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Let me offer one reason and one misconception as an answer to your question.The main reason that it is easier to write (seemingly) correct mathematical proofs is that they are written at a very high level. Suppose that you could write a program like this:function MaximumWindow(A, n, w):using a sliding window, calculate (in O(n)) the sums of all ...
(I am probably risking a few downvotes here, as I have no time/interest to make this a proper answer, but I find the text quoted (and the rest of the article cited) below to be quite insightful, also considering they are written by a well-known mathematician. Perhaps I can improve the answer later.)The idea, which I suppose isn't particularly distinct from ...
A common error I think is to use greedy algorithms, which is not always the correct approach, but might work in most test cases.Example: Coin denominations, $d_1,\dots,d_k$ and a number $n$,express $n$ as a sum of $d_i$:s with as few coins as possible.A naive approach is to use the largest possible coin first,and greedily produce such a sum.For ...
I immediately recalled an example from R. Backhouse (this might have been in one of his books). Apparently, he had assigned a programming assignment where the students had to write a Pascal program to test equality of two strings. One of the programs turned in by a student was the following:issame := (string1.length = string2.length);if issame thenfor ...
Allow me to start by quoting E. W. Dijkstra:"Programming is one of the most difficult branches of applied mathematics; the poorer mathematicians had better remain pure mathematicians." (from EWD498)Although what Dijkstra meant with `programming' differs quite a bit from the current usage, there is still some merit in this quote. The other answers have ...
Here is an algorithm for the identity function:Input: $n$Check if the $n$th binary string encodes a proof of $0 > 1$ in ZFC, and if so, output $n+1$Otherwise, output $n$Most people suspect this algorithm computes the identity function, but we don't know, and we can't prove it in the commonly accepted framework for mathematics, ZFC.
Lamport provides some ground for disagreement on prevalence of errors in proofs in How to write a proof (pages 8-9):Some twenty years ago, I decided to write a proof of the Schroeder-Bernsteintheorem for an introductory mathematics class. The simplest proof I couldfind was in Kelley’s classic general topology text. Since Kelleywas writing for a ...
One big difference is that programs typically are written to operate on inputs, whereas mathematical proofs generally start from a set of axioms and prior-known theorems. Sometimes you have to cover multiple corner cases to get a sufficiently general proof, but the cases and their resolution is explicitly enumerated and the scope of the result is implicitly ...
The best example I ever came across is primality testing:input: natural number p, p != 2output: is p a prime or not?algorithm: compute 2**(p-1) mod p. If result = 1 then p is prime else p is not.This works for (almost) every number, except for a very few counter examples, and one actually needs a machine to find a counterexample in a realistic period ...
They say the problem with computers is that they do exactly what you tell them.I think this might be one of the many reasons.Notice that, with a computer program, the writer (you) is smart but the reader (CPU) is dumb.But with a mathematical proof, the writer (you) is smart and the reader (reviewer) is also smart.This means you can never afford to get ...
There are indeed programs like this. To prove this, let's suppose to the contrary that for every machine that doesn't halt, there is a proof it doesn't halt.These proofs are strings of finite length, so we can enumerate all proofs of length less than $s$ for some integer $s$.We can then use this to solve the halting problem as follows:Given a Turing ...
Ultimately, you'll need a mathematical proof of correctness. I'll get to some proof techniques for that below, but first, before diving into that, let me save you some time: before you look for a proof, try random testing.Random testingAs a first step, I recommend you use random testing to test your algorithm. It's amazing how effective this is: in my ...
One issue that I think was not addressed in Yuval's answer, is that it seems you are comparing different animals.Saying "the code is correct" is a semantic statement, you mean to say that the object described by your code satisfies certain properties, e.g. for every input $n$ it computes $n!$. This is indeed a hard task, and to answer it, one has to look ...
First, let us make two maybe obvious, but important assumptions:_.random_item can choose the last position._.random_item chooses every position with probability $\frac{1}{n+1}$.In order to prove correctness of your algorithm, you need an inductive argument similar to the one used here:For the singleton list there is only one possibility, so it is ...
Here's one that was thrown at me by google reps at a convention I went to. It was coded in C, but it works in other languages that use references.Sorry for having to code on [cs.se], but it's the only to illustrate it.swap(int& X, int& Y){X := X ^ YY := X ^ YX := X ^ Y}This algorithm will work for any values given to x and y, ...
What is so different about writing faultless mathematical proofs and writing faultless computer code that makes it so that the former is so much more tractable than the latter?I believe that the primary reasons are idempotency (gives the same results for the same inputs) and immutability (doesn't change).What if a mathematical proof could give different ...
There is a whole class of algorithms that is inherently hard to test: pseudo-random number generators. You can not test a single output but have to investigate (many) series of outputs with means of statistics. Depending on what and how you test you may well miss non-random characteristics.One famous case where things went horribly wrong is RANDU. It ...
This is not a secure encryption scheme. It is similar to a Hill cipher, and vulnerable to similar attacks. For instance, it is vulnerable to known-plaintext attacks: an attacker who observes a ciphertext E and knows the corresponding message M can recover the secret key and thus decrypt all other messages that were encrypted with the same key.The ...
I will use the following simple sorting algorithm as an example:repeat:if there are adjacent items in the wrong order:pick one such pair and swapelsebreakTo prove the correctness I use two steps.First I show that the algorithm always terminates.Then I show that the solution where it terminates is the one I want.For the first point,...
We are indeed assuming $P(k)$ holds for all $k < n$. This is a generalization of the "From $P(n-1)$, we prove $P(n)$" style of proof you're familiar with.The proof you describe is known as the principle of strong mathematical induction and has the formSuppose that $P(n)$ is a predicate defined on $n\in \{1, 2, \dotsc\}$. If we can show that$...
I agree with what Yuval has written. But also have a much simpler answer: In practice softwares engineers typically don't even try to check for correctness of their programs, they simply don't, they typically don't even write down the conditions that define when the program is correct.There are various reasons for it. One is that most software engineers ...
You need the line C(i:j) = 0 just before the innermost loop; otherwise, the code is incorrect.Assuming that line is in place, here is the (strongest possible) invariant just before the assignment in the innermost loop:\begin{align*}C_{IJ} &= \sum_{k=1}^{n} A_{Ik}B_{kJ} & \text{for all $I$ and $J$ such that $1\le I < i$ and $1\le J\le n$}\\\...
No, your algorithm doesn't work. Consider if the array A isA = [1 1 1 1 1 2 2 3 3 3 3 3 3].Then the array B will beB = [5 5 5 5 5 2 2 6 6 6 6 6 6].The sum of B will be 65, and the length of B will be 13, so after division, we'll get the number 5. This is equal to the first element of B, so your algorithm will output "Yes". Nonetheless, not all ...
There are a lot of good answers already but there are still more reasons math and programming aren't the same.1Mathematical proofs tend to be much simpler than computer programs. Consider the first steps of a hypothetical proof:Let a be an integerLet b be an integerLet c = a+bSo far the proof is fine. Let's turn that into the first ...
There is no (one) formal definition of "optimal substructure" (or the Bellman optimality criterion) so you can not possibly hope to (formally) prove you have it.You should do the following:Set up your (candidate) dynamic programming recurrence.Prove it correct by induction.Formulate the (iterative, memoizing) algorithm following the recurrence.
Cryptosystems which are algebraic in nature are amenable to algebraic cryptanalysis.If you are trying to design a secure cryptosystem for actual use, there is one important maxim that you should keep in mind:Don't design your own cryptosystem!It is easy to design weak cryptosystems. Off-the-shelf cryptosystems have withstood breaking attempts by the ...
I like Yuval's answer, but I wanted to riff off of it for a bit. One reason you might find it easier to write Math proofs might boil down to how platonic Math ontology is. To see what I mean, consider the following:Functions in Math are pure (the entire result of calling a function is completely encapsulated in the return value, which is deterministic and ...
Proving that a program is "thread safe" is hard. It is possible, however, to concretely and formally define the term "data race." And it is possible to determine whether an execution trace of a specific run of a program does or does not have a data race in time proportional to the size of the trace. This type of analysis goes back at least to 1988: ...
2D local maximuminput: 2-dimensional $n \times n$ array $A$output: a local maximum -- a pair $(i,j)$ such that $A[i,j]$ has no neighboring cell in the array that contains a strictly larger value.(The neighboring cells are those among $A[i, j+1], A[i, j-1], A[i-1, j], A[i+1, j]$ that are present in the array.) So, for example, if $A$ is$$\begin{...
First off to answer your main question: there is no flaw in the proof.The point where you are not following the reasoning of the proof is that:E belongs to a cycle in the graph and to the MST.Not every other edge of the cycle are in the MST.If you look at your drawing, then either one of the edges towards the rest of the MST is redundant. Because ... |
$g(x) = x^2$ is indeed a parabola and thus has just one optimum.
However, the $\text{MSE}(\boldsymbol{x}, \boldsymbol{y}) = \sum_i (y_i - f(x_i))^2$, where $\boldsymbol{x}$ are the inputs, $\boldsymbol{y}$ the corresponding labels and the function $f$ is the model (e.g. a neural network), is
not necessarily a parabola. In general, it is only a parabola if $f$ is a constant function and the sum is over one element.
For example, suppose that $f(x_i) = c, \forall i$, where $c \in \mathbb{R}$. Then $\text{MSE}(\boldsymbol{x}, \boldsymbol{y}) = \sum_i (y_i - c)^2$ will only change as a function of one variable, $\boldsymbol{y}$, as in the case of $g(x) = x^2$, where $g$ is a function of one variable, $x$. In that case, $(y_i - c)^2$ will just be a shifted version (either to the right or left depending on the sign of $c$) of $y_i^2$, so, for simplicity, let's ignore $c$. So, in the case $f$ is a constant function, then $\text{MSE}(\boldsymbol{x}, \boldsymbol{y}) = \sum_i y_i^2$, which is a sum of parabolas $y_i^2$, which is called a
paraboloid. In this case, the paraboloid corresponding to $\text{MSE}(\boldsymbol{x}, \boldsymbol{y}) = \sum_i y_i^2$ will only have one optimum, just like a parabola. Furthermore, if the sum is just over one $y_i$, that is, $\text{MSE}(\boldsymbol{x}, \boldsymbol{y}) = \sum_i y_i^2 = y^2$ (where $\boldsymbol{y} = y$), then the MSE becomes a parabola.
In other cases, the MSE might not be a parabola or have just one optimum. For example, suppose that $f(x) = x^2$, $y_i = 1$ ($\forall i$), then $h(x) = (1 - x^2)^2$ looks as follows
which has two minima (and one maximum): at $x=0$ and $x=1$. We can find the two minima of this function $h$ using calculus: $h'(x) = -4x(1 - x^2)$, which becomes zero when $x=0$ and $x=1$.
In this case, we only considered one term of the sum. If we considered the sum of terms of the form of $h$, then we could even have more "complicated" functions.
To conclude, given that $f$ can be arbitrarily complex, then also $\text{MSE}(\boldsymbol{x}, \boldsymbol{y})$, which is a function of $f$, can also become arbitrarily complex and have multiple minima. Given that neural networks can implement arbitrarily complex functions, then $\text{MSE}(\boldsymbol{x}, \boldsymbol{y})$ can easily have multiple minima. Moreover, the function $f$ (e.g. the neural network) changes during the training phase, which might introduce more complexity, in terms of which functions the MSE can be and thus which (and how many) optima it can have. |
Will it be correct to use
Mean absolute scaled error in non time series data? I've got a set which contains a lot of zeros, so errors like MAPE can not be used here. MASE based on difference between yt and yt-1 (formula like here), but I don't have time series, just observations of some activity of 200 people (independent), so maybe I can just use difference based on second and firstperson and so on? If not, do you know any percentage measures that I can use in that kind of set?
Will it be correct to use
(The standard version of) MASE only works for time series. The purpose of the $Y_t$ - $Y_{t-1}$ term in the denominator is that it represents a naive forecast, that it is your best guess for $Y_t$ is $Y_{t-1}$.
In your case, your data is non-sequential and more importantly it is independent, so the idea of using $I_{n-1}$ as an estimator for $I_n$ doesn't really make sense (why $I_{n-1}$? Why not $I_{n-2}$ or $I_{n-199}$?)
To answer your 2nd question, it depends on what error you are trying to measures exactly ?
For dealing with the issue of zeros, a possible alternative for MAPE is sMAPE (Symmetric Mean Absolute Percentage Error). That might be a good option for you.
Addressing the comment:
When $Y_{emp}$ and $Y_{pred}$ are both equal to zero, than you set the sMAPE to 0 (since they are equal to each other, there is no error) (add an if clause to your calculation). Many software packages do that already, even with MAPE. For the second problem, you can use the modified version of the sMAPE:
$$\text{sMAPE} = \frac{1}{N}\sum \frac{|y_{emp} -y_{pred}|}{{|y_{emp}|+|y_{pred}|}}$$
This gives you a 100% error in cases like you mentioned ($Y_{emp} = 0$ and $Y_{pred}=0.005$)
So the modified version of sMAPE could be:
$$\text{sMAPE} = \begin{cases} 0, & \text{ if } Y_{emp} = 0 & and & Y_{pred}=0 \\ \frac{1}{N}\sum \frac{|y_{emp} -y_{pred}|}{{|y_{emp}|+|y_{pred}|}} & \text{otherwise} \\ \end{cases} $$
If even 100% is too high a threshold, you can apply some sort of log transformation.
Since I first wrote this answer, I noticed elsewhere in this forum that Rob Hyndman (the original author of MASE) proposes a modified version of MASE for non-time series data. See here
If $e_j$ denotes a prediction error on the test data, then the scaled errors are $$ q_{j} = \frac{\displaystyle e_{j}}{\displaystyle\frac{1}{N}\sum_{i=1}^N |Y_i-\bar{Y}|}. $$ where $y_1,\dots,y_N$ denotes the training data.
Note here that $Y_{t-1}$ is replaced with $\bar{Y}$.
In the time series case, this would mean that the prediction is being compared to the predictions produced by a mean forecast instead of a naive forecast (as it is in the standard formulation of MASE). |
Frequent Links Tisserand's parameter Tisserand's parameter (or Tisserand's invariant) is a value calculated from several orbital elements (semi-major axis, orbital eccentricity, and inclination) of a relatively small object and a larger "perturbing body". It is used to distinguish different kinds of orbits. It is named after French astronomer Félix Tisserand, and applies to restricted three-body problems, in which the three objects all differ greatly in size. Definition
For a small body with semimajor axis <math>a\,\!</math>, eccentricity <math>e\,\!</math>, and inclination <math>i\,\!</math>, relative to the orbit of a perturbing larger body with semimajor axis <math>a_P</math>, the parameter is defined as follows:
[1] <math>T_P\ = \frac{a_P}{a} + 2\cdot\sqrt{\frac{a}{a_P} (1-e^2)} \cos i</math>
The quasi-conservation of Tisserand's parameter is a consequence of Tisserand's relation.
Applications T J, Tisserand’s parameter with respect to Jupiter as perturbing body, is frequently used to distinguish asteroids (typically <math>T_J > 3</math>) from Jupiter-family comets (typically <math>2< T_J < 3</math>). The roughly constant value of the parameter before and after the interaction (encounter) is used to determine whether or not an observed orbiting body is the same as one previously observed in Tisserand's Criterion. The quasi-conservation of Tisserand's parameter constrains the orbits attainable using gravity assist for outer Solar system exploration. T N, Tisserand's parameter with respect to Neptune, has been suggested to distinguish near-scattered (affected by Neptune) from extended-scattered trans-Neptunian objects (not affected by Neptune; e.g. 90377 Sedna). Tisserand's parameter could be used to infer the presence of an intermediate-mass black hole at the center of the Milky Way using the motions of orbiting stars. [2] Related notions
The parameter is derived from one of the so-called Delaunay standard variables, used to study the perturbed Hamiltonian in a 3-body system. Ignoring higher-order perturbation terms, the following value is conserved:
<math> \sqrt{a (1-e^2)} \cos i</math>
Consequently, perturbations may lead to the resonance between the orbital inclination and eccentricity, known as Kozai resonance. Near-circular, highly inclined orbits can thus become very eccentric in exchange for lower inclination. For example, such a mechanism can produce sungrazing comets, because a large eccentricity with a constant semimajor axis results in a small perihelion.
See also Tisserand's relation for the derivation and the detailed assumptions |
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.
\[\Delta{T_f} = T_f(solvent) - T_f (solution) = K_f \times m\]
where \(\Delta{T_f}\) is the freezing point depression, \(T_f\) (solution) is the freezing point of the solution, \(T_f\) (solvent) is the freezing point of the solvent, \(K_f\) is the freezing point depression constant, and
m is the molality. Introduction
Nonelectrolytes are substances with no ions, only molecules. Strong electrolytes, on the other hand, are composed mostly of ionic compounds, and essentially all soluble ionic compounds form electrolytes. Therefore, if we can establish that the substance that we are working with is uniform and is not ionic, it is safe to assume that we are working with a nonelectrolyte, and we may attempt to solve this problem using our formulas. This will most likely be the case for all problems you encounter related to freezing point depression and boiling point elevation in this course, but it is a good idea to keep an eye out for ions. It is worth mentioning that these equations work for both volatile and nonvolatile solutions. This means that for the sake of determining freezing point depression or boiling point elevation, the vapor pressure does not effect the change in temperature. Also remember that a pure solvent is a solution that has had nothing extra added to it or dissolved in it. We will be comparing the properties of that pure solvent with its new properties when added to a solution.
Adding solutes to an ideal solution results in a positive ΔS, an increase in entropy. Because of this, the newly altered solution’s chemical and physical properties will also change. The properties that undergo changes due to the addition of solutes to a solvent are known as colligative properties. These properties are dependent on the amount of solutes added, not on their identity. Two examples of colligative properties are boiling point and freezing point: due to the addition of solutes, the boiling point tends to increase and freezing point tends to decrease.
The freezing point and boiling point of a pure solvent can be changed when added to a solution. When this occurs, the freezing point of the pure solvent may become lower, and the boiling point may become higher. The extent to which these changes occur can be found using the formulas:
\[\Delta{T}_f = -K_f \times m\]
\[\Delta{T}_f = K_b \times m\]
where \(m\) is the solute
molality and \(K\) values are proportionality constants; (\(K_f\) and \(K_b\) for freezing and boiling, respectively.
If solving for the proportionality constant is not the ultimate goal of the problem, these values will most likely be given. Some common values for \(K_f\) and \(K_b\) respectively, are:
Solvent \(K_f\) \(K_b\) Water 1.86 .512 Acetic acid 3.90 3.07 Benzene 5.12 2.53 Phenol 7.27 3.56
Molality is defined as the number of moles of solute per kilogram
solvent. Be careful not to use the mass of the entire solution. Often, the problem will give you the change in temperature and the proportionality constant, and you must find the molality first in order to get your final answer.
The solute, in order for it to exert any change on colligative properties, must fulfill two conditions. First, it must not contribute in the vapor pressure of the solution and second, it must remain suspended in the solution even during phase changes. Because the solvent is no longer pure with the addition of solutes, we can say that the
chemical potential of the solvent is lower. Chemical potential is the molar Gibb's energy that one mole of solvent is able to contribute to a mixture. The higher the chemical potential of a solvent is, the more it is able to drive the reaction forward. Consequently, solvents with higher chemical potentials will also have higher vapor pressures.
Boiling point is reached when the chemical potential of the pure solvent, a liquid, reaches that of the chemical potential of pure vapor. Because of the decrease of the chemical potential of mixed solvents and solutes, we observe this intersection at higher temperatures. In other words, the boiling point of the impure solvent will be at a higher temperature than that of the pure liquid solvent. Thus,
boiling point elevation occurs with a temperature increase that is quantified using
\[\Delta{T_b} = K_b b_B\]
where
\(K_b\) is known as the ebullioscopic constantand \(m\) is the molality of the solute.
Freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent. Again, since we are dealing with mixtures with decreased chemical potential, we expect the freezing point to change. Unlike the boiling point, the chemical potential of the impure solvent requires a colder temperature for it to reach the chemical potential of the solid pure solvent. Therefore, a
freezing point depression is observed.
Example \(\PageIndex{1}\)
2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from 5.53 to 4.90 \(^{\circ}C\). What is the molar mass of the compound?
SOLUTION
First we must compute the molality of the benzene solution, which will allow us to find the number of moles of solute dissolved.
\[ \begin{align*} m &= \dfrac{\Delta{T}_f}{-K_f} \\[4pt] &= \dfrac{(4.90 - 5.53)^{\circ}C}{-5.12^{\circ}C / m} \\[4pt] &= 0.123 m \end{align*}\]
\[ \begin{align*} \text{Amount Solute} &= 0.07500 \; kg \; benzene \times \dfrac{0.123 \; m}{1 \; kg \; benzene} \\[4pt] &= 0.00923 \; m \; solute \end{align*}\]
We can now find the molecular weight of the unknown compound:
\[ \begin{align*} \text{Molecular Weight} =& \dfrac{2.00 \; g \; unknown}{0.00923 \; mol} \\[4pt] &= 216.80 \; g/mol \end{align*}\]
The freezing point depression is especially vital to aquatic life. Since saltwater will freeze at colder temperatures, organisms can survive in these bodies of water.
Applications
Road salting takes advantage of this effect to lower the freezing point of the ice it is placed on. Lowering the freezing point allows the street ice to melt at lower temperatures. The maximum depression of the freezing point is about −18 °C (0 °F), so if the ambient temperature is lower, \(\ce{NaCl}\) will be ineffective. Under these conditions, \(\ce{CaCl_2}\) can be used since it dissolves to make three ions instead of two for \(\ce{NaCl}\).
Problems
Benzophenone has a freezing point of 49.00
oC. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59 oC. Find the freezing point depression constant for the solvent.(answ.: 9.80oC/m) References Atkins, Peter and de Paula, Julio. Physical Chemistry for the Life Sciences. New York, N.Y.: W. H. Freeman Company, 2006. (124-136). Contributors Lorigail Echipare (UCD), Zachary Harju (UCD) |
You don't use a pre-generated list of primes. That would make it easy to crack as you note. The algorithm you want to use would be something like this (see note 4.51 in HAC, see also an answer on crypto.SE):Generate a random $512$ bit odd number, say $p$Test to see if $p$ is prime; if it is, return $p$; this is expected to occur after testing about $Log(p)...
Let's assume for an instant that you could build a large table of all primes. Then... what ? How would you use it ? What would you look up ?If you "just" scan the table and try to divide the number to factor by each prime, then this is known as trial division; there is no need to store the primes (they can be regenerated on-the-fly; that's the division ...
Primes are important because the security of many encryption algorithms are based on the fact that it is very fast to multiply two large prime numbers and get the result, while it is extremely computer-intensive to do the reverse. When you have a number which you know is the product of two primes, finding these two prime numbers is very hard. This problem is ...
The question to answer is "Is N the product of P*Q?" I believe that the easiest way to understand Shor is to imagine two sine waves, one length P and one length Q. Assuming that P and Q are co-prime, then the question above can also be answered "At what point does the harmony of P overlapped with Q repeat itself?" And the answer can be determined quickly, ...
The premise "we don't have a way of generating and verifying a 2048-bit prime number with 100% accuracy" is wrong (if we trust the computers performing the operations): it has long been known practicable ways to generate randomly-seeded provable primes, and it is a (somewhat marginal) practice in RSA key generation (see FIPS 186-4 appendix B.3.2). We can ...
Is this number specified anywhere?It was formally specified in this RFC as the 1536 bit MODP group (although its use predates that RFC). However, from what I've seen, the 2048 bit MODP group from that same document is actually more popular.Why was this particular number picked?Well, it's a safe prime; in addition, the leading 64 bits and the ...
However, factoring a large integer is extremely difficult, even for a computer using known factoring algorithms.Not categorically. Factoring a large integer is trivial if it is only composed of small factors.A fairly naive algorithm for factoring N is the following:while N > 1:for p in increasing_primes:while p divides N:N = N / p...
No, it is not at all feasible to build an index of prime factors to break RSA.Even if we consider 384-bit RSA, which was in use but breakable two decades ago, the index would need to include a sizable portion of the 160 to 192-bit primes, so that the smallest factor of the modulus has a chance to be in the index. Per the Prime number theorem there are in ...
Short answer: Yes.The discrete logarithm can be attacked in a multitude of ways: Baby-step giant-step (BSGS), Pollard's Rho, Pohlig-Hellman, and the several variants of Index Calculus, the best of which currently is the Number Field Sieve.Let $n$ be the order of the generator of our field $\mathbb{F}_p$; it is $n = p-1$. We are trying to find $x$ given $...
mpz_nextprime states in the documentation and source (file: mpz/nextprime.c) that it simply finds the next prime larger than the provided input. There are various methods of doing so (depending on how efficient it tries to be), but they should all produce the same answer. Looking at the code, mpz_nextprime first tests a number against a large quantity of ...
This procedure is known as incremental search and his described in the Handbook of Applied Cryptography (note 4.51, page 148). Although some primes are being selected with higher probability than others, this allows no known attacks on RSA; roughly speaking, incremental search selects primes which could have been selected anyway and there are still ...
A Mersenne prime is a prime number that can be written in the form $M_p = 2^n-1$, and they’re extremely rare finds. Of all the numbers between 0 and $2^{25,964,951}-1$ there are 1,622,441 that are prime, but only 42 are Mersenne primes.The second sentence is wrong.What they meant to say is that there are 1,622,441 numbers of the form they mentioned in ...
The main reasons we usually choose $p$ an $q$ prime numbers are:For a given size of $N=pq$, that makes $N$ harder to factor, hence RSA safer. Although efficient factoring algorithms do not find factors by trial division, it remains much easier to find very small prime factors than large ones. If we chose $p$ and/or $q$ at random without consideration for ...
In the general case, for proper security with Diffie-Hellman, we need a value $g$ such that the order of $g$ (the smallest integer $n \geq 1$ such that $g^n = 1 \mod p$) is a multiple of a large enough prime $q$. "Large enough" means "of length at least $2t$ bits if we target $t$-bit security". Since $n$ necessarily divides $p-1$, $q$ divides $p-1$.We ...
It has to do with optimizing RSA.It turns out that using the Chinese Remainder Theorem with $p$, $q$, $d\pmod{p-1}$, and $d\pmod{q-1}$ (i.e., prime1, prime2, exponent1, exponent2 from the data structure in the question) to run the decryption operation faster than if you only had $d,n$.For more information on how it is done, I found this reference http://...
The algorithm you quote is usually called textbook RSA and is not used in practice for numerous security reasons (the problem you pointed out, is just one of them).In practice, you have to pad (or armor) your message. This should be done using the RSA-OAEP (also called PKCS#1 v2.0) scheme. It transforms your message (1) into a pseudorandom block (not 1) ...
I can think of two places where we use a Mersenne Prime within cryptogaphy:As a modulus within a prime elliptic curve. $2^{521}-1$ is a prime, and so we can define an elliptic curve using $GF(2^{521}-1)$, which is in moderately common use. One reason we use such a modulus (rather than another prime of approximately the same size) is that the special form ...
The critical facts enabling to find such $p$ in practice are:We can easily tell with practical certainty if an integer with many thousand bits is prime or not, using a primality test such as Miller-Rabin, even though we are typically unable to tell all its factors when it is not prime.About $1.4/b$ integers of $b$ bits are prime. Thus it is more likely ...
Generating your own group for Diffie-Hellman is not a tough issue; but it is somewhat expensive (it depends on the context, but a 25 MHz ARM device would not like to do it often) and it is not really needed: a good point of DH (and DSA) is that the group parameters can be shared between many users, with no ill effect on the confidentiality of their ...
What we really need is a number $\lambda$ satisfying $x^{\lambda+1} \equiv x \pmod n$ for all integers $x$ (which, by induction, then implies that $x^{k\lambda+1} \equiv x \pmod n$ for any $k$).Given such a $\lambda$, and an arbitrary encryption exponent $e$ which is coprime to it, we can then find the multiplicative inverse of $e$ modulo $\lambda$, i.e. a ...
There are two approaches to such a validation:Test: you can look at the number and decide without involving the person who gave it to you.Proof: The person who generated the number can also give you additional information that will convince you it is a correct RSA number.There are no tests for RSA numbers.There are proofs for RSA numbers, including "...
Well, to answer your questions in order:How big should $p$ be? Well, it should be large enough to defend against the known attacks against it. The most efficient attack is NFS; that has been used against numbers on the order of $2^{768}$ (a 232 digit number). It would appear wise to pick a $p$ that's considerably bigger than that; around 1024 bits at a ...
"I understand that the typical approach is to use pre-generated lists of large primes."This is what I also thought. But I had not considered how many primes we might choose from. As it turns out you choose from ~2.8x10^147 primes with a 1024 bit RSA key and from about ~7.0x10^613 with a 4096 bit RSA key. Then you have up to 4.9x10^1227 possible pairs of ...
In order to generate a RSA key pair, you are to find a public exponent $e$ and a private exponent $d$ such that, for all $m \in \mathbb Z_n^*$, i.e. $m$ is relatively prime to $n$, $(m^e)^d \equiv m \pmod n$. It is a consequence of Euler's theorem that if $e, d$ satisfy the equation $ed \equiv 1 \pmod {\phi(n)}$, they are such a valid public/private exponent ...
RSA moduli are generally of the form $N = pq$ for two primes $p$ and $q$. It is also important that $p$ and $q$ have (roughly) the same size. The main reason is that the security of RSA is related to the factoring problem. The most difficult numbers to factor are numbers that are the product of two primes of similar size.Note. There are basically two ...
Wiener's result has been improved several times, and it is hard to tell how big the private exponent must be to be safe from further progress.Also, the proposed technique, assuming $d>n^{1/3}$, requires a minimum of ${1\over3}\cdot log_2(n)$ modular multiplications for the sparsest $d$ conceivable (a power of two), compared to say ${7\over6} \cdot log_2(...
We want a non-trivial factorization of a moderate odd integer $n$ into positive integers $p$ and $q$, knowing that such factorization with $|p-q|$ suitably small exists.Perhaps the most elementary method answering the question is trial division by integers starting at $\lfloor\sqrt n\rfloor$, going down. This succeeds after checking divisibility of $n$ by ...
There is no more efficient way of generating a safe prime. Even in OpenSSL's optimized code, it can take a long time to generate a safe prime (30 seconds, a minute, 2 minutes). Run "openssl gendh 1024" on your computer to see (on my 2015 MacBook pro it can take a long time, but the variance is really high so try a few times).The comments talk about safe ... |
Reading this paper which is itself an exposition of Parikh and Wilczek's paper, I get to a point where I fail to be able to follow the calculation. Now this is undoubtably because my calculational skills have been affected by decades of atrophy, so I wonder if someone can help. The paper computes the tunneling transmission coefficient for a particle (part of a particle/antiparticle pair) created just inside the horizon. The WKB transmission coefficient is given by $$T = exp({-\frac{2}{\hbar} Im(S)})$$ where "Im" is the imaginary part, and the action, $S$, is evaluated over the classically forbidden region. Using Painleve-Gullstrand coordinates for the black hole, the paper derives, fairly straightforwardly, $$Im(S) = Im \int_{2M}^{2(M-\omega)}{dr \int_{0}^{\omega}d\omega' {\frac{1}{1-\sqrt{\frac{2(M-\omega')}{r}}}}}$$ (eq 41). $\omega$ is the energy carried out by the tunneling particle. Now the next step is where I get stuck. The following equation (42) suggests that the r integration has been performed to get from (41)->(42). Here's what happens when I try to do it:
Presumably the contour integral being referred to is in the complexified r variable. The way the "r" appears isn't very nice, so we make a substitution $z=r^{\frac{1}{2}}$, giving $$Im \int_{2M}^{2(M-\omega)}dz{\frac{2z^2}{z-\sqrt{2(M-\omega')}}}$$. The talk of deforming the contour "in the E plane" suggests we make the energy slightly imaginary - add $i\delta$ to $\omega$ where $\delta$ is small. The expression $\sqrt{2(M-\omega+i\delta)}$ can then be expanded in powers of delta, leaving $\sqrt{2(M-\omega)}+i\delta$ (after the expansion I've redefined delta to take out the constant factors - this doesn't matter because we're going to contour-integrate round the pole anyway, so moving the pole up and down a bit makes no difference). So the z-plane looks like the first diagram here, where the large X is the displaced pole. We ultimately want to evaluate the integral between the two points on the real axis.
Well, the one thing I can do is integrate round the green contour shown in the second figure. The answer is just $2\pi i$ times the residue at the simple pole, i.e. in this case $$2\pi i\cdot\lim_{z \to {\sqrt{2(M-\omega')}+i\delta}}(z-\sqrt{2(M-\omega')}+i\delta)\cdot \frac{2z^2}{(z-\sqrt{2(M-\omega')}+i\delta} $$, $$=2\pi i\cdot2\cdot(\sqrt{2(M-\omega')}+i\delta)^2$$ $$=8\pi i \cdot(M-\omega')$$ approx.
This looks similar to what I want, namely $4\pi i \cdot(M-\omega')$ in order to get equation (42) (apart from a factor of 2). However
(1) The next step would be to relate the closed contour integral to the integral along the real axis. Unfortunately, this relies on the integrand vanishing for large $|z|$ in the positive half plane, but this appears not to be the case here
(2) And anyway we want the integral between $\sqrt{2(M-\omega)}$ and $\sqrt{2M}$, so how would I deal with the integral along other parts on the real axis (i.e. outside of the classically forbidden region)?
Any hints would be welcome (or an alternative way of computing the tunneling transmission coefficient).This post imported from StackExchange Physics at 2014-03-22 17:31 (UCT), posted by SE-user twistor59 |
Difference between revisions of "Higher-dimensional Fujimura"
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Let [math]\overline{c}^\mu_{n,4}[/math] be the largest subset of the tetrahedral grid:
[math] \{ (a,b,c,d) \in {\Bbb Z}_+^4: a+b+c+d=n \}[/math]
which contains no tetrahedrons [math](a+r,b,c,d), (a,b+r,c,d), (a,b,c+r,d), (a,b,c,d+r)[/math] with [math]r \gt 0[/math]; call such sets
tetrahedron-free.
These are the currently known values of the sequence:
n 0 1 2 3 4 5 6 [math]\overline{c}^\mu_{n,4}[/math] 1 3 7 14 24 37 55 n=0
[math]\overline{c}^\mu_{0,4} = 1[/math]:
There are no tetrahedrons, so no removals are needed.
n=1
[math]\overline{c}^\mu_{1,4} = 3[/math]:
Removing any one point on the grid will leave the set tetrahedron-free.
n=2
[math]\overline{c}^\mu_{2,4} = 7[/math]:
Suppose the set can be tetrahedron-free in two removals. One of (2,0,0,0), (0,2,0,0), (0,0,2,0), and (0,0,0,2) must be removed. Removing any one of the four leaves three tetrahedrons to remove. However, no point coincides with all three tetrahedrons, therefore there must be more than two removals.
Three removals (for example (0,0,0,2), (1,1,0,0) and (0,0,2,0)) leaves the set tetrahedron-free with a set size of 7.
General n
A lower bound of 2(n-1)(n-2) can be obtained by keeping all points with exactly one coordinate equal to zero.
You get a non-constructive quadratic lower bound for the quadruple problem by taking a random subset of size [math]cn^2[/math]. If c is not too large the linearity of expectation shows that the expected number of tetrahedrons in such a set is less than one, and so there must be a set of that size with no tetrahedrons. I think [math] c = \frac{24^{1/4}}{6} + o(\frac{1}{n})[/math].
With coordinates (a,b,c,d), take the value a+2b+3c. This forms an arithmetic progression of length 4 for any of the tetrahedrons we are looking for. So we can take subsets of the form a+2b+3c=k, where k comes from a set with no such arithmetic progressions. [This paper] gives a complicated formula for the possible number of subsets.
One upper bound can be found by counting tetrahedrons. For a given n the tetrahedral grid has [math]\frac{1}{24}n(n+1)(n+2)(n+3)[/math] tetrahedrons. Each point on the grid is part of n tetrahedrons, so [math]\frac{1}{24}(n+1)(n+2)(n+3)[/math] points must be removed to remove all tetrahedrons. This gives an upper bound of [math]\frac{1}{8}(n+1)(n+2)(n+3)[/math]. |
Let $p$ be prime, let $q$ be a power of $p$ and let $E/\mathbf{F}_q$ be an elliptic curve defined over the finite field $\mathbf{F}_q$. Let $\overline{\mathbf{F}_q}$ be the algebraic closure of $\mathbf{F}_q$.
Main Question:How might one conceive of $E/\overline{\mathbf{F}_q}$ both geometrically and as a group ? (And is there any benefit to having such a mental picture ?)
If my limited understanding is correct (and most of what follows I don't yet actually understand),
$E/\overline{\mathbf{F}_q}$ is all torsion : it ought to be the union (colimit) of the $E/\mathbf{F}_{q^d}$ since $\overline{\mathbf{F}_q}$ is the union (colimit) of the $\mathbf{F}_{q^d}$, if $\gcd(m,p)=1$ then the $m$-torsion $(E/\overline{\mathbf{F}_q})[m]\simeq (\mathbb{Z}/m\mathbb{Z})\times(\mathbb{Z}/m\mathbb{Z})$, as one climbs up the ladder of powers of $q$, more points of the elliptic curve $E/\overline{\mathbf{F}_q}$ reveal themselves; it mayhappen that for some power $q^d$ the $m$-torsion of the curve $E/\mathbf{F}_{q^d}$, where $m$ is a power of a prime $\neq p$, forms a cyclic group $\mathbb{Z}/m\mathbb{Z}$ and only at a later stage the full $m$-torsion will appear, and $p$-torsion behaves differently : either ((for all $\nu$ the groups $(E/\overline{\mathbf{F}_q})[p^\nu]$ are trivial)), or ((for all $\nu$ the groups $(E/\overline{\mathbf{F}_q})[p^\nu]$ are cyclic $\simeq\mathbb{Z}/p^\nu\mathbb{Z}$)).
ProposalIs it a good mental picture to think of $E/\overline{\mathbf{F}_q}$ as a kind of quotient of $(\mathbb{Q}/\mathbb{Z})\times(\mathbb{Q}/\mathbb{Z})$ that would squeeze $(\mathbb{Z}[\frac1p]/\mathbb{Z})\times(\mathbb{Z}[\frac1p]/\mathbb{Z})$ either to a point or to a copy of $(\mathbb{Z}[\frac1p]/\mathbb{Z}$ ? That is, somewhat in analogy to complex elliptic curves, a kind of strange rational torus with some compression of the $p$-torsion?
Auxiliary Question:If such a mental picture is correct, how does one picture the process of successive unveiling of the points that are rational over the intermediary fields $\mathbf{F}_{q}\subset\mathbf{F}_{q^{a}}\subset\mathbf{F}_{q^{ab}}\subset\overline{\mathbf{F}_{q}}$ ? (And is there any point to it?) |
Waecmaths
Title waecmaths question Question 1
If
Question 2
Evaluate $\frac{{{(3.2)}^{2}}-{{(4.8)}^{2}}}{3.2+4.8}$
Question 3
Find the product of 0.0409 and 0.0021 leaving your answer in the standard form
Question 4
Simplify $\sqrt{50}+\frac{10}{\sqrt{2}}$
Question 5 Question 6
Convert 42
Question 7
A trader bought 100 tubers of yam at 5 for N350.00. She sold them in set of 4 for N290.00. Find her gain percent
Question 8
If $p-2g+1=g+3p$ and $p-2=0$ , find
Question 9
Simplify $\frac{\tfrac{1}{x}+\tfrac{1}{y}}{x+y}$
Question 10
Simplify $\sqrt[3]{27{{x}^{3}}{{y}^{9}}}$
Question 11
Given that
Question 12
Factorize $5{{y}^{2}}-2ay-3{{a}^{2}}$
Question 13
If 4
Question 14
If $p=\tfrac{3}{5}\sqrt{\tfrac{q}{r}}$ express q in term of p and r
Question 15
Simplify $\frac{2{{x}^{2}}-5x-12}{4{{x}^{2}}-9}$
Question 16
In the diagram $\angle QPR={{90}^{\circ }}$ , if ${{q}^{2}}=25-{{r}^{2}}$ find the value of
Question 17 Question 18
If the volume of a cube is 343cm
Question 19
In the diagram PQRS is a rhombus$\left| PR \right|=10cm\text{ and }\left| QS \right|=24cm$. Calculate the perimeter of the rhombus.
Question 20
In the diagram \[\left| PR \right|=\left| QR \right|\] and $\left| PR \right|=\left| RS \right|=\left| SP \right|$ , calculate the size of $\angle QRS$
Question 21
In the figure shown,
Question 22
In the diagram, $PQ\parallel RS,\text{ }QU\parallel PT\text{ and }\angle PSR={{42}^{{}^\circ }}$. Find the angle
Question 23
The angle of a quadrilateral are ${{(x+10)}^{o}},2{{y}^{o}},{{90}^{o}}\text{ and (100}-y{{)}^{o}}$. Find
Question 24
In the figure $\left| PX \right|=\left| XQ \right|,\text{ }\left| PQ \right|\parallel \left| YZ \right|$ and $XY\parallel QR$>What is the ratio of the area of XYZQ to $\Delta YZR$ ?
Question 25
In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of x
Question 26
If ${{x}^{o}}$ is obtuse, which of the following is true?
Question 27
if $\tan x=1$evaluate $\sin x+\cos x$ leaving your answer in surd form
Question 28
If $\cos {{(x+25)}^{o}}=\sin {{45}^{o}}$, find the value of
Question 29 Question 30
If ${{2}^{n}}=128$ , find the value of $({{2}^{n-1}})({{5}^{n-2}})$
Question 31
If $x\propto (45+\tfrac{1}{2}y)$ which of the following is true
Question 32
Simplify $\frac{\log \sqrt{8}}{\log 4-\log 2}$
Question 33
Every staff in an office owns either a Mercedes or Toyota car. 20 owns Mercedes, 15 own Toyota and 5 own both. How many staff are there in the office?
Question 34
A train travel 60km in
Question 35
An arc of a circle, radius 14cm is 18.33cm long Calculate correct to the nearest degree, the angle which the arc subtends at the centre of the circle Take$\pi=\tfrac{22}{7}$
Question 36
What is the length of an edge of a cube whose total surface is
Question 37 Question 38
If the perimeter of Δ
PRS? Question 39
If $p=\frac{1}{2}$ and $\frac{1}{p-1}=\frac{2}{p+x}$ , find the value of
Question 40
Find the quadratic equation whose roots are |
Mean Field Models
Interactions in real matters are complicated and hard to solve exactly since this problem involves many-body interactions. A number of physics-chemical systems which undergo phase transitions have long-range orders. One example is ferromagnetism. Magnetic moments of individual paritcles in a ferromagnet align themselves under a certain critical temperature even without external magnetic field.
The ordering indicates the interactions in these systems ara not completely random. The main idea of
Mean Field Models is treating the interactions on any one particle as an average interaction [2]. This model effectively turns a many-body problem to a solvable one-body problem. It also provides a theoretical basis for understanding a variaty of phenomena such as ferromagnetism, gas-liquid transitions, and order-disorder transitions in alloys [3]. Example-Ferromagnetism
In the following example we can see how ferromagnetism arises from Ising model in the zeroth approximaiton [3].Consider the Ising model on an N-dimensional cubic lattice under an external magnetic field
B. The Hamiltonian of the system in configuration {<math>\sigma_1,\sigma_2,...,\sigma_N</math>} is given by
<math>\; H</math>{<math>\sigma_i</math>}=<math>-J \sum_{n.n}\sigma_i\sigma_j-\mu B \sum_i\sigma_i</math>
, where <math>\sigma_i</math>=1 for an "up" spin and -1 for a "down" spin, <math>J</math> is the exchange interaction, and <math>\mu</math> is the magnetic dipole moment. A long-range order parameter <math>L</math> can be defined as
<math>L=1/N \sum_i\sigma_i = (N_+ - N_-)/N </math>
, where <math>N_+</math> (<math>N_-</math>) is the total number of "up" ("down") spins, and numbers <math>N_+</math> and <math>N_-</math> must satisfy the relation
<math>N_+ + N_- =N</math>
The magentization <math>M</math> is then given by
<math>M= (N_+ - N_-) \mu = N \mu L </math>
, the parameter <math>L</math> is a direct measure the magnetizaiton in the system.
In the spirit of mean-field model, the first part of the Hamiltonian can be replaced by the expression <math>-J (1/2q<\sigma>) \sum_i \sigma_i</math>, i.e. for a given <math>\sigma_i</math> with <math>q</math> nearest-neighbors, each of the <math>q \sigma_j</math> is replaced by <<math>\sigma</math>>.
