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Background
Let's say you own a convenience store
Each day $N$ customers will visit you, where $N \in (1,7 \, billion \, humans)$
$X_i$ is the amount that each customer will spend
All the $X_i$'s are iid and also independent of $N$
Let's say you want to know the expected value of total revenue made in a day. I know the following is true (Wald's equation),
$$E \bigg[ \sum_{i=1}^{N} X_i \bigg] = E[N]E[X]$$
My question
Can you explain why it is illegal to do,
$$E \bigg[ \sum_{i=1}^{N} X_i \bigg] = \sum_{i=1}^{N} E[X_i] = NE[X]$$
I know this is illegal because $NE[X] \neq E[N]E[X]$. I know the LOE doesn't always hold for infinite sums but in this situation $N$ is capped by the human population. If you could explain in lay terms I would greatly appreciate it (versus going into sigma algebras and IUT). Thanks. |
Premise: Let $A := [ k(1), k(2), ..., k(n) ]$ and $B:=[ l(1), l(2), ..., l(n) ]$ be two Arrays where $k$ and $l$ are permutations. (What I'm trying to express: $A$ and $B$ contain the same elements in different orders.)
$A$ and $B$ are of different types, thus it is not allowed to mix them in the same data structure.
It is not possible to compare two elements that are from the same array with each other. However, given is an ordering $A \times B \rightarrow \{<, >, =\}$ (in words: it is possible to compare an $a \in A$ with a $b \in B$ which will tell whether $a > b$, $a = b$ or $a<b$).
There are no restrictions on space being used.
The
task is to sort both arrays.
I think I have found a solution analogous to QuickSort that also has $\Theta(n~\,log\,n)$ best case and $\Theta(n^2)$ worst case time complexity.
My
question: Is there an algorithm that runs in $\Theta(n~\,log\,n)$ in the worst case aswell?
My approach:
Facilitate $A[1] = B[1]$ by iterating through $B[2, ..., n]$ and swapping the element that is equal to $A[1]$ to $B[1]$.
Use that element as a Pivot and, in the fashion of QuickSort, go through $A[2..n]$ comparing the elements with $B[1]$ and vice versa for $B[2..n]$.
Rearrange the pivot so that to the left of it there are only elements that are smaller than the Pivot and all others to the right of them. At this point it is guaranteed that
Do this recursively until reaching a base case in which there is only a single element to be sorted.
Now, as far as I can see, similarly to QuickSort, in the best case we require $log~n$ layers of recursion and in each we have a total of $n$ required for recombination.
However, this has the weakness that if the input is already sorted, we will end up with quadratic runtime.
When using a D&C approach, one has to take care that the two arrays are divided so that the resulting smaller arrays still contain the same elements. |
The usual justification for regarding POVMs as fundamental measurements is via Neumark's theorem, i.e., by showing that they can always be realized by a projective measurement in a larger Hilbert space.
That justification is sometimes problematic because for some applications is important
not to enlarge the Hilbert space, so as to guarantee that the result you proved via POVMs is really about a Hilbert space of that dimension, and not just a shadow of a larger Hilbert space.
So, my question is, how to implement POVMs without enlarging the Hilbert space?
The only strategy I know is doing PVMs stochastically and grouping outcomes; for instance, the POVM $$\bigg\{\frac{1}{2}|0\rangle\langle0|,\frac{1}{2}|1\rangle\langle1|,\frac{1}{2}|+\rangle\langle+|,\frac{1}{2}|-\rangle\langle-|\bigg\}$$ can be implemented by measuring either $\{|0\rangle\langle0|,|1\rangle\langle1|\}$ or $\{|+\rangle\langle+|,|-\rangle\langle-|\}$ with probability $1/2$; by grouping the outcomes one can then measure the POVM $$\bigg\{\frac{1}{2}|0\rangle\langle0|+\frac{1}{2}|+\rangle\langle+|,\frac{1}{2}|1\rangle\langle1|+\frac{1}{2}|-\rangle\langle-|\bigg\}$$ or $\{I/2,I/2\}$
But this class can't be the whole set of POVMs; the Mercedes-Benz POVM (which has three outcomes proportional to $|0\rangle$ and $\frac{1}{2}|0\rangle \pm \frac{\sqrt{3}}{2}|1\rangle$) clearly can't be implemented this way. Is there a neat characterization of this class? Is there published research on it? Even better, is there another (more powerful) way of implementing POVMs without enlarging the Hilbert space?This post has been migrated from (A51.SE) |
Unfortunately, I couldn't solve this question. Actually, I am very suspicious about whether the question is correct or not.
An experiment has a sample space that consists of 8 equally likely outcomes. $ S = \{x_1, x_2, ..., x_8\} $. Three events are defined as $ A = \{x_2, x_4, x_6, x_8\}, B = \{x_1, x_3, x_6, x_7\}, C = \{x_1, x_2, x_3, x_5\} $ . Find the probabilities of the following events.
a)$ A \cap B$
b) $ \overline{A \setminus B}$
c)$ A \cap (B\cup\overline{A})$
d $\overline{A \cap B \cap C} $
I thought for question in a $ A \cap B = \{x_6\} $ so $ P(A \cap B) = 1/8 $ but i am not sure |
I'm trying to find the quotient of a free $\mathbb{Z}$-module, but somehow I don't really find the right procedure on how to get the right quotient (nor have I found any sources). I've read What does it mean here to describe the structure of this quotient module? already, but I didn't manage to reproduce the second matrix from that specific post. In my specific example I have $\mathbb{Z}^3$ as a left-$\mathbb{Z}$ module. Lets call the generators $e_1, e_2, e_3$. My submodule is given by
$$K:=\operatorname{span}_{\mathbb{Z}}\left<i_1:=\left(\matrix{0\\-2\\1}\right), i_2:=\left(\matrix{-2\\0\\1}\right)\right>$$
My idea to get the quotient module $\mathbb{Z}^3/K$ thus far was to use the equalities
$$i_1 = -2e_2+e_3\\ i_2 = -2e_1+e_3 $$
Which yields (for $z_1,z_2,z_3,l_1,l_2\in\mathbb{Z}$):
$$ z_1e_1+z_2e_2+z_3e_3 \equiv z_1e_1+z_2e_2+z_3e_3+l_1i_1+l_2i_2\\ = z_1e_1+z_2e_2+z_3e_3+l_1(-2e_2+e_3)+l_2(-2e_1+e_3)\\ = (z_1-2l_1)e_1 + (z_2-2l_2)e_2 + (z_3+l_1+l_2)e_3$$
So for $z_1$ even, the first summand should be identical to zero. For $z_1$ uneven it should be identical to one. Same goes for $z_2$ and the second summand, whereas the third summand should be identical to zero in any case (though i'm not really sure about how to deal with that one, since I've already fixed $l_1$ and $l_2$). Hence I'd say my quotient should be $\mathbb{Z}_2^2$. However, letting sage do the calculations, it claims that the result should be $\mathbb{Z}\oplus\mathbb{Z}_2$
So my question is: Is my result wrong? If so: Where's my mistake and is there a better way to do it? |
...or not...
The black hole firewall saga continues. The original paper by AMPS has collected 35 citations according to SPIRES. Most of the recent 10 are papers that are primarily about issues that different than black hole firewalls.
However, Stephen Hsu who is also a blogger (article about the topic; don't confuse him with Steven Chu although there may be some similarities here, too) just posted a new, 3-page preprint that attacks the essential error done by AMPS:
The most important fact of this sort is that for a fixed pure initial state of the star, Alice (the infalling observer) always has some probability amplitudes (and some overall probability) that she falls into a black hole and some probability amplitudes (and some overall probability) that she doesn't. There are many possible microstates (possible evolution) in the first group and many possible microstates (possible evolution) in the second group and the overall state vector is the sum of all the pieces from both groups. I discussed the presence of the "Yes" as well as "No" branches here (click).
I found this cartoon by a rudimentary Alice-Bob search on Google Images but the content of the cartoon is actually exactly what I need. Alice correctly tells Bob who wants to marry her that in a quantum world, both possibilities – she will marry him and she won't, she will escape him to a black hole interior or she won't – have nonzero probability amplitudes. As the article below discusses many times, Bob much like Polchinski et al. seem to misunderstand this basic point about quantum mechanics. Well, this misunderstanding may be shared by most men as well because they often interpret the quantum mechanical women's "Maybe" as "Yes". :-)
Only the overall wave function must evolve unitarily from the initial state, only the total wave function (including pieces with Alice inside and pieces with Alice never inside any black hole) are subject to the constraint that the information should be preserved. This accuracy is very important because, as Papadodimas and Raju emphasized, even exponentially tiny corrections to the operators (I mean the matrix entries are exponentially tiny) may turn a pure state into a mixed one and vice versa.
Hsu says that Alice observes no firewalls but it's important to discuss "which Alice" we mean when we say that she sees no firewall; and "which Alice[s]" we need to preserve the information. The answers to both questions differ.
The answer to the first question is that
a particular Alicewith a particular history, one in which she's sure that she's falling into a black hole (a single term in the overall wave function), will experience no firewalls. If we imagined that the whole quantum evolution (evolved state) contains this Alice (term), we could argue – following AMPS – that the information isn't conserved.
However, the unitary evolution of the initial state contains, as we have emphasized, also other "Alices", including "Alices" who avoid the black hole interior at all times. The word "contain" in the previous sentence is meant to represent the mathematical inclusion of a term in a sum; the correct "physical" interpretation says that the laws of quantum mechanics imply that it's possible for Alice to evolve differently and, perhaps, avoid the black hole throughout her life. The total wave function is a state similar to Schrödinger's cat, one composed of macroscopically distinct states that quickly decohere from each other. This state evolves unitarily from the initial state and will evolve unitarily to the final state of the Hawking radiation that remembers the initial state.
There's no need for the information to be preserved on the "branch" of one particular Alice. One particular Alice who fell into a black hole just made some measurements of certain observables \(A_i\) that identify her "branch" of the wave function. One of these observables \(A_0\) is the qubit that determines whether she fell into a black hole or not.
These observables \(A_i\), including \(A_0\), refuse to commute with the observables \(B_k\) that observers outside the black hole (spatially separated from the particular Alice who is inside) may perform. There's nothing wrong about this fact; the fact that these sets of operators \(\{A_i\},\{B_k\}\) don't commute with each other is the essence of black hole complementarity. As long as the probabilities of the measured values of \(A_i\) are nonzero in the eigenstates corresponding to the measured values of \(B_k\), there is no contradiction.
Polchinski et al. made the mistake of trying to restrict the discussion of all the observations (including those outside) to the subspace of eigenvalues \(A_i\) that a particular infalling Alice has measured. But that's just wrong: most of the states identified by other observations (e.g. those done outside the hole, for example by Bob) fail to belong to this low-dimensional space because the relevant states onto which Bob projects are eigenstates of operators such as \(B_k\) that don't commute with \(A_i\), so they can't possibly belong to a particular subspace of shared \(A_i\) eigenstates.
Steve Hsu also discusses one interesting point showing that the fact that the black hole evolves into a superposition of macroscopically distinct states isn't just a formality that affects Alice in the same sense as it usually affects Schrödinger's cat. Instead, the black hole itself is evolving into a superposition of macroscopically distinct states, especially when it comes to the location of the black hole.
When the black hole is sending the Hawking quanta to random directions, their total momentum more or less averages out. But it doesn't average out exactly. The momentum of a Hawking particle goes like the temperature \(T\sim 1/R\) and each particle reduces the entropy of the remaining black hole by something like \(\Delta S\sim -1\). After the Page time (half of the initial entropy has been evaporated away), the black hole has sent something like \(S/2\) Hawking quanta. The momentum of each was \(1/R\) with a random sign/direction. If we add them, we may see by the maths of random walks that the total momentum deposited to the recoiling black hole shooting the Hawking quanta is of order \(\pm\sqrt{S/2}/R\sim\pm 1/\sqrt{G}\) (the last expression only holds for \(d=4\) while the previous expressions held for a general \(d\)) so the average velocity is of order \(P/M=\pm 1/M\sqrt{G}\). Yes, that's smaller than \(c=1\) because the mass of the black hole is larger than the Planck mass \(1/\sqrt{G}\).
When you finally multiply this velocity by the Page time \(t\sim G^2 M^3\), you get the estimate for the total distance that the black hole has traveled after the long Page time due to the recoils as it was shooting the Hawking bullets:\[
\Delta x \sim \pm \frac{1}{M\sqrt{G}} \cdot G^2 M^3\sim\pm M^2 G^{3/2}.
\] Now, all the formulae are only OK in \(d=4\). In the Planck units, the change of the location of the black hole goes like \(M^2\) which is still a very high number i.e. long distance! It is much longer, \(M \sqrt{G}\) times, than the black hole radius. (I wrote the derivation directly into the blog and haven't seen it previously; thank God, the result agreed with Hsu's underived claim.) So the Hawking quanta are pretty random but they easily combine to an uncertainty of the black hole location that may become macroscopic – in fact, much longer than the black hole radius – after a sufficiently long time such as the Page time.
(If you want to know, after the fraction \(f\) of the black hole mass evaporates away, \(\Delta x\) is generalized to \(f^{3/2}G^{3/2}M^2\); the simplified expression above was for the Page fraction \(f=1/\sqrt{2}\). It's not hard to derive the \(f^{3/2}\) dependence, I think: the velocity goes like \(f^{1/2}\) for the same reason why the random walk \(\Delta x\) goes like \(t^{1/2}\); the power \(f^{3/2}\) is just the indefinite integral of that which adds \(1\) to the exponent. Also, note that the "branches" with different locations of the black hole inevitably decohere from each other if we decide to trace over the degrees of freedom in the outgoing Hawking radiation.)
It's very important not to overlook the other branches of the evolution, including the inevitably nonzero branches in which Alice never falls into a black hole, if we want to verify that the information is conserved. AMPS failed in this test of accuracy. They were imagining that the full exact quantum evolution contains "one particular Alice with one particular life story" which is one of the typical errors in which people incorrectly use a classical reasoning in the quantum realm or, if I locate the blunder even more precisely, the error in which people "erase" the other terms in the wave function ("collapse" the wave function) prematurely because they just feel uncomfortable about the superpositions and want to return to the classical world (where all properties are objectively and uniquely determined) as soon as possible. But in this case (much like others), the "collapse" is premature, indeed, because the preservation of the information may depend on the interference of the many branches in which Alice fell into a black hole and the branches in which she hasn't.
(In fact, AMPS – and others – are not only making the error of assuming that Alice's being inside a black hole is a classical fact. They often want to determine Alice's position relatively to the horizon with quite some amazing precision. This contradicts the inevitably inaccuracy resulting from the random-walk character of the Hawking evaporation. The stunning similarity of this mistake to Einstein's mistake during his debates with Bohr – those about Einstein's box – suggest that physicists have learned almost nothing from these debates.)
So yes, Hsu's paper is another one that helps to settle my original suspicion that the error done by Polchinski et al. ultimately boils down to their misunderstanding of the foundations of quantum mechanics, something that they (much like others) try to "reshape" far too classically.
And that's the memo.
P.S.: I can't resist to mention that in 2005, when Stephen Hawking admitted that the information wasn't lost after all, he also offered his own proof that it doesn't. When properly reorganized, most of the histories that contribute to the final Hawking radiation are histories in which the black hole is completely avoided, he argued. This 2005 claim due to Hawking (which wasn't uncritically accepted as "yet another full proof", just to be sure) is not quite the same thing as the claim in Hsu's paper – that it's important that Alice herself may avoid the black hole interior – but it also has a similar spirit because it emphasizes the importance of histories and observers that never see a/the black hole interior. One may only derive the paradoxical "information is lost" conclusion if he assumes that the chances are 100% that the black hole is formed and an observer sees the interior but the probability that this ain't so, while it may be small, is always important to fix the qualitative conclusion and to see that the information is preserved. |
In Calculus I, Apostol appeals to the definition of an inductive set given in previous pages to prove the PMI (it's somewhat funny because the proof is just two lines).
I have been trying to do it in another way. I would like to know whether there is something wrong with it.
Theorem (I.36): Principle of Mathematical Induction: Let $S$ be a set of positive integers which has the following properties: (a)The number $1$ is in the set $S$. (b)If an integer $k$ is in $S$, then so is $k+1$. Proof:
Let suppose that there is a set $S \subset \mathbb{N}$ which contains $1$ and an integer $k > 1$. Let assume that $k+1 \notin S$ (i.e. property
(b) does not hold). Then it follows that if $k$ is in $S$, then so is $k-1 = m$, which is the same as saying that, for every integer $m \geq 1$ in $S$, there is an integer $m+1 \in S$. Hence properties (a) and (b) are fulfilled, $\implies S = \mathbb{N}$. |
What you did wrong was to start with the goal of the proof, and then follow it with implications. If you are trying to prove statement X, then the only informative logic is $\text{something} \rightarrow X$, not $X \rightarrow \text{something}$. An exception is when you try to make a statement of the form $X \rightarrow \text{false}$ in order to prove $\text{not } X$.
For example, suppose I try to prove that "y is divisible by 12", that is $12\mid y$. If something is divisible by 12, then it is also divisible by 4, so we could say $12\mid y \rightarrow 4 \mid y$,
but that doesn't help because even if we prove that $4 \mid y$ then we still don't know that $12 \mid y$. On the other hand, if we could prove $24 \mid y$, that would help because $24 \mid y \rightarrow 12 \mid y$.
I suggest approaching your problem another way, consider well known the function:
$$\text{sgn}(x) = \begin{cases} -1 & \text{ for } x < 0 \\ 0 & \text{ for } x = 0 \\ 1 & \text{ for } x > 0 \end{cases}$$
sgn just tells us if x is positive, negative, or zero.
We know a,b, and c aren't zero.
If $ab > 0$, then $\text{sgn}(a) = \text{sgn}(b)$. If $bc < 0$, then $\text{sgn}(b) = -\text{sgn}(c)$. Together we get that $\text{sgn}(a) = -\text{sgn}(c)$. That implies that $ac < 0$, which implies that $b^2 - 4ac > 0$. |
You asked for a qualitative picture, so here goes.
Consider a simplified example: the quantum harmonic oscillator.
Its ground state is given by
$$ \Psi(x) = \text{const} \cdot \exp \left( - m \omega_0 x^2 / 2 \hbar \right). $$
Now suppose that we are measuring the position of this oscillator in the ground state. We could get any real value, with probability density $|\Psi|^2$. In reality, because of the exponential decay, most of the values are distributed within the window of width
$$ \Delta x \sim \sqrt{\frac{2 \hbar}{m \omega_0}}, $$
with the mean concentrated at $x = 0$.
Because measuring an individual oscillator is a complicated process which results in it getting entangled with the measurement device, let's simplify the problem – say we have an ensemble of non-interacting oscillators all in ground states, and we measure them all independently. The distribution of values $\{x_i\}$ is expected to mostly lie within the mentioned above window, but the actual values are unknown. We usually say that those are due to
quantum fluctuations of the position operator.
The same thing happens with the quantum field, which upon inspection is nothing more than a collection of weakly interacting harmonic oscillators. If we take an ensemble of vacuum quantum field configurations (say, independent experiments at a particle accelerator), and we measure a value of the field at a point, we will see that it is not equal to zero (as it would be in the classical theory), but instead the values are distributed within an error window and are otherwise random. This are quantum fluctuations of the QFT vacuum.
These fluctuations are sometimes attributed to "virtual particles", or "virtual pairs", which are said to be "born from the vacuum". Sometimes it is also said that they can "borrow energy from vacuum for a short period of time". AFAIK these are just analogies, appealing to the consequence of Erenfest's theorem (the so-called time-energy uncertainty relation).
But the fluctuations undisputably have very real, measurable effects. Qualitatively, those effects come from a difference between the physical picture of the same thing painted by classical fields and quantum fields. You can say that quantum fields reproduce classical fields on certain scales (measured in the field value), which are much greater than the size of the error window. But once the precision with which you measure field values becomes comparable to the size of the error window, quantum effects kick in. Those who like painting intuitive pictures in their heads say that this is caused by quantum fluctuations, or virtual particles.
UPDATE
Belief that observed Casimir effect has something to do with vacuum fluctuations of the
fundamental QFT is misguided. In fact, in the calculation of the Casimir force we use an effective field theory – free electromagnetism in the 1D box, bounded by the two plates. Then we look at the effective vacuum state of this effective QFT, and we interpret the Casimir force as a consequence of the dependence of its properties on the displacement between the plates, $d$.
From the point of view of the fundamental QFT however (Standard Model, etc.)
there is no external conducting plates in the first place. If there were, it would violate Lorentz invariance. Real plates used in real experiments are made of the same matter described by the fundamental QFT, thus the state of interest is extremely complicated. What we observe as Casimir force is really just a complicated interaction of the fundamental QFT, which describes the time evolution of the complicated initial state (which describes the plates + electromagnetic field in between).
It is hopeless to try to calculate this in the fundamental QFT, just like it is hopeless to calculate the properties of the tennis ball by studying directly electromagnetic interactions holding its atoms together. Instead, we turn to the effective description, which captures all the interesting properties of our setup. In this case it is free electromagnetic effective QFT in the 1D box.
So to summarize: we are looking at the vacuum state of the effective QFT and the dependence of its properties on $d$. Alternatively, we are observing an extremely complicated fundamental system in a state which we can't hope to describe. |
Since I just finished optimizing a lot of them in a software, DifferentialEquations.jl, I decided to just lay out a comparison of the main Order 4/5 methods. The Fehlberg method was left out because it's commonly known to be less efficient than the DP5 method.
Backstories Dormand-Prince 4/5
The Dormand-Prince method was developed to be accurate as a 4/5 pair with local extrapolation usage (i.e. step with the order 5 pair. This is because it was designed to be close to optimal (i.e. minimal) principle truncation error coefficient (under the restraint of also having the minimal number of steps to achieve order 5). It has an order 4 interpolation which is free, but needs extra steps for an order 5 interpolation.
Cash-Karp 4/5
The Cash-Karp method was developed to satisfy different constraints, namely to deal with non-smooth problems better. They chose the $c_i$, the percentage of the timestep in the $i$th step (i.e. $t+c_i \Delta t$ is the time the $i$th step is calculated at) to be as uniform as possible, yet still achieve order 5. Then it also was derived to have embedded 1st, 2nd, 3rd, and 4th order methods with this uniformity of the $c_i$. They are spaced in such a manner that you can find out where a stiff part starts by which difference is large. Moreover, note that the more stiff the equation, the worse a higher order method does (because it needs bounds on higher derivatives). So they develop a strategy which uses the 5 embedded methods to "quit early": i.e. if you detect stiffness, stop at stage $i<6$ to decrease the number of function calls and save time. So in the end, this "pair" was developed with a lot of other constraints in mind, and so there's no reason to expect it would be "more accurate", at least as a 4/5 pair. If you add all of this other machinery then, on (semi-)stiff problems, it will be more accurate (but in that case you may want to use a different method like a W-Rosenbrock method). This is one reason why this pair hasn't become standard over the DP5 pair, but it still can be useful (maybe it would be good for a hybrid method which switches to a stiff solver when stiffness is encountered?).
Bogacki & Shampine 4/5
To round out the answer, let's discuss the Bogacki & Shampine pair that was mentioned in the comment. The BS5 method drops the constraint of "using the least function calls" (it uses 8 instead of 6) in order to do 2 things:
Get really low principle truncation error coefficients.
Produce an order 5 interpolation with lower error coefficients.
These coefficients are so low that for many problems with tolerances that users likely use, it measures as though its 6th order. Their paper shows that for cheap function calls, this can be more efficient than DP5 by about the same amount at DP5 was over RKF5 (the Fehlberg method).
You might put two-and-two together and see: wait a second, Shampine is the same person who developed the MATLAB ODE suite, this was after the BS5 pair paper was published, why doesn't MATLAB's
ode45 use the BS5 pair? One reason is, it was mostly done before the BS5 pair was relaesed. The other reason is because the
ode45 function was developed to minimize time. While the BS5 pair is more efficient (i.e. gets lower accuracy), the purpose of
ode45 is to have good enough error to make a good enough plot. This means that, in order to deal with the large steps, it also produces two extra interpolated solutions between every step. For the DP5 method, there is a "free" order 4 interpolation, and so this is much faster than using BS5. Since it is also "accurate enough" at moderate tolerances, this method is set as the standard because it gives a better standard user experience than BS5 when doing interactive computing (so this choice was context specific).
Tsitorous 4/5
Here's one less people know about. It's derived in this paper. It's derived using less assumptions than the DP5 method, and tries to get a pair with lower principle truncation error coefficients. In its tests, it states that it achieves this. It also has a free order 4 interpolation like the DP5 method.
Numerical tests
I wrote the numerical package DifferentialEquations.jl to be a pretty comprehensive set of solvers for Julia. Along the warpath, I implemented over 100 Runge-Kutta methods, and hand-optimized plenty. Three of the hand optimized integrators are the DP5, BS5, and Tsit5 methods (I did not do CK5 because, as noted in the backstory, it's main case is for problems which are kind of stiff. I think the better way to handle them is to use DP5/BS5 and switch to stiff solvers as necessary in a manner like LSODE, but that's a story for a different time) (one way to see they are close to optimal is that these methods are faster than the Hairer
dopri5 implementations, so they are at least decent implementations). Tests between a lot of Runge-Kutta methods on nonstiff equations can be found in the benchmarks folder. I am adding more as I go along, but you can see from the linear ODE and the Three-Body problem work precision diagrams, I measure the DP5 and Tsit5 methods to have almost identical efficiency, beating out the BS5 method in the linear ODE, while it's DP5 and BS5 that are almost identical on the Three-Body problem with Tsit5 behind. From this information, at least for now, I have settled on the DP5 method as the default, matching previous recommendations. That may change with future tests (or you could add benchmarks! Feel free to contribute, or Star the repo to give this effort more support).
Conclusion
In conclusion, the Order 5 pairs go like this:
The Dormand-Prince 4/5 pair is a good go-to pair since it's well-optimized in terms of principle truncation error coefficient and has a cheap order 4 interpolation, which makes it fast for producing decent plots.
The Cash-Karp pair has more constraints on it to better handle stiff equations. However, to get the full benefit you'll want to use the full algorithm with the 5 embedded methods.
Bogacki & Shampine Order 5 method may be the most efficient in terms of producing error per function calls (it has a double error estimator, so in harder problems it probably does better), but that allows it to take larger timesteps. However, if you're just wanting to produce a smooth plot, you then have to counter-act this method: use a lower tolerance (so it will take longer than DP5 but with less error) or use more interpolated steps. In the end, this meant that it might not be better for interactive applications, although it might be better for some scientific computing applications.
The Tsitorous 4/5. Was developed fairly recently (2011) to beat out the DP5 in a head-to-head comparison. My tests don't give me a reason to believe that it's so much better than DP5 that it should now be considered as the new standard method, but future tests may begin to side in its favor.
Edit
I did improve the Tsit5 implementation. It now does better than DP5 on most tests, both the DifferentialEquations.jl and the Hairer dopri implementations (though one might be surprised that the DifferentialEquations.jl implementations are actually faster, which of course helps the Tsit5 implementation). I now recommend it as the default order 4/5 method.
If you are interested in comparing two integrators solve
$$\frac{dq}{dt} = p \\ \frac{dp}{dt} = -q$$
with initial values $(q,p) = (1,0)$ and $dt = 0.1$ for both of them. Then plot the errors
$$dq = q - \cos(t) ; dp = p + \sin(t)$$
for a reasonable range of $t$ (0 to 100, one thousand time steps in all) and choose the integrator with the smaller error. This harmonic oscillator test will show you the phase and amplitude errors with very little effort.
P J Channell and C Scovel. Symplectic integration of Hamiltonian systems. Journal of Computational Physics 73, 468 (1973). |
As part of my fine exercise of Pioneer orbit calculations, I am using a polynomial extrapolation technique (actually, I believe it is called an acceleration technique) called Richardson extrapolation.
The basic idea is this: we compute something (for instance, a numerical integral) using an iteration step of size $h_1$. We repeat the computation several times using integration steps $h_2$, $h_3$, ..., $h_n$. We label successive results as $p_1$, $p_2$, ..., $p_n$. We now seek the polynomial $P$ of order $n-1$ such that $P(h_i)=p_i$. We then evaluate $P(0)$, which gives an estimate of the result that we would get, had we used an iteration step of 0.
In other words, we seek $a_1$, $a_2$, ..., $a_n$ such that
\begin{equation}
\begin{pmatrix} 1&h_1&h_1^2&...&h_1^{n-1}\\ 1&h_2&h_2^2&...&h_2^{n-1}\\ .&.&.&...&.\\ 1&h_n&h_n^2&...&h_n^{n-1} \end{pmatrix} \begin{pmatrix} a_1\\ a_2\\~\\ \cdot\cdot\cdot\\ a_n \end{pmatrix} = \begin{pmatrix} p_1\\ p_2\\~\\ \cdot\cdot\cdot\\ p_n \end{pmatrix}. \end{equation}
Naturally, the solution of this equation can be obtained by straight matrix inversion. However, matrix inversion is computationally costly. Under some circumstances, it may be possible to avoid matrix inversion altogether. One such case is when successive intervals are halved (or multiplied by any other factor $t$), such that $h_1=th_0$, $h_2=th_1$, etc. In this case, we have (writing $h=h_0$):
\begin{equation}
\begin{pmatrix} 1&th&(th)^2&...&(th)^{n-1}\\ 1&t^2h&(t^2h)^2&...&(t^2h)^{n-1}\\ .&.&.&...&.\\ 1&t^nh&(t^nh)^2&...&(t^nh)^{n-1} \end{pmatrix} \begin{pmatrix} a_1\\ a_2\\~\\ \cdot\cdot\cdot\\ a_n \end{pmatrix} = \begin{pmatrix} p_1\\ p_2\\~\\ \cdot\cdot\cdot\\ p_n \end{pmatrix}. \end{equation}
The matrix $M$ on the left hand side of this equation can be triangularized. The $i$th row, $j$th column of this matrix is
\begin{equation}
M_{ij}=(t^ih)^{j-1}. \end{equation}
We can now form the rows of a triangularized matrix $M^*$ as follows:
\begin{align}
M'_{kj}&=M_{kj}-M_{1j}~~~&(k=2,...,n),\\ M''_{kj}&=M'_{kj}-\frac{M'_{k2}}{M'_{22}}M'_{2j}~~~&(k=3,...,n),\\ M'''_{kj}&=M''_{kj}-\frac{M''_{k3}}{M''_{33}}M''_{3j}~~~&(k=4,...,n),\\ &...,\\ M^{(l)}_{kj}&=M^{(l-1)}_{kj}-\frac{M^{(l-1)}_{kl}}{M^{(l-1)}_{ll}}M^{(l-1)}_{lj}~~~&(k=l+1,...,n). \end{align}
where we used the notation $M^{(0)}=M$, $M^{(1)}=M'$, $M^{(2)}=M''$, etc. Then:
\begin{equation}
M^*_{kj}=M^{(k-1)}_{kj}. \end{equation}
Correspondingly, we must also adjust the right-hand side. using similar notation for $p^{(0)}=p$, $p^{(1)}=p'$, etc.:
\begin{align}
p^{(l)}_k&=p^{(l-1)}_k-\frac{M^{(l-1)}_{kl}}{M^{(l-1)}_{ll}}p^{(l-1)}_l,\\ p^*_k&=p^{(k-1)}_k. \end{align}
The solution for $a_i$ can then be written in the form:
\begin{align}
a_n&=\frac{p^*_n}{M^*_{nn}},\\ a_{n-1}&=\frac{p^*_{n-1}-M^*_{(n-1)n}a_n}{M^*_{(n-1)(n-1)}},\\ a_k&=\frac{p^*_k-\sum\limits_{l=k+1}^nM^*_{kl}a_l}{M^*_{kk}}. \end{align}
Finally, we have
\begin{equation}
P(0)=a_1, \end{equation}
which is the result we sought. Simultaneously, we could also evaluate $P(h)$, which could then be compared against the next numerical iteration to see how well our polynomial approximation predicts its value.
These calculations can be trivially implemented using computer algebra code, such as the following lines in Maxima:
N:5; t:2; M:zeromatrix(N,N); for i thru N do for j thru N do M[i,j]:(t^i*h)^(j-1); P:zeromatrix(N,1); for i thru N do P[i,1]:concat(p,i); for k from 2 thru N do for i from k thru N do (P[i]:P[i]-P[k-1]*M[i,k-1]/M[k-1,k-1],M[i]:M[i]-M[k-1]*M[i,k-1]/M[k-1,k-1]); M; factor(P); factor((M^^-1).P); |
I am a newbie in finite difference methods, so I apologize in advance if the question is trivial.
I am trying to solve the advenction equation, i.e. $\frac{\partial \phi(x,t)}{\partial t} + v \frac{\partial \phi(x,t)}{\partial x}=0$, where $v$ is a velocity, with the finite difference method. The system is of course also solvable analytically, and it turns out that every function of the kind $f(x,t) = f(x-vt)$ is a solution. Therefore, given an arbitrary initial condition $\phi(x,0)$, the "shape" of $\phi(x,0)$ will move towards positive x (for $v>0$) or negative x (for $v<0$).
I first formulated the problem with the backward formula for the spatial derivative,
$\phi(x_i,t_{j+1}) = \phi(x_i,t_j) - v\cdot\Delta T\cdot(\phi(x_i,t_j)-\phi(x_{i-1},t_j))/\Delta x$.
The initial condition is given by a Gaussian packet centered at the origin,
$\phi(x_i,t_1) = exp(-(x_i/2\sigma)^2)$.
With $v>0$ the gaussian packet does indeed move towards right with increasing times, and maintaining always the same shape. However, if I set $v<0$, I obtain large instabilities, with an exponentially increasing amplitude. If I reformulate the problem in terms of the forward formula for the spatial derivative
$\phi(x_i,t_{j+1}) = \phi(x_i,t_j) - v\cdot\Delta T\cdot(\phi(x_{i+1},t_j)-\phi(x_{i},t_j))/\Delta x$
I obtain the opposite behavior: with a negative $v$, the packet propagates nicely towards left, but a positive $v$ leads to a large divergence of the system.
Finally, I tried to use the centered formula for the spatial derivative, $\phi(x_i,t_{j+1}) = \phi(x_i,t_j) - v\cdot\Delta T\cdot(\phi(x_{i+1},t_j)-\phi(x_{i-1},t_j))/(2\Delta x)$
and in this case I obtain divergences for both signs of $v$.
Is there a particular reason for this kind of behaviour? What should be the "most correct" way to implement this problem?
I attach here a short Matlab script that I made, and that reproduces what I described.
L = 1; %Length v = 1; %Speed i=sqrt(-1); DeltaT = ((L/abs(v))/1000) DeltaX = 1*abs(v)*DeltaT x=-L:DeltaX:L; t = 0:DeltaT:(L/abs(v)); sigma = L/50; Tbreak = 1e-5; %Parameter for animation %% v = 1, backward formula v=1; phi = zeros(length(x),length(t));phi(:,1) = exp(-((x)./(2*sigma)).^2); for n=1:(length(t)-1) for k=2:(length(x)-1) phi(k,n+1)=phi(k,n) - v*DeltaT*(phi(k,n)-phi(k-1,n))/(DeltaX); end end figure(1); for n=1:(length(t)) plot(x,phi(:,n)) pause(Tbreak) end %% v = -1, backward formula v=-1; phi = zeros(length(x),length(t));phi(:,1) = exp(-((x)./(2*sigma)).^2); for n=1:(length(t)-1) for k=2:(length(x)-1) phi(k,n+1)=phi(k,n) - v*DeltaT*(phi(k,n)-phi(k-1,n))/(DeltaX); end end figure(1); for n=1:(length(t)) plot(x,phi(:,n)) pause(Tbreak) end %% v = 1, forward formula v=1; phi = zeros(length(x),length(t));phi(:,1) = exp(-((x)./(2*sigma)).^2); for n=1:(length(t)-1) for k=2:(length(x)-1) phi(k,n+1)=phi(k,n) - v*DeltaT*(phi(k+1,n)-phi(k,n))/(DeltaX); end end figure(1); for n=1:(length(t)) plot(x,phi(:,n)) pause(Tbreak) end %% v = -1, forward formula v=-1' phi = zeros(length(x),length(t));phi(:,1) = exp(-((x)./(2*sigma)).^2); for n=1:(length(t)-1) for k=2:(length(x)-1) phi(k,n+1)=phi(k,n) - v*DeltaT*(phi(k+1,n)-phi(k,n))/(DeltaX); end end figure(1); for n=1:(length(t)) plot(x,phi(:,n)) pause(Tbreak) end |
Generalized linear models for population dynamics in two juxtaposed habitats
1.
Normandie Univ, UNIHAVRE, LMAH, FR-CNRS-3335, 76600 Le Havre, France
2.
USTHB, LAMNEDP, Faculté de Mathématiques, BP.32, El Alia, Bab Ezzouar, 16111 Alger, Algérie
3.
Normandie Univ, UNIHAVRE, LMAH, FR-CNRS-3335, 76600 Le Havre, France
In this work we introduce a generalized linear model regulating the spread of population displayed in a $ d $-dimensional spatial region $ \Omega $ of $ \mathbb{R}^{d} $ constituted by two juxtaposed habitats having a common interface $ \Gamma $. This model is described by an operator $ \mathcal{L} $ of fourth order combining the Laplace and Biharmonic operators under some natural boundary and transmission conditions. We then invert explicitly this operator in $ L^{p} $-spaces using the $ H^{\infty } $-calculus and the Dore-Venni sums theory. This main result will lead us in a later work to study the nature of the semigroup generated by $ \mathcal{L} $ which is important for the study of the complete nonlinear generalized diffusion equation associated to it.
Keywords:Population dynamics, diffusion equation, semigroups, Landau-Ginzburg free energy functional. Mathematics Subject Classification:35B65, 35C15, 35J40, 35R20, 47A60, 47D06, 92D25. Citation:Rabah Labbas, Keddour Lemrabet, Stéphane Maingot, Alexandre Thorel. Generalized linear models for population dynamics in two juxtaposed habitats. Discrete & Continuous Dynamical Systems - A, 2019, 39 (5) : 2933-2960. doi: 10.3934/dcds.2019122
References:
[1]
W. Arendt, C. J. K. Batty, M. Hieber and F. Neubrander,
[2] [3]
D. L. Burkholder,
A geometrical characterisation of Banach spaces in which martingale dif-ference sequences are unconditional,
[4] [5] [6] [7]
A. Favini, R. Labbas, S. Maingot, K. Lemrabet and H. Sidibé,
Resolution and Optimal Re-gularity for a Biharmonic Equation with Impedance Boundary Conditions and Some Gene-ralizations,
[8]
D. Gilbarg and N. Trudinger,
[9] [10] [11] [12] [13]
R. Labbas, S. Maingot, D. Manceau and A. Thorel,
On the regularity of a generalized diffusion problem arising in population dynamics set in a cylindrical domain,
[14]
K. Limam, R. Labbas, K. Lemrabet, A. Medeghri and M. Meisner,
On Some Transmission Problems Set in a Biological Cell, Analysis and Resolution,
[15] [16]
A. Lunardi,
[17] [18] [19] [20] [21]
J. L. Rubio de Francia, Martingale and integral transforms of Banach space valued functions,
[22]
H. Triebel,
show all references
References:
[1]
W. Arendt, C. J. K. Batty, M. Hieber and F. Neubrander,
[2] [3]
D. L. Burkholder,
A geometrical characterisation of Banach spaces in which martingale dif-ference sequences are unconditional,
[4] [5] [6] [7]
A. Favini, R. Labbas, S. Maingot, K. Lemrabet and H. Sidibé,
Resolution and Optimal Re-gularity for a Biharmonic Equation with Impedance Boundary Conditions and Some Gene-ralizations,
[8]
D. Gilbarg and N. Trudinger,
[9] [10] [11] [12] [13]
R. Labbas, S. Maingot, D. Manceau and A. Thorel,
On the regularity of a generalized diffusion problem arising in population dynamics set in a cylindrical domain,
[14]
K. Limam, R. Labbas, K. Lemrabet, A. Medeghri and M. Meisner,
On Some Transmission Problems Set in a Biological Cell, Analysis and Resolution,
[15] [16]
A. Lunardi,
[17] [18] [19] [20] [21]
J. L. Rubio de Francia, Martingale and integral transforms of Banach space valued functions,
[22]
H. Triebel,
[1]
Shujuan Lü, Hong Lu, Zhaosheng Feng.
Stochastic dynamics of 2D fractional Ginzburg-Landau equation
with multiplicative noise.
[2]
Hong Lu, Shujuan Lü, Mingji Zhang.
Fourier spectral approximations to the dynamics of 3D fractional complex Ginzburg-Landau equation.
[3]
Feng Zhou, Chunyou Sun.
Dynamics for the complex Ginzburg-Landau equation on non-cylindrical domains I: The diffeomorphism case.
[4]
Leonid Berlyand, Volodymyr Rybalko, Nung Kwan Yip.
Renormalized Ginzburg-Landau energy and location of near boundary vortices.
[5]
Ko-Shin Chen, Peter Sternberg.
Dynamics of Ginzburg-Landau and Gross-Pitaevskii vortices on manifolds.
[6] [7] [8] [9]
Satoshi Kosugi, Yoshihisa Morita, Shoji Yotsutani.
A complete bifurcation diagram of the Ginzburg-Landau equation with periodic boundary conditions.
[10] [11]
Noboru Okazawa, Tomomi Yokota.
Subdifferential operator approach to strong wellposedness of the
complex Ginzburg-Landau equation.
[12] [13]
Sen-Zhong Huang, Peter Takáč.
Global smooth solutions of the complex Ginzburg-Landau equation and their dynamical properties.
[14]
Hassen Aydi, Ayman Kachmar.
Magnetic vortices for a Ginzburg-Landau type energy with discontinuous constraint. II.
[15]
Robert Denk, Yoshihiro Shibata.
Generation of semigroups for the thermoelastic plate equation with free boundary conditions.
