We examine nature of Asteroid’sorbit with Lyapunov Characteristic Exponents (LCEs) over a finite intervals of time. LCE of the system confirms that the motion of asteroid is chaotic in nature. With the effect of radiation pressure and oblateness the length of curve varies in both the planes. Oblateness factor is found to be more perturbative than radiation pressure. To see the precision of result obtain from numerical integration we show the error propagation and the numerical stability is assured around the singularity by applying regularized equations of motion for precise long-term study.
Article ID 0030 June 2018
We have reconsidered the simultaneous and homogeneous optical–IR light curves and the corresponding spectral indices curve of the blazarPKS0537–441 from January 2011 toMay2015. All the curves show significant fluctuations on various timescales, and the flux variations seem to be more pronounced towards the IR bands. The relation between average fluxes and spectral indices reveals the existence of redder-when-brighter (RWB) and bluer-when-brighter (BWB) trends at different flux levels, along with a long-term achromatic trend and a mild RWB trend on short-term timescales. Cross-correlation analyses present an energy-dependent time delay that the lower-frequency variations follow higher-frequency ones by a few weeks and a hysteresis pattern between spectra and fluxes. Our analysis reveals some potential coherence between low-energy-peaked BL Lacs (LBLs) and FSRQs, and indicates that the observed flux variability and spectral changes could be due to the superposition of a dominant jet emission, an underlying thermal contribution from a more slowly varyingdisk and/or other geometric effects under the shock-in-jet scenario.
Article ID 0031 June 2018
We investigate the dependence of physical properties of galaxies on small- and large-scale density environment. The galaxy population consists of mainly passively evolving galaxies in comparatively lowdensity regions of Sloan Digital Sky Survey (SDSS). We adopt (i) local density, $\rho_{20}$, derived using adaptive smoothing kernel, (ii) projected distance, $r_p$, to the nearest neighbor galaxy and (iii) the morphology of the nearest neighbor galaxy as various definitions of environment parameters of every galaxy in our sample. In orderto detect long-range interaction effects, we group galaxy interactions into four cases depending on morphology of the target and neighbor galaxies. This study builds upon an earlier study by Park and Choi (2009) by including improved definitions of target and neighbor galaxies, thus enabling us to better understand the effect of “the nearest neighbor” interaction on the galaxy. We report that the impact of interaction on galaxy properties is detectable at least up to the pair separation corresponding to the virial radius of (the neighbor) galaxies. Thisturns out to be mostly between 210 and 360 $h^{-1}$ kpc for galaxies included in our study.We report that early type fraction for isolated galaxies with $r_p \ge r_{vir,nei}$ is almost ignorant of the background density and has a very weakdensity dependence for closed pairs. Star formation activity of a galaxy is found to be crucially dependent on neighbor galaxy morphology. We find star formation activity parameters and structure parameters of galaxies to be independent of the large-scale background density.We also exhibit that changing the absolute magnitude of the neighbor galaxies does not affect significantly the star formation activity of those target galaxies whose morphology and luminosities are fixed.
Article ID 0032 June 2018
We have attempted to examine the ability of coronal mass ejections to cause geoeffectiveness. To that end, we have investigated total 571 cases of higher-speed (>1000 km/s) coronal mass ejection events observed during the years 1996–2012. On the basis of angular width (W) of observance, events of coronal mass ejection were further classified as front-side or halo coronal mass ejections (W $=$ 360$^{\circ}$); back-side halo coronal mass ejections (W = 360$^{\circ}$); partial halo (120$^{\circ}$ < W < 360$^{\circ}$) and non-halo (W < 120$^{\circ}$). From further analysis, we found that front halo coronal mass ejections were much faster and more geoeffective in comparison of partial halo and non-halo coronal mass ejections. We also inferred that the front-sided halo coronal mass ejections were 67.1% geoeffective while geoeffectiveness of partial halo coronal mass ejections and non-halo coronal mass ejections were found to be 44.2% and 56.6% respectively. During the same period of observation, 43% ofback-sided CMEs showed geoeffectiveness. We have also investigated some events of coronal mass ejections having speed >2500 km/s as a case study. We have concluded that mere speed of coronal mass ejection and their association with solar flares or solar activity were not mere criterion for producing geoeffectiveness but angular width of coronal mass ejections and their originating position also played a key role.
Article ID 0033 June 2018
We demonstrate that extremely rapid and weak periodic and non-periodic signals can easily be detected by using the autocorrelation of intensity as a function of time. We use standard radio-astronomical observations that have artificial periodic and non-periodic signals generated by the electronics of terrestrial origin. The autocorrelation detects weak signals that have small amplitudes because it averages over long integration times. Another advantage is that it allows a direct visualization of the shape of the signals, while it isdifficult to see the shape with a Fourier transform. Although Fourier transforms can also detect periodic signals,a novelty of this work is that we demonstrate another major advantage of the autocorrelation, that it can detect non-periodic signals while the Fourier transform cannot. Another major novelty of our work is that we use electric fields taken in a standard format with standard instrumentation at a radio observatory and therefore no specialized instrumentation is needed. Because the electric fields are sampled every 15.625 ns, they therefore allow detection of very rapid time variations. Notwithstanding the long integration times, the autocorrelationdetects very rapid intensity variations as a function of time. The autocorrelation could also detect messages from Extraterrestrial Intelligence as non-periodic signals.
Article ID 0034 June 2018
To investigate possible factors related to OH megamaser formation (OH MM, $L_{{\rm H}_2{\rm O}} > 10L_{\odot}$), we compiled a large HCN sample from all well-sampled HCN measurements so far in local galaxies and identifiedwith the OH MM, OH kilomasers ($L_{{\rm H}_2{\rm O}} > 10L_{\odot}$, OH kMs), OH absorbers and OH non-detections (non-OH MM). Through comparative analysis on their infrared emission, CO and HCN luminosities (good tracers for the low-density gas and the dense gas, respectively), we found that OH MM galaxies tend to have stronger HCN emission and no obvious difference on CO luminosity exists between OH MM and non-OH MM. This implies that OH MM formation should be related to the dense molecular gas, instead of the low-density molecular gas. It can be also supported by other facts: (1) OH MMs are confirmed to have higher mean molecular gas density and higher dense gas fraction ($L_{\rm HCN}/L_{\rm CO}$) than non-OH MMs. (2) After taking the distance effect into account, the apparent maser luminosity is still correlated with the HCN luminosity, while no significant correlation can be found at all between the maser luminosity and the CO luminosity. (3) The OH kMs tend to have lower values than those of OH MMs, including the dense gas luminosity and the dense gas fraction. (4) From analysis of known data of another dense gas tracer HCO$^+$, similar results can also be obtained. However, from our analysis,the infrared radiation field can not be ruled out for the OH MM trigger, which was proposed by previous works on one small sample (Darling in ApJ 669:L9, 2007). On the contrary, the infrared radiation field should play one more important role. The dense gas (good tracers of the star formation) and its surrounding dust are heated by the ultra-violet (UV) radiation generated by the star formation and the heating of the high-density gas raises the emission of the molecules. The infrared radiation field produced by the re-radiation of the heated dust inturn serves for the pumping of the OH MM.
Article ID 0035 June 2018
The rest-frame $(g–r) /M_r$ color–magnitude relations of 12 Abell-type clusters are analyzed in the redshift range ($0.02\lesssim z \lesssim 0.10$) and within a projected radius of 0.75 Mpc using photometric data from SDSS-DR9. We show that the color–magnitude relation parameters (slope, zero-point, and scatter) do not exhibit significant evolution within this low-redshift range. Thus, we can say that during the look-back time of $z \sim 0.1$ all red sequence galaxies evolve passively, without any star formation activity.
Article ID 0036 June 2018
A geomagnetic storm affects the dynamics and composition of the ionosphere and also offers an excellent opportunity to study the plasma dynamics. In the present study, we have used the VHF scintillations data recorded at low latitude Indian station Varanasi (Geomag. latitude $=$ 14$^{\circ}$55$'$N, long. $=$ 154$^{\circ}$E) which is radiated at 250 MHz from geostationary satellite UFO-02 during the period 2011–2012 to investigate the effects of geomagnetic storms on VHF scintillation.Various geomagnetic and solar indices such as Dst index, Kpindex, IMF Bz and solar wind velocity (Vx) are used to describe the geomagnetic field variation observed during geomagnetic storm periods. These indices are very helpful to find out the proper investigation and possible interrelation between geomagnetic storms and observed VHF scintillation. The pre-midnight scintillation is sometimes observed when the main phase of geomagnetic storm corresponds to the pre-midnight period. It is observed that for geomagnetic storms for which the recovery phase starts post-midnight, the probability ofoccurrence of irregularities is enhanced during this time and extends to early morning hours.
Article ID 0037 June 2018
In the present paper, the numerical simulation of Inertial Alfven wave (IAW) in low-$\beta$ plasma applicable to the auroral region at 1700 km was studied. It leads to the formation of localized structures when the nonlinearity arises due to ponderomotive effect and Joule heating. The effect of perturbation and magnitude of pump IAW, formed the localized structures of magnetic field, has been studied. The formed localized structures at different times and average spectral index scaling of power spectrum have been observed. Resultsobtained from simulation reveal that spectrum steepens with power law index $\sim −$3.5 for shorter wavelength. These localized structures could be a source of particle acceleration and heating by pump IAW in low-$\beta$ plasma.
Article ID 0038 June 2018
A hypothesis put forward in late 20th century and subsequently substantiated experimentally posited the existence of optical vortices (twisted light). An optical vortex is an electromagnetic wave that in addition to energy and momentum characteristic of flatwaves also possesses angular momentum. In recent yearsoptical vortices have found wide-ranging applications in a number of branches including cosmology. The main hypothesis behind this paper implies that the magnitude of gravitational redshift for an optical vortex will differ from the magnitude of gravitational redshift for flat light waves. To facilitate description of optical vortices, we have developed the mathematical device of gravitational interaction in seven-dimensional time-space that we apply to the theory of electromagnetism. The resulting equations are then used for a comparison of gravitational redshift in optical vortices with that of normal electromagnetic waves. We show that rotating bodies creating weak gravitational fields result in a magnitude of gravitational redshift in optical vortices that differs from themagnitude of gravitational redshift in flat light waves. We conclude our paper with a numerical analysis of the feasibility of detecting the discrepancy in gravitational redshift between optical vortices and flat waves in the gravitational fields of the Earth and the Sun.
Article ID 0039 June 2018
The effect of density and velocity gradients on the Kelvin–Helmholtz instability (KHI) of two superimposed finite-thickness fluid layers are analytically investigated. The linear normalized frequency and normalized growth rate are presented. Then, their behavior as a function of the density ratio of the lightfluid to the heavy one $(r)$ was analyzed and compared to the case of two semi-infinite fluid layers. The results showed that the values of normalized frequency of KHI for two finite-thickness fluid layers are less than their counterparts for two semi-infinite fluid layers. The behavior of normalized growth rate as a functionof the velocity and density gradients capitulates to the effect of velocity gradient at the large values of $(r)$.
Current Issue
Volume 40 | Issue 5 October 2019
Since January 2016, the Journal of Astrophysics and Astronomy has moved to Continuous Article Publishing (CAP) mode. This means that each accepted article is being published immediately online with DOI and article citation ID with starting page number 1. Articles are also visible in Web of Science immediately. All these have helped shorten the publication time and have improved the visibility of the articles.
Click here for Editorial Note on CAP Mode |
Fix a natural number $n$. Moreover fix your favorite field and category to work in - algebraic, complex, real, analytic, smooth, topological, whatever.
For a vector space $V$ let Gr$_n(V)$ be the Grassmanian of all $n$-dimensional subspaces of $V$.
Every morphism $f:X\to$ Gr$_n(V)$ gives rise to a vector bundle on $X$. It also gives rise to a family $(V_x)_{x\in X}$ of $n$-dimensional subspaces of $V$.
Let us impose on this family the condition that for any nonzero $v\in V$ there is a
unique $x\in X$ with $v\in V_x$.
Call a vector bundle "special" (just to call it something) if it can be obtained from some morphism $f$ (for some $V$) with the property that the corresponding family $(V_x)_{x\in X}$ satisfies the above condition.
The only examples of "special" vector bundles that I know are related to that question I link to.
Let $k\subset K$ be a finite field extension of degree $n$. Then, for $V$ a $K$-vector space, we have embedding
P$(V)\hookrightarrow$ Gr$_n(V_k)$
(view $K$-lines in $V$ as $n$-dimensional $k$-subspaces of the $k$-vector space $V$). The corresponding family is the tautological one, and the $k$-vector bundle on P$(V)$ so obtained is the tautological $K$-line bundle viewed as an $n$-dimensional $k$-bundle.
More generally, $K$ might be a non-commutative (associative) division ring, or a (non-associative) octonion algebra (but then dimension must be restricted).
Are there any other examples?
Does for every $X$ exist a vector bundle on $X$ that is "special" in this sense?
In particular, which manifolds have the property that their tangent bundles are "special"?
For $n=1$, and only imposing uniqueness of $x$ with $v\in V_x$ without its existence, this more or less amounts to very ample line bundles. However for $n>1$ it is essentially more than requiring $f$ to be an embedding into the Grassmanian, not only because of existence, but also because there is additional restriction that all $V_x$ are not only pairwise different but also must have pairwise zero intersections. |
Article ID 0040 August 2018
We calculate the transport coefficients of low-density nuclear matter, especially the nuclear pasta phase, using quantum molecular dynamics simulations. The shear viscosity as well as the thermal and electricalconductivities are determined by calculating the static structure factor of protons for all relevant density, temperature and proton fractions, using simulation data. It is found that all the transport coefficients have similar orders of magnitude as found earlier without considering the pasta phase. Our results are thus in contrast to the common belief that the pasta layer is highly resistive and therefore have important astrophysical consequences.
Article ID 0041 August 2018 Review
Here we outline some recent activities in the theory and phenomenology of Galactic cosmic rays, in the light of the great precision of
direct cosmic ray measurements reached in the last decade. In the energy domain of interest, ranging from a few GeV/nucleon to tens of TeV/nucleon, data have revealed some novel features requiring an explanation.We shall emphasize the importance of a more refined modeling, of achieving a better assessment of theoretical uncertainties associated to the models, and of testing key predictions specific of different models against the rich datasets available nowadays. Despite the still shaky theoretical situation, several hints have accumulated suggesting the need to go beyond the approximation of a homogeneous and non-dynamical diffusion coefficient in the Galaxy.
Article ID 0042 August 2018
The interior of neutron stars consists of the densest, although relatively cold, matter known in the universe. Here, baryon number densities might reach values close to ten times the nuclear saturation density. These suggest that the constituents of neutron star cores not only consist of nucleons, but also of more exotic baryons like hyperons or a phase of deconfined quarks.We discuss the consequences of such exotic particles on the gross properties and phenomenology of neutron stars. In addition, we determine the general phase structure of dense and also hot matter in the chiral parity-doublet model and confront model results with the recent constraints derived from the neutron star merger observation.
Article ID 0043 August 2018 Review
HAGAR, an array of seven atmospheric Cherenkov telescopes located at Hanle in Himalayas, has been observing VHE gamma ray sources since September 2008. Taking advantage of the high altitude location, HAGAR could achieve an energy threshold of about 200 GeV. Several astronomical sources, mostly pulsars and blazar class active galactic nuclei, have been observed in the last nine years. Pulsations from Crab pulsar and emission from blazars Mkn 421 and Mkn 501 has been detected successfully. Details of HAGAR telescopearray will be given and some important results will be discussed. Also the future plans will be described briefly.
Article ID 0044 August 2018
Article ID 0045 August 2018
The long awaited event of the detection of a gravitational wave from a binary neutron star merger and its electromagnetic counterparts marked the beginning of a newera in observational astrophysics. The brandnew field of gravitational wave astronomy combined with multi-messenger observations will uncover violent, highly energetic astrophysical events that could not be explored before by humankind. This article focuses on the presumable appearance of a hadron–quark phase transition and the formation of regions of deconfined quark matter in the interior of a neutron star merger product. The evolution of density and temperature profiles inside the inner region of the produced hypermassive/ supramassive neutron star advises an incorporation of a hadron–quark phase transition in the equation of state of neutron star matter. The highly densed and hot neutron star matter of the remnant populate regions in the QCD phase diagram where a non neglectable amount of deconfined quark matter is expected to be present. If a strong hadron–quark phase transition would happenduring the post-merger phase, it will be imprinted in the spectral properties of the emitted gravitational wave signal and might give an additional contribution to the dynamically emitted outflow of mass.
Article ID 0046 August 2018 Review
In the recent past, measurements of $\sigma_8$ from large scale structure observations have shown some discordance with its value obtained from Planck CMB within the $\Lambda$CDM frame. This discordance naturally leads to a mismatch in the value of $H_0$ also. Under the presumption that these discordances are not due to systematics, several attempts have been made to ameliorate the tensions. In this article, we describe the methods of determination of σ8 from large scale as well as CMB observations.We discuss that these discrepancies vanish if we consider the energy momentum tensor for an imperfect fluid which could arise due to self-interaction of dark matter or in an effective description of large scale structure.We demonstrate how the presence of viscosities in cold dark fluid on large scales ameliorate the problem elegantly than other solutions. We also estimate the neutrino mass in the viscous cosmological setup.
Article ID 0047 August 2018 Review
Fast Radio Bursts (FRBs) are short duration highly energetic dispersed radio pulses.We developed a generic formalism (Bera
et al. 2016, MNRAS, 457, 2530) to estimate the FRB detection rate for any radiotelescope with given parameters. By using this model, we estimated the FRB detection rate for two Indian radio telescope; the Ooty Wide Field Array (OWFA) (Bhattacharyya et al. 2017, J. Astrophys. Astr., 38, 17)and the upgraded Giant Metrewave Radio Telescope (uGMRT) (Bhattacharyya et al. 2018, J. Astrophys. Astr.) with three beam-forming modes. Here, we summarize these two works.We considered the energy spectrum ofFRBs as a power law and the energy distribution of FRBs as a Dirac delta function and a Schechter luminosity function.We also considered two scattering models proposed by Bhat et al. (2004, Astrophys. J. Suppl. Series,206, 1) and Macquart & Koay (2013, ApJ, 776, 125) for these works and we consider FRB pulse without scattering as a special case. We found that the future prospects of detecting FRBs by using these two Indian radio telescopes is good. They are capable to detect a significant number of FRBs per day. According to our prediction, we can detect $\sim$10$^5$−10$^8$, $\sim$10$^3$−10$^6$ and $\sim$10$^5$−10$^7$ FRBs per day by using OWFA, commensal systems of GMRT and uGMRT respectively. Even a non detection of the predicted events will be very useful in constraining FRB properties.
Article ID 0048 August 2018 Review
A review of observational properties of the Vela pulsar across a wide energy spectrum is given. Then current approaches to the modelling of pulsars and their wind zones are briefly presented. The challenges posed to the models by the diversity of Vela’s light curves in different energy ranges are discussed.
Article ID 0049 August 2018 Review
In extremely dense neutrino environments like in supernova core, the neutrino-neutrino refraction may give rise to self-induced flavor conversion. These neutrino flavor oscillations are well understood from the idea of the exponentially growing modes of the interacting oscillators in the flavor space. Until recently, the growth rates of these modes were found to be of the order of the vacuum oscillation frequency $\Delta m^2/2E$ $[O(1 {\rm km}^{−1})]$ and were considered slow growing. However, in the last couple of years it was found that if the system was allowed to have different zenith-angle distributions for the emitted $\nu_e$ and $\bar{\nu}_e$ beams then the fastest growing modes of the interacting oscillators grew at the order of $\mu =\sqrt{2G_{\rm F}n_ν}$ , a typical $\nu–\nu$ interaction energy $[O(10^5 {\rm km}^{−1})]$. Thus the growth rates are very large in comparison to the so-called ‘slow oscillations’ and canresult in neutrino flavor conversion on a much faster scale. In fact, the point that the growth rates are no longer dependent on the vacuum oscillation frequency $\Delta m^2/2E$, makes these ‘fast flavor conversions’ independentof $\Delta m^2$ (thus mass) and energy. This is a surprising result as neutrino flavor conversions are considered to be the ultimate proof of massive neutrinos. However, the importance of this effect in the realistic astrophysical scenarios still remains to be understood.
Article ID 0050 August 2018
Imaging Atmospheric Cherenkov Telescopes have revealed more than 100 TeV sources along the galactic plane, around 45% of them remain unidentified. However, radio observations revealed that dense molecular clumps are associated with 67% of 18 unidentified TeV sources. In this paper, we propose that an electron–positron magnetospheric accelerator emits detectable TeV gamma-rays when a rapidly rotating black hole enters a gaseous cloud. Since the general-relativistic effect plays an essential role in this magnetosphericlepton accelerator scenario, the emissions take place in the direct vicinity of the event horizon, resulting in a point-like gamma-ray image.We demonstrate that their gamma-ray spectra have two peaks around 0.1 GeV and 0.1 TeV and that the accelerators become most luminous when the mass accretion rate becomes about 0.01% of the Eddington accretion rate. We compare the results with alternative scenarios such as the cosmic-ray hadron scenario, which predicts an extended morphology of the gamma-ray image with a single power-law photonspectrum from GeV to 100 TeV.
Article ID 0051 August 2018 Review
BHAL CHANDRA JOSHI PRAKASH ARUMUGASAMY MANJARI BAGCHI DEBADES BANDYOPADHYAY AVISHEK BASU NEELAM DHANDA BATRA SURYARAO BETHAPUDI ARPITA CHOUDHARY KISHALAY DE L. DEY A. GOPAKUMAR Y. GUPTA M. A. KRISHNAKUMAR YOGESH MAAN P. K. MANOHARAN ARUN NAIDU RANA NANDI DHRUV PATHAK MAYURESH SURNIS ABHIMANYU SUSOBHANAN
Radio pulsars show remarkable clock-like stability, which make them useful astronomy tools in experiments to test equation of state of neutron stars and detecting gravitational waves using pulsar timing techniques. A brief review of relevant astrophysical experiments is provided in this paper highlighting thecurrent state-of-the-art of these experiments. A program to monitor frequently glitching pulsars with Indian radio telescopes using high cadence observations is presented, with illustrations of glitches detected in this program, including the largest ever glitch in PSR B0531 $+$ 21. An Indian initiative to discover sub-$\mu$Hz gravitational waves, called Indian Pulsar Timing Array (InPTA), is also described briefly, where time-of-arrival uncertainties and post-fit residuals of the order of $\mu$s are already achievable, comparable to other international pulsar timing array experiments. While timing the glitches and their recoveries are likely to provide constraints on the structure of neutron stars, InPTA will provide upper limits on sub-$\mu$Hz gravitational waves apart from auxiliary pulsarscience. Future directions for these experiments are outlined.
Article ID 0052 August 2018 Review
During the last decade, very high energy astrophysics emerged as a new branch of astronomy with major discoveries achieved by the present ground-based gamma-ray Cherenkov telescopes. The sample of cosmic sources firmly detected at very high energy (VHE) now exceeds two hundred objects, includingactive galactic nuclei (AGN), pulsar wind nebulae, and several other types of sources of which a significant number are unidentified ones. The scientific return from recent VHE data is particularly interesting for AGN science, shedding new light on particle acceleration and emission processes around supermassiveblack holes, and probing the intergalactic space by the analysis of VHE photons propagating from bright remote sources to the Earth. The perspectives of this research field are promising with new generation VHE instruments such as CTA, a project of open observatory at extreme energies at the horizon 2023,allowing a deep analysis of the sky in the highest part of the electromagnetic spectrum, from 20 GeV to 300 TeV.
Current Issue
Volume 40 | Issue 5 October 2019
Since January 2016, the Journal of Astrophysics and Astronomy has moved to Continuous Article Publishing (CAP) mode. This means that each accepted article is being published immediately online with DOI and article citation ID with starting page number 1. Articles are also visible in Web of Science immediately. All these have helped shorten the publication time and have improved the visibility of the articles.
Click here for Editorial Note on CAP Mode |
Article ID 0053 October 2018
The aim of this paper is to investigate the association of the geomagnetic storms with themagnitude of interplanetary magnetic field IMF ($B$), solar wind speed ($V$), product of IMF and wind speed ($V\cdot B$), Ap index and solar wind plasma density ($n_{\rm p}$) for solar cycles 23 and 24. A Chree analysis by the superposed epoch method has been done for the study. The results of the present analysis showed that $V\cdot B$ is more geoeffective when compared to V or B alone. Further the high and equal anti-correlation coefficient is found between Dst and Ap index ($−$0.7) for both the solar cycles. We have also discussed the relationship between solar wind plasma density (np) and Dst and found that both these parameters are weakly correlated with each other. Wehave found that the occurrence of geomagnetic storms happens on the same day when IMF, $V$, Ap and $V\cdot B$ reach their maximum value while 1 day time lag is noticed in case of solar wind plasma density with few exceptions. The study of geomagnetic storms with various solar-interplanetary parameters is useful for the study of space weather phenomenon.