Noting that <math><\sigma>=1/N<\sum_i \sigma_i>= <L></math>, the total confiqurational energy of the system can be written as
<math>E=-1/2 (q J <L>) NL - (\mu B) N L</math>
The expectation valve of <math>E</math> is then given by
<math>U=-1/2 q J N <L>^2 - \mu B N L</math>
The energy expended in changing any "up" spin into a "down" on is given by
<math>\Delta \epsilon = -J (q <\sigma>) \Delta \sigma -\mu B \Delta \sigma =2 \mu (q J/ \mu <\sigma > + B)</math>
, for <math>\Delta \sigma =-2</math> here.
The relative values of the equilibrium numbers <<math>N_+</math>> and <<math>N_-</math>> then follow the Boltzmann distribution,
<math><N_-> / <N_+> = </math> exp<math>( -\Delta \epsilon / k T) = </math> exp<math>(- 2 \mu (B' +B) /k T)</math> , where <math>B'</math> denotes the internal molecular field
<math>B'=q J <\sigma> / \mu = qJ (<M> / N \mu^2)</math>
Given the ratio of <math>N_-</math> and <math>N_+</math> and defnition of order parameter <math>L</math>, we can obtain
<math>(q j <L> + \mu B) / k T = 1/2 </math>ln <math>(1 +<L>)/(1-<L>) = </math> tanh<math>^{-1} <L></math>
To see the possibility of spontaneous magnetization, we can let <math>B \rightarrow 0</math>, which leads to the relationship
<math><L_0>=</math> tanh(<math>q j <L_0> / k T</math>)
The above equatin has solutions when
<math>q J /kT > 1</math>.
This also means spontaneous magnetization can appear when
<math>T < q j/ k = T_c</math>
The exact solution of <math><L_0></math> can be solved numerically. Fig. 2 shows <math><L_0></math> as a function of temperature.
References
[3] R. K. Pathria, "Statistical Mechanics", 2nd ed., Butterworth-Heinemann, 1996 |
Let's consider an arbitrary infinitesimal transformation of the fields and their coordinates :
$$x'^{\mu}= x^{\mu} + \delta x^{\mu} = x^{\mu} + \frac{\delta x^{\mu}}{\delta{\omega}^a}{\omega}^a\tag{1}$$
$${\Phi}'(x') = {\Phi}(x) + \delta\Phi= {\Phi}(x) + \frac{\delta\Phi}{\delta\omega^a}{\omega}^a\tag{2}$$
where ${\{\omega}^a\}$ is a set of infinitesimal parameters. The corresponding variation of the Action is:
$$\delta S= \delta (\int{dx \,\mathcal{L}[\Phi,\partial^{\mu}\Phi]})=\int{(dx \, \delta \mathcal{L}} +\mathcal{L}\,\delta dx)= \int{dx (\delta \mathcal{L}+\mathcal{L}\,\partial_{\mu}\delta x^{\mu})}\tag{3}$$
Now if
$$\delta=\delta_0 + \delta x^{\mu}\partial_{\mu},\tag{4}$$
where:
$\delta_0 \phi (x)=\phi'(x)-\phi(x)$ at first order, then the variation becomes:
$$\delta S=\int{dx\,(\delta\mathcal{L}+\mathcal{L}\,\partial_{\mu}\delta x^{\mu})}=\int{dx(\delta_0\mathcal{L}+\delta x^{\mu}\partial_{\mu}\mathcal{L}+\mathcal{L}\,\partial_{\mu}\delta x^{\mu})}$$ $$=\int{dx\,(\delta_0\mathcal{L}+ \partial_\mu(\mathcal{L}\delta x^{\mu}))}=\int{dx\,(\frac{\partial\mathcal{L}}{\partial{\Phi}}\delta_0\Phi}+\frac{\partial{\mathcal{L}}}{\partial[\partial_\mu\Phi]}\delta_0\partial_\mu\Phi+ \partial_\mu(\mathcal{L}\delta x^{\mu})).\tag{5}$$
Using the Euler-Lagrange equations and commuting $\delta_0$ and $\partial$, the last integral becomes:
$$\delta S=\int{dx\,\partial_\mu(\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta_0\Phi+\mathcal{L}\delta x^\mu)}.\tag{6}$$
Now we go back to $\delta$ from $\delta_0$:
$$\delta S =\int{dx\,\partial_\mu(\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta\Phi-\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta x^\nu\partial_\nu\Phi+\mathcal{L}\delta x^\mu)}\tag{7}$$
and finally:
$$\delta S =\int{dx\,\partial_\mu(\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta\Phi-\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta x^\nu\partial_\nu\Phi+\mathcal{L}\delta x^\mu)} \\ =\int{dx\,\partial_\mu[\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta\Phi+(\mathcal{L}\delta^\mu_\nu-\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\partial_\nu\Phi)\delta x^\nu]}\\ =\int{dx\,\partial_\mu\{[\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\frac{\delta\Phi}{\delta\omega^a}+(\mathcal{L}\delta^\mu_\nu-\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\partial_\nu\Phi)\frac{\delta x^\nu}{\delta\omega^a}]\delta\omega^a\}}.\tag{8}$$
Now defining:
$$j^{a\mu} = (\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\partial_\nu\Phi-\mathcal{L}\delta^\mu_\nu)\frac{\delta x^\nu}{\delta\omega^a} - \frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\frac{\delta\Phi}{\delta\omega^a}\tag{9} $$
we eventually get:
$$\delta S = - \int{dx\,\partial_\mu(j^{a\mu}\delta\omega^a)}.\tag{10}$$
Let's suppose for now that $\{\omega^a\}$ are $x$-independent, such that:
$$\delta S = - \int{dx\,\partial_\mu(j^{a\mu})\delta\omega^a}.\tag{11}$$
Here's my first question: do I need to ask that $S$ is invariant under my transformations in order to get $$\partial_\mu\,j^{a\mu}=0~?\tag{12}$$ Or can the transformation be arbitrary given that in every case $$\delta S = 0\tag{13}$$ for all infinitesimal variations of the fields satisfying Euler-Lagrange dynamical equations?
Then it comes my second question. My book$^{(1)}$ says that, even not taking EL equations into account, the variation of the action under infinitesimal arbitrary transformation $(1),(2)$ is: $$\delta S=-\int{dx\,j^{\mu a}\partial_{\mu}\omega^a}\tag{14}.$$ Now I don't know how $(14)$ could be derived without taking EL equations into account beacause if $j^{\mu a}$ does not have $\frac{\partial\mathcal{L}}{\partial\Phi}$ terms is just thanks to those equations.
$^{(1)}$Philippe Di Francesco, Pierre Mathieu, David Sénéchal - Conformal Field Theory, page 41. |
Maxwell's equations are elementary geometric identities that apply to all thrice differentiable vector fields. This statement appears to be at odds with Proca's equation of a massive vector field, and with the concept of magnetic monopoles. This brief note explains how Proca's equation can be obtained by redefining the four-current, whereas the existence of monopoles is a consequence of topological defects.
Maxwell's equations of the electromagnetic field are, in fact, identities that apply to any thrice differentiable vector field on a smooth manifold. This statement is seen most easily using the language of differential forms. We begin with a real-valued one-form $A$ (i.e., a vector field). This field is none other than the electromagnetic four-potential, comprising the electrostatic scalar potential and the magnetic vector potential.
Forming the exterior derivative of $A$, we obtain the electromagnetic field tensor, a real-valued two-form:
\[F=\mathrm{d}A.\]
As it is well known, if $F$ is written in component form in four dimensions, its six independent components are the components of the three-dimensional electric field and magnetic field vectors $\vec{E}$ and $\vec{B}$.
A basic property of the exterior derivative is that it is nilpotent, i.e., ${\rm d}^2=0$. (This can be proven easily using the explicit definition of the derivative operator, but it also follows from the fact that the algebra of differential forms is a Grassman-algebra.) From this, the following equation trivially follows:
\[{\rm d}F=0.\]
Written out in explicit form, this equation splits into two of Maxwell's equations: Faraday's law and Gauss's law for magnetism.
It is also possible to form the codifferential of $F$. Up to sign
1, it reads:
\[J=\star{\rm d}{\star{F}}=\star{\rm d}G,\]
where $\star$ represents the Hodge dual of a differential form. (Note that definition of the Hodge dual requires a metric. Alternatively, we could treat $F$ and $G$—corresponding to the three-vector fields $\vec{E}$, $\vec{B}$, and $\vec{D}$, $\vec{H}$, respectively—as principal quantities and forego the requirement to define a metric. In this case, however, ${\rm d}F=0$ must be imposed, as it is no longer an identity that is automatically satisfied.) In four dimensions (and only in four dimensions!), the codifferential of the two-form $F$ is a one-form, i.e., another vector field: it is called the four-current, and the defining equation, when written out in component form, splits into the remaining two of Maxwell's equations, Gauss's law and Ampère's law.
The codifferential is also nilpotent (${\star{\rm d}\star}{\star{\rm d}\star}={\star{\rm d}}^2\star=0$), hence
\[{\star{\rm d}}{\star J}=0,\]
which is the equation of charge and current conservation.
As the above results are all trivially obtainable identities that apply to any thrice differentiable vector field $A$ on an arbitrary (curved!) smooth (or at least thrice diffentiable) manifold, there seems to be little room to generalize Maxwell's equations to the case of massive fields, as was done by Alexandru Proca. It also doesn't appear possible to introduce magnetic monopoles: their non-existence seems to be an assured fact.
Yet appearances can be deceiving. To introduce Proca's theory into the picture, all one must do is to recognize that the definition of $J$ represents a somewhat arbitrary choice. We are, in fact, free to define $J$ differently, for instance by
\[J-\mu^2A={\star{\rm d}}{\star F}.\]
This is just Proca's equation for a massive vector field with mass $\mu$. (If the Lorenz gauge is used (${\star{\rm d}}{\star A}=0$), this is just $\nabla^2A+\mu^2A=J$, and we have the theory of a massive vector field with conserved current.)
So what about magnetic monopoles? Again, the trivial identities obtained above seem to preclude their existence. However, we must not forget the underlying assumption that the vector field $A$ was everywhere thrice differentiable. If it isn't, the picture changes. In particular, Dirac's magnetic monopole can exist precisely because the vector field $A$ is
not differentiable along the Dirac string even though $F$ and $G$ remain both continuous and differentiable along it. Such topological defects in the vector field or the underlying manifold can lead to many interesting consequences. 1A sign ambiguity is introduced by the signature of the metric, but this is not relevant to the present discussion. |
I will try to give the motivation behind this problem and later the math formality.
Given a grayscale image (1 Channel - M by N Matrix). Someone marks some pixels as anchors. Now, you need to interpolate the other pixels (Which are not anchors) by minimizing a given cost function s.t. the end result is an image which has the original image values at the anchors and interpolated values else were s.t. it minimizes the cost function.
Given an $M$ by $N$ matrix (A 1 channel image for that matter) $ I $.
Subset of the elements (Pixels) in the matrix are marked as reference and their location is a group marked as $ S $.
The optimization cost function is given by:
$$ \sum_{\mathbf{r}} \left( E(\mathbf{r}) - \sum_{\mathbf{s} \in N(\mathbf{r})} {w}_{\mathbf{rs}} E(\mathbf{s}) \right)^2, \text{ s.t. } \forall p \in S \; E(\mathbf{p}) = I(\mathbf{p})\,, $$
where a bold letter $ \mathbf{p}, \mathbf{r}, \mathbf{s} $ means an element (Pixel) location.
The group $ N(r) $ is the neighborhood of $ \mathbf{r} $, which is size $ k $ namely, a $ k $ by $ k $ rectangle where $ \mathbf{r} $ is in the middle.
The weights $ {w}_{\mathbf{rs}} $ are defined as following:
$$ {w}_{\mathbf{rs}} \propto \exp\left(-\frac{(I(\mathbf{r}) - I(\mathbf{s}))^2}{2\sigma^2_r} \right )\ \ \text{ s.t. } \sum_{\mathbf{s} \in N(\mathbf{r})} w_{\mathbf{rs}} = 1\,. $$
Namely, the weights are normalized to 1 within the neighborhood. The variance is calculated on the matrix $ I $ in the neighborhood (You can assume it is given).
So the problem is to find a matrix $ E $ which is equal to $ I $ on all reference points and interpolates other places by bringing the cost function to minimum.
It looks like a weighted least squares per neighborhood (The inner brackets).
I couldn't formalize it (For the whole matrix) a way that can be easily calculated and solved in e.g. MATLAB.Is there a way to formalize it as classic Weighted LS problem?
Or any other form which using the classic tools will bring solution (Get the interpolated matrix)?
Could any one help with that?
P.S.I tried using the Weighted LS per neighborhood disregarding the rest didn't yield the expected results (Obviously).
Tried it just to see how far the real solution is from this naive solution. |
In Gaussian clustering (i.e. General Mixture Models) we model the data with some clusters. For example, in the below figure, we have two clusters $C_1, C_2$, each of which are modeled with a Gaussian (Normal) distribution: $p_{C_1} \sim \mathcal{N}(\mu_1, \Sigma_1)$ and $p_{C_2} \sim \mathcal{N}(\mu_2, \Sigma_2)$ with proportions $\pi_1$ and $\pi_2$.
The probability that a given point $x$ belongs to $C_1$ (is generated from $C_1$) is: $$ p(C_1 | x ) = \frac { \pi_1 \mathcal{N}(\mu_1, \Sigma_1)} {\sum_{i=1}^2{ \pi_i \mathcal{N}(\mu_i, \Sigma_i) } } $$ The probability that $x$ belongs to $C_2$ can be similarly calculated.
Now, assume that we have a group of points that are generated (sampled) from another Gaussian distribution $q(x) \sim \mathcal{N}(\mu_q, \Sigma_q)$.
Obviously, we can compute the probability of each point belonging to each cluster. However, we instead want to compute
the expected probability of a sample belonging to each cluster. In other words, if we generate infinite samples from the $q(x)$, what is the average (expected) value of $p(C_1 | x )$ for all points?
**P.S. ** I need the solution in closed-form. |
Consider the one-dimensional Ising model with constant magnetic field and node-dependent interaction on a finite lattice, given by
$$H(\sigma) = -\sum_{i = 1}^N J_i\sigma_i\sigma_{i + 1} - h\sum_{i = 1}^N\sigma_i$$
where $\sigma = \{\sigma_i\}_{i = 1,\dots, N}\in\Omega := \{\pm 1\}^N$, $\{J_i\}_{i = 1,\dots, N}$ are nearest neighbor interaction strength couplings, and $h \in \mathbb{R}$ is the magnetic field. Let's consider the ferromagnetic case, that is, $J_i \geq 0$ for $i = 1, \dots, N$, and for the sake of simplicity (though this doesn't matter in the thermodynamic limit), take periodic boundary conditions. Neither in the finite volume, nor in the thermodynamic limit does this model exhibit critical behavior for finite temperatures.
On the other hand, as soon as we allow $h$ to be complex (and fix the temperature), even in the finite volume $N$, the partition function has zeros as a function of $h$. In the thermodynamic limit these zeros accumulate on some set on the unit circle in the complex plane (Lee-Yang circle theorem).
Now the question: let's consider information geometry of the Ising model, as described above, when $h$ is real. In this case the induced metric is defined and the curvature does not develop singularities (obviously, since there are no phase transitions). Now, what about information geometry of the Ising model when $h$ is complex? This is a bit puzzling to me, since then the partition function attains zeros in the complex plane, so that the logarithm of the partition function is not defined everywhere on the complex plane, and the definition of metric doesn't extend directly to this case (the metric involves the log of the partition function), let alone curvature.
Is anyone aware of any literature in this direction? I thought it would be a good idea to ask before I try to develop suitable methods from scratch.
One could of course try to define the metric and curvature first for real $h$ and only then extend the final formulas to complex $h$. This seems a bit unnatural to me, and even dangerous.
EDIT: Allow me to elaborate on what I mean by "information geometry." Let us consider for simplicity the finite volume model (i.e., $N < \infty$ above). The Gibbs state (i.e., the probability distribution on $\Omega$ of maximal entropy), given by
$$P(\sigma) = \frac{e^{\beta H(\sigma)}}{\sum_{\sigma\in\Omega} e^{\beta H(\sigma)}}$$
where $\beta$ is inverse temperature, obviously depends on the temperature and the magnetic field. So it is convenient to write, for example, $P := P_{(\beta, h)}$ to make this dependence explicit.
Now, the Gibbs states can be identified with points in the parameter space
$$M := \{(\beta, h): \beta, h\in (0,\infty)\}$$
On this space one can define a metric, the so-called Fisher information metric, that naturally measures the distance between two Gibbs states $P_{(\beta, h)}$ and $P_{(\beta', h')}$. The definition of this metric (as you may have guessed!) involves the partition function. This metric then induces a geometry on the parameter space $M$, the so-called statistical manifold (see, for example, http://en.wikipedia.org/wiki/Information_geometry for more details). The curvature (induced by the metric) is an interesting quantity to study. As it turns out, the curvature develops singularities at phase transitions (also something you may have guessed, since the metric involves the partition function, and hence so does the curvature tensor).
Actually, the This post has been migrated from (A51.SE)
definition of the metric involves the logarithm of the partition function. All is fine so far, since $h$ is real (assume nonzero), and everything is well-defined. However, as soon as we pass to complex $h$, the partition function admits zeros, and it is no longer clear (at least to me) how the above constructions should generalize. See, for example, Section 4 in http://eprints.nuim.ie/268/1/0207180.pdf). |
Think about a setting where there are $n$ tasks and $m$ machines. We are interested in task-machine assignment. Let $p_i$ be a non-negative completion time of job $i$. Also, $x_i$ denotes the machine where task $i$ is scheduled. A machine processes the task having the smallest completion time, then the second etc. So for simplicity we assume the tasks available are ordered such that $p_1\leq p_2 \leq \ldots \leq p_n$.
The completion time of each task is given with:
$C_j(x) = \sum_{k \leq j,x_k=x_j} p_k $.
And the social cost is defined as:
$C(x)= \sum_{j \in [n]}C_j(x)$.
An optimal solution $x^*$ is a strategy that minimizes $C(x)$.
How can we prove that there exists an efficient polynomial time algorithm that computes the optimal solution? |
The ramification index is multiplicative in towers. That is:
If $K\subseteq E\subseteq L$ is a tower of extensions, $P$ is a prime in $K$, $Q$ is a prime in $E$ with $Q\cap K=P$, and $Q'$ is a prime in $L$ with $Q'\cap E=Q$, then letting $e(Q|P)$ be the ramification index of $Q$ over $P$, $e(Q'|Q)$ the ramification index of $Q'$ over $Q$, and $e(Q'|P)$ the ramification index of $Q'$ over $P$, we have have
$$e(Q'|P) = e(Q'|Q)e(Q|P).$$
To see this, you do the factoring: if we factor $P$ into primes in $E$, then $Q$ occurs with exponent $e(Q|P)$; then we factor in $L$, the exponent of $Q'$ will be its exponent in the factorization of $Q$, $e(Q'|Q)$, times the exponent to which $Q$ occurs in that of $P$, $e(Q|P)$, since $(\mathfrak{P}^a)^b = \mathfrak{P}^{ab}$.
If all extensions are Galois (that is, $L$ and $E$ are both Galois over $K$), then the ramification index depends only on $P$ and not on $Q$ or $Q'$, but it's still multiplicative, but here you get the further information that the ramification index also divides the order of the extension.
All of that information it may give you enough leverage to determine the ramification index in $E_i/K$ in some circumstances (if you know the ramification in $L/K$, then the multiplicativity means that knowing the index in $E_i/K$ is equivalent to knowing it in $L/E_i$). But it may not. For example, if you have a biquadratic extension $L/K$ whose Galois group is the Klein $4$-group, with a prime $P$ that has ramification index $2$, then for the two intermediate fields $E_1$ and $E_2$ you could have that $P$ already ramifies in $E_1$, but the ramification index in $L/E_1$ is $1$; or the other way around.
Added. To answer your final question without reference to discriminants:
No rational prime can have ramification degree $4$ in $K=\mathbb{Q}(\sqrt{-5},\sqrt{-1})$: that would require it to ramify in both $\mathbb{Q}(\sqrt{5})$ and in $\mathbb{Q}(\sqrt{-1})$ by the multiplicativity of the ramification degree, but no prime ramifies in both. So the ramification degree of any ramified rational prime is $2$.
Let $L_1 = \mathbb{Q}(\sqrt{5})$, $L_2=\mathbb{Q}(\sqrt{-5})$, $L_3=\mathbb{Q}(\sqrt{-1})$. These are all the intermediate fields of the extension.
Let $p$ be a prime that ramifies in $K$. Let $Q$ be a prime of $K$ lying over $p$, and let $P_i$ be the prime of $L_i$ lying under $Q$. If $I_p$ is the inertia group of $p$, and $L_i$ is the fixed field of $I_p$, then $e(Q|P_i) = e(Q|p) = 2$, and $e(Q|P_j)=1$ with $j\neq i$. Again, multiplicativity of the ramification tells you that $p$ would then need to ramify in both the "other two" subextensions.
In particular, if $L_i = L_2 = \mathbb{Q}(\sqrt{-5})$ (so that there is a prime of $\mathbb{Q}(\sqrt{-5})$ that ramifies in $K$), then $p$ would have to ramify in
both $\mathbb{Q}(\sqrt{5})$ and in $\mathbb{Q}(\sqrt{-1})$. But no rational prime ramifies in both; so we can never have the fixed field of $I_p$ equal to $\mathbb{Q}(\sqrt{-5})$, which means $K/\mathbb{Q}(\sqrt{-5})$ must be unramified everywhere, as claimed. |
Taking this as my original question: Do we know if there exists a RHS and an initial (unlucky) guess that will require $\Theta(\sqrt{\kappa})$steps?
The answer to the question is "no". The idea of this answer comes from the comment from Guido Kanschat.
Claim: For any given condition number $k$, there exists a matrix $A$, with that condition number for which the CG algorithm will terminate in at most two steps (for any given RHS and initial guess).
Consider $A\in \mathbb{R}^{n\times n} $ where $A=\mathrm{diag}(1,\kappa,\kappa,\ldots, \kappa)$. Then the condition number of $A$ is $\kappa$. Let $b\in \mathbb{R}^n$ be the RHS, and denote the eigenvalues of $A$ as $\lambda_i$ where$$\lambda_i = \left\{\begin{array}{ll}1 & i=1\\\kappa & i\not= 1 \end{array} \right. .$$
We first consider the case where $x^{(0)} \in \mathbb{R}^n$, the initial guess, is zero. Denote $x^{(2)}\in \mathbb{R}^n$ as the second estimate of $A^{-1}b$ from the CG algorithm. We show that $x^{(2)} =A^{-1}b$ by showing $\langle x^{(2)}-A^{-1}b, A(x^{(2)}-A^{-1}b)\rangle =0$. Indeed, we have
\begin{align*} \langle x^{(2)}-A^{-1}b, A(x^{(2)}-A^{-1}b)\rangle &= \left\| x^{(2)}-A^{-1}b \right\|_A^2 \\&=\min_{p\in \mathrm{poly}_{1} } \left\| (p(A)-A^{-1}) b \right\|_A^2\\&=\min_{p\in \mathrm{poly}_{1} } \sum_{i=1}^n (p(\lambda_i) - \lambda_i^{-1})^2 \lambda_i b_i^2 \\&\le \sum_{i=1}^n (\widehat{p}(\lambda_i) - \lambda_i^{-1})^2 \lambda_i b_i^2= 0\end{align*}
Where we use the first order polynomial $\widehat{p}$ defined as $\widehat{p}(x)= (1+\kappa-x)/\kappa$. So we proven the case for $x^{(0)}= 0$.
If $x^{(0)} \not = 0$, then $x^{(2)}= \overline{x^{(2)}}+ x^{(0)}$ where $\overline{x^{(2)} }$ is the second estimate of the CG algorithm with $b$ replaced with $\overline{b} = b-A x^{(0)}$. So we have reduced this case to the previous one. |
[NOTE: I'm using $n$ instead of $M$ as the key dimension here, sorry... it's too much trouble to go back and change everything...]
Well, $X$ is actually going to be an $2n+1\times 2n+1$ matrix with this block structure when you are done:$$\begin{bmatrix} 1 & \beta^T \\ \beta & B \\ & & \beta_1 \\ & & & \beta_2 \\ & & & & \ddots \\ & & & & & \beta_n \end{bmatrix}$$So in addition to your single equality constraint you need to make sure that $X_{1,1}=1$ and $X_{k+1,1}=X_{n+k+1,n+k+1}$ for $k=1,2,\dots,n+1$. So you have $n+2$ equality constraints to build. So I think it looks something like this:$$\bar{C} = \begin{bmatrix} 0 & 0_{1\times n} \\ 0_{n\times 1} & C \\ & & 0\\ & & & 0\\ & & & & \ddots \\ & & & & & 0 \end{bmatrix}$$$$\bar{A}_1 = \begin{bmatrix} 0 & 0_{1\times n} \\ 0_{n\times 1} & A \\ & & 0\\ & & & 0\\ & & & & \ddots \\ & & & & & 0 \end{bmatrix}, \quad a_1 = 1$$$$\bar{A}_2 = \begin{bmatrix} 1 & 0_{1\times n} \\ 0_{n\times 1} & 0_{n\times n} \\ & & 0\\ & & & 0\\ & & & & \ddots \\ & & & & & 0 \end{bmatrix}, \quad a_2 = 1$$$$\bar{A}_{k+2} = \begin{bmatrix} 0 & \tfrac{1}{2}e_k^T \\ \tfrac{1}{2}e_k & 0_{n\times n} \\ & & -\mathop{\textrm{diag}}(e_k) \end{bmatrix}, \quad a_{2+k} = 0, \quad k=1,2,\dots, n$$where $e_k$ is the $n$-vector with $1$ in the $k$th position and zeros elsewhere. See the section "Using the subroutine interface to CSDP" in the CSDP user guide for information about how to translate this block structure into CSDP's preferred format.
But, since I have you here: it would be much easier for you if you did not have to plug this into CSDP. If running in MATLAB is acceptable to you, you could use my toolbox CVX to solve this, and the model would look like this:
cvx_begin sdp
variables B(n,n) b(n)
minimize(trace(C*B))
subject to
trace(A*B) == 1;
[ 1, b' ; b, B ] >= 0;
b >= 0;
cvx_end
Of course, if you're stuck in C, I understand! CSDP is a very nice solver particularly for standalone applications. I hope I have been helpful either way. |
Several years ago, I became one of the developers of the open-source computer algebra system (CAS) called Maxima.
Maxima is the open source descendant of the first ever computer algebra system, MACSYMA. Initially developed by the US Department of Defense, it was later commercialized. The company behind commercial MACSYMA has since disappeared, but in the late 1990s, the DOE agreed to permit the original version to be released under an open-source license.
My interest in Maxima is due to my interest in general relativity. Perhaps more than most other areas of physics, general relativity relies heavily on computer algebra tools, due to the complexity involved with defining and analyzing metrics in curved spacetime.
Maxima has two packages related to general relativity work. They both perform complicated tensor calculations. One package,
itensor, is a general-purpose package that deals with indexed objects: specifically, objects with covariant, contravariant, and derivative indices. The package knows about contraction rules, the raising and the lowering of indices, ordinary and covariant differentiation, and Christoffel symbols. The other package,
ctensor, is really a collection of subroutines that are designed to compute tensor components used mainly in general relativity, including the components of the Christoffel symbols, the Riemann tensor, the Ricci tensor, and the Weyl tensor. The two packages neatly complement each other: many problems can be solved by writing up and deriving indicial tensor equations using
itensor, and then using a special function provided by the
itensor package to convert the result into component form that can then be processed by
ctensor.
The main problem with the tensor packages was, simply put, that they were broken! This is where I come in: having been able to fix the core functionality in
itensor, I decided to offer my services to the Maxima development team.
I'm working on more than mere fixes, however. I have also extended the functionality of the tensor packages. On the one hand, I improved the algebraic power of
itensor, introducing a new notation that helps preserve index ordering in more complicated tensor equations. On the other hand, I added to both
ctensor and
itensor the capability to deal with not just the standard metric formalism, but also with rigid frames, torsion, and conformal nonmetricity. Time permitting, I'd also like to add more capabilities in the future, to make Maxima "competitive" with other well-known tensor packages, such as SHEEP, CLASSI, and grTensorII.
Meanwhile, I added a third tensor package:
atensor is a package that can deal with generalized (tensor) algebras, including Clifford, Grassmann, and Lie-algebras. I also fixed the
cartan package, a package that deals with with differential forms. Last but not least, I changed these four packages so that their naming conventions now conform to that of commercial MACSYMA.
I have drafted a paper that summarizes the work I've done. For additional reference, here's a link to the tensor package manuals (snapshot from the current development version of Maxima) and some demos:
Algebraic tensor manipulation (
atensor)
Component tensor manipulation (
ctensor)
Indicial tensor manipulation (
itensor)
Tensor package demos (text output, ~370 kB; as of March, 2008)
The following are two more complete examples that demonstrate some of the new capabilities that I added to these packages:
The Kaluza-Klein metric
In 1919, Theodor Kaluza proposed an extension to general relativity: using an appropriately constructed fifth dimension, he was able to incorporate electromagnetism into Einstein's theory of gravity. Recently, I endeavored to replicate the most basic of Kaluza's results: the equation of motion for a particle in empty five-dimensional space, as seen from a four-dimensional perspective.
Now that I am working with Maxima, the question arose: can the same result be reproduced using this computer algebra system? Surprisingly, the answer is a yes. With only minor changes to the current Maxima code base, I was able to complete the derivation.
The Petrov classification
One of the common problems in general relativity is determining the equivalence of two metrics. Because the same manifold can be mapped using drastically different coordinate systems, it is usually not at all evident whether or not two metrics describe the same manifold. A set of routines that I am working on make it possible to derive the Petrov class for a metric specified using an orthonormal tetrad base.
Differentiation with respect to field variables
The newest addition (March 2008) to Maxima is the ability to differentiate indexed expressions with respect to indexed variables, most notably the metric. This feature makes it possible to use Maxima to derive the Euler-Lagrange equations from a field Lagrangian; indeed, Maxima can now deduce the field equations of general relativity from the Einstein-Hilbert action, and also can deal with other theories based on a modified Lagrangian, such as Brans-Dicke theory.
The power of the package can now be demonstrated by showing how Maxima, starting from the Lagrangian density of the gravitational field (the Einstein-Hilbert Lagrangian) can derive, and solve, the field equations in the spherically symmetric case, a plot of the celebrated Schwarzschild gravity well. In other words, starting with
\[{\cal L}=\frac{1}{16\pi G}(R+2\Lambda)\sqrt{-g},\]
and running the following code:
if get('ctensor,'version)=false then load(ctensor); if get('itensor,'version)=false then load(itensor); remsym(g,2,0); remsym(g,0,2); remsym(gg,2,0); remsym(gg,0,2); remcomps(gg); imetric(gg); icurvature([a,b,c],[e])*gg([d,e],[])$ contract(rename(expand(%)))$ %,ichr2$ contract(rename(expand(%)))$ canform(%)$ contract(rename(expand(%)))$ components(gg([a,b],[]),kdels([a,b],[u,v])*g([u,v],[])/2); components(gg([],[a,b]),kdels([u,v],[a,b])*g([],[u,v])/2); %th(4),gg$ contract(rename(expand(%)))$ contract(canform(%))$ imetric(g); contract(rename(expand(%th(2))))$ remcomps(R); components(R([a,b,c,d],[]),%th(2)); g([],[a,b])*R([a,b,c,d])*g([],[c,d])$ contract(rename(canform(%)))$ contract(rename(canform(%)))$ components(R([],[]),%); decsym(g,2,0,[sym(all)],[]); decsym(g,0,2,[],[sym(all)]); ishow(1/(16*%pi*G)*((2*L+'R([],[])))*sqrt(-determinant(g)))$ L0:%,R$ canform(contract(canform(rename(contract(expand(diff(L0,g([],[m,n]))- idiff(diff(L0,g([],[m,n],k)),k)+idiff(rename(idiff(contract( diff(L0,g([],[m,n],k,l))),k),1000),l)))))))$ ishow(e([m,n],[])=canform(%*16*%pi/sqrt(-determinant(g))))$ EQ:ic_convert(%)$ ct_coords:[t,r,u,v]; lg:ident(4); lg[2,2]:-a^2/(1-k*r^2); lg[3,3]:-a^2*r^2; lg[4,4]:-a^2*r^2*sin(u)^2; dependencies(a(t)); cmetric(); derivabbrev:true; christof(false); e:zeromatrix(4,4); ev(EQ); expand(radcan(ug.e)); lg:ident(4); lg[1,1]:B; lg[2,2]:-A; lg[3,3]:-r^2; lg[4,4]:-r^2*sin(u)^2; kill(dependencies); dependencies(A(r),B(r)); cmetric(); christof(false); e:zeromatrix(4,4); ev(EQ); E:expand(radcan(ug.e)); exp:findde(E,2); solve(ode2(exp[1],A,r),A); %,%c=-2*M; a:%[1],%c=-2*M; ode2(ev(exp[2],a),B,r); b:ev(%,%c=rhs(solve(rhs(%)*rhs(a)=1,%c)[1])); factor(ev(ev(exp[3],a,b),diff)); lg:ev(lg,a,b),L=0$ ug:invert(lg)$ block([title: "Schwarzschild Potential for Mass M=2",M:2.], plot3d([r*cos(th),r*sin(th),1-ug[1,1]],[r,5.,50.],[th,-%pi,%pi], ['grid,20,30],['z,-2,0],[psfile],['legend,title]));
we end up with this plot:
I believe this capability is unique to the Maxima tensor package. |
What is Geometric Optics?
When a ray of light is reflected by some angle by a barrier in its pathway, it is said that the light beam is rebounded and this procedure is known as reflection of light.
When the beam of light crosses from one medium to the other, it diverges slightly from its route. This deviation hinges over the medium where the light is crossing through. This procedure is termed as the refraction of light.
Snell’s Law (Laws of Refraction and Reflection):
When light is refracted or reflected, one can check the basis of procedure for refraction and reflection along with the things linked to them.
The Laws of Reflection in Geometric Optics:
The conduct of the ray of light that is rebounded by a smooth mirror is judged based on the laws of reflection. But primarily, one needs to know the basic terms.
Light approaching a mirror is termed as incident ray and the beam of light that is reflected by the mirror is termed as reflected ray. At the spot of reflection, the perpendicular that is drawn is termed as a normal. The angle amidst the incident ray and normal is the angle of incidence, while the angle amidst the reflected ray and normal is the angle of reflection.
Conferring to the laws of reflection, the angle of incidence is always equivalent to the angle of reflection.
The law of refraction is also termed as Snell’s Law. Conferring to this law, “
the ratio between the sine of the angle of incidence and sine of the angle of refraction in two separate mediums wherein the light is travels equal to the ratio of the mediums phase velocities.” It is also equivalent to the reciprocal ratio between the indices of refraction .i.e,
\(\frac{sin\; \Theta _{1}}{sin\; \Theta _{2}}=\frac{v_{1}}{v_{2}}=\frac{\lambda _{1}}{\lambda _{2}}=\frac{n_{1}}{n_{2}}\)
Refractive Index of the Medium in Geometric Optics:
The refractive index is denoted by ‘n’ and described as a dimensionless number which shows that the light or any radiation travels through a medium. It is mathematically expressed as:
n = c/v
Where,
‘v’ indicates phase velocity of light in a medium, and
‘c’ indicates speed of light in a vacuum.
Refraction through a Glass Prism:
The ray of light comprises of different colors and also called as white light. When the white light crosses through a prism, it is disintegrated into the colors it comprises of, that is violet, indigo, blue, green, yellow, orange and red. This procedure is also termed as dispersion that occurs sue to different wavelength of each color. Because of these discrete wavelengths, the dispersed ray from the prism refract into diverse angles and therefore displaying all the colors that are present in the white light. This color configuration is also termed as VIBGYOR titled with the first letter in the order of diminishing deviations.
Stay tuned with BYJU’S to learn more about Geometric Optics. |
A full review of each paper will be released after the seminar is finished. Meanwhile, here are questions that have been asked.
Table of Contents Deep Reinforcement Learning with Double Q-learning (Double DQN) Prioritized Experience Replay (PER) Dueling Network Architectures for Deep Reinforcement Learning (Dueling DQN) Deep Reinforcement Learning with Double Q-learning
Double DQN
Q1. Why must states -6 and 6 always be included to avoid extrapolation?
Suppose that -6 is not sampled. Then, its value must be estimated from or some other close values. However, since values less than -6 cannot be sampled, it must be
extrapolated only from one side: values more than -6. We want to test the amount of overestimation fairly, so we want all values to be intrapolated. If -6 and 6 is always included, any other value without samples can be intrapolated from some samples from below and some samples from above. Prioritized Experience Replay
PER
Q1. Why do the authors mention the power-law distribution while explaining the rank-based variant?
The power-law distribution is monotonic, which might not be obvious with the equation $p_i = \frac{1}{\text{rank}(i)}$.
Q2. How does the Sum-Tree implementation work for proportional-based variant?
To choose with priority, we need to first sum all the priority weights $W = \sum_i w_i$. Then, we select a random number $r$ from $[0, W]$. Finally, if $r \in [0, w_1]$, choice with weight $w_1$ is chosen. If $r \in [w_1, w_1 + w_2]$, choice with weight $w_2$ is chosen. In other words, if $r \in [\sum_{i=1}^k w_i, \sum_{i=1}^{k+1} w_i]$, choice with weight $w_{k+1}$ is chosen. The simplist possible data structure is an array $[w_1\;w_2\;w_3\;\ldots\;w_n]$. Then, to choose an element, we need to iteratively check if $r < \sum w_i$ for all $i$. This is inefficient, so we instead use the
sum-tree implementation.
The sum-tree can be thought of as performing a binary search on the discrete CDF $\sum w_i$. In a tree structure, each parent node contains the sum of all its children. Then, to sample, we just need $O(\log n)$ inequality checks. This implementation does suffer from worse update complexity of $O(\log n)$, since the value of all parents and grandparents of the updated weight must also be updated. However, as a whole, it is faster than the naive array implementation.
Dueling Network Architectures for Deep Reinforcement Learning
Dueling DQN
Q1. What is gradient clipping?
Gradient clipping is a technique used most frequently for training deep neural networks or recurrent neural networks (RNNs) to prevent exploding gradients. When the error gradient is backpropagated long distances, the gradient can either “vanish” or “explode”. A vanishing gradient results in neural network weights not changing, while an exploding gradient results in instability or divergence. To mitigate such instability, it is a common technique to artificially limit the size of gradients. This is called “gradient clipping.”
Q2. Why is the naive aggregation module $Q = V + A$ unidentifiable?
Suppose that for some state $s$, $V_1(s; \theta, \beta) = x$ and $A_1(s, a_i; \theta, \alpha) = y_i$. Then, using the naive aggregating module $Q = V + A$, $Q_1(s, a_i; \theta, \alpha, \beta) = x + y_i$. Now, consider a different set of value estimate functions $V_2$ and $A_2$: $V_2(s; \theta, \beta) = x + \epsilon$ and $A_2(s, a_i; \theta, \alpha) = y_i - \epsilon$. Then $Q_2(s, a_i; \theta, \alpha, \beta) = x + y_i = Q_1(s, a_i; \theta, \alpha, \beta)$. Thus, different $V$ and $A$ can be aggregated to have same $Q$, so we cannot recover $V$ and $A$ from values of $Q$. Thus, this aggregation module is unidentifiable. |
I'm going to teach calculus for the first time to undergraduate students. I would like to know if there is some book about how to teach the concepts of calculus (e.g. limits, derivatives, etc.).
There are so many Calc 1, 2, and Multi video lectures online now. Watch a few... take notes on what you think are some good traits of the person teaching. MIT's Calc course (super fast) might be helpful if you're teaching that style and speed.
Some resources
This text, How to Teach Mathematics - S. Krantz, has many useful guides for the beginning person. I don't agree with all, but it's a great resource. Suzanne Kelton wrote a guide for the AMS. Its pdf is here: teaching Mathematics at the college level - pg. 31 goes into Calc stuff! Also, do not be ashamed to search for AP Calculus teachers' resources. It may be a "high school" course, but some of the teachers out there are amazing and offer a lot of guidance. Specific topic choice may vary but good teachers are that no matter what. Lin McCullin's website is an prime example. Watch those videos yourself from his site :) Ignore test-specific things. Pedagogy! If you are in the US, go to your MAA sectional meeting in the Fall! Please, please, please incorporate technology as a reasonable tool that you actually teach how to use in the right situations. This is vital and super valuable.
Numbers and Functions: Steps into Analysis, R.P. Burn, Cambridge University Press, ISBN 0 521 78836 6.
This book takes a constructivist approach to the subject. It takes the form of a sequence of problems, introducing the key ideas step by step.
In his preface Burn identifies common problems student often encounter: the apparently needlessly formal approach of topics that are familiar from school as well as the insistent use of arbitrary and contrived definitions. Accordingly the book builds on high school concepts and naturally introduces key definitions.
Personally it saved me in my first year of uni.
Below is a review from 'The Mathematical Gazette'
Mathematics should essentially be fun to do and to read and I think that generally students only love to do those things which interests them. So, one basic (and I think mandatory) component of your teaching should be about making the subject interesting for the students.
Presuming that by
Calculus you mean Real Analysis I have the following suggestions for you.
First, let the students feel themselves the importance of a rigorous study of calculus. For this, I think that the introductory chapters of Tom M. Apostol's Calculus I and Terence Tao's Analysis I are two excellent references. Also I would like to recommend reading the last three sections of Part II of the book The Princeton Companion to Mathematics.
Second, when you will be giving assignments to your students try to be scientific in choosing the problems. To be precise, I don't recommend giving difficult (it's a relative term, though) problems to the beginners, nor do I recommend giving a set of problems which doesn't have any relation(s) among them. Though, it's difficult to express precisely what I mean by the above statements, I will nevertheless try to give an example which is intended to make the matter a bit more clear.
Consider the following problems,
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(x)=0$ for all $x\in\mathbb{Q}$. Show that, $f(x)=0$ for all $x\in\mathbb{R}$.
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Let $T=\left\{\dfrac{m}{2^n}\mid m,n\in\mathbb{Z}\right\}$ such that we have $f(x)=0$ for all $x\in T$. Can we say that $f(x)=0$ for all $x\in\mathbb{R}$?
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Let $S=\left\{\dfrac{a}{2^n}\mid a, n\in\mathbb{Z}\right\}$ ($a$ is a constant) such that we have $f(x)=0$ for all $x\in S$. Can we say that $f(x)=0$ for all $x\in\mathbb{R}$?
These problems have a very curious property. To solve the first problem, you need to show that $\mathbb{Q}$ is dense in $\mathbb{R}$. The second problem you need to show that a certain subset of $\mathbb{Q}$ is dense in $\mathbb{R}$. Finally, the third problem asks you to prove that a subset of the subset of Problem 2 is dense in $\mathbb{R}$. So you can see that the first problem is follows from the second one, whereas the second problem follows from the third one. Can you appreciate the order in which the exercises are arranged ? Can you intuitively grasp what I meant in the above lines ?
Third, make your students familiar with some basic ideas of Mathematical Logic at the very beginning. For example, the concept of conjunction, disjunction, negation etc.
Finally, I think that for teaching calculus L. V. Tarasov's book Calculus: Basic Concepts for High Schools is an invaluable guide. |
Rajesh: the next, tenth season of TBBT may be the last one Stops: CMS sees an excess over 2 sigma in their search for top squarks, see page 3, Figure 2, lower left. \(600\GeV\) stops may be "indicated" by that picture.
Adam Falkowski often acts as a reasonable (and educated) man. But I think that his title
What's going on? Both ATLAS and CMS have shown some weak hints of a possible "flavor violation", namely the possible decay of the \(125\GeV\) Higgs boson to a strange lepton pair\[
h \to \mu^\pm \tau^\mp
\] Note that the muon and the tau are "similar" but so far, we've always created a muon-antimuon or tau-antitau pair. The individual lepton numbers \(L_e, L_\mu,L_\tau\) for the generations have been preserved. And the Standard Model makes such a situation natural (although one may predict some really tiny flavor-violating processes even in the Standard Model).
Because the muon of one sign is combined with the tau with another sign – with a particle from a different generation of leptons – the process above, if possible, would be one of the so-called (and so far unseen) flavor-violating processes.
ATLAS and CMS have seen excesses in the 2012 run. They have large error margins so nothing is conclusive at all but the branching ratio (the percentage of Higgses that decay according to the template) for \(h\to \mu\tau\) was measured "somewhat positive" by ATLAS and CMS.
In 2012, ATLAS and CMS had \[
\eq{
{\rm ATLAS:} & B(h\to \tau\mu) = 0.53\%\pm 0.51\%\\
{\rm CMS:} & B(h\to \tau\mu) = 0.84\%\pm 0.37\%
}
\] which are 1-sigma and 2.3-sigma excesses, respectively, combining to 2.5 sigma or so (that's Falkowski's figure). Now, in the 2015 dataset which is 6 times smaller or so, CMS found\[
{\rm CMS:} B(h\to \tau\mu) = -0.76\%\pm 0.81\%
\] The mean value of the branching ratio is reported to be negative. It's a sign of a deficit except that we
knowthat the branching ratio cannot be negative. So a treatment that acknowledges the asymmetry of the distribution and the error margins would probably be highly appropriate here.