[16]
Yun Lan, Ji Shu.
Dynamics of non-autonomous fractional stochastic Ginzburg-Landau equations with multiplicative noise.
[17]
Henri Berestycki, Jean-Michel Roquejoffre, Luca Rossi.
The periodic patch model for population dynamics with fractional diffusion.
[18]
Shujuan Lü, Chunbiao Gan, Baohua Wang, Linning Qian, Meisheng Li.
Traveling wave solutions and its stability for
3D Ginzburg-Landau type equation.
[19]
Hongzi Cong, Jianjun Liu, Xiaoping Yuan.
Quasi-periodic solutions for complex
Ginzburg-Landau equation of nonlinearity $|u|^{2p}u$.
[20]
2018 Impact Factor: 1.143
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Is there any formal definition about the average height of a binary tree?
I have a tutorial question about finding the average height of a binary tree using the following two methods:
The natural solution might be to take the average length of all possible paths from the root to a leaf, that is
$\qquad \displaystyle \operatorname{avh}_1(T) = \frac{1}{\text{# leaves in } T} \cdot \sum_{v \text{ leaf of } T} \operatorname{depth}(v)$.
Another option is to define it recursively, that is the average height for a node is the average over the average heights of the subtrees plus one, that is
$\qquad \displaystyle \operatorname{avh}_2(N(l,r)) = \frac{\operatorname{avh}_2(l) + \operatorname{avh}_2(r)}{2} + 1$
with $\operatorname{avh}_2(l) = 1$ for leafs $l$ and $\operatorname{avh}_2(\_) = 0$ for empty slots.
Based on my current understanding, for example the average height of the tree $T$
1 / \ 2 3 /4
is $\operatorname{avh}_2(T) = 1.25$ by the second method, that is using recursion.
However, I still don't quite understand how to do the first one. $\operatorname{avh}_1(T) = (1+2)/2=1.5$ is not correct. |
$\phi(n)$ being Euler's totient function.
Regarding effort put into the problem:
In the case that $n$ is a prime $p$, then it is given that $\phi(p) = p-1$. It is also given that $n\ne 2$, so the fact that $p-1 \ge \sqrt p$, or $(p-1)^2 \ge p$ is easy enough to prove in it of itself.
Considering the case were $n$ is some non-prime number, then $n$ may be written as its power prime factorization $n=p^{a_1}_1p^{a_2}_2 \ldots p^{a_k}_k$. It is given that $\phi$ is multiplicative, so applying this to the original problem we have $\phi(n)=\phi(p^{a_1}_1) \phi(p^{a_2}_2) \ldots \phi(p^{a_k}_k)$.
It is given that $\phi(p^{a_j}_j)=p^j_j-p^{a_j-1}_j=p^{a_j}_j(1-\frac{1}{p_j})$, so the above can be rewritten as
$\phi(n)=p^{a_1}_1p^{a_2}_2 \ldots p^{a_k}_k(1-\frac{1}{p_1})(1-\frac{1}{p_2}) \ldots (1-\frac{1}{p_k})=n(1-\frac{1}{p_1})(1-\frac{1}{p_2}) \ldots (1-\frac{1}{p_k})$.
Which is where I am stuck, as I don't quite see how to prove that the above product (or product squared) is greater than $\sqrt n$ (or $n$ in the product squared case). Is this the wrong angle to be approaching the problem from, or are one of my inferences altogether mistaken? |
In the last section of chapter 3 (page 57) in
Probability and Computing: Randomized Algorithms and Probabilistic Analysis by Mitzenmacher and Upfal, a randomized algorithm is discussed for finding the median of a set $S$ of distinct elements in $O(n)$ time. The algorithm discussed is a Monte Carlo algorithm. As such if we would like to run this variation of algorithm till we find a solution, how do we show that it still has a linear running time?
The Algorithm (source, starts on slide 66, page 53):
Input: A set $S$ of $n$ elements over a
totally ordered universe$^{***}$.
Output: The median of set $S$, denoted by $m$.
Pick a (multi-)set $R$ of $\lceil n^\frac{3}{4} \rceil$ elements in $S$ chosen independently and uniformly at random
withreplacement.
Sort the set $R$.
Let $d$ be the $(\lfloor \frac{1}{2} n^\frac{3}{4} - \sqrt{n} \rfloor)$-th smallest element in sorted set $R$.
Let $d$ be the $(\lceil \frac{1}{2} n^\frac{3}{4} + \sqrt{n} \rceil)$-th smallest element in sorted set $R$.
By comparing every element in $S$ to $d$ and $u$, compute the set: $$C = \left\{ x \in S : d \leq x \leq u \right\}$$ and the numbers $$\ell_d = |\{x \in S : x < d\}|$$ and the numbers $$\ell_u = |\{x \in S : x > u\}|$$
If $\ell_d > \frac{n}{2}$ or $\ell_u > \frac{n}{2}$ then $FAIL$.
If $|C| \leq 4n^\frac{3}{4}$ then sort the set $C$, otherwise $FAIL$.
Output the $(\lfloor \frac{n}{2} \rfloor - \ell_d + 1)$th element in the sorted order of $C$. |
The key step in Kontsevich's proof of deformation quantization of Poisson manifolds is the so-called formality theorem where 'a formal complex' means that it admits a certain condition. I wonder why it is called 'formal'. I only found the definition of Sullivan in Wikipedia: 'formal manifold is one whose real homotopy type is a formal consequence of its real cohomology ring'. But still I am confused because most of articles I found contain the same sentence only and I cannot understand the meaning of 'formal consequence'. Does anyone know the history of this concept?
I would guess that the terminology goes back to the work of Sullivan and Quillen on rational homotopy theory. You should probably also look at the paper of Deligne-Griffiths-Morgan-Sullivan on the real homotopy theory of Kähler manifolds. Actually, I think that at least some familiarity with the DGMS paper is an important prerequisite for understanding many of Kontsevich's papers.
I am not totally sure, but I believe that the definitions are as follows:
A differential graded algebra $(A,d)$ is called formal if it is quasi-isomorphic (in general, if we work in the category of dg algebras and not, say, the category of A-infinity algebras, we need a "zig-zag" of quasi-isomorphisms) to $H^\ast(A,d)$ considered as a dg algebra with zero differential.
A space X is called formal (over the rationals resp. the reals) if its cochain dg algebra $C^\ast(X)$ (with rational resp. real coefficients) with the standard differential is a formal dg algebra.
One of the things I'm not sure about is whether in the definition we should require $H^\ast(A,d)$ to be commutative; but for spaces this is not an issue since $H^\ast(X)$ is always (graded-)commutative.
The DGMS paper proves that if X is a compact Kähler manifold, then the de Rham dg algebra consisting of (real, $C^\infty$) differential forms on X with the standard de Rham differential is a formal dg algebra.
The phrase "the real (resp. rational) homotopy type of X is a formal consequence of the real (resp. rational) cohomology ring of X", which appears in e.g. the DGMS paper, simply means that the real (resp. rational) homotopy theory of X is determined by (and is probably explicitly and algorithmically computable from?) the cohomology ring of X. In other words, if X and Y are formal (over the rationals resp. the reals) and have isomorphic (rational resp. real) cohomology rings, then their respective (rational resp. real) homotopy theories are the same (and are explicitly computable, if we know the cohomology ring(s)?). For example, the ranks of their homotopy groups will be equal.
Actually I am not totally sure whether what I said in the last paragraph is true. I think it's true when X and Y are simply connected. I'm not sure about what happens more generally.
In the context of rational homotopy theory, I think the term "formal" is fine, for the reasons I've explained above. Perhaps in the more general context of dg algebras, the use of the term "formal" makes less sense. However, I think that it is still reasonable, for the following reasons. Let me use the more "modern" language of A-infinity algebras. In general, it is not true that a dg algebra $(A,d)$ is quasi-isomorphic to $H^\ast(A,d)$ considered as a dg algebra with zero differential. However, it is a "standard" fact (Kontsevich-Soibelman call this the "homological perturbation lemma" (for example, it's buried somewhere in this paper), and you can find it in the operads literature as the "transfer theorem") that you can put an A-infinity structure on $H^\ast(A,d)$ which makes $A$ and $H^\ast(A,d)$ quasi-isomorphic as A-infinity algebras. The A-infinity structure manifests itself as a series of $n$-ary products satisfying various compatibilities. Intuitively at least, these $n$-ary products should be thought of as being analogous to Massey products in topology. So $H^\ast(A,d)$ with this A-infinity structure does carry some "homotopy theoretic" information. In this language then, a dg algebra $(A,d)$ is formal if it is quasi-isomorphic, as an A-infinity algebra, to $H^\ast(A,d)$ with all higher products zero. In other words, all of the "Massey products" vanish*, and thus the only remaining "homotopy theoretic" information is that coming from the ordinary ring structure on $H^\ast(A,d)$.
*Don Stanley notes correctly that vanishing of Massey products is weaker than formality. However, I believe that triviality of the A-infinity structure is equivalent to formality. In the language of the DGMS paper, which does not use the A-infinity language, they say that formality is equivalent to the vanishing of Massey products "in a uniform way". I believe this uniform vanishing is the same as triviality of A-infinity structure. From the paper:
... a minimal model is a formal consequence of its cohomology ring if, and only if, all the higher order products vanish in a uniform way.
and also
[Choosing a quasi-isomorphism from a minimal dg algebra to its cohomology] is a way of saying that
one may make uniform choices so that the forms representing all Massey products and higher order Massey products are exact. This is stronger than requiring each individual Massey product or higher order Massey product to vanish. The latter means that, given one such product, choices may be made to make the form representing it exact, and there may be no way to do this uniformly.
(Sorry for the proliferation of parentheses, and sorry for my lack of certainty on all of this, I have not thought about this in a while. People should definitely correct me if I'm wrong on any of this.)
Paraphrasing Groucho Marx: if you don't like my first answer..., well I have another one. :-)
Here it is: let $X$ be a simply connected differentiable manifold.
Rational homotopy theory tells us that the
rational homotopy type of $X$ (that is, its homotopy type modulo torsion) is contained in its minimal model, $M_X$, which is a commutative differential graded (cdg) algebra.
By definition, this means that you have a quasi-isomorphism (
quis, a morphism of cdg algebras inducing an isomorphism in cohomology)
$$ M_X \longrightarrow \Omega^*(X) \ . $$
Here, $\Omega^* (X)$ is the algebra of differential forms of $X$ and the
minimality of $M_X$ means that, in a certain, but precise, sense, it is the smallest cdg algebra for which such a quis exists.
The fact that $M_X$
contains the rational homotopy type of $X$ implies, for instance, that you can obtain the ranks of the homotopy groups of $X$ from it:
rank $\pi_n(X) =$ number of degree n generators (as an algebra) of $M_X$, for $n \geq 2$.
Nice, isn't it? :-)
The problem is that the algebra $\Omega^*(X)$ is, in general, not computable, so you can not obtain from it the minimal model $M_X$. And here is where formality comes to help you.
Almost by definition, $X$ is a
formal space if there exists two quis
$$ \Omega^*(X) \longleftarrow M_X \longrightarrow H^*(X;\mathbb{Q}) $$
Hence, if $X$ is formal you can compute its minimal model $M_X$, and hence its rational homotopy type, directly from the cohomology algebra $H^*(X; \mathbb{Q})$, which is nicer (smaller, more computable) than $\Omega^*(X)$.
And the final point is that there are plenty of examples of spaces which are known to be formal.
(Final remark: Actually, you'd have to put $A_{PL}^*(X;\mathbb{Q})$ instead of $\Omega^*(X)$ to work over the rationals, but this you can find it explained in the references we have provided for you.)
Formal can mean slightly different things in different contexts.
A commutative differential graded algebra (CDGA) is formal if it is quasi-isomorphic to it's homology. This is stronger than having all the higher Massey products equal to 0 (I think there are such examples in the Halperin-Stasheff paper).
To a space you can associate a CDGA (via Sullivan's $A_{pl}$ functor) which is basically the deRham complex when the space is a manifold. In nice cases this functor induces an equivalence from the rational homotopy category to the homotopy category of CDGA. Quasi-isormorphic CDGA correspond to (rationally) homotopy equivalent spaces. You can also tensor with the reals to get real CDGA.
If A is a CDGA which is quasi-isomorphic to $A_{pl}(X)$ for a space $X$ then A is often called a model of X. A space is formal if $A_{pl}$ of it is formal. So a formal space is modeled by its cohomology. In that sense its rational homotopy type is a formal consequence of its cohomology.
I think you have to be slightly careful with using $C^*$. This functor lands in differential graded algebra which are not commutative, so possibly the notion of formality could be different. In particular if you consider two CDGA there may be more strings of quasi-isomorphisms between them as DGAs then as CDGAs. I believe it is unknown if two CDGA that are quasi-isomorphic as DGA have to be quasi-isomorphic as CDGA.
Maybe you could take a look at
Y. Félix, J. Oprea, D. Tanré; Algebraic models in Geometry, Oxford Graduate Text in Math. 17 (2008)
where they talk about formality in the context of rational homotopy theory, RHT, (for instance, in sections 2.7 and 3.1.4). Also the more classical, but excellent little book
D. Lehmann; Théorie homotopique des formes différentielles, Astérisque 45
is worth reading (section V.9).
As for formality in the context of operads, allow me a little self-promotion :-) :
F. Guillén, V. Navarro, P. Pascual, Agustí Roig, Moduli spaces and formal operads; Duke Math. J. 129, 2 (2005).
In this work, we translate some classical results concerning formality in RHT to chain operads. For instance, the Deligne-Griffiths-Morgan-Sullivan theorem about formality of Kähler manifolds, formality's independence of the ground field... And extend them also to modular operads. |
Godbole, Rohini M and Pancheri, G (2001)
$\sigma^{tot}_{ee \gamma \gamma}$ at $e^+e^-$ colliders. [Preprint]
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Abstract
In this talk I briefly summarize different models for $\sigma^{tot}_{2 \gamma}$ ($e^+e^- \to \gamma \gamma \to$ hadrons) and contrast model predictions with the data. I will then discuss the capability of the future $e^+e^-$ and $\gamma \gamma$ colliders to distinguish between various models and end with an outlook for future work.
Item Type: Preprint Department/Centre: Division of Physical & Mathematical Sciences > Centre for Theoretical Studies (Ceased to exist at the end of 2003) Depositing User: Ramnishath A Date Deposited: 31 Jul 2004 Last Modified: 19 Sep 2010 04:13 URI: http://eprints.iisc.ac.in/id/eprint/812 Actions (login required)
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Is there a "simple" mathematical proof that is fully understandable by a 1st year university student that impressed you because it is beautiful?
closed as primarily opinion-based by Daniel W. Farlow, Najib Idrissi, user91500, LutzL, Jonas Meyer Apr 7 '15 at 3:40
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.
Here's a cute and lovely theorem.
There exist two irrational numbers $x,y$ such that $x^y$ is rational.
Proof. If $x=y=\sqrt2$ is an example, then we are done; otherwise $\sqrt2^{\sqrt2}$ is irrational, in which case taking $x=\sqrt2^{\sqrt2}$ and $y=\sqrt2$ gives us: $$\left(\sqrt2^{\sqrt2}\right)^{\sqrt2}=\sqrt2^{\sqrt2\sqrt2}=\sqrt2^2=2.\qquad\square$$
(Nowadays, using the Gelfond–Schneider theorem we know that $\sqrt2^{\sqrt2}$ is irrational, and in fact transcendental. But the above proof, of course, doesn't care for that.)
How about the proof that
$$1^3+2^3+\cdots+n^3=\left(1+2+\cdots+n\right)^2$$
I remember being impressed by this identity and the proof can be given in a picture:
Edit: Substituted $\frac{n(n+1)}{2}=1+2+\cdots+n$ in response to comments.
Cantor's Diagonalization Argument, proof that there are infinite sets that can't be put one to one with the set of natural numbers, is frequently cited as a beautifully simple but powerful proof. Essentially, with a list of infinite sequences, a sequence formed from taking the diagonal numbers will not be in the list.
I would personally argue that the proof that $\sqrt 2$ is irrational is simple enough for a university student (probably simple enough for a high school student) and very pretty in its use of proof by contradiction!
Prove that if $n$ and $m$ can each be written as a sum of two perfect squares, so can their product $nm$.
Proof: Let $n = a^2+b^2$ and $m=c^2+d^2$ ($a, b, c, d \in\mathbb Z$). Then, there exists some $x,y\in\mathbb Z$ such that
$$x+iy = (a+ib)(c+id)$$
Taking the magnitudes of both sides are squaring gives
$$x^2+y^2 = (a^2+b^2)(c^2+d^2) = nm$$
I would go for the proof by contradiction of an infinite number of primes, which is fairly simple:
Assume that there is a finite number of primes. Let $G$ be the set of allprimes $P_1,P_2,...,P_n$. Compute $K = P_1 \times P_2 \times ... \times P_n + 1$. If $K$ is prime, then it is obviously notin $G$. Otherwise, noneof its prime factors are in $G$. Conclusion: $G$ is notthe set of allprimes.
I think I learned that both in high-school and at 1st year, so it might be a little too simple...
By the concavity of the $\sin$ function on the interval $\left[0,\frac{\pi}2\right]$ we deduce these inequalities: $$\frac{2}\pi x\le \sin x\le x,\quad \forall x\in\left[0,\frac\pi2\right].$$
The first player in Hex has a winning strategy.
There are no draws in hex, so one player must have a winning strategy. If player two has a winning strategy, player one can steal that strategy by placing the first stone in the center (additional pieces on the board never hurt your position) then using player two's strategy.
You cannot have two dice (with numbers $1$ to $6$) biased so that when you throw both, the sum is uniformly distributed in $\{2,3,\dots,12\}$.
For easier notation, we use the equivalent formulation "You cannot have two dice (with numbers $0$ to $5$) biased such that when you throw both, the sum is uniformly distributed in $\{0,1,\dots,10\}$."
Proof:Assume that such dice exist. Let $p_i$ be the probability that the first die gives an $i$ and $q_i$ be the probability that the second die gives an $i$. Let $p(x)=\sum_{i=0}^5 p_i x^i$ and $q(x)=\sum_{i=0}^5 q_i x^i$.
Let $r(x)=p(x)q(x) = \sum_{i=0}^{10} r_i x^i$. We find that $r_i = \sum_{j+k=i}p_jq_k$. But hey, this is also the probability that the sum of the two dice is $i$. Therefore, $$ r(x)=\frac{1}{11}(1+x+\dots+x^{10}). $$ Now $r(1)=1\neq0$, and for $x\neq1$, $$ r(x)=\frac{(x^{11}-1)}{11(x-1)}, $$ which clearly is nonzero when $x\neq 1$. Therefore $r$ does not have any real zeros.
But because $p$ and $q$ are $5$th degree polynomials, they must have zeros. Therefore, $r(x)=p(x)q(x)$ has a zero. A contradiction.
Given a square consisting of $2n \times 2n$ tiles, it is possible to cover this square with pieces that each cover $2$ adjacent tiles (like domino bricks). Now imagine, you remove two tiles, from two opposite corners of the original square. Prove that is is now no longer possible to cover the remaining area with domino bricks.
Proof:
Imagine that the square is a checkerboard. Each domino brick will cover two tiles of different colors. When you remove tiles from two opposite corners, you will remove two tiles with the
samecolor. Thus, it can no longer be possible to cover the remaining area.
(Well, it may be
too "simple." But you did not state that it had to be a university student of mathematics. This one might even work for liberal arts majors...)
One little-known gem at the intersection of geometry and number theory is Aubry's reflective generation of primitive Pythagorean triples, i.e. coprime naturals $\,(x,y,z)\,$with $\,x^2 + y^2 = z^2.\,$ Dividing by $\,z^2$ yields $\,(x/z)^2+(y/z)^2 = 1,\,$ so each triple corresponds to a rational point $(x/z,\,y/z)$ on the unit circle. Aubry showed that we can generate all such triples by a very simple geometrical process. Start with the trivial point $(0,-1)$. Draw a line to the point $\,P = (1,1).\,$ It intersects the circle in the
rational point $\,A = (4/5,3/5)\,$ yielding the triple $\,(3,4,5).\,$ Next reflect the point $\,A\,$ into the other quadrants by taking all possible signs of each component, i.e. $\,(\pm4/5,\pm3/5),\,$ yielding the inscribed rectangle below. As before, the line through $\,A_B = (-4/5,-3/5)\,$ and $P$ intersects the circle in $\,B = (12/13, 5/13),\,$ yielding the triple $\,(12,5,13).\,$ Similarly the points $\,A_C,\, A_D\,$ yield the triples $\,(20,21,29)\,$ and $\,(8,15,17),\,$ We can iterate this process with the new points $\,B,C,D\,$ doing the same we did for $\,A,\,$ obtaining further triples. Iterating this process generates the primitive triples as a ternary tree
$\qquad\qquad$
Descent in the tree is given by the formula
$$\begin{eqnarray} (x,y,z)\,\mapsto &&(x,y,z)-2(x\!+\!y\!-\!z)\,(1,1,1)\\ = &&(-x-2y+2z,\,-2x-y+2z,\,-2x-2y+3z)\end{eqnarray}$$
e.g. $\ (12,5,13)\mapsto (12,5,13)-8(1,1,1) = (-3,4,5),\ $ yielding $\,(4/5,3/5)\,$ when reflected into the first quadrant.
Ascent in the tree by inverting this map, combined with trivial sign-changing reflections:
$\quad\quad (-3,+4,5) \mapsto (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$
$\quad\quad (-3,-4,5) \mapsto (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$
$\quad\quad (+3,-4,5) \mapsto (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$
See my MathOverflow post for further discussion, including generalizations and references.
I like the proof that there are infinitely many Pythagorean triples.
Theorem:There are infinitely many integers $ x, y, z$ such that $$ x^2+y^2=z^2 $$ Proof:$$ (2ab)^2 + ( a^2-b^2)^2= ( a^2+b^2)^2 $$
One cannot cover a disk of diameter 100 with 99 strips of length 100 and width 1.
Proof: project the disk and the strips on a semi-sphere on top of the disk. The projection of each strip would have area at most 1/100th of the area of the semi-sphere.
If you have any set of 51 integers between $1$ and $100$, the set must contain some pair of integers where one number in the pair is a multiple of the other.
Proof: Suppose you have a set of $51$ integers between $1$ and $100$. If an integer is between $1$ and $100$, its largest odd divisor is one of the odd numbers between $1$ and $99$. There are only $50$ odd numbers between $1$ and $99$, so your $51$ integers can’t all have different largest odd divisors — there are only $50$ possibilities. So two of your integers (possibly more) have the same largest odd divisor. Call that odd number $d$. You can factor those two integers into prime factors, and each will factor as (some $2$’s)$\cdot d$. This is because if $d$ is the largest divisor of a number, the rest of its factorization can’t include any more odd numbers. Of your two numbers with largest odd factor $d$, the one with more $2$’s in its factorization is a multiple of the other one. (In fact, the multiple is a power of $2$.)
In general, let $S$ be the set of integers from $1$ up to some even number $2n$. If a subset of $S$ contains more than half the elements in $S$, the set must contain a pair of numbers where one is a multiple of the other. The proof is the same, but it’s easier to follow if you see it for a specific $n$ first.
The proof that an isosceles triangle ABC (with AC and AB having equal length) has equal angles ABC and BCA is quite nice:
Triangles ABC and ACB are (mirrored) congruent (since AB = AC, BC = CB, and CA = BA), so the corresponding angles ABC and (mirrored) ACB are equal.
This congruency argument is nicer than that of cutting the triangle up into two right-angled triangles.
Parity of sine and cosine functions using Euler's forumla:
$e^{-i\theta} = cos\ (-\theta) + i\ sin\ (-\theta)$
$e^{-i\theta} = \frac 1 {e^{i\theta}} = \frac 1 {cos\ \theta \ + \ i\ sin\ \theta} = \frac {cos\ \theta\ -\ i\ sin\ \theta} {cos^2\ \theta\ +\ sin^2\ \theta} = cos\ \theta\ -\ i\ sin\ \theta$
$cos\ (-\theta) +\ i\ sin\ (-\theta) = cos\ \theta\ +i\ (-sin\ \theta)$
Thus
$cos\ (-\theta) = cos\ \theta$
$sin\ (-\theta) = -\ sin\ \theta$
$\blacksquare$
The proof is actually just the first two lines.
I believe Gauss was tasked with finding the sum of all the integers from $1$ to $100$ in his very early schooling years. He tackled it quicker than his peers or his teacher could, $$\sum_{n=1}^{100}n=1+2+3+4 +\dots+100$$ $$=100+99+98+\dots+1$$ $$\therefore 2 \sum_{n=1}^{100}n=(100+1)+(99+2)+\dots+(1+100)$$ $$=\underbrace{101+101+101+\dots+101}_{100 \space times}$$ $$=101\cdot 100$$ $$\therefore \sum_{n=1}^{100}n=\frac{101\cdot 100}{2}$$ $$=5050.$$ Hence he showed that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}.$$
If $H$ is a subgroup of $(\mathbb{R},+)$ and $H\bigcap [-1,1]$ is finite and contains a positive element. Then, $H$ is cyclic.
Fermat's little theorem from noting that modulo a prime p we have for $a\neq 0$:
$$1\times2\times3\times\cdots\times (p-1) = (1\times a)\times(2\times a)\times(3\times a)\times\cdots\times \left((p-1)\times a\right)$$
Proposition (No universal set): There does not exists a set which contain all the sets (even itself)
Proof: Suppose to the contrary that exists such set exists. Let $X$ be the universal set, then one can construct by the axiom schema of specification the set
$$C=\{A\in X: A \notin A\}$$
of all sets in the universe which did not contain themselves. As $X$ is universal, clearly $C\in X$. But then $C\in C \iff C\notin C$, a contradiction.
Edit: Assuming that one is working in ZF (as almost everywhere :P)
(In particular this proof really impressed me too much the first time and also is very simple)
Most proofs concerning the Cantor Set are simple but amazing.
The total number of intervals in the set is zero.
It is uncountable.
Every number in the set can be represented in ternary using just
0 and
2. No number with a
1 in it (in ternary) appears in the set.
The Cantor set contains as many points as the interval from which it is taken, yet itself contains no interval of nonzero length. The irrational numbers have the same property, but the Cantor set has the additional property of being closed, so it is not even dense in any interval, unlike the irrational numbers which are dense in every interval.
The Menger sponge which is a 3d extension of the Cantor set
simultaneously exhibits an infinite surface area and encloses zero volume.
The derivation of first principle of differentiation is so amazing, easy, useful and simply outstanding in all aspects. I put it here:
Suppose we have a quantity $y$ whose value depends upon a single variable $x$, and is expressed by an equation defining $y$ as some specific function of $x$. This is represented as:
$y=f(x)$
This relationship can be visualized by drawing a graph of function $y = f (x)$ regarding $y$ and $x$ as Cartesian coordinates, as shown in Figure(a).
Consider the point $P$ on the curve $y = f (x)$ whose coordinates are $(x, y)$ and another point $Q$ where coordinates are $(x + Δx, y + Δy)$.
The slope of the line joining $P$ and $Q$ is given by:
$tanθ = \frac{Δy}{Δx} = \frac{(y + Δy ) − y}{Δx}$
Suppose now that the point $Q$ moves along the curve towards $P$.
In this process, $Δy$ and $Δx$ decrease and approach zero; though their ratio $\frac{Δy}{Δx}$ will not necessarily vanish.
What happens to the line $PQ$ as $Δy→0$, $Δx→0$? You can see that this line becomes a tangent to the curve at point $P$ as shown in Figure(b). This means that $tan θ$ approaches the slope of the tangent at $P$, denoted by $m$:
$m=lim_{Δx→0} \frac{Δy}{Δx} = lim_{Δx→0} \frac{(y+Δy)-y}{Δx}$
The limit of the ratio $Δy/Δx$ as $Δx$ approaches zero is called the derivative of $y$ with respect to $x$ and is written as $dy/dx$.
It represents the slope of the tangent line to the curve $y=f(x)$ at the point $(x, y)$.
Since $y = f (x)$ and $y + Δy = f (x + Δx)$, we can write the definition of the derivative as:
$\frac{dy}{dx}=\frac{d{f(x)}}{dx} = lim_{x→0} [\frac{f(x+Δx)-f(x)}{Δx}]$,
which is the required formula.
This proof that $n^{1/n} \to 1$ as integral $n \to \infty$:
By Bernoulli's inequality (which is $(1+x)^n \ge 1+nx$), $(1+n^{-1/2})^n \ge 1+n^{1/2} > n^{1/2} $. Raising both sides to the $2/n$ power, $n^{1/n} <(1+n^{-1/2})^2 = 1+2n^{-1/2}+n^{-1} < 1+3n^{-1/2} $.
Can a Chess Knight starting at any corner then move to touch every space on the board exactly once, ending in the opposite corner?
The solution turns out to be childishly simple. Every time the Knight moves (up two, over one), it will hop from a black space to a white space, or vice versa. Assuming the Knight starts on a black corner of the board, it will need to touch 63 other squares, 32 white and 31 black. To touch all of the squares, it would need to end on a white square, but the opposite corner is also black, making it impossible.
The Eigenvalues of a skew-Hermitian matrix are purely imaginary.
The Eigenvalue equation is $A\vec x = \lambda\vec x$, and forming the vector norm gives $$\lambda \|\vec x\| = \lambda\left<\vec x, \vec x\right> = \left<\lambda \vec x,\vec x\right> = \left<A\vec x,\vec x\right> = \left<\vec x, A^{T*}\vec x\right> = \left<\vec x, -A\vec x\right> = -\lambda^* \|\vec x\|$$ and since $\|\vec x\| > 0$, we can divide it from left and right side. The second to last step uses the definition of skew-Hermitian. Using the definition for Hermitian or Unitarian matrices instead yields corresponding statements about the Eigenvalues of those matrices.
I like the proof that not every real number can be written in the form $a e + b \pi$ for some integers $a$ and $b$. I know it's almost trivial in one way; but in another way it is kind of deep. |
The correct formulation of the result is called spectral theorem. It encompasses all type of spectra for self-adjoint operators, and thus both (1) and (2).
Given a self adjoint operator $A$ with only discrete spectrum (you have such situations for compact operators, and operators with compact resolvent) the spectral theorem has a nice and simple form:$$A=\sum_{i}\lambda_i P_i\; ,$$where the sum ranges over all possible eigenvalues $\lambda_i$ and $P_i$ is the projection on the eigensubspace corresponding to $\lambda_i$ (that if the eigenvalue has multiplicity one has the form $\lvert \psi_i\rangle\langle \psi_i\rvert$ where $\psi_i$ is the normalized eigenvector). The spectral theorem gives also the so-called functional calculus (I will not be completely precise mathematically but to give the idea will be enough): given a suitable function $f:\mathbb{R}\to \mathbb{C}$, the operator $f(A)$ is defined by$$f(A)=\sum_i f(\lambda_i)P_i\; .$$The particular case of $e^{-itA}$ is the one that gives you quantum evolution.
Now if $B$ is an operator with general spectrum (i.e. a discrete part and a continuum one) the theorem still holds, but instead of having a simple discrete family of projections $P_i$ we have what is called a spectral family $\{P_\Omega\}_{\Omega\in \{X, X\subseteq \mathbb{R}\}}$ of projections. This family is uncountable, so you cannot take the sum, but it defines a projection valued measure $dP_\lambda$ (a function from sets to self-adjoint projections that satisfies the axioms of a measure) and it is possible to integrate functions with respect to it. So the spectral theorem for (measurable) functions $f$ of $B$ becomes:$$B=\int_{\mathbb{R}}\lambda dP_\lambda\; ,$$$$f(B)=\int_{\mathbb{R}}f(\lambda) dP_\lambda\; .$$Or, in the case of the evolution operator $e^{-itB}$ applied to $u\in \mathscr{H}$ (the Hilbert space):$$u(t)=e^{-itB}u=\int_{\mathbb{R}}e^{-it\lambda} dP_\lambda u \; .$$
Observe that if $B$ has only discrete spectrum, the corresponding measure is just the sum of delta measures pointed at each discrete eigenvalue $\lambda_i$, so it can be written $dP_\lambda = \sum_i \delta(\lambda-\lambda_i)P_\lambda d\lambda$. Therefore the discrete spectrum is just a special case of the spectral theorem.
Anyways, what you are writing is an operator on the right hand side, and a function on the left hand side, so it is not correct as written ;-) |
My questions is related to combining text and math fonts. I am a real fan of Georgia-like "dense" and "no so thin" fonts. It seems that books published by AMS use similar one.
My problem is that I cannot find a good math font to fit the text:
newtxmath seems too thin for Georgia, it fits Times New Roman more. I have tried using
STIX Math Two, but the
\bm package won't work with it. Also,
mathbb, mathcal and
mathscr styled letters look much better in
newtxmath.
I am searching for a solution of at least one of these problems:
Can I somehow load the symbols I like from
newtxmathpackage and make
bmwork?
What is the font that looks good with Georgia and has good support for math symbols, proper spacing and works with other packages (most prefferable -- loadable with
unicode-mathpackage)?
Some examples:
MWE contains a short example of a formula and text. I am including some packages into MWE that sometimes conflict with the font-loading stuff.
I use LuaLaTeX to compile.
\documentclass[a4paper,10pt,openany]{book}\usepackage{geometry}\geometry{margin=1in}%\usepackage{mathtools}\usepackage{amsthm}\usepackage{thmtools}\usepackage{wasysym}%\usepackage{newtxmath}%\usepackage[notext,not1,notextcomp]{stix}%\let\coloneqq\relax%\let\Coloneqq\relax%\let\eqqcolon\relax\usepackage[math-style=ISO]{unicode-math}\setmathfont{STIX Two Math}%\setmathfont{XITS Math}\usepackage{bm}\usepackage{mathrsfs}\usepackage{lipsum}\usepackage[utf8]{inputenc}\usepackage{polyglossia}\defaultfontfeatures{Ligatures=TeX}\setmainfont{Georgia}\setmainlanguage{english}\DeclareFontFamily{U}{skulls}{}\DeclareFontShape{U}{skulls}{m}{n}{ <-> skull }{}\newcommand{\skull}{\text{\usefont{U}{skulls}{m}{n}\symbol{'101}}}%\begin{document}If $\omega$ is a positive linear functional on a $C^{\ast}$-algebra~$A$,then we can construct a unique (up to unitary equivalence)representation~$\pi_\omega$ of algebra~$A$ in some Hilbertspace~$H_\omega$ over field of scalars $\mathbb{C}$ anda vector~$\xi_\omega$ such that $$ \omega(a)=\left(\pi_{\omega(a)}\xi_\omega,\,\xi_\omega\right). $$\end{document} |
I think that there is no formula. The best one can do is to estimate. Here is a simpler problem of the same sort: suppose you have a parametrization of the boundary of a simply connected region, andsuppose that 0 is inside. Consider the Riemann mapping f of this region sending 0 to 0.The problem is to find |f'(0)|. There is no formula in any reasonable case.
Of course this is not a theorem, because one cannot define what a "formula" is.Both quantities, the modulus of a ring, and |f'(0)| in the simplified problemare solutions of certain extremal problems. So one can write a "formula" involving supover some class of functions.
Added on 9.19: I don't know why the question about a "formula" is important. There arereasonably good converging algorithms for finding moduli of rings,of course. The closest thing to a "formula" for a conformal map of a simply connected region thatI know is described in the papers of Wiegmann and Zabrodin, for example, MR1785428.Perhaps this can be modified to make a formula for the modulus of a ring.
Added on the same day: Here is a "formula". Let $\mu$ and $\nu$ be two probability measures,one sitting on each boundary component. Let $\rho=\mu-\nu$. Then$$\log r=-2\pi\sup\int\int\log|z-w|d\rho(z)d\rho(w),$$where the $\sup$ is taken over all such measures. If your boundaries are smooth, the measures arealso smooth, and can be described by smooth densities.
Explanation. Think of the boundaries as bases of metal cyinders, and put unit charges on them,one positive another negative. Then allow the charges to flow according to Coulomb Law.they will occupy the equilibrium position (minimizing the energy). This minimal energy is$\log r/\(2\pi)$ and it is conformally invariant. It is the so-called capacity of a condenser.
This was given as an example of what I meant by a formula containing a sup over a set offunctions. |
Show that the spectral radius of the matrix A is less than or equal any natural norm, i.e:
$$\rho(A) \leq ||A||=\max_{||x||=1}{||Ax||}$$
where $\rho(A)=\max\{|\lambda|:\lambda \text{ is a eigenvalue of $A$}\}$.
I'm a little confused with this, because I'm not sure about the definition of natural norm of $A$. If $\rho(A)=\lambda^*$ and $u$ is such that $||u||=1$, then: \begin{align*} ||A||&\geq ||Au||\\ &=||\lambda^* u||\\ &=|\lambda^*| \end{align*} Is this ok?
Thanks. |
I’ve encountered the following difficult question that I don’t know how to solve.
$L_1$ and $L_2$ are regular languages over the same $\Sigma$. $$\begin{align}L^\wedge=&\{σ_1σ_2...σ_n\mid n\ge1, \sigma_1, \sigma_2, \cdots, \sigma_n\in \Sigma, \\ &\exists \mu_1, \mu_2,\cdots,\mu_n,\zeta_1,\zeta_2,\cdots\, \zeta_n \in\Sigma, \\ &\mu_1\mu_2\cdots\mu_n\in L_2,\sigma_1\mu_1\zeta_1\sigma_2\mu_2\zeta_2...\sigma_n\mu_n\zeta_n\in L_1\}\\ &\cup\{a\mid a\in L_1, a\in L_2, a=\epsilon\}\end{align}$$ where the second set on the right hand side just says that $\epsilon\in L^\wedge\Leftrightarrow \epsilon\in L_1\land \epsilon \in L_2$.
How can we prove $L^\wedge$ is regular using closure properties or a product automaton?
What i tried to do is:
Since $n \geq 0$, we can build $\Sigma^* = \{ \sigma^* | \sigma \in \Sigma \} $ $(\Sigma \cup \Sigma) \to \Sigma^*$, then for some function h we can assign $h(\sigma)=\sigma$ (also $h(\sigma')=\sigma$), so because of homomorphism $h^{-1}(L_1 \cup L_2) = \{ \sigma_1...\sigma_n |\forall 1 \leq i \leq n \to \sigma_i \in \{ \mu_i,\mu_i' \}, \mu_1...\mu_n \in L_1 \cup L_2 \} $,
So if $(L_1 \cup L_2)' = h^{-1}(L_1 \cup L_2) \cap ( \sigma'\Sigma \Sigma')*=\{ \sigma_1'\sigma_2\sigma_3'...\sigma_{n-2}'\sigma_{n-1}\sigma{n}' | \sigma_1... \sigma_n \in (L_1 \cup L_2)$ is also regular
So if for some function $f$, $f(\sigma)=\sigma$ and $f(\sigma')=\epsilon$ $(f|f:(\Sigma' \cup \Sigma) \to \Sigma^*)$
Then $f(L_1' \cup L_2') = \{ \mu_1...\mu_n | σ_1μ_1ξ_1...σ_nμ_nξ_n∈L_1 \cup L_2 = L^∧$ and if regular because of regular languages closure to homomorphism
I know it’s complicated and would appreciate help with it, seeing how to do it correctly (pretty sure I've made some mistakes along the way). |
In this paper we consider a telegraph equation with time-dependent coefficients, governing the persistent random walk of a particle moving on the line with a time-varying velocity $c(t)$ and changing direction at instants distributed according to a non-stationary Poisson distribution with rate $\lambda (t)$. We show that, under suitable assumptions, we are able to find the exact form of the probability distribution. We also consider the space-fractional counterpart of this model, finding the characteristic function of the related process. A conclusive discussion is devoted to the potential applications to run-and-tumble models.
In this paper the fractional Cox–Ingersoll–Ross process on ${\mathbb{R}_{+}}$ for $H<1/2$ is defined as a square of a pointwise limit of the processes ${Y_{\varepsilon }}$, satisfying the SDE of the form $d{Y_{\varepsilon }}(t)=(\frac{k}{{Y_{\varepsilon }}(t){1_{\{{Y_{\varepsilon }}(t)>0\}}}+\varepsilon }-a{Y_{\varepsilon }}(t))dt+\sigma d{B^{H}}(t)$, as $\varepsilon \downarrow 0$. Properties of such limit process are considered. SDE for both the limit process and the fractional Cox–Ingersoll–Ross process are obtained.
We present a generalization of the Yule model for macroevolution in which, for the appearance of genera, we consider point processes with the order statistics property, while for the growth of species we use nonlinear time-fractional pure birth processes or a critical birth-death process. Further, in specific cases we derive the explicit form of the distribution of the number of species of a genus chosen uniformly at random for each time. Besides, we introduce a time-changed mixed Poisson process with the same marginal distribution as that of the time-fractional Poisson process.
For a class of non-autonomous parabolic stochastic partial differential equations defined on a bounded open subset $D\subset {\mathbb{R}^{d}}$ and driven by an ${L^{2}}(D)$-valued fractional Brownian motion with the Hurst index $H>1/2$, a new result on existence and uniqueness of a mild solution is established. Compared to the existing results, the uniqueness in a fully nonlinear case is shown, not assuming the coefficient in front of the noise to be affine. Additionally, the existence of moments for the solution is established.
The problem of European-style option pricing in time-changed Lévy models in the presence of compound Poisson jumps is considered. These jumps relate to sudden large drops in stock prices induced by political or economical hits. As the time-changed Lévy models, the variance-gamma and the normal-inverse Gaussian models are discussed. Exact formulas are given for the price of digital asset-or-nothing call option on extra asset in foreign currency. The prices of simpler options can be derived as corollaries of our results and examples are presented. Various types of dependencies between stock prices are mentioned.