Article ID 0054 October 2018
We used binary octahedrons to investigate the dynamical behaviors of binary asteroid systems. The mutual potential of the binary polyhedron method is derived from the fourth order to the sixth order. The irregular shapes, relative orbits, attitude angles, as well as the angular velocities of the binary asteroid system are included in the model. We investigated the relative trajectory of the secondary relative to the primary, the total angular momentum and total energy of the system, the three-axis attitude angular velocity of the binary system,as well as the angular momentum of the two components. The relative errors of the total angular momentum and the total energy indicate that the calculation has a high precision. It is found that the influence of the orbital and attitude motion of the primary from the gravitational force of the secondary is obvious. This study is useful in understanding the complicated dynamical behaviors of the binary asteroid systems discovered in our Solar system.
Article ID 0055 October 2018
IUE has made very successful long term and intense short time-scale monitoring spectroscopic study of NGC 4151, a Seyfert 1 galaxy for over nearly 18 years from its launch in 1978 to 1996. The long-term observations have been useful in understanding the complex relation between UV continuum and emission line variability Seyfert galaxies. In this paper, we present the results of our studies on the short-timescale intense monitoring campaign of NGC 4151 undertaken during December 1–15, 1993. A most intense monitoring observation of NGC 4151 was carried out by IUE in 1993, when the source was at its historical high flux state with a shortest interval of 70 min between two successive observations.We present our results on emission line and continuum variability amplitudes characterized by $F_{\rm var}$ method.We found highest variability of nearly 8.3% at 1325 å continuum with a smallest amplitude of 4% at 2725 å. The relative variability amplitudes ($R_{\rm max}$) have been found to be 1.372, 1.319, 1.302 and 1.182 at 1325, 1475, 1655 and 2725 å continuum respectively. The continuum and emission line variability characteristics obtained in the present analysis are in very good agreement with the results obtained by Edelson
et al. (1996) and Crenshaw et al. (1996) from the analysis of the same observational spectral data. The large amplitude rapid variability characteristics obtained in our study have been attributed to the continuum reprocessing of X-rays absorbed by the material in the accretion disk as proposed by Shakura and Sunyaev (1973). The continuum and emission light curves have shown four distinct high amplitude events of flux maxima during the intense monitoring campaign of 15 days, providing a good limit on the amplitude of UV variability and the BLR size in low luminosity Seyfert galaxies and are useful for constraining the continuum emission models. The decreasing $F_{\rm var}$ amplitude of UV continuum with respect to increasing wavelength obtained in the present study and consistent with similar observations by Edelson et al. (1996) and Crenshaw et al. (1996) is a significant result of the intense monitoring observations.
Article ID 0056 October 2018
Contemporary piece of writing devotes to the investigation of plane symmetric cosmological model with quark and strange quark matter in the deformations of the Einstein’s theory of General Relativity (GR). At small or large scales (ultraviolet or infrared gravity), deformations of the Einstein’s theory could provide a better handling of cosmic acceleration without magnetism (along with singularities). In particular, a proper deformation of GR in the ultraviolet regime could play the role of describing the transition between GR and quantum gravity. As a matter of fact, although with a different purpose in mind, it was Einstein himself who proposed in the 30’s the reformulation of GRby taking the field of orthonormal frames or tetrads as the dynamical variable instead of the metric tensor (Einstein,
Phys. Math. Kl 217, 401, 1928). As per the observation, pressure and energy density of the model approaches the bag constant in negative and positive ways at $t\to \infty$, i.e. $p \to −B_c$ and $\rho \to B_c$, the negative pressure due to the Dark Energy (DE) in the context of accelerated expansion of the universe. So the strange quark matter gives an idea of existence of dark energy in the universe and supports the observations of the SNe-I (Riess et al., Astron. J. 116,1009, 1998; Perlmutter et al. Astrophys. J. 517, 565, 1999). Also these results agree with the study of Aktas and Aygun ( Chinese J. Phys. 55, 71, 2017) and Sahoo et al. ( New. Astron. 60, 80, 2018).
Article ID 0057 October 2018
The circular restricted three-body problem, where two primaries are taken as heterogeneous oblate spheroid with three layers of different densities and infinitesimal body varies its mass according to the Jeans law, has been studied. The system of equations of motion have been evaluated by using the Jeans law and hence the Jacobi integral has been determined. With the help of system of equations of motion, we have plotted the equilibrium points in different planes (in-plane and out-of planes), zero velocity curves, regions of possible motion, surfaces (zero-velocity surfaces with projections and Poincaré surfaces of section) and the basins of convergence with the variation of mass parameter. Finally, we have examined the stability of the equilibrium points with the help of Meshcherskii space–time inverse transformation of the above said model and revealed that all the equilibrium points are unstable.
Article ID 0058 October 2018
We present the stellar parameters of the individual components of the two old close binary systems HIP14075 and HIP14230 using synthetic photometric analysis. These parameters are accurately calculated based on the best match between the synthetic photometric results within three different photometric systems with the observed photometry of the entire system. From the synthetic photometry,we derive the masses and radii of HIP14075 as ${\mathcal M}^A = 0.99\pm 0.19{\mathcal M}^{\odot}$, $R_A = 0.877\pm 0.08R_{\odot}$ for the primary and ${\mathcal M}^B = 0.96\pm 0.15{\mathcal M}_{\odot}$,$R_B = 0.821\pm 0.07R_{\odot}$ for the secondary, and of HIP14230 as ${\mathcal M}^A = 1.18\pm 0.22{\mathcal M}_{\odot}$, $R_A = 1.234\pm 0.05R_{\odot}$ for the primary and ${\mathcal M}^B = 0.84 \pm 0.12{\mathcal M}_{\odot}$, $R^B = 0.820\pm 0.05R_{\odot}$ for the secondary. Both systems depend on Gaia parallaxes. Based on the positions of the components of the two systems on a theoretical Hertzsprung–Russell diagram, we find that the age of HIP 14075 is $11.5\pm 2.0$ Gyr and of HIP 14230 is $3.5\pm 1.5$ Gyr.Our analysis reveals that both systems are old close binary systems ($\approx$>4 Gyr). Finally, the positions of the components of both the systems on the stellar evolutionary tracks and isochrones are discussed.
Article ID 0059 October 2018
In this paper, we have constructed the cosmological model of the universe in $f (R, T)$ theory of gravity in a Bianchi type VI$_h$ universe for the functional $f (R, T)$ in the form $f (R, T) = \mu R + \mu T$, where $R$ and $T$ are respectively Ricci scalar and trace of energy momentum tensor and $\mu$ is a constant. We have made use of the hyperbolic scale factor to find the physical parameters and metric potentials defined in the space-time. The physical parameters are constrained from different representative values to build up a realistic cosmological model aligned with the observational behaviour. The state finder diagnostic pair is found to be in the acceptable range. The energy conditions of the model are also studied.
Article ID 0060 October 2018
We report the observations of the solar chromosphere from a newly commissioned solar telescope at the incursion site near Pangong Tso lake in Merak (Leh/Ladakh). This new Hα telescope at the Merak site is identical to the Kodaikanal H$_{\alpha}$ telescope. The telescope was installed in the month of August 2017 at the Merak site. The telescope consists of a 20-cm doublet lens with additional re-imaging optics. A Lyot filter with 0.5 å passband isolates the Balmer line of the hydrogen spectra to make the observations of the solarchromosphere. The observations made in Hα wavelength delineates the magnetic field directions at the sunspot and the quiet regions. A CCD detector records the images of the chromosphere with a pixel resolution of 0.27$''$and covers 9.2$'$ field-of-view. This telescope has a good guiding system that keeps the FoV in the intended position. We report the development of control software for tuning the filter unit, control detector system, observations and calibration of the data to make it useful for the scientific community. Some preliminary results obtained from the Merak H$_{\alpha}$ telescope are also presented. This high altitude facility is a timely addition to regularly obtain H$_{\alpha}$ images around the globe.
Article ID 0061 October 2018
We have classified a sample of 37,492 objects from SDSS into QSOs, galaxies and stars using photometric data over five wave bands ($u, g, r , i$ and $z$) and UV
GALEX data over two wave bands (near-UV and far-UV) based on a template fitting method. The advantage of this method of classification is that it does not require any spectroscopic data and hence the objects for which spectroscopic data is not available can also be studied using this technique. In this study, we have found that our method is consistent by spectroscopic methods given that their UV information is available. Our study shows that the UV colours are especially important for separating quasars and stars, as well as spiral and starburst galaxies. Thus it is evident that the UVbands play a crucial role in the classification and characterization of astronomical objects that emit over a wide range of wavelengths, but especially for those that are bright at UV. We have achieved the efficiency of 89% for the QSOs, 63% for the galaxies and 84% for the stars. This classification is also found to be in agreement with the emission line diagnostic diagrams.
Article ID 0062 October 2018
We have studied the evolution of cosmological parameters by considering exponential harmonic field with collisional matter. A comparison has been made with the behavior of these parameters in the presence of ordinary matter and the model $\Lambda CDM$. We have also compared the evolution of these parameters with the ones obtained in the modified gravity $f (R)$ and $f (R, T)$ theory case. The results are in line with those of the modified gravity so that the harmonic exponential field can be used to explain why the Universe has gone from the deceleration phase to the acceleration phase.
Article ID 0063 October 2018
In this paper, computation of the halo orbit for the KS-regularized photogravitational circular restricted three-body problem is carried out. This work extends the idea of Srivastava
et al. ( Astrophys. Space Sci. 362: 49, 2017) which only concentrated on the (i) regularization of the 3D-governing equations of motion, and (ii) validation of the modeling for small out-of-plane amplitude ($A_z = 110000$ km) assuming the third order analytical approximation as an initial guess with and without differential correction. This motivated us to compute the halo orbits for the large out-of-plane amplitudes and to study their stability analysis for the regularized motion. The stability indices are described as a function of out-of-plane amplitude, mass reduction factor and oblateness coefficient. Three different Sun–planet systems: the Sun–Earth, Sun–Mars and the Sun–Jupiter are chosen in this study. Stable halo orbits do not exist around the $L_1$ point, however, around the $L_2$ point stable halo orbits are found for the considered systems.
Current Issue
Volume 40 | Issue 5 October 2019
Since January 2016, the Journal of Astrophysics and Astronomy has moved to Continuous Article Publishing (CAP) mode. This means that each accepted article is being published immediately online with DOI and article citation ID with starting page number 1. Articles are also visible in Web of Science immediately. All these have helped shorten the publication time and have improved the visibility of the articles.
Click here for Editorial Note on CAP Mode |
A jewelry company requires for its products to pass three tests before they are sold at stores. For gold rings, 90 % passes the first test, 85 % passes the second test, and 80 % passes the third test. If a product fails any test, the product is thrown away and it will not take the subsequent tests. If a gold ring failed to pass one of the tests, what is the probability that it failed the second test?
Let $F$ be the event that a gold ring fails one of the three tests. Let $F_2$ be the event that it fails the second test. Then what we need to compute is the conditional probability\[P(F_2 \mid F) = \frac{P(F_2 \cap F)}{P(F)}.\]
The numerator is\[P(F_2 \cap F) = P(F_2) = 0.9 \cdot 0.15.\](A gold ring passes the first test with probability $0.9$ and fails the second test with probability $1-0.85=0.15$.)
The complement $F^c$ of $F$ is the event that a gold ring passes all the tests. Thus\[P(F) = 1- P(F^c) = 1 – 0.9 \cdot 0.85 \cdot 0.8.\]It follows that the desired probability is\begin{align*}P(F_2 \mid F) &= \frac{0.9 \cdot 0.15}{1 – 0.9 \cdot 0.85 \cdot 0.8}= \frac{135}{388} \approx 0.348\end{align*}
Therefore, given that a gold ring failed to pass one of the tests, the probability that it failed the second test is about 34.8 %.
Independent Events of Playing CardsA card is chosen randomly from a deck of the standard 52 playing cards.Let $E$ be the event that the selected card is a king and let $F$ be the event that it is a heart.Prove or disprove that the events $E$ and $F$ are independent.Definition of IndependenceEvents […]
Pick Two Balls from a Box, What is the Probability Both are Red?There are three blue balls and two red balls in a box.When we randomly pick two balls out of the box without replacement, what is the probability that both of the balls are red?Solution.Let $R_1$ be the event that the first ball is red and $R_2$ be the event that the […]
Complement of Independent Events are IndependentLet $E$ and $F$ be independent events. Let $F^c$ be the complement of $F$.Prove that $E$ and $F^c$ are independent as well.Solution.Note that $E\cap F$ and $E \cap F^c$ are disjoint and $E = (E \cap F) \cup (E \cap F^c)$. It follows that\[P(E) = P(E \cap F) + P(E […]
Overall Fraction of Defective Smartphones of Three FactoriesA certain model of smartphone is manufactured by three factories A, B, and C. Factories A, B, and C produce $60\%$, $25\%$, and $15\%$ of the smartphones, respectively.Suppose that their defective rates are $5\%$, $2\%$, and $7\%$, respectively. Determine the overall fraction of […]
Conditional Probability Problems about Die RollingA fair six-sided die is rolled.(1) What is the conditional probability that the die lands on a prime number given the die lands on an odd number?(2) What is the conditional probability that the die lands on 1 given the die lands on a prime number?Solution.Let $E$ […]
Independent and Dependent Events of Three Coins TossingSuppose that three fair coins are tossed. Let $H_1$ be the event that the first coin lands heads and let $H_2$ be the event that the second coin lands heads. Also, let $E$ be the event that exactly two coins lands heads in a row.For each pair of these events, determine whether […]
Probability Problems about Two DiceTwo fair and distinguishable six-sided dice are rolled.(1) What is the probability that the sum of the upturned faces will equal $5$?(2) What is the probability that the outcome of the second die is strictly greater than the first die?Solution.The sample space $S$ is […] |
$\delta(t)$ is an impulse that's infinitely thin and infinitely high. The area under it is 1 though. $\delta(t - nT)$ places the impulse at time $nT$.
Now this is being multiplied by some function, in your case $e^{-st}$ and that product is integrated.
Integrating is just a glorified sum. Mathematicians try to disguise it as something fancy by using a special symbol, but in the end it's really just plain old addition. What is summed up in an integral? The values of the function to be integrated within the given range.
The result of an integral is
the area under the plot of a function. $\delta(t - nT)$ has area 1 and is so extremely thin, it can be thought of to exist only at one position ($nT$) and to be 0 elsewhere. Kind of like rectangle in a histogram representing one value, but so thin, it could almost work at the fashion week.
Multiplying a function by a scalar value basically means scaling the function (stretching it along the y axis). You have a rectangle of area 1 and you scale one dimension by a factor. This factor is the value of the function you are multiplying it with: $e^{-st}$ and the value is evaluated at the time the impulse occurs: $nT$, so the area of the rectangle is $1 \cdot e^{-snT}$, which is the result of the integral.
If you have been doing programming with embedded systems you are probably familiar with how masking individual bits works with bitwise operators:
0101100101110
& 0000000100000
---------------
0000000100000
You can think about the impulse as the single bit that masks out all values of a function except the single value at its position.
$$\int^\infty _{0^-}\delta(t-nT)e^{-st}dt = e^{-nsT}$$
Is just a special form of
$$\int^{+\infty}_{-\infty}\delta(t-T)f(t)~dt = f(T)$$
which is sometimes called
the sifting property or the sampling property. The delta function is said to "sift out" the value at $t = T$. |
Statistical model for a Completely Randomized Design:
$$y_{ij}=\mu+\tau_{i}+\epsilon_{ij}\quad i=1,2,\ldots,a; j=1,2,\ldots,n$$
where,
$y_{ij}$ is the $j$th observation of $i$th treatment effect
$\mu$ is the overall mean
$\tau_{i}$ is the $i$th treatment effect
$\epsilon_{ij}$ random error term and $\epsilon_{ij}\sim NID(0,\sigma^2)$
For testing the hypothesis of no difference in treatment means
$$H_0:\tau_1=\tau_2=\ldots=\tau_a = 0 \quad vs.\quad H_1:\tau_i\ne 0\quad \text{for some } i.$$
Under
null hypothesis , the ratio$$F=\frac{\frac{SS_{treatments}}{a-1}}{\frac{SS_{error}}{N-a}}$$is distributed as F with $a-1$ and $N-a$ degrees of freedom .
My question is: Why do i have to compute the F-ratio under
nullhypothesis, why do not under
alternativehypothesis ? |
dyld: Library not loaded: /sw/lib/libpng12.0.dylib Referenced from: /sw/bin/latex Reason: Incompatible library version: latex requires version 30.0.0 or later, but libpng12.0.dylib provides version 26.0.0 sh: line 1: 81964 Trace/BPT trap latex -output-directory=/var/folders/71/71wgK3o7FtGClPSEzjU3v++++TM/-Tmp-/inkscape-9yWR6s -halt-on-error /var/folders/71/71wgK3o7FtGClPSEzjU3v++++TM/-Tmp-/inkscape-9yWR6s/eq.tex > /var/folders/71/71wgK3o7FtGClPSEzjU3v++++TM/-Tmp-/inkscape-9yWR6s/eq.out invalid LaTeX input: \(\displaystyle\frac{\pi^2}{6}=\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{k^2}\) temporary files were left in: /var/folders/71/71wgK3o7FtGClPSEzjU3v++++TM/-Tmp-/inkscape-9yWR6s
Curiously, when I print out the version number of "/sw/lib/libpng12.0.dylib" via otool, it tells me that I already use the most recent version, which is 32.0.0, so there should be no problems.
I just found out what goes wrong: The Inkscape .dmg application package which I got from the Inkscape homepage comes with an own version of libpng12.0.dylib, which lies in "Inkscape.app/Contents/Resources/lib/", that has version number 26.0.0. It seems that Inkscape uses this file although the error message above explicitely shows the Fink path "/sw/lib/" !!
So I just renamed the libpng12.0.dylib which comes with the application package so that Inkscape doesn't find / use this file anymore and now the LaTeX effect works fine.
Just wanted to tell you in case someone else has a similar problem...
bluefloyd
P.S.: My version of Inkscape is 0.46.
P.P.S.: The identical problem occurs when using the textext extension. The solution is the same. An additional comment concerning textext: On a Mac there are actually two places where you can put the files of the textext extension:
1) ~/inkscape/extensions/
2)/Applications/Inkscape.app/Contents/Resources/extensions/
I propose putting the files in 2), since the extension is looking for inkex.py which has to be in the same directory as the extensions itself; inkex.py lies in 2), not in 1). Another possibility would be to use 1) and copy inkex.py (and probably some other needed files?) also from 2) to 1). |
A graph is
almost Hamiltonian if it contains a cycle that visits every node at least once and at most twice.
Is the problem of determining whether a graph is almost Hamiltonian NP-complete?
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Yes it is NP-complete. Membership in NP is trivial. Thus, we must only show NP-hardness. To do so, we use a reduction from the original Hamilton cycle problem.
Given a graph $G=(\{v_1,\dots,v_n\},E)$, we can construct a new Graph $G'=(\{v_1,\dots,v_n\}\cup\{v_1',\dots,v_n'\},E \cup \{(v_i,v_i')\mid 1\leq i\leq n\} \cup \{(v_i',v_i)\mid 1\leq i\leq n\})$. Note that in order to reach one of the nodes $v_i'$, one must visit $v_i$ twice on a cycle $v_i - v_i' - v_i$. Apart from that, the edges in $G'$ are the same as in $G$. Thus, $G'$ is almost Hamiltonian if and only if $G$ is Hamiltonian, which completes the proof. |
Let $K_\alpha(t,x)$ be the (generalised or fractional)
Heat kernel which corresponds to the fractional Heat equation (I'm not sure that's the right name) in $\mathbb R^n$$$u_t=(-\Delta)^\alpha u, \quad \alpha\in(0,1].$$
A fact, which appears to be very well-known, states that: $$ K_\alpha(t,x)>0, \quad \text{for all $x\in\mathbb R^n,\, t>0$ and $\alpha\in(0,1]$ }. $$ It is so well-known that no paper where I have encountered this fact gives any reference.
I am looking for a reference where I could see the proof of this fact. |
Multi-scan Experimental Data (BBM data)
Measurement data obtained in-situ can be easily imported and loaded to OptiLayer using
In the lower part of the window
Using
you can plot discrepancies between theoretical (or model) spectral characteristic and corresponding measurement scan.
Discrepancies window represent partial discrepancies \(D_j, j=1,...,m\), \(m\) is the number of design layers for the multi-scan measurement mode:
\[ D_j=\frac 1L\sum\limits_{i=1}^L \left(T(d_1,...,d_j;\lambda_i)-\hat{T}(\lambda_i)\right)^2, \]
where \(\lambda_1,...,\lambda_L\) are measurement spectral points, \(\hat{T}(\lambda_i)\) are measurement values.
The scan discrepancies are represented as a diagram (see below) or in a numerical form in
Efficient numerical algorithms provided by OptiRE enable performing reliable reverse engineering of multilayer coatings. We demonstrated reliability of the reverse engineering results obtained on the basis of multiscan measurements in our publications.
Recent version of OptiLayer allows you to analyze BBM movies.
\[ DF^2=\frac 1M\frac 1L\sum\limits_{i=1}^M\sum\limits_{j=1}^L\left(T(d_1(1+\delta_1),...,d_m(1+\delta_m))-\hat{T}(\lambda_j)\right)^2 \]
Initial fitting of
Achieved fitting of
Estimated relative errors in layer thicknesses. Two errors in layers number 3 and 12 are reproduced with high accuracy.
Evolutionary Reverse Engineering Algorithm
Solve --> Random Errors
Solve --> Quasi-Random Errors
Solve --> Quasi-Random Inhomogeneities
Solve --> Random Errors (scientific mode)
Solve --> Quasi-Random Errors (scientific mode)
In comparison with the previous versions, there is a choice between two algorithms in "Method" field:
\[ DF^2=\frac{1}{m}\sum\limits_{k=1}^m\frac 1L\sum\limits_{j=1}^L\left[\frac{T(d_1,...,d_m;\lambda_j)-\hat{T}^{(k)}(\lambda_j)}{\Delta T_j}\right]^2, \]
where \(m\) is the number of design layers, \(\{\lambda_j\}, \; j=1,...,L\) is the wavelength grid, \(\hat{T}^{(k)}(\lambda_j)\) transmittance scan recorded after the deposition of \(k-\)th layer, \(\Delta T_j\) are measurement tolerances.
In the case of "Triangular algorithm", partial discrepancy functions \(DF(i)\) are minimized with respect to all "Active" layer thicknesses among \(d_1,...,d_i\):
\[ DF^2(i)=\frac{1}{i}\sum\limits_{k=1}^i\frac 1L\sum\limits_{j=1}^L\left[\frac{T(d_1,...,d_i;\lambda_j)-\hat{T}^{(k)}(\lambda_j)}{\Delta T_j}\right]^2,\]
where \(m\) is the number of design layers, \(\{\lambda_j\}, \; j=1,...,L\) is the wavelength grid, \(\hat{T}^{(k)}(\lambda_j)\) transmittance scan recorded after the deposition of \(k-\)th layer, \(\Delta T_j\) are measurement tolerances. |
Suppose there is a tutorial session at a university. We have a set of $k$ questions $Q = \{ q_1 \ldots q_k \}$ and a set of $n$ students $S = \{ s_1 \ldots s_n \}$. Each student has a doubt in a certain subset of questions, i.e. for each student $s_j$, let $Q_j \subseteq Q$ be the set of questions that a student has a doubt it. Assume that $\forall 1 \leq j \leq n: Q_j \neq \phi$ and $\bigcup_{1\leq j\leq n}Q_j = Q$.
All students enter the tutorial session in the beginning (at $t = 0$). Now, a student leaves the tutorial session as soon as all the questions in which he has a doubt in have been discussed. Suppose that the time taken to discuss each question is equal, say 1 unit$^*$. Let $t_j$ be the time spent by $s_j$ in the tutorial session. We want to find out an optimal permutation $\sigma$ in which questions are discussed $(q_{\sigma(1)} \ldots q_{\sigma(n)})$ such the the quantity $T_\sigma = \Sigma_{1\leq j \leq n}t_j$ is minimized.
I have not been able to design a polynomial time algorithm, or prove $\mathsf{NP}$-hardness.
We can define a decision version of the problem $$ \mathsf{TUT} = \{\langle k, n, \mathcal{F}_Q, C \rangle \mid \exists \sigma : T_{\sigma} \leq C\} $$
where $\mathcal{F}_Q$ is the set of $Q_j$'s.