But OK, let us ignore this complaint and just combine the excesses and deficits from the assumed Gaussians blindly.
If you combine the 2012 and 2015 data, you get something like\[
{\rm combo:} B(h\to \tau\mu) = +0.55\%\pm 0.30\%
\] or so. I calculated it using my "brain analog computer". When switching to the combo, the mean value has dropped from 2012 and the significance of its being nonzero is some 1.8 sigma. It's less than 2.5 sigma but it's still an excess, a well over 90% confidence level that something is there.
When the significance level decreases from 2.5 to 1.8 sigma, it's a decrease but it's in no way a decisive decrease. 1.8 sigma will be found attractive by a smaller number of physicists than 2.5 sigma but not a "dramatically smaller" number (perhaps 3 times smaller?). In particular, the title
CMS: Higgs to mu tau is going awayis just rubbish. CMS has recorded a deficit but in a smaller amount of data. In this small dataset, no confirmation or refutation of the previous excess could have been expected, and it wasn't given. If this small amount of data were enough for a (5-sigma) discovery of the flavor-violating processes, then the 2015 data would pretty much sharply contradict the 2012 data. If the 2015 data were enough to "safely rule out" the flavor-violating decays around 0.5%-1%, then they would need a big deficit that would disagree with the 2012 data (and the Standard Model), too.
It just isn't possible to decide about the tantalizing signal quickly, and it couldn't have been decided in this way.
What's really untrue about the title is the tense, "is going away". This title pretty much explicitly says that (according to Adam Falkowski), there has been an established downward trend that will continue. But this is just a plain lie. In the short run, the confidence level behaves as a random walk
So whether the latest changes in a very small dataset (or amount of time) have increased or decreased the confidence level is a matter of chance. Both possibilities are equally likely for a short enough period of time – and this was clearly an example of a short enough period of time. The decrease in this short period of time does notimply the decrease in the future because Nature's random generator in individual collisions (or their small enough sets) acts independently.
I am not saying that the flavor-violating decay is there. I really believe it's unlikely, perhaps it has a 10% probability at this point. But even if the claim is untrue – or reasonably believed by an informed physicist to be untrue – I can still see that somewhat isn't describing the evidence honestly.
The point is that the 2.5-sigma hint is a weak one, and Falkowski has all the good reasons to be skeptical because flavor violation is a somewhat extraordinary claim (although not as extraordinary as some people like to suggest). However, the decrease from 2.5 sigma to 1.8 sigma is a decrease by 0.7 sigma which is even (much) smaller than 2.5 sigma.
If someone is laughing at 2.5 sigma but a change by 0.7 sigma is enough for him to say that the "debate is over", he is simply not acting fairly.
Needless to say, the misinterpretation of some random wiggles in a random walk as some "long term trend" or a "law of physics" isn't typical for particle physics. An identical discussion has repeatedly – and much more characteristically – taken place in the weather data. In a 1-year or 10-year or 40-year period, a change of the global mean temperature was observed and some people misinterpret it as some "long-term trend that has to continue".
For a long enough period of time, there could be
somereason for such an interpretation. But if you pick a 6 times shorter period and start to pretend that the trend in this 6 times shorter period may be trusted as much as the trend from the longer dataset, you are simply a demagogue. The shorter periods of time (or smaller collections of collisions) you consider, the more likely it is for the excesses or deficits to be due to chance.
Falkowski's "is going away" spin is virtually identical to the stupid pissing contests of low-brow climate skeptics and low-brow climate alarmists who saw some weather in a recent week and use it as a prediction of the weather in 2100.
The CMS 2015 data are formally a 1-sigma deficit relatively to the Standard Model – which is almost certainly due to chance. If you believe that the branching ratio is some 0.8% as (optimistically) indicated in 2012, the deficit shown in the 2015 data is 1.8 sigma relatively to the flavor-violating extension of the Standard Model. That's larger but not "dramatically different from 1 sigma", a deficit that is there, anyway, and it's not the same 1.8 as the 1.8 sigma excess in the bigger dataset – simply because you know that a larger dataset measures the excesses or deficits more accurately than the smaller one.
The smaller datasets and the shorter periods are unavoidably more affected by the random numbers than their larger siblings.
To 2-sigma exclude the nonzero branching ratios indicated by the mild 2-sigma excesses in 2012, we will probably need the amount of collisions that is at least as large as the 2012 dataset. To "decide" much earlier than that is simply statistically indefensible. And you need at least to 2-sigma exclude the reasonable theory with a nonzero FLV branching ratio to claim to have evidence that "the signal is going away". Falkowski doesn't have it. He has basically used a 0.7-sigma evidence to "settle" a question – and that's much worse than using some 2.5-sigma to do the same thing. |
first of all I'm sorry for my bad English and second I'm sorry for my mistakes of understanding the following topic, I still going to school and learning this for interest.
The topic is Myhill-Nerode and the equivalence classes of a regular or non regular language.
I know that every element of a equivalence class by Myhill-Nerode fulfills this property:
$ \equiv_{A} \triangleq\{(x, y) | \forall z \in \Sigma^{*} \cdot(x z \in A \leftrightarrow y z \in A)\} $
If I understand this right, than a equivalence class consist of element (words) $x$ which we can expand with a word $y$ but for all words $x$ and $y$ of the same class must apply, adding a word $z$ to them both must be in or out of the language.
Hope that is right until now.
Now I will show you my problem:
I have the language (its from a book):
$ \mathrm{B} \triangleq\left\{73 \mathrm{a}^{n} 7 \mathrm{b}^{\mathrm{m}} | \mathrm{n}, \mathrm{m} \in \mathbb{N} \wedge \mathrm{n}=\mathrm{m}+2\right\} $ with $ \Sigma_{\mathrm{M}} \triangleq\{\mathrm{a}, \mathrm{b}, 3,7\} $
And a complete solution:
$1: [\lambda]\equiv_{B}=\{\lambda\} $
$2: [7]\equiv_{B}=\{7\} $
$3: \left[73 a^{k}\right] \equiv_{B}=\left\{73 a^{k}\right\} $ für $ k \in \mathbb{N} $
$4: \left[73 a^{l+2} 7\right] \equiv_{B} =\left\{73 \mathrm{a}^{\imath+2+n} 7 \mathrm{b}^{\mathrm{n}} | \mathrm{n} \in \mathbb{N}\right\} \quad $ für $ l \in \mathbb{N} $
$5: [3]_{\equiv \mathrm{B}}=\Sigma^{*} \backslash\left([\lambda]_{\equiv \mathrm{B}} \cup[7]_{\equiv \mathrm{B}}\right. \cup\left(\bigcup_{k \in \mathbb{N}}\left[73 a^{k}\right] \equiv_{B}\right) \cup \left(\bigcup_{\mathfrak{l} \in \mathbb{N}}\left[73 \mathrm{a}^{\ l+ 2} 7\right] \equiv_{\mathrm{B}}\right) ) $
So in $1$ they build a class of the empty word $\lambda$ and $z = B$ has to be the language by her self to be in the language?
In $2$ they build they build the class of $7$ and z has to be something like this $ z = \left\{3 \mathrm{a}^{n} 7 \mathrm{b}^{\mathrm{m}} | \mathrm{n}, \mathrm{m} \in \mathbb{N} \wedge \mathrm{n}=\mathrm{m}+2\right\}$ to be in the language.
In $3$ they build a class or better infinitely many classes. But here is my problem. I cannot find a $z$ which is working for all classes.
For example we have the words $x_i$ and $z_i$
$x_1 = 73 \to z_1= {a}^{n+2}7b^n$ with $n\in N$
$x_2 = 73a \to z_2= {a}^{n+1}7b^n$ with $n\in N$
$x_3 = 73a^2 \to z_3= {a}^{n}7b^n$ with $n\in N$
$x_4 = 73a^3 \to z_4= {a}^{n}7b^n+1$ with $n\in N$
and so on.
But why is this ok? I Mean they are 2 words in this class for example $x_1$ and $x_2$ who $x_2$ would not be in $B$ with $z_1$.
I hope you can tell me on a simple and understanding way how those classes by Myhill work and how i can find them without making big mistakes. |
4th generation and B decays (in session "Flavour physics")
A stringent test of $\mu-e$ universality in $K\rightarrow l\nu$ decays by NA62 at CERN (in session "Flavour physics, top properties, lepton universality")
Antimatter cosmic rays : backgrounds or signals ? (in session "Flavour Physics - Dark Matter")
ATIC (in session "Flavour Physics - Dark Matter")
Beyond SM : noncommutative approach (in session "SM and beyond at colliders")
BSW Higgs (Tevatron) (in session "SM and beyond at colliders")
Can LHC disprove leptogenesis? (in session "Neutrino Physics - Leptogenesis")
CAST results and Axion Review (in session "Dark Matter - Astroparticle Physics")
CKM elements from squark-gluino loops (in session "Flavour physics: the future")
CKM fits as of winter 2009 and sensitivity to New Physics (in session "Flavour physics: the future")
CLEO's Impact on CKM (in session "Flavour physics, top properties, lepton universality")
Collective flavor transitions of supernova neutrinos (in session "Neutrino Physics")
Current Kamland Results (in session "Neutrino Physics")
Dark Matter and the PAMELA data (in session "Flavour Physics - Dark Matter")
Dark matter in view of recent experiments (in session "Dark Matter")
Dark matter: the MiMac project (in session "Dark Matter")
Diboson physics (in session "SM and beyond at colliders")
Direct determination of neutrino mass parameters at future colliders (in session "Young Scientists Forum 3")
Early Measurement of the W and Z Cross Sections in the Electron Decay Channels with CMS (in session "Young Scientists Forum 1")
Electromagnetic Leptogenesis (in session "Neutrino Physics - Leptogenesis")
EW physics at HERA (in session "SM and beyond at colliders")
Experimental Summary (in session "Summaries")
Final Results from the HiRes Experiment/Telescope Array Status (in session "Dark Matter - Astroparticle Physics")
First evidence of WW,WZ->l nu qq at Tevatron (in session "Young Scientists Forum 1")
Flavor violation in SUSY (in session "Flavour physics, top properties, lepton universality")
Heavy scalar dark matter : constraints and observability (in session "Flavour Physics - Dark Matter")
Hierarchical Soft Terms and Flavor Physics (in session "Young Scientists Forum 1")
Higgs dependent leptogenesis (in session "Neutrino Physics - Leptogenesis")
Higgs in two photons (in session "Young Scientists Forum 2")
Higgs Physics and Beyond the SM at ATLAS/CMS (in session "SM and beyond at colliders")
High energy gamma rays observations with the Fermi gamma-ray Telescope (in session "Dark Matter - Astroparticle Physics")
High mass SM Higgs (Tevatron) (in session "SM and beyond at colliders")
Hot topic Belle (in session "Flavour physics")
Hot Topics in BaBar (in session "Flavour physics")
Initial Results from Electron-Neutrino Appearance in MINOS (in session "Neutrino Physics - Dark Matter")
Instability in coupled dark sectors (in session "Neutrino Physics - Dark Matter")
Is antigravity possible? (in session "Young Scientists Forum 3")
Latest Results from the IceCube Neutrino Observatory (in session "Dark Matter - Astroparticle Physics")
Less-dimensions and the origin of Dark Matter (in session "Dark Matter - Astroparticle Physics")
Lifetimes and rare decays (Tevatron) (in session "Flavour physics")
Looking for New Physics - Prospects of B-Physics at LHCb (in session "Flavour physics: the future")
Low mass SM Higgs (Tevatron) (in session "SM and beyond at colliders")
Measurement of the relative fraction of the gluon-gluon fusion in top-antitop production process at 1.96 TeV proton-antiproton collisions using CDF (in session "Young Scientists Forum 3")
Measurement of top quark properties at the Tevatron (in session "Flavour physics, top properties, lepton universality")
Measurement of topquark mass (in session "Young Scientists Forum 3")
Measurement of Z → mumu cross section in LHC (in session "Young Scientists Forum 1")
MiniBoone (fermilab neutrino oscillations) (in session "Neutrino Physics - Dark Matter")
Mixing and DeltaGamma_s, Tevatron (in session "Flavour physics")
NA62: New Opportunities in Rare Kaon Decays (in session "Flavour physics: the future")
Neutrino data and implications for theta_13 (in session "Neutrino Physics")
Neutrino dipole moments and solar experiments (in session "Neutrino Physics")
New perspectives for heavy flavour physics from the lattice (in session "Flavour physics")
Non Standard Neutrino Interactions (in session "Neutrino Physics - Dark Matter")
Pamela and ATIC - an Astrophysicist View (in session "Flavour Physics - Dark Matter")
Particle Dark Matter in the galactic halo: results from DAMA/LIBRA (in session "Dark Matter")
Particle production in an early supersymmetric universe (in session "Young Scientists Forum 2")
QCD and backgrounds for LHC (in session "SM and beyond at colliders")
Radiative and electroweak penguin measurements at B-factories (in session "Flavour physics")
Readiness of the ATLAS Experiment for First Data (in session "SM and beyond at colliders")
Readiness of the CMS Detector for First Data (in session "SM and beyond at colliders")
Readiness of the LHCb experiment for first data (in session "Flavour physics")
Reconciling dark matter and neutrino masses in mSUGRA (in session "Neutrino Physics")
Results from KLOE (in session "Flavour Physics - Dark Matter")
Results from the Borexino experiment (in session "Neutrino Physics")
Results from the Pierre Auger Observatory (in session "Dark Matter - Astroparticle Physics")
Results of Cuoricino and review of neutrinoless double beta searches (in session "Neutrino Physics - Leptogenesis")
Review of direct searches of dark matters and status of EdelweissII (in session "Dark Matter")
Review of phi1,phi2,ph3 measurements (in session "Flavour physics")
Review of rare decays (Babar & Belle) (in session "Flavour physics")
Review of Vub and Vcb (in session "Flavour physics")
Review on flavor in a warped extra dimension (in session "Flavour physics: the future")
Revisiting the Global Electroweak Fit of the Standard Model and Beyond with Gfitter (in session "SM and beyond at colliders")
S4 and A4 -based flavour models (in session "Flavour physics")
Scalar Dark Matter and DAMA (in session "Young Scientists Forum 3")
SciBoone (in session "Neutrino Physics - Dark Matter")
Search for an Ultra Light Higgs Boson in the Rare Decay $K_{L}\rightarrow\pi^{0}\pi^{0}\mu^{+}\mu^{-}$ (in session "Flavour Physics - Dark Matter")
Search for new Physics at HERA (in session "Flavour physics")
Search for nucleon decay in Super-Kamiokande (in session "Neutrino Physics")
Searches for new physics at the
Tevatron in photon and jet final states (in session "SM and beyond at colliders")
Searches in lepton final states (in session "SM and beyond at colliders")
Single Top Physics at the Tevatron (in session "Flavour physics, top properties, lepton universality")
Solar system dark matter and constraints (in session "Dark Matter")
Solar system dark matter and simulations (in session "Dark Matter")
Standard Model Physics with ATLAS and CMS (in session "SM and beyond at colliders")
Status and first results of the ANTARES neutrino telescope (in session "Dark Matter - Astroparticle Physics")
Status of the Unitarity Triangle analysis (in session "Flavour physics: the future")
Status report on Double Chooz (in session "Neutrino Physics")
SU(5) and A(4) (in session "Young Scientists Forum 2")
SUSY Gauge Singlets and Dualities (in session "Young Scientists Forum 2")
SUSY searches in Leptonic Final States with CMS (in session "Young Scientists Forum 2")
Symmetries and Asymmetries of B -> K* mu+ mu- Decays in the Standard Model and Beyond (in session "Flavour physics")
t-tbar production, etc (Tevatron) (in session "Flavour physics, top properties, lepton universality")
Tevatron SM Higgs Combined Result (low and high mass regions) (in session "Latest news from the Tevatron")
TGC Limits and Search for Resonances in WZ Final State (in session "Young Scientists Forum 2")
The OPERA experiment (in session "Neutrino Physics - Dark Matter")
The PAMELA Space Experiment (in session "Flavour Physics - Dark Matter")
The SuperB factory project (in session "Flavour physics: the future")
Theory Summary (in session "Summaries")
To Measure Theta_13: The Daya Bay Reactor Neutrino Experiment (in session "Neutrino Physics")
UHECR Source correlations and the Auger analysis (in session "Dark Matter - Astroparticle Physics")
Unification in Higher Dimensional Scenario (in session "Young Scientists Forum 1")
W and Z masses and production properties (in session "SM and beyond at colliders")
ZZ → l l nu nu production ( D0 ) (in session "Young Scientists Forum 1")
Include materials from selected contributions |
> Input
> Input
>> 1²
>> (3]
>> 1%L
>> L=2
>> Each 5 4
>> Each 6 7
>> L⋅R
>> Each 9 4 8
> {0}
>> {10}
>> 12∖11
>> Output 13
Try it online!
Returns a set of all possible solutions, and the empty set (i.e. \$\emptyset\$) when no solution exists.
How it works
Unsurprisingly, it works almost identically to most other answers: it generates a list of numbers and checks each one for inverse modulus with the argument.
If you're familiar with how Whispers' program structure works, feel free to skip ahead to the horizontal line. If not: essentially, Whispers works on a line-by-line reference system, starting on the final line. Each line is classed as one of two options. Either it is a
nilad line, or it is a operator line.
Nilad lines start with
>, such as
> Input or
> {0} and return the exact value represented on that line i.e
> {0} returns the set \$\{0\}\$.
> Input returns the next line of STDIN, evaluated if possible.
Operator lines start with
>>, such as
>> 1² or
>> (3] and denote running an operator on one or more values. Here, the numbers used do not reference those explicit numbers, instead they reference the value on that line. For example,
² is the square command (\$n \to n^2\$), so
>> 1² does not return the value \$1^2\$, instead it returns the square of line
1, which, in this case, is the first input.
Usually, operator lines only work using numbers as references, yet you may have noticed the lines
>> L=2 and
>> L⋅R. These two values,
L and
R, are used in conjunction with
Each statements.
Each statements work by taking two or three arguments, again as numerical references. The first argument (e.g.
5) is a reference to an operator line used a function, and the rest of the arguments are arrays. We then iterate the function over the array, where the
L and
R in the function represent the current element(s) in the arrays being iterated over. As an example:
Let \$A = [1, 2, 3, 4]\$, \$B = [4, 3, 2, 1]\$ and \$f(x, y) = x + y\$. Assuming we are running the following code:
> [1, 2, 3, 4]
> [4, 3, 2, 1]
>> L+R
>> Each 3 1 2
We then get a demonstration of how
Each statements work. First, when working with two arrays, we zip them to form \$C = [(1, 4), (2, 3), (3, 2), (4, 1)]\$ then map \$f(x, y)\$ over each pair, forming our final array \$D = [f(1, 4), f(2, 3), f(3, 2), f(4, 1)] = [5, 5, 5, 5]\$
Try it online!
How this code works
Working counter-intuitively to how Whispers works, we start from the first two lines:
> Input
> Input
This collects our two inputs, lets say \$x\$ and \$y\$, and stores them in lines
1 and 2 respectively. We then store \$x^2\$ on line 3 and create a range \$A := [1 ... x^2]\$ on line 4. Next, we jump to the section
>> 1%L
>> L=2
>> Each 5 4
>> Each 6 7
The first thing executed here is line
7,
>> Each 5 4, which iterates line
5 over line 4. This yields the array \$B := [i \: \% \: x \: | \: i \in A]\$, where \$a \: \% \: b\$ is defined as the modulus of \$a\$ and \$b\$.
We then execute line
8,
>> Each 6 7, which iterates line
6 over \$B\$, yielding an array \$C := [(i \: \% \: x) = y \: | \: i \in A]\$.
For the inputs \$x = 5, y = 2\$, we have \$A = [1, 2, 3, ..., 23, 24, 25]\$, \$B = [0, 1, 2, 1, 0, 5, 5, ..., 5, 5]\$ and \$C = [0, 0, 1, 0, 0, ..., 0, 0]\$
We then jump down to
>> L⋅R
>> Each 9 4 8
which is our example of a dyadic
Each statement. Here, our function is line
9 i.e
>> L⋅R and our two arrays are \$A\$ and \$C\$. We multiply each element in \$A\$ with it's corresponding element in \$C\$, which yields an array, \$E\$, where each element works from the following relationship:
$$E_i =\begin{cases}0 & C_i = 0 \\A_i & C_i = 1\end{cases}$$
We then end up with an array consisting of \$0\$s and the inverse moduli of \$x\$ and \$y\$. In order to remove the \$0\$s, we convert this array to a set (
>> {10}), then take the set difference between this set and \$\{0\}\$, yielding, then outputting, our final result. |
A
supposedly fast method to find the sum of a geometric series is the following one.
Let $$S = \sum_{n = 0}^{+\infty} q^n$$ then $$S = 1 + q\left[\sum_{n = 0}^{+\infty} q^n\right] = 1 + qS.$$ Hence $$(1 - q)S = 1 \implies S = \frac1{1 - q}$$
This result, apparently correct, is actually wrong, since we have never required $|q| < 1$, and so it would seem that the geometric series converges for every value of $q$, which is not true.
I cannot find where the error lies in the derivation of the above result. Any hints? |
For a normed linear space $(X,||\cdot||)$ I want to show
a)For a continuous map $f : X \to \mathbb{R}$ the set $$f^{-1}(r):=\{x\in X: f(x)=r\} \subset X$$ is closed.
b) $$S(0,r)=\{x\in X: ||x||=r\}\subset X$$ is closed and bounded.
c) Provide an example that $S(0,r)$ is in general not compact.
I believe these questions lead on from eachother but I have a couple of things which confuse me.
My attempt at a):
Suppose we let $y$ be a limit point of $f^{-1}(r)$. So there is a sequence $\{y_n\}$ such that $y_n \in \{ x \in X : f(x)=r\}$ for all $n$ and $\lim_{n \to \infty} y_n = y$. Since $f$ is continuous we then have $f(y)= \lim_{n \to \infty}f(y_n)=\lim_{n \to \infty} r= r.$ Hence $ y \in f^{-1}(r)$, so $f^{-1}$ contains all of its limit points and is closed.
I understand this is a proof and generally 'show that' questions have a more direct approach so perhaps this is the wrong approach. Any feedback on my attempt is much appreciated.
Also in regards to b) and c), as far as I understood a compact space is one which is closed and bounded. But then b) and c) would contradict eachother so is this not the case all the time? |
Gouy-Chapman Debye layer
The Debye layer is the ionic concentration and potential distribution structure that appears on the surface of a charged electrode in contact with solvents in which are dissolved ionic species. Louis Georges Gouy and David Leonard Chapman at the beginning of the XX century proposed a model of the Debye layer resulting from the combined effect of its thermal diffusion and its electrostatic attraction or repulsion. In effect, in a stationary situation and assuming fluid at rest, the Poisson-Nernst-Planck equations are,
\displaystyle 0 = \nabla \cdot (e \omega_i Z_i c_i \nabla \phi) + \nabla \cdot (\omega_i k_B T \nabla c_i) \quad \mathrm{with} \quad \nabla \cdot (\epsilon \nabla \phi) = \sum_i e c_i
where \phi is the electric potential and c_i is the number of i-ions per volume. \omega_i and Z_i are the i-ion mobility and valence. k_B is the Boltzmann constant, e is the electron charge, \epsilon the electrical permittivity and T the temperature.
The above equations, written in dimensionless form, reduces in the case of a fully dissolved binary system in a planar geometry to,
\displaystyle \hat{c}_+ = exp (-\hat{\phi}), \, \hat{c}_- = exp (\hat{\phi}) \quad \mathrm{with} \quad (\hat{\phi})_{xx} = 2 \sinh (\hat{\phi}).
#include "grid/multigrid1D.h"#include "diffusion.h"#include "run.h"#include "ehd/pnp.h"#define Volt 1.0#define DT 0.01
We assume a fully dissolved binary system labelling the positive ion as Cp and the counterion as Cm. The valence is one, (|Z|=1).
scalar phi[];scalar Cp[], Cm[];int Z[2] = {1,-1};scalar * sp = {Cp, Cm};
Ions are repelled by the electrode due to its positive volume conductivity while counterions are attracted (negative conductivity).
On the left the charged planar electrode is set to a constant potential \phi =1. The concentrations of the positive and negative ions depend exponentially on the voltage electrode.
phi[left] = dirichlet(Volt);Cp[left] = dirichlet (exp(-Volt));Cm[left] = dirichlet (exp(Volt));
In the bulk of the liquid, on the right boundary, the electrical potential is zero and the ion concentrations match the bulk concentration i.e
phi[right] = dirichlet (0.);Cp[right] = dirichlet (1.);Cm[right] = dirichlet (1.);
Initially, we set the ion concentration to their bulk values together with a linear decay of the electric potential \phi.
At each instant, the concentration of each species is updated taking into account the ohmic transport.
#if 1 ohmic_flux (sp, Z, dt, K);#else ohmic_flux (sp, Z, dt); // fixme: this does not work yet#endif
Then, the thermal diffusion is taken into account.
for (scalar s in sp) diffusion (s, dt);
The electric potential \phi has to be re-calculated since the net bulk charge has changed.
Results
We compare the numerical results (symbols) with the analytical solution (lines).
set xlabel 'x'gamma = tanh(0.25)fi(x) = 2*log((1+gamma*exp(-sqrt(2)*x))/(1-gamma*exp(-sqrt(2)*x)))nplus(x) = exp(-fi(x))nminus(x) = exp(fi(x))plot 'log' u 1:2 notitle, fi(x) t '{/Symbol f}',\ 'log' u 1:3 notitle, nplus(x) t 'n+',\ 'log' u 1:4 notitle, nminus(x) t 'n-' lt 7 |
If we take a spacetime with one spatial dimension, we can write a vector as $A^\mu=(t, x)$. This is a
contravariant vector, and we can calculate the covariant vector by multiplying it with the Minkowski metric:$$A_\mu=\begin{bmatrix} 1& 0\\0&-1\end{bmatrix}A^\mu=\begin{bmatrix} t\\-x\end{bmatrix}$$
(I'm not sure if this is notationally valid, since I might be mixing up tensor notation and matrix notation, but you get the idea. )
We can then calculate the
proper distance using the einstein summation convention: $A^\mu A_\mu$.
We can see that coindidentally, the conjugate of a complex number transforms in the same way. If we have a complex number written as a $2\times 1$ matrix $\alpha = \begin{bmatrix} a\\bi \end{bmatrix}$
Then the complex conjugate of $\alpha$ is found by
$$ \alpha^*= \begin{bmatrix} 1& 0\\0&-1 \end{bmatrix} \alpha $$
So we see that the
relation between contravariant and covariant vectors in spacetime is similar to the relation between complex conjugates. Is there a deeper connection between the two, or is this just a mathematical coincidence with no significance? |
Abstract : We study a class of multi-species birth-and-death processes going almost surely to extinction and
admitting a unique quasi-stationary distribution (qsd for short). When rescaled by $K$ and in the limit $K\to+\infty$,
the realizations of such processes get close, in any fixed finite-time window, to the trajectories of a dynamical system whose vector field is defined by the birth and death rates.
Assuming that this dynamical has a unique attracting fixed point, we analyzed in a previous work what happens for large but finite $K$,
especially the different time scales showing up. \newline
In the present work, we are mainly interested in the following question:
Observing a realization of the process, can we determine the so-called engineering resilience?
To answer this question, we establish two relations which intermingle the resilience, which is a
macroscopic quantity defined for the dynamical system, and the fluctuations of the process, which are
microscopic quantities. Analogous relations are well known in nonequilibrium statistical mechanics.
To exploit these relations, we need to introduce several estimators which we control for times between $\log K$
(time scale to converge to the qsd) and $\exp(K)$ (time scale of mean time to extinction).
We also provide variance estimates.
Along the way, we prove moment estimates of independent interest for the process started either from an arbitrary state or
from the qsd. We also obtain weak convergence of the rescaled qsd to a Gaussian measure. |
I have been reading "Generalized Additive Models an Introduction with R" by Simon Wood and have come across a section I'm having trouble with. On page 13 it is stated that the model or design matrix
$X$ can always be decomposed into $$X = Q \left[ \begin{matrix}{R}\\{0}\end{matrix}\right] = Q_fR$$
where $R$ is a $p \times p$ upper triangular matrix and $Q$ is an $ n \times n $ orthogonal matrix, the first p columns of which form $Q_f$. Recall that orthogonal matrices rotate vectors, but do not change their length. Orthogonality also means that $QQ^t = Q^tQ=I_n$. Applying $Q^t$ to $y-X\beta$ implies that $$\left\|y-X\beta\|^2=\right\|Q^ty-Q^tX\beta\|^2=\left\|Q^ty-\left[\begin{matrix}R\\0\end{matrix}\right]\beta\right\|^2 $$
Writing $Q^ty=\left[\begin{matrix}f\\r\end{matrix}\right]$, where $f$ is vector of dimension $p$, and hence $r$ is a vector of dimension $n − p$, yields
$$ \|y-X\beta\|^2=\left\|\left[\begin{matrix}f\\r\end{matrix}\right] - \left[\begin{matrix}R\\0\end{matrix}\right]\beta \right\|^2=\|f-R\beta\|^2+\|r\|^2 $$
The length of $r$ does not depend on $\beta$ , while $\|f − R\beta\|^2$ can be reduced to zero by choosing $\beta$ so that $R$ equals $f$ . Hence $$ \hat\beta=R^{-1}f $$ is the least squares estimator of $\beta$. Notice that $\|r\|^2=\|y-X\hat\beta\|^2$, the residual sum of squares of the model fit.
I find this a bit confusing for a number of reasons. First, it seems to me that $Q$ will only be $n\times n$ if X is $n\times n$. However in general $Q$ will be $n\times p$ where the number of parameters $p$ is less than the number of subjects/cases $n$. It is also stated that the first $p$ columns of $Q$ make up $Q_f$. In order to multiply $Q$ and $R$, $Q$ must contain $p$ columns. Second, if we had a simple design matrix of dimension 2x4 (i.e. we have an intercept and slope paramater and 4 subjects) y would be 4x1 and $Q^t$ 2x4. $Q^ty$ will be 2x1. Thus it is not clear to me how $Q^ty =\left[\begin{matrix}f\\r\end{matrix}\right]$ where $f$ is vector of dimension $p$, and hence $r$ is a vector of dimension $n − p$. In this case $f = 2$ and $n-p=2$ which would give $Q^ty$ a dimension of 4x1.
I have checked the books errata and there are no mistakes listed for this section, so I must not be thinking about this quite right. If anyone could shed some light on what piece of this puzzle I'm missing, I'd greatly appreciate some enlightenment. |
A two-step mixed inpainting method with curvature-based anisotropy and spatial adaptivity
1.
Instituto de Investigación en Señales, Sistemas e Inteligencia Computacional, sinc(ⅰ), FICH-UNL/CONICET, Argentina
2.
Ciudad Universitaria, CC 217, Ruta Nac. No 168, km 472.4, (3000) Santa Fe, Argentina
3.
Instituto de Matemática Aplicada del Litoral, IMAL, CONICET-UNL, Centro Científico
4.
Tecnológico CONICET Santa Fe, Colectora Ruta Nac. 168, km 472, Paraje "El Pozo", 3000, Santa Fe, Argentina and Departamento de Matemática
5.
Facultad de Ingeniería Química, Universidad Nacional del Litoral, Santa Fe, Argentina
The image inpainting problem consists of restoring an image from a (possibly noisy) observation, in which data from one or more regions are missing. Several inpainting models to perform this task have been developed, and although some of them perform reasonably well in certain types of images, quite a few issues are yet to be sorted out. For instance, if the image is expected to be smooth, the inpainting can be made with very good results by means of a Bayesian approach and a maximum a posteriori computation [
In this work we present a two-step inpainting process. The first step consists of using a CDD inpainting to build a pilot image from which to infer
a- priori structural information on the image gradient. The second step is inpainting the image by minimizing a mixed spatially variant anisotropic functional, whose weight and penalization directions are based upon the aforementioned pilot image. Results are presented along with comparison measures in order to illustrate the performance of this inpainting method. Mathematics Subject Classification:Primary: 94A08, 68U10; Secondary: 65F22. Citation:Francisco J. Ibarrola, Ruben D. Spies. A two-step mixed inpainting method with curvature-based anisotropy and spatial adaptivity. Inverse Problems & Imaging, 2017, 11 (2) : 247-262. doi: 10.3934/ipi.2017012
References:
[1] [2] [3] [4] [5]
F. Ibarrola and R. Spies,
Image restoration with a half-quadratic approach to mixed weighted smooth and anisotropic bounded variation regularization,
[6] [7] [8]
G. L. Mazzieri, R. D. Spies and K. G. Temperini,
Mixed spatially varying $ L^2$-$ BV$ regularization of inverse ill-posed problems,
[9]
R. Rockafellar,
Convex Analysis, Princeton University Press, 1970.
Google Scholar
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L. I. Rudin, S. Osher and E. Fatemi,
Nonlinear total variation based noise removal algorithms (proceedings of the 11th annual international conference of the center for nonlinear studies),
show all references
References:
[1] [2] [3] [4] [5]
F. Ibarrola and R. Spies,
Image restoration with a half-quadratic approach to mixed weighted smooth and anisotropic bounded variation regularization,
[6] [7] [8]
G. L. Mazzieri, R. D. Spies and K. G. Temperini,
Mixed spatially varying $ L^2$-$ BV$ regularization of inverse ill-posed problems,
[9]
R. Rockafellar,
Convex Analysis, Princeton University Press, 1970.
Google Scholar
[10]
L. I. Rudin, S. Osher and E. Fatemi,
Nonlinear total variation based noise removal algorithms (proceedings of the 11th annual international conference of the center for nonlinear studies),
Isotropic T1 CDD Anisotropic T1-TV 20.140 35.497 36.329
Isotropic T1 CDD Anisotropic T1-TV 20.140 35.497 36.329
Isotropic T1 CDD Anisotropic T1-TV 29.127 29.952 30.868
Isotropic T1 CDD Anisotropic T1-TV 29.127 29.952 30.868
Isotropic T1 CDD Anisotropic T1-TV 20.568 21.360 22.112
Isotropic T1 CDD Anisotropic T1-TV 20.568 21.360 22.112
Gray CKS Gray A T1-TV Color CKS Color A T1-TV 29.770 32.421 21.464 24.284
Gray CKS Gray A T1-TV Color CKS Color A T1-TV 29.770 32.421 21.464 24.284
[1]
Bernadette N. Hahn.
Dynamic linear inverse problems with moderate movements of the object: Ill-posedness and regularization.
[2] [3] [4] [5] [6] [7]
G. Fonseca, G. Rodríguez-Blanco, W. Sandoval.
Well-posedness and ill-posedness results for the regularized Benjamin-Ono equation in weighted Sobolev spaces.
[8] [9]
Yonggeun Cho, Gyeongha Hwang, Soonsik Kwon, Sanghyuk Lee.
Well-posedness and ill-posedness for the cubic fractional Schrödinger equations.
[10] [11]
Tsukasa Iwabuchi, Kota Uriya.
Ill-posedness for the quadratic nonlinear Schrödinger
equation with nonlinearity $|u|^2$.
[12]
Shunlian Liu, David M. Ambrose.
Sufficiently strong dispersion removes ill-posedness in truncated series models of water waves.
[13] [14] [15]
Marcel Braukhoff.
Semiconductor Boltzmann-Dirac-Benney equation with a BGK-type collision operator: Existence of solutions vs. ill-posedness.
[16]
Chao Deng, Xiaohua Yao.
Well-posedness and ill-posedness for the 3D generalized Navier-Stokes equations in $\dot{F}^{-\alpha,r}_{\frac{3}{\alpha-1}}$.
[17]
Felix Lucka, Katharina Proksch, Christoph Brune, Nicolai Bissantz, Martin Burger, Holger Dette, Frank Wübbeling.
Risk estimators for choosing regularization parameters in ill-posed problems - properties and limitations.
[18]
Sergiy Zhuk.
Inverse problems for linear ill-posed differential-algebraic equations with uncertain parameters.
[19] [20]
Thorsten Hohage, Mihaela Pricop.
Nonlinear Tikhonov regularization in Hilbert scales for inverse boundary value problems with random noise.
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I want to post a message to say, after applying Hoot's answer, you're still 'not done yet' (this problem is quite technical).
Let's recall the situation. You can reduce to $f:X\rightarrow Y=\operatorname{Spec} B$ dominant morphism of integral schemes of finite type (we no longer need the generically finite assumption) inducing a finite extension $K(Y)\rightarrow K(X)$. This means that $X$ is covered by $U_i=\operatorname{Spec} A_i$ open such that $B\rightarrow A_i$ is finitely generated, and $A_i$ is generated by elements algebraic over $B$ as Hoot said. Note that in fact $B\hookrightarrow A_i$ is mono ($\forall i$) since $U_i$ is dense in $X$ since $X$ irred., and $f$ is dominant, so $U_i$ is dense in $Y$. Let $x_{ij}$ generate $A_i$ over $B$. Then $x_{ij}$ satisfies an equation of algebraic dependence over $B$ with leading coefficient $b_{ij}$. Put $U=\{b_{ij}\}_{i,j}$; then $W:=\operatorname{Spec} B[U^{-1}]$ is an open affine subset of $Y$ and again since $U_i\rightarrow Y$ is dense, $f^{-1}(W)\cap U_i\ne\emptyset\forall i$. Moreover, now $B[U^{-1}]\hookrightarrow A_i[U^{-1}]$ has $A_i[U^{-1}]$ finitely generated by integral elements; hence is finite over $B[U^{-1}]$.
We have reduced to $f : X\rightarrow Y=\operatorname{Spec} B$ with $U_i=\operatorname{Spec} A_i$ covering $X$ and $B\hookrightarrow A_i$ finite. But we need to find a nonempty open subset $V\subset Y$ such that $f:f^{-1}(V)\rightarrow V$ is finite (in particular, affine). Put $W:=\bigcap U_i$ (nonempty open since $X$ irred.). Let $\mathfrak a_i\subset A_i$ be such that $U_i\supset V(\mathfrak a_i)=U_i-W$. $B\subset A_i$ are integral domains, and $K(A_i)$ is algebraic over $K(B)$. It is then a fact that every nonzero ideal of $A_i$ intersects $B$ nontrivially. $\mathfrak a_i$ is nonzero since $W\ne\emptyset$. Hence $\exists f_i\in B\subset A_i$ such that $D(f_i)\subset W\subset U_i$. Put $U:=\{f_i\}_i$. Then $V:=\operatorname{Spec} B[U^{-1}]$ is nonempty affine open in $Y$ and so $\emptyset\ne f^{-1}(V)\subset W$. Note that $f^{-1}(V)$ coincides with $\bigcap_i X_{f_i}$, where $f_i$ is the image of $f_i$ in $\Gamma(X,\mathscr O_X)$ ($B$ injects into global sxns since $f$ dominant). Moreover $f^{-1}(V)\subset W$ hence coincides with $\bigcap_i X_{f_i}\cap U_j=\bigcap_i D(f_i)\cap U_j\forall j=\operatorname{Spec} A_j[U^{-1}]$, where as always we identify $B$ with its image as appropriate. Hence $f:f^{-1}(V)\rightarrow V$ is our desired finite morphism.
One last thing: here is a different (probably functionally equivalent) way to prove that $K(X)$ is a finite extension of $K(Y)$. Reduce to $f:\operatorname{Spec} A=X\rightarrow Y=\operatorname{Spec} B$ dominant generically finite morphism of integral affine schemes of finite type. Identifying $B$ with its monomorphic image we have $B\subset A$ domains with only finitely many primes of $A$ lying over $(0)\in\operatorname{Spec} B$. Replacing $B$ by $K(B)$ we have $A$ finitely generated over a field hence Noetherian and Jacobson (by the Nullstellensatz). A Noetherian ring $A$ is Jacobson iff $\forall\mathfrak p\in\operatorname{Spec} A$ of dimension one, $R/\mathfrak p$ has infinitely many ideals. $A$ has only finitely many ideals after replacing $B$ by $K(B)$, so all primes of $A$ must be maximal. Thus $A$ is Artinian (and Noetherian) hence finite over $K(B)$. Therefore $K(A)=K(X)$ is a finite extension of $K(B)=K(Y)$. |
Find the Laurent series expansion of $f(z)=\frac{z^2-1}{(z+2)(z+3)^2}$ at $0<|z+3|<1$
I have a couple of doubts in how to handle this problem: First of all, should I do it with partial fractions? Because now I have a polynomial in the numerator, so I don't know how to proceed if that's the case.
If not, can I use the general term for the Laurent series?
$$a_n=\frac{1}{2\pi i}\int \frac{f(\zeta)d\zeta}{(\zeta +3)^{n+1}}$$
But what happens with the $(z+3)^2$ in the numerator? Im pretty mixed up. |
The radius of circular disc is increasing at the rate of 0.5cm/sec. At what rate is the area of the disc increasing when its radius is 6cm
Jamb Maths 2004
A farmer planted 5000 grains of maize and harvested 5000cobs, each bearing 500 grains. What is the ratio of the number of grains sowed to the number harvested?
Simplify $\frac{1}{\sqrt{3}+2}$
The shaded region in the venn diagram above is
Evaluate $\frac{\tfrac{1}{10}\times \tfrac{2}{3}+\tfrac{1}{4}}{\tfrac{1}{2}\div \tfrac{3}{5}-\tfrac{1}{4}}$
Given that $\sqrt[3]{{{4}^{2x}}}=16$ find the value of
x
In a class of 40 students,each student offers at least one of physics and chemistry. If the number of students that offers physics is 3 times the number that offer both subject and the number that offer chemistry is twice the number that offer physics. Find the number of students that offers physics only
Find
P , if 451 6 – P 7 =305 6
If $6{{\log }_{x}}2-3{{\log }_{x}}3=3{{\log }_{5}}0.2$, find
x
Three teacher shared a packet of chalk. The first teacher got $\frac{2}{5}$of the chalk and the second teacher received $\frac{2}{15}$of the remainder. What fraction did the third teacher received ? |
Many novel ideas are found on the Internet. One not so novel notion is that Einstein was wrong and that the "lightspeed limit" is really just some international conspiracy of conservative "establishment" scientists. Those who make this point neglect the fact, however, that the deduction about the speed of light is not a result of some exotic assumptions or blind speculation, but a fairly simple consequence of some fundamental assumptions about nature: in other words, if you wish to prove that Einstein was wrong, you have to show that either elementary logic is incorrect, or that some of our basic assumptions about nature are outright false.
Here is why.
Symmetries in Nature
To begin, here are our assumptions, some of which are generic in nature, while others are specifically about electricity and magnetism. First, the generic ones:
Space is homogeneous. The equations of physics work the same in New York and Los Angeles, on Earth or on Mars, in the Milky Way or in the Andromeda Galaxy. Space is isotropic. The equations of physics don't change just because you turn around and look in a different direction. Space is symmetric under time translation. The equations of physics are the same today as they were yesterday, and as they will be tomorrow and the day after that. Space is symmetric under a "boost". The equations of physics work the same in moving coordinate systems: your watch, computer, or your body for that matter won't cease to function just because you're moving on a train, an airplane, or spacecraft. Maxwell's Equations
The specific assumptions about electricity and magnetism are the culmination of 100 years of research and experiment, and were first put into modern form by Maxwell in the 1860s. In plain English, this is what they say:
The sum total of the electric fieldaround a volume of space is proportional to the charges contained within. The sum total of the magnetic fieldaround a volume of space is always zero, indicating that there are no magnetic charges (monopoles). (With a bar magnet, the number of field lines "going in" and those "going out" cancel each other out exactly, so there is no deficit that would show up as a net magnetic charge.) A change over time in the electric field or a movement of electric charges (current) induces a proportional vorticity in the magnetic field. A change over time in the magnetic field induces a proportional vorticity in the electric field, but in the opposite direction.
These four assumptions can also be stated exactly using mathematical language: specifically, the language of vector calculus. But before we continue, it is important to make note of the fact that
we are done with the assumptions: what follows is rigorous logic. In other words, if one wishes to argue that Einstein's conclusion is wrong, one either has to throw logic out the window, or find fault in one or more of the assumptions above. Empty Space
To examine the speed of light in free space, we can simplify two of our assumptions. In free space there are no charged bodies or particles about, and therefore Maxwell's first assumption reads as:
The sum total of the electric fieldaround a volume of empty space is zero, indicating there is no electric charge contained within.