The insurance model when the amount of claims depends on the state of the insured person (healthy, ill, or dead) and claims are connected in a Markov chain is investigated. The signed compound Poisson approximation is applied to the aggregate claims distribution after $n\in \mathbb{N}$ periods. The accuracy of order $O({n^{-1}})$ and $O({n^{-1/2}})$ is obtained for the local and uniform norms, respectively. In a particular case, the accuracy of estimates in total variation and non-uniform estimates are shown to be at least of order $O({n^{-1}})$. The characteristic function method is used. The results can be applied to estimate the probable loss of an insurer to optimize an insurance premium.
The discrete time risk model with two seasons and dependent claims is considered. An algorithm is created for computing the values of the ultimate ruin probability. Theoretical results are illustrated with numerical examples. |
If a rocket is going 50,000 kilometers per hour and wanted to go 75,000 kilometers per hour wouldn't the rocket just need an additional thrust equaling 25,000 kilometers per hour? Or is it something different?
In some sense, you're correct. It is just that much thrust. The difficulty is that rockets work by expelling mass, so the mass of the rocket is variable, making the thrust required for a specific acceleration (or delta-v) variable as well.
If the rocket wer mostly non-fuel, then the mass wouldn't change much and you might be able to ignore the difference. But such a rocket can't accelerate much and is much less useful than a rocket that is mostly fuel.
Let's say you want to have a specific configuration arrive at the final velocity (the rocket, plus enough fuel to do other things). For this example, I'll make that a mass of 25000kg. Further, let's assume we carry cryogenic hydrogen and oxygen for fuel. If so, the engine produces $v_e$ of $\sim 4400\text{m/s}$.
Now we have everything to determine how much additional fuel is required.
$$\Delta v = v_e \ln \frac{m_0}{m_f}$$ $$\ln \frac{m_0}{m_f} = \frac{\Delta v}{v_e}$$ $$\frac{m_0}{m_f} = e^{\frac {\Delta v}{v_e}}$$ $$m_0 = m_f e^{\frac {\Delta v}{v_e}}$$ $$m_0 = (25000\text{kg}) e^{\frac {6944 \text{m/s}}{4400 \text{m/s}}}$$ $$m_0 = 121000\text{kg}$$
So for this rocket, with this mass, you'd have to carry and burn 96,000kg of fuel to perform that acceleration. More efficient engines (those with greater $v_e$) would require less fuel.
If a rocket is going 50,000 kilometers per hour and wanted to go 75,000 kilometers per hour wouldn't the rocket just need an additional thrust equaling 25,000 kilometers per hour?
Yes, but ...
The problem is that the material needed to attain that extra 25,000 kph is just dead weight during the interval over which the rocket gains the initial 50,000 kph velocity. This carrying of dead weight that eventually becomes fuel is what results in the ideal rocket equation. |
I stumbled upon this link while googling for an answer to the same question.
Under point number 3, the sole answer mentions an example where it is said $\mathbb{R}/\mathbb{Q}$ is not Hausdorff, because the topology generated is trivial, assuming $\mathbb{R}$ has the standard topology. I agree with the conclusion but I think the reason given is wrong. I just wanted to cross-check my argument.
If I take $ (\mathbb{R}/\mathbb{Q}) \backslash [e] $ where e is the equivalence class of the Euler number $e$,
then the inverse of the projection map $\pi: \mathbb {R}\rightarrow \mathbb{R}/\mathbb{Q}$ is $\mathbb{R}\backslash \{e\}$ which is an open set in $\mathbb{R}$.
And so the topology consists of open sets other than $X$ and $\phi$, which means the topology is not trivial.
Is there something wrong with the above argument?
NOTE: I agree with the conclusion though. Let $U \neq \phi$ be an open set in $\mathbb{R}/\mathbb{Q}$ . Let $q\in \mathbb{Q}$ then if $[q] \notin U$,$\pi^{-1}(U)$ will consist of only irrationals which cannot be open,and so the space is not hausdorff, since any two non-empty open sets have non-empty intersection containing atleast $[q]$. |
$$ \newcommand{\bsth}{{\boldsymbol\theta}} \newcommand{\va}{\textbf{a}} \newcommand{\vb}{\textbf{b}} \newcommand{\vc}{\textbf{c}} \newcommand{\vd}{\textbf{d}} \newcommand{\ve}{\textbf{e}} \newcommand{\vf}{\textbf{f}} \newcommand{\vg}{\textbf{g}} \newcommand{\vh}{\textbf{h}} \newcommand{\vi}{\textbf{i}} \newcommand{\vj}{\textbf{j}} \newcommand{\vk}{\textbf{k}} \newcommand{\vl}{\textbf{l}} \newcommand{\vm}{\textbf{m}} \newcommand{\vn}{\textbf{n}} \newcommand{\vo}{\textbf{o}} \newcommand{\vp}{\textbf{p}} \newcommand{\vq}{\textbf{q}} \newcommand{\vr}{\textbf{r}} \newcommand{\vs}{\textbf{s}} \newcommand{\vt}{\textbf{t}} \newcommand{\vu}{\textbf{u}} \newcommand{\vv}{\textbf{v}} \newcommand{\vw}{\textbf{w}} \newcommand{\vx}{\textbf{x}} \newcommand{\vy}{\textbf{y}} \newcommand{\vz}{\textbf{z}} \DeclareMathOperator*{\argmin}{argmin} \DeclareMathOperator\mathProb{\mathbb{P}} \renewcommand{\P}{\mathProb} % need to overwrite stupid paragraph symbol \DeclareMathOperator\mathExp{\mathbb{E}} \newcommand{\E}{\mathExp} \DeclareMathOperator\Uniform{Uniform} \DeclareMathOperator\poly{poly} \DeclareMathOperator\diag{diag} \newcommand{\pa}[1]{ \left({#1}\right) } \newcommand{\ha}[1]{ \left[{#1}\right] } \newcommand{\ca}[1]{ \left\{{#1}\right\} } \newcommand{\norm}[1]{\left\| #1 \right\|} \newcommand{\nptime}{\textsf{NP}} \newcommand{\ptime}{\textsf{P}} \newcommand{\R}{\mathbb{R}} \newcommand{\card}[1]{\left\lvert{#1}\right\rvert} \newcommand{\abs}[1]{\card{#1}} \newcommand{\sg}{\mathop{\mathrm{SG}}} \newcommand{\se}{\mathop{\mathrm{SE}}} \newcommand{\mat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \DeclareMathOperator{\var}{var} \DeclareMathOperator{\cov}{cov} \newcommand\independent{\perp\kern-5pt\perp} \newcommand{\CE}[2]{ \mathExp\left[ #1 \,\middle|\, #2 \right] } \newcommand{\disteq}{\overset{d}{=}} $$
BERT Prerequisite 2: The Transformer
In the last post, we took a look at deep learning from a very high level (Part 1). Here, we’ll cover the second and final prerequisite for setting the stage for discussion about BERT, the Transformer.
The Transformer is a novel sequence-to-sequence architecture proposed in Google’s Attention is All You Need paper. BERT builds on this significantly, so we’ll discuss here why this architecture was important.
The Challenge
Recall the language of the previous post applied to supervised learning. We’re interested in a broad class of settings where the input \(\textbf{x}\) has some shared structure with the output \(\textbf{y}\), which we don’t know ahead of time. For instance, \(\textbf{x}\) might be an English sentence and \(\textbf{y}\) might be a German sentence with the same context.
For a parameterized model \(M(\theta)\) which might just be a function over \(\textbf{x}\), we recall the \(L\)-layer MLP from last time, where \(\theta=\mat{\theta_1& \theta_2&\cdots&\theta_L}\), \[ M(\theta)= x\mapsto f_{\theta_L}^{(L)}\circ f_{\theta_{L-1}}^{(L-1)}\circ\cdots\circ f_{\theta_1}^{(1)}(x)\,, \] and we define each layer as \[ f_{\theta_i}=\max(0, W_ix+b_i)\,,\,\,\, \mat{W_i & b_i} = \theta_i\,. \]
Most feed-forward neural nets (FFNNs) are just variants on this architecture, with some loss typically like \(\norm{M(\theta)(\textbf{x}) - \textbf{y}}^2\).
One issue with this, and typical FFNNs, is that they’re mappings from some fixed size vector space \(\mathbb{R}^m\) to another \(\mathbb{R}^k\). When your inputs are variable-length sequences like sentences, this doesn’t make sense for two reasons:
Sentences can be longer than the width of your input space (not a fundamental issue, you could just make \(m\) really large). The inputs don’t respect the semantics of the input dimensions.
For typical learning tasks, the \(i\)-th input dimension corresponds to a meaningful position in the input space. E.g., for images, this is the \(i\)-th pixel in the space of fixed size \(64\times 64\) images. It’s next to the \((i-1)\)-th and \((i+1)\)-th pixels, and every \(64\times 64\) image \(\textbf{x}\) will also have its \(i\)-th pixel in the \(i\)-th place.
Not so for sentences. In sentences, the subject may the first or second or third word. It might be preceded by an article, or it might not. If you look at a fixed offset for many different sentences, you’d be hard pressed to find a robust semantics for the word or letter that you see there. So it’s unreasonable to assume a model could extract relevant structure with such a representation.
Recursive Neural Networks (RNNs)
The typical resolution to this problem in deep learning is to use RNNs. For an overview, see Karpathy’s blog post.
To resolve this issue, we can view our input as a variable-length list of fixed length vectors \(\{\textbf{x}_i\}_{i}\). Next, we modify our FFNN to accept two fixed-length parameters at a time step \(i\), a hidden state \(\textbf{h}_i\) and input \(\textbf{x}_i\). It’s the green box in the diagram above.
This retains essential properties of FFNNs that allow it to optimize well (backprop still works). But, from a perspective of input semantics, we’ve resolved our problem by assuming the hidden state at timestep \(\textbf{h}_i\) tells the FFNN how to interpret the \(i\)-th sequence element (which could be a word or word part or character in the sentence). The FFNN is then also responsible for updating how the \((i+1)\)-th sequence element is to be interpreted, by returning \(\textbf{h}_{i+1}\) on the evaluation in timestep \(i\).
We might want to wait until the network reads the entire input if the entire variable-length output may change depending on all parts of the input (the second to last diagram above). This is the case in translation, where words at the end of the source language may end up at the beginning in the target language.
Alternatively, we might do something like try to classify off of the hidden state after reading the sentence, like identifying the sentiment of a text-based review.
RNN challenges
Consider the task of translating English to Spanish. Let’s suppose our inputs are sequences of words, like
I arrived at the bank after crossing the {river,road}.
The proper translation might be either:
Llegué a la orilla después de cruzar el río.
or:
Llegué al banco después de cruzar la calle.
Notice how we need to look at the
whole sentence to translate it correctly. The choice of “river” or “road” affects the translation of “bank”.
This means that the RNN needs to store information about the entire sentence when translating. For longer sentences, we’d definitely need to use a larger hidden state, but also we’re assuming the network would even be able to train to a parameter setting that properly recalls whole-sentence information.
The Transformer
The problem we faced above is one of
context: to translate “bank” properly we need the full context of the sentence. This is what the Transformer architecture addresses. It inspects each word in the context of others.
Again, let’s view each word in our input sequence as some embedded vector \(\textbf{e}_i\) (for context on word embeddings, check out the Wikipedia page).
Our goal is to come up with a new embedding for each word, \(\textbf{a}_i\), which contains context from all other words. This is done through a mechanism called attention. For a code-level explanation, see The Annotated Transformer, though I find that focusing on a particular word (the one at position \(i\)) helped me understand better.
The following defines (one head of) a Transformer block. A transformer block just contextualizes embeddings. They can be stacked on top of each other and then handed off to the transformer decoder, which is a more complicated kind of transformer that includes attention over both the inputs and outputs. Luckily, we don’t need that for BERT.
Remember, at the end of the day, we’re trying to take one sequence \(\{\textbf{e}_i\}_i\) and convert it into another sequence \(\{\textbf{a}_i\}_i\) which is then used as input for another stage that does the actual transformation. The point is that the representation \(\{\textbf{a}_i\}_i\) is broadly useful for many different decoding tasks.
Apply an FFNN pointwise to each of the inputs \(\{\textbf{e}_i\}_i\) to get \(\{\textbf{x}_i\}_i\).
Now consider a fixed index \(i\). How do we contextualize the word at \(\textbf{x}_i\) in the presence of other words \(\textbf{x}_1,\cdots,\textbf{x}_{i-1},\textbf{x}_{i+1},\cdots,\textbf{x}_s\)?
We attend to the sequence itself. Attention tells us how much to pay attention to each element when coming up with a fixed-width context for the \(i\)-th element. This is done with the inner product.
After computing how important each element \(\textbf{x}_j\) is to the element in question \(\textbf{x}_i\) as \(\alpha_j\), we combine the weighted sum of each of the \(\textbf{x}_j\) themselves.
After doing this for every index \(i\in[s]\), we get a new sequence \(\textbf{a}_i\). That’s it!
This glosses over a couple normalization, multiple heads, and computational details, but it’s the gist of self-attention and the Transformer block.
One thing worth mentioning is the positional encoding, which makes sure that information about a word being present in the \(i\)-th position is present before the first Transformer block is applied.
After possibly many transformer blocks, we get our \(L\)-th sequence of embeddings, \(\{\textbf{a}^{(L)}_i\}_i\). We plug this as input to another model, the transformer decoder, which uses a similar process to eventually get a loss based on some input-output pair of sentences (e.g., in translation, the decoder converts the previous sequence into \(\{\textbf{b}_j\}_j\), which is compared with the actual translation \(\{\textbf{y}^{(L)}_j\}_j\)
So What?
On the face of it, this all sounds like a bunch of hand-wavy deep learning nonsense. “Attention”, “embedding”, etc. all look like fancy words to apply to math that is operating on meaningless vectors of floating-point numbers. Layer on top of this (lol) the other crap I didn’t cover, like multiple heads, normalization, and various knobs pulled during training, and the whole thing looks suspect.
It’s not clear which parts are essential, but something is doing its job:
And self-attention looks like it’s doing something like what we think it should.
Regardless how much of a deep learning believer you are, this architecture solves problems which require contextualizing our representation of words, and it picks the right things to attend to in examples.
Next time
We’ll see how BERT uses the context-aware Transformer to come up with a representation without any supervision. |
In basic type inference for 𝜆-calculus with parametric polymorphism à la Hindley–Milner, when can we say that we cannot give a type to a lambda constructor? For example $$(λx.λy.y(x\ y))(λz.z)$$
A term $M$ is well-typed if and only if there is a type derivation that leads to a judgement of the form $\Gamma \vdash M : \tau$ for some context $\Gamma$ and some type $\tau$. (I use the word “type” in its general sense which can include quantified variables; in the terminology commonly used with Hindler-Milner, that's a
type scheme.) So, to prove that you cannot give a type to a lambda term from first principles, you prove that there is no type derivation that leads to such a judgement.
Hindley-Milner has a nice property in this respect: it's
syntax-directed, i.e. it's presented as a set of deduction rules such that for any term, there is a single rule that can be used to end a deduction of this term. (Being syntax-directed is actually a property of a presentation of the type system, not a property of the type system. In this post, I use the classical syntax-directed presentation which builds generalization into the let rule, rather than having a separate rule for generalization.)
Another nice property (shared by almost every type system) is that unused variables can always be removed from the context. So for example, if $(\lambda x y.y (x y)) (\lambda z.z)$ is well-typed, it must have a type derivation ending with the App rule and, since the term has no free variable, an empty context: $$ \dfrac{\vdash (\lambda x y.y (x y)) : \tau_1 \to \tau_0 \qquad \vdash (\lambda z. z) : \tau_1} {\vdash (\lambda x y.y (x y)) (\lambda z.z) : \tau_0} $$
You can of course make use of any theorem that you already know. In particular, the
principal typing theorem can be useful: if a term is well-typed then all of its types are instances of a particular one. For example, the identity function $(\lambda z.z)$'s possible types are all instances of $\forall \alpha. \alpha \to \alpha$. Similarly, it's easy to see that the function $(\lambda x y.y (x y))$ has the most general type $\forall \beta \gamma. ((\beta \to \gamma) \to \beta) \to ((\beta \to \gamma) \to \gamma)$.
In order to find a type for $(\lambda x y.y (x y)) (\lambda z.z)$, we saw above that we need to find types $\tau_0$ and $\tau_1$ such that $\tau_1 \to \tau_0$ is an instance of $\forall \beta \gamma. ((\beta \to \gamma) \to \beta) \to ((\beta \to \gamma) \to \gamma)$ and $\tau_1$ is an instance of $\forall \alpha. \alpha \to \alpha$. This means that there must exist base types $T_1$, $T_2$ and $T_3$ which satisfy the equations $$ \begin{align} \tau_1 &= \forall \alpha \beta \gamma. (T_2 \to T_3) \to T_2 \\ \tau_0 &= \forall \alpha \beta \gamma. (T_2 \to T_3) \to T_3 \\ \tau_0 &= \forall \alpha \beta \gamma. T_1 \to T_1 \\ \end{align} $$ Hence $T_3 = T_1 = (T_2 \to T_3)$. But this is impossible due to the structure of types: a type can't be a strict subterm of itself. Therefore the premise of was wrong: the term $(\lambda x y.y (x y)) (\lambda z.z)$ is not well-typed.
chi's answer shows how to use another theorem about Hindley-Milner to shorten the proof:
subject reduction, i.e. the property that if a term is well-typed and it reduces to another term then that other term also has the same type. Since $(\lambda x y.y (x y)) (\lambda z.z) \to_\beta^* \lambda y. y y$, if the original term has a type then so does $\lambda y. y y$. By contraposition, if $\lambda y. y y$ is not well-typed then neither is the original term. You can show that $\lambda y. y y$ is not well-typed by a shorter application of the methodology above: note that if it is well-typed, it must be an application of the lambda typing rule, which leads to a type equation of the form $T_1 \to T_1 = T_1$, which has no solution for the same reason as before.
There are, of course, other type systems where the term in question is well-typed. These type systems must have some additional ways of typing terms that Hindley-Milner doesn't have. I'll give two examples:
If you extend the syntax of types to allow them to be recursive (as with
ocaml -rectypes), then the reasoning about the type equation having no solution because a type can't contain itself breaks down, and indeed both $\lambda y. y y$ and the original term are well-typed in this system. For example, the most general type of $\lambda y. y y$ is $\forall \alpha \beta [\alpha = \alpha \to \beta]. \alpha$.
If you add a rule that allows
intersection types: $$ \dfrac{\Gamma \vdash M : \tau_1 \qquad \Gamma \vdash M : \tau_2} {\Gamma \vdash M : \tau_1 \wedge \tau_2} $$ then the reasoning above breaks down because the presentation of the type system is not type-directed. The intersection rule can be used with any language construct. With this type system, $y : \alpha \wedge (\alpha \to \beta) \vdash_{\wedge} y : \alpha$ and $y : \alpha \wedge (\alpha \to \beta) \vdash_{\wedge} y : \alpha \to \beta$ and so $\lambda y. y y$ has the type $\forall \alpha \beta. (\alpha \wedge (\alpha \to \beta)) \to \beta$. The original term is also well-typed. With a rule that departs so radically from being syntax-directed, you might think that it's difficult to prove that a term is not typable, and you'd be right: the intersection rule is so strong that it causes all strongly normalizing terms to be well-typed. Conversely, intersection types with base types (but without quantifiers or a top type) only allow strong normalizing terms, so to prove that a term is not typable in that system, it's sufficient to prove that a term is not strongly normalizing. See Does there exist a Turing complete typed lambda calculus?) for more information about intersection types.
No, that term can not be typed in Hindley Milner, or any other "standard" type system. Here's a rough sketch of a proof.
Suppose by contradiction it had a type. Since type is preserved under beta reduction (by the subject reduction theorem) we would get that all these terms also have the same type
$$ \begin{array}{l} (λx.λy.y(x\ y))(λz.z) \\ (λy.y((λz.z)\ y) \\ (λy.y\ y) \\ \end{array} $$
The last term is however not typeable, since $y$ should have a function type $\tau_1 \to \tau_2$ because it is applied as a function, but since $y$ itself is the argument we should also have that $y$ has type $\tau_1$. Since $\tau_1 = \tau_1\to\tau_2$ is impossible, we obtain a contradiction.
Another way to see the same fact is running the type inference algorithm: doing so at a certain point would require to solve the equation $\tau_1 = \tau_1\to\tau_2$, triggering the failure of type inference.
Let me add that applying a function to itself, directly or indirectly (as done above), is the archetypal way to produce a term having no type in typed lambda calculi. |
When solving an equation of the following form:
$$ \begin{aligned} \frac {\partial A}{\partial t} &= EB - A \\ \frac {\partial B}{\partial t} &= EA - B \\ \frac {\partial E}{\partial t} &= u \\ \frac {\partial u}{\partial t} &= u + \frac{1}{10^4}\frac {\partial E}{\partial z} - \frac{1}{10^8}\frac {\partial^2 E}{\partial z^2} - \frac {\partial^2 B}{\partial t^2} - \frac {\partial B}{\partial t} \end{aligned} $$
I am able to produce the desired output that matches figures in this text when applying a Cash-Karp method and approximating the spatial derivatives by central differences. (Backward difference approximations of order 5 for the spatial derivatives appear to work, too). The domain I am considering in the above case is from $t = [0,1000] \times z = [0,1000]$ and my step-size in the $z$-direction is $\Delta z \sim $ 1m. Once again, this code seems to work correctly.
To my understanding, explicit methods like the ones I've implemented and finite difference approximations are more accurate when using smaller step-sizes. Of course there should be a precision limit based on floating point error, but I can re-scale my variables to avoid this problem.
Now I am using the exact same initial/boundary conditions as for the above, but wanting to simply consider the domain $t = [0,1000]$ and $z = [0,10^{-4}]$. It should correspond to my solution above but simply on a reduced domain, right? Well when I run the code now, $\Delta z \sim 10^{-7}$ and it appears there are very fast and large oscillations in only the $z$-component for $A, B,$ and $E$ (larger than any oscillations than on the original larger domain). I have an adaptive step based on the Cash-Karp method and the steps seem to work perfectly fine on this domain. But it appears my solutions on $t = [0,1000] \times z = [0,10^{-4}]$ and $t = [0,1000] \times z = [0,1000]$ do not correspond. When reducing the domain of $t$, the oscillations are still present, but just not as large. To my knowledge, it's arising because of the spatial derivative approximations where a small $\Delta z$ results in a large $\frac {\partial E}{\partial z}$. But is this to be expected when using such precise values with finite difference approximations? Are these large oscillations I am observing characteristics of the true solution or simply numerical errors? |
This question already has an answer here:
I have the same problem as in this question.
But can someone elaborate on the answer? The poster says that:
Solving this system of 4 ODE's with rk4 will solve for all your state variables simultaneously. position and velocity do not need to be handled seperately (and shouldn't be).
How is this actually implemented? There is still no expression that I can see for $v$, so I do not see how we can obtain the RK coefficients, and I cannot see how to handle position and velocity ‘not seperately’.
For a system of 2 coupled oscillators we have:
$m_1\ddot{x}_1=-k_1x_1+k_2(-x_1+x_2)$
$m_2\ddot{x}_2=-k_3x_2+k_2(x_1-x_2)$
where $x_1$ and $x_2$ are the displacements of masses 1 and 2 from their equilibrium positions. If we then substitute $v=\dot{x}$, we get:
$\dot{v}_1m_1=-k_1x_1+k_2(-x_1+x_2)$
and similar for $\dot{v}_2$. The answer to the question I linked suggested we combine the system to give (they have used a slightly different functional form but the underlying mechanics is the same):
$$ \left(\begin{array}{c} \dot{x}_1 \\ \dot{x}_2 \\ \dot{v}_1 \\ \dot{v}_2 \end{array}\right) = \left(\begin{alignat}{1} & v_1 \\ & v_2 \\ -\omega_1^2(x_1-R_1) &+ \omega_2^2(x_2-x_1-w-R_2)\\ &-\omega_2^2(x_2-x_1-w-R_2) \end{alignat}\right) $$
However I'm unsure how to actually proceed from here. RK4 algorithm needs to evaluate the RK coefficients at different timesteps, and my issue is $\dot{v}=f(x_1,x_2)$ instead of $\dot{v}=f(v)$ in normal (uncoupled) RK4. I can't find $v$ at a new timestep as I need $x$ to do that, and I can't find $x$ without $v$. Is it a matter of writing some operator matrix equation and solving that? |
Let $a \in \Bbb{R}$. Let $\Bbb{Z}$ act on $S^1$ via $(n,z) \mapsto ze^{2 \pi i \cdot an}$. Claim: The is action is not free if and only if $a \Bbb{Q}$. Here's an attempt at the forward direction: If the action is not free, there is some nonzero $n$ and $z \in S^1$ such that $ze^{2 \pi i \cdot an} = 1$. Note $z = e^{2 \pi i \theta}$ for some $\theta \in [0,1)$.
Then the equation becomes $e^{2 \pi i(\theta + an)} = 1$, which holds if and only if $2\pi (\theta + an) = 2 \pi k$ for some $k \in \Bbb{Z}$. Solving for $a$ gives $a = \frac{k-\theta}{n}$...
What if $\theta$ is irrational...what did I do wrong?
'cause I understand that second one but I'm having a hard time explaining it in words
(Re: the first one: a matrix transpose "looks" like the equation $Ax\cdot y=x\cdot A^\top y$. Which implies several things, like how $A^\top x$ is perpendicular to $A^{-1}x^\top$ where $x^\top$ is the vector space perpendicular to $x$.)
DogAteMy: I looked at the link. You're writing garbage with regard to the transpose stuff. Why should a linear map from $\Bbb R^n$ to $\Bbb R^m$ have an inverse in the first place? And for goodness sake don't use $x^\top$ to mean the orthogonal complement when it already means something.
he based much of his success on principles like this I cant believe ive forgotten it
it's basically saying that it's a waste of time to throw a parade for a scholar or win he or she over with compliments and awards etc but this is the biggest source of sense of purpose in the non scholar
yeah there is this thing called the internet and well yes there are better books than others you can study from provided they are not stolen from you by drug dealers you should buy a text book that they base university courses on if you can save for one
I was working from "Problems in Analytic Number Theory" Second Edition, by M.Ram Murty prior to the idiots robbing me and taking that with them which was a fantastic book to self learn from one of the best ive had actually
Yeah I wasn't happy about it either it was more than $200 usd actually well look if you want my honest opinion self study doesn't exist, you are still being taught something by Euclid if you read his works despite him having died a few thousand years ago but he is as much a teacher as you'll get, and if you don't plan on reading the works of others, to maintain some sort of purity in the word self study, well, no you have failed in life and should give up entirely. but that is a very good book
regardless of you attending Princeton university or not
yeah me neither you are the only one I remember talking to on it but I have been well and truly banned from this IP address for that forum now, which, which was as you might have guessed for being too polite and sensitive to delicate religious sensibilities
but no it's not my forum I just remembered it was one of the first I started talking math on, and it was a long road for someone like me being receptive to constructive criticism, especially from a kid a third my age which according to your profile at the time you were
i have a chronological disability that prevents me from accurately recalling exactly when this was, don't worry about it
well yeah it said you were 10, so it was a troubling thought to be getting advice from a ten year old at the time i think i was still holding on to some sort of hopes of a career in non stupidity related fields which was at some point abandoned
@TedShifrin thanks for that in bookmarking all of these under 3500, is there a 101 i should start with and find my way into four digits? what level of expertise is required for all of these is a more clear way of asking
Well, there are various math sources all over the web, including Khan Academy, etc. My particular course was intended for people seriously interested in mathematics (i.e., proofs as well as computations and applications). The students in there were about half first-year students who had taken BC AP calculus in high school and gotten the top score, about half second-year students who'd taken various first-year calculus paths in college.
long time ago tho even the credits have expired not the student debt though so i think they are trying to hint i should go back a start from first year and double said debt but im a terrible student it really wasn't worth while the first time round considering my rate of attendance then and how unlikely that would be different going back now
@BalarkaSen yeah from the number theory i got into in my most recent years it's bizarre how i almost became allergic to calculus i loved it back then and for some reason not quite so when i began focusing on prime numbers
What do you all think of this theorem: The number of ways to write $n$ as a sum of four squares is equal to $8$ times the sum of divisors of $n$ if $n$ is odd and $24$ times sum of odd divisors of $n$ if $n$ is even
A proof of this uses (basically) Fourier analysis
Even though it looks rather innocuous albeit surprising result in pure number theory
@BalarkaSen well because it was what Wikipedia deemed my interests to be categorized as i have simply told myself that is what i am studying, it really starting with me horsing around not even knowing what category of math you call it. actually, ill show you the exact subject you and i discussed on mmf that reminds me you were actually right, i don't know if i would have taken it well at the time tho
yeah looks like i deleted the stack exchange question on it anyway i had found a discrete Fourier transform for $\lfloor \frac{n}{m} \rfloor$ and you attempted to explain to me that is what it was that's all i remember lol @BalarkaSen
oh and when it comes to transcripts involving me on the internet, don't worry, the younger version of you most definitely will be seen in a positive light, and just contemplating all the possibilities of things said by someone as insane as me, agree that pulling up said past conversations isn't productive
absolutely me too but would we have it any other way? i mean i know im like a dog chasing a car as far as any real "purpose" in learning is concerned i think id be terrified if something didnt unfold into a myriad of new things I'm clueless about
@Daminark They key thing if I remember correctly was that if you look at the subgroup $\Gamma$ of $\text{PSL}_2(\Bbb Z)$ generated by (1, 2|0, 1) and (0, -1|1, 0), then any holomorphic function $f : \Bbb H^2 \to \Bbb C$ invariant under $\Gamma$ (in the sense that $f(z + 2) = f(z)$ and $f(-1/z) = z^{2k} f(z)$, $2k$ is called the weight) such that the Fourier expansion of $f$ at infinity and $-1$ having no constant coefficients is called a cusp form (on $\Bbb H^2/\Gamma$).
The $r_4(n)$ thing follows as an immediate corollary of the fact that the only weight $2$ cusp form is identically zero.
I can try to recall more if you're interested.
It's insightful to look at the picture of $\Bbb H^2/\Gamma$... it's like, take the line $\Re[z] = 1$, the semicircle $|z| = 1, z > 0$, and the line $\Re[z] = -1$. This gives a certain region in the upper half plane
Paste those two lines, and paste half of the semicircle (from -1 to i, and then from i to 1) to the other half by folding along i
Yup, that $E_4$ and $E_6$ generates the space of modular forms, that type of things
I think in general if you start thinking about modular forms as eigenfunctions of a Laplacian, the space generated by the Eisenstein series are orthogonal to the space of cusp forms - there's a general story I don't quite know
Cusp forms vanish at the cusp (those are the $-1$ and $\infty$ points in the quotient $\Bbb H^2/\Gamma$ picture I described above, where the hyperbolic metric gets coned off), whereas given any values on the cusps you can make a linear combination of Eisenstein series which takes those specific values on the cusps
So it sort of makes sense
Regarding that particular result, saying it's a weight 2 cusp form is like specifying a strong decay rate of the cusp form towards the cusp. Indeed, one basically argues like the maximum value theorem in complex analysis
@BalarkaSen no you didn't come across as pretentious at all, i can only imagine being so young and having the mind you have would have resulted in many accusing you of such, but really, my experience in life is diverse to say the least, and I've met know it all types that are in everyway detestable, you shouldn't be so hard on your character you are very humble considering your calibre
You probably don't realise how low the bar drops when it comes to integrity of character is concerned, trust me, you wouldn't have come as far as you clearly have if you were a know it all
it was actually the best thing for me to have met a 10 year old at the age of 30 that was well beyond what ill ever realistically become as far as math is concerned someone like you is going to be accused of arrogance simply because you intimidate many ignore the good majority of that mate |
This vignette focuses on MCMC diagnostic plots, in particular on diagnosing divergent transitions and on the
n_eff and
Rhat statistics that help you determine that the chains have mixed well. Plots of parameter estimates from MCMC draws are covered in the separate vignette
Plotting MCMC draws, and graphical posterior predictive model checking is covered in the Graphical posterior predictive checks vignette.
Note that most of these plots can also be browsed interactively using the shinystan package.
In addition to
bayesplot we’ll load the following packages:
Before we delve into the actual plotting we need to fit a model to have something to work with. In this vignette we’ll use the eight schools example, which is discussed in many places, including Rubin (1981), Gelman et al. (2013), and the RStan Getting Started wiki. This is a simple hierarchical meta-analysis model with data consisting of point estimates
y and standard errors
sigma from analyses of test prep programs in
J=8 schools. Ideally we would have the full data from each of the previous studies, but in this case we only have the these estimates.
The model is: \[ \begin{align*} y_j &\sim {\rm Normal}(\theta_j, \sigma_j), \quad j = 1,\dots,J \\ \theta_j &\sim {\rm Normal}(\mu, \tau), \quad j = 1, \dots, J \\ \mu &\sim {\rm Normal}(0, 10) \\ \tau &\sim {\rm half-Cauchy}(0, 10), \end{align*} \] with the normal distribution parameterized by the mean and standard deviation, not the variance or precision. In Stan code:
// Saved in 'schools_mod_cp.stan'data { int<lower=0> J; vector[J] y; vector<lower=0>[J] sigma;}parameters { real mu; real<lower=0> tau; vector[J] theta;}model { mu ~ normal(0, 10); tau ~ cauchy(0, 10); theta ~ normal(mu, tau); y ~ normal(theta, sigma);}
This parameterization of the model is referred to as the centered parameterization (CP). We’ll also fit the same statistical model but using the so-called non-centered parameterization (NCP), which replaces the vector \(\theta\) with a vector \(\eta\) of a priori
i.i.d. standard normal parameters and then constructs \(\theta\) deterministically from \(\eta\) by scaling by \(\tau\) and shifting by \(\mu\): \[\begin{align*}\theta_j &= \mu + \tau \,\eta_j, \quad j = 1,\dots,J \\\eta_j &\sim N(0,1), \quad j = 1,\dots,J.\end{align*}\] The Stan code for this model is:
// Saved in 'schools_mod_ncp.stan'data { int<lower=0> J; vector[J] y; vector<lower=0>[J] sigma;}parameters { real mu; real<lower=0> tau; vector[J] eta;}transformed parameters { vector[J] theta; theta = mu + tau * eta;}model { mu ~ normal(0, 10); tau ~ cauchy(0, 10); eta ~ normal(0, 1); // implies theta ~ normal(mu, tau) y ~ normal(theta, sigma);}
The centered and non-centered are two parameterizations of the same statistical model, but they have very different practical implications for MCMC. Using the
bayesplot diagnostic plots, we’ll see that, for this data, the NCP is required in order to properly explore the posterior distribution.
To fit both models we first translate the Stan code to C++ and compile it using the
stan_model function.
We then fit the model by calling Stan’s MCMC algorithm using the
sampling function (the increased
adapt_delta param is to make the sampler a bit more “careful” and avoid false positive divergences),
Warning: There were 110 divergent transitions after warmup. Increasing adapt_delta above 0.9 may help. Seehttp://mc-stan.org/misc/warnings.html#divergent-transitions-after-warmup
Warning: There were 1 chains where the estimated Bayesian Fraction of Missing Information was low. Seehttp://mc-stan.org/misc/warnings.html#bfmi-low
Warning: Examine the pairs() plot to diagnose sampling problems
Warning in throw_sampler_warnings(nfit): Bulk Effective Samples Size (ESS) is too low, indicating posterior means and medians may be unreliable.Running the chains for more iterations may help. Seehttp://mc-stan.org/misc/warnings.html#bulk-ess
Warning in throw_sampler_warnings(nfit): Tail Effective Samples Size (ESS) is too low, indicatingposterior variances and tail quantiles may be unreliable.Running the chains for more iterations may help. Seehttp://mc-stan.org/misc/warnings.html#tail-ess
and extract a
iterations x chains x parameters array of posterior draws with
as.array,
You may have noticed the warnings about divergent transitions for the centered parametrization fit. Those are serious business and in most cases indicate that something is wrong with the model and the results should not be trusted. For an explanation of these warnings see Divergent transitions after warmup. We’ll have a look at diagnosing the source of the divergences first and then dive into some diagnostics that should be checked even if there are no warnings from the sampler.
The No-U-Turn Sampler (NUTS, Hoffman and Gelman, 2014) is the variant of Hamiltonian Monte Carlo (HMC) used by Stan and the various R packages that depend on Stan for fitting Bayesian models. The
bayesplot package has special functions for visualizing some of the unique diagnostics permitted by HMC, and NUTS in particular. See Betancourt (2017), Betancourt and Girolami (2013), and Stan Development Team (2017) for more details on the concepts. Documentation:
help("MCMC-nuts")
The special
bayesplot functions for NUTS diagnostics are
bayesplot MCMC module:(matching pattern '_nuts_') mcmc_nuts_acceptance mcmc_nuts_divergence mcmc_nuts_energy mcmc_nuts_stepsize mcmc_nuts_treedepth
Those functions require more information than simply the posterior draws, in particular the log of the posterior density for each draw and some NUTS-specific diagnostic values may be needed. The
bayesplot package provides generic functions
log_posterior and
nuts_params for extracting this information from fitted model objects. Currently methods are provided for models fit using the
rstan, rstanarm and brms packages, although it is not difficult to define additional methods for the objects returned by other R packages. For the Stan models we fit above we can use the
log_posterior and
nuts_params methods for stanfit objects:
Iteration Value Chain1 1 -7.289305 12 2 -9.230824 13 3 -6.660970 14 4 -4.504500 15 5 -3.310762 16 6 -2.131661 1
Iteration Parameter Value Chain1 1 accept_stat__ 0.9772738 12 2 accept_stat__ 0.8734830 13 3 accept_stat__ 0.9566386 14 4 accept_stat__ 0.5040941 15 5 accept_stat__ 0.5647954 16 6 accept_stat__ 0.7317091 1
In addition to the NUTS-specific plotting functions, some of the general MCMC plotting functions demonstrated in the
Plotting MCMC draws vignette also take optional arguments that can be used to display important HMC/NUTS diagnostic information. We’ll see examples of this in the next section on divergent transitions.
When running the Stan models above, there were warnings about divergent transitions. Here we’ll look at diagnosing the source of divergences through visualizations.
The parallel coordinates plot (
mcmc_parcoord) is probably the first plot to have look at if you have no idea, where the divergences in your model might be coming from. This function works in general without including information about the divergences, but if the optional
np argument is used to pass NUTS parameter information, then divergences will be colored in the plot (by default in red).
The
mcmc_parcoord plot shows one line per iteration, connecting the parameter values at this iteration. This lets you see any global patterns in the divergences. Here, you may notice that divergences in the centered parameterization happen exclusively when
tau, the hierarchical standard deviation, goes near zero and the values of the
thetas are essentially fixed. This makes
tau immediately suspect. See Gabry et al. (2019) for another example of the parallel coordinates plot.
The
mcmc_pairs function can be also be used to look at multiple parameters at once, but unlike
mcmc_parcoord (which works well even when including several dozen parameters)
mcmc_pairs is more useful for up to ~8 parameters. It shows univariate histograms and bivariate scatter plots for selected parameters and is especially useful in identifying collinearity between variables (which manifests as narrow bivariate plots) as well as the presence of multiplicative non-identifiabilities (banana-like shapes).
Let’s look at how
tau interacts with other variables, using only one of the
thetas to keep the plot readable:
Note that each bivariate plot is present twice – by default each of those contain half of the chains, so you also get to see if the chains produced similar results (see the documentation for the
condition argument for other options). Here, the interaction of
tau and
theta[1] seems most interesting, as it concentrates the divergences into a tight region.
Further examples of pairs plots and instructions for using the various optional arguments to
mcmc_pairs are provided via
help("mcmc_pairs").
Using the
mcmc_scatter function (with optional argument
np) we can look at a single bivariate plot to investigate it more closely. For hierarchical models, a good place to start is to plot a “local” parameter (
theta[j]) against a “global” scale parameter on which it depends (
tau).
We will also use the
transformations argument to look at the log of
tau, as this is what Stan is doing under the hood for parameters like
tau that have a lower bound of zero. That is, even though the draws for
tau returned from Stan are all positive, the parameter space that the Markov chains actual explore is unconstrained. Transforming
tau is not strictly necessary for the plot (often the plot is still useful without it) but plotting in the unconstrained is often even more informative.
First the plot for the centered parameterization:
The shape of this bivariate distribution resembles a funnel (or tornado). This one in particular is essentially the same as an example referred to as Neal’s funnel (details in the Stan manual) and it is a clear indication that the Markov chains are struggling to explore the tip of the funnel, which is narrower than the rest of the space.
The main problem is that large steps are required to explore the less narrow regions efficiently, but those steps become too large for navigating the narrow region. The required step size is connected to the value of
tau. When
tau is large it allows for large variation in
theta (and requires large steps) while small
tau requires small steps in
theta.
The non-centered parameterization avoids this by sampling the
eta parameter which, unlike
theta, is
a priori independent of
tau. Then
theta is computed deterministically from the parameters
eta,
mu and
tau afterwards. Here’s the same plot as above, but with
eta[1] from non-centered parameterization instead of
theta[1] from the centered parameterization:
We can see that the funnel/tornado shape is replaced by a somewhat Gaussian blob/cloud and the divergences go away. Gabry et al. (2019) has further discussion of this example.
Ultimately we only care about
eta insofar as it enables the Markov chains to better explore the posterior, so let’s directly examine how much more exploration was possible after the reparameterization. For the non-centered parameterization we can make the same scatterplot but use the values of
theta[1] = mu + eta[1] * tau instead of
eta[1]. Below is a side by side comparison with the scatterplot of
theta[1] vs
log(tau) from the centered parameterization that we made above. We will also force the plots to have the same \(y\)-axis limits, which will make the most important difference much more apparent:
# A function we'll use several times to plot comparisons of the centered # parameterization (cp) and the non-centered parameterization (ncp). See# help("bayesplot_grid") for details on the bayesplot_grid function used here.compare_cp_ncp <- function(cp_plot, ncp_plot, ncol = 2, ...) { bayesplot_grid( cp_plot, ncp_plot, grid_args = list(ncol = ncol), subtitles = c("Centered parameterization", "Non-centered parameterization"), ... )}scatter_theta_ncp <- mcmc_scatter( posterior_ncp, pars = c("theta[1]", "tau"), transform = list(tau = "log"), np = np_ncp, size = 1)compare_cp_ncp(scatter_theta_cp, scatter_theta_ncp, ylim = c(-8, 4))
Once we transform the
eta values into
theta values we actually see an even more pronounced funnel/tornado shape than we have with the centered parameterization. But this is precisely what we want! The non-centered parameterization allowed us to obtain draws from the funnel distribution without having to directly navigate the curvature of the funnel. With the centered parameterization the chains never could make it into the neck of funnel and we see a clustering of divergences and no draws in the tail of the distribution.