We can then find out the minimum $T_\sigma$ using binary search on $C$ and find out the optimal $\sigma$ using partial assignments to $\sigma$ in polynomial time using an oracle for $\mathsf{TUT}$. Also, $\mathsf{TUT} \in \mathsf{NP}$ because the optimal $\sigma$ can be used as a certificate which we can verify easily in polynomial time.
My question: Is $\mathsf{TUT}$ $\mathsf{NP}$-complete or can we design a polynomial time algorithm for it?
Sidenote: By the way, I thought of this question after an actual tutorial session, in which the TA discussed the questions in the normal order $q_1 \ldots q_n$ because of which many students had to wait until the end.
Example Let $k=3$ and $n=2$. $Q_1 = \{q_3\}$ and $Q_2 = \{q_1, q_2, q_3\}$. We can see that an optimal $\sigma = \langle 3, 1, 2 \rangle$ because in that case, $s_1$ leaves after $t_1 = 1$ and $s_2$ leaves after $t_2 = 3$, so sum is 4. However, if we discuss the questions in the order $\langle 1, 2, 3\rangle$, then $s_1$ and $s_2$ both have to wait till the end and $t_1 = t_2 = 3$, so sum is 6.
$^*$You are free to solve the more general case where each question $q_i$ takes $x_i$ units to discuss! |
Why exactly is X(0) the DC component of a signal?
How is it equal to N times x(n)'s average value and why it is at X(0)?
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Why exactly is X(0) the DC component of a signal?
How is it equal to N times x(n)'s average value and why it is at X(0)?
Follows from the DFT definition. It's defined as
\begin{equation} X(k) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{kn}{N}} \end{equation}
So $X(0)$ is
\begin{equation} X(0) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{0 \cdot n}{N}} \end{equation}
Having $k=0$ gives $e^0=1$ all the time so that
\begin{equation} X(0) = \sum_{n=0}^{N-1} x(n) 1 \end{equation}
Comparing this to the average
\begin{equation} \overline{x} = \frac{1}{N} \sum_{n=0}^{N-1} x(n) \end{equation}
shows that $X(0) = N \overline{x}$ |
My problem can be describe by following image:
I know coordinates of an example
P point. Say, they are equal to
(8,8). I also know the length of
a,
b and
c sides of the triangle which are equal to
10. Now, how one can calculate the coordinates of
ABC points?
SOLVED (11/3/2016)
Thanks to @EmilioNovati answer to my other question ("How to calculate $B(x_1,y_1)$ when $\alpha$ and $A(x_0,y_0)$ are known?"), I found a solution to my problem with coordinates of triangle vertices.
First lets look at following drawing:
Now, in order to calculate $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$, one can do:
$x_C = x_P + a*cos(\alpha)$, $y_C = y_P + a*sin(\alpha)$;
$x_A = x_P + a*cos(\alpha+\beta)$, $y_A = y_P + a*sin(\alpha+\beta)$;
$x_B = x_P + a*cos(\alpha+\beta+\gamma)$, $y_B = y_P + a*sin(\alpha+\beta+\gamma)$. |
As I was studying about sufficiency I came across your question because I also wanted to understand the intuition about From what I've gathered this is what I come up with (let me know what you think, if I made any mistakes, etc).
Let $X_1,\ldots,X_n$ be a random sample from a Poisson distribution with mean $\theta>0$.
We know that $T({\bf{X}})=\sum_{i=1}^{n} X_i$ is a sufficient statistic for $\theta$, since the conditional distribution of $X_1,\ldots,X_n$ given $T({\bf{X}})$ is free of $\theta$, in other words, does not depend on $\theta$.
Now, statistician $A$
knows that $X_1,\ldots,X_n \overset{i.i.d}{\sim} Poisson(4)$ and creates $n=400$ random values from this distribution:
n<-400
theta<-4
set.seed(1234)
x<-rpois(n,theta)
y=sum(x)
freq.x<-table(x) # We will use this latter on
rel.freq.x<-freq.x/sum(freq.x)
For the values statistician $A$ has created, he takes the sum of it and asks statistician $B$ the following:
"I have these sample values $x_1,\ldots,x_n$ taken from a Poisson distribution. Knowing that $\sum_{i=1}^{n} x_i = y = 4068$, what can you tell me about this distribution?"
So, knowing only that $\sum_{i=1}^{n} x_i = y = 4068$ (and the fact that the sample arose from a Poisson distribution) is sufficient for statistician $B$ to say anything about $\theta$? Since we know that this is a sufficient statistic we know that the answer is "yes".
To gain some intution about the meaning of this, let's do the following (taken from Hogg & Mckean & Craig's "Introduction to Mathematical Statistics", 7th edition, exercise 7.1.9):
"$B$ decides to create some fake observations, which he calls $z_1,z_2,\ldots,z_n$ (as he knows they will probably not be equal the original $x$-values) as follows. He notes that the conditional probability of independent Poisson random variables $Z_1,Z_2\ldots,Z_n$ being equal to $z_1,z_2,\ldots,z_n$, given $\sum z_i = y$, is
$$\cfrac{\frac{\theta^{z_1}e^{-\theta}}{z_1!} \frac{\theta^{z_2}e^{-\theta}}{z_2!} \cdots \frac{\theta^{z_n}e^{-\theta}}{z_n!}}{\frac{n \theta^{y}e^{-n\theta}}{y!}}=\frac{y!}{z_1!z_2! \cdots z_n!} \left(\frac{1}{n}\right)^{z_1} \left(\frac{1}{n}\right)^{z_2} \cdots \left(\frac{1}{n}\right)^{z_n}$$
since $Y=\sum Z_i$ has a Poisson distribution with mean $n \theta$. The latter distribution is multinomial with $y$ independent trials, each terminating in one of $n$ mutually exclusive and exhaustive ways, each of which has the same probability $1/n$. Accordingly, $B$ runs such a multinomial experiment $y$ independent trials and obtains $z_1,\ldots,z_n$."
This is what the exercise states. So, let's do exactly that:
# Fake observations from multinomial experiment
prob<-rep(1/n,n)
set.seed(1234)
z<-as.numeric(t(rmultinom(y,n=c(1:n),prob)))
y.fake<-sum(z) # y and y.fake must be equal
freq.z<-table(z)
rel.freq.z<-freq.z/sum(freq.z)
And let's see what $Z$ looks like (I'm also plotting the real density of Poisson(4) for $k=0,1,\ldots,13$ - anything above 13 is pratically zero -, for comparison):
# Verifying distributions
k<-13
plot(x=c(0:k),y=dpois(c(0:k), lambda=theta, log = FALSE),t="o",ylab="Probability",xlab="k",
xlim=c(0,k),ylim=c(0,max(c(rel.freq.x,rel.freq.z))))
lines(rel.freq.z,t="o",col="green",pch=4)
legend(8,0.2, legend=c("Real Poisson","Random Z given y"),
col = c("black","green"),pch=c(1,4))
So, knowing nothing about $\theta$ and knowing only the sufficient statistic $Y=\sum X_i$ we were able to recriate a "distribution" that looks a lot like a Poisson(4) distribution (as $n$ increases, the two curves become more similar).
Now, comparing $X$ and $Z|y$:
plot(rel.freq.x,t="o",pch=16,col="red",ylab="Relative Frequency",xlab="k",
ylim=c(0,max(c(rel.freq.x,rel.freq.z))))
lines(rel.freq.z,t="o",col="green",pch=4)
legend(7,0.2, legend=c("Random X","Random Z given y"), col = c("red","green"),pch=c(16,4))
We see that they are pretty similar, as well (as expected)
So, "for the purpose of making a statistical decision, we can ignore the individual random variables $X_i$ and base the decision entirely on the $Y=X_1+X_2+\cdots+X_n$" (Ash, R. "Statistical Inference: A concise course", page 59). |
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1988, Série Lendo, 125 p. --
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2. Temperature and Density in the Foot Points of the Molecular Loops in the Galactic Center; Analysis of Multi-J Transitions of 12CO (J = 1–0, 3–2, 4–3, 7–6), 13CO (J = 1–0), and C18O (J = 1–0)
Publications of the Astronomical Society of Japan, ISSN 0004-6264, 6/2010, Volume 62, Issue 3, pp. 675 - 695
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The European Physical Journal. C, Particles and Fields, ISSN 1434-6044, 2016, Volume 76
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The elliptic flow of prompt and non-prompt $$J/\psi $$ J/ψ was measured in the dimuon decay channel in Pb+Pb collisions at $$\sqrt{s_{_\text {NN}}}=5.02$$...
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8. Observation of a near-threshold omega J/psi mass enhancement in exclusive B -> K omega J/psi decays
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We report the observation of a near-threshold enhancement in the omega J/psi invariant mass distribution for exclusive B -> K omega J/psi decays. The results...
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A measurement of $$J/\psi $$ J/ψ and $$\psi (2\mathrm {S})$$ ψ(2S) production is presented. It is based on a data sample from Pb+Pb collisions at...
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postmarketing registration | thrombolysis | tissue plasminogen activator | alteplase | acute ischemic stroke | CONTROLLED-TRIAL | MANAGEMENT | J-ACT | GUIDELINES | ECASS II | CLINICAL NEUROLOGY | TISSUE-PLASMINOGEN-ACTIVATOR | DOUBLE-BLIND | ATLANTIS | PERIPHERAL VASCULAR DISEASE | OUTCOMES | ASSOCIATION | Humans | Japan | Middle Aged | Male | Tissue Plasminogen Activator - administration & dosage | Treatment Outcome | Stroke - drug therapy | Tissue Plasminogen Activator - therapeutic use | Time Factors | Brain Ischemia - drug therapy | Fibrinolytic Agents - therapeutic use | Thrombolytic Therapy - methods | Female | Registries | Aged | Brain Ischemia - mortality | Stroke - mortality
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Physical Review C - Nuclear Physics, ISSN 0556-2813, 10/2015, Volume 92, Issue 3
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Journal Article |
Tagged: subspace Problem 131
Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$.
\[V:=\left\{ \quad\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \in \R^4 \quad \middle| \quad x_1-x_2+x_3-x_4=0 \quad\right\}.\] Find a basis of the subspace $V$ and its dimension. Problem 125
Let $S$ be the following subset of the 3-dimensional vector space $\R^3$.
\[S=\left\{ \mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, x_1, x_2, x_3 \in \Z \right\}, \] where $\Z$ is the set of all integers. Determine whether $S$ is a subspace of $\R^3$. Problem 121
Let $A$ be an $m \times n$ real matrix. Then the
null space $\calN(A)$ of $A$ is defined by \[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\] That is, the null space is the set of solutions to the homogeneous system $A\mathbf{x}=\mathbf{0}_m$.
Prove that the null space $\calN(A)$ is a subspace of the vector space $\R^n$.
(Note that the null space is also called the kernel of $A$.) Read solution Problem 119
Let $\mathbf{a}$ and $\mathbf{b}$ be fixed vectors in $\R^3$, and let $W$ be the subset of $\R^3$ defined by
\[W=\{\mathbf{x}\in \R^3 \mid \mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0\}.\]
Prove that the subset $W$ is a subspace of $\R^3$.
Read solution Problem 79
Let $V$ be the set of all $n \times n$ diagonal matrices whose traces are zero.
That is,
\begin{equation*}
V:=\left\{ A=\begin{bmatrix} a_{11} & 0 & \dots & 0 \\ 0 &a_{22} & \dots & 0 \\ 0 & 0 & \ddots & \vdots \\ 0 & 0 & \dots & a_{nn} \end{bmatrix} \quad \middle| \quad \begin{array}{l} a_{11}, \dots, a_{nn} \in \C,\\ \tr(A)=0 \\ \end{array} \right\} \end{equation*}
Let $E_{ij}$ denote the $n \times n$ matrix whose $(i,j)$-entry is $1$ and zero elsewhere.
(a) Show that $V$ is a subspace of the vector space $M_n$ over $\C$ of all $n\times n$ matrices. (You may assume without a proof that $M_n$ is a vector space.) (b) Show that matrices \[E_{11}-E_{22}, \, E_{22}-E_{33}, \, \dots,\, E_{n-1\, n-1}-E_{nn}\] are a basis for the vector space $V$.
Add to solve later
(c) Find the dimension of $V$. Read solution Problem 61
Let $V$ and $W$ be subspaces of $\R^n$ such that $V \cap W =\{\mathbf{0}\}$ and $\dim(V)+\dim(W)=n$.
(a) If $\mathbf{v}+\mathbf{w}=\mathbf{0}$, where $\mathbf{v}\in V$ and $\mathbf{w}\in W$, then show that $\mathbf{v}=\mathbf{0}$ and $\mathbf{w}=\mathbf{0}$. (b) If $B_1$ is a basis for the subspace $V$ and $B_2$ is a basis for the subspace $W$, then show that the union $B_1\cup B_2$ is a basis for $R^n$. (c) If $\mathbf{x}$ is in $\R^n$, then show that $\mathbf{x}$ can be written in the form $\mathbf{x}=\mathbf{v}+\mathbf{w}$, where $\mathbf{v}\in V$ and $\mathbf{w} \in W$.
Add to solve later
(d) Show that the representation obtained in part (c) is unique. Problem 60
Let $T: \R^3 \to \R^3$ be the linear transformation given by orthogonal projection to the line spanned by $\begin{bmatrix}
1 \\ 2 \\ 2 \end{bmatrix}$. (a) Find a formula for $T(\mathbf{x})$ for $\mathbf{x}\in \R^3$. (b) Find a basis for the image subspace of $T$. (c) Find a basis for the kernel subspace of $T$. (d) Find the $3 \times 3$ matrix for $T$ with respect to the standard basis for $\R^3$. (e) Find a basis for the orthogonal complement of the kernel of $T$. (The orthogonal complement is the subspace of all vectors perpendicular to a given subspace, in this case, the kernel.) (f) Find a basis for the orthogonal complement of the image of $T$. (g) What is the rank of $T$?
(
Johns Hopkins University Exam) Problem 33
Suppose that $S$ is a fixed invertible $3$ by $3$ matrix. This question is about all the matrices $A$ that are diagonalized by $S$, so that $S^{-1}AS$ is diagonal. Show that these matrices $A$ form a subspace of $3$ by $3$ matrix space.
(
MIT-Massachusetts Institute of Technology Exam) Problem 15
Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?
(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ for $i=1,2,3,4$. (b) At $0$ each of the polynomials has the value $1$. Namely $p_i(0)=1$ for $i=1,2,3,4$.
(
University of California, Berkeley) |
Equilibrium Le Chatelier's Principle and Factors Effecting Equilibrium
Le Chatelier's Principle :
Applicable for only reversible reaction
Effect of concentration : If concentration of reactant ↑ / product ↓ → forward reaction If concentration of reactant ↓ / product ↑ → backward reaction Effect of pressure : Pressure(P) ↑ no. of moles ↓ (or) volume ↓ Pressure(P) ↓ no. of moles ↑ (or) volume ↑
ex :
↑ of P → favours forward reaction ↓ of P → favours backward reaction Effect of temperature : ↑ of temperature → favours endothermic reaction. ↓ of temperature → favours exothermic reaction.
ex :
↑ T → B.R → Reddish brown ↑ ↓ T → exothermic → intensity RB ↓ Haber's process : N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} ; \Delta H = -92kJ Temperature (T) : Low T ∴ 725 - 775 K Catalyst : Fe (or) FeO promoter : K_{2}O + Al_{2}O_{3} (or) Molybdenum Contact process : 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} ; \Delta H = -189kJ Pressure : High pressure 1 - 2 atm Temperature : Optimum temperature, 673 - 723 K Catalyst : Pt (or) V 2O 5 Degree of dissociation (α) using vapour density : \alpha = \frac{No. of \ particles \ dissociated}{No.of \ particles \ taken} \frac{D}{d} = 1 + (n - 1)x \ \Rightarrow \frac{D}{d} - 1 = (n - 1)x \frac{D - d}{(n - 1)d} = x \ (or) \ \frac{M - m}{m(n - 1)} = x D = initial vapour density d = equilibrium vapour density. View the Topic in this Video from 0:11 to 6:23
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1. Degree of Dissociation \alpha = \frac{D - d}{d(y - 1)}
where, y = number of moles of product from one mole of reactant, D = theoretical vapour density and d = observed vapour density
2. \alpha = \frac{M_{c} - M_{o}}{M_{o}}
where, M c = calculated molecular weight and M o = observed molecular weight. |
This tag is for questions involving the Dirac delta function, either in the informal sense, or in the distribution sense. The Dirac delta function is a mathematical construct which is called a generalized function or a distribution and was originally introduced by the British theoretical physicist Paul Dirac.
Mathematically, the delta function is not a function, because it is too singular. Instead, it is said to be a “distribution.” It is a generalized idea of functions, but can be used only inside integrals.
In fact, $~\int \delta~(x)~dx~$ can be regarded as an “operator” which pulls the value of a function at zero. Put it this way, it sounds perfectly legitimate and well-defined. But as long as it is understood that the delta function is eventually integrated, we can use it as if it is a function.
Definition:The Dirac delta functionor, delta function$~(~\delta~(x)~)~$ is defined by the properties $$\delta~(x) = \begin{cases} 0 \quad \text{if} \ x\not=0\\ \infty \quad \text{if} \ x=x\end{cases}\qquad \text{and}\qquad \int_0^1 \delta~(x)~dx=1$$
This function is very useful as an approximation for a tall narrow spike function, namely an impulse. For example, to calculate the dynamics of a baseball being hit by a bat, approximating the force of the bat hitting the baseball by a delta function is a useful device. The delta function not only enables the equations to be simplified, but it also allows the motion of the baseball to be calculated by only considering the total impulse of the bat against the ball, rather than requiring the details of how the bat transferred energy to the ball.
There are three main properties of the Dirac Delta function that we need to be aware of. These are,
$1.\quad$ $$~\delta \left( {t - a} \right) = 0~~~~~~~t \ne a~$$ $2.\quad$ $$~\displaystyle \int_{{a - \varepsilon }}^{{ a + \varepsilon }}{{\delta \left( {t - a} \right)~dt}} = 1,\hspace{0.25in}\varepsilon > 0~$$ $3.\quad$ $$~\displaystyle \int_{{ a - \varepsilon }}^{{ a + \varepsilon }}{{f\left( t \right)\delta \left( {t - a} \right)~dt}} = f\left( a \right),\hspace{0.25in}~~~~~~~~~~\varepsilon > 0~$$
References: |
I would like to maximize the following definite double integral with respect to $0<r<\frac{1}{2}$:
$\int _0^{\frac{d}{1-r}}\int _{\frac{d+r x-x}{r}}^s [(1-r) x+r y -d]dydx$
under the constraints: $0\leq d \leq 1$, $2d < s \leq 4$, $0 < r < \frac{1}{2}$, $0 \leq x \leq s$, $0\leq y \leq s$.
Note that the variables of integration, i.e. $x$ and $y$, are both bounded by $0$ and $s$; for this reason, I have to be careful in my code regarding the limits of integration regions.
Here is my code:
maxR = Flatten[Table[{d, s, r /. Last@Maximize[{Integrate[(1 - r)*x + r*y - d, {x, 0, Min[s, d/(1 - r)]}, {y, Max[0, Min[s, (d + r*x - x)/r]], s}], 0 <= d <= 1, 2 d < s, 0 < r < 1/2}, {r}]}, {d, 0, 1, .1}, {s, 1, 4, .1}], 1];ListPlot3D[maxR, PlotRange -> {0, 1}, AxesLabel -> {"d", "s", "r"}]
And this is running forever. Can anyone help please?
By the way, this is an extension of Ulich's answer to this question. While the latter shows how to get the solution of the integral, my question here regards a maximization of the integral. |
I'm trying to calculate heat loss when a human body is simply suspended in $20 \sideset{^{\circ}}{}{\mathrm{C}}$ water, but I keep getting absurdly high results. It should be a simple application of this equation$$ H = \frac{K \cdot A \cdot \Delta T}{\Delta x} \,,$$where:
$H$ is heat loss,
$K$ is the thermal conductivity of skin ($0.3\frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}$, according to this page),
$A$ is the surface area of a human ($1.7 \, {\mathrm{m}}^{2}$, according to this page),
$\Delta T$ is the temperature difference $\left(37 - 20 = 17\right) ,$ and
$\Delta x$ is the thickness of the skin ($\sim 1 \, \mathrm{mm}$, according to this site).
Plugging those numbers in, we get$$H~=~ \frac{K \cdot A \cdot \Delta T}{\Delta x}~ \approx ~ \frac{0.3 \cdot 1.7 \cdot 17}{0.001}~ \approx ~ 8670 \, \mathrm{W}\,.$$Now, that seemed fine until I realized that $8500 \, \mathrm{W}$ is a
lot of energy – essentially 2 food calories per second. That seems highly inaccurate and implies that you're incapable of surviving more than 20 minutes in $20 \sideset{^{\circ}}{}{\mathrm{C}}$ water before using up an entire day's worth of calories.
Is there an error in my calculations somewhere, or an error in the numbers I'm using?
I also found this source which suggests that the thermal conductivity of water is around $83\frac{\mathrm{W}}{{\mathrm{m}}^2 \cdot \mathrm{K}},$ which is both a very different number and has different units.
I don't think that this discrepancy is due to a protective layer of warmer water forming near the body, because swimmers also survive longer than half an hour and they're constantly disrupting any warm boundary layer that may form.
This question is similar, but I can't figure out how to reverse-engineer it. Perhaps someone else can explain the logic there more clearly? |
The travelling salesman problem (TSP) asks the following question: Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the origin city? It is an NP-hard problem in combinatorial optimization, important in operations research and theoretical computer science.
TSP can be formulated as an integer linear program. Label the cities with the numbers $0, \ldots, n$ and define: $$ x_{ij} = \begin{cases} 1 & \text{the path goes from city } i \text{ to city } j \\ 0 & \text{otherwise} \end{cases} $$ For $i = 1, \ldots, n$, let $u_i$ be an artificial variable, and finally take $c_{ij}$ to be the distance from city $i$ to city $j$. Then TSP can be written as the following integer linear programming problem:
\begin{align} \min &\sum_{i=0}^n \sum_{j\ne i,j=0}^nc_{ij}x_{ij} && \\ & 0 \le x_{ij} \le 1 && i,j=0, \cdots, n \\ & u_{i} \in \mathbf{Z} && i=0, \cdots, n \\ & \sum_{i=0,i\ne j}^n x_{ij} = 1 && j=0, \cdots, n \\ & \sum_{j=0,j\ne i}^n x_{ij} = 1 && i=0, \cdots, n \\ &u_i-u_j +nx_{ij} \le n-1 && 1 \le i \ne j \le n \end{align}
The first set of equalities requires that each city be arrived at from exactly one other city, and the second set of equalities requires that from each city there is a departure to exactly one other city. The last constraints enforce that there is only a single tour covering all cities, and not two or more disjointed tours that only collectively cover all cities. To prove this, it is shown below (1) that every feasible solution contains only one closed sequence of cities, and (2) that for every single tour covering all cities, there are values for the dummy variables $u_i$ that satisfy the constraints.
To prove that every feasible solution contains only one closed sequence of cities, it suffices to show that every subtour in a feasible solution passes through city 0 (noting that the equalities ensure there can only be one such tour). For if we sum all the inequalities corresponding to $x_{ij}=1$ for any subtour of $k$ steps not passing through city 0, we obtain: $$ nk \leq (n-1)k, $$ which is a contradiction.
It now must be shown that for every single tour covering all cities, there are values for the dummy variables $u_i$ that satisfy the constraints. Without loss of generality, define the tour as originating (and ending) at city 0. Choose $u_{i}=t$ if city $i$ is visited in step $t$ $(i, t = 1, 2, \ldots, n)$. Then $$ u_i-u_j\le n-1, $$ since $u_i$ can be no greater than $n$ and $u_j$ can be no less than 1; hence the constraints are satisfied whenever $x_{ij}=0$. For $x_{ij}=1$, we have $$u_{i} - u_{j} + nx_{ij} = (t) - (t+1) + n = n-1,$$ satisfying the constraint.
The problem has been shown to be NP-hard (more precisely, it is complete for the complexity class FPNP; see function problem), and the decision problem version (“given the costs and a number x, decide whether there is a round-trip route cheaper than x”) is NP-complete.
In the general case, finding a shortest travelling salesman tour is NPO-complete. If the distance measure is a metric and symmetric, the problem becomes APX-complete and Christofides’s algorithm approximates it within 1.5.
If the distances are restricted to 1 and 2 (but still are a metric) the approximation ratio becomes 8/7. In the asymmetric, metric case, only logarithmic performance guarantees are known, the best current algorithm achieves performance ratio $0.814 log(n)$; it is an open question if a constant factor approximation exists. |
The catenary is a plane curve, whose shape corresponds to a hanging homogeneous flexible chain supported at its ends and sagging under the force of gravity.