Similarly, we can drop the bit about electric current from the third assumption:
A change over time in the electric field in empty space induces a proportional vorticity in the magnetic field. Divergence
Mathematically, the first two assumptions are expressed through the concept of divergence. If we imagine the electric field with lines of force, as in a high-school physics textbook, divergence basically tells us how the lines are "spreading out". For the lines to spread out, there must be something, intuitively speaking, to "fill the gaps": these things would be particles of charge. But there are no such things in empty space, so we can say that the divergence of the electric field in empty space is identically zero:
\[{\rm div}~{\bf\mathrm{E}}=0.\]
The electric field is a vector field: the force it produces has a strength as well as a direction. The divergence of a vector field in a given coordinate system is computed through partial derivatives of the vector components:
\[{\rm div}~{\bf\mathrm{v}}=\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}.\]
So far so good. What we said about the electric field also applies to the magnetic field of course:
\[{\rm div}~{\bf\mathrm{B}}=0.\]
Vorticity
What about the vorticity? The vorticity of a vector field is also computed through partial derivatives:
\[{\rm curl}~{\bf\mathrm{v}}=\begin{pmatrix} \frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}\\ \frac{\partial v_x}{\partial z}-\frac{\partial v_z}{\partial x}\\ \frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\end{pmatrix}.\]
Unlike the divergence of a vector field, which is a number field (called a scalar field), the vorticity of a vector field is another vector field. Intuitively what it means is that a vortex not only has strength, but it also has an axis pointing in a specific direction.
In this mathematical formalism, the second pair of Maxwell's equations in empty space can be expressed as:
\begin{align}{\rm curl}~{\bf\mathrm{B}}&=\epsilon_0\mu_0\partial{\bf\mathrm{E}}/\partial t,~~~{\rm and}\\ {\rm curl}~{\bf\mathrm{E}}&=-\partial{\bf\mathrm{B}}/\partial t,~~~~~\end{align}
where $\epsilon_0$ is the vacuum's electrical permittivity, $\mu_0$ its magnetic permeability.
To further simplify calculations, we'll assume that the field depends only on one spatial coordinate, say, \(x\). Feynman offers the example of a large (infinite?) charged sheet in the \(y\)-\(z\) plane that moves in a direction perpendicular to its surface as a source of this field. The same computation can be performed in the general case, but it is a lot more complicated (and a lot less instructive.)
Solution in One Spatial Dimension
In this case, the first pair of Maxwell's equations tells us that \(E_x\) and \(B_x\) must be constant functions.
The second pair of Maxwell's equations reduces to the following simple set:
\begin{align} -\frac{\partial B_z}{\partial x}&=\epsilon_0\mu_0\frac{\partial E_y}{\partial t},\\ \frac{\partial B_y}{\partial x}&=\epsilon_0\mu_0\frac{\partial E_z}{\partial t},\\ -\frac{\partial E_z}{\partial x}&=-\frac{\partial B_y}{\partial t},\\ \frac{\partial E_y}{\partial x}&=-\frac{\partial B_z}{\partial t}. \end{align}
Using the first and the fourth equation, for instance, we can find a solution for \(B_z\) (or \(E_y\)). Consider:
\[-\frac{\partial^2B_z}{\partial x^2}=\epsilon_0\mu_0\frac{\partial^2E_y}{\partial x \partial t},\]
and
\[\frac{\partial^2E_y}{\partial x \partial t}=-\frac{\partial^2B_z}{\partial t^2},\]
or
\[\epsilon_0\mu_0\frac{\partial^2B_z}{\partial t^2}-\frac{\partial^2B_z}{\partial x^2}=0.\]
This can be rewritten as:
\[\left(\sqrt{\epsilon_0\mu_0}\frac{\partial}{\partial t}-\frac{\partial}{\partial x}\right) \left(\sqrt{\epsilon_0\mu_0}\frac{\partial}{\partial t}+\frac{\partial}{\partial x}\right)B_z=0.\]
Solutions to this equation can be found in the form:
\[B_z=f_1(ct-x)+f_2(ct+x),\]
where \(f_1\) and \(f_2\) are arbitrary functions and $c=1/\sqrt{\epsilon_0\mu_0}$. The same solution exists for \(B_y\), \(E_y\), and \(E_z\).
If we set \(f_2=0\), then
\[B_z=f_1(ct-x),\]
which is a legitimate solution to Maxwell's equations. What this means is that if the field has a certain value at \(t=0\), \(x=0\), then it'll have the same value at \(t=t_0\), \(x=t_0\). Similarly, if we set \(f_1=0\), a field that has a certain value at \(t=0\), \(x=0\), then it'll have the same value at \(t=t_0\), \(x=-t_0\). Thus we can say that the electromagnetic field represented by this solution is moving at velocity $c$ along the \(x\) axis in either of two directions. The value of $c$ is observer independent, unless the properties of the vacuum ($\epsilon_0$ and $\mu_0$) are themselves dependent on the motion of the observer.
If, on the other hand, we assume that the vacuum is the same for all observers, the observed speed will be the
same to all observers. Same regardless of where they are. Regardless of when they make their measurements. Regardless of how fast they themselves are moving, and in which direction they are facing. Whether you move towards a light source or away from it, the speed appears the same. Special Relativity
This of course makes no sense in ordinary Euclidean spacetime: when you are running ahead of a moving train, it'll appear slower (i.e., take longer to hit you) than when you're running towards it.
Special relativity is simply the most economical way to solve this dilemma. The idea is to find the simplest geometry in which all our initial assumptions can be simultaneously true.
Why geometry? If you think about it, when you switch from a stationary coordinate system to a moving one (i.e., from a coordinate system fixed to the clock of a railway station to one that is fixed to the main axis of your steam engine) it's really just a simple coordinate transformation: \(t'=t\), \(x'=x-vt\). And herein lies the problem: after this coordinate transformation, in the new coordinate system a ray of light no longer satisfies the conditions that we derived previously. If, in the old coordinate system, an electromagnetic field had the same value at \(t=0\), \(x=0\) and \(t=t_0\), \(x=t_0\), in the new coordinate system, it'll have the same values at \(t'=0\), \(x'=0\) and \(t'=t_0\), \(x'=t_0-vt_0\), and this contradicts what we just learned about Maxwell's equations as \(x'\) won't be equal to \(t'\).
The simple geometry of special relativity, Minkowski spacetime, is built around the assumption that the quantity \(dt^2-dx^2-dy^2-dz^2\) remains constant under a "boost", i.e., when you change from one moving coordinate system to another. In our simple scenario with only one spatial coordinate, this reduces to \(dt^2-dx^2\) remaining constant when you switch from a stationary to a moving system. For rays of light moving in either direction, \(dt^2-dx^2\) remains 0 regardless whether you measure it from a moving or stationary system, which is precisely what we want in order to remain consistent with Maxwell's equations.
This assumption leads to a new form of coordinate transformation, the Lorentz transformation. To see why, compare the values for the station and the train in the diagram above. For the station, \(dt=t_0\), \(dx=x_0=vt_0\) (this, after all, is how we define the train's velocity \(v\)) and therefore, \(dt^2-dx^2\) is \(t_0^2-v^2t_0^2\). For the train, \(dx'=0\) and thus \((dt')^2-(dx')^2\) is \((t_0')^2\). We want the values for the station and the train to be equal:
\begin{align}(t_0')^2&=t_0^2-v^2t_0^2,\\ (t_0')^2&=t_0^2(1-v^2),\\ t_0'&=t_0\sqrt{1-v^2},\end{align}
and this, of course, is the fabled Lorentz transform.
Any other approach would either have to use a more complicated geometry (the late 19
th century concept of "ether" can be viewed as an attempt to do just this) or it would require giving up at least some of our initial assumptions. And what's wrong with that, you ask? Well, those assumptions are supported by an enormous number of physical observations, not the least of which is the observation that this computer in front of me is functioning as expected, even though it is moving about at a not altogether inconsiderable velocity as the Earth spins, moves around the Sun and, along with the Sun, moves about in the Universe... References
Feynman, Richard P., The Feynman Lectures on Physics II.(chapter 20), Addison-Wesley, 1977 |
The real problem I am facing is the following. INSTANCE: I have sets $N:=\{1,\ldots,n\}$ and $K:=\{1,\ldots,k\}$ and matrix $a_{ij}>0$ for all $i\in K$ and $j\in N$. QUESTION: I need to find a subset $S$ of $N$ of size as small as possible and partition the set $K$ into $|S|$ disjoint sets $K_j$ whose union equals $K$ such that for all $j\in S$, I have $$\sum_{j'\in S\\j'\neq j} a_{ij'} \leqslant a_{ij}-1,$$ for all $i\in K_j$. Example:
Given $n=k=3$ and the matrix $$ \begin{bmatrix} 0.6 & 2.7 & 1.2\\ 1.3 & 2.6 & 0.8\\ 1.5 & 0.4 & 0.6 \end{bmatrix}. $$
In this example, $S$ should be equal to $S=\{1, 2\}$ and $K_1=\{3\}$ and $K_2=\{1,2\}$.
I noticed two facts:
If there exists some $j\in N$ such that $a_{ij}\geqslant 1$ for all $i\in K$ then $S=\{j\}$ and $K_j=K$; and If there exists some $i\in K$ such that $a_{ij}<1$ then $S=\emptyset$. My question:Is it possible to solve this optimization problem in polynomial time (at least with approximation algorithm)?
The first thing I tried to do is to transform it into a known problem and then applied a known algorithm for that. I thought about transforming it to a
set cover or bin packing but I failed and also I do not think that this is interesting. The problem I tried to formulate.
I have sets $N:=\{1,\ldots,n\}$ and $K:=\{1,\ldots,k\}$ and matrix $a_{ij}>0$ for all $i\in K$ and $j\in N$. Also, I have $n$ disjoints sets $K_j\subset K$ for each $j\in N$, (I added $K_j$ as inputs because I could not formulate it otherwise.)
Finally, I get this: $$ \begin{align*} & {\underset{S}{\text{minimize}}} & & |S|\\[3pt] & \text{subject to} & & \sum_{j'\in S\\j'\neq j} a_{ij'} \leqslant a_{ij}-1,\forall\, j\in S,i\in K_j,\\[3pt] & & & S\subseteq N.\\ \end{align*} $$
Thanks. |
It is well-known that the Mapping Class Group of a closed surface of genus $g$ surjects onto $Sp(2g, \mathbb{Z})$ (see, for example the Farb-Margalit book). However, I was wondering if there is a simple proof that the set of pseudo-Anosov elements in MCG surjects onto $Sp(2g, \mathbb{Z})$ (and also a reference for where this might have been stated first) -- I can construct a somewhat sophisticated proof, but this should be easier.
Here's an outline of a proof. Consider a pseudo-Anosov mapping class $\phi$ such that the infinite cyclic group $C = \langle \phi \rangle$ is malnormal. Using the methods of Ivanov and/or McCarthy one can show that for any mapping class $\psi \not\in C$, the mapping class $\psi \phi^k$ is pseudo-Anosov for sufficiently large $k$. Now apply this with $\phi$ in the Torelli group and $\psi$ representing any element of $SP(2g,\mathbb{Z})$.
Some details added later: First part: There is a simple direct construction of a pseudo-Anosov in the Torelli group: use Penner's recipe, which says that if $c,d$ are a filling pair of curves then $\tau_c \tau_d^{-1}$ is pseudo-Anosov. Take $c,d$ to be separating. The malnormality requirement can also be obtained by choosing $c,d$ so that no mapping class fixes both $c,d$ nor interchanges them. Second part: Since $\psi \not\in C$ then (as in Sam's answer) $\psi$ fixes neither the attracting nor repelling measured geodesic laminations of $\phi$, and neither does it interchange them; thie follows from the assumption that $C$ is malnormal. Now pick a tiny compact neighborhood $K$ of the attracting lamination of $\phi$, so tiny that $\psi(K)$ does not contain the repelling lamination. Pick a high power $\phi^k$ so that $\phi^k(\psi(K))$ is in the interior of $K$ and so that $\phi^k\psi$ is contracting on $K$; this is possible because, in projective train track coordinates, $\phi$ is contracting near its attracting lamination. By this means (and working similarly with the inverse) one shows that $\phi^k\psi$ has north-south dynamics (and hence so does its conjugate $\psi\phi^k$). Finally, no reducible mapping class has north south dynamics, so $\psi\phi^k$ is pseudo-Anosov.
Suppose that $f$ is a pseudo-Anosov and let $\lambda^\pm$ be its stable and unstable laminations. Suppose that $g$ is any mapping class with the following property: $g(\lambda^+) \neq \lambda^-$. Then, for all sufficiently large $n$, the composition $f^n g$ is also pseudo-Anosov. This is proved using the north-south dynamics of $f$.
Now, as in Mosher's answer, choose $f$ in Torelli and let $g$ be a lift of the desired element of $Sp$. If needed, compose $g$ with a bounding pair map to arrange the side-condition. We are done.
Added later: Here are some of the details of the dynamical argument. I've edited this several times to try and make it correct. Full disclosure - I first heard this from Yair Minsky.
Pick $U$, a small neighborhood of $\lambda^+$, chosen so that $\lambda^-$ is not contained in $g(U)$. I'll also want $f|U$ to be strictly contracting. Using north-south dynamics, there is an $m^+ > 0$ so that for all $n > m^+$ we have $f^{m^+} g(U) \subset U$. The contracting property implies that the map $f^n g$ has a unique fixed point in $U$.
On the other hand, pick a small neighborhood $V$ of $g^{-1}(\lambda^-)$, so that $\lambda^+$ is not in $V$. Also, we require $f^{-1}|g(V)$ to be strictly contracting There is a power $m^- > 0$ so that for all $n > m^-$ we have $V \subset f^n g(V)$. Deduce that the map $f^n g$ has a unique fixed point in $V$.
We place one more restriction on $n$. We need $f^n g(V^c)$ to be contained in $U$. I now claim that $h = f^n g$ has no other fixed points, and so is pseudo-Anosov. |
1) Let us for simplicity assume that the Hilbert space is finite-dimensional. Let $\hat{A}:H\to H$ and $\hat{B}:H\to H$ be two Hermitian operators (a.k.a. quantum observables). Their two spectra must then be finite sets of eigenvalues
$${\rm Spec}(A)~=~\{a_1, \ldots, a_n\} \qquad {\rm and} \qquad {\rm Spec}(B)~=~\{b_1, \ldots, b_m\}.$$
2) In the question(v1) OP specifically asks what happens if the spectra are
degenerate? We can decompose the Hilbert space
$$H~=~\oplus_{i=1}^n E^{(\hat{A})}_i$$
in orthogonal eigenspaces
$$E^{(\hat{A})}_i~:=~ {\rm ker}(\hat{A}-a_i{\bf 1}) ~\subseteq~H$$for $\hat{A}$. Let us define corresponding projection operators $\hat{P}^{(\hat{A})}_i:H\to E^{(\hat{A})}_i$.Then we may decompose $$\hat{A}~=~\sum_{i=1}^n a_i\hat{P}^{(\hat{A})}_i. $$
3) Let us for simplicity assume that the initial state is a pure state given by a ket $\mid\psi\rangle$. (For the case of a mixed state, see this answer.) A measurement of the observable $\hat{A}$, with the outcome $a_i$, collapses the initial state $\mid\psi\rangle$ into a new state
$$ \mid\psi^{\prime}\rangle~=~\frac{\hat{P}^{(\hat{A})}_i\mid\psi\rangle}{\sqrt{\langle\psi\mid\hat{P}^{(\hat{A})}_i\mid\psi\rangle}}.$$
4) Similarly, we can do the same thing with the other operator,$$H~=~\oplus_{j=1}^m E^{(\hat{B})}_j,$$$$E^{(\hat{B})}_j~:=~ {\rm ker}(\hat{B}-b_j{\bf 1}) ~\subseteq~H,$$
$$\hat{B}~=~\sum_{j=1}^m b_j\hat{P}^{(\hat{B})}_j. $$
5) Since the two operators $\hat{A}$ and $\hat{B}$ commute $[\hat{A},\hat{B}]=0$, i.e., are compatible, there exists a common set of orthogonal eigenspaces
$$E_{ij} ~:=~ E^{(\hat{A})}_i \cap E^{(\hat{B})}_j~\subseteq~H, \qquad\qquad H~=~\oplus_{i=1}^n\oplus_{j=1}^m E_{ij}.$$
Note that some of the subspaces $E_{ij}$ could be trivial (zero-dimensional). All the projection operators $\hat{P}^{(\hat{A})}_i$ and $\hat{P}^{(\hat{B})}_j$ will also commute.
6) If we next perform a measurement of the observable $\hat{B}$, with result $b_j$, on the state $\mid\psi^{\prime}\rangle$ from section 3, the state collapses into
$$ \mid\psi^{\prime\prime}\rangle~=~\frac{\hat{P}^{(\hat{B})}_j\mid\psi^{\prime}\rangle}{\sqrt{\langle\psi^{\prime}\mid\hat{P}^{(\hat{B})}_j\mid\psi^{\prime}\rangle}},$$
which after some straightforward algebra reduces to
$$ \mid\psi^{\prime\prime}\rangle~=~\frac{\hat{P}^{(\hat{B})}_j\hat{P}^{(\hat{A})}_i\mid\psi\rangle}{\sqrt{\langle\psi\mid\hat{P}^{(\hat{B})}_j\hat{P}^{(\hat{A})}_i\mid\psi\rangle}}. $$
7) The latter expression is symmetric in $\hat{A} \leftrightarrow \hat{B}$,and therefore performing the measurements in opposite order, i.e., measuring first $\hat{B}$, with outcome $b_j$, and then $\hat{A}$, with result $a_i$ would produce the same final state $\mid\psi^{\prime\prime}\rangle.$
8) So to answer the question, in the degenerate case there may still be a collapse associated with the second measurement. However, if the spectrum of the first observable is non-degenerated, then there is no second collapse. |
Formally, I see it as emerging from the trace necessary to compute observables in quantum mechanics,
$$ \langle O(t) \rangle = \text{Tr}\left[\rho O(t) \right] \, .$$
I use Heisenberg's picture, $O(t) = \text{e}^{\text{i} t H} O \, \text{e}^{-\text{i} t H}$ and $\hbar = 1$. The path integral representation of the time evolution operator gives
$$\left. \langle x \right| \text{e}^{\text{i} t H} \left| x \right. \rangle = \int \text{D}x \, \text{e}^{i S} \, ,$$
with $S = \int_0^t \text{d}t [...]$ an action that contains a simple time integral from $t=0$ to $t=t$. In thermal equilibrium, you can set $\text{i}t = -\beta$ and compute observables from
$$ \langle O \rangle = \text{Tr}\left[\text{e}^{-\beta H} O \right] \, .$$
There is only one time path. Out-of-equilibrium however the
two terms $ \text{e}^{\pm \text{i} t H}$ must be taken into account. Each leads to the same action. Only the direction of time changes. The full expression for the time dependent observable is
\begin{align*} \langle O(t) \rangle & = \int \text{d}x \, \text{d}x'\, \text{d}x'' \, \text{d}x''' \, \\ & \qquad \times \left. < x \right| \rho \left| x' > \right. \left. < x' \right| \text{e}^{\text{i} t H} \left| x'' > \right. \left. < x'' \right| O \left| x''' > \right. \, \left. < x''' \right| \text{e}^{-\text{i} t H} \left| x > \right. \, ,\\ & = \int Dx \, \text{e}^{\text{i} \left[ S_{+} + S_{-}\right]} \, \left. \, \left. < x_-(0) \right| \rho \left| x_+(0) > \right. \, < x_+(t) \right| O \left| x_-(t) \right. > \, ,\end{align*}
where $S_+ = \int_0^t \text{d}t[...]$, $S_- = \int_t^0 \text{d}t[...]$ and $x_{\pm}(t)$ are the paths on the forward and backwards paths respectively. The integrals over the boundaries, $x=x(0)$, $x'=x(0)$ $x''=x(t)$ and $x'''=x(t)$ have been absorbed into the functional integral, $\int Dx$.
This is a very quick answer. Tell me if you have trouble filling in the gaps. The generalisation from quantum mechanics to quantum field theory is straightforward. |
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Production of $K*(892)^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$ =7 TeV
(Springer, 2012-10)
The production of K*(892)$^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$=7 TeV was measured by the ALICE experiment at the LHC. The yields and the transverse momentum spectra $d^2 N/dydp_T$ at midrapidity |y|<0.5 in ...
Transverse sphericity of primary charged particles in minimum bias proton-proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV
(Springer, 2012-09)
Measurements of the sphericity of primary charged particles in minimum bias proton--proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV with the ALICE detector at the LHC are presented. The observable is linearized to be ...
Pion, Kaon, and Proton Production in Central Pb--Pb Collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012-12)
In this Letter we report the first results on $\pi^\pm$, K$^\pm$, p and pbar production at mid-rapidity (|y|<0.5) in central Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV, measured by the ALICE experiment at the LHC. The ...
Measurement of prompt J/psi and beauty hadron production cross sections at mid-rapidity in pp collisions at root s=7 TeV
(Springer-verlag, 2012-11)
The ALICE experiment at the LHC has studied J/ψ production at mid-rapidity in pp collisions at s√=7 TeV through its electron pair decay on a data sample corresponding to an integrated luminosity Lint = 5.6 nb−1. The fraction ...
Suppression of high transverse momentum D mesons in central Pb--Pb collisions at $\sqrt{s_{NN}}=2.76$ TeV
(Springer, 2012-09)
The production of the prompt charm mesons $D^0$, $D^+$, $D^{*+}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at the LHC, at a centre-of-mass energy $\sqrt{s_{NN}}=2.76$ TeV per ...
J/$\psi$ suppression at forward rapidity in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012)
The ALICE experiment has measured the inclusive J/ψ production in Pb-Pb collisions at √sNN = 2.76 TeV down to pt = 0 in the rapidity range 2.5 < y < 4. A suppression of the inclusive J/ψ yield in Pb-Pb is observed with ...
Production of muons from heavy flavour decays at forward rapidity in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012)
The ALICE Collaboration has measured the inclusive production of muons from heavy flavour decays at forward rapidity, 2.5 < y < 4, in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV. The pt-differential inclusive ...
Particle-yield modification in jet-like azimuthal dihadron correlations in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012-03)
The yield of charged particles associated with high-pT trigger particles (8 < pT < 15 GeV/c) is measured with the ALICE detector in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV relative to proton-proton collisions at the ...
Measurement of the Cross Section for Electromagnetic Dissociation with Neutron Emission in Pb-Pb Collisions at √sNN = 2.76 TeV
(American Physical Society, 2012-12)
The first measurement of neutron emission in electromagnetic dissociation of 208Pb nuclei at the LHC is presented. The measurement is performed using the neutron Zero Degree Calorimeters of the ALICE experiment, which ... |
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Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV
(Elsevier, 2013-04-10)
The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (|y| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Elliptic flow of muons from heavy-flavour hadron decays at forward rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(Elsevier, 2016-02)
The elliptic flow, $v_{2}$, of muons from heavy-flavour hadron decays at forward rapidity ($2.5 < y < 4$) is measured in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The scalar ...
Centrality dependence of the pseudorapidity density distribution for charged particles in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2013-11)
We present the first wide-range measurement of the charged-particle pseudorapidity density distribution, for different centralities (the 0-5%, 5-10%, 10-20%, and 20-30% most central events) in Pb-Pb collisions at $\sqrt{s_{NN}}$ ...
Beauty production in pp collisions at √s=2.76 TeV measured via semi-electronic decays
(Elsevier, 2014-11)
The ALICE Collaboration at the LHC reports measurement of the inclusive production cross section of electrons from semi-leptonic decays of beauty hadrons with rapidity |y|<0.8 and transverse momentum 1<pT<10 GeV/c, in pp ... |
Siril processing tutorial Convert your images in the FITS format Siril uses (image import) Work on a sequence of converted images Pre-processing images Registration (Global star alignment) → Stacking Stacking
The final step to do with Siril is to stack the images. Go to the "stacking" tab, indicate if you want to stack all images, only selected images or the best images regarding the value of FWHM previously computed. Siril proposes several algorithms for stacking computation.
Sum Stacking
This is the simplest algorithm: each pixel in the stack is summed using 32-bit precision, and the result is normalized to 16-bit. The increase in signal-to-noise ratio (SNR) is proportional to [math]\sqrt{N}[/math], where [math]N[/math] is the number of images. Because of the lack of normalisation, this method should only be used for planetary processing.
Average Stacking With Rejection Percentile Clipping: this is a one step rejection algorithm ideal for small sets of data (up to 6 images). Sigma Clipping: this is an iterative algorithm which will reject pixels whose distance from median will be farthest than two given values in sigma units ([math]\sigma_{low}[/math], [math]\sigma_{high}[/math]). Median Sigma Clipping: this is the same algorithm except than the rejected pixels are replaced by the median value of the stack. Winsorized Sigma Clipping: this is very similar to Sigma Clipping method but it uses an algorithm based on Huber's work [1] [2]. Linear Fit Clipping: this is an algorithm developed by Juan Conejero, main developer of PixInsight [2]. It fits the best straight line ([math]y=ax+b[/math]) of the pixel stack and rejects outliers. This algorithm performs very well with large stacks and images containing sky gradients with differing spatial distributions and orientations.
These algorithms are very efficient to remove satellite/plane tracks.
Median Stacking
This method is mostly used for dark/flat/offset stacking. The median value of the pixels in the stack is computed for each pixel. As this method should only be used for dark/flat/offset stacking, it does not take into account shifts computed during registration. The increase in SNR is proportional to [math]0.8\sqrt{N}[/math].
Pixel Maximum Stacking
This algorithm is mainly used to construct long exposure star-trails images. Pixels of the image are replaced by pixels at the same coordinates if intensity is greater.
Pixel Minimum Stacking
This algorithm is mainly used for cropping sequence by removing black borders. Pixels of the image are replaced by pixels at the same coordinates if intensity is lower.
In the case of NGC7635 sequence, we first used the "Winsorized Sigma Clipping" algorithm in "Average stacking with rejection" section, in order to remove satellite tracks ([math]\sigma_{low}=4[/math] and [math]\sigma_{high}=3[/math]).
The output console thus gives the following result:
22:26:06: Pixel rejection in channel #0: 0.215% - 1.401%
22:26:06: Pixel rejection in channel #1: 0.185% - 1.273% 22:26:06: Pixel rejection in channel #2: 0.133% - 1.150% 22:26:06: Integration of 12 images: 22:26:06: Normalization ............. additive + scaling 22:26:06: Pixel rejection ........... Winsorized sigma clipping 22:26:06: Rejection parameters ...... low=4.000 high=3.000 22:26:09: Saving FITS: file NGC7635.fit, 3 layer(s), 4290x2856 pixels 22:26:19: Background noise value (channel: #0): 10.013 (1.528e-04) 22:26:19: Background noise value (channel: #1): 6.755 (1.031e-04) 22:26:19: Background noise value (channel: #2): 6.621 (1.010e-04)
After that, the result is saved in the file named below the buttons, and is displayed in the grey and colour windows. You can adjust levels if you want to see it better, or use the differe1nt display mode. In our example the file is the stack result of all files, i.e., 12 files.
The images above picture the result in Siril using the Auto-Stretch rendering mode. Note the improvement of the signal-to-noise ratio regarding the result given for one frame in the previous step (take a look to the sigma value). The increase in SNR is of [math]19.7/6.4 = 3.08 \approx \sqrt{12} = 3.46[/math] and you should try to improve this result adjusting [math]\sigma_{low}[/math] and [math]\sigma_{high}[/math].
Now should start the process of the image with crop, background extraction (to remove gradient), and some other processes to enhance your image. To see processes available in Siril please visit this page. Here an example of what you can get with Siril:
Peter J. Huber and E. Ronchetti (2009), Robust Statistics, 2nd Ed., Wiley Juan Conejero, ImageIntegration, Pixinsight Tutorial |
Rationalize $\frac{2}{6-5\sqrt{3}}$
Jamb Maths 2006
If $(k{{2}_{6}})\times 6={{3}_{5}}{{(k4)}_{5}}$ what is the value of
k
In a small village of 500 people, 350 speak the local language, while 200 speak Pidgin English. What is the percentage of the population speak both.
Simplify $\left( \tfrac{7}{9}-\tfrac{2}{3} \right)\div \left( \tfrac{1}{3}+\tfrac{2}{5}\div \tfrac{4}{5} \right)$
If $E\subseteq G\subseteq U,$where
U is the universal set, then the shaded venn diagram representing U – E or E c
Compute 110011
2 + 11111 2
Simplify ${{(25)}^{-\tfrac{1}{3}}}\times {{(27)}^{\tfrac{1}{3}}}+{{(121)}^{-\tfrac{1}{4}}}\times {{(625)}^{-\tfrac{1}{4}}}$
Find the tax on th income of N20,000. IF no tax is paid on the first N10,000 and tax is paid N50 in N1000 on the next N5000 and at N55 in N1000 on the remainder.
Calculate the logarithm to base 9 of ${{3}^{-4}}\times {{9}^{2}}\times {{(81)}^{-1}}$
Convert 2232
4 to a number in base six |
During my PhD thesis, I have partly worked on the problem of the automatic accurate test data generation. In order to be complete and self-contained, I have addressed all kinds of data types, including strings. This article is the first one of a little series that aims at showing how to generate accurate and relevant strings under several constraints.
What is a regular expression?
We are talking about formal language theory here. In the known world, there are four kinds of languages. More formally, in 1956, the Chomsky hierarchy has been formulated, classifying grammars (which define languages) in four levels:
unrestricted grammars, matching langages known as Turing languages, no restriction, context-sensitive grammars, matching contextual languages, context-free grammars, matching algebraic languages, based on stacked automata, regular grammars, matching regular languages.
Each level includes the next level. The last level is the “weaker”, which must not sound negative here. Regular expressions are used often because of their simplicity and also because they solve most problems we encounter daily.
A regular expression is a small language with very few operators and, most of the time, a simple semantics. For instance
ab(c|d) means: a word (a data) starting by
ab and followed by
c or
d. We also have quantification operators (also known as repetition operators), such as
?,
* and
+. We also have
{ to define a repetition between
x, y}
and
x
. Thus,
y
? is equivalent to
{0,1},
* to
{0,} and
+ to
{1,}. When
is missing, it means \displaystyle +\infty , so unbounded (or more exactly, bounded by the limits of the machine). So, for instance
y
ab(c|d){2,4}e? means: a word starting by
ab, followed 2, 3 or 4 times by
c or
d (so
cc,
cd,
dc,
ccc,
ccd,
cdc and so on) and potentially followed by
e.
The goal here is not to teach you regular expressions but this is kind of a tiny reminder. There are plenty of regular languages. You might know POSIX regular expression or Perl Compatible Regular Expressions (PCRE). Forget the first one, please. The syntax and the semantics are too much limited. PCRE is the regular language I recommend all the time.
Behind every formal language there is a graph. A regular expression is compiled into a Finite State Machine (FSM). I am not going to draw and explain them, but it is interesting to know that behind a regular expression there is a basic automaton. No magic.
Why focussing regular expressions?
This article focuses on regular languages instead of other kind of languages because we use them very often (even daily). I am going to address context-free languages in another article, be patient young padawan. The needs and constraints with other kind of languages are not the same and more complex algorithms must be involved. So we are going easy for the first step.
Understanding PCRE: lex and parse them
The
Hoa\Compiler library provides both \displaystyle LL(1) and \displaystyle LL(k) compiler-compilers. The documentation describes how to use it. We discover that the \displaystyle LL(k) compiler comes with a grammar description language called PP. What does it mean? It means for instance that the grammar of the PCRE can be written with the PP language and that
Hoa\Compiler\Llk will transform this grammar into a compiler. That’s why we call them “compiler of compilers”.
Fortunately, the
Hoa\Regex library provides the grammar of the PCRE language in the
hoa://Library/Regex/Grammar.pp file. Consequently, we are able to analyze regular expressions written in the PCRE language! Let’s try in a shell at first with the
hoa compiler:pp tool:
$ echo 'ab(c|d){2,4}e?' | hoa compiler:pp hoa://Library/Regex/Grammar.pp 0 --visitor dump> #expression> > #concatenation> > > token(literal, a)> > > token(literal, b)> > > #quantification> > > > #alternation> > > > > token(literal, c)> > > > > token(literal, d)> > > > token(n_to_m, {2,4})> > > #quantification> > > > token(literal, e)> > > > token(zero_or_one, ?)
We read that the whole expression is composed of a single concatenation of two tokens:
a and
b, followed by a quantification, followed by another quantification. The first quantification is an alternation of (a choice betwen) two tokens:
c and
d, between 2 to 4 times. The second quantification is the
e token that can appear zero or one time. Pretty simple.
The final output of the
Hoa\Compiler\Llk\Parser class is an Abstract Syntax Tree (AST). The documentation of
Hoa\Compiler explains all that stuff, you should read it. The \displaystyle LL(k) compiler is cut out into very distinct layers in order to improve hackability. Again, the documentation teach us we have four levels in the compilation process: lexical analyzer, syntactic analyzer, trace and AST. The lexical analyzer (also known as lexer) transforms the textual data being analyzed into a sequence of tokens (formally known as lexemes). It checks whether the data is composed of the good pieces. Then, the syntactic analyzer (also known as parser) checks that the order of tokens in this sequence is correct (formally we say that it derives the sequence, see the Matching words section to learn more).
Still in the shell, we can get the result of the lexical analyzer by using the
--token-sequence option; thus:
$ echo 'ab(c|d){2,4}e?' | hoa compiler:pp hoa://Library/Regex/Grammar.pp 0 --token-sequence # … token name token value offset----------------------------------------- 0 … literal a 0 1 … literal b 1 2 … capturing_ ( 2 3 … literal c 3 4 … alternation | 4 5 … literal d 5 6 … _capturing ) 6 7 … n_to_m {2,4} 7 8 … literal e 12 9 … zero_or_one ? 13 10 … EOF 15
This is the sequence of tokens produced by the lexical analyzer. The tree is not yet built because this is the first step of the compilation process. However this is always interesting to understand these different steps and see how it works.
Now we are able to analyze any regular expressions in the PCRE format! The result of this analysis is a tree. You know what is fun with trees? Visiting them.
Visiting the AST
Unsurprisingly, each node of the AST can be visited thanks to the
Hoa\Visitor library. Here is an example with the “dump” visitor:
use Hoa\Compiler;use Hoa\File;// 1. Load grammar.$compiler = Compiler\Llk\Llk::load( new File\Read('hoa://Library/Regex/Grammar.pp'));// 2. Parse a data.$ast = $compiler->parse('ab(c|d){2,4}e?');// 3. Dump the AST.$dump = new Compiler\Visitor\Dump();echo $dump->visit($ast);
This program will print the same AST dump we have previously seen in the shell.
How to write our own visitor? A visitor is a class with a single
visit method. Let’s try a visitor that pretty print a regular expression, i.e. transform:
ab(c|d){2,4}e?
into:
ab( c | d){2,4}e?
Why a pretty printer? First, it shows how to visit a tree. Second, it shows the structure of the visitor: we filter by node ID (
#expression,
#quantification,
token etc.) and we apply respective computations. A pretty printer is often a good way for being familiarized with the structure of an AST.
Here is the class. It catches only useful constructions for the given example:
use Hoa\Visitor;class PrettyPrinter implements Visitor\Visit { public function visit ( Visitor\Element $element, &$handle = null, $eldnah = null ) { static $_indent = 0; $out = null; $nodeId = $element->getId(); switch($nodeId) { // Reset indentation and… case '#expression': $_indent = 0; // … visit all the children. case '#quantification': foreach($element->getChildren() as $child) $out .= $child->accept($this, $handle, $eldnah); break; // One new line between each children of the concatenation. case '#concatenation': foreach($element->getChildren() as $child) $out .= $child->accept($this, $handle, $eldnah) . "\n"; break; // Add parenthesis and increase indentation. case '#alternation': $oout = []; $pIndent = str_repeat(' ', $_indent); ++$_indent; $cIndent = str_repeat(' ', $_indent); foreach($element->getChildren() as $child) $oout[] = $cIndent . $child->accept($this, $handle, $eldnah); --$_indent; $out .= $pIndent . '(' . "\n" . implode("\n" . $cIndent . '|' . "\n", $oout) . "\n" . $pIndent . ')'; break; // Print token value verbatim. case 'token': $tokenId = $element->getValueToken(); $tokenValue = $element->getValueValue(); switch($tokenId) { case 'literal': case 'n_to_m': case 'zero_or_one': $out .= $tokenValue; break; default: throw new RuntimeException( 'Token ID ' . $tokenId . ' is not well-handled.' ); } break; default: throw new RuntimeException( 'Node ID ' . $nodeId . ' is not well-handled.' ); } return $out; }}
And finally, we apply the pretty printer on the AST like previously seen:
$compiler = Compiler\Llk\Llk::load( new File\Read('hoa://Library/Regex/Grammar.pp'));$ast = $compiler->parse('ab(c|d){2,4}e?');$prettyprint = new PrettyPrinter();echo $prettyprint->visit($ast);
Et voilà !
Now, put all that stuff together!
Isotropic generation
We can use
Hoa\Regex and
Hoa\Compiler to get the AST of any regular expressions written in the PCRE format. We can use
Hoa\Visitor to traverse the AST and apply computations according to the type of nodes. Our goal is to generate strings based on regular expressions. What kind of generation are we going to use? There are plenty of them: uniform random, smallest, coverage based…
The simplest is isotropic generation, also known as random generation. But random says nothing: what is the repartition, or do we have any uniformity? Isotropic means each choice will be solved randomly and uniformly. Uniformity has to be defined: does it include the whole set of nodes or just the immediate children of the node? Isotropic means we consider only immediate children. For instance, a node
#alternation has \displaystyle c^1 immediate children, the probability \displaystyle C to choose one child is:
\displaystyle P(C) = \frac{1}{c^1}
Yes, simple as that!
We can use the
Hoa\Math library that provides the
Hoa\Math\Sampler\Random class to sample uniform random integers and floats. Ready?
Structure of the visitor
The structure of the visitor is the following:
use Hoa\Visitor;use Hoa\Math;class IsotropicSampler implements Visitor\Visit { protected $_sampler = null; public function __construct ( Math\Sampler $sampler ) { $this->_sampler = $sampler; return; } public function visit ( Visitor\Element $element, &$handle = null, $eldnah = null ) { switch($element->getId()) { // … } }}
We set a sampler and we start visiting and filtering nodes by their node ID. The following code will generate a string based on the regular expression contained in the
$expression variable:
$expression = '…';$ast = $compiler->parse($expression);$generator = new IsotropicSampler(new Math\Sampler\Random());echo $generator->visit($ast);
We are going to change the value of
$expression step by step until having
ab(c|d){2,4}e?.
Case of
#expression
A node of type
#expression has only one child. Thus, we simply return the computation of this node:
case '#expression': return $element->getChild(0)->accept($this, $handle, $eldnah); break;
Case of
token
We consider only one type of token for now:
literal. A literal can contain an escaped character, can be a single character or can be
. (which means everything). We consider only a single character for this example (spoil: the whole visitor already exists). Thus:
case 'token': return $element->getValueValue(); break;
Here, with
$expression = 'a'; we get the string
a.
Case of
#concatenation
A concatenation is just the computation of all children joined in a single piece of string. Thus:
case '#concatenation': $out = null; foreach($element->getChildren() as $child) $out .= $child->accept($this, $handle, $eldnah); return $out; break;
At this step, with
$expression = 'ab'; we get the string
ab. Totally crazy.
Case of
#alternation
An alternation is a choice between several children. All we have to do is to select a child based on the probability given above. The number of children for the current node can be known thanks to the
getChildrenNumber method. We are also going to use the sampler of integers. Thus:
case '#alternation': $childIndex = $this->_sampler->getInteger( 0, $element->getChildrenNumber() - 1 ); return $element->getChild($childIndex) ->accept($this, $handle, $eldnah); break;
Now, with
$expression = 'ab(c|d)'; we get the strings
abc or
abd at random. Try several times to see by yourself.
Case of
#quantification
A quantification is an alternation of concatenations. Indeed,
e{2,4} is strictly equivalent to
ee|eee|eeee. We have only two quantifications in our example:
? and
{. We are going to find the value for
x, y}
and
x
and then choose at random between these bounds. Let’s go:
y
case '#quantification': $out = null; $x = 0; $y = 0; // Filter the type of quantification. switch($element->getChild(1)->getValueToken()) { // ? case 'zero_or_one': $y = 1; break; // {x,y} case 'n_to_m': $xy = explode( ',', trim($element->getChild(1)->getValueValue(), '{}') ); $x = (int) trim($xy[0]); $y = (int) trim($xy[1]); break; } // Choose the number of repetitions. $max = $this->_sampler->getInteger($x, $y); // Concatenate. for($i = 0; $i < $max; ++$i) $out .= $element->getChild(0)->accept($this, $handle, $eldnah); return $out; break;
Finally, with
$expression = 'ab(c|d){2,4}e?'; we can have the following strings:
abdcce,
abdc,
abddcd,
abcde etc. Nice isn’t it? Want more?
for($i = 0; $i < 42; ++$i) echo $generator->visit($ast), "\n";/** * Could output: * abdce * abdcc * abcdde * abcdcd * abcde * abcc * abddcde * abddcce * abcde * abcc * abdcce * abcde * abdce * abdd * abcdce * abccd * abdcdd * abcdcce * abcce * abddc */
Performance
This is difficult to give numbers because it depends of a lot of parameters: your machine configuration, the PHP VM, if other programs run etc. But I have generated 1 million ( \displaystyle 10^6 ) strings in less than 25 seconds on my machine (an old MacBook Pro), which is pretty reasonable.
Conclusion and surprise
So, yes, now we know how to generate strings based on regular expressions! Supporting all the PCRE format is difficult. That’s why the
Hoa\Regex library provides the
Hoa\Regex\Visitor\Isotropic class that is a more advanced visitor. This latter supports classes, negative classes, ranges, all quantifications, all kinds of literals (characters, escaped characters, types of characters —
\w,
\d,
\h…—) etc. Consequently, all you have to do is:
use Hoa\Regex;// …$generator = new Regex\Visitor\Isotropic(new Math\Sampler\Random());echo $generator->visit($ast);
This algorithm is used in Praspel, a specification language I have designed during my PhD thesis. More specifically, this algorithm is used inside realistic domains. I am not going to explain it today but it allows me to introduce the “surprise”.
Generate strings based on regular expressions in atoum
atoum is an awesome unit test framework. You can use the
Atoum\PraspelExtension extension to use Praspel and therefore realistic domains inside atoum. You can use realistic domains to validate
and to generate data, they are designed for that. Obviously, we can use the
Regex realistic domain. This extension provides several features including
sample,
sampleMany and
predicate to respectively generate one datum, generate many data and validate a datum based on a realistic domain. To declare a regular expression, we must write:
$regex = $this->realdom->regex('/ab(c|d){2,4}e?/');
And to generate a datum, all we have to do is:
$datum = $this->sample($regex);
For instance, imagine you are writing a test called
test_mail and you need an email address:
public function test_mail ( ) { $this ->given( $regex = $this->realdom->regex('/[\w\-_]+(\.[\w\-\_]+)*@\w\.(net|org)/'), $address = $this->sample($regex), $mailer = new \Mock\Mailer(…), ) ->when($mailer->sendTo($address)) ->then ->…}
Easy to read, fast to execute and help to focus on the logic of the test instead of test data (also known as fixtures). Note that most of the time the regular expressions are already in the code (maybe as constants). It is therefore easier to write and to maintain the tests.
I hope you enjoyed this first part of the series :-)! This work has been published in the International Conference on Software Testing, Verification and Validation: Grammar-Based Testing using Realistic Domains in PHP. |
Suppose that I have two charged particles in the configuration below.
Let us assume the following:
We apply a constant force $f$ to the the bottom particle so that it has a constant acceleration $a(t)=f/m$. The velocity of the bottom particle is negligible to simplify the calculation of the electric field. The top particle is initially stationary with a large mass $M$. The distance $r$ is large enough so that the Coulomb repulsion between the particles, which is inversely proportional to $r^2$, is negligible.
Under these conditions the Lienard-Wiechert retarded electric field due to the bottom particle, accelerating at time $t=0$, produces a force $F$ on the top particle, at a later time $t=r/c$, given by:
$$F(t=r/c)=\frac{qQa(t=0)}{4\pi\epsilon_0c^2r}.$$
Let us say that in a time interval $\Delta t$ the top particle gains a momentum $F\Delta t$ towards the left.
My question is the following: How is this momentum change balanced?
The conventional answer is to say that the EM field gains an opposite momentum to the right.
But the only way I can see the EM field changing is if the top particle accelerates to the left, under the action of $F$, producing its own counter electric field towards the right. However this won't work as the mass $M$ of the top particle is assumed to be large so that its acceleration, and thus its induced electric field, is negligible.
In summary I can see how the electric field in the vicinity of the top particle can transfer momentum to it but I can't see any mechanism whereby the top particle transfers momentum back to the field if we are free to make the assumption that it is so heavy that it can absorb the momentum without changing its motion appreciably.