Another useful diagnostic plot is the trace plot, which is a time series plot of the Markov chains. That is, a trace plot shows the evolution of parameter vector over the iterations of one or many Markov chains. The
np argument to the
mcmc_trace function can be used to add a rug plot of the divergences to a trace plot of parameter draws. Typically we can see that at least one of the chains is getting stuck wherever there is a cluster of many red marks.
Here is the trace plot for the
tau parameter from the centered parameterization:
The first thing to note is that all chains seem to be exploring the same region of parameter values, which is a good sign. But the plot is too crowded to help us diagnose divergences. We may however zoom in to investigate, using the
window argument: |
Try the thrifty mobile template, too. Physicists Roger Cohen, William Happer, and Richard Lindzen wrote a calm and sensible text about weather extremes in the Wall Street Journal:
But the bulk of the article is dedicated to clarifications of the fact that there's no unusual trend in the weather extremes.
Tornadoes, droughts, hurricanes, and wildfires are discussed as the most important examples of the fact that these largely unpredictable events are doing what they have always been doing. They randomly appear and disappear.
Despite the fact that these topics have been discussed for years and one could say decades, I think that most people – and maybe even most people who have science degrees – remain remarkably ignorant about the basic logic of all these things, especially the statistical properties of weather events. In particular, people don't distinguish "red noise" (or similar "pink noise") from the "white noise" (or quantities well approximated by it).
The global mean temperature has been changing during the glaciation cycles; the maximum and the minimum differ by nearly 10 °C which is significant. A cool world or a warm world persists for thousands of years. Smaller but qualitatively similar changes may be seen at shorter timescales such as several centuries (compare the medieval warm period and the little ice age). There are numerous drivers that influence the temperatures and allow them to remain "elevated" or "depressed". So it is surely incorrect to say that the annual global mean temperatures for each year are independent from those in the previous year or two. That would be needed for us to say that they behave as the "white noise".
Still, the white noise is the only "type of random data" that most people are familiar with. The white noise would look like this:
White noise. Good enough for tornadoes, hurricanes, or even earthquakes as functions of the year.
At every "moment", whatever its tiny length is, the quantity is distributed according to some fixed distribution and it is independent from the value at any previous moment. For that reason, the "jump" of the quantity from one moment to another is almost always infinite. In other words, the huge frequencies are heavily represented in the Fourier decomposition of the white noise.
We could say that tornadoes, hurricanes, wildfires, and other things are more or less described by the white noise. Every year is different, the previous year is forgotten and the "qualitative" weather events and natural catastrophes arrive randomly according to this most widely understood type of a random distribution. Each year, you throw dice and the number you get (reparameterized by a fixed function) gives you an estimate how many natural catastrophes of a certain type you get during the year.
Everything we know indicates that extreme weather events that depend on dramatic changes of temperature, pressure, or humidity as a function of space or time are nicely described as the white noise. That's why we don't observe any trend in those graphs. The graphs are highly variable but each year, the previous year (or two) are forgotten. It's always more sensible to build your expectation upon the ensemble of many (or thousands of) previous years, regardless of the fact that most of them are very far.
The climate alarmists often say that the number of extreme events such as tornadoes and hurricanes increases and they like to pretend that such claims are scientifically justified and that such claims are a part of the "greenhouse effect science" package. But these propositions are complete lies that are unfortunately often bought by a large fraction of the public. Once you look at extreme things driven by rapid change or large gradients such as tornadoes, there exists no empirical reason to think that anything is changing about them; and there exists no theoretical explanation why it should be happening. According to everything that science may say, such changes are simply not happening. Tornadoes are as random and as frequent as they were decades or centuries ago.
On the other hand, the temperature is more accurately modeled by the "red noise" or at least some kind of "pink noise" which is something in between the red noise and the white noise. While the "white noise" is a collection of random numbers that are independent from the previous ones, "red noise" is what you get by integrating (or partial summing of) the "white noise". In other words, "red noise" is the graph of a random walk. To emphasize the relationship between the two, let me also say that the derivative of the red noise is the white noise.
Graphs of red noise. Good enough model for temperatures or prices of stocks.
Each new year or new moment or new datapoint isn't independent from the previous one. Instead, it stands on the "shoulders" of the previous one and adds a random deviation away from it. Consequently, as you can see on the picture above, the graphs of red noise are much more continuous than the graphs of white noise we started with. In other words, the Fourier decomposition of the red noise suppresses the contribution from the very high-frequency oscillations and it adds strong low-frequency wiggles.
(If the graphs depict pressure as a function of time, you will hear a lower-pitch noisy sound than in the case of the white noise. If the graphs depict the electromagnetic field, the resulting noisy "light" will look redder because the lower-frequency red light will be overrepresented relatively to the white noise which is considered "color neutral" here; that's why it's called "red noise".)
These low-frequency wiggles inevitably look like "trends" over arbitrarily long timeframes.
Another concept in physics that is almost synonymous with "red noise" or "random walk" is the Brownian motion. Albert Einstein's 1905 paper – and an independent Polish mountaineer Marian Smoluchowski's 1906 paper – derived some basic mathematical patterns about the Brownian motion. The typical distance \(s(t)\) that the particle moves after time \(t\) goes like\[
\overline{s(t)} \sim C\cdot \sqrt{t}.
\] So it increases with \(t\) but it increases less quickly than the direct proportionality would predict. The graph of \(s(t)\) may be drawn and you may calculate the apparent trend between \(t=0\) and \(t=t\):\[
\overline {|v(t)|} = \frac{s(t)}{t} \sim \frac{C}{\sqrt{t}}.
\] You see that the "apparent velocity" or "apparent rate" goes to zero as you increase the period of time over which you look at the quantity. But it goes to zero very slowly. You need to increase the timescale 9 times for the apparent rate to decrease 3 times.
The global mean temperature is at least morally closer to the red noise, much like the prices of stocks. The price of a stock is pretty random and if you look at its evolution over a few years, it's often twice as high at one moment than another moment. But it's very unlikely for the stock to jump 100 percent or drop 50 percent overnight. Instead, it's the change of the price from the previous day to the new day that is random so to see large enough changes, you usually have to wait for a longer time. (That wasn't the case of the white noise.)
For this reason, it's totally normal for the temperature in a year (or few years) after a warm year to be comparably warm, too. The year 1998 was warm and due to the basic properties of the red noise or pink noise, it's very unlikely for nearby years to among the coolest ones. That's simply prohibited by the continuity of the temperature graph. And the temperature graph is continuous (although not smooth) because the Earth has a finite heat capacity and an infinitely abrupt change of the temperature would require an infinite amount of energy.
That's why all the comments of the type "most of the years in the last decade are among the top 10 warmest years" are not surprising at all. That's exactly what the natural theory of the noisy graphs – even without any human or other "special" influences – predicts. It predicts such a clustering of warm years simply because the temperature is a continuous function of time; the noise apparent in the temperature graphs is closer to the red noise than the white noise.
Being excited about this clustering is exactly as silly as saying "Look, 4 out of the history's 5 best days for the Apple stock price occurred in April 2012 (the only exception is yesterday, isn't it a stunning proof of something?". No, it's not shocking at all. There simply had to be a maximum near $640 and the moments near the maximum are likely to have similarly high values. (I am not claiming that the Apple stock won't get above $632 in the future; I am not claimiing that it will, either. Instead, we just talk about the known history.)
On the other hand, there are many other quantities that depend on time and that have no reason to be continuous – e.g. the number of tornadoes in the year Y – and those things don't have to remember and don't remember the previous year or two. So you will find no trends in them.
Things related to the drought are somewhere in between. The precipitation itself is more or less "white noise" (especially above the ocean) because whether it rains today is mostly independent on whether it was raining a month or a year ago. However, the soil and other entities on the surface of the globe have some inertia. They store water or they may run out of water (and this also influences the evaporation rate and precipitation) and if they do so, chances are that such conditions will continue for the following year or two or three, too. This inertia means that the dry years will tend to be clumped more than what you would expect from the white noise, i.e. from the complete independence. But the inertia diminishes if you look at timescales longer than a decade. If there are some unlucky coincidences that create drought in some region, chances are high that in 10 years, the situation returns to the "normal" (calculable from recent centuries, for example). That was the case of the Dust Bowl in the 1930s.
The actual things that have changed are things such as the global mean temperature that are expected to change and "oscillate around the new values" because their behavior is closer to the red noise (random walk), much like the behavior of the stock prices or the position of a particle undergoing the Brownian motion. But the global mean temperature – which may only be calculated with the desired accuracy if you have accurate thermometers everywhere and at all times and you're extremely careful about the statistical procedures used to compute the average – has only changed by something like 0.7 °C per century, something a human would probably not be able to detect at all even if the surrounding temperature was stable (the real one has additional oscillations by dozens of degrees at all time so the human detection of the underlying "global trend" is completely impossible).
If one focuses on the precise values of the temperature, it is totally inevitable – a trivial consequence of the red noise – that he gets some "apparent trends". That's what the red noise always does. But the actual temperature changes coming from this stuff are tiny and they affect neither humans nor tornadoes. All things such as tornadoes depend on properties of the climate system that are not changing. And if they're changing, the changes are negligible.
You know, one could naturally expect that there is some residual dependence. For example, the number of tornadoes could be proportional to a power of the absolute temperature, \(T^k\). I am talking about the absolute temperature because it is the more natural quantity describing temperature in physics (for example, the volume of the ideal gas at fixed pressure is proportional to the absolute temperature).
Now, the global mean temperature has changed from 288 kelvins to 288.7 kelvins in a century (or some numbers of this kind, no one knows because there is not even any precise universal definition of the "global mean temperature"). That's a change by 0.25 percent. Consequently, \(T^k\) has changed by \(k\) times 0.25 percent per century. Unless \(k\) is much greater than one, and it would be unnatural if it were much greater (so it is unlikely), we see that change in the predicted number of tornadoes per year will be smaller than one percent or so (relatively to the average predicted for the year 1912). That's obviously undetectable even if such an effect existed. And by the way, if it existed, it's likely that a warmer world will lead to smaller, and not greater, amount of similar activity because this activity depends on temperature gradients and they're predicted to decrease, and not increase, in a warming world (because the equator-to-pole temperature difference drops as the poles are claimed to be warming faster).
There's an immense layer of irrationality underlying the whole climate panic. Every rational person knows that a 0.7 °C increase of some carefully-averaged temperature is undetectable by humans and irrelevant for all other phenomena that matter. It's just a tiny change of the temperature. Nature routinely deals with much larger and harsher changes every day – larger changes per day than the centennial change someone claims to be dangerous. For the same reason, things like tornadoes and hurricanes have no good reason to behave as anything else than the "white noise", so there won't be any trend and there isn't any statistically significant trend in any of these quantities. So it's trivial to see that given the estimates for the rates that we have measured, there can't exist absolutely any reason for concern when it comes to the climate change.
Saying that people talking about a dangerous climate change are good scientists is preposterous beyond imagination. They're dimwits incapable of understanding (or unwilling to understand) basic concepts of mathematics, statistics, and natural sciences.
And that's the memo. |
Equilibrium of a droplet suspended in an electric field
A droplet suspended in a fluid subjected to a uniform electric field deforms due to the competing effect of electrical forces and surface tension. If the electrification level is moderate an equilibrium shape is reached. It was observed experimentally that droplets deforms as prolate or oblate spheroids (i.e. larger elongation aligned with the external electric field or viceversa). The analytical analysis of O’Konski & Thacher (1953) unexpectedly predicted prolate forms whereas the experiments showed oblate ones (and viceversa).
It was the genius of Geoffrey Taylor (1966) who shed light on the problem. The work of O’Konski & Thacher assumed that both fluids (inner and outer) were perfect dielectrics, given the low conductivity of the fluids involved. Taylor realized that the conductivity could be very low but it was not zero, so that the charges could migrate through the “leaky” media, thus accumulating at the fluid interface and altering radically the pattern of electrical forces.
Moreover, approximating the electrostatic Maxwell equations with the Taylor–Melcher leaky dielectric model and assuming small deformation Taylor predicted recirculating velocities and provided analytical expressions for the radial and azimuthal velocities as functions of the dimensionless radius r and the azimuthal coordinate \theta. For r < 1 this gives, \displaystyle v_r = Ar(1-r^2)(3\cos^2 \theta -1) \quad \mathrm{and} \quad v_\theta = 3Ar/2 \left( 1-5r^2/3 \right)\cos 2\theta and for r \ge 1, \displaystyle v_r = A(r^{-4}-r^{-2})(3\cos^2 \theta -1) \quad \mathrm{and} \quad v_\theta = -Ar^{-4} \sin 2\theta with \displaystyle A = -\frac{9}{10} \frac{E_\infty^2 R_d \varepsilon_o}{ \mu_o} \frac{1}{1+\lambda}\frac{R-Q}{(R+2)^2}, where R, Q and \lambda are the ratio of inner to outer conductivity, permittivity and viscosity, respectively. E_\infty is the imposed electrid field, R_d the droplet radius, \varepsilon_o is the outer permittivity and \mu_2 is the outer viscosity. The electrical forces induces recirculations in both (viscous) fluids.
This test case is also discussed in Lopez-Herrera et al, 2011.
The problem is assumed to be axisymmetric.
#include "axi.h"#include "navier-stokes/centered.h"#include "ehd/implicit.h"#include "ehd/stress.h"#include "vof.h"#include "tension.h"
We need to track the interface with the volume fraction field
f. The viscosity is constant but the coefficients will vary due to the axisymmetric metric terms.
scalar f[], * interfaces = {f};face vector muv[];
The maximum level of resolution, LEVEL, will be varied from 8 to 10.
int LEVEL = 8;#define Ef 1.34 // External electric field#define R0 0.1 // Radius of the droplet #define F 50. #define R 5.1 // Conductivity ratio#define Q 10.0 // permittivity ratio#define CMU 0.1 // Outer viscosity #define theta (M_PI/4.)#define LAM 1. // Viscosity ratio#define VC (sq(Ef)*R0/CMU) // characteristic velocity#define A (-9./10.*(R - Q)/sq(R + 2.)/(1. + LAM))
F is the conductivity of the outer medium. F has no influence on the steady solution but decreases the characteristic electrical relaxation time and consequently the electrical transient.
#define cond(T) (F*((1. - (T)) + R*(T)))#define perm(T) ((1. - (T)) + Q*(T))
The electric potential is linear.
phi[top] = dirichlet(Ef*x);phi[left] = dirichlet(Ef*x);phi[right] = dirichlet(Ef*x);
We make sure there is no flow through the boundaries, otherwise the compatibility condition for the Poisson equation can be violated.
uf.n[left] = 0.;uf.n[right] = 0.;uf.n[top] = 0.;uf.n[bottom] = 0.;
The domain spans [0:2]. We will compute only a quarter of the droplet, making use of axisymmetry and right-left symmetry. The surface tension coefficient is unity. The viscosity coefficients are variable.
We initialize the volume fraction field corresponding to a circular interface of radius R0.
fraction (f, sq(R0) - sq(x) - sq(y));
We initialize the electrical potential.
Permittivity and electrical conductivity are face values and also incorporate the metric factors. The viscosity is constant but the viscosity coefficients need to incorporate the metric factors.
event properties (i++){ foreach_face() { double ff = (f[] + f[-1])/2.; epsilon.x[] = perm(ff)*fm.x[]; K.x[] = cond(ff)*fm.x[]; muv.x[] = CMU*fm.x[]; } boundary ((scalar *){epsilon, K, muv});}
Convergence
We store the horizontal component of the velocity to check its convergence with time.
We monitor the variation in the horizontal component of the velocity and the convergence of the multigrid solvers every 20 timesteps.
If the change is small enough (i.e. the solution has converged for this level of refinement), we increase the level of refinement. If the simulation has converged and the level of refinement is 10, we stop the simulation.
if (i > 0 && du < 1e-5) return (LEVEL++ == 10); /* stop */}
Results
At the end of the simulation we create two files:
log (standard error) will contain the dimensionless radial and azimuthal velocities and their theoretical values as functions of the dimensionless radial coordinate along the line \theta= 45^o. vector.svg displays the velocity field, interface position and isopotential lines, as displayed by gfsview-batch.
event result (t = end) { double h = 0.35*L0/(2*99); for (int i = 1; i <= 100; i++) { double x = i*h, y = i*h, r = sqrt(sq(x) + sq(y))/R0; double ux = interpolate (u.x, x, y)/VC; // dimensionless velocities double uy = interpolate (u.y, x, y)/VC; double vrt, vtt; // theoretical radial and azimuthal velocities; if (r < 1.) { vrt = A*r*(1. - sq(r))*(3.*sq(sin(theta)) - 1.); vtt = 3*A/2*r*(1. - 5./3.*sq(r))*sin(2.*theta); } else { vrt = A/sq(r)*(1/sq(r) - 1.)*(3.*sq(sin(theta)) - 1.); vtt = - A*1./sq(sq(r))*sin(2.*theta); } fprintf (stderr, "%g %g %g %g %g\n", r, (ux*x + uy*y)/(R0*r), vrt, (-uy*x + ux*y)/(R0*r), vtt); } FILE * fp = popen ("gfsview-batch2D taylor.gfv", "w"); output_gfs (fp); fprintf (fp, "Save vectors.svg { format = SVG }\n"); pclose (fp);}
The mesh is adapted according to interpolation errors on the volume fraction, charge density and velocity fields.
event adapt (i += 20) { adapt_wavelet ({f, rhoe, u.x, u.y}, (double[]){1e-3, 1, 2e-4, 2e-4}, maxlevel = LEVEL);}
set xlabel 'r'set ylabel 'v'plot 'log' u 1:2 notitle, 'log' u 1:3 w l t "v_r",\ 'log' u 1:4 notitle, 'log' u 1:5 w l t "v_{/Symbol q}"
Bibliography
[konski1953]
Chester T O’Konski and Henry C Thacher Jr. The distortion of aerosol droplets by an electric field.
[taylor1966]
Geoffrey Ingram Taylor. Studies in electrohydrodynamics. i. the circulation produced in a drop by an electric field. |
Does any know how I can combine itemize and aligned (with argument [t]) -- or another multi-line mathematical environment -- in order to obtain a multi-line equation whose first line is aligned to the left and the remaining lines are aligned to the right?
I tried to illustrate what I would like to obtain in the picture below.
My current latex-code is the following:
\documentclass[letterpaper]{article}\usepackage{amsthm}\usepackage{mathtools}\usepackage{bm}\begin{document}$\bm{f}^{\cup}(B^1,B^2), \bm{f}^{\cap}(B^1,B^2) \in \mathcal{B}(\Sigma_1\cup\Sigma_2, w\cdot w^{\prime})$, respectively, where\begin{itemize}\small \item $\begin{aligned}[t] \ell(\bm{f}^{\cup}(B^1,B^2)) = \{w^{\prime}\cdot(\mathfrak{p}-1) + \mathfrak{p}^{\prime} \colon \mathfrak{p}\in\ell(B_i),\\\mathfrak{p}^{\prime}\in\ell(B_i)\}\text{;} \end{aligned}$ \item $r(B_i^{*}) = \{w^{\prime}\cdot(\mathfrak{q}-1) + \mathfrak{q}^{\prime} \colon \mathfrak{q} \in r(B_i), \mahfrak{q}^{\prime}\in r(B_i)\}$ ;\end{itemize}\end{document}
which results in something like the picture below.
Thanks!! |
Given:
the receiving antenna's gain in dBi, $G_\mathrm{dBi}$ the power received by the antenna in dBm, $P_\mathrm{dBm}$ the frequency in Hz, $f$ the impedance of free space in ohms, $Z_0$ the speed of light in meters per second, $c$
then the field strength in V/m is:
$$ \begin{align*}V &=\:\sqrt{ 0.004\, \pi\, Z_0 }\:f/c\:10^{(P_\mathrm{dBm}-G_\mathrm{dBi}) / 20}\\&\approx 7.258\times 10^{-9}\: f\:10^{(P_\mathrm{dBm}-G_\mathrm{dBi}) / 20}\end{align*}$$
If the frequency is in MHz and the result is in µV/m then the constant becomes more manageable:
$$ V_\mathrm{\mu V/m}\approx 7258\: f_\mathrm{MHz}\: 10^{(P_\mathrm{dBm}-G_\mathrm{dBi}) / 20}$$
Note that the receiving antenna must match the polarization of the voltage field being measured. In situations where the polarization isn't known, an isotropic receiver can be used, which is usually an orthogonal arrangement of short dipoles in 3 axes. Such an antenna can measure a voltage gradient in any direction.
I find it's easier to understand why this is true by thinking about power density rather than field strength.
If you have a 1 watt isotropic antenna at the center of a sphere, the entire 1 watt of that transmitted power must pass through the sphere. The power density is then the transmitted power, divided by the surface area of the sphere. The result is a number in units of watts per square meter.
The receiving antenna, assuming it is identically polarized, is then a "net" for this energy flux. The size of the net is the antenna aperture, which can be calculated from its gain.
$$ A = {\lambda^2 \over 4 \pi} 10^{G_\text{dBi}/10} $$
For more detail, see What does the Friis transmission equation represent and how is it derived?
In the far field, power density and the field strength are related by the free space impedance, which is approximately 377Ω. Converting between power density and field strength is very much like Ohm's law. This can be derived from Poynting's theorem:
$$ \text{power density} = {\text{field strength}^2 \over \text{free space impedance}} $$
Or solved for field strength:
$$ \text{field strength} = \sqrt{\text{power density} \times \text{free space impedance}} $$
Thus, if you can measure the power density at some point, say by measuring the feedpoint power from an antenna of known gain, then you can calculate the field strength.
For example, say the antenna has a gain of 3 dBi at a wavelength of 2 meters, and you measure a power at the feedpoint of 0.1μW.
The aperture is:
$$ {(2\:\mathrm{m})^2 \over 4 \pi} 10^{3\:\mathrm{dBi}/10} = 0.635\:\mathrm{m}^2 $$
The power density is then:
$$ {0.1\:\mathrm{\mu W} \over 0.635\:\mathrm{m}^2} = 0.157\:\mathrm{\mu W}/\mathrm{m}^2 $$
And the corresponding field strength:
$$ \sqrt{0.157\:\mathrm{\mu W/m^2} \times 377\:\Omega} = 7705\:\mathrm{\mu V/m}$$
Put these three equations together and simplify to get the equation at the top.
For more detail, Semtech has a good paper on field strength and many related topics. |
Suppose we have the initial-value problem on $(0,L)$:
$$ \frac{d u(x)}{d x} = f(x) u(x),\, \qquad x\in\Omega,\,~~ u(0) = u_0, $$
I am reading a claim that says if we multiply the ODE by $u$ and integrate over $(0,L)$, we have
$$ \frac{1}{2}u^2(L) - \frac{1}{2} u^2_0 = \int_{0}^{L} f(x)u^2(x) \,dx $$
"from which the $L^\infty$-stability of the solution follows." I agree that the equation is correct, but:
Why does this guarantee stability? What exactly is meant by $L^\infty$-stability? I interpret stability in the context that the numerical solution will remain bounded as the the step size is reduced, but here, we do not have a discretization yet...
This discussion is given in the context of discontinuous Galerkin methods. |
A
constellation diagram is a diagram that shows how the information in a digital modulation scheme such as phase shift keying (PSK) is represented in a modulated carrier. This article is a bit of an experiment; instead of introducing the constellation diagram by simply presenting it, I’m going to try to have it emerge from the formulas of PSK modulation.
I’ve already mentioned the general waveform for
Multiple Phase-Shift Keying ( MPSK) at the end of the article mentioned above, as
\[s_m(t)=A\cos\left(2\pi f_c t+\frac{\pi(2m+1)}{M}\right),\]
for \(m=0,\ldots,M-1\). In this equation, \(A\) is the amplitude of the carrier, \(f_c\) is the frequency of the carrier, and \(M\) is the number of phase offsets. This formula represents a carrier wave of which the phase is suddenly shifted for each new symbol (if that symbol is different from the previous one, of course). This is a
digital modulation scheme, because the phase is changed in fixed steps instead of continuously as for analog modulation techniques.
To get to the constellation diagram, we first switch to the
analytic form of the waveform (see What is an Analytic Signal? for details), as follows.
\[\tilde{s}_m(t)=s_m(t)+i\mathcal{H}(s_m(t))\]
This results in
\[\tilde{s}_m(t)=A\cos\left(2\pi f_c t+\frac{\pi(2m+1)}{M}\right)+iA\sin\left(2\pi f_c t+\frac{\pi(2m+1)}{M}\right)\]
when applying the Hilbert transform, and finally in
\[\tilde{s}_m(t)=Ae^{\textstyle i\left(2\pi f_ct+\frac{\pi(2m+1)}{M}\right)},\]
by applying Euler’s formula. Because the waveform for MPSK is simply a cosine with a phase offset, the calculation of the analytic signal is the same as in the mentioned article on analytic signals.
The common part for the expression for each of the \(\tilde{s}_m(t)\) with \(m=0,\ldots,M-1\) is a vector of length \(A\) rotating with a constant frequency of \(f_c\), given by
\[Ae^{i2\pi f_ct}=A\cos(2\pi f_ct)+iA\sin(2\pi f_ct).\]
If you plot this on the complex plane, then you get a circle. This is not a very informative thing to do, but we
are getting closer to the constellation diagram.
The key insight to get to the constellation diagram is that the analytic signal allows putting the carrier frequency \(f_c\) to zero. Of course, to do that, you have to be willing to use complex signals throughout your modem instead of real ones. This is often referred to as
inphase ( I) and quadrature ( Q) processing, where I represents the real axis of the complex plane and Q represents the imaginary axis.
For the specific example of MPSK, setting \(f_c=0\) results in the waveform
\[\tilde{s}_m(t)=Ae^{\textstyle i\frac{\pi(2m+1)}{M}}=A\cos\left(\frac{\pi(2m+1)}{M}\right)+iA\sin\left(\frac{\pi(2m+1)}{M}\right),\]
still for \(m=0,\ldots,M-1\).
For the even more specific example of MSPK with \(M=4\), and for \(A=1\), this leads to the constellation diagram of Figure 1. Because \(A=1\), the coordinates of the top-right point, for example, are \((\sqrt{2}/2,\sqrt{2}/2)\).
For
binary phase shift keying (BPSK), so for \(M=2\), this leads to the diagram of Figure 2.
In application such as
cable or DSL modems for broadband internet, more elaborate constellations such as Quadrature Amplitude Modulation ( QAM) are used. In QAM, both the phase and the amplitude of the signal are adapted in order to be able to send more bits per symbol. An example is 16-QAM (4 bits per symbol), as shown in Figure 3. Cable modems typically use 64-QAM or 256-QAM, and, for DSL, constellations as large as 32768-QAM (15 bits per symbol) are used routinely.
In conclusion, a constellation diagram is a way to represent the waveforms of a digital modulation scheme in a concise way. Of course, the examples shown above are the perfect constellations. In practice, the constellation diagram is used to detect problems with the received signal, such as excessive noise or timing problems. |
Let $G$ be a group and $H$ be a subgroup of $G$. Let $A$ be the set of all left cosets of $H$ in $G$. We know that $G$ can acts on $A$ by $$g\cdot xH=gxH.$$
For any $g\in G$, define that $\sigma_{g}:A\to A, \sigma_g(xH)=g\cdot xH$. Then $\sigma_g$ is a permutation on $A$.
Let $S_A$ be the set of all permutation on $A$. Define the permutation representation $\theta:G\to S_A$. Then $\theta$ is a homomorphism and $$\begin{align} \ker{\theta}&=\text{core }H\\ &=\{g\in G\mid g\in aHa^{-1}\text{ for all }a\in G\}\\ &=\bigcap_{a\in G}aHa^{-1}. \end{align}$$
The dual notion of the $\text{core}$ is the normal closure (or conjugate closure) of a subset $X$ of $G$. Which is defined by $$\begin{align} \bar{X}&=\langle \{gxg^{-1}\mid g\in G, x\in X\}\rangle\\ &=\bigcap_{X\subseteq N\lhd G}N. \end{align}$$
Is there a group action whose kernel of the permutation representation is the normal closure $\bar{X}$ of $X$?
Thanks for any kind of tips. |
The number of storms in the upcoming rainy season is Poisson distributed but with a parameter value that is uniformly distributed between (0,5). That is Λ is uniformly distributed over (0,5), and given Λ = λ, the number of storms is Poisson with mean λ. Find the probability there are at least three storms this season
Ok, I feel like I can get started with this problem, but I'm having a really difficult time wrapping my head around the idea of conditioning on a random variable. and I would appreciate if somebody could explain to me, in simple terms, what is going on when we condition on a random variable. I haven't been able to find a good explanation of this concept and I've seen this problem done both by taking the expectation of $\Lambda$ and placing it into the PMF of the Poisson distribution and by multiplying the Poisson the PDF of $\Lambda$, $\frac{1}{5}$ over the interval [0,5] and integrating the product from 0 to 5 w.r.t. $\lambda$.
Anyway, here goes nothing
Obviously, the problem starts out quite simple:
Let $X =$ number of storms in the season
Let $\Lambda =$ the rate of the Poisson distribution, which varies as a uniform r.v. over [0,5].
We want the following:
$Pr(X \geq 3|\Lambda = \lambda) = 1- Pr(X < 3|\Lambda = \lambda) $
And then of course we can count down to zero storms in a season in a summation. But that's where I get lost. I don't know how to handle the random parameter and more-so how to handle these type of problems in general and I'd appreciate some feedback. Please note that I am not looking to solve this problem using the definition of conditional probability. I am trying to solve this by conditioning on $\Lambda$, that is, treating it as a constant.
Please note that this is NOT homework but rather for independent study. |
I use this decomposition all the time, but I have never read a paper solely devoted to the topic. From my experience a complete characterization of the constraints on $t_{i_{1}, t_{2},..t_{n}}$ is tricky, and so if you want to be sure $\rho$ is physical you should calculate the density matrix and its eigenvalues.
However, there are a lot of necessary conditions that have a useful form in this decomposition. For example, for a positive unit-trace Hermitian operator $\rho$ is follows that
$|t_{i_{1}, i_{2},.. i_{n}}| \leq 1$
$tr ( \rho^{2} ) =\frac{1}{2^{n}} \sum_{i_{1}, i_{2},.. i_{n}} t_{i_{1}, i_{2},.. i_{n}}^{2} \leq 1 $
The above condition tells us that if we think of $t$ as a vector in a real vector space, then the physical states live within the unit sphere. This is a bit like the Bloch sphere for 1 qubit but for many qubits we have some other constraints that take the form of hyperplanes. For every $\vert \psi \rangle$ expressed in the same form$\vert \psi \rangle \langle \psi \vert = \frac{1}{2^{n}} \sum_{i_{1},i_{2},... i_{n}} Q_{i_{1},i_{2},... i_{n}}\sigma_{i1} \otimes \sigma_{i2}... \sigma_{in}$ we require that
$\langle \psi \vert \rho \vert \psi \rangle \geq 0 $ and so$\sum Q_{i_{1},i_{2},... i_{n}}t_{i_{1},i_{2},... i_{n}}\geq 0$which defines a hyperplane.
The problem is you have a hyperplane for every $\psi$ so that requiring $t$ to satisfy every inequality one of the infinite hyperplanes is impossible to check by brute force. If you want sufficient conditions for positivity of $\rho$ I suspect you have to calculate eigenvalues.This post has been migrated from (A51.SE) |
Fujimura's problem
Let [math]\overline{c}^\mu_n[/math] the largest subset of the triangular grid
[math]\Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c=n \}[/math]
which contains no equilateral triangles [math](a+r,b,c), (a,b+r,c), (a,b,c+r)[/math] with [math]r \gt 0[/math]; call such sets
triangle-free. (It is an interesting variant to also allow negative r, thus allowing "upside-down" triangles, but this does not seem to be as closely connected to DHJ(3).) Fujimura's problem is to compute [math]\overline{c}^\mu_n[/math]. This quantity is relevant to a certain hyper-optimistic conjecture.
n 0 1 2 3 4 5 [math]\overline{c}^\mu_n[/math] 1 2 4 6 9 12 Contents n=0
[math]\overline{c}^\mu_0 = 1[/math]:
This is clear.
n=1
[math]\overline{c}^\mu_1 = 2[/math]:
This is clear.
n=2
[math]\overline{c}^\mu_2 = 4[/math]:
This is clear (e.g. remove (0,2,0) and (1,0,1) from [math]\Delta_2[/math]).
n=3 [math]\overline{c}^\mu_3 = 6[/math]:
For the lower bound, delete (0,3,0), (0,2,1), (2,1,0), (1,0,2) from [math]\Delta_3[/math].
For the upper bound: observe that with only three removals each of these (non-overlapping) triangles must have one removal:
set A: (0,3,0) (0,2,1) (1,2,0) set B: (0,1,2) (0,0,3) (1,0,2) set C: (2,1,0) (2,0,1) (3,0,0)
Consider choices from set A:
(0,3,0) leaves triangle (0,2,1) (1,2,0) (1,1,1) (0,2,1) forces a second removal at (2,1,0) [otherwise there is triangle at (1,2,0) (1,1,1) (2,1,0)] but then none of the choices for third removal work (1,2,0) is symmetrical with (0,2,1) n=4 [math]\overline{c}^\mu_4=9[/math]:
The set of all [math](a,b,c)[/math] in [math]\Delta_4[/math] with exactly one of a,b,c =0, has 9 elements and is triangle-free. (Note that it does contain the equilateral triangle (2,2,0),(2,0,2),(0,2,2), so would not qualify for the generalised version of Fujimura's problem in which [math]r[/math] is allowed to be negative.)
Let [math]S\subset \Delta_4[/math] be a set without equilateral triangles. If [math](0,0,4)\in S[/math], there can only be one of [math](0,x,4-x)[/math] and [math](x,0,4-x)[/math] in S for [math]x=1,2,3,4[/math]. Thus there can only be 5 elements in S with [math]a=0[/math] or [math]b=0[/math]. The set of elements with [math]a,b\gt0[/math] is isomorphic to [math]\Delta_2[/math], so S can at most have 4 elements in this set. So [math]|S|\leq 4+5=9[/math]. Similar if S contain (0,4,0) or (4,0,0). So if [math]|S|\gt9[/math] S doesn’t contain any of these. Also, S can’t contain all of [math](0,1,3), (0,3,1), (2,1,1)[/math]. Similar for [math](3,0,1), (1,0,3),(1,2,1)[/math] and [math](1,3,0), (3,1,0), (1,1,2)[/math]. So now we have found 6 elements not in S, but [math]|\Delta_4|=15[/math], so [math]S\leq 15-6=9[/math].
Remark: curiously, the best constructions for [math]c_4[/math] uses only 7 points instead of 9. n=5 [math]\overline{c}^\mu_5=12[/math]:
The set of all (a,b,c) in [math]\Delta_5[/math] with exactly one of a,b,c=0 has 12 elements and doesn’t contain any equilateral triangles.
Let [math]S\subset \Delta_5[/math] be a set without equilateral triangles. If [math](0,0,5)\in S[/math], there can only be one of (0,x,5-x) and (x,0,5-x) in S for x=1,2,3,4,5. Thus there can only be 6 elements in S with a=0 or b=0. The set of element with a,b>0 is isomorphic to [math]\Delta_3[/math], so S can at most have 6 elements in this set. So [math]|S|\leq 6+6=12[/math]. Similar if S contain (0,5,0) or (5,0,0). So if |S| >12 S doesn’t contain any of these. S can only contain 2 point in each of the following equilateral triangles:
(3,1,1),(0,4,1),(0,1,4)
(4,1,0),(1,4,0),(1,1,3)
(4,0,1),(1,3,1),(1,0,4)
(1,2,2),(0,3,2),(0,2,3)
(3,2,0),(2,3,0),(2,2,1)
(3,0,2),(2,1,2),(2,0,3)
So now we have found 9 elements not in S, but [math]|\Delta_5|=21[/math], so [math]S\leq 21-9=12[/math].
n=6 [math]15 \leq \overline{c}^\mu_6 \leq 17[/math]:
[math]15 \leq \overline{c}^\mu_6[/math] from the bound for general n.
Note that there are eight extremal solutions to [math] \overline{c}^\mu_3 [/math]:
Solution I: remove 300, 020, 111, 003
Solution II: remove 030, 111, 201, 102 Solution III (and 2 rotations): remove 030, 021, 210, 102 Solution III' (and 2 rotations): remove 030, 120, 012, 201
Also consider the same triangular lattice with the point 020 removed, making a trapezoid. Solutions based on I-III are:
Solution IV: remove 300, 111, 003
Solution V: remove 201, 111, 102 Solution VI: remove 210, 021, 102 Solution VI': remove 120, 012, 201
Suppose we can remove all equilateral triangles on our 7×7x7 triangular lattice with only 10 removals.
The triangle 141-411-114 must have at least one point removed. Remove 141, and note because of symmetry any logic that follows also applies to 411 and 114.
There are three disjoint triangles 060-150-051, 240-231-330, 042-132-033, so each must have a point removed.
(Now only six removals remaining.)
The remainder of the triangle includes the overlapping trapezoids 600-420-321-303 and 303-123-024-006. If the solutions of these trapezoids come from V, VI, or VI', then 6 points have been removed. Suppose the trapezoid 600-420-321-303 uses the solution IV (by symmetry the same logic will work with the other trapezoid). Then there are 3 disjoint triangles 402-222-204, 213-123-114, and 105-015-006. Then 6 points have been removed. Therefore the remaining six removals must all come from the bottom three rows of the lattice.
Note this means the "top triangle" 060-330-033 must have only four points removed so it must conform to solution either I or II, because of the removal of 141.
Suppose the solution of the trapezoid 600-420-321-303 is VI or VI'. Both solutions I and II on the "top triangle" leave 240 open, and hence the equilateral triangle 240-420-222 remains. So the trapezoid can't be VI or VI'.
Suppose the solution of the trapezoid 600-420-321-303 is V. This leaves an equilateral triangle 420-321-330 which forces the "top triangle" to be solution I. This leaves the equilateral triangle 201-321-222. So the trapezoid can't be V.
Therefore the solution of the trapezoid 600-420-321-303 is IV. Since the disjoint triangles 402-222-204, 213-123-114, and 105-015-006 must all have points removed, that means the remaining points in the bottom three rows (420, 321, 510, 501, 312, 024) must be left open. 420 and 321 force 330 to be removed, so the "top triangle" is solution I. This leaves triangle 321-024-051 open, and we have reached a contradiction.
n = 7 n = 8
[math]\overline{c}^\mu_{8} \geq 22[/math]:
008,026,044,062,107,125,134,143,152,215,251,260,314,341,413,431,440,512,521,620,701,800
n = 9
[math]\overline{c}^\mu_{9} \geq 26[/math]:
027,045,063,081,126,135,144,153,207,216,252,270,315,342,351,360,405,414,432,513,522,531,603,630,720,801
n = 10
[math]\overline{c}^\mu_{10} \geq 29[/math]:
028,046,055,064,073,118,172,181,190,208,217,235,262, 316,334,352,361,406,433,442,541,550,604,613,622, 721,730,901,1000
General n
A lower bound for [math]\overline{c}^\mu_n[/math] is 2n for [math]n \geq 1[/math], by removing (n,0,0), the triangle (n-2,1,1) (0,n-1,1) (0,1,n-1), and all points on the edges of and inside the same triangle. In a similar spirit, we have the lower bound
[math]\overline{c}^\mu_{n+1} \geq \overline{c}^\mu_n + 2[/math]
for [math]n \geq 1[/math], because we can take an example for [math]\overline{c}^\mu_n[/math] (which cannot be all of [math]\Delta_n[/math]) and add two points on the bottom row, chosen so that the triangle they form has third vertex outside of the original example.
An asymptotically superior lower bound for [math]\overline{c}^\mu_n[/math] is 3(n-1), made of all points in [math]\Delta_n[/math] with exactly one coordinate equal to zero.
A trivial upper bound is
[math]\overline{c}^\mu_{n+1} \leq \overline{c}^\mu_n + n+2[/math]
since deleting the bottom row of a equilateral-triangle-free-set gives another equilateral-triangle-free-set. We also have the asymptotically superior bound
[math]\overline{c}^\mu_{n+2} \leq \overline{c}^\mu_n + \frac{3n+2}{2}[/math]
which comes from deleting two bottom rows of a triangle-free set and counting how many vertices are possible in those rows.
Another upper bound comes from counting the triangles. There are [math]\binom{n+2}{3}[/math] triangles, and each point belongs to n of them. So you must remove at least (n+2)(n+1)/6 points to remove all triangles, leaving (n+2)(n+1)/3 points as an upper bound for [math]\overline{c}^\mu_n[/math].
Asymptotics
The corners theorem tells us that [math]\overline{c}^\mu_n = o(n^2)[/math] as [math]n \to \infty[/math].