The catenary is similar to parabola (Figure \(1\)).
So it was believed for a long time. In the early \(17\)th century Galileo doubted that a hanging chain is actually a parabola. However, a rigorous proof was obtained only half a century later after Isaak Newton and Gottfried Leibniz developed a framework of differential and integral calculus.
The solution of the problem about the catenary was published in \(1691\) by Christiaan Huygens, Gottfried Leibniz, and Johann Bernoulli.
Below we derive the equation of catenary and some its variations.
Suppose that a heavy uniform chain is suspended at points \(A, B,\) which may be at different heights (Figure \(2\)).
Consider equilibrium of a small element of the chain of length \(\Delta s.\) The forces acting on the section of the chain are the distributed force of gravity
\[\Delta P = \rho g\Delta s,\]
where \(\rho\) is the density of the chain material, \(g\) is the acceleration of gravity, \(A\) is the cross sectional area of the thread, and the tension forces \(T\left( x \right)\) and \(T\left( {x + \Delta x} \right),\) respectively, at points \(x\) and \({x + \Delta x}.\)
The equilibrium conditions of the element of the length \(\Delta s\) for projections on the axes \(Ox\) and \(Oy\) are written as
\[ {- T\left( x \right)\cos \alpha \left( x \right) }+{ T\left( {x + \Delta x} \right)\cos \alpha \left( {x + \Delta x} \right) }={ 0,}\]
\[ {- T\left( x \right)\sin\alpha \left( x \right) }+{ T\left( {x + \Delta x} \right)\sin\alpha \left( {x + \Delta x} \right) }-{ \Delta P = 0.} \]
It follows from the first equation that the horizontal component of the tension force \(T\left( x \right)\) is always a constant:
\[{T\left( x \right)\cos \alpha \left( x \right) = {T_0} }={ \text{const}.}\]
Using differentials in the second equation we can rewrite it as
\[{d\left( {T\left( x \right)\sin\alpha \left( x \right)} \right) }={ dP\left( x \right).}\]
As \(T\left( x \right) =\) \( {\large\frac{{{T_0}}}{{\cos\alpha \left( x \right)}}\normalsize},\) we have
\[
{d\left( {{T_0}\tan\alpha \left( x \right)} \right) = dP\left( x \right),\;\;}\Rightarrow {{T_0}d\left( {\tan\alpha \left( x \right)} \right) = dP\left( x \right).} \]
Take into account that \(\tan \alpha \left( x \right) =\) \({\large\frac{{dy}}{{dx}}\normalsize} = y’,\) so the equilibrium equation is written in the differential form as
\[
{{T_0}d\left( {y’} \right) = dP\left( x \right),\;\;}\Rightarrow {{T_0}d\left( {y’} \right) = \rho gAds.} \]
The chain element of the length \(\Delta s\) can be expressed by the formula
\[ds = \sqrt {1 + {{\left( {y’} \right)}^2}} dx.\]
As a result we obtain the differential equation of the catenary:
\[
{{T_0}\frac{{dy’}}{{dx}} = \rho gA\sqrt {1 + {{\left( {y’} \right)}^2}} ,\;\;}\Rightarrow {{T_0}y^{\prime\prime} = \rho gA\sqrt {1 + {{\left( {y’} \right)}^2}} .} \]
The order of this equation can be reduced. By denoting \(y = z,\) we can represent it as the first order equation:
\[{T_0}z’ = \rho gA\sqrt {1 + {z^2}} .\]
The last equation can be solved by separating variables.
\[
{{T_0}dz = \rho gA\sqrt {1 + {z^2}} dx,\;\;}\Rightarrow {\frac{{dz}}{{\sqrt {1 + {z^2}} }} = \frac{{\rho gA}}{{{T_0}}}dx,\;\;}\Rightarrow {\int {\frac{{dz}}{{\sqrt {1 + {z^2}} }}} = \frac{{\rho gA}}{{{T_0}}}\int {dx} ,\;\;}\Rightarrow {\ln \left( {z + \sqrt {1 + {z^2}} } \right) }={ \frac{x}{a} + {C_1}.} \]
Here we denoted \(\large\frac{{\rho gA}}{{{T_0}}}\normalsize\) as \(\large\frac{1}{a}\normalsize.\)
The tangent to the catenary at the lowest point is parallel to the \(x\)-axis. Hence,
\[{z\left( {x = 0} \right) = y’\left( {x = 0} \right) }={ 0.}\]
We can determine the constant \({C_1}\) from here:
\[{\ln 1 = 0 + {C_1},\;\; }\Rightarrow{ {C_1} = 0.}\]
Thus, we get the following equation:
\[z + \sqrt {1 + {z^2}} = {e^{\large\frac{x}{a}\normalsize}}.\]
Multiplying both sides of the equation by the conjugent expression \(z – \sqrt {1 + {z^2}} \) gives
\[
{{\left( {z + \sqrt {1 + {z^2}} } \right) \cdot}\kern0pt{ \left( {z – \sqrt {1 + {z^2}} } \right) }={ \left( {z – \sqrt {1 + {z^2}} } \right){e^{\large\frac{x}{a}\normalsize}},\;\;}}\Rightarrow {{{z^2} – \left( {1 + {z^2}} \right) }={ \left( {z – \sqrt {1 + {z^2}} } \right){e^{\large\frac{x}{a}\normalsize}},\;\;}}\Rightarrow { – 1 = \left( {z – \sqrt {1 + {z^2}} } \right){e^{\large\frac{x}{a}\normalsize}},\;\;}\Rightarrow {z – \sqrt {1 + {z^2}} = – {e^{ – \large\frac{x}{a}\normalsize}}.} \]
Adding to the previous equation, we find the expression for \(z = y’:\)
\[\require{cancel}
{{z + \cancel{\sqrt {1 + {z^2}}} }+{ z – \cancel{\sqrt {1 + {z^2}}} }={ {e^{\large\frac{x}{a}\normalsize}} – {e^{ – \large\frac{x}{a}\normalsize}},\;\;}}\Rightarrow {{z = \frac{{{e^{\large\frac{x}{a}\normalsize}} – {e^{ – \large\frac{x}{a}\normalsize}}}}{2} }={ \sinh \frac{x}{a},\;\;}}\Rightarrow {y’ = \sinh \frac{x}{a}.} \]
Integrating once more gives the final nice expression for the shape of the catenary:
\[y = a\cosh \frac{x}{a}.\]
Thus, the catenary is described by the hyperbolic cosine function. Its shape is uniquely determined by the parameter \(a = {\large\frac{{{T_0}}}{{\rho gA}}\normalsize}\) as shown in Figure \(3.\)
Catenaries are often found in nature and technology. For example, the square sail under the pressure of the wind takes the form of a catenary (this problem has been considered by Jacob Bernoulli).
The archs in the form of an inverted catenary (such as Saarinen’s Gateway Arch in St.Louis shown in Figure \(4\)) are often used in architecture and construction.
They have a high stability because the internal compression forces are ideally compensated and do not cause sagging.
The catenary has another interesting feature. When revolved about the
x-axis, the catenary gives the surface called catenoid. This surface has minimum surface area, i.e. any part of the catenoid will be less than any other surface bounded by the same contour. In particular, the soap film between two circles trying to minimize the free energy takes the form of a catenoid. Solved Problems
Click a problem to see the solution. |
The typical thing to do is visual inspection of the pre-treatment trends for the control and treatment group. This is particularly easy if you only have those two groups given a single binary treatment. Ideally the pre-treatment trends should look something like this:
This graph was taken from a previous answer to the question why we need the common trends assumption. This includes also an explanation of the blue-dashed line which is the counterfactual outcome for the treated that can be assumed if we can reasonably verify the parallel trends assumption.
A formal test which is also suitable for multivalued treatments or several groups is to interact the treatment variable with time dummies. Suppose you have 3 pre-treatment periods and 3 post-treatment periods, you would then regress$$y_{it} = \lambda_i + \delta_t + \beta_{-2}D_{it} + \beta_{-1}D_{it} + \beta_1 D_{it} + \beta_2 D_{it} + \beta_3 D_{it} + \epsilon_{it}$$
where $y$ is the outcome for individual $i$ at time $t$, $\lambda$ and $\delta$ are individual and time fixed effects (this is a generalized way of writing down the diff-in-diff model which also allows for multiple treatments or treatments at different times).
The idea is the following. You include the interactions of the time dummies and the treatment indicator for the first two pre-treatment periods and you leave out the one interaction for the last pre-treatment period due to the dummy variable trap. Also now all the other interactions are expressed relative to the omitted period which serves as the baseline. If the outcome trends between treatment and control group are the same, then $\beta_{-2}$ and $\beta_{-1}$ should be insignificant, i.e. the difference in differences is not significantly different between the two groups in the pre-treatment period.
An attractive feature of this test is that also the interactions of the time dummies after the treatment with the treatment indicator is informative. For instance, $\beta_{1}, \beta_2, \beta_3$ show you whether the treatment effect fades out over time, stays constant, or even increases. An application of this approach is Autor (2003).
Note that the literature generally refers to $\beta_{-2}, \beta_{-1}$ as "leads" and $\beta_{1}, \beta_2, \beta_3$ as "lags", even though they are merely interactions of the treatment indicator with time dummies and are not actually leads and lags of the treatment indicator in a time-series jargon sense. A more detailed explanation of this parallel trends test is provided in the lecture notes by Steve Pischke (here on page 7, or here on page 9). |
Consider a consumer whose preferences can be represented by the following utility function: $$u(x_1,x_2)=\dfrac{x_2}{(1+x_1)^2}.$$
Assume the agent's income is $y=5$. The price of one unit of good $1$ is $p_1=1$. For each unit of good $1$ the agent buys, he qualifies to buy up to one unit of good $2$ at an additional price of $p_2=1$. In other words, to buy one unit of good $2$ the agent has to first buy one unit of good $1$. The agent must consume everything he buys. Using this information, sketch the feasible set. Is it convex? Derive the utility maximizing bundle.
How does your answers to question 4. change if the agent does not have to consume everything he buys ("free disposal")?
For Q 4 :
Utility maximization problem of the consumer is :
\begin{eqnarray*} \max_{x_1, x_2} & \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} & \ \ x_1+x_2 \leq 5 \\ \text{and} & \ \ 0 \leq x_2 \leq x_1 \end{eqnarray*}
Here is the constraint set of the consumer, along with a few indifference curves:
Observe that the constraint set is convex and the consumer does not spend all his income in optimum. His optimal consumption bundle is $(x_1, x_2) = (1,1)$.
For Q 5 :
Utility maximization problem (with free disposal) of the consumer is :
\begin{eqnarray*} \max_{x_1, x_2, b_1, b_2} & \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} & \ \ b_1+b_2 \leq 5 \\ & \ \ 0 \leq b_2 \leq b_1 \\ \text{and} & \ \ 0 \leq x_1 \leq b_1, 0 \leq x_2 \leq b_2\end{eqnarray*}
Here $b_1$, $b_2$ denotes the amount of the two commodities bought by the consumer, and $x_1$, $x_2$ denotes the amount consumed. In this case, the consumer will try and maximize his consumption of commodity 2 $(x_2)$ by buying as much amount of commodity 2 $(b_2)$ as he can. Clearly, the solution to this utility maximization problem is$b_1 = b_2 = x_2 = 2.5, x_1 = 0$.
For Q 4, here is a way to solve the optimization problem using Lagrangian method :
Given the utility maximization problem of the consumer :
\begin{eqnarray*} \max_{x_1, x_2} & \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} & \ \ x_1+x_2 \leq 5 \\ \text{and} & \ \ 0 \leq x_2 \leq x_1 \end{eqnarray*}
We set up the Lagrangian as follows:
$\mathcal{L}(x_1, x_2) = \dfrac{x_2}{(1+x_1)^2} - \lambda(x_1+x_2-5) +\mu_1(x_1-x_2)+ \mu_2x_2 $
Necessary conditions for optimality are as follows :
$\dfrac{\partial \mathcal{L}}{\partial x_1} = \dfrac{-2x_2}{(1+x_1)^3} - \lambda + \mu_1 = 0$
$\dfrac{\partial \mathcal{L}}{\partial x_2} = \dfrac{1}{(1+x_1)^2} -\lambda - \mu_1 + \mu_2 = 0$
$x_1+x_2 \leq 5$, $\lambda \geq 0$ and $\lambda(x_1+x_2-5) = 0$
$x_1 \geq x_2$, $\mu_1 \geq 0$ and $\mu_1(x_1-x_2) = 0$
$x_2 \geq 0$, $\mu_2 \geq 0$ and $\mu_2x_2 = 0$
Solving the above system, we get
$x_1 = 1$, $x_2 = 1$, $\mu_1 = \frac{1}{4}$, $\mu_2=0$, $\lambda = 0$
Alternatively for Q 4, the problem can also be converted to a single variable optimization problem. This is because the individual will always consume equal amounts of $x_1$ and $x_2$ in equilibrium. Substituting $x_2=x_1$ yields :
\begin{eqnarray*} \max_{x_2} & \ \ \frac{x_2}{(1+x_2)^2} \\ \text{s.t.} & \ \ 0 \leq x_2 \leq 2.5 \end{eqnarray*}
Differentiating $\dfrac{x_2}{(1+x_2)^2}$ with respect to $x_2$ gives the following necessary condition
$\dfrac{(1+x_2)^2 - 2(1+x_2)x_2}{(1+x_2)^4} = \dfrac{1-x_2^2}{(1+x_2)^4} = 0$
which yields $x_2 = 1$, and the corresponding value of $x_1 = 1$. |
Remember that by the definition of the limit supremum, we have$$\limsup_{t\to\infty} f(t) \leq M \iff \left(\forall \epsilon > 0, \exists T : t > T \implies f(t) \leq M + \epsilon\right)$$In particular, note that we cannot conclude $f(t) \leq M$ for any $t.$ (e.g. note that $\limsup_{n\to\infty} 1 + \frac{1}{n} \leq 1$)In your context, this means ...
Assuming you are talking about probability measures on the reals, and convolution with respect to $(\mathbb R, +)$, the answer is No. A convolution semigroup need not be continuous anywhere.Let $f$ be any discontinuous solution of Cauchy's functional equation $f(x+y)=f(x)+f(y)$ (these are discontinuous everywhere, and unbounded in every neighborhood) and ...
For $i\in\{1,\dots,6\}$, let $m_i = E(T_B | Z_t = i)$ be the expected number of steps until reaching $\{3,4,5\}$, starting from state $i$. Trivially, $m_3=m_4=m_5=0$.To obtain a system of linear constraints, apply first-step analysis (conditioning on the first step out of state $i$). For $i=1$, we have\begin{align}m_1 &= \sum_{j\in\{2,6\}} E(T_B | ...
Taking two steps in the Markov chain can lead to one of two things, with equal probability:$1 \to 2 \to 3$ or $1 \to 6 \to 5$ and we're done.$1 \to 2 \to 1$ or $1 \to 6 \to 1$ and we're back where we started.We took $2$ steps. In the first case, we have $0$ steps left, and in the second case, we have $\mathbb E[T_B]$ more steps left in expectation. ...
You can actually compute the integral with the formula $d(tY_t) = Y_t \, dt + t \, dY_t$. Fixing $Y_t = B_t^2$, you have by Itô formula that $dY_t = 2B_t dB_t + dt$ and therefore\begin{align}\int_0^1 B_t^2 d t &= \int_0^1 d(tB_t^2) - \int_0^1 td(B_t^2)\\&= B_1^2 - \int_0^12tB_tdB_t - \int_0^1tdt\\\&= B_1^2 - \frac12 - \int_0^12tB_tdB_t.\end{...
You have the right mean and variance for$$ \int_0^1 B_t^2 \; dt \; ?$$But, its distribution is not of any known functions. In fact, it is rather complex. You may find some research results in the link below.https://projecteuclid.org/download/pdf_1/euclid.aop/1020107767
This question has been asked before in this forum. My answer in Determining distribution of $X_t = \int_0^t W_s^2 \mathrm{d} s$ gives a couple of references to the original solution of Cameron & Martin, and a different approach of M. Kac.
Well, in these sorts of things there's not really much to do except try to exploit the independence of the increments. So here, $W_{t+2}=W_{t}+W_{t+2}-W_{t}$ and $W_{t+1}=W_t+W_{t+1}-W_t$. Thus, we get$$\mathbb{E}W_t(W_{t+1}+W_{t+2})=2\mathbb{E} W_t^2+\mathbb{E}W_t(W_{t+2}-W_{t})+\mathbb{E}W_t(W_{t+1}-W_t)$$And by independence and Gaussianity, we thus ...
As an alternative approach in the spirit of your initial attempt, you can condition on the number $2k$ of steps to reach $\{3,4,5\}$ from $1$:\begin{align}E(T_B)&= \sum_{k=1}^\infty 2k\ P(\text{$2k$ steps}) \\&= \sum_{k=1}^\infty 2k\ 2^k \left(\frac{1}{2}\right)^{2k} \\&= \sum_{k=1}^\infty k \left(\frac{1}{2}\right)^{k-1} \\&= \frac{1}{...
Hints:$\ S_n\ $ is odd when $\ n\ $ is odd, and even when $\ n\ $ is even.If $\ S_n = m\ $ then $\ \frac{m+n}{2}\ $ of the first $\ n\ $ steps must have been to the right and $\ \frac{n-m}{2}\ $ of them must have been to the left.The number of rightward steps among the first $\ n\ $ follows a binomial distribution with parameters $\ n\ $ and $\frac{1}{2}\...
Your sollution is overly complicated. The easy path to finding the probability generating function of the product of independent random variables is to use the definition.$$\mathsf G_X(t)=\mathsf E(t^X)$$Let $Y$ be the random variable resulting in $1$ if the coin toss is head and $2$ if it is tails. So $Z=YX$ and $Y\sim\mathcal U\{1,2\}$.$$\begin{...
I actually found a way to solve my problem.The rows of the limiting transition matrix are linear combinations of the stationary distributions, $\pi_k$, $k=1,...,K$ of the $K$ strongly connected components that correspond to a leaf of the condensation graph (i.e. the absorbing states).The coefficients associated with each $\pi_k$ for each row are given by ...
This is a birth-death process, so we may instead use the detailed balance equations$$\lambda(1-\mu)\pi_{n-1} = (1-\lambda)\mu \pi_n,\quad n\geqslant 1.$$This yields the recurrence$$\pi_n = \left(\frac{\lambda(1-\mu)}{\mu(1-\lambda)}\right)^n\pi_0,\quad n\geqslant 0.$$Now, assuming that $\frac{\lambda(1-\mu)}{\mu(1-\lambda)}<1$, we have convergence ...
Let S be the set of all the states of the Markov chain. Consider a matrix $A$ and $B$ such that $A$=$P^{r-1}$=$(a_{ij})_{i,j\in S}$ and $B$=$P^r$=$(b_{ij})_{i,j\in S}$ respectively, where $P$=$(p_{ij})_{i,j\in S}$ is the transition matrix of the Markov Chain.Then,$P^{r+1} = BP = \Biggl($$\sum_{k} b_{ik}p_{kj} \Biggr)_{i,j\in S}$As per the question, $... |
In Zee's Group Theory in a Nutshell book, he says that the antisymmetric tensor $A^{ij}$ furnishes a 6 dimensional representation of $SO(4)$. He further argues that this 6 dimensional representation can be broken up into two 3 dimensional representations by listing the following grouping ($A^{14}$, $A^{24}$, $A^{34}$) and ($A^{12}$, $A^{23}$, $A^{31}$). Why does this make sense? Why do the elements in each group only transform amongst themselves?
OP is correct: The Lie algebra isomorphism $$so(4)~\cong~ so(3)_+~\oplus~ so(3)_-\tag{1}$$ uses (anti)self-dual real antisymmetric $4\times 4$ matrices, respectively, cf. e.g. this Math.SE post.
However Ref. 1 is trying to make another point. Instead Ref. 1 is discussing branching rules for the subgroup $$H:=SO(3)~\ni R ~\mapsto~ \begin{pmatrix} R & \vec{0}\cr \vec{0}^t & 1 \end{pmatrix}_{4\times 4}~\in ~ G:=SO(4).\tag{2}$$ In this specific embedding (2) of the subgroup, the explicit splitting of the $G$-representation into irreducible $H$-representations $${\bf 6}_G\cong {\bf 3}_H\oplus {\bf 3}_H\tag{3}$$ is as Zee writes: $$ A^{\prime 4i}~=~R^i{}_j A^{4j}, \qquad A^{\prime k\ell} ~=~ R^k{}_iA^{ij} R^{\ell}{}_j, \qquad i,j,k,\ell~\in\{1,2,3\}.\tag{4} $$
References:
A. Zee, Group Theory in a Nutshell for Physicists,2016; p. 197.
The Lie algebras of SO(4) and SU(2)$\times$SU(2) are isomorphic, so the you can get representations of SO(4) former by taking the tensor product of two representations of SU(2). |
I was a little confused by the answer above, hence I'll give it another shot. I think the question is not actually about 'classical' linear regression but about the style of that particular source. On the classical regression part:
However, the linearity assumption by itself does not put any structure on our model
That is absolutely correct. As you have stated, $\epsilon$ might as well kill the linear relation and add up something completely independent from $X$ so that we cannot compute any model at all.
Is Greene being sloppy? Should he actually have written: $E(y|X)=Xβ$
I do not want to answer the first question but let me sum up the assumptions you need for usual linear regression:
Let us assume that you observe (you are given) data points $x_i \in \mathbb{R}^d$ and $y_i \in \mathbb{R}$ for $i=1,...,n$. You need to assume that the data $(x_i, y_i)$ you have observed comes from independently, identically distributed random variables $(X_i, Y_i)$ such that ...
There exists a fixed (independent of $i$) $\beta \in \mathbb{R}^d$ such that $Y_i = \beta X_i + \epsilon_i$ for all $i$ and the random variables $\epsilon_i$ are such that
The $\epsilon_i$ are iid as well and $\epsilon_i$ is distributed as $\mathcal{N}(0, \sigma)$ ($\sigma$ must be independent of $i$ as well)
For $X = (X_1, ..., X_n)$ and $Y = (Y_1, ..., Y_n)$ the variables $X, Y$ have a common density, i.e. the single random variable $(X, Y)$ has a density $f_{X,Y}$
Now you can run down the usual path and compute
$$f_{Y|X}(y|x) = f_{Y,X}(y,x)/f_X(x) = \left(\frac{1}{\sqrt{2\pi d}}\right)^n \exp{\left( \frac{-\sum_{i=1}^n (y_i - \beta x_i)^2}{2\sigma}\right)} $$
so that by the usual 'duality' between machine learning (minimalization of error functions) and probability theory (maximization of likelihoods) you maximize $-\log f_{Y|X}(y|x)$ in $\beta$ which in fact, gives you the usual "RMSE" stuff.
Now as stated: If the author of the book you are quoting wants to make this point (which you have to do if you ever want to be able to compute the 'best possible' regression line in the basic setup) then yes, he
must make this assumption on the normalicity of the $\epsilon$ somewhere in the book.
There are different possibilities now:
He does not write this assumption down in the book. Then it is an error in the book.
He does write it down in form of a 'global' remark like 'whenever I write $+ \epsilon$ then the $\epsilon$ are iid normally distributed with mean zero unless stated otherwise'. Then IMHO it is bad style because it causes exactly the confusion that you feel right now. That is why I tend to write the assumptions in some shortened form in
every Theorem. Only then every building block can be viewed cleanly in its own right. He does write it down closely to the part you are quoting and you/we just did not notice it (also a possibility :-))
However, also in a strict mathematical sense, the normal error is something canonical (the distribution with the highest entropy [once the variance is fixed], hence, producing the strongest models) so that some authors tend to skip this assumption but use in nontheless. Formally, you are absolutely right: They are using mathematics in the "wrong way". Whenever they want to come up with the equation for the density $f_{Y|X}$ as stated above then they need to know $\epsilon$ pretty well, otherwise you just have properties of it flying around in every senseful equation that you try to write down. |
I need help to make a diagram(square), someone can teach me how to do?
I know that I could look at the posts to see a model, but I am stopped for 7 days to edit questions
Thanks in advance.
$$\begin{array}{ccc}A&\to& B\\\downarrow &\nearrow&\uparrow\\A'&\to& B'\end{array}$$
$$\begin{array}{ccc}A&\to& B\\\downarrow &\nearrow&\uparrow\\A'&\to& B'\end{array}$$
Even if you are stopped from editing questions for seven days, you can still see the MathJax code!