P.S. My hypothesis is that a balancing momentum $F\Delta t$ to the right is transmitted backwards in time from the top particle at time $t=r/c$ to the bottom particle at time $t=0$ using an advanced electromagnetic interaction. This momentum then has the effect of reducing the effective mass of the bottom particle (less external force has to be supplied to produce a given acceleration $a$). |
A tweet has got me musing again. @ColinTheMathmo sent out a tweet making a seemingly audacious claim that from just two data points he could deduce all the coefficients and powers in a polynomial that you had defined as long as the coefficients and powers were zero or positive integers
You pick a polynomial with coeffs in Z & >=0. I give you x, you tell me the value. We do that again. Then I can name your poly. How?
— Colin Wright (@ColinTheMathmo) October 31, 2012
i.e. you pick a polynomial of the form:
\begin{equation} y = \sum {a_i \cdot x^i} \qquad for \qquad i \in Z^*, a_i \in Z^* \qquad (1)\end{equation}
He then supplies an \(x\) value, you return the corresponding \(y\) value, this procedure is repeated
once only and then he can tell you all the \(a_i\) values in your polynomial.
I looked at this tweet on the way to work one day and starting thinking about it. I wondered whether you cleverly supplied some complex numbers for the \(x\) values which could indicate the powers of x in use, but Colin assured me that the \(x\) values could be in \(Z\) too. In my head, I started drawing graphs. Let’s call the first number asked for \(x_1\). I could see that, given \(y_1\) for that value, there are a finite number of graphs for polynomials satisfying \((1) \) that go through that point and that for the constraints given, these graphs were all monotonically increasing in \(x\).
I could see all this in my mind’s eye, but couldn’t quite work out how that would let me uniquely identify each one – i.e. at what point all those lines no longer cross. So, when I had a minute, I wrote a little python program to spit some graphs out. Below is one graph for a particular sample pair (\(x_1=2\),\(y_1=41\)) showing all the possible polynomials satisfying \((1)\) that go through that point. For those who would like to look at the graphs in a bit more detail – there’s a gallery at the end of the post with various \(x,y\) pairs and zoom levels. Scrappy Python code to generate these is also available on request.
Click the graph to see a full size, better resolution, image
From the picture above and a bit of reasoning that any higher powers of \(x\) will cross all curves at lower values of \(x\), I’m pretty convinced that for any point you ask for, if you work out the shallowest polynomial of order 2 through the point and the steepest straight line and then find their intersection, all the graphs beyond that point are distinct, will never cross and are monotonically increasing in \(x\).
With this insight, if I define
\begin{aligned} b & ={\lfloor}{\frac{y_1}{x_1}}{\rfloor} \\ c & = y_1-{x_1}^2 – (y_1 mod x_1) \end{aligned} The intersection is at \begin{equation}x = \frac{b + \sqrt{b^2 – 4c}}{2} \end{equation} So, we ask for \(x_2\) greater than that value and the polynomial is uniquely identified. For instance, in the example above, this works out to \(x_2 \geq 18\) How to then get back to each co-efficient is interesting, but basically a search problem.
Now, Colin has reliably informed me that this is
too complicated and also challenged me as to whether I can prove it will always work – which I’m not sure I can, but I think the above sort of way showsit.
I’m going to have a think about a few of my other ideas, which are (in no particular order)
ask for 1 as the first x (\(x_1 = 1\)), thus giving you at least the sum of all co-efficients. I have a feeling this might be useful put in the value that you get back from the first answer as the second value you ask for (not sure why I think that might be good – I doubt it really) find a value of \(x_2\) which is \(f(x_1)\) but lower than the value outlined above which can similarly always have the graphs distinct. I’m not hopeful. Gallery of graphs |
Problem
When train my
linear chain CRF with annotated observations, I feed it with a number of sequences containing observation values and a "ground-truth" label for each observation. I'm currently using the hCRF Matlab interface. (see 1)
In my case I have 4 continuous observation values (some in the interval [-1,1], some around [140, 200]; none outside [-10, 250] though). The label is an integer between 1 and 17, giving a total of 17 possible labels.
Question
After training, my model consists of 17*17 = 289 edge weights and 17*4 = 68 window weights. Can I understand the edge weights as some sort of probabilities for state transitions (even though they are not actual probabilities in the range [0,1])? And what exactly do the edge weights tell me?
Reference
For reference, the CRF produces conditional probabilities of the form
$$p(y|x) = \frac{1}{Z(x)} \exp{ \left\{ \sum_{k=1}^{K} \lambda_k f_k(y, x) \right\}}$$
where to my understanding, the $\lambda_k$ are my edge weights and the $Z(x)$ are the window weights I get.
I read some tutorials about CRFs but I still do not quite understand, how the feature functions $f_k(y, x)$ are looking if I simply put in sequences of 4 tuples (the 4 observation values).
Thanks in advance for any pointers. |
One way to look at this is to enumerate several sets.
Firstly, how many permutations of
california contain
aa and
ii? Let's call that set $C$, and its size is $|C|$.
Secondly, how many contain
aa? And how many
ii? Let's call these sets $A$ and $I$.
Let's call the total number of permutations $U$. The set of permutations which contain
no consecutive letters is therefore $U - A - I + C$. And so the number we are looking for is $|U| - |A| - |I| + |C|$.
We must add back $C$ because it's the overlapping area in the Venn diagram. That is to say, both set $A$ and set $I$ contain $C$, and so if we subtract them from $U$ via set difference, we end up subtracting $C$ twice. $A$ contains $C$ because the set of all permutations of
california which contain
aa includes all permutations that contain
ii also, and that subset of $A$ corresponds to $C$.
Let's deal with set C:
This can be answered as: given an array of ten places, how many ways can we place the digraph
aa and
ii into that array? Each of these ways leaves six spaces, which we then fill with permutations of
cflnor: the remaining letters in
california. We multiply together these possibilities: that is to say $6!$ times the number of ways of filling those digraphs.
Now, there are 9 ways to place
aa into the array, obviously. Out of these 9 ways, the edge placements
aa........ and
........aa, support 7 ways of adding a
ii. All 7 interior placements support 6 ways of introducing
ii. So the possibilities are $2\times 7 + 7\times 6 = 56$.
Hence $|C| = 56\times 6! = 56\times 720 = 40320$.
Some diagrams:
Edge
aa-fill:
aaii...... 1
aa.ii..... 2
aa..ii.... 3
aa...ii... 4
aa....ii.. 5
aa.....ii. 6
aa......ii 7
Interior
aa-fill:
.aaii..... 1
.aa.ii.... 2
.aa..ii... 3
.aa...ii.. 4
.aa....ii. 5
.aa.....ii 6
iiaa...... 1
..aaii.... 2
..aa.ii... 3
..aa..ii.. 4
..aa...ii. 5
..aa....ii 6
Now let's deal with A and I. They are easy mirror cases. Basically, we have nine ways to place
aa into ten spaces, leaving eight spaces, which we then fill with permutations of the remaining letters. Thus $|A| = 9\times 8! = 9! = 362880$ and also $|I| = 362880$.
Of course $|U| = 10! = 3628800$.
We can now calculate $|U| - |A| - |I| + |C| = 3628800 - 2\times 362880 + 40320 = 2943360$.
Oops: does not validate by machine:
$ txr -i
1> (count-if (notf (op search-regex @1 #/aa|ii/)) (perm "california"))
2338560
The correct answer is 2338560.
To be continued ...
(Math has cliffhangers too!)
And here is where I screwed up! In calculating the cardinalities $|C|$ and $|A| = |I|$, I assumed that there is only one distinct
aa and
ii: that both
a-s are the same object. But in fact they are distinct: there are two
aa digraphs and two
ii digraphs (I'm dealing with permutations which need not be distinct). Therefore, set $|C|$ is actually four times larger, due the four combinations of digraphs, hence $4\times 56\times 720 = 161280$. Similarly, $|A|$ and $|I|$ are twice as large: $2\times 9!$. With these corrections, $|U| - |A| - |I| + |C| = 2338560$.
Now suppose we want to consider only distinct permutations, so that there is exactly one digraph
aa and one digraph
ii. The cardinality of the new $|U|$ is reduced by a factor of four, since strings that differ only in exchanges of one
a with the other, and one
i with the other are equivalent. It is $|U| = \frac{10!}{2} = 907200$. The original $|C| = 40320$ value is correct, since that calculation considered both
i-s and
a-s as indistinct. The correct $|A|$ calculation is $|A| = |I| = 9!\div 2 = 181440$: we must divide by two because the remaining letters after the digraph is positioned include two which are indistinct. And so $|U| - |A| - |I| + |C| = 907200 - 2\times 181440 + 40320 = 584640$.
Check by machine:
3> (count-if (notf (op search-regex @1 #/aa|ii/)) (uniq (perm "california")))
584640
So those are the answers: if the
i-s and
a-s are considered distinct, then $2338560$ permutations don't contain a digraph. If they are considered indistinct, then $584640$ permutations don't contain a digraph. |
How to prove that space $\mathbb{R}_w$, the countably infinite product of $\mathbb{R}$ in the box topology, is not metrizable? I have tried finding a solution to this problem, but failed. Kindly help me find this answer.
At no point of this product in the box topology, the space is first countable: e.g. for $p=(p_1, p_2,p_3,\ldots)$: suppose $\{U_n: n \in \mathbb{N}\}$ is a local base at $p$. Every $U_n$ contains an open box around $p$ and as all open sets in $\mathbb{R}$ are unions of open intervals, for each $m$:
$$\exists r^{(m)}_1,r^{(m)}_2, \ldots,r^{(m)}_i,\ldots: p \in \prod_i (p_i - r^{(m)}_i, p_i + r^{(m)}_i) \subseteq U_m$$.
Then define $O = \prod_i (p_i - \frac{1}{2} r^{(i)}_i, p_i + \frac{1}{2}r^{(i)}_i)$ which is box-open, and contains $p$. But no $U_n$ can be a subset of $O$ ($U_n$ fails at the $n$-th coordinate), and so the $U_n$ cannot form a local base at $p$.
All metric spaces do have local countable bases everywhere: $\{B(p, \frac{1}{n}):n \in \mathbb{N}\}$ will do.
So this box product is not metrisable.
Hint: Do it by contradiction method. You must have learnt the sequence lemma:It says " If A a subset of a topological space X and there is a sequence in A converging to a point x then $x\in cl (A) $. The converse holds if X is metrizable". So assuming $\mathbb {R} ^{\omega}$ is metrizable try to find a contradiction to this theorem(i.e., prove the converse of the above Lemma isn't true.)
Try it with the set $$A=\{(x_1,x_2,\cdots)| x_i>0 for\, all \, i\in \mathbb {N}\} $$. $\mathbf {0} $ is a limit point of this set but there is no sequence of points in A that converges to $\mathbf {0} $ in box topology. Hope it helps? |
I'm trying to find functions $f,g \in L^p(\mathbb{R})$ with $$||f+g||_p > ||f||_p+||g||_p$$ where $p \in (0,1)$. All my ideas failed so fair, any help and hints appreciated!
Take $f$ supported on $[0,1]$ and equal to $1$, $g$ supported on $[1,2]$ and equal to $a>0$.
Then $\|f+g\|_p=(1+a^p)^{1/p}$, $\|f\|_p=1$, $\|g\|_p=a$.
We want $1+a^p > (1+a)^p$. By binomial expansion, the RHS is only $1+a/p+O(a^2)$ which is less than $1+a^p$ for $a>0$ small enough.
For who doesn't like big $O$ estimates: it suffices to prove that $$\lim_{a \to 0+}\frac{(1+a)^p-1}{a^p} < 1$$ We have by definition of derivative: $$\lim_{a \to 0+}\frac{(1+a)^p-1}{a} = p$$ and so for $p<1$: $$\lim_{a \to 0+}\frac{(1+a)^p-1}{a^p} = p \cdot \lim_{a \to 0+}a^{1-p}=0$$
The two are essentially the same, the only difference being that Taylor expansion gives the error $O(a^2)$ while the definition of derivative gives $o(a)$, which is sufficient. |
Waecmaths
Title waecmaths question Question 24
If $\sin x=\tfrac{5}{13}$ and ${{0}^{\circ }}\le x\le {{90}^{\circ }}$ find the value of $(\cos x-\tan x)$
Question 25
An object is 6m away from a mast. The angle of depression of the object from the top of the mast is 50
Question 26
The bear of
Question 29
A ship sails
Question 33
Given that $\cos x=\tfrac{12}{13}$evaluate $\frac{1-\tan x}{\tan x}$
Question 38
A man’s eye level is 1.7m above the horizontal ground and 13m from a vertical pole. If the pole is 8.3m high, calculate, correct to the nearest degree, the angle of elevation of the top of the pole from his eyes
Question 18
In diagram \[\left| QR \right|=10m,\text{ }\left| SR \right|=8m,\]\[\angle QPS={{30}^{\circ }},\angle QRP={{90}^{\circ }}\text{ and }\left| PS \right|=x\]find
Question 24
If cos(
Question 25
A kite on a taut string of length 50m inclined at an angle of 54
Question 26
The position of three ships
Question 3
Given that $\cos {{x}^{\circ }}=\tfrac{1}{r}$ express tan
Question 12
Esther was facing S 20
Question 49
A boy looks through a window of a building and sees a man fruits on the ground 50
Question 25
Given that $\tan x=1$ where ${{0}^{\circ }}\le x\le {{90}^{\circ }}$ evaluate $\frac{1-{{\sin }^{2}}x}{\cos x}$
Question 26
If $\sin 3y=\cos 2y$ and ${{0}^{\circ }}\le y\le {{90}^{\circ }}$, find the value of
Question 46
In the diagram $\angle WOX={{60}^{\circ }}$, $\angle YOE={{50}^{\circ }}$and $\angle OXY={{30}^{\circ }}$what is the bearing of
Question 21
Kweku walked 8
Question 26
In the diagram
Question 27
If $\cos \theta =x$and $\sin {{60}^{\circ }}=x+0.5$ ${{0}^{\circ }}<\theta <{{90}^{\circ }}$find correct to the nearest degree, the value of
Question 49
An object is 6m from the base of a mast. If the angle of depression of the object from the top of the mast is 50
Question 25
Given that $\sin {{60}^{\circ }}=\tfrac{\sqrt{3}}{2}\text{ and }\cos {{60}^{\circ }}=\tfrac{1}{2}$ evaluate $\frac{1-\sin {{60}^{\circ }}}{1+\cos {{60}^{\circ }}}$
Question 26
Find the bearing of
Question 27
If $\left| XY \right|=50m$ how far east of
Question 16
In the diagram $\angle QPR={{90}^{\circ }}$ , if ${{q}^{2}}=25-{{r}^{2}}$ find the value of
Question 26
If ${{x}^{o}}$ is obtuse, which of the following is true?
Question 27
if $\tan x=1$evaluate $\sin x+\cos x$ leaving your answer in surd form
Question 28
If $\cos {{(x+25)}^{o}}=\sin {{45}^{o}}$, find the value of
Question 17
In the diagram $\left| LN \right|=4cm,\angle LNM={{90}^{\circ }}$ and $\tan y=\tfrac{2}{3}$What is the area of $\triangle LMN$
Question 30 Question 31
In the diagram$\left| PQ \right|=4cm$, $\left| QR \right|=6cm,\text{ }\left| RS \right|=12cm$ $\angle QRS={{90}^{\circ }}$. Find the value of
Question 33
The bearing of a point
Question 40
$\triangle PQR$ is a equilateral triangle with sides $2\sqrt{3}cm$ calculate its height
Question 10
If tan
Question 13
In the diagram $\left| QR \right|=5cm$ $\angle PQR={{60}^{\circ }}$ and $\angle PSR={{45}^{\circ }}$. Find $\left| PS \right|$ leaving your answer in surd form
Question 27
In the diagram,
Question 50
A ladder 16m leans against an electric pole. If the ladder makes an angle of 65
Question 23
Given that $\tan x=\tfrac{2}{3}$, where ${{0}^{\circ }}\le x\le {{90}^{\circ }}$, find the value of $2\sin x$
Question 24 Question 25
The angle of elevation of an aircraft from a point
Question 7
From a point |
We prove that for a compact toric manifold whose anti-canonical divisor isnumerically effective, the Lagrangian Floer superpotential defined byFukaya-Oh-Ohto-Ono is equal to the superpotential written down by using thetoric mirror map under a convergence assumption. This gives a method to computeopen Gromov-Witten invariants using mirror symmetry.
Given a smooth target curve $X$, we explore the relationship betweenGromov-Witten invariants of $X$ relative to a smooth divisor and orbifoldGromov-Witten invariants of the $r$-th root stack along the divisor. We provedthat relative invariants are equal to the $r^0$-coefficient of thecorresponding orbifold Gromov-Witten invariants of $r$-th root stack for $r$sufficiently large. Our result provides a precise relation between relative andorbifold invariants of target curves generalizing the result ofAbramovich-Cadman-Wise to higher genus invariants of curves. Moreover, when $r$is sufficiently large, we proved that relative stationary invariants of $X$ areequal to the orbifold stationary invariants in all genera. Our results lead to some interesting applications: a new proof of genus zeroequality between relative and orbifold invariants of $X$ via localization; anew proof of the formula of Johnson-Pandharipande-Tseng for double Hurwitznumbers; a version of GW/H correspondence for stationary orbifold invariants.
We show that the product in the quantum K-ring of a generalized flag manifold$G/P$ involves only finitely many powers of the Novikov variables. In contrastto previous approaches to this finiteness question, we exploit the finitedifference module structure of quantum K-theory. At the core of the proof is abound on the asymptotic growth of the $J$-function, which in turn comes from ananalysis of the singularities of the zastava spaces studied in geometricrepresentation theory.
We study the higher genus equivariant Gromov-Witten theory of the Hilbertscheme of n points of the plane. Since the equivariant quantum cohomology issemisimple, the higher genus theory is determined by an R-matrix via theGivental-Teleman classification of Cohomological Field Theories (CohFTs). Weuniquely specify the required R-matrix by explicit data in degree 0. As aconsequence, we lift the basic triangle of equivalences relating theequivariant quantum cohomology of the Hilbert scheme and theGromov-Witten/Donaldson-Thomas correspondence for 3-fold theories of localcurves to a triangle of equivalences in all higher genera. The proof uses thepreviously determined analytic continuation of the fundamental solution of theQDE of the Hilbert scheme. The GW/DT edge of the triangle in higher genusconcerns new CohFTs defined by varying the 3-fold local curve in the modulispace of stable curves. The equivariant orbifold Gromov-Witten theory of the symmetric product of theplane is also shown to be equivalent to the theories of the triangle in allgenera. The result establishes a complete case of the crepant resolutionconjecture.
We consider the question of how geometric structures of a Deligne-Mumfordstack affect its Gromov-Witten invariants. The two geometric structures studiedhere are {\em gerbes} and {\em root constructions}. In both cases, we explainconjectures on Gromov-Witten theory for these stacks and survey some recentprogress on these conjectures.
We study open-closed orbifold Gromov-Witten invariants of 3-dimensionalCalabi-Yau smooth toric DM stacks with respect to Lagrangian branes ofAganagic-Vafa type. We prove an open mirror theorem for toric Calabi-Yau3-orbifolds, which expresses generating functions of orbifold disk invariantsin terms of Abel-Jacobi maps of the mirror curves. This generalizes aconjecture by Aganagic-Vafa [arXiv:hep-th/0012041] and Aganagic-Klemm-Vafa[arXiv:hep-th/0105045], proved by the first and the second authors in[arXiv:1103.0693], on disk invariants of smooth toric Calabi-Yau 3-folds.
We introduce K-theoretic Gromov-Witten invariants of algebraic orbifoldtarget spaces. Using the methods developed by Givental-Tonita we characterizeGiventals Lagrangian cone of quantum K theory of orbifolds in terms of thecohomological cone.
For a Fermat quasi-homogeneous polynomial $W$, we study a family ofK-theoretic quantum invariants parametrized by a positive rational number$\epsilon$. We prove a wall-crossing formula by showing the generatingfunctions lie on the Lagrangian cone of the permutation-equivariant K-theoreticFJRW theory of $W$.
We study Givental's Lagrangian cone for the quantum orbifold cohomology oftoric stack bundles and prove that the I-function gives points in theLagrangian cone, namely we construct an explicit slice of the Lagrangian conedefined by the genus $0$ Gromov-Witten theory of a toric stack bundle.
Let $X$ be a compact toric K\"ahler manifold with $-K_X$ nef. Let $L\subsetX$ be a regular fiber of the moment map of the Hamiltonian torus action on $X$.Fukaya-Oh-Ohta-Ono defined open Gromov-Witten (GW) invariants of $X$ as virtualcounts of holomorphic discs with Lagrangian boundary condition $L$. We prove aformula which equates such open GW invariants with closed GW invariants ofcertain $X$-bundles over $\mathbb{P}^1$ used to construct the Seidelrepresentations for $X$. We apply this formula and degeneration techniques toexplicitly calculate all these open GW invariants. This yields a formula forthe disc potential of $X$, an enumerative meaning of mirror maps, and adescription of the inverse of the ring isomorphism of Fukaya-Oh-Ohta-Ono.
Using the mirror theorem [CCIT15], we give a Landau-Ginzburg mirrordescription for the big equivariant quantum cohomology of toric Deligne-Mumfordstacks. More precisely, we prove that the big equivariant quantum D-module of atoric Deligne-Mumford stack is isomorphic to the Saito structure associated tothe mirror Landau-Ginzburg potential. We give a GKZ-style presentation of thequantum D-module, and a combinatorial description of quantum cohomology as aquantum Stanley-Reisner ring. We establish the convergence of the mirrorisomorphism and of quantum cohomology in the big and equivariant setting.
We present a generalization of the Bogomolov-Miyaoka-Yau inequality toDeligne-Mumford surfaces of general type.
We derive a formula for the virtual class of the moduli space of rubber mapsto $[\mathbb{P}^1/G]$ pushed forward to the moduli space of stable maps to$BG$. As an application, we show that the Gromov-Witten theory of$[\mathbb{P}^1/G]$ relative to $0$ and $\infty$ are determined by knowncalculations.
In this note we prove that the crepant transformation conjecture for acrepant birational transformation of Lawrence toric DM stacks studied in\cite{CIJ} implies the monodromy conjecture for the associated wall crossing ofthe symplectic resolutions of hypertoric stacks, due to Braverman, Maulik andOkounkov.
For each positive rational number $\epsilon$, we define $K$-theoretic$\epsilon$-stable quasimaps to certain GIT quotients $W\sslash G$. For$\epsilon>1$, this recovers the $K$-theoretic Gromov-Witten theory of $W\sslashG$ introduced in more general context by Givental and Y.-P. Lee. For arbitrary $\epsilon_1$ and $\epsilon_2$ in different stability chambers,these $K$-theoretic quasimap invariants are expected to be related bywall-crossing formulas. We prove wall-crossing formulas for genus zero$K$-theoretic quasimap theory when the target $W\sslash G$ admits a torusaction with isolated fixed points and isolated one-dimensional orbits.
For a gerbe $\Y$ over a smooth proper Deligne-Mumford stack $\B$ banded by afinite group $G$, we prove a structure result on the Gromov-Witten theory of$\Y$, expressing Gromov-Witten invariants of $\Y$ in terms of Gromov-Witteninvariants of $\B$ twisted by various flat $U(1)$-gerbes on $\B$. This isinterpreted as a Leray-Hirsch type of result for Gromov-Witten theory ofgerbes.
We study Gromov-Witten theory of hypertoric Deligne-Mumford stacks from twopoints of view. From the viewpoint of representation theory, we calculate theoperator of small quantum product by a divisor, following \cite{BMO},\cite{MO}, \cite{MS}. From the viewpoint of Lawrence toric geometry, we compareGromov-Witten invariants of a hypertoric Deligne-Mumford stack with those ofits associated Lawrence toric stack.
We propose a conjectural determination of the Gromov-Witten theory of a rootstack along a smooth divisor. We verify our conjecture under an additionalassumption.
We show that the Virasoro conjecture in Gromov--Witten theory holds for thethe total space of a toric bundle $E \to B$ if and only if it holds for thebase $B$. The main steps are: (i) we establish a localization formula thatexpresses Gromov--Witten invariants of $E$, equivariant with respect to thefiberwise torus action, in terms of genus-zero invariants of the toric fiberand all-genus invariants of $B$; and (ii) we pass to the non-equivariant limitin this formula, using Brown's mirror theorem for toric bundles.
For a toric Calabi-Yau (CY) orbifold $\mathcal{X}$ whose underlying toricvariety is semi-projective, we construct and study a non-toric Lagrangian torusfibration on $\mathcal{X}$, which we call the Gross fibration. We apply theStrominger-Yau-Zaslow (SYZ) recipe to the Gross fibration of $\mathcal{X}$ toconstruct its mirror with the instanton corrections coming from genus 0 openorbifold Gromov-Witten (GW) invariants, which are virtual counts of holomorphicorbi-disks in $\mathcal{X}$ bounded by fibers of the Gross fibration. We explicitly evaluate all these invariants by first proving an open/closedequality and then employing the toric mirror theorem for suitable toric(partial) compactifications of $\mathcal{X}$. Our calculations are then appliedto (1) prove a conjecture of Gross-Siebert on a relation between genus 0 openorbifold GW invariants and mirror maps of $\mathcal{X}$ -- this is called theopen mirror theorem, which leads to an enumerative meaning of mirror maps, and (2) demonstrate how open (orbifold) GW invariants for toric CY orbifoldschange under toric crepant resolutions -- an open analogue of Ruan's crepantresolution conjecture.
We use the mirror theorem for toric Deligne-Mumford stacks, proved recentlyby the authors and by Cheong-Ciocan-Fontanine-Kim, to compute genus-zeroGromov-Witten invariants of a number of toric orbifolds and gerbes. We prove amirror theorem for a class of complete intersections in toric Deligne-Mumfordstacks, and use this to compute genus-zero Gromov-Witten invariants of anorbifold hypersurface.
We prove a Givental-style mirror theorem for toric Deligne--Mumford stacks X.This determines the genus-zero Gromov--Witten invariants of X in terms of anexplicit hypergeometric function, called the I-function, that takes values inthe Chen--Ruan orbifold cohomology of X.
We construct an integrable hierarchy in the form of Hirota quadraticequations (HQE) that governs the Gromov--Witten (GW) invariants of the Fanoorbifold projective curve $\mathbb{P}^1_{a_1,a_2,a_3}$. The vertex operators inour construction are given in terms of the $K$-theory of$\mathbb{P}^1_{a_1,a_2,a_3}$ via Iritani's $\Gamma$-class modification of theChern character map. We also identify our HQEs with an appropriateKac--Wakimoto hierarchy of ADE type. In particular, we obtain a generalizationof the famous Toda conjecture about the GW invariants of $\mathbb{P}^1$ .
We investigate the relationship between the Lagrangian Floer superpotentialsfor a toric orbifold and its toric crepant resolutions. More specifically, westudy an open string version of the crepant resolution conjecture (CRC) whichstates that the Lagrangian Floer superpotential of a Gorenstein toric orbifold$\mathcal{X}$ and that of its toric crepant resolution $Y$ coincide afteranalytic continuation of quantum parameters and a change of variables. Relatingthis conjecture with the closed CRC, we find that the change of variableformula which appears in closed CRC can be explained by relations between open(orbifold) Gromov-Witten invariants. We also discover a geometric explanation(in terms of virtual counting of stable orbi-discs) for the specialization ofquantum parameters to roots of unity which appears in Y. Ruan's original CRC["The cohomology ring of crepant resolutions of orbifolds", Gromov-Wittentheory of spin curves and orbifolds, 117-126, Contemp. Math., 403, Amer. Math.Soc., Providence, RI, 2006]. We prove the open CRC for the weighted projectivespaces $\mathcal{X}=\mathbb{P}(1,\ldots,1,n)$ using an equality between openand closed orbifold Gromov-Witten invariants. Along the way, we also prove anopen mirror theorem for these toric orbifolds.
We construct a new effective orbifold $\widehat{\Y}$ with an $S^1$-gerbe $c$to study an $S^1$-gerbe $\mathfrak{t}$ on a $G$-gerbe $\Y$ over an orbifold$\B$. We view the former as the relative dual, relative to $\B$, of the latter.We show that the two pairs $(\Y, \mathfrak{t})$ and $(\widehat{\Y}, c)$ haveisomorphic categories of sheaves, and also the associated twisted groupoidalgebras are Morita equivalent. As a corollary, the K-theory and cohomologygroups of $(\Y, \mathfrak{t})$ and $(\widehat{\Y}, c)$ are isomorphic. |
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ and $a+b+c=1$.
Prove that: $$\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\leq\frac{31}{27}$$
The equality occurs for $(a,b,c)=\left(\frac{19}{27},\frac{8}{27},0\right)$.
This inequality is similar to the following inequality, which was proposed by Walther Janous.
For all non-negatives $x$, $y$ and $z$ such that $xy+xz+yz\neq0$ prove that: $$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq\frac{5}{4}\sqrt{x+y+z}$$
My proof:
By Cauchy-Schwarz $$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$ Id est, it remains to prove that $$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or $$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$ $$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or $$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$ $$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.
If we want to use a similar way for the starting inequality, we need to use Holder, which gives very big numbers.
Maybe there is another reasoning?
Thank you! |
In the book "Reflections on Relativity" by Kevin Brown, there is a chapter called "Relatively Straight", in which he derives the geodesic equations using the Euler equation. Online version
Just after the second mention of the Euler equation (about 80% down), there is the following text: "Therefore, we can apply Euler's equation to immediately give the equations of geodesic paths on the surface with the specified metric
$$ \frac{\partial F}{\partial x^\sigma} - \frac{d}{d\lambda}\frac{\partial F}{\partial \dot{x}^\sigma}$$
For an n-dimensional space this represents n equations, one for each of the coordinates $x_1, x_2, ..., x_n$. Letting $w = (\frac{ds}{dl})^2 = F^2 = g_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta$ this can be written as
$$\frac{\partial w^{1/2}}{\partial x^\sigma} - \frac{d}{d\lambda}\frac{\partial w^{1/2}}{\partial \dot{x}^\sigma} = \frac{d}{d\lambda}\frac{\partial w}{\partial \dot{x}^\sigma} - \frac{\partial w}{\partial x^\sigma} - \frac{1}{2w}\frac{dw}{d\lambda} \frac{\partial w}{\partial \dot{x}^\sigma} = 0 $$
I get the substitution of sqrt(w) for F on the LHS, but can't see how he obtains the middle expression. I have tried using the product/chain rules as is usual with these things, but just cannot see what he is doing here.
I can usually follow Kevin's work, with a bit of effort, but this one seems a little trickier than I am used to. Can anyone help me to understand the trick? |
Difference between revisions of "Abstract.tex"
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The Hales-Jewett theorem asserts that for every $r$ and every $k$ there exists $n$ such that every colouring of the $n$-dimensional grid $
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The Hales-Jewett theorem asserts that for every $r$ and every $k$ there exists $n$ such that every colouring of the $n$-dimensional grid $k^n$ contains a combinatorial line. This result is a generalization of van der Waerden's theorem, and it is one of the fundamental results of Ramsey theory. The van der Waerden's theorem has a famous density version, conjectured by Erd\H os and Tur\'an in 1936, proved by Szemer\'edi in 1975 and given a different proof by Furstenberg in 1977. The Hales-Jewett theorem has a density version as well, proved by Furstenberg and Katznelson in 1991 by means of a significant extension of the ergodic techniques that had been pioneered by Furstenberg in his proof of Szemer\'edi's theorem. In this paper, we give the first proof of the theorem of Furstenberg and Katznelson. proof is surprisinglysimple: indeed, it gives what is probably the simplest known proof of Szemer\'edi's theorem.
\end{abstract}
\end{abstract}
Revision as of 17:08, 11 June 2009
\begin{abstract}The Hales--Jewett theorem asserts that for every $r$ and every $k$ there exists $n$ such that every $r$-colouring of the $n$-dimensional grid $\{1, \dotsc, k\}^n$ contains a combinatorial line. This result is a generalization of van der Waerden's theorem, and it is one of the fundamental results of Ramsey theory. The van der Waerden's theorem has a famous density version, conjectured by Erd\H os and Tur\'an in 1936, proved by Szemer\'edi in 1975 and given a different proof by Furstenberg in 1977. The Hales--Jewett theorem has a density version as well, proved by Furstenberg and Katznelson in 1991 by means of a significant extension of the ergodic techniques that had been pioneered by Furstenberg in his proof of Szemer\'edi's theorem. In this paper, we give the first elementary proof of the theorem of Furstenberg and Katznelson, and the first to provide a quantitative bound on how large $n$ needs to be. In particular, we show that a subset of $[3]^n$ of density $\delta$ contains a combinatorial line if $n \geq 2 \upuparrows O(1/\delta^3)$. Our proof is surprisingly\noteryan{``reasonably
, maybe} simple: indeed, it gives what is probably the simplest known proof of Szemer\'edi's theorem. \end{abstract} |
2013-07-01
Central Limit Theorems for Markov-modulated infinite-server queues Publication Publication
This paper studies an infinite-server queue in a Markov environment, that is, an infinite-server queue with arrival rates and service times depending on the state of an independently evolving Markovian background process. Scaling the arrival rates $\lambda_i$ by a factor $N$ and the rates $q_{ij}$ of the background process by a factor $N^\alpha$, with $\alpha \in \mathbb R^+$, we establish a central limit theorem as $N$ tends to $\infty$. We find different scaling regimes, which depend on the specific value of $\alpha$. Remarkably, for $\alpha<1$, we find a central limit theorem in which the centered process has to be normalized by $N^{{1-}\alpha/2}$ rather than $\sqrt{N}$; in the expression for the variance deviation matrices appear.
Additional Metadata THEME Life Sciences (theme 5) Publisher CWI Series Life Sciences [LS] Citation
Blom, J.G, deTurck, K.E.E.S, & Mandjes, M.R.H. (2013).
Central Limit Theorems for Markov-modulated infinite-server queues. Life Sciences [LS]. CWI. |
Let $\mathcal G= \{G_\alpha: \alpha < \kappa\}$ be a family of topological groups. My question is this:
Is the topological product $\Pi \{G_\alpha: \alpha < \kappa\} $ a topological group?
Thanks ahead.
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It is, and it follows from the universal property of continuity of maps into products, and its corollary of continuity of product maps, see my answer here. The universal property I mean is stated as:
For any space $Y$ and any product $X= \prod_{\alpha < \kappa} X_\alpha$: a function $f: Y \to X$ is continuous w.r.t the product topology on $X$ iff $\forall \alpha < \kappa$: $\pi_\alpha \circ f$ is continuous, where $\pi_\alpha : X \to X_\alpha$ are the projections.
To introduce some notation: Let $\mu_\alpha: G_\alpha \times G_\alpha \to G_\alpha$ be the continuous multiplication of $G_\alpha$, and $i_\alpha: G_\alpha \to G_\alpha$ be its inversion.
Then let $G = \prod_{\alpha < \kappa} G_\alpha$ be the product in the product group structure $(\mu_G,e_G, i_G)$ (defined coordinatewise), and in the product topology (i.e. the initial topology with respect to the projections $\pi_\alpha: G \to G_\alpha$), then $i_G = \prod_{\alpha < \kappa} i_\alpha$, so is continuous, using the universal property:
$$\forall \alpha < \kappa: \pi_\alpha \circ i_G = i_\alpha$$ and the right hand is continuous by assumption.
Then the coordinatewise multiplication rule says $(\mu_G(g,h))_\alpha = \mu_\alpha(g_\alpha, h_\alpha)$ for all $\alpha < \kappa, g,h \in G$, so
$$\forall \alpha < \kappa : \pi_\alpha \circ \mu_G = \mu_\alpha \circ (\pi_\alpha \times \pi_\alpha)$$ where the right hand side is continuous as the composition of a continuous product map and the continuous $\mu_\alpha$. So the universal property says that $\mu_G$ is also continuous.
So $G$ is indeed a topological group in this group law/topology combination. |
Many thanks to Jens Seeber for comments and corrections!
Phrases like “
Terminal objects are unique up to isomorphism” are everywhere in category theory. In this post, I’ll explain the concept of “uniqueness up to isomorphism”, and its best buddy, “uniqueness up to unique isomorphism”.
I’ll also talk about the philosophy behind these concepts, and how they allow category theorists to define objects in terms of their
relationship to other objects rather than by some internal properties. In other words, how can we avoid asking “what is this object?”, but instead ask “how does this object behave?”.
I will assume only a minimal background in category theory, but you should at least know what a category is, and what is meant by “objects” and “morphisms”.
We’ll begin with some intuition for how isomorphism mathematically means “same shape”. Next, we’ll make that intuition precise, before examining the example of terminal objects, which are both unique up to isomorphism, and up to
unique isomorphism. 1 Intuition
What made the concept click for me was when someone explained the meaning of the sentence “
terminal objects are unique up to isomorphism”. They said:
A category might contain multiple terminal objects, but if it does, then there must exist an isomorphism between those objects.
Take the category of sets for example, where “the” terminal object is the singleton set. Just believe me for now; we’ll prove it later. Clearly there are many singleton sets - \(\{foo\}\), \(\{\pi\}\) and \(\{💅\}\) to name a few - but we talk about “the” terminal object of a category - why?
1.1 Isomorphic means “same shape” - literally and formally
The reason is that all singleton sets are isomorphic in the category of sets, and the existence of an isomorphism between two objects literally and formally means they have the
same shape, in the sense that they have the same relationship to other objects.
Mathematically, we capture this idea by talking about the
morphisms of an object. In particular, the existence of an isomorphism between two objects \(A\) and \(B\) implies that there is a bijection between their morphisms. That is, if objects \(A\) and \(B\) are isomorphic, and \(A\) has a morphism \(f : A \to C\), then \(B\) must have a “twin” morphism \(f' : B \to C\).
We’ll state this formally and prove it in the next section, but the intuition you should have is that when two objects \(A\) and \(B\) are
isomorphic they have the same relationship with other objects in the category. This is the sense in which they behave the same way. 1.2 “ Unique up to isomorphism” is with respect to a property
Before we get formal, I want to clarify something that really tripped me up.
When we say “
unique up to isomorphism”, we mean crucially with respect to some property. What kind of property? Well for example, the property of “being a terminal object”.
I’m going to repeat this using a definition from stackoverflow,
Unique up to isomorphism means that all the objects
satisfying a given definitionare isomorphic
Emphasis mine. Again, examples of “a given definition” might be the definition of “terminal object”, or the definition of “initial object”.
2 Isomorphisms 2.1 Definition: Isomorphism
An isomorphism is defined as a morphism with an inverse. That is, if you compose an isomorphism \(\alpha : A \to B\) with its inverse \(\alpha^{-1} : B \to A\), they cancel each other out and become the identity morphism \(id_A\).
Graphically
1, we can follow the arrow \(\alpha\) from \(A\) to \(B\), then follow the inverse \(\alpha^{-1}\) back, and it’s the same as just following the identity morphism \(id_A\).
The reverse is also true - we can follow \(\alpha^{-1}\), then \(\alpha\), and it’s the same as just following \(id_B\).
Formally, if \(\alpha : A \to B\) is an isomorphism then it has an inverse \(\alpha^{-1}\) such that \(\alpha^{-1} \circ \alpha = id_A\), and also \(\alpha \circ \alpha^{-1} = id_B\). We also say that \(A\) and \(B\) are
isomorphic, and write \(A \cong B\).
Note that there can be more than one isomorphism, so \(A\) and \(B\) can be isomorphic in more than one way! We will come back to this when we talk about universal properties.
2.2 Isomorphic means “same shape”
So what do isomorphisms have to do with how objects relate to each other?
Precisely, if \(A \cong B\), then given an arbitrary object \(Y\), there is a bijection mapping “outgoing” morphisms \(f : A \to Y\) to outgoing morphisms \(f' : B \to Y\).
This also works for “incoming” morphisms- given an arbitrary object \(X\), there is a bijection from morphisms \(g : X \to A\) to morphisms \(g' : X \to B\).
Graphically:
We’ll focus on proving this correspondence for the lower half of the diagram- i.e., that the bijection exists for the “output” morphisms. The proof for the “input” morphisms is almost identical, so I’ll omit it.
Without further ado, let’s formally state and prove this proposition.
2.2.1 Proposition: There is a bijection between sets of morphisms of isomorphic objects
Suppose \(A \cong B\) with isomorphism \(\alpha\). Then for all objects \(Y\), there is a function \(\phi : \mathscr{C}(A,Y) \to \mathscr{C}(B,Y)\). Furthermore, \(\phi\) is a bijection, and therefore has an inverse, \(\phi^{-1}\).
The notation
2 \(\mathscr{C}(A,Y)\) means the set of morphisms from \(A\) to \(Y\), and is indeed a set 3. Therefore \(\phi\) is a function and takes a morphism \(A \to Y\) as input, and produces a morphism \(B \to Y\) as an output.
Now let’s prove the proposition.
2.2.2 Proof of Proposition 2.2.1
First we need to construct \(\phi\) and its inverse, \(\phi^{-1}\). Then, we’ll show that \(\phi^{-1} \circ \phi\) is the identity function, i.e. that \(\phi^{-1}\) is indeed the inverse of \(\phi\). Showing the existence of an inverse is the same as showing \(\phi\) is bijective, so once we’ve done this, we’ll be finished!
Let’s start by constructing \(\phi\). Our job is to take a morphism \(f : A \to Y\) and turn it into a morphism \(f' : B \to Y\). Thinking graphically, we need a way to start at \(B\), and eventually follow \(f\) to reach \(Y\).
The obvious choice is to just use \(\alpha^{-1}\), and indeed that is what we do:
\[ \phi : \mathscr{C}(A,Y) \to \mathscr{C}(B, Y) \] \[ \phi(f) = f \circ \alpha^{-1} \]
Similarly, the inverse function does the same thing, but starting from \(A\):
\[ \phi^{-1} : \mathscr{C}(B,Y) \to \mathscr{C}(A, Y) \] \[ \phi^{-1}(f') = f' \circ \alpha \]
So we’ve constructed \(\phi\), now we have to prove that it has an inverse. Equivalently, we show that \(\phi^{-1} \circ \phi : \mathscr{C}(A,Y) \to \mathscr{C}(A,Y)\) is the identity function:
\[ \begin{align} (\phi^{-1} \circ \phi) (f) &= \phi^{-1} ( \phi (f) ) \\ &= \phi^{-1} ( f \circ \alpha^{-1} ) \\ &= f \circ \alpha^{-1} \circ \alpha \\ &= f \circ id_B \\ &= f \end{align} \]
Here we simply apply the definitions of \(\phi\) and \(\phi^{-1}\), and then rely on the fact that \(\alpha^{-1} \circ \alpha = id_B\) by the definition of isomorphism.
Strictly speaking we should also show that \(\phi \circ \phi^{-1}\) is also the identity function, but the proof is almost identical so we’ll omit it.
So we have shown that there is a bijection between sets of morphisms of isomorphic objects, and therefore they are
formally of the same shape \(\blacksquare\). 2.3 Recap
I’ll pause now and recap what we’ve covered so far.
We’ve defined what an isomorphism is, and shown how its existence means that two objects are
formally of the same shape, by which we mean there is a bijection between their morphisms.
In the next section, we’ll look at the example of
terminal objects. Objects satisfying this definition are unique up to isomorphism, which we will prove after going through an example in the category of sets. 3 Terminal Objects
We’ll first examine the category of sets, and discover that there is a unique function mapping any set to the singleton set. We’ll see how that also implies all singleton sets are isomorphic.
Next, we’ll see how these properties are a consequence of the singleton set being the terminal object of \(\textbf{Set}\). In particular, we’ll define terminal objects, and show that \(\textbf{Set}\) has them.
3.1 The Category of Sets
The category of sets is called \(\textbf{Set}\). Its objects are sets, and its morphisms are functions.
By way of example, here is a diagram of a tiny fragment of \(\textbf{Set}\). Obviously not all objects and morphisms are pictured, because there are an infinite number of sets and functions!
This diagram shows some of the objects and morphisms of \(\textbf{Set}\). For example, the morphism \(f : \mathbb{N} \to \{0,1,4,9,..\}\) maps natural numbers to their squares.
Also shown are two suggestively-named morphisms \(\alpha\) and \(\alpha^{-1}\), which denote an isomorphism. We’ll come back to these shortly!
Finally, notice that all the functions whose codomain is a singleton set have the same form. In particular, they all look like \(f(x) = t\), where \(t\) is the single element of the codomain set.
3.2 Unique function from any set to the singleton set
The functions/morphisms to singleton sets in the diagram above all look “the same” because there is
only one possible function with a given domain whose codomain is the singleton set.
To convince you, think of a function from some set \(A\) to \(\{\bullet\}\) as the set of input/output pairs:
input output \(a_0\) \(\bullet\) \(a_1\) \(\bullet\) \(a_2\) \(\bullet\) \(\vdots\) \(\vdots\)
Clearly there is only one choice of output for each input, and so there is only one possible function- it’s like we’re just tagging each element of \(A\) with \(\bullet\).