By looking at those triples (a,b,c) with a+2b inside a Behrend set, one can obtain the lower bound [math]\overline{c}^\mu_n \geq n^2 \exp(-O(\sqrt{\log n}))[/math]. |
Main Page The Problem
Let [math][3]^n[/math] be the set of all length [math]n[/math] strings over the alphabet [math]1, 2, 3[/math]. A
combinatorial line is a set of three points in [math][3]^n[/math], formed by taking a string with one or more wildcards [math]x[/math] in it, e.g., [math]112x1xx3\ldots[/math], and replacing those wildcards by [math]1, 2[/math] and [math]3[/math], respectively. In the example given, the resulting combinatorial line is: [math]\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}[/math]. A subset of [math][3]^n[/math] is said to be line-free if it contains no lines. Let [math]c_n[/math] be the size of the largest line-free subset of [math][3]^n[/math]. [math]k=3[/math] Density Hales-Jewett (DHJ(3)) theorem: [math]\lim_{n \rightarrow \infty} c_n/3^n = 0[/math]
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be considered by the Polymath project is to explore a particular combinatorial approach to DHJ, suggested by Tim Gowers.
Useful background materials
Some background to the project can be found here. General discussion on massively collaborative "polymath" projects can be found here. A cheatsheet for editing the wiki may be found here. Finally, here is the general Wiki user's guide
Threads (1-199) A combinatorial approach to density Hales-Jewett (inactive) (200-299) Upper and lower bounds for the density Hales-Jewett problem (final call) (300-399) The triangle-removal approach (inactive) (400-499) Quasirandomness and obstructions to uniformity (inactive) (500-599) Possible proof strategies (active) (600-699) A reading seminar on density Hales-Jewett (active) (700-799) Bounds for the first few density Hales-Jewett numbers, and related quantities (arriving at station)
We are also collecting bounds for Fujimura's problem.
Here are some unsolved problems arising from the above threads.
Here is a tidy problem page.
Bibliography M. Elkin, "An Improved Construction of Progression-Free Sets ", preprint. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem for k=3“, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–241. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem“, J. Anal. Math. 57 (1991), 64–119. B. Green, J. Wolf, "A note on Elkin's improvement of Behrend's construction", preprint. K. O'Bryant, "Sets of integers that do not contain long arithmetic progressions", preprint. R. McCutcheon, “The conclusion of the proof of the density Hales-Jewett theorem for k=3“, unpublished. |
I've been meaning to create a simple example that demonstrates how a likelihood ratio can be calculated. A few days ago, I did. The inspiration was a plot that appeared in many recent publications, which compared the various observational channels at the LHC against the standard model prediction vs. a "null hypothesis", which was the standard model without the Higgs boson. The two hypothesis were represented by two values, 0 and 1, and the various channels were marked on the vertical axis, their values plotted with horizontal error bars.
So for this example, I started with ten random data points and associated error bars, shown below.
The null hypothesis is that the values are centered on the vertical axis at 0. The alternative hypothesis is that the values are centered on 1.
The average of these values is 0.57. Their error bars are large. The (normalized) $\chi^2$ values per degrees of freedom associated with the two hypotheses do not suggest a significant difference:
\begin{align} \chi^2_0&=\frac{1}{n-1}\sum\limits_{i=1}^n\frac{a_i^2}{\sigma_i^2}=2.80,\\ \chi^2_1&=\frac{1}{n-1}\sum\limits_{i=1}^n\frac{(a_i-1)^2}{\sigma_i^2}=1.86. \end{align}
However, when we compute the likelihood ratio, the picture changes. The likelihoods are calculated as
\begin{align} L_0&=\prod\limits_{i=1}^n{\cal L}(a_i,\sigma_i)=3.73\times 10^{-7},\\ L_1&=\prod\limits_{i=1}^n{\cal L}(a_i-1,\sigma_i)=2.61\times 10^{-5}, \end{align} where ${\cal L}(x,\sigma)$ is the Gaussian distribution of $x$ with standard deviation $\sigma$.
The ratio of $L_1$ and $L_0$ is 70.1; i.e., the '1' hypothesis is more than 70 times as likely as the '0' hypothesis.
Another way of expressing this is that if the only two competing hypotheses are the '0' and '1' cases described here, then we can normalize the likelihoods and express them as probabilities:
\begin{align} {\cal P}_0&=\frac{{\cal L}_0}{{\cal L}_0+{\cal L}_1}=1.4\%,\\ {\cal P}_1&=\frac{{\cal L}_1}{{\cal L}_0+{\cal L}_1}=98.6\%, \end{align} which actually looks like a nearly $2.5\sigma$ significance.
We can also interpret this result as a prediction that as more data are collected in an experiment that yielded these results, under the assumption that either the '0' or the '1' hypothesis must be true, the data will converge on the '1' hypothesis with a probability of 98.6%.
I guess this demonstrates the statement that I found on Wikipedia, which says that ``use of the likelihood ratio test can be justified by the Neyman–Pearson lemma, which demonstrates that such a test has the highest power among all competitors''. |
Is there a closed-form expression for this nested sum?
$$s(n)=\sum_{i=1}^n\;\; \sum_{j=i+1}^n \sum_{k=i+j-1}^n1$$
If yes, what is it and how can it be derived?
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Is there a closed-form expression for this nested sum?
$$s(n)=\sum_{i=1}^n\;\; \sum_{j=i+1}^n \sum_{k=i+j-1}^n1$$
If yes, what is it and how can it be derived?
The usual convention with the sum sign $\sum$ is that $\sum_{k=p}^q a_k:=0$ if $q<p$. This has the following effect: Given an $i$ with $1\leq i\leq n$ the next index variable $j$ has to satisfy $$i+1\leq j\leq\min\{n, n+1-i\}=n+1-i\ .$$ This then implies that the variable $i$ in fact has to satisfy $2i\leq n$, or $$1\leq i\leq\left\lfloor{n\over2}\right\rfloor\ .$$ Given $i$ and $j$ with these constraints the innermost sum becomes $$\sum_{k=i+j-1}^n 1=n+2-i-j\ .$$ The next sum (over $j$) then becomes $$A_i:=\sum_{j=i+1}^{n+1-i}(n+2-i-j)$$ and has $n+1-2i$ terms. It follows that $$A_i=(n+1-2i)\>{1\over2}\>(n+2-2i)\qquad\bigl(1\leq i\leq\lfloor n/2\rfloor\bigr)\ .$$ Here we have used that the sum of a finite arithmetic series is the number of terms times the arithmetic mean of its outermost terms. From now on we have to distinguish the cases of even and odd $n$.
If $n=2m$ then Mathematica produces $$s(n)=\sum_{i=1}^m A_i={4m^3+3m^2-m\over6}={n(2n^2+3n-2)\over 24}\ .$$ If $n=2m+1$ we similarly obtain $$s(n)=\sum_{i=1}^m A_i={4m^3+9m^2+5m\over6}={2n^3+3n^2-2n-3\over 24}\ .$$
By choosing experimentally a small value of $n$ and writing out by hand in a simple $i-j$ grid the values of the innermost summation, it becomes clear that:
$$\begin{align}S(n)=S(2m)&=\sum_{s=1}^m \sum_{r=1}^{2s-1}r=\sum_{s=1}^m\binom {2s}2\\ &=\frac 16m(m+1)(4m-1)\end{align}$$
$$\begin{align}S(n)=S(2m+1)&=\sum_{s=1}^m \sum_{r=1}^{2s}r=\sum_{s=1}^m\binom {2s+1}2\\ &=\frac 16m(m+1)(4m+5)\end{align}$$
The above can also be derived as follows:
By considering each $i,j$ combination in turn and the corresponding limits on $k$ (or as pointed out by Christian Blatter in his solution), it is clear that the applicable limits of $i,j$ are narrower $\color{red}{\text{(shown below in red})}$ than in the original question, as the innermost summation cannot be negative, as specified by the condition in the Iverson brackets $\color{lightblue}{\text{(shown below in light blue)}}$.
The case for $n=2m+1$ can be shown using a similar method.
Hint: the general solution strategy is to solve first the innermost sum, expand that, and repeat the procedure.
For solving the sums use
$\sum_{k=1}^n k = (n(n+1))/2$,
$\sum_{k=1}^n k^2 = (n(n+1)(2n+1))/6$,
$\sum_{k=1}^n k^3 = (n^2(n+1)^2)/4$. |
That is not very difficult actually. Let $\Lambda\subset\Bbb R^n$ be the lattice on which you sampled your points, and let $\Phi:\Bbb R^n\to\Bbb R^n$ be a linear transformation under which $\Bbb Z^n\to\Lambda$ isomorphically (i.e. the columns of the matrix of $\Phi$ generate $\Lambda$). Let $f:\Bbb R^n\to\Bbb C$ be the function you are interested in, and of which you are interested in its values in $\Lambda$. From now on, we will alternatingly identify $f$ with a function on $\Lambda$ taking the value $f(\lambda)$ at $\lambda\in\Lambda$, and with a function (or rather distribution)
$$\sum_{\lambda\in\Lambda}f(\lambda)\delta(\lambda - x)$$
on $\Bbb R^n$, i.e. zero outside $\Lambda$, and a delta function at lattice points.
We can obtain the properties of $f$ in terms of its Fourier transform from the Fourier transform of $f\circ\Phi$, which is a function on $\Bbb Z^n$:
$$\mathscr F(f\circ\Phi)(s) \equiv \int f\circ\Phi(x)e^{-2\pi i\langle s,x\rangle}dx = \frac1{|\det\Phi|}\int f(y)e^{-2\pi i\langle\Phi^{-T}s,x\rangle}dy = \frac1{|\det\Phi|}\mathscr F(f)(\Phi^{-T}s)$$
Here a substitution $y = \Phi^{-1}x$ was made, and it was used that $\langle s,\Phi^{-1}y\rangle = \langle\Phi^{-T}s,y\rangle$. Note that your conventions w.r.t. the sign of the exponent and factors of $\sqrt{2\pi}$ may differ.
Now what does the formula mean that we just wrote down? $f\circ\Phi$ is just our set of samples of $f$ arranged in an $n$-dimensional array, whose values have to be interpreted as lying on a standard grid. Its Fourier transform gives an array of the same shape, but the interpretation is different: the value at index $s$ in reality is the value at $\Phi^{-T}s$ in $\Bbb R^n$, multiplied by $|\det\Phi|$.
In particular, if the function $f$ is supported on $\Lambda\subset\Bbb R^n$, then its Fourier transform is supported on the image of $\Phi^{-T}$, the span of the columns of the inverse transposed matrix of $\Phi$. |
Logical AND: Use the linear constraints $y_1 \ge x_1 + x_2 - 1$, $y_1 \le x_1$, $y_1 \le x_2$, $0 \le y_1 \le 1$, where $y_1$ is constrained to be an integer. This enforces the desired relationship. (Pretty neat that you can do it with just linear inequalities, huh?)Logical OR: Use the linear constraints $y_2 \le x_1 + x_2$, $y_2 \ge x_1$, $y_2 \ge x_2$, ...
The logical AND relation can be modeled in one range constraintinstead of three constraints (as in the other solution).So instead of the three constraints$$y_1\geq x_1+x2−1,\qquad y_1\leq x_1,\qquad y_1\leq x_2\,,$$it can be written using the single range constraint$$0 \leq x_1 + x_2-2y_1 \leq 1\,.$$Similarly, for logical OR:$$0 \leq 2y_1 - x_1 - x_2 ≤...
If a problem is NP-Hard it means that there exists a class of instances of that problem whose are NP-Hard.It is perfectly possible for other specific classes of instances to be solvable in polynomial time.Consider for example the problem of finding a 3-coloration of a graph. It is a well-known NP-Hard problem. Now imagine that its instances are restricted ...
No, special cases can be easier.Consider this IP, for example, given $a_i \geq 0$ for $i \in [1..n]$:$\qquad\displaystyle \min \sum_{i=1}^n x_ia_i$s.t. $\quad\displaystyle\sum_{i=1}^n x_i \geq 1$and $\ \displaystyle x_i \in \mathbb{N}$ for $i \in [1..n]$.It finds the minimum among $a_1, \dots, a_n$ (that for which, inevitably, $x_i=1$ in an optimal ...
This problem is NP-hard by reduction from Vertex Cover.In the Vertex Cover problem, we are given a graph $G = (V, E)$ and a number $r$, and our task is to determine whether there is some subset $U$ of at most $r$ vertices from $V$ such that every edge in $E$ is incident on at least one vertex in $U$. (Equivalently: Is it possible to kill every edge in $G$ ...
A classical result of Berlekamp, McEliece, and van Tilborg shows that the following problem, maximum likelihood decoding, is NP-complete: given a matrix $A$ and a vector $b$ over $\mathbb{F}_2$, and an integer $w$, determine whether there is a solution to $Ax = b$ with Hamming weight at most $w$.You can reduce this problem to your problem. The system $Ax = ...
All constraints in a linear program are convex (if $x,y$ satisfy the constraints, then $tx+(1-t)y$ also does for all $0 \leq t \leq 1$). The constraint $|a|+b > 3$ is not convex, since $(4,0)$ and $(-4,0)$ are both solutions while $(0,0)$ is not. It is also not closed, which is another reason why you cannot use it in a linear program (change $>$ to $\...
Some ILPs can be solved rapidly (to an exact solution) in practice; some cannot. Usually when we are talking about solving an ILP, we are looking for an exact solution, though some ILP solvers can find an approximate solution as well (find the best solution they can within the time constraints). There are no hard-and-fast rules.Of course, ILP is NP-hard, ...
Yes, there is a polynomial-time algorithm for your problem. There's no need to use a LP or ILP; you can solve it directly using combinatorial, graph-based methods. In particular, we can solve your problem by a reduction to the assignment problem, i.e., to computing a maximum weight matching in a weighted bipartite graph.Suppose we have a bipartite graph ...
I've found out a very interesting document that answers my question: http://lpsolve.sourceforge.net/5.5/absolute.htm It's about integer programming and it covers all possible cases I think. See section >= minimum to handle abs(X) >= minimum. Here is another one with more tricks: http://orinanobworld.blogspot.de/2012/07/modeling-absolute-values.htmlThere ...
In $\mathbb R$, one can simply round down or round up to obtain an element of $\mathbb Z$. Only two choices!However, in $\mathbb R^n$, one has $2^n$ ways of rounding to obtain an element of the integer lattice $\mathbb Z^n$. For example, if $n=100$, one has $2^{100} \approx 10^{30}$ possible ways of rounding.Of course, one can round all entries of an $n$...
There is unlikely to be any efficient algorithm.Your first class of constraints are monotone exactly-1 CNF clauses. Your second class of constraints are monotone CNF clauses. The monotone part indicates that negated literals aren't allowed (you can't have $x_1 - x_3 = 1$ or $x_1 - x_4 \ge 1$).Thus, in the special case where you have only type-2 ...
First of all, let me start by making clear that the notion of 'solvable in polynomial time' is something defined on a class of problem instances. It makes no sense to speak of polynomial time for a single problem as any single problem can be solved in $O(1)$!That said, there is a notable class of ILP's that is known to be polynomial time solvable. This ...
Sure, of course you can create random linear programming problems. Why not?Yes, in general, you can usually verify the solution to a linear programming problem faster than you can find the solution. Verifying the solution just involves plugging into the equations and checking that all equations hold. Finding the solution requires a bit more work (e.g., ...
All four problems are NP-complete.Let me start by proving two easy results:Your problem #3 is NP-complete, even when the matrix has only a single row, as explained here: Complexity of a subset sum variant.This implies that #4 is NP-complete as well, as #3 is a special case of #4.Now for the strong result: Your problem #1 is NP-complete. The proof of ...
Simplifying the problem:Givena positive integer $r$positive integers $m, n \geq 2$a partial binary matrix $\mathrm A \in \{*, 0,1\}^{m \times n}$ (where $*$ denotes an unknown entry)determine whether it is possible to complete the given partial matrix $\mathrm A$ with values in $\{0,1\}$ such that the resulting completed matrix ...
It's the same was for all NP problems; the optimisation problem isFind a valid solution $s$ that minimises¹ $f(s)$!and the corresponding decision problem isIs there a valid solution $s$ with $f(s) \leq k$?You see that the former immediately solves the latter, and you can solve the former by using the latter with binary search over the set of ...
If there are only constraints that place a lower bound on the number of trucks, but no constraints that place an upper limit on the number of trucks, then of course you can round up. That will still give you a solution.However, there are multiple caveats:First, this isn't always possible. Sometimes there are both constraints that place lower limits and ...
You could add a constant $0<A<M$ and then you add this constraint:$$A-y(A+M)\leqslant x_1-x_2\leqslant M(1-y).$$If $y=1$ then you are left with$$-M\leqslant x_1-x_2\leqslant 0,$$ which says that $x_1\leqslant x_2$.And if $y=0$ then you will be left with$$A\leqslant x_1-x_2\leqslant M,$$ which says that $x_1>x_2$ (since $0<A<M$).
This can be expressed with just the equation $X=Y$. Since $X,Y$ are zero-or-one variables, the only possible assignments that are consistent with your condition are $X=Y=0$ and $X=Y=1$.See Express boolean logic operations in zero-one integer linear programming (ILP) for many more boolean conditions and how to express them as linear inequalities.
The notion of duality suggested by the discrete Farkas lemma found in Lasserre's paper does not correspond to an integer program. There is a notion of duality for integer programs (see for example here), but strong duality does not hold, so I doubt it can be used along the lines you suggest.
The decision problem version of integer linear programming is known to be in NP. In particular, determining whether an integer linear program has a feasible solution is in NP. It is known that if an integer linear program has a feasible solution, then it has a feasible solution whose length (in bits) is at most a polynomial in the length (in bits) of the ...
One-in-three 3SAT is NP-complete. Looking at the reduction, it inherits the APX-hardness of 3SAT. You can formulate one-in-three 3SAT as a binary integer program with binary entries, so you problem is APX-hard.
I have a method for you that will help you find valid solutions (matrices) for many possible values of $m,n$. However, it is not a complete answer to your question. It can try to find a matrix for a particular value of $m,n$, but it might fail, and if it fails, you've learned nothing; my method cannot prove that no such matrix exists.The method is based ...
Yes, some IP formulations are less useful than others. The technique used to show that an LP relaxation can only be so good is showing integrality gaps. For a minimization problem, an integrality gap of $k$ in an instance in which the optimal integer solution has value $V$ but the LP has a solution of value at most $V/k$. This shows that any rounding ...
The existence of such a reduction follows immediately from the Cook-Levin theorem, which guarantees that you can reduce any problem in NP to SAT and describes explicitly one way to do it. Just work through the steps of the proof of the theorem (which can be found in any textbook) and you'll obtain a valid reduction from 0-1 ILP to SAT.
You can do this by introducing the two inequalities$$x_1 \le x_2 + M (1-y)$$and$$x_1 > x_2 - M y.$$The former encodes the requirement $y=1 \implies x_1 \le x_2$ (you can see that if $y=1$, then the $M(1-y)$ term disappears; if $y=0$, then $M(1-y)$ becomes something huge and the inequality is automatically satisfied). The latter encodes the ...
Maybe there is something I do not understand in your question, but the way it is formulated it seems that the set of solutions is:set $b_i=1$ if $x_i > 0$set $b_i=0$ if $x_i < 0$all others $b_i$ can either be zero or oneIt is clear that any other assignment will give you a lower value (either by adding a positive value or not adding a positive ... |
I am trying to understand the equation (122)
in http://xxx.lanl.gov/pdf/hep-th/0509097.pdf
Making my proper computation I am obtaining
$$-\int_{M^4}da \wedge d\omega_{2} ^{1}=-\int_{M^4}[d^2a \wedge\omega_{2} ^{1}-d(da \wedge \omega_{2} ^{1}) ]=$$
$$-\int_{M^4}d^2a \wedge\omega_{2} ^{1}+\int_{M^4}d(da \wedge \omega_{2} ^{1}) =$$
$$-\int_{M^4}d^2a \wedge\omega_{2} ^{1}+\int_{\partial M^4}da \wedge \omega_{2} ^{1}= -\int_{M^4}d^2a \wedge\omega_{2} ^{1}+0= -\int_{M^4}d^2a \wedge\omega_{2} ^{1}$$
My question is: if my computation is correct and then a minus sign is missing in the last integral of (122)? |
The runtime for Dijkstra's algorithm implemented with a priority queue on a sparse graph is $O((E+V)\log V)$. For a dense graph such as a complete graph, there can be $V(V-1)/2$ edges.
Since $E \sim V^2$, is the runtime $O((V+V^2)\log V)$?
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The runtime for Dijkstra's algorithm implemented with a priority queue on a sparse graph is $O((E+V)\log V)$. For a dense graph such as a complete graph, there can be $V(V-1)/2$ edges.
Since $E \sim V^2$, is the runtime $O((V+V^2)\log V)$?
The runtime of Dijkstra's algorithm (with Fibonacci Heaps) is $O(|E|+|V|\log|V|)$, which is different from what you were posting.
If $|E|\in \Theta(|V|^2)$, that is your graph is very dense, then this gives you runtime of $O(|V|^2+|V|\log|V|)=O(|V|^2)$. A better runtime would be "surprising", since you have to look at every edge at least once.
When using binary heaps, you get a runtime of $O((|E|+|V|)\log|V|)$ which for $|E|\in \Theta(|V|^2)$ gives $O(|V|^2\log |V|)$.
You can also get rid of the log factor in dense graphs, and do it in $O(V^2)$. See here.
They might give you around $V=5000$ nodes but $O(V^2) = O(2.5 \cdot 10^7)$ edges which will give the leading edge over normal $O(E\log V)$ Dijkstra. |
$\def\Hadw{\mathop{\rm Hadw}}$This is true for finite graphs, and false for (not necessarily connected) infinite graphs. Right now I do not know what happens for infinite connected graphs.
1. Each component $G_1\subseteq G$ corresponds to an isolated vertex $v_{G_1}$ in $\Hadw(G)$ and a component $\Hadw(G_1)\setminus \{v_{G_1}\}$; this component is empty if $G_1$ consists of an isolated vertex. For finite graphs, this means that we find the number of isolated vertices and the Hadwiger graphs of other components; so the problem reduces to the same problem for finite connected graphs. For infinite graphs, this leads to the counterexample. Let $G$ be a union of a countable number of components none of which is an isolated vertex, and let $H$ be $G$ augmented with an isolated vertex; then $G\not\cong H$ but $\Hadw(G)\cong \Hadw(H)$.
The rest part is devoted to the reconstruction of a connected graph $G$ by its Hadwiger graph.
2. In this case, we will show a bit more, namely:
Knowing $\Hadw(G)$, we can find all the vertices in $\Hadw(G)$ corresponding to the vertices of $G$ (then the induced graph is isomorphic to $G$).
We proceed by the induction on $|V(G)|$. If $|V(G)|=1$ then the statement is obvious.
Any vertex $P\in V(\Hadw(G))$ has degree 1 exactly if $P=V(G)\setminus \{v\}$, where $v\in V(G)$ is not a cut vertex; in this case $\{v\}$ is its only neighbor. Thus we may reconstruct all the vertices of $\Hadw(G)$ which correspond to non-cut vertices of $G$ (notice that there is at least one such vertex!).
Let $T=\{v\}\in V(\Hadw(G))$ be one of such vertices. Denote by $N$ the set of all neighbors of $v$ in $G$; denote $$ L=\{X\in V(\Hadw(G)): (\{v\}\cup N)\subseteq X\}.$$Notice that $L$ is nonempty.
Consider now the distances $d(S,T)$ from every vertex $S\in V(\Hadw(G))$ (distinct from $T$ and $V(G)$) to $T$.
(i) If $v\notin S$ but $S\cap N\neq\varnothing$ then $d(S,T)=1$.
(ii) If $S\cap(\{v\}\cup N)=\varnothing$ then $d(S,T)=2$ due to a path in $G$ from $v$ to $S$; moreover, in this case $S$ has a neighbor in $L$.
(iii) If $v\in S$ but $N\not\subseteq S$ (that is, $v\in S$ but $S\notin L$) then $d(S,T)=2$ due to any vertex in $N\setminus S$; but in this case $S$ has no neighbor in $L$.
(iv) If $S\in L$ then $d(S,T)=3$ since the distance from every neighbor of $S$ to $T$ is 2.
Thus we can reconstruct the set $L$ (due to the distance 3 from $T$), and then set of all $S$ such that $v\in S$ but $S\notin L$ (due to the distance 2 from $T$ and non-existence of a neighbor in $L$). Thus we have reconstructed all $S\in V(\Hadw(G))$ containing $v$. Now we can remove all these vertices obtaining the graph $\Hadw(G-\{v\})$ for which the induction assumption is applicable. |
1D arterial flow
A 1D model for arterial flows can be derived from the Navier-Stokes equations, in terms of the cross sectional area A and flow rate Q, we have \displaystyle \partial_t A +\partial_x Q = 0 \displaystyle \partial_t Q +\partial_x (Q^2/A) = - A \partial_x p/\rho - f_r where p(A) models the wall properties of the arteries, \rho is the blood density and f_r stands for the wall shear stress. For a simple linear wall relation, p = K A with K a constant, we can write the flux as F = (Q,Q^2/A + 2 e_1 A) and the source term as S = (0,-e_2 Q/A) using two parameters e_1 and e_2.
Before including the conservation solver, we need to overload the default
update function of the predictor-corrector scheme in order to add our source term.
#include "grid/cartesian1D.h"#include "predictor-corrector.h"static double momentum_source (scalar * current, scalar * updates, double dtmax);event defaults (i = 0) update = momentum_source;#include "conservation.h"
Variables
We define the conserved scalar fields a and q which are passed to the generic solver through the
scalars list. We don’t have any conserved vector field.
scalar a[], q[];scalar * scalars = {a,q};vector * vectors = NULL;
The other parameters are specific to the example.
double e1, e2, omega, Amp;
Functions
We define the
flux function required by the generic solver.
void flux (const double * s, double * f, double e[2]){ double a = s[0], q = s[1], u = q/a; f[0] = q; f[1] = q*q/a + e1*a*a; // min/max eigenvalues double c = sqrt(2.*e1*a); e[0] = u - c; // min e[1] = u + c; // max}
We need to add the source term of the momentum equation. We define a function which, given the current states, fills the
updates with the source terms for each conserved quantity.
static double momentum_source (scalar * current, scalar * updates, double dtmax){
We first compute the updates from the system of conservation laws.
double dt = update_conservation (current, updates, dtmax);
We recover the current fields and their variations from the lists…
scalar a = current[0], q = current[1], dq = updates[1];
We add the source term for
q.
foreach() dq[] += - e2*q[]/a[]; return dt;}
Boundary conditions
We impose a sinusoidal flux Q(t) at the left of the domain.
q[left] = dirichlet(Amp*sin(2.*pi*omega*t));
Parameters
For small amplitudes Amp = 0.01 at the input boundary condition the system has analytical solutions for e1 < e2, in this case the spatial envelope of the flux rate behaves like Q=Amp\times e^{-e2/2x} [Wang at al., 2013].
Initial conditions
The initial conditions are A=1 and Q=0.
Outputs
We print to standard error the spatial profile of the flow rate Q.
We get the following comparison between the numerical solution and the linear theory for the flow rate Q.
Amp = 0.01e2 = 0.1set yrange [0.008:]set ylabel 'Q'set xlabel 'x'plot 'log' w l t 'numerical', Amp*exp(-e2/2.*x) t 'linear theory' |
On a differentiable manifold, the
exterior derivative extends the concept of the differential of a function to differential forms of higher degree. The exterior derivative was first described in its current form by Élie Cartan; it allows for a natural, metric-independent generalization of Stokes' theorem, Gauss's theorem, and Green's theorem from vector calculus.
If a
k-form is thought of as measuring the flux through an infinitesimal k-parallelepiped, then its exterior derivative can be thought of as measuring the net flux through the boundary of a ( k + 1)-parallelepiped.
Contents Definition 1 Axioms for the exterior derivative 1.1 Exterior derivative in local coordinates 1.2 Invariant formula 1.3 Stokes' theorem on manifolds 2 Examples 3 Further properties 4 Closed and exact forms 4.1 de Rham cohomology 4.2 Naturality 4.3 Exterior derivative in vector calculus 5 Gradient 5.1 Divergence 5.2 Curl 5.3 Invariant formulations of grad, curl, div, and Laplacian 5.4 See also 6 References 7 Definition
The exterior derivative of a differential form of degree
k is a differential form of degree k + 1.
If
f is a smooth function (a 0-form), then the exterior derivative of f is the differential of f . That is, d f is the unique 1-form such that for every smooth vector field X, d f ( X) = d X f , where d X f is the directional derivative of f in the direction of X.
There are a variety of equivalent definitions of the exterior derivative of a general
k-form. Axioms for the exterior derivative
The exterior derivative is defined to be the unique
R-linear mapping from k-forms to ( k + 1)-forms satisfying the following properties: d f is the differential of f for smooth functions f . d(d f ) = 0 for any smooth function f . d( α ∧ β) = d α ∧ β + (−1) ( p α ∧ d β) where α is a p-form. That is to say, d is an antiderivation of degree 1 on the exterior algebra of differential forms.
The second defining property holds in more generality: in fact, d(d
α) = 0 for any k-form α; more succinctly, d 2 = 0. The third defining property implies as a special case that if f is a function and α a k-form, then d( fα) = d( f ∧ α) = d f ∧ α + f ∧ d α because functions are 0-forms, and scalar multiplication and the exterior product are equivalent when one of the arguments is a scalar. Exterior derivative in local coordinates
Alternatively, one can work entirely in a local coordinate system (
x 1, ..., x). First, the coordinate differentials d n x 1, ..., d x form a basic set of one-forms within the coordinate chart. The formulas in this section rely on the Einstein summation convention. Given a multi-index n I = ( i 1, ..., i ) with 1 ≤ k i ≤ p n for 1 ≤ p ≤ k (and an abuse of notation d x), the exterior derivative of a I k-form \omega = f_I \mathrm{d} x^I = f_{i_1,i_2\cdots i_k}\mathrm{d}x^{i_1}\wedge \mathrm{d}x^{i_2}\wedge\cdots\wedge \mathrm{d}x^{i_k}
over
R is defined as n \mathrm{d}{\omega} = \sum_{i=1}^n \frac{\partial f_I}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d} x^I.
For a general
k-form \omega = \sum_I f_I \mathrm{d}x^I,
where the components of the multi-index
I run over all the values in {1, ..., n}, the definition of the exterior derivative is extended linearly. Note that whenever i is one of the components of the multi-index I then d x ∧ d i x = 0 (see wedge product). I
The definition of the exterior derivative in local coordinates follows from the preceding definition. Indeed, if
ω = f d I x ∧ ... ∧ d i 1 x, then i k \begin{align} \mathrm{d}{\omega} &= \mathrm{d} \left (f_I \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} \right ) \\ &= \mathrm{d}f_I \wedge \left (\mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} \right ) + f_I \mathrm{d} \left ( \mathrm{d} x^{i_1}\wedge \cdots \wedge \mathrm{d}x^{i_k} \right ) \\ &= \mathrm{d}f_I \wedge \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} + \sum_{p=1}^k (-1)^{(p-1)} f_I \mathrm{d} x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_{p-1}} \wedge \mathrm{d}^2x^{i_p} \wedge \mathrm{d}x^{i_{p+1}} \wedge \cdots \wedge\mathrm{d} x^{i_k} \\ &= \mathrm{d}f_I \wedge \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} \\ &= \sum_{i=1}^n \frac{\partial f_I}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} \\ \end{align}
Here, we have interpreted
f as a 0-form, and then applied the properties of the exterior derivative. I Invariant formula
Alternatively, an explicit formula can be given for the exterior derivative of a
k-form ω, when paired with k + 1 arbitrary smooth vector fields V 0, V 1, ..., V k: \mathrm{d}\omega(V_0,...,V_k) = \sum_i(-1)^{i} V_i \left( \omega \left (V_0, \ldots, \hat V_i, \ldots,V_k \right )\right) +\sum_{i
where [
V, i V] denotes the Lie bracket and a hat denotes the omission of that element: j \omega \left (V_0, \ldots, \hat V_i, \ldots,V_k \right ) = \omega \left (V_0, \ldots, V_{i-1}, V_{i+1}, \ldots, V_k \right ).
In particular, for 1-forms we have: d
ω( X, Y) = X(ω( Y)) − Y(ω( X)) − ω([ X, Y]), where X and Y are vector fields, X(ω( Y)) is the scalar field defined by the vector field X∈Γ( T M) applied as a differential operator ("directional derivative along X") to the scalar field defined by applying ω∈Γ *( T M) as a covector field to the vector field Y∈Γ( T M) and likewise for Y( ω( X)). Stokes' theorem on manifolds
If
M is a compact smooth orientable n-dimensional manifold with boundary, and ω is an ( n − 1)-form on M, then the generalized form of Stokes' theorem states that: \int_M \mathrm{d}\omega = \int_{\partial{M}} \omega
Intuitively, if one thinks of
M as being divided into infinitesimal regions, and one adds the flux through the boundaries of all the regions, the interior boundaries all cancel out, leaving the total flux through the boundary of M. Examples
Example 1. Consider σ = u d x 1 ∧ d x 2 over a 1-form basis d x 1, ..., d x. The exterior derivative is: n \begin{align} \mathrm{d} \sigma &= \mathrm{d}(u) \wedge \mathrm{d}x^1 \wedge \mathrm{d}x^2 \\ &= \left(\sum_{i=1}^n \frac{\partial u}{\partial x^i} \mathrm{d}x^i\right) \wedge \mathrm{d}x^1 \wedge \mathrm{d}x^2 \\ &= \sum_{i=3}^n \left( \frac{\partial u}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x^1 \wedge \mathrm{d}x^2 \right ) \end{align}
The last formula follows easily from the properties of the wedge product. Namely, d
x ∧ d i x = 0. i
Example 2. Let σ = u d x + v d y be a 1-form defined over R 2. By applying the above formula to each term (consider x 1 = x and x 2 = y) we have the following sum, \begin{align} \mathrm{d} \sigma &= \left( \sum_{i=1}^2 \frac{\partial u}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x \right) + \left( \sum_{i=1}^2 \frac{\partial v}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}y \right) \\ &= \left(\frac{\partial{u}}{\partial{x}} \mathrm{d}x \wedge \mathrm{d}x + \frac{\partial{u}}{\partial{y}} \mathrm{d}y \wedge \mathrm{d}x\right) + \left(\frac{\partial{v}}{\partial{x}} \mathrm{d}x \wedge \mathrm{d}y + \frac{\partial{v}}{\partial{y}} \mathrm{d}y \wedge \mathrm{d}y\right) \\ &= 0 - \frac{\partial{u}}{\partial{y}} \mathrm{d}x \wedge \mathrm{d}y + \frac{\partial{v}}{\partial{x}} \mathrm{d}x \wedge \mathrm{d}y + 0 \\ &= \left(\frac{\partial{v}}{\partial{x}} - \frac{\partial{u}}{\partial{y}}\right) \mathrm{d}x \wedge \mathrm{d}y \end{align} Further properties Closed and exact forms
A
k-form ω is called closed if d ω = 0; closed forms are the kernel of d. ω is called exact if ω = d α for some ( k − 1)-form α; exact forms are the image of d. Because d 2 = 0, every exact form is closed. The Poincaré lemma states that in a contractible region, the converse is true. de Rham cohomology
Because the exterior derivative d has the property that d
2 = 0, it can be used as the differential (coboundary) to define de Rham cohomology on a manifold. The k-th de Rham cohomology (group) is the vector space of closed k-forms modulo the exact k-forms; as noted in the previous section, the Poincaré lemma states that these vector spaces are trivial for a contractible region, for k > 0. For smooth manifolds, integration of forms gives a natural homomorphism from the de Rham cohomology to the singular cohomology over R. The theorem of de Rham shows that this map is actually an isomorphism, a far-reaching generalization of the Poincaré lemma. As suggested by the generalized Stokes' theorem, the exterior derivative is the "dual" of the boundary map on singular simplices. Naturality
The exterior derivative is natural in the technical sense: if
f : M → N is a smooth map and Ω is the contravariant smooth functor that assigns to each manifold the space of k k-forms on the manifold, then the following diagram commutes
so d(
f ∗ ω) = f ∗d ω, where f ∗ denotes the pullback of f . This follows from that f ∗ ω(·), by definition, is ω( f ∗(·)), f ∗ being the pushforward of f . Thus d is a natural transformation from Ω to Ω k . k+1 Exterior derivative in vector calculus
Most vector calculus operators are special cases of, or have close relationships to, the notion of exterior differentiation.
Gradient
A smooth function
f : R → n R is a 0-form. The exterior derivative of this 0-form is the 1-form \mathrm{d}f = \sum_{i=1}^n \frac{\partial f}{\partial x^i}\, \mathrm{d}x^i = \langle \nabla f,\cdot \rangle.
That is, the form d
f acts on any vector field V by outputting, at each point, the scalar product of V with the gradient ∇ f of f .
The 1-form d
f is a section of the cotangent bundle, that gives a local linear approximation to f in the cotangent space at each point. Divergence
A vector field
V = ( v 1, v 2, ... v) on n R has a corresponding ( n n − 1)-form \begin{align} \omega_V &= v_1 \left (\mathrm{d}x^2 \wedge \mathrm{d}x^3 \wedge \cdots \wedge \mathrm{d}x^n \right) - v_2 \left (\mathrm{d}x^1 \wedge \mathrm{d}x^3 \cdots \wedge \mathrm{d}x^n \right ) + \cdots + (-1)^{n-1}v_n \left (\mathrm{d}x^1 \wedge \cdots \wedge \mathrm{d}x^{n-1} \right) \\ &=\sum_{p=1}^n (-1)^{(p-1)}v_p \left (\mathrm{d}x^1 \wedge \cdots \wedge \mathrm{d}x^{p-1} \wedge \widehat{\mathrm{d}x^{p}} \wedge \mathrm{d}x^{p+1} \wedge \cdots \wedge \mathrm{d}x^n \right ) \end{align}
where \widehat{\mathrm{d}x^{p}} denotes the omission of that element.
(For instance, when
n = 3, in three-dimensional space, the 2-form ω is locally the scalar triple product with V V.) The integral of ω over a hypersurface is the flux of V V over that hypersurface.
The exterior derivative of this (
n − 1)-form is the n-form \mathrm{d} \omega _V = \operatorname{div}(V) \left (\mathrm{d}x^1 \wedge \mathrm{d}x^2 \wedge \cdots \wedge \mathrm{d}x^n \right ). Curl
A vector field
V on R also has a corresponding 1-form n \eta_V = v_1 \mathrm{d}x^1 + v_2 \mathrm{d}x^2 + \cdots + v_n \mathrm{d}x^n.,
Locally,
η is the dot product with V V. The integral of η along a path is the work done against − V V along that path.
When
n = 3, in three-dimensional space, the exterior derivative of the 1-form η is the 2-form V \mathrm{d} \eta_V = \omega_{\operatorname{curl}(V)}. Invariant formulations of grad, curl, div, and Laplacian
On any Riemannian manifold, the standard vector calculus operators can be written in coordinate-free notation as follows:
\begin{array}{rcccl} \operatorname{grad}(f) &=& \nabla f &=& \left( \mathrm{d} f \right)^\sharp \\ \operatorname{div}(F) &=& \nabla \cdot F &=& \star \mathrm{d} \left( \star F^\flat \right) \\ \operatorname{curl}(F) &=& \nabla \times F &=& \left[ \star \left( \mathrm{d} F^\flat \right) \right]^\sharp, \\ \Delta f &=& \nabla^2 f &=& \star \mathrm{d} \left( \star \mathrm{d} f \right) \\ \end{array}
where \star is the Hodge star operator and \flat and \sharp are the musical isomorphisms.
See also References
Flanders, Harley (1989). Differential forms with applications to the physical sciences. New York: Dover Publications. p. 20. Ramanan, S. (2005). Global calculus. Providence, Rhode Island: American Mathematical Society. p. 54. Conlon, Lawrence (2001). Differentiable manifolds. Basel, Switzerland: Birkhäuser. p. 239. Darling, R. W. R. (1994). Differential forms and connections. Cambridge, UK: Cambridge University Press. p. 35.
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First, look at the bits that are set in $M$ and write it as $M = m_1+m_2+\dots$ where each $m_i$ is a power of two and such that $m_1$ is the smallest power of two, and so on.
If the numbers $x_i \in X$ are
distinct powers of two and you can use each one zero or one times, the problem is trivial: If all $m_i$ are found in $X$, then $M$ can be formed by them, otherwise it can't. This is because missing powers of two can't be formed by a) any combination of larger powers of two or b) distinct smaller powers of two.
If the numbers $x_i$ aren't guaranteed to be distinct, it's a bit more complex. Missing powers of two still can't be formed by larger powers of two, but they can be formed by non-distinct smaller powers of two.
Let's look at how smaller powers of two can form larger powers of two. If you have powers of two $A = [a_1, \dots, a_n]$ that are smaller than a power of two $N$, and $S=\sum_{i=0}^n a_i$, then $N$ can be formed by them iff $S \ge N$. Let $b$ be the smallest power of two in $A$ and $c$ its number of occurences in $A$. If $c$ is odd, one $a_i=b$ can't combine with anything to form a higher power of two and the corresponding bit will be set in $S$. The others can be replaced with $\lfloor \frac{c}{2} \rfloor$ copies of the number $2b$ in $A$. If we continue this with higher powers of two until we get to $N$, we find that we can form $\lfloor \frac{S}{N} \rfloor$ copies of $N$ with $A$.
So, start with $m_1$. If it's found in $X$, we remove it and move to the next power of two. If it's not, we calculate the sum $S$ of all $x_i$ smaller than $m_1$ and remove them from $X$. If $S \lt N$, then $M$ can't be formed by $X$. Otherwise, we add $\lfloor \frac{S}{m_1} \rfloor - 1$ copies of $m_1$ to $X$ and continue in the same manner with the next smallest power of two in $M$. |
Given a graph with $n \leq 50 $ vertices. Count all $k$-cliques of this graph, where $k = 1, \ldots , n$.
I need the most efficient algorithm.
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Your problem (assuming I understand your English correctly) is counting the number of cliques in a graph of size $n = 50$. Counting the number of cliques in a graph is #P-complete (see this paper, which shows that counting the number of independent sets in a graph is #P-complete even for bipartite graphs).