Right click on a math expression (for instance, that in Asaf's answer), in the context menu, select "show math as" -> "tex commands". This will bring up a pop-up window showing the MathJax expression that went into constructing that particular display. |
$\displaystyle y^{'} = {\frac{y}{x}} + 1$
cannot be solved for $y$ as a power series $x$. Solve this equation for $y$ as a power series in powers for $ x-1 $.: Introduce $t=x-1$ as a new independent variable and solve the resulting equation for y as a power series in $t$.)
So far I have proved that it cannot be solved for y as power series in terms of x because you get negative exponents. I did solve for x, which equal t+1.
So, the equation looks like $\displaystyle y^{'} = {\frac{y}{t+1}} + 1$.
Can I multiply both sides by $(t+1)$ so the equation looks like $ \displaystyle (t+1)y^{'} = y + (t+1)$?
My solution so far:
$y^{'} = \displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n-1}}$
$y = \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}}$
$ \displaystyle (t+1)y^{'} = y + (t+1)$
$(t+1)\displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n-1}} = \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} + (t+1)$
$\displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n}} - \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} =(t+1)$
$\displaystyle \sum^{\infty}_{n=0}{n}{c_n}{(t+1)^{n}} - \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} =(t+1)$
$\displaystyle \sum^{\infty}_{n=0}[nc_n-c_n]*(t+1)^n = (t+1)$
$\displaystyle nc_n-c_n = (t+1)$
$\displaystyle c_n(n-1) = (t+1)$
$\displaystyle c_n = \frac {(t+1)}{n-1} $
Substituing x back in we get:
$\displaystyle c_n = \frac {(x)}{n-1} $
Did I do this correctly? |
Let\[\mathbf{v}_{1}=\begin{bmatrix}1 \\ 1\end{bmatrix},\;\mathbf{v}_{2}=\begin{bmatrix}1 \\ -1\end{bmatrix}.\]Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?
For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]
For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]
(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?
(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.
(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.
(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]
(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?
(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.
Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis\[B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.\]Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.
Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.\]
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,\[T (f) (x) = x f(x).\]
Prove that $T$ is a linear transformation, and find its range and nullspace.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$.
(a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.
Let $V$ be a vector space and $B$ be a basis for $V$.Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form\[\begin{bmatrix}1 & 0 & 2 & 1 & 0 \\0 & 1 & 3 & 0 & 1 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0\end{bmatrix}.\]
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.Let\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}a & b\\c& -a\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\]be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.
(c) Prove that\[B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}\]is a basis for $V$. |
I wanted to better understand dfa. I wanted to build upon a previous question:Creating a DFA that only accepts number of a's that are multiples of 3But I wanted to go a bit further. Is there any way we can have a DFA that accepts number of a's that are multiples of 3 but does NOT have the sub...
Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show th...
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? )I wonder about counting the irreducible elements bounded by a lower...
How would you make a regex for this? L = {w $\in$ {0, 1}* : w is 0-alternating}, where 0-alternating is either all the symbols in odd positions within w are 0's, or all the symbols in even positions within w are 0's, or both.
I want to construct a nfa from this, but I'm struggling with the regex part |
I wanted to better understand dfa. I wanted to build upon a previous question:Creating a DFA that only accepts number of a's that are multiples of 3But I wanted to go a bit further. Is there any way we can have a DFA that accepts number of a's that are multiples of 3 but does NOT have the sub...
Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show th...
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? )I wonder about counting the irreducible elements bounded by a lower...
How would you make a regex for this? L = {w $\in$ {0, 1}* : w is 0-alternating}, where 0-alternating is either all the symbols in odd positions within w are 0's, or all the symbols in even positions within w are 0's, or both.
I want to construct a nfa from this, but I'm struggling with the regex part |
Let $\phi$ be a scalar field in an interacting theory ($\phi^3$ or $\phi^4$, for example). If $|0\rangle$ is the vacuum of the interacting theory and $P^\mu$ is the four-momentum operator, we have that
$$\langle 0 | \phi(x) | 0 \rangle = \langle 0 | e^{-iPx} \phi(0) e^{iPx} | 0 \rangle = \langle 0 | \phi(0) | 0 \rangle $$
In chapter 5 of his QFT book, Srednicki says that
We would like $ \langle 0 | \phi(0) | 0 \rangle$ to be zero. This is because we would like $a_1^\dagger (\pm \infty)$, when acting on $|0\rangle$, to create a single particle state. We do
notwant $a_1^\dagger (\pm \infty)$ to create a linear combination of a single particle state and the ground state.
Here $a_1^\dagger (\pm \infty)$ is the creation operator $a^\dagger$ for a momentum $\mathbf{k}_1$ taken at time $\pm \infty$, which (according to the book) guarantees that the particle is located away from the origin. In other words, we define $|k\rangle = \lim\limits_{t \to -\infty} a^\dagger(\mathbf{k}, t) |0\rangle$.
It seems to me that Srednicki wants $\langle 0 | k \rangle = 0$, which sounds reasonable. But applying the LSZ formula for the special case of one initial particle and zero final particles, I get $\langle 0 | k \rangle = i (2\pi)^4 m^2 \langle 0 | \phi(0) | 0 \rangle \delta^4(k)$ (the $2\pi$'s might be off). This is nonzero only when $k^\mu=0$, a.k.a. never, since $k$ must be on shell. So why must we ask that the fields's VEV be zero, when it seems that $\langle 0 | k \rangle = 0$ anyway? |
As far as I know the electroweak theory only holds for some energy scale (probably the one we use at the LHC). For that scale there is a gauge theory for the electroweak interaction. Under this scale symmetry breaking occurs and this unification does no longer hold. Then we get again electromagnetism (QED). What theory then describes the weak interaction? I thought we need this unification because there is no gauge theory for the weak interaction that is conistent with QED.
I believe you're mistakenly thinking that we have a theory (electroweak theory) that works above a 'certain scale' and then below that scale that theory goes out the window, leaving us having to define a new theory. electroweak theory is always the same, it is a unified framework for the observed electric and weak forces. It is based on the gauge group $SU(2)\times U(1)_Y$. (Y is hypercharge).
Let's see if I can explain without writing long equations and entering into much technicalities..
A theory is given by writing down a Lagrangian. This will contain several fields: in our case fermion matter fields, gauge bosons fields for the above symmetry group, and a Higgs field. Initially this lagrangian is invariant under local $SU(2)\times U(1)_Y$ symmetry transformations. Under some special scale you mentioned the symmetry is broken really just because we have had to redefine what are the physical fields in the electroweak lagrangian for some reason. (I'll explain this better below, but leaping ahead this is, loosely speaking, because the system as a whole has collapsed into a new state). So the lagrangian is rewritten in terms of the new fields and written in this new way the lagrangian is no longer invariant under gauge transformations: 'the symmetry has been broken'. In our case $SU(2)\times U(1)_Y$ breaks down to $U(1)_{EM}$ which is related to good old electromagnetic gauge theory. In the mean time however in our redefined lagrangian we have new interaction terms involving what was 'left over' (in our redefinitions) by the gauge boson fields of the bigger $SU(2)\times U(1)_Y$ group. These are now identified with the $Z, \:W^{\pm}$ bosons. The interaction terms involve products of these fields and matter fields. Assuming you know enough about Feynman diagrams, you'll know that interaction terms of this kind in the lagrangian amount to processes with vertices joining various particles and $Z, \:W^{\pm}$ bosons. We say that these processes occur 'via the weak force'. For example, the first part of the weak decay of a muon to an electron diagram as in figure comes from a $\sim \mu \nu_{\mu} W^-$ term in the lagrangian.
The point is the electroweak theory is a mathematical framework which allows us to treat weak and electric forces together ('unifying them'). It amounts to writing down the initial Lagrangian with the $SU(2)\times U(1)_Y$ symmetry. Although we then make a sort of 'change of variables' in the lagrangian, it is always inside the same framework. At the end all physical processes are a direct consequence of the terms you have in your lagrangian.
Finally, a little more about what happens at the symmetry breaking critical scale. Firstly, when we say energy 'scale' here we really mean 'temperature' of the entire system (e.g. the universe) as a whole. (Mind you this is not the same as firing a 'few' protons at the LHC with energy above the critical scale). As the system cools down the form of the Higgs field part of the lagrangian changes. Above the critical scale the vacuum sits at a nice stable minima of the Higgs potential. As the system cools down this point becomes unstable, while stable minima develop elsewhere (see the wine bottle example). The system thus is forced to collapse into one of the new stable minima, warranting all the business of redefining what the fields are in order to QFT properly. This is part of the so-called Higgs mechanism.
I am not sure what level of description you are looking for. I will start off at a very basic level and maybe modify this if needed. The $SU(2)$ for the weak interactions has $W^\pm$ and $Z$ gauge bosons. The first of these are flavor changing charged gauge bosons and $Z$ is neutral. These have a correspondence with the $\sigma_\pm$ and $\sigma_3$ matrix representation of $SU(2)$. The old theory of weak interactions involved massive gauge bosons, which have a longitudinal degree of freedom. The problem is that at high energy this degree of freedom can propagate in ways that are causality violating.
Peter Higgs proposed a scalar field $\phi$ with a Ginsburg-Landau quartic potential $V(\phi)~=~-\mu|\phi|^2~+~\lambda|\phi|^4$ and at minimum energy can have zero kinetic energy and has a degeneracy of possible states. Along the Mexican hat potential trough the boson does not need to move to maintain its potential configuration, and there are a large number of them. I leave it to the reader to evaluate $\partial V(|\phi|^2)/\partial\phi~=~0$ to find this.
The scalar fields exist in doublets $(H^+,~H^-)$ and $(H^0,~h_0)$. These fields are coupled to the gauge interactions by the standard covariant formalism $$ {\cal D}_\mu\phi~=~\partial_\mu\phi~+~ie/\hbar A_\mu~+~igB^a_\mu\phi^a, $$ where $A$ is the QED gauge potential and $B^a_\mu$ are the $W^\pm$ and $Z$ the charge neutral $Z$, and the index values $a~=~+,~-,~0$ indicate this and couple to the field components correspondingly. There is some abuse of notation here, and I am going to not focus on the matter of hypercharge and the weak (Weinberg) angle.
Gauge bosons have two degrees of freedom corresponding to the electric and magnetic field. There is no longitudinal field if they are massless. We may think of this as there being no rest frame where $m~=~0$ exists, and only $m~=~\pm 1$ exist. However at lower energy where the components of the Higgs field $H^\pm,~H^0$ are minimal they couple with the $B^a_\mu$ weak interaction gauge potentials and are absorbed. The loss of these three degrees of freedom in the Higgs field is then in the longitudinal components of the $W^\pm,~Z$ particles. The remaining $h_0$ is the Higgs particle the LHC detected in 2012.
At very high energy the QED and weak interactions are massless bosons with the meaning of charge mixed into hypercharge according to the weak angle. They have the same interaction strength. However, as transverse momenta of interactions or energy approaches the EW breaking scale this Higgs mechanism kicks in so the weak interacting bosons are massive, as seen by having longitudinal degrees of freedom, and they decouple from the remaining photon of QED that remains massless. |
Among the worst of barabarisms is that of introducing symbols which are quite new in mathematical, but perfectly understood in common, language. Writers have borrowed from the Germans the abbreviation n! to signify 1.2.3.(n – 1).n, which gives their pages the appearance of expressing surprise and admiration that 2, 3, 4, &c. should be found in mathematical results.
You know, I’d always thought the formula for “
n choose k” was a little, well, enthusiastic:
[tex]\left(\begin{array}{c} n \\ k \end{array}\right) = \frac{n!}{k!(n-k)!}.[/tex]
“
n! k! n minus k!! You gotta believe me, guys!”
Still, though, if I saw de Morgan’s way of writing the factorial of
n, I’d read it as
[tex]1 \cdot 2 \cdot 3 \cdot (n-1) \cdot n,[/tex]
which is only the factorial of
n when n is 5. I guess there’s little point in pleasing the dead. . . . |
Translating Code to Mathematics
Given a (more or less) formal operational semantics you can translate an algorithm's (pseudo-)code quite literally into a mathematical expression that gives you the result, provided you can manipulate the expression into a useful form. This works well for
additive cost measures such as number of comparisons, swaps, statements, memory accesses, cycles some abstract machine needs, and so on. Example: Comparisons in Bubblesort
Consider this algorithm that sorts a given array
A:
bubblesort(A) do 1
n = A.length; 2
for ( i = 0 to n-2 ) do 3
for ( j = 0 to n-i-2 ) do 4
if ( A[j] > A[j+1] ) then 5
tmp = A[j]; 6
A[j] = A[j+1]; 7
A[j+1] = tmp; 8
end 9
end 10
end 11
end 12
Let's say we want to perform the usual sorting algorithm analysis, that is count the number of element comparisons (line 5). We note immediately that this quantity does not depend on the content of array
A, only on its length $n$. So we can translate the (nested)
for-loops quite literally into (nested) sums; the loop variable becomes the summation variable and the range carries over. We get:
$\qquad\displaystyle C_{\text{cmp}}(n) = \sum_{i=0}^{n-2} \sum_{j=0}^{n-i-2} 1 = \dots = \frac{n(n-1)}{2} = \binom{n}{2}$,
where $1$ is the cost for each execution of line 5 (which we count).
Example: Swaps in Bubblesort
I'll denote by $P_{i,j}$ the subprogram that consists of lines
i to
j and by $C_{i,j}$ the costs for executing this subprogram (once).
Now let's say we want to count
swaps, that is how often $P_{6,8}$ is executed. This is a "basic block", that is a subprogram that is always executed atomically and has some constant cost (here, $1$). Contracting such blocks is one useful simplification that we often apply without thinking or talking about it.
With a similar translation as above we come to the following formula:
$\qquad\displaystyle C_{\text{swaps}}(A) = \sum_{i=0}^{n-2} \sum_{j=0}^{n-i-2} C_{5,9}(A^{(i,j)})$.
$A^{(i,j)}$ denotes the array's state before the $(i,j)$-th iteration of $P_{5,9}$.
Note that I use $A$ instead of $n$ as parameter; we'll soon see why. I don't add $i$ and $j$ as parameters of $C_{5,9}$ since the costs do not depend on them here (in the uniform cost model, that is); in general, they just might.
Clearly, the costs of $P_{5,9}$ depend on the content of $A$ (the values
A[j] and
A[j+1], specifically) so we have to account for that. Now we face a challenge: how do we "unwrap" $C_{5,9}$? Well, we can make the dependency on the content of $A$ explicit:
$\qquad\displaystyle C_{5,9}(A^{(i,j)}) = C_5(A^{(i,j)}) + \begin{cases} 1 &, \mathtt{A^{(i,j)}[j] > A^{(i,j)}[j+1]} \\ 0 &, \text{else} \end{cases}$.
For any given input array, these costs are well-defined, but we want a more general statement; we need to make stronger assumptions. Let us investigate three typical cases.
The worst case
Just from looking at the sum and noting that $C_{5,9}(A^{(i,j)}) \in \{0,1\}$, we can find a trivial upper bound for cost:
$\qquad\displaystyle C_{\text{swaps}}(A) \leq \sum_{i=0}^{n-2} \sum_{j=0}^{n-i-2} 1 = \frac{n(n-1)}{2} = \binom{n}{2}$.
But
can this happen, i.e. is there an $A$ for this upper bound is attained? As it turns out, yes: if we input an inversely sorted array of pairwise distinct elements, every iteration must perform a swap¹. Therefore, we have derived the exact worst-case number of swaps of Bubblesort.
The best case
Conversely, there is a trivial lower bound:
$\qquad\displaystyle C_{\text{swaps}}(A) \geq \sum_{i=0}^{n-2} \sum_{j=0}^{n-i-2} 0 = 0$.
This can also happen: on an array that is already sorted, Bubblesort does not execute a single swap.
The average case
Worst and best case open quite a gap. But what is the
typical number of swaps? In order to answer this question, we need to define what "typical" means. In theory, we have no reason to prefer one input over another and so we usually assume a uniform distribution over all possible inputs, that is every input is equally likely. We restrict ourselves to arrays with pairwise distinct elements and thus assume the random permutation model.
Then, we can rewrite our costs like this²:
$\qquad\displaystyle \mathbb{E}[C_{\text{swaps}}] = \frac{1}{n!} \sum_{A} \sum_{i=0}^{n-2} \sum_{j=0}^{n-i-2} C_{5,9}(A^{(i,j)})$
Now we have to go beyond simple manipulation of sums. By looking at the algorithm, we note that every swap removes exactly one inversion in $A$ (we only ever swap neighbours³). That is, the number of swaps performed on $A$ is exactly the number of inversions $\operatorname{inv}(A)$ of $A$. Thus, we can replace the inner two sums and get
$\qquad\displaystyle \mathbb{E}[C_{\text{swaps}}] = \frac{1}{n!} \sum_{A} \operatorname{inv}(A)$.
Lucky for us, the average number of inversions has been determined to be
$\qquad\displaystyle \mathbb{E}[C_{\text{swaps}}] = \frac{1}{2} \cdot \binom{n}{2}$
which is our final result. Note that this is
exactly half the worst-case cost. Note that the algorithm was carefully formulated so that "the last" iteration with
i = n-1 of the outer loop that never does anything is not executed.
"$\mathbb{E}$" is mathematical notation for "expected value", which here is just the average. We learn along the way that no algorithm that only swaps neighbouring elements can be asymptotically faster than Bubblesort (even on average) -- the number of inversions is a lower bound for all such algorithms. This applies to e.g. Insertion Sort and Selection Sort. The General Method
We have seen in the example that we have to translate control structure into mathematics; I will present a typical ensemble of translation rules. We have also seen that the cost of any given subprogram may depend on the current
state, that is (roughly) the current values of variables. Since the algorithm (usually) modifies the state, the general method is slightly cumbersome to notate. If you start feeling confused, I suggest you go back to the example or make up your own.
We denote with $\psi$ the current state (imagine it as a set of variable assignments). When we execute a program
P starting in state $\psi$, we end up in state $\psi / \mathtt{P}$ (provided
P terminates).
Individual statements
Given just a single statement
S;, you assign it costs $C_S(\psi)$. This will typically be a constant function.
Expressions
If you have an expression
E of the form
E1 ∘ E2 (say, an arithmetic expression where
∘ may be addition or multiplication, you add up costs recursively:
$\qquad\displaystyle C_E(\psi) = c_{\circ} + C_{E_1}(\psi) + C_{E_2}(\psi)$.
Note that
the operation cost $c_{\circ}$ may not be constant but depend on the values of $E_1$ and $E_2$ and evaluation of expressions may change the state in many languages,
so you may have to be flexible with this rule.
Sequence
Given a program
P as sequence of programs
Q;R, you add the costs to
$\qquad\displaystyle C_P(\psi) = C_Q(\psi) + C_R(\psi / \mathtt{Q})$.
Conditionals
Given a program
P of the form
if A then Q else R end, the costs depend on the state:
$\qquad\displaystyle C_P(\psi) = C_A(\psi) + \begin{cases} C_Q(\psi/\mathtt{A}) &, \mathtt{A} \text{ evaluates to true under } \psi \\ C_R(\psi/\mathtt{A}) &, \text{else} \end{cases}$
In general, evaluating
A may very well change the state, hence the update for the costs of the individual branches.
For-Loops
Given a program
P of the form
for x = [x1, ..., xk] do Q end, assign costs
$\qquad\displaystyle C_P(\psi) = c_{\text{init_for}} + \sum_{i=1}^k c_{\text{step_for}} + C_Q(\psi_i \circ \{\mathtt{x := xi\}})$
where $\psi_i$ is the state before processing
Q for value
xi, i.e. after the iteration with
x being set to
x1, ...,
xi-1.
Note the extra constants for loop maintenance; the loop variable has to be created ($c_{\text{init_for}}$) and assigned its values ($c_{\text{step_for}}$). This is relevant since
computing the next
xi may be costly and
a
for-loop with empty body (e.g. after simplifying in a best-case setting with a specific cost) does not have zero cost if it performs iterations.
While-Loops
Given a program
P of the form
while A do Q end, assign costs
$\qquad\displaystyle C_P(\psi) \\\qquad\ = C_A(\psi) + \begin{cases} 0 &, \mathtt{A} \text{ evaluates to false under } \psi \\ C_Q(\psi/\mathtt{A}) + C_P(\psi/\mathtt{A;Q}) &, \text{ else} \end{cases}$
By inspecting the algorithm, this recurrence can often be represented nicely as a sum similar to the one for for-loops.
Example: Consider this short algorithm:
while x > 0 do 1
i += 1 2
x = x/2 3
end 4
By applying the rule, we get
$\qquad\displaystyle C_{1,4}(\{i := i_0; x := x_0\}) \\\qquad\ = c_< + \begin{cases} 0 &, x_0 \leq 0 \\ c_{+=} + c_/ + C_{1,4}(\{i := i_0 + 1; x := \lfloor x_0/2 \rfloor\}) &, \text{ else} \end{cases}$
with some constant costs $c_{\dots}$ for the individual statements. We assume implicitly that these do
not depend on state (the values of
i and
x); this may or may not be true in "reality": think of overflows!
Now we have to solve this recurrence for $C_{1,4}$. We note that neither the number of iterations not the cost of the loop body depend on the value of
i, so we can drop it. We are left with this recurrence:
$\qquad\displaystyle C_{1,4}(x) = \begin{cases} c_> &, x \leq 0 \\ c_> + c_{+=} + c_/ + C_{1,4}(\lfloor x/2 \rfloor) &, \text{ else} \end{cases}$
This solves with elementary means to
$\qquad\displaystyle C_{1,4}(\psi) = \lceil \log_2 \psi(x) \rceil \cdot (c_> + c_{+=} + c_/) + c_>$,
reintroducing the full state symbolically; if $\psi = \{ \dots, x := 5, \dots\}$, then $\psi(x) = 5$.
Procedure Calls
Given a program
P of the form
M(x) for some parameter(s)
x where
M is a procedure with (named) parameter
p, assign costs
$\qquad\displaystyle C_P(\psi) = c_{\text{call}} + C_M(\psi_{\text{glob}} \circ \{p := x\})$.
Note again the extra constant $c_{\text{call}}$ (which might in fact depend on $\psi$!). Procedure calls are expensive due to how they are implemented on real machines, and sometimes even dominate runtime (e.g. evaluating the Fibonacci number recurrence naively).
I gloss over some semantic issues you might have with the state here. You will want to distinguish global state and such local to procedure calls. Let's just assume we pass only global state here and
M gets a new local state, initialized by setting the value of
p to
x. Furthermore,
x may be an expression which we (usually) assume to be evaluated before passing it.
Example: Consider the procedure
fac(n) do
if ( n <= 1 ) do 1
return 1 2
else 3
return n * fac(n-1) 4
end 5
end
As per the rule(s), we get:
$\qquad\displaystyle\begin{align*} C_{\text{fac}}(\{n := n_0\}) &= C_{1,5}(\{n := n_0\}) \\ &= c_{\leq} + \begin{cases} C_2(\{n := n_0 \}) &, n_0 \leq 1 \\ C_4(\{n := n_0 \}) &, \text{ else} \end{cases} \\ &= c_{\leq} + \begin{cases} c_{\text{return}} &, n_0 \leq 1 \\ c_{\text{return}} + c_* + c_{\text{call}} + C_{\text{fac}}(\{n := n_0 - 1\}) &, \text{ else} \end{cases}\end{align*}$
Note that we disregard global state, as
fac clearly does not access any. This particular recurrence is easy to solve to
$\qquad\displaystyle C_{\text{fac}}(\psi) = \psi(n) \cdot (c_{\leq} + c_{\text{return}}) + (\psi(n) - 1) \cdot (c_* + c_{\text{call}})$
We have covered the language features you will encounter in typical pseudo code. Beware hidden costs when analysing high-level pseudo code; if in doubt, unfold. The notation may seem cumbersome and is certainly a matter of taste; the concepts listed can not be ignored, though. However, with some experience you will be able to see right away which parts of the state are relevant for which cost measure, for instance "problem size" or "number of vertices". The rest can be dropped -- this simplifies things significantly!
If you think now that this is far too complicated, be advised: it
is! Deriving exact costs of algorithms in any model that is so close to real machines as to enable runtime predictions (even relative ones) is a tough endeavour. And that's not even considering caching and other nasty effects on real machines.
Therefore, algorithm analysis is often simplified to the point of being mathematically tractable. For instance, if you don't need exact costs, you can over- or underestimate at any point (for upper resp. lower bounds): reduce the set of constants, get rid of conditionals, simplify sums, and so on.
A note on asymptotic cost
What you will usually find in literature and on the webs is the "Big-Oh analysis". The proper term is
asymptotic analysis which means that instead of deriving exact costs as we did in the examples, you only give costs up to a constant factor and in the limit (roughly speaking, "for big $n$").