There’s also a special case of this unique function: when both domain and codomain of the function are singleton sets.
For example, the unique function between \(\{pi\}\) and \(\{\bullet\}\) is the function \(\alpha = \pi \mapsto \bullet\).
In this case, the function is also a
bijection, and its inverse is obviously \(\alpha^{-1} = \bullet \mapsto \pi\). 3.3 All singleton sets are isomorphic
Recall that in \(\textbf{Set}\), morphisms are functions. Isomorphisms, then, are functions with an inverse, which means that in \(\textbf{Set}\), isomorphisms are
bijective functions.
We just found bijection between \(\{\bullet\}\) and \(\{\pi\}\), and so we’ve shown that those two objects (sets) are isomorphic.
This isomorphism is clearly a consequence of the property of singleton sets that there is a unique function from every set, but the isomorphism itself is quite boring: it merely amounts to a renaming of the singleton set’s only element.
The idea of isomorphism is that of a “weaker equality”, where we can ignore this tedious renaming of elements, and instead look at objects in terms of their relationships to other objects.
That is really the idea behind “unique up to isomorphism”- ignoring irrelevant internal details, and instead
focusing on relationships.
The particular property we saw here - that each singleton set has a unique function mapping to it - is generalised by the definition of “terminal object”, which we’ll look at next.
3.4 Definition: Terminal Objects
Given a category \(\mathscr{C}\), we say an object \(T\) is
terminal 4 if, for all objects \(A \in ob(\mathscr{C})\), there exists a unique morphism \(!_A : A \to T_{}\) (pronounced “bang A”).
We usually draw this as the following diagram, where the dotted line denotes a unique morphism.
Note that this implies that \(!_T {}_{} = id_T\), by uniqueness. By the definition of terminal object, there can be only one morphism \(T \to T\), and by the definition of a category, there must be an identity morphism of that type, so they are equal.
Now let’s show that \(\textbf{Set}\) has terminal objects.
3.5 Terminal objects in \(\textbf{Set}\)
To show that \(\textbf{Set}\) has terminal objects, we must show there is a set \(T\) with a unique function \(!_A {}_{} : A \to T\) for every set \(A\).
Of course, we’ve already done this: \(T\) is any singleton set, and the unique morphism \(!_A {}_{}\) is the function \(f(a) = t\) where \(t\) denotes the single element of \(T\).
So therefore, \(\textbf{Set}\) has terminal objects. \(\blacksquare\).
3.6 Terminal objects are unique up to isomorphism
Now we’ll step up a level of generality. Earlier, we proved that singleton sets are all isomorphic. Now we’ll prove that terminal objects in
any category are unique up to isomorphism: a much more general result.
The proof goes as follows:
Suppose we have two arbitary terminal objects, \(T\) and \(T'\). Then by the definition of terminal object, there must exist unique morphisms \(!_{T'} {}_{} : T' \to T\) and \(!'_T {}_{} : T \to T'\).
Proving \(T \cong T'\) means constructing an isomorphism. Taking \(!_{T'}\) and \(!'_T\) as the isomorphism and its inverse, we must show both that \(!_{T'} \circ !'_T = id_T\), and that \(!'_T \circ !_{T'} = id_{T'} {}_{}\)
This is fairly straightforward, and we use the same reasoning as before. Because \(!_{T'} \circ !'_T\) goes from \(T \to T\), it must equal \(id_T\) by uniqueness- there must be exactly one function of this type by the definitions of terminal object and category.
The same reasoning applies to the second case, so \(T\) and \(T'\) are isomorphic. Because \(T\) and \(T'\) were arbitrary, we have shown that any two terminal objects are isomorphic, and we are done. \(\blacksquare\).
3.7 Terminal Objects are unique up to unique isomorphism
Not only this, but the isomorphism \(!_{T'} {}_{}\) is unique (again, by the definition of terminal object). Therefore, terminal objects are also
unique up to unique isomorphism!
This is because the property of “terminality” is a
universal property. 4 Universal Properties
One of the triumphs of category theory is its ability to
define objects by their relationships, rather than their internal properties. For example, the definition of terminal object allows us to define the singleton set without even mentioning elements!
This idea is captured by the notion of
Universal Properties, of which the definition of “terminal object” is an example.
While universal properties can be defined and studied formally in category theory, we’ll only talk informally about them here; in particular, to say that if a property is universal, one consequence is that the objects it defines are
unique up to unique isomorphism.
While “uniqueness up to isomorphism” says that objects satisfying a given definition are isomorphic, they may be isomorphic in
more than one way.
In contrast,
uniqueness up to unique isomorphism means that any two objects satisfying the definition have a unique isomorphism between them, and are therefore isomorphic in exactly one way.
This captures more closely the idea of objects being
indistinguishable by their relationships, so we can view objects satisfying a universal property as “ equal, if we ignore irrelevant and boring details”. 5 Summary
I’ll finish by summarising what we’ve covered:
“unique up to isomorphism” is with respect to some property, e.g. “terminal objects are unique up to isomorphism” Category theory talks about objects in terms of their relationship to other objects, not by the internal properties of objects. unique up to (unique) isomorphismgives us a way to talk formally about how a certain property makes a collection of objects indistinguishable in terms of their relationship to other objects.
So to wrap it up with an example, when we say “terminal objects are unique up to isomorphism”, what we mean is:
There might be multiple terminal objects, but… … if there are, they have an isomorphism, and that means … … all those terminal objects have the exact same relationship to all other objects, and so … … they are interchangeable, or in a sense “indistinguishable”
Cheers!
If you haven’t seen this kind of diagram before, the nodes are objects and the arrows morphisms. A path around the graph corresponds to a composition of morphisms.↩
The letter \(\mathscr{C}\) is a calligraphic \(C\) - \(C\) for “Category”. You can type it in latex as
\mathscr{C}.↩
In general, the collection of morphisms between two objects \(A\) and \(B\) is not required to be a set. In fact, by requiring this, we are restricting ourselves to working with locally small categories, but I hope this restriction makes this blog post more clear!↩
Usually authors will denote terminal objects as \(1\), but we’ll use \(T\) to avoid confusion.↩ |
Nuclear reactions are going on all around us. Using correctly balanced equations is important whetting to understand nuclear reactions. All equations need to be balance to conform to two conservation laws: The mass number and the electrical charge.
Learning Objectives Using the conservation laws to find an unknowns in a nuclear reaction equation Write a balanced nuclear equation for an natural transmutations Success Criteria Balance nuclear reactions Identify the products of nuclear reactions Recognize the three primary modes of radioatciity Identify stability via the Belt of Stability Nuclear Reactions
Nuclear chemistry is the subfield of chemistry dealing with radioactivity, nuclear processes, such as nuclear transmutation, and nuclear properties. It is the chemistry of radioactive elements such as the actinides, radium and radon together with the chemistry associated with equipment (such as nuclear reactors) which are designed to perform nuclear processes. This includes the corrosion of surfaces and the behavior under conditions of both normal and abnormal operation (such as during an accident). An important area is the behavior of objects and materials after being placed into a nuclear waste storage or disposal site.
As you remember from Chem 2A, an atom is composed of a nucleus and 1 or more electrons. While, the nucleus has 1/100,000 the radius of the atom, it has nearly all its mass. Nuclei are composed of positively charged protons and neutral neutrons. Elements are distinguished by the
atomic number \(Z\) which quantifies the number of protons. Similarly, the \(N\) refers number of neutrons in a nucleus. The mass number (\(A\)) is the total number of protons \(Z\) and neutrons \(N\)
\[A = Z + N\]
An isotope is a nucleus with the same \(Z\) as another nucleus, but with a different neutron number \(N\). Isotopes are nearly identical and have the same number of protons, e.g. U-235, U-236, and U-238 refer to three isotopes of uranium with mass numbers of 235, 236 and 238. A nuclear nomenclature can be adopted to describe an specific nucleus that explicitly indicates both \(A\) and \(Z\):
\[ \ce{^{A}_{Z} Element\, Symbol}\]
Within this nomenclature, the U-235, U-236, and U-238 nuclei would be written as \( \ce{^{235}_{92} U}\), \( \ce{^{236}_{92} U}\), and \( \ce{^{238}_{92} U}\), respectively. Note that \(Z\)
does not change for isotopes of the same element.
The nuclear nomenclature can be used to describe more than isotopes. for example, the \(\ce{^4_2He}\) is a helium nucleus and is also known as an
alpha particle. Similarly, an electron is written as \(\ce{^0_{-1}e}\) and is known as a beta particle when emitted by a nucleus. In a nuclear decay reaction, also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. Q1
The following are two nuclear decay reactions
\[\ce{^{220}_{87}Fr} \rightarrow \ce{^{4}_{2}He } + \ce{^{216}_{85}At} \label{eq1}\]
\[\ce{^{16}_{7}N} \rightarrow \ce{^{0}_{-1}e} + \ce{^{16}_{8}O} \label{eq2}\]
It is common that the electric charges is not indicated (but can be).
What are the products of the decay of Francium-220? What are the products of the decay of Nitrogen-16? Is an alpha particle gained or released in the decay of Francium-220? Is an electron gained or released in the decay of Nitrogen-16? What is the mass number of an alpha particle? What is the charge of an alpha particle? What is the mass number of a beta particle? What is the charge of an beta particle? Q2
By examining Equations \(\ref{eq1}\) and \(\ref{eq2}\), what is the mathematical relationship between the total
mass number of the reactants and the total mass number of the products. Show work. Q3
By examining Equations \(\ref{eq1}\) and \(\ref{eq2}\), what is the mathematical relationship between the total
charge of the reactants and the total charge of the products. Show work.
Radioactivity
A nucleus that is not permanently stable is radioactive and eventually decays into another. Although the decay of a particular radioactive nucleus is random, 50% of a collection of radioactive nuclei decays in
one half-life (\(t_{1/2}\)), just like in chemical chemistry. There are various ways in which nuclei can decay through emitting radiation Alpha Decay: Alpha decay only occurs in the heaviest nuclei with \(Z > 83\) or \(A ≥ 200\). As an alpha particle is a helium nucleus 4He, the mass number of the daughter nucleus is reduced by 4 and the atomic number (Z) is reduced by 2. Examples of alpha decay are
\[\ce{^{238}_{92}U} → \ce{^{234}_{90}Th} + \alpha\]
with a \(t_{1/2} = 4 \times 10^9\, \text{years}\) or
\[\ce{^{216}_{88}Ra} →\ce{^{212}_{84}Po} + \alpha\]
with \(t_{1/2} = 45 \times 10^{-6}\, \text{sec}\).
Alpha particles are stopped by a few cm of air, a sheet of paper or human skin.
Beta Decay: Beta decay occurs for all values of mass number. It is the emission of an electron (known as \(β^-\) decay) or a positron (known as \(β^+\) decay) with a resulting change in atomic number, \(Z\), but not mass number, \(A\). Neutrons or protons in the nucleus decay by emitting electrons via the weak interaction. This decay is always accompanied by emission of a neutrino or anti-neutrino (a massless particle with no charge). \(β^-\) Decay
\[\ce{^{1}_0 n} → \ce{^{1}_1p^+} + \ce{^{0}_{-1} β^{-}} + \text{anti-neutrino}\]
\[\ce{^{27}_{12}Mg} → \ce{^{27}_{13}Al}^{*} + \ce{^{0}_{-1} β^{-}} + \text{anti-neutrino}\]
The \(*\) is meant to signify that the nucleus is unstable (e. g., in the excited-state) and will also decay (e. g., via gamma decay).
\(β^+\) Decay
\[\ce{^{27}_{14}Si} → \ce{^{27}_{13}Al}^* + \ce{^{0}_{+1} β^{+}} + \text{neutrino}\]
Beta radiation is stopped by a thin piece of wood or plastic.
Gamma Decay: Gamma radiation is high energy electromagnetic particles/waves. They are emitted by nuclei in an excited state attempting to return to their ground state.
\[\ce{^{27}_{13}Al}^{*} \rightarrow \ce{^{27}_{13}Al} + \gamma\]
There is no accompanying change in their mass number, A, or the atomic number, Z, with a half-llfe usually less than \(1\,µs\). Gamma radiation can only be stopped by a substantial thickness of heavy material such as lead.
Q4
Use the laws of conservation of \(A\), \(N\) and \(Z\) and charge to determine the identify of \(X\) in the nuclear reaction equations below (include both \(Z\) and \(A\) for \(X\). You will need to refer to a periodic table.
\(\ce{^{222}_{86} Rn} \rightarrow \ce{^{4}_{2} He} + X\) \(\ce{^{14}_{6}C} \rightarrow \ce{^{0}_{-1} e^{-}} + X\) \(\ce{X} \rightarrow \ce{^{0}_{+1} e^{+}} + \ce{^{19}_{9} F} \) Q5
Write the balanced equation for the beta decay of Sr-90.
Belt of Stability
he graph of stable elements is commonly referred to as the
Band (or Belt) of Stability. The graph consists of a y-axis labeled neutrons, an x-axis labeled protons, and a nuclei. At the higher end (upper right) of the band of stability lies the radionuclides that decay via alpha decay, below is positron emission or electron capture, above is beta emissions and elements beyond the atomic number of 83 are only unstable radioactive elements. Stable nuclei with atomic numbers up to about 20 have an neutron:proton ratio of about 1:1 (solid line). Q6
Find the reactions from Q4 and Q5 on the graph above. Where are they located on the belt (above, below or on the belt)?
Q7
The nuclei on the belt are stable and those off the belt are unstable. Hence to move a nucleus from on the belt to off requires energy (i.e., not spontaneous) and vice versa. Are the decays in Q4 and Q5 spontaneous or non-spontanous?
Additional Questions Q8
Identify the following as Alpha, beta, gamma or neutron:
\(\ce{^1_0 n}\) \(\ce{^0_{-1} e}\) \(\ce{^4_2 He^{2+}}\) \(\ce{^0_0 \gamma}\) Nuclear decay with no mass nor charge: An electron Least penetrating nuclear decay Most damaging nuclear decay to the human body Nuclear decay that can be stopped by skin paper Nuclear decay that can be stopped by aluminum Q9
How many protons, neutrons, and electrons are in \(\ce{^{195}Pt^{+2}}\)?
Q10
\(\ce{^{62}Cr}\) decays via beta emission. Which statement is correct concerning \(\ce{^{62}Cr}\)?
The number of electrons decrease. The number of protons decrease. The number of protons increase. The number of neutrons decrease. The number of protons increase and the number of neutrons decrease. Q11
Complete the following nuclear equations (the question marks)
\(\ce{^{42}_{19}K \rightarrow ^{0}_{-1}e^- + ?}\) \(\ce{^{239}_{94}Pu \rightarrow ^{4}_{2}He^{2+} + ?}\) \(\ce{^{9}_{4}Be \rightarrow ^{9}_{4}Be} \,+ \, ?\) \(\ce{^{235}_{92}U \rightarrow ? + ^{231}_{90}Th}\) \(\ce{^{6}_{3}Li \rightarrow ^{4}_{2}He^{2+}}\, + \, ?\) \(\ce{? \rightarrow ^{142}_{56}Ba + ^{91}_{36}Kr + 3\; ^{1}_{0}n}\) Q12
Write equations for the following nuclear decay reactions. Make sure that both mass numbers and atomic numbers are balanced on each side
Decay of polonium-218 by alpha emission. Decay of carbon-14 by beta \(\beta^-\) emission. The alpha decay of radon-198 The beta \(\beta^-\) decay of uranium-237 Decay Chain
In nuclear science, the decay chain refers to the radioactive decay of different discrete radioactive decay products as a chained series of transformations. They are also known as "radioactive cascades". Most radioisotopes do not decay directly to a stable state, but rather undergo a series of decays until eventually a stable isotope is reached (i.e., a nucleus on the belt of stability). Decay stages are referred to by their relationship to previous or subsequent stages. A parent isotope is one that undergoes decay to form a daughter isotope. One example of this is uranium (atomic number 92) decaying into thorium (atomic number 90). The daughter isotope may be stable or it may decay to form a daughter isotope of its own. The daughter of a daughter isotope is sometimes called a granddaughter isotope.
Q13
The figure below maps the radioactive decay of \(\ce{^{238}U}\) into \(\ce{^{206}Pb}\). Use this figure to answer the following three questions:
How many alpha particles are produced as one atom of \(\ce{^{238}U}\) decays to one atom of \(\ce{^{206}Pb}\)? Draw the decay pathway you used for this calculation? Does it change if you picked a different pathway? How many beta particles are produced as one atom of \(\ce{^{238}U}\) decays to one atom of \(\ce{^{206}Pb}\)? Draw the decay pathway you used for this calculation? Does it change if you picked a different pathway? What is the final product in the decay series of \(\ce{^{238}U}\)? Q14 Write the nuclear equation showing that when \(\ce{^{229}Pa}\) goes through two consecutive alpha decays to form \(\ce{^{221}Fr}\). Write the nuclear equation showing that when \(\ce{^{210}Po}\) goes through two consecutive alpha decays and then a beta decay and then another alpha decay. Q15
Throium-232 undergoes radioactive decay until a stable isotope is reached. Write the nuclear reaction for each of the 11 steps in the decay of \(\ce{^{238}Th}\) with each product becoming the reactant of the next decay. What is the final stable isotope?
Step 1: Alpha decay Step 2: Beta decay Step 3: Beta decay Step 4: Alpha decay Step 5: Alpha decay Step 6: Alpha decay Step 7: Alpha decay Step 8: Beta decay Step 9: Beta decay Step 10: Alpha decay Step 11: Beta decay |
I'm solving the 3D Diffusion equation $$u_t=k(u_{xx}+u_{yy}+u_{zz})$$
in MATLAB using Fourier techniques. I assume a 3D Fourier expansion $(e^{-ipx},e^{-imy},e^{-imz})$of the solution.
Physical space: $u(x,y,z,t)$. Fourier Space: $c(m,n,p,t)$.
Substitution and differentiation result in: $$c(m,n,p)^{N+1}-c(m,n,p)^{N} = -\frac{k\Delta T}{2} (p^2+m^2+n^2)(c(m,n,p)^{N+1}+c(m,n,p)^{N})$$ after applying a Crank-Nicolson scheme.
I'm using
fftn() and
ifftn() to forward my coefficients in time and bring them back to physical space. However I achieve universal decay from the initial condition to zero with no heat flux in any direction, for all time. Typical time step:
0.00001. Typical
k=0.005.
Is the problem with my application of
fftn() or the stability of my finite-difference?
Edit: I've taken an initial condition of $50 \sin(2x)$. I merely took the boundaries of that initial condition and imposed them as the boundary conditions for all time. No good I'm guessing? Edit 2: I forgot the wavenumber squared on the rhs. Thanks James! |
As multigrid methods are known to have grid independent convergence rates with $O(N)$ computational cost, then why would one be interested in using Krylov subspace methods at all, for which convergence rates deteriorate with grid refinement? Are there any problems where Krylov solvers perform better than MG methods?
The answer depends somewhat on the discretization; for example, some boundary integral discretizations result in very well-conditioned matrices, for which a Krylov solver works just fine without having to introduce the multi-level machinery of multigrid.
For ill-conditioned matrices arising from finite difference or finite element methods, Krylov solvers are almost always paired with a preconditioner, for which multigrid methods can work very well.
This question is pretty well discussed in literature. However, there are lots of questions concerning multigrid on SciComp, so I decided to compose more or less detailed answer.
I. When multigrid DOES work fine as a stand–alone solver
MG works fine for rather simple elliptic problems. In fact, both MG and PCG (with MG cycles as inner iterations) usually give similar results for such problems in terms of iterations / working time and stay robust as you refine the mesh.
I solved diffusion–reaction 2D problem $$ -\epsilon \, \Delta u + u = f, \quad \mathbf x \in \Omega, \\ u|_{\Gamma_D} = g_D, \\ \left( \epsilon \, \nabla u \cdot \hat{\mathbf n} + u \right)|_{\Gamma_R} = g_R $$
with CATS’ PDEs on hierarchy of triangular meshes $\Omega_1 \subset \Omega_2 \subset \dots \subset \Omega_8$ of unit square with P1 FEs. I obtained the following results:
$\text{MG(V)}$ and $\text{MG(W)}$ are V– / W–cycles as stand–alone solvers; $\mathbf P_{\text{MG(V)}}\text{CG}$ is preconditioned conjugate gradient with one V–cycle as an inner iteration; $\mathbf P_{\text{ILU(0)}}\text{CG}$ is preconditioned conjugate with incomplete $(\mathbf L + \mathbf I) \, \mathbf D \, (\mathbf I + \mathbf L^T)$ preconditioner (zero fill–in). I used 2 pre– and post–smoothing SSOR iterations inside MG–cycles. Numb of DOFs on the coarsest grid: 17; numb of DOFs on the finest grid: 656 897. In the table: “numb of iterations, $m$” / “solving time (in seconds), $t$.” I always started with a random initial guess $\mathbf x_0$ in order to introduce components of different frequencies in the initial error vector $(\mathbf x - \mathbf x_0)$; I stopped solving when $||\mathbf r_m||/||\mathbf r_0|| < 10^{-12}$.
Here’s comparison in terms of time $t$ as system size $n$ grows:
Indeed, you can see that $\text{MG}$ allows you to solve this problem in $O(n)$ time; so does $\mathbf P_{\text{MG}}\text{CG}$, but there is no difference between them in this case. I think this makes people ask questions like
As multigrid methods are known to have grid independent convergence rates with $O(n)$ computational cost, then why would one be interested in using Krylov subspace methods at all […]?
II. When multigrid DOES NOT work fine as a stand–alone solver
Consider the following cases.
II.1 More complex elliptic problems
It is usual to have jumping coefficients. For one, consider the following magnetostatic Poisson problem $$ \nabla \cdot \left( \frac{1}{\hat{\mu}} \, \nabla A_z \right) = \mu_0 \, J_z. $$
It is natural to have different values for magnetic permeability $\hat{\mu}$ for a magnet (gray box on the picture above) and air; current density $J_z$ is only nonzero within wires (red and blue squares on the picture above).
MG as a stand–alone solver loses its robustness (or even diverges) with respect to jump sizes for such problems, although the configuration is very simple.
However, it still works nice as a preconditioner for Krylov solvers. If I have some free time, I will add a similar table for this problem.
There are lots of simple elliptic problems for which MG does not work. It is useful to look through a paper I mentioned in this answer in order to get a deeper understanding.
II.2 “Real–life” problems
Complex real–life problems involve different physics, several unknown fields and so forth. From the algebraic point of view it usually means block structure of the resulting (non-)linear systems. (Consider, for one, Stokes or Oseen problems, fluid–structure–interaction models typical for haemodynamics, thermoelasticity etc.)
It turns out that MG is in general a poor choice for such kind of problems. However, these problems usually “contain” some kind of “diffusion” (read: elliptic / Laplace blocks in the matrix), and one usually keeps this in mind while constructing preconditioners. Indeed, one often uses some kind of MG–cycles to approximate actions of inverses of elliptic blocks. One example I mentioned in this answer.
There are multiple answers to your question.
For one, as others have said, multigrid (MG) and krylov based solvers can work really well
together, so this is a good reason to still use krylov solvers.
A second good reason to use krylov solvers is if the matrix to be inverted is ill conditioned ( for example due to discontinuous material properties ). In this case, MG does not guarantee convergence whereas Congugate Gradient method (e.g.) will. |
Difference between revisions of "Timeline of prime gap bounds"
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[http://math.mit.edu/~drew/admissible_35146_395122.txt 395,122]? [m=2] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comments Sutherland])
−
473,244,502? [m=4] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comments Sutherland])
+
473,244,502? [m=4] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comments
+ + + +
Sutherland])
| A numerical precision issue was discovered in the earlier m=4 calculations
| A numerical precision issue was discovered in the earlier m=4 calculations
|}
|}
Revision as of 17:10, 22 December 2013 [math]H = H_1[/math] is a quantity such that there are infinitely many pairs of consecutive primes of distance at most [math]H[/math] apart. Would like to be as small as possible (this is a primary goal of the Polymath8 project). [math]k_0[/math] is a quantity such that every admissible [math]k_0[/math]-tuple has infinitely many translates which each contain at least two primes. Would like to be as small as possible. Improvements in [math]k_0[/math] lead to improvements in [math]H[/math]. (The relationship is roughly of the form [math]H \sim k_0 \log k_0[/math]; see the page on finding narrow admissible tuples.) More recent improvements on [math]k_0[/math] have come from solving a Selberg sieve variational problem. [math]\varpi[/math] is a technical parameter related to a specialized form of the Elliott-Halberstam conjecture. Would like to be as large as possible. Improvements in [math]\varpi[/math] lead to improvements in [math]k_0[/math], as described in the page on Dickson-Hardy-Littlewood theorems. In more recent work, the single parameter [math]\varpi[/math] is replaced by a pair [math](\varpi,\delta)[/math] (in previous work we had [math]\delta=\varpi[/math]). These estimates are obtained in turn from Type I, Type II, and Type III estimates, as described at the page on distribution of primes in smooth moduli.
In this table, infinitesimal losses in [math]\delta,\varpi[/math] are ignored.
Date [math]\varpi[/math] or [math](\varpi,\delta)[/math] [math]k_0[/math] [math]H[/math] Comments 10 Aug 2005 6 [EH] 16 [EH] ([Goldston-Pintz-Yildirim]) First bounded prime gap result (conditional on Elliott-Halberstam) 14 May 2013 1/1,168 (Zhang) 3,500,000 (Zhang) 70,000,000 (Zhang) All subsequent work (until the work of Maynard) is based on Zhang's breakthrough paper. 21 May 63,374,611 (Lewko) Optimises Zhang's condition [math]\pi(H)-\pi(k_0) \gt k_0[/math]; can be reduced by 1 by parity considerations 28 May 59,874,594 (Trudgian) Uses [math](p_{m+1},\ldots,p_{m+k_0})[/math] with [math]p_{m+1} \gt k_0[/math] 30 May 59,470,640 (Morrison)
58,885,998? (Tao)
59,093,364 (Morrison)
57,554,086 (Morrison)
Uses [math](p_{m+1},\ldots,p_{m+k_0})[/math] and then [math](\pm 1, \pm p_{m+1}, \ldots, \pm p_{m+k_0/2-1})[/math] following [HR1973], [HR1973b], [R1974] and optimises in m 31 May 2,947,442 (Morrison)
2,618,607 (Morrison)
48,112,378 (Morrison)
42,543,038 (Morrison)
42,342,946 (Morrison)
Optimizes Zhang's condition [math]\omega\gt0[/math], and then uses an improved bound on [math]\delta_2[/math] 1 Jun 42,342,924 (Tao) Tiny improvement using the parity of [math]k_0[/math] 2 Jun 866,605 (Morrison) 13,008,612 (Morrison) Uses a further improvement on the quantity [math]\Sigma_2[/math] in Zhang's analysis (replacing the previous bounds on [math]\delta_2[/math]) 3 Jun 1/1,040? (v08ltu) 341,640 (Morrison) 4,982,086 (Morrison)
4,802,222 (Morrison)
Uses a different method to establish [math]DHL[k_0,2][/math] that removes most of the inefficiency from Zhang's method. 4 Jun 1/224?? (v08ltu)
1/240?? (v08ltu)
4,801,744 (Sutherland)
4,788,240 (Sutherland)
Uses asymmetric version of the Hensley-Richards tuples 5 Jun 34,429? (Paldi/v08ltu) 4,725,021 (Elsholtz)
4,717,560 (Sutherland)
397,110? (Sutherland)
4,656,298 (Sutherland)
389,922 (Sutherland)
388,310 (Sutherland)
388,284 (Castryck)
388,248 (Sutherland)
387,982 (Castryck)
387,974 (Castryck)
[math]k_0[/math] bound uses the optimal Bessel function cutoff. Originally only provisional due to neglect of the kappa error, but then it was confirmed that the kappa error was within the allowed tolerance.
[math]H[/math] bound obtained by a hybrid Schinzel/greedy (or "greedy-greedy") sieve
6 Jun 387,960 (Angelveit)
387,904 (Angeltveit)
Improved [math]H[/math]-bounds based on experimentation with different residue classes and different intervals, and randomized tie-breaking in the greedy sieve. 7 Jun
26,024? (vo8ltu)
387,534 (pedant-Sutherland)
Many of the results ended up being retracted due to a number of issues found in the most recent preprint of Pintz. Jun 8 286,224 (Sutherland)
285,752 (pedant-Sutherland)
values of [math]\varpi,\delta,k_0[/math] now confirmed; most tuples available on dropbox. New bounds on [math]H[/math] obtained via iterated merging using a randomized greedy sieve. Jun 9 181,000*? (Pintz) 2,530,338*? (Pintz) New bounds on [math]H[/math] obtained by interleaving iterated merging with local optimizations. Jun 10 23,283? (Harcos/v08ltu) 285,210 (Sutherland) More efficient control of the [math]\kappa[/math] error using the fact that numbers with no small prime factor are usually coprime Jun 11 252,804 (Sutherland) More refined local "adjustment" optimizations, as detailed here.
An issue with the [math]k_0[/math] computation has been discovered, but is in the process of being repaired.
Jun 12 22,951 (Tao/v08ltu)
22,949 (Harcos)
249,180 (Castryck) Improved bound on [math]k_0[/math] avoids the technical issue in previous computations. Jun 13 Jun 14 248,898 (Sutherland) Jun 15 [math]348\varpi+68\delta \lt 1[/math]? (Tao) 6,330? (v08ltu)
6,329? (Harcos)
6,329 (v08ltu)
60,830? (Sutherland) Taking more advantage of the [math]\alpha[/math] convolution in the Type III sums Jun 16 [math]348\varpi+68\delta \lt 1[/math] (v08ltu) 60,760* (Sutherland) Attempting to make the Weyl differencing more efficient; unfortunately, it did not work Jun 18 5,937? (Pintz/Tao/v08ltu)
5,672? (v08ltu)
5,459? (v08ltu)
5,454? (v08ltu)
5,453? (v08ltu)
60,740 (xfxie)
58,866? (Sun)
53,898? (Sun)
53,842? (Sun)
A new truncated sieve of Pintz virtually eliminates the influence of [math]\delta[/math] Jun 19 5,455? (v08ltu)
5,453? (v08ltu)
5,452? (v08ltu)
53,774? (Sun)
53,672*? (Sun)
Some typos in [math]\kappa_3[/math] estimation had placed the 5,454 and 5,453 values of [math]k_0[/math] into doubt; however other refinements have counteracted this Jun 20 [math]178\varpi + 52\delta \lt 1[/math]? (Tao)
[math]148\varpi + 33\delta \lt 1[/math]? (Tao)
Replaced "completion of sums + Weil bounds" in estimation of incomplete Kloosterman-type sums by "Fourier transform + Weyl differencing + Weil bounds", taking advantage of factorability of moduli Jun 21 [math]148\varpi + 33\delta \lt 1[/math] (v08ltu) 1,470 (v08ltu)
1,467 (v08ltu)
12,042 (Engelsma) Systematic tables of tuples of small length have been set up here and here (update: As of June 27 these tables have been merged and uploaded to an online database of current bounds on [math]H(k)[/math] for [math]k[/math] up to 5000). Jun 22 Slight improvement in the [math]\tilde \theta[/math] parameter in the Pintz sieve; unfortunately, it does not seem to currently give an actual improvement to the optimal value of [math]k_0[/math] Jun 23 1,466 (Paldi/Harcos) 12,006 (Engelsma) An improved monotonicity formula for [math]G_{k_0-1,\tilde \theta}[/math] reduces [math]\kappa_3[/math] somewhat Jun 24 [math](134 + \tfrac{2}{3}) \varpi + 28\delta \le 1[/math]? (v08ltu)
[math]140\varpi + 32 \delta \lt 1[/math]? (Tao)
1,268? (v08ltu) 10,206? (Engelsma) A theoretical gain from rebalancing the exponents in the Type I exponential sum estimates Jun 25 [math]116\varpi+30\delta\lt1[/math]? (Fouvry-Kowalski-Michel-Nelson/Tao) 1,346? (Hannes)
1,007? (Hannes)
10,876? (Engelsma) Optimistic projections arise from combining the Graham-Ringrose numerology with the announced Fouvry-Kowalski-Michel-Nelson results on d_3 distribution Jun 26 [math]116\varpi + 25.5 \delta \lt 1[/math]? (Nielsen)
[math](112 + \tfrac{4}{7}) \varpi + (27 + \tfrac{6}{7}) \delta \lt 1[/math]? (Tao)
962? (Hannes) 7,470? (Engelsma) Beginning to flesh out various "levels" of Type I, Type II, and Type III estimates, see this page, in particular optimising van der Corput in the Type I sums. Integrated tuples page now online. Jun 27 [math]108\varpi + 30 \delta \lt 1[/math]? (Tao) 902? (Hannes) 6,966? (Engelsma) Improved the Type III estimates by averaging in [math]\alpha[/math]; also some slight improvements to the Type II sums. Tuples page is now accepting submissions. Jul 1 [math](93 + \frac{1}{3}) \varpi + (26 + \frac{2}{3}) \delta \lt 1[/math]? (Tao)
873? (Hannes)
Refactored the final Cauchy-Schwarz in the Type I sums to rebalance the off-diagonal and diagonal contributions Jul 5 [math] (93 + \frac{1}{3}) \varpi + (26 + \frac{2}{3}) \delta \lt 1[/math] (Tao)
Weakened the assumption of [math]x^\delta[/math]-smoothness of the original moduli to that of double [math]x^\delta[/math]-dense divisibility
Jul 10 7/600? (Tao) An in principle refinement of the van der Corput estimate based on exploiting additional averaging Jul 19 [math](85 + \frac{5}{7})\varpi + (25 + \frac{5}{7}) \delta \lt 1[/math]? (Tao) A more detailed computation of the Jul 10 refinement Jul 20 Jul 5 computations now confirmed Jul 27 633 (Tao)
632 (Harcos)
4,686 (Engelsma) Jul 30 [math]168\varpi + 48\delta \lt 1[/math]# (Tao) 1,788# (Tao) 14,994# (Sutherland) Bound obtained without using Deligne's theorems. Aug 17 1,783# (xfxie) 14,950# (Sutherland) Oct 3 13/1080?? (Nelson/Michel/Tao) 604?? (Tao) 4,428?? (Engelsma) Found an additional variable to apply van der Corput to Oct 11 [math]83\frac{1}{13}\varpi + 25\frac{5}{13} \delta \lt 1[/math]? (Tao) 603? (xfxie) 4,422?(Engelsma)
12 [EH] (Maynard)
Worked out the dependence on [math]\delta[/math] in the Oct 3 calculation Oct 21 All sections of the paper relating to the bounds obtained on Jul 27 and Aug 17 have been proofread at least twice Oct 23 700#? (Maynard) Announced at a talk in Oberwolfach Oct 24 110#? (Maynard) 628#? (Clark-Jarvis) With this value of [math]k_0[/math], the value of [math]H[/math] given is best possible (and similarly for smaller values of [math]k_0[/math]) Nov 19 105# (Maynard)
5 [EH] (Maynard)
600# (Maynard/Clark-Jarvis) One also gets three primes in intervals of length 600 if one assumes Elliott-Halberstam Nov 20 Optimizing the numerology in Maynard's large k analysis; unfortunately there was an error in the variance calculation Nov 21 68?? (Maynard)
582#*? (Nielsen])
59,451 [m=2]#? (Nielsen])
42,392 [m=2]? (Nielsen)
356?? (Clark-Jarvis) Optimistically inserting the Polymath8a distribution estimate into Maynard's low k calculations, ignoring the role of delta Nov 22 388*? (xfxie)
448#*? (Nielsen)
43,134 [m=2]#? (Nielsen)
698,288 [m=2]#? (Sutherland) Uses the m=2 values of k_0 from Nov 21 Nov 23 493,528 [m=2]#? Sutherland Nov 24 484,234 [m=2]? (Sutherland) Nov 25 385#*? (xfxie) 484,176 [m=2]? (Sutherland) Using the exponential moment method to control errors Nov 26 102# (Nielsen) 493,426 [m=2]#? (Sutherland) Optimising the original Maynard variational problem Nov 27 484,162 [m=2]? (Sutherland) Nov 28 484,136 [m=2]? (Sutherland Dec 4 64#? (Nielsen) 330#? (Clark-Jarvis) Searching over a wider range of polynomials than in Maynard's paper Dec 6 493,408 [m=2]#? (Sutherland) Dec 19 59#? (Nielsen)
10,000,000? [m=3] (Tao)
1,700,000? [m=3] (Tao)
38,000? [m=2] (Tao)
300#? (Clark-Jarvis)
182,087,080? [m=3] (Sutherland)
179,933,380? [m=3] (Sutherland)
More efficient memory management allows for an increase in the degree of the polynomials used; the m=2,3 results use an explicit version of the [math]M_k \geq \frac{k}{k-1} \log k - O(1)[/math] lower bound. Dec 20
55#? (Nielsen)
36,000? [m=2] (xfxie)
175,225,874? [m=3] (Sutherland)
27,398,976? [m=3] (Sutherland)
Dec 21 1,640,042? [m=3] (Sutherland) 429,798? [m=2] (Sutherland) Optimising the explicit lower bound [math]M_k \geq \log k-O(1)[/math] Dec 22 1,628,944? [m=3] (Castryck)
75,000,000? [m=4] (Castryck)
3,400,000,000? [m=5] (Castryck)
395,154? [m=2] (Sutherland)
1,523,781,850? [m=4] (Sutherland)
82,575,303,678? [m=5] (Sutherland)
A numerical precision issue was discovered in the earlier m=4 calculations Legend: ? - unconfirmed or conditional ?? - theoretical limit of an analysis, rather than a claimed record * - is majorized by an earlier but independent result # - bound does not rely on Deligne's theorems [EH] - bound is conditional the Elliott-Halberstam conjecture [m=N] - bound on intervals containing N+1 consecutive primes, rather than two strikethrough - values relied on a computation that has now been retracted
See also the article on
Finding narrow admissible tuples for benchmark values of [math]H[/math] for various key values of [math]k_0[/math]. |
Suppose we have a function $g$ and two random variables $\tilde{X} = (\tilde{X}_1, \tilde{X}_2, \tilde{X}_3)$ and $X = (X_1, X_2, X_3)$ which are iid. Furthermore, $\tilde{X}_1, \tilde{X}_2, \tilde{X}_3$ and $X_1, X_2, X_3$ are independent random vectors. I am interested in the expectation $\mathbb{E}(g(\tilde{X}_1, X_2, X_3)) $.
According to Serfling, Robert J. (1980), I could use the V-statistic (or U-statistic) by defining the kernel$$ h(\tilde{X}, X) = h((\tilde{X}_1, \tilde{X}_2, \tilde{X}_3), (X_1, X_2, X_3)) := g(\tilde{X}_1, X_2, X_3)$$ and using $\mathbb{E}( h(\tilde{X}, X))$.
But I would like to do this differently, for example: \begin{align*} \mathbb{E}(g(\tilde{X}_1, X_2, X_3)) &= \mathbb{E}_{(X_2, X_3)}(\mathbb{E}_{\tilde{X_1} \mid (X_2, X_3)} (g(\tilde{X}_1, X_2, X_3)) \\ &= \mathbb{E}_{(X_2, X_3)}(\mathbb{E}_{\tilde{X_1}} (g(\tilde{X}_1, X_2, X_3)) \\ &= \mathbb{E}_{(X_2, X_3)}(\mathbb{E}_{X_1} (g(X_1, X_2, X_3)) \\ \end{align*}
My idea was to 1) use the law of total expectation (first line), then remove the condition due to independence (second line) and finally use$$ \mathbb{E}_{X_1} (g(X_1, X_2, X_3) = \mathbb{E}_{\tilde{X}_1} (g(\tilde{X}_1, X_2, X_3)$$ since $X_1$ and $\tilde{X}_1$ are iid.
Except that I get a biased estimation analogous to the V-statistic, does someone see a problem in my approach? Is it mathematically correct to do this? |
I have implemented a pretty straightforward finite element solver for the following Poisson equation. For the purposes of this question we can assume the source term and the Dirichlet data both vanish. I assume that $g\in H^{-1/2}(\partial \Omega_N) \cap L^2(\partial \Omega_N)$ (instead of just $g\in H^{-1/2}(\partial\Omega_N)$)
\begin{aligned} \mathbf{q} + \mathbf{\nabla}u &= 0 & \mathrm{in\,} \Omega,\\ \mathbf{\nabla} \cdot \mathbf{q} &= 0& \mathrm{in\ } \Omega,\\ u& = 0 &\mathrm{on\ } \partial\Omega_D,\\ \mathbf{q}\cdot \mathbf{\eta} &= g & \mathrm{on\ } \partial\Omega_N. \end{aligned}
I accounted for Neumann boundary conditions on all or some of the boundary by using Lagrange multipliers as follows.
Let the The domain $\Omega$ have a triangulation $T$ made up of simplexes $K$. We denote $$\mathbf{W}_h= \{\mathbf{p}\in \mathbf{H}^{\mathrm{div}}(\Omega)\colon \forall K\in T \quad \mathbf{p}\big|_K \in \mathrm{RT}^k(K))$$ $$V_h = \{v\in L^2(\Omega)\colon \forall K \quad v\big|_K \in P^k(K)\}$$ $$X_h = \{\mu \in L^2(\partial \Omega_N)\colon \forall K\in T\,\, \forall \mathrm{edge}\subset\partial K\cap \partial\Omega_N\quad \mu \big|_{\mathrm{edge}} \in P^k(\mathrm{edge})\}$$
Where $\mathrm{RT}^k(K)$ is the standard Raviart Thomas space and $P^k(K)$ is the space of polynomials of total degree no more than $k$.
The problem is to find the triple $(\mathbf{q}_h,u_h,\lambda_h) \in \mathbf{W}_h \times V_h \times X_h$ so that for all test functions $(\mathbf{p}, v, s) \in (W_h \times V_h \times X_h)$ we have \begin{aligned} 0 &= \int_\Omega \mathbf{p}\cdot \mathbf{q}_h\, \mathrm{d} x - \int_\Omega (\mathbf{\nabla} \cdot \mathbf{p}) u_h\, \mathrm{d} x + \int_{\partial\Omega_N} \mathbf{p} \lambda_h \cdot \mathbf{\eta}\, \mathrm{d} x \\ 0 &= \int_\Omega v (\mathbf{\nabla} \cdot \mathbf{q}_h)\, \mathrm{d} x\\ \int_{\partial\Omega_N}s g \, \mathrm{d}x&= \int_{\partial\Omega_N} s\mathbf{q}_h \cdot \mathbf{\eta}\, \mathrm{d} x \end{aligned}
The scheme works (well defined, converges to true solution, etc.) and is (I think) close to standard practice. However, in all the references I have seen in the literature (On the Mixed Finite Element Method with Lagrange Multipliers Babuska and Gatica, 2003 Numerical Methods for Partial Differential Equation and section 4.4 of A simple Introduction to the Mixed Finite Element Method by Gatica) require that the trial and test space for the Lagrange multiplier, $X_h$, have elements that are continuous across elements of the "boundary mesh".
Is this scheme close to standard practice? Are there any references I could read that discuss this scheme? What problems am I allowing by choosing a piecewise discontinuous space for the Lagrange multiplier? A disadvantage of the scheme is that it requires extra regularity on $g$. Are there more disadvantages? |
$$ \newcommand{\bsth}{{\boldsymbol\theta}} \newcommand{\va}{\textbf{a}} \newcommand{\vb}{\textbf{b}} \newcommand{\vc}{\textbf{c}} \newcommand{\vd}{\textbf{d}} \newcommand{\ve}{\textbf{e}} \newcommand{\vf}{\textbf{f}} \newcommand{\vg}{\textbf{g}} \newcommand{\vh}{\textbf{h}} \newcommand{\vi}{\textbf{i}} \newcommand{\vj}{\textbf{j}} \newcommand{\vk}{\textbf{k}} \newcommand{\vl}{\textbf{l}} \newcommand{\vm}{\textbf{m}} \newcommand{\vn}{\textbf{n}} \newcommand{\vo}{\textbf{o}} \newcommand{\vp}{\textbf{p}} \newcommand{\vq}{\textbf{q}} \newcommand{\vr}{\textbf{r}} \newcommand{\vs}{\textbf{s}} \newcommand{\vt}{\textbf{t}} \newcommand{\vu}{\textbf{u}} \newcommand{\vv}{\textbf{v}} \newcommand{\vw}{\textbf{w}} \newcommand{\vx}{\textbf{x}} \newcommand{\vy}{\textbf{y}} \newcommand{\vz}{\textbf{z}} \DeclareMathOperator*{\argmin}{argmin} \DeclareMathOperator\mathProb{\mathbb{P}} \renewcommand{\P}{\mathProb} % need to overwrite stupid paragraph symbol \DeclareMathOperator\mathExp{\mathbb{E}} \newcommand{\E}{\mathExp} \DeclareMathOperator\Uniform{Uniform} \DeclareMathOperator\poly{poly} \DeclareMathOperator\diag{diag} \newcommand{\pa}[1]{ \left({#1}\right) } \newcommand{\ha}[1]{ \left[{#1}\right] } \newcommand{\ca}[1]{ \left\{{#1}\right\} } \newcommand{\norm}[1]{\left\| #1 \right\|} \newcommand{\nptime}{\textsf{NP}} \newcommand{\ptime}{\textsf{P}} \newcommand{\R}{\mathbb{R}} \newcommand{\card}[1]{\left\lvert{#1}\right\rvert} \newcommand{\abs}[1]{\card{#1}} \newcommand{\sg}{\mathop{\mathrm{SG}}} \newcommand{\se}{\mathop{\mathrm{SE}}} \newcommand{\mat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \DeclareMathOperator{\var}{var} \DeclareMathOperator{\cov}{cov} \newcommand\independent{\perp\kern-5pt\perp} \newcommand{\CE}[2]{ \mathExp\left[ #1 \,\middle|\, #2 \right] } \newcommand{\disteq}{\overset{d}{=}} $$
Beating TensorFlow Training in-VRAM
In this post, I’d like to introduce a technique that I’ve found helps accelerate mini-batch SGD training in my use case. I suppose this post could also be read as a public grievance directed towards the TensorFlow Dataset API optimizing for the large vision deep learning use-case, but maybe I’m just not hitting the right incantation to get
tf.Dataset working (in which case, drop me a line). The solution is to TensorFlow
harder anyway, so this shouldn’t really be read as a complaint.