Several efficient exponential time algorithms for the problem (efficient in the sense that their running time is $O(c^n)$ for $c<2$) are described in Jeff Erickson's lecture notes - see section 4.2 on page 4. The simplest algorithm, whose running time is $O(\phi^n)$ (where $\phi$ is the golden ratio), is as follows. For an arbitrary vertex $v \in G$ with neighborhood $N(v)$, we have the following recurrence for the number $I(G)$ of independent sets in $G$ (to get the number of cliques, complement $G$): $$ I(G) = \begin{cases} 2I(G-v), & N(v) = \emptyset, \\ I(G-v) + I(G\setminus(N(v)+v)), & \text{otherwise}. \end{cases} $$
This algorithm might be practical for $n=50$. If it isn't, you can try the other algorithms described in the lecture notes. Some of them require some modification for your case, and therefore the running times listed there might not hold in your case. |
$|V_{cb}|$ from $\bar{B}^0\rightarrow D^{*+} \ell^- \bar{\nu} $ zero-recoil form factor using 2$+$1$+$1 flavour HISQ and NRQCD
Pre-published on: 2017 February 13
Published on: 2017 March 24
Abstract
We present the status of our ongoing calculation of the zero-recoil form factor for the semileptonic decay $\bar{B}^0\rightarrow D^{*+}l^-\bar{\nu}$ using lattice QCD with 2+1+1 flavours of highly improved staggered quarks in the sea (the MILC HISQ configurations) and using non-relativistic QCD for the bottom quark. We combine our result for $ F(1)$ with the latest HFAG average of $\eta_{EW} F(1)|V_{cb}|$ to get a preliminary value for $|V_{cb}|$. |
I'm working on a high-dimensional inference problem (around 2000 model parameters) for which we are able to robustly perform MAP estimation by finding the global maximum of the log-posterior using a combination of gradient-based optimisation and a genetic algorithm.
I'd very much like to be able to make some estimate of the uncertainties on the model parameters in addition to finding the MAP estimate.
We are able to efficiently calculate the gradient of the log-posterior with respect to the parameters, so long-term we're aiming to use Hamiltonian MCMC to do some sampling, but for now I'm interested in non-sampling based estimates.
The only approach I know of is to calculate the inverse of the Hessian at the mode to approximate the posterior as multivariate normal, but even this seems infeasible for such a large system, since even if we calculate the $\sim 4\times10^{6}$ elements of the Hessian I'm sure we couldn't find its inverse.
Can anyone suggest what kind of approaches are typically used in cases like this?
Thanks!
EDIT - additional information about the problem Background This is an inverse-problem related to a large physics experiment. We have a 2D triangular mesh which describes some physical fields, and our model parameters are the physical values of those fields at each vertex of the mesh. The mesh has about 650 vertices, and we model 3 fields, so that's where our 2000 model parameters come from.
Our experimental data is from instruments which do not measure these fields directly, but quantities that are complicated non-linear functions of the fields. For each of the different instruments we have a forward-model which maps the model parameters to predictions of the experimental data, and a comparison between the prediction and the measurement yields a log-likelihood.
We then sum up the log-likelihoods from all these different instruments, and also add some log-prior values which apply some physical constraints to the fields.
Consequently I doubt this 'model' falls neatly into a category - we don't have a choice of what the model is, it is dictated by how the actual instruments function that collect our experimental data.
Data set The data set is composed of 500x500 images, and there is one image for each camera so total data points is 500x500x4 = $10^6$. Error model We take all errors in the problem to be Gaussian at the moment. At some point I might try to move over to a student-t error model just for some extra flexibility, but things still seem to function well with just Gaussians. Likelihood example This is a plasma physics experiment, and the vast majority of our data comes from cameras pointed at the plasma with particular filters in front of the lenses to look only at specific parts of the light spectrum.
To reproduce the data there are two steps; first we have to model the light that comes from the plasma on the mesh, then we have to model that light back to a camera image.
Modelling the light that comes from the plasma unfortunately depends on what are effectively rate coefficients, which say how much light is emitted by different processes given the fields. These rates are predicted by some expensive numerical models, so we have to store their output on grids, and then interpolate to look up values. The rate function data is only ever computed once - we store it then build a spline from it when the code starts up, and then that spline gets used for all the function evaluations.
Suppose $R_1$ and $R_2$ are the rate functions (which we evaluate by interpolation), then the emission at the $i$'th vertex of the mesh $\mathcal{E}_i$ is given by $$ \mathcal{E}_i = R_1(x_i, y_i) + z_i R_2(x_i, y_i) $$ where $(x,y,z)$ are the 3 fields we model on the mesh. Getting the vector of emissions to a camera image is easy, it's just multiplication with a matrix $\mathbf{G}$ which encodes what parts of the mesh each camera pixel looks through.
Since the errors are Gaussian the log-likelihood for this particular camera is then $$ \mathcal{L} = -\frac{1}{2} (\mathbf{G}\vec{\mathcal{E}} - \vec{d})^{\top}\mathbf{\Sigma}^{-1} (\mathbf{G}\vec{\mathcal{E}} - \vec{d}) $$
where $\vec{d}$ is the camera data. The total log-likelihood is a sum of 4 of the above expressions but for different cameras, which all have different versions of the rate functions $R_1, R_2$ because they're looking at different parts of the light spectrum.
Prior example We have various priors which effectively just set certain upper and lower bounds on various quantities, but these tend not to act too strongly on the problem. We do have one prior that acts strongly, which effectively applies Laplacian-type smoothing to the fields. It also takes a Gaussian form: $$ \text{log-prior} = -\frac{1}{2}\vec{x}^{\top}\mathbf{S}\vec{x} -\frac{1}{2}\vec{y}^{\top}\mathbf{S}\vec{y} -\frac{1}{2}\vec{z}^{\top}\mathbf{S}\vec{z} $$ |
Let $M$ be a compact finite-dimensional manifold and $f\colon M\to M$ be a diffeomrphism. By $P_n(f)$ we denote the number of periodic points of $f$ with period $n$, that is, the number of fixed points of $f^n$.Katok Lyapunov exponents, entropy and periodic orbits for diffeomorphisms, Publications Mathématiques de l'IHÉS 51, no. 1 (1980), pp. 137-173 showed that the topological entropy of a $\mathcal{C}^{1+\varepsilon}$ diffeomorphism of a compact surface obeys the following inequality: $$\limsup_{n\to\infty} \frac{\log P_n(f)}{n}\ge h(f).$$He also conjectured (see paragraph 5 page 141 of the article cited above) that this inequality holds in any dimension generically in the $\mathcal{C}^1$-topology. In other words:
There is a dense $G_\delta$ set $\mathcal{G} \subset \text{Diff}^1(M)$ such that for any $f\in\mathcal{G}$ one has$$\limsup_{n\to\infty} \frac{\log P_n(f)}{n}\ge h(f).$$Is this conjecture still open? I would be grateful for any references.
Let $M$ be a compact finite-dimensional manifold and $f\colon M\to M$ be a diffeomrphism. By $P_n(f)$ we denote the number of periodic points of $f$ with period $n$, that is, the number of fixed points of $f^n$.Katok Lyapunov exponents, entropy and periodic orbits for diffeomorphisms, Publications Mathématiques de l'IHÉS 51, no. 1 (1980), pp. 137-173 showed that the topological entropy of a $\mathcal{C}^{1+\varepsilon}$ diffeomorphism of a compact surface obeys the following inequality: $$\limsup_{n\to\infty} \frac{\log P_n(f)}{n}\ge h(f).$$He also conjectured (see paragraph 5 page 141 of the article cited above) that this inequality holds in any dimension generically in the $\mathcal{C}^1$-topology. In other words:
Nice question. I think it has not been directly adressed in the literature, but combining known results it seems that a positive answer to that question can be given. Still some details must be carried out (which I did not) to be sure.
The main point is that $C^1$-
far from homoclinic tangencies the result must be true due to the existence of a maximal entropy measure (see here, here, here and references therein) where the splitting has one dimensional center (and so Katok type results can be carried out, see here). There is a subtelty here which is that the central exponent may be a priori equal to zero, in which case one would like to use the genericity hypothesis to show that the estimate holds (I think the concept of principal simbolic extension in the above references should be helpful for this). Here there might be some work to be done.
In the open set where homoclinic tangencies are dense, results of Kaloshin and from Bonatti-Diaz-Fisher suggest that there should be superexponential growth of periodic orbits, but some care must be taken since the precise statements do not work for this case. On the other hand, since what is needed is way less than superexponential growth, I don't see this part as being the hard one. |
At last I think I understand what this strange business with spinors is
really all about. Everybody knows that it has to do with SU(2) being a double cover of SO(3) of course, but I think I finally comprehend why.
SO(3) is the group of proper rotations in 3D space. For instance, a rotation around the $z$-axis by angle $\theta$ can be represented by the following matrix product:
\[\begin{pmatrix}x'\\y'\\z'\end{pmatrix} = \begin{pmatrix} \cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1 \end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix}.\]
Yet there's another way to represent the same rotation using matrix algebra. We can arrange the coordinates into a 2×2 complex matrix and perform the following multiplication:
\[\begin{pmatrix} iz'&y'+ix'\\-y'+ix'&-iz' \end{pmatrix}= \begin{pmatrix}e^{i\theta/2}&0\\0&e^{-i\theta/2}\end{pmatrix} \cdot \begin{pmatrix}iz&y+ix\\-y+ix&-iz\end{pmatrix}\cdot \begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}.\]
If you carry out the computation (not too difficult) you'll find that this representation of a spatial rotation is equivalent to the previous representation. And it can be easily generalized to an arbitrary rotation (though the calculations tend to get more tedious). What we find is that the complex matrix used to represent a rotation will be a unitary complex matrix (unitary meaning that when multiplied by its own conjugate transpose, the result is the unit matrix) of determinant +1. These matrices form a group under matrix multiplication: the group is called SU(2). In other words, SU(2) is equivalent to SO(3).
Well, almost. Whereas in the first representation, there's precisely one matrix that corresponds to a particular spatial rotation, in the second representation there are two:
\[\begin{pmatrix}e^{i\theta/2}&0\\0&e^{-i\theta/2}\end{pmatrix}{\rm~~and~~} \begin{pmatrix}-e^{i\theta/2}&0\\0&e^{-i\theta/2}\end{pmatrix} = \begin{pmatrix}e^{i\theta/2+\pi}&0\\0&e^{-i\theta/2-\pi}\end{pmatrix}\]
both correspond to the same rotation, since the minus sign, occurring twice, will cancel. Another way of putting this is that SU(2) is a
double cover of SO(3). If $\vec{B}$ is an element of SU(2) and $\vec{V}$ is the 2×2 complex matrix representation of a set of coordinates, the rotation can be written up as $\vec{B}\cdot\vec{V}\cdot\overline{\vec{B}}$.
So far so good. These matrices can be used to represent a rotation in 3D space; i.e., a coordinate transformation that corresponds to a proper rotation of the coordinate system. In which case it is customary to talk about covariant and contravariant quantities, i.e., quantities that transform differently under a coordinate transformation.
What spinors make possible is to define a third type of quantity. One that transforms according to $\vec{B}\cdot\vec{V}$, i.e., "one half" of a rotation if you wish. Or (I think Penrose called it that) the "square root" of a rotation. A quantity like this is called a
spinor. The most notorious property of a spinor of course is that when $\theta$ is 360º, the spin matrix is not the identity matrix, so it takes a rotation of 720º in order for a spinor to return to itself.
Needless to say, all this stuff can be generalized to higher dimensions, but not arbitrarily. The correspondence between SU(2) and SO(3) is accidental. In general, the dimensionality of SU($n$) is $n^2-1$, and the dimensionality of SO($m$) is $m(m-1)/2$, so corresponding pairs of $n$ and $m$ can be found by solving the Diophantine equation $2(n^2-1)=m(m-1)$. Solutions include the trivial SU(1) = SO(1) (both groups consisting of the single 1×1 "matrix" (1)), $n=2$ and $m=3$, and $n=4$, $m=6$ (SU(4) and SO(6) are isomorphic, i.e., one is
not a double cover of the other). There are also solutions to this equation that do not result in a covering at all.
In other words, these kinds of spinors are a rather special counterpart to 3D rotations. They may, in fact, provide a glimpse into that great unanswered mystery: why is the space we live in 3-dimensional?
So what are spinors for anyway? Quantities that transform like a spinor appear in quantum mechanics: the coefficients in the Dirac-equation form a spinor, for instance, because when you rewrite the equation under a change of coordinates, in order for it to remain invariant, the Dirac-coefficients must transform as spinors do. |
We show that the canonical random-cluster measure associated to isoradial graphs is critical for all q \geq 1. Additionally, we prove that the phase transition of the model is of the same type...
Pages
We calculate the scaling dimensions of operators with large global charge and spin in 2+1 dimensional conformal field theories. By the state-operator correspondence, these operators correspond...
The Drude weight is a central quantity for the transport properties of quantum spin chains. The canonical definition of Drude weight is directly related to Kubo formula of conductivity. However...
We study metastable motions in weakly damped Hamiltonian systems. These are believed to inhibit the transport of energy through Hamiltonian, or nearly Hamiltonian, systems with many degrees of...
We consider a system of N bosons interacting through a singular two-body potential scaling with N and having the form N^{3\beta-1} V (N^\beta x), for an arbitrary parameter \beta \in (0,1). We...
In an earlier paper, we proved the validity of large deviations theory for the particle approximation of quite general chemical reaction networks (CRNs). In this paper, we present a more...
This paper initiates the study of semitoric integrable systems with two degrees of freedom and with proper momentum-energy map, but with possibly nonproper S 1-momentum map. This class of...
Oleg Viro introduced an invariant of rigid isotopy for real algebraic knots and links in RP
3 which can be viewed as a first order Vassiliev invariant. In this paper we classify real...
Scrambling in interacting quantum systems out of equilibrium is particularly effective in the chaotic regime. Under time evolution, initially localized information is said to be scrambled as it...
We introduce a notion of weakly log-canonical Poisson structures on positive varieties with potentials. Such a Poisson structure is log-canonical up to terms dominated by the potential. To a...
We propose a class of {\rm AdS}_3/{\rm CFT}_2 dualities with N=(2,2) supersymmetry. These dualities relate string theory on {\rm AdS}_3 \times ({\rm S}^3\times \mathbb{T}^4)/{\rm G} to marginal...
We prove that the K-theoretic Nekrasov instanton partition functions have a finite radius of convergence in the instanton counting parameter and are holomorphic functions of the Coulomb... |
I was inspired in the shape of the integrals for $\zeta(2)$ in
A. Córdoba, Encounters at the interface between Number Theory and Harmonic Analysis, Proceedings of the Segundas Jornadas de Teoría de Números, page 102 (2007) Biblioteca de la Revista Matemática Iberoamericana, to ask to Wolfram Alpha online calculator about integrals of a different kind, to obtain integrals that get zeta values.
Example 1.After a lot of trials with Wolfram AlphaI've found ( here was fixed a typo, see the comments from the users, thanks them) $$\int_{-\infty}^\infty\frac{x^2}{-1+\cosh (2x)}dx=\frac{\pi^2}{6}.$$
When I did the change of variables $u=e^{2x}$ I can show (using the change of variables and the evaluation from Wolfram Alpha) that previous integral is evaluated as $$\int_0^\infty\frac{\log^2 u}{-8u+4u^2+4}du=\frac{\pi^2}{6},$$
see it:
integrate (log x)^2/(-8x+4x^2+4) dx from x=0 to x=infinite
Example 2.After a lot experiments, I would like to get $\frac{\pi^4}{90}$ as an integral of previous kind, I say a double integral, and not a multiple of an integral of this kind, some of my attempts were, for example (in comments you can see an ample variety) this $$\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{x^2y^2}{(-1+\cosh (2x))(1+\cosh(y)))}dxdy=\frac{\pi^4}{9},$$ or this $$\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{x^2y^2}{(1+\cosh (2x))(1+\cosh( \sqrt{5}y)))}dxdy=\frac{\pi^4}{90\sqrt{5}}.$$ In comments are more of my attempts to get a similar integral for $\zeta(4)=\frac{\pi^4}{90}$.
These are
the codes of previous Example 2:
integrate x^2y^2/((-1+cosh(2x))(1+cosh(y)))dx dy from x=-infty to x=infty, from y=-infty to y=infty
integrate x^2y^2/((1+cosh(2x))(1+cosh(sqrt(5)y)))dx dy from x=-infty to x=infty, from y=-infty to y=infty
I would like to know
Question.Can you provide to me hints to finish the calculation $\int_0^\infty\frac{\log u}{-8u+4u^2+4}du=\frac{\pi^2}{6}$? Can you obtain an integral, not a multiple of an integral of previous kind (see also the comments) to get $\zeta(4)$? Many thanks. |
I have an expectation given as:
$\mathbb{E}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$
where $K$ is just an arbitrary number (i.e. the strike price, but that's unimportant) and $S$ can be modelled by the equation $S_{t} = \exp((r-\frac{1}{2})t + \sigma W_{t})$. Also, the expectation is under the $\mathbb{P}$-measure, not the $\mathbb{Q}$-measure, so effectively this expectation is $\mathbb{E}^{\mathbb{P}}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$
Now, when trying to evaluate $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right)$ under the $\mathbb{Q}$-measure, then solving this expectation is fairly easy, since you can integrate the SDE and take the $\log$ of $S$ to get $\log(S_{T}) = \log(S_{0}) + (r-\frac{1}{2})T + \sigma\sqrt{T}N(0,1)$ (since $W_{t}\approx \sqrt{T}N(0,1)$) and thus we have for $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right) = \mathbb{P}({S_{T}\geq K})$:
$\log(S_{T}) = (r-\frac{1}{2})T + \sigma\sqrt{T}N(0,1) > \log(K)$ and thus rearranging this equation gives
$N(0,1) > \frac{\log(K/S_{0}) - (r-\frac{1}{2}\sigma^{2})T}{\sigma\sqrt{T}}$
and through a little bit more rearrangement (i.e. knowing that $N(0,1)<-x = 1 - N(x)$ we get finally $d_{2} = \frac{(r-\frac{1}{2}\sigma^{2})T + \log(S_{0}/K)}{\sigma\sqrt{T}}$ and thus $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right) = N(d_{2})$.
The problem I have now however is that I'm a bit unsure how to find $\mathbb{E}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$, I honestly don't really know how to calculate it, so if someone could help me out I'd really appreciate it. Thanks in advance. |
Show that $L = L_\phi \cup L_{\{\sum^*\}} \notin RE$ with Rice theorem.
Well I did show that with reduction, by using $HP'$.
Simply by creating a function from $f(\langle M \rangle, x) = (M')$
Thus, if $M$ stops on $x$ then $M'$ simulates the machine that stops on the empty word and accepts/rejects like it.
$(\langle M \rangle, x) \in HP' \rightarrow M$ rejects every input$ \rightarrow \langle M' \rangle \in L_2$
$(\langle M \rangle, x) \notin HP' \rightarrow M$ halts on $x \rightarrow L(M') = \phi $(it stops only on the empty word) $ \rightarrow \langle M' \rangle \notin L_2$
EDIT: In other words, can I just say that there're languages that contains $L$ thus $\in RE$ and languages that don't contain $L$ thus $\notin RE$. For example $L_{\phi \cup \sum^*\cup L_1}$ or just $L_1$. And $L_1$ is defined as a language that accepts all the languages with at least 200 words for example. Will this work for the Rice's theorem? |
The time hierarchy theorem states that
$DTIME(f(n)) \neq DTIME(f(n)\log(f(n)))$
(Let me acknowledge that this statement isn't 100 percent accurate because f must be "time constructible" and potentially other things that I'm missing)
If we were to take some function f, we can consider
$DTIME(f(n)) \subset DTIME(f(n)\sqrt{\log(f(n))})\subset DTIME(f(n)\log(f(n)))$
Where we know that at least one of these inclusions is strict. Do we know of examples where one of these inclusions isn't?
In general we might want to prove that $DTIME(f(n)) \neq DTIME(f(n)g(f(n)))$ for a function $g$ which is as small as possible. Do we have better upper bounds on $g$ (like $g(n) = \log\log n$)? do we have non-trivial lower bounds on $g$ in the form of some (time constructible) $f$ for which $DTIME(f(n) = DTIME(f(n)g(f(n)))$? |
I have been strugling with this LL(1) transformation for far longer than expected and am getting quite desperate. We are given a grammar $ G = \left\{V_N, V_T, S, \Phi \right\}$, with the non-terminal set $V_N = \left\{S\right\}$, terminal set $V_T = \left\{\underline{o}, \underline{l}, \underline{r}, \underline{i}\right\}$, start symbol $ S $ and the production rules in $\Phi$:
$$ \begin{equation}\begin{split} S \rightarrow & \ S \underline{o} S \\ S \rightarrow & \ S \underline{l} S \underline{r} \\ S \rightarrow & \ \underline{i}\end{split}\end{equation}$$
I have tried the standard tricks like left factoring, left recursion removing, which lead to an infinite transformation loop. Trying to fix first and follow set clashes also did not help. Here are some results after applying the transformations:
$$ \begin{equation}\begin{split} S \rightarrow & \ \underline{i} A & \ \ \ \ \ \ \ \ \ \ \ S \rightarrow & \ S A \\ A \rightarrow & \ \underline{o} S A & \ \ \ \ \ \ \ \ \ \ \ S \rightarrow & \ \underline{i} \\ A \rightarrow & \ \underline{l} S \underline{r} A & \ \ \ \ \ \ \ \ \ \ \ A \rightarrow & \ \underline{o} S \\ A \rightarrow & \ \epsilon & \ \ \ \ \ \ \ \ \ \ \ A \rightarrow & \ \underline{l} S \underline{r} \end{split}\end{equation} $$
However I go about the problem, I cannot seem to get even close to eliminating the double entries in my LL(1) table. |
I have attempted to find a definition of a monoidal category which incorporates $n$-fold tensor products instead of just binary tensor products.
Definition. A "multi-monoidal category" consists of a category $\mathcal{C}$, for every $n \geq 0$ a functor $T_n : \mathcal{C}^n \to \mathcal{C}$, denoted by $(A_1,\dotsc,A_n) \mapsto A_1 \otimes \dotsc \otimes A_n$, an isomorphism $\eta : T_1 \cong \mathrm{id}_{\mathcal{C}}$, for all $n_1,\dotsc,n_k \geq 0$ an isomorphism $$\mu_{n_1,\dotsc,n_k} : T_k \circ (T_{n_1} \times \dotsc \times T_{n_k}) \cong T_{n_1+\dotsc+n_k}.$$
The following coherence conditions should hold:
Coherence of $\eta$ with $\mu$: We have $\mu_{n_1} = \eta \circ T_{n_1} : T_1 \circ T_{n_1} \to T_{n_1}$.
Coherence of $\mu$ with $\mu$: The square
$$\begin{array}{cc} T_k \circ \bigl(T_{n_1} \circ (T_{m_{11}} \times \dotsc \times T_{m_{1n_1}}) \times \dotsc \times T_{n_k} \circ (T_{m_{k1}} \times \dotsc \times T_{m_{k n_k}})\bigr) & \rightarrow & T_k \circ (T_{m_{11}+\dotsc+m_{1 n_1}} \times \dotsc \times T_{m_{k1}+\dotsc+m_{k n_k}}) \\ \downarrow && \downarrow \\ T_{n_1+\dotsc+n_k} \circ (T_{m_{11}} \times \dotsc \times T_{m_{k n_k}}) & \rightarrow & T_{m_{11}+\dotsc+m_{k n_k}} \end{array}$$ commutes.
Notice that when $\mathcal{C}$ is discrete, this is the monadic definition of a monoid (as compared to the usual definition).
Questions. (1) Did I forget some coherence condition?(2) Is this concept already known? Does it have a name? It really looks like the most natural thing in the world, especially when you think "operadic" or "monadic". (3) Most important for me: Is this concept equivalent to the definition of a monoidal category? If yes, what is a reference for this? The idea for the equivalence is straight forward (the $n$-fold tensor product is an iteration of binary tensor products etc.), but I believe that it will probably require much work to check the coherence conditions in both directions. |
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Now showing items 1-10 of 32
The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC
(Elsevier, 2018-01)
This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ...
First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
(Elsevier, 2018-06)
The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ...
D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV
(American Physical Society, 2018-03)
The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ...
Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV
(Elsevier, 2018-05)
We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2018-01)
We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ...
Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV
(Springer Berlin Heidelberg, 2018-07-16)
Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range |η| < 0.8 ... |
Your problem is NP-hard. There is a reduction from Independent Set to its decision version.Consider an instance $G=(V,E)$ of Independent Set, you construct a network with vertices $\{s,t\}\cup V\cup V'$ where each vertex in $V'$ corresponds to a pair of vertices in $V$. For example, if $V=\{1,2,3\}$, then $V'=\{v_{12},v_{23},v_{13}\}$. Then we construct ...
Your post-traversal numbering of the component $C$ is incorrect. The vertex labeled as $1,6$ was assigned its post-traversal number too early; a depth-first search will also explore the vertices of $D$ before assigning the post-traversal number to this vertex.
The problem statement needs to be specified more precisely to be meaningful.If I'm being pedantic, I would say that the minimum number of propositional variables is zero: you can solve the graph coloring problem (taking exponential time if necessary), then either use the formula $\text{True}$ or $\text{False}$ according to whether the graph is $C$-...
I suggest you use a search algorithm on the following statespace: the state is a set of adjacent tiles whose numbers sum to 7 or less; it is possible to transition from state $s$ to $s'$ if $s'$ is obtained by adding to $s$ one tile that is adjacent to some tile in $s$. The initial state has the empty set $\emptyset$, and it can transition to any state ...
Your conjecture is false. There are regular graphs with an even number of vertices yet without a 1-regular subgraph. See this question on Mathematics.The complement of such a graph gives a counterexample to your claim that you can always add a perfect matching to increase the regularity (when the number of vertices is even).In the bipartite case, however,...
While your description of the problem is not super clear. I can guess that you're looking at a maximum bipartite matching problem. Which by the way, can also be viewed as a degenerate case of maxflow problem. Both problems are classic, you should have no problem finding resources on them. For example, this.
The minimum cost perfect matching problem is formally defined as an optimization problem where:The set of instances is the set of all edge-weighted undirected graphs $G=(V, E, w)$, where $w : E \to \mathbb{R}$.Given an instance $G=(V, E, w)$, the set of feasible solutions are the perfect matchings $M$ of $G$, i.e., all the sets $M \subseteq E$ such that, $...
The problem was considered by Kahn and Saks in their paper Balancing poset extensions. Earlier related works are Fredman, How good is the information theory bound in sorting? and Linial, The information-theoretic bound is good for merging.Kahn and Saks showed that partially sorted lists can be sorted in $O(\log M)$, where $M$ is the number of possible ... |
First important thing to understand is that vortices and vorticity are not the same thing, despite the similarity of the words. A vortex is a region in a flow with spinning features (at a rather large scale if you wish), but it may be irrotational (zero vorticity). Vorticity is a local property of the fluid, the rate of rotation of an imaginary particle fluid at that point. A viscous flow between two plates is actually rotational (non-zero vorticity), although it is laminar (layered).
To answer your question regarding potential flow. A potential flow is a flow where the velocity field derives from a potential. In 2D for example we'd have, given the velocity potential $\phi$:$$ u = \frac{\partial}{\partial x} \phi\\ v = \frac{\partial}{\partial y} \phi$$Such a flow is necessarily irrotational ($\nabla \wedge \vec{u} = 0$), since$$ -\frac{\partial}{\partial y}u+\frac{\partial}{\partial x}v = -\frac{\partial}{\partial y}\frac{\partial}{\partial x}\phi + \frac{\partial}{\partial x}\frac{\partial}{\partial y}\phi = 0$$Note that it doesn't tell us anything directly about vortices.
But how do we know if we're supposed to suspect a potential flow in the first place? A clear no-go are viscous flows with no-slip boundaries. A no-slip boundary (in a reference frame where the boundary is at rest) is characterized by$$ \vec{u}\vert_\text{boundary} = 0$$The fluid in contact with the no-slip boundary (hereafter just "wall") does not move, but the one slightly above does. If the wall is parallel to the $x$-axis, then this means that the variation of the flow in the $y$-axis must be non-zero:$$ \frac{\partial}{\partial y} u \neq 0$$else the fluid would never move. But as soon as you have that, then of course $\omega= \nabla \wedge \vec{u} \neq 0$ too! For a viscous fluid, there is always vorticity generated at a wall.
For non-viscous flows, walls are given by the condition$$\vec{u}\cdot\vec{n}\vert_\text{boundary} = 0$$
i.e., no flow may go through the wall, but it might well slip tangentially without any loss of speed. Given a reasonable geometry the flow can move around, there should be no vorticity. I wouldn't bet on what happens if you blow a perfect fluid into a closed box: hard to believe that the flow impacting with box walls (what with the corners and all) wouldn't be rotational, but I'd have to check.
There's still another mechanism that can introduce $\partial_y u\neq0$ or $\partial_x v \neq 0$: very particular body forces. That's actually one of the means by which you can study turbulence in a periodic box of fluid (
i.e., without walls): you generate shear by a random body force. But you won't see those in most cases.
In applications, potential flow is particularly used in the inviscid flow approximation around profiles, with the flow extending to infinity all around (avoiding the "box scenario" I mentioned above). At most you'd have the gravity force, but that's a nice uniform field in most approximations.
Moreover, in 2D, an irrotational inviscid flow stays irrotational in time by the Kelvin circulation theorem. Good stuff. Ever noticed that we always study airfoils in 2D? Interesting.
(There's another reason for studying airfoils in 2D: the lift of a wing section per unit length is actually quite well predicted, whereas in 3D inviscid flows, the lift is always zero! See this other question. That's where more advanced methods kick in if you need to calculate stuff in 3D anyway.)
If I missed something, ask me some questions through the comments and I'll try to update the answer. |
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Now showing items 1-10 of 27
Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ... |
Charge Stabilization in Nonpolar Solvents
Original Entry by Michelle Borkin, AP225 Fall 2009
Overview
M. F. Hsu, E. R. Dufresne, and D. A. Weitz, Langmuir 21, 4881-4887 (2005).
Keywords Summary
This paper investigates the use of surfactants to control charges is nonpolar solvents (<math>\epsilon \approx 2</math> versus for an aqueous solution <math>\epsilon \approx 80</math>) where the electrostatic charge barrier is much greater than
kT. Understanding nonpolar solvents and colloid interactions are important for industrial and commercial applications such as electrophoretic ink or the stabilization of soot particles in oil. Surfactants play the key role in creating in such solutions the charge-stabilizing aggregates. The research presented in this paper focuses on nanometer sized reverse micelles in nonpolar solvents and investigates the electrokinetics and thermodynamic properties to explain how the micelles effect charge. As opposed to simple salt ions, these large reverse micelles have low ionization energies and charge surfaces by stabilizing counterions. They find very strong surface potentials (2.0-4.4 kT) and Debye screening lengths (0.2-1.4 <math>\mu m</math>) that strongly depend on the concentration of reverse micelles in the system. Soft Matter
For the experiments, the reverse micelles were created using aerosol-OT (AOT) which forms nanometer sized reversed micelles (contains ~30 surfactant molecules) above its 1mM in dodecane critical micelle concentration (cmc). The colloid particles in the system are PMMA particles with PHSA grafted to their surface for steric stabilization (radius = 780 nm). The solution is contained in cells between glass plates thus the model and results presented are based on 2D descriptions and analysis. As shown in
Figure 1, when AOT (i.e. the reverse micelles) is added to the solution (b) the colloidal particles evenly disperse due to the electrostatic forces.
They observe a strong dependence on AOT concentration in controlling the interactions, specifically by controlling the range of interaction among the particles. As shown in
Figure 2 (b), is "soft" and long ranged - the interparticle repulsion is greater than the thermal energy for 5x the particle radius and the interactions become "stiffer" and short ranged as the micelle concentration is increased. However, the surface potential barley changes with micelle concentration. Other striking observations include that the measured zeta potential is comporable to those measured in water with highly charged particles (the Debye screening lengths are also much larger than those measured in such a solution).
Finally, to further investigate screening lengths the ionic strength of conductivity was measured. The measured conductivity (
<math>\sigma</math>) as a function of AOT concentration is shown in Figure 3 (a). The conductivity values span more than two orders of magnitude and the fraction of ionized micelles is independent of concentration. The derived inverse screening lengths are also plotted in Figure 3 (b). Also, instead of following a standard charge exchange description as found in weak electrolytes, a neutral micelle reversibly exchanges charge through collision <math>A + A \rightleftharpoons A^{+} + A^{-}</math> leading to an ionization fraction independent of concentration (schematic in Figure 3). |
I assume you start with a massive Klein-Gordon QFT:
$$ S[\phi] = \int d^4 x \left( \frac{1}{2} (\partial \phi)^2 - \frac{1}{2}m^2 \phi^2 \right), $$
and you want to treat the massive term as a perturbation around the massless Klein-Gordon QFT just for the heck of it (even though an exact nonperturbative solution exists in this case).
Deriving Feynman rules shouldn't be a big problem. The easiest way is to use path integrals. The derivation then goes in three steps:
Step 1: you write down a generic path integral for some expectation functional $\Omega[\phi]$:$$ \left< \Omega \right> = \int D\phi e^{i S[\phi]} \Omega[\phi]. $$
Now (even though this integral can be computed rigorously) you wish to split $S[\phi]$ into the free (massless) and interaction parts, ending up with$$ \left< \Omega \right> = \int D\phi e^{i S_0[\phi]}\left( 1 + i\int d^4 x \left( - \frac{1}{2} m^2 \phi^2 \right) + \frac{1}{2!}i^2 \left[ \int d^4 x \left( - \frac{1}{2} m^2 \phi^2 \right)\right]^2 + \dots \right) \Omega[\phi]. $$
You then observe that the evaluation of the expectation $\Omega$ in the full theory is equivalent to the evaluation of the infinite series of the complex expectations containing powers of the space-time integrals in the "free" theory with action $S_0[\phi] = \int d^4 x \frac{1}{2} (\partial \phi)^2$.
You come up with a graphical way to represent the terms in your expansion. Since each of the functionals in the series is evaluated in a free theory, the Wick theorem holds, meaning that it is a sum of contractions. In the contractions, two types of vertices contribute: the external vertices coming from $\Omega$, and internal vertices coming from the integrals containing $m$. Moreover, the factor of $\frac{1}{2}$ cancels out $2!=2$ ways to contract a 2-valent vertex, so the expression for the internal (interaction) vertex is:
$$ A(v) = \int d^4 x_v i m^2. $$
Different terms in the series then correspond to diagrams with different number of interaction vertices.
Step 2: one remembers that the expectation can be normalized to give $\left<1\right> = 1$, if one does not consider vacuum bubble diagrams. This is standard QFT diagrammatics, textbook authors usually refer to it as "bubble graph exponentiation".
Step 3: one writes propagators of the free theory in terms of the momentum integrals:
$$ \Delta(x, y) = \int \frac{d^4 p}{(2\pi)^4} \frac{i e^{i p x}}{p^2 + i \varepsilon}.$$
Notice the absense of $m$.
Now we can perform the vertex integrals, ending up with the momentum-space Feynman rules:
For every propagator with momentum $p$ we associate $$\frac{i}{p^2 + i \varepsilon},$$ For every vertex we associate $$i m^2,$$ We integrate over undetermined loop momenta, and impose momemtum conservation.
Now one can calculate the actual 2-point function using the rules which she just derived. Consider $\Omega[\phi] = \phi(x) \phi(y)$ for $x, y$ spacetime points. The only contributing diagram is a bamboo graph with $n$ interacting vertices. Its contribution is given by
$$ \left< \phi(x) \phi(y) \right> = \sum_n \int \frac{d^4 p}{(2\pi)^4} e^{i p x} \left( \frac{i}{p^2 + i \varepsilon} \right)^{n+1} \cdot \left( i m^2 \right)^n = \int \frac{d^4 p}{(2\pi)^4} e^{i p x} \frac{i}{p^2 - m^2 + i \varepsilon},$$
where the last equation is the sum of the geometric series. Thus, we obtain the usual massive propagator. Perturbative and nonperturbative methods agree in this case, which is kinda nice. |
Surface energies Contents Introduction
"The comprehension of the laws which govern any material system is greatly facilitated by considering the energy and entropy of the system in the various states of which it is capable. As the difference of the values of the energy for any two states represents the combined amount of work and heat received or yielded by the system when it is brought from one state to another, and the difference of entropy is the limit of all the possible values of the integral <math>\int{\frac{dQ}{t}}</math>, (
dQ denoting the element of the heat received from external sources, and t the temperature of the part of the system receiving it,) the vary values of the energy and entropy characterize in all that is essential the effects producible by the system in passing from one state to another." J. Willard Gibbs. "On the equilibrium of heterogeneous substances." Trans. Conn. Acad., III, pp. 108 - 248, Oct 1875 - May 1876, and pp. 343 - 524, May 1877 - July, 1878.
The Gibbsian layer: "...in the transition zone between two phases there exits a field of electric and molecular forces that decreases with distance into each of the contiguous phases. We shall refer to these forces as surface forces (not to be confused with the term 'surface forces' used in mechanics to denote forces applied to a surface)." B.V. Derjaguin and N.V. Churaev, Surface forces, 1987, pp. 4-5. The non-Gibbsian layer: "In contrast to Gibbs' interpretation, the excess of entropy and mass of the components related to the interlayer depend on h (with small values of h), and here it is impossible to distinguish the excesses on each interface." B.V. Derjaguin, Theory of stabiity of colloids and thin films, 1989, p.23-24. Surface energy and measurements
Surface energy quantifies the disruption of intermolecular bonds that occurs when a surface is created. For a liquid, the surface tension (force per unit length) and the surface energy density are identical. Water, a special case, has a surface energy density of 0.08 J/m
2 and a surface tension of 0.08 N/m.
As first described by Thomas Young in 1805 in the Philosophical Transactions of the Royal Society of London, it is the interaction between the forces of cohesion and the forces of adhesion which determines whether or not wetting, the spreading of a liquid over a surface, occurs. If complete wetting does not occur, then a bead of liquid will form, with a contact angle which is a function of the surface energies of the system. Surface energy is most commonly quantified using a contact angle goniometer and a number of different methods.
Thomas Young described surface energy as the interaction between the forces of cohesion and the forces of adhesion which, in turn, dictate if wetting occurs. If wetting occurs, the drop will spread out flat. In most cases, however, the drop will bead to some extent and by measuring the contact angle formed where the drop makes contact with the solid the surface energies of the system can be measured.
Young established the well-regarded Young's Equation which defines the balances of forces caused by a wet drop on a dry surface. If the surface is hydrophobic then the contact angle of a drop of water will be larger. Hydrophilicity is indicated by smaller contact angles and higher surface energy. (Water has rather high surface energy by nature; it is polar and forms hydrogen bonds). The Young equation gives the following relation,
<math>\gamma_{\mathrm{SL}}+\gamma_{\mathrm{LV}}\cos{\theta_\mathrm{c}}=\gamma_{\mathrm{SV}}\,</math>
where
<math>
\theta_\mathrm{c}=\arccos\left(\frac{r_\mathrm{A}\cos{\theta_\mathrm{A}}+r_\mathrm{R}\cos{\theta_\mathrm{R}}}{r_\mathrm{A}+r_\mathrm{R}}\right)~;~~r_\mathrm{A}=\left(\frac{\sin^3{\theta_\mathrm{A}}}{2-3\cos{\theta_\mathrm{A}}+\cos^3{\theta_\mathrm{A}}}\right)^{1/3}~;~~r_\mathrm{R}=\left(\frac{\sin^3{\theta_\mathrm{R}}}{2-3\cos{\theta_\mathrm{R}}+\cos^3{\theta_\mathrm{R}}}\right)^{1/3}</math>where <math>\gamma_{\mathrm{SL}}</math>, <math>\gamma_{\mathrm{LV}}</math>, and <math>\gamma_{\mathrm{SV}}</math> are the interfacial tensions between the solid and the liquid, the liquid and the vapor, and the solid and the vapor, respectively. The equilibrium contact angle that the drop makes with the surface is denoted by <math>\theta_\mathrm{c}</math>. Even in a perfectly smooth surface a drop will assume a wide spectrum of contact angles ranging from the so called
advancing contact angle, <math>\theta_\mathrm{A}</math>, to the so called receding contact angle, <math>\theta_\mathrm{R}</math>. The equilibrium contact angle (<math>\theta_\mathrm{c}</math>) can be calculated from <math>\theta_\mathrm{A}</math> and <math>\theta_\mathrm{R}</math>. Electrostatic and induction energies
Charge-charge Angle and temperature independent <math>\frac{q_{a}q_{b}}{4\pi \varepsilon _{0}r}\,\!</math> Charge-dipole Averaged over all orientations <math>-\frac{kT}{3}\frac{q_{a}^{2}\mu _{b}^{2}}{(4\pi \varepsilon _{0})^{2}r^{4}}\,\!</math> At the maximum <math>-\frac{q_{a}\mu _{b}}{4\pi \varepsilon _{0}r^{2}}\,\!</math> Charge-quadrupole Averaged over all orientations <math>-\frac{kT}{20}\frac{q_{a}^{2}Q_{b}^{2}}{(4\pi \varepsilon _{0})^{2}r^{6}}\,\!</math> Dipole-dipole Averaged over all orientations <math>-\frac{2kT}{3}\frac{\mu _{a}^{2}\mu _{b}^{2}}{(4\pi \varepsilon _{0})^{2}r^{6}}\,\!</math> At the maximum <math>-\frac{2\mu _{a}\mu _{b}}{4\pi \varepsilon _{0}r^{3}}\,\!</math> Charge-induced dipole Angle and temperature independent <math>-\frac{q_{a}^{2}\alpha _{b}}{8\pi \varepsilon _{0}r^{4}}\,\!</math> Dipole-induced dipole Averaged over all orientations <math>-\frac{\mu _{a}^{2}\alpha _{b}}{4\pi \varepsilon _{0}r^{6}}\,\!</math> where charge on particle a (C) = <math>q_{a}\,\!</math> interparticle distance (m) <math>r\,\!</math> dipole moment of particle a (C-m) = <math>\mu _{a}\,\!</math> polarizability of particle a (m3) = <math>\alpha _{a}\,\!</math> quadrupole moment of particle a (C-m3) = <math>Q_{a}^{{}}\,\!</math> permittivity of free space = <math>\varepsilon _{0}\,\!</math> Molecular polarizabilities are often estimated from the sum of bond polarizabilities. Representative molecular properties Short and long range forces
Dispersion attraction:Long range; primarily dependent on particle properties. Dispersion attractions arise when charge fluctuations in a molecule give rise to instantaneous dipole moments even when the time averaged dipole moment is zero. Such induced dipole-induced dipole interactions are known as London dispersion forces. Such interactions are even present between hydrocarbons which may not interact via the other interactions described in this list. It is interesting to note that these are the only interactions that occur between noble gas atoms. Without dispersive attraction between noble gas atoms, noble gases would never be found in liquid form. Dispersive attractions become larger as the atom or molecule gets larger in size. Larger molecules tend to be easier to polarize. Electrostatic repulsion: Intermediate range; heavily dependent on solution properties. Electrostatic repulsion arises when particles with like charges repel each other. Steric (entropic) repulsion: Short range, primarily dependent on solution properties. This repulsion arises from the volume that a particle occupies. Debye shielding is an example of such a force. This interaction is particularly important is pharacology and biology where molecules may be very large. Such shielding may influence what reactions may occur or limit the circumstances under which the reaction may take place. Two particles; two kinds of forces; over two distance scales. In many systems, there are multiple interactions that have differing effects depending on the length scale at which particles are observed. While long range forces may be observed at short scales, in this region short range forces tend to be larger and dominate the interactions between the particles. Derjaguin approximation for spheres
Israelachvili's text did a fantastic job illustrating this, the interaction forces versus energies, starting at 10.5 in "Intermolecular and Surface Forces". The final product of the Derjaguin approximation is to show the "force between two spheres in terms of the energy per unit area of two flat surfaces at the same separation D" (p.163).