This is (often) fair since abstract statements have
some (generally unknown) costs in reality, depending on machine, operating system and other factors, and short runtimes may be dominated by the operating system setting up the process in the first place and whatnot. So you get some perturbation, anyway.
Here is how asymptotic analysis relates to this approach.
Identify
dominant operations (that induce costs), that is operations that occur most often (up to constant factors). In the Bubblesort example, one possible choice is the comparison in line 5.
Alternatively, bound all constants for elementary operations by their maximum (from above) resp. their minimum (from below) and perform the usual analysis.
Perform the analysis using execution counts of this operation as cost. When simplifying, allow estimations. Take care to only allow estimations from above if your goal is an upper bound ($O$) resp. from below if you want lower bounds ($\Omega$).
Make sure you understand the meaning of Landau symbols. Remember that such bounds exist for all three cases; using $O$ does not imply a worst-case analysis.
Further reading
There are many more challenges and tricks in algorithm analysis. Here is some recommended reading.
There are many questions tagged algorithm-analysis around that use techniques similar to this. |
I know that if $f_\mathrm{m}$ is the
"Nyquist frequency" (max frequency) and $f_\mathrm{s}$ sampling rate then $f_\mathrm{s}>2f_\mathrm{m}$. Am I correct so far?
I have a signal $x(t)$ with max frequency $f_1$ and $h(t)$ with $f_2$ and we define $y(t)= h(t)*x(t)$ ($*$ for convolution) and we need to find the sampling frequency/Nyquist frequency of this function.
So the Nyquist frequency of $x(t)$ is $f_{\mathrm{s}_{x}} >2f_1$ and of $h(t)$ is $f_{\mathrm{s}_{h}} >2f_2$.
Now I saw that someone wrote that using the convolution theorem we get $Y(f)=H(f)X(f)$, so there must be that $f_\mathrm{s} \leq \min\{2f_1,2f_2 \}$ stating that this is an upper bound because frequency may cancel each other.
Why is that true?
I must mention that he wrote $Nq(x)$ instead of $f_{\mathrm{s}_{x}}$ (I just understood that he meant the same), also it's not supposed to be $f_\mathrm{s} \geq \min\{2f_1,2f_2 \}$ ?
I'm also not sure about is that $X(f)=0$ if $|f|>\frac{Nq(x)}{2}$.
Why is that? |
If I toss a coin, then according to the many worlds interpretation of QM, in half those worlds I'll get a head. If I then toss again, then in a quarter I will have got two heads. And so on. There will therefore be some extreme worlds where I always get heads. What happens to the normal distribution of probabilities say in a world where I always get a head, you always toss a six on a die and all electrons are spin up? Similar extreme outcomes of will not happen in only one world, but an infinite number. In these worlds the normal distribution of events will not occur. What an I missing about the many-world interpretation of QM?
Without entering the quantum mechanics of the situation, we can see that each toss is a new world. The next toss is another world, so the series of heads do not add in the way you think to make a world of all heads.
Each world deserted by each new toss will have the usual probabilities of heads or tails.
A world of all heads is possible with sequential tossings making a history of all heads, but not in the way you think:
There will therefore be some extreme worlds where I always get heads
The "always get heads" assumes that you have freedom to keep tossing in the same world. You can only "always have gotten heads" in one world line.
The many worlds interpretation is just mathematics made visual, in my opinion.
Of course to even register that such a world line exists innumerable numbers of worlds will have been created so as to have the history in your world line that such a world existed ! Thinking mathematically is much simpler.
Your observation is correct. If tossing a coin were a quantum measurement, in the many-worlds interpretation of quantum mechanics (MW), there would be a branch of "worlds" in which the outcome was always heads.
This would not violate anything we know about probability or quantum mechanics. The full ensemble of worlds would have the expected binomial distribution of outcomes. e.g. for two coin tosses, there would be one world with $HH$, two with $HT$ and $TH$, and one with $TT$. The probability that "you" live in the $TTT\ldots$ branch is exactly equal to the probbility of there being a single Universe in which you toss a coin and see $TTT\ldots$ You wouldn't be able to infer anything about the Universe/MWI if you saw $TTT\ldots$, other than that you had witnessed something remarkably unlikely.
I will add a few comments regarding MWI and the most important aspect of probability in the MWI. This discussion is all a bit misleading without them.
The "branching of worlds" is an macroscopic phenomena that emerges around the time that a pure state is significantly de-cohered. It cannot be understood from the microscopic theory, from the bottom-up.
In the MWI, there is no collapse of the wavefunction, just unitary evolution of the wavefunction by Schrodinger's equation. The MWI doesn't contain a Born rule for probabilities!, $$ P_a = |\langle a | \psi \rangle|^2 $$ That's very problematic. The branching occurs around the time a pure state has decohered, such that the basis states don't interefere. There is one branch per basis state. e.g. $$ |\psi\rangle = \frac35|\phi\rangle + \frac45|\chi\rangle $$ would not result in 25 Universes, 9 in state $|\phi\rangle$ and 16 in state $|\chi\rangle$, but two branches, one $|\phi\rangle$ and one $|\phi\rangle$. The correct probabilities do not emerge from the branching.
Probability in MWI is an outstanding problem. There have been attempts to derive the Born rule and arguments that it emerges, but I'm not sure whether any solution is widely accepted even in the MWI community.
As far as I am aware, we are only able to speculate about these other "Universes" in the sense that they represent an aggregate of potential (indeterminate) wave functions relating to what events may occur in the time direction known as "future". Or might have occured in the "past" but didn't in this universe. Every time you toss a coin and look at it, that wave function collapses and you are able to determine which universe you are in. Each added toss of the coin adds to the number of collapsed wave functions - and thus potential alternative wave functions which did not occur in
this universe.
Reversing time's arrow would seem to decrease the number of alternative possible wave functions, and thus decrease the number of "Universes".
Assuming that you are not an expert coin tosser and can engineer tails every time through skill, then any random distribution of toss sequences can occur, including a lifetime of tails or a normal distribution of tails V's heads, as you say, not forgetting that an infinite run of heads lies within the normal distribution, albeit at a probability approaching zero. You may discover that you live in the world where you decide never to toss a coin again, that mere decision will not collapse the wave function of any possible future toss, as you may change your mind, and take up the old hobby again.
What you are missing is that it is not a viable strategy to get the probability by counting worlds. The only viable way of calculating the probability in MWI is the Born rule. See this paper especially section 9: http://arxiv.org/abs/0906.2718. |
Electromagnetic Induction The Experiments of Faraday and Henry, Magnetic Flux The magnetic flux through a surface (area \tt \overrightarrow{A}) placed in a uniform magnetic field \tt \overrightarrow{B} is defined as \tt \phi \overrightarrow{B} \cdot \overrightarrow{A}= BA \cos \theta
Where θ is the angle between \tt \overrightarrow{B} and \tt \overrightarrow{A}
If there is a change of magnetic flux in a circuit, there will be an induced emf and this will last until the change persists, is FARADAY’S LAW. The magnitude of induced emf is equal to the time rate of change of magnetic flux \tt e = \frac{d \phi}{dt} The polarity of the induced emf tends to produce a current which opposes the change that produces it. This is known as LENZ’S LAW
φ = BAN cos θ, the change in flux can be caused by changing B (or) A or N Only when ‘B’ is changed, the average induced emf is, \tt e = - AN \cos \theta \frac{(B_{2} - B_{1})}{(t_{2} - t_{1})} Only when ‘A’ is changed, the average induced emf is given by, \tt e = - BN \cos \theta \frac{(A_{2} - A_{1})}{(t_{2} - t_{1})} Only when ‘θ’ is change, the average induced emf is, \tt e = - BAN \frac{(\cos \theta_{2} - \cos \theta_{1})}{(t_{2} - t_{1})} When the coil is rotated with constant angular velocity ‘ω’, then \tt e = - BAN \frac{d(\cos \omega t)}{dt} = BAN \omega \ sin \omega t The change in flux by changing the number of turns is practically difficult. The current induced in a coil is given by \tt i = \frac{e}{R} = \frac{-1}{R} \frac{\Delta \phi}{\Delta t} \tt q = - \frac{\Delta \phi}{R} \left(i = \frac{q}{\Delta t}\right) If a rod of length ‘l’ is moved with a velocity ‘v’ perpendicular to its length in a uniform magnetic field B which is perpendicular to both its length as well as its velocity, then e = Blv If the rod is moved making an angle ‘θ’ with its length, then e = Blv sin θ FLEMINGS RIGHT HAND RULE : Stretch the thumb, fore finger and middle finger such that they all are mutually perpendicular to each other. If Thumb represents – direction of velocity
Forefingers represents – direction of field
Middle finger represents – direction of induced current
If ends of the rod A and B are connected by an external resistance R, then the current in the rod is given as, \tt i = \frac{Blv}{R} The force acting on the rod due to current carrying conductor, \tt F = Bil \Rightarrow F = \frac{B^{2} l^{2} v}{R} \left(i = \frac{Blv}{R}\right) To move the rod with constant velocity, the power applied by external agent is \tt P = F V = \frac{B^{2} l^{2} v^{2}}{R} A rod of length l is rotated about O perpendicular to its length in an uniform magnetic field B, then emf, \tt e = \frac{1}{2} B l^{2} \omega
A rod is rotated about an axis passing through its centre, O and perpendicular to its length then emf across the ends is zero.
emf across \tt OB = \frac{-1}{8} B l^{2} \omega emf across \tt OA = + \frac{1}{8} B l^{2} \omega The flow patterns of induced currents resemble the swirling eddies in water and these currents are called EDDY CURRENT (OR) FOUCAULT CURRENTS The eddy currents are reduced by using laminations of metal to make a metal core. Insulating material like lacquer is used to separate laminations. When the current flowing in one coil changes, magnetic flux linked with the second coil placed near to it also changes. The emf induced in secondary coil is called mutually induced emf. This phenomenon is known as MUTUAL INDUCTION View the Topic in this video From 1:03 To 56:32
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1. The magnetic Net flux through the surface \phi = \oint \overrightarrow{B}. d \overrightarrow{A} = BA \cos \theta |
A second order differential equation is written in general form as
\[F\left( {x,y,y’,y^{\prime\prime}} \right) = 0,\]
where \(F\) is a function of the given arguments.
If the differential equation can be resolved for the second derivative \(y^{\prime\prime},\) it can be represented in the following explicit form:
\[y^{\prime\prime} = f\left( {x,y,y’} \right).\]
In special cases the function \(f\) in the right side may contain only one or two variables. Such incomplete equations include \(5\) different types:
\[
{y^{\prime\prime} = f\left( x \right),\;\;}\kern-0.3pt {y^{\prime\prime} = f\left( y \right),\;\;}\kern-0.3pt {y^{\prime\prime} = f\left( {y’} \right),\;\;}\kern-0.3pt {y^{\prime\prime} = f\left( {x,y’} \right),\;\;}\kern-0.3pt {y^{\prime\prime} = f\left( {y,y’} \right).} \]
With the help of certain substitutions, these equations can be transformed into first order equations.
In the general case of a second order differential equation, its order can be reduced if this equation has a certain symmetry. Below we discuss two types of such equations (cases \(6\) and \(7\)):
The function \(F\left( {x,y,y’,y^{\prime\prime}} \right)\) is a homogeneous function of the arguments \(y,y’,y^{\prime\prime};\) The function \(F\left( {x,y,y’,y^{\prime\prime}} \right)\) is an exact derivative of the first order function \(\Phi\left( {x,y,y’} \right).\)
Consider these \(7\) cases of reduction of order in more detail.
Case \(1.\) Equation of type \(y^{\prime\prime} = f\left( x \right)\)
For an equation of type \(y^{\prime\prime} = f\left( x \right),\) its order can be reduced by introducing a new function \(p\left( x \right)\) such that \(y’ = p\left( x \right).\) As a result, we obtain the first order differential equation
\[p’ = f\left( x \right).\]
Solving it, we find the function \(p\left( x \right).\) Then we solve the second equation
\[y’ = p\left( x \right)\]
and obtain the general solution of the original equation.
Case \(2.\) Equation of type \(y^{\prime\prime} = f\left( y \right)\)
The right-hand side of the equation depends only on the variable \(y.\) We introduce a new function \(p\left( y \right),\) setting \(y’ = p\left( y \right).\) Then we can write:
\[{y^{\prime\prime} = \frac{d}{{dx}}\left( {y’} \right) }={ \frac{{dp}}{{dx}} }={ \frac{{dp}}{{dy}}\frac{{dy}}{{dx}} }={ \frac{{dp}}{{dy}}p,}\]
so the equation becomes:
\[\frac{{dp}}{{dy}}p = f\left( y \right).\]
Solving it, we find the function \(p\left( y \right).\) Then we find the solution of the equation \(y’ = p\left( y \right),\) that is, the function \(y\left( x \right).\)
Case \(3.\) Equation of type \(y^{\prime\prime} = f\left( {y’} \right)\)
In this case, to reduce the order we introduce the function \(y’ = p\left( x \right)\) and obtain the equation
\[{y^{\prime\prime} = p’ }={ \frac{{dp}}{{dx}} }={ f\left( p \right),}\]
which is a first order equation with separable variables \(p\) and \(x.\) Integrating, we find the function \(p\left( x \right),\) and then the function \(y\left( x \right).\)
Case \(4.\) Equation of type \(y^{\prime\prime} = f\left( {x,y’} \right)\)
Here we use the substitution \(y’ = p\left( x \right),\) where \( p\left( x \right)\) is a new unknown function. As a result, we obtain the first order equation:
\[p’ = \frac{{dp}}{{dx}} = f\left( {x,p} \right).\]
By integrating, we find the function \( p\left( x \right).\) Next, we solve one more equation of the \(1\)st order
\[y’ = p\left( x \right)\]
and find the general solution \( y\left( x \right).\)
Case \(5.\) Equation of type \(y^{\prime\prime} = f\left( {y,y’} \right)\)
To solve this equation, we introduce a new function \( p\left( y \right),\) setting \( y’ = p\left( y \right),\) similar to case \(2.\) Differentiating this expression with respect to \(x\) leads to the equation
\[
{y^{\prime\prime} = \frac{{d\left( {y’} \right)}}{{dx}} = \frac{{dp}}{{dx}} } = {\frac{{dp}}{{dy}}\frac{{dy}}{{dx}} }={ \frac{{dp}}{{dy}}p.} \]
As a result, our original equation is written as an equation of the \(1\)st order
\[p\frac{{dp}}{{dy}} = f\left( {y,p} \right).\]
Solving it, we find the function \( p\left( y \right).\) Then we solve another first order equation
\[y’ = p\left( y \right)\]
and determine the general solution \( y\left( x \right).\)
The above \(5\) cases of reduction of order are not independent. Based on the structure of the equations, it is clear that case \(2\) follows from the case \(5\) and case \(3\) follows from the more general case \(4.\)
Case \(6.\) Function \(F\left( {x,y,y’,y^{\prime\prime}} \right)\) is homogeneous with respect to the arguments \(y, y’, y^{\prime\prime}\)
If the left side of the differential equation
\[F\left( {x,y,y’,y^{\prime\prime}} \right) = 0\]
satisfies the condition of homogeneity, that is the relationship
\[{F\left( {x,ky,ky’,ky^{\prime\prime}} \right) }={ {k^m}F\left( {x,y,y’,y^{\prime\prime}} \right)}\]
is valid for any \(k\), the order of the equation can be reduced by substitution
\[y = {e^{\int {zdx} }}.\]
After the function \( z\left( x \right)\) is found, the original function \( y\left( x \right)\) is determined by the integration formula
\[y\left( x \right) = {C_2}{e^{\int {zdx} }},\]
where \({C_2}\) is the constant of integration.
Case \(7.\) Function \(F\left( {x,y,y’,y^{\prime\prime}} \right)\) is an exact derivative
If one can find a function \(\Phi\left( {x,y,y’} \right),\) which does not contain the second derivative \(y^{\prime\prime}\) and satisfies the equation
\[{F\left( {x,y,y’,y^{\prime\prime}} \right) }={ \frac{d}{{dx}}\Phi \left( {x,y,y’} \right),}\]
then the solution of the original equation is given by the integral
\[\Phi \left( {x,y,y’} \right) = C.\]
Using this way the second order equation can be reduced to first order equation.
In some cases, the left part of the original equation can be transformed into an exact derivative, using an integrating factor.
Solved Problems
Click a problem to see the solution.
Example 1Solve the equation \(y^{\prime\prime} = \sin x + \cos x.\) Example 2Solve the equation \(y^{\prime\prime} = {\large\frac{1}{{4\sqrt y }}\normalsize}.\) Example 3Solve the equation \(y^{\prime\prime} = \sqrt {1 – {{\left( {y’} \right)}^2}}.\) Example 4Solve the equation \(\sqrt x y^{\prime\prime} = {\left( {y’} \right)^2}.\) Example 5Solve the equation \(y^{\prime\prime} = \left( {2y + 3} \right){\left( {y’} \right)^2}.\) Example 6Solve the equation \(yy^{\prime\prime} = \) \({\left( {y’} \right)^2} – {\large\frac{{3{y^2}}}{{\sqrt x }}\normalsize}.\) Example 7Solve the equation \(yy^{\prime\prime} + {\left( {y’} \right)^2} =\) \( 2x + 1.\) Example 1.Solve the equation \(y^{\prime\prime} = \sin x + \cos x.\)
Solution.
This example relates to the Case \(1.\) Consider the function \(y’ = p\left( x \right).\) Then \(y^{\prime\prime} = p’.\) Consequently,
\[p’ = \sin x + \cos x.\]
Integrating, we find the function \(p\left( x \right):\)
\[
{\frac{{dp}}{{dx}} = \sin x + \cos x,\;\;}\Rightarrow {dp = \left( {\sin x + \cos x} \right)dx,\;\;}\Rightarrow {{\int {dp} }={ \int {\left( {\sin x + \cos x} \right)dx} ,\;\;}}\Rightarrow {{p = – \cos x }+{ \sin x + {C_1}.}} \]
Given that \(y’ = p,\) we integrate one more equation of the \(1\)st order:
\[
{{y’ = – \cos x }+{ \sin x + {C_1},\;\;}}\Rightarrow {{\int {dy} }={ \int {\left( { – \cos x + \sin x + {C_1}} \right)dx} ,\;\;}}\Rightarrow {{y = – \sin x }-{ \cos x + {C_1}x + {C_2}.}} \]
The latter formula gives the general solution of the original differential equation. |
Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals Marks: 10 Marks Year: May 2016 Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals Marks: 10 Marks Year: May 2016
Parameters IC741 Values Differential input Resistance 2MΏ Input capacitance 1-4 pF Open Loop Voltage Gain 200,000 CMRR 90 dB Output Voltage Swing ±13 to ±15 V Output Resistance 75 Ώ Input Voltage Range ±12 to ±13 V Power Supply Rejection Ratio 30µV/V Power Consumption 85 mW Gain-Bandwidth Product 1MHz Average Temperature Coefficient of Offset Parameters $12 pA/C^0$ Supply Current 2.8 mA Slew Rate 15 µA
It is the ratio of differential voltage gain $A_d$ to the common mode voltage gain $A_C$.
$$CMRR=\dfrac{A_d}{A_C}$$
Now $A_d$ is nothing but open loop voltage gain $A_{OL}$. And $A_c$ is measured by using the circuit as shown in figure
The common mode input $V_C$ is applied to both the input terminals of OP-amp. Then the output $V_{OC}$ is measured. Then common mode gain $A_C$ can be obtained as,
$$A_c=\dfrac{V_{OC}}{V_c}$$
It is generally very small and not specified in the data sheet. The CMRR is generally specified for the op-amp and is expressed in dB. For op-amp 741C it is 90dB.
Slew Rate
The slew rate is defined as the maximum rate of change of output voltage with time.
The slew rate is specified in V/μsec. Thus,
$$Slew \ \ Rate=S=\dfrac{dV_0}{dt}\bigg| \max$$
The slew rate is caused due to limited charging rate of the compensating capacitor and current limiting and saturation of the internal stages of an op-amp, when a high frequency, large amplitude signal is applied.
The internal capacitor voltage cannot change instantaneously. It is given by $(dV_c)/dt=I/C$. For large charging rate, the capacitor should be small or charging current should be large. Hence the slew rate for the op-amp whose maximum internal capacitor charging current is known, can be obtained as
$$S=\dfrac{I_{|max}}{C}$$
For example, for IC 741 the charging current is 15 μA and the internal capacitor is 30 pF, hence its slew rate is
$$S=\dfrac{15 \times 10^{-6}}{30 \times 10^{-12}}=\dfrac{0.5}{10^{-6}}V/\sec \\ =0.5 V/\mu \sec$$ |
Last time, we found that the problem of the hydrogen atom could be split into a radial part and an angular part. Thanks to spherical symmetry, the angular part could be studied using angular momentum operators and spherical harmonics. We found that the 3D behavior of the electron could be reinterpreted as a 1D wavefunction of a particle in an effective potential which was the two-body interaction potential plus a “barrier” term which depended upon the angular momentum quantum number. Today, we’re going to solve the radial part of the problem and thereby find the eigenstates and eigenenergies of the hydrogen atom.
The technique we’ll employ has a certain charm, because we solved the first part, the angular dependence, using commutator relations, while as we shall see, the
radial dependence can be solved with anticommutator relations. THE FAMILY OF COULOMB HAMILTONIANS
We ended up with a family of Hamiltonians labeled by the angular momentum quantum number:
[tex]H_l^{\rm Coul} = -\frac{\hbar^2}{2\mu} \partial_r^2 -\frac{Ze^2}{r} + \frac{l(l + 1) \hbar^2}{2\mu r^2}.[/tex]
For each eigenstate of any [tex]H_l^{\rm Coul}[/tex], the hydrogen atom has a set of states labeled by the quantum number [tex]m[/tex], which takes the values
[tex]m = 0, \pm 1, \pm 2, \ldots, \pm l.[/tex]
However, because the original problem was spherically symmetric, the operator [tex]L_3[/tex] does not appear in the overall Hamiltonian (there’s no preferred axis), and so the energy will not depend upon [tex]m[/tex]. Geometrical symmetry implies that states of different [tex]m[/tex] but the same [tex]l[/tex] are
degenerate; as we proceed, we’ll see an additional example of symmetry implying degeneracy unfold before us.
Carrying around all those constants in [tex]H_l^{\rm Coul}[/tex] would be a pain and not particularly useful, so let’s get rid of them. By appropriately rescaling our variables (craftily choosing “better rulers” for distance and energy) we can rewrite [tex]H_l^{\rm Coul}[/tex] in a cleaner form, preserving the essential features of the Hamiltonians without all that dimensionful baggage:
[tex]H_l^{\rm Coul} = -\partial_x^2 + \frac{l(l+1)}{x^2} – \frac{1}{x}.[/tex]
[I originally screwed this bit up; see the comments below.]
Now, we have a family of related Hamiltonians, labeled by a nonnegative integer. Why does this situation sound familiar? When we studied supersymmetry and shape invariance, we saw just such a hierarchy of Hamiltonians appear! So, let’s see if those ideas are applicable to this problem.