Nonetheless, if you are working with a new-ish GPU that has enough memory to hold a decent portion of your data alongside your neural network, you may find the final training approach I present here useful. The experiments I’ve run fall exactly in line with this “in-VRAM” use case (in particular, I’m training deep reinforcement learning value and policy networks on semi-toy environments, whose training profile is many iterations of training on a small replay buffer of examples). For some more context, you may want to check out an article on the TensorForce blog, which suggests that RL people should be building more of their TF graphs like this.
Briefly, if you have a dataset that fits into a GPU’s memory, you’re giving away a lot of speed with the usual TensorFlow pipelining or data-feeding approach, where the CPU delivers mini-batches whose forward/backward passes are computed on GPUs. This gets worse as you move to pricier GPUs, whose relative CPU-GPU bandwidth-to-GPU-speed ratio drops. Pretty easy change for a 2x.
Punchline
Let’s get to it. With numbers similar to my use case, 5 epochs of training take about
16 seconds with the standard
feed_dict approach,
12-20 seconds with the TensorFlow Dataset API, and 8 seconds with a custom TensorFlow control-flow construct.
This was tested on an Nvidia Tesla P100 with a compiled TensorFlow 1.4.1 (CUDA 9, cuDNN 7), Python 3.5. Here is the test script. I didn’t test it too many times (exec trace). Feel free to change the data sizes to see if the proposed approach would still help in your setting.
Let’s fix the toy benchmark supervised task we’re looking at:
Vanilla Approach
This is the (docs-discouraged) approach that everyone really uses for training. Prepare a mini-batch on the CPU, ship it off to the GPU.
Note code here and below is excerpted (see the test script link above for the full code). It won’t work if you just copy it.
This drops whole-dataset loss from around 4500 to around 4, taking around
16 seconds for training. You might worry that random-number generation might be taking a while, but excluding that doesn’t drop the time more than 0.5 seconds. Dataset API Approach
With the dataset API, we set up a pipeline where TensorFlow orchestrates some dataflow by synergizing more buzzwords on its worker threads. This should constantly feed the GPU by staging the next mini-batch while the current one is sitting on the GPU. This might be the case when there’s a lot of data, but it doesn’t seem to work very well when the data is small and GPU-CPU latency, not throughput, is the bottleneck.
Another unpleasant thing to deal with is that all those orchestrated workers and staging areas and buffers and shuffle queues need magic constants to work well. I tried my best, but it seems like performance is very sensitive with this use case. This could be fixed if Dataset detected (or could be told) it could be placed onto the GPU, and then it did so.
For a small
bufsize, like
1000, this trains in around
12 seconds. But then it’s not actually shuffling the data too well (since all data points can only move by a position of 1000). Still, the loss drops from around 4500 to around 4, as in the
feed_dict case. A large
bufsize like
1000000, which you’d think should effectively move the dataset onto the GPU entirely, performs
worse than
feed_dict at around
20 seconds.
I don’t think I’m unfair in counting
it.initializer time in my benchmark (which isn’t that toy, either, since it’s similar to my RL use case size). All the training methods need to load the data onto the GPU, and the data isn’t available until run time.
Using a TensorFlow Loop
This post isn’t a tutorial on
tf.while_loop and friends, but this code does what was promised: just feed everything once into the GPU and do all your epochs without asking for permission to continue from the CPU.
This one crushes at around
8 seconds, dropping loss again from around 4500 to around 4. Discussion
It’s pretty clear Dataset isn’t feeding as aggressively as it can, and its many widgets and knobs don’t help (well, they do, but only after making me do more work). But, if TF wants to invalidate this blog post, I suppose it could add yet another option that plops the dataset into the GPU. |
The Hartree method discussed previously is useful as an introduction to the solution of many-particle system and to the concepts of self-consistency and of the self-consistent- eld, but its importance is con ned to the history of physics. In fact the Hartree method is not just approximate: it is wrong, by construction, since its wavefunction is not antisymmetric to electron permutation! The Hartree-Fock approach discussed below is a better approach, which correctly takes into account the antisymmetric character of the trial wavefunctions.
Although these
Hartree equations are numerically tractable via the self-consistent field method, it is not surprising that such a crude approximation fails to capture elements of the essential physics. The Pauli exclusion principle demands that the many-body wavefunction be antisymmetric with respect to interchange of any two electron coordinates, e.g.
\[\Psi(\mathbf{r}_{1},\mathbf{r}_{2}, \ldots, \mathbf{r}_{N}) = - \Psi(\mathbf{r}_{2},\mathbf{r}_{1}, \ldots, \mathbf{r}_{N}) \label{2.7}\]
which clearly cannot be satisfied by the multi-electron wavefunctions of the form used in the Hartree Approximation (Equation \(\ref{2.3}\)).
\[\Psi(\mathbf{r}_1,\mathbf{r}_2, \ldots, \mathbf{r}_N) \approx \psi_{1}(\mathbf{r}_1)\psi_{2}(\mathbf{r}_2) \ldots \psi_{N}(\mathbf{r}_N) \label{2.3}\]
This indistinguishability condition can be satisfied by forming a Slater determinant of single-particle orbitals
\[\Psi(\mathbf{r}_{1}, \mathbf{r}_{2}, \ldots, \mathbf{r}_{N})= \dfrac{1}{\sqrt{N}} \left \vert\psi(\mathbf{r}_{1})\psi(\mathbf{r}_{2}) \ldots \psi(\mathbf{r}_{N}) \right\vert \label{2.8}\]
This decouples the electrons resulting in single-particle
Hartree-Fock equations:
\[-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r}) + V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r}) + V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r}) - \sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} = \epsilon_{i}\psi_{i}(\mathbf{r}). \label{2.9}\]
As with the
Hartree equations, the first term is the kinetic energy of the \(i^{th}\) electron
\[-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r}) \]
and the second term is the electron-nuclear potential between the \(i^{th}\) electron and nucleus
\[V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r}) \]
The third term (sometimes called the “Hartree” term) is the electrostatic potential between the \(i^{th}\) electron and the
average charge distribution of the other N-1 electrons.
\[V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r}) = J_{j,k} = \int |\phi_j(r)|^2 |\phi_k(r’)|^2 \dfrac{e^2}{r-r'} dr dr’ \label{8.3.9}\]
These three terms are identical to Hartree Equations with the product wavefunction ansatz (i.e., orbital approximation). The fourth term of Equation \(\ref{2.9}\) is not in the Hartree Equations:
\[\sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert}\]
and is the
exchange term. This term resembles the direct Coulomb term, but for the exchanged indices. It is a manifestation of the Pauli exclusion principle, and acts so as to separate electrons of the same spin. This “exchange” term acts only on electrons with the same spin and comes from the Slater determinant form of the wavefunction. Physically, the effect of exchange is for like-spin electrons to avoid each other. The exchange term adds considerably to the complexity of these equations.
The Hartree-Fock Equations in Equation \(\ref{2.9}\) can be recast as series of Schrödinger-like equations:
\[ \hat {F} | \varphi _i \rangle = \epsilon _i| \varphi _i \rangle \label {8.7.2}\]
where \(\hat {F}\) is called the
Fock operator and \( \{| \varphi_i \rangle \}\) are the Hatree-Fock orbitals with corresponding energies \(\epsilon_i\).
The Fock operator is a one-electron operator and solving a Hartree-Fock equation gives the energy and Hartree-Fock orbital for one electron. For a system with 2N electrons, the variable i will range from 1 to N; i.e there will be one equation for each orbital. The reason for this is that only the spatial wavefunctions are used in Equation \(\ref{8.7.2}\). Since the spatial portion of an orbital can be used to describe two electrons, each of the energies and wavefunctions found by solving Equation \(\ref{8.7.2}\) will be used to describe two electrons.
The nature of the Fock operator reveals how the Hartree-Fock (HF) or Self-Consistent Field (SCF) Method accounts for the electron-electron interaction in atoms and molecules while preserving the idea of
independent atomic orbitals. The wavefunction written as a Slater determinant of spin-orbitals is necessary to derive the form of the Fock operator, which is
\[\hat {F} = \hat {H} ^0 + \sum _{j=1}^N ( 2 \hat {J} _j - \hat {K} _j ) = -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {Ze^2}{4 \pi \epsilon _0 r} + \sum _{j=1}^N (2\hat {J}_j - \hat {K} _j ) \label {8.7.3}\]
As shown by the expanded version on the far right, the first term in this equation, \(\hat {H}^0\), is the familiar hydrogen-like operator that accounts for the kinetic energy of an electron and the potential energy of this electron interacting with the nucleus. The next term accounts for the potential energy of one electron in an average field created by all the other electrons in the system. The \(\hat {J}\) and \(\hat {K}\) operators result from the electron-electron repulsion terms in the full Hamiltonian for a multi-electron system. These operators involve the one-electron orbitals as well as the electron-electron interaction energy.
The Fock operator (Equation \(\ref{8.7.3}\) depends on
all occupied orbitals (because of the exchange and Coulomb operators). Therefore, a specific orbital can only be determined if all the others are known. One must use iterative methods to solve the HF equations like the Self-consistent field method discussed previously for the Hartree Approximation.
Exchange Energy
The exchange interaction is a quantum mechanical effect that only occurs between identical particles. Despite sometimes being called an exchange force in analogy to classical force,
it is not a true force, as it lacks a force carrier. The effect is due to the wavefunction of indistinguishable particles being subject to exchange symmetry, that is, either remaining unchanged (symmetric) or changing its sign (antisymmetric) when two particles are exchanged. Both bosons and fermions can experience the exchange interaction. For fermions, it is sometimes called Pauli repulsion and related to the Pauli exclusion principle. For bosons, the exchange interaction takes the form of an effective attraction that causes identical particles to be found closer together, as in Bose–Einstein condensation.
Example \(\PageIndex{1}\): Hartree-Fock Energy of Helium
For example, the electron 1 in helium (with \(Z=2\), then
\[\hat {H}^0 (1) = - \dfrac {\hbar ^2}{2m} \nabla ^2_1 - \dfrac {2e^2}{4 \pi \epsilon _0 r_1} \label {8.7.4}\]
The Fock operator is couched in terms of the coordinates of the one electron whose perspective we are taking (which we will call electron 1 throughout the following discussion), and the
average field created by all the other electrons in the system is built in terms of the coordinates of a generic “other electron” (which we’ll call electron 2) that is considered to occupy each orbital in turn during the summation over the N spatial orbitals.
As with the Hartree Equations, solving the Hartree-Fock Equations is mathematically equivalent to assuming each electron interacts only with the average charge cloud of the other electrons. This is how the electron-electron repulsion is handled. This also why this approach is also called the Self-Consistant Field (SCF) approach.
The best possible one-electron wavefunctions, by definition, will give the
lowest possible total energy for a multi-electron system used with the complete multielectron Hamiltonian to calculate the expectation value for the total energy of the system. These wavefunctions are called the Hartree-Fock wavefunctions and the calculated total energy is the Hartree-Fock energy of the system. Hartree-Fock Energy
The Hartree-Fock equations \(h_e \phi_i = \epsilon_i \phi_i\) imply that the orbital energies \(\epsilon_i\) can be written as:
\[ \begin{align} \epsilon_i &= \langle \phi_i | h_e | \phi_i \rangle \\[4pt] &= \langle \phi_i | T + V | \phi_i \rangle + \sum_{j({\rm occupied})} \langle \phi_i | J_j - K_j | \phi_i \rangle \label{8.7.6} \\[4pt] &= \langle \phi_i | T + V | \phi_i \rangle + \sum_{j({\rm occupied})} [ J_{i,j} - K_{i,j} ],\label{8.7.7} \end{align}\]
where \(T + V\) represents the kinetic (\(T\)) and nuclear attraction (\(V\)) energies, respectively. Thus, \(\epsilon_i\) is the average value of the kinetic energy plus Coulombic attraction to the nuclei for an electron in \(\phi_i\) plus the sum over all of the spin-orbitals occupied in \(\psi\) of Coulomb minus Exchange interactions of these electrons with the electron in \(\phi_i\).
If \(\phi_i\) is an occupied spin-orbital, the \(j = i\) term \([ J_{i,i} - K_{i,i}]\) disappears in the above sum and the remaining terms in the sum represent the Coulomb minus exchange interaction of \(\phi_i\) with all of the \(N-1\) other occupied spin-orbitals. If \(\phi_i\) is a virtual spin-orbital, this cancelation does not occur because the sum over \(j\) does not include \(j = i\). So, one obtains the Coulomb minus exchange interaction of \(\phi_i\) with all \(N\) of the occupied spin-orbitals in \(\psi\). Hence the energies of occupied orbitals pertain to interactions appropriate to a total of \(N\) electrons, while the energies of virtual orbitals pertain to a system with \(N+1\) electrons. This difference is very important to understand and to keep in mind.
To give an idea of how well HF theory can predict the ground state energies of several atoms, consider Table \(\PageIndex{1}\) below:
Atom Hartree-Fock Energy Experiment \(He\) \(-5.72\) \(-5.80\) \(Li\) \(-14.86\) \(-14.96\) \(Ne\) \(-257.10\) \(-257.88\) \(Ar\) \(-1053.64\) \(-1055.20\) Koopmans' Theorem
Koopmans' theorem states that the first ionization energy is equal to the negative of the orbital energy of the highest occupied molecular orbital. Hence, the ionization energy required to generated a cation and detached electron is represented by the removal of an electron from an orbital without changing the wavefunctions of the other electrons. This is called the "frozen orbital approximation." Let us consider the following model of the detachment or attachment of an electron in an \(N\)-electron system.
In this model, both the parent molecule and the species generated by adding or removing an electron are treated at the single-determinant level. The Hartree-Fock orbitals of the parent molecule are used to describe both species. It is said that such a model neglects orbital relaxation (i.e., the re-optimization of the spin-orbitals to allow them to become appropriate to the daughter species).
Within this model, the energy difference between the daughter and the parent can be written as follows (\(\phi_k\) represents the particular spin-orbital that is added or removed:
for electron detachment (vertical ionization energies) \[ \color{red} E_{N-1} - E_N = - \epsilon_k \label{8.7.8}\] and for electron attachment (electron affinities) \[ \color{red} E_N - E_{N+1} = - \epsilon_k .\label{8.7.9}\]
Example \(\PageIndex{2}\): Electron Affinity
Let’s derive this result for the case in which an electron is added to the \(N+1^{st}\) spin-orbital. The energy of the \(N\)-electron determinant with spin-orbitals \(\phi_1\) through \(f_N\) occupied is
\[E_N = \sum_{i=1}^N \langle \phi_i | T + V | \phi_i \rangle + \sum_{i=1}^{N} [ J_{i,j} - K_{i,j} ] \nonumber\]
which can also be written as
\[E_N = \sum_{i=1}^N \langle \phi_i | T + V | \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N} [ J_{i,j} - K_{i,j} ].\nonumber\]
Likewise, the energy of the \(N+1\)-electron determinant wavefunction is
\[E_{N+1} = \sum_{i=1}^{N+1} \langle \phi_i | T + V | \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N+1} [ J_{i,j} - K_{i,j} ]. \nonumber\]
The difference between these two energies is given by
\[ \begin{align*} E_{N} – E_{N+1} = &- \langle \phi_{N+1} | T + V | \phi_{N+1} \rangle - \frac{1}{2} \sum_{i=1}^{N+1} [ J_{i,N+1} - K_{i,N+1} ] \\[4pt] &- \frac{1}{2} \sum_{j=1}^{N+1} [ J_{N+1,j} - K_{N+1,j} ] \\[4pt] &= - \langle \phi_{N+1} | T + V | \phi_{N+1} \rangle - \sum_{i=1}^{N+1} [ J_{i,N+1} - K_{i,N+1} ] \\[4pt] &= - \epsilon_{N+1}. \end{align*}\]
That is, the energy difference is equal to minus the expression for the energy of the \(N+1^{st}\) spin-orbital, which was given earlier.
The Hartree-Fock equations deal with exchange exactly; however, the equations neglect more detailed correlations due to many-body interactions. The effects of electronic correlations are not negligible; indeed the failure of Hartree-Fock theory to successfully incorporate correlation leads to one of its most celebrated failures.
Advanced: Electron Correlation and the "Exchange Hole"
In the Copenhagen Interpretation, the squared modulus of the wavefunction gives the probability of finding a particle in a given place. The many-body wavefunction gives the N-particle distribution function, i.e. \(|Φ(r_1, ..., r_N )|^2\) is the probability density that particle 1 is at \(r_1\), ..., and particle \(N\) is at \(r_N\). However, when trying to work out the interaction between electrons, what we want to know is the probability of finding an electron at \(r\), given the positions of all the other electrons \(\{r_i\}\). This implies that the electron behaves quantum mechanically when we evaluate its wavefunction, but as a classical point particle when it contributes to the potential seen by the other electrons.
The contributions of electron-electron interactions in N-electron systems within the Hartree and Hartree-Fock methods are shown in Figure \(\PageIndex{2}\). The conditional electron probability distributions \(n(r)\) of \(N-1\) electrons around an electron with given spin situated at \(r=0\). Within the Hartree approximation, all electrons are treated as independent, therefore \(n(r)\) is structureless. However, within the Hartree-Fock approximation, the \(N\)-electron wavefunction reflects the Pauli exclusion principle and near the electron at \(r=0\) the exchange hole can be seen where the the density of spins equal to that of the central electron is reduced. Electrons with opposite spins are
unaffected (not shown). Summary
So, within the limitations of the HF, frozen-orbital model, the ionization potentials (IPs) and electron affinities (EAs) are given as the negative of the occupied and virtual spin-orbital energies, respectively. This statement is referred to as Koopmans’ theorem; it is used extensively in quantum chemical calculations as a means of estimating ionization potentials (Equation \(\ref{8.7.8}\)) and Electron Affinities (Equation \(\ref{8.7.9}\)) and often yields results that are qualitatively correct (i.e., ± 0.5 eV). In general Hartree-Fock theory gives a great first order solution (99%) to describing multi-electron systems, but that last 1% is still too great for quantitatively describing many aspects of chemistry and more sophisticated approaches are necessary. These are discussed elsewhere. |
Trying to prove or disprove this (pretty sure it's correct): Let $\mathcal{F}=\{f\mid f:\mathbb{N}\to\mathbb{R}^+\}$
$$\forall f\in\mathcal{F}: \left\lfloor \sqrt{\lfloor f(n)\rfloor }\right\rfloor \in O\left(\sqrt{f(n)}\right)\;.$$
How can I get rid of both of the floor functions? Because I think that'll make it simpler
Progress: Now I am trying to prove $\lfloor \sqrt{\lfloor x)\rfloor }\rfloor = \lfloor \sqrt{x} \rfloor$ which will then make the rest of proof pretty easy.
Right now I have: $$Let \ k = \lfloor \sqrt{\lfloor x\rfloor }\rfloor$$
$$Then \ k \le \sqrt{\lfloor x\rfloor } < k +1$$ $$Then \ k^2 \le \lfloor x\rfloor < (k+1)^2$$ But I am not too sure how to reach a point where it says $$k \le \sqrt[]{x} < k+1$$ Which is the definition of floor function and conclude they are equal |
This question already has an answer here:
I recently learned that $\left(-\frac{1}{2}\right)! = \sqrt{\pi}$ but I don't understand how that makes sense. Can someone please explain how this is possible?
Thanks!
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This question already has an answer here:
I recently learned that $\left(-\frac{1}{2}\right)! = \sqrt{\pi}$ but I don't understand how that makes sense. Can someone please explain how this is possible?
Thanks!
In order to extend the factorial function to any real number, we introduce the Gamma Function, which is a strange object defined as follows:
$$ \Gamma(s)=\int_0^\infty t^{s-1}e^{-t} \, dt $$
The gamma function comes with the special property that $n!=\Gamma(n+1)$ for
natural numbers $n$, so to evaluate $(-1/2)!$, (which by itself is not technically defined) we define it to be $\Gamma(1/2)$ and hence we evaluate the integral
$$ (-1/2)!:=\Gamma(1/2)=\int_0^\infty\frac{e^{-t}}{\sqrt{t}} \,dt $$ To evaluate this integral, we make the substitution $u=\sqrt{t}$, which results in the well known Gaussian integral:
$$ \int_0^\infty \frac{e^{-t}}{\sqrt{t}}dt=2\int_0^\infty \frac{e^{-u^2}}{u}u \, du=\int_{-\infty}^\infty e^{-u^2} \, du=\sqrt{\pi} $$
I don't know where you've seen this notation. One thing you surely know is the socalled Gamma Function $\Gamma (z)$. This function is a complex function with a host of properties. One of its properties is that if evaluated on $\mathbb{N}$ it coincides with the factorial function, this is, $\Gamma (n+1) = n! $ if $n\in \mathbb{N}$. Also, one can find that $\Gamma (1/2) = \sqrt{\pi}$. So maybe by using a notation I don't know about someone could write $(-1/2)!$ instead of $ \Gamma (1/2)$.
I posted this very question and an answer to it: Why is $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ ?
What follows is the answer I posted there. Several others also posted good answers.
If there's any justice in the universe, someone must have asked here how to show that $$ \int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt{2\pi}. $$ Let's suppose that has been answered here. Let (capital) $X$ be a random variable whose probability distribution is $$ \frac{e^{-x^2/2}}{\sqrt{2\pi}}\,dx. $$ Consider the problem of finding $\mathbb E(X^2)$. It is $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty x^2 e^{-x^2/2}\,dx = \text{(by symmetry)} \frac{2}{\sqrt{2\pi}} \int_0^\infty x^2 e^{-x^2/2}\,dx $$ $$ \sqrt{\frac2\pi}\int_0^\infty xe^{-x^2/2}\Big(x\,dx\Big) = \sqrt{\frac2\pi}\int_0^\infty \sqrt{u}\ e^{-u}\,du = \sqrt{\frac2\pi}\ \Gamma\left(\frac32\right) = \frac12\sqrt{\frac2\pi} \Gamma\left(\frac12\right). $$ So it is enough to show that this expected value is $1$. That is true if the sum of two independent copies of it has expected value $2$. So: $$ \Pr\left(X^2+Y^2<w\right) = \frac{1}{2\pi}\iint\limits_\mathrm{disk} e^{-(x^2+y^2)/2}\,dx\,dy $$ where the disk has radius $\sqrt{w}$. This equals $$ \frac{1}{2\pi}\int_0^{2\pi}\int_0^{\sqrt{w}} e^{-\rho^2/2} \,\rho\,d\rho\,d\theta = \int_0^{\sqrt{w}} e^{-\rho^2/2} \,\rho\,d\rho. $$ This last equality holds because we are integrating with respect to $\theta$ something not depending on $\theta$. Differentiating this with respect to $w$ gives the probability density function of the random variable $X^2+Y^2$: $$ e^{-w/2}\sqrt{w}\frac{1}{2\sqrt{w}} = \frac{e^{-w/2}}{2}\text{ for }w>0. $$ So $$ \mathbb E(X^2+Y^2) = \int_0^\infty w \frac{e^{-w/2}}{2}\,dw =2. $$
factorial of negative numbers are $defined$ in terms of Gamma function, i.e. $(x!) := \Gamma(x+1)\forall x\in\mathbb R$, except for $x$ to be negative integer. this is because the two of them agree on positive integers, and so this is just a convention. |
In a lot of plots they use dimensionless quantities, why really we don't let quantities in their physical dimension and plot the curves normally.
The reason why this is done is because dimensionless data usually follows a single curve rather than multiple curves for different values of the parameters. This is nice because the figures become nicer and the data more organized.
This dimensional reduction of data occurs because, by the Buckingham-$\Pi$ theorem, there are fewer combined dimensionless parameters than there are individual dimensional parameters. These dimensionless parameters give more information about the underlying phenomena than the dimensional parameters do by themselves.
Let's take an example; below is a figure of transient heat transfer through a semi-infinite solid. If you look at the top figure it shows that at different times the cross-sectional temperature profiles are different; this makes sense because the solid is heating up. However if you properly scale the temperature and spatial coordinate and make them dimensionless: $$\theta = \frac{T-T_i}{T_s-T_i} \quad \eta = \frac{x}{\sqrt{4\pi t}}$$ then all profiles (for limited short time scale) collapse onto a single curve, the figure below.
I think this single curve in general is much more useful because it no longer depends on the values of the individual parameters (e.g. initial and wall temperatures) but on a dimensionless number which i can replicate if i ever have a similar situation but with different values for the initial and wall temperatures. I can then still apply this graph.
In my opinion most scientific figures should be presented in dimensionless form. Then it also doesn't matter anymore what units system the author worked in; he can use imperial units and i can work in metric units, it doesn't matter as long as the combination of parameters into dimensionless parameters is the same value.
Dimensionless constants have a special significance in physics. This is well described in the answers to the question Dimensionless Constants in Physics.
Having said this, whether or not you use dimensionless quantities in data analysis is to some extent a matter of taste. In my time as an industrial colloid scientist I generally didn't manipulate plots to be of dimensionless quantities, but then this was very much applied physics. If I were studying fundamental issues, e.g. at CERN, I might be more inclined to work with dimensionless quantities. |
I have a nonlinear inequality constrained optimization problem of the form $$\begin{array}{cc} \min & f(x) \\ \textrm{s.t.} & g(x) \ge 0 \end{array}$$ where $x \in \mathbb{R}^n$, $f : \mathbb{R}^n \to \mathbb{R}$, $g : \mathbb{R}^n \to \mathbb{R}^m$. I am currently solving this system with pseudotransient continuation (3) applied to a semismooth Newton's method (1,2). The semismooth Newton's method encodes the KKT conditions for a local solution to the constrained optimization into a single semismooth equation. Pseudotransient continuation is an unnecessarily fancy name for adding a diagonal term to the gradient of this equation (which includes the Hessian of the energy $f$) and running Newton's method.
Unfortunately, pseudotransient continuation globally convergent only for sufficiently large diagonal adjustments, and my current problem is converging only down to a nasty period two oscillation between two states. Without the constraints, global convergence could be enforced using line search using the original energy function $f$. However, the KKT conditions depend only on the gradient of $f$, not its value, and the original energy is somewhat obscured in the passage to the semismooth equation.
Question: Are there natural ways to perform line search in the context of constrained optimization? Note that it is not sufficient to apply line search to the residual norm, as described by Jed Brown here.
References: |
As known that the lower incomplete gamma function can be written as $\gamma(a,x) = x^{a}e^{-x}\sum_k^\infty{{x^{k}}\over a^{k+1}}.$ What is the format for $\sum_j^\infty{\gamma(v/p-j,rx^{p})} $ in series representation? (where $v$, $p$ and $r$ are parameters)
$\gamma(a,x)$ in R is zipfR::Igamma(a, x, lower=TRUE, log=FALSE). If $\sum_j^\infty{\gamma(v/p-j,rx^{p})} $, what is the $a$ and $b$ should be in Igamma()?
closed as off-topic by jbowman, kjetil b halvorsen, Peter Flom♦ Nov 8 '18 at 12:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
" Self-study questions(including textbook exercises, old exam papers, and homework) that seek to understand the concepts are welcome, but those that demand a solution need to indicate clearly at what step help or advice are needed. For help writing a good self-study question, please visit the meta pages." – Peter Flom
If your question is about the generalized incomplete gamma function that you mention in an earlier question, then note that the path you try to follow here has been done before (I have now added this also to my previous answer to your previous question):
They eventually express the function in terms of a Kampé de Fériet function (a two-variable generalization of the generalized hypergeometric series)
$$\begin{array}{rl} \Gamma \left( -v, \frac{1}{x};\frac{z^2}{4} \right) = & \Gamma\left( - v , \frac{1}{x} \right) \Gamma(1+c) \left( \frac{2}{z} \right)^v I_v(z) \\ & - \frac{z^2}{4} \frac{x^{1+v}}{1+v} e^{-1/x} F \mathstrut_{2:0;0}^{0:2;1} \left[ \begin{array}{ r r r } \dfrac{\phantom{2,2+v}}{\phantom{2,2+v}} : & 1,1+v ; & 1 ; \\ 2, 2+v : & \dfrac{\phantom{1,1+v}}{\phantom{1,1+v}} ; & \dfrac{\phantom{1}}{\phantom{1}} ; & \end{array} -x \dfrac{z^2}{4}, \dfrac{z^2}{4} \right] \end{array}$$ where now $v \neq -1, -2, -3, ...$
and for non negative integers
$$\begin{array}{rl} \Gamma \left( n, x; z \right) = & \frac{(-z)^n}{n!} \lbrace \sum_{k=1}^n \frac{(-n)_k}{z^k} \Gamma(k,x) + _0F_1[-;n+1;z]\Gamma(0,x) \\& - \frac{z}{n+1} \frac{e^{-x}}{x} F \mathstrut_{2:0;0}^{0:2;1} \left[ \begin{array}{ r r r } \dfrac{\phantom{2,n+2}}{\phantom{2,2+v}} : & 1,1 ; & 1 ; \\ 2, n+2 \;: & \dfrac{\phantom{1,1}}{\phantom{1,1}} ; & \dfrac{\phantom{1}}{\phantom{1}} ; & \end{array} -x^{-1} z, z \right] \rbrace \end{array}$$ |
Electronic Journal of Probability Electron. J. Probab. Volume 13 (2008), paper no. 33, 980-999. Large-$N$ Limit of Crossing Probabilities, Discontinuity, and Asymptotic Behavior of Threshold Values in Mandelbrot's Fractal Percolation Process Abstract
We study Mandelbrot's percolation process in dimension $d \geq 2$. The process generates random fractal sets by an iterative procedure which starts by dividing the unit cube $[0,1]^d$ in $N^d$ subcubes, and independently retaining or discarding each subcube with probability $p$ or $1-p$ respectively. This step is then repeated within the retained subcubes at all scales. As $p$ is varied, there is a percolation phase transition in terms of paths for all $d \geq 2$, and in terms of $(d-1)$-dimensional ``sheets" for all $d \geq 3$.
For any $d \geq 2$, we consider the random fractal set produced at the path-percolation critical value $p_c(N,d)$, and show that the probability that it contains a path connecting two opposite faces of the cube $[0,1]^d$ tends to one as $N \to \infty$. As an immediate consequence, we obtain that the above probability has a discontinuity, as a function of $p$, at $p_c(N,d)$ for all $N$ sufficiently large. This had previously been proved only for $d=2$ (for any $N \geq 2$). For $d \geq 3$, we prove analogous results for sheet-percolation.
In dimension two, Chayes and Chayes proved that $p_c(N,2)$ converges, as $N \to \infty$, to the critical density $p_c$ of site percolation on the square lattice. Assuming the existence of the correlation length exponent $\nu$ for site percolation on the square lattice, we establish the speed of convergence up to a logarithmic factor. In particular, our results imply that $p_c(N,2)-p_c=(\frac{1}{N})^{1/\nu+o(1)}$ as $N \to \infty$, showing an interesting relation with near-critical percolation.
Article information Source Electron. J. Probab., Volume 13 (2008), paper no. 33, 980-999. Dates Accepted: 12 June 2008 First available in Project Euclid: 1 June 2016 Permanent link to this document https://projecteuclid.org/euclid.ejp/1464819106 Digital Object Identifier doi:10.1214/EJP.v13-511 Mathematical Reviews number (MathSciNet) MR2413292 Zentralblatt MATH identifier 1191.60109 Subjects Primary: 60K35: Interacting random processes; statistical mechanics type models; percolation theory [See also 82B43, 82C43] Secondary: 60D05: Geometric probability and stochastic geometry [See also 52A22, 53C65] 28A80: Fractals [See also 37Fxx] 82B43: Percolation [See also 60K35] Rights This work is licensed under aCreative Commons Attribution 3.0 License. Citation
Broman, Erik; Camia, Federico. Large-$N$ Limit of Crossing Probabilities, Discontinuity, and Asymptotic Behavior of Threshold Values in Mandelbrot's Fractal Percolation Process. Electron. J. Probab. 13 (2008), paper no. 33, 980--999. doi:10.1214/EJP.v13-511. https://projecteuclid.org/euclid.ejp/1464819106 |
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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 ... |
The closest positive answers to your question that I could find is for sparse diagonal perturbations (see below).
With that said, I do not know of any algorithms for the general case, though there is a generalization of the technique you mentioned for scalar shifts from SPD matrices to all square matrices:
Given any square matrix $A$, there exists a Schur decomposition $A=U T U^H$, where $U$ is unitary and $T$ is upper triangular, and $A+\sigma I = U (T + \sigma I) U^H$ provides a Schur decomposition of $A + \sigma I$. Thus, your precomputation idea extends to all square matrices through the algorithm:
Compute $[U,T]=\mathrm{schur}(A)$ in at most $\mathcal{O}(n^3)$ work. Solve each $(A+\sigma I) x = b$ via $x := U (T +\sigma I)^{-1} U^H b$ in $\mathcal{O}(n^2)$ work (the middle inversion is simply back substitution).
This line of reasoning reduces to the approach you mentioned when $A$ is SPD since the Schur decomposition reduces to an EVD for normal matrices, and the EVD coincides with the SVD for Hermitian positive-definite matrices.
Response to update: Until I have a proof, which I do not, I refuse to claim that the answer is "no". However, I can give some insights as to why it's hard, as well as another subcase where the answer is yes.
The essential difficulty is that, even though the update is diagonal, it is still in general full rank, so the primary tool for updating an inverse, the Sherman-Morrison-Woodbury formula, does not appear to help. Even though the scalar shift case is also full rank, it is an extremely special case since it commutes with every matrix, as you mentioned.
With that said, if each $D$ was sparse, i.e., they each had $\mathcal{O}(1)$ nonzeros, then the Sherman-Morrison-Woodbury formula yields an $\mathcal{O}(n^2)$ solve with each pair $\{D,b\}$. For example, with a single nonzero at the $j$th diagonal entry, so that $D=\delta e_j e_j^H$:
$$[A^{-1}+\delta e_j e_j^H]^{-1} = A^{-1} - \frac{\delta A^{-1} e_j e_j^H A^{-1}}{1+\delta (e_j^H A^{-1} e_j)},$$
where $e_j$ is the $j$th standard basis vector.
Another update: I should mention that I tried the $A^{-1}$ preconditioner that @GeoffOxberry suggested on a few random SPD $1000 \times 1000$ matrices using PCG and, perhaps not surprisingly, it seems to greatly reduce the number of iterations when $||D||_2/||A||_2$ is small, but not when it is $\mathcal{O}(1)$ or greater. |
I am working to solve Poisson's equation in 2D axisymmetric cylindrical coordinates using the Jacobi method. The $L^2$ norm decreases from $\sim 10^3$ on the first iteration (I have a really bad guess) to $\sim 0.2$ very slowly. Then, the $L^2$ norm begins to increase over many iterations.
My final matrix is weakly diagonally dominate, except for the 2nd order Neumann condition at $r = 0$.
Can I make a small tweek to make this work, is it numeric or do I need a new algorithm?
My geometry is parallel plates with sharp points at $r = 0$ on both plates.
My boundary conditions are $$\left. \frac{\partial V}{\partial r} \right|_{r=0} = 0$$
Although I would like my second radial BC to be $$\left. \frac{\partial V}{\partial r} \right|_{r=\infty} = 0$$ I settled for $$\left. \frac{\partial V}{\partial r} \right|_{r=a} = 0$$
Then Dirichlet conditions at the upper and lower boundaries $$V(r, L(r) ) = V_0$$ $$V(r, U(r) ) = V_L$$
where $$L(r) = \begin{cases} & 0 \text{ if } r \geq R_L \\ & H_L (1 - \frac{r}{R_L} ) \text{ if } r \leq R_L \end{cases}$$
and
$$U(r) = \begin{cases} & H \text{ if } r \geq R_U \\ & H + H_U (\frac{r}{R_U} - 1 ) \text{ if } r \leq R_U \end{cases}$$ |
I am trying to implement a simple normal-inverse-Wishart conjugate prior distribution for a multivariate normal with unknown mean and covariance in numpy/scipy such that it can take a data vector and construct a posterior. I'm using the update equations specified by Wikipedia for a NIW: http://en.wikipedia.org/wiki/Conjugate_prior
My distribution class is as follows:
import numpy as npfrom scipy.stats import chi2class NormalInverseWishartDistribution(object): def __init__(self, mu, lmbda, nu, psi): self.mu = mu self.lmbda = float(lmbda) self.nu = nu self.psi = psi self.inv_psi = np.linalg.inv(psi) def sample(self): sigma = np.linalg.inv(self.wishartrand()) return (np.random.multivariate_normal(self.mu, sigma / self.lmbda), sigma) def wishartrand(self): dim = self.inv_psi.shape[0] chol = np.linalg.cholesky(self.inv_psi) foo = np.zeros((dim,dim)) for i in range(dim): for j in range(i+1): if i == j: foo[i,j] = np.sqrt(chi2.rvs(self.nu-(i+1)+1)) else: foo[i,j] = np.random.normal(0,1) return np.dot(chol, np.dot(foo, np.dot(foo.T, chol.T))) def posterior(self, data): n = len(data) mean_data = np.mean(data, axis=0) sum_squares = np.sum([np.array(np.matrix(x - mean_data).T * np.matrix(x - mean_data)) for x in data], axis=0) mu_n = (self.lmbda * self.mu + n * mean_data) / (self.lmbda + n) lmbda_n = self.lmbda + n nu_n = self.nu + n psi_n = self.psi + sum_squares + self.lmbda * n / float(self.lmbda + n) * np.array(np.matrix(mean_data - self.mu).T * np.matrix(mean_data - self.mu)) return NormalInverseWishartDistribution(mu_n, lmbda_n, nu_n, psi_n)
I am running a simple sanity check to see if the posterior converges to the true distribution:
x = NormalInverseWishartDistribution(np.array([0,0])-3,1,3,np.eye(2))samples = [x.sample() for _ in range(100)]data = [np.random.multivariate_normal(mu,cov) for mu,cov in samples]y = NormalInverseWishartDistribution(np.array([0,0]),1,3,np.eye(2))z = y.posterior(data)print 'mu_n: {0}'.format(z.mu)print 'psi_n: {0}'.format(z.psi)
The mean is appropriately converging, but the scale matrix appears to be converging to incorrectly large values along the diagonal, rather than the true value of 1.
As far as I can tell, I'm copying the update rule exactly. Am I implementing something inappropriately here?
Edit: It looks like in fact the posterior is converging, but the sample routine is returning biased samples. Am I doing something wrong in my sampling method? Edit2: I've confirmed that the same phenomenon happens in the MCMCpack riwish function in R:
> library(MCMCpack)> samples <- replicate(100000, riwish(3, matrix(c(1,0,0,1),2,2)))> mean(samples[1,1,])[1] 4.889211
This leads me to believe I must be misunderstanding something. From the Wikipedia page (http://en.wikipedia.org/wiki/Inverse-Wishart_distribution), we have:
$\newcommand{\E}{\mathrm{E}}$ $\E[{\Sigma}] = \frac{\Psi}{\nu - p - 1}$
However, in my test case, $\nu = p + 1$, so $\E[{\Sigma}] = \Psi$. Thus, if I sample $\Sigma$ a bunch of times, shouldn't I recover $\Psi$ on average?
Edit 3: Realized my problem was simple algebraic oversight: I needed to set $\nu = p+2$. Problem solved |
I think you may be asking a few related questions, so here's my attempt to answer some of them:
Are there examples of scalar Goldstone bosons?
Yes, at least at a hypothetical level. One simple example in particle physics is the
dilaton, the Goldstone boson of spontaneous breaking of scale symmetry.
Remarks:
When scale symmetry is broken by strong dynamics, the dilaton is often related to the radion, the radial excitation in the size of an extra dimension, which is related to the dilaton through the holographic principle.
Scale symmetry (and the larger conformal symmetry that contains it) is anomalous in our universe. Renormalization violates scale invariance, which is why we have things like anomalous dimensions---literally the scaling dimensions of our fields and operators appear to change at different length scales. This means that the dilaton is at best a pseudo-Goldstone boson.
I believe that in this case you do get a Coulomb potential.
Remark: in your question you ask if you get a Coulomb potential, which is a very interesting question. But part of the reason why it is interesting is that a pseudoscalar particle does not mediate a Coulomb potential, but rather a spin-dependent $1/r^3$ potential. Why are our favorite Goldstone bosons pseudoscalars?
Our "favorite Goldstone bosons" in particle physics are the longitudinal modes of the $W$ and $Z$ (pseudoscalars) and the neutral pions/kaons (pseudoscalar mesons!). Nothing in Goldstone's theorem says that the Goldstone mode has to be pseudoscalar, so what's going on?
This is a little subtle and I'm not sure if I have the whole story, so I invite more knowledgeable readers to correct me!
The parity of a spin-0 particle---that is, whether it is a scalar or pseudoscalar---is not really well defined in a field theory until you have a sense of what (charge--)parity means in that theory. In most particle theories we play with, the sense of parity comes from how the particle interacts with fermions. This boils down to: (in 4-component fermion notation) whether there's a $i\gamma^5$ in the interaction or (in 2-component fermion notation) whether the coupling is imaginary. (These two are equivalent because the hermitian conjugate of the latter term gives the negative term in the $\gamma^5$.)
You can parameterize the Goldstone $\pi(x)$ in the usual Coleman--Callan--Wess--Zumino way as $$U(x) = \exp(i\pi^a(x) T^a/v)\, v \ ,$$where $v$ is the vev (the order parameter of spontaneous symmetry breaking), and $T^a$ are the generators of the broken symmetry. $U(x)$ is a
linear representation under the whole symmetry group. For the Higgs this is like saying$$H(x) = \exp(i\pi^a T^a/v)\, \begin{pmatrix}0\\v+h(x)\end{pmatrix}$$where we've added in the radial mode $h(x)$ for clarity. The $\pi^a$ here are the Goldstones that are eaten by the $W$ and $Z$. $H(x)$ transforms as an SU(2) doublet, but we've pushed the Goldstone degrees of freedom into the exponential. What this boils down to is the following rule:
To identify the Goldstones: transform the vev by a general broken generator, parameterized by some direction $\epsilon^a$. Promote this parameter to a dynamical field, $\epsilon^a \to \pi^a(x)/v$. That is the Goldstone.
When you have a
compact symmetry, the generators are always Hermitian. This is simply the generalization of $e^{i\theta}$ which is compact and lies on the unit circle versus $e^{x}$ which may take any value in (0,$\infty$). This, in turn, tells you something about the factors of $i$ floating around when you couple the Goldstones to fermions. As we mentioned in (1), the factors of $i$ when coupling to fermions have to do with whether a spin-0 is scalar of pseudoscalar.
In simple cases, we assume that the fermions are also linear representations of the global symmetry. Then you may use the global symmetry to write down the allowed interaction terms with respect to the order parameter of symmetry breaking (e.g. writing the Yukawa interactions with respect to the Higgs doublet in the Standard Model).
When you turn the crank, I believe this gives you that the Goldstone of a spontaneously broken internal symmetry interacts with fermions as a pseudoscalar. Now there are lots of caveats. Here are two:
We've used one definition of CP:
how the field interacts with fermions. There are other sense of CP that pop out purely from group theory. See, for example this recent thesis.