First, Israelachvili relates force F(D) between two curved surfaces with the interaction free energy W(D) of two flat plates. Using a previously derived equation for the free energy of a sphere-surface interaction, he shows that the force value for a sphere near a flat surface is:
<math>F(D)= -\frac{\partial W(D)}{\partial D} = -\frac{4\pi^2 C\rho^2 R}{(n-2)(n-3)(n-4)D^{(n-4)}}</math>
Which is simplified to just the interaction free energy of two planar surfaces (flat disks!) per unit area:
<math>F(D)_{sphere}=2\pi RW(D)_{planes}</math>
He goes on to the derive the Derjaguin approximation to show that this relationship is valid for any type of force law (incredible) if the separation distance is D is much less than the radii of the spheres.
Calculate the force between two spheres, by integrating over force between thin disks: <math>F(D)=\int\limits_{Z=D}^{\infty }{2\pi xdxf\left( Z \right)}</math>
Now, looking at the diagram and remembering the Chord Theorem,
<math>x^2 \approx 2R_1z_1 = 2R_2z_2</math>
<math>Z = D + z_1 + z_2 = D + \frac{x^2}{2} (\frac{1}{R_1}+\frac{1}{R_2})</math>
<math>dZ= (\frac{1}{R_1}+\frac{1}{R_2})xdx</math>
<math>F\left( D \right)\approx 2\pi \left( \frac{R_{1}R_{2}}{R_{1}+R_{2}} \right)W\left( D \right)</math>
Where W(D) is the energy of interaction of two flat plates.
Destabilization of colloidal mixtures
Colloidal mixtures can become unstable, resulting in particle movement towards one another and aggregating in flocs. Flocculation is a process in which a solute comes out of the solution in the form of flakes or floc. This process can result in creation of photonic glasses using one of the following methods:
- Removing the electrostatic force that keeps the forces separate. This can be accomplished by changing the pH of the solution to effectively neutralize the charge in the solution. In this manner, Van Der Waals attractive force will dominate the electrostatic repulsion and the particles will flocculate.
- Physically deforming particles in the mixture. Doing so may allow Van Der Waals attractive force to "win" over other repulsive forces that we mentioned earlier (electrostatic, steric ...)
- Adding polymer of opposite charge compared to the colloids in the mixture. Added polymers can "bridge" two colloids, attracting two or more towards each other and thus aggregating.
- Adding nonadsorbed polymers (depletants) to the mixture can cause aggregation due to steric effects.
Below is an SEM image of a sample of "photonic glass" on the left and regular disordered media on right.
Current Research: An additonal way to potentially nucleate the formation of colloidal crystals from a suspension is to dope it with packed sphere structures as described in Building Materials by Packing Spheres Colloidal Stability
The long-term colloidal stability of a dispersion will be of great importance in a number of industries such as pharmaceutical, ceramic, paints and pigments. The term “stability” can have different connotations to different applications. When applied to colloids, a stable colloidal system is one in which the particles resist flocculation or aggregation and exhibits a long shelf-life. This will depend upon the balance of the repulsive and attractive forces that exist between particles as they approach one another. If all the particles have a mutual repulsion then the dispersion will remain stable. However, if the particles have little or no repulsive force then some instability mechanism will eventually take place e.g. flocculation, aggregation etc.
In certain circumstances, the particles in a colloidal disperson may adhere to one another and form aggregates of successively increasing size that may settle out under the influence of gravity. An initially formed aggregate is called a floc and the process of its formation flocculation. The floc may or may not separate out. If the aggregate changes to a much denser form, it is said to undergo coagulation. An aggregate usually separates out either by sedimentation (if it is more dense than the medium) or by creaming (if it less dense than the medium). The term’s flocculation and coagulation have often been used interchangeably. Usually coagulation is irreversible whereas flocculation can be reversed by the process of deflocculation. The following figure schematically represents some of these processes.
Cheese Making
Colloidal chemistry is connected with the cheese making process. Colloidal suspensions of casein micelles aggregate during a process called acidifcation where calcium ion concentrations increase thereby decreasing the net negative electric charge on the casein micelles causing them to collapse. In their new form they aggregate instead of being dispersed and form a gel related to the curd part of cheese!!!! So the calcium ions cause changes in surface charge and surface forces along with decreased repulsion between micelles allowing us to enjoy the wonders of dairy in the soft matter form. Tons of information including plots sort of like phase diagrams defining cheese making processes are available online in the two links below. |
Genus two partition functions of 2d chiral conformal field theories are given by Siegel modular forms. We compute their conformal blocks and use them to perform the conformal bootstrap. The...
Pages
We study correlation functions of a conserved spin-1 current J_\mu in three dimensional Conformal Field Theories (CFTs). We investigate the constraints imposed by permutation symmetry and...
We construct a Hennings type logarithmic invariant for restricted quantum sl(2) at a 2p-th root of unity. This quantum group U is not braided, but factorizable. The invariant is defined for a...
We prove an inequality on decision trees on monotonic measures which generalizes the OSSS inequality on product spaces. As an application, we use this inequality to prove a number of new results...
We describe a large family of nonequilibrium steady states (NESS) corresponding to forced flows over obstacles. The spatial structure at large distances from the obstacle is shown to be...
The modern conformal bootstrap program often employs the method of linear functionals to derive the numerical or analytical bounds on the CFT data. These functionals must have a crucial "...
We investigate the behaviour of the bosonic string on AdS3 with H-flux at stringy scales, looking in particular for a `tensionless` limit in which there are massless higher spin gauge fields. We...
It was recently shown that the CFT dual of string theory on {\rm AdS}_3 \times {\rm S}^3 \times T^4, the symmetric orbifold of T4, contains a closed higher spin subsector. Via holography, this...
We compute genus two partition functions in two dimensional conformal field theories at large central charge, focusing on surfaces that give the third Renyi entropy of two intervals. We compute...
Let K\subset S^3 be a knot, X:= S^3\setminus K its complement, and T the circle group identified with R/Z. To any oriented long knot diagram of K, we associate a quadratic polynomial in...
We consider inhomogeneous Erd\H{o}s-R\'enyi graphs. We suppose that the maximal mean degree d satisfies d \ll \log n. We characterize the asymptotic behavior of the n
1−o(1) largest...
We show that the maximal extension sl(2) times psl(2|2) times C3 of the sl(2|2) superalgebra can be obtained as a contraction limit of the semi-simple superalgebra d(2,1;epsilon) times sl(2). We... |
Suppose you were trying to determine the mean rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion.
We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics.
The sample data help us to make an estimate of a population parameter. We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, called confidence intervals.
In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student’s-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed.
If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from iTunes. If so, you could conduct a survey and calculate the sample mean, [latex]\displaystyle\overline{x}[/latex], and the sample standard deviation,
s. You would use [latex]\displaystyle\overline{x}[/latex] to estimate the population mean and s to estimate the population standard deviation. The sample mean, [latex]\displaystyle\overline{x}[/latex], is the point estimate for the population mean, μ. The sample standard deviation, s, is the point estimate for the population standard deviation, σ.
Each of [latex]\displaystyle\overline{x}[/latex] and
s is called a statistic.
A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is a range of values calculated from a given set of sample data. The confidence interval is likely to include an unknown population parameter.
Suppose, for the iTunes example, we do not know the population mean
μ, but we do know that the population standard deviation is σ = 1 and our sample size is 100. Then, by the central limit theorem, the standard deviation for the sample mean is
[latex]\displaystyle\frac{{\sigma}}{{\sqrt{n}}} = \frac{{1}}{{\sqrt{100}}}=0.1[/latex]
The empirical rule, which applies to bell-shaped distributions, says that in approximately 95% of the samples, the sample mean, [latex]\displaystyle\overline{x}[/latex] , will be within two standard deviations of the population mean
μ. For our iTunes example, two standard deviations is (2)(0.1) = 0.2. The sample mean [latex]\displaystyle\overline{x}[/latex] =0.1 is likely to be within 0.2 units of μ.
Because [latex]\displaystyle\overline{x}[/latex] is within 0.2 units of
μ, which is unknown, then μ is likely to be within 0.2 units of [latex]\displaystyle\overline{x}[/latex] in 95% of the samples. The population mean μ is contained in an interval whose lower number is calculated by taking the sample mean and subtracting two standard deviations (2)(0.1) and whose upper number is calculated by taking the sample mean and adding two standard deviations. In other words, μ is between [latex]\displaystyle\overline{x}[/latex]− 0.2 and [latex]\displaystyle\overline{x}[/latex] + 0.2 in 95% of all the samples.
For the iTunes example, suppose that a sample produced a sample mean [latex]\displaystyle\overline{x}[/latex]= 2. Then the unknown population mean
μ is between [latex]\displaystyle\overline{x}[/latex]−0.2=2−0.2=1.8 and [latex]\displaystyle\overline{x}[/latex] +0.2=2+0.2=2.2
We say that we are
95% confident that the unknown population mean number of songs downloaded from iTunes per month is between 1.8 and 2.2. The 95% confidence interval is (1.8, 2.2).
The 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean
μ or our sample produced an [latex]\displaystyle\overline{x}[/latex] that is not within 0.2 units of the true mean μ. The second possibility happens for only 5% of all the samples (95–100%).
Remember that a confidence interval is created for an unknown population parameter like the population mean,
μ. Confidence intervals for some parameters have the form: (point estimate – margin of error, point estimate + margin of error )
The margin of error depends on the confidence level or percentage of confidence and the standard error of the mean.
When you read newspapers and journals, some reports will use the phrase “margin of error.” Other reports will not use that phrase, but include a confidence interval as the point estimate plus or minus the margin of error. These are two ways of expressing the same concept.
Although the text only covers symmetrical confidence intervals, there are non-symmetrical confidence intervals (for example, a confidence interval for the standard deviation). |
Some theories posit that a neutron star's magnetic field is residual, left over from the remnant's creation. Strong arguments are given for this in Flowers & Ruderman (1977)
1: Dynamo mechanisms, which are responsible for the magnetic fields of many celestial bodies, cannot exist in mature neutron stars, because damping of fluid motion would have removed any significant movement of conductive fluid inside the remnant. Permanent magnets cannot exist inside neutron stars, according to the present-day models (this, admittedly, is from 1977; since then, we have made more strides in understanding neutron stars).
Therefore, the magnetic field of a neutron star must be "fossilized". That's not to say these fields can't flair up again - in fact, arguments are presented in Price & Rosswog (2006) that during neutron star mergers shortly before a catastrophic event (e.g. a gamma-ray burst) the magnetic fields can be amplified considerably. However, the resulting merger may destroy the two bodies.
I should also mention a second major mechanism for magnetic field formation, discussed in (among others) (Spruit). In this mechanism, magnetic fields are generated through core collapse, through convection, field generation in "stable zones", and neutrino convection.
However, not everyone believes that magnetic fields cannot grow after neutron star formation. This overview discusses several models that say that thermal processes can occur within ~100,000 years of the neutron star's formation that can build up a magnetic field. There are two main models:
The battery model - originally proposed for "normal" stars like the Sun - states that different ionized components near the object's core behave differently due to different gravitational masses, with electrons wandering outwards a bit due to gravity and partial pressure. This acts like a battery, generating currents that in turn produce a magnetic field. However, matter in a neutron star is degenerate, and so a straightforward version of this mechanism is impossible. It is possible that temperature-dependent pressure could solve this, but the model is still not favored.
The thermoelectric mechanism solves the problem of degenerate components that arises from the battery model. It requires a non-zero vertical temperature gradient (which is present) and an existing "seed" magnetic field. The gradient brings "hotter" electrons up and "cooler" electrons down, which create a horizontal temperature gradient. This gradient requires pressure changes to go hand in hand with it, which thus brings about a thermoelectric field. The thermoelectric field helps the "seed" field grow.
Here, the basic equation is$$\frac{\partial\vec{B}}{\partial t}=\overbrace{\vec{\nabla}\times\left(\vec{V}\times\vec{B}\right)}^{\text{Field convection term}}-\overbrace{\vec{\nabla}Q_0\times\vec{\nabla}T}^{\text{Battery term}}-\overbrace{\vec{\nabla}\times\left[\frac{\vec{\nabla}\times\vec{B}}{4\pi\sigma_0}\right]}^{\text{Ohmic decay term}}$$
There are various theories out there about why neutron stars have magnetic fields - some even say that neutron stars are giant magnets
2. The thing is, there is no consensus at the moment as to what the actual reason is. It doesn't seem like what you propose is correct, but we can't be sure because modeling convective processes under these conditions (as well as the interiors of neutron stars) isn't easy. For example, differential rotation needs to be accounted for. It appears that the answer to your question is "no", though.
1 They also explain that at birth, the spin axis and magnetic dipole are nearly aligned, and that the dipole drifts over time, leading to a gap like the displayed in the picture linked to in the question. 2 The accompanying paper is Hansson & Ponga (2011). |
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Latest revision Your text Line 6: Line 6:
:<math>p = B \left[ \left( \frac{\rho}{\rho_0} \right)^\gamma -1 \right]</math>
:<math>p = B \left[ \left( \frac{\rho}{\rho_0} \right)^\gamma -1 \right]</math>
−
In it, <math>\rho_0</math> is a reference density around which the density varies
+
In it, <math>\rho_0</math> is a reference density around which the density varies
−
<math>\gamma</math> is the [[Heat capacity#Adiabatic index | adiabatic index]]
+
<math>\gamma</math> is the [[Heat capacity#Adiabatic index | adiabatic index]] and <math>B</math> is a pressure parameter.
Usually, the equation is used to model a nearly incompressible system. In this case,
Usually, the equation is used to model a nearly incompressible system. In this case,
Line 22: Line 22:
Therefore, if <math>B=100 \rho_0 v^2 / \gamma</math>, the relative density fluctuations
Therefore, if <math>B=100 \rho_0 v^2 / \gamma</math>, the relative density fluctuations
−
will be about 0.01.
+
will be about 0.01.
If the fluctuations in the density are indeed small, the
If the fluctuations in the density are indeed small, the |
Nikolaj Kunst asked: As I understand it from the threads I read, D-branes are viewed as somewhat secondary to strings: If I know what all the open strings do, then I know what the D-branes do as well. But if the D-brane is a "big" and higherdimensional object, then there should also be a part of its surface, which is not covered by strings ending there. Why or how is the whole thing still characterized by the strings? Do the moving string ends drag the D-branes with them? Is there something like a Lagrangian and dynamical rules for the D-branes alone? Do I only need the D-brane after quantization and/or is there a classical concept of it? In the former case, does a classical limit of a D-brane and an effective action for it make sense? What is the precise reason one doesn't write down Lagrangians for non-point particle/non-string/non-field quantities? L.M. answers:
Much like in the case of the spacetime itself, you may write low-energy effective quantum field theories for D-branes. The most popular effective action for a D-brane that one encounters in the literature is a non-linear generalization of Yang-Mills theory known as the Dirac-Born-Infeld action which is also coupled to extra scalars, and so on. It describes essentially everything at long distances, regardless of the shape of the D-brane, as long as the curvature is low etc.
\[ S_p = -T_p \int_{\mathcal{M}_{p+1}}\!\!\!\!\!d^{p+1}\xi\, e^{-\Phi} \sqrt{ \det ( G_{ab} + B_{ab} + 2\pi\alpha' F_{ab} ) } \] Well, this simple action with the induced metric \(G_{ab}\) already depends on the transverse coordinates as well (and while doing so, it mostly generalizes the Nambu-Goto action) but one should also add the fermionic superpartners if any.
But this action breaks down at short distances or high energies, comparable to the string scale, because there are other massive excitations living on the D-brane and their interactions which modify the dynamics. You may try to describe all those things accurately (at least to all orders in perturbation theory in \(g_s\)) by string field theory in the D-brane world volume but even if you do so, you're brought to my main point:
D-branes' degrees of freedom are not as fundamental as the strings' degrees of freedom; they're actually "composed" of strings.
So it is really incorrect to imagine that the degrees of freedom of D-branes are independent from the stringy degrees of freedom, i.e. that they're independent objects. They're not. D-branes are topological defects which play the same role as solitons in quantum field theory as long as the degrees of freedom carried by strings ("string fields") are considered analogous to the normal fields in a quantum field theory. A TV example of a soliton is the magnetic monopole which may be represented as a classical configuration – a solution – in which the electromagnetic fields have certain values (and a topologically nontrivial "charge"). In this sense, the magnetic monopole is "composed" out of excitations of the electromagnetic field. Such objects, solitons, are usually much heavier than the "fundamental" objects when the coupling constant is weak; and the same thing holds for a low \(g_s\) and D-branes (solitons of string theory).
In the same way, D-branes are solitons in the closed string fields, including the Ramond-Ramond \(p\)-form generalizations of the electromagnetic field. And all their excitations are (paradoxically, by the open-closed duality which boils down to the fact that the annulus and cylinder shapes are equivalent) equivalent to open string states.
So let me try to convince you that you should adopt a very different philosophical picture of what the D-branes actually are. First, consider gravity. It is a part of string theory and in perturbative string theory, gravitons always arise from the massless states of closed strings, right?
In this sense, all the degrees of freedom \(g_{\mu\nu}\) in the spacetime fail to be "independent objects" or "independent fields" or "independent degrees of freedom" from strings. They are literally made out of closed strings. Any gravitational wave, a classical perturbation of \(g_{\mu\nu}\), is equivalent to the coherent state of closed strings in the graviton mode. All these fields are "built from" closed string degrees of freedom. Assuming that you choose the name "vacuum" for the appropriate state in the Hilbert space describing spacetime in string theory, any other state in the Hilbert space may be obtained as a linear superposition of extra strings' creation operators acting on this vacuum (and it doesn't have to be "the right" vacuum).
The D-brane situation is fully analogous. Imagine a flat D8-brane at \(X^9=0\). You may change its shape by making \(X^9\) slightly varied as a function of the remaining 8+1 spacetime coordinates. Such a wave will propagate by the speed of light. The field \(X^9\) is a scalar field on the brane.
However, this field and its excitations – quanta of the waves – are not independent objects or independent fields or independent degrees of freedom from fundamental strings. In fact, the wave of changing \(X^9\) which propagates by the speed of light in the 8+1 world volume dimensions of the D8-brane I randomly chose is literally and physically made out of open strings. A coherent state of open strings is how you make the D-brane classically wiggle. A coherent state of open strings in another state changes the electromagnetic fields inside the D-brane (or the stack). All other degrees of freedom living in the D-brane are literally composed of open strings as well! If you're at the quantum level and you want to consider the most general states of the D-branes in its Hilbert space, any superposition of open strings created upon the background is allowed and nothing else about the D-brane exists.
This picture created by Steuard Jensen and brought to the broader world by David Tong suggests that aside from open strings connecting the D-branes, the D-branes may also independently wiggle. But the D-brane wiggles (and everything else occurring inside them) are actually made out of the same open strings.
That's why only closed (and open, in the presence of D-branes) strings are fundamental degrees of freedom in perturbative string theory. Everything else is made out of them: spacetime closed-string-like fields such as the metric tensor are made out of closed strings; all degrees of freedom describing the state of the D-branes are made out of open strings. This fact is why in the perturbative description, strings produce a theory of everything.
This hegemony of fundamental strings is only violated if you try to go beyond the perturbative expansions. Because of S-dualities, one may be sure that in some sense, all D-branes and other branes are equally fundamental as strings (e.g. F-strings are exchanged with D1-branes in type IIB string theory). But there is no known explicit calculation-ready formalism that would make this "brane democracy" manifest. Instead of this democratic description, we possess various desciptions such as expansions around various limiting points. The expansion around \(g_s=0\) is always dominated by strings and their degrees of freedom and everything else seems to be composed out of them. (The CFT description of an AdS/CFT system has different, more braney fundamental degrees of freedom; in this case, the excited stringy states are made out of these gauge-theoretical, D-brane-like degrees of freedom.)
That's also why D-branes don't bring any ambiguity or uncertainty to the dynamical rules of string theory: there is only one string theory and it is unique. The addition of a D-brane means that the background is a non-trivial solution for the closed string degrees of freedom and all the dynamics around this solution is dictated by stringy diagrams. That's surely a good property of string theory – and a necessary for one a truly unifying theory.
If you tried to find independent D-brane Lagrangians describing everything that have nothing to do with strings, something I just told you doesn't exist, you would face another problem: all such hypothetical world volume theories would suffer from the same short-distance (UV) problems as quantized gravity without strings. Strings whose world volume is 1+1-dimensional are unique among all objects you can think of because their dimensionality is high enough so that the shape of the world sheet may create topologically nontrivial and therefore interacting histories (pants diagrams etc.); but it is also low enough so that all potential UV problems in the world volume (namely world sheet of the string) are under control.
That's not true for higher-dimensional objects (whose UV problems are analogous to those of quantized general relativity) so there can't be any "membrane or higher brane theory" that would share the same consistency standards with string theory. As we know, other objects of various dimensions exist in string theory but whenever strings are used as fundamental ones, the others can't be fundamental at the same moment. The hegemony of strings in the perturbative description is not only a symptom of the uniqueness of string theory and its being a "theory of everything" but it is also necessary for the finiteness of the theory. Whatever isn't made out of strings would create UV problems comparably severe to those in non-stringy quantum gravity. |
According to many sources (including Wikipedia, Stephani&Kluge, D.J. Acheson) a steady state ist:
In systems theory, a system in a steady state has numerous properties that are unchanging in time. This means that for those properties $p$ of the system, the partial derivative with respect to time is zero:
$$ \frac{\partial p}{\partial t} = 0 $$
But why is it defined like that? Why not ${d \over dt} p=0$ ? If only $ \frac{\partial p}{\partial t} = 0 $ then there will still be change in time if $p=p(\vec r(t))$ !
Since people seem to disagree that this is even a legitimate question here is a motivation for that, Wikipedia about total and partial derivative:
The total derivative of a function is different from its corresponding partial derivative ($\partial$). Calculation of
the total derivative of f with respect to t does not assume that the other arguments are constantwhile t varies; instead, it allows the other arguments to depend on t.
So why can we assume the other variables are constant? If we are using $ \frac{\partial p}{\partial t} = 0 $ to determine if a steady state is present, why should we be able to assume no indirect dependence on time exists?
My try to explain this, if it is nonsense of course an explanation why would be great but just a good answer to my question would be too:
Reasons why ${\partial p \over \partial t} =0$ might make more sense than the total derivative to equal zero.
1. ${d p \over dt} =0 $ on it's own already indicates a conservation of the flux $j$ and density $\rho$ associated with $p$ by Reynolds theorem:$${d \over dt} p={d \over dt}\int_V \rho dV= \int_V \frac{\partial \rho}{\partial t} dV+ \int_{\partial V} \underbrace{\rho \vec v}_{= j} \cdot \vec n dA$$
since ${d p\over dt} =0$ normally holds independent of V if we use Gauss theorem $$\Rightarrow \frac{\partial \rho}{\partial t}+\nabla j=0$$
We see the total derivative being zero gives us a continuity equation.
2.
${ d p \over dt} =0 $ implies if we write it out $${dp \over dt } = { \partial p \over \partial t} + ( v \nabla ) p=0$$ i.e. $${ \partial p \over \partial t} =- ( v \nabla ) p$$ So by the definition of the partial derivative: if we hold all other variables and only look at the change in $t$, we see, this derivative does not vanish and we even have a time dependend $p$. |
Boltzmann Distribution Edited by Pichet Adstamongkonkul, AP225, Fall 2011 Introduction
Widely known to be one of the most important fundamental concepts in Chemistry, Quantum Mechanics, Statistical Mechanics and Thermodynamics, the Boltzmann Distribution is an exponential distribution law that describes the probability distribution of states based on the underlying energy levels.[2] For instance, if we are looking at the states of the configurations of a molecule, the Boltzmann distribution function will be the fraction of molecules that are in each conformation.
Definition
Given a system with N particles, if we let <math>p_i = \frac{n_i}{N}</math> represent the fraction of a molecule in the state i, which has energy Ei associated with it, then the Boltzmann distribution has a general form of:
<math>\frac{n_i}{N}=p_i^*=\frac{g_i e^{\frac{-E_i}{k_BT}}}{Q}</math> , where <math>Q=\sum_{i=1}^N g_i e^{\frac{-E_i}{k_BT}}</math> is the partition function and gi is the degeneracy. KB is the Boltzmann’s constant, which equals to 1.38 x 10-23 J/K. N is a total number of molecules or particles in the system (<math>N=\sum_{i=1}^N n_i</math>) [1]
The degeneracy of the state sometimes occurs when there are more than one
microstates (the different states of molecules which have the same energy; each microstate is like a snapshot of the system) for each macrostate (a collection of microstates). The degeneracy becomes important in the partition of rotational energy of molecules.[1]
For each molecule or particle, there is some
internal energy, due to its intrinsic property, such as rotational or vibrational states of the molecule. The sum total of all the internal and interaction energies for a particular state i of the system gives the value of each energy level, <math>E_i</math>. Question: Is "intrinsic" the same as "internal"? What about entropy? Is this definition of energy exact enough?
One can also find the relative populations of particles in two different microstates I and j at equilibrium by
<math>\frac{p_i^*}{p_j^*}=e^{\frac{-(E_i-E_j)}{kT}}</math>
The Boltzmann distribution only applies to particles at high enough temperature and low enough density that the quantum effects can be ignored. It predicts more particles, molecules, or configurations will be in the low energy states, while a few will have high energies, given that all energy levels are equivalent. If each molecule in the system takes a small fraction of energy of the system, there will be more ways to distribute the remaining energy to other molecules in the system.[3]
Application
The Boltzmann distribution can be applied to find the number of molecules, or the particle densities, in each layer of the atmosphere according to the gravitational field.
For the Newtonian particles with velocities and kinetic energy, the Boltzmann distribution provides the basis of another distribution function, the Maxwell-Boltzmann distribution, which gives a fundamental relationship between temperature and the velocities of the gas molecules.
<math>P(v_x)=\frac{m}{\sqrt{2 \pi kT}} e^{\frac{-m v_x^2}{2kT}}</math>[1]
Classical particles with this energy distribution are said to obey Maxwell-Boltzmann statistics.
References
[1] Ken A. Dill, and Sarina Bromberg. Molecular Driving Forces: Statistical Thermodynamics in Chemistry and Biology. New York: Garland Science, 2003. 171-77.
[2] Wikipedia contributors. “Boltzmann distribution.” Wikipedia, The Free Encyclopedia. 9 Dec 2011.
[3] Edmund L. J. Tisko. “Boltzmann Distribution.” University of Nebraska at Omaha. Lecture. |
I received requests to write about two types of universal motor: conductively and inductively compensated. This post can be considered as the second part of universal motor operation.
Purpose of compensating winding
In an electric motor, armature winding are rotor’s wires. When motor is on, electric current which pass through conductor wires in rotor produces a magnetic field. This generated field is the armature reaction, it weakens and distorts stator’s magnetic field.
If stator’s magnetic field get weaker, the motor loses torque. Because torque \vec{\tau} on a coil turn is directly proportional to magnetic flux density \vec{B}. \vec{\mu} is magnetic dipole moment and N is number of turns.
\vec{\tau}=N\vec{\mu }\times \vec{B}
The universal motor uses compensating winding to reduce armature reaction. \times is vector multiplication, it’s different from escalar multiplication between two numbers.
Conductive compensation
In this configuration, compensating winding is connected in series with main and armature winding. In addition to receive current directly from power supply.
Magnetic field of compensating winding (C) is put in slots of main poles of stator (S), it’s involve the rotor (R) and almost cancel armature reaction field. (F) and (A) are field and armature winding respectively.
About speed control, universal motor has the same feature of DC series motor, too high speed when there is no load. A rheostat (variable resistor) in series or electronic circuits are used to have a bigger speed control range.
Inductively compensation
In this configuration, compensating winding is short-circuited.
Armature winding acts as primary of a transformer, while compensator is the secondary which receives magnetic field generated in rotor. Why compensating is winding in short-circuit? When it receives armature’s magnetic flux, receives an induced voltage, in short-circuit, there is current circulation. Compensating winding’s magnetic field is in opposite polarity to armature winding. The result is that both fields are in opposite phases and cancel each other. |
We describe level-line decompositions of the two-dimensional Gaussian Free Field (GFF) with free boundary conditions. In particular, we point out a simple way to couple the GFF with free...
Pages
The d-dimensional long-range Ising model, defined by spin-spin interactions decaying with the distance as the power 1/r^{d+s}, admits a second order phase transition with continuously varying...
We survey here the construction and the basic properties of descendent invariants in the theory of stable pairs on nonsingular projective 3-folds. The main topics covered are the rationality of...
Descent equations play an important role in the theory of characteristic classes and find applications in theoretical physics, e.g. in the Chern-Simons field theory and in the theory of...
We define an elliptic version of the stable envelope of Maulik and Okounkov for the equivariant elliptic cohomology of cotangent bundles of Grassmannian. It is a version of the construction...
We discuss, in the context of energy flow in high-dimensional systems and Kolmogorov-Arnol'd-Moser (KAM) theory, the behavior of a chain of rotators (rotors) which is purely Hamiltonian, apart...
We study the fundamental relationship between stable quotient invariants and the B-model for local CP2 in all genera. Our main result is a direct geometric proof of the holomorphic anomaly...
The classical quadratic formula and some of its lesser known variants for solving the quadratic equation are reviewed. Then, a new formula for the roots of a quadratic polynomial is presented....
In this paper, we study positivity phenomena for the e-coefficients of Stanley's chromatic function of a graph. We introduce a new combinatorial object: the {\em correct} sequences of unit...
The goal of the present paper is to explain, based on properties of the conformal loop ensembles CLE_\kappa (both with simple and non-simple loops, i.e., for the whole range \kappa \in (8/3, 8)...
The relation between the bosonic higher spin {\cal W}_\infty[\lambda] algebra, the affine Yangian of \mathfrak{gl}_{1}, and the SHc algebra is established in detail. For generic \lambda we find...
We study pre-Lie pairs, by which we mean a pair of a homotopy Lie algebra and a pre-Lie algebra with a compatible pre-Lie action. Such pairs provide a wealth of algebraic structure, which in... |
The topics we'll look at in this article are:
Friction Scene Collision Jump Table
I highly recommended reading up on the previous two articles in the series before attempting to tackle this one. Some key information in the previous articles is built upon within this article.
Note: Although this tutorial is written using C++, you should be able to use the same techniques and concepts in almost any game development environment. Video Demo
Here's a quick demo of what we're working towards in this part:
Friction
Friction is a part of collision resolution. Friction always applies a force upon objects in the direction opposite to the motion in which they are to travel.
In real life, friction is an incredibly complex interaction between different substances, and in order to model it, vast assumptions and approximations are made. These assumptions are implied within the math, and are usually something like "the friction can be approximated by a single vector" - similarly to how rigid body dynamics simulates real life interactions by assuming bodies with uniform density that cannot deform.
Take a quick look at the video demo from the first article in this series:
The interactions between the bodies are quite interesting, and the bouncing during collisions feels realistic. However, once the objects land on the solid platform, they just sort of all press away and drift off the edges of the screen. This is due to a lack of friction simulation.
Impulses, Again?
As you should recall from the first article in this series, a particular value,
j, represented the magnitude of an impulse required to separate two objects penetration during a collision. This magnitude can be referred as
jnormal or
jN as it is used to modify velocity along the collision normal.
Incorporating a friction response involves calculating another magnitude, referred to as
jtangent or
jT. Friction will be modeled as an impulse. This magnitude will modify the velocity of an object along the negative tangent vector of the collision, or in other words along the friction vector. In two dimensions, solving for this friction vector is a solvable problem, but in 3D the problem becomes much more complex.
Friction is quite simple, and we can make use of our previous equation for
j, except we will replace all instances of the normal
n with a tangent vector
t.
\[ Equation 1:\\
j = \frac{-(1 + e)(V^{B}-V^{A})\cdot n)} {\frac{1}{mass^A} + \frac{1}{mass^B}}\]
Replace
n with
t:
\[ Equation 2:\\
j = \frac{-(1 + e)((V^{B}-V^{A})\cdot t)} {\frac{1}{mass^A} + \frac{1}{mass^B}}\]
Although only a single instance of
n was replaced with
t in this equation, once rotations are introduced a few more instances must be replaced besides the single one in the numerator of Equation 2.
Now the matter of how to calculate
t arises. The tangent vector is a vector perpendicular to the collision normal that is facing more towards the normal. This might sound confusing - don't worry, I have a diagram!
Below you can see the tangent vector perpendicular to the normal. The tangent vector can either point to the left or the right. To the left would be "more away" from the relative velocity. However, it is defined as the perpendicular to the normal that is pointing "more towards" the relative velocity.
As stated briefly earlier, friction will be a vector facing opposite to the tangent vector. This means that the direction in which to apply friction can be directly computed, since the normal vector was found during the collision detection.
Knowing this, the tangent vector is (where
n is the collision normal):
\[ V^R = V^{B}-V^{A} \\
t = V^R - (V^R \cdot n) * n \]
All that is left to solve for
jt, the magnitude of the friction, is to compute the value directly using the equations above. There are some very tricky pieces after this value is computed that will be covered shortly, so this isn't the last thing needed in our collision resolver:
// Re-calculate relative velocity after normal impulse // is applied (impulse from first article, this code comes // directly thereafter in the same resolve function) Vec2 rv = VB - VA // Solve for the tangent vector Vec2 tangent = rv - Dot( rv, normal ) * normal tangent.Normalize( ) // Solve for magnitude to apply along the friction vector float jt = -Dot( rv, t ) jt = jt / (1 / MassA + 1 / MassB)
The above code follows Equation 2 directly. Again, it's important to realize that the friction vector points in the opposite direction of our tangent vector, and as such we must apply a negative sign when we dot the relative velocity along the tangent to solve for the relative velocity along the tangent vector. This negative sign flips the tangent velocity and suddenly points in the direction in which friction should be approximated as.
Coulomb's Law
Coulomb's law is the portion of friction simulation that most programmers have trouble with. I myself had to do quite a bit of studying to figure out the correct way of modeling it. The trick is that Coulomb's law is an inequality.
Coulomb friction states:
\[ Equation 3: \\
F_f <= \mu F_n \]
In other words, the force of friction is always less than or equal to the normal force multiplied by some constant
μ (whose value depends on the materials of the objects).
The normal force is just our old
j magnitude multiplied by the collision normal. So if our solved
jt (representing the force of friction) is less than
μ times the normal force, then we can use our
jt magnitude as friction. If not, then we must use our normal force times
μ instead. This "else" case is a form of clamping our friction below some maximum value, the max being the normal force times
μ.
The whole point of Coulomb's law is to perform this clamping procedure. This clamping turns out to be the most difficult portion of friction simulation for impulse-based resolution to find documentation on anywhere - until now, at least! Most white papers I could find on the subject either skipped friction entirely, or stopped short and implemented improper (or non-existent) clamping procedures. Hopefully by now you have an appreciation for understanding that getting this part right is important.
Lets just dish out the clamping all in one go before explaining anything. This next code block is the previous code example with the finished clamping procedure and friction impulse application all together:
// Re-calculate relative velocity after normal impulse // is applied (impulse from first article, this code comes // directly thereafter in the same resolve function) Vec2 rv = VB - VA // Solve for the tangent vector Vec2 tangent = rv - Dot( rv, normal ) * normal tangent.Normalize( ) // Solve for magnitude to apply along the friction vector float jt = -Dot( rv, t ) jt = jt / (1 / MassA + 1 / MassB) // PythagoreanSolve = A^2 + B^2 = C^2, solving for C given A and B // Use to approximate mu given friction coefficients of each body float mu = PythagoreanSolve( A->staticFriction, B->staticFriction ) // Clamp magnitude of friction and create impulse vector Vec2 frictionImpulse if(abs( jt ) < j * mu) frictionImpulse = jt * t else { dynamicFriction = PythagoreanSolve( A->dynamicFriction, B->dynamicFriction ) frictionImpulse = -j * t * dynamicFriction } // Apply A->velocity -= (1 / A->mass) * frictionImpulse B->velocity += (1 / B->mass) * frictionImpulse
I decided to use this formula to solve for the friction coefficients between two bodies, given a coefficient for each body:
\[ Equation 4: \\
Friction = \sqrt[]{Friction^2_A + Friction^2_B} \]
I actually saw someone else do this in their own physics engine, and I liked the result. An average of the two values would work perfectly fine to get rid of the use of square root. Really, any form of picking the friction coefficient will work; this is just what I prefer. Another option would be to use a lookup table where the type of each body is used as an index into a 2D table.
It is important that the absolute value of
jt is used in the comparison, since the comparison is theoretically clamping raw magnitudes below some threshold. Since
j is always positive, it must be flipped in order to represent a proper friction vector, in the case dynamic friction is used.
Static and Dynamic Friction
In the last code snippet static and dynamic friction were introduced without any explanation! I'll dedicate this whole section to explaining the difference between and necessity of these two types of values.
Something interesting happens with friction: it requires an "energy of activation" in order for objects to start moving when at complete rest. When two objects are resting upon one another in real life, it takes a fair amount of energy to push on one and get it moving. However once you get something sliding it is often easier to keep it sliding from then on.
This is due to how friction works on a microscopic level. Another picture helps here:
As you can see, the small deformities between the surfaces are really the major culprit that creates friction in the first place. When one object is at rest on another, microscopic deformities rest between the objects, interlocking. These need to be broken or separated in order for the objects to slide against one another.
We need a way to model this within our engine. A simple solution is to provide each type of material with two friction values: one for static and one for dynamic.
The static friction is used to clamp our
jt magnitude. If the solved
jt magnitude is low enough (below our threshold), then we can assume the object is at rest, or nearly as rest and use the entire
jt as an impulse.
On the flipside, if our solved
jt is above the threshold, it can be assumed that the object has already broken the "energy of activation", and in such a situation a lower friction impulse is used, which is represented by a smaller friction coefficient and a slightly different impulse computation.
Scene
Assuming you did not skip any portion of the Friction section, well done! You have completed the hardest part of this entire series (in my opinion).
The
Scene class acts as a container for everything involving a physics simulation scenario. It calls and uses the results of any broad phase, contains all rigid bodies, runs collision checks and calls resolution. It also integrates all live objects. The scene also interfaces with the user (as in the programmer using the physics engine).
Here is an example of what a scene structure may look like:
class Scene { public: Scene( Vec2 gravity, real dt ); ~Scene( ); void SetGravity( Vec2 gravity ) void SetDT( real dt ) Body *CreateBody( ShapeInterface *shape, BodyDef def ) // Inserts a body into the scene and initializes the body (computes mass). //void InsertBody( Body *body ) // Deletes a body from the scene void RemoveBody( Body *body ) // Updates the scene with a single timestep void Step( void ) float GetDT( void ) LinkedList *GetBodyList( void ) Vec2 GetGravity( void ) void QueryAABB( CallBackQuery cb, const AABB& aabb ) void QueryPoint( CallBackQuery cb, const Point2& point ) private: float dt // Timestep in seconds float inv_dt // Inverse timestep in sceonds LinkedList body_list uint32 body_count Vec2 gravity bool debug_draw BroadPhase broadphase };
There is not anything particularly complex about the
Scene class. The idea is to allow the user to add and remove rigid bodies easily. The
BodyDef is a structure that holds all information about a rigid body, and can be used to allow the user to insert values as a sort of configuration structure.
The other important function is
Step(). This function performs a single round of collision checks, resolution and integration. This should be called from within the timestepping loop outlined in the second article of this series.
Querying a point or AABB involves checking to see which objects actually collide with either a pointer or AABB within the scene. This makes it easy for gameplay-related logic to see how things are placed within the world.
Jump Table
We need an easy way to pick out which collision function should be called, based on the type of two different objects.
In C++ there are two major ways that I am aware of: double dispatch and a 2D jump table. In my own personal tests I found the 2D jump table to superior, so I'll go into detail about how to implement that. If you're planning to use a language other than C or C++ I am sure an array of functions or functor objects can be constructed similarly to a table of function pointers (which is another reason I chose to talk about jump tables rather than other options that are more specific to C++).