We should note one complication right away: when we explored SUSY QM and shape-invariant partner potentials, we worked with an original Hamiltonian whose ground-state energy was
zero. Here, however, we’re looking at an electron bound by an atom, so the energies of the eigenstates will be negative. Shifts of the energy scale aren’t going to pose profound difficulties, but we should be on our guard. LIVING UP TO OUR SUPERPOTENTIAL
Earlier, we took a
superpotential and from it derived two partner potentials. Thus, if [tex]W[/tex] was the superpotential, then
[tex]A = \partial_x + W(x)[/tex]
and
[tex]A^\dag = -\partial_x + W(x),[/tex]
so our partner Hamiltonians would be
[tex]H^{(1)} = A^\dag A = -\partial_x^2 + W^2(x) – \partial_x W(x)[/tex]
and
[tex]H^{(2)} = AA^\dag = -\partial_x^2 + W^2(x) + \partial_x W(x).[/tex]
To study [tex]H_l^{\rm Coul}[/tex] in this framework, we’ll need a superpotential which, when put through these steps, yields both the [tex]1/x[/tex] Coulomb law and the angular momentum barrier. The superpotential will, therefore, depend upon [tex]l[/tex]. We choose the following:
[tex]W_l = -\frac{(l + 1)}{x} + \frac{1}{2(l + 1)}.[/tex]
This gives us a pair of operators for each value of [tex]l[/tex]:
[tex]A_l = \partial_x – \frac{l + 1}{x} + \frac{1}{2(l + 1)},[/tex]
and
[tex]A^\dag_l = -\partial_x – \frac{l + 1}{x} + \frac{1}{2(l + 1)}.[/tex]
The first Hamiltonian is given by one ordering,
[tex]H_l^{(1)} = A_l^\dag A_l,[/tex]
which works out to be
[tex]H_l^{(1)} = -\partial_x^2 + \frac{l(l+1)}{x^2} – \frac{1}{x} + \frac{1}{4(l+1)^2},[/tex]
and when we define a Hamiltonian with the other ordering,
[tex]H_l^{(2)} = A_l A^\dag_l,[/tex]
we get as the partner Hamiltonian
[tex]H_l^{(2)} = -\partial_x^2 + \frac{(l+1)(l+2)}{x^2} – \frac{1}{x} + \frac{1}{4(l+1)^2}.[/tex]
By comparing these two expressions, we get that
[tex]H_l^{(2)} = H_{l + 1}^{(1)} + \frac{1}{4(l + 1)^2} – \frac{1}{4(l + 2)^2}.[/tex]
Here, we see the shape-invariance criterion: [tex]H_l^{(2)}[/tex] is a Hamiltonian which depends upon the parameter [tex]l[/tex], and we can reach [tex]H_l^{(2)}[/tex] by taking its partner [tex]H_l^{(1)}[/tex], tweaking a parameter and adding an offset which does not depend upon [tex]x[/tex]. We can see the energy-axis shift mentioned above if we rewrite these operators in terms of the Coulomb Hamiltonian. First, we see that
[tex]H_l^{(1)} = H_l^{\rm Coul} + \frac{1}{4(l + 1)^2},[/tex]
so the operators [tex]H_l^{(1)}[/tex] and [tex]H_l^{\rm Coul}[/tex] will have the same eigenstates, although their eigenvalues will be separated by a term dependent upon [tex]l[/tex]. We also have
[tex]H_l^{(2)} = H_{l + 1}^{\rm Coul} + \frac{1}{4(l + 1)^2}.[/tex]
BUILDING UP THE STATES
We are now in a position to employ the diagrammatic approach we built up earlier. Let’s label the eigenstates of a particular Hamiltonian in the hierarchy by [tex]\nu[/tex], such that [tex]\nu = 0[/tex] for the ground state. Each state in our diagram will then be labeled by [tex]l[/tex] and [tex]\nu[/tex], which we write in Dirac notation as [tex]|\nu, l\rangle[/tex]. The isospectrality between partner potentials implies that state [tex]\nu[/tex] of Hamiltonian [tex]H_l^{(1)}[/tex] has the same energy as state [tex]\nu – 1[/tex] of [tex]H_l^{(2)}[/tex]. The state annihilated by [tex]A_l[/tex] is defined by
[tex]A_l |0, l\rangle = 0,[/tex]
which tells us that
[tex]H_l^{(1)} |0, l\rangle = 0.[/tex]
In turn, because of the energy shift, the Coulomb Hamiltonian satisfies
[tex]H_l^{\rm Coul} |0, l\rangle = -\frac{1}{4(l + 1)^2} |0, l\rangle.[/tex]
Likewise, for [tex]l + 1[/tex],
[tex]H_{l+1}^{(1)} |0, l + 1\rangle = 0.[/tex]
Using the shape-invariance condition yields
[tex]H_l^{(2)} |0, l + 1\rangle = \left[\frac{1}{4(l + 1)^2} – \frac{1}{4(l + 2)^2}\right]|0, l + 1\rangle.[/tex]
By the other key property, isospectrality, we know that [tex]H_l^{(1)}[/tex] has a state of the same energy, one notch higher in [tex]\nu[/tex]:
[tex]H_l^{(1}} |1, l\rangle = \left[\frac{1}{4(l + 1)^2} – \frac{1}{4(l + 2)^2}\right]|1, l\rangle.[/tex]
Again, the physical energy of the hydrogen atom can be found by adding a constant:
[tex]H_l^{\rm Coul} |1, l\rangle = -\frac{1}{4(l + 2)^2} |1, l\rangle.[/tex]
Notice how as we moved from [tex]\nu = 0[/tex] to [tex]\nu = 1[/tex], the value in the denominator went from [tex]l + 1[/tex] to [tex]l + 2[/tex]? The rest of the states can be built by working “leftwards” in the state diagram:
[tex]|\nu, l\rangle = A^\dag_l A^\dag_{l + 1} \cdots A^\dag_{l + \nu – 1} |0, l + \nu\rangle.[/tex]
This just means repeating the process we went through earlier, meaning that the denominator of the eigenvalue will depend on [tex]l + \nu + 1[/tex]. Note that the energy of the state labeled by [tex]\nu[/tex] and [tex]l[/tex] depends only upon their sum. If we restore the constants we scaled away earlier, we get that
[tex]E_{\nu l} = -\frac{\mu e^4 Z^2}{2\hbar^2} \frac{1}{(l + \nu + 1)^2},[/tex]
so we can call [tex]l + \nu + 1[/tex] by the new name [tex]n[/tex] and write
[tex]E_n = -\frac{\mu e^4 Z^2}{2\hbar^2} \frac{1}{n^2}.[/tex]
This is the energy of the [tex]n{\rm th}[/tex] eigenstate of the hydrogenic Hamiltonian. We could work out explicit expressions for states of our choice, but before we plug-and-chug any specific examples, let’s take a breather and look at a picture.
THE SLIGHTLY BIGGER PICTURE
The diagram we drew before now looks like this:
The eigenstate diagram shows
more symmetry than we had expected on geometrical considerations alone: the only quantum number necessary for determining the energy is [tex]n[/tex], which corresponds to the energy-level index in Bohr’s model of the atom. If our interaction potential were not shape invariant, this degeneracy would be broken.
This is the way I learned to solve the hydrogenic atom in the misty days of my undergraduacy. The only textbook I know of which takes an approach like this is Ohanian’s
Principles of Quantum Mechanics; other than a handful of universities, most schools attack the problem by plowing into Schrödinger’s second-order differential equation and eventually finding a recursion relation for the Laguerre polynomials. Prof. Rajagopal‘s lecture notes call the standard method “much more painful,” and as for why most textbooks follow that route, “Go figure.” I suspect that too many teachers of quantum mechanics have been bitten by the Matrix Zombie and think that mathematics beyond differential equations is just too hard for introductory classes. Rather than making the time investment necessary to use “more advanced” techniques, they solve problems in laborious and rather unilluminating ways.
Unfortunately, MIT’s OpenCourseWare project doesn’t provide the lecture notes we used, or any later editions thereof; the site for 8.05 Quantum Physics II just lists the sections of textbooks which should be read, instead of providing actual juicy PDFs. This post, in particular, was based on the 8.05 material, while my earlier overview of the general superalgebra machinery mostly follows Fred Cooper, Avinash Khare and Uday Sukhatme’s review article, “Supersymmetry and Quantum Mechanics” (1994). As that review explains, Schrödinger himself solved for the hydrogen atom eigenstates with a method rather like this, in 1940; many years later, the supersymmetric context of that “factorization” method was discovered.
From here, we can go in several directions. After perhaps working a few examples, we can head towards the relativistic regime and find SUSY-based solutions to the Dirac Equation. Also, we can look back at classical mechanics and relate these ideas to the Laplace-Runge-Lenz vector, an avenue which will eventually lead us to
superalgebras with central charge and BPS bounds. I’m also strongly tempted to look at the application of SUSY to diffusion problems via the Fokker-Planck Equation. SUSY QM SERIES: |
Based on your terminology and comments, you seem to be having an internal versus external/syntax versus semantics confusion.
Syntactically, the Law of the Excluded Middle (LEM) is the axiom schema $P\lor\neg P$. Even in type theory, this would be something like $\Pi P:\mathsf{Prop}.P\lor (P\to\bot)$.
Externally, i.e. semantically (and particularly for categorical semantics), LEM means subobjects are complemented (or, particularly, subobjects of $1$ are). Propositions being "empty" or the "universe" sounds like semantic terminology. In type theory, $\top$ is simply the unit type having exactly one value. Calling it "the universe" doesn't make a lot of sense (and also conflicts with other uses of the term "universe" in type theory). Calling $\bot$ the "empty type" is reasonable enough, but calling it "empty" begins to suggest a little too much but isn't completely unreasonable. In the
semantics of (single-sorted) first-order logic, people often talk about "true" (unary) predicates being interpreted as the "universe (of discourse)" or "domain" and "false" predicates being interpreted as the empty set. This "universe" terminology doesn't work that well even for multi-sorted first-order logic and is an even poorer fit (in this sense) for type theoretic work.
As a concrete example, $\mathbf{Set}\times\mathbf{Set}$ is a Boolean topos (so LEM holds) that is not two-valued. In particular, it has four truth values, $\bot=(0,0)$, $\top=(1,1)$, and also $(1,0)$ and $(0,1)$. All the operations are defined component-wise. You can easily show $P\lor\neg P$ and even $(P=\top)\lor(P=\bot)$ are valid with respect to this semantics. The latter can
internally be read as saying "every proposition is either $\bot$ or $\top$". Externally, if we assign the truth value $(1,0)$ to $P$, then we're asking for the truth value of $(1,0)=(1,1)$ and $(1,0)=(0,0)$ which have the truth values $(1,0)$ and $(0,1)$ respectively. Of course, the join of these, i.e. the interperation of $\lor$, is $(1,1)$, i.e. the truth value of $\top$ as LEM requires.
Diaconescu's proof (which is only three pages long!) is talking about toposes, i.e. categorical semantics for constructive set theories. Unsurprisingly, it focuses on complemented subobjects.
$\neg\neg(P\lor\neg P)$ holds constructively. The only way to
internally say "a proposition is neither empty nor the universe" in the way you appear to be using the words is to say something that is equivalent 1 to $\neg(P\lor\neg P)$, e.g. $\neg(P=\top)\land\neg(P=\bot)$, but this is contradictory in constructive logic itself even before we talk about LEM or the Axiom of Choice. Externally, we can easily say "if we have a subobject that's not complemented, then the Axiom of Choice does not hold" which is exactly what happens in Diaconescu's proof. If, however, you state (externally) that "if $0_1:0\rightarrowtail 1$ and $id_1:1\rightarrowtail 1$ aren't the only subobjects of $1$, then the Axiom of Choice does not hold" then this is just false. Any Boolean topos that is not two-valued, such as $\mathbf{Set}\times\mathbf{Set}$, would be a counter-example.
1 In the presence of propositional extensionality which Diaconescu's theorem requires. Here's a formal proof in Agda that $\neg(P\equiv\top)\land\neg(P\equiv\bot)$ is contradictory:
data ⊥ : Set where
record ⊤ : Set where constructor tt
record _∧_ {ℓ}(A B : Set ℓ) : Set ℓ where
constructor _,_
field
fst : A
snd : B
¬ : ∀{ℓ} → Set ℓ → Set ℓ
¬ A = A → ⊥
data _≡_ {ℓ}{A : Set ℓ} (a : A) : A → Set ℓ where
Refl : a ≡ a
isProp : Set → Set
isProp P = (x y : P) → x ≡ y
isProp-⊤ : isProp ⊤
isProp-⊤ tt tt = Refl
isProp-⊥ : isProp ⊥
isProp-⊥ () y
_↔_ : Set → Set → Set
P ↔ Q = (P → Q) ∧ (Q → P)
postulate propExt : {P Q : Set} → isProp P → isProp Q → (P ↔ Q) → P ≡ Q
thm1 : {P : Set} → isProp P → P → P ≡ ⊤
thm1 prf p = propExt prf isProp-⊤ ((λ _ → tt) , λ _ → p)
thm2 : {P : Set} → P ≡ ⊤ → P
thm2 Refl = tt
thm3 : {P : Set} → isProp P → ¬ P → P ≡ ⊥
thm3 prf ¬p = propExt prf isProp-⊥ (¬p , λ{ () })
thm4 : {P : Set} → P ≡ ⊥ → ¬ P
thm4 Refl p = p
thm5 : {P : Set} → isProp P → ¬ (P ≡ ⊤) ∧ ¬ (P ≡ ⊥) → ⊥
thm5 prf (¬P≡⊤ , ¬P≡⊥) with (λ p → ¬P≡⊤ (thm1 prf p)) | (λ ¬p → ¬P≡⊥ (thm3 prf ¬p))
... | ¬P | ¬¬P = ¬¬P ¬P |
Today, on a very special episode of Science After Sunclipse, Mary Sue discovers that for a quantum-mechanical system with a central potential, the eigenfunctions of the Hamiltonian can be separated into radial and angular factors, and the angular dependence can be understood using angular momentum operators.
ROTATION GENERATORS AND ANGULAR MOMENTUM Way back, we showed that the momentum operator was the generator of spatial translations. Thanks to the time we spent studying 3D rotations, we can deduce an analogous fact about angular momentum. Recall that the commutator of two rotations about axes [tex]i[/tex] and [tex]j[/tex] is
[tex][J_i,J_j] = i\epsilon_{ijk}J_k.[/tex]
As Prof. Freedman would say, we learned in elementary school that in classical mechanics, the angular momentum of a point particle can be written
[tex] \vec{L} = \vec{x} \times \vec{p}.[/tex]
Let’s invent the corresponding expression for a quantum particle, by re-interpreting the variables [tex]\vec{x}[/tex] and [tex]\vec{p}[/tex] as operators which satisfy the canonical commutation relation
[tex][x_i,p_j] = i\hbar \delta_{ij}.[/tex]
To work with the angular momentum, we’ll express the cross product using the Levi-Civita symbol:
[tex]L_k = \epsilon_{ijk} x_i p_j.[/tex]
Now, we’ll show that the components of the angular momentum satisfy the same commutation relation as the generators of spatial rotations (up to a constant we don’t really have to care about). To see what this means, let’s start by finding the commutator of [tex]L_1[/tex] and [tex]L_2[/tex]. By the definition of angular momentum, we get
[tex][L_1,L_2] = [x_2 p_3 – x_3 p_2,x_3 p_1 – x_1 p_3],[/tex]
the right-hand side of which can be simplified to
[tex][x_2 p_3,x_3 p_1] + [x_3 p_2,x_1 p_3].[/tex]
Position and momentum operators commute as long as they do not have the same index, so we can pull some of the operators outside of the brackets,
[tex]x_2 [p_3,x_3] p_1 + x_1 [x_3,p_3] p_2,[/tex]
and then apply the canonical commutator to find
[tex]i\hbar(x_1 p_2 – x_2 p_1).[/tex]
Comparing this with the definition of angular momentum we wrote above, we see that
[tex][L_1,L_2] = i\hbar L_3.[/tex]
Notice that we have the commutator of two operators giving us some constants times the
third operator. Swapping the indices on the left-hand side changes the sign of the commutator, and (of course) repeating an index makes the commutator work out to zero. By running through the above argument with different subscripts, we can show that in general, the commutator takes the familiar form
[tex][L_i,L_j] = i\hbar \epsilon_{ijk} L_k.[/tex]
This is just the relation we had before when we studied rotation generators, except for Planck’s constant (and when we work in “natural units,” we set Planck’s constant to 1 anyway).
When we looked at spatial translations, we worked with Taylor expansions and “infinitesimal generators.” We can do the analogous thing here, although the derivation will not be necessary for the work which follows. Take a particle whose state is specified by a scalar wavefunction [tex]\psi(\vec{x})[/tex] and perform a rotation
[tex]\vec{x} \rightarrow \vec{x}^\prime = R\vec{x},[/tex]
where
[tex]R(\theta,\hat{n}) = R(\vec{\omega})[/tex]
is a rotation of some small angle [tex]\theta[/tex] around the axis defined by [tex]\hat{n}[/tex]. The wave function transforms as
[tex]\psi(\vec{x}) \rightarrow \tilde{\psi}(\vec{x}) [/tex]
such that
[tex]\tilde{\psi}(\vec{x}^\prime) = \psi(\vec{x}).[/tex]
It may not be obvious at first glance, but for a small rotation, the transformed wavefunction can be written as a first-order correction to the original, plus terms higher-order in the angle [tex]\theta[/tex]:
[tex]\tilde{\psi}(\vec{x}) = \psi(\vec{x}) – (\vec{\omega}\times\vec{x})\cdot\grad\psi(\vec{x}) + \mathcal{O}(\theta^2).[/tex]
After some algebraic manipulation, one gets that
[tex]\ket{\tilde{\psi}} = \idmat – \frac{i}{\hbar}\vec{\omega} \cdot \vec{L}}\ket{\psi}[/tex]
where [tex]\vec{L} = \vec{x} \times \vec{p}[/tex].
SPHERICAL COORDINATES
Our problem of interest has an intrinsic spherical symmetry: the interaction potential does not depend upon direction, only distance. However, we’ve been working in Cartesian coordinates. Let’s shift ourselves into a coordinate system which reflects the symmetry of the problem, so we can get at its essence more easily. Mary Sue!
“Yes?”
How can we turn the Cartesian coordinates [tex]x_i[/tex] into spherical coordinates?
“Um. . . Like this?”
[tex]x_1 = r\sin\theta\cos\phi,[/tex]
[tex]x_2 = r\sin\theta\sin\phi,[/tex]
[tex]x_3 = r\cos\theta.[/tex]
Excellent, although if we were mathematicians, our [tex]\theta[/tex] and [tex]\phi[/tex] would be defined the other way round. Now, Mary Sue, I’d like you to show for homework —
“Awwww.”
— for homework, show that with this change of variables, the Cartesian components of the angular momentum can be written in the following way:
[tex]L_1 = i\hbar\left(\sin\phi \partial_\theta + \frac{\cos\phi}{\tan\theta}\partial_\phi\right).[/tex]
[tex]L_2 = i\hbar\left(-\cos\phi\partial_\theta + \frac{\sin\phi}{\tan\theta}\partial_\phi\right).[/tex]
[tex]L_3 = -i\hbar\partial_\phi.[/tex]
In addition, we’ll be working with the square of the magnitude of the angular momentum,
[tex]\vec{L}^2 = L_1^2 + L_2^2 + L_3^2,[/tex]
which in spherical coordinates works out to be
[tex]\vec{L}^2 = -\hbar^2\left(\partial_\theta^2 + \frac{1}{\tan\theta}\partial_\theta + \frac{1}{\sin^2\theta}\partial_\phi^2\right).[/tex]
ANGULAR MOMENTUM EIGENFUNCTIONS
“Excuse me, Professor Science?”
Yes, Mary Sue?
“This is where we pick one component of the angular momentum, like say [tex]L_3[/tex], and then simultaneously diagonalize [tex]L_3[/tex] and [tex]\vec{L}^2[/tex], right?”
Exactly. And what are the eigenfunctions we find?
“They’re the spherical harmonics — we did this in high school. They’re labeled by two integers, the quantum numbers [tex]l[/tex] and [tex]m[/tex].”
So, drawing on our high-school education, we can write
[tex]\vec{L}^2 Y_l^m = l(l+1) \hbar^2 Y_l^m,[/tex]
[tex]L_3 Y_l^m = m\hbar Y_l^m.[/tex]
Recall, in addition, that [tex]l[/tex] can take the values 0, 1, 2, 3, …, while for any value of [tex]l[/tex], the number [tex]m[/tex] ranges from [tex]-l[/tex] to [tex]+l[/tex].
Spherical harmonics occur in classical electromagnetism as well as in quantum mechanics, since the Laplacian also appears in equations for the electrical potential. The detailed investigation of the spherical harmonics is written up in lots of places, so we’ll move on for now.
SEPARATING THE HAMILTONIAN
So far, we have manipulated our original Schrödinger Equation for a two-particle system into the following form, where [tex]\psi[/tex] is the wavefunction which encapsulates the relative degrees of freedom of the two particles; for the hydrogen atom, this is almost, but not exactly, the “wavefunction of the electron.”
[tex]\left(-\frac{\hbar^2}{2\mu} \grad^2 + V(r)\right)\psi = E\psi.[/tex]
The spherical symmetry of our problem leads us to work in spherical coordinates. In that coordinate system, the Laplacian takes the following form:
[tex]\grad^2 = \frac{1}{r} \partial_r^2 r + \frac{1}{r^2}\left(\partial_\theta^2 + \frac{1}{\tan\theta}\partial_\theta + \frac{1}{\sin^2\theta}\partial_\phi^2\right).[/tex]
Comparing this to the expression for [tex]\vec{L}^2[/tex] whose derivation we sketched above, we can rewrite our Hamiltonian in terms of the “electron’s” angular momentum:
[tex]H = -\frac{\hbar^2}{2\mu} \frac{1}{r}\partial_r^2 r + \frac{1}{2\mu r^2}\vec{L}^2 + V(r).[/tex]
Note that the only term which acts on the angular dependency of the wavefunction is the one containing [tex]\vec{L}^2[/tex]. If we separate variables and write the wavefunction as the product of a radial function and a spherical harmonic,
[tex]\psi(\vec{x}) = R_l(r) Y_l^m(\theta,\phi),[/tex]
then the total function is an eigenfunction of the Hamiltonian
[tex]H\psi = E\psi,[/tex]
but it is also an eigenfunction of [tex]\vec{L}^2[/tex]:
[tex]\vec{L}^2\psi = l(l + 1)\hbar^2 \psi.[/tex]
For a given value of [tex]l[/tex], then, our Schrödinger Equation can be written
[tex]\left[-\frac{\hbar^2}{2\mu} \frac{1}{r}\partial_r^2 r + \frac{l(l + 1)\hbar^2}{2\mu r^2} + V(r)\right]R_l(r) = ER_l(r).[/tex]
We can clean this up a little by scaling by [tex]r[/tex], thusly:
[tex]R_l(r) = \frac{1}{r} u_l(r).[/tex]
With this transformation, the Schrödinger Equation takes the form
[tex]\left[-\frac{\hbar^2}{2\mu} \partial_r^2 + \frac{l(l + 1)\hbar^2}{2\mu r^2} + V(r)\right]u_l(r) = Eu_l(r).[/tex]
This is just what we’d get if we wrote a Schrödinger Equation for a single particle moving in one dimension with the potential
[tex]V_{\rm eff.} (r) = V(r) + \frac{l(l + 1) \hbar^2}{2\mu r^2}.[/tex]
We call this the
effective potential. It is just the interaction potential we wrote originally, plus a contribution which depends upon the angular momentum quantum number [tex]l[/tex], a contribution which is known as the angular momentum barrier. This is the quantum analogue of our classical intuition that a particle with large angular momentum spinning around inside a potential well is less likely to be found near the center of the well than is a particle with smaller angular momentum. In other words, the effective potential is where “centrifugal force” sneaks in.
All the work we’ve done so far applies equally well to any two-body system where the interaction potential depends upon the separation distance. We could, for example, use this machinery to attack the 3D harmonic oscillator (two massive particles connected by a massless spring). Instead of doing that, however, we’re going to attack the hydrogenic atoms, where a single electron moves in the Coulomb field of the nucleus. For a nucleus containing [tex]Z[/tex] protons, the interaction potential will be
[tex]V(r) = -\frac{Ze^2}{r}.[/tex]
Therefore, when the electron is in an eigenstate of angular momentum quantum number [tex]l[/tex], its behavior will obey the Schrödinger Equation with the Hamiltonian
[tex]H_l = -\frac{\hbar^2}{2\mu} \partial_r^2 -\frac{Ze^2}{r} + \frac{l(l + 1) \hbar^2}{2\mu r^2}.[/tex]
Next time, we will apply the machinery of SUSY QM and shape invariance to this family of Hamiltonians and deduce both the eigenvalues and the eigenfunctions of the hydrogenic atom.
SUSY QM SERIES |
Given the following equations for the achievable rate of the Minimum Mean-Squared Error Receiver [1]:
$$\mu = \frac{1}{K-1} \sum_{i=1, i \neq k}^{K}{\frac{1}{Mpd_{i}\left(1 - \frac{K-1}{M}+ \frac{K-1}{M}\mu \right) + 1}}$$
and
$$\kappa \left(1 + \sum_{i=1, i \neq k}^{K}{\frac{pd_{i}}{ \left( Mpd_{i}\left(1 - \frac{K-1}{M}+ \frac{K-1}{M}\mu \right) + 1 \right)^2} } \right) = \sum_{i=1, i \neq k}^{K}{\frac{pd_{i}\mu + 1}{ \left( Mpd_{i}\left(1 - \frac{K-1}{M}+ \frac{K-1}{M}\mu \right) + 1 \right)^2} }$$
I've got to find, $\mu$ and $\kappa$. I'm trying to solve these equations with matlab, firstly I tried with solve and then vpasolve, however, when $d$ is a vector with different numbers (i.e., random) it seems matlab never finds a solution. For the case when $d$ is a vector with all values equal to 1 for example, it finds the solution quickly.
This is the script I'm using to find the solution: https://pastebin.com/AWDwyR0U
Am I doing something wrong? Is there a faster way to find a solution to these equations?