If you're not too attached to the Goldstone being a
true Goldstone, but are allowing for explicit breaking terms, then even the Goldstones of a spontaneously broken compact approximate symmetry may have scalar interactions. This is the case in models of a pseudo-Goldstone Higgs. The standard theory breaks $SO(5)\to SO(4)$, where $SO(4)$ contains the four degrees of freedom of a Higgs doublet. These are all identified as Goldstones of the breaking, but the global $SO(5)$ is broken explicitly by small terms. At loop level, these generate the usual Higgs properties (electroweak symmetry breaking potential, scalar interactions to Standard Model fermions). Can $SU(3)_L\times SU(3)_R$ chiral symmetry break to $SU(3)_V$?
I do not think this is possible. What you're proposing is that $$SU(3)_L\times SU(3)_R\to SU(3)_A , $$however, the axial part of chiral symmetry simply is not a group. If you compute the commutation relations of $SU(3)_A$ generators, you get elements of $SU(3)_V$. So the algebra of $SU(3)_A$ does not close.
I hope that helps! |
Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!! |
Most methods for oscillatory integrals I know about deal with integrals of the form $$ \int f(x)e^{i\omega x}\,dx $$ where $\omega$ is large.
If I have an integral of the form $$ \int f(x)g_1(x)\cdots g_n(x)\,dx, $$ where $g_k$ are oscillatory functions whose roots are only known approximately, but some kind of asymptotic form $$ g_k(x) \sim e^{i\omega_k x} $$ is known, with the frequencies $\omega_k$ all different (and $\mathbb{Q}$-linearly independent), then how can I evaluate this integral?
Unlike in the case of $e^{i\omega x}$, the polynomial integrals $\int x^a \prod g_k(x)$ are not known, so I can't construct a set of polynomial interpolants for $f(x)$ and integrate the interpolants exactly.
In my exact problem, $g_k$'s are Bessel functions $J_0(\omega_k x)$, and $f(x)=x^\alpha$, and the region of integration is $[0,\infty)$. The method I am using now is to sum up integral contributions over intervals $[x_{k-1},x_k]$ between roots up to some cutoff $M$, then use the asymptotic expansion for $g_k(x)$ for large $x$. This algorithm's time complexity is exponential in $n$ because it involves expanding the product $g_1\ldots g_n$, each of which has a number $r$ of asymptotic terms, giving $r^n$ total terms; pruning terms that are too small doesn't reduce the run time enough to make this feasible for large $n$.
Heuristic non-rigorous answers, suggestions and references are all welcome. |
Given $a_0=1$ and:$$a_n=a_{\frac{n}{2}}+a_{\frac{n}{3}}+a_{\frac{n}{6}}$$Find convergence or divergence of $\frac{a_n}{n}$.
Such a weird problem; I don't know how to attack it. I'm also fairly certain I typed it out correctly.
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On the assumption that the question is really $$a_n=a_{\lfloor\frac{n}2\rfloor}+a_{\lfloor\frac{n}3\rfloor}+a_{\lfloor\frac{n}6\rfloor}$$ starting with $a_0=1$ (otherwise how do you work out $a_1$?), this is OEIS A007731 and was covered by P. Erdos, A. Hildebrand, A. Odlyzko, P. Pudaite and B. Reznick, The asymptotic behavior of a family of sequences, Pacific J. Math., 126 (1987), pp. 227-241.
They showed that $\dfrac{a_n}{n}$ converges to $\dfrac{12}{\log_e 432} \approx 1.97744865$, though the coveregence is very slow: $\frac{a_n}{n}$ is about $1.8430$ when $n=432^5-1$ but about $2.1175$ when $n=432^5$.
Caveat: This answer is concerned with the problem as stated by the OP, that is, with sequence(s) defined by $a_0=1$ and $a_n=a_{n/2}+a_{n/3}+a_{n/6}$ for every $n\geqslant1$ such that the RHS of the equality makes sense.
user27572: Are you absolutely sure that this problem does not have a solution as given?
Absolutely sure. As explained by @Antonio in the comments the initial condition $a_0=1$ is not enough to determine $(a_n)_{n\geqslant0}$ since one needs to choose the value of $a_n$ for every $n$ not a multiple of $6$.
Obviously, each sequence defined by $a_0=1$ and $a_n=\beta n$ for some fixed $\beta$ and for every nonzero $n$, is a solution. For this choice, the sequence $(b_n)_{n\geqslant1}$ defined by $b_n=a_n/n$ is convergent, to $\beta$.
Here is a different, explicit, choice of $a_n$ for $n$ a power of $2$ and for $n$ a power of $3$, which yields a
divergent subsequence $(b_n)_n$ when $n$ is restricted to the products of a power of $2$ by a power of $3$, and, a fortiori, a divergent complete sequence $(b_n)_{n\geqslant1}$. Define$$
a_n=r^{i+j}\quad\text{if}\quad n=2^i3^j,
$$for some fixed $r$.A one-line computation shows that this solves the desired recursion as soon as $r$ solves $r^2=2r+1$, hence, for example, for $r=1+\sqrt2$.
Now, $b_n\to+\infty$ when $n\to\infty$ while $n$ is a power of $2$ because $r\gt2$ and $b_n\to0$ when $n\to\infty$ while $n$ is a power of $3$ because $r\lt3$. Furthermore, for every finite positive $\beta$, there exists a sequence $(n_k)_k$ such that every $n_k$ is the product of a power of $2$ by a power of $3$ and such that $b_{n_k}\to\beta$ when $k\to\infty$. Thus, the set of limit points of the sequence $(b_n)_{n}$ when $n$ is restricted to the products of a power of $2$ by a power of $3$ is $[0,+\infty]$.
More generally, $a_n=\beta\cdot\prod\limits_pr_p^{i_p}$ for $n=\prod\limits_pp^{i_p}$, where the products are over the set of primes $p$, yields a solution for every family of weights $(r_p)_p$ such that $r_2r_3=r_2+r_3+1$.
The (admissible) choice $r_p=p$ for every prime $p$ yields $a_n=\beta n$. This is the only case when $(b_n)_{n\geqslant1}$ converges. The (admissible) choice $r_2=r_3=1+\sqrt2$ yields the solution above.
Note: The recursion $a_n=a_{\lfloor n/2\rfloor}+a_{\lfloor n/3\rfloor}+a_{\lfloor n/6\rfloor}$ is a different story... |
I have some non-hermitian matrix $A$, that I have the left and right eigenvectors. (Calculated using SLEPc, by finding the eigenvectors of $A$ and $A^H$).
I'm not sure how to orthogonalize them however. I know that they must obey the relation:
$$L^HR = 1$$
But I'm not sure how to enforce this. The normalization of the vectors isn't clear to me either, since $\left<l|r\right> = 1$, does $\left<l|l\right> = 1$? (And similarly $\left<r | r\right>$?).
Grahm-Schmidt (tildes represent non-orthogonalized quantities):
$$ \tilde{q}_i = \frac{\tilde{r}_i}{\left<\tilde{r}_i | \tilde{r}_i \right>}$$ $$ r_i = \tilde{q}_i - \sum_{j\neq i}\left<\tilde{l}_j|\tilde{q}_i\right> \tilde{l}_j^H$$
seems like it might work, but unlike grahm-schmidt for self-orthogonalizing, the first step of normalization doesn't feel right. And what about $l_i$? Is $l_i$ found by doing the same procedure? i.e.:
$$ l_i = \tilde{q}_i - \sum_{j\neq i}\left<\tilde{q}_j|r_i\right>^H r_j^H$$
Is there some resource that can give me more information about non-hermitian eigenvalue problems? |
Given the linear system $\dot{z} = Az$.
(a) Assume all the eigenvalues of $A$ have negative real part. Give a counterexample to this statement: every solution of $\dot{z} = Az$ satisfies $|z(t)|\leq |z(s)| \ \forall\ t> s $
(b) Assume A is symmetric and all the eigenvalues of $A$ have negative real part. Prove that every solution satisfies $|z(t)|\leq |z(s)|\ \forall\ t> s$.
My thought: I was trying many different types of matrices (the ones which has $2$ real eigenvalues with $2$ negative real parts or $2$ complex eigenvalues with $2$ negative real part), but they all satisfy the inequality. Can someone please help me with an example?
For part (b), I think of using $A$ as symmetric must have all real eigenvalues. Thus, by applying Lemma A, which is following:
Consider $\dot{x} = Ax + f(t,x) + f_0(t)$ with $A,\ f$ and $f_0$ are continuous and $f(t,0) = 0 \ \forall\ t\in R$. Assume there exists constants $K\geq 1$, $M,L\geq 0$ and $\theta > \lambda + KL$ ($\lambda < 0$ is the negative real part). Then if we have:
(a) $||e^{At}||\leq Ke^{\lambda t}$ for $t\geq 0$
(b) $||f(t,x) - f(t,y)||\leq L||x-y||$ for all $L,x,y\in R$.
(c) $||f_0(t)||\leq Me^{\theta t}$ for $t\geq \tau$ ($\tau$ is some fixed constant).
Then if $u(t)$ is a solution, then for all $t\geq \tau$, we have: $||u(t)||e^{-\theta t}\leq K||u(\tau)||e^{-\theta \tau} + \frac{KM}{\theta - \lambda - KL}$.
Applying the result above into part (b) with sufficiently large $K\geq 1,\ M=L=\theta = 0$ (since $\lambda < 0$), we have: $||u(t)||\leq K||u(\tau)||$. But how do we "cancel" the constant $K$ here? |
I've been wondering about the following question: Suppose that $P$ is a polynomial of degree $n$ with complex coefficients. Assume that $a_0, a_1, \dots, a_n \in \mathbb{C}$ are distinct. Are the polynomials $$P(x + a_0), P(x + a_1), \dots, P(x + a_n)$$ linearly independent?
Consider the matrix of coefficients, where each row corresponds to an evaluation at a point, and each column to a power of $x$. Any linear dependency among the columns amounts to some degree~$n$ polynomial in the point, which is supposed to have $n+1$ roots. Since this can't happen, the polynomials are independent.
Example (almost Vandermonde) - $P(x) = x^2$: $$\begin{pmatrix} 1 & 2a & a^2 \\ 1 & 2b & b^2 \\ 1 & 2c & c^2 \end{pmatrix}$$ A linear dependency of the columns is a quadratic polynomial having three different roots $a,b,c$.
EDIT: Just to clarify what happens in the general case. Here's the matrix corresponding to $P(x) = x^2 + \alpha x + \beta$: $$\begin{pmatrix} 1 & 2a + \alpha & a^2 + \alpha a + \beta \\ 1 & 2b + \alpha & b^2 + \alpha b + \beta \\ 1 & 2c + \alpha & c^2 + \alpha c + \beta \end{pmatrix}$$ Since the degree of the polynomial in column $t$ is $t$ (if we start counting from zero), it's easy to see that the any non-zero combination of columns results in a non-zero polynomial (consider the rightmost non-zero column).
Here a fancy proof that uses operators and differentiation.
Let $T_a$ be the translation operator which maps polynomials to their translates
$$ T_a p(x) = p(x + a) .$$
Restricting attention to polynomials of degree $n$ or lower, this is a map from an $n+1$-dimensional vector space to itself. The question to be resolved is whether the $n+1$ translation operators $T_{a_0}, T_{a_1}\dots, T_{a_n}$ are linearly independent (when applied to polynomials of degree $n$).
It is well-known that the translation operator can be written as the exponentiation of the differentiation operator $Dp(x) := p'(x)$ as follows
$$ T_a = e^{aD} = 1 + \frac1{1!} aD + \frac1{2!} a^2D^2 + \dots + \frac1{n!} a^n D^n .$$
This is Taylor's theorem. Note that since $D$ is nilpotent on the space of polynomials of degree $n$, $D^{n+1} = 0$, the series is actually a finite sum.
Now, the point is that for any polynomial $p(x)$ of degree $n$, we know that its derivatives $p(x), Dp(x), D^2p(x), \dots, D^n p(x)$ are
linearly independent because the degree always decreases by 1. Hence, they form a basis of our vector space of polynomials. The expansion of the $T_a$ in terms of this new basis is given by the expression above.
To show that the translates $T_{a_k} p(x)$ are linearly independent, we only have to check that their matrix with respect to the new basis is nonsingular. It reads
$$\begin{pmatrix} 1 & 1 & \dots & 1 \\ a_0 & a_1 & \dots & a_n \\ \frac1{2!} a_0^2 & \frac1{2!} a_1^2 & \dots & \frac1{2!} a_n^2 \\ \vdots \\ \frac1{n!} a_0^n & \frac1{n!} a_1^n & \dots & \frac1{n!} a_n^n \\ \end{pmatrix}$$
But its determinant is clearly a non-zero multiple of the Vandermonde determinant, which is non-zero if and only if the values $a_0,a_1,\dots,a_n$ are pairwise distinct.
(Another way to see that this matrix is non-singular is to note that the Vandermonde determinant is related to the interpolation problem, whose solution is unique.)
If you write $P(x) = \sum_{j=0}^n p_j x^j$ and set $ \sum_{i=0}^n c_i P(x+a_i) = 0$, the full expression is written $\sum_{i=0}^n \sum_{k=0}^n \sum_{j=0}^k c_i p_k {k \choose j} a_i^{k-j} x^j$. If we find the coefficient of each power $x_j$ as a matrix times the vector $(c_0, ..., c_n)$, and then setting the system equal to zero, we can deduce the determinant of the matrix is zero. The matrix is $A_{ij} = \sum_{k=j}^n p_k {k \choose j} a_i^{k-j}$.
Setting the determinant equal to zero gives a polynomial in all $a_i$. I don't know if this polynomial is simpler than the matrix hints at, but it is certainly possible to find coefficients $a_i$ such that the polynomials are linearly independent, though for "most" values it won't be the case, as the solution set must be of at least one dimension less than $C^{n+1}$ (seen by fixing all $a_1, ..., a_n$ arbitrarily and thereby determining a finite set of possible $a_0$).
Here a fancy proof by induction that uses differentiation.
Case $n=0$.
A polynomial of degree $0$ has the form $P(x)=c_0$ with $c_0\neq 0$ which is clearly linearly independent.
Case $n=k+1$.
Assume that there is a linear dependence
$$\lambda_0 P(x+a_0) + \lambda_1 P(x+a_1) + \dots + \lambda_n P(x+a_n) = 0.$$
Since this relation holds for all $x$, we can differentiate this relation. After differentiating $n$ times, we obtain a relation for the $n$-th derivative
$$\lambda_0 P^{(n)}(x+a_0) + \lambda_1 P^{(n)}(x+a_1) + \dots + \lambda_n P^{(n)}(x+a_n) = 0.$$
But since $P$ has degree $n$, we know that $P^{(n)}(x+a_i)$ is equal to a constant $c\neq 0$, which implies
$$ \lambda_0 + \lambda_1 + \dots + \lambda_n = 0 .$$
By expressing $\lambda_0$ in terms of the other coefficients, we obtain a relation
$$ \lambda_1 (P(x+a_1) - P(x+a_0)) + \dots + \lambda_n (P(x+a_n) - P(x+a_0)) = 0$$
of polynomials of degree $n-1$. By induction, we conclude that all the $\lambda_i$ must vanish. |
Currently, when one clicks the \(\TeX\) button while writing a post, one is presented with the example of a solution to a second-order polynomial equation.
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
PolarKernel asked on the thread;
Public Beta Feature Requests whether we want a more physics-ish or high-level example equation instead.
I see that we could have one of the following kinds of example equation:
Functional ones; common examples, like our current example equation.
Physics-ish fun ones, say, something like \(\left\langle \mathcal{T}\left\{ \exp\left(\int \mathrm{d}x^D J_{4D}(x)\mathcal{O}(x)\right) \right\} \right\rangle_{\mathrm{CFT}} = Z_{\mathrm{AdS}}\left[\lim_{\mathrm{boundary}} J \omega^{\Delta-D+n } = J_{4D}\right]\) (from AdS/CFT)
Commonly used, useful equations, e.g. Dirac Field Lagrangian, or Polyakov Action, or something like that.
Solely to show some of the features of MathJax; e.g. \( {\mathcal{L}} = \left[ {\begin{array}{*{20}{c}}{2k\sum\limits_\pi ^\infty {{\Gamma ^{\int {\rm{d}} {\mathbb{F}}}}} }&0\\0&1\end{array}} \right] \)
Something creative and fun; here's a dumb thing I thought of: \(\bf{{MaThJaX}^{-1}\lvert p\hbar0\rangle}=0\).
Commonly used equations may be advantageous because they may reduce equation-typesetting time. But with different conventions used, etc., it may not turn out to be very useful eventually.
Physics-ish equations may help ward off low-level questions, and give the message that "
this is not the place for your question", which is good.
The creative and fun things are useless but fun.
So, please propose example equations, or say "I'd rather stick to our current, functional, example equation." if you prefer. |
Search
Now showing items 1-8 of 8
Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV
(Springer, 2015-01-10)
The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (|y|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ...
Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV
(Springer Berlin Heidelberg, 2015-04-09)
The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ...
Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV
(American Physical Society, 2015-06)
The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ...
K*(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2015-02)
The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ...
Production of inclusive $\Upsilon$(1S) and $\Upsilon$(2S) in p-Pb collisions at $\mathbf{\sqrt{s_{{\rm NN}}} = 5.02}$ TeV
(Elsevier, 2015-01)
We report on the production of inclusive $\Upsilon$(1S) and $\Upsilon$(2S) in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV at the LHC. The measurement is performed with the ALICE detector at backward ($-4.46< y_{{\rm ...
Elliptic flow of identified hadrons in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Springer, 2015-06-29)
The elliptic flow coefficient ($v_{2}$) of identified particles in Pb--Pb collisions at $\sqrt{s_\mathrm{{NN}}} = 2.76$ TeV was measured with the ALICE detector at the LHC. The results were obtained with the Scalar Product ...
Measurement of electrons from semileptonic heavy-flavor hadron decays in pp collisions at s =2.76TeV
(American Physical Society, 2015-01-07)
The pT-differential production cross section of electrons from semileptonic decays of heavy-flavor hadrons has been measured at midrapidity in proton-proton collisions at s√=2.76 TeV in the transverse momentum range ...
Multiplicity dependence of jet-like two-particle correlations in p-Pb collisions at $\sqrt{s_NN}$ = 5.02 TeV
(Elsevier, 2015-02-04)
Two-particle angular correlations between unidentified charged trigger and associated particles are measured by the ALICE detector in p–Pb collisions at a nucleon–nucleon centre-of-mass energy of 5.02 TeV. The transverse-momentum ... |
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Now showing items 1-10 of 167
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV
(Elsevier, 2013-04-10)
The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (|y| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ... |
$\newcommand{\psym}{\text{Psym}_n}$ $\newcommand{\sym}{\text{sym}}$ $\newcommand{\Sym}{\operatorname{Sym}}$ $\newcommand{\Skew}{\operatorname{Skew}}$ $\newcommand{\SO}{\operatorname{SO}_n}$ $\renewcommand{\skew}{\operatorname{skew}}$ $\newcommand{\GLp}{\operatorname{GL}_n^+}$
Let $\psym$ be the space of real symmetric positive-definite $n \times n$ matrices, and $\GLp$ be the group of real $n \times n$ invertible matrices with positive determinant.
Let $O:\GLp \to \SO$ be the orthogonal polar factor map, defined by requiring $A= O(A)P$ for some symmetric positive-definite $P$. Note that $O(A)=A(\sqrt{A^TA})^{-1}$.
Is there a nice "closed-form
algebraic formula" for the differential $dO_A$? If not, perhaps there is a formula for $\langle dO_A(B),C \rangle $? (similarly to what happens with the Levi-Civita connection, where we have an implicit characterization of $\nabla_XY$ in terms of the Koszul formula). I am fine with using positive matrix square roots and inverses, but not with using integral formulas or vectorization operations like here or here. (I also don't want to use explicitly the singular values of $A$). Here are some partial results:
Since $dO_{QA}(QB)=QdO_A(B)$ (and $dO_{AQ}(BQ)=dO_A(B)Q$), the question can be reduced to the case where $A \in \psym$. (The "dual" orthogonal case is easy: $dO_{Id}(B)=\skew(B)$, and for every $Q \in \SO$, $dO_{Q}(QB)=Q\skew B$).
For every $B \in \skew, dO_A(OBP)=OB$, i.e. $dO_A(X)=XP^{-1}$ if $O^TXP^{-1} \in \skew$: Set $\alpha(t)=Oe^{tB}P$. Then $O(\alpha(t))=Oe^{tB}$, so $dO_A(OBP)=OB$.
The problem with calculating $dO_A(OBP)$ when $B \in \sym$, is that $e^{tB}P$ does not need to be symmetric, even though $e^{tB},P$ are both positive-definite. If this was the case, then it would imply $dO_A(OBP)=O\skew B$, which is false in general (see below).
One could conjecture that perhaps $dO_A(OBP)=OB$ holds also for $B \in \sym$, or equivalently that $dO_A(X)=XP^{-1}$ for every $X \in M_n$. This is false since $dO_A$ is not injective, due to dimensional incompatibility.
Another possible conjecture is $dO_A(OBP)=O\skew B$. However, this is also false:
Indeed, for $A=P \in \psym$, this reduces to $dO_P(BP)=\skew B$. Suppose that $C:=BP \in \sym$. Then must have $dO_P(BP)=0$. Indeed, by differentiating $A=OP$ we obtain $$ \dot A=\dot O P+O\dot P, \tag{1}$$
which for $A=P,O=Id$ becomes $ \dot A=\dot O P+\dot P$. Note that $C \in \sym \Rightarrow dP_P(C)=C$ (since $P_{\psym}=Id_{\psym}$ and $C \in T_P{\psym}$), i.e. $\dot P=\dot A$, which implies $\dot O=0$.
Thus, we proved that $BP \in \sym$ implies $dO_P(BP)=0$. However, this is incompatible with $dO_P(BP)=\skew B$ in general: $BP \in \sym \iff BP=PB^T$.
For $P=\text{diag}(\sigma_1,\sigma_2), B=\begin{pmatrix} 1 & b \\\ c & 1 \end{pmatrix}$, this happens if and only if $\sigma_2b=\sigma_1c$, so if $\sigma_1 \neq \sigma_2$, then $B$ is not symmetric. Thus, $dO_P(BP)=0 \neq \skew B$.
Comment: Equation $(1)$ implies that one can equivalently focus upon the "dual" problem, of computing $dP_A$, instead of $dO_A$. Here is a previous attempt of mine to go in this direction. |
General question
I work on the plane where I have a two-dimensional shape $V$ that is cut in a collection of parts $\{V_i\}$ that do not overlap
$ V_i ~~\text{s.t.}~~ \bigcup_i \overline{V}_i = \overline{V} ~~\text{and}~~ \bigcap_i V_i = \{\emptyset\} $
I know the value of the integral of a scalar field $q(\pmb{x})$ over each part $V_i$
$ \int_{V_i} q(\pmb{x}) \, \pmb{dx} ~~ \text{known for all \(i\)} $
and I would like to get an estimation of the field $q(\pmb{x})$ for any point $\pmb{x}$ of space.
What methods can do that?
My particular case
where the colors in the plot corresponds to the value of a packing factor $P(V_i)$ calculated from a characteristic function $\chi$ as $ P(V_i) := \frac{\int_{V_i} \chi(\pmb{x})}{\int_{V_i} 1} $
I would like to find a function $p$ such that$ P(V_i) = \int_{V_i} p(\pmb{x}) ~~ \forall i$
Of course, being in a general 2D case, there is no such thing as a primitive function… How can I estimate function $p$? |
Let $$f(x)=\sum_{n=1}^{\infty}\dfrac{\cos{nx}}{\sqrt{n^3+n}}$$ and $F(x)=\int_{0}^{x}f(t)\,\mathrm dt,F(0)=0$.
Show that: $$\dfrac{\sqrt{2}}{2}-\dfrac{1}{15}<F\left(\dfrac{\pi}{2}\right)<\dfrac{\sqrt{2}}{2}$$
Find the value $F\left(\dfrac{\pi}{2}\right)$.
Since the series $f$ converges normally on $\mathbb{R}$, we can integrate term-by-term. This gives $$ F\left(\frac{\pi}{2}\right)=\sum_{n\geq 1}\frac{1}{\sqrt{n^3+n}}\int_0^{\pi/2}\cos (nx)dx=\sum_{k\geq 0}\frac{(-1)^k}{\sqrt{2}(2k+1)^\frac{3}{2}(2k^2+2k+1)^\frac{1}{2}}. $$ The series converges absolutely to $S$. But this is an alternating series which satisfies Leibniz criterion (the relevant function is indeed decreasing on $(0,+\infty)$). So the sequence of partial sums $S_n$ alternates about $S$ and satisfies $$ S_{2n+1}< S< S_{2n} \qquad \forall n\geq 0. $$ In particular, we get $$ \frac{\sqrt{2}}{2}-\frac{1}{15}<\frac{\sqrt{2}}{2}-\frac{1}{3\sqrt{5}\sqrt{6}}=S_1<S<S_0=\frac{\sqrt{2}}{2}. $$ This answers 1.
I don't know if there is a closed form for $S$ and I am not alone.
I was delayed in posting this, so it is little more than a comment to julien's answer.
Since the series $\sum\limits_{n=1}^\infty\frac1{\sqrt{n^3+n}}$ is convergent, we can integrate term by term $$ F(x)=\sum_{n=1}^\infty\frac{\sin(nx)}{n\sqrt{n^3+n}} $$ $\sin((2k)\pi/2)=0$ and $\sin((2k+1)\pi/2)=(-1)^k$; therefore, $$\begin{align} F(\pi/2) &=\frac1{\sqrt2}\sum_{k=0}^\infty\frac{(-1)^k}{\sqrt{(2k+1)^3(2k^2+2k+1)}}\\ &=\frac1{\sqrt2}-\frac1{\sqrt{270}}+\dots \end{align} $$ By the alternating series test, the final sum is between $\frac1{\sqrt2}$ and $\frac1{\sqrt2}-\frac1{\sqrt{270}}$. That is, $$ \frac1{\sqrt2}-\frac1{15}\lt\frac1{\sqrt2}-\frac1{\sqrt{270}}\lt F(\pi/2)\lt\frac1{\sqrt2} $$
This evaluates to $0.6587329279592957$, but the ISC does not find a closed form. |
Waecmaths
Title waecmaths question Question 1
Evaluate (0.13)
Question 2
Simplify 11011
Question 3
Simplify $\frac{{{25}^{\tfrac{2}{3}}}\div {{25}^{\tfrac{1}{6}}}}{{{({1}/{5}\;)}^{-\tfrac{7}{6}}}\times {{({1}/{5}\;)}^{\tfrac{1}{6}}}}$
Question 4
Simplify $\frac{x-4}{4}-\frac{x-3}{6}$
Question 5
Given that $\frac{1-2x}{4x-3}$, find the value of
Question 6 Question 7
A fair coin is tossed three times. Find the probability of getting two head and one tail
Question 8
If 30% of
Question 9
A baker used 40% of a 50kg of flour. If $\tfrac{1}{8}$ of the amount used was for cake, how many kilogram of flour was used for the cake?
Question 10
If tan
Question 11
In the diagram
Question 12
Which of this diagram show(s) the correct method of constructing a perpendicular line
Question 13
In the diagram $\left| QR \right|=5cm$ $\angle PQR={{60}^{\circ }}$ and $\angle PSR={{45}^{\circ }}$. Find $\left| PS \right|$ leaving your answer in surd form
Question 14
Represent the solution of the inequality $\frac{x-2}{2}\ge \frac{x+3}{4}$ on the number line
Question 15
In the diagram find
Question 16
The ratio of boys to girls in a class is 5:3, find the probability of selecting at random a girl from the class
Question 17
In the diagram
Question 18
The table below satisfies the relation $y=k\sqrt{x}$ , where
Question 19
The table below satisfies the relation $y=k\sqrt{x}$ , where
Question 20
What must be added to ${{x}^{2}}-3x$ to make it a perfect square?
Question 21
In the diagram
Question 22
Given that $(2x-1)(x+5)=2{{x}^{2}}-mx-5$, what is the value of
Question 23
What is the place value of 9 in the number 3.0492?
Question 24
If the simple interest on a sum of money invested at 3% per annum for 2½ years in N123. Find the principal
Question 25
A machine valued at N20,000 depreciates by 10% every year. What will be the value of the machine at the end of two years
Question 26
If $2x+y=10$ and $y\ne 0$ which of the following is not a possible value of
Question 27
In the diagram,
Question 28
If
Question 29 Question 30 Question 31
If ${{\log }_{a}}270-{{\log }_{a}}10+{{\log }_{a}}\tfrac{1}{3}=2$ what is the value of
Question 32
A conical water–jug is 7cm in diameter and 6cm deep. Find the volume of water it can hold [Take $\pi =\tfrac{22}{7}$ ]
Question 33
In the diagram, the two circles have a common a centre O, if the area of the larger circle is 100
Question 34
If $P={{\left[ \frac{Q(R-T)}{15} \right]}^{\tfrac{1}{3}}}$ Make
Question 35
In the diagram,
Question 36
A seller allows 20% discount for cash payment on a marked price of his goods. What is the ratio of the cash payment to the marked price
Question 37 Question 38 Question 39
Find the
Question 40
The diagram shows an arc |
Some tricks I've seen:
Tricks with notable products
$(a + b)^2 = a^2 + 2ab + b^2$
This formula can be used to compute squares. Say that we want to compute $46^2$. We use $46^2 = (40+6)^2 = 40^2+2\cdot40\cdot6 +6^2 = 1600 + 480 + 36 = 2116$. You can also use this method for negative $b$:$ 197^2 = (200 - 3)^2 = 200^2 - 2\cdot200\cdot3 + 3^2 = 40000 - 1200 + 9 = 38809 $
The last subtraction can be kind of tricky: remember to do it right to left, and take out the common multiples of 10:$ 40000 - 1200 = 100(400-12) = 100(398-10) = 100(388) = 38800 $The hardest thing here is to keep track of the amount of zeroes, this takes some practice!
Also note that if we're computing $(a+b)^2$ and a is a multiple of $10^k$ and $b$ is a single digit-number, we already know the last $k$ digits of the answer: they are $b^2$, then the rest (going to the right) are zeroes. We can use this even if a is only a multiple of 10: the last digit of $(10 * a + b)^2$ (where $a$ and $b$ consist of a single digit) is $b$. So we can write (or maybe only make a mental note that we have the final digit) that down and worry about the more significant digits.
Also useful for things like $46\cdot47 = 46^2 + 46 = 2116 + 46 = 2162$. When both numbers are even or both numbers are uneven, you might want to use:
$(a+b)(a-b) = a^2 - b^2$Say, for example, we want to compute $23 \cdot 27$. We can write this as $(25 - 2)(25 + 2) = 25^2 - 2^2 = (20 + 5)^2 = 20^2 + 2\cdot20\cdot5 + 5^2 - 4 = 400 + 200 + 25 - 4 = 621$.
Divisibility checks
Already covered by Theodore Norvell. The basic idea is that if you represent numbers in a base $b$, you can easily tell if numbers are divisible by $b - 1$, $b + 1$ or prime factors of $b$, by some modular arithmetic.
Vedic math
A guy in my class gave a presentation on Vedic math. I don't really remember everything and there probably are a more cool things in the book, but I remember with algorithm for multiplication that you can use to multiplicate numbers in your head.
This picture shows a method called lattice or gelosia multiplication and is just a way of writing our good old-fashioned multiplication algorithm (the one we use on paper) in a nice way. Please notice that the picture and the Vedic algorithm are not tied: I added the picture because I think it helps you appreciate and understand the pattern that is used in the algorithm. The gelosia notation shows this in a much nicer way than the traditional notation.
The algorithm the guy explained is essentially the same algorithm as we would use on paper. However, it structures the arithmetic in such a way that we never have remember too many numbers at the same time.
Let's illustrate the method by multiplying $456$ with $128$, as in the picture. We work from left to right: we first compute the least significant digits and work our way up.
We start by multiplying the least significant digits:
$6 \cdot 8 = 48$: the least significant digit is $8$, remember the $4(0)$ for the next round (of course, I don't mean zero times four here but four, or forty, whatevery you prefer: be consistent though, if you include the zero here to make forty, you got do it everywhere).$ 8 \cdot 5(0) = 40(0) $
$ 2(0) \cdot 6 = 12(0) $ $ 4(0) + 40(0) + 12(0) = 56(0) $: our next digit (to the left of the $8$) is $6$: remember the $5(00)$
$ 8 \cdot 4(00) = 32(00) $
$ 2(0) \cdot 5(0) = 10(00) $ $ 1(00) \cdot 6 = 6(00) $ $ 5(00) + 32(00) + 10(00) + 6(00) = 53(00) $: our next digit is a $3$, remember the $5(000)$
Pfff... starting with 2-digit numbers is a better idea, but I wanted to this longer one to make the structure of the algorithm clear. You can do this much faster if you have practiced, since you don't have to write it all down.
$ 2(0) \cdot 4(00) = 8(000) $
$ 1(00) \cdot 5(0) = 5(000)$ $ 5(000) + 8(000) + 5(000) = 18(000)$: next digit is an $8$, remember the $1(0000)$
$ 1(00) \cdot 4(00) = 4(0000) $
$ 1(0000) + 4(0000) = 5(0000) $: the most significant digit is a $5$.
So we have $58368$.
Quadratic equations
There are multiple ways to solve a quadratic equation in your head. The easiest are quadratic with integer coefficients. If we have $x^2 + ax + c = 0$, try to find $r_{1, 2}$ such that $r_1 + r_2 = -a$ and $r_1r_2 = c$. It is also possible to solve for non-integer solutions this way, but it is usually too hard to actually come up with solutions this way.
Another way is just to try divisors of the constant term. By the rational root theorem (google it, I can't link anymore
sigh) all solutions to $x^n + ... + c = 0$ need to be divisors of $c$. If $c$ is a fraction $\frac{p}{q}$, the solutions need to be of the form $\frac{a}{b}$ where $a$ divides $p$ and $b$ divides $q$.
If this all fails, we can still put the abc-formula in a much easier form:
$ ux^2 + vx + w = 0 $
$ x^2 + \frac{v}{u}x + \frac{w}{u} = 0 $ $ x^2 + \frac{v}{u}x + \frac{w}{u} = 0 $ $ x^2 - ax - b = 0 $
$ x^2 = ax + b $
(This is the form that I found easiest to use!) $ (x - \frac{a}{2})^2 = (\frac{a}{2})^2 + b $ $ x = \frac{a\pm\sqrt{a^2 + 4b}}{2} = \frac{a}{2} \pm \sqrt{(\frac{a}{2})^2 + b} $
I'm sure there are also a lot of techniques for estimating products and the like, but I'm not really familiar with them.
Tricks that aren't really usable but still pretty cool
See this excerpt from Feynman's "Surely you're joking, Mr. Feynman!" about how he managed to amaze some of his colleagues, and also this video from Numberphile. |
I am trying to solve an APSP (All-Pair Shortest Path) problem on a weighted graph. This graph is actually a 1, 2 or 3 dimensional grid, and the weights on each edge represent the distance between its two vertices. What I want to have is the geodesic graph distance (shortest path through the graph), for every pair of vertices.
I want a diffusion-based method, because it is faster than a Dijkstra, or a Floyd-Warshall algorithm. I'm trying to achieve this using the heat equation: $$\frac{du}{dt} = \Delta u.$$ In the end, my application needs a kernel of the form $\exp(-d^2/\gamma)$ with $d$ the graph geodesic distance.
My hope is that since the solution is supposed to be the Green function for diffusion:
$$ u(t,x,y) = \left(\frac{1}{4\pi t}\right)^{-\frac{dim}{2}}\exp\left(\frac{-d^2(x,y)}{4t}\right),$$
then I can directly use that solution (with a few adjustements to get rid of the factor in front) as my kernel, and the parameter $\gamma$ will be adjusted by adjusting $t$.
I haven't been able to do something that works yet, and I would love some help. I have tried many things so far, and there are multiple problems that arise. It is difficult and long to explain all of them in one question, so I will first explain what I think is the beginning of a good approach, and then ask a few general questions.
In the same way it is done in the first step of the Geodesic in Heat algorithm by Crane et al., with a backward Euler step, I can solve the linear system: $$(Id - tL)u = u_0 \tag{1}$$ with $t$ the diffusion step, $L$ the laplacian matrix, and $u_0$ a Dirac at one of the vertices.
Solving equation (1) actually gives a kernel of the form $\exp(-d/\gamma)$, which is not desired. Therefore I have to do K subiterations in time, and solve K times: $$(Id - \frac{t}{k}L)u = u_0 \iff Mu = u_0$$ which gives $u = M^{-1}...M^{-1}u_0 \iff u = M^{-K}u_0$.
As K increases, the kernel is supposed to converge to a square one $\exp(-d^2/\gamma)$.
Now here come the questions :
Should I use a graph Laplacian, or a finite differences Laplacian ?AFAIU, a graph laplacian is normalized to have 1 in the diagonal, whereas a FE Laplacian has the degree in the diagonal, and is multiplied by $\frac{1}{h^2}$ How do I embed the graph weights in the Laplacian, so that the distance I get in the solution is the graph geodesic distance ?I want to be able to predict what will be the range of values of $d(x,y)$ in the solution, with regard to the range of the weights, and the parameters $t$, $K$, and $n$ the number of points in one direction (total domain size: $N = n^{dim}$). Which boundary conditions should I use in the Laplacian ?I feel like I shouldn't impose the function values (Dirichlet) at the boundary, because that would mean imposing the highest distance. Or should I ? I tried homogeneous Neumann conditions, and homogeneous Dircihlet conditions, but in either cases I get some distortion near the boundaries of the parabola $d(x,y)^2$ (which I check by computing the log of the solution $u(t)$, and substracting the minimum). Should I use a diffusion equation instead ?: $\frac{\partial u}{\partial t}=\nabla\cdot\left(\kappa\nabla u\right)$, since the diffusion is spatially dependent ?
References :
Crane et al. 2013 : Geodesics in heat: A new approach to computing distance based on heat flow |
The
impulse response of a system is, perhaps not entirely unexpectedly, the response of a system to an impulse. The concepts of signals and systems, in the context of discrete-time signal processing, are introduced in the article Discrete-Time Signal Processing. This article introduces the all important impulse response, and shows how knowing only the impulse response of an LTI system can be used to determine the output of that system for any given input. Impulse
Defining an
impulse in continuous-time signal processing is quite tricky. There, it is defined through the Dirac delta, which can be described as
\[\delta(x)=\begin{cases}+\infty,&x=0\\0,&x\ne 0\end{cases}\mathrm{,\quad with }\int_{-\infty}^\infty\delta(x)\,dx=1.\]
For discrete-time signal processing, things are much easier. There, an impulse is defined through the signal
\[\delta[n]=\begin{cases}1,&n=0\\0,&n\ne 0\end{cases},\]
which is called the
unit sample function or, simply, impulse. The center part (since its length is infinite, albeit zero everywhere except for the origin) of this signal is shown below, with the zero time point in the middle of the graph. Impulse Response
As already noted in Discrete-Time Signal Processing, an LTI system is completely characterized by its impulse response. This means that knowing the impulse response is sufficient to completely predict what the system will output for any possible input. The impulse response is often written using the letter \(h\), as \(h[n]\). For example, the
identity system has \(h[n]=\delta[n]\).
As a somewhat more complex example, consider the system for which the impulse response is shown below.
A system with this impulse response computes a
moving average. We know the output of this system when the input is an impulse, since we know the impulse response. What is the output of the moving average system for the following signal?
The output is shown below. Even with these very simple signals, the relation between the input and the output is not obvious to the eye, although small examples like this could still be computed manually.
An example with audio signals is shown in How to Record a Lion in a Concert Hall. That article shows how the impulse response of a concert hall makes it possible to compute how the roar of a lion would sound in that concert hall.
Convolution
For the specific case of an LTI system, the reaction of the system to an impulse can be computed by
convolution. The convolution of two (discrete and one-dimensional) signals is defined by the infinite sum
\[(x*h)[n]\equiv\sum_{m=-\infty}^\infty\!x[m]\,h[n-m],\]
where, in this application, \(x\) is the original signal and \(h\) is the impulse response.
Computing a convolution by following this definition closely is a very expensive operation (in terms of computation). However, the ubiquitous use of convolution is enabled by the
Fast Fourier Transform (FFT) algorithm. |
The Annals of Statistics Ann. Statist. Volume 4, Number 4 (1976), 779-787. Gauss-Markov Estimation for Multivariate Linear Models with Missing Observations Abstract
In this note we discuss multivariate linear models from the coordinate-free point of view, as earlier done by Eaton (1970). We generalize the result of Eaton by allowing for missing observations. This leads to models of the kind $EY \in L, Cov Y \in\{P(I \otimes \sum)P'\}$ where $P$ is a diagonal mapping. The paper starts by deriving the conditions for existence of Gauss-Markov estimators (GME) of $EY$ in models where the covariance-mappings are not necessarily nonsingular. These conditions are then applied to the above models if $\Sigma$ runs either over all PSD-mappings or over all diagonal PSD-mappings. In the latter case $L$ must be of the form $L = L_1 \times \cdots \times L_p$ while in the general case some further conditions on the $L_i$ must be met. (If $P = I$, then $L_i = L_j$ must hold for all $i, j$; this is equivalent to the result obtained by Eaton). Examples show that these conditions are satisfied only under rather exceptional conditions.
Article information Source Ann. Statist., Volume 4, Number 4 (1976), 779-787. Dates First available in Project Euclid: 12 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aos/1176343551 Digital Object Identifier doi:10.1214/aos/1176343551 Mathematical Reviews number (MathSciNet) MR411060 Zentralblatt MATH identifier 0336.62052 JSTOR links.jstor.org Citation
Drygas, Hilmar. Gauss-Markov Estimation for Multivariate Linear Models with Missing Observations. Ann. Statist. 4 (1976), no. 4, 779--787. doi:10.1214/aos/1176343551. https://projecteuclid.org/euclid.aos/1176343551 |
Waecmaths
Title waecmaths question Question 31
If $P=\{y:2y\ge 6\}$and $Q=\{y:y-3\le 3\}$ when
Question 44
Using the Venn diagram, find$n(X\cap Y')$
Question 11
If $X=\{0,2,4,6\}$, $Y=\{1,2,3,4\}$ and$Z=\{1,3\}$ are subset of $U=\{x:0\le x\le 6\}$, find $X\cap (Y\cup Z)$
Question 12
Find the truth set of the equation ${{x}^{2}}=3(2x+9)$
Question 37
If
Question 34
Given that
Question 44
The venn diagram show the number of students in a class who like reading(
Question 13
Given that $P=\{x:1\le x\le 6\}$ and $Q=\{x:2<x<10\}$ , where
Question 15
Given that $T=\{x:-2<x\le 9\}$ where
Question 38
Detemine $M'\cap N$from the Venn diagram
Question 7
Two sets are disjoints if
Question 36
If
Question 48
Which shaded region in the following diagram represents$(P\cup Q)\cap R$?
Question 33
Every staff in an office owns either a Mercedes or Toyota car. 20 owns Mercedes, 15 own Toyota and 5 own both. How many staff are there in the office?
Question 38
The Venn diagram shows the choice of food of a number of visitors to a canteenthere were 35 visitors in all, find the value of
Question 39
The Venn diagram shows the choice of food of a number of visitors to a canteen. How many people took at least two kinds of food?
Question 42
If
Question 42
Describe the shaded portion in the diagram
Question 49
Consider the following statements:
X : Locally manufactured tyres are attractive
Y : Many locally manufactured tyres do not last long
Denoting locally manufactured by M, attractive tyres by R and lost lasting tyres tyres by L.
Which of these Venn diagrams illustrates the statements?
Question 34 Question 6
If $M=\{x:3\le x<8\}$ and $N=\{x:8<x<12\}$ , which of the following is true
I. $8\in M\cap N$
II. $8\in M\cup N$
III. $M\cap N=\varnothing $
Question 46 Question 6
If
$\mu =\{x:0<x\le 10\text{ and }x\text{ is an integer }\!\!\}\!\!\text{ }$find $(T'\cap M')$ |
The asymptotic sampling distribution, after taking plim, of the OLS estimator is given by $\sqrt{N}(\hat{\beta}-\beta) = E[X_iX_i^T]^{-1} \left(1/\sqrt{N} \sum_{I=1}^NX_ie_i \right) $
It must be shown that the asymptotic variance can be written as: $ Var[E[X_iX_i^T]^{-1} 1/\sqrt{N} \sum_{I=1}^NX_ie_i] = \left( E[X_iX_i^T]^{-1} \right) \left( E[X_iX_i^T e_i^2]\right) \left( E[X_i X_i^T]^{-1} \right) $
Because the variance is linear in parameters, my idea was to start like this:
$V[\sqrt{N}(\hat{\beta}-\beta)] = \sqrt{N}Var[\hat{\beta}] + 0$
Because the $Var[\beta]=0$. Thus one could use the fact that $Var[\hat{\beta}]=E[(\hat{\beta}-\beta)(\hat{\beta}-\beta)^T]$.
However, while there seems to be a relationship I stuck at this point, mostly because of the $\sqrt{N}$. Thanks for help. |
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