A jump table in C or C++ is a table of function pointers. Indices representing arbitrary names or constants are used to index into the table and call a specific function. The usage could look something like this for a 1D jump table:
enum Animal { Rabbit Duck Lion }; const void (*talk)( void )[] = { RabbitTalk, DuckTalk, LionTalk, }; // Call a function from the table with 1D virtual dispatch talk[Rabbit]( ) // calls the RabbitTalk function
The above code actually mimics what the C++ language itself implements with
virtual function calls and inheritance. However, C++ only implements single dimensional virtual calls. A 2D table can be constructed by hand.
Here is some psuedocode for a 2D jump table to call collision routines:
collisionCallbackArray = { AABBvsAABB AABBvsCircle CirclevsAABB CirclevsCircle } // Call a collsion routine for collision detection between A and B // two colliders without knowing their exact collider type // type can be of either AABB or Circle collisionCallbackArray[A->type][B->type]( A, B )
And there we have it! The actual types of each collider can be used to index into a 2D array and pick a function to resolve collision.
Note, however, that
AABBvsCircle and
CirclevsAABB are almost duplicates. This is necessary! The normal needs to be flipped for one of these two functions, and that is the only difference between them. This allows for consistent collision resolution, no matter the combination of objects to resolve.
Conclusion
By now we have covered a huge amount of topics in setting up a custom rigid body physics engine entirely from scratch! Collision resolution, friction, and engine architecture are all the topics that have been covered thus far. An entirely successful physics engine suitable for many production-level two dimensional games can be constructed with the knowledge presented in this series so far.
Looking ahead into the future, I plan to write one more article devoted entirely to a very desirable feature: rotation and orientation. Oriented objects are exceedingly attractive to watch interact with one another, and are the final piece that our custom physics engine requires.
Resolution of rotation turns out to be quite simple, though collision detection takes a hit in complexity. Good luck until next time, and please do ask questions or post comments below! |
I've been trying to solve the following problem from Stewart's Calculus Textbook for a while without any success. My answer makes sense, but I'm looking for a way to solve it analytically.
The problem concerns a pulley that is attached to the ceiling of a room at a point
C by a rope of length r. At another point B on the ceiling, at a distance d from C (where d > r), a rope of length l is attached and passed through the pulley at F and connected to a weight W. The weight is released and comes to rest at its equilibrium position D. This happens when the distance | ED| is maximized. Show that when the system reaches equilibrium, the value of x is:
$$\frac{r}{4d}(r+\sqrt{r^2+8d^2})$$
Here is what I've done. First, I expressed |
DE| as a function of x
$$|DE|(x)={a}_{2}+{a}_{3}=l-{a}_{1}+\sqrt{{r}^{2}-{x}^{2}}=l-\sqrt{{a}_{3}^2+y^2}+\sqrt{r^2-x^2}$$
from what follows that $$|DE|(x)=l-\sqrt{r^2+d^2-2xd}+\sqrt{r^2-x^2}$$ defined for $$0\leq x \leq r$$
...and it works since $$|DE|(0)=l+r-\sqrt{r^2+d^2}$$ and $$|DE|(r)=l-|r-d|$$
To find the maximum of this function, I calculated
|DE|'(x)
$$|DE|'(x)=\frac{d}{\sqrt{r^2+d^2-2xd}}-\frac{x}{\sqrt{r^2-x^2}}$$
I proved the two radicals at the denominator are defined for $$0\leq x < r$$
...so basically I'm interested in finding when |DE|'(x) equals zero, more specifically the roots of
$$d\sqrt{r^2-x^2}-x\sqrt{r^2+d^2-2xd}=0$$ that becomes
$$2dx^3-(r^2+2d^2)x^2+d^2r^2=0$$
I graphed
|DE|(x) and |DE|'(x) (using l = 15, r = 3, and d = 4), they are consistent with the problem. |DE|'(x) has only one root at about 2.76 and at the same point |DE|(x) has its maximum. Moreover, if you substitute my test numbers for l, r, and d in the given formula for |DE|(x) maximum you get the same numerical result.
So, how was the author of the problem able to find an analytical solution to the problem?
Thanks! |
So the question that I'm working on is the following.
Show that $\Pi_{p\leq z}(1-\dfrac{1}{p})=\dfrac{C(1+\mathcal{o}(1))}{\log z}$.
First off I take logs and just work with the sum and thisis what I get. $$\sum_{p\leq z}\log (1-\dfrac{1}{p})=-\sum_{p\leq z}\sum_{n=1}^{\infty}\dfrac{1}{np^{n}} $$
now so now I want to show that this sum is $-\log\log z +C' +\mathcal{o}(\log x^{-1})$.
But I'm not sure on who to do this, anyone got any ideas? |
I would like to know whether a universally-quantified type $T_a$: $$T_a = \forall X: \left\{ a\in X,f:X→\{T, F\} \right\}$$ is a sub-type, or special case, of an existentially-quantified type $T_e$ with the same signature: $$T_e = \exists X: \left\{ a\in X,f:X→\{T, F\} \right\}$$
I'd say "yes": If something is true "for all X" ($\forall X$), then it must also be true "for some X" ($\exists X$). That is, a statement with '$\forall$' is simply a more restricted version of the same statement with '$\exists$': $$∀X, P(X) \overset?\implies ∃X, P(X).$$
Am I wrong somewhere?
Background: Why am I asking this?
I am studying existential types in order to understand why and how "Abstract [Data] Types Have Existential Type". I cannot get a good grasp of this concept from theory alone; I need concrete examples, too.
Unfortunately, good code examples are hard to find because most programming languages have only limited support for existential types. (For instance, Haskell's
forall, or Java's
?wildcards.) On the other hand, universally-quantified types are supported by many recent languages via "generics".
What's worse,
generics seems to easily get mixed up with existential types, too, making it even harder to tell apart existential from universal types. I'm curious why this mix-up occurs so easily.An answer to this question might explain it: If universal types are indeed only a special case of existential types, then it's no wonder that generic types, e.g. Java's
List<T>, can be interpreted either way. |
I have a question about a proof in Rosenberg and Schochet's paper "the Künneth theorem and the Universal Coefficient Theorem for Kasparov's generalized K-functor", proposition 2.6. First of all, the setting:
Def.: Let $N$ be the bootstrap class of $C^*$-algebras, it's the smallest full subcategory of the seperable nuclear $C^*$-algebras which contains the separable Type I $C^*$-algebras and and is closed under strong Morita equivalence, inductive limits, extensions, and crossed products by $\mathbb{R}$ and by $\mathbb{Z}$. And if $J$ is an ideal in $A$ and $J$ and $A$ are in $N$, then so is $A/J$. And if $A$ and $A/J$ are in $N$ then so is $J$.
Theorem (2.1): Let $A\in N$ and let $B$ be a $\sigma$-unital $C^*$-algebra such that $K_*(B)$ is an injective $\mathbb{Z}$-module. Then the map $$\gamma(A,B):KK_*(A,B)\to Hom(K_*(A),K_*(B))$$ is an isomorphism.
One step to prove this theorem is the following
Proposition (2.6): If $K_*(B)$ is injective and if $\gamma(A,B)$ is an isomorphism, then $\gamma(A\rtimes_{\rho} \mathbb{R},B)$ is an isomorphism for any continuous action of $\mathbb{R}$ on $A$.
Proof: The Thom isomorphism theorems of Connes yield natural isomorphisms
$$Hom(K_i(A\rtimes_{\rho} \mathbb{R}),K_j(B))\cong Hom(K_{i+1}(A),K_j(B))$$ and $$KK_i(A\rtimes_{\rho} \mathbb{R},B)\cong KK_{i+1}(A,B).$$The proposition follows immediately. $\Box$
The argument above is the following: $$\require{AMScd}\begin{CD} KK_i(A\rtimes_{\rho} \mathbb{R},B) @>\gamma(A\rtimes_{\rho} \mathbb{R},B)>> Hom(K_i(A\rtimes_{\rho} \mathbb{R}),K_j(B)) \\ @VV \eta V @VV \sigma V \\ KK_{i+1}(A,B) @>\gamma(A,B)>> Hom(K_{i+1}(A),K_j(B)) \\ \end{CD}$$ is a commutative diagram, where $\eta , \sigma$ and $\gamma(A,B)$ are isomorphisms, therefore $\gamma(A\rtimes_{\rho} \mathbb{R},B)$ is an isomorphism.
My question is: why is this diagram commutative?
For this, I tried to figure out how to write down the maps explicitely. The map $\gamma(A,B)$ comes from the Kasparov-product
$$KK(\mathbb{C},A)\times KK(A,B)\to KK(\mathbb{C},B)\; (\epsilon_1 , \epsilon_2)\mapsto \epsilon_1 \otimes \epsilon_1 ,$$ because it is $KK_i(\mathbb{C},A)\cong K_i(A)$, $KK_i(\mathbb{C},B)\cong K_i(B)$. Hence the Kasparov-product induces a homomorphism (which should be $\gamma(A,B)$ I think) $$KK(A,B)\to Hom(K(A), K(B)),$$ $$\epsilon_2 \mapsto (\epsilon_1 \mapsto \epsilon_1 \otimes \epsilon_2).$$
The map $\sigma$ is the contravariant $Hom(-,K_j(B))$-functor apllied to the (Connes-Thom)-isomorphism $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$.
But I don't know how to write down the isomorphisms $\sigma$ and $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ explicitely such that I'm stuck to prove that the diagram commutes.
Can you help me to prove that the diagram commutes, or do you know how to write down the maps $\sigma$ and $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ explicitely?
Best |
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Glia, ISSN 0894-1491, 01/2019, Volume 67, Issue 1, pp. 182 - 192
Epidemiologic studies have provided compelling evidence that prenatal stress, through excessive maternal glucocorticoids exposure, is associated with...
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2007, Colecção "a IELTsar se vai ao longe", ISBN 9789727727681, Volume 19, 275
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7. Observation of Higgs boson production in association with a top quark pair at the LHC with the ATLAS detector
Physics Letters B, ISSN 0370-2693, 09/2018, Volume 784, Issue C, pp. 173 - 191
The observation of Higgs boson production in association with a top quark pair ( ), based on the analysis of proton–proton collision data at a centre-of-mass...
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8. Combinations of single-top-quark production cross-section measurements and $|f_{\rm LV}V_{tb}|$ determinations at $\sqrt{s}=7$ and 8 TeV with the ATLAS and CMS experiments
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JHEP 05 (2019) 088 This paper presents the combinations of single-top-quark production cross-section measurements by the ATLAS and CMS Collaborations, using...
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9. Mesenchymal stem cells secretome as a modulator of the neurogenic niche: Basic insights and therapeutic opportunities
Frontiers in Cellular Neuroscience, ISSN 1662-5102, 07/2015, Volume 9, Issue JULY, p. 249
Neural stem cells (NSCs) and mesenchymal stem cells (MSCs) share few characteristics apart from self-renewal and multipotency. In fact, the neurogenic and...
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The European Physical Journal C, ISSN 1434-6044, 3/2019, Volume 79, Issue 3, pp. 1 - 41
The efficiency of the photon identification criteria in the ATLAS detector is measured using $$36.1\hbox { fb}^1$$ 36.1fb1 to $$36.7\hbox { fb}^1$$ 36.7fb1 of...
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Tectonics, ISSN 0278-7407, 12/2007, Volume 26, Issue 6, pp. TC6009 - n/a
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Physical review letters, ISSN 0031-9007, 07/2019, Volume 123, Issue 4, p. 042001
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14. Electron reconstruction and identification in the ATLAS experiment using the 2015 and 2016 LHC proton-proton collision data at $\sqrt{s}$ = 13 TeV
European Physical Journal. C, Particles and Fields, ISSN 1434-6044, 02/2019, Volume 79, Issue 8
Eur. Phys. J. C 79 (2019) 639 Algorithms used for the reconstruction and identification of electrons in the central region of the ATLAS detector at the Large...
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15. Search for Higgs boson decays into a pair of light bosons in the $bb\mu\mu$ final state in $pp$ collision at $\sqrt{s} = $13 TeV with the ATLAS detector
07/2018
Phys. Lett. B 790 (2019) 1 A search for decays of the Higgs boson into a pair of new spin-zero particles, $H \to aa$, where the $a$-bosons decay into a...
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Phys. Lett. B 788 (2019) 96 This Letter presents a search for heavy charged long-lived particles produced in proton-proton collisions at $\sqrt{s} = 13$ TeV at...
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05/2018
Phys. Lett. B 790 (2019) 108 Measurements of the yield and nuclear modification factor, $R_\mathrm{ AA}$, for inclusive jet production are performed using 0.49...
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18. Measurements of gluon-gluon fusion and vector-boson fusion Higgs boson production cross-sections in the $H \to WW^{\ast} \to e\nu\mu\nu$ decay channel in $pp$ collisions at $\sqrt{s}=13$ TeV with the ATLAS detector
08/2018
Phys. Lett. B 789 (2019) 508 Higgs boson production cross-sections in proton-proton collisions are measured in the $H \to WW^{\ast} \to e\nu\mu\nu$ decay...
Physics - High Energy Physics - Experiment
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06/2018
Phys. Rev. Lett. 121, 152002 (2018) This Letter presents a normalized differential cross-section measurement in a fiducial phase-space region where...
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I was reading about the Crank-Nicolson method, and it is often said that it can produce "spurious oscillations" or that this method is prone to "ringing", especially for large time step and stiff initial value.
I'm curious about that phenomena but I can't find any satisfying answer. I read on this website that this could be because the Crank-Nicolson scheme was not L-stable.
What is the definition of L-stable ? How this is related to the production of oscillations ?
I found in this paper a beginning of an explanation thanks to a Fourier transform (see eq. 9, p. 255), but I really can't understand why the frequency $\varphi$ is only between $-\pi$ and $\pi$, and why $\varphi \approx \pm \pi$ should correspond to "high frequencies". Can you help me with that, too ?
Finally, I would like to know if this spurious oscillations problem always appear with a Crank-Nicolson scheme, or does it depend on the equation you discretize ?
To sum up, my questions are :
What is the origin of the spurious oscillations when using a Crank-Nicolson scheme ? Does it depend on the equation we consider ? What is the L-stability and in what it is related with the oscillations ? (if it is) Can you help me to understand eq. 9 of this paper, why $-\pi < \varphi < \pi$ and not $\varphi \in \mathbb{R}$ ? |
This online article is a supplement to our paper arXiv:1412.7152.
The object of study is a self-gravitating Bose-Einstein condensate, which hypothetically may form a dark matter star or a dark matter core inside a normal star.
The dimensionless GPE-Poisson system
The simulations presented on this page use a dimensionless form of the GPE, together with Poisson's equation for gravity:
\begin{align}i\frac{\partial\psi}{\partial t}&=-\frac{1}{2}\nabla^2\psi+V\psi+c|\psi|^2\psi-\mu\psi,\\ \nabla^2 V&=4\pi G|\psi|^2.\end{align}
This form of the GPE-Poisson system is possible if we choose units such that the reduced Planck constant is $\hbar=1$ and the particle mass is $m=1$.
These choices fix two out of three units of measurement (e.g., the units of mass and action), leaving the choice of the third unit (e.g., length) arbitrary.
We normalize the wavefunction such that $\int |\Psi|^2~dV=N$ is the total particle number. That is, after restoring units, $M=Nm$ is the total condensate mass.
Typically, for small $m$ and large $M$, $N$ becomes very large, which is computationally inconvenient. However, we note that the GPE is invariant under the following set of transformations:
\begin{align}t\rightarrow\lambda^2 t,\qquad r\rightarrow\lambda r,&\qquad V\rightarrow\lambda^{-2}V,\qquad\mu\rightarrow\lambda^{-2}\mu,\\ c\rightarrow\lambda^{-2}\kappa^{-2}c,\qquad G&\rightarrow\lambda^{-4}\kappa^{-2} G,\qquad\psi\rightarrow\kappa\psi.\end{align}
As a result of these rescalings the GPP system will read
\begin{align}\frac{\kappa}{\lambda^2}i\frac{\partial\psi}{\partial t}&=-\frac{1}{2}\frac{\kappa}{\lambda^2}\nabla^2\psi+\frac{\kappa}{\lambda^2}V\psi+\frac{\kappa}{\lambda^2}c|\psi|^2\psi-\frac{\kappa}{\lambda^2}\mu\psi,\\ \lambda^{-4}\nabla^2V&=\lambda^{-4}4\pi G|\psi|^2. \end{align}
In particular, choosing a value of $\kappa\ne 1$ amounts to rescaling $N\rightarrow\kappa^2N$ and $G\rightarrow\kappa^{-2}G$. We shall use this later on to normalize the values of $G$ and $N$ for computationa l convenience.
A Bose star
We wish to simulate a BEC star. Its mass has to be roughly the same as one solar mass, or about $2\times 10^{30}$ kg. The Schwarzschild radius corresponding to this mass is roughly 3 km. Our BEC star must be significantly larger than the Schwarzschild radius in order for the non-relativistic GPE to work. We choose $R=50$ km.
The simulation volume must be significantly larger than the simulated star, to ensure that the star does not interact too much with the boundaries of the volume. After experimentation, we choose a simulation box with coordinates that extend from $-440$ to $+440$ km.
Our simulation code divides space into cells. Too many cells slow down the simulation unacceptably. As a practical limit, we cannot use much more than 100 cells per one side, as the computation time increases as the cube of this number. For practical simulations, we use 80 cells, so the cell size is $11\times 11\times 11$ km.
If we take $11~{\rm km}=\frac{1}{2}\lambda$ to be the Compton half-wavelength of a particle, the corresponding mass is $m=h/c\lambda\simeq 10^{-46}~\rm{kg}\simeq 5.6\times 10^{-11}$ eV. This is the maximum Bose particle mass that we can simulate using a spatial resolution of 11 km.
The time that corresponds to a Compton wavelength of 22 km is approximately 70 μs; this determines the maximum allowable simulation timestep. In most simulations, we shall use τ = 10 μs as our simul ation timestep.
With these considerations in mind, we choose the following units for the purpose of simulating a BEC stellar remnant or stellar core:
We set our unit of mass, $[{\rm M}]=10^{-46}$ kg, the mass of a $5.6\times 10^{-11}$ eV Bose-particle. We set $\hbar=1$. Therefore, $10^{-34}~{\rm m}^2{\rm kg}/{\rm s}=1~[{\rm L}]^2[{\rm M}]/[{\rm L}]=10^{-46}~[{\rm L}]^2{\rm kg}/[{\rm T}]$, or $1~{\rm m}^2/{\rm s}=10^{-12}[{\rm L}]^2/[{\rm T}]$. We choose our unit of length: $[{\rm L}]=1~{\rm km}=1000~{\rm m}$. So $1~{\rm m}^2/{\rm s}=10^{-12}[1000~{\rm m}]^2/[{\rm T}]=10^{-6}{\rm m}^2/[{\rm T}]$. Therefore, $[{\rm T}]=10^{-6}~{\rm s}$.
So we measure masses, lengths and time in the following units:
\begin{align} 1~[{\rm M}]&=10^{-46}~{\rm kg}\simeq 5.61\times 10^{-11}~{\rm eV},\\ 1~[{\rm L}]&=1~{\rm km}=1000~{\rm m},\\ 1~[{\rm T}]&=10^{-6}~{\rm s}=1~\mu{\rm s}.\end{align}
In particular, the
size of our simulation volume is measured in kilometers; the simulation step is measured in microseconds; and $N=2\times 10^{76}$ simulated particles constitute one solar mass.
In these units, $G=6.67\times 10^{-78}$. Furthermore, a Bose star with one solar mass, $M=2\times 10^{30}$ kg, contains $N=2\times 10^{76}$ particles.
We now choose $\kappa=10^{-38}$, allowing us to rescale $G\rightarrow 0.0667$ and $N\rightarrow 2$. It must be emphasized that the physical conclusions are unaffected by this rescaling. (However, this rescali ng not only ensures that intermediate calculations do not exceed the double-precision range, it allows all calculations to stay within range of single-precision floating point, making it possible in the future t o improve performance by a GPU-based single-precision reimplementation.)
Angular velocity is measured in radians per microsecond. For an $R=50~{\rm km}$ sphere, the equatorial circumference is $2\pi R\simeq 314~{\rm km}$. A tangential velocity of 300,000 km/s (speed of light) corresponds to 955 revolutions per second, or 6,000 radians per second. That is, 0.006 radians per microsecond (our unit of time). In order to remain within the approximately nonrelativistic regime, the refore, angular velocities must be less than $10^{-3}~[{\rm T}]^{-1}$. Numerical stability
As we perform a simulation using a finite simulation volume, finite spatial grid size, and finite time step, it is important to characterize our understanding of the limitations of our simulation tool. Specifically, we need to understand how nonphysical parameters of the simulation, specifically the finite size of the simulation volume element, the finite timestep, and the finite extent of the simulation volume, affect the stability of the simulation. For this purpose, we ran four very long simulations, varying these nonphysical parameters only.
The parameters in these simulations were identical (the physical parameters described a compact one solar mass Bose star with a 50 km radius, simulated in a cubical volume with 880 km long edges), except for the nonphysical values of the time step and spatial resolution, which are summarized in the table below:
Run Time step Spatial grid Case 1 10 μs 80 × 80 × 80 Case 2 20 μs 80 × 80 × 80 Case 3 5 μs 80 × 80 × 80 Case 4 10 μs 60 × 60 × 60
Below is a set of brief videos that show the density and phase evolution of the four cases.
Case 1 Case 2 Case 3 Case 4
High resolution simulations and rotation
To investigate the physics of a Bose-star, we ran four cases. In three cases, we investigated the effects of varying the rotation rate Ω; in the fourth case, we ran a simulation with a modest rotation rate, but much higher spatial resolution (at the cost of reducing the overall simulation volume.) All four cases ran for 1 second of simulated time, using a $60\times 60\times 60$ spatial grid.
The table below summarizes the four cases that were investigated in this manner:
Run Box size (km) Ω (rad/s) Case 5 880 km $~~~~~~10^{-3}$ Case 6 880 km $3\times 10^{-4}$ Case 7 880 km no rotation Case 8 280 km $~~~~~~10^{-4}$
Videos for these four simulation runs can be viewed below:
Case 5 Case 6 Case 7 Case 8
Finally, here is a video of of the phase evolution of 36 Bose stars, as described in our paper (arXiv:1412.7152). These simulations were performed using a simulation volume of 480×480×480 km and a spatial grid of 60×60×60, while varying $N$, $\Omega$, $R$ and $a$:
$N={}$ $2\times 10^{76}$ $1\times 10^{76}$ $4\times 10^{76}$ $a/a_0$ $\Omega$ $R={}$ 50 km 80 km 50 km 80 km 50 km 80 km 1 0 case 11 case 12 case 13 case 14 case 15 case 16 0.0001 case 21 case 22 case 23 case 24 case 25 case 26 0.5 0 case 31 case 32 case 33 case 34 case 35 case 36 0.0001 case 41 case 42 case 43 case 44 case 45 case 46 2 0 case 51 case 52 case 53 case 54 case 55 case 56 0.0001 case 61 case 62 case 63 case 64 case 65 case 66 |
Thermal noise Johnson–Nyquist noise ( thermal noise, Johnson noise, or Nyquist noise) is the electronic noise generated by the thermal agitation of the charge carriers (usually the electrons) inside an electrical conductor at equilibrium, which happens regardless of any applied voltage. The generic, statistical physical derivation of this noise is called the fluctuation-dissipation theorem, where generalized impedance or generalized susceptibility is used to characterize the medium.
Thermal noise in an idealistic resistor is approximately white, meaning that the power spectral density is nearly constant throughout the frequency spectrum (however see the section below on extremely high frequencies). When limited to a finite bandwidth, thermal noise has a nearly Gaussian amplitude distribution.
[1] Contents History Noise voltage and power
Thermal noise is distinct from shot noise, which consists of additional current fluctuations that occur when a voltage is applied and a macroscopic current starts to flow. For the general case, the above definition applies to charge carriers in any type of conducting medium (e.g. ions in an electrolyte), not just resistors. It can be modeled by a voltage source representing the noise of the non-ideal resistor in series with an ideal noise free resistor.
\bar {v_{n}^2} = 4 k_B T R
where
k B is Boltzmann's constant in joules per kelvin, Tis the resistor's absolute temperature in kelvins, and Ris the resistor value in ohms (Ω). Use this equation for quick calculation, at room temperature:
\sqrt{\bar {v_{n}^2}} = 0.13 \sqrt{R} ~\mathrm{nV}/\sqrt{\mathrm{Hz}}.
For example, a 1 kΩ resistor at a temperature of 300 K has
\sqrt{\bar {v_{n}^2}} = \sqrt{4 \cdot 1.38 \cdot 10^{-23}~\mathrm{J}/\mathrm{K} \cdot 300~\mathrm{K} \cdot 1~\mathrm{k}\Omega} = 4.07 ~\mathrm{nV}/\sqrt{\mathrm{Hz}}.
For a given bandwidth, the root mean square (RMS) of the voltage, , is given by
v_{n} = \sqrt{\bar {v_{n}^2}}\sqrt{\Delta f } = \sqrt{ 4 k_B T R \Delta f }
where Δ
f is the bandwidth in hertz over which the noise is measured. For a 1 kΩ resistor at room temperature and a 10 kHz bandwidth, the RMS noise voltage is 400 nV. [5] A useful rule of thumb to remember is that 50 Ω at 1 Hz bandwidth correspond to 1 nV noise at room temperature.
A resistor in a short circuit dissipates a noise power of
P = {v_{n}^2}/R = 4 k_B \,T \Delta f.
The noise generated at the resistor can transfer to the remaining circuit; the maximum noise power transfer happens with impedance matching when the Thévenin equivalent resistance of the remaining circuit is equal to the noise generating resistance. In this case each one of the two participating resistors dissipates noise in both itself and in the other resistor. Since only half of the source voltage drops across any one of these resistors, the resulting noise power is given by
P = k_B \,T \Delta f
where
P is the thermal noise power in watts. Notice that this is independent of the noise generating resistance. Noise current
The noise source can also be modeled by a current source in parallel with the resistor by taking the Norton equivalent that corresponds simply to divide by
R. This gives the root mean square value of the current source as:
i_n = \sqrt - 1}
f \ll \frac{k_B T}{h}
(this assumption is valid until few terahertz at room temperature) then the exponential can be expressed in terms of its Taylor series. The relationship then becomes:
\Phi (f) \approx 2 R k_B T.
In general, both
R and T depend on frequency. In order to know the total noise it is enough to integrate over all the bandwidth. Since the signal is real, it is possible to integrate over only the positive frequencies, then multiply by 2.Assuming that R and T are constants over all the bandwidth , then the root mean square (RMS) value of the voltage across a resistor due to thermal noise is given by
v_n = \sqrt { 4 k_B T R \Delta f },
that is, the same formula as above.
See also References
This article incorporates MIL-STD-188).
Amplifier noise in RF systems Thermal noise (undergraduate) with detailed math Johnson–Nyquist noise or thermal noise calculator – volts and dB Derivation of the Nyquist relation using a random electric field, H. Sonoda |
I am using Fourier series to find the analytical solution to the 2D heat equation. The problem is that the integrals which are used to calculate the coefficients of the series cannot be solved analytically.
I am currently using GSL's Monte Carlo integration to compute them an this introduces error to the series. My question is will the solution converge to the analytical solution or not? If not, How can I get an estimate of the error?
EDIT
This is my initial condition:
\begin{align} &C_0(x,y) = \left\lbrace \begin{array}{lll} 0 & &r(x,y) \geq 15\\ 4- \dfrac{4}{15} r(x,y) & &r(x,y) < 15 \end{array}\right. \label{ch3-diffusion-2d-circle-init-cond}\\ &r(x,y) = \sqrt{(x-75)^2 + (y-50)^2} \end{align}
and this is the integral:
$$A_{mn}={4\over l_x l_y}\int_0^{l_y} \int_0^{l_x} C_0(x,y)\sin\left({m\pi\, x\over l_x}\right)\sin\left({n\pi\, y\over l_y}\right)\,dx\,dy, \quad \forall m,n \in \aleph$$
where $l_x=l_y = 100$.
This is what I get after 500 iterations:
Which hasn't really improved from iteration 100. |
I've looked at this and it doesn't help because I don't know anything about SVD. Can someone dumb it down for me please?
It is not exactly true that non-square matrices can have eigenvalues. Indeed, the definition of an eigenvalue is for square matrices. For non-square matrices, we can define singular values:
Definition: The singular values of a $m \times n$ matrix $A$ are the positive square roots of the nonzero eigenvalues of the corresponding matrix $A^{T}A$. The corresponding eigenvectors are called the singular vectors.
Of course, these have certain properties, that may or may not be useful for what you are trying to study.
Eigenvalues and eigenvectors of a matrix, say $A$, help us find subspaces which are invariant under $A$ (when $A$ is seen as a linear transformation). If $A$ is non-square, then $A:\mathbb{R}^m\rightarrow \mathbb{R}^n$, where $m\neq n$. Hence $Av=\lambda v$ makes no sense, since $Av\notin\mathbb{R}^m$.
Non-square matrices do not have eigenvalues. If the matrix X is a real matrix, the eigenvalues will either be all real, or else there will be complex conjugate pairs.
Source: Wikipedia. |
Let $\mathcal{C}$ be a category. We can construct a category of morphisms $\mathcal{M}$ by letting the objects be morphisms in $\mathcal{C}$ and morphisms be appropriate pairs of morphisms that give a commutative diagram.
Let $f:A \to B$, $g:C \to D$, $h:E \to F$, and $k:G \to H$. A morphism $(r,s) \in \mathrm{Hom}(f,g)$ consists of a map $r:A \to C$ and a map $s:B \to D$, and a morphism $(t,u) \in \mathrm{Hom}(h,k)$ consists of a map $t:E \to G$ and a map $u:F \to H$, all so that the appropriate diagrams commute.
Part of the requirement of a category is that hom-sets be disjoint. Suppose $(r,s)=(t,u)$ for some $(t,u) \in \mathrm{Hom}(h,k)$. Then we must have $r:E \to G$ and $s:F \to H$, so it must be that $A=E$, $C=G$, $B=F$, and $D=H$.
Since the morphisms $r:A \to C$ and $s:B \to D$ form a commutative diagram with $f:A \to B$ and $g:C \to D$, we have the relations $k \circ r=s \circ h$ and $g \circ r=s \circ f$. I'm not sure how to show from here that $r=t$ and $s=u$, which would show that if two hom-sets are not equal then they are disjoint. I see that they must have the same domain and codomain, but I feel like this isn't enough.
How do we show that hom-sets in a morphism category are disjoint? |
A quantum system has Hamiltonian $H$ with normalised eigenstates $\psi_n$ and corresponding energies $E_n$ ($n = 1,2,3...$). A linear operator $Q$ is defined by its action on these states: $$ Q\psi_1 = \psi_2 $$ $$ Q\psi_2 = \psi_1 $$ $$ Q\psi_n = 0, n>2 $$ Show that $Q$ has eigenvalues 1 and -1 and find the corresponding normalised eigenstates $\zeta_1$ and $\zeta_2$, in terms of energy eigenstates. Calculate $\langle H \rangle$ in each of the states $\zeta_1$ and $\zeta_2$.
A measurement of $Q$ is made at time=0, and the result 1 is obtained. The system is then left undisturbed for a time $t$, at which instant another measurement of $Q$ is made. What is the probability that the result will again be 1? Show that the probability is 0 if the measurement is made after a time $T = \pi \hbar/(E_2 - E_1)$, assuming $E_2 - E_1> 0$.
I found $$ \zeta_1 = (\psi_1 + \psi_2)/\sqrt{2} $$ $$ \zeta_2 = (\psi_1 - \psi_2)/\sqrt{2} $$ and $\langle H \rangle = (E_1 + E_2)/2$ for both.
I have trouble doing the second part.
Doesnt the system collapse into $ \zeta_1$ given we know this is the state at time =0? so probability will be 1? |
A ladder 16m leans against an electric pole. If the ladder makes an angle of 65
o with the ground. How far up the electric pole does its top reach WAEC MAY/ JUNE 2006
A ladder 16m leans against an electric pole. If the ladder makes an angle of 65
Simplify $\frac{20}{5\sqrt{28}-2\sqrt{63}}$
The temperature in a chemical plant was –5
oC at 2.00am . The temperature fell by 6 oC and then rose again by 7 oC. What was the final temperature?
Which of the following is not a measure of central tendency?
In the diagram $\left| XY \right|=12cm,\left| XZ \right|=9cm,\left| ZN \right|=3cm$ and $\left| ZY \right|=$ calculate $\left| MY \right|$
A rectangular carpet 2.6m long and 2.4 wide covers 5% of rectangular floor. Calculate the area of the floor.
The value of three angles at a point are 3
y - 45 o, y + 25 o and y o. Find the value of y
In the diagram, what is
x+ y in term of z ?
If
P = { 1,3, 5, 7,9} and Q = {2, 4, 6, 8, 10} are subsets of a universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, what are the elements of $P'\cap Q'$
The wheel of a tractor has diameter 1.4m, what distance does it cover in 100complete revolution? [Take $\pi =\tfrac{22}{7}$] |
Instead of arguing with other people's answers in the comments I thought it might be more productive to present my own point of view. I find myself completely unable to understand why anyone would take off points for this student's answer.
Just to be clear, this isn't because I'm being somehow lax or generous as a grader. My opinion is that
this is a model solution to the problem, written clearly and well, and I can imagine writing exactly what this student wrote as part of homework solution or exam solution that I distribute to a class. In the context of Calculus I, it's also how I would do this problem on the board during class if a student asked me about it. On the Status of Infinity
Some of the other calculus teachers here have mentioned that they teach their students that "infinity isn't a number".
I find this statement very strange, and I suppose that my position is that
infinity is a number. It certainly isn't a real number, since it's not included in the usual real number system. But neither is the imaginary unit $i$, and I don't think many people would argue that $i$ isn't a number. The number $i$ is included in the system of complex numbers, and the number $\infty$ is included in the system of extended real numbers, which is the set $\mathbb{R}\cup\{-\infty,\infty\}$. I don't see the difference.
Of course, there's no standard definition of "number" in mathematics, so there's no objective truth either way. This is part of why it strikes me as so odd that a teacher would say that "$\infty$ isn't a number".
It's possible that what they mean is that "you can't do arithmetic with $\infty$". But of course
you can do arithmetic with $\infty$. For example,$$\infty + \infty = \infty,\qquad \infty \cdot \infty = \infty,\qquad\text{and}\qquad 3\cdot \infty = \infty.$$These definitions are absolutely standard in mathematics, and I would feel free to use them in a conference talk or journal article without comment. I would hope that most calculus students would know how to do basic arithmetic with $\infty$ by the end of a first calculus course, but apparently this varies by instructor.
There are also arithmetic operations involving $\infty$ that are undefined, such as$$\infty - \infty,\qquad \frac{\infty}{\infty},\qquad\text{and}\qquad 0\cdot\infty.$$The last is sometimes defined to be zero (e.g. in the theory of Lebesgue integration), but in the context of calculus it's better to leave it undefined.
As far as I know, all of this is completely standard, and in my experience arithmetic involving $\infty$ and $-\infty$ is commonly used by mathematicians without further explanation or comment. I've seen lots of examples of this, but to cite a specific one it's certainly the case that Rudin's
Real & Complex Analysis textbook (an extremely standard choice for a graduate analysis course) uses the extended real number system throughout. On the Student's Answer
The student's answer depends primarily on the following theorem
Theorem. Let $f\colon \mathbb{R}\to\mathbb{R}$ and $g\colon\mathbb{R}\to\mathbb{R}$ be functions, and let $a\in [-\infty,\infty]$. If$$\lim_{x\to a} f(x) = L\qquad\text{and}\qquad \lim_{x\to a} g(x) = M$$ for some $L,M\in[-\infty,\infty]$ and the product $LM$ is defined, then$$\lim_{x\to a} f(x)\,g(x) = LM.$$
This is a well-known and standard theorem in analysis. In the context of this theorem, the student's work constitutes a perfectly good proof of the fact that$$\lim_{x\to\infty} \bigl(x-\sqrt{x}\bigr) = \infty.$$It is no more or less correct than something like$$\lim_{x\to 0} \frac{x\sin x + 2 \sin x}{x} = \lim_{x\to 0} \,\bigl(x+2\bigr)\!\left(\frac{\sin x}{x}\right) = (2)(1) = 2.$$I don't see why this proof would require any more explanation or rigor, in either a calculus or real analysis course, and I feel the same way about the student's proof. I suppose it
might be reasonable for an analysis professor to always require students to cite the theorems that they are using, as opposed to using theorems implicitly as part of a calculation. I certainly don't think this would be a reasonable requirement for student answers in a calculus course. Should we teach arithmetic with infinity to calculus students?
I do, and I would certainly hope that most other calculus instructors do as well. Dealing with the concept of infinity is a major theme of calculus, and the rules for arithmetic involving infinity ultimately derive from the idea of a limit. How does it help to avoid talking about this?
Actually, it seems to me that it would be difficult to cover the idea of an "indeterminate form" without covering this material. I guess at least some of the teachers here manage to avoid saying that "infinity plus infinity equals infinity" by always saying "the sum of two quantities that are both approaching infinity again approaches infinity", but what's the purpose of being so obtuse? If there's a simple way to say something, just say it that way.
And in any case, the reality is that you
can do arithmetic with infinity. Saying that $\infty+\infty$ is undefined or indeed anything other than $\infty$ is just wrong, both at an intuitive level and from the point of view of standard notation and terminology. Students will figure out that it's true on their own, and will try to guess what other arithmetic rules you're not telling them. If you tell students that $\infty + \infty$ isn't $\infty$, you lose your credibility, and they won't believe you later when you tell them that $\infty - \infty$ isn't $0$. Okay, but should we mark the student wrong?
Even if you don't talk about arithmetic involving infinity in your calculus class, the fact remains that it is absolutely standard mathematical notation. Students often seek help from mathematics tutors, other math professors, online videos, and so forth, and any one of those sources might be teaching your students about how to use infinity in this fashion. Can you really justify deducting points from students who don't write their mathematics the way that
you want it written? I feel like one of the most basic principles of grading is that correct answers should receive full credit, unless the answer explicitly violates the instructions for the question. This student's answer is completely correct, and in my opinion giving it anything less than 5/5 is just arbitrary and unfair. |
In the diagram
O is the centre of circle and PQ is a diameter. Triangle of RSQ is an equilateral triangle of side 4cm. Find the area of the shaded region WAEC MAY/ JUNE 2007
In the diagram
In the diagram, $\left| XR \right|=4cm$, $\left| RZ \right|=12cm,\left| SR \right|=n,\text{ }\left| YZ \right|=m$ and $\overline{SR}\parallel \overline{YZ}$. Find
m in terms of n
The volume of a cylinder is 1200cm
3 and the area of its base is 150cm 2. Find the height of the cylinder
Two circles have radii 16cm and 23cm. What is the difference between their circumferences? (take $\pi =\tfrac{22}{7}$)
Simplify $\frac{4}{2x}-\frac{2+x}{x}$
Find the smallest value of
k such that ${{2}^{2}}\times {{3}^{3}}\times 5\times k$ is a perfect square
For what range of values of
x is $4x-3(2x-1)>1$
Expand the expression $(3a-xy)(3a+xy)$
Make
w the subject of the relation $\frac{a+bc}{wd+f}=g$
Which of the following is a factor of $2-x-{{x}^{2}}$ |
The definition of Pohlmeyer invariants in flat-space (as per eq-2.16 in Urs Schreiber's DDF and Pohlmeyer invariants of (super)string) is the following:
$ Z^{\mu_1...\mu_N} (\mathcal{P}) = \frac{1}{N} \int\limits_0^{2\pi} d\sigma^1 \int\limits_{\sigma_1}^{\sigma_1+2\pi} d\sigma^2... \int\limits_{\sigma_{N-1}}^{\sigma_1} d\sigma^N \ \mathcal{P}^{\mu_1}(\sigma_1)\ \mathcal{P}^{\mu_2}(\sigma_2)...\mathcal{P}^{\mu_N}(\sigma_N) $
Another reference is eq-3.29 in Thomas Thiemann's The LQG -- String: Loop Quantum Gravity Quantization of String Theory I. Flat Target Space. These formulations of the Pohlmeyer invariants is used to show its equivalence to DDF invariants of light-cone gauge fixed strings in flat-space (cf. eq-2.53 in Urs Schreiber's paper and chap-2.3.2 in Green, Schwarz & Witten vol-I)
Now, what is the corresponding invariant for the Pohlmeyer reduction in AdS case (ref: Miramontes' "Pohlmeyer reduction revisited" arXiv:0808.3365)? I cant seem to find one that resembles the flat-space expression.
This post has been migrated from (A51.SE) |
Find the quadratic equation whose roots are – ½ and 3
WAEC MAY/ JUNE 2009
Simplify ${{(x-3y)}^{2}}-{{(x+3y)}^{2}}$
Which shaded region in the following diagram represents$(P\cup Q)\cap R$?
The angles of a triangle are (
x +10) o, (2 x – 40) o and (3 x – 90) o. Which of the following accurately described the triangle
Find the value of
x in the diagram
From the diagram, which of the following statements is/are correct?
I.
is a radius of a centre XQ Q
II. $\left| XQ \right|=\left| QY \right|$
III. $\left| QX \right|=\left| XY \right|$
Simplify $\tfrac{1}{2}\sqrt{32}-\sqrt{18}+\sqrt{2}$
If a student is picked at random from the four Departments, what is the probability that he is
not from the Education Department
The pie shows that the distribution of 4320 students who graduated from foru Department in a university.
How many students from the Science?
A box contains black, white and red identical balls. The probability of picking a black ball at random from the box is $\tfrac{3}{10}$ and the probability of picking a white ball at random is $\tfrac{2}{5}$. If there are 30 balls in the box, how many of them are red? |
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