Reference:
[1] Hien Quoc Ngo, Erik G. Larsson, and Thomas L. Marzetta, "Energy and Spectral Efficiency of Very Large Multiuser MIMO Systems", IEEE Transactions on Communications, vol. 61, no. 4, April 2013. |
Bourguignon-Lawson's 1981 paper.
The Hilbert function is a classical invariant of a variety (with a given embedding) that is easy to compute. It determines some properties of the variety (such as degree, dimension, and arithmetic genus), but it cannot determine more sophisticated invariants. A minimal free resolution determines more sophisticated properties of the variety while still being easily computable. For example, any set of seven points in P^3 in linearly general position has the same Hilbert function, but minimal free resolutions can distinguish whether the points lie on a rational normal curve. Furthermore, a minimal free resolution retains all the information of the Hilbert function. In this talk, we will define a minimal free resolution and associated invariants. With the definitions in place, we will show that minimal free resolutions retain all the information of the Hilbert function then explain the above example. If time permits, we will additionally show that minimal free resolutions are well behaved when restricting a variety to a hypersurface.
briefly recall a combinatorial approach to the description and quantization of Teichmuller spaces of Riemann surfaces $\Sigma_{g,s}$ of genus $g$ with $s$ holes and algebras of geodesic functions on these surfaces. We describe sets of geodesic functions in W.Thurston shear coordinates based on an ideal triangle decomposition of Riemann surfaces with holes and demonstrate the polynomiality and positivity properties of the corresponding geodesic functions. In the algebraic setting, these sets are related to traces of monodromies of $SL_2$ connection on $\Sigma_{g,s}$, and Darboux-type Poisson and quantum relations on shear coordinates were proven to generate Goldman brackets on geodesic functions. I will describe these structures and their recent generalizations to $SL_2$ and $SL_n$ (decorated) character varieties on Riemann surfaces $\Sigma_{g,s,n}$ with holes and $n$ marked points on hole boundaries and how it is interlaced with cluster algebras, reflection equations, and groupoids of upper triangular matrices. [Based on work in collaboration with M.Mazzocco, V.Roubtsov, and M.Shapiro.]
We are interested in the topology of some spaces obtained by relaxing total positivity in the real Grassmannian. We define two families of subsets of the Grassmannian each of which include both the totally nonnegative Grassmannian and the whole Grassmannian. In this initial study of such subsets of the Grassmannian we focus of subsets of real projective space where interesting topology already appears. We we are able to find a regular CW complex which can be leveraged to compute some invariants like the fundamental group and Euler characteristic. We also conjecture some "ball-like" properties (e.g. Cohen-Macualayness).
The purpose is to investigate temporal clusters of extremes defined as subsequent exceedances of high thresholds in a stationary time series. Two meaningful features of these clusters are the probability distribution of the cluster size and the ordinal patterns within a cluster. The latter have been introduced in order to handle data sets with several thousand data points appearing in medicine, biology, finance and computer science. Since these patterns take only the ordinal structure of consecutive data points into account, the method is robust under monotone transformations and measurement errors. We verify the existence of the corresponding limit distributions in the framework of regularly varying time series, develop non-parametric estimators and show and their asymptotic normality under appropriate mixing conditions. (This is joint work with Marco Oesting.)
In this talk, I will present a new data-driven paradigm on how to quantify the structural uncertainty (model-form uncertainty) and learn the physical laws hidden behind the noisy data in the complex systems governed by partial differential equations. The key idea is to identify the terms in the underlying equations and to approximate the coefficients of the terms with error bars using Bayesian machine learning algorithms on the available noisy measurement. In particular, Bayesian sparse feature selection and parameter estimation are performed. Numerical experiments show the robustness of the learning algorithms with respect to noisy data and size, and its ability to learn various candidate equations with error bars to represent the quantified uncertainty.
Talk II on Bourguignon-Lawson's 1978 paper
The stable parametrized h-cobordism theorem provides a critical link in the chain of homotopy theoretic constructions that show up in the classification of manifolds and their diffeomorphisms. For a compact smooth manifold M it gives a decomposition of Waldhausen's A(M) into QM_+ and a delooping of the stable h-cobordism space of M. I will talk about joint work with Malkiewich on this story when M is a smooth compact G-manifold.
We show $C^\infty$ local rigidity for a broad class of new examples of solvable algebraic partially hyperbolic actions on ${\mathbb G}=\mathbb{G}_1\times\cdots\times \mathbb{G}_k/\Gamma$, where $\mathbb{G}_1$ is of the following type: $SL(n, {\mathbb R})$, $SO_o(m,m)$, $E_{6(6)}$, $E_{7(7)}$ and $E_{8(8)}$, $n\geq3$, $m\geq 4$. These examples include rank-one partially hyperbolic actions. The method of proof is a combination of KAM type iteration scheme and representation theory. The principal difference with previous work
that used KAM scheme is very general nature of the proof: no specific information about unitary representations of ${\mathbb G}$ or ${\mathbb G}_1$ is required.
This is a continuation of the last talk.
A classical problem in knot theory is determining whether or not a given 2-dimensional diagram represents the unknot. The UNKNOTTING PROBLEM was proven to be in NP by Hass, Lagarias, and Pippenger. A generalization of this decision problem is the GENUS PROBLEM. We will discuss the basics of computational complexity, knot genus, and normal surface theory in order to present an algorithm (from HLP) to explicitly compute the genus of a knot. We will then show that this algorithm is in PSPACE and discuss more recent results and implications in the field.
We show that the three-dimensional homology cobordism group admits an infinite-rank summand. It was previously known that the homology cobordism group contains an infinite-rank subgroup and a Z-summand. The proof relies on the involutive Heegaard Floer homology package of Hendricks-Manolescu and Hendricks-Manolescu-Zemke. This is joint work with I. Dai, M. Stoffregen, and L. Truong.
There is a close analogy between function fields over finite fields and number fields. In this analogy $\text{Spec } \mathbb{Z}$ corresponds to an algebraic curve over a finite field. However, this analogy often fails. For example, $\text{Spec } \mathbb{Z} \times \text{Spec } \mathbb{Z} $ (which should correspond to a surface) is $\text{Spec } \mathbb{Z}$ (which corresponds to a curve). In many cases, the Fargues-Fontaine curve is the natural analogue for algebraic curves. In this first talk, we will give the construction of the Fargues-Fontaine curve.
Consider a collection of particles in a fluid that is subject to
a standing acoustic wave. In some situations, the particles tend to
cluster about the nodes of the wave. We study the problem of finding a
standing acoustic wave that can position particles in desired locations,
i.e. whose nodal set is as close as possible to desired curves or
surfaces. We show that in certain situations we can expect to reproduce
patterns up to the diffraction limit. For periodic particle patterns, we
show that there are limitations on the unit cell and that the possible
patterns in dimension d can be determined from an eigendecomposition of a
2d x 2d matrix.
No abstract available.
Although Seiberg Witten invariant originally introduced for four manifolds, but its three dimensional version is also interesting .After a brief discussion on the definition of the Seiberg Witten invariant on three manifolds we will see some results from literature equating this invariant to some known invariants of three manifolds.
I'll discuss joint work with Rita Gitik (UM) on the problem of recognizing quasi-positive elements of a group G defined by
a finite presentation (X ; R). An element of G is quasi-positive if it can be represented by a word that is a product of conjugates of positive powers of letters in X. The recognition problem is to determine whether or not a given word (using both positive and negative powers of letters in X) represents an element of G that is quasi-positive. This problem was solved by Orevkov when G is free with basis X or when G is the 3-strand braid group with its standard generating set. I'll present a new solution to the recognition problem for free groups and discuss some of the challenges posed by braid groups and related groups.
We develop a version of cluster algebra extending the ring of Laurent polynomials by adding Grassmann variables. These algebras can be described in terms of “extended quivers” which are oriented hypergraphs. We describe mutations of such objects and define a corresponding commutative superalgebra. Our construction includes the notion of weighted quivers that has already appeared in different contexts. This project is a step towards understanding the notion of cluster superalgebra.
We discuss a number of $h$-principle phenomena which were recently discovered in the field of contact and symplectic geometry. In generality, an $h$-principle is a method for constructing global solutions to underdetermined PDEs on manifolds by systematically localizing boundary conditions. In symplectic and contact geometry, these strategies typically are well suited for general constructions and partial classifications. Some of the results we discuss are the characterization of smooth manifolds admitting contact structures, high dimensional overtwistedness, the symplectic classification of flexible Stein manifolds, and the construction of exotic Lagrangians in $C^n$.
This talk is the second part of the talk that I gave last semester in the dynamics seminar. I will present sketch of the proof of the main result of the joint work with Dmitry Kleinbock in which we found an upper bound for the set of points in a homogeneous space whose orbit miss a fixed open subset.
The comparison principle for second order elliptic equations is one of the cornerstone of the theory of viscosity solutions. Works of Sayah and more recently by Jakobsen-Karlsen and Barles-Imbert have expanded it to many important subfamilies of nonlocal elliptic equations. In joint work with Chenchen Mou and Andrzej Święch we show how optimal transportation methods can be used to couple Lévy measures, which encode the integro-differential part of nonlocal operators, which allow us to obtain comparison principles for new families of nonlocal equations. Our method puts all previous subfamilies of nonlocal operators (Levy-Ito operators, operators of order less than 1) in a single framework, while also yielding results for new families.
I will describe an infinite family of integer-valued concordance homomorphisms defined using knot Floer homology. These invariants have topological applications to concordance genus, concordance unknotting number, and bridge index. This is joint work with Irving Dai, Jen Hom, and Matt Stoffregen.
A long standing problem in geometry, conjectured by Yau in 1982, is that any any $3$-manifold admits an infinite number of distinct minimal surfaces. The analogous problem for geodesics on surfaces led to the discovery of deep interactions between dynamics, topology, and analysis. The last couple of years brought dramatic developments to Yau’s conjecture, which has now been settled due to the work of Marques-Neves and Song. I will survey the history of the problem and the several contributions made.
Networks are widely recognized as important for phenomena that span multiple disciplines
including the diffusion of information (communication), the contagion of illness (public health), the
passage of legislation (political science), the emergence of norms (sociology), the coordination of economic activities (economics & geography), the formation of relationships (developmental psychology), and detection of national security threats (intelligence). However, the challenges of collecting network data have limited the feasibility of network research in these areas. In this project, we will explore and refine the use of bipartite projections to validly infer otherwise unobserved or unobservable networks. Several formal methods for making such inferences have been proposed, however it remains unknown which models lead to valid inferences and under what circumstances. We use bipartite network data to systematically test, compare, and extend available models for inferring networks from bipartite projections.
Department of Mathematics
Michigan State University
619 Red Cedar Road
C212 Wells Hall
East Lansing, MI 48824
Phone: (517) 353-0844
Fax: (517) 432-1562
College of Natural Science |
So to be clear: We are proving the binomial theorem by mathematical induction?
For speed of typing, I will use $\dbinom n k$ rather than ${^n\mathrm C_k}$
It rather sounds like you are at this stage:
Assume for any $n\geq 1$ that: $(1+x)^n = \sum\limits_{k=0}^n \dbinom nk x^k$
Then it would be that: $(1+x)^{n+1} ~{= (1+x)\sum\limits_{k=0}^n \dbinom nk x^k \\ = \dbinom n 0 +\sum\limits_{k=1}^n \left({\dbinom nk+\dbinom n{k-1}}\right)x^k + \dbinom nn x^{n+1} }$
Well, there is an identity: $\dbinom{n+1}{k}=\dbinom n k+\dbinom n{k-1}$
You should now be able to continue.
Is there a way to do this without putting it in summation form, and only using the last two terms listed as the sum?
That would be:
$$\begin{align}& (1+x)^{n+1}\\ =~& (1+x)^n + (1+x)^n x \\ = ~& (\tbinom n0+\tbinom n1x+\ldots+\tbinom nkx^k+\ldots+\tbinom nn x^n)+(\tbinom n0x+\ldots+\tbinom n{k-1}x^k+\ldots+\tbinom n{n-1}x^n+\tbinom nn x^{n+1})\\ = ~& \tbinom n0+(\tbinom n1+\tbinom n0)x+\ldots+(\tbinom nk+\tbinom n{k-1})x^k+\ldots+(\tbinom nn+\tbinom n{n-1}) x^n+\tbinom nn x^{n+1}\\ \vdots ~&\end{align}$$ |
I am having some troubles understanding the algorithm.
My algorithm looks like this.
how come am i able to compute the DFT coefficient using this algoritm. As far as i know $$ X[k] = \sum_{i=0}^{N-1}x(n)W_N^k $$
which using Goertzel is the same as $$X(k) = y_k(N) $$ and since $$y_k(n) = v_k(n) -W_n^kv_k(n-1)$$ and $$v_k(n) =2cos(\frac{2 \pi k}{N}v_k(n-1)-v_k(n-2)+x(n)$$ which means $$X(k) = y_k(N) = v_k(N) -W_n^kv_k(N-1)$$ which isn't the same as $$X[k] = \sum_{i=0}^{N-1}x(n)W_N^k$$ ??
how come is Goertzel using $$|X(k)^2|$$ to compute the amplitude spectrum, where DFT uses $$|X(k)|$$ to compute the amplitude spectrum. ?? |
Associations to the word «Von» Noun Neumann Graf Humboldt Braun Erich Bis Friedrich Philipp Johann Ritter Bismarck Nach Heinrich Zu Albrecht Goethe Ludwig Trier Gottfried Wilhelm Christoph Georg Walther Eduard Wolfgang Franz Hermann Herr Konrad Frau Schiller Moritz Theodor Dietrich Der Bernhard Dem Ulrich Ernst Auf Otto Joachim Aus Maximilian Eugen Oskar Leben Stephan Adolf Karl Hartmann Jakob Mainz Oder Rudolf Manfred Baroness Gustav Hans Werner Ferdinand Leopold Gerhard Mueller Deutschland Gesellschaft Das Constantin Siemens Horst Josef Elisabeth Adolph Rosen Habsburg Essen Sigismund Helmut Lorenz Anton Prussian Claus Carl Welt Johannes Galen Den Fischer Fritz Abbess Baron Archduke Heinz Emil Clemens Holstein Wiktionary
VON HIPPEL-LINDAU DISEASE, noun. Angiomatosis
VON KÁRMÁN VORTEX STREET, noun. (physics) a double row of vortices in a fluid sometimes found in the wake of a cylindrical body (such as in a river, downstream of a bridge support), eddies being produced from alternate sides of the body
VON KARMAN VORTEX STREET, noun. Alternative form of Kármán vortex street
VON KÁRMÁN VORTEX STREETS, noun. Plural of von Kármán vortex street
VON MANGOLDT FUNCTION, noun. (math) A certain arithmetic function that is neither multiplicative nor additive. It is denoted by Λ(n) and defined as :\(\Lambda(n) = \begin{cases} \log p & \mbox{if }n=p^k \mbox{ for some prime } p \mbox{ and integer } k \ge 1, \\ 0 & \mbox{otherwise.} \end{cases}\)
VON NEUMANN ENTROPY, noun. (quantum physics) The entropy of a quantum state. If the state is expressed as a quantum density matrix \(\rho\), then this entropy can be expressed mathematically as \(S = -\text{tr}(\rho \log \rho)\) where \(\text{tr}\) is the trace operator and the logarithm is natural.
VON NEUMANN MACHINE, noun. (computing) Any computer that has a central processing unit and can store programs in memory.
VON NEUMANN MACHINE, noun. (science fiction) Any machine that is capable of self-replication.
VON NEUMANN MACHINES, noun. Plural of von Neumann machine
VON NEUMANN PROBE, noun. (science fiction) A von Neumann machine, or machine capable of self-replication, that is sent on interstellar voyages to investigate or terraform.
VON NEUMANN PROBES, noun. Plural of von Neumann probe
VON RECKLINGHAUSEN DISEASE, proper noun. Neurofibromatosis type I
VON RESTORFF EFFECT, noun. The effect that distinctive things are easier to remember
VON WILLEBRAND DISEASE, noun. (medicine) An hereditary disease, characterized by a tendency to hemorrhage, and caused by a defect in blood platelet activity.
VON WILLEBRAND'S DISEASE, noun. (medicine) An hereditary disease, characterized by a tendency to hemorrhage, and caused by a defect in blood platelet activity.
Wise words
Always aim at complete harmony of thought and word and deed. Always aim at purifying your thoughts and everything will be well. |
I am trying to check the conditions of the Phragmén Lindelöf Principle (see below) in order to use it on the following corollary:
CorollaryLet $u$ be a subharmonic function on an unbounded proper domain subdomain $D$ of $\mathbb{C}$ such that \begin{eqnarray*} \limsup_{z \to \zeta} u(z) \leq 0 \quad \text{for} \quad \zeta \in \partial D \backslash \{\infty\} \quad \text{and} \quad \limsup_{z \to \infty} \frac{u(z)}{\log|z|} \leq 0. \end{eqnarray*} Then $u \leq 0$ on $D$.
Proof. Take $w \in \partial D$ and aplly Theorem 2.3.2 with $v(z)=\log |z-w|$. $\square$
This is the theorem that the corollary uses:
Theorem 2.3.2 (Phragmén Lindelöf Principle)Let $u$ be a subharmonic function on a unbounded domain D in $\mathbb{C}$ such that \begin{eqnarray*} \limsup_{z \to \zeta} u(z) \leq 0 \end{eqnarray*} for all $\zeta \in \partial D \backslash \{\infty\}$. Suppose also taht there exists a finite-valued superharmonic function $v$ on $D$ such that \begin{eqnarray*} \liminf_{z \to \infty} v(z) > 0 \quad \text{and} \quad \limsup_{z \to \infty} \frac{u(z)}{v(z)} \leq 0. \end{eqnarray*} Then $u \leq 0$ on $D$.
So we need $v$ to be finite-valued and superharmonic, which is the case since $z \neq w$ (right?). But I am having trouble understanding why the limit conditions are fulfilled: \begin{eqnarray*}\liminf_{z \to \infty} \log |z-w|>0 \\\limsup _{z \to \infty} \frac{u(z)}{\log |z-w|} \leq 0\end{eqnarray*} What if $w=\infty$? (I should mention that limits are taken with respect to $\infty$.) And why do we need the subdomain to be
proper?
Thanks! |
If you have a harmonic oscillator with damping $D$ (e.g. small angle pendulum)
$$\ddot{\theta}+D\dot{\theta} + \theta=0$$
then the solution I get in the underdamped case ($D^2-4<0$) is: $$\theta=\theta_0e^{-\gamma t}\cos{(\omega t)}$$ where $\gamma=\frac{D}{2}$ and $\omega=\frac{1}{2}\sqrt{4-D^2}$, and $\theta_0$ is the inital angular displacement. We have assumed the inital angular velocity $\dot{\theta}$ is zero.
The solution I get for the critically damped case ($D^2-4=0$) is: $$\theta=\theta_0(1+t)e^{-\gamma t}$$
Fine. Except that they don't match. When $D^2-4=0$, the underdamped case becomes $\theta=\theta_0e^{-\gamma t}$ which is not the same as the critically damped case.
And if there is no damping, the resulting cosine wave is exactly the same as the simulation.
So does this mean the analytic solution for an underdamped pendulum is actually an approximation only valid for small damping, or have I done something wrong? |
This class logs the out of bag risk for a specific loss function. It is also possible to use custom losses to log performance measures. For details see the use case or extending compboost vignette.
S4 object.
1 2
use_as_stopper [
logical(1)]
Boolean to indicate if the logger should also be used as stopper.
used_loss [
Loss object]
The loss used to calculate the empirical risk by taking the mean of the returned defined loss within the loss object.
eps_for_break [
numeric(1)]
This argument is used if the loss is also used as stopper. If the relativeimprovement of the logged inbag risk falls above this boundary the stopperreturns
TRUE.
oob_data [
list]
A list which contains data source objects which corresponds to the source data of each registered factory. The source data objects should contain the out of bag data. This data is then used to calculate the prediction in each step.
oob_response [
numeric]
Vector which contains the response for the out of bag data given withinthe
list.
This logger computes the risk for a given new dataset
\mathcal{D}_\mathrm{oob} = \{(x^{(i)},\ y^{(i)})\ |\ i \in I_\mathrm{oob}\}and stores it into a vector. The OOB risk \mathcal{R}_\mathrm{oob} foriteration m is calculated by: \mathcal{R}_\mathrm{oob}^{[m]} = \frac{1}{|\mathcal{D}_\mathrm{oob}|}∑\limits_{(x,y) \in \mathcal{D}_\mathrm{oob}} L(y, \hat{f}^{[m]}(x)) Note:
If
m=0 than \hat{f} is just the offset.
The implementation to calculate
\mathcal{R}_\mathrm{emp}^{[m]} isdone in two steps:
Calculate vector
risk_temp of losses for every observation forgiven response
y^{(i)} and prediction \hat{f}^{[m]}(x^{(i)}).
Average over
risk_temp.
This procedure ensures, that it is possible to e.g. use the AUC or anyarbitrary performance measure for risk logging. This gives just onevalue for
risk_temp and therefore the average equals the lossfunction. If this is just a value (like for the AUC) then the value isreturned.
This class is a wrapper around the pure
C++ implementation. To seethe functionality of the
C++ class visithttps://schalkdaniel.github.io/compboost/cpp_man/html/classlogger_1_1_oob_risk_logger.html.
This class doesn't contain public fields.
summarizeLogger()
Summarize the logger object.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
# Define data:X1 = cbind(1:10)X2 = cbind(10:1)data.source1 = InMemoryData$new(X1, "x1")data.source2 = InMemoryData$new(X2, "x2")oob.list = list(data.source1, data.source2)set.seed(123)y.oob = rnorm(10)# Used loss:log.bin = LossBinomial$new()# Define logger:log.oob.risk = LoggerOobRisk$new(FALSE, log.bin, 0.05, oob.list, y.oob)# Summarize logger:log.oob.risk$summarizeLogger()
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Lenz’s law named after Emil Lenz who formulated it during 1834. It depends on the principle of conservation of energy and Newton’s third law and is the most convenient method to determine the direction of the induced current. It states that the direction of an induced current is always such as to oppose the change in the circuit or the magnetic field that produces it.
Lenz Law Definition
Lenz’s law states that induced electromotive force with different polarities induces a current whose magnetic field opposes the change in magnetic flux through the loop in order to ensure that original flux is maintained through the loop when current flows in it.
Lenz Law Formula
The Lenz Law is reflected in the formula of Faraday’s law. Here the negative sign is contributed from Lenz law. The expression is –\(Emf=-N\left ( \frac{\Delta \phi }{\Delta t} \right )\)
Where,
Emf is the induced voltage (also known as electromotive force).
N is the number of loops.\(\Delta \phi \) Change in magnetic flux. \(\Delta t\) Change in time.
Following is the table with links of other
Physics related laws: Lenz Law Applications
Lenz law applications are plenty. Some of them are listed below-
Eddy current balances Metal detectors Eddy current dynamometers Braking systems on train AC generators Card readers Microphones Lenz Law Experiment
To find the direction of the induced electromotive force and current we look to Lenz’s law. Some experiments were proved by Lenz’s in accordance with his theory.
First Experiment
In the first experiment, he concluded that when the current in the coil flows in the circuit the magnetic field lines are produced. As the current flows through the coil increases, the magnetic flux will increase. The direction of the flow of induced current would be such that it opposes when the magnetic flux increases.
Second Experiment
In the second experiment, he concluded that when the current carrying coil is wound on an iron rod with its left end behaving as N-pole and is moved towards the coil S, an induced current will be produced.
Third Experiment
In the third experiment, he concluded that when the coil is pulled towards the magnetic flux, the coil linked with it goes on decreasing that means that the area of the coil inside the magnetic field decreases. According to Lenz’s law, the motion of the coil is opposed when the induced current is applied in the same direction.
To produce current force is exerted by the magnet in the loop. To oppose the change a force must be exerted by the current on the magnet.
Practice Questions on Lenz’s Law Q1: State Lenz’s law Ans: Lenz’s law states that induced electromotive force with different polarities induces a current whose magnetic field opposes the change in magnetic flux through the loop in order to ensure that original flux is maintained through the loop when current flows in it. Q2: Write the Lenz law formula and explain the terms. Ans: Lenz law formula is given by- \(Emf=-N\left ( \frac{\Delta \phi }{\Delta t} \right )\) Where,
Emf is the induced voltage (also known as electromotive force).
N is the number of loops.\(\Delta \phi \) Change in magnetic flux. \(\Delta t\) Change in time.
Q3: What is the prime importance of Lenz law? Ans: Lenz law is used to determine the direction of the induced current. Q4: What are the applications of Lenz law? Ans: The applications of Lenz law are Eddy current balances Metal detectors Eddy current dynamometers Braking systems on train Q5: Where is Lenz law used? Ans:Lenz’s law is used to explain how electromagnetic circuits obey the conservation of energy and Newton’s third law